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5 Steps to a 5

AP Calculus AB

MCGRAW-HILL

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5 Steps to a 5

AP Calculus AB

William Ma

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We hope you enjoy this McGraw-Hill eBook! If you’d like more

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Dedication

To my mother, Lai-ping , who borrowed money

to hire a tutor for me when I was in 7th grade.

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Contents

Preface / xiiiAcknowledgments / xv

PART I HOW TO USE THIS BOOKHow Is This Book Organized? / 3Introduction to the Five-Step Program / 3Three Approaches to Preparing for the AP Calculus AB Exam / 4

Overview of the Three Plans / 4Calendar for Each Plan / 5

Graphics Used in the Book / 9

PART II WHAT YOU NEED TO KNOW ABOUT THEAP CALCULUS AB EXAMBackground on the AP Exam / 13

What Is Covered in the AP Calculus AB Exam? / 13What Is the Format of the AP Calculus AB Exam? / 13What Are the Advanced Placement Exam Grades? / 13How Is the AP Calculus AB Exam Grade Calculated? / 14Which Graphing Calculators Are Allowed for the Exam? / 14

Calculators and Other Devices Not Allowed for the AP Calculus Exam / 14Other Restrictions on Calculators / 15How Much Work Do I Need to Show When I Use a Graphing Calculator in

Section II, Free-Response Questions? / 15What Do I Need to Bring to the Exam? / 15

Tips for Taking the Exam / 15Getting Started! / 17Diagnostic Test / 19Answers and Solutions / 27Scoring and Interpretation / 39

PART III COMPREHENSIVE REVIEWChapter 1 Review of Pre-Calculus / 43

1.1 Lines / 43Slope of a Line / 43Equations of Lines / 44Parallel and Perpendicular Lines / 45

1.2 Absolute Values and Inequalities / 47Absolute Values / 47Inequalities and the Real Number Line / 48Solving Absolute Value Inequalities / 49

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Solving Polynomial Inequalities / 50Solving Rational Inequalities / 51

1.3 Functions / 53Definition of a Function / 53Operations on Functions / 53Inverse Functions / 56Trigonometric and Inverse Trigonometric Functions / 58Exponential and Logarithmic Functions / 61

1.4 Graphs of Functions / 64Increasing and Decreasing Functions / 64Intercepts and Zeros / 65Odd and Even Functions / 67Shifting, Reflecting, and Sketching Graphs / 68

1.5 Rapid Review / 721.6 Practice Problems / 731.7 Solutions to Practice Problems / 74

Chapter 2 Limits and Continuity / 76

2.1 The Limit of a Function / 76Definition and Properties of Limits / 76Evaluating Limits / 77One-Sided Limits / 78Squeeze Theorem / 81

2.2 Limits Involving Infinities / 83Infinite Limits (as x → a) / 83Limits at Infinity as (x → ±∞) / 85Vertical and Horizontal Asymptotes / 87

2.3 Continuity of a Function / 90Theorems on Continuity / 90

2.4 Rapid Review / 92

2.5 Practice Problems / 932.6 Cumulative Review Problems / 952.7 Solutions to Practice Problems / 952.8 Solutions to Cumulative Review Problems / 97

Chapter 3 Differentiation / 993.1 Derivatives of Algebraic Functions / 99

Definition of the Derivative of a Function / 99Power Rule / 102The Sum, Difference, Product, and Quotient Rules / 103The Chain Rule / 104

3.2 Derivatives of Trigonometric, Inverse Trigonometric, Exponential, andLogarithmic Functions / 105

Derivatives of Trigonometric Functions / 105Derivatives of Inverse Trigonometric Functions / 107Derivatives of Exponential and Logarithmic Functions / 108

3.3 Implicit Differentiation / 1103.4 Approximating a Derivative / 1123.5 Derivatives of Inverse Functions / 1133.6 Higher Order Derivatives / 1153.7 Rapid Review / 116

viii • Contents

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3.8 Practice Problems / 1173.9 Cumulative Review Problems / 118

3.10 Solutions to Practice Problems / 1183.11 Solutions to Cumulative Review Problems / 121

Chapter 4 Graphs of Functions and Derivatives / 1234.1 Rolle’s Theorem, Mean Value Theorem, and Extreme Value Theorem / 123

Rolle’s Theorem and Mean Value Theorem / 123Extreme Value Theorem / 1264.2 Determining the Behavior of Functions / 127

Test for Increasing and Decreasing Functions / 127First Derivative Test and Second Derivative Test for Relative Extrema / 130Test for Concavity and Points of Inflection / 133

4.3 Sketching the Graphs of Functions / 138Graphing without Calculators / 138Graphing with Calculators / 140

4.4 Graphs of Derivatives / 1414.5 Rapid Review / 1454.6 Practice Problems / 148

4.7 Cumulative Review Problems / 1504.8 Solutions to Practice Problems / 1504.9 Solutions to Cumulative Review Problems / 155

Chapter 5 Applications of Derivatives / 1575.1 Related Rate / 157

General Procedure for Solving Related Rate Problems / 157Common Related Rate Problems / 157Inverted Cone (Water Tank) Problem / 158Shadow Problem / 160Angle of Elevation Problem / 161

5.2 Applied Maximum and Minimum Problems / 162General Procedure for Solving Applied Maximum and

Minimum Problems / 162Distance Problems / 162Area and Volume Problems / 164Business Problems / 166

5.3 Rapid Review / 1675.4 Practice Problems / 1685.5 Cumulative Review Problems / 1705.6 Solutions to Practice Problems / 1705.7 Solutions to Cumulative Review Problems / 176

Chapter 6 More Applications of Derivatives / 1786.1 Tangent and Normal Lines / 178

Tangent Lines / 178Normal Lines / 183

6.2 Linear Approximations / 186Tangent Line Approximation / 186Estimating the nth Root of a Number / 188Estimating the Value of a Trigonometric Function of an Angle / 188

6.3 Motion Along a Line / 189Instantaneous Velocity and Acceleration / 189

Contents • ix

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x • Contents

Vertical Motion / 190Horizontal Motion / 191


Chapter 7 Integration / 2007.1 Evaluating Basic Integrals / 200

Antiderivatives and Integration Formulas / 200Evaluating Integrals / 202

7.2 Integration by U-Substitution / 205The U-Substitution Method / 205U-Substitution and Algebraic Functions / 205U-Substitution and Trigonometric Functions / 207U-Substitution and Inverse Trigonometric Functions / 208U-Substitution and Logarithmic and Exponential Functions / 209



Chapter 8 Definite Integrals / 2188.1 Riemann Sums and Definite Integrals / 218

Sigma Notation or Summation Notation / 218Definition of a Riemann Sum / 219Definition of a Definite Integral / 220

Properties of Definite Integrals / 2218.2 Fundamental Theorems of Calculus / 222First Fundamental Theorem of Calculus / 222Second Fundamental Theorem of Calculus / 224

8.3 Evaluating Definite Integrals / 225Definite Integrals Involving Algebraic Functions / 226Definite Integrals Involving Absolute Value / 227Definite Integrals Involving Trigonometric, Logarithmic, and

Exponential Functions / 228Definite Integrals Involving Odd and Even Functions / 229

8.4 Rapid Review / 2308.5 Practice Problems / 231

8.6 Cumulative Review Problems / 2328.7 Solutions to Practice Problems / 2328.8 Solutions to Cumulative Review Problems / 235

Chapter 9 Areas and Volumes / 2379.1 The Function F (x) = ∫ a

x

f (t )dt / 2379.2 Approximating the Area under a Curve / 241

Rectangular Approximations / 241Trapezoidal Approximations / 244

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Contents • xi

9.3 Area and Definite Integrals / 245Area under a Curve / 245Area between Two Curves / 249

9.4 Volumes and Definite Integrals / 252Solids with Known Cross Sections / 252The Disc Method / 256The Washer Method / 261



Chapter 10 More Applications of Definite Integrals / 27510.1 Average Value of a Function / 275

Mean Value Theorem of Integrals / 275Average Value of a Function on [a,b] / 276

10.2 Distance Traveled Problems / 27710.3 Definite Integral as Accumulated Change / 280

Business Problems / 280Temperature Problems / 281Leakage Problems / 282Growth Problems / 282

10.4 Differential Equations / 283Exponential Growth/Decay Problems / 283Separable Differential Equations / 285


PART IV PRACTICE MAKES PERFECTPractice Exam 1 / 299

Answers / 310Solutions / 312Scoring Sheet / 321

Practice Exam 2 / 323Answers / 335Solutions / 337Scoring Sheet / 346

APPENDIXESAppendix I: Formulas and Theorems / 349Appendix II: Special Topic: Slope Fields / 353Appendix III: Bibliography / 358Appendix IV: Websites / 359

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Preface

Congratulations! You are an AP Calculus student. Not too shabby! As you know, APCalculus is one of the most challenging subjects in high school. You study mathematicalideas that helped change the world. Not that long ago, calculus was taught in graduateschools. Today, smart young people like yourself study calculus in high school. Most col-leges will give you credit if you score a 3 or more on the AP Calculus Exam.

So how do you do well on the AP Exam? How do you get a 5? Well, you’ve alreadytaken the first step. You’re reading this book. The next thing you need to do is to makesure that you understand the materials and do the practice problems. In recent years, theAP Calculus exams have gone through many changes. For example, today the questionsno longer stress long and tedious algebraic manipulations. Instead, you are expected tobe able to solve a broad range of problems including problems presented to you in the

form of a graph, a chart, or a word problem. For many of the questions, you are alsoexpected to use your calculator to find the solutions.

After having taught AP Calculus for many years and having spoken to students andother calculus teachers, I understand some of the difficulties that students might encounterwith the AP Calculus exams. For example, some students have complained about notbeing able to visualize what the question was asking and others students said that evenwhen the solution was given, they could not follow the steps. Under these circumstances,who wouldn’t be frustrated? In this book, I have addressed these issues. Whenever pos-sible, problems are accompanied by diagrams and solutions are presented in a step-by-step manner. The graphing calculator is used extensively whenever it is permitted. Thebook also begins with a big chapter on review of precalculus. The purpose is to make thebook self-contained so that if a student needs to look up a formula, a definition, or a con-

cept in precalculus, it is right there in the book. You might skip this chapter and beginwith Chapter 2.

So how do you get a 5 on the AP Calculus AB exam?

Step 1: Pick one of the study plans from the book.Step 2: Study the chapters and do the practice problems as scheduled.Step 3: Take the Diagnostic Test and Practice Tests.Step 4: Get a good night’s sleep the day before the exam.

As an old martial artist once said, “First you must understand. Then you must prac-tice.” Have fun and good luck!

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Acknowledgments

I could not have written this book without the help of the following people:

My high school calculus teacher, Michael Cantor, who taught me calculus.

Professor Leslie Beebe who taught me how to write.

David Pickman who fixed my computer and taught me Equation Editor.

Jennifer Tobin, who was a senior at Herricks High School and is now attending TheCollege of New Jersey, who tirelessly edited many parts of the manuscript and with whomI look forward to co-author a math book in the future.

Robert Teseo and his calculus students who field-tested many of the problems.All the students in my BC Calculus class at Herricks for their comments and support.

Mark Reynolds who proofread part of the manuscript.

Robert Main who meticulously edited the entire manuscript.

Maxine Lifshitz who offered many helpful comments and suggestions.

Don Reis whose patience and encouragement kept me writing.

Barbara Gilson, the sponsoring editor, Grace Freedson, the project editor, MaureenWalker, the editing supervisor, and Betsy Winship, the production manager, for all theirassistance.

Sam Lee and Derek Ma who were on 24-hour call for technical support.

My older daughter Janet for not killing me for missing one of her concerts.

My younger daughter Karen who helped me with many of the computer graphics.

My wife Mary who gave me many ideas for the book and who often has more confi-dence in me than I have in myself.

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5 Steps to a 5

AP Calculus AB

MCGRAW-HILL

http://dx.doi.org/10.1036/0071406247

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HOW TO USE THIS BOOK

PART I

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HOW IS THIS BOOK ORGANIZED?

Part I contains an introduction to the Five-Step Program and three study plans for prepar-ing for the AP Calculus AB exam.

Part II contains a full-length diagnostic test. The diagnostic test contains the samenumber of multiple-choice and free-response questions as an AP Calculus AB exam.

Part III (Comprehensive Review) contains 10 chapters. Chapter 1 is a review of pre-

calculus. It is included in the book so that you do not have to look for a pre-calculus bookin the event you need to refer to a previous concept. You may skip this chapter if you wishand begin with Chapter 2. At the end of each chapter you will find a set of 20 practice prob-lems and, beginning with Chapter 2, a set of 5 cumulative review problems. These prob-lems have been created to allow you to practice your skills. They have also been designedto avoid unnecessary duplications. At the end of each chapter, a Rapid Review gives yousome of the highlights of the chapter.

Part IV contains two full-length practice tests as well as the answers, explanations,and worksheets to compute your scores.

Note: The exercises in this book are done with the TI-89 graphing calculator.

INTRODUCTION TO THE FIVE-STEP PROGRAM

The Five-Step Program is designed to provide you with the skills and strategies vital tothe exam and the practice that can help lead you to that perfect 5. Each of the five stepswill provide you with the opportunity to get closer and closer to the “Holy Grail” 5.

Step One leads you through a brief process to help determine which type of exampreparation you want to commit yourself to.

1. Month-by-month: September through May.

2. The calendar year: January through May.3. Basic training: Six weeks prior to the exam.

Step Two helps develop the knowledge you need to succeed on the exam.

1. A comprehensive review of the exam.2. One “Diagnostic Test” which you can go through step-by-step and question-by-

question to build your confidence level.3. A summary of formulas related to the AP Calculus AB exam.4. A list of interesting and related websites and a bibliography.

Step Three develops the skills necessary to take the exam and do well.

1. Practice multiple-choice questions.2. Practice free-response questions.

Step Four helps you develop strategies for taking the exam.

1. Learning about the test itself.2. Learning to read multiple choice questions.3. Learning how to answer multiple choice questions, including whether or not to guess.4. Learning how to plan and write the free-response questions.

How to Use This Book • 3

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Step Five will help you develop your confidence in using the skills demanded on theAP Calculus AB exam.

1. The opportunity to take a diagnostic exam.2. Time management techniques/skills.3. Two practice exams that test how well-honed your skills are.

THREE APPROACHES TO PREPARING FORTHE AP CALCULUS AB EXAM

Overview of the Three Plans

No one knows your study habits, likes and dislikes better than you. So, you are the onlyone who can decide which approach you want and/or need to adopt to prepare for theAdvanced Placement Calculus AB exam. Look at the brief profiles below. These mayhelp you to place yourself in a particular prep mode.

You are a full-year prep student (Approach A) if:

1. You are the kind of person who likes to plan for everything far in advance . . . andI mean far . . . ;

2. You arrive at the airport 2 hours before your flight because, “you never know whenthese planes might leave early . . .”;

3. You like detailed planning and everything in its place;4. You feel you must be thoroughly prepared;5. You hate surprises.

You are a one-semester prep student (Approach B) if:

1. You get to the airport 1 hour before your flight is scheduled to leave;2. You are willing to plan ahead to feel comfortable in stressful situations, but are okay

with skipping some details;3. You feel more comfortable when you know what to expect, but a surprise or two

is cool;4. You’re always on time for appointments.

You are a 6-week prep student (Approach C) if:

1. You get to the airport just as your plane is announcing its final boarding;2. You work best under pressure and tight deadlines;

3. You feel very confident with the skills and background you’ve learned in your APCalculus AB class;

4. You decided late in the year to take the exam;5. You like surprises;6. You feel okay if you arrive 10–15 minutes late for an appointment.

4 • How to Use This Book

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CALENDAR FOR EACH PLAN

A Calendar for Approach A:A Year-Long Preparation for the AP Calculus AB Exam

Although its primary purpose is to prepare you for the AP Calculus AB Exam you will

take in May, this book can enrich your study of calculus, your analytical skillsand your problem solving techniques.

SEPTEMBER–OCTOBER (Check off the

activities as you complete them.)

Determine into which student mode youwould place yourself.Carefully read Parts I and II.Get on the web and take a look at the AP

website(s).Skim the Comprehensive Review section.(These areas will be part of your year-longpreparation.)Buy a few highlighters.Flip through the entire book. Break thebook in. Write in it. Toss it around a littlebit . . . Highlight it.Get a clear picture of what your ownschool’s AP Calculus AB curriculum is.Begin to use the book as a resource to sup-plement the classroom learning.

Read and study Chapter 1—Review of Pre-Calculus.Read and study Chapter 2—Limits andContinuity.Read and study Chapter 3—Differentiation.

NOVEMBER (The first 10 weeks have elapsed.)

Read and study Chapter 4—Graphs of Functions and Derivatives.Read and study Chapter 5—Applicationsof Derivatives.

DECEMBER

Read and study Chapter 6—MoreApplications of Derivatives.Review Chapters 1–3.

JANUARY (20 weeks have now elapsed.)

Read and study Chapter 7—Integration.Review Chapters 4–6.

FEBRUARY

Read and study Chapter 8—DefiniteIntegrals.Read and study Chapter 9—Areas andVolumes.Take the Diagnostic Test.

Evaluate your strengths and weaknesses.Study appropriate chapters to correctweaknesses.

MARCH (30 weeks have now elapsed.)

Read and study Chapter 10—MoreApplications of Definite Integrals.Review Chapters 7–9.

APRIL

Take Practice Exam 1 in first week of April.Evaluate your strengths and weaknesses.Study appropriate chapters to correctweaknesses.Review Chapters 1–10.

MAY—First Two Weeks (THIS IS IT!)

Take Practice Exam 2.Score yourself.

Study appropriate chapters to correctweaknesses.Get a good night’s sleep the night beforethe exam. Fall asleep knowing you arewell prepared.

GOOD LUCK ON THE TEST!

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A Calendar for Approach B:A Semester-Long Preparation for the AP Calculus AB Exam

Working under the assumption that you’ve completed one semester of calculus studies,the following calendar will use those skills you’ve been practicing to

prepare you for the May exam.


JANUARY

Carefully read Parts I and II.Read and study Chapter 1—Review of Pre-Calculus.Read and study Chapter 2—Limits andContinuity.Read and study Chapter 3—Differentiation.Read and study Chapter 4—Graphs of Functions and Derivatives.

FEBRUARY

Read and study Chapter 5—Applicationsof Derivatives.Read and study Chapter 6—MoreApplications of Derivatives.Read and study Chapter 7—Integration.Take the Diagnostic Test.Evaluate your strengths and weaknesses.Study appropriate chapters to correctweaknesses.

Review Chapters 1–4.

MARCH (10 weeks to go.)

Read and study Chapter 8—DefiniteIntegrals.

Read and study Chapter 9—Areas andVolumes.Read and study Chapter 10—MoreApplications of Definite Integrals.Review Chapters 5–7.

APRIL

Take Practice Exam 1 in first week of April.

Evaluate your strengths and weaknesses.Study appropriate chapters to correctweaknesses.Review Chapters 1–10.


Take Practice Exam 2.Score yourself.Study appropriate chapters to correctweaknesses.Get a good night’s sleep the night beforethe exam. Fall asleep knowing you arewell prepared.


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A Calendar for Approach C:A Six-Week Preparation for the AP Calculus AB Exam

At this point, we are going to assume that you have been building your calculus knowledgebase for more than six months. You will, therefore, use this book primarily

as a specific guide to the AP Calculus AB Exam.Given the time constraints, now is not the time to try to expand your

AP Calculus AB curriculum. Rather, it is the time to limit and refinewhat you already do know.


APRIL 1st–15th

Skim Parts I and II.Skim Chapters 1–5.Carefully go over the “Rapid Review”sections of Chapters 1–5.Take the Diagnostic Test.

Evaluate your strengths and weaknesses.Study appropriate chapters to correctweaknesses.

APRIL 16th–May 1st

Skim Chapters 6–10.Carefully go over the “Rapid Review”sections of Chapters 6–10.

Complete Practice Exam 1.Score yourself and analyze your errors.Study appropriate chapters to correctweaknesses.


Complete Practice Exam 2.Score yourself and analyze your errors.Study appropriate chapters to correctweaknesses.Get a good night’s sleep. Fall asleep know-ing you are well prepared.


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Summary of the Three Study Plans

Approach A: Approach B: Approach C:Month September Plan January Plan 6-Week Plan

September–October Chapters 1, 2 & 3

November 4 & 5

December 6Review 1–3

January 7 ChaptersReview 4–6 1, 2, 3 & 4

5, 6 & 7February 8 & 9 Diagnostic Test

Diagnostic Test Review 1–4

March 10 8, 9 & 10Review 7–9 Review 5–7

Diagnostic TestReview 1–5April Practice Exam 1 Practice Exam 1 Practice Exam 1

Review 1–10 Review 1–10 Review 6–10

May Practice Exam 2 Practice Exam 2 Practice Exam 2


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GRAPHICS USED IN THE BOOK

To emphasize particular skills, strategies, and practice, we use four sets of icons through-out this book. You will see these icons in the margins of Parts I, II, and III.

The first icon is an hourglass. We’ve chosen this to indicate the passage of time dur-ing the school year. This hourglass icon will be in the margin next to an item which maybe of interest to one of the three types of students who are using this book.

For the student who plans to prepare for the AP Calculus exam during the entireschool year, beginning in September through May, we use an hourglass which is full onthe top.

For the student who decides to begin preparing for the exam in January of the cal-endar year, we use an hourglass which is half full on the top and half full on the bottom.

For the student who chooses to prepare during the final 6 weeks before the exam,we use an hourglass which is empty on the top and full on the bottom.

The second icon is a clock that indicates a timed practice activity or a time man-agement strategy. It indicates on the face of the dial how much time to allow for a givenexercise. The full dial will remind you that this is a strategy that can help you learn tomanage your time on the test.

The third icon is an exclamation point that points to a very important idea, concept,or strategy point you should not pass over.

The fourth icon, a sun, indicates a tip that you might find useful.


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WHAT YOU NEED TOKNOW ABOUT THEAP CALCULUS AB EXAM

PART II


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BACKGROUND ON THE AP EXAM

What Is Covered in the AP Calculus AB Exam?

The AP Calculus AB exam covers the following topics:

• Functions, Limits and Graphs of Functions, Continuity

• Definition and Computation of Derivatives, Second Derivatives, Relationshipbetween the Graphs of Functions and their Derivatives, Applications of Derivatives

• Finding Antiderivatives, Definite Integrals, Applications of Integrals, FundamentalTheorem of Calculus, Numerical Approximations of Definite Integrals, and SeparableDifferential Equations.

Students are expected to be able solve problems that are expressed graphically, numer-ically, analytically, and verbally. For a more detailed description of the topics coveredin the AP Calculus AB exam, visit the College Board website at: www.collegeboard.org/ ap/calculus.

What Is the Format of the AP Calculus AB Exam?The AP Calculus AB exam has 2 sections:

Section I contains 45 multiple-choice questions with 105 minutes.

Section II contains 6 free-response questions with 90 minutes.

The time allotted for both sections is 3 hours and 15 minutes. Below is a summary of thedifferent parts of each section.

What You Need to Know About the AP Calculus AB Exam • 13

Part A 28 questions No Calculator 55 Minutes

Part B 17 questions Calculator 50 Minutes

Part A 3 questions Calculator 45 Minutes

Part B 3 questions No Calculator 45 Minutes

Section I

Multiple-Choice

Section IIFree-Response

During the time allotted for Part B of Section II, students may continue to work on ques-tions from Part A of Section II. However, they may not use a calculator at that time.Please note that you are not expected to be able to answer all the questions in order toreceive a grade of 5. If you wish to see the specific instructions for each part of the test,visit the College Board website at: www.collegeboard.org/ap/calculus.

What Are the Advanced Placement Exam Grades?

Advanced Placement Exam grades are given in a 5-point scale with 5 being the highestgrade. The grades are described below:

5 Extremely Well Qualified4 Well Qualified3 Qualified2 Possibly Qualified1 No Recommendation

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How Is the AP Calculus AB Exam Grade Calculated?

• The exam has a total raw score of 108 points: 54 points for the multiple-choice ques-tions in Section I and 54 points for the free-response questions for Section II.

• Each correct answer in Section I is worth 1.2 points, an incorrect answer is worth(1 ⁄ 4)(−1.2) points, and no points for unanswered questions. For example, supposeyour result in Section I is as follows:

Correct Incorrect Unanswered

36 4 5

Your score for Section I would be:

36 × 1.2 − 4 × (1 ⁄ 4)(1.2) = 43.2 − 1.2 = 42. Not a bad score!

• Each complete and correct solution for Section II is worth 9 points.• The total raw score for both Section I and II is converted to a 5-point scale. The cut-off

points for each grade (1–5) vary from year to year. Visit the College Board website at:www.collegeboard.com/ap for more information. Below is a rough estimate of theconversion scale:

Total Raw Score Approximate AP Grade

75–108 560–74 445–59 331–44 20–30 1

Remember, these are approximate cut-off points.

Which Graphing Calculators Are Allowed for the Exam?The following calculators are allowed:

Texas Instruments Hewlett-Packard Casio Sharp

TI-82 HP-28 series FX-9700 series EL-9200 seriesTI-83/TI-83 Plus HP-38G FX-9750 series EL-9300 seriesTI-85 HP-39G CFX-9800 series EL-9600 seriesTI-86 HP-40G CFX-9850 seriesTI-89 HP-48 series CFX-9950 series

HP-49 series CFX-9970 seriesAlgebra FX 2.0 series

For a more complete list, visit the College Board website at: www.collegeboard.com/ap.If you wish to use a graphing calculator that is not on the approved list, your teachermust obtain written permission from the ETS before April 1st of the testing year.

Calculators and Other Devices Not Allowed for the AP Calculus Exam

• TI-92, HP-95 and devices with QWERTY keyboards• Non-graphing scientific calculators• Laptop computers

• Pocket organizers, electronic writing pads or pen-input devices

14 • What You Need to Know About the AP Calculus AB Exam

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Other Restrictions on Calculators

• You may bring up to 2 (but no more than 2) approved graphing calculators to the exam.• You may not share calculators with another student.• You may store programs in your calculator.• You are not required to clear the memories in your calculator for the exam.• You may not use the memories of your calculator to store secured questions and take

them out of the testing room.

How Much Work Do I Need to Show When I Use a GraphingCalculator in Section II, Free-Response Questions?

• When using a graphing calculator in solving a problem, you are required to write thesetup that leads to the answer. For example, if you are finding the volume of a solid,you must write the definite integral and then use the calculator to compute the numer-

ical value, e.g., Simply indicating the answer without

writing the integral would only get you one point for the answer but no other credit

for the work.• You may not use calculator syntax to substitute for calculus notations. For example,you may not write “ ” instead of “

”

• You are permitted to use the following 4 built-in capabilities of your calculator toobtain an answer: plotting the graph of a function, finding the zeros of a function, find-ing the numerical derivative of a function, and evaluating a definite integral. All othercapabilities of your calculator can only be used to check your answer. For example,you may not use the built-in Inflection function of your calculator to find points of inflection. You must use calculus using derivatives and showing change of concavity.

What Do I Need to Bring to the Exam?

• Several Number 2 pencils• A good eraser and a pencil sharpener• Two black or blue pens• One or two approved graphing calculators with fresh batteries. (Be careful when you

change batteries so that you don’t lose your programs.)• A watch• An admissions card or a photo I.D. card if your school or the test site requires it.• Your Social Security number• Your school code number if the test site is not at your school• A simple snack if the test site permits it. (Don’t try anything you haven’t eaten before.

You might have an allergic reaction.)• A light jacket if you know that the test site has strong air conditioning• Do not bring Wite Out or scrap paper.

TIPS FOR TAKING THE EXAM

General Tips

• Write legibly.

• Label all diagrams.

π π5 2252

0

3

x dx( ) =∫ .

Volume =Volume = ( ) ∗( ) =∫ π π5 2 0 3 225x x ˆ , , ,

Volume = ( ) =∫ π π5 2252

0

3

x dx .


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• Organize your solution so that the reader can follow you line of reasoning.• Use complete sentences whenever possible. Always indicate what the final answer is.

More Tips

• Do easy questions first.• Write out formulas and indicate all major steps.

• Guess if you can eliminate some of the choices in a multiple-choice question.• Leave a multiple-choice question blank if you have no clue what the answer is.• Be careful to bubble in the right grid, especially if you skip a question.• Move on. Don’t linger on a problem too long.• Go with your first instinct if you are unsure.

Still More Tips

• Indicate units of measure.• Simplify numeric or algebraic expressions only if the question asks you to do so.

• Carry all decimal places and round only at the end.• Round to 3 decimal places unless the question indicates otherwise.

• Watch out for different units of measure, e.g., the radius, r, is 2 feet, find in inchesper second.

• Use calculus notations and not calculator syntax, e.g., write and not

• Use only the four specified capabilities of your calculator to get your answer: plot-ting graph, finding zeros, calculating numerical derivatives, and evaluating definiteintegrals. All other built-in capabilities can only be used to check your solution.

• Answer all parts of a question from Section II even if you think your answer to anearlier part of the question might not be correct.

Enough Already . . . Just 3 More Tips

• Be familiar with the instructions for the different parts of the exam before the day of the exam. Visit the College Board website at: www.collegeboard.com/ap for moreinformation.

• Get a good night sleep the night before.• Have a light breakfast before the exam.

x x ˆ , .2( )x dx2

dr

dt


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GETTING STARTED!

Answer Sheet for Diagnostic Test—Section I


PART A

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

PART B

76.

77.

78.

79.

80.

81.

82.

83.

84.

85.

86.

87.

88.

89.

90.

91.

92.

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Number of Questions Time Use of Calculator

28 55 Minutes No


DIAGNOSTIC TEST

Section I—Part A

Directions:Use the answer sheet provided in the previous page. All questions are given equal weight. There is no penaltyfor unanswered questions. However, 1 ⁄ 4 of the number of the incorrect answers will be subtracted from the

number of correct answers. Unless otherwise indicated, the domain of a function f is the set of all realnumbers. The use of a calculator is not permitted in this part of the test.

1. Evaluate

2. If f(x) = −2csc(5x), find

3. Evaluate

4. Given the equation what is anequation of the normal line to the graph at x = 5?

5. The graph of f is shown in Figure D-1. Draw apossible graph of f ′ on (a, b).

y x= − 1,

lim .x

x

x→−∞

−2 4

2

′

f

π6

.

11

4

x

dx

∫ .

a c 0 d e f b x

f

y

0

10

–10

20

2 4 6 8

V

V (t )

( f e e t / s e c o n d )

t

(seconds)

y

f

x –1 0 1 2 3 4 5

6. If what is the value of

7. Evaluate

8. If

9. The graph of the velocity function of a movingparticle is shown in Figure D-2. What is the total

distance traveled by the particle during 0 ≤ t ≤ 6?

2 3 61

x dx kk

−( ) =−∫ , .find

1 2

2

−∫

x

x dx.

lim ?x

f x→ +

′( )0

f x xex( ) = 2 ,

10. If find h′(π).

11. What is the average value of the function y = e−4x

on [−ln 2, ln 2]?

12. The graph of f consists of four line segments,for −1 ≤ x ≤ 5 as shown in Figure D-3. What is

the value of f x dx( )−∫ 15

?

h x t dt x

( ) = ∫ sin ,π2

Figure D-1

Figure D-2

Figure D-3

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a 0 b

g

y

x

Figure D-4

13. Given the equation y = (x + 1)(x − 3)2, what isthe instantaneous rate of change of y at x = −1?

14. If

15. What is the slope of the tangent to the curve y =cos (xy) at x = 0?

16. If f ′(x) = g (x) and g is a continuous function for

all real values of x, then is

(A)

(B) f (2) − f (0)

(C) f (6) − f (0)

(D)

(E) 3f (6) − 3f (0)

17. What is

18. The graph of the function g is shown inFigure D-4. Which of the following is true for

g on (a, b)?

I. g is monotonic on (a, b).

II. g ′ is continuous on (a, b).III. g ″ > 0 on (a, b).

lim

tan tan

?∆

∆

∆x

x

x→

+

−

0

4 4

π π

1

30

1

36f f ( ) − ( )

1

36

1

30f f ( ) − ( )

g x dx30

2

( )∫

h x x xx x

h xx

( ) = >− ≤

( )→

ifif

find the4

412 42

, lim .

21. A function f is continuous on [−2, 0] and someof the values of f are shown below.

If f (x) = 2 has no solution on [−2, 0], then bcould be

(A) 3(B) 2(C) 1(D) 0(E) −2

22. Find the area of the region enclosed by thegraph of y = x2 −x and the x-axis.

23. If for all real values of k, then

which of the graphs in Figure D-5 (see page 21)could be the graph of f ?

24. If f (x) is an antiderivative of and f (0) =ln 2, find f (ln 2).

25. If = 2 sin x and at x = π, y = 2, find a

solution to the differential equation.

26. When the area of a square is increasing fourtimes as fast as the diagonals, what is the length

of a side of the square?

27. The graph of f is shown in Figure D-6 onpage 21 and f is twice differentiable, which of the following statements is true?

(A) f (10) < f ′(10) < f ″ (10)

(B) f ″ (10) < f ′(10) < f (10)

(C) f ′(10) < f (10) < f ″ (10)

(D) f ′(10) < f ″ (10) < f (10)

(E) f ″ (10) < f (10) < f ′(10)

28. If a function f is continuous for all valuesof x, which of the following statements isalways true?

I.

II.

III. f x dx f x dx f x dxb

a

b

c

c

a

( ) = ( ) − ( )∫ ∫ ∫

f x dx f x dx f x dxa

c

a

b

c

b

( ) = ( ) − ( )∫ ∫ ∫


b

a

c

b

c

( ) = ( ) + ( )∫ ∫ ∫

dy

dx

e

e

x

x + 1

f x dxk

k

( ) =−∫ 0

x −2 −1 0

f 4 b 4

19. The velocity function of a moving particle on thex-axis is given as v(t ) = t 2 − t. For which values of t is the particle’s speed decreasing?

20. Evaluate sin .2t dt x

( )∫ π

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0

0 0

y

y y

x x

0 0

y y

x x x

(A)

(D) (E)

(B) (C)

Figure D-5

0 10

f

y

x

Figure D-6

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Section I—Part B

Directions:Use the same answer sheet as for Part A. Please note that the questions begin with number 76.This is not an

error. It is done to be consistent with the numbering system of the actual AP Calculus AB Exam.All questions are given equal weight. There is no penalty for unanswered questions. However,1 ⁄ 4 of the number of incorrect answers will be subtracted from the number of correct answers.

Unless otherwise indicated, the domain of a function f is the set of all real numbers.The use of a calculator is permitted in this part of the test.

76. The area under the curve from x = 1 to

x = k is 8. Find the value of k.

77. If find the

value(s) of x where g has a local minimum.

78. If which of the following

statements about g are true?

I. g has a relative maximum at x = 2.II. g is differentiable at x = 6.III. g has a point of inflection at x = −2.

79. The graph of f ′, the derivative of f is shown inFigure D-7. At which value(s) of x is graph of f concave up?

g x x x( ) = − −2 4 12 ,

g x t dt x

( ) =

∫ 22

5

22

sin , ,π

π π on

y x= 81. The velocity function of a moving particle is

for 0 ≤ t ≤ 6. What is the

maximum acceleration of the particle on theinterval 0 ≤ t ≤ 6?

82. Water is leaking from a tank at the rate off (t ) = 10 ln(t + 1) gallons per hour for 0 ≤ t ≤ 10,where t is measured in hours. How manygallons of water has leaked from the tank atexactly after 5 hours?

83. Write an equation of the normal line to the

graph of y = x3 for x ≥ 0 at the point wheref ′(x) = 12.

84. If and the graph of f is shown

in Figure D-8, which of the graphs in Figure D-9on page 23 is a possible graph of g ?

g x f t dt a

x

( ) = ( )∫

v t t t ( ) = − +3

2

32 5


17 50 Minutes Yes

x 2 x 3 x 4 x 1 0

y

x

Figure D-7

a b0

f

y

x

Figure D-8

80. How many points of inflection does the graph of y = sin (x2) have on the interval [−π, π]?

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a 0 b

y

x

a b0 a b0

y y

x x

a

0 0a b

b

y y

x x

(A)

(D) (E)

(B) (C)

Figure D-9


85. For 0 ≤ x ≤ 3π, find the area of the regionbounded by the graphs of y = sin x and y = cos x.

86. Carbon-14 has a half-life of 5730 years. If y isthe amount of carbon-14 present and y decays

according to the equation where k is

a constant and t is measured in years, find thevalue of k.

87. Given a differentiable function f with

and Using a tangent line to the

graph of what is an approximate value

of

88. Find the volume of the solid generated byrevolving about the x-axis the region boundedby the graph of y = sin 2x for 0 ≤ x ≤ π and the

line y =1

2.

f π π2 180

+

?

x = π2

,

′

= −f

π2

1.

f π2

3

=

dy

dt ky= ,

89. The graphs of f ′, g ′, p ′ and q ′ are shown inFigure D-10 on page 24. Which of the functions

f, q, p or q have a point of inflection on (a, b)?

90. At what value(s) of x do the graphs of

have perpendicular

tangent lines?

91. Let f be a continuous function on [0, 6] and hasselected values as shown below:

Using three midpoint rectangles of equal lengths,

what is the approximate value of

92. What is the volume of the solid whose base is theregion enclosed by the graphs of y = x2 andy = x + 2 and whose cross sections perpendicularto the x-axis are squares?

f x dx( )∫ ?0

6

f x x

x y x( ) = = −

lnand 2

x 0 1 2 3 4 5 6f(x) 1 2 5 10 17 26 37

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a 0 b

y

y

x

a a0 0b b x x

f '

p'

0a b

y

y

x

g'

q'

Figure D-10

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Section II—Part A

Directions:Show all work. You may not receive any credit for correct answers without supporting work. You may use an

approved calculator to help solve a problem. However, you must clearly indicate the setup of your solutionusing mathematical notations and not calculator syntax. Calculators may be used to find the derivative of a

function at a point, compute the numerical value of a definite integral, or solve an equation. Unless otherwiseindicated, you may assume the following: (a) the numeric or algebraic answers need not be simplified,(b) your answer, if expressed in approximation, should be correct to 3 places after the decimal point,

and (c) the domain of a function f is the set of all real numbers.

1. The slope of a function f at any point (x, y) is

The point (2, 4) is on the graph of f.

(a) Write an equation of the line tangent to thegraph of f at x = 2.

(b) Use the tangent line in part (a) to approxi-mate f (2.1).

(c) Solve the differential equation

with the initial conditionf (2) = 4.

(d) Use the solution in part (c) and find f (2.1).

2. A drum containing 100 gallons of oil is puncturedby a nail and begins to leak at the rate of

where t is measured

in minutes and 0 ≤ t ≤ 10.

(a) How much oil to the nearest gallon leakedout after t = 6 minutes?

(b) What is the average amount of oil leaked outper minute from t = 0 to t = 6 to the nearestgallon?

1012

sin πt

gallons

minute

dy

dx

x

y=

+4 1

2

4 1

2

x

y

+.

(c) Write an expression for f (t ) to represent thetotal amount of oil in the drum at time t ,where 0 ≤ t ≤ 10.

(d) At what value of t to the nearest minute willthere be 40 gallons of oil remaining in thedrum?

3. Given the function f (x) = xe2x.

(a) At what value(s) of x, if any, is f ′(x) = 0?(b) At what value(s) of x, if any, is f ″ (x) = 0?(c) Find and

(d) Find the absolute extrema of f and justifyyour answer.

(e) Show that if f (x) = xeax where a > 0, the

absolute minimum value of f is−1

ae.

lim .x

f x→−∞

( )limx

f x→∞

( )


3 45 Minutes Yes

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Section II—Part B

Directions:The use of a calculator is not permitted in this part of the test. When you have finished this part of the test,

you may return to the problems in Part A of Section II and continue to work on them. However, youmay not use a calculator. You should show all work. You may not receive any credit for correctanswers without supporting work. Unless otherwise indicated, the numeric or algebraic answers

need not be simplified, and the domain of a function f is the set of all real numbers.

4. The graph of f ′, the derivative of the function f,for −4 ≤ x ≤ 6 is shown in Figure D-11.

(a) At what value(s) of x does f have a relative

maximum value? Justify your answer.(b) At what value(s) of x does f have a relative

minimum value? Justify your answer.(c) At what value(s) of x does f ″ (x) > 0? Justify

your answer.(d) At what value(s) of x, if any, does the graph

of f have a point of inflection? Justify youanswer.

(e) Draw a possible sketch of f (x), if f (−2) = 3.

5. Let R be the region enclosed by the graph ofy = x3, the x-axis and the line x = 2.

(a) Find the area of region R.(b) Find the volume of the solid obtained by

revolving region R about the x-axis.

(c) The line x = a divides region R into tworegions such that when the regions arerevolved about the x-axis, the resulting solidshave equal volume. Find a.

(d) If region R is the base of a solid whose crosssections perpendicular to the x-axis aresquares, find the volume of the solid.

6. Given the equation x2y2 = 4,

(a) Find

(b) Write an equation of the line tangent to thegraph of the equation at the point (1, −2).

(c) Write an equation of the line normal to thecurve at point (1, −2).

(d) The line is tangent to the curve

at the point P. Find the coordinates of point P.

y x= +12

2

dy

dx.


3 45 Minutes No

y

x

y= f ' ( x )

–4 –3 –2 –1 1 2 3 4 5 60

Figure D-11

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ANSWERS AND SOLUTIONS

Answers to Diagnostic Test—Section I

Part A

1. 2

2.

3. −1/2

4. y = −4x + 22

5. See solution.

6. 2

7.

8. {−2, 5}

9. 50 feet

10. 0

11.255

128 2ln

−− +

1

x

x c

−20 3

12. 2

13. 16

14. Does not exist.

15. 0

16. A

17. 2

18. II & III

19.

20.

21. A

22.1

6

−( ) +

1

22

1

2cos x

1

2

1,

23. D

24. ln 3

25. y = −2 cos x

26. 4

27. C

28. I & III

Part B

76. 132/3

77. 2π

78. I

79. x < x2

80. 8

81. 12

82. 57.506

83.

84. A

85. 5.657

86.

87. 2.983

88. 1.503

89. q

90. 1.370

91. 76

92.81

10

− ln 2

5730

yx

= −

+1

12

49

6

Answers to Diagnostic Test—Section II

Part A

1. (a) (3 pts.)

(b) 4.113 (1 pt.)

(c) (4 pts.)

(d) 4.113 (1 pt.)

2. (a) 38 gallons (3 pts.)(b) 6 gallons (2 pts.)

(c) (1 pt.)

(d) 8 minutes (3 pts.)

3. (a) −0.5 (1 pt.)(b) −1 (1 pt.)(c) (2 pts.)

(d) (3 pts.)

(e) See solution. (2 pts.)

−1

2e

lim & limx x

f x f x→∞ →−∞

( ) = ∞ ( ) = 0

f t x dxt

( ) = − ( )∫ 100 10120

sin π

y x x= + +2 62

y x= −( ) +9

82 4

Part B

4. (a) x = 2 (2 pts.)(b) x = 5 (2 pts.)(c) (−4, 0) and (4, 6) (1 pt.)(d) x = 0 and x = 4 (2 pts.)(e) See solution. (2 pts.)

5. (a) 4 (2 pts.)

(b) (2 pts.)

(c) (3 pts.)

(d) (2 pts.)

6. (a) (3 pts.)

(b) y = 2x − 4 (2 pts.)

(c) (2 pts.)

(d) (2, −1) and (−2, 1) (2 pts.)

y x= − −1

2

3

2

dy

dx

y

x=

−

128

7

26

7

128

7

π

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Solutions to Diagnostic Test—Section I

Part A—No calculators.

1. From Chapter 8

2. From Chapter 3

3. From Chapter 2

4. From Chapter 6

dy

dxx =

=−( )

=( )

=5

12

12

1

2 5 1

1

2 4

1

4

y x x dy

dx x

x

= − = −( ) = −( )

=−( )

−

1 11

21

1

2 1

12

12

12

;

lim lim

,

lim

lim

x x

x

x

x

x

xx

xx

x x x

xx

x

→−∞ →−∞

→−∞

→−∞

−=

−−

→ −∞ = −( )

=− −

=− −

= − = −

2

2

2

2

2

2

2

4

2

4

2

4

2

1 1

2

1

2

1

2

note: as

f x x

f x x x

x x

f

( ) = − ( )

′( ) = − −( ) ( )[ ]( )

= ( ) ( )

′

=

= ( ) −( ) = −

2 5

2 5 5 5

10 5 5

610

5

6

5

6

10 2 3 20 3

csc

csc cot

csc cot

csc cotπ π π

11

2

2

2 4 2 1 4 2 2

1

4

1

4

1

4

1

412

12 1

2

12

12

xdx x dx x x∫ ∫ = =

= ]

= ( ) − ( ) = − =

−

5. From Chapter 4

See Figure DS-1.

x y x

y x y x

y x

= = − = − = ( )

=

= −

− = − −( ) ⇒ = − −( ) +

= − +

5 1 5 1 2 5 2

1

44

2 4 5 4 5 2

4 22

At

Slope of normal line negative reciprocal

of

Equation of normal line:

or

, ; ,

.

[

[

[

[

a

a d b

e0 b

+'

"

'

+00 –

+–

incr. f

f

incr.decr.

incr.decr.

Concave downwardConcave

upward

a e x

y

bd 0

A possible graph of f

Based on the graph of f :

Figure DS-1

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11. From Chapter 10

12. From Chapter 9

13. From Chapter 3

y x x

dy

dx x x x

x x x

dy

dxx

= +( ) −( )

= ( ) −( ) + −( ) +( )

= −( ) + −( ) +( )

= − −( ) + − −( ) − +( )

= −( ) + =

=−

1 3

1 3 2 3 1

3 2 3 1

1 3 2 1 3 1 1

4 0 16

2

2

2

1

2

2

;

f x dx f x dx f x dx( ) = ( ) + ( )

= − ( )( ) + +( )( )

= − + =

− −∫ ∫ ∫ 1

5

1

5

1

1

1

22 1

1

22 4 1

1 3 2

Average value =−

=−

− −

= ( )

− +

( )

=−

+

=

−

−

− − −( )

−

−

1

2 2 4

1

2 2 4 4

1

2 2 4 4

1

2 2

2

4

2

4

1

4 2

4 2 4 2

2 4 2 4

4 4

ln

ln

ln

ln

ln

ln

ln ln

ln ln

e

e e

e e

x

x

22 2

1

644

1

2 2

255

64

255

128 2

ln

ln

ln

− +

=

=

Average value

Let or

=− −( )

= − = − −

=

=

−

= + =

−

+

−

−

− −

∫

∫ ∫

1

2 2

4 44

4

1

4

1

4

4

2

2

4 4

ln ln

;

ln

ln

e dx

u x du dx du

dx

e dx e

du

e c e c

x

x u u x

6. From Chapters 2 and 3

7. From Chapter 7

You can check the answer by differentiating

your result.

8. From Chapter 8

You can check your answer by evaluating

9. From Chapter 9

10. From Chapter 8

h x t dt h x x

h

x

( ) = ⇒ ′( ) =

′( ) = = =

∫ sin sin

sin

π

π π

2

0 0

Total distance

feet

= ( ) + ( )

= ( )( ) + ( ) −( )

= + =

∫ ∫ v t v t dt 4

5

0

4

12

4 20 12

2 10

40 10 50

2 3 2 31

2

1

6

x x dx−( ) −( )−

−

−∫ ∫ and

Set

5 or

k k k k

k k k k

2 23 4 6 3 10 0

2 0 5 2

− − = ⇒ − − =⇒ −( ) +( ) = ⇒ = = −

2 3 3

3 1 3 1

3 1 3

3 4

2

1 1

2 2

2

2

x dx x x

k k

k k

k k

k k

−( ) = − ]= −( ) − −( ) − −( )( )= − − +( )

= − −

− −∫

1 1 11

11

1

2

2 2

2

2 2

2

1

−= −

= −

= −( ) =−

− +

= − − +

∫ ∫ ∫

∫ −−

x

x dx

x

x

x dx

x dx

x dx x

x c

x x c

f x xe xe x

xe x

x f x e e x e xe

f x e xe e

xx

x

x x x x

x x

x x

( ) = = ≥− <

≥ ′( ) = + ( ) = +

′( ) = +( ) = + =→ →+ +

22 02 0

0 2 2 2 2

2 2 2 0 20 0

0

ifif

If ,

lim lim


7/23/2019 007137714 x


[0

12

0

1

0V(t )

a(t )

t

– – – – – – – – – – – – – – – – – – – – – – + + + + +

– – – – – – – – – + + + + + + + + + + + + + + + + + +

0


17. From Chapter 3

18. From Chapter 4

I. Since the graph of g is decreasing and thenincreasing, It is not monotonic.

II. Since the graph of g is a smooth curve, g ′ iscontinuous,

III. Since the graph of g is concave upward, g ″ >0.

Thus only statements II and III are true.

19. From Chapter 6

See Figure DS-2.

v t t t

v t t t

t t

a t v t t

a t t t

v t a t

( ) = −

( ) = ⇒ −( ) = ⇒

= =

( ) = ′( ) = −

( ) = ⇒ − = =

( ) < ( ) >

2

0 1 0

0 1

2 1

0 2 1 01

2

0 0 1

1

21

Set

or

Set or

Since and on1

2

the speed of the particle is decreasing on

.

, ,

, .

+

−

= ( )

=

= ( ) =

→

x

x

d

dx x

x

x

0

4

22

4 4

42 2

lim

tan tan

tan

sec

∆

∆

∆Thus

π π

π

′( ) = +( ) − ( )

→f x

f x x f x

xx1

0

1 1lim∆

∆∆

Figure DS-2

14. From Chapter 2

15. From Chapter 6

16. From Chapter 8

Let or

Thus, the correct choice is A

u x du dx du

dx

g x dx g u du

g u du

f u c f x c

g x dx f x f f

= = =

( ) = ( ) = ( )

= ( ) + = ( ) +

( ) = ( )[ ] = ( ) − ( )

( )

∫ ∫ ∫

∫

3 33

33

1

3

1

3

1

33

3 13 3 13 6 13 00

2

0

2

;

.

y xy dy

dx xy y x

dy

dx

dy

dx y xy x xy

dy

dx

dydx

x xy dydx

y xy

dy

dx x xy y xy

dy

dx

y xy

x xy

x y xy

= ( ) = − ( )[ ] +

= − ( ) − ( )

+ ( ) = − ( )

+ ( )[ ] = − ( )

= − ( )

+ ( )

= = ( ) = ( ) =

cos ; sin

sin sin

sin sin

sin sin

sin

sin

, cos cos

1

1

1

0 0At 11 0 1

1 0

1 0 0

0

10

0 0

0 1

; ,

sin

sin

.

,

( )

= −( ) ( )

+ ( ) = =

=

= =

dy

dx

x

x y

Thus the slope of the tangent at is

h x x x

x x

h x x

h x x

h x h x h x

x x

x x

x x x

( ) = >

− ≤

( ) = = =

( ) = −( ) = −( ) =

( ) ≠ ( ) ( )

→ →

→ →

→ → →

+ +

− −

+ −

if

if

Since thus

does not exist.

4

12 4

4 2

12 4 12 4

2

4 4

4 4

2 2

4 4 4

lim lim

lim lim

lim lim , lim

7/23/2019 007137714 x


23. From Chapter 9

Thus the graph in choice (D) is the only oddfunction.

24. From Chapter 7

25. From Chapter 10

26. From Chapter 5

Let be the diagonal of a square.

Area of a square

Since

z

A z

dA

dt

z dz

dt z

dz

dt

dA

dt

dz

dt

dz

dt z

dz

dt z

=

= =

= = ⇒ =

2

2

2

2

4 4 4;

dy

dx x dy xdx

dy xdx y x c

x y c

c

c c

y x

= ⇒ =

= ⇒ = − +

= = ⇒ = − +

⇒ = −( ) −( ) +

⇒ = + =

= −

∫ ∫

2 2

2 2

2 2 2

2 2 1

2 2 0

2

sin sin

sin cos

, cos

.

cos

At

or

Thus

π π

Thus andf x e f

e

x( ) = +( ) ( )

= +( ) = +( ) =

ln ln

ln ln lnln

1 2

1 2 1 32

Let

and

u e du e dx

f x ee

dxu

du

u c e c

f e c c

f c

c

x x

x

x

x

= + =

( ) =+

=

= + = + +

( ) = + + = ( ) +

( ) = ⇒ ( ) + =

⇒ =

∫ ∫

1

11

1

0 1 2

0 2 2 2

0

0

;

ln ln

ln ln

ln ln ln

f x dx f x

f x f x

k

k

( ) = ⇒ ( )

( ) = − −( )

−∫ 0 is odd function

i.e.,

,

Area = −( ) = −

= −

− = −

=

∫ x x dx x x2

0

13 2

0

1

3 2

1

3

1

20

1

6

1

6


20. From Chapter 8

21. From Chapter 1

See DS-3.

sincos cos

cos

cos

22

2

2

2

2

2

1

22

1

2

t dt t x

x

x

x

( ) = − ( )

=

− ( )

− − ( )

= − ( ) +

∫ ππ

π

–2

–1

0–1–2

1

4

2

3

(–2,4) (0,4)

y

x

y = 2

Figure DS-3

0 1

y

x

y= x 2– x

Figure DS-4

If b = 2, then x = −1 would be a solution forf (x) = 2.

If b = 1, 0 or −2, f (x) = 2 would have two solu-tions for f (x) = 2.

Thus, b = 3, choice (A).

22. From Chapter 9

To find points of intersection, set y = x2 − x = 0⇒ x(x − 1) = 0 ⇒ x = 0 or x = 1. See Figure DS-4.

7/23/2019 007137714 x


78. From Chapter 5

See Figure DS-6.


Let s be a side of the square. Since the diagonal z= 4, s2 + s2 = z2 or 2s2 = 16. Thus, s2 = 8 or s =

27. From Chapter 4

The graph indicates that (1) f (10) = 0, (2) f ′(10)< 0, since f is decreasing; and (3) f ″ (10) > 0,since f is concave upward.

Thus f ′(10) < f (10) < f ″ (10), choice (C).

28. From Chapter 8

I.

The statement is true, since the upper andlower limits of the integrals are in sequence,i.e., a → c = a → b → c.

II.

The statement is not always true.

III.

The statement is true.

Thus only statements I and III are true.

Part B—Calculators are permitted.

76. From Chapter 9

77. From Chapter 8

Since g (x) = then g ′(x) = 2 sin x

Set g ′(x) = 0 ⇒ 2 sin x = 0 ⇒ x = π, or 2π

g ″ (x) = 2 cos x and g ″ (π) = 2 cos π = −2 and

g ″ (2π) = 1

22

sin ,t dt x

π∫

Area

Since set

or

= = =

=

= − ( )

= − = −( )

= −( ) = ⇒ −

= ⇒ = =

∫ ∫ xdx x dx x

x k

k k

A k k

k k

kk

k

k

12

32

11

1

32

1

32

32

32

32

32

32

32

2

32

2

3

2

3

2

31

2

3

2

3

2

31

82

31 8 1

12 13 13

,

33

1

13

82

3

You can check your result by evaluating

to obtainxdx .∫

f x dx f x dx f x dx

f x dx f x dx

b

c

c

a

b

a

a

c

b

a

( ) = ( ) − ( )

= ( ) + ( )

∫ ∫ ∫

∫ ∫


f x dx f x dx

a

b

c

b

a

c

b

c

a

c

( ) = ( ) − ( )

= ( ) + ( )

∫ ∫ ∫

∫ ∫


c

b

c

a

b

( ) = ( ) + ( )∫ ∫ ∫

2 2× .

+

–

+

y=25int

3π 2π2

5π2

π π20

–2

2

y

t

Figure DS-5

Figure DS-6

x 2

incr.'

"

decr.

+ –

Concave

upward

Concave

downward

Figure DS-7

The graph of g indicates that a relative maxi-mum occurs at x = 2, g is not differentiable atx = 6, since there is a cusp at x = 6 and g doesnot have a point of inflection at x = −2, sincethere is no tangent line at x = −2.

Thus, only statement I is true.

79. From Chapter 4

See Figure DS-7.

Thus, g has a local minimum at x = 2π.

You can also approach the problem geometri-cally by looking at the area under the curve. SeeFigure DS-5.

The graph of f is concave upward for x < x2.

7/23/2019 007137714 x


Slope of normal = negative reciprocal of slope of

tangent

At x = 2, y = x3 = 23 = 8; (2, 8)


84. From Chapter 4

See Figure DS-10.

Since g x f t dt g x f xa

x

( ) = ( ) ′( ) = ( )∫ , .

⇒ = − −( ) +

= − +

y x

y x

1

122 8

1

12

49

6

or

.

y x− = − −( )81

122

= −1

2

f x dy

dx x x

x

′( ) = ⇒ = = ⇒

= ⇒ =

2 212 3 12

4 2


80. From Chapter 4

See Figure DS-8.

Figure DS-8

Figure DS-9

[ [a b0

rel. max.

0+ –

incr. decr.

g' ( x )= f ( x )

g( x )

Figure DS-10

Figure DS-11

Enter y1 = sin(x2)

Using the Inflection function of your calculator,you obtain four points of inflection on [0, π].The points of inflection occur at x = 0.81, 1.81,2.52, and 3.07. Since y1 = sin(x2), is an evenfunction, there is a total of eight points of inflec-

tion on [−π, π]. An alternate solution is to enterThe graph of y2 indicates

that there are eight zeros on [−π, π].

81. From Chapter 6

See Figure DS-9.

v t t

t

a t v t t t

( ) = − +

( ) = ′( ) = −

3

2

2

32 5

4

y d

dx y x x2

2

2 1 2= ( )( ), , .

The graph indicates that the maximum accelera-tion occurs at the endpoint t = 6. a(t ) = t 2 − 4t and a(6) = 62 − 4(6) = 12

82. From Chapter 10

Amount of Water Leaked =

Using your calculator, you obtain 10(6 ln 6 − 5)which is approximately 57.506 gallons.

83. From Chapter 6

y x x dy

dx x= ≥ =3 20 3, ;

10 10

5

ln t dt +( )∫

The only graph that satisfies the behavior of g ischoice (A)

85. From Chapter 9

See Figure DS-11.

Using the Intersection function of the calculator,you obtain the intersection points atx = 0.785398, 3.92699, and 7.06858.

Area = −( )

+ −( )

= + ≈

∫

∫

sin cos

cos sin

. . .

.

.

.

.

x x dx

x x dx

0 785398

3 92699

3 92699

7 06858

2 82843 2 82843 5 65685

7/23/2019 007137714 x


You can also find the area by:

86. From Chapter 10

87. From Chapter 6

88. From Chapter 7

See Figure DS-12.

Thus f π π π π π

π

2 180 2 180 23

3180

2 98255 2 983

+

≈ − +

+ +

≈ − ≈ ≈. .

f

f s

x

y x

y x

π π

π

π

π

π

23

23

21

21

3 12

23

= ⇒

′

= − ⇒

= −

− = − −

= − + +

,

.

:

is on the graph

lope of the tangent at

is

Equation of tangent line or

dydx

ky y y e

y y t

y y e e

e k

k k

k

kt

k k

k

= ⇒ =

= ⇒ = =

= ⇒ =

= ( ) ⇒

=

− = ⇒ − =

= −

( )

0

0

0 0

5730 5730

5730

57301

25730

1

2

1

2

1

2

1

25730

1 2 5730 2 5730

2

5730

Half-life when

Thus,

ln ln ln

ln ln ln

ln

Area = − ≈

≈

∫ sin cos .

. .

.

.

x x dx785398

7 06858

5 65685

5 657

Using your calculator, you obtain:

Volume of solid ≈ (0.478306)π ≈ 1.50264≈ 1.503

89. From Chapter 4

See Figure DS-13.

Volume of solid = ( ) −

∫ π π

π

sin 21

2

2

2

2

52

x dx

To find the points of intersection, set

or or

sin sin21

22

1

2

26

25

6 12

5

12

1x x

x x x x

= ⇒ =

⇒ = = ⇒ = =

−

π π π π


Figure DS-12

[ [0 ba

incr incr

+ –

Concave

upward

Concave

downward

q'

q"

q

Figure DS-13

A change of concavity occurs at x = 0 for q.Thus, q has a point of inflection at x = 0. None

of the other functions has a point of inflection.

90. From Chapter 6

Using the Solve function on your calculator, youobtain x ≈ 1.37015 ≈ 1.370.

91. From Chapter 9

Length of a rectangle = −

=6 0

32

Perpendicular tangents ⇒ ′( )( )

= −

⇒

−

−( ) = −

f x dy

dx

x

x

x x

1

12 1

2 2

ln

f x x

x f x x

x x

x

x

x

x

y x dy

dx x

( ) = ′( ) = ( )( ) − ( )

= −

= − = −

ln;

ln

ln

;

1 1

1

2

2

2 2

2

7/23/2019 007137714 x



10–1 2 x

y y= x+2

y= x 2

Figure DS-14

Midpoints are x = 1, 3 and 5 and f (1) = 2, f (3) =10 and f (5) = 26

92. From Chapter 10

See Figure DS-14.

To find points of intersection, set x2 = x + 2 ⇒x2 − x − 2 = 0 ⇒ x = 2 or x = −1

Volume of solid,

Using your calculator, you obtain: V =81

10

V x x dx= + −( )−∫ 2 2 2

1

2

Area of cross section = +( ) −( )x x2 2 2

f x dx( ) ≈ + +( ) ≈ ( ) =∫ 2 2 10 26 2 38 760

6

7/23/2019 007137714 x


(c) The amount of oil in the drum at t

(d) Let a be the value of t:

3. (a)

(b)

Set f ″ (x) = 0 ⇒ 4e2x (1 + x) = 0

Since e2x > 0, thus 1 + x = 0 or x = −1.

(c) since xe2x increases without

bound as x approaches ∞.

lim limx

x

x xxe

x

e→−∞ →−∞ −=2

2

lim ,x

xxe→−∞

= ∞2

′( ) = +

( ) = ( ) + + ( )( )

= + + = += +( )

f x e xe

f x e e x e

e e xe e xe

e x

x x

x x x

x x x x x

x

2 2

2 2 2

2 2 2 2 2

2

2

2 2 2 2

2 2 4 4 4

4 1

″

f x xe

f x e x e e xe

e x

f x e x

e x x

x

x x x x

x

x

x

( ) =′( ) = + ( )( ) = +

= +( )

′( ) = ⇒ +( ) =

> + = = −

2

2 2 2 2

2

2

2

2 2

1 2

0 1 2 0

0 1 2 0 0 5

Set

Since thus or

.

, .

cos

cos .

cos . .

. .

a

a

a

a

a

ππ

π

π π

π

π

12120 60

12

12

2

20 570796

120 570796 2 17827

2 17827 12 8 32038

8

1

( ) = −( )( )

( ) = − +( ) ≈ −

≈ −( ) ≈

= ( )( ) ≈

≈

−

minutes

100 10 12 40

100 12012

40

100 12012

120 0 40

100 12012

120 40

12012

120 40 100

0

0

− ( ) =

− −( ) ( )[ ] =

− −( ) ( ){− −( ) ( )[ ]} =

+ ( ) ( ) − ( ) =

( ) ( ) = ( ) + −

∫ sin

cos

cos

cos

cos

cos

π

π π

π π

π

π π

π

π π

π

t dt

t

a

a

a

a

a

f t x dxt

( ) = − ( )∫ 100 10120

1

sin πPart A—Calculators are permitted.

1. (a)

Equation of tangent line:

or

(b)

(c)

Since the point (2, 4) is on the graph of f,

(d)

2. (a) The amount of oil leaked out after 6 minutes

(b) Average amount of oil leaked out per minutefrom t = 0 to t = 6:

=− ( ) = ( )

= ≈

∫ 1

6 010

12

1

6120

6 3662 6

0

6

sin

.

ππ

t dt

gallons.

= ( )

=− ( )

= − ( )[ ]= ≈ ≈

∫ 1012

1012

12120

12

120 38 1972 38

0

6

0

6

0

6

sin

cos

cos

.

π

π

π

π π

π

t dt

t

t

gallons gallons.

y x x

f

= + +

( ) = ( ) + + =

≈ ≈

2 6

2 1 2 2 1 2 1 6 16 92

4 11339 4 113

2

2

. . . .

. .

y x x= + +2 62

2 4 1

2 4 1

2 2 4

4 2 2 2 6

2 6 2 6

2 2

2 2

2 2 2

ydy x dx

ydy x dx

y x x c f

c c

y x x y x x

= +( )= +( )

= + + ( ) =

= ( ) + + ⇒ =

= + + = ± + +

∫ ∫ ;

Thus or

f 2 19

82 1 2 4

0 9

84

4 1125 4 113

. ..

. .

( ) ≈ −( ) + ≈ +

≈ ≈

y x= −( ) +9

82 4

y x− = −( )49

82

dy

dx

x

y

dydx

= +

( )

= ( ) +( ) =( )

4 1

22 4

4 2 12 4

98

2 4

; ,

,


Solutions to Diagnostic Test—Section II

7/23/2019 007137714 x



(b) Since f decreases on (−4, −2) and increases on(−2, 2), f has a relative minimum at x = −2.And f decreases on (2, 5) and increases on(5, 6), f has a relative minimum at x = 5.

(c) See Figure DS-16.

Since f ′ is increasing on the intervals (−4, 0)and (4, 6), f ″ > 0 on (−4, 0) and (4, 6).

(d) A change of concavity occurs at x = 0 and atx = 4. (See Figure DS-16).

Thus f has a point of inflection at x = 0 and atx = 4.

as x → −∞, the numerator → −∞

as x → −∞, the denominator e−2x → ∞

However, the denominator increases at amuch greater rate and thus

(d) Since as x → ∞, xe2x increases without bound,f has no absolute maximum value. From part(a) f (x) has one critical point at x = −0.5.Since f ′(x) = e2x (1 + 2x), f ′(x) < 0 for x < −0.5and f ′(x) > 0 for x > −0.5, thus f has a relativeminimum at x = −0.5, and it is the absoluteminimum because x = −0.5 is the only criticalpoint on an open interval. The absolute

minimum value is

(e)

Thus is the only critical point, and

f has an absolute minimum at .

The absolute minimum value of f is forall a > 0.

Part B—No calculators

4. (a) See Figure DS-15.

−1

ae

f a a

ea e

aea a− = − = − = −−

−1 1 1 11

1 .

xa

= −1

xa

= −1

f x xe a

f x e x e a e axe

e ax

f x e ax xa

xa

f x xa

f x

ax

ax ax ax ax

ax

ax

( ) = >

′( ) = + ( )( ) = +

= +( )

′( ) = ⇒ +( ) = = −

< − ′( ) < − ′( ) >

,

.

, , .

0

1

0 1 01

10

10

Set or If

and if >

− = −−( )0 51

2

2 0 5.

.e

e

lim .x

xxe→−∞

=2 0

[ [

+0 0 0 +––'

incr. incr.decr. decr.

rel. min rel. minrel. max

–4 –2 2 5 6

Figure DS-15

[ [0 4 6–4

incr.'

"

decr. incr.

+ +–

Concave

upward

Concave

upward

Concave

downward

Figure DS-16

(–2,3)

0 1 2 3 4 5 6–1–2–3–4

y

f

x

Figure DS-17

(e) See Figure DS-17.

5. See Figure DS-18 on page 38.

(a)

(b) Volume of solid = ( ) =

= ( )

=

∫ π π

π π

x dx x3 2

7

0

2

0

2

7

7

2

7

128

7

Area of R x dx x

= =

= − =∫ 3

4

0

2

0

2 4

4

2

40 4

Since f increases on (−2, 2) and decreases on(2, 5), f has a relative maximum at x = 2

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(c)

(d)

6. (a) x2y2 = 4

Differentiating using product and chain rules:

(b)

(c) Slope of normal

slope of tangent

= −

= −1 1

2

Equation of tangent

or

:

.

y x

y x

y x

− −( ) = −( )

⇒ + = −

= −

2 2 1

2 2 2

2 4

dy

dx1 2

2

12

,−( )

= − −( )

=

2 2 0

2 2

22

2 2

2 2

2

2

xy x y dy

dx

x y dy

dx xy

dydx

xyx y

yx

+ ( ) =

= −

= − = −

Volume of solid = =

=∫ x dx x6

7

0

2

0

2

7

128

7.

Area of cross section = ( ) =x x3 2 6

π π

π π π π

x dx

x a

a a

a

a

3 2

0

7

0

7

7 6

1

2

128

7

7

64

7 7

64

7

64 2 26

7

( ) =

= =

= = =

∫

; ;

;

(d)

Possible Points for P are: (2, 1), (2, −1), (−2, 1),and (−2, −1).

Since the only points to which the

tangent line has a slope of are (2, −1) and

(−2, 1), since the x and y coordinates must

have opposite signs.

1

2

dy

dx

y

x=

−,

If y x y x

x x

= − = ⇒ −( ) = ⇒

= ⇒ = ±

1 4 1 4

4 2

2 2 2 2

2

,

y x m dy

dx

y

x

y

x y x

x y x y

y y y y

y y

y

y x y x

x x

= + = = −

− = ⇒ − =

= = −

−( ) = =

= =

= ±

= = ⇒ ( ) = ⇒

= ⇒ = ±

1

22

1

2

1

22

4 2

2 4 4 4

4 4 1

1

1 4 1 4

4 2

2 2

2 2 2 2

4 4

2 2 2 2

2

;

;

; ;

;

,

and

Set

substitute

If

Equation of normal

or

: y x

y x

y x

− −( ) = − −( )

⇒ + = − +

= − −

21

21

21

2

1

2

1

2

3

2

0 2

y

x

y= x 3

x =2

Figure DS-18

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Section I—Part A

× 1.2 =No. Correct Subtotal A

× (0.25) × 1.2 =No. Incorrect Subtotal B

Part A (Subtotal A − Subtotal B) =Subtotal C

Section I—Part B

× 1.2 =No. Correct Subtotal D

× (0.25) × 1.2 =No. Incorrect Subtotal E

Part B (Subtotal D − Subtotal E) =Subtotal F

Section II—Part A (Each question is worth 9 points.)

+ + =Q#1 Q#2 Q#3 Subtotal G

Section II—Part B (Each question is worth 9 points.)

+ + =Q#1 Q#2 Q#3 Subtotal H

Total Raw Score (Subtotals C F G H) =


SCORING AND INTERPRETATION

Scoring Sheet for Diagnostic Test

Approximate Conversion Scale:

Total Raw Score Approximate AP Grade

75–108 5

60–74 4

45–59 3

31–44 2

0–30 1

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COMPREHENSIVE REVIEW

PART III


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1.1 LINES

Main Concepts: Slope of a Line, Equations of a Line, Parallel and Perpendicular Lines

Slope of a Line

Given two points A (x1, y1) and B (x2, y2), the slope of the line passing through the twogiven points is defined as

Note that if (x2 − x1) = 0, then x2 = x1 which implies that points A and B are on a verticalline parallel to the y-axis and thus, the slope is undefined.

Example 1

Find the slope of the line passing through the points (3, 2) and (5, −4). Using the definition

, the slope of the line is

Example 2

Find the slope of the line passing through the points (−5, 3) and (2, 3). The slope

This implies that the points (−5, 3) and (2, 3) are

on a horizontal line parallel to the x-axis.

m = −

− −( ) =

+ = =

3 3

2 5

0

2 5

0

70.

m = − −

− =

−= −

4 2

5 3

6

23.m

y y

x x=

−−

2 1

2 1

m y y

x x x x=

−−

−( ) ≠2 1

2 1

2 1 0where

Chapter 1

Review of Pre-Calculus

43Copyright 2002 The McGraw-Hill Companies, Inc. Click Here for Terms of Use.

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44 • Comprehensive Review

Equations of Lines

y = mx + b Slope-intercept form of a line where m is its slope and b isy-intercept.

y − y1 = m(x − x1) Point-slope form of a line where m is the slope and (x1, y1) isa point on the line.

Ax + By + C = 0 General form of a line where A, B and C are constants andA and B are not both equal to 0.

Example 1

Write an equation of the line through the points (−2, 1) and (3, −9). The slope of line

passing through (−2, 1) and (3, −9) is Using the point-

slope form and the point (−2, 1),

y − 1 = −2[x − (−2)]y − 1 = −2(x + 2) or y = −2x − 3

An equation of the line is y = −2x − 3.

Example 2

An equation of a line l is 2x + 3y = 12. Find the slope, the x-intercept and the y-interceptof line l.

Begin by expressing the equation 2x + 3y = 12 in slope-intercept form.

Therefore, m, the slope of line l, is and b, the y-intercept, is 4. To find the x-intercept,

set y = 0 in the original equation 2x + 3y = 12. Thus, 2x + 0 = 12 and x = 6. Thex-intercept of line l is 6.

Example 3

Equations of vertical and horizontal lines involve only a single variable. Here are sev-eral examples:

−2

3

2 3 12

3 2 12

2

34

x y

y x

y x

+ =

= − +

= −

+

m = − −

− −( ) =

−= −

9 1

3 2

10

52.

y

0

y

x 0

y

x 0

y

x 0

m>0 m<0 m=0 m is undefined.

Horizontal line Vertical line

Parallel to x-axis Parallel to y-axis

Figure 1.1-1

Example 3

Here is a summary of 4 different orientations of lines and their slopes:

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Parallel and Perpendicular Lines

Given two non-vertical lines l 1 and l 2 with slopes m1and m2 respectively.

Lines l 1 and l 2 are parallel if and only if m1 = m2.

Lines l 1 and l 2 are perpendicular if and only ifm1m2 = −1.

Example 1

Write an equation of the line through the point (−1, 3) and parallel to the line 3x − 2y = 6.

Review of Pre-Calculus • 45

Figure 1.1-2

2 x –4

y

0

x=2 x= – 4

y

x0

2

– 3

y=2

y= – 3

l1

y

l2

x0

Figure 1.1-3

l2

0 x

y

l1

Figure 1.1-4

Figure 1.1-5

3 2 6

2 3 6

3

2

6

2

3

23

x y

y x

y x

y x

− =− = − +

= −−

+−

= −

Begin by expressing 3x − 2y = 6 inslope-intercept form.

–3

x

y

0

3x – 2y=6

(–1,3) .

• Write clearly. If the reader (the person grading your test) can’t read your hand-writing, you’re in trouble.

Therefore, the slope of the line 3x − 2y = 6 is m = and the slope of the line par-3

2

allel to the line 3x − 2y = 6 is also . Since the line parallel to 3x − 2y = 6 passes32

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Figure 1.1-6

Figure 1.1-7

Example 3

Write an equation of the circle with center at C(−2, 1) and tangent to the line l havingthe equation x + y = 5.

.

y

x

y=x+1

(1,2)

A(3,0)

B( – 1,4)

.

.

0

Begin by finding the midpoint of AB— .

Midpoint = The

slope of AB—

is Therefore,

the perpendicular bisector of AB—

has a slopeof 1. Since the perpendicular bisector of AB

—

passes through the midpoint, you could use the point-slope formand obtain y − 2 = 1 (x − 1) ory − 2 = x − 1 or y = x + 1.

m = −− −

= −4 0

1 31.

3 1

2

0 4

21 2

+ −( ) +

= ( ), , .

M

x

y

0

C( – 2,1)

x + y=5Let M be the point of tangency. Express theequation x + y = 5 in slope-intercept form andobtain y = −x + 5. Thus, the slope of line l is

−1. Since is a radius drawn to the point of

tangency, it is perpendicular to line l, and theslope of is 1. Using the point-slope for-

mula, the equation of is y − 1 = 1 [x −(−2)] or y = x + 3. To find the coordinates of point M, solve the two equations y = −x + 5and y = x + 3 simultaneously. Thus, −x + 5 =x + 3 which is equivalent to 2 = 2x or x = 1.

CM

CM

CM

Substituting x = 1 into y = x + 3, you have y = 4. Therefore, the coordinates of M are

(1, 4). Since is the radius of the circle, you should find the length of by

using the distance formula

Thus, Now that you know both the radius

of the circle and its center, (−2, 1), use the formula (x − h)2 + (y − k)2 = r2 to

find an equation of the circle. Thus, an equation of the circle is (x − (−2))2 + (y − 1)2 = 18or (x + 2)2 + (y − 1)2 = 18.

• Even functions are symmetrical with respect to the y-axis. That means you onlyneed to do one side of the graph and then reflect it to get the other side.

r =( )18

CM = − −( )[ ] + −( ) =1 2 4 1 182 2

.

d x x y y= −( ) + −( )2 1

2

2 1

2

.

CMCM

through the point (−1, 3), you can use the point-slope form and obtain the equation

Example 2

Write an equation of the perpendicular bisector of the line segment joining the pointsA(3, 0) and B(−1, 4).

y x y x− = − −( )[ ] − = +( )33

21 3

3

21or .

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1.2 ABSOLUTE VALUES AND INEQUALITIES

Main Concepts: Absolute Values, Inequalities and The Real Number Line, Solving Absolute Value Inequalities, Solving Polynomial Inequalities,Solving Rational Inequalities

Absolute Values

Let a and b be real numbers.

1. 3.

2. 4.

Example 1

Solve for x: 3x − 12 = 18.Depending on whether the value of (3x − 12) is positive or negative, the equation3x − 12 = 18 could be written as 3x − 12 = 18 or 3x − 12 = −18. Solving both equa-tions, you have x = 10 or x = −2. (Be sure to check both answers in the original equa-tion.) The solution set for x is {−2, 10}.

Example 2

Solve for x: 2x − 12 = 4x + 24 .The given equation implies that either 2x − 12 = 4x + 24 or 2x − 12 = −(4x + 24).

Solving both equations, you have x = −18 or x = −2. Checking x = −18 with the originalequation: 2(−18) − 12 = 4(−18) + 24 or −36 − 12 = −72 + 24 or −48 = −48.Checking x = −2 with the original equation, you have −16 = 16. Thus, the solution set

for x is {−18, −2}.

Example 3

Solve for x: 11 − 3x = 1 − x.Depending on the value of (11 − 3x)—whether it is greater than or less than 0—the

given equation could be written as 11 − 3x = 1 − x or 11 − 3x = −(1 − x). Solving bothequations, you have x = 5 and x = 3. Checking x = 5 with the original equation yields11 − 3(5) = 1 − 5 or −4 = −4 which is not possible. Checking x = 3 with the originalequation shows that 11 − 3(3) = 1 − 3 or −2 = −2 which is also not possible. Thus,the solution for x is the empty set { }. You could also solve the equation using a graph-ing calculator.

a a a a

a a2 0

0= = ≥

− <

,,

ifif

ab a b=

a b b a− = −a a a

a a= ≥

− <

,,

ifif

00


[–10,10] by [–10,10]

Figure 1.2-1

Enter y1 = 11 − 3x and y2 = 1 − x. The twographs do not intersect, thus, no commonsolution.

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Interval Notation Set Notation Graph

Inequalities and the Real Number Line

Properties of Inequalities:Let a, b, c, d, and k be real numbers:

1. If a < b and b < c, then a < c. 3. If a < b and k > 0, then ak < bk.e.g., −7 < 2 and 2 < 5 ⇒ −7 < 5 e.g., 3 < 5 and 2 > 0 ⇒ 3(2) < 5(2)

2. If a < b and c < d, then a + c < b + d. 4. If a < b and k < 0, then ak > bk.

e.g., 5 < 7 and 3 < 6 ⇒ 5 + 3 < 7 + 6 e.g., 3 < 5 and 2 < 0 ⇒ 3(−2) > 5(−2)

Example 1

Solve the inequality 6 − 2x ≤ 18 and sketch the solution on the real number line.

Solving the inequality 6 − 2x ≤ 18−2x ≤ 12

x ≥ −6

Therefore, the solution set is the interval [−6, ∞) or expressing the solution set in setnotation: {xx ≥ −6}.

Example 2

Solve the double inequality −15 ≤ 3x + 6 < 9 and sketch the solution on the real numberline.

Solving the double inequality −15 ≤ 3x + 6 < 9−21 ≤ 3x < 3

−7 ≤ x < 1

Therefore, the solution set is the interval [−7, 1) or expressing the solution in set notation:{x −7 ≤ x < 1}.

Example 3

Here is a summary of the different types of intervals on a number line:


[–7 1

)

[ ]a b

( )a b

[ )a b

( ]a b

[a

(a

]b

)b

{x a ≤ x ≤ b}

{x a < x < b}

{x a ≤ x < b}

{x a < x ≤ b}

{x x ≥ a}

{x x > a}

{x x ≤ b}

{x x < b}

{x x is a real number}

[a, b]

(a, b)

[a, b)

(a, b]

[a, ∞)

(a, ∞)

(−∞, b]

(−∞, b)

(−∞, ∞)

[–6 0

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Solving Absolute Value Inequalities

Let a be a real number such that a ≥ 0

x ≥ a ⇔ (x ≥ a or x ≤ −a) and x > a ⇔ (x > a or x < a)

x ≤ a ⇔ (−a ≤ x ≤ a) and x < a ⇔ (−a < x < a)

Example 1

Solve the inequality 3x − 6 ≤ 15 and sketch the solution on the real number line. Thegiven inequality is equivalent to

−15 ≤ 3x − 6 ≤ 15−9 ≤ 3x ≤ 21−3 ≤ x ≤ 7

Therefore, the solution set is the interval [−3, 7] or written in set notation {x −3 ≤x ≤ 7}.

Example 2

Solve the inequality 2x + 1 > 9 and sketch the solution on the real number line. Theinequality 2x + 1 > 9 implies that

2x + 1 > 9 or 2x + 1 < −9

Solving the two inequalities on the above line, you have x > 4 or x < −5. Therefore, thesolution set is the union of the two disjoint intervals (4, ∞) ∪ (−∞, −5) or writing the solu-tion in set notation: {x (x > 4) or (x < −5)}.

Example 3

Solve the inequality 1 − 2x ≤ 7 and sketch the solution on the real number line. Theinequality 1 − 2x ≤ 7 implies that

−7 ≤ 1 − 2x ≤ 7−8 ≤ −2x ≤ 6

4 ≥ x ≥ −3−3 ≤ x ≤ 4

Therefore, the solution set is the interval [−3, 4] or writing the solution in set notation:{x −3 ≤ x ≤ 4}.


[–7.9,7.9] by [–5,10]

Figure 1.2-2

Note: You can solve an absolute value inequality byusing a graphing calculator. For instance, in example 3,enter y1 = 1 − 2x and y2 = 7. The graphs intersect atx = −3 and 4, and y1 is below y2 on the interval (−3, 4).Since the inequality is ≤, the solution set is [−3, 4].

[ ]–3 7

( )–5 4

[ ]–3 4

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• Label everything in a diagram including axes, origin, function, lines, intercepts(if appropriate), intersection points, and special points.

Example 2

Solve the inequality x3 − 9x < 0 using a graphing calculator.

Solving Polynomial Inequalities

1. Write the given inequality in standard form with the polynomial on the left and zerothe right.

2. Factor the polynomial if possible.3. Find all zeros of the polynomial.4. Using the zeros on a number line, determine the test intervals.

5. Select an x-value from each interval and substitute it in the polynomial.6. Check the endpoints of each interval with the inequality.7. Write the solution to the inequality.

Example 1

Solve the inequality x2 − 3x ≥ 4.

Steps: 1. Write in standard form: x2 − 3x − 4 ≥ 02. Factor polynomial: (x − 4)(x + 1)3. Find zeros: (x − 4)(x + 1) = 0 implies that x = 4 and x = −1

4. Determine intervals:(−∞, −1) and (−1, 4) and (4, ∞)

5. Select an x-value in each interval and evaluated polynomial at that value:


Figure 1.2-3

Selected Factor Factor PolynomialInterval x-value ( x + 1) ( x − 4) ( x − 4)( x + 1)

(−∞, −1) −2 − − +

(−1, 4) 0 + − −

(4, ∞) 6 + + +

Therefore the intervals (−∞, −1) and (4, ∞) make (x − 4)(x + 1) > 0.

6. Check end-points:Since the inequality x2 − 3x − 4 ≥ 0 is greater than or equal to 0, both end-points x = −1 and x = 4 are included in the solution.

7. Write solution: The solution is (−∞, −1] ∪ [4, ∞).

[–8,8] by [–5,5]

Note: The inequality x2 − 3x ≥ 4 could have beensolved using a graphing calculator. Enter y1 = x2 − 3xand y2 = 4. The graph of y1 is above y2 on (−∞, −1)and (4, ∞). Since the inequality is ≥, the solution set is

(−∞, −1] or [4, ∞).

–1 4

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Therefore, the intervals (−∞, −3) and (0, 3) make x(x − 3)(x + 3) < 0.

8. Check end-points:Since the inequality x3 − 9x < 0 is strictly less than 0, none of the endpointsx = −3, 0 and 3 are included in the solution.

9. Write the solution: The solution is (−∞, −3) ∪ (0, 3).

Solving Rational Inequalities

1. Rewrite the given inequality so that all the terms are on the left and only zero is onthe right.

2. Find the least common denominator and combine all the terms on the left into asingle fraction.

3. Factor the numerator and the denominator if possible.4. Find all x-values for which the numerator or the denominator is zero.5. Putting these x-values on a number line; determine the test intervals.6. Select an x-value from each interval and substitute it in the fraction.7. Check the endpoints of each interval with the inequality.8. Write the solution to the inequality.


Selected Factor Factor Factor PolynomialInterval x-value x ( x + 3) ( x − 3) x( x − 3)( x + 3)

(−∞, −3) −5 − − − −

(−3, 0) −1 − + − +

(0, 3) 1 + + − −(3, ∞) 6 + + + +

Figure 1.2-4

[–10,10] by [–15,15]

Steps: 1. Enter y = x3 − 9x into yourgraphing calculator.

2. Find the zeros of y: x = −3, 0and 3.

3. Determine the intervals onwhich y < 0: (−∞, −3) and (0, 3).

4. Check if the end-points satisfythe inequality. Since the inequal-

ity is strictly less than 0, the end-points are not included in thesolution.

5. Write the solution to the in-equality. The solution is (−∞, −3)∪ (0, 3).

Example 3

Solve the above inequality x3 − 9x < 0 algebraically.

Steps: 1. Write in standard form: x3 − 9x < 0 is already in standard form.2. Factor polynomial: x(x − 3)(x + 3)

3. Find zeros: x(x − 3)(x + 3) = 0 implies that x = 0, x = 3 and x = −3.4. Determine intervals:

(−∞, −3), (−3, 0), (0, 3) and (3, ∞)

5. Select an x-value and evaluate polynomial:

0 3–3

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Example 3

Solve the inequality ≥ x using a graphing calculator.1

x

Example 1

Solve the inequality

Steps: 1. Rewrite:

2. Combine:

3. Set numerator and denominator equal to 0 and solve for x: x = 2 or 3.

4. Determine intervals:(−∞, 2), (2, 3) and (3, ∞)

5. Select x-value and evaluate fraction:

2 5 3

3

02

3

0x x

x

x

x

− − +

−

≤ ⇔ −

−

≤

2 5

31 0

x

x

−−

− ≤

2 5

31

x

x

−−

≤ .


Selected

Fraction

Interval x-value ( x − 2) ( x − 3)

(−∞, 2) 0 − − +(2, 3) 2.5 + − −

(3, ∞) 6 + + +

x

x

−−

2

3

Therefore, the interval (2, 3) makes the fraction < 0.

6. Check endpoints:At x = 3, the fraction is undefined. Thus the only endpoint is x = 2. Sincethe inequality is less than or equal to 0, x = 2 is included in the solution.

7. Write solution: The solution is the interval [2, 3).

Example 2

Solve the inequality ≤ 1 using a graphing calculator.2 5

3

x

x

−−

[–7.9,79] by [–3.8,3.8]

Figure 1.2-5

1. Enter y1 = and y2 = 1.

2. Find the intersection points: x = 2. Note that atx = 3, y1 is undefined.)

3. Determine the intervals on which y1 is belowy2: The interval is (2, 3)

4. Check if the endpoints satisfy the inequality.Since the inequality is less than or equal to 1,the endpoint at x = 2 is included in the solution.

5. Write the solution to the inequality. The solutionis the interval [2, 3).

2 5

3

x

x

−−

2 3

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[–8,8] by [–4,4]

Figure 1.2-6

3

0 3 x

y

–3

Figure 1.3-1

1. Enter y1 = and y2 = x.

2. Find the intersection points: at x = −1and 1. Note that x = 0, y1 is undefined.)

3. Determine the intervals on which y1 ≥y2. The interval are (−∞, −1) and (0, 1).

4. Check if the end-points satisfy the in-equality. Since the inequality is greater

than or equal to x, the endpoints at x =−1 and 1 are included in the solution.5. Write the solution to the inequality. The

solution is the interval (−∞, −1] ∪ (0, 1].

1

x

1.3 FUNCTIONS

Main Concepts: Definition of a Function, Operations on Functions, Inverse Functions,Trigonometric and Inverse Trigonometric Functions, Exponential and Logarithmic Functions

Definition of a Function

A function f is a set of ordered pairs (x, y) in which for every x coordinate there is oneand only one corresponding y coordinate. We write f (x) = y.

The domain of f is the set of all possible values of x and the range of f is the set of allvalues of y.

Vertical Line Test

If all vertical lines pass through the graph of an equation in at most one point, then the

equation is a function.

Example 1

Given y = , where √ denotes the positive square root, sketch the graph of theequation, determine if the equation is a function and find the domain and range of theequation.

9 2− x

Since the graph of y = passes the verti-cal line test, the equation is a function. Let y =

f (x). The expression implies that 9 − x2

≥ 0. By inspection, note that −3 ≤ x ≤ 3. Thus thedomain is [−3, 3]. Since f (x) is defined for all val-ues of x ∈ [−3, 3] and f (−3) = 0 is the minimumvalue and f (0) = 3 is the maximum value, therange of f (x) is [0, 3].

9

2

− x

9 2− x

Example 2

Given f (x) = x2 − 4x, find f (−3), f (−x) andf x h f x

h

+( ) − ( ).

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f x h f x

h

x h x h x x

h

x hx h x h x x

hhx h h

h x h

+( ) − ( )=

+( ) − +( ) − −( )

= + + − − − +

= + −

= + −

2 2

2 2 2

2

4 4

2 4 4 4

2 42 4.

f

f x x x x x

−( ) = −( ) − −( ) = + =

−( ) = −( ) − −( ) = +

3 3 4 3 9 12 21

4 4

2

2 2


• Given the function is discontinuous at x = 3,

but the lim .x

f x→

( ) =3

6

f x

x

x if x

if x

f x( ) =−−

≠

=

( )

2 9

33

0 3

,

Operations on FunctionsLet f and g be two given functions. Then for all x in the intersection of the domains of f and g, the sum, difference, product, and quotient of f and g are defined as:

The composition of f with g is:

(f g )(x) = f [ g (x)]

where the domain of f g is the set containing all x in the domain of g for which g (x) isin the domain of f.

Example 1

Given f (x) = x2 − 4 and g (x) = x − 5, find

(a) (f g )(−1) (b) ( g f )(−1) (c) (f + g )(−3) (d) (f − g )(1)

(e) (fg )(2) (f ) (0) (g) (5) (h) (4).

(a) (f g (x) = f [ g (x)] = f (x − 5) = (x − 5)2 −4 = x2 − 10x + 21.Thus (f g )(−1) = (−1)2 − 10(−1) + 21 = 1 + 10 + 21 = 32.

Or (f g )(−1) = f [ g (−1)] = f (−6) = 32

(b) ( g f )(x) = g [f (x)] = g (x2 − 4) = (x2 − 4) − 5 = x2 − 9.Thus ( g f )(−1) = (−1)2 − 9 = 1 − 9 = −8.

(c) (f + g )(x) = (x2 − 4) + (x − 5) = x2 + x − 9. Thus (f + g )(−3) = −3.

(d) (f − g )(x = (x2 − 4) − (x − 5) = x2 − x + 1. Thus (f − g )(1) = 1.

(e) (fg (x) = (x2 − 4)(x − 5) = x3 − 5x2 − 4x + 20. Thus (fg )(2) = 0.

g

f

f

g

f

g

f g x f x g x

f g x f x g x

fg x f x g x

f

g x

f x

g x g x

+( )( ) = ( ) + ( )

−( )( ) = ( ) − ( )

( )( ) = ( ) ( )

( ) =

( )( )

( ) ≠, 0

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(f) (x) = , x ≠ 5. Thus (0) =

(g) Since g (5) = 0, x = 5 is not in the domain of , (5) is undefined.

(h) (x) = , x ≠ 2 or −2. Thus (4) =

Example 2

Given h(x) = and k(x) =

(a) find (x) and indicate its domain; and

(b) find (x) and indicate its domain.

(a) (x) =

The domain of h(x) is [0, ∞) and the domain of k(x) is [−3, 3].

The intersection of the two domains is [0, 3]. However, k(3) = 0.

Therefore the domain of is [0, 3).

Note that is not equivalent to outside of the domain [0, 3).

(b) (x) =

The intersection of the two domains is [0, 3]. However, h(0) = 0.

Therefore the domain of is (0, 3].

Example 3

Given the graphs of functions f (x) and g (x):

h

k

9 2− x

x

h

k

x

x9 2−x

x9 2−

h

k

x

x92

−

h

k

h

k

h

k

9 2− x :x

−1

12.

g

f

x

x

−−

5

42

g

f

f

g

f

g

4

5.

f

g

x

x

2 4

5

−−

f

g


1 0

1

2

2

3

3

x

f(x)

Figure 1.3-2

1 0

1

2

2

3

3

x

g(x)

Figure 1.3-3

Find (a) (f + g )(1) (b) (fg )(0) (c) (0) (d) f [ g (3)]

(a) (f + g )(1) = f (1) + g (1) = 3. (b) (fg )(0) = f (0) g (0) = 3(0) = 0

f

g

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(c) (0) = undefined.

(d) f [ g (3)] = f (1) = 1.

Inverse Functions

Given a function f, the inverse of f (if it exists) is a function g such that f ( g (x)) = x forevery x in the domain of g and g (f (x)) = x for every x in the domain of f. The function

g is written as f −1. Thus: f [f −1(x)] = x and f −1[f (x)] = x.

The graphs of f and f −1 are reflections of each other inthe line y = x. The point (a, b) is on the graph of f if andonly if the point (b, a) is on the graph of f −1.

A function f is one-to-one if for any two points x1 and x2 in the domain such that x1 ≠ x2,then f (x1) ≠ f (x2).

Equivalent Statements: Given a function f,1. The function f has an inverse.2. The function f is one-to-one.3. Every horizontal line passes through the graph of f at most once.

Finding the inverse of a function f:1. Check if f has an inverse, i.e., f is one-to-one or passes the horizontal line test.2. Replace f (x) by y.3. Interchange the variables x and y.

4. Solve for y.5. Replace y by f −1(x).6. Indicate the domain of f −1(x) as the range of f (x).7. Verify f −1(x) by checking if f [f −1(x)] = f −1[f (x)] = x.

Example 1

Given the graph of f (x), find (a) f −1(0), (b) f −1(1) and (c) f −1(3).

(a) By inspection, f (3) = 0. Thus, f −1(0) = 3.(b) Since f (1) = 1, f −1(1) = 1.(c) Since f (0) = 3, f −1(3) = 0.

f

g

0

0

3

0

( )( )

=f

g


(b,a) x

y y=x f

f – 1

. .

(a,b)

Figure 1.3-4

1 0

1

2

2

3

3

x

f(x)

Figure 1.3-5

Example 2

Determine if the given function has an inverse:(a) f (x) = x3 + x − 2 (b) f (x) = x3 − 2x + 1

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Example 3

Find the inverse function of f (x) =

1. Since f (x) a strictly increasing function,the inverse function exists.

2. Let y = f (x). Thus, y =

3. Interchange x and y. You have

x =

4. Solve for y. Thus, y =

5. Replace y by f −1(x). You have

f −1(x) =

6. Since the range of f (x) is [0, ∞), the domain of f −1(x) is [0, ∞).

7. Verify f −1(x) by checking:

Since x > 0,

f −1[f (x)] = f x x

x− −( ) = − +

=1

2

2 12 1 1

2

f f x f x x

x− ( )[ ] = +

= +

− =1

2 21

22

1

21 ;x x2 = .

x2 1

2

+.

x2 1

2

+.

2 1y − .

2 1x − .

2 1x − .

(a) By inspection, the graph of f (x) = x3 + x − 2is strictly increasing which implies that f (x)is one-to-one. (You could also use the hor-izontal line test.) Therefore, f (x) has aninverse function.


[–10,10] by [10,10]

Figure 1.3-6

[–8,8] by [–4,3]

Figure 1.3-7

Figure 1.3-8

(b) By inspection, the graph of f (x) = x3 − 2x+ 1 fails the horizontal line test. Thus,f (x) has no inverse function.

0.5

f

y

x0

y=x f

-1

0.5

• Organize your solution so that the reader can follow your line of reasoning.Write all formulas used and indicate all major steps taken.

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Trigonometric and Inverse Trigonometric Functions


Domain:{ –∞ < x < ∞}

Range: {-1 ≤ y ≤ 1} Amplitude: 1

Frequency: 1 Period: 2π

y = sin x

Figure 1.3-9

Domain: {–∞ < x < ∞}

Range: {–1 ≤ y ≤ 1} Amplitude: 1


y = cos x

Figure 1.3-11

Domain: {all x ,...2

3,2

π π ±±≠ }

Range: {–∞ < y < ∞}

Frequency: 1 Period: π

y = tan x

Figure 1.3-13

y = cos–1 x

Domain: {–1 ≤ x ≤ 1}

Range: {0 ≤ y ≤ π}

0. 5

π

Figure 1.3-12

Domain: {-∞ < x < ∞}

Ra < y <nge: { }2

π

2π −

–0.5

0.5π

y = tan–1 x

Figure 1.3-14

Domain: {-1 ≤ x ≤ 1}

Range: { }22

π π ≤≤− y

y = sin–1 x

0. 5

–0. 5

Figure 1.3-10

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y = sec x

Domain: {all x ±≠2

π , ±2

3π , …}

Range: { y ≤ –1 and y ≥ 1}


–1

1

Figure 1.3-15

y = csc x

Domain: {all x ≠ 0, ±π, ±2π}Range:{ y ≤ –1 and y ≥ 1}


–1

1

Figure 1.3-17

y = cot x

Domain: {all x ≠ 0, ±π, ±2π}

Range: {–∞ < y < ∞}

Frequency: 1 Period: π

Figure 1.3-19

y = csc–1 x

Domain: { x ≤ –1 or x ≥ 1}

Range: { }0,22

≠≤≤− y y π π

2π

–2

π

Figure 1.3-18

y = cot–1 x

Domain: {–∞ < x < ∞}

Range: {0 < y < π}

π

Figure 1.3-20

Domain: { x ≤ –1 or x ≥ 1}

Range: {0 ≤ y ≤ π, y ≠2

π }

y = sec–1 x

π

2π

Figure 1.3-16

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Formulas for using a calculator to get sec−1x, csc−1x, and cot−1x:

sec−1x = cos−1(1/ x)csc−1x = sin−1(1/ x)cot−1x = π /2 − tan−1x

Example 1

Sketch the graph of the function y = 3sin2x. Indicate its domain, range, amplitude,

period and frequency.The domain is all real numbers. The range is [−3, 3]. The amplitude is 3, which isthe coefficient of sin2x. The frequency is 2, the coefficient of x, and the period is (2π) ÷(the frequency), thus (2π) ÷ 2 = π. (See Figure 1.3-21.)


Figure 1.3-21

π /3

Cos x < 0

π /3

π /2

3π /2

Cos x < 0

√3

12π /3

π /6(0 or 2π)

Figure 1.3-22

Example 2

Solve the equation cos x = −0.5 if 0 ≤ x ≤ 2π.Note that cos (π /3) = 0.5 and that cosine is negative in the 2nd and 3rd quadrants.

Since cos x = −0.5, x must be in the 2nd or 3rd quadrants with a reference angle of π /3.In the 2nd quadrant, x = π − (π /3) = 2π /3 and in the 3rd quadrant, x = π + (π /3) = 4π /3.

Thus x = 2π /3 or 4π /3. (See Figure 1.3-22.)

Example 3

Evaluate tan−1 (3).Using your graphing calculator, enter tan−1 (3) and the result is 1.2490457724.

Note that the range of tan−10 x is (−π /2, π /2) and −π /2 ≤ 1.2490457724 ≤ π /2. Thustan−1 (3) ≈ 1.2490457724.

Example 4

Evaluate .sin cos−

1 32

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Note that = π /6, and thus, sin (π /6) = 0.5. Or you could use a calculator

and enter sin and get 0.5.

Exponential and Logarithmic FunctionsExponential Function with base a: f (x) = ax where a > 0 and a ≠ 1.Domain: {all real numbers}. Range: {y y > 0}. Y -intercept: (0, 1). Horizontal Asymptote:x-axis. Behavior: strictly increasing. (See Figure 1.3-23.)

cos−

1 32

cos−

1 32


(0,1)

y

x

0

.

y=a x , a >1

0

(0,1)

y

x

y=a x , 0< a<1

.

Figure 1.3-23

y

x0

y=loga x, a >1

.(1,0)

y

x0

y=loga x,0<a<1

.(1,0)

Figure 1.3-24

Properties of Exponents:Given a > 0, b > 0 and x and y are real numbers, then

ax ay = a(x+y)

ax ÷ ay = a(x− y)

(ax)y = axy

(ab)x = ax bx

Logarithmic Function with base a: y = loga x if and only if ay = x where x > 0, a > 0 anda ≠ 1. (See Figure 1.3-24.)

a

b

a

b

x x

x

=

Domain: {xx > 0}. Range: {all real numbers}. X -intercept: (1, 0). Vertical Asymptote:y-axis. Behavior: strictly increasing.

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Note that y = loga x and y = ax are inverse functions (i.e., loga (ax) = a(logax) = x.) (SeeFigure 1.3-25.)


y=a x , a > 1

(0,1)

y

x0

. .(1,0)

y=loga x, a > 1

y=x

Figure 1.3-25

. y=e

x

(0,1)

y

x0

.

.(1,0)

y=ln x

y=x

.(1,e)

(e,1)

Figure 1.3-26

Properties of Logarithms:Given x, y, and a are positive numbers with a ≠ 1 and n is a real number, then

loga (xy) = loga x + loga y

loga = loga x − loga y

loga xn = n loga x

Note that loga 1 = 0, loga a = 1 and loga ax = x.

The Natural Base e:

e ≈ 2.71828182846 . . .

The expression approaches the number e as x gets larger and larger.

An equivalent expression is . The expression also approaches e as happroaches 0.

Exponential Function with base e: f (x) = ex

The Natural Logarithmic Function: f (x) = ln x = loge x where x > 0.Note that y = ex and y = ln x are inverse functions: (eln x = ln (e x) = x)Also note that e0 = 1, ln 1 = 0, and ln e = 1. (See Figure 1.3-26.)

1

1

+( )h h

1

1

+( )h h

11

+

x

x

x

y

Properties of the Natural Logarithmic and Exponential Functions:Given x and y are real numbers, then

ex ey = e(x+y)

ex ÷ ey = e(x− y)

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Example 2

Evaluate (a) log2 8, (b) log5 , and (c) ln e5.

(a) Let n = log2 8 and thus, 2n = 8 = 23. Therefore n = 3.

(b) Let n = log5 , and thus, 5n = = 5−2. Therefore, n = −2.

(c) You know that y = e x and y = ln x are inverse functions. Thus, ln e5 = 5.

Example 3

Express ln [x(2x + 5)3] as the sum and multiple of logarithms.

ln [x(2x + 5)3] = ln x + ln (2x + 5)3

= ln x + 3 ln (2x + 5)

Example 4

Solve 2e x+1 = 18 to the nearest thousandth.

2e x+1 = 18

e x+1 = 9

ln (e x+1) = ln 9

x + 1 = ln 9

x = 1.197

1

25

1

25

1

25


Figure 1.3-27

y

x0

y=ln(x-2)

x=2

.(3,0)

(ex)y = exy

ln (xy) = ln x + ln y

ln = ln x − ln y

ln x n = n ln x

Change of Base Formula:

Example 1

Sketch the graph of f (x) = ln (x − 2)Note that the domain of f (x) is {x x > 2} and that f(3) = ln(1) = 0 and thus, the

x-intercept is 3. (See figure 1.3-27.)

logln

lna x

x

a a a= > ≠where and0 1

x

y

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Example 5

Solve 3 ln 2x = 12 to the nearest thousandth.

3 ln 2x = 12

ln 2x = 4

eln 2x = e4

2x = e4

x = e4 /2 = 27.299

1.4 GRAPHS OF FUNCTIONS

Main Concepts: Increasing and Decreasing Functions; Intercepts and Zeros; Odd and Even Functions; Shifting, Reflecting, and Stretching Functions

Increasing and Decreasing Functions

Given a function f defined on an interval:

• f is increasing on an interval if f (x1) < f (x2) whenever x1 < x2 for any x1 and x2 in theinterval

• f is decreasing on an interval if f (x1) > f (x2) whenever x1 < x2 for any x1 and x2 in theinterval

• f is constant on an interval if f (x1) = f (x2) for any x1 and x2 in the interval.

A function value f (c) is called a relative minimum of f if there exists an interval (a,b) inthe domain of f containing c such that f (c) ≤ f (x) for all x ∈ (a,b).

A function value f (c) is called a relative maximum of f if there exists an interval (a,b)in the domain of f containing c such that f (c) ≥ f (x) for all x ∈ (a,b) (See Figure 1.4-1.)


Decreasing

f(x)

x

y

0

.

. Increasing

Decreasing

Constant

Relative

Maximum

Relative Minimum

Figure 1.4-1

In the following examples, using your graphing calculator, determine the intervalsover which the given function is increasing, decreasing or constant. Indicate any relativeminimum and maximum values of the function.

Example 1: f (x ) = x 3 − 3x + 2

The function f (x) = x3 − 3x + 2 is increasing on (−∞, −1) and (1, ∞), decreasing on (−1, 1).

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Intercepts and Zeros

Given a function f, if f (a) = 0, then the point (a, 0) is an x-intercept of the graph of thefunction, and the number a is called a zero of the function.


Figure 1.4-2

[–8,8] by [–5,5]

Figure 1.4-3

[–5,5] by [–4,4]

Figure 1.4-4

[–5,5] by [–4,4]

Example 2: g (x ) = (x − 1)3

Note that g (x) = (x − 1)3 is increasing for the entire domain (−∞, ∞) and it has no rela-tive minimum or relative maximum values. (See Figure 1.4-3.)

Example 3:

The function f is decreasing on the intervals (−∞, 2) and (2, ∞) and it has no relative

minimum or relative maximum values. (See Figure 1.4-4.)

f x x

x 2 ( ) =

−

• The is equivalent to which is cos π or −1.d

dx x xsin =πlim

sin sin

h

h

h→

+( ) −0

π π

A relative minimum value of the function is 0 occurring at the point (1,0) and a rel-ative maximum value is 4 located at the point (−1,4). (See Figure 1.4-2.)

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If f (0) = b, then b is the y-intercept of the graph of the function. (See Figure 1.4-5.)


.

y

x0

b

c ed

f(x)

. . . x-intercepts: f(c)=0, f(d)=0, f(e) = 0

y-intercept: f(0) = b

Figure 1.4-5

[–5,5] by [–4,4]

Figure 1.4-6

[–10,10] by [–10,20]

Figure 1.4-7

Note that to find the x-intercepts or zeros of a function, you should set f (x) = 0 andto find the y-intercept, let x be 0 (i.e., find f (0).)

In the following examples, find the x-intercepts, y-intercept, and zeros of the given

function if they exist.

Example 1: f (x ) = x 3 − 4x

Using your graphing calculator, note that the x-intercepts are −2, 0, 2 and the y-interceptis 0. The zeros of f are −2, 0 and 2. (See Figure 1.4-6.)

Example 2: f (x ) = x 2 − 2x + 4

Using your calculator, you see that the y-intercept is (0, 4) and the function f has nox-intercept or zeros. (See Figure 1.4-7.)

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Odd and Even Functions

A function f is an even function if f (−x) = f (x) for all x in the domain. The graph of aneven function is symmetrical with respect to the y-axis. If a point (a,b) is on the graph,so is the point (−a,b). If a function is a polynomial with only even powers, then it is aneven function. (See Figure 1.4-8.)


0

..

y

x

(a,b)( – a,b)

f(x)

Figure 1.4-8

x

y

0

..

(a,b)

( – a, – b)

f(x)

Figure 1.4-9

A function f is an odd function if f (−x) = −f (x) for all x in the domain. The graph of an odd function is symmetrical with respect to the origin. If a point (a, b) is on the graph,so is the point (−a, −b). If a function is a polynomial with only odd powers and a zeroconstant, then it is an odd function. (See Figure 1.4-9.)

In the following example, determine if the given functions are even, odd or neither.

Example 1: f (x ) = x 4 − x 2

Begin by examining f (−x). Since f (−x) = (−x)4 − (−x)2 = x4 − x2, f (−x) = f (x). Therefore,f (x) = x4 − x2 is an even function. Or using your graphic calculator, you see that the graphof f (x) is symmetrical with respect to the y-axis. Thus, f (x) is an even function. Or, sincef has only even powers, it is an even function. (See Figure 1.4-10.)

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Example 2: g (x ) = x 3 + x

Examine g (−x). Note that g (−x) = (−x)3 + (−x) = −x3 − x = − g (x). Therefore, g (x) = x3 +x is an odd function. Or looking at the graph of g (x) in your calculator, you see thatthe graph is symmetrical with respect to the origin. Therefore, g (x) is an odd function.Or, since g (x) has only odd powers and a zero constant, it is an odd function. (SeeFigure 1.4-11.)


[–4,4] by [–3,3]

Figure 1.4-10

[–4,4] by [–3,3]

Figure 1.4-11

[–4,4] by [–3,3]

Figure 1.4-12

Example 3: h (x ) = x 3 + 1

Examine h(−x). Since h(−x) = (−x)3 + 1 = −x3 + 1, h(−x) ≠ h(x) which indicates that h(x)is not even. Also, −h(x) = −x3 − 1; therefore, h(−x) ≠ −h(x) which implies that h(x) is not

odd. Using your calculator, you notice that the graph of h(x) is not symmetrical respectto the y-axis or the origin. Thus, h(x) is neither even nor odd. (See Figure 1.4-12.)

Shifting, Reflecting, and Stretching Graphs

Vertical and Horizontal Shifts

Given y = f (x) and a > 0, the graph of

y = f (x) + a is a vertical shift of the graph of y = f (x) a units upward.

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y = f (−x) is a reflection of the graph of y = f (x) about the y-axis. (See Figure 1.4-16.)


x

y

0

y=f(x) y=f( – x)

a – a

Figure 1.4-16

(a,b)

y

x0 ..

y=f(x)

y= – f( – x)

( – a, – b)

Figure 1.4-17

y

x0

f(x)af(x),a>0

af(x),a<0

.

Figure 1.4-18

y = −f (−x) is a reflection of the graph of y = f (x) about the origin. (See Figure 1.4-17.)

Stretching Graphs

Given y = f (x), the graph of

y = af (x) where a > 1 is a vertical stretch of the graph of y = f (x), and

y = af (x) where 0 < a < 1 is a vertical shrink of the graph of y = f (x). (See Figure 1.4-18.)

Example 1

Sketch the graphs of the given functions and verify your results with your graphingcalculator: f (x) = x2, g (x) = 2x2 and p(x) = (x − 3)2 + 2.

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Note that g (x) is a vertical stretch of f (x) and that p(x) is a horizontal shift of f (x)3 units to the right followed by a vertical shift of 2 units upward. (See Figure 1.4-19.)


[–5,8] by [–5,10]

g(x) f(x)

p(x)

Figure 1.4-19

[–5,5] by [–10,10]

g(x) f(x)

h(x)

Figure 1.4-20

6

5

4

1 2 3 4 5 6

3

2

1

x

y

0

f(x)

Figure 1.4-21

Example 2

Figure 1.4-20 contains the graphs of f (x) = x3, h(x), and g (x). Find an equation for h(x)and an equation for g (x).

The graph of h(x) is a horizontal shift of the graph of f (x) one unit to the right.

Therefore, h(x) = (x − 1)3. The graph of g (x) is a reflection of the graph of f (x) about thex-axis followed by a vertical shift 2 units upward. Thus, g (x) = −x3 + 2.

Example 3

Given f (x) as shown in Figure 1.4-21, sketch the graph of f (x − 2), f (x) + 1 and 2f (x).

Note that (a) f (x − 2) is a horizontal shift of f (x) 2 units to the right, (b) f (x) + 1 is avertical shift of f (x) 1 unit upward, and (c) 2f (x) is a vertical stretch of f (x) by a factor of 2. (See Figure 1.4-22.)

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1.5 RAPID REVIEW

1. If line l is parallel to the line y − 3x = 2, find ml .

Answer: ml = 3

2. If line l is perpendicular to the line 2y + x = 6, find ml .

Answer: ml = 23. If x2 + y = 9, find the x-intercepts and y-intercepts.

Answer: The x-intercepts are ±3 (by setting y = 0) and the y-intercept is 9 (by set-ting x = 0).

4. Simplify (a) ln(e3x) and (b) eln (2x).

Answer: Since y = lnx and y = ex are inverse functions, thus ln(e3x) = 3x andeln (2x) = 2x

5. Simplify ln .

Answer: Since ln = ln(a) − ln(b), thus ln = ln(1) − ln x = −ln x.

6. Simplify ln(x3).

Answer: Since ln(ab) = b ln a, thus ln(x3) = 3 ln x.

7. Solve the inequality x2 − 4x > 5, using your calculator.

Answer: Let y1 = x2 − 4x and y2 = 5. Look at the graph and see where y1 is abovey2. Solution is {x:x < −1 or x > 5}. (See Figure 1.5-1.)

1

x

a

b

1

x


f(x – 2)

2f(x)6

5

4

1 2 3 4 5 6

3

2

1

x

y

0

f(x)+1

Figure 1.4-22

8. Evaluate sin , tan , and cos .

Answer: sin = , tan = 1, and cos = .3

2π6 π4 12π6

π6

π4

π6

[–2,6] by [–5,10]

Figure 1.5-1

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1.7 SOLUTIONS TO PRACTICE PROBLEMS


1. Rewrite the equation 3x − 4y + 12 = 0 in y = mx

+ b form: y = x + 3. Thus, theslope of the line

is . Since line l is parallel to this line, the slope

of line l must also be . Line l also passes through

the point (−2, 5). Therefore, an equation of line l

is y − 5 = (x + 2).

2. Let M be the midpoint of . Using the midpointformula, you will find the coordinates of M to be

(2, 3). The slope of median is . Thus, an

equation of is y − 3 = (x − 2).

3. Since the circle is tangent to the line y = −1, theradius of the circle is 2 units. Therefore, theequation of the circle is (x − 2)2 + (y + 3)2 = 4.

4. The two derived equations are x − 2 = 2x + 5and x − 2 = −2x − 5. Form x − 2 = 2x + 5, x = −7and from x − 2 = −2x − 5, x = −1. However,substituting x = −7 into the original equationx − 2 = 2x + 5 results in 9 = −9 which is notpossible. Thus the only solution is −1.

5. The inequality 6 − 3x < 18 is equivalentto −18 < 6 −3x < 18. Thus, −24 < −3x < 12.Dividing through by −3 and reversing theinequality sign, you have 8 > x > −4 or− 4 < x < 8.

6. Since f (x + h) = (x + h)2 + 3(x + h), the expression

is equivalent to

7. The graph of equation (2) 4x2 + 9y2 = 36 is anellipse and the graph of (4) y2 − x2 = 4 is ahyperbola intersecting the y-axis at two distinctpoints. Both of these graphs fail the vertical linetest. Only the graphs of equations (1) xy = −8

= + +

= + +2 3

2 32xh h h

h x h .

x h x h x x

hx xh h x h x x

h

+( ) + +( )[ ] − +[ ]

= + + + +( ) − −

2 2

2 2 2

3 3

2 3 3 3

f x h f x

h

+( ) − ( )

3

4

AM

3

4AM

BC

3

4

34

3

4

3

4

(a parabola) and (3) 3x2 − y = 1 (a hyperbola inthe 2nd and 4th quadrant) pass the vertical linetest. Thus, only (1) xy = −8 and (3) 3x2 − y = 1are functions.

8. The domain of g (x) is −5 ≤ x ≤ 5 and the domainof f (x) is the set of all real numbers. Therefore, the

domain of (f g )(x) = = 25 − x2 is theinterval −5 ≤ x ≤ 5.

9. From the graph, (f − g )(2) = f (2) − g (2) = 1 − 1 = 0,(f g )(1) = f [ g (1)] = f (0) = 1, and ( g f )(0) = g [f (0)]= g (1) = 0.

10. Let y = f (x) and thus y = x3 + 1. Switch x and yand obtain x = y3 + 1. Solve for y and you will

have y =

. Thus f −1(x)=

.

11. The amplitude is 3, frequency is 1 ⁄ 2, and period is4π. (See Figure 1.7-1.)

x −( )

11

3x −( )

11

3

25 22

−( )x

Figure 1.7-1

y = lnxy = ln(–x) y = –ln(x+2)

Figure 1.7-2

13. Enter into your calculator y1 = 2x + 4 and y2 =10. Locate the intersection points. They occur at

12. Note that (1) y = ln x is the graph of the naturallogarithmic function. (2) y = ln(−x) is the reflec-tion about the y-axis. (3) y = −ln(x + 2) is a hori-zontal shift 2 units to the left followed by areflection about the x-axis. (See Figure 1.7-2.)

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14. Enter into your calculator y1 = x3 − 2x and y2 =1. Find the intersection points. The points arelocated at x = −1, −0.618 and 1.618. Since y1 isabove y2 in the intervals −1 < x < −0.618 and x> 1.618 excluding the end points, the solution tothe inequality are the intervals −1 < x < −0.618and x > 1.618. (See Figure 1.7-4.)

19. Examine f (−x) and f (−x) = −2(−x)4 + (−x)2 + 5 =−2x4 + x2 + 5 = f (x). Therefore, f (x) is an evenfunction. Note that the graph of f (x) is symmet-rical with respect to the y-axis; thus, f (x) is aneven function. (See Figure 1.7-6.)


[–10,10] by [–10,15]

Figure 1.7-3

[–2,2] by [–2,2]

Figure 1.7-4

[–4,4] by [–4,7]

Figure 1.7-5

(−1, 2). Since the inequality is ≤ which includesthe end point at x = 2, the solution is (−1, 2].(See Figure 1.7-5.)

[–4,4] by [–4,7]

Figure 1.7-6

20. Enter y1 = x4 − 4x3 into your calculator and

examine the graph. Note that the graph isdecreasing on the interval (−∞, 3) and increasingon (3, ∞). The function crosses the x-axis at 0and 4. Thus, the zeros of the function are 0 and4. There is one relative minimum point at (3,−27). Thus, the relative minimum value for thefunction is −27. There is no relative maximum.(See Figure 1.7-7.)

[–2,5] by [–30,10]

Figure 1.7-7

15. Enter into your calculator andtan arccos2

2

x = −7 and 3. Note that y1 is below y2 from x =−7 to 3. Since the inequality is less than or equalto, the solution is −7 ≤ x ≤ 3. (See Figure 1.7-3.)

obtain 1. (Note that = π /4 and tan(π /4) = 1.)

16. Factor e2 x − 6e x + 5 = 0 as (e x − 5)(e x − 1) = 0.Thus (ex − 5) = 0 or (e x − 1) = 0 resulting in ex =5 and e x = 1. Taking the natural log of bothsides yields ln (ex) = ln 5 ≈ 1.609 and ln (e x) =ln 1 = 0. Therefore to the nearest thousandth,x = 1.609 or 0. [Note that ln (e x) = x.]

17. The equation 3 ln 2x − 3 = 12 is equivalent toln 2x = 5. Therefore, eln 2x = e5, 2x = e5 ≈148.413159 and x ≈ 74.207.

18. Enter y1 = and y2 = 1 into your calcula-

tor. Note that y1 is below y2 = 1 on the interval

2 1

1

x

x

−+

arccos2

2

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Chapter 2

Limits and Continuity

2.1 THE LIMIT OF A FUNCTION

Main Concepts: Definition and Properties of Limits, Evaluating Limits, One-sided Limits, Squeeze Theorem

Definition and Properties of Limits

Definition of Limit

Let f be a function defined on an open interval containing a, except possibly at a itself.Then (read as the limit of f (x) as x approaches a is L) if for any ε > 0, there

exists a δ > 0 such that f (x) − L < ε whenever x − a < δ.

Properties of Limits

Given and and L, M, a, c and n are real numbers, then

1.

2.

3.

4.

5.

6. lim limx a

n

x a

nnf x f x L

→ →( )[ ] = ( )( ) =

limlim

lim,

x a

x a

x a

f x

g x

f x

g x

L

M M

→

→

→

( )( )

= ( )

( ) = ≠ 0

lim lim limx a x a x a

f x g x f x g x L M→ → →

( ) ( )[ ] = ( ) ( ) =

lim lim limx a x a x a

f x g x f x g x L M→ → →

( ) ± ( )[ ] = ( ) ± ( ) = +

lim limx a x a

cf x c f x cL→ →

( )[ ] = ( ) =

limx a

c c→

=

limx a

g x M→

( ) =limx a

f x L→

( ) =

limx a

f x L→

( ) =


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Limits and Continuity • 77

Evaluating Limits

If f is a continuous function on an open interval containing the number a, then

Common techniques in evaluating limits are:

1. substituting directly2. factoring and simplifying3. multiplying the numerator and denominator of a rational function by the conjugate

of either the numerator or denominator4. using a graph or a table of values of the given function.

Example 1

Find the limit:

Substituting directly:

Example 2Find the limit:

Using the product rule, = (3π)(sin π) = (3π)(0) = 0.

Example 3

Find the limit:

Factoring and simplifying: = (2 − 1)

= 1. (Note that had you substituted t = 2 directly in the original expression, you wouldhave obtained a zero in both the numerator and denominator.)

Example 4

Find the limit:

Factoring and simplifying:

Example 5

Find the limit:

Multiplying both the numerator and the denominator by the conjugate of the numerator,

=+ +( )

=+ +( )

=+ +

=→ →

lim limt t

t

t t t 0 0

2 2

1

2 2

1

0 2 2

1

2 2

t t

t

t

t

t

t t t t

+ +( ) + − + +

+ +

=

+ −

+ +( )→ →2 2

2 2 2 2

2 2

2 2

2 20 0

, lim limyields

limt

t

t →

+ −0

2 2

=+

=+

=→

limx b x b b b b

1 1 1

25 5 5 5 5

lim limx b x b

x b

x b

x b

x b x b→ →

−−

= −

−( ) +( )

5 5

10 10

5 5

5 5 5 5

lim .x b

x b

x b→

−−

5 5

10 10

lim lim limt t t

t t

t

t t

t t

→ → →

− +−

= −( ) −( )

−( ) = −( )

2

2

2 2

3 2

2

1 2

21

limt

t t

t →

− +−2

2 3 2

2

lim sin lim lim sinx x x

x x x x→π →π →π

= ( )( )3 3

lim sinx

x x→π

3

lim .x

x→

+ = ( ) + =5

3 1 3 5 1 4

limx

x→

+5

3 1

lim .x a

f x f a→

( ) = ( )

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(Note that substituting 0 directly into the original expression

would have produced a 0 in both the numerator and denominator.)

Example 6

Find the limit:

Enter in the calculator. You see that the graph of f (x) approaches 3 as x

approaches 0. Thus, the (Note that had you substituted x = 0 directly

in the original expression, you would have obtained a zero in both the numerator anddenominator.) (See Figure 2.1-1.)

limsin

.t

x

x→=

0

3 2

23

y xx

1 3 22

= sin

limsin

x

x

x→0

3 2

2

=

=

1

2 2

2

2

2

4.

[–10,10] by [–4,4]

Figure 2.1-1

• Always indicate what the final answer is, e.g., “The maximum value of f is 5.”Use complete sentences whenever possible.

One-Sided Limits

Let f be a function and a is a real number. Then the right-hand limit: f (x) represents

the limit of f as x approaches a from the right, and left-hand limit: f (x) represents the

limit of f as x approaches a from the left.

limx a→ −

limx a→ +

Example 7

Find the limit:

Enter into your calculator. You notice that as x approaching 3 from the

right, the graph of f (x) goes higher and higher, and that as x approaching 3 from the

left, the graph of f (x) goes lower and lower. Therefore, is undefined. (See Fig

ure 2.1-2.)

limx x→ −3

13

yx

11

3=

−

limx x→ −3

1

3

[–2,8] by [–4,4]

Figure 2.1-2

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Existence of a Limit

Let f be a function and let a and L be real numbers. Then the two-sided limit f (x) =

L if and only if the one-sided limits exist and f (x) = f (x) = L.

Example 1

Given find the limits (a) f (x), (b) f (x), and (c) f (x).

Substituting x = 3 into f (x) leads to a 0 in both the numerator and denominator. Factor

f (x) as which is equivalent to (x + 1) where x ≠ 3. Thus, (a) f (x) =

(x + 1) = 4, (b) f (x) = (x + 1) = 4, and (c) since the one-sided limits exist and

are equal, f (x) = f (x) = 4, therefore the two-sided limit f (x) exists and

f (x) = 4. (Note that f (x) is undefined at x = 3, but the function gets arbitrarily close to 4as x approaches 3. Therefore the limit exists. (See Figure 2.1-3.)

limx→3

limx→3

limx→ −3

limx→ +3

limx→ −3

limx→ −3

limx→ +3

limx→ +3

x x

x

−( ) +( )−( )

3 1

3

limx→3limx→ −3limx→ +3f x

x x

x( ) = − −

−

2 2 3

3 ,

limx a→ −

limx a→ +

limx a→

[–8,8] by [–6,6]

Figure 2.1-3

Example 2

Given f (x) as illustrated in the accompanying diagram (Figure 2.1-4.) Find the limits(a) f (x), (b) f (x), and (c) f (x).lim

x→0limx→ −0

limx→ +0

[–8,8] by [–10,10]

Figure 2.1-4

(a) As x approaches 0 from the left, f (x) gets arbitrarily close to 0. Thus, f (x) = 0.

(b) As x approaches 0 from the right, f (x) gets arbitrarily close to 2. Therefore, f (x)

= 2. Note that f (0) ≠ 2.

(c) Since f (x) ≠ f (x), f (x) does not exist.

Example 3

Given the greatest integer function f (x) = [x], Find the limits (a) f (x), (b) f (x),

and (c) f (x).limx→1

limx→ −1

limx→ +1

limx→0

limx→ −0

limx→ +0

limx→ +0

limx→ −0

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a. Enter y1 = int(x) in your calculator. You see that as x approaches 1 from the right, thefunction stays at 1. Thus, [x] = 1. Note that f (1) is also equal 1. (b) As x approaches

1 from the left, the function stays at 0. Therefore, [x] = 0. Notice that [ x] ≠

f (1). (c) Since [x] ≠ [x], therefore, [x] does not exist. (See Figure 2.1-5.)limx→1

limx→ +1

limx→ −1

limx→ −1

limx→ −1

limx→ +1

Figure 2.1-5

–1

1

2

–2

y

x

–2 –1 2 310

• Remember ln(e) = 1 and eln3 = 3 since y = ln x and y = ex are inverse functions.

Example 4

Given x ≠ 0, find the limits (a) f (x), (b) f (x), and (c) f (x).

(a) From inspecting the graph, = 1, (b) = −1, and (c) since

≠ , therefore, does not exist. (See Figure 2.1-6.)limx

x

x→

=0

limx

x

x→ −

0

limx

x

x→ +0

limx

x

x→ −=

0

limx

x

x→ +=

0

limx→0

limx→ −0

limx→ +0

f x x

x( ) = ,

[–4.4] by [–4,4]

Figure 2.1-6

Example 5

If

Thus f (x) does not exist.limx→0

lim lim lim lim .x x

x

x x

xf x xe f x e→ → → →+ + − −

( ) = = ( ) = =0 0 0 0

20 1and

f x e x

xe xf x

x

x x( ) =

− ≤ <

≤ ≤

( )→

2

0

4 0

0 4

for

forfind, lim .

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Squeeze Theorem

If f, g, and h are functions defined on some open interval containing a such that g (x) ≤f (x) ≤ h(x) for all x in the interval except possibly at a itself, and g (x) = h(x) = L,then f (x) = L.

Theorems on Limits

(1) and (2)

Example 1

Find the limit if it exists:

Substituting 0 into the expression would lead to Rewrite as and

thus, As x approaches 0, so does 3x. There-

fore, (Note that is equivalent to

by replacing 3x by x.) Verify your result with a calculator. (See Figure 2.1-7.)limsin

x

x

x→0

limsin

3 0

3

3x

x

x→3

3

33

3

33 1 3

0 3 0lim

sinlim

sin.

x x

x

x

x

x→ →= = ( ) =

lim

sin

lim

sin

lim

sin

.x x x

x

x

x

x

x

x→ → →= =0 0 0

3 3 3

3 3

3

3

3

3

3

sin x

x

sin 3x

x

0

0.

limsin

x

x

x→0

3

limcos

x

x

x→

−=

0

10lim

sinx

x

x→=

01

limx a→

limx a→

limx a→

[–10,10] by [–4,4]

Figure 2.1-7

Example 2

Find the limit if it exists:

Rewrite As h approaches 0, so do 3h and 2h. Therefore,

(Note that substituting h = 0 into the original

expression would have produced ) Verify your result with a calculator. (See

Figure 2.1-8.)

00

.

limsin

sin

limsin

limsin

.h

h

h

h

h

h

hh

h

→

→

→

= = ( )

( ) =

0

3 0

2 0

3

2

33

32

2

3 1

2 1

3

22

sin

sin

sin

sin.

3

2

3 33

2

2

h

h

hh

h

h

as

2

limsin

sin.

h

h

h→0

3

2

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Example 3

Find the limit if it exists: limcos

.y

y

y→ −0

2

1


[–3,3] by [–3,3]

Figure 2.1-8

[–8,8] by [–2,10]

Figure 2.1-9

Example 4

Find the limit if it exists: limcos

.x

x

x→0

3

[–10,10] by [–30,30]

Figure 2.1-10

Substituting 0 in the expression would lead to Multiplying both the numerator and

denominator by the conjugate (1 + cos y) produces

(Note that ) Verify your result with a

calculator. (See Figure 2.1-9.)

limsin

limsin

lim

limsin

.y y

y

y

y

y y

y

y

y

→ →

→

→

= =( )

= =0 0

0

0

11

1

11

limsin

lim cos limsin

lim cos .y y y y

yy

y yy

y→ → → →

+( ) = +( ) = ( ) +( ) =0

2

0

2

0

2

0

2 21 1 1 1 1 2

limcos

coslim

cos

sinlim

sinlim cos

y y y y

y y

y

y y

y

y

y y

→ → → →

+( )−

= +( ) = +( ) =

0

2

2 0

2

2 0

2

2 0

21

1

11

limcos

cos

cosy

y

y

y

y→ −+( )+( )

=0

2

1

1

1

00

.

Using the quotient rule for limits, you have Verify

your result with a calculator. (See Figure 2.1-10.)

limcos

lim

lim cos.

x

x

x

x

x

x

x→

→

→

= ( )

( ) = =

0

0

0

3 3 0

10

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2.2 LIMITS INVOLVING INFINITIES

Main Concepts: Infinite Limits (as x → a ), Limits at Infinity (as x → ∞ ), Horizontal and Vertical Asymptotes

Infinite Limits (as x → a)If f is a function defined at every number in some open interval containing a, except pos-sibly at a itself, then

(1) means that f (x) increases without bound as x approaches a,

(2) means that f (x) decreases without bound as x approaches a.

Limit Theorems

(1) If n is a positive integer, then

(a)

(b)

(2) If the then

(3) If the

(Note that limit theorems 2 and 3 hold true for x → a+ and x → a−.)

Example 1

Evaluate the limit: (a) and (b)

The limit of the numerator is 5 and the limit of the denominator is 0 through positive val-

ues. Thus, (b) The limit of the numerator is 5 and the limit of the denom-

inator is 0 through negative values. Therefore, Verify your result with

a calculator. (See Figure 2.2-1.)

lim .x

x

x→ −

−−

= −∞2

3 1

2

lim .x

x

x→ +

−−

= ∞2

3 1

2

limx

x

x→ −

−−2

3 1

2limx

x

x→ +

−−2

3 1

2

limx a

f x

g x

g x

g x→

( )( )

= −∞ ( )

∞ ( )

if approaches 0 through positive values

if approaches 0 through negative values

lim , , lim ,x a x a

f x c c g x→ →

( ) = < ( ) =0 0and then

limx a

f x g x

g x g x→ ( )( ) = ∞ ( )

−∞ ( )

if approaches 0 through positive valuesif approaches 0 through negative values

lim , , lim ,x a x a

f x c c g x→ →

( ) = > ( ) =0 0and

limx

nx

n

n→ −=

∞

−∞

0

1 if is even

if is odd

limx

nx→ += ∞

0

1

limx a

f x→

( ) = −∞

limx a

f x→

( ) = ∞

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Example 2

Find:

Factor the denominator obtaining The limit of the

numerator is 9 and the limit of the denominator is (0)(6) = 0 through negative values.

Therefore, Verify your result with a calculator. (See Figure 2.2-2.)lim .x

x

x→ − − = −∞3

2

2 9

lim lim .x x

x

x

x

x x→ →− −− =

−( ) +( )3

2

23

2

9 3 3

limx

x

x→ − −3

2

2 9


[–5,7] by [–40,20]

Figure 2.2-1

[–10,10] by [–10,10]

Figure 2.2-2

Example 3

Find:

Substituting 5 into the expression leads to Factor the numerator into

As x → 5−, (x − 5) < 0. Rewrite (x − 5) as −(5 − x). As x → 5−, (5 − x)

> 0 and thus, you may express (5 − x) as Therefore,

(x − 5) = −(5 − x) = Substituting these equivalent expressions into

the original problem, you have

The limit of the numerator is 10 and the

limit of the denominator is 0 through positive values. Thus, the lim .x

x

x→ −

−−

= −∞5

225

5

= − −( ) +( )

−( ) −( ) = −

+( )−( )→ →− −

lim lim .x x

x x

x x

x

x5 5

5 5

5 5

5

5

lim limx x

x

x

x x

x x→ →− −

−−

= −( ) +( )

−( ) −( )5

2

5

25

5

5 5

5 5

− −( ) −( )5 5x x .

5 5 52

−( ) = −( ) −( )x x x .

5 5−( ) +( )x x .

25 2− x00

.

limx

x

x→ −

−−5

225

5

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Example 4

Find: where [x] is the greatest integer value of x.

As x → 2−, [x] = 1. The limit of the numerator is (1 − 2) = −1. As x → 2−, (2 − x) = 0 through

positive values. Thus, lim .x

x x

x→ −

[ ] −−

= −∞2 2

lim ,x

x x

x→ −

[ ] −−2 2


[–10,30] by [–5,10]

Figure 2.2-3

• Do easy questions first. The easy ones are worth the same number of points asthe hard ones.

Limits at Infinity (as x → ±∞)

If f is a function defined at every number in some interval (a, ∞), then

means that L is the limit of f (x) as x increases without bound.

If f is a function defined at every number is some interval (−∞, a), then

means that L is the limit of f (x) as x decreases without bound.

Limit Theorem

If n is a positive integer, then

(a)

(b)

Example 1

Evaluate the limit:

Divide every term in the numerator and denominator by the highest power of x, and inthis case, it is x and obtain:

Verify your result with a calculator. (See Figure 2.2-3.)

lim limlim lim

lim lim

lim lim

lim limx x

x x

x x

x x

x x

x

xx

x

x

x

x

x

→∞ →∞

→∞ →∞

→∞ →∞

→∞ →∞

→∞ →∞

−+

=−

+=

( ) −

( ) +

=( ) −

( ) +

= − ( )

+ ( )

6 13

2 5

613

25

613

25

6 131

2 51

6 13 0

2 5 0== 3.

limx

xx→∞ −+6 13

2 5

limx nx→−∞

=1

0

limx nx→∞

=1

0

limx

f x L→−∞

( ) =

limx

f x L→∞

( ) =

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Example 2

Evaluate the limit:

Divide every term in the numerator and denominator by the highest power of x and in

this case, it is x3. Thus, Verify your result

with a calculator. (See Figure 2.2-4.)

lim lim .x x

x

xx x

x

→−∞ →−∞

−+

=−

+=

−+

=3 10

4 5

3 10

45

0 0

4 00

3

2 3

3

limx

x

x→−∞

−+

3 10

4 53


[–4,4] by [–20,10]

Figure 2.2-4

[–10,30] by [–5,3]

Figure 2.2-5

Example 3

Evaluate the limit:

Divide every term in the numerator and denominator by the highest power of x and in this

case, it is x2. Therefore, The limit

of the numerator is −1 and the limit of the denominator is 0. Thus,


lim .x

x

x→∞

−+

= −∞1

10 7

2

lim lim

lim lim

lim lim

.x x

x x

x x

x

x

x

x x

x

x x→∞ →∞

→∞ →∞

→∞ →∞

−

+ =

−

+=

− ( )

+

1

10 7

11

10 7

11

10 7

2 2

2

2

2

limx

x

x→∞

−+

1

10 7

2

Example 4

Evaluate the limit: limx

x

x→−∞

+

+

2 1

32

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As x → −∞, x < 0 and thus, x = − Divide the numerator and denominator by

x (not x2 since the denominator has a square root.) Thus, you have

Replacing the x below you have


=+

+

−

=+

− +

=( ) −

− ( ) +

=−

= −→−∞ →−∞

→−∞ →−∞

→−∞ →−∞

lim limlim lim

lim lim

.x x

x x

x x

x

x

x

x

x

x

x

x

2 1

3

21

13

21

13

2

12

2

22 2

limx

x

x→−∞

+

+

2 1

32x x2 23+ −( )by ,=

+

+→−∞lim .x

x

x

x

x

2 1

32

limx

x

x→−∞

+

+

2 1

32

x2 .


[–4,10] by [–4,4]

Figure 2.2-6

Vertical and Horizontal Asymptotes

A line y = b is called a horizontal asymptote for the graph of a function f if either

A line x = a is called a vertical asymptote for the graph of a function f if either

Example 1

Find the horizontal and vertical asymptotes of the function

To find the horizontal asymptotes, examine the

The andlim lim lim ,x x x

f x x

xx

x

→∞ →∞ →∞( ) =

+−

=+

−= =

3 5

2

35

12

3

13

lim lim .x x

f x f x→∞ →−∞

( ) ( )and the

f x x

x( ) =

+−

3 5

2.

limx a a

f x f x→ →+ −

( ) = +∞ ( ) = +∞or limx

lim lim .x x

f x b f x b→∞ →−∞

( ) = ( ) =or

• Remember that ln ln ln ln .1

11

x x x y e

xx

x

= ( ) − = − = =− and

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you notice that and Therefore, the line x = 0 (or the

y-axis) is a vertical asymptote. (See Figure 2.2-8.)

lim .x

f x→ −

( ) = −∞0

lim ,x

f x→ +

( ) = ∞0


[–5,5] by [–30,30]

Figure 2.2-8

[–20,20] by [–3,3]

Figure 2.2-9

Relationship between the limits of rational functions as x → ∞ and horizontal asymptotes:

Given

(1) if the degree of p(x) is same as the degree of q(x), thenwhere a is the coefficient of the highest power of x in p(x) and b is the coefficient

of the highest power of x in q(x). The line is a horizontal asymptote. See

example 1, above.(2) if the degree of p(x) is smaller than the degree of q(x), then

The line y = 0 (or x-axis) is a horizontal asymptote. See example 2, above.

(3) if the degree of p(x) is greater than the degree of q(x), then and

Thus, f (x) has no horizontal asymptote. See example 3, above.

Example 4Using your calculator, find the horizontal asymptotes of the function

Enter The graph shows that f (x) oscillates back and forth about the x-axis.

As x → ±∞, the graph of gets closer and closer to the x-axis which implies that f (x)approaches 0. Thus, the line y = 0 (or the x-axis) is a horizontal asymptote. (SeeFigure 2.2-9.)

y x

x1

2=

sin.

f x x

x( ) =

2 sin.

lim .x

f x→−∞

( ) = ±∞

limx

f x→∞

( ) = ±∞

lim .x

f x→−∞

( ) = 0limx

f x→∞

( ) =

y a

b=

lim limx xf x f x a

b→∞ →−∞( ) = ( ) =

f x p x

q x( ) =

( )( )

, then

• When entering a rational function into a calculator, use parentheses for both thenumerator and denominator, e.g., (x−2) ÷ (x+3).

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2.3 CONTINUITY OF A FUNCTION

Continuity of a Function at a Number: A function f is said to be continuous at a num-ber a if the following three conditions are satisfied:

1. f (a) exists

2. exists

3.

The function f is said to be discontinuous at a if one or more these three conditions arenot satisfied and a is called the point of discontinuity.

Continuity of a Function over an Interval:A function is continuous over an interval if itis continuous at every point in the interval.

Theorems on Continuity

1. If the functions f and g are continuous at a, then the functions f + g, f − g, f − g, and

f/g, g (a) ≠ 0, are also continuous at a.2. A polynomial function is continuous everywhere.3. A rational function is continuous everywhere, except at points where the denominator

is zero.4. Intermediate-Value Theorem: If a function f is continuous on a closed interval [a, b]

and k is a number with f (a) ≤ k ≤ f (b), then there exists a number c in [a, b] such thatf (c) = k.

Example 1

Find the points of discontinuity of the function

Since f (x) is a rational function, it is continuous everywhere, except at points where thedenominator is 0. Factor the denominator and set it equal to 0: (x − 2)(x + 1) = 0. Thusx = 2 or x = −1. The function f (x) is undefined at x = −1 and at x = 2. Therefore, f (x) isdiscontinuous at these points. Verify your result with a calculator. (See Figure 2.3-1.)

f x x

x x( ) =

+− −

5

22.

limx a f x f a→ ( ) = ( )

limx a

f x→

( )


[–5,5] by [–10,10]

Figure 2.3-1

Example 2

Determine the intervals on which the given function is continuous:

f x

x x

x x

x

( ) =+ −

− ≠

=

2 3 10

22

10 2

,

,

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Check the three conditions of continuity at x = 2:

Condition 1: f (2) = 10. Condition 2:

Condition 3: f (2) ≠ f (x). Thus, f (x) is discontinuous at x = 2. The

function is continuous on (−∞, 2) and (2, ∞). Verify your result with a calculator. (SeeFigure 2.3-2.)

limx→2

= +( ) =→

lim .x

x2

5 7

lim limx x

x x

x

x x

x→ →

+ −−

= +( ) −( )

−2

2

2

3 10

2

5 2

2


Remember thatd

dx x x x dx x c

1 12

= − = +∫ and

1ln .

[–8,12] by [–3,17]

Figure 2.3-2

Example 3

For what value of k is the function continuous at x = 6?

For f (x) to be continuous at x = 6, it must satisfy the three conditions of continuity.

Condition 1: f (6) = 62 − 2(6) = 24. Condition 2: thus

must also be 24 in order for the to equal 24. Thus, which

implies 2(6) + k = 24 and k = 12. Therefore, if k = 12, condition (3) f (6) = is

also satisfied.

Example 4

Given f (x) as shown in Figure 2.3-3, (a) find f (3) and and (b) determine if f (x)is continuous at x = 3? Explain why.

lim ,x

f x→

( )3

limx

f x→

( )6

limx x k→ − +( ) =6 2 24limx f x→ ( )6

limx

x k→ −

+( )6

2lim ;x

x x→ −

−( ) =6

2 2 24

f xx x x

x k x( ) =

− ≤

+ >

2 2 6

2 6

,

,

[–3,8] by [–4,8]

Figure 2.3-3

The graph of f (x) shows that f (3) = 5 and the Since f (3) ≠ there-fore f (x) is discontinuous at x = 3.

lim ,x

f x→

( )3

lim .x

f x→

( ) =3

1

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x 0 3 5

f −4 b −4

Example 5

If g (x) = x2 − 2x − 15, using the Intermediate Value Theorem show that g (x) has a rootin the interval [1, 7].

Begin by finding g (1) and g (7), and g (1) = −16 and g (7) = 20. If g (x) has a root, then g (x) crosses the x-axis, i.e., g (x) = 0. Since −16 ≤ 0 ≤ 20, by the Intermediate ValueTheorem, there exists at least one number c in [1, 7] such that g (c) = 0. The number cis a root of g (x).

Example 6

A function f is continuous on [0, 5] and some of the values of f are shown below.


y

x

f ( x ) = –2

3

2

1

1 2 3 4 50

–1

–2

–3

–5

(3,3)

(3,1)

(3,0)

(3,–2)

(3,–5) (5,–4)(0,–4)

Figure 2.3-4

If f (x) = −2 has no solution on [0, 5] then b could be

(A) 3 (B) 1 (C) 0 (D) −2 (E) −5If b = −2, then x = 3 would be a solution for f (x) = −2.

If b = 0, 1, or 3, f (x) = −2 would have two solutions for f (x) = −2.Thus, b = −5, choice (E). (See Figure 2.3-4.)

2.4 RAPID REVIEW

1. Find f (2) and f (x) and determine if f is continuous at x = 2. (See Figure 2.4-1 on

page 93.)

Answer: f (2) = 2, f (x) = 4, and f is discontinuous at x = 2.

2. Evaluate lim .x a

x a

x a→

−−

2 2

limx→2

limx→2

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Answer:

3. Evaluate

Answer: The limit is −3, since the polynomials in the numerator and denominator havethe same degree.

4. Determine if is continuous at x = 3.

Answer: The function f is continuous, since f (3) = 9, and

5. If

Answer:

6. Evaluate

Answer: The limit is

7. Evaluate

Answer: The limit is −∞, since (x2 − 25) approaches 0 through negative values.

8. Find the vertical and horizontal asymptotes of f (x) =

Answer: The vertical asymptotes are x = ±5, and the horizontal asymptote is y = 0,since lim .

xf x

→±∞ ( ) = 0

1

252x −.

lim .x

x

x→ − −5

2

2 25

6

23 1

0= =

→, lim

sin.since

x

x

x

lim sin

sin.

x

x

x→0

6

2

lim , lim .x x x

f x f x f x→ → →

( ) = ( ) = ( ) =+ −0 0 0

1 1since lim

f xe x

xf x

x

x( ) =

≠

=

( )→

for

forfind

0

5 0 0, lim .

f f xx

33

( ) = ( )→

lim .

lim limx x

f x f x→ →+ −

( ) = ( ) =3 3

9

f xx x

x x( ) =

+ <

≥

6 3

32

for

for

lim .x

x

x x→∞

−+ +1 3

100 99

2

2

lim .x a

x a x a

x a a

→

+( ) −( )−

= 2


2.5 PRACTICE PROBLEMS

2. If b ≠ 0, evaluate

3. limx

x

x→

− −0

2 4

limx b

x b

x b→

−−

3 3

6 6

20

2

4

(4,2)

y

x

f(x)

(2,2)

Figure 2.4-1

Part A—The use of a calculator is not allowed.

Find the limits of the following:

1. lim cosx

x x→

− )0

5

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4.

5.

6.

7.

8.

9.

10.

11.

12. The graph of a function f is shown in Figure 2.5-1.

limt

t

t → +

−−3

2 9

3

limsin

sinx

x

x→0

3

4

limx

xe

x→∞ −1 3

If for

forf x

e x

x e xfind f x

x

x x( ) =

≤ <

≤ ≤

( )→

0 1

1 52 1, lim .

limx

x

x→−∞ −

3

42

limx

x

x→∞ +3

5 8

2

limx

x x

x x→−∞

+ −+

2

3 2

2 3

2

limx

x

x→∞

−+

5 6

2 1114. Find the limit: when [x] is the greatest

integer of x.

15. Find the points of discontinuity of the function

16. For what value of k is the function

continues at x = 3?

17. Determine if

is continuous at x = 2. Explain why or why not.

18. Given f (x) as shown in Figure 2.5-2, find

f x

x x

x x

x

( ) =+ − −

− ≠

=

2 5 14

22

12 2

,

,

if

if

g x x x

x k x( ) =

+ ≤

− >

2 5 3

2 3

,

,

f x x

x x( ) =

++ −

1

4 122.

limx

x

x→ +

+ [ ]−5

5

5

x −2 0 2

f(x) 3 b 4

8

7

6

5

43

2

1

0 1 2 3 4 5 6 7 8 9

y

x

f

Figure 2.5-1

[–2,8] by [–4,7]

Figure 2.5-2

Which of the following statements is/are true?

Ι.

ΙΙ.

ΙΙΙ. x = 4 is not in the domain of f

Part B—Calculators are allowed.

13. Find the horizontal and vertical asymptotes of the

graph of the function f xx x

( ) =+ −

1

22.

limx

f x→

) =4

2

limx

f x→ −

( ) =4

5

a) f (3)b) f (x)

c) f (x)

d) f (x)

e) Is f (x) continuous at x = 3? Explain why orwhy not.

19. A function f is continuous on [−2,2] and some of the values of f are shown below:

limx→3

limx→ −3

limx→ +3

If f has only one root, r, on the closed interval[−2,2], and r ≠ 0, then a possible value of b is

(A) −3 (B) −2 (C) −1 (D) 0 (E) 1

20. Evaluate limcos

sin.

x

x

x→

−0 2

1

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Part A—No calculators are permitted.

1. Using the product rule,

(Note that cos 0 = 1.)

2. Rewrite

Substitute x = b and obtain

3. Substituting x = 0 into the expression

leads to which is an indeterminate form. Thus,

multiply both the numerator and denominator by

the conjugate and obtain

limx

x

x

x

x→

− − + −

+ −

0

2 4 2 4

2 4

2 4+ −( )x

00

2 4− − x

x

1 1

23 3 3b b b+ = .

lim lim .x b x b

x b

x b x b x b→ →

−−( ) +( )

=+

3 3

3 3 3 3 3 3

1

limx b

x b

x b→

−−

3 3

6 6as

= −( )( ) = −( )( ) = −0 5 0 5 1 5cos .

lim cos lim lim cosx x x

x x x x→ → →

−( )( ) = −( ) 0 0 05 5


21. Write an equation of the line passing throughthe point (2, −4) and perpendicular to the line3x − 2y = 6.

22. The graph of a function f is shown in Figure 2.6-1.

Which of the following statements is/are true?Ι.ΙΙ. x = 4 is not in the domain of f .ΙΙΙ. does not exist.

23. Evaluate

24. Find

25. Find the horizontal and vertical asymptotes of f (x)

=+

x

x2 4.

limtan

.x

x

x→0

limx

x

x→

−−0

3 4

2

limx

f x→

( )4

lim .x

f x→ −

( ) =4

3

2.6 CUMULATIVE REVIEW PROBLEMS

8

7

6

5

4

3

2

1

0 1 2 3 4 5 6 7 8 9

y

x

f

Figure 2.6-1


4. Since the degree of the polynomial in thenumerator is the same as the degree of thepolynomial in the denominator,

5. Since the degree of the polynomial in thenumerator is 2 and the degree of thepolynomial in the denominator is 3,

6. The degree of the monomial in the numerator is 2and the degree of the binomial in the denominator

is 1. Thus,

7. Divide every term in both the numerator anddenominator by the highest power of x and in

lim .x

x

x→∞ + = ∞

3

5 8

2

lim .x

x x

x x→∞

+ −+

=2

3 2

2 3

20

lim .x

x

x→∞

−+

= − = −5 6

2 11

6

23

=+ −( )

=+ − ( )( )

=→

limx x0

1

2 4

1

2 4 0

1

4

= − −( )

+ −( ) =

+ −( )→ →lim limx x

x

x x

x

x x0 0

4 4

2 4 2 4

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12. The graph of f indicates that:

Ι. is true.

ΙΙ.

ΙΙΙ. “x = 4 is not in the domain of f ” is false sincef (4) = 2.


13. Examining the graph in your calculator, you noticethat the function approaches the x-axis as x → ∞or as x → −∞. Thus, the line y = 0 (the x-axis) is ahorizontal asymptote. As x approaches 1 fromeither side, the function increases or decreaseswithout bound. Similarly, as x approaches −2from either side, the function increases ordecreases without bound. Therefore, x = 1 andx = −2 are vertical asymptotes. (See Figure 2.7-1.)

lim lim .x x

f x f x→ →

( ) = ( ) =( )4 4

2 is false. The 5

limx

f x→ −

( ) =4

5


this case, it is x. Thus, you have

As x → −∞, x = Since the denominatorinvolves a radical, rewrite the expression as

8.

9. However, asx → ∞; the rate of increase of ex is much greaterthan the rate of decrease of (1 − x3 ). Thus

10. Divide both numerator and denominator by x and

obtain Now rewrite the limit as

As x approaches

0, so do 3x and 4x. Thus, you have

11. As t → 3+, (t − 3) > 0 and thus (t − 3) =

Rewrite the limit as

The limit of the numerator is

and the denominator is approaching 0 through

positive values. Thus, lim .t

t

t → +

−−

= ∞3

2 9

3

6

= +( )

−( )→ +lim .t

t

t 3

3

3

limt

t t

t → +

−( ) +( )

−( )3 2

3 3

3

t −( )32

.

3

4

3

34

4

3 1

4 1

3

4

3 0

4 0

limsin

limsin

.x

x

x

xx

x

→

→

= ( )

( ) =

lim

sin

sinlim

sin

sin.

x x

xx

x

x

xx

x

x

→ →=

0 0

3 33

44

4

3

4

33

4

4

lim

sin

sin.

x

x

xx

x

→0

3

4

lim .x

xe

x→∞ − = −∞

1 3

lim lim .x

x

xe x→∞ →∞= ∞ − ) = ∞and 13

= ) = ) =→ →−

lim . lim .x

x

xe e f x e

1 1Thus

lim lim limx x

x

x

f x x e e f x→ → →+ + −

( ) = ( ) = ( )1 1

2

1

and

=− −

= −3

1 03

lim limx x

x

xx

x x

→−∞ →−∞−

−

=

− −

3

43

142

2 2

− x2 .

lim .x

x

x

x

x

→−∞ −

3

42

[–6,5] by [–3,3]

Figure 2.7-1

14. As x → 5+, the limit of the numerator (5 + [5]) is10 and as x → 5+, the denominator approaches 0

through negative values. Thus, the= −∞.

15. Since f (x) is a rational function, it iscontinuous everywhere except at values wherethe denominator is 0. Factoring and setting thedenominator equal to 0, you have (x + 6)(x − 2) = 0. Thus, the points of discontinuityare at x = −6 and x = 2. Verify your result witha calculator. (See Figure 2.7-2.)

limx

x

x→ +

+ [ ]−5

5

5

[–8,8] by [–4,4]

Figure 2.7-2

16. In order for g (x) to be continuous at x = 3, itmust satisfy the three conditions of continuity:

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Thus, the line y = 1 is a

horizontal asymptote. The

As x → −∞, x = − . Thus,

Therefore, the line

y = −1 is a horizontal asymptote. As for verticalasymptotes, f (x) is continuous and defined for allreal numbers. Thus, there is no verticalasymptote.

=

− +

= −→−∞

lim .x

x

1

14

1

2

lim limx x

x

x

x

x

x

x

→−∞ →−∞+=

+

−

2 2

2

4 4

x2=+→−∞

lim .x

x

x2 4

limx

f x→−∞

( )

=

+

=→∞

lim .x

x

1

14

1

2


25. To find horizontal asymptotes, examine the

f (x) and the f (x). The

Dividing by the highest

power of x (and in this case, it’s x), you obtain

As x → ∞, Thus, you have

lim limx x

x

x

x

x

x

x

→∞ →∞+=

+2

2

2

2

4

1

4

x2lim .x

x

x

x

x

→∞ +2 4

=

+→∞

lim .x

x

x

2

4

limx

f x→∞

( )limx→−∞

limx→∞

limsin

limcos

.x x

x

x x→ →= ( )( ) =

0 0

11 1 1

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3.1 DERIVATIVES OF ALGEBRAIC FUNCTIONS

Main Concepts: Definition of the Derivative of a Function; Power Rule; The Sum, Difference, Product, and Quotient Rules; The Chain Rule

Definition of the Derivative of a Function

The derivative of a function f, written as f ′, is defined as

if this limit exists. (Note that f ′(x) is read as f prime of x.)

Other symbols of the derivative of a function are:

Let mtangent be the slope of the tangent to a curve y = f (x) at a point on the curve. Then

(See Figure 3.1-1.)

Given a function f, if f ′(x) exists at x = a, then the function f is said to be differentiableat x = a. If a function f is differentiable at x = a, then f is continuous at x = a. (Note thatthe converse of the statement is not necessarily true, i.e., if a function f is continuous atx = a, then f may or may not be differentiable at x = a.) Here are several examples of functions that are not differentiable at a given number x = a). (See Figures 3.1-2 to3.1-5 on page 100.)

m f x f x h f x

h

m x a f a f a h f a

h

f x f a

x a

h

h x a

tangent

tangent at or

= ′( ) = +( ) − ( )

=( ) = ′( ) = +( ) − ( ) ( ) − ( )

−

→

→ →

lim

lim lim .

0

0

D f d

dx f x y f x y

dy

dx D yx x, , , , , .( ) = ( ) ′and if and

′( ) = +( ) − ( )

→f x

f x h f x

hhlim ,

0

Chapter 3

Differentiation


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Example 1

If f (x) = x2 − 2x − 3, find (a) f ′(x) using the definition of derivative, (b) f ′(0), (c) f ′(1),and (d) f ′(3).

(a) Using the definition of derivative,

= + − + − − − −[ ]

= + + − − −[ ] − − −[ ] = + −

→

→ →

lim[( ) ( ) ]

lim lim

h

h h

x h x h x x

h

x xh h x h x xh xh h hh

0

2 2

0

2 2 2

0

2

2 3 2 3

2 2 2 3 2 3 2 2

′( ) = +( ) − ( )

→f x

f x h f x

hhlim

0

f(x) y

x

0

tangent

.(a, f(a))

Slope of tangent to f(x)at x=a is m = f ' (a)

y

x0

f(x)

x=a

a

f is discontinuousat x=a

f has a cornerat x=a

y

x0

.

f(x)

(a, (f(a))

a

f has a cusp at x=a

y

x0

.

a

(a, (f(a))

f(x)

f(x)

x=a

f has a verticaltangent at x=a

y

x0

.(a, (f(a))

a

Figure 3.1-1

Figure 3.1-2 Figure 3.1-3

Figure 3.1-4 Figure 3.1-5

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Differentiation • 101

[–3.14,6.28] by [–3,3]

Figure 3.1-6

Example 3

If the function f (x) = x2/3 + 1, find all points where f is not differentiable.The function f (x) is continuous for all real numbers and the graph of f (x) forms a

“cusp” at the point (0, 1). Thus, f (x) is not differentiable at x = 0. See Figure 3.1-7.

[–5,5] by [–1,6]

Figure 3.1-7

(b) f ′(0) = 2(0) − 2 = −2 (c) f ′(1) = 2(1) − 2 = 0 and (d) f ′(3) = 2(3) − 2 = 4.

Example 2

Evaluate

The expression is equivalent to the derivative of the func-

tion f (x) = cos x at x = π, i.e., f ′(π). The derivative of f (x) = cos x at x = π is equivalentto the slope of the tangent to curve of cos x at x = π. The tangent is parallel to the

x-axis. Thus, the slope is 0 or

Or, using an algebraic method, note that cos(a + b) = cos(a) cos(b) − sin(a) sin(b).

Then rewrite

(See Figure 3.1-6.)

= − ( ) − −( ) = − ( ) + = − ( ) −[ ] = − ( ) −[ ] =→ → → →

lim cos lim cos lim cos lim cos .h h h h

hh

hh

hh

hh0 0 0 0

1 1 1 1 0

limcos cos

limcos cos sin sin cos

h h

h

h

h h

h→ →

+( ) − ( )=

( ) ( ) − ( ) ( ) − ( )0 0

π π π π π

limcos cos

.h

h

h→

+( ) − ( )=

00

π π

limcos cos

h

h

h→

+( ) − ( )0

π π

limcos cos

h

h

h→

+( ) − ( )0

π π

= + −( ) = + −( ) = −

→ →lim lim .h h

h x h

h x h x

0 0

2 22 2 2 2

Example 4

Using a calculator, find the derivative of f (x) = x2 + 4x at x = 3.There are several ways to find f ′(3), using a calculator. One way is to use the nDeriv

function of the calculator. From the main (Home) screen, select F3-Calc and then selectnDeriv. Enter nDeriv (x2 + 4x, x)x = 3. The result is 10.

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Derivative withFunction Written in cxn form Derivative Positive Exponents

3x 3x1 3x0 = 3 3

−5x7 −5x7 −35x6 −35x6

x−2 −2x−3

− −1

32

3x x or

1

3 23−−1

3

2

3x−x

1

3−x1

3

13

23

x x or

1

x−3

2

−= −

−21

2

1

2

x

x2−2

x

−23x

12x

4 41

2

x x

or

4

1

2

x

−

8

1

2

x8 x


Power Rule

If f (x) = c where c is a constant, then f ′(x) = 0.If f (x) = xn where n is a real number, then f ′(x) = nxn−1.If f (x) = cxn where c is a constant and n is a real number, then f ′(x) = cnxn−1.

Summary of Derivatives of Algebraic Functions:

Example 1

If f (x) = 2x3, find (a) f ′(x), (b) f ′(1) and (c) f ′(0)

Note that (a) f ′(x) = 6x2, (b) f ′(1) = 6(1)2 = 6, and (c) f ′(0) = 0.

Example 2

Note that does not exista and thus, and b2( ) = = = − = −

( )− −

=

yx

x dy

dx x

x

dy

dx x

12

22

3

30

If find a and b which represents atyx

dy

dx

dy

dx

dy

dx x

x

= ( ) ( ) =

=

102

0

, .

d

dx c

d

dx x nx

d

dx c cnxn n n n( ) = ( ) = ( ) =− −0 1 1, , and

• Always write out all formulas in your solutions.

because the expression is undefined.

Example 3

Here are several examples of algebraic functions and their derivatives:

−2

0

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Example 4

Using a calculator, find f ′(x) and f ′(3) if f (x) =

There are several ways of finding f ′(x) and f ′(9) using a calculator. One way to usethe d (differentiate) function. Go to Home screen. Select F3-Calc and then select

d (differentiate). Enter The result is To find f ′(3), enter

The result is

The Sum, Difference, Product, and Quotient Rules

If u and v are two differentiable functions, then

Summary of Sum, Difference, Product and Quotient Rules:

Example 1

Find f ′(x) if f (x) = x3 − 10x + 5.Using the sum and difference rules, you can differentiate each term and obtain f ′(x)= 3x2 − 10. Or using your calculator, select the d (differentiate) function and enterd (x3 − 10x + 5, x) and obtain 3x2 − 10.

Example 2

If y = (3x − 5)(x4 + 8x − 1), find

Using the product rule let u = (3x − 5) and v = (x4 + 8x − 1).

Then = (3)(x4 + 8x − 1) + (4x3 + 8)(3x − 5) = (3x4 + 24x − 3) + (12x4 − 20x3 +

24x − 40) = 15x4 − 20x3 + 48x − 43. Or you can use your calculator and enter d ((3x − 5)(x4 + 8x − 1), x) and obtain the same result.

Example 3

Using the quotient rule , let u = 2x − 1 and v = x + 5. Then f ′(x) =u

v

u v v u

v

′=

′ − ′2

If findf x x

x f x( ) =

−+

′( )2 1

5, .

dydx

d

dx uv v

du

dx u

dv

dx( ) = + ,

dy

dx.

u v u v uv u v v u u

v

u v v u

v±( )′ = ′ ± ′ ( )′ = ′ + ′

′

= ′ − ′

&2

d

dx u v

du

dx

dv

dx

d

dx uv v

du

dx u

dv

dx

d

dx

u

v

v du

dx u

dv

dxv

v

±( ) = ±

( ) = +

=

−≠

Sum & Difference Rules

Product Rule

Quotient Rule2

0,

′( ) = −f 3 154

.d x x x1 ( )( ) =, .3

′( ) = −

f x

x

1

23

2

.d x x1 ( )( ), .

1

x.


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Or you can use

your calculator and enter d ((2x − 1)/(x + 5), x) and obtain the same result.

Example 4

Using your calculator, find an equation of the tangent to the curve f (x) = x2 − 3x + 2 atx = 5.

Find the slope of the tangent to the curve at x = 5 by entering d (x2 − 3x + 2, x)x = 5.The result is 7. Compute f (5) = 12. Thus, the point (5,12) is on the curve of f (x). Anequation of the line whose slope m = 7 and passing through the point (5,12) is y − 12 =7(x − 5).

2 5 1 2 1

5

2 10 2 1

5

11

55

2 2 2

( ) +( ) − ( ) −( )

+( ) =

+ − +

+( ) =

+( ) ≠ −

x x

x

x x

x xx, .

• Remember that The integral

formula is not usually tested in the AB exam.

d

dx x

x x dx x x x cln ln ln .= = − +∫

1and

The Chain Rule

If y = f (u) and u = g (x) are differentiable functions of u and x respectively, then

Example 1

Using the chain rule, let u = 3x − 5 and thus, y = u10. Then,

Since Or you can use

your calculator and enter d ((3x − 5)10, x) and obtain the same result.

Example 2

If findf x x x f x( ) = − ′( )5 25 2 , .

dy

dx u x x= ( )( ) = −( ) ( ) = −( )10 3 10 3 5 3 30 3 59 9 9

.dy

dx

dy

du

du

dx= ,

dy

du u

du

dx= =10 39 and .

If findy x dy

dx= −( )3 5

10

, .

d

dx f g x f g x g x

dy

dx

dy

du

du

dx( )( )[ ] = ′ ( )( ) ′( ) = or

Using the product rule,

To find use the chain rule and let u = 25 − x2.

Thus, Substituting this quantity

back into f ′(x), you have

′( ) = −( ) + ( )

−

−( )

=

−( ) −

−( ) =

−

−( )f x x x

x

x

x x

x

x

x5 25 5 25

5 25 5

25

125 10

25

21

2

212

2 2

212

2

212 .

d

dx x x x

x

x

251

225 2

25

21

2 2 1 2

21

2

−( ) = −( ) −( ) = −

−( )

−.

d

dx x25 2

1

2−( ) ,

′( ) = −( ) ( ) + ( ) −( ) = −( ) + ( ) −( )f x x d

dx x x

d

dx x x x

d

dx x25 5 5 25 5 25 5 252

1

2 21

2 21

2 21

2 .

Rewrite asf x x x f x x x( ) = − ( ) = −( )5 25 5 252 21

2 .

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Or you can use your calculator and enter and obtain the same result.

Example 3

Using the chain rule, let

To find use the quotient rule.

Thus, Substituting this quantity

back into

An alternate solution is to use the product rule and rewrite as

y = and use the quotient rule. Another approach is to express

y = (2x − 1)3(x−6) and use the product rule. Of course, you can always use your calcula-tor if you are permitted to do so.

3.2 DERIVATIVES OF TRIGONOMETRIC, INVERSETRIGONOMETRIC, EXPONENTIAL, ANDLOGARITHMIC FUNCTIONS

Main Concepts: Derivatives of Trigonometric Functions, Derivatives of InverseTrigonometric Functions, Derivatives of Exponential and

Logarithmic Functions

Derivatives of Trigonometric Functions

Summary of Derivatives of Trigonometric Functions:

Note that the derivatives of cosine, cotangent, and cosecant all have a negative sign.

Example 1

If findy x x dy

dx

dydx x x x

= +

= +

6 3

12 3

2 sec , .

sec tan .

d

dx

x x d

dx

x x

d

dx x x

d

dx x x

d

dx x x x

d

dx x x x

sin cos sin

tan sec csc

sec sec tan csc cot .

( ) = ( ) = −

( ) = ( ) = −

( ) = ( ) = −

cos

cot

csc

2 2

= −( )2 1

3

6

x

x

2 13

2 3

x

x

−( )

( )

y x

x= −

2 1

2

3

dy

dx

x

x

d

dx

x

x

x

x

x x

x

x x

x= −

−

= −

− +

= − −( ) −( )

32 1 2 1

32 1 2 2 6 1 2 1

2

2

2 2

2 2

4

2

7.

d

dx

x

x

x x x

x

x x

x

2 1 2 2 2 1 2 22

2

2 2

2

4

−

= ( )( ) − ( ) −( )

( ) =

− +.

d

dx

x

x

2 12

−

,

u x

x

dy

dx

x

x

d

dx

x

x

= −

= −

−

2 13

2 1 2 12 2

2

2. .Then

If findy x

x

dy

dx= −

2 1

2

3

, .

d x x x5 25 2−( ),


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Example 2

Find f ′(x) if f (x) = cot (4x − 6).Using the chain rule, let u = 4x − 6. Then f ′(x) = [−csc2 (4x − 6)] [4] = −4 csc2 (4x − 6).

Or using your calculator, enter d (1/tan(4x − 6), x) and obtain which is an

equivalent form.

Example 3

Find f ′(x) if f (x) = 8 sin(x2).Using the chain rule, let u = x2. Then f ′(x) = [8 cos(x2)] [2x] = 16x cos(x2).

Example 4If y = sin x cos(2x), find

Using the product rule, let u = sin x and v = cos(2x).

Then = cos x cos(2x) + [−sin(2x)] (2) (sin x) = cos x cos(2x) − 2 sin x sin(2x).

Example 5

If y = sin [cos(2x)], find

Using the chain rule, let u = cos(2x). Then

To evaluate , use the chain rule again by making another u-substitution,

this time for 2x. Thus, Therefore,

Example 6

Find f ′(x) if f (x) = 5x csc x.Using the product rule, let u = 5x and v = csc x. Thenf ′(x) = 5 csc x + (−csc x cot x) (5x) = 5 csc x − 5x (csc x) (cot x).

Example 7

Rewrite Using the chain rule, let u = sin x. Thus,

Example 8

Using the quotient rule, let u = tan x and v = (1 + tan x). Then,

dy

dx

x x x x

x

x x x x x

x=

( ) +( ) − ( )( )

+( ) =

+ ( )( ) − ( )( )

+( )

sec tan sec tan

tan

sec sec tan sec tan

tan

2 2

2

2 2 2

2

1

1 1

If findy x

x

dy

dx=

+tan

tan, .

1

dy

dx x x

x

x

x

x= ( ) ( ) =

( )=

−1

22 2

1

21

2

sin coscos

sin

cos

sin.

asy x y x= = ( )sin sin .1 2

If findy x dy

dx= sin , .

dy

dx

x x x xcos cos sin sin cos cos .2 2 2 2 2 2( )[ ] − ( )( ) = − ( ) ( )[ ]

d

dx x x xcos sin sin .2 2 2 2 2( )[ ] = − ( )[ ] = − ( )

d

dx xcos 2( )[ ]

dy

dx

dy

du

du

dx x

d

dx x= = ( )[ ] ( )[ ] cos cos cos .2 2

dydx

.

dy

dx

dy

dx.

−−( )

4

4 62sin x


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Note: For all of the above exercises, you can find the derivatives by using a calculatorprovide that you are permitted to do so.

Derivatives of Inverse Trigonometric Functions

Summary of Derivatives of Inverse Trigonometric Functions:

Let u be a differentiable function of x, then

Note that the derivatives of cos−1 x, cot−1 x and csc−1 x all have a “−1” in their numerators.

Example 1

If y = 5 sin−1 (3x), find

Let u = 3x. Then

Or using a calculator, enter d [5 sin−1 (3x), x] and obtain the same result.

Example 2

Example 3

Let u = 3x2. Then

dy

dx x x

du

dx x xx

x x=

( ) −=

− ( ) =

−

1

3 3 1

1

3 9 16

2

9 12 2 2 2 2.

If findy x dy

dx= ( )−sec , .1 23

Let Thenu x f x x

du

dx x x x x

x x

= ′( ) = + ( ) = +

= +

=+( )

−

.

.

1

1

1

1

1

2

1

1

1

2

1

2 1

2

1

2

Find if ′( ) ( ) = −f x f x xtan .1

dy

dx x

du

dx x x= ( )

− ( )=

− ( )( ) =

−5

1

1 3

5

1 33

15

1 92 2 2.

dydx

.

d

dx u

u

du

dx u

d

dx u

u

du

dx u

d

dx u

u

du

dx

d

dx u

u

du

dx

d

dx u

u u

du

dx u

d

dx u

u u

du

dx u

sin , cos ,

tan cot

sec , csc , .

− −

− −

− −

=−

< = −−

<

=+

= −

+

=−

> = −

−>

1

2

1

2

1

2

1

2

1

2

1

2

1

11

1

11

1

1

1

1

1

11

1

11

=+( )

( )

+

= ( )

+

=

+( )

sec

tan,

cos

sin

cos

cos

cos sincos

cos sin

.

2

2

2

2

2

2 2

1

1

1

1

1

x

x

x

x

x

x

x xx

x x

which is equivalent to


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Example 4

Note: For all of the above exercises, you can find the derivatives by using a calculator

provided that you are permitted to do so.

Derivatives of Exponential and Logarithmic Functions

Summary of Derivatives of Exponential and Logarithmic Functions:

Let u be a differentiable function of x, then

For the following examples, find and verify your result with a calculator.

Example 1

Example 2

y = xex − x2ex

Using the product rule for both terms, you have

Example 3

y = 3sin x

Let u = sin x. Then, dy

dx

du

dx x xx x x= ( )( ) = ( )( ) = ( )( )3 3 3 3sin sin sincos cos .ln ln 3 ln 3

dy

dx e e x x e e x e xe xe x e e xe x e

x e xe e e x x

x x x x x x x x x x x

x x x x

= ( ) + ( ) − ( ) + ( )[ ] = + − − = − −

= − − + = − − +( )

1 2 2

1

2 2 2

2 2 .

y e xe e

dy

dx e e e e e

x

x x

= + +

= ( )( ) + + = +

3 3 3

3 3 3 3 3

5

3 5 0 3 5 (Note that is a constant.)

dy

dx

d

dx e e

du

dx

d

dx a a a

du

dx a a

d

dx u

u

du

dx u

d

dx u

u a

du

dx a a

u u u u

a

( ) = ( ) = ≠

( ) =

( ) = ≠

ln , &

ln , logln

, & .

>

> >

0 1

10

10 1

=−

1

12x x.

Therefore, dy

dx

x

du

dx

x

x x

x x

x x

x

= −

−

= −

−

−=

−( )

=− ( )

1

11

1

11

1 1

1

1

12 22

2

2

22 2

Rewrite as Thenux

u x du

dx x

x=

= = − = −− −1

111 2

2. .

Let Thenux

dy

dx

x

du

dx=

= −

−

1 1

11

2. .

If findyx

dy

dx=

−cos , .1 1


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Example 4

Let u = x3. Then,

Example 5

y = (ln x)5

Let u = ln x. Then,

Example 6

y = ln(x2 + 2x − 3) + ln5

Let u = x2 + 2x − 3. Then,

(Note that ln5 is a constant. Thus the derivative of ln5 is 0.)

Example 7y = 2xln x + x

Using the product rule for the first term,

you have

Example 8

y = ln (ln x)

Let u = ln x. Then,

Example 9

y = log5 (2x + 1)

Let u = 2x + 1. Then

Example 10

Write an equation of the line tangent to the curve of y = ex at x = 1.The slope of the tangent to the curve y = ex at x = 1 is equivalent to the value of the

derivative of y = ex evaluated at x = 1. Using your calculator, enter d (e^(x), x)x = 1 andobtain e. Thus, m = e, the slope of the tangent to the curve at x = 1. At x = 1, y = e1 = e,and thus the point on the curve (1, e). Therefore, the equation of the tangent is y − e =e(x − 1) or y = ex. (See Figure 3.2-1.)

dy

dx x

du

dx x x=

+( ) =

+( ) ( ) =

+( )1

2 1 5

1

2 1 52

2

2 1 5ln ln ln.

dy

dx x

du

dx x x x x= =

=

1 1 1 1

ln ln ln.

dy

dx x

x x x x= ( ) +

( ) + = + + = +2

12 1 2 2 1 2 3ln ln ln .

= ++ −2 2

2 32

x

x x.

dy

dx x x

du

dx x x x=

+ − + =

+ − +( )

1

2 30

1

2 32 2

2 2

dydx

x dudx

xx

xx

= ( ) = ( )

= ( )5 5 1 54 4

4

ln ln ln .

dy

dx e

du

dx e x x ex x

x

= [ ] = [ ] =( ) ( ) ( )3 33

3 32 2 .

y e x= ( )3

[–1,3] by [–2,8]

Figure 3.2-1

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3.3 IMPLICIT DIFFERENTIATION

Procedure for Implicit Differentiation: Given an equation containing the variables x

and y for which you cannot easily solve for y in terms of x, you can find by:

Steps: 1. Differentiate each term of the equation with respect to x.

2. Move all terms containing to the left side of the equation and all other

terms to the right side.

3. Factor out on the left side of the equation.

4. Solve for .

Example 1

Step 1: Differentiate each term of the equation with respect to x. (Note y is treated as

a function of x.)

Step 2: Move all terms containing to the left side of the equation and all other

terms to the right:

Step 3: Factor out

Step 4: Solve for

Example 2

Step 1: Differentiate each term with respect to x:

Step 2: Move all

Step 3: Factor out

Step 4: Solve for

Example 3

Find ifdydx

x y x y x y+( ) − −( ) = +2 2 5 5

dy

dx

dy

dx

y x

y x

y x

y x: =

−−

= −

−6 3

3 6

2

2

2

2

2

2

dy

dx

dy

dx y x y x: 23 6 6 3 2−( ) = −

dy

dx y

dy

dx x

dy

dx y xterms to the left side: 23 6 6 3 2− = −

3 3 6 6x y dydx

y dydx

x2 2+ = ( ) + ( )

Given , find3 3x y xy dy

dx+ = 6 .

dy

dx

dy

dx

x

y: =

− +−

2 4

2 7

dy

dx

dy

dx y x: 2 7 2 4−( ) = − +

2 7 2 4y dydx

dydx

x− = − +

dy

dx

2 7 2 4 0y dy

dx

dy

dx x− + − =

Find if 2 2dy

dx y y x x− + − =7 4 10.

dy

dx

dy

dx

dy

dx

dy

dx

• Guess, if you can eliminate some of the choices in a multiple-choice question.

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Step 1: Differentiate each term with respect to x:

Distributing 2(x + y) and −2(x − y), you have

Step 2: Move all terms to the left side:

Step 3: Factor out :

Step 4: Solve for

Example 4

Write an equation of the tangent to the curve x2 + y2 + 19 = 2x + 12y at (4, 3)

The slope of the tangent to the curve at (4, 3) is equivalent to the derivative at (4, 3).

Using implicit differentiation, you have

Thus the equation of the tangent is y − 3 = (1)(x − 4) or y − 3 = x − 4.

Example 5

Find if

cos

dy

dx x y x

x y dy

dx

dy

dx x y

dydx x y

, sin

cos

cos

+( ) =

+( ) +

=

+ =+( )

= +( ) −

2

1 2

12

2 1

2 2 2 12

2 12 2 2

2 12 2 2

2 2

2 12

1

6

1 4

3 61

4 3

x y dy

dx

dy

dx

y dy

dx

dy

dx x

dy

dx y x

dy

dx

x

y

x

y

dy

dx

+ = +

− = −

−( ) = −

= −

− =

−−

= −

− =

( )

and,

dy

dx

dy

dx

dy

dx

x y

x y: =

−−

5 4

4 5

4

4

dy

dx x y x y y x x y x y

dy

dx x y x y y x y

dy

dx x y x y

2 2 5 5 2 2 2

2 2 2 2 5 5 4

4 5 5 4

4 4

4 4

+( ) + −( ) −[ ] = − − + −

+ + − −[ ] = −

−[ ] = −

4 4 2

dy

dx

2 2 5 5 2 24 4x y dy

dx x y

dy

dx y

dy

dx x x y x y+( ) + −( ) − = − +( ) + −( )

dy

dx

2 2 2 5 5 4x y x y dy

dx x y x y

dy

dx x y

dy

dx+( ) + +( ) − −( ) + −( ) = +2 4

2 1 2 1 5 5x y dy

dx x y

dy

dx x y

dy

dx+( ) +

− −( ) −

= +4 4


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3.4 APPROXIMATING A DERIVATIVE

Given a continuous and differentiable function, you can find the approximate value of a derivative at a given point numerically. Here are two examples.

Example 1

The graph of a function f on [0, 5] is shown in Figure 3.4-1. Find the approximate value

of f ′(3). (See Figure 3.4-1.)

0

1

2

3

4

5

6

7

8

1 2 3 4 5 6 7

y

x

f

Figure 3.4-1

Since f ′(3) is equivalent to the slope of the tangent to f (x) at x = 3, there are severalways you can find its approximate value.

Method 1: Using slope of the line segment joining the points at x = 3 and x = 4.

Method 2: Using the slope of the line segment joining the points at x = 2 and x = 3.

Method 3: Using the slope of the line segment joining the points at x = 2 and x = 4.

Note that is the average of the results from methods 1 and 2.3

2

f f

m f f

2 2 4 5

4 2

4 2

5 2

4 2

3

2

( ) = ( ) =

= ( ) − ( )

− =

−−

=

and

f f

m f f

2 2 3 3

2 2

3 2

3 2

3 21

( ) = ( ) =

= ( ) − ( )−

= −−

=

and

f f

m f f

3 3 4 5

4 3

4 3

5 3

4 32

( ) = ( ) =

= ( ) − ( )

− =

−−

=

and

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• Remember that the lim sinsin lim sin .x xxx xx→ →= = =0 062 62 3 1because the

Thus depending on which line segment you use.

Example 2

Let f be a continuous and differentiable function. Selected values of f are shown below.Find the approximate value of f ′ at x = 1.

′( ) ≈f 3 1 2, or3

2

[–2,4] by [–2,4]

Figure 3.4-2

You can use the difference quotient to approximate f ′(a).

Or, you can use the symmetric difference quotient to approximatef ′(a).

Thus, f ′(3) ≈ 0.49, 0.465, 0.54 or 0.63 depending on your method.Note that f is decreasing on (−2, −1) and increasing on (−1, 3). Using the symmet-

ric difference quotient with h = 3 would not be accurate. (See Figure 3.4-2.)

Let

Let

h f f f

h f f f

= ′( ) ≈ ( ) − ( )

− ≈

−≈

= ′( ) ≈ ( ) − −( )

− −( ) ≈

−≈

1 12 0

2 0

2 08 1

20 54

2 13 1

3 1

2 52 0

40 63

;.

.

;.

.

f a h f a hh+( ) − −( )2

Let

Let

h f f f

h f f f

= ′( ) ≈ ( ) − ( )

− ≈

−≈

= ′( ) ≈ ( ) − ( )

− ≈

−≈

1 12 1

2 1

2 08 1 59

10 49

2 13 1

3 1

2 52 1 59

20 465

;. .

.

;. .

.

f a h f a

h

+( ) − ( )

x −2 −1 0 1 2 3

f 1 0 1 1.59 2.08 2.52

3.5 DERIVATIVES OF INVERSE FUNCTIONS

Let f be a one-to-one differentiable function with inverse function f −1.If f ′(f −1(a)) ≠ 0, then the inverse function f −1 is differentiable at a and

(See Figure 3.5-1.)f af f a

−−( )′ ( ) =

′ ( )( )1

1

1.

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Example 1

If f (x) = x3 + 2x − 10, find (f −1)′(x)

Step 1: Check if (f −1)′(x) exists. f ′(x) = 3x2 + 2 and f ′(x) > 0 for all real values of x.Thus f (x) is strictly increasing which implies that f (x) is 1 − 1. Therefore(f −1)′(x) exists.

Step 2: Let y = f (x) and thus y = x3 + 2x − 10.Step 3: Interchange x and y to obtain the inverse function x = y3 + 2y − 10.

Step 4: Differentiate with respect to

Step 5: Apply formula

Example 2

Example 1 could have been done by using implicit differentiation.

Step 1: Let y = f ′(x), and thus y = x3 + 2x − 10.Step 2: Interchange x and y to obtain the inverse function x = y3 + 2y − 10.Step 3: Differentiate each term implicitly with respect to x.

1 3 2 0= + −y dy

dx

dy

dx2

d

dx x

d

dx y

d

dx y

d

dx( ) = ( ) + ( ) − −( )3 2 10

dx

dy dxdy

y f x

y= =

+ ( )′ ( ) =

+−1 1

3 2

1

3 22

1

2. Thus,

dx

dy dxdy

=1

.

y dxdy

y: 2= +3 2

If so that then withy f x x f y dy

dx dxdy

dx

dy= ( ) = ( ) = ≠−1 1

0, .

( ) ( )( ( )) a f f '

a f –1

–1 ' 1

(a, (f –1

(a))

y

x

0

y=x

f

f -1

..(f

– 1(a),a)

m=(f – 1

)'

(a)

m=f'(

=

f – 1

(a))

Figure 3.5-1

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Step 4: Solve for

Example 3

If f (x) = 2x5 + x3 + 1, find (a) f (1) and f ′(1) and (b) (f −1)(4) and (f −1)′(4).Enter y1 = 2x5 + x3 + 1. Since y1 is strictly increasing, thus f (x) has an inverse.

(a) f (1) = 2(1)5 + (1)3 + 1 = 4f ′(x) = 10x4 + 3x2

f ′(1) = 10(1)4 + 3(1)2 = 13

(b) Since f (1) = 4 implies the point (1,4) is on the curve f (x) = 2x5 + x3 + 1, thereforethe point (4,1) (which is the reflection of (1,4) on y = x) is on the curve (f −1)(x).Thus (f −1)(4) = 1

Example 4

If f (x) = 5x3 − x + 8, find (f −1)′(8).Enter y1 = 5x3 − x + 8. Since y1 is strictly increasing near x = 8, f (x) has an inverse nearx = 8.Note that f (0) = 5(0)3 − 0 + 8 = 8 which implies the point (0,8) is on the curve of f (x).Thus, the point (8,0) is on the curve of (f −1)(x).

f ′(x) = 15x2 − 1f ′(0) = −1

Therefore f f

−( )′( ) =′( )

=−

= −1 81

0

1

11.

f f

−( )′ ( ) =′( )

=1 41

1

1

13

dy

dx y f x

y=

+ ( )′ ( ) =

+−1

3 2

1

3 22

1

2. Thus

1 3 22= +( )dy

dx y

dy

dx.

3.6 HIGHER ORDER DERIVATIVES

If the derivatives f ′ of a function f is differentiable, then the derivative of f ′ is the sec-ond derivative of f represented by f ″ (reads as f double prime). You can continue todifferentiate f as long as there is differentiability.

Some of the Symbols of Higher Order Derivatives:

f ′(x), f ″ (x), f ′″ (x), f (4)(x)

dy

dx

d y

dx

d y

dx

d y

dx,

,,

2

2

3

3

4

4

• Leave a multiple-choice question blank, if you have no clue. You don’t have toanswer every question to get a 5 on the AP Calculus AB exam.

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y ′ , y ″ , y ′″ , y (4)

Dx(y), D2x(y), D3

x(y), D4x(y)

Note that

Example 1If y = 5x3 + 7x − 10, find the first four derivatives.

Example 2

Differentiate again:

Example 3

If y = x cos x, find y″ .Using the product rule, y′ = (1)(cos x) + (x)(−sin x) = cos x − xsin x

y″ = −sin x − [(1)(sin x) + (x)(cos x)] = −sin x − sin x −xcos x

= −2sin x − xcos x

Or, you can use a calculator and enter d (x*cosx, x, 2) and obtain the same result.

′′( ) = − = −

= −

′′( ) = −

= −−

f x xx x

f 1

4

1

4

1

44

1

4 4

1

32

32

32 3 3

and

Rewrite: and differentiate:f x x x f x x( ) = = ′( ) = −1

21

21

2

If find 4f x x f ( ) = ′′( ), .

dy

dx x

d y

dx x

d y

dx

d y

dx= + = = =15 7 30 30 0

2

2

3

3

4

4

2 ; ; ;

d y

dx

d

dx

dy

dx

dy

dx

2

2 =

′ or .

3.7 RAPID REVIEW

1. If

Answer: Using the chain rule,

2. Evaluate

Answer: The limit is equivalent to

3. Find f ′(x) if f (x) = ln(3x).

Answer: ′( ) = ( ) =f x

x x

1

3

31

.

d

dx x

xcos sin .

== −

= −π

π6 6

1

2

limh

h

h→

+

−

0

6 6cos cos

.

π π

dy

dx e xx

= ( )( )

2

2 .

y e dy

dxx=

3

, .find

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4. Find the approximate value of f ′(3). (See Figure 3.7-1.)



Find the derivative of each of the following functions.

1. y = 6x5 − x + 10

2.

3.

4.

5. f (x) = (3x − 2)5

(x2

− 1)

y x

x=

−

2

65 1

y x

x=

−5 16

2

f xx x

( ) = +1 1

23

6.

7. y = 10 cot(2x − 1)

8. y = 3xsec(3x)

9.

10. y = 8 cos−1 (2x)

11. y = 3e5 + 4xe x

12. y = ln(x2

+ 3)

y x= −( ) 10 42cos sin

y

x

x=

+

−

2 1

2 1

y

x

f

(4,3)

(2,1)0

Figure 3.7-1

Answer: Using the slope of the line segment joining (2,1) and

5. Find if xy = 5x2.

Answer: Using implicit differentiation,

6. If

Answer: Rewrite y = 5x−2. Then

7. Using a calculator, write an equation of the line tangent to the graph f (x) = −2x4 atthe point where f ′(x) = −1.

Answer: f ′(x) = −8x3

. Using a calculator, enter Solve (−8x^3 = −1, x) and

obtain Using your calculator Thus tangent

is y x+ = − −

1

81

1

2.

f 1

2

1

8

= − .x f = ⇒ ′

= −

1

2

1

21.

dy

dx x

d y

dx x

x= − = =− −10 30

303

2

2

4

4and .

yx

d y

dx=

52

2

2, .find

1 1010

y x dy

dx x

dy

dx

x y

x+ = =

−. .Thus

dy

dx

4 3 33 1

4 21, , .( ) ′( ) ≈

−−

=f

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25. (Calculator) If

determine if f (x) is continuous at x = 3.)Explain why or why not.

f xx

x x

x

( ) =−−

≠

=

2 9

33

3 3

,

,

,


15. Let f be a continuous and differentiable function.Selected values of f are shown below. Find theapproximate value of f ′ at x = 2.

below. Find the approximate value of f ′ atx = 1.


13.

14. The graph of a function f on [1, 5] is shown inFigure 3.8-1. Find the approximate value of f ′(4).

Find , ifdy

x x y xy

310 52 3+ = − .

x −1 0 1 2 3

f 6 5 6 9 14

0 1 2 3 4 5 6

1

2

3

4

5

6

y

f

x

Figure 3.8-1

x 0 1 2 3 4 5f 3.9 4 4.8 6.5 8.9 11.8

(“Calculator” indicates that calculators are permitted)

21. Find

22. If f (x) = cos2(π − x), find f ′(0).

23. Find

24. (Calculator) Let f be a continuous and differen-tiable function. Selected values of f are shown

limx

x

x x→∞

−+ −

25

10 2 2.

limh o

h

h→

+

−

sin sin

.

π π2 2

16. If f (x) = x5 + 3x − 8, find (f −1)′(−8).

17. Write an equation of the tangent to the curve y =ln x at x = e.

18. If y = 2xsin x, find

19. If the function f (x) = (x − 1)2/3 + 2, find all pointswhere f is not differentiable.

20. Write an equation of the normal line to the curve

x cos y = 1 at 23

, .π

d y

dx x

2

2 2at =

π.




1. Applying the power rule,

2. Rewrite

as f x x x( ) = +− −

12

3 .

f xx x

( ) = +1 1

23

dy

dx x= −30 14 .

Differentiate,

3. Rewrite

y x

x y

x

x x x x=

−= − = − −5 1 5 1

56

2

6

2 2

4 2 as .

′( ) = − − = − −− −

f x x xx x

25

3

2 53

2

3

1 2

3

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Differentiate,

An alternate method is to differentiate

directly, using the quotient rule.

4. Applying the quotient rule,

5. Applying the product rule, u = (3x − 2)5 andv = (x2 − 1) and then the chain rule,

6. Rewrite

Applying first the chain rule and then the quo-tient rule,

dy

dx

x

x

x x

x

x

x

x

x

x

x

= +−

( ) −( ) − ( ) +( )

−( )

=+−

−−( )

=+( )

−( )

−

−

−1

2

2 1

2 1

2 2 1 2 2 1

2 1

1

2

1

2 1

2 1

4

2 1

1

2

1

2 1

2 1

4

2 1

12

2

12

2

12

12

(( )

= −

+( ) −( )

2

12

32

2

2 1 2 1x x

y x

x y

x

x=

+−

= +−

2 1

2 1

2 1

2 1

12

as .

′( ) = − − + −

= −( ) −( ) + −( )

= −( ) −( ) + −( )[ ]

= −( ) − + −[ ]

= −( )

f x x x x x

x x x x

x x x x

x x x x

x

[ ( ) ( )][ ] [ ][( ) ]5 3 2 3 1 2 3 2

15 1 3 2 2 3 2

3 2 15 1 2 3 2

3 2 15 15 6 4

3 2

4 2 5

2 4 5

4 2

42 2

44 221 4 15x x− −( )

= − −

−( )

= − +( )

−( )

20 2

5 1

2 10 1

5 1

7

6 2

6

6 2

x x

x

x x

x

dy

dx

x x x x

x

x x x

x

= ( ) −( ) − ( )( )

−( )

= − −

−( )

2 5 1 30

5 1

10 2 30

5 1

6 5 2

6 2

7 7

6 2

y x

x=

−5 16

2

dy

dx x x x

x= − −( ) = +−20 2 20

23 3 3

3.

Note:

An alternate method of solution is to write

and use the quotient rule.

Another method is to write

and use the product rule.

7. Let

8. Using the product rule,

= 3 sec(3x) + 9x sec(3x) tan(3x) = 3 sec(3x)

[1 + 3x tan(3x)]

9. Using the chain rule, let u = sin(x2

− 4).

10. Using the chain rule, let u = 2x.

11. Since 3e5 is a constant, thus its derivative is 0.

= 4ex + 4xex = 4e x(1 + x)

12. Let

=+

2

32

x

x

u x dy

dx xx= +( ) =

+

( )2

23

1

32,

dy

dx e e xx x= + ( ) ( ) + ( ) ( )0 4 4

dy

dx x x=

−

− ( )

( ) =

−

−8

1

1 22

16

1 42

2

dy

dx x x x

x x x

= − −( )[ ]( ) −( )[ ] ( )

= − −( ) −( )

10 4 4 2

20 4 4

2 2

2 2

sin sin cos

cos sin sin

dy

dx x x x x= ( )[ ]( ) + ( ) ( )[ ]( )[ ]3 3 3 3 3 3sec sec tan

= − −( )20 2 12csc x

u x dy

dx

x= − = − −( ) ( )2 1 10 2 1 22, csc

2 11

2x −( )

y x= +( )2 11

2

y x

x=

+−

2 1

2 1

if which implies

or

2 1

2 10

1

2

1

2

x

x x

x

+−

−> <

> .

2 1

2 1

2 1

2 1

12

12

12

x

x

x

x

+−

=

+( )

−( ),

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13. Using implicit differentiation, differentiate eachterm with respect to x.

14. Since f ′(4) is equivalent to the slope of thetangent to f (x) at x = 4, there are several waysyou can find its approximate value.

Method 1: Using the slope of the line segmentjoining the points at x = 4 andx = 5.

f (5) = 1 and f (4) = 4

Method 2: Using the slope of the line segmentjoining the points at x = 3 andx = 4.

f (3) = 5 and f (4) = 4

Method 3: Using the slope of the line segment

joining the points at x = 3 andx = 5.

f (3) = 5 and f (5) = 1

Note that −2 is the average of the results frommethods 1 and 2. Thus f ′(4) ≈ −3, −1 or −2depending on which line segment you use.

m f f =

( ) − ( )−

= −

− = −

5 3

5 3

1 5

5 32

m f f =

( ) − ( )−

= −

− = −

4 3

4 3

4 5

4 31

m f f =

( ) − ( )

− =

−= −

5 4

5 4

1 4

13

dy

dx

y x

y x

dy

dx

x y

x y=

− −+

= − +( )

+5 2

3 5

2 5

5 32 2or

dy

dx y x y x3 5 5 22 +( ) = − −

3 5 5 22y dy

dx x

dy

dx y x+ = − −

2 3 5 52x y dydx

y x dydx

+ = − −

2 3 0 5 52x y dy

dx y

dy

dx x+ = − ( )( ) + ( )

15. You can use the difference quotient

to approximate f ′(a).

Let

Or, you can use the symmetric difference quotient


Let

Thus, f ′(2) ≈ 4 or 5 depending on your method.

16. Enter y1 = x5 + 3x − 8. The graph of y1 is strictlyincreasing. Thus f (x) has an inverse. Note thatf (0) = −8. Thus the point (0, −8) is on the graphof f (x) which implies that the point (

−8, 0) is on

the graph of f −1(x).

f ′(x) = 5x4 + 3 and f ′(0) = 3

Since

17.

Thus the slope of the tangent to y = ln x at

At x = e, y = ln x = ln e = 1, which

means the point (e, 1) is on the curve of y = ln x.

Therefore, an equation of the tangent is

(See Figure 3.10-1.)ye

x e y x

e− = −( ) =1

1or .

x ee

= is1

.

dy

dx x

dy

dx ex e

= ==

1 1 and

f f

f − −( )′ −( ) =′( )

( )′ −( ) =1 181

08

1

3, .thus

h f f f

= ′( ) ≈ ( ) − ( )

− ≈

−≈1 2

3 1

2 0

14 6

24;

f a h f a h

h

+( ) − −( )2

h f f f

= ′( ) ≈ ( ) − ( )

− ≈

−−

≈1 23 2

3 2

14 9

3 25;

f a h f a

h

+( ) − ( )

Figure 3.10-1

18. dy

dx x x x x x x= ( )( ) + ( )( ) = +2 2 2 2sin cos sin cos

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3.11 SOLUTIONS TO CUMULATIVE REVIEW PROBLEMS

Or, using a calculator, enter d (2x − sin(x), x, 2)

and obtain −π.

19. Enter + 2 in your calculator. The

graph of y1 forms a cusp at x = 1. Therefore, f isnot differentiable at x = 1.

y x1 12

3= −( )

x = π2

d y

dx x

2

2

2

4

2

2

2 2

0 22

1

=

=

−

= −

( ) = −

π

π π π

ππ

cos sin

d y

dx x x x x

x x x x

x x x

2

22 2 2

2 2 2

4 2

= + ( )( ) + −( )( )[ ]

= + −

= −

cos cos sin

cos cos sin

cos sin

20. Differentiate with respect to x:

Thus, the slope of the tangent to the curve at

The slope of the normal line to the curve at

Therefore an equation of the normal line isy x− = − −( )π

32 3 2 .

23

2 3

12 3, .π( ) = − = −is m

23

1

2 3, .π( ) =is m

dy

dxx y= =

= ( )

( ) ( ) =

=2

3

3

23

12

2 32

1

2 3,

cos

sinπ

π

π

dy

dx

y

x y=

cos

sin

cos siny x y dy

dx− = 0

1 0( ) + −( )

( ) =cos siny y

dy

dxx

21. The expression is the

derivative of sin x at which is the slope

of the tangent to sin x at The tangent

to sin x at is parallel to the x-axis.

Therefore the slope is 0, i.e.,

An alternate method is to expand sin as

= ( ) + − ( )

= ( ) −[ ] +

→

→

limsin cos cos sin sin

limsin cos cos sin

h

h

h h

h

h h

h

0

0

2 2 2

21

2

π π π

π π

sin cos cos sin .

limsin sin

π π

π π

2 2

2 20

( ) + ( )+( ) − ( )

→

h h

h

hhThus,

π2

+( )h

limsin sin

.h

h

h→

+( ) − ( )=

0

2 2 0

π π

x = π2

x = π2

.

x =

π2

limsin sin

h

h

h→

+( ) − ( )0

2 2π π

22. Using the chain rule, let u = (π − x).Then, f ′(x) = 2 cos(π − x)[−sin(π − x)](−1)

= 2 cos(π − x) sin(π − x)f ′(0) = 2 cos π sin π = 0.

23. Since the degree of the polynomial in thedenominator is greater than the degree of the

polynomial in the numerator, the limit is 0.

24. You can use the difference quotient


Let h f f f

= ′( ) ≈ ( ) − ( )−

≈ −

≈

1 23 2

3 2

6 5 4 8

11 7

;

. .. .

f a h f a

h

+( ) − ( )

= ( )[ ] + ( ) ( ) = ( ) =sin cos cosπ π π2

02

12

0

= ( ) −

− ( )

→ →sin lim

coscos lim

sin

π π2 20 0h h

h

h

h

h

1

= ( ) −

− ( )

→ →

lim sincos

lim cossin

h h

h

h

h

h0 021

2π π

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Or, you can use the symmetric difference quo-

tient to approximate f ′(a).

Thus, f ′(2) ≈ 1.7, 2.05 of 1.25 depending onyour method.

Let

Let

h f f f

h f f f

= ′( ) ≈ ( ) − ( )

−

≈ −

≈

= ′( ) ≈ ( ) − ( )

−

≈ − ≈

1 23 1

3 1

6 5 4

21 25

2 24 0

4 0

8 9 3 9

4 1 25

;

..

;

. ..

f a h f a hh

+( ) − −( )2

Let h f f f

= ′( ) ≈ ) − )

−

≈ −

≈

2 24 2

4 2

8 9 4 8

22 05

;

. .. .


25. (See Figure 3.11-1.) Checking the three conditionsof continuity:

[–10,10] by [–10,10]

Figure 3.11-1

1 3 3

29

3

3 3

3

3 3 3 6

3

3

2

3

3

3

( ) ( ) =

( ) −

−

= +( ) −( )

−( )

= +( ) = ( ) + =

( ) ( ) ≠ ( ) ( )=

→ →

→

→

f

x

x

x x

x

x

f f x f x

x

x x

x

x

lim lim

lim .

lim ,3 Since is discontinous

at 3.

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4.1 ROLLE’S THEOREM, MEAN VALUE THEOREM, ANDEXTREME VALUE THEOREM

Main Concepts: Rolle’s Theorem, Mean Value Theorem, Extreme Value Theorem

Chapter 4

Graphs of Functions andDerivatives

• Set your calculator to Radians and change it to Degrees if/when you need to.Don’t forget to change it back to Radians after you’re finished using it in Degrees.

Rolle’s Theorem and Mean Value Theorem

Rolle’s Theorem: If f is a function that satisfies the following three conditions:

1. f is continuous on a closed interval [a, b]2. f is differentiable on the open interval (a, b)3. f (a) = f (b) = 0

then there exists a number c in (a, b) such that f ′(c) = 0. (See Figure 4.1-1.)Note that if you change condition 3 from f (a) = f (b) = 0 to f (a) = f (b), the conclu-

sion of Rolle’s Theorem is still valid.

123

(c, f (c))

f

f ' (c)=0

y

a c b x

0

Figure 4.1-1


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(c,f(c))

(a,f(a))

(b,f(b))

c x

y

f

0

f' (c) = f(b) – f(a)

b – a

Figure 4.1-2

Figure 4.1-3

Example 1

If f (x) = x2 + 4x − 5, show that the hypotheses of Rolle’s Theorem are satisfied on theinterval [−4, 0] and find all values of c that satisfy the conclusion of the theorem. Checkthe three conditions in the hypothesis of Rolle’s Theorem:

(1) f (x) = x2 + 4x − 5 is continuous everywhere since it is polynomial.(2) The derivative f ′(x) = 2x + 4 is defined for all numbers and thus is differentiable on

(−4, 0).(3) f (0) = f (−4) = −5. Therefore, there exists a c in (−4, 0) such that f ′(c) = 0. To find c,

set f ′(x) = 0. Thus 2x + 4 = 0 ⇒ x = −2, i.e., f ′(−2) = 0. (See Figure 4.1-3.)

[–5,3] by [–15,10]

Example 2

Let Using Rolle’s Theorem, show that there exists a number

c in the domain of f such that f ′(c) = 0. Find all values of c.

f x x x

x( ) = − − +3 2

3 22 2.

Mean Value Theorem: If f is a function that satisfies the following conditions:

1. f is continuous on a closed interval [a, b]2. f is differentiable on the open interval (a, b)

then there exists a number c in (a, b) such that

. (See Figure 4.1-2.)′( ) = ( ) − ( )

−

f c f b f a

b a

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Note f (x) is a polynomial and thus f (x) is continuous and differentiable everywhere.

Enter The zero’s of y1 are approximately −2.3, 0.9 and 2.9

i.e. f (−2.3) = f (0.9) = f (2.9) = 0. Therefore, there exists at least one c in the interval(−2.3, 0.9) and at least one c in the interval (0.9, 2.9) such that f ′(c) = 0. Use d (differen-tiate) to find f ′(x): f ′(x) = x2 − x − 2. Set f ′(x) = 0 ⇒ x2 − x − 2 = 0 or (x − 2)(x + 1) = 0.

Thus x = 2 or x = −1, which implies f ′(2) = 0 and f ′(−1) = 0. Therefore the values of c are −1 and 2. (See Figure 4.1-4.)

y x x

x13 2

2 23 2

= − − + .

Graphs of Functions and Derivatives • 125

Figure 4.1-4

[–8,8] by [–4,4]

Figure 4.1-5

[–4,4] by [–20,40]

Example 3

The points P(1, 1) and Q(3, 27) are on the curve f (x) = x3. Using the Mean Value Theorem,

find c in the interval (1, 3) such that f ′(c) is equal to the slope of the secant .

The slope of secant is Since f (x) is defined for all real

numbers, f (x) is continuous on [1, 3]. Also f ′(x) = 3x2 is defined for all real numbers.Thus f (x) is differentiable on (1,3). Therefore, there exists a number c in (1, 3) such that

f ′(c) = 13. Set f ′(c) = 13 ⇒ 3(c)2 = 13 or c2 = c = ± . Since only is in the

interval (1, 3), thus c = . (See Figure 4.1-5.)13

3

13

3

13

3

13

3

m = −

− =

27 1

3 113.PQ

PQ

Example 4

Let f be the function Determine if that the hypotheses of the Mean

Value Theorem are satisfied on the interval [0,2] and if so, find all values of c that satisfythe conclusion of the theorem.

Enter The graph y1 shows that there is a cusp at x = 1. Thus, f (x)

is not differentiable on (0, 2) which implies there may or may not exists a c in (0,2) such

y x1 12

3= −( ) .

f x x( ) = −( )12

3 .

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that The derivative and

Set . Note f is not differentiable (a + x = 1). There-

fore c does not exist. (See Figure 4.1-6.)

2

31 0 1

13x x−( ) = ⇒ =

1 1

20

−= .

f f 2 0

2 0

( ) − ( )

− =′( ) = −( )

−

f x x2

31

13′( ) =

( ) − ( )

−f c

f d 2 0

2 0.


[–8,8] by [–4,4]

Figure 4.1-6

[–3,3] by [–4,20]

Figure 4.1-7

Extreme Value Theorem

Extreme Value Theorem: If f is a continuous function on a closed interval [a,b], then f has both a maximum and a minimum value on the interval.

Example 1

If f (x) = x3 + 3x2 − 1, find the maximum and minimum values of f on [−2,2]. Since (f (x) isa polynomial, it is a continuous function everywhere. Enter y1 = x3 + 3x2 − 1. The graphof y1 indicates that f has a minimum of −1 at x = 0 and a maximum value of 19 at x = 2.(See Figure 4.1-7.)

Example 2

If find any maximum and minimum values of f on [0,3]. Since f (x) is a ratio-

nal function, it is continuous everywhere except at values where the denominator is 0.

f xx

( ) =1

2,

• The formula for finding the area of an equilateral triangle is area

where s is the length of a side. You might need this to find the volume of a solidwhose cross sections are equilateral triangles.

= s2 34

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4.2 DETERMINING THE BEHAVIOR OF FUNCTIONS

Main Concepts: Test for Increasing and Decreasing Functions, First Derivative Test and Second Derivative Test for Relative Extrema, Test for Concavityand Points of Inflection

Test for Increasing and Decreasing Functions

Let f be a continuous function on the closed interval [a,b] and differentiable on the open

interval (a,b).1. If f ′(x) > 0 on (a, b), then f is increasing on [a,b]2. If f ′(x) < 0 on (a, b), then f is decreasing on [a,b]3. If f ′(x) = 0 on (a, b), then f is constant on [a,b]

Definition: Let f be a function defined at a number c. Then c is a critical number of f if either f ′(c) = 0 or f ′(c) does not exist. (See Figure 4.2-1.)


[–1,4] by [–1,6]

Figure 4.1-8

f(x)

x

y

0

.

..

0=′ f

f ′ > 0

f increasing

0=′ f

constant

0=′ f

f ′ < 0

f decreasing

f ′ < 0 f decreasing

Figure 4.2-1

In this case, at x = 0, f (x) is undefined. Since f (x) is not continuous on [0,3], the Extrema

Value Theorem may not be applicable. Enter The graph of y1 shows that as

x → 0+, f (x) increases without bound (i.e., f (x) goes to infinity). Thus f has no maximum

value. The minimum value occurs at the endpoint x = 3 and the minimum value is

(See Figure 4.1-8.)

1

9.

yx

11

2= .

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Example 1

Find the critical numbers of f (x) = 4x3 + 2x2.To find the critical numbers of f (x), you have to determine where f ′(x) = 0 and where

f ′(x) does not exist. Note f ′(x) = 12x2 + 4x, and f ′(x) is defined for all real numbers. Letf ′(x) = 0 and thus 12x2 + 4x = 0 which implies 4x(3x + 1) = 0 ⇒ x = −1/3 or x = 0.Therefore the critical numbers of f are 0 and −1/3. (See Figure 4.2-2.)


[–1,1] by [–1,1]

Figure 4.2-2

[–3,8] by [–4,4]

Figure 4.2-3

y

3

3 4 5 6 x 210

f '

Figure 4.2-4

Example 2

Find the critical numbers of

Note that f ′(x) is undefined at x = 3 and that f ′(x)

≠ 0. Therefore, 3 is the only critical number of f. (See Figure 4.2-3.)

′( ) = −( ) =−( )

−

f x xx

2

53

2

5 3

35

35

.

f x x( ) = −( )325 .

Example 3

The graph of f ′ on (1,6) is shown in Figure 4.2-4. Find the intervals on which f is increas-ing or decreasing.

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Solution: See Figure 4.2-5.


decr. decr.incr.

– + –

21 6

5

'

x

Figure 4.2-5

Figure 4.2-6

Thus, f is decreasing on [1,2] and [5,6] and increasing on [2,5].

Example 4

Find the open intervals on which is increasing or decreasing.

Step 1: Find the critical numbers of f.

Set f ′(x) = 0 ⇒ 4x = 0 or x = 0.Since f ′(x) is a rational function, f ′(x) is undefined at values where the denomi-nator is 0. Thus, set x2 − 9 = 0 ⇒ x = 3 or x = −3. Therefore the critical numbersare −3,0, and 3.

Step 2: Determine intervals.

Intervals are (−∞,−3), (−3,0), (0,3) and (3,−∞).

Step 3: Set up a table.

–3 0 3

′( ) = −( ) ( ) =

−( )

−

−f x x x

x

x

2

39 2

4

3 9

21

3

21

3

f x x( ) = −( )22

39

Interval (,3) (3,0) (0,3) (3,)

Test Point −5 −1 1 5

f ′(x) − + − +

f (x) decr incr decr incr

Step 4: Write a conclusion. Therefore f (x) is increasing on [−3,0] and [3,∞) anddecreasing on (−∞,−3] & [0,3]. (See Figure 4.2-6.)

[–8,8] by [–1,5]

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Example 5

The derivative of a function f is given as f ′(x) = cos(x2). Using a calculator, find the values

of x on such that f is increasing. (See Figure 4.2-7.)−

π π2 2

,


[-π,π] by [-2,2]

Figure 4.2-7

–1.2533π 1.25332

π2

–

– f '

f

–+

[ ]

decr. incr. decr.

x

Figure 4.2-8

.

.

. f ′ > 0 f ′ > 0 f ′ < 0

rel. max. f ′ = 0

rel. max

f ′ undefined

f ′ = 0

rel. min.

f ′ = undefined

rel. min.

f ′ > 0 f ′ > 0

f ′ < 0 f ′ < 0 f ′ < 0

.

Figure 4.2-9

Using the zero function of the calculator, you obtain x = 1.25331 is a zero of f ′ on

Since f ′(x) = cos(x2) is an even function, x = −1.25331 is also a zero on

(See Figure 4.2-8.)

−

π

2

0, .0

2

, .π

Thus f is increasing on [−1.2533, 1.2533].

• Bubble in the right grid. You have to be careful in filling in the bubblesespecially when you skip a question.

First Derivative Test and Second DerivativeTest for Relative Extrema

First Derivative Test for Relative Extrema:

Let f be a continuous function and c be a critical number of f. (Figure 4.2-9.)

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1. If f ′(x) changes from positive to negative at x = c (f ′ > 0 for x < c and f ′ < 0 for x >c), then f has a relative maximum at c.

2. If f ′(x) changes from negative to positive at x = c (f ′ < 0 for x < c and f ′ > 0 for x >c), then f has a relative minimum at c:

Second Derivative Test for Relative Extrema:

Let f be a continuous function at a number c.

1. If f ′(c) = 0 and f ″ (c) < 0, then f (c) is a relative minimum.2. If f ′(c) = 0 and f ″ (c) > 0, then f (c) is a relative minimum.3. If f ′(c) = 0 and, f ″ (c) = 0, then the test is inconclusive. Use the first Derivative Test.

Example 1

See Figure 4.2-10. The graph of f ′, the derivative of a function f is shown in Figure4.2-10. Find the relative extrema of f.


0 3

y

x

f '

–2

Figure 4.2-10

–2 3

incr. incr.decr.

+ – +

rel. min rel. max

x

'

Figure 4.2-11

Solution: See Figure 4.2-11.

Thus f has a relative minimum at x = −2, and a relative maximum at x = 3.

Example 2

Find the relative extrema for the function

Step 1: Find f ′(x).

f ′(x) = x2 − 2x − 3

f x x

x x( ) = − −3

2

33 .

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Step 2: Find all critical numbers of f (x).Note that f ′(x) is defined for all real numbers.Set f ′(x) = 0: x2 − 2x − 3 = 0 ⇒ (x − 3)(x + 1) = 0 ⇒ x = 3 or x = −1.

Step 3: Find f ″ (x): f ″ (x) = 2x − 2.

Step 4: Apply the Second Derivative Test.f ″ (3) = 2(3) − 2 = 4 ⇒ f (3) is a relative minimum.f ″ (−1) = 2(−1) − 2 = −4 ⇒ f (−1) is a relative maximum.

Therefore, −9 is a relative minimum value of f and is a relative maximum

value. (See Figure 4.2-12.)

5

3

f f 3 33

3 3 3 9 1 53

32

( ) = − ( ) − ( ) = − −( ) =and .


[–5,7] by [–10,10]

Figure 4.2-12

Example 3

Find the relative extrema for the function

Using the First Derivative Test:

Step 1: Find f ′(x).

Step 2: Find all critical numbers of f.Set f ′(x) = 0. Thus 4x = 0 or x = 0.Set x2 − 1 = 0. Thus f ′(x) is undefined at x = 1 and x = −1. Therefore the criti-cal numbers are −1, 0, 1.


The intervals are (−∞,−1), (−1,0), (0,1), and (1,∞).


0 1–1

′( ) = −( ) ( ) =−( )

−

f x x x x

x

2

31 2

4

3 1

21

3

21

3

.

f x x( ) = −( )22

31 .

Interval (,1) x 1 (1,0) x 0 (0,1) x 1 (1,)

Test Point −2 −1/2 1/2 2

f ′(x) − undefined + 0 − undefined +

f (x) decr rel min incr rel max decr rel min incr

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Step 5: Write a conclusionUsing the First Derivative Test, note that f (x) has a relative maximum at x = 0and relative minimum at x = −1 and x = 1.

Note that f (−1) = 0, f (0) = 1 and f (1) = 0. Therefore, 1 is a relative maximum value and 0is a relative minimum value. (See Figure 4.2-13.)


[–3,3] by [–2,5]

Figure 4.2-13

concave

downward

f ″ < 0 f ″ < 0

Figure 4.2-14

concave

upward

f ″ > 0 f ″ > 0

Figure 4.2-15

• Don’t forget the constant, C, when you write the antiderivative after evaluatingan indefinite integral, e.g., cos sin .xdx x C= +

Test for Concavity and Points of Inflection

Test for Concavity:

Let f be a differentiable function.

1. If f ″ > 0 on an interval I, then f is concave upward on I.

2. If f ″ < 0 on an interval I, then f is concave downward on I.

See Figures 4.2-14 and 4.2-15.

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A point P on a curve is a point of inflection if

1. the curve has a tangent line at P, and2. the curve changes concavity at P (from concave upward to downward or from con-

cave downward to upward).

See Figures 4.2-16 to 4.2-18.


f ″ < 0

f ″ > 0

pt. of inflection.

Figure 4.2-16

. f ″ < 0

f ″ > 0

pt. of inflection

Figure 4.2-17

f ″ < 0 f ″ > 0 .CUSP

not a pt. of

inflection

Figure 4.2-18

Note that if a point (a, f (a)) is a point of inflection, the f ″ (c) = 0 or f ″ (c) does notexist. (The converse of the statement is not necessarily true.)

Note: There are some textbooks that define a point of inflection as a point wherethe concavity changes and do not require the existence of a tangent at the point of

inflection. In that case, the point at the cusp in Figure 4.2-18 would be a point of inflection.

Example 1

See Figure 4.2-19. The graph of f ′, the derivative of a function f is shown in Figure 4.2-19on page 135. Find the points of inflection of f and determine where the function f is con-cave upward and where it is concave downward on [−3,5].

Solution: See Figure 4.2-20 on page 135.Thus f is concave upward on [−3,0) and (3,5], and is concave downward on (0,3).There are two points of inflection: one at x = 0 and the other at x = 3.

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Example 2

Using a calculator, find the values of x at which the graph of y = x2e x changes concavity.Enter y1 = x^2 e^x and y2 = d (y1(x), x,2). The graph of y2, the second derivative

of y, is shown in Figure 4.2-21. Using the Zero function, you obtain x = −3.41421 andx = −0.585786. (See Figures 4.2-21 and 4.2-22.)


[ ]–3 0 3 5

incr. decr. incr.

Concave

Upward

Concave

Upward

Concave

Downard

pt. of

infl.

pt. of

infl.

+ – +

x

f '

"

f

Figure 4.2-20

[–4,1] by [–2,5]

Figure 4.2-21

1

1

–1

–2

–3

2

3

4

–1–2–3 2 3 4 50

y

x

f '

Figure 4.2-19

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Thus, f changes concavity at x = −341421 and x = −0.585786.

Example 3

Find the points of inflection of f (x) = x3 − 6x2 + 12x − 8 and determine the intervalswhere the function f is concave upward and where it is concave downward.

Step 1: Find f ′(x) and f ″ (x).

f ′(x) = 3x2 − 12x + 12

f ″ (x) = 6x − 12

Step 2: Set f ″ (x) = 0

6x − 12 = 0

x = 2

Note that f ″ (x) is defined for all real numbers.


The intervals are (−∞,2) and (2,∞).


2


+ + –

–3.41421 –0.585786

x

f "

Concave

upward

Concave

downward

Concave

upward

Change of

concavity

Change of

concavity

Figure 4.2-22

Interval (,2) x 2 (2,)

Test Point 0 5

f ″ (x) − 0 +

f (x) concave point of concavedownward inflection upward

Since f (x) has change of concavity at x = 2, the point (2, f (2)) is a point of inflection. f (2) = (2)3 − 6(2)2 + 12(2) −8 = 0.

Step 5: Write a conclusion.

Thus f (x) is concave downward on (−∞,2), concave upward on (2,∞) and f (x)has a point of inflection at (2,0). (See Figure 4.2-23.)

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Example 4

Find the points of inflection of and determine the intervals where the

function f is concave upward and where it is concave downward.

Step 1: Find f ′(x) and f ″ (x).

Step 2: Find values of x where f ″ (x) = 0 or f ″ (x) is undefined.Note that f ″ (x) ≠ 0 and that f ″ (1) is undefined.


The intervals are (−∞,1), and (1,∞).


1

′( ) = −( ) =−( )

( ) = − −( ) = −

−( )

−

−

f x xx

f x xx

2

31

2

3 1

2

91

2

9 1

13

13

13

43

″

f x x( ) = −( )12

3


[–1,5] by [–5,5]

Figure 4.2-23

Interval (,1) x 1 (1,)

Test Point 0 2

f ″ (x) − undefined −

f (x) concave no change concavedownward of inflection downward

Note that f (x) has no change of concavity at x = 1, f does not have a point of inflection.

Step 5: Write a conclusion.Therefore f (x) is concave downward on (−∞,∞) and has no point of inflection.(See Figure 4.2-24.)

[–3,5] by [–1,4]

Figure 4.2-24

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Example 5The graph of f is shown in Figure 4.2-25 and f is twice differentiable, which of the fol-lowing statements is true:


y

f

x 0 5

Figure 4.2-25

(A) f (5) < f ′(5) < f ″ (5)

(B) f ″ (5) < f ′(5) < f (5)

(C) f ′(5) < f (5) < f ″ (5)

(D) f ′(5) < f ″ (5) < f (5)

(E) f ″ (5) < f (5) < f ′(5)

The graph indicates that (1) f (5) = 0, (2) f ′(5) < 0, since f is decreasing; and (3) f ″ (5)< 0, since f is concave upward. Thus f ′(5) < f (5) < f ″ (5), choice (C).

• Move on. Don’t linger on a problem too long. You can earn many more pointsfrom other problems.

4.3 SKETCHING THE GRAPHS OF FUNCTIONS

Main Concepts: Graphing without Calculators, Graphing with Calculators

Graphing without Calculators

General Procedure for Sketching the Graph of a Function

Steps:

1. Determine the domain and if possible the range of the function f (x).

2. Determine if the function has any symmetry i.e., if the function is even (f (x) = f (−x));odd (f (x) = −f (−x)) or periodic (f (x + p) = f (x))

3. Find f ′(x) and f ″ (x).

4. Find all critical numbers (f ′(x) = 0 or f ′(x) is undefined) and possible points of inflection (f ″ (x) = 0 or f ″ (x) is undefined).

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5. Using the numbers in Step 4, determine the intervals on which to analyze f (x).

6. Set up a table using the intervals, to(a) determine where f (x) is increasing or decreasing.(b) find relative and absolute extrema.(c) find points of inflection.(d) determine the concavity of f (x) on each interval.

7. Find any horizontal, vertical, or slant asymptotes.

8. If necessary, find the x-intercepts, the y-intercepts, and a few selected points.

9. Sketch the graph.

Example

Sketch the graph of

Step 1: Domain: all real numbers x ≠ ±5.

Step 2: Symmetry: f (x) is an even function (f (x) = f (−x)); symmetrical with respect toy-axis.

Step 3:

Step 4: Critical numbers:f ′(x) = 0 ⇒ −42x = 0 or x = 0f ′(x) is undefined at x = ±5 which are not in the domain.

Possible points of inflection:f ″ (x) ≠ 0 and f ″ (x) is undefined at x = ±5 which are not in the domain.

Step 5: Determine Intervals:

Intervals are (−∞,−5), (−5,0), (0,5) & (5,∞)

Step 6: Set up a table:

–5 0 5

′( ) = ( ) −( ) − ( ) −( )−( )

= −−( )

( ) = − −( ) − −( )( ) −( )

−( )=

+( )

−( )

f x x x x xx

xx

f x x x x x

x

x

x

2 25 2 425

4225

42 25 2 25 2 42

25

42 3 25

25

2 2

2 2 2 2

2 2 2

2 4

2

2 3″

f x x

x( ) =

−−

2

2

4

25.


Intervals (,5) x 5 (5,0) x 0 (0,5) x 5 (5,)

f (x) undefined 4/25 undefined

f ′(x) + undefined + 0 − undefined −

f ″ (x) + undefined − − − undefined +

conclusion incr incr rel decr decrconcave concave max concave concaveupward downward downward upward

Step 7: Vertical asymptote: x = 5 and x = −5Horizontal asymptote: y = 1

Step 8: y-intercept:

x-intercept: (−2,0) and (2,0)

0 425

,( )

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See Figure 4.3-1.


[–8,8] by [–4,4]

Figure 4.3-1

[–2,4] by [–4,4]

Figure 4.3-2

Graphing with Calculators

Example 1

Using a calculator, sketch the graph of indicating all relative extrema,points of inflection, horizontal and vertical asymptotes, intervals where f (x) is increas-ing or decreasing, and intervals where f (x) is concave upward or downward.

1. Domain: all real numbers; Range: all real numbers2. No symmetry3. Relative Minimum (0,0); Relative Maximum (1.2,2.03); Points of Inflection

(−0.6,2.56)4. No asymptote.5. f (x) is decreasing on (−∞,0], [1.2,∞); and increasing on (0,1.2)6. Evaluating the f ″ (x) on either side of the point of inflection (−0.6,2.56)

⇒ f (x) is concave upward on (−∞,−0.6) and concave downward on (−0.6,∞) (SeeFigure 4.3-2.)

d x x x x

d x x x x

− ∧

( ) + ∧

( )( ) = − →

− ∧ ( ) + ∧ ( )( ) = − → −

5

3

3 2

3

2 2 0 19

53

3 23

2 1 4 66

, , .

, , .

f x x x( ) = − +

53

23

3

Example 2

Using a calculator, sketch the graph of f (x) = e−x2 / 2; indicating all relative minimum andmaximum points; points of inflection, vertical and horizontal asymptotes, intervals onwhich f (x) is increasing, decreasing, concave upward or concave downward.

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1. Domain: all real numbers; Range (0,1]2. Symmetry: f (x) is an even function, and thus is symmetrical with respect to the y-axis.3. Relative maximum: (0,1)

No relative minimumPoints of inflection: (−1,0.6) and (1,0.6)

4. y = 0 is a horizontal asymptote; no vertical asymptote5. f (x) is increasing on (−∞,0] and decreasing on [0,∞)6. f (x) is concave upward on (−∞,−1) and (1,∞); and concave downward on (−1,1)

See Figure 4.3-3.


[–4,4] by [–1,2]

Figure 4.3-3

0

y

f

ba x

Figure 4.4-1

• When evaluating a definite integral, you don’t have to write a constant C,

e.g., Notice, no C.2 82

1

3

1

3

xdx x= =∫ .

4.4 GRAPHS OF DERIVATIVES

The functions f, f ′, and f ″ are interrelated, and so are their graphs. Therefore, you can

usually infer from the graph of one of the three functions (f, f ′, or f ″ ) and obtain infor-mation about the other two. Here are some examples.

Example 1

The graph of a function f is shown in Figure 4.4-1. Which of the following is true for f on (a,b)?

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(a) Summarize the information of f ′ on a number line:

The function f has a relative maximum at x = −4 and at x = 4; and a relative mini-mum at x = 0.

(b) The function f is increasing on interval (−∞,−4] and [0,4], and f is decreasing on[−4,0] and [4,∞).

(c) Summarize the information of f ′ on a number line:

A change of concavity occurs at x = −2 and at x = 2 and f ′ exists at x = −2 and atx = 2, which implies that there is a tangent line to the graph of f at x = −2 and atx = 2. Therefore, f has a point of inflection at x = −2 and at x = 2.

decr incr decr

concave

downward upward downward

concave concave

–2 2

f '

" + – –

–4 0 4

incrincr

'

decr decr

+ – + –

I. f ′ ≥ 0 on (a,b)II. f ″ > 0 on (a,b)

Solution:

I. Since f is strictly increasing, f ′ ≥ 0 on (a,b) is true.II. The graph is concave downward on (a,0) and upward on (0,b). Thus f ″ > 0 on (0,b)

only. Therefore only statement I is true.

Example 2

Given the graph of f ′ in Figure 4.4-2, find where the function f (a) has its relative maxi-mum(s) or relative minimums, (b) is increasing or decreasing, (c) has its point(s) of inflec-tion, (d) is concave upward or downward, and (e) if f (−2) = f (2) = 1 and f (0) = −3, drawa sketch of f.


x

y

0–4 –2 2 4

f ′

Figure 4.4-2

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(a) f ′(x) = 0 at x = −4, 2, 4, 8. Thus f has a horizontal tangent at these values.

(b) Summarize the information of f ′ on a number line:

The first Derivative Test indicates that f has a relative maximum at x = −4 and 4;and f has a relative minimum at x = 2 and 8.

(c) The function f is increasing on (−8,−4], [2,4] and [8,∞) and is decreasing on [−4,2]and [4,8].

incr decr incr decr incr

–4 2 4 8

+ f '

f

– + – +

(d) The graph of f is concave upward on the interval (−2,2) and concave downward on(−∞,−2) and (2,∞).

(e) A sketch of the graph of f is shown in Figure 4.4-3.


rel. min.

.

..

.

–3

2 4

y

x

0–4 –2

.rel. max. rel. max.

pts. of

inflection

Figure 4.4-3

y

x

0–4 –1 2 3 4 6 8

f ′

Figure 4.4-4

Example 3

Given the graph of f ′ in Figure 4.4-4, find where the function f (a) has a horizontal tan-gent, (b) has its relative extrema, (c) is increasing or decreasing, (d) has a point of inflec-tion, and (e) is concave upward or downward.

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(d) Summarize the information of f ′ on a number line:

–1 3 6

decr' incr decr incr


f ″ − + − +f concave concave concave concave

downward upward downward upward

A change of concavity occurs at x = −1, 3, and 6. Since f ′(x) exists, f has a tangentat every point. Therefore, f has a point of inflection at x = −1, 3, and 6.

(e) The function f is concave upward on (−1,3) and (6,∞) and concave downward on(−∞,−1) and (3,6).

Example 4

A function f is continuous on the interval [−4,3] with f (−4) = 6 and f (3) = 2 and the fol-lowing properties:

Intervals (4,2) x 2 (2,1) x 1 (1,3)

f ′ − 0 − undefined +

f ″ + 0 − undefined −

(a) Find the intervals on which f is increasing or decreasing.

(b) Fund where f has its absolute extrema.

(c) Find where f has the points of inflection.

(d) Find the intervals on where f is concave upward or downward.

(e) Sketch a possible graph of f.

Solution:

(a) The graph of f is increasing on [1,3] and decreasing on [−4,−2] and [−2,1].

(b) At x = −4, f (x) = 6. The function decreases until x = 1 and increases back to 2 atx = 3. Thus, f has its absolute maximum at x = −4 and its absolute minimum at x = 1.

(c) A change of concavity occurs at x = −2, and since f ′(−2) = 0 which implies a tangentlines exists at x = −2, thus f has a point of infection at x = −2.

(d) The graph of f is concave upward on (−4,−2) and concave downward on (−2,1) and(1,3).

(e) A possible sketch of f is shown in Figure 4.4-5.

(–4,6)

(3,2)

f (x)

y

x 10–1–2–3–4 2 3

Figure 4.4-5

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Example 5

If (See Figure 4.4-6.)f x x find f xx

( ) = +( ) ′( )→ −

ln , lim .10


[–2,5] by [–2,4]

Figure 4.4-6

–2 0

–4

2

y

x

f ( x )

Figure 4.5-1

The domain of f is (−1, ∞).

4.5 RAPID REVIEW

1. If f ′(x) = x2 − 4, find the intervals where f is decreasing. (See Figure 4.5-1.)

Thus,

if

if

Therefore,

′( ) = +

>

−+

>

′( ) = −+

= −

→ →− −

f x x

x

x x

f xxx x

1

10

1

10

1

11

0 0

lim lim .

f

f x xx x

x x

0 0 1 1 0

11 0

1 0

( ) = +( ) = ( ) =

( ) = +( ) =+( ) >

− +( ) >

ln ln

lnln

ln.

if

if

Answers: Since f ′(x) < 0 if −2 < x < 2, f is decreasing on (−2,2).

2. If f ″ (x) = 2x − 6 and f ′ is continuous, find the values of x where f has a point of inflec-tion. (See Figure 4.5-2.)

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Answers: Thus f has a point of inflection at x = 3.

3. See Figure 4.5-3. Find the values of x where f has a change of concavity.


3

0–

concave

downward

concave

upward

Point of Inflection

x

+

Figure 4.5-2

0–2 2

5

y

x

f ( x )

Figure 4.5-3

incr. decr.

+ –

concave

upward

concave

downward

x

0

Figure 4.5-4

1

–2 0

f ( x )

x

y

Figure 4.5-5

Answers: f has a change of concavity at x = 0. See Figure 4.5-4.

4. See Figure 4.5-5. Find the values of x where f has a relative minimum.

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Answers: Thus, f has a relative minimum at x = −2. See Figure 4.5-6.


x –2

0

incr. decr.

– +

Figure 4.5-6

0 10

y

x

Figure 4.5-7

y

x

f

–3 0 3

4

Figure 4.5-8

incr. decr.

+ –

concave

upward

concave

downward

x

0

Figure 4.5-9

5. See Figure 4.5-7. Given f is twice differentiable, arrange f (10), f ′(10), f ″ (10) fromsmallest to largest.

Answers: f (10) = 0, f ′(10) > 0 since f is increasing and f ″ (10) < 0 since f is concavedownward. Thus the order is f ″ (10), f (10), f ′(10).

6. See Figure 4.5-8. Find the values of x where f ′ is concave up.

Answers: Thus, f ′ is concave upward on (−∞,0). See Figure 4.5-9.

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1. If f (x) = x3 − x2 − 2x, show that the hypothesesof Rolle’s Theorem are satisfied on the interval[−1,2] and find all values of c that satisfy the

conclusion of the theorem.

2. Let f (x) = ex. Show that the hypotheses of theMean Value Theorem are satisfied on [0, 1] andfind all values of c that satisfy the conclusion of the theorem.

3. Determine the intervals in which the graph of

is concave upward or

downward.

4. Given f (x) = x + sin x 0 ≤ × ≤ 2π, find any pointsof inflection of f.

5. Show that the absolute minimum of

on [−5,5] is 0 and the

absolute maximum is 5.

6. Given the function f in Figure 4.6-1, identify thepoints where:

(a) f ′ < 0 and f ″ > 0, (b) f ′ < 0 and f ″ < 0,

(c) f ′ = 0 (d) f ″ does not exist.

f x x( ) = −25 2

f x x

x( ) =

+−

2

2

9

25

9. The graph of f ′ on [−3,3] is shown in Figure 4.6-3.

Find the values of x on [−3,3] such that (a) f isincreasing and (b) f is concave downward.


y

f

x

A

B

0

C

D

E

Figure 4.6-1

f ″

y

x0

.. .c

ba

Figure 4.6-2

y

x –3 3210

f '

Figure 4.6-3

7. Given the graph of f ″ in Figure 4.6-2, determinethe values of x at which the function f has apoint of inflection. (See Figure 4.6-2.)

8. If f ″ (x) = x2(x + 3)(x − 5), find the values ofx at which the graph of f has a change ofconcavity.

10. The graph of f is shown in Figure 4.6-4 on page149 and f is twice differentiable. Which of thefollowing has the largest value:

(A) f (−1)

(B) f ′(−1)

(C) f ″ (−1)

(D) f (−1) and f ′(−1)

(E) f ′(−1) and f ″ (−1)

Sketch the graphs of the following functionsindicating any relative and absolute extrema,points of inflection, intervals on which the func-tion is increasing, decreasing, concave upwardor concave downward.


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15. A function f is continuous on the interval [−2,5]with f (−2) = 10 and f (5) = 6 and the followingproperties:

11. f (x) = x4 − x2

12.


13. Given the graph of f ′ in Figure 4.6-5, determineat which of the four values of x (x1, x2, x3, x4)does f have

f x x

x( ) =

+−

4

4


–1 0

f

y

x

Figure 4.6-4

–1

y

x

1 2 3 4 50

f

Figure 4.6-6

f ′

x 4 x 1 x

0 x 2 x 3

Figure 4.6-5

y

0 3 6

f ′

Figure 4.6-7

(a) the largest value,

(b) the smallest value,

(c) a point of inflection,

(d) and at which of the four values of x does f ″ have the largest value.

14. Given the graph of f in Figure 4.6-6, determineat which values of x is

(a) f ′(x) = 0

(b) f ″ (x) = 0

(c) f ′ a decreasing function.

Intervals (2,1) x 1 (1,3) x 3 (3,5)

f ′ + 0 − undefined +

f ″ − 0 − undefined +

(a) Find the intervals on which f is increasing ordecreasing.

(b) Find where f has its absolute extrema.

(c) Find where f has points of inflection.

(d) Find the intervals on where f is concaveupward or downward.


16. Given the graph of f ′ in Figure 4.6-7, find wherethe function f (a) has its relative extrema (b) isincreasing or decreasing (c) has its point(s) of inflection, (d) is concave upward or downward,and (e) if f (0) = 1 and f (6) = 5, draw a sketch of f.

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17. If f (x) = x2 − 6x − 7 , which of the followingstatements about f are true?

I. f has a relative maximum at x = 3.

II. f is differentiable at x = 7.

III. f has a point of inflection at x = −1.

18. How many points of inflection does the graph of

y = cos (x2

) have on the interval [−π,π]?


Sketch the graphs of the following functionsindicating any relative extrema, points of inflec-tion, asymptotes, and intervals where the func-tion is increasing, decreasing, concave upwardor downward.

19.

20. f (x) = cos x sin2

x [0,2π]

f x ex

( ) = −

32

2


“Calculator” indicates that calculators are permitted.

21. Find

22. Evaluate

23. Find

24. (Calculator) Determine the value of k such thatthe function

is continuous for all

real numbers.

25. A function f is continuous on the interval [−1,4]with f (−1) = 0 and f (4) = 2 and the followingproperties:

f xx x

x k x( ) =

− ≤

+ >

2 1 1

2 1

,

,

d y

dx y x x

2

2

22 3 1if = ( ) + −cos

limx

x

x→

+ −0

9 3

dy

dx x y xyif 2 2 2

10+( ) =



(c) Find where f has points of inflection.

(d) Find where intervals on where f is concaveupward or downward.


Intervals (1,0) x 0 (0,2) x 2 (2,4)

f ′ + undefined + 0 −

f ″ + undefined − 0 −



1. Condition 1: Since f (x) is a polynomial, it is con-

tinuous on [−1,2].Condition 2: Also, f (x) is differentiable on [−1,2]because f ′(x) = 3x2 − 2x − 2 is defined for allnumbers in [−1,2].

Condition 3: f (−1) = f (2) = 0. Thus f (x) satisfiesthe hypotheses of Rolle’s Theorem which meansthere exists a c in [−1, 2] such that f ′(c) = 0. Setf ′(x) = 3x2 − 2x − 2 = 0. Solve 3x2 − 2x − 2 = 0,using the quadratic formula and obtain

Thus x ≈ 1.215 or −0.549 andx = ±1 7

3.

both values are in the interval (−1,2). Therefore

2. Condition 1: f (x) = ex is continuous on [0,1].

Condition 2: f (x) is differentiable on [0,1] sincef ′(x) = e x is defined for all numbers in [0,1].Thus, there exists a number c in [0,1] such

that . Set f ′(x) =

ex = (e − 1). Thus ex = (e − 1). Take ln of bothsides. ln(e x) = ln(e − 1) ⇒ x = ln(e − 1). Thusx ≈ 0.541 which is in the (0,1). Thereforec = ln(e − 1).

′( ) = −

− = −( )f c

e ee

1 0

1 01

c = ±1 7

3.

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3.

Set f ″ > 0. Since (3x2 + 25) > 0, ⇒ (x2 − 25)3 > 0⇒ x2 − 25 > 0, 50 x < −5 or x > 5. Thus f (x) isconcave upward on (−∞,−5) and (5,∞) and con-cave downward on (−5,5).

4. Step 1: f (x) = x + sin x, f ′(x) = 1 + cos x, f ″ (x) =−sin x.

Step 2: Set f ″ (x) = 0 ⇒ −sin x = 0 or x = 0, π, 2π

Step 3: Check intervals.

f x x

x f x

x x x x

x

x

x

f x x x x x

x

x

x

( ) = +

− ′( )

= −( ) − ( ) +( )

−( )

= −

−( )

( ) = − −( ) − −( )( ) −( )

−( )

= +( )

−( )

2

2

2 2

2 2

2 2

2 2 2

2 4

2

2 3

9

25

2 25 2 9

25

68

25

68 25 2 25 2 68

25

68 3 25

25

,

and

″


f ′(0) = 0 and f ″ (0) = (and f (0) = 5) ⇒ (0, 5)

is a relative maximum. Since f (5) and f (−5) areboth undefined, use the First Derivative Test.The domain of f is [−5, 5].

−1

5

2ππ0

[ ]

concave

downward downward

concave

+ – f "

f

5–5 0 f

f ' + –

incr. decr.

Step 4: Check for tangent line:

At x = π, f ′(x) = 1 + (−1) = 0 ⇒ there is atangent line at x = π.

Step 5: Thus (π, π) is a point of inflection.

5. Step 1:

Step 2:

Step 3: Find critical numbers. f ′(x) = 0; at x = 0;and f ′(x) is undefined at x = ±5.

Step 4:

f x

x

x x x

x

x

x

x

″ ( ) =

−( ) −( )

− −( ) −( ) −( )

−( )

= −

−( )−

−( )

−

1 25

1

225 2

25

1

25 25

2 2

21

2

2

12

2

23

2

′( ) = −( ) −( )

= −

−( )

−

f x x x

x

x

1

225 2

25

21

2

21

2

Rewrite asf x f x x( ) ( ) = −( )25 21

2 .

Since x = −5 and x = 5 are endpoints of [−5,5],and f (−5) = f (5) = 0, and there is no other rela-tive minimum points, thus 0 is the absolute min-imum value. Similarly, since there is only onerelative maximum value and it is greater thanf (x) at the endpoints, the point (0,5) is theabsolute maximum point and 5 is the absolutemaximum value.

6. (a) Point A f ′ < 0 ⇒ decreasing and f ″ > 0 ⇒concave upward.

(b) Point E f ″ < 0 ⇒ decreasing and f ″ < 0 ⇒concave downward.

(c) Point B and D f ′ = 0 ⇒ horizontal tangent.

(d) Point C f ″ does not exist ⇒ vertical tangent.

7. A change in concavity ⇒ a point of inflection.At x = a, there is a change of concavity; f ″ goesfrom positive to negative ⇒ concavity changes

from upward to downward. At x = c, there is achange of concavity; f ″ goes from negative topositive ⇒ concavity changes from downwardto upward. Therefore f has two points of inflec-tion, one at x = a and the other at x = c.

8. Set f (x) = 0. Thus x2(x + 3)(x − 5) = 0 ⇒ x = 0,x = −3 or x = 5. (See Figure 4.8-1.)

–3 0 5

Change of

concavity

Concave

upward

Concave

downward

Concave

downward

Concave

upward

Change of

concavity

x

" + – – +

Figure 4.8-1

Thus f has a change of concavity at x = −3 andat x = 5.

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9. See Figure 4.8-2.


f '

f "

f ' incr. decr. incr.

f Concave

upward

Concave

downward

Concave

upward

x

x

[

[

]

]

0

0

–3

–3

2 3

3

decr. decr. incr.

– – +

+ +–

Figure 4.8-2

(0,0)rel. max

y

f

x

Abs. min Abs. min

pt of

infl.

pt of

infl.

( )√12 4

1– –,

( )√16 36

–5– , ( )√16 36

–5,

( )√12 4

1–,

Figure 4.8-3

Intervals x 0

f (x) 0 −5/36 −1/4

f ′(x) 0 − − − 0 +

f ″ (x) − − 0 + + +

conclusion rel decr decr decr rel incrmax concave pt. of concave min concave

downward inflection upward upward

1 2, ∞( ) x 1 21 6 1 2,( ) x 1 60 1 6,( )

The function has an absolute minimum value of (−1/4) and no absolute maximum value.

Thus f is increasing on [2,3] and concave down-ward on (0,1)

10. The correct answer is (A)

f (−1) = 0; f ′(−1) < 0 since f is decreasing andf ″ (−1) < 0 since f is concave downward. Thusf (−1) has the largest value.

11. Step 1: Domain: all real numbers.

Step 2: Symmetry: Even function (f (x) = f (−x));symmetrical with respect to the y-axis.

Step 3: f ′(x) = 4x3

− 2x and f ″ (x) = 12x2

− 2.Step 4: Critical numbers:

f ′(x) is defined for all real numbers. Setf ′(x) = 4x3 − 2x = 0 ⇒ 2x(2x2 − 1) = 0 ⇒

x = 0 or x =

Possible points of inflection:

f ″ (x) is defined for all real numbers. Setf ″ (x) = 12x2 − 2 = 0 ⇒ 2(6x2 − 1) = 0 ⇒

x = ± 1 6.

± 1 2.

Step 5: Determine intervals:

1/21/2– – 1/61/6 0√ √ √ √

Intervals are: ,

Since f ′(x) is symmetrical with respect to they-axis, you only need to examine half of theintervals.

Step 6: Set up a table (See Table 4.8-1.)

Step 7: Sketch the graph. (See Figure 4.8-3.)

1 2, .∞

( )1 6 1 2, ,( )0 1 6, ,

( )−( )1 6 0, ,

− −( )1 2 1 6, ,−∞ −( ), ,1 2

Table 4.8-1

12. Step 1: Domain: all real numbers x ≠ 4.

Step 2: Symmetry: noneStep 3: Find f ′ and f ″ .

Step 4: Critical numbers:

f ′(x) ≠ 0 and f ′(x) is undefined at x = 4.

′( ) = ( ) −( ) − ( ) +( )

−( )

= −

−( ) ( ) =

−( )

f x x x

x

xf x

x

1 4 1 4

4

8

4

16

4

2

2 3, ″

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–4–1

y

1

4 x

x = 4

f

y = 1

Figure 4.8-4

x 4 x 3 x 2 x 1

f

x

y

0

. .

Figure 4.8-5

+ – +

incr f

f '

x x decr incr1 4

rel. max. rel. min.

concave pt. of inflection

concavedownward upward

x 3 – +

f '

f "

f

decr incr


Intervals are (−∞, 4) and (4, ∞).

Step 6: Set up table as below:

4

13. (a)

(b) And f has the smallest value at x = x4.

(c)

A change of concavity occurs at x = x3, andf ′(x3) exists which implies there is a tangentto f at x = x3. Thus, at x = x3, f has a pointof inflection.

(d) The function f ″ represents the slope of thetangent to f ′. The graph indicates that theslope of the tangent to f ′ is the largest atx = x4. (See Figure 4.8-5.)

14. (a) Since f ′(x) represents the slope of the tan-gent, f ′(x) = 0 at x = 0, and x = 5.

(b) At x = 2, f has a point of inflection whichimplies that if f ″ (x) exists, f ″ (x) = 0. Sincef ′(x) is differentiable for all numbers in thedomain, f ″ (x) exists, and f ″ (x) = 0 at x = 2.

(c) Since the function f is concave downwardson (2,∞), f ″ < 0 on (2,∞) which implies f ′ isdecreasing on (2,∞).

15. (a) The function f is increasing on the intervals[−2,1] and [3,5].

(b) The absolute maximum occurs at x = 1,since it is a relative maximum, f (1) > f (−2)and f (5) < f (−2). Similarly, the absolute min-imum occurs at x = 3, since it is a relativeminimum, and f (3) < f (5) < f (−2).

(c) No point of inflection. (Note that at x = 3 f has a cusp.)

Note: Some textbooks define a point of

inflection as a point where the concavitychanges and do not require the existence of

Step 7: Horizontal asymptote:

Thus, y = 1 is a

horizontal asymptote.

Vertical asymptote:

Thus, x = 4 is a vertical asymptote.

Step 8: x-intercept: Set f ′(x) = 0 ⇒ x + 4 = 0;x = −4

y-intercept: Set x = 0 ⇒ f (x) = −1.

Step 9: Sketch the graph. (See Figure 4.8-4.)

lim lim ;x x

x

x

x

x→ →+ −

+−

= ∞ +−

= −∞4 4

4

4

4

4and

lim ;x

x

x→±∞

+−

=4

41

Interval (,4) (4,)

f ′ − −f ″ − +

conclusion Decr Concave Incr Concavedownward upward

The function f has the largest value (of the fourchoices) at x = x1.

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The function f has its relative minimum at x = 0and its relative maximum at x = 6.

(b) The function f is increasing on [0,6] anddecreasing on (−∞,0] and [6,∞).

(c)


incr decr

+ –

concave

upward

concave

downward

pt. of

inflection

3

f '

f "

f

Since f ′(3) exists and a change of concavity occursat x = 3, f has a point of inflection at x = 3.

(d) Concave upward on (−∞,3) and downwardon (3,∞).

(e) Sketch a graph. (See Figure 4.8-7.)


The graph of f indicates that a relative maximumoccurs at x = 3, f is not differentiable at x = 7,since there is a cusp at x = 7 and f does not have apoint of inflection at x = −1, since there is no tan-gent line at x = −1. Thus, only statement I is true.

decr decrincr

rel. min. rel. max.

+ ––

0 6 f

f '

f

..

y

x0

. .

[ ]

–2 1 3 5

(5,6)(–2,10)

Figure 4.8-6

f. y

x0

..3 6

(6,5)

(0,1)

Figure 4.8-7

[–5,10] by [–5,20]

Figure 4.8-8

[–π,π] by [–2,2]

Figure 4.8-9

a tangent. In that case, at x = 3, f has a pointof inflection.

(d) Concave upward on (3, 5) and concavedownward on (−2, 3).

(e) A possible graph is shown in Figure 4.8-6.

16. (a)


Enter y1 = cos(x2)

Using the Inflection function of your calculator,you obtain three points of inflection on [0 , π].The points of inflection occur at x = 1.35521,2.1945 and 2.81373. Since y1 = cos(x2), is aneven function; there is a total of 6 points of inflection on [−π, π]. An alternate solution is to

enter The graph of y2

indicates that there are 6 zero’s on [−π, π].

y d

dx y x x2

2

2 1 2= ( )( ), , .

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19. Enter y1 = 3 e ∧ (−x ∧ 2/2). Note that thegraph has a symmetry about the y-axis. Usingthe functions of the calculator, you will find:

(a) a relative maximum point at (0,3), which isalso the absolute maximum point;

(b) points of inflection at (−1,1.819) and(1,1.819);

(c) y = 0 (the x-axis) a horizontal asymptote;

(d) y1 increasing on (−∞,0] and decreasing on[0,∞); and

(e) y1 concave upward on (−∞,−1) and (1,∞)and concave downward on (−1,1). (SeeFigure 4.8-10.)


[–4,4] by [–1,4]

Figure 4.8-10

[–1,9.4] by [–1,1]

Figure 4.8-11

20. (See Figure 4.8-11.) Enter y1 = cos(x) (sin(x))∧ 2. A fundamental domain of y1 is [0,2π]. Usingthe functions of the calculator, you will find:

(a) relative maximum points at (0.955,0.385),(π,0) and (5.328,0.385) and relative mini-mum points at (2.186,−0.385) and(4.097,−0.385);

(b) points of inflection at (0.491,0.196),

(2.651,−0.196), (3.632,−0.196),

and (5.792,0.196);

(c) no asymptote;

(d) function is increasing on intervals (0,0.955),(2.186,π) and (4.097,5.328) and decreasingon intervals (0.955,2.186), (π,4.097) and(5.328,2π);

(e) function is concave upwards on intervals

(0,0.491), and

(5.792, 2π) and concave downward on the

intervals , (2.651,3.632) and

3

25 792

π, . .

0 491 2. , ,π

π π2

2 651 3 6323

2, . , . , ,

3

20

π,

π2

0, ,


21. (x2 + y2)2 = 10xy

2 2 2 10 10

4 4 10 10

4 10 10 4

4 10 10 4

2 2

2 2 2 2

2 2 2 2

2 2

x y x y

dy

dx y x

dy

dx

x x y y x y dy

dx y x

dy

dx

y x y dy

dx x

dy

dx y x x y

dy

dx y x y x y x x

+( ) +

= + ( )

+( ) + +( ) = + ( )

+( ) − ( ) = − +( )

+( ) −( ) = − 22 2

2 2

2 2

2 2

2 2

10 4

4 10

5 2

2 5

+( )

= − +( )

+( ) − =

− +( )+( ) −

y

dy

dx

y x x y

y x y x

y x x y

y x y x

22. lim lim

lim

lim

lim

.

x x

x

x

x

x

x

x

x

x

xx

x x

x

x x

x

→ →

→

→

→

+ −=

+ −( ) + +( )

+ +( )=

+( ) −

+ +( )=

+ +( )=

+ +=

+ +

=+

=

0 0

0

0

0

9 3 9 3 9 3

9 39 9

9 3

9 3

1

9 3

1

0 9 3

1

3 3

1

6

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23.

24. (Calculator) The function f is continuous every-

where for all values of except possibly at x = 1.Checking with the three conditions of continuityat x = 1:

(1) f (1) = (1)2 − 1 = 0

(2)

thus 2 + k = 0 ⇒ k = −2. Sincetherefore

(3) f (1) = Thus, k = −2.

25. (a) Since f ′ > 0 on (−1,0) and (0,2), the function f is increasing on the intervals [−1,0] and[0,2]. And f ′ < 0 on (2,4), f is decreasing on[2,4].

(b) The absolute maximum occurs at x = 2,since it is a relative maximum and it is theonly relative extremum on (−1,4). Theabsolute minimum occurs at x = −1, since

lim .x

f x→

) =1

0

lim .x

f x→

( ) =1

0

lim lim ,x x

f x f x→ →+ −

) = ) =1 1

0

lim , lim ;x x

x k k x→ →+ −

+( ) = + −( ) =1 1

22 2 1 0

y x x

dy

dx x x x x

d y

dx x x

= ( ) + −

= − ( )[ ]( ) + = − ( ) +

= − ( )( )( ) + = − ( ) +

cos

sin sin

cos cos .

2 3 1

2 2 6 2 2 6

2 2 2 6 4 2 6

2

2

2


f (−1) < f (4) and the function has no relativeminimum on [−1,4].

(c) A change of concavity occurs at x = 0.However, f ′(0) is undefined, which implies f may or may not have a tangent at x = 0.Thus f may or may not have a point of inflection at x = 0.

(d) Concave upward on (−1,0) and concavedownward on (0,4).

(e) A possible graph is shown in Figure 4.9-1.

(–1,0)

0

1

2

1–1 2 3 4

possible

point of

reflection

y f

(4,2)

Figure 4.9-1

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5.1 RELATED RATE

Main Concepts: General Procedure for Solving Related Rate Problems, Common Related Rate Problems, Inverted Cone (Water Tank) Problem,Shadow Problem, Angle of Elevation Problem

General Procedure for Solving Related Rate Problems

1. Read the problem and if appropriate, draw a diagram.2. Represent the given information and the unknowns by mathematical symbols.3. Write an equation involving the rate of change to be determined. (If the equation

contains more than one variable, it may be necessary to reduce the equation to onevariable.)

4. Differentiate each term of the equation with respect to time.5. Substitute all known values and known rates of change into the resulting equation.6. Solve the resulting equation for the desired rate of change.7. Write the answer and, if given, indicate the units of measure.

Common Related Rate Problems

Example 1

When area of a square is increasing twice as fast as its diagonals, what is the length of a side of the square?

Let z represent the diagonal of the square. The area of a square is

SincedA

dt

dz

dt

dz

dt z

dz

dt z= = ⇒ =2 2 2,

dA

dt z

dz

dt z

dz

dt =

=2

1

2.

A z=

2

2.

Chapter 5

Applications of Derivatives


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Let s be a side of the square. Since the diagonal z = 2, then s2 + s2 = z2

Example 2

Find the surface area of a sphere at the instant when the rate of increase of the volumeof the sphere is nine times the rate of increase of the radius.

Volume of a sphere: Surface area of a sphere: S = 4πr2

Since

Since S = 4πr2, thus the surface area is S = 9 square units.

Note: At 9 = 4πr2, you could solve for r and obtain You could

then substitute into the formula for surface area s = 4πr2 and obtain 9. These

steps are of course correct but not necessary.

Example 3

The height of a right circular cone is always three times the radius. Find the volume of the cone at the instant when the rate of increase of the volume is twelve times the rateof increase of the radius.

Let r, h be the radius and height of the cone respectively.

Since h = 3r, the volume of the cone

When

Thus V r= =

=

=π ππ

ππ π π

3

3

2 8 8.

dV

dt

dr

dt

dr

dt r

dr

dt r r= = ⇒ = ⇒ =12 12 3 4

22 2, .π ππ

V r dV

dt r

dr

dt = =π π3 23; .

V r h r r r= = ( ) =1

3

1

332 2 3π π π .

r =3

2

1

π

r r2 9

4

3

2

1= =

π πor .

dV

dt

dr

dt

dr

dt r

dr

dt r= = =9 9 4 9 42 2, .you have orπ π

V r dV

dt r

dr

dt = =

4

343 2π ;

V r=4

33

π ;

⇒ = ⇒ = ⇒ = =2 4 4 2 22 2 2s s s sor .


• Go with your first instinct if you are unsure. Usually that’s the correct one.

Inverted Cone (Water Tank) Problem

A water tank is in the shape of an inverted cone. The height of the cone is 10 metersand the diameter of the base is 8 meters as shown in Figure 5.1-1. Water is beingpumped into the tank at the rate of 2 m3 /min. How fast is the water level rising whenthe water is 5 meters deep? (See Figure 5.1-1 on page 159.)

Solution:

Step 1: Define the variables. Let V be the volume of water in the tank; h be the heightof the water level at t minutes; r be the radius of surface of the water at t min-utes; and t be the time in minutes.

Step 2: Given: Height = 10 m, diameter = 8 m

Find: .dhdt hat = 5

dV

dt = 2 3m min.

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Step 3: Set up an equation

Using similar triangles, you have See

Figure 5.1-2.

4

104 10

2

5= ⇒ = =

r

h h r r

h; .or

V r h=1

32π .

Applications of Derivatives • 159

10 m

5m

8m

Figure 5.1-1

r

4

h

10

Figure 5.1-2

Thus, you can reduce the equation to one variable:

Step 4: Differentiate both sides of the equation with respect to t .

Step 5: Substituting known values.

Step 6: Thus, the water level is rising at when the water is 5 m high.12π m min

Evaluating at m

m

dh

dt h

dh

dt h

= =

( )

=

=

525

2

1

5

1

2

5

2;

.

π

π

min

min

24

25

25

2

12

2= =

π

πh

dh

dt

dh

dt h; m min

dV dt

h dhdt

h dhdt = ( ) =4

753 4

252 2π π

V h

h h=

=1

3

2

5

4

75

2

3π π

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Shadow Problem

A light on the ground 100 feet from a building is shining at a 6-foot tall man walkingaway from the streetlight and towards the building at the rate of 4 ft/sec. How fast ishis shadow on the building growing shorter when he is 40 feet from the building? SeeFigure 5.1-3.


Solution:

Step 1: Let s be the height of the man’s shadow; x be the distance between the manand the light; and t be the time in seconds.

Step 2: Given: man is 6 ft tall; distance between light and building =

100 ft. Find

Step 3: See Figure 5.1-4. Write an equation using similar triangles, you have:

ds

dt xat = 60.

dx

dt = 4 ft sec;

100 ft

Building

Light6 ft

Figure 5.1-3

100

6

x

s

Figure 5.1-4

Step 4: Differentiate both sides of the equation with respect to t .

Step 5: Evaluate

Note: when the man is 40 ft from the building, x (distance from the light)is 60 ft.

ds

dt at x = 60.

ds

dt x

dx

dt x

dx

dt x x= −( )( ) =

−=

−( ) =

−−1 600600 600

424002

2 2 2ft sec

6100

600 600 1

sx s

x x= = = −;

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Step 6: The height of the man’s shadow on the building is changing at −2

3ft sec .

ds

dt x =

= −

( )= −

60

2

2400

60

2

3ft sec ft sec


Angle of Elevation Problem

A camera on the ground 200 meters away from a hot air balloon records the balloonrising into the sky at a constant rate of 10 m/sec. How fast is the camera’s angle of ele-vation changing when the balloon is 150 m in the air? See Figure 5.1-5.

• Indicate units of measure, e.g., the velocity is 5 m/sec or the volume is 25 in3.

200 m

x

x

y

θ

Camera

Balloon

Figure 5.1-5

Step 1: Let x be the distance between the balloon and the ground; θ be the camera’sangle of elevation; and t be the time in seconds.

Step 2: Given: distance between camera and the point on the grounddx

dt = 10 m sec;

where the balloon took off is 200 m,

Step 3: Find

Step 4: Differentiate both sides with respect to t.

Step 5:

Using the Pythagorean Theorem: y2 = x2 + (200)2

y2 = (150)2 + (200)2

y = ± 250

Since y > 0, then y = 250. Thus, sec .θ = =250

200

5

4

sec θ = =y

x200

150and at

sec ; sec sec2

2 21200 1200 1 10 120θ θ θ θ θd dt dxdt d dt = = ( ) =

d

dt x

θat m= 150 .

tan .θ = x

200

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Step 6: The camera’s angle of elevation changes at approximately 1.833 deg/secwhen the balloon is 150 m in the air.

5.2 APPLIED MAXIMUM AND MINIMUM PROBLEMS

Main Concepts: General Procedure for Solving Applied Maximum and Minimum

Problems, Distance Problem, Area and Volume Problems, Business Problems

General Procedure for Solving Applied Maximum andMinimum Problems

Steps:1. Read the problem carefully and if appropriate, draw a diagram.2. Determine what is given and what is to be found and represent these quantities by

mathematical symbols.

3. Write an equation that is a function of the variable representing the quantity to bemaximized or minimized.

4. If the equation involves other variables, reduce the equation to a single variablethat represents the quantity to be maximized or minimized.

5. Determine the appropriate interval for the equation (i.e., the appropriate domainfor the function) based on the information given in the problem.

6. Differentiate to obtain the first derivative and to find critical numbers.7. Apply the First Derivative Test or the Second Derivative Test by finding the second

derivative.8. Check the function values at the end points of the interval.9. Write the answer(s) to the problem and, if given, indicate the units of measure.

Distance Problems

Find the shortest distance between the point A (19, 0) and the parabola y = x2 − 2x + 1.

Solution:

Step 1: Draw a diagram. See Figure 5.2-1.Step 2: Let P(x, y) be the point on the parabola and let Z represents the distance

between points P(x, y) and A(19, 0).Step 3: Using the distance formula,

=

=

= =

≈

1

205

4

1

2025

16

1

125

4

4

125

032

1 833

2radian

radian

deg

sec

. sec

.

or

sec

Evaluating radiand

dt x

θθ=

= =

150

2 2

1

20

1

205

4

secsec


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(Special case: In distance problems, the distance and the square of the dis-tance have the same maximum and minimum points). Thus, to simplify com-

putations, let L = Z2

= (x − 19)2

+ (x − 1)4

. The domain of L is (−∞,∞).

Step 4:

is defined for all real numbers.

Set = 0; 2x3 − 6x2 + 7x − 21 = 0. The factors of 21 are ±1, ±3, ±7 and ±21.

Using Synthetic Division, 2x3 − 6x2 + 7x − 21 = (x − 3)(2x2 + 7) = 0 ⇒ x = 3.Thus the only critical number is x = 3.(Note: Step 4 could have been done using a graphing calculator.)

Step 5: Apply the First Derivative Test.

dL

dx

dL

dx

Differentiate dL

dx x x

x x x x x x x

x x x

= −( )( ) + −( ) ( )

= − + − + − = − + −

= − + −( )

2 19 1 4 1 1

2 38 4 12 12 4 4 12 14 42

2 2 6 7 21

3

3 2 3 2

3 2

Z x y x x x

x x x x

= −( ) + −( ) = −( ) + − + −( )

= −( ) + −( )( ) = −( ) + −( )

19 0 19 2 1 0

19 1 19 1

2 2 2 2 2

2 2 2 2 4


• Simplify numeric or algebraic expressions only if the question asks you to do so.

Figure 5.2-1

0 3

0 – L '

L

+

[

decr incr

rel. min

Step 6: Since x = 3 is the only relative minimum point in the interval, it is the absoluteminimum.

Step 7: At

Thus, the shortest distance is

x Z= = −( ) + − ( ) +( ) = −( ) + ( )

= = =

3 3 19 3 2 3 1 16 4

272 16 17 4 17 4 17

2 2 2 2 2

,

. .

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Area and Volume Problems

Example—Area Problem

The graph of encloses a region with the x-axis and y-axis in the first

quadrant. A rectangle in the enclosed region has a vertex at the origin and the opposite

vertex on the graph of . Find the dimensions of the rectangle so that itsarea is a maximum.

Solution:

Step 1: Draw a diagram. (See Figure 5.2-2.)

y x= − +

1

2 2

y x= − +1

22


y = – x + 2

y

y

x x

P( x,y)

0

1

2

Figure 5.2-2

Step 2: Let P(x, y) be the vertex of the rectangle on the graph of .

Step 3: Thus the area of the rectangle is:

The domain of A is [0,4].

Step 4: Differentiate.

Step 5:

A(x) has one critical number x = 2.

Step 6: Apply Second Derivative Test

has a relative maximum point at x = 2; A(2) = 2.

Since x = 2 is the only relative maximum, it is the absolute maximum. (Noteat the endpoints: A(0) = 0 & A(4) = 0)

Step 7: At x = 2,

Therefore the length of the rectangle is 2, and its width is 1.

y = − ( ) + =1

22 2 1

d A

dx A x

2

21= − ⇒ ( )

Set dAdx

x x= ⇒ − + = =0 2 0 2;

dA

dxis defined for all real numbers.

dA

dx x= − + 2

A xy A x x x x= = − +

= − +or

1

22

1

222

y x= − +1

22

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Example—Volume Problem (with calculator)

If an open box is to be made using a square sheet of tin, 20 inches by 20 inches, by cut-ting a square from each corner and folding the sides up; find the length of a side of thesquare being cut so that the box will have a maximum volume.

Solution:

Step 1: Draw a diagram. (See Figure 5.2-3.)

Step 2: Let x be the length of a side of the square to be cut from each corner.

Step 3: The volume of the box is V = x(20 − 2x)(20 − 2x).

The domain of V is [0, 10].

Step 4: Differentiate V (x).

Entering d (x (20 − 2x) (20 − 2x), x) = 4(x − 10)(3x − 10).

Step 5: V ′(x) is defined for all real numbers:

Set V ′(x) = 0 by entering: solve (4(x − 10)(3x − 10) = 0, x), and obtain x = 10

or The critical numbers of V (x) are x = 10 and V (10) = 0

and = 592.59. Since V (10) = 0, you need to test only x =10

3.V

10

3

x =10

3.x =

10

3.

x x

x

x

x

x

x

x

x

20

20–2 x

20–2 x

2

0 – 2

x

20

20–2 x

Figure 5.2-3

Step 6: Using the Second Derivative Test, d (x (20 − 2x) (20 − 2x), x, 2) x =10

3.

and obtain −80. Thus, is a relative maximum. And since it is the onlyV 10

3

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Solution:

Step 1: Write an equation.

Profit = Revenue − Cost

P = R − C

Revenue = (Units Sold)(Price Per Unit)

R = xp(x) = x(10) = 10xP = 10x − (500 + 3x + 0.01x2)

Step 2: Differentiate.

Enter d (10x − (500 + 3x + 0.01x^2, x) and obtain 7 − 0.02x.

Step 3: Find critical numbers.

Set 7 − 0.02x = 0 ⇒ x = 350.

Critical number is x = 350.

Step 4: Apply Second Derivative Test.

Enter d (10x − (500 + 3x + 0.01x^2), x, 2)x = 350 and obtain −02.Since x = 350 is the only relative maximum, it is the absolute maximum.

Step 5: Write a Solution

Thus, producing 350 units will lead to maximum profit.

5.3 RAPID REVIEW

1. Find the instantaneous rate of change at x = 5 of the function

2. If the diameter of a circle h is increasing at a constant rate of 0.1 cm/sec, find the rateof change of the area of the circle when the diameter is 4 cm.

3. The radius of a sphere is increasing at a constant rate of 2 inches per minute. In termsof the surface area, what is the rate of change of the volume of the sphere?

Answer: since

4. Using your calculator, find the shortest distance between the point (4, 0) and the liney = x. (See Figure 5.3-1.)

S r dV

dt S= =π 2 32, in. min.V r

dr

dt r

dr

dt = =

4

343 2π π;

Answer A r h

h

dA

dt h

dh

dt

:

cm

= =

=

= = ( )( ) =

π π

π π π

2

2

2

2

2

1

4

1

2

1

24 0 1 0 2. . sec .

Answer f x x x

f x x x

f

: ( ) = − = −( )

( ) = −( ) ( ) = −( )

′( ) =

−

2 1 2 1

1

22 1 2 2 1

51

3

12

12

12

f x x( ) = −2 1.


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Answer:

Enter y1 = ((x − 4)^2 + x^2)^(.5) and y2 = d(y1(x), x)

Use the Zero function for y2 and obtain x = 2. Use the Value function for y1 atx = 2 and obtain y1 = 2.82843. Thus the shortest distance is approximately 2.828.


S x y x x= −( ) + −( ) = −( ) +4 0 42 2 2 2


[–6.3,10] by [–2,6]

Figure 5.3-1

13 ft

Wall

ground

Figure 5.4-1


1. A spherical balloon is being inflated. Find thevolume of the balloon at the instant when therate of increase of the surface area is eight timesthe rate of increase of the radius of the sphere.

2. A 13-foot ladder is leaning against a wall. If the

top of the ladder is sliding down the wall at 2 ft/ sec, how fast is the bottom of the ladder movingaway from the wall, when the top of the ladder is5 feet from the ground? (See Figure 5.4-1.)

5. A water tank in the shape of an inverted conehas an altitude of 18 feet and a base radius of 12 feet. If the tank is full and the water isdrained at the rate of 4 ft3 /min, how fast isthe water level dropping when the water level is6 feet high?

6. Two cars leave an intersection at the same

time. The first car is going due east at the rateof 40 mph and the second is going due south atthe rate of 30 mph. How fast is the distancebetween the two cars increasing when the firstcar is 120 miles from the intersection?

7. If the perimeter of an isosceles triangle is 18 cm,find the maximum area of the triangle.

8. Find a number in the interval (0,2) such that thesum of the number and its reciprocal is theabsolute minimum.

9. If an open box is to be made using a cardboard8 cm by 15 cm by cutting a square from eachcorner and folding the sides up. Find the lengthof a side of the square being cut so that the boxwill have a maximum volume.

10. What is the shortest distance between the point

and the parabola y = −x2?

11. If the cost function is C(x) = 3x2 + 5x + 12, find

the value of x such that the average cost is aminimum.

21

2, −

3. Air is being pumped into a spherical balloon atthe rate of 100 cm3 /sec. How fast is the diameterincreasing when the radius is 5 cm.

4. A man 5 feet tall is walking away from a street-light hung 20 feet from the ground at the rate

of 6 ft/sec. Find how fast is his shadow islengthening.

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12. A man with 200 meters of fence plans to enclosea rectangular piece of land using a river on oneside and a fence on the other three sides. Findthe maximum area that the man can obtain.


13. A trough is 10 meters long and 4 meters wide.(See Figure 5.4-2.)

The two sides of the trough are equilateral tri-angles. Water is pumped into the trough at1 m3 / min. How fast is the water level risingwhen the water is 2 meters high?

14. A rocket is sent vertically up in the air with theposition function s = 100t 2 where s is measured inmeters and t in seconds. A camera 3000 m away

is recording the rocket. Find the rate of change of the angle of elevation of the camera 5 sec afterthe rocket went up.

15. A plane lifting off from a run away at an angleof 20°. If the speed of the plane is 300 mph,how fast is the plane gaining altitude?

16. Two water containers are being used. (SeeFigure 5.4-3.)

One container is in the form of an inverted right

circular cone with a height of 10 feet and a radiusat the base of 4 feet. The other container is a rightcircular cylinder with a radius of 6 feet and aheight of 8 feet. If water is being drained from theconical container into the cylindrical container atthe rate of 15 ft3 /min, how fast is the water levelfalling in the conical tank when the water level inthe conical tank is 5 feet high? How fast is thewater level rising in the cylindrical container?

17. The wall of a building has a parallel fence that is6 feet high and 8 feet from the wall. What is the

length of the shortest ladder that passes over thefence and leans on the wall? (See Figure 5.4-4.)


4 m

1 0 m

Figure 5.4-2

10 m

4 ft

6 ft

8 ft

Figure 5.4-3

Ladder

8 ft

Wall

Fence

6 ft

Figure 5.4-4

7/23/2019 007137714 x



18. Given the cost function C(x) = 2500 + 0.02x +0.004x2, find the product level such that theaverage cost per unit is a minimum.

19. Find the maximum area of a rectangle inscribedin an ellipse whose equation is 4x2 + 25y2 = 100.

20. A right triangle is in the first quadrant with avertex at the origin and the other two verticeson the x- and y-axes. If the hypotenuse passesthrough the point (0.5, 4), find the vertices of the triangle so that the length of the hypotenuseis the shortest.



21.

22.

23. The graph of f ′ is shown Figure 5.5-1. Findwhere the function f: (a) has its relative extremaor absolute extrema; (b) is increasing or decreas-ing; (c) has its point(s) of inflection; (d) is con-cave upward or downward; and (e) if f (3) = −2,draw a possible sketch of f. (See Figure 5.5-1.)

24. (Calculator) At what values(s) of x does the tan-gent to the curve x2 + y2 = 36 have a slope of −1.

Evaluate lim .x

xx x→∞ − + +

100

4 2

If findy x dy

dx= −( )( )sin cos , .2 6 1


1.

2. Pythagorean Theorem yields x2 + y2 = (13)2.

Differentiate: 2 2 0x dx

dt y

dy

dt

dy

dt

x

y

dx

dt

+ = ⇒

= −

.

At cubic unitsr V = =

=

1 4

3

1 4

3

3

2π π

π π, .

Since

or

dS

dt

dr

dt

dr

dt r

dr

dt r

r

= = ⇒ =

=

8 8 8 8 8

1

,

.

π π

π

dS

dt r

dr

dt = 8π

Volume: Surface Area:V r S r= =4

343 2π π;

At x = 5, (5)2 + y2 = 13 ⇒ y = ±12, since y > 0,y = 12.

Therefore,

The ladder is moving away from the wall at

when the top of the ladder is 5 feet

from the ground.

3. Volume of a sphere is

Differentiate:

Substitute:

Let x be the diameter. Since

Thus Thedx

dt r =

=

=5

21 2

π πcm cmsec sec .

x r dx

dt

dr

dt = =2 2, .

100 4 512

= ( ) ⇒ =ππ

dr

dt

dr

dt cm sec .

dV

dt r

dr

dt r

dr

dt =

( ) =

4

33 42 2π π .

V r=4

33

π .

5

6ft sec

dy

dt = − −( ) =

5

122

5

6ft ftsec sec .


y

f '

x 0 3

Figure 5.5-1

25. (Calculator) Find the shortest distance betweenthe point (1, 0) and the curve y = x3.

7/23/2019 007137714 x


diameter is increasing at when the

radius is 5 cm.

4. See Figure 5.6-1. Using similar triangles, with ythe length of the shadow you have:

2

πcm sec Differentiate: Substituting

known values:

The water level is dropping at when

h = 6 ft.

6. See Figure 5.6-3. Step 1. Using the PythagoreanTheorem, you have x2 + y2 = z2. You also havedx

dt

dy

dt = =40 30and .

1

4π

ft min

⇒ − = = −4 161

4π

πdh

dt

dh

dt or ft min

− = ( )44

96

2π dh

dt

dV

dt h

dh

dt =

4

92π .


5 ft

20 ft

y x

Light

Figure 5.6-1

Differentiate:

5. See Figure 5.6-2. Volume of a cone V r h=1

32π .

dy

dt

dx

dt

dy

dt = ⇒ = ( )

=

1

3

1

36

2 ft sec.

520

20 5 5 15

53

=+

⇒ = + ⇒

= =

yy x

y y x y

x y x

or .

5 m

h

r 18

12

Figure 5.6-2

Using similar triangles, you have

thus reducing the equation to

V h h h=

( ) =1

3

2

3

4

27

2

3π π

.

= =32

3r r hor ,

12

182= ⇒

r

h h

y

x

z

N

S

W E

Figure 5.6-3

Step 2. Differentiate:

At x = 120, both cars have traveled 3hours and thus, y = 3(30) = 90. By thePythagorean Theorem, (120)2 + (90)2 =z2 ⇒ z = 150.

Step 3. Substitute all known values into theequation:

2(120)(40) + 2(90)(30) = 2(150)

Thus = 50 mph.

Step 4. The distance between the two cars isincreasing at 50 mph at x = 120.

7. See Figure 5.6-4. Step 1. Applying the Pytha-gorean Theorem, you have x2 = y2 + (9 − x)2 ⇒ y2

= x2 − (9 − x)2 = x2 − (81 − 18x + x2) = 18x − 81 =

9(2x − 9), or y = = ± −( )3 2 9x± −( )9 2 9x

dz

dt

dzdt

2 2 2x dx

dt y

dy

dt z

dz

dt + = .

7/23/2019 007137714 x


triangle (18 − 2x) =

(9 − x) = (9 − x).

Step 2.

Step 3. Set

is undefined at x = . The critical92

dAdx

dA

dx x x= ⇒ − = =0 54 9 0 6;

dA

dx x x

x

x x

x

x

x

= −( ) ( ) −( )

+ −( )( ) −( )

= −( ) − −( )

−

= −

−

−3

2 2 9 2 9

1 3 2 9

3 9 3 2 9

2 9

54 9

2 9

12

12

3 2 91

2x −( )

3 2 9x −( )A x= −( )1

23 2 9

8. See Figure 5.6-5. Step 1. Let x be the number

and be its reciprocal.1

x


+undef undef 0 –

69/2incr decr

A'

A

Thus at x = 6, the area A is a relative

maximum.

Step 5. Check endpoints. The domain of A is

and A(9) = 0.

Therefore, the maximum area of anisosceles triangle with the perimeter of

18 cm is (Note, at x = 6, thetriangle is an equilateral triangle.)

9 3 cm.

92 9

92

0,

;

( ) =A

A 6 3 2 6 9 9 6 9 3( ) = ( ) − −( ) = .

y

x

9 – x 9 – x

x

Figure 5.6-4

x

x

x x

x

x

x

x x

x x

x x

x

x

15–2 x

8–2 x

Figure 5.6-5

Step 2.

Step 3.

Step 4. Set

⇒ x = ±1, since the domain is(0,2), thus x = 1.

is defined for all x in (0,2).

Critical number is x = 1.

Step 5. Second Derivative Test: and

Thus at x = 1, s is a relative minimum.Since it is the only relative extremum,thus, at x = 1, it is the absolute minimum.

9. Step 1. Volume V = x(8 − 2x)(15 − 2x) with 0 ≤x ≤ 4.

Step 2. Differentiate: Rewrite asV = 4x3 − 46x2 + 120x

Step 3. Set V = 0 ⇒ 12x2 − 92x + 120 = 0⇒ 3x2 − 23x + 30 = 0. Using the qua-dratic formula, you have x = 6 or

. And is defined for all real

numbers.

dV

dxx =

5

3

dV

dx x x= − +12 92 1202

d s

dxx

2

2

1

2=

= .

d s

dx x

2

2 3

2=

ds

dx

⇒ − =11

02x

ds

dx = 0

ds

dx x

x= + −( ) = −−1 1 1

12

2

S x

x

x= + < <1

0 1with .

since y > 0, y = The area of the3 2 9x −( ).

numbers are , 6.

Step 4. First Derivative Test

9

2

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Step 4. Second Derivative Test.

Thus at is a relative maximum.

Step 5. Check endpoints.

At x = 0, V = 0 and at x = 4, V = 0.

Therefore, at , V is the absolute

maximum.

10. See Figure 5.6-6. Step 1. Distance Formula:

x =5

3

x =

5

3

d V

dx x

d V

dx

d V

dx

x

x

2

2

2

2

6

2

25

3

24 92

52 52

= −

= = −

=

=

;

and

minimum and since it is the only relativeextrema, it is the absolute minimum

Step 5. At x = 1,

Therefore, the shortest distance is

11. Step 1. Average Cost

Step 2.

Step 3. Set

Since x > 0, x = 2 and

= 17. is undefined at x = 0 which isnot in the domain.

Step 4. Second Derivative Test:

Thus at x = 2, the average cost is aminimum

12. See Figure 5.6-7. Step 1. Area A = x(200 − 2x) =200x − 2x2 with 0 ≤ x ≤ 100

d C

dx x

d C

dxx

2

2 3

2

2

2

243= =

=

and

dC

dx

C 2( )⇒ ±x 2.

dC

dx x x= ⇒ − = ⇒ −0 3

120 3

122 2

dC

dx x

x= − = −−3 12 3

122

2

C C xx

x xx

xx

= ( ) = + +

= + +

3 5 12

3 512

2

5

4.

Z = ( ) − ( ) + =1 4 117

4

5

4

4

.

Figure 5.6-6

x x

River

(200–2 x )

Figure 5.6-7

Step 2. A′(x) = 200 − 4x

Step 3. Set A′(x) = 0 ⇒ 200 − 4x = 0; x = 50

Step 4. Second Derivative Test

A″ (x) = −4; Thus at x = 50, the area is arelative maximum.

A(50) = 5000 m2.

Step 5. Check endpoints

A(0) = 0 and A(100) = 0; Therefore atx = 50, the area is the absolute maximumand 5000 m2 is the maximum area.

Step 2. Let S = Z2, since S and Z have the same

maximums and minimums.

Step 3. Set = 0; x = 1 and is defined for

all real numbers.

Step 4. Second Derivative Test:

and Thus at x = 1, Z is ad S

dxx

2

2

1

12=

= .

d S

dx x

2

2

212=

dS

dx

dS

dx

S x x dS

dx x= − + = −4 34

17

44 4;

Z x y

x x

x x x x

x x

= −( ) + − −

= −( ) + − +

= − + + − +

= − +

2

1

2

21

2

4 41

4

417

4

2

2

2 2

2

2 4 2

4

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13. Step 1. Let h be the height of the trough and4 be a side of one of the two equilateraltriangles. Thus, in a 30–60 right triangle,

Step 2. Volume V = (area of the triangle) 10

=

Step 3. Differentiate with respect to t.

Step 4. Substitute known values

The water level is rising

when the water level is 2 m high.

14. See Figure 5.6-8. Step 1. tan θ = S /3000

3

40m min

120

32

3

40= ( ) =

dh

dt

dh

dt ; min .m

dV

dt h

dh

dt =

( )

10

32

1

2

2

310

10

3

2h h h( )

=

h = 2 3.

The angle of elevation is changing at0.197 radian/sec, 5 seconds after lift off.

15. See Figure 5.6-9. Sin 20300

° = ht

d

dt

θ= 0 197. radian sec

since sec .θ = Z

3000

h = (sin 20°) 300t ; = (sin 20°)(300) ≈

102.61 mph. The plane is gaining altitude at102.61 mph.

16.

Similar triangles:

Substitute known values:

The water level in the cone is falling at

≈ −1.19 ft/min when the water level

is 5 feet high.

V cyclinder = πR2 H = π(6)2 H = 36 πH .

≈ 0.1326 ft/min or 1.592 in./min

The water level in the cylinder is rising at

≈ 0.1326 ft/min.512π ft min

dV

dt

dH

dt

dH

dt

dV

dt

dH

dt − =

= ( ) =

361

36

1

3615

5

12

ππ

π π

; ;

minft

−15

4πft min

− = = −

≈ −15 415

41 19π

πdh

dt

dh

dt ; . minft

− = ( )154

255

2π dh

dt ;

V h

h h dV

dt h

dh

dt come =

= = ( )

1

3

2

5

4

75

4

753

2

3 2π π π

; .

4

10

5 22

5

= ⇒ = =r

h

r h r h

or

V r hcome =1

32π

dh

dt

3000 m

S

Z

θ

Camera

Figure 5.6-8

Step 2. Differentiate with respect to t.

Step 3. At t = 5; S = 100(5)2 = 2500;

Thus Z2 = (3000)2 + (2500)2 =

15,250,000. Therefore Z = ±500 61,

sec ;

sec

sec

2

2

2

1

3000

1

3000

1

1

3000

1200

θ θ θ

θ

θ

d

dt

dS

dt

d

dt

dS

dt

t

=

=

=

( )

since Z > 0, Substituteknown values into the question:

d

dt

θ=

( )1

3000

1

500 613000

1000

2

,

Z = 500 61.

h

20°

Figure 5.6-9

7/23/2019 007137714 x



17. Step 1. Let x be the distance of the foot of theladder from the higher wall. Let y be the heightof the point where the ladder touches the higher

wall. The slope of the ladder is or

Thus ⇒ (y − 6)(8 − x) = −48

⇒ 8y − xy − 48 + 6x = −48 ⇒ y(8 − x) = −6x

⇒ y =

Step 2. Pythagorean Theorem: l 2 = x2 + y2 = x2 +

Since

Step 3. Enter y1 =

The graph of y1 is continuous on theinterval x > 8. Use the minimum func-tion of the calculator and obtain x =14.604; y = 17.42. Thus the minimumvalue of l is 19.731 or the shortest lad-der is approximately 19.731 feet.

18. Step 1. Average Cost

+ 0.02 + .004x

Step 2. Enter y1 + .02 + .004 x

Step 3. Use the Minimum function in the calcu-lator and obtain x = 790.6.

Step 4. Verify the result with the First DerivativeTest. Enter y2 = d (2500/ x + .02 + 004x,x); Use the Zero function and obtain

x = 790.6. Thus at x = 790.6.

Apply the First Derivative Test:

dC

dx = 0;

=2500

x

=2500

x

= + +2500 0 02 0 004 2. .x x

x

C C

x C x= ( )

; Thus

x x x ˆ ˆ .2 6 8 2+ −( ) −( )[ ]{ }

l l x x

x x

> = +

−

−

>0

6

882

2

, ,

−−

6

8

2

x

x

−−6

8

x

x

yx−− = −6

86

8

mx

= −

−6 0

8.

m y=

−−

6

0 8

Thus the minimum average cost perunit occurs at x = 790.6 (The graph of the average cost function is shown inFigure 5.6-10.)

' – 0 +

decr

rel. min

incr

0 790.6

Figure 5.6-10

19. See Figure 5.6-11. Step 1. Area A = (2x)(2y); 0 ≤x ≤ 5 and 0 ≤ y ≤ 2.

( x,y)

y

2

–2

–5 5 x x

y

Figure 5.6-11

Step 2. 4x2 + 25y2 = 100; 25y2 = 100 − 4x2

Step 3.

Step 4. Enter

Use the Maximum function and obtainx = 3.536 and y1 = 20.

Step 5. Verify the result with the First DerivativeTest.

y x

x124

5100 4= −

A x x

xx

= ( )

−( )

= −

22

5100 4

4

5100 4

2

2

y x

y x

y

y x x

2

2 2

2 2

100 4

25

100 4

25

0

100 4

25

100 4

5

= −

⇒ = ± −

≥

= −

= −

Since

7/23/2019 007137714 x


Since f has only one relative extremum,it is the absolute extremum.

Step 6. Thus at x = 2.5, the length of thehypotenuse is the shortest. At x = 2.5,

The vertices of the

triangle are (0,0), (2.5,0) and (0,5).

y = )

− =

4 2 5

2 5 0 55

.

. ..


Enter . Use

Zero function and obtain x = 3.536.

Note that:

y d x

x x224

5100 4= −

,

Step 3.

Step 4. Enter and use

the minimum function of the calculatorand obtain x = 2.5.

Step 5. Apply the First Derivative Test.

Enter y2 = d (y1(x), x) and use the zerofunction and obtain x = 2.5.

Note that:

y x x

x1

2

2

4

0 5= +

−

.

Since l l x x

x> = +

−

0

4

0 52

2

,.

l x x

x

l x x

x

2 2

2

2

2

4

0 5

4

0 5

= +−

= ± +−

.;

.

⇒ − − +

= −( ) =

=−

xy y x

y x x

y x

x

0 5 4 2

2 0 5 4

4

0 5

.

; .

.

f '

f incr

rel. max

decr

0 3.536

+ –0

The function f has only one relativeextrema. Thus it is the absolute extrema.Therefore, at x = 3.536, the area is 20and the area is the absolute maxima.

20. See Figure 5.6-12. Step 1. Distance formula: l 2 =x2 + y2; x > 0.5 and y > 4

y

y

x ( x,0)

(0 ,y) (0.5,4)

x

l

0

Figure 5.6-12

Step 2. The slope of the hypotenuse:

m y

x

y x

= −

− =

−−

⇒ −( ) −( ) =

4

0 0 5

4

0 5

4 0 5 2

. .

.

21. Rewrite:

Thus dy

dx x x

x

x x

x

= −( )( )[ ] −( )( )[ ]

− −( )[ ]( )

= − −( ) −( )( )[ ]

−( )( )[ ]

2 6 1 6 1

6 1 6

12 6 1 6 1

6 1

sin cos cos cos

sin

sin sin cos

cos cos

y x= −( )( )[ ]sin cos 6 12

22. As x → ∞, the numerator approaches 0

and the denominator increases without bound

(i.e. ∞). Thus the

23. (a) Summarize the information of f ′ on anumber line.

lim .x

xx x→∞ − + +

=100

40

2

100

x

.5 2.5

decr. incr.

rel. min

l ' – +0

l


7/23/2019 007137714 x



Since f has only one relative extremum, it is

the absolute extremum. Thus at x = 3, it isan absolute minimum.

(b) The function f is decreasing as the interval(−∞, 3) and increasing on (3, ∞).

(c)

3

f ' – 0 +

f decr

rel. min

incr

f '

f "

f

– +3

incr

concave

upward

concave

upward

0 incr

no change of concavity ⇒ no point of inflection.

(d) The function f is concave upward for theentire domain (−∞, ∞).

(e) Possible sketch of graph for f (x). SeeFigure 5.7-1.

3

(3,–2)

x 0

y

f

Figure 5.7-1

24. (Calculator) See Figure 5.7-2.

Step 1. Differentiate: 2 2 0x y dy

dx

dy

dx

x

y

+ = ⇒

= −

Figure 5.7-2

Figure 5.7-3

Step 1. Distance formula:

= −( ) +x x12 6

z x x= −( ) + ( )12 3 2

Step 2. Enter y1 = ((x − 1) ∧ 2 + x ∧ 6). Usethe Minimum function of the calculatorand obtain x = .65052 and y1 = .44488.Verify the result with the First DerivativeTest. Enter y2 = d (y1(x), x) and use theZero Function and obtain x = .65052.

0 0.65052

z' – +0

z decr

rel min

incr

Thus the shortest distance is approxi-mately 0.445.

Step 3. Solve for y: x2 + y2 = 36 ⇒ y2 = 36 − x2;

y =

Step 4. Thus, y = x ⇒ = x ⇒ 36 − x2

= x2 ⇒ 36 = 2x2 or x =

25. (Calculator) See Figure 5.7-3.

±3 2

± −36 2x

± −36 2x

Step 2. Setdy

dx

x

y y x= − ⇒

−= − ⇒ =1 1

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Chapter 6

More Applicationsof Derivatives

6.1 TANGENT AND NORMAL LINES

Main Concepts: Tangent Lines, Normal Lines

Tangent Lines

If the function y is differentiable at x = a, then the slope of the tangent line to the graph

of y at x = a is given as

Types of Tangent Lines:

Horizontal Tangents (See Figure 6.1-1.)dy

dx =

0 .

m dydx

x a

x a

tangent at =( )=

= .

178

Figure 6.1-1


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More Applications of Derivatives • 179

Figure 6.1-2

Figure 6.1-3

[.5π,π] by [4,4]

Figure 6.1-4

Vertical Tangents (See Figure 6.1-2.)dy

dx

dx

dydoes not exist but =

0 .

Parallel Tangents (See Figure 6.1-3.)dy

dx

dy

dxx a x c= =

=

.

Example 1

Write an equation of the line tangent to the graph of y = −3 sin 2x at(See Figure 6.1-4.) x =

π2 .

y x dy

dx x x

x dy

dx

x y x

x x

= − = − ( )[ ] = − ( )

=

= − ( )[ ] = − =

= = − ( ) = − ( )[ ] = − ( ) =

=

3 2 3 2 2 6 2

26 2

26 1

23 2 3 2

23 0

2

sin ; cos cos

: cos cos .

, sin sin sin .

slope of tangent at

Point of tangency: At

π π π

π π π

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Example 2

If the line y = 6x + a is tangent to the graph of y = 2x3, find the value(s) of a.

Solution:

(See Figure 6.1-5.)y x dy

dx x= =2 63 2; .

Therefore2

is the point of tangency.

Equation of Tangent: or

π

π π

,

.

0

0 12 2

− = −( ) = −y x y x


[2,2] by [6,6]

Figure 6.1-5

The slope of the line y = 6x + a is 6.

Since y = 6x + a is tangent to the graph of y = 2x3, thus for some values of x.

Set 6x2 = 6 ⇒ x2 = 1 or x = ± 1.

At x = −1, y = 2x3 = 2(−1)3 = −2; (−1,−2) is a tangent point. Thus, y = 6x + a ⇒ −2= 6(−1) + a or a = 4.At x = 1, y = 2x3 = 2(1)3 = 2; (1,2) is a tangent point.Thus y = 6x + a ⇒ 2 = 6(1) + a or a = −4.Therefore, a = ± 4.

Example 3

Find the coordinates of each point on the graph of y2 − x2 − 6x + 7 = 0 at which the tan-gent line is vertical. Write an equation of each vertical tangent. (See Figure 6.1-6.)

dy

dx = 6

–7 0 1 x

y

y2 – x 2 – 6 x + 7 = 0

x = –7 x = 1

Figure 6.1-6

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Step 1: Find

Step 2: Find

Step 3: Find points of tangency.At y = 0, y2 − x2 − 6x + 7 = 0 becomes −x2 − 6x + 7 = 0 ⇒ x2 + 6x − 7 = 0

⇒ (x + 7)(x − 1) = 0 ⇒ x = −7 or x = 1.Thus the points of tangency are (−7,0) and (1,0)

Step 4: Write equation for vertical tangents:x = −7 and x = 1.

Example 4

Find all points on the graph of y = xex at which the graph has a horizontal tangent.

Step 1: Find

Step 2: Find the x-coordinate of points of tangency.

If x ≥ 0, set ex + xex = 0 ⇒ ex (1 + x) = 0⇒ x = −1 but x ≥ 0, therefore, no solution.

If x = 0, set −ex − xex = 0 ⇒ −ex (1 + x) = 0⇒ x = −1.

Step 3: Find points of tangency.

Thus at the point the graph has a horizontal tangent.

(See Figure 6.1-7.)

−( )1 1, ,e

At x y xe ee

x= − = − = − −( ) =1 111, .

Horizontal Tangent ⇒ =dy

dx0

y xexe x

xe x

dy

dx

e xe x

e xe x

x

x

x

x x

x x

= = ≥

− <

= + ≥

− + <

if

if

if

if

0

0

0

0

dydx

Vertical tangent

Set

⇒ =

= =+( )

=+

= ⇒ =

dx

dy

dx

dy dydx

x

y

y

x

dx

dy y

0

1 1

3 3

0 0.

dy

dx

y x x

y dy

dx x

dy

dx

x

y

x

y

2 2 6 7 0

2 2 6 0

2 6

2

3

− − + =

− − =

= +

= +

dy

dx


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Example 5

Using your calculator, find the value(s) of x to the nearest hundredth at which the slope

of the line tangent to the graph of y = 2 ln(x2 + 3) is equal to (See Figure 6.1-8

and 6.1-9.)

−1

2.


[3,1] by [0.5,1.25]

Figure 6.1-7

[5,5] by [1,7]

Figure 6.1-8

[10,3] by [1,10]

Figure 6.1-9

Step 1: Enter y1 = 2 ln(x^2 + 3)

Step 2: Enter y2 = d (y1(x), x) and enter

Step 3: Using the Intersection function of the calculator for y2 and y3, you obtainx = −7.61 or x = −0.39.

Example 6

Using your calculator, find the value(s) of x at which the graphs of y = 2x2 and y = ex haveparallel tangents.

Step 1:

Step 2: Find the x-coordinate of the points of tangency. Parallel tangents ⇒ slopes areequal.

Set 4x = e x ⇒ 4x − ex = 0

Find for both anddy

dx y x y e

y x dy

dx x

y e dy

dx e

x

x x

= =

= =

= =

2

2 4

2

2 ;

;

y3 1

2= −

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Normal Lines

The normal line to the graph of f at the point (x1, y1) is the line perpendicular to the tan-gent line at (x1, y1). (See Figure 6.1-10.)


• Watch out for different units of measure, e.g., the radius, r, is 2 feet, find in

inches per second.

dr

dt

( x 1, y1)

Normal Line

Tangent f

Figure 6.1-10

Note that the slope of the normal line and the slope of the tangent line at any point

on the curve are negative reciprocals provided that both slopes exist.

(mnormal line)(mtangent line) = −1.

Special Cases: See Figure 6.1-11.

Figure 6.1-11

Tangent Tangent Tangent

Normal

f

NormalNormal

f f

At these points, mtangent = 0; but mnormal does not exist.See Figure 6.1-12.

Using the Solve function of the calculator, enter Solve (4x − e^ (x) = 0,x) andobtain x = 2.15 and x = 0.36.

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At these points, mtangent does not exist; however mnormal = 0.

Example 1

Write an equation for each normal to the graph of y = 2 sin x for 0 ≤ x ≤ 2π that has a

slope of

Step 1: Find mtangent

Step 2: Find mnormal

⇒ x = cos−1 (−1) or x = π. (See Figure 6.1-13.)

mm x

m

x

x

normal

tangent

normalSet

= − = −

= ⇒ − = ⇒ = −

1 1

2

1

2

1

2

1

2

1

cos

cos

cos

y x dy

dx x= =2 2sin ; cos

12

.


Figure 6.1-12

Normal

f

f Tangent Tangent

Normal

[1.5π,2.5π] by [3,3]

Figure 6.1-13

Step 3: Write equation of normal line.At x = π, y = 2 sin x = 2(0) = 0; (π, 0)

Since equation of normal is:

y x y x− = −( ) = −01

2

1

2 2π

π or .

m =1

2,

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Example 2

Find the point on the graph of y = ln x such that the normal line at this point is parallelto the line y = −ex − 1.



slope of y = −ex − 1 is −e.Since normal is parallel to the line y = −ex − 1, set mnormal = −e⇒ −x = −e or x = e.

Step 3: Find point on graph. At x = e, y = ln x = ln e = l . Thus the point of the graph ofy = ln x at which the normal is parallel to y = −ex − 1 is (e, 1). (See Figure 6.1-14.)

mm

x

xnormal

tangent

= −

= −

= −1 1

1

y x dy

dx x= =ln ;

1


[6.8,9.8] by [5,3]

Figure 6.1-14

Example 3

Given the curve : (a) write an equation of the normal to the curve at the

point and (b) does this normal intersect the curve at any other point? If yes, find

the point.



Step 3: Write equation of normal

m

y x y x

normal

Equation of normal: or

= ( )

− = −( ) = −

4 2 12

1

24 2 4

15

2

; ,

,

mm

x

x

m

normal

tangent

normalAt

= −

= −−

=

( ) = =

1 1

1

2 12

2 4

2

2

2, , .

yx

dy

dx x

x= = −( )( ) = −−1

112

2;

2 12

,( )

yx

=1

yx

=1

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Step 4: Find other points of intersection.

Using the Intersection function of your calculator,enter

and obtain x = −0.125 and y = −8. Thus, the normal line intersect

the graph of at the point (−0.125, −8) as well.y x=1

y x2 415

2= −

yx

1

1= and

yx

y x= = −1

415

2;


(a, f (a))

y = f (a) + f' (a)( x – a)

y

f ( x )

Figure 6.2-1

Tangent Line Approximation (or Linear Approximation):

Since the curve of f (x) and the tangent line are close to each other for points near x = a,f (x) ≈ f (a) + f ′(a)(x − a)

Example 1

Write an equation of the tangent line to f (x) = x3 at (2,8). Use the tangent line to findthe approximate values of f (1.9) and f (2.01).

Differentiate f (x): f ′(x) = 3x2; f ′(2) = 3(2)2 = 12. Since f is differentiable at x = 2, thus anequation of the tangent at x = 2 is:

• Remember that 1 1 0dx x C d

dx= + ( ) =∫ and .

6.2 LINEAR APPROXIMATIONS

Main Concepts: Tangent Line Approximation, Estimating the nth Root of a Number, Estimating the Value of a Trigonometric Function of an Angle

Tangent Line Approximation

An equation of the tangent line to a curve at the point (a, f (a)) is:

y = f (a) + f ′(a)(x − a); providing that f is differentiable at a. See Figure 6.2-1.

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y = f (2) + f ′(2)(x − 2)y = (2)3 + 12(x − 2) = 8 + 12x − 24 = 12x − 16f (1.9) ≈ 12(1.9) − 16 = 6.8f (2.01) ≈ 12(2.01) − 16 = 8.12. (See Figure 6.2-2.)


y

f ( x ) = x 3

(2,8)

0 x

1.9 2 2.01

Tangent line

Not to Scale

y = 12 x – 16

Figure 6.2-2

Example 2

If f is a differentiable function and f (2) = 6 and , find the approximation

value of f (2.1).

Using tangent line approximation, you have

(a) f (2) = 6 ⇒ the point of tangency is (2,6)

(b) ⇒ the slope of the tangent at x = 2 is m = −1

2.′( ) = −f 2

1

2

′( ) = −f 21

2

(c) the equation of the tangent is

(d) thus

Example 3

The slope of a function at any point (x, y) is The point (3,2) is on the graph

of f. (a) Write an equation of the line tangent to the graph of f at x = 1. (b) Use the tan-gent line in part (a) to approximate f (3.1).

(a)

Equation of tangent: y − 2 = −2(x − 3) or y = −2x + 8

(b) f (3.1) ≈ −2(3.1) + 8 ≈ 1.8.

Let theny f x dy

dx

x

y

dy

dxx y

= ( ) = − +

= − +

= −= =

,

.,

1

3 1

22

3 2

− +x

y

1.

f 2 11

22 1 7 5 95. . . .( ) ≈ − ( ) + ≈

y x y x− = − −( ) = − +61

22

1

27or

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Estimating the n th Root of a Number

Another way of expressing the tangent line approximation is:

f (a + ∆x) ≈ f (a) + f ′(a)∆x; where ∆x is a relatively small value.

Example 1

Find the approximation value of using linear approximation.

Using f (a + ∆x) ≈ f (a) + f ′(a)∆x, let f (x) = a = 49 and ∆x = 1.

Thus f (49 + 1) ≈ f (49) + f ′(49)(1) ≈

Example 2

Find the approximate value of using linear approximation.

Let Since and

you can use Thus

f x x a x f x xx

f f a x f a f a x

f f f f

( ) = = = − ′( ) = =

′( ) =( )

= +( ) ≈ ( ) + ′( )

( ) = −( ) ≈ ( ) + ′( ) −( ) ≈ + −( )

−13

23

23

23

64 21

3

1

3

64 1

3 64

148

62 64 2 64 64 2 41

482

, , .

, .

∆

∆ ∆

≈≈ 3 958. .

623

491

249 1 7

1

147 0714

12+ ( ) ( ) ≈ + ≈

−

. .

x ;

50


• Use calculus notations and not calculator syntax, e.g., write and notx x∧ )2, .

x dx2

Estimating the Value of a Trigonometric Function of an Angle

Example

Approximate the value of sin 31°.

Note: You must express the angle measurement in radians before applying linear

approximations.

Since you can use linear approximations:

f f f π π π π π

π π π

π

6 180 6 6 180

6 6 180

1

2

3

2 180

0 515

+

≈

+ ′

≈ +

≈ +

≈

sin cos

. .

′( ) = ′

=

=f x x f cos cos ,andπ π6 6

3

2

Let andf x x a x

( ) = = =sin , .

π π

6 180∆

306

1180

° °= =π π

radians and radians.

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6.3 MOTION ALONG A LINE

Main Concepts: Instantaneous Velocity and Acceleration, Vertical Motion, Horizontal Motion

Instantaneous Velocity and Acceleration

Position Function: s(t )

Instantaneous Velocity:

If particle is moving to the right →, then v(t ) > 0.If particle is moving to the left ←, then v(t ) < 0.

Acceleration:

Instantaneous speed: v(t )

Example 1

The position function of a particle moving on a straight line is s(t ) = 2t 3 − 10t 2 + 5.Find (a) the position, (b) instantaneous velocity, (c) acceleration and (d) speed of theparticle at t = 1.

Solution

(a) s(1) = 2(1)3 − 10(1)2 + 5 = −3(b) v(t ) = s′(t ) = 6t 2 − 20t

v(1) = 6(1)2 − 20(1) = −14(c) a(t ) = v′(t ) = 12t − 20

a(1) = 12(1) − 20 = −8(d) Speed = v(t ) = v(1) = 14

Example 2

The velocity function of a moving particle is − 4t 2 + 16t − 64 for 0 ≤ t ≤ 7.

What are the minimum and maximum acceleration of the particle on 0 ≤ t ≤ 7?

See Figure 6.3-1. The graph of a(t ) indicates that:

(1) the minimum acceleration occurs at t = 4 and (a)(4) = 0.(2) The maximum acceleration occurs at t = 0 and a(0) = 16.

v t t

t t

a t v t t t

( ) = − + −

( ) = ′( ) = − +

3

2

2

34 16 64

8 16

v t t

( ) =3

3

a t v t dv

dt a t s t

d s

dt ( ) = ′( ) = ( ) = ( ) =or ″

2

2

v t s t ds

dt ( ) = ′( ) =


[1,7] by [2.20]

Figure 6.3-1

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Example 3

The graph of the velocity function is shown in Figure 6.3-2.


0

1

–1

–2

–3

2

3

4

1 2 3 4

v(t )

v

t

Figure 6.3-2

(a) when is the acceleration 0?(b) when is the particle moving to the right?(c) when is the speed the greatest?

Solution:

(a) a(t ) = v′(t ) and v′(t ) is the slope of tangent to the graph of v.At t = 1 and t = 3, the slope of the tangent is 0.

(b) For 2 < t < 4, v(t ) > 0. Thus the particle is moving to the right during 2 < t < 4.(c) Speed = v(t ) at t = 1, v(t ) = −4.

Thus speed at t = 1 is −4 = 4 which is the greatest speed for 0 ≤ t ≤ 4.

• Use only the four specified capabilities of your calculator to get your answer: plot-ting graph, finding zeros, calculating numerical derivatives, and evaluating definiteintegrals. All other built-in capabilities can only be used to check your solution.

Vertical Motion

Example

From a 400-foot tower, a bowling ball is dropped. The position function of the bowlingball s(t ) = −16t 2 + 400, t ≥ 0 is in seconds. Find:

(a) the instantaneous velocity of the ball at t = 2 sec(b) the average velocity for the first 3 sec(c) when the ball will hit the ground

Solution

(a) v(t) = s′(t ) = −32t v(2) = −32(2) = −64 ft/sec

(b) Average velocity = ( ) − ( )

− =

− ( ) +( ) − +( )

= −

s s3 0

3 0

16 3 400 0 400

3

48

2

ft sec

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(c) When the ball hits the ground, s(t ) = 0.Thus set s(t ) = 0 ⇒ −16t 2 + 400 = 0; 16t 2 = 400; t 2 = 25; t = ±5Since t ≥ 0, t = 5. The ball hits the ground at t = 5 sec.


t

0 1

0v(t ) + – – – – – – – – – – – –++++++ +++++++0

3

RightDirection

of MotionStopped Stopped

RightLeft

t

v(t )

a(t )

+ + + + + + +

+ + + + + + + + + + +

+ + + + +0 0– – – – – – – – – – – –

– – – – – – – – – – – 0

1

1

0 3

t 20

t 20 3

ParticleSlowing

downSpeeding

up

Stopped Stopped

Speedingup

Slowingdown

Figure 6.3-3

Figure 6.3-4

• Remember that the volume of a sphere is and the surface area is

s = 4πr2. Note that v ′ = s.

v r=4

33π

Horizontal Motion

Example

The position function of a particle moving on a straight line is s(t ) = t 3 − 6t 2 + 9t − 1, t ≥ 0.Describe the motion of the particle.

Step 1. Find v(t ) and a(t ). v(t ) = 3t 2 − 12t + 9a(t ) = 6t − 12

Step 2. Set v(t ) and a(t ) = 0.Set v(t ) = 0 ⇒ 3t 2 − 12t + 9 = 0 ⇒ 3(t 2 − 4t + 3) = 0⇒ 3(t − 1)(t − 3) = 0 or t = 1 or t = 3.

Set a(t ) = 0 ⇒ 6t − 12 = 0 ⇒ 6(t − 2) = 0 or t = 2.

Step 3. Determine the directions of motion. See Figure 6.3-3.

Step 4. Determine acceleration. See Figure 6.3-4.

Step 5. Draw the motion of the particle. See Figure 6.3-5.s(0) = −1, s(1) = 3, s(2) = 1 and s(3) = −1

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At t = 0, the particle is at −1 and moving to the right. It slows down and stops at t = 1 andat t = 3. It reverses direction (moving to the left) and speeds up until it reaches 1 at t = 2.It continues moving left but slows down and it stops at −1 at t = 3. Then it reverses direc-tion (moving to the right) again and speeds up indefinitely. (Note: “Speeding up” isdefined as when v(t ) increases and “slowing down” is defined as when v(t ) decreases.)

6.4 RAPID REVIEW

1. Write an equation of the normal line to the graph y = ex at x = 0.

At x = 0, y = e0 = 1 ⇒ you have the point (0,1).Equation of normal y − 1 = −1(x − 0) or y = −x + 1.

2. Using your calculator, find the values of x at which the functions y = −x2 + 3x andy = ln x have parallel tangents.

3. Find the linear approximation of f (x) = x3 at x = 1 and use the equation to find f (1.1).

Answer: f (1) = 1 ⇒ (1,1) is on the tangent line and f ′(x) = 3x2 ⇒ f ′(1) = 3.y − 1 = 3(x − 1) or y = 3x − 2f (1.1) ≈ 3(1.1) − 2 ≈ 1.3.


(a) When is the acceleration zero? (b) is the particle moving to the right or left?

Answer y x x dy

dx x

y x dy

dx x

xx

xx

x x x

:

ln

.

, .

= − + ⇒ = − +

= ⇒ =

− + =

− + =

= =

2 3 2 3

1

2 31

2 31

11

2

Set Using the Solve function on your calculator, enter

Solve and obtain or

Answer dydx

e e e mx

x x

x: normal

=== = = ⇒ = −

0

0

0 1 1.


t = 3

t = 0

t = 2

t = 1

–1 0 1 3Position s(t )

Figure 6.3-5

0 2 4

y

t

v(t )

Figure 6.4-1

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t seconds. Draw a sketch of the correspondingvelocity function.

9. The position function of a moving particle is shownin Figure 6.5-2. For which value(s) of t (t 1, t 2, t 3) is:

(a) the particle moving to the left?

(b) the acceleration negative?

(c) the particle moving to the right and slowingdown?



1. Find the linear approximation of at x = 0 and use the equation to approximatef (0.1).

2. Find the approximate value of using linearapproximation.

3. Find the approximation value of cos 46° usinglinear approximation.

4. Find the point on the graph of y = x3 suchthat the tangent at the point is parallel to theline y − 12x = 3.

5. Write an equation of the normal to the graph of y = e x at x = ln 2.

6. If the line y − 2x = b is tangent to the graph

y = −x2

+ 4, find the value of b.

7. If the position function of a particle is

− 3t 2 + 4, find the velocity and position of particlewhen its acceleration is 0.

8. The graph in Figure 6.5-1 represents thedistance in feet covered by a moving particle in

s t t

( ) =3

3

283

f x x( ) = +( )11

4

Answer: (a) a(t ) = v′(t ) and v′(t ) is the slope of the tangent. Thus, a(t ) = 0 at t = 2.(b) Since v(t ) ≥ 0, the particle is moving to the right.

5. Find the maximum acceleration of the particle whose velocity function is v(t ) = t 2 + 3on the interval 0 ≤ t ≤ 4.

Answer: a(t ) = v′(t ) = 2(t ) on the interval 0 ≤ t ≤ 4, a(t ) has its maximum value att = 4. Thus a(t ) = 8. The maximum acceleration is 8.


5

4

3

2

1

0 1 2 3 4 5

s(t )

t Seconds

Feet s

Figure 6.5-1

t 1 t 2

t 3

s(t )

s

t

Figure 6.5-2

10. The velocity function of a particle is shown inFigure 6.5-3.

(a) when does the particle reverse direction?

(b) when is the acceleration 0?

(c) When is the speed the greatest?

11. A ball is dropped from the top of a 640-footbuilding. The position function of the ball iss(t ) = −16t 2 + 640, where t is measured in secondsand s(t ) is in feet. Find:

(a) The position of the ball after 4 seconds.

(b) The instantaneous velocity of the ball at t = 4.

(c) The average velocity for the first 4 seconds.

(d) When the ball will hit the ground.

(e) The speed of the ball when it hits the ground.

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Intervals (−1,2) x = 2 (2,5) x = 5 (5,8)

f ′ + 0 − − −

f ″ − − − 0 +

(c) When is the particle standing still?

(d) When does the particle have the greatest speed?


13. The position function of a particle moving on aline is s(t ) = t 3 − 3t 2 + 1, t ≥ 0 where t is measuredin seconds and s in meters. Describe the motion

of the particle.

14. Find the linear approximation of f (x) = sin x atx = π. Use the equation to find the approximate

value of

15. Find the linear approximation of f (x) = ln(1 + x)at x = 2.

16. Find the coordinates of each point on the graphof y2 = 4 − 4x2 at which the tangent line is vertical.

Write an equation of each vertical tangent.

17. Find the value(s) of x at which the graphs ofy = ln x and y = x2 + 3 have parallel tangents.

18. The position functions of two moving particles ares1(t ) = ln t , and s2(t ) = sin t, a and the domain of both functions is 1 ≤ t ≤ 8. Find the values of t suchthat the velocities of the two particles are the same.

19. The position function of a moving particle on aline is s(t ) = sin(t ) for 0 ≤ t ≤ 2π. Describe themotion of the particle.

20. A coin is dropped from the top of a tower andhits the ground 10.2 seconds later. The positionfunction is given as s(t ) = −16t 2 + v0t + s0,where s is measured in feet, t in seconds andv0 is the initial velocity and s0 is the initialposition. Find the approximate height of thebuilding to the nearest foot.

f 181

180

π

.


12. The graph of the position function of a movingparticle is shown. See Figure 6.5-4.

(a) What is the particle’s position at t = 5?

(b) When is the particle moving to the left?

10

1

–1

–2

–3

–4

–5

2

3

4

5

2 3 4

v(t )

v

t

Figure 6.5-3

1

1

0

2

3

4

2 3 4 5 6 7

s(t )

s

(seconds)

( f e e t )

t

Figure 6.5-4



21. Find

22. Given f (x) = x3 − 3x2 + 3x − 1 and the point(1,2) is on the graph of f −1(x). Find the slope of the tangent line to the graph of f −1(x) at (1,2).

23. Evaluate limx 100→

−

−

x

x

100

10.

dy

dx y x xif = ( )−sin .1 2

24. A function f is continuous on the interval [−1,8]with f (0) = 0, f (2) = 3, and and thefollowing properties:

f 8 12( ) =

7/23/2019 007137714 x


Then f ′(x) = −sin x and f ′(45°) =

4. Step 1: Find mtangent

Step 2: Set mtangent = slope of line y − 12x = 3.Since y − 12x = 3 ⇒ y = 12x + 3,then m = 12.Set 3x2 = 12 ⇒ x ±2 since x ≥ 0, x = 2.Set −3x2 = 12 ⇒ x2 = −4. Thus φ.

Step 3: Find the point on the curve.(See Figure 6.7-1.)

y xx x

x x

dy

dx

x x

x x

= = ≥

− <

= >

− <

3

3

3

2

2

0

0

3 0

3 0

if

if

if

if

′

= −

( ) =

= =

+

≈

+ ′

≈

+

≈ −

f

f f f

f f f

π

π π π

π π π π π

π π

4

2

2

4623

90 4 180

4 180 4 4 180

2

2

2

2 180

2

2

2

360

°

f π π4 4

2

2

=

=cos




(c) Find where f has the points of inflection.

(d) Find the intervals on where f is concaveupward or downward.


25. The graph of the velocity function of a movingparticle for 0 ≤ t ≤ 8 is shown in Figure 6.6-1.Using the graph:

(a) estimate the acceleration when v(t ) = 3 ft/sec.

(b) the time when the acceleration is a minimum.

8

7

6

5

4

3

2

1

0 1 2 3 4 5 6 7 8 9t

v

v(t )

( f e e t / s e c )

(seconds)

Figure 6.6-1



1. Equation of tangent line: y = f (a) + f ′(a)(x − a)

2. f (a + ∆x) ≈ f (a) + f ′(a)∆x

3. f (a + ∆x) ≈ f (a) + f ′(a)∆x. Convert to radians:

and 1° = ; 45° =

Let f (x) = cos x and f (45°) =

π4

.π

180

46

180

23

90= ⇒ =

aa

ππ

Let and

Then

and and

f x x f f

f x x

f f

f f f

( ) = ( ) = +( )

′( ) = ( )

′( ) = ( ) =

+( ) ≈ ( ) + ′( )( ) ≈ +

( ) ≈

−

3

23

28 27 1

1

3

27

1

27 27 3

27 1 27 27 1 3

1

271 3 037

.

.

. .

′( ) = +( ) ( ) = +( )

′( ) = ( ) =

= + −( ) = +

( ) = + ( ) =

− −

f x x x

f f

y x x

f

1

41 1

1

41

01

40 1

1 14

0 1 14

0 1 11

40 1 1 025

34

34

and

Thus

;

, .

. . . .

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8. On the interval (0, 1), the slope of the line segmentis 2. Thus the velocity v(t ) = 2 ft/sec. On (1, 3), v(t)= 0 and on (3, 5), v(t ) = −1. See Figure 6.7-2.


At x = 2, y = x3 = 23 = 8.Thus the point is (2,8).



Step 3: Write equation of normalAt x = ln 2, y = ex = eln 2 = 2. Thus thepoint of tangency is (ln 2, 2).The equation of normal:


Step 2: Find the slope of line y − 2x = by − 2x = b ⇒ y = 2x + b or m = 2.

Step 3: Find point of tangency. Set mtangent = slope

of line y − 2x = b − 2x = 2 ⇒ x = −1.At x = −1, y = −x2 + 4 = −(−1)2

+ 4 = 3; (−1, 3).

Step 4: Find b.Since the line y − 2x = b passes through thepoint (−1, 3), thus 3 − 2 (−1) = b or b = 5.

7. v(t ) = s′(t ) = t 2 − 6t ; a(t ) = v′(t ) = s″ (t ) = 2t − 6Set a(t ) = 0 ⇒ 2t − 6 = 0 or t = 3.

v(3) = (3)2 − 6(3) = −9; s(3) = − 3(3)2 + 4 = −14.3

3

3

( )

y x dy

dx x= − + = −2 4 2; .

y x

y x

− = − −( )

= − −( ) +

2 12

2

1

22 2

ln

ln .

or

At normal

tangent

x mm

= = −

= −ln ,2

1 1

2

y e dy

dx e

dy

dx e

x x

x

= =

= ==

;

ln

ln

2

2 2

[3,4] by [5,15]

Figure 6.7-1

1

–1

–2

10 2 3 4 5

2

v

t

Figure 6.7-2

9. (a) At t = t 2, the slope of the tangent is negative.Thus, the particle is moving to the left.

(b) At t = t 1, and at t = t 2, the curve is concave

downward acceleration is negative.

(c) At t = t 1, the slope > 0 and thus the particle ismoving to the right. The curve is concavedownward ⇒ the particle is slowing down.

10. (a) At t = 2, v(t ) changes from positive to negative,and thus the particle reverses its direction.

(b) At t = 1, and at t = 3, the slope of the tangentto the curve is 0. Thus the acceleration is 0.

(c) At t = 3, speed is equal to −5 = 5 and 5 isthe greatest speed.

11. (a) s(4) = −16(4)2 + 640 = 384 ft

(b) v(t ) = s′(t) = −32t v(4) = −32(4) ft/sec = −128 ft/sec

(c) Average Velocity

−64 ft/sec

(d) Set s(t ) = 0 ⇒ −16t 2 + 640 = 0 ⇒ 16t 2 = 640

or t =Since t ≥ 0, t = or t ≈ 6.32 sec.

(e) v 2 10 32 2 10 64 10

202 39

( ) = − ( ) = −

≈

ft sec

or ft sec.

+2 10

±2 10.

= −

=384 640

4

= ( ) − ( )

−

s s4 0

4 0

⇒ =d s

dt

2

2

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The particle is initially at 1 (t = 0). It moves tothe left speeding up until t = 1, when it reaches−1. Then it continues moving to the left, butslowing down until t = 2 at −3. The particlereverses direction, moving to the right andspeeding up indefinitely.

14. Linear approximation: y = f (a) + f ′(a)(x − a)a = πf (x) = sin x and f (π) = sin π = 0

f ′(x) = cos x and f ′(π) = cos π = −1.Thus y = 0 + (−1)(x − π) or y = −x + π.

15. y = f (a) + f ′(a)(x − a)f (x) = ln (1 + x) and f (2) = ln (1 + 2) = ln 3

16. Step 1:

Step 2:

Step 3: Find points of tangencyAt y = 0, y2 = 4 − 4x2 becomes 0 = 4 − 4x2

⇒ x = ±1.Thus points of tangency are (1,0) and(−1,0)

Find

Set or

dx

dy

dx

dy dydx y

y

x

dx

dy

y

x y

x

.

.

= =−

= −

= ⇒ −

= =

1 1

4 4

04

0 0

Find dy

dx

y x

y dy

dx x

dy

dx

x

y

.

2 24 4

2 84

= −

= − ⇒ = −

′( ) =+

′( ) =+

=

= + −( )

f xx

f

y x

1

12

1

1 2

1

3

31

32

and

Thus

.

ln .

f

y

181

180

181

180 1800 0175

π

ππ

π

= −

+ =

−≈ −

is approximately

or

:

. .


12. (a) At t = 5, s(t ) = 1

(b) For 3 < t < 4, s(t ) decreases. Thus, theparticle moves to the left when 3 < t < 4.

(c) When 4 < t < 6, the particle stays at 1.

(d) When 6 < t < 7, speed = 2 ft/sec, the greatestspeed, which occurs where s has the greatest

slope.


13. Step 1. v(t ) = 3t 2 − 6t a(t ) = 6t − 6

Step 2. Set v(t ) = 0 ⇒ 3t 2 − 6t = 0 ⇒ 3t (t − 2) = 0,or t = 0 or t = 2Set a(t ) = 0 ⇒ 6t − 6 = 0 or t = 1.

Step 3. Determine the directions of motion. See

Figure 6.7-3.

[

0 2

Left

Stopped Stopped

RightDirectionof Motion

t

v(t ) 0 0– – – – – – – – + + + + + + +

Figure 6.7-3


[

[

t

t

t

– – – –

– – – +0 + + + + + + + + +– –

– – – 00v(t ) + + + + + + +

0

0

[

0

2

2

1 2

Speedingup

Speedingup

Slowingdown

Stopped Stopped

Motion of Particle

a(t )

Figure 6.7-4

Step 5. Draw the motion of the particle. SeeFigure 6.7-5. s(0) = 1, s(1) = −1 ands(2) = −3.

t = 2t = 1

t = 0

t > 2

s(t )

–3 –1 0 1

Figure 6.7-5

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Step 4: Write equations of vertical tangentsx = 1 and x = −1.

17. Step 1:

Step 2: Find the x-coordinate of point(s) oftangency.Parallel tangents ⇒ slopes are equal

Set

Using the Solve function of your calculator,

enter solve and obtain

or Since for y = ln x,

x > 0, thus .

18.

s2(t ) = sin(t ) and s2′(t ) = cos(t ); 1 ≤ t ≤ 8

Enter and y2 = cos(x). Use the Intersection

function of the calculator and obtain t = 4.917 andt = 7.724.

19. Step 1. s(t ) = sin t v(t ) = cos t a(t ) = −sin t

Step 2. Set v(t ) = 0 ⇒ cos t = 0;

Set a(t ) = 0 ⇒ −sin t = 0; t = π and 2π.

Step 3. Determine the directions of motion. SeeFigure 6.7-6.

t = π π2

3

2and .

yx

1

1=

s t t s t t

t 1 1

11 8( ) = ( ) = ≤ ≤ln ;and ′

x =2

2

x = − 22

.x = 22

12

x x x=

,

Set1

2x

x= .

Find for anddy

dx y x y x

y x dy

dx x

y x dydx

x

= = +

= =

= + =

ln

ln ;

;

2

2

3

1

3 2

[ [

0 2ππ2

3π2

RightDirectionof Motion

Stopped Stopped

RightLeft

t

v(t ) 0+ + + + + + + ++ – – – – – – – 0

Figure 6.7-6

[ [

[[

[

0 2π

0 π 2π

[

2ππ0

3π2

3π2

π2

π2

t

t

t

Slowingdown

Slowingdown

Speedingup

Speedingup

Stopped Stopped

v(t )

a(t )

Motion ofParticle

+ + + + + + + ++ + – – – – – – – – –

– – – – – – – – – + + + + + + + +0

0 0

Figure 6.7-7

Step 5. Draw the motion of the particle. SeeFigure 6.7-8.

t =3π2

t = π2

z = 0

t = π

t = 2π

–1 0

5(t )1

Figure 6.7-8

The particle is initially at 0, s(0) = 0. It moves tothe right but slows down to a stop at 1 when

It then turns and moves to the

left speeding up until it reaches 0, when t = π,s(π) = 0 and continues to the left but slowing down

to a stop at −1 when It then

turns around again, moving to the right, speedingup to 1 when t = 2π, s(2π) = 0.

20. s(t ) = −16t 2 + v0t + s0.s0 = height of building and v0 = 0.Thus s(t ) = −16t 2 + s0.When the coin hits the ground, s(t ) = 0, t = 10.2.Thus, set s(t ) = 0 ⇒ −16t 2 + s0 = 0 ⇒ −16(10.2)2

+ s0 = 0s0 = 1664.64 ft. The building is approximately1665 ft tall.

t s= = −3

2

3

21

π π, .

t s=

=

π π2 2

1, .

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24. (a) f ′ > 0 on (−1,2), f is increasing on (−1,2)f ′ < 0 on (2,8), f is decreasing on (2,8)

(b) At x = 2, f ′ = 0 and f ″ < 0, thus at x = 2,f has a relative maximum. Since it is the onlyrelative extremum on the interval, it is an

absolute maximum. Since f is a continuousfunction on a closed interval and at its

endpoints f (−1) < 0 and f (8) = thus

f has an absolute minimum at x = −1.

(c) At x = 5, f has a change of concavity andf ′ exists at x = 5.

(d) f ″ < 0 on (−1,5), f is concave downward on(−1,5).f ″ > 0 on (5,8), f is concave upward on (5,8).

(e) A possible graph of f is given in Figure 6.8-1.

12 ,



21. Using product rule, let u = x; v = sin−1 (2x).

22. Let y = f (x) ⇒ y = x3 − 3x2 + 3x − 1.To find f −1 (x), switch x and y: x = y3 − 3y2 + 3y − 1

23. Substituting x = 0 into the expression

would lead to Multiply both numerator and

denominator by the conjugate of the denominator

An alternative solution is to factor the numerator:

lim .x

x x

x→

−( ) +( )−( ) =100

10 10

10 20

lim

lim

lim .

x

x

x

x

x

x

x

x x

x

x

→

→

→

−( )−( )

+( )+( )

=

−( ) +( )−( )

+( ) = + =

100

100

100

100

10

10

10

100 10

100

10 10 10 20

x +( )10 :

0

0.

x

x

−

−

100

10

dx

dy y y

dy

dx dxdy

y y

dy

dxy

= − +

= =− +

= ( ) − ( ) + ==

3 6 3

1 1

3 6 3

1

3 2 6 2 3

1

3

2

2

2

2

;

dy

dx x

xx

x

x

x

= ( ) ( ) +− ( )

( ) ( )

= ( ) − −

−

−

1 21

1 22

2

2

1 4

1

2

1

2

sin

sin

10–1 2 3 4 5 6 7 8

(2,3)

(8,1 ⁄ 2)

3 f

y

x

Figure 6.8-1

25. (a) v(t ) = 3 ft/sec at t = 6. The tangent line tothe graph of v(t ) at t = 6 has a slope of approximately m = 1. (The tangent passesthrough the points (8,5) and (5,0); thusm = 1). Therefore the acceleration is 1 ft/sec2.

(b) The acceleration is a minimum at t = 0,since the slope of the tangent to the curve of v(t ) is the smallest at t = 0.

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Chapter 7

7.1 EVALUATING BASIC INTEGRALS

Main Concepts: Antiderivatives and Integration Formulas, Evaluating Integrals

Integration

• Answer all parts of a question from Section II even if you think your answer toan earlier part of the question might not be correct. Also, if you don’t know theanswer to part one of a question, and you need it to answer part two, just makeit up and continue.

200

Antiderivatives and Integration Formulas

Definition: A function F is an antiderivative of another function f if F ′(x) = f (x) for all xin some open interval. Any two antiderivatives of f differ by an additive constant C. Wedenote the set of antiderivatives of f by ∫ f (x)dx, called the indefinite integral of f.

Integration Rules:

1.

2.

3.

4.

Differentiation Formulas:

1.

2. d

dx ax a( ) =

d

dx x( ) = 1

f x g x dx f x dx g x dx( ) ± ( )[ ] = ( ) ± ( )∫ ∫ ∫

− ( ) = − ( )∫ ∫ f x dx f x dx

a f x dx a f x dx( ) = ( )∫ ∫

f x dx F x C F x f x( ) = ( ) + ⇔ ′( ) = ( )∫

Integration Formulas:

1.

2. a dx ax c= +∫

1dx x c= +∫


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Integration • 201

Differentiation Formulas (cont.):

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

More Integration Formulas:

16.

17.

18.

19.

20.

21.

22.

23.1 1

2 2

1 1

x x adx

a

x

a c

a

a

x c

+= + +− −∫ sec cosor

1

1 12 2

1

a x dx

a

x

a c

+ =

+−∫ tan

12 2

1

a xdx

x

a c

−=

+

−∫ sin

ln lnx dx x x x c= − +∫ csc ln csc cotx dx x x c= − +∫

sec ln sec tanx dx x x c= + +∫

cot ln sin ln cscx dx x c x c= + − +∫ or

tan ln sec ln cosx dx x c x c= + − +∫ or

d

dx x

x xsec−( ) =

−1

2

1

1

d

dx

x

x

tan−( ) =

+

1

2

1

1

d

dx x

xsin−( ) =

−1

2

1

1

d

dx a a ax x( ) = ( )ln

d

dx e ex x( ) =

d dx

xx

ln( ) = 1

d

dx x x xcsc csc cot( ) = − ( )

d

dx x x xsec sec tan( ) =

d

dx x xcot csc( ) = − 2

d

dx x xtan sec( ) = 2

d

dx

x xsin cos( ) =

d

dx x xcos sin( ) = −

d

dx x nxn n( ) = −1

Integration Formulas (cont.):

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.1

12

1

x xdx x c

−= +−∫ sec

1

12

1

+

= +−

∫ x

dx x ctan

1

1 2

1

−= +−∫

xdx x csin

a dx a

a c a ax

x

= + > ≠∫ ln,0 1

e dx e cx x= +∫

1x

dx x c∫ = +ln

csc cot cscx x dx x c( ) = − +∫

sec tan secx x dx x c( ) = +∫

csc cot2 x dx x c= − +∫

sec tan2 x dx x c= +∫

cos sinx dx x c= +∫

sin cosxdx x c= − +∫

x dx x

n c nn

n

=+

+ ≠ −+

∫ 1

11,

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24.

Note: After evaluating an integral, always check the result by taking the derivative of the answer (i.e., taking the derivative of the antiderivative).

sinsin

. sincos2 2

2

2

4

1 2

2x dx

x xc x

x= −

( )+ =

−∫ Note:

• Remember that the volume of a right-circular cone is where r is the

radius of the base and h is the height of the cone.

v r h= 13

2π

Evaluating Integrals

Integral Rewrite Antiderivative

x dx3∫

dx

x dx

∫

5dx

x dx5

2∫

12x dx∫

123

xdx∫

x

x dx

+∫

1

x x dx5 1+( )∫

1dx∫

x dx1

2

∫

x dx−∫ 2

12

3

23

xdx x dx∫ ∫ =

−

11

+

∫ x dx

x x dx6 +( )∫

xc

4

4+

xc

xc

32

32

32

2

3+ +or

x + c

5x + c

xc

xc

72

72

72

2

7+ +or

xc

x c

−

− +

−+

1

1

1or

xc x c

133

13

3+ +or

x x c+ +ln

x xc

7 2

7 2+ +

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Integration • 203

Example 1

Example 2

Example 3

If and the point (0,−1) lies on the graph of y, find y.

Since then y is an antiderivative of . Thus

The point (0,−1) is on the graph of y.

Thus y = x3 + 2x + c becomes −1 = 03 + 2(0) + c or c = −1. Therefore, y = x3 + 2x − 1.

Example 4

Example 5

Example 6

Evaluate x x dx∫ −( )2 3

Evaluate

Rewrite as

3 1

31 1

31

31

31

2

2

2

2

1

x xx

dx

x x dx

x x dx

x x x

c x xx

c

+ −

+ −

= + −

= + −−

+ = + + +

∫

∫ ∫ −

−

ln ln

Evaluate

Rewrite as

11

11

1

13

3

43

43

43

13

3

−

−

= −

= −−

+ = + +

∫

∫ ∫ −

−

xdx

xdx x dx

x x

c xx

c

y x dx x x c= +( ) = + +∫ 3 2 22 3 .

dy

dx

dy

dx x= +3 22 ,

dy

dx x= +3 22 ,

Evaluate

Rewrite as

xx

dx

xx

dx x x dx x x

c

xx

c

+

+

+( ) = +

− +

= − +

∫

∫ ∫ −−

1

1

32

2

2

3

1

2

3

3

12 3

32 2

32

2

Evaluate

Apply the formula

x x x dx

x dx x

n c n

x x x dx x

x x

x c

n

n

5 2

1

5 2

6

3

2

6 1

11

6 16

22

− + −( )

=+

+ ≠

− + −( ) = − + − +

∫

∫

∫

+

, .

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Example 7

Example 8

Example 9

Example 10

Example 11

Example 12

Example 13

Reminder: You can always check the result by taking the derivative of the answer.

Evaluate 7

77

7

x

x

x

dx

dx c

∫

∫ = +ln

Evaluate

Rewrite as

1

9

1

3 3

2

2 2

1

−

−=

+

∫

∫ −

xdx

xdx

xcsin

Evaluate

Rewrite as 3

3

1

1

13

2

2

1

+

+ = +

∫

∫ −

x dx

x dx x ctan

Evaluate

Rewrite the integral as

ee

dx

e dx e c

x

x

x x

2

∫

∫ = +

Evaluate

Rewrite

or

sin

cos

sin

cos costan sec tan sec

ln sec ln sec tan lnsec

sec tan

ln sin

x

x dx

x

x x dx x x dx x dx x dx

x x x c x

x x c

x c

−

−

= −( ) = −

= − + + =+

+

− + +

∫

∫ ∫ ∫ ∫

1

1

1

Evaluate 4

4 4

cos cot

cos cot sin ln sin

x x dx

x x dx x x c

−( )

−( ) = − +

∫ ∫

Evaluate x x dx

x x dx x

x c

3

3

4

4

44

4

−( )

−( ) = + +

∫

∫

sin

sin cos

Rewrite x x dx x x dx

x xc x x c

∫ ∫ −( ) = −( )

= − + = − +

12 2

52

12

72

32 7

2 3

3 3

72

3

32

2

72


7/23/2019 007137714 x


7.2 INTEGRATION BY U-SUBSTITUTION

Main Concepts: The U-Substitution Method, U-Substitution and Algebraic Functions,U-Substitution and Trigonometric Functions, U-Substitutionand Inverse Trigonometric, U-Substitution and Logarithmic and

Exponential Functions

The U-Substitution Method

The Chain Rule for Differentiation:

The Integral of a Composite Function:If f ( g (x)) and f ′ are continuous and F ′ = f, then

Making a U-Substitution:

Let u = g (x); then du = g ′(x)dx

Procedure for Making a U-Substitution:

Steps:1. Given f ( g (x)); Let u = g (x)2. Differentiate: du = g ′(x)dx3. Rewrite the integral in terms of u.4. Evaluate the integral.5. Replace u by g (x).6. Check your result by taking the derivative of the answer.

U-Substitution and Algebraic Functions

Another Form of the Integral of a Composite Function:

If f is a differentiable function, then

Making a U-Substitution:

Let u = f (x); then du = f ′(x) dx.

Example 1

Evaluate ∫ x(x + 1)10 dx

f x f x dx u du u

n c

f x

n c n

n n

n n

( )( ) ′( ) = =+

+ = ( )( )

+ + ≠ −∫ ∫

+ +

11

1 11,

f x f x dx f x

n c n

n

n

( )( ) ′( ) = ( )( )

+ + ≠ −∫ +1

11,

f g x g x dx f u du F u c F g x c( )( ) ′( ) = ( ) = ( ) + = ( )( ) +∫ ∫

f g x g x dx F g x c( )( ) ′( ) = ( )( ) +∫

d

dx F g x f g x g x F f ( )( ) = ( )( ) ′( ) ′ =where

Integration • 205

• Be familiar with the instructions for the different parts of the exam before theday of exam. Visit the College Board website at: www.collegeboard.com/ap formore information.

7/23/2019 007137714 x


Step 1. Let u = x + 1; then x = u − 1Step 2. Differentiate: du = dxStep 3. Rewrite: ∫ (u − 1)u10 du = ∫ (u11 − u10) du

Step 4. Integrate:

Step 5. Replace u:

Step 6. Differentiate and Check:

Example 2

Evaluate

Step 1. Let u = x − 2; then x = u + 2Step 2. Differentiate du = dx

Step 3. Rewrite:

Step 4. Integrate:

Step 5. Replace:


Example 3

Evaluate

Step 1. Let u = 2x − 5


Step 3. Rewrite:

Step 4. Integrate:

Step 5. Replace u:

Step 6. Differentiate and Check:3

10

5

32 5 2 2 5

23

23

−( ) ( ) = −( )x x

3 2 5

10

53x

c−( )

+

1

2 53

3

10

53

53u

c u

c

+ = +

u du u du2

32

3

212∫ ∫ =

du dx du

dx= ⇒ =22

2 52

3x dx−( )∫

5

2

2 2

5

3

2

4 2

3

2 2 2

2 2 2 2 2

32

12

3

2

1

2

12

12

−( )+

−( )

= −( ) + −( )

= −( ) −( ) +[ ] = −( ) −

x x

x x

x x x x x xor

2 2

5

4 2

3

52

32x x

c−( )

+ −( )

+

u uc

52

32

52

2

32

+ +

u udu u u du u u du+( ) = +( ) = +( )∫ ∫ ∫ 2 2 212 32 12

x x dx−∫ 2

12 1

12

11 1

111 1

1 1 1 1 1

11 10

11 10

10 10 10

x xx x

x x x x x x

+( )−

+( )= +( ) − +( )

= +( ) + −( ) = +( ) +( )or

x xc

+( )−

+( )+

1

12

1

11

12 11

u uc

12 11

12 11− +


7/23/2019 007137714 x


Example 4

Evaluate

Step 1. Let u = x3 − 8


Step 3. Rewrite:

Step 4. Integrate:

Step 5. Replace u:


U-Substitution and Trigonometric Functions

Example 1

Evaluate ∫ sin 4x dx

Step 1. Let u = 4x


Step 3. Rewrite:

Step 4. Integrate:

Step 5. Replace u:


Example 2

Evaluate

Step 1. Let u = tan x

Step 2. Differentiate: du = sec2 x dx

Step 3. Rewrite:

Step 4. Integrate

Step 5. Replace u:

Step 6. Differentiate and Check: 23

23

12 2 2( )

( )( ) = ( )tan sec sec tanx x x x

2 23

23

2tan tanx c x c( ) + +or

33

2

2

32 3

2u

c u c+ = +

3 31

2 21

2tan secx x dx u du( ) =∫ ∫

3 2sec tanx x dx( )∫

−

−( )( ) =

1

44 4 4sin sinx x

− ( ) +1

44cos x c

14

14

−( ) + = − +cos cosu c u c

sin sinu du

u du4

1

4= ∫ ∫

du dx du

dx= =44

or

−

−( ) −( ) ( ) =

−( )

−1

124 8 3

8

3 5 2

2

3 5x x

x

x

1

128

1

12 8

3 4

3 4− −( ) +

−

−( )+

−x c

xcor

1

3 4

4uc

−

−

+

13

13

1 135 5

5

udu

u du u du∫ ∫ ∫ = = −

du x dx du

x dx= ⇒ =33

2 2

x

xdx

2

3 5

8−( )∫

Integration • 207

7/23/2019 007137714 x


• Remember that the area of a semi-circle is Don’t forget the If the

cross sections of a solid are semi-circles, the integral for the volume of the solid

will involve which is1

4.

1

2

2

12

.12

2πr

Example 3

Evaluate

Step 1. Let u = x3

Step 2. Differentiate

Step 3. Rewrite:

Step 4. Integrate:

Step 5. Replace u:

Step 6. Differentiate & Check:2

33 23 2 2 3cos cosx x x x( )[ ] = ( )

2

33sin x c( ) +

2

3sin u c+

2 23

23

3 2cos cos cosx x dx u du u du( )[ ] = =∫ ∫ ∫

du x dx du

x dx= ⇒ =33

2 2

2 2 3x x dxcos( )∫


U-Substitution and Inverse Trigonometric Functions

Example 1Evaluate

Step 1. Let u = 2x

Step 2. Differentiate

Step 3. Rewrite:

Step 4. Integrate:

Step 5. Replace u:

Step 6. Differentiate and Check:1

2

1

1 2 3

23

1

3

1

1 4 9

1

9

1

1 4 9

1

9 1 4 9

1

9 4

2 2

2 2 2

− ( )=

−

=−

=−( )

=−

x x

x x x

1

2

2

31sin−

+

xc

1

2 31

sin−

+u

c

1

9 2

1

2 32 2 2−=

−∫ ∫

u

du du

u

du x du

dx= =22

;

dx

x9 4 2−∫

7/23/2019 007137714 x


Example 2

Evaluate

Step 1. Rewrite:

Let u = x + 1

Step 2. Differentiate: du = dx

Step 3. Rewrite:

Step 4. Integrate:

Step 5. Replace u:

Step 6. Differentiate and Check: 1

2

1 12

1 1 2

1

4

1

1 1 4

1

4

4

4 1

1

2 5

2 2

2 2

( )+ +( )[ ]

=

+ +( )

+ +( )=

+ +

x x

x x x.

1

2

1

21tan− +

+

xc

1

2 21tan−

+

uc

1

22 2+∫ u du

1

2 1 4

1

1 2

1

2 12 2 2 2 2x x x

dxx

dx+ +( ) +

=+( ) +

=+ +( )

∫ ∫ ∫

1

2 52x x dx

+ +∫

Integration • 209

U-Substitution and Logarithmic and Exponential Functions

Example 1

Evaluate

Step 1. Let u = x4 − 1


Step 3. Rewrite:

Step 4. Integrate:

Step 5. Replace u:

Step 6. Differentiate & Check:1

4

1

14

14

3

3

4

− ( ) =

−x x

x

x.

1

414ln x c− +

1

4ln u c+

1

4

1

4

1

u

du

u du∫ ∫ =

du x dx du

x dx= ⇒ =44

3 3

x

x dx

3

4 1−∫

• If the problem gives you the diameter of a sphere is 6 and you are using formulas

such as or s = 4πr2, don’t forget that r = 3.v r=4

33π

7/23/2019 007137714 x


Example 2

Evaluate

Step 1. Let u = cos x + 1Step 2. Differentiate: du = −sin x dx ⇒ −du = sin x dx

Step 3. Rewrite:

Step 4. Integrate:

Step 5. Replace u:


Example 3

Evaluate

Step 1. Rewrite by dividing (x2 + 3) by (x − 1).

Let u = x − 1.

Step 2. Differentiate: du = dx

Step 3. Rewrite:

Step 4. Integrate:

Step 5. Replace u:


Example 4

Evaluate

Step 1. Let u = ln x


Step 3. Rewrite:1

3u dx∫

dux

dx=1

ln x

x dx

3∫

2

21 4

1

11

4

1

3

1

2x

x c x

x

x

x+ +

−

+ = + +

− =

+−

.

x

x dx

xx x c

2 23

1 24 1

+−

= + + − +∫ ln

4 1ln x c− +

4 ln u c+

4

1

u du∫

x

x dx x

x dx x dx

x dx

xx

x dx

2

2

3

11

4

11

4

1

24

1

1

+−

= + +−

= +( ) +

−

= + +−

∫ ∫ ∫ ∫

∫

x

x x

x

2 3

11

4

1

+−

= + +−

;

x

x dx

2 3

1

+−∫

−+

−( ) =

+1

1 1cossin

sin

cos.

x x

x

x

− + +ln cos x c1

− +ln u c

−= −

∫ ∫

du

u

du

u

sin

cos

x

x dx

+∫ 1


7/23/2019 007137714 x


Step 4. Integrate

Step 5. Replace u:


Example 5

Evaluate

Step 1. Let u = 2x − 5


Step 3. Rewrite:

Step 4. Integrate:

Step 5. Replace u:


Example 6

Evaluate

Step 1. Let u = ex + 1

Step 2. Differentiate: du = ex dx

Step 3. Rewrite:

Step 4. Integrate:

Step 5. Replace u:


Example 7

Evaluate

Step 1. Let u = 3x2


Step 3. Rewrite: e du

e duu u

6

1

6= ∫ ∫

du x dx du

x dx= ⇒ =66

xe dxx3 2

∫

1

1 1e e

e

ex

x

x

x+ =

+ .

ln e cx + +1

ln u c+

1

u du∫

e

e dx

x

x +∫ 1

1

222 5 2 5e ex x− −( ) = .

1

2

2 5e c

x −( ) +

12

e cu +

e du

e duu u

2

1

2

= ∫ ∫

du dx du

dx= ⇒ =22

e dxx2 5−( )∫

1

62

1

3( )( )

=ln

ln.x

x

x

x

1

6

2

ln x c( ) +

1

3 2

1

6

2

2

+ = +

uc u c

Integration • 211

7/23/2019 007137714 x


Step 4. Integrate:

Step 5. Replace u:


Example 8Evaluate

Step 1. Let u = 2x


Step 3. Rewrite:

Step 4. Integrate:

Step 5. Replace u:


Example 9

Evaluate

Step 1. Let u = x4


Step 3. Rewrite:

Step 4. Integrate:

Step 5. Replace u:


Example 10

Evaluate

Step 1. Let u = cos πx

Step 2. Differentiate: du = −π sin πx dx;

Step 3. Rewrite:

Step 4. Integrate: − +1

π e cu

e du

e duu u−

= − ∫ ∫ π π

1

− =du

xdxπ

πsin

sin cosπ πx e dxx( )∫

5 4 5 4 5 54 4

3 3x x

x x( ) ( )

( ) =ln ln .

5

4 5

4x

cln

+

1

45 5u c( ) +ln

54

1

45u udu

du= ∫ ∫

du x dx du

x dx= ⇒ =44

3 3

x dxx35

4( )∫

5 2 5 2 5 52 2x x( )( ) =ln ln .

5

2 5

2x

cln

+

1

25 5 5 2 5u uc c( ) + = +ln ln

52

1

25u udu

du= ∫ ∫

du dx du

dx= ⇒ =22

52x

dx( )∫

1

663 32 2

e x xex x( )( ) = .

1

63 2

e cx +

1

6e cu +


7/23/2019 007137714 x


Step 5. Replace u:


7.3 RAPID REVIEW

1. Evaluate

Answer: Rewrite as

2. Evaluate

Answer: Rewrite as

3. Evaluate

Answer: Rewrite as Let u = x2 − 1.

4. Evaluate ∫ sin x dx

Answer: −cos x + c.

5. Evaluate

Answer: Let u = 2x and obtain

6. Evaluate

Answer: Let u = ln x;

7. Evaluate

Answer: Let u = x2;du

xdx e

cx

2 2

2

= +and obtain .

xe dxx2

∫

dux

dx x

c= ( )

+1

2

2

and obtainln

.

ln x

x dx∫

1

22sin .x c+

cos 2x dx( )∫

Thusdu

xdx u du u

c x c2

1

2

1

2 32

1

31

12

32

23

2= ⇒ = + = −( ) +∫ .

x x dx21

21−( )∫ .

x x dx2 1−∫ .

xx

dx x

x c2

31

3−

= − +∫ ln .

x

x dx

3 1−∫ .

x dx x

cx

c−−

=−

+ = − +∫ 2

1

1

1.

12x dx.∫

− ( ) −( ) = ( )1

π π π ππ πe x x ex xcos cossin sin .

− +1

ππe cxcos


Integration • 213

Evaluate the following integrals in problems 1 to 20.No calculators allowed. (However, you may use cal-culators to check your results.)

1. x x x dx5 23 1+ − +( )∫

2.

3.

4. x x dx3 2 1+∫

x x dx3 4 5

10−( )∫

xx

dx−

∫

12

7/23/2019 007137714 x




5.

6.

7.

8.

9.

10.

11.

12. 1x x

dxln∫

e e dxx x2 4( )( )∫

1 12

2

x x dxsec

∫

1

2 102x x dx

+ +∫

sin

cos

x

x dx

3∫

x x dxcsc2 2( )∫

tan x

dx2

∫

x

xdx

2 5

1

+

−∫

13.

14.

15.

16.

17. If and the point (0,6) is on the

graph of y, find y.

18.

19.

20. If f (x) is the antiderivative of and f (1) = 5,find f (e).

1

x

e e

e e dx

x x

x x

−+

−

−∫

− ( )∫ 3e e dxx xsin

dy

dx ex= + 2

x x dx∫ +( )1

3

2

4

9 2−( )∫ x x dx

e

e dx

x

x

4 1−∫

ln e dxx5 1+( )∫

“Calculator” indicates that calculators arepermitted.

21. The graph of the velocity function of a movingparticle for 0 ≤ t ≤ 10 is shown in Figure 7.5-1.

(b) At what time is the particle moving to theright?

22. Air is pumped into a spherical balloon, whosemaximum radius is 10 meters. For what value of r is the rate of increase of the volume a hundredtimes that of the radius?

23. Evaluate

24. (Calculator) The function f is continuous anddifferentiable on the interval [0,2] with f ″ (x)> 0 for all x in the interval [0,2]. Some of thepoints on the graph are shown below.

ln3 x

x dx

( )∫

0

1

1 2 3 4 5 6 7 8 9 10–1

–2

–3

–4

–5

2

3

4

5

t

v(t )

Figure 7.5-1

(a) At what value of t is the speed of the parti-cle the greatest?

x 0 0.5 1 1.5 2

f (x) 1 1.25 2 3.25 5

Which of the following is the best approxima-tion for f ′(1)?

(a) f ′(1) < 2

(b) 0.5 < f ′(1) < 1

(c) 1.5 < f ′(1) < 2.5

7/23/2019 007137714 x


Integration • 215

0 1 2 3 4 5 6 7 8

f " (t )

y

t

Figure 7.5-2

(d) 2.5 < f ′(1) < 3.5

(e) f ′(1) > 2

25. The graph of the function f ″ on the interval[1,8] is shown in Figure 7.5-2. At what value(s)of t on the open interval (1,8), if any, does thegraph of the function f ′:

(a) have a point of inflection?

(b) have a relative maximum or minimum

(c) concave upward?

5. Let u = x − 1; du = dx and (u + 1) = x

Rewrite:

6. Let or 2du = dx

7. Let u = x2; du = 2x dx or

Rewrite: csc csc

cot

cot

2 2

2

2

1

2

1

2

1

2

u du

u du

u c

x c

∫ ∫ =

= − +

= − ( ) +

dux dx

2=

Rewrite: tan tan

ln cos

ln cos

u du u du

u c

xc

2 2

2

22

( ) =

= − +

= − +

∫ ∫

u x

du dx= =2

1

2;

= + +( )

= + + +

= −( )

+ −( )

+ −( ) +

−

∫ u u u du

u u uc

x x

x c

32

12

12

52

32

12

52

32

12

2 6

5 2

2

3 2

6

1 2

2 1

5

4 1

3

12 1

u

udu

u u

udu

+( ) +=

+ +∫ ∫

1 5 2 62

2

12


No calculators except for verifying your results.

1.

2. Rewrite:

3. Let u = x4 − 10 du = 4x3 dx or

Rewrite:

4. Let

Rewrite: x x x dx u u du

u u du u u du

u uc

u uc

x xc

2 2

12

32

12

52

32

52

32

25

2 23

2

1 12

1

21

1

2

1

2 52

32

5 3

1

5

1

3

+ ( ) = −( )

= −( ) = −( )

= −

+ = − +

= +( )

− +( )

+

∫ ∫

∫ ∫

du x dx du

x dx= =22

or

u x u x= + ⇒ −( ) =2 21 1 and

xc

4 6

10

24=

−( )+

u du

u du u

c5 5

6

4

1

4

1

4 6= =

+∫ ∫

dux dx

43= .

x x dx x x

c x

x c

12 2

32 1

32

32

1

2

3

1−( ) = −

− + = + +−

−

∫

xx

xx c

6

3

2

6 2+ − + +

7/23/2019 007137714 x


8. Let u = cos x; du = −sin x dx or −du = sin x dx

9. Rewrite

Let u = x + 1; du = dx

Rewrite

10.

11.

Let u = 6x; du = 6 dx or

12. Let u = ln x;

13. Since e x and ln x are inverse functions:

14.

Let u = 3x; du = 3 dx;

Rewritee

e e dx e e dx

e dx e dx

x

x

x x

x x

4

2

3

3

1−

= −( )

= −

−

−

∫ ∫

∫ ∫

ln e dx x dx x

x cx5 1

2

5 15

2+( ) = +( ) = + +∫ ∫

Rewrite1

u

du u c x c= + = +∫ ln ln ln

dux

dx=1

Rewrite e du

e du e c

e c

u u u

x

6

1

6

1

6

1

66

∫ ∫ = = +

= +

du dx6

=

Rewrite e dx e dxx x x2 4 6+( ) = ∫ ∫

Rewrite: sec sec

tan

– tan

2 2

1

u du u du

u c

x c

−( ) = −

= − +

=

+

∫ ∫

Let orux

dux

dx dux

dx= = − − =1 1 12 2

;

1

3

1

3 3

1

3

1

3

2 2

1

1

u du

uc

xc

+ =

+

= +

+

−

−

∫ tan

tan

1

2 1 9

1

1 3

2

2 2

x x

dx

xdx

+ +( ) +

=+( ) +

∫ ∫

Rewrite: −

= − = −−

+

= +

∫ ∫ −du

u

du

u

uc

x c

3 3

2

2

2

1

2 cos

Let v = −x; dv = −dx;

Note: c1 and c2 are arbitrary constants, andthus c1 + c2 = c.

15.

16.

17.

The point (0, 6) is on the graph of y. Thus, 6 = e0

+ 2(0) + c ⇒ 6 = 1 + c or c = 5. Therefore, y = ex

+ 2x + 5.

18. Let u = ex; du = ex dx

19. Let u = ex + e−x; du = (e x − e−x) dx

Rewrite:

or

1

1

1

1

1

2

2

2

u du u c e e c

ee

c

e

e c

e e c

e x c

x x

x

x

x

x

x x

x

∫ = + = + +

= + +

= +

+

= + − +

= + − +

−ln ln

ln

ln

ln ln

ln

Rewrite: − ( ) = − −( ) +

= ( ) +

∫ 3 3

3

sin cos

cos

u du u c

e cx

Since thendy

dx e y e dx

e x c

x x

x

= + = +( )

= + +

∫ 2 2

2

,

.

Rewrite: u du u du u

c

xc

4 4

5

32

5

2

3

2

3

2

3 5

2 1

15

∫ ∫

= =

+

=+( )

+

Let oru x du x dx

du x dx xdx

= + =

= =

13

22

3

32

12

12

;

Rewrite 9 −( ) = −( )

= − +

= − +

∫ ∫ x x dx x x dx

x xc

x x

c

21

21

25

2

32

72

32

72

9

9

3 2 7 2

62

7

Thus e dx e dx e e cx x x x3 31

3− = + +− −∫ ∫

e dx e dv e c e cx v v x− −= −( ) = − + = − +∫ ∫ 2 2

e dx e du

e c e cx u u x3

1

3

3

1

3

1

3∫ ∫ =

= + = +


7/23/2019 007137714 x


Given f (1) = 5; thus ln(1) + c = 5 ⇒ 0 + c = 5 orc = 5.

Thus, f (x) = ln x + 5 and f (e) = ln(e) + 5= 1 + 5 = 6.

Integration • 217

Tangent

0 0.5 1 1.5 2

AB C

D

D f

y

x

Not to Scale

Figure 7.7-1

20. Since f (x) is the antiderivative of

1f x

x d x cln .( ) = = +∫

1

x,



21. (a) At t = 4, speed is 5 which is the greatest on0 ≤ t ≤ 10.

(b) The particle is moving to the right when6 < t < 10.

22.

23.

24. Label given points as A, B, C, D and E.

Since f ″ (x) > 0 ⇒ f is concave upward for all xin the interval [0,2].

Let

Rewrite:

u x dux

dx

u du u

c x

c

xc

= =

= + = ( )

+

= ( )

+

∫

ln ;

ln

ln

1

4 4

4

3

44

4

or

Since

r

r r meters

= ± = ±

≥ =

25 5

05

π π

π

.

, .

If then 100dV

dt

dr

dt

dr

dt

r dr

dt r

=

= ⇒ =

100

4 100 42 2

,

π π

V r dV

dt r

dr

dt r

dr

dt = =

( ) =

4

3

4

33 43 2 2π π π;

Therefore 1.5 < f ′(1) < 2.5, choice (c). SeeFigure 7.7-1.

Thus andm f m m

mBC CD BC

CD

< ′( ) < ==

1 1 5

2 5

; .

.

25. (a) f ″ is decreasing on [1,6) ⇒ f ′ ″ < 0 ⇒ f ′ is con-cave downward on [1,6) and f ″ is increasing on(6,8] ⇒ f ′ ″ is concave upward on (6,8].

Thus, at x = 6, f ′ has a change of concavity.Since f ′ exists at x = 6 (which implies thereis a tangent to the curve of f ′ at x = 6), f ′has a point of inflection at x = 6.

(b) f ″ > 0 on [1,4) ⇒ f ′ is increasing and f ″ < 0on (4,8] ⇒ f ′ is decreasing. Thus at x = 4,f ′ has a relative maximum at x = 4. There isno relative minimum.

(c) f ″ is increasing on (6,8] ⇒ f ′″ > 0 ⇒ f ′ isconcave upward on (6,8].

7/23/2019 007137714 x


8.1 RIEMANN SUMS AND DEFINITE INTEGRALS

Main Concepts: Sigma Notation, Definition of a Riemann Sum, Definition of a Definite Integral, and Properties of Definite Integrals

Sigma Notation or Summation Notation

where i is the index of summation, l is the lower limit and n is the upper limit of sum-mation. (Note: The lower limit may be any non-negative integer ≤ n.)

Examples

Summation Formulas

If n is a positive integer, then:

1. a ani

n

==∑

1

i

k

i

k

i

k

i

k

k

2 2 2 2

5

7

0

3

1

3

1

4

5 6 7

2 2 0 2 1 2 2 2 3

2 1 1 1 3 5 7

1 1 2 3 4

= + +

= ( ) + ( ) + ( ) + ( )

+( ) = − + + + +

−( ) ( ) = − + − +

=

=

=−

=

∑

∑

∑

∑

a a a an

i

n

1 2 3

1

+ + + +=

∑ L

Chapter 8

Definite Integrals


7/23/2019 007137714 x


Definite Integrals • 219

2.

3.

4.

5.

Example

(Note: This question has not appeared in an AP Calculus AB Exam in recent years).

Evaluate

Rewrite as

i i

n

i i

n n i i

n i i

n

n n n n n

n

n n n n n

i

n

i

n

i

n

i

n

i

n

+( )

+( )+( ) = +

= +( ) +( )

+ +( )

= +( ) +( ) + +( )

=

= = = =

∑

∑ ∑ ∑ ∑

1

1 1 1

1 1 2 1

6

1

2

1 1 2 1 3 1

6

1

1

2

1

2

1 1

=

+( ) +( ) + +( )

= +( ) +( ) +[ ]

= +( ) +( )

= +( ) +( )

n n n

n n n n

n n

1 2 1 3 1

6

1 2 1 3

6

1 2 4

6

1 2

3

i n n n n n

i

n

4

3 2

1

1 6 9 1

30= +( )

+ + −( )=∑

i n n

i

n3

2 2

1

1

4=

+( )=∑

i n n n

i

n2

1

1 2 1

6=

+( ) +( )=∑

i n n

i

n

= +( )

=∑

1

21

• Remember in exponential growth/decay problems, the formulas are

dy

dx ky y y ekt = =and 0 .

Definition of a Riemann Sum

Let f be defined on [a, b] and x1’s be points on [a, b] such that x0 = a, xn = b and a < x1

< x2 < x3 . . . < xn−1 < b. The points a, x1, x2, x3, . . . xn+1, b form a partition of f denotedas ∆ on [a, b]. Let ∆xi be the length of the ith interval [xi−1, xi] and ci be any point in the

ith

interval. Then the Riemann sum of f for the partition is

Example 1

Let f be a continuous function defined on [0, 12] as shown below.

f c xi i

i

n

( )=∑ ∆1.

x 0 2 4 6 8 10 12

f (x) 3 7 19 39 67 103 147

Find the Riemann sum for f (x) over [0,12] with 3 subdivisions of equal length and themidpoints of the intervals ci’s.

7/23/2019 007137714 x



Length of an interval (See Figure 8.1-1.)∆xi = −

=12 0

34.

y

x c1

x 1 x 2 x 3 x 4

c2 c3 c4

43.532.521.510.5 x 0 = 0

Figure 8.1-1

y

x c1

x 1 x 2 x 3

c2 c3

12108642 x 0 = 0

Figure 8.1-2

Riemann sum

The Riemann sum is 596.

Example 2

Find the Riemann sum for f (x) = x3 + 1 over the interval [0,4] using 4 subdivisions of

equal length and the midpoints of the intervals as ci’s. (See Figure 8.1-2.)

= ( ) = ( ) + ( ) + ( )

= ( ) + ( ) + ( ) =

=∑ f c x f c x f c x f c xi i

i

∆ ∆ ∆ ∆1

3

1 1 2 2 3 3

7 4 39 4 103 4 596

The Riemann sum is 66.

Definition of a Definite Integral

Let f be defined on [a, b] with the Riemann sum for f over [a, b] written as .

If max ∆xi is the length of the largest subinterval in the partition and theexists, then the limit is denoted by:

is the definite integral of f from a to b.

Example 1

Using a midpoint Riemann sum with three subdivisions of equal length to find the approx-

imate value of .x dx2

0

6

∫

f x dxa

b

( )∫

limmax ∆

∆x

i i

i

n

a

b

i

f c x f x dx→

=( ) = ( )∑ ∫ 0

1

limmax ∆ ∆x

i i

i

n

if c x

→=

( )∑01

f c xi i

i

n

( )=∑ ∆

1

Length of an interval

Riemann sum

Enter

∆

∆

x b a

n c i i

f c x i i

i i

i i

i i

i i i

= −

= −

= = + =( ) = −

= ( ) = −( ) +[ ] = −( ) +

−( ) +( ) =

= = =∑ ∑ ∑

∑

4 0

41 0 5 1 0 5

0 5 11 0 5 1

0 5 1 1 4 66

1

43

1

43

1

4

3

; . .

. .

. , , , .

7/23/2019 007137714 x



midpoints are x = 1, 3 and 5.

Example 2

Using the limit of the Riemann sum, find .

Using n subintervals of equal lengths, the length of an interval

Let ci = xi; max ∆xi → 0 ⇒ n → ∞

Thus .

(Note: This question has not appeared in an AP Calculus AB Exam in recent years.)

Properties of Definite Integrals

1. If f is defined on [a, b], and the limit exists, then f is integrableon [a, b].

2. If f is continuous on [a, b], then f is integrable on [a, b].

If f (x), g (x), h(x) are integrable on [a, b], then

3.

4.

5. when c is a consonant.

6.

7. provided f (x) ≥ 0 on [a, b]

8. provided f (x) ≥ g (x) on [a, b]

9. f x dx f x dxa

b

a

b

( ) ≤ ( )∫ ∫

f x dx g x dxa

b

a

b

( ) ≥ ( )∫ ∫

f x dxa

b

( ) ≥∫ 0

f x g x dx f x dx g x dxa

b

a

b

a

b

( ) ± ( )[ ] = ( ) ± ( )∫ ∫ ∫

cf x dx c f x dxa

b

a

b

( ) = ( )∫ ∫

f x dx f xa

b

b

a

( ) = − ( )∫ ∫

f x dxa

a

( ) =∫ 0

limmax ∆

∆x

i i

i

n

i

f x x→

=( )∑

01

3 361

5

xdx∫ =

3 14 4

3 14 4

121

4 12 4 1

2

1

5

1 1

1

xdx f i

n n

i

n n

n

i

n n n

n n

n

ni

n

ni

n

ni

n

n

∫ ∑ ∑

∑

= +

= +

= +

= +

+

→∞=

→∞=

→∞=

→∞

lim lim

lim lim

= + +( )( ) = +( ) = +

=

→∞ →∞ →∞lim lim limn n nn

n nn

nn

122 1

123 2 36

2436

∆

∆∆

xn n

xn

i

xdx f c x

i i

x i i

i

n

i

= −

= = +

= ( )∫ ∑→=

5 1 41

4

31

5

01

;

limmax

31

5

xdx∫

x dx f x f x f x2

0

6

1 3 5 1 2 9 2 25 2

70

≈ ( ) + ( ) + ( ) ≈ ( ) + ( ) + ( )

≈

∫ ∆ ∆ ∆

.

∆x f x x= −

= ( ) =6 0

32 2,

7/23/2019 007137714 x



10. provided g (x) ≤ f (x) ≤ h(x) on [a, b]

11. provided m ≤ f (x) ≤ M on [a, b]

12. provided f (x) is integrable on an interval contain-

ing a, b, c.

Examples

1.

2.

3.

4.

5.

note: Or

do not have to be arranged from smallest to largest.

The remaining properties are best illustrated in terms of the area under the curve of thefunction as discussed in the next section.

= + ∫ ∫ ∫ a b cb

c

a

b

a

c

, ,

xdx xdx xdx1

3

1

5

5

3

∫ ∫ ∫ = +

xdx xdx xdx1

5

1

3

3

5

∫ ∫ ∫ = +

x x dx x dx xdx dx3

0

43

0

4

0

4

0

4

2 1 2 1− +( ) = − +∫ ∫ ∫ ∫

5 52 2

2

7

2

7

x dx x dx=−− ∫ ∫

x dx x dx4 4

5

1

1

5

= −∫ ∫

cos xdx =∫ 0π

π


c

a

b

b

c

( ) = ( ) + ( )∫ ∫ ∫ ;

m b a f x dx M b aa

b

- -( ) ≤ ( ) ≤ ( )∫ ;

g x dx f x dx h x dxa

b

a

b

a

b

( ) ≤ ( ) ≤ ( )∫ ∫ ∫ ;

• Don’t forget that f x dx f x dx( ) = − ( )−

−∫ ∫ 0

3

3

0

.

8.2 FUNDAMENTAL THEOREMS OF CALCULUS

Main Concepts: The First Fundamental Theorem of Calculus, The Second Fundamental Theorem of Calculus

First Fundamental Theorem of Calculus

If f is continuous on [a, b] and F is an antiderivative of f on [a, b], then

Example 1

Evaluate

4 14

4 2 2

22

22 0 16

3

0

24 2

0

2 4

2

0

2

4

2

x x dx x x

x x x

x+ −( ) = + − ] = + − ]

= + −

− ( ) =

∫

4 13

0

2

x x dx+ −( )∫

f x dx F b F a

F b F a F x

a

b

a

b

( ) = ( ) − ( )

( ) − ( ) ( )]∫ .

.Note is often denoted as

7/23/2019 007137714 x



Example 2

Example 3

Example 4

If f ′(x) = g (x) and g is a continuous function for all real values of x, expressin terms of f.

Let u = 3x; du = 3dx or

Example 5

Evaluate

Cannot evaluate using the First Fundamental Theorem of Calculus since

is discontinuous at x = 1.

Example 6

Using a graphing calculator, evaluate

Using a TI-89 graphing calculator, enter and obtain 2π.4 2 2 2−( ) −( )∫ x x^ , , ,

4 2

2

2

−−∫ x dx.

f xx

( ) =−1

1

1

10

4

x dx−∫

g x dx g u du

g u du f u c

f x c

g x dx f x f f

f f

3

3

1

3

1

31

33

31

33

1

33 5

1

33 2

1

315

1

36

2

5

2

5

( ) = ( ) = ( ) = ( ) +

= ( ) +

( ) = ( )] = ( )( ) − ( )( )

= ( ) − ( )

∫ ∫ ∫

∫

.

dudx

3=

g x dx32

5

( )∫

If find

Set

or or

Since

4 1 30 0

4 1 2 2 2 2 2

2 6

2 6 30 2 36 0

2 9 4 09

24

2

2

2

2

2 2

2

2 2

x dx k k

x dx x x k k

k k

k k k k

k k k k

k

k k

+( ) = >

+( ) = + ] = +( ) − −( ) −( )= + −

+ − = ⇒ + − =

⇒ +( ) −( ) = = − =

−

− −

∫

∫

, , .

.

kk k> =0 4, .

Evaluate sin

sin cos cos cos

xdx

xdx x

−

− −

∫

∫ = − ] = −[ ] − − −( )[ ]

= − −( )[ ] − − −( )[ ] = ( ) − ( ) =

π

π

π

π

π

π

π π

1 1 1 1 0

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Second Fundamental Theorem of Calculus

If f is continuous on [a, b] and , then F ′(x) = f (x) at every point x

in [a, b].

Example 1

Example 2

Example 3

Example 4

Find if

Rewrite:

dy

dx y t dt

y t dt

x

x

; sin .

sin

=

= −

∫

∫

2

2

1

1

Find if

Let then

Rewrite:

dy

dx y

t dt

u x du

dx

y t dt

dy

dx

dy

du

du

dx u x x

x

u

; .

; .

=

= =

=

= = ( ) =( )

=

∫

∫

1

2 2

1

12

1

22

1

4

31

2

31

3 3 3

If findh x t dt h

h x x h

x

( ) = + ′( )

′( ) = + ′( ) = + =∫ 1 8

1 8 8 1 3

3 , .

; .

cos cos cos

sin sin

cos sin

sin sin

sin sin

sin

22

1

2

1

2

1

22

21

22

1

22

1

22

4

1

22

1

2 2

1

22

1

4 4

t dt u du

u du

u c t c

t dt t

x

x

x

x x

( ) = =

= + = ( ) +

( ) = ( )]

= ( ) −

= ( ) −

= ( ) −

∫ ∫ ∫

∫ π π

π

π

22

Evaluate

Let or

cos

;

2

2 22

4

t dt

u t du dt du

dt

x

( )

= = =

∫ π

F x f t dt a

x

( ) = ( )∫

7/23/2019 007137714 x



Example 5

Example 6

, integrate to find F (x) and then differentiate to find F ′(x).

8.3 EVALUATING DEFINITE INTEGRALS

Main Concepts: Definite Integrals Involving Algebraic Functions; Definite Integrals Involving Absolute Volume; Definite Integrals Involving Trigonometric, Logarithmic, and Exponential Functions; Definite Integrals Involving Odd and Even Functions

F x t t x x

xx

F x x

x

x

( ) = − ] = −

− − ( )

= − +

′( ) =

− = −

3

1

3 3

3

2

2

34

34 1

34 1

34

11

3

33

4 4.

F x t dt x

( ) = −( )∫ 2

14

Since y e dt e dt

dy

dx

d

dx e dt d

dx e dt

e d

dx x e

x e e

t x

t x

t

x

t

x

x x

x x

= + − +

= + − +

= +

( ) − +( )

= + − +

∫ ∫

∫ ∫

1 1

1 1

1 1

2 1 1

0 0

0 0

2

2

2

2

2

Find ifdy

dx y e dt

y e dt e dt y e dt e dt

e dt e dt

t

x

x

t

x

t x

t x

t x

t x

t x

; .= +

= + + + = − + + +

= + − +

∫

∫ ∫ ∫ ∫

∫ ∫

1

1 1 1 1

1 1

2

2 2

2

0

0 0 0

0 0

Let then

Rewrite:

u x du

dx x

y tdt

dy

dx

dy

du

du

dx u x x x x x

u

;

sin

sin sin sin

= =

= −

= = −( ) = −( ) = − ( )

∫

2

1

2 2

2

2 2 2

• If the problem asks you to determine the concavity of f ′ (not f ), you need toknow if f ″ is increasing or decreasing or if f ′″ is positive or negative.

7/23/2019 007137714 x



Definite Integrals Involving Algebraic Functions

Example 1

Verify your result with a calculator.

Example 2

Evaluate:

Begin by evaluating the indefinite integral

Let u = x2 − 1; du = 2x dx or = x dx


Example 3


Rewrite: y y dy y y dy

y y y y

13

13

8

1 13

13

8

1

43

23

8

1 43

23

8

1

43

23

43

1

4 3 2 3

3

4

3

2

3 1

4

3 1

2

3 8

4

3 8

+

= +( )

= +

= +

= −( )

+ −( )

− −( )

+ −( )

−

− −

−

−

−

−

−

−

∫ ∫

/ /

223

2

3

4

3

212 6

63

4

= +

− +( ) =

−

Evaluate yy

dy3

38

1 1+

−

−

∫

Rewrite:

Thus the definite integral

u duu du

uc

uc

xc

x x dx x

7

7

8 8

2 8

0

22 7

2 8

0

2

2 8 2 88

88

2

1

2

1

2 8 16

1

16

11

16

2 1

16

0 1

16

3

16

1

16

3 1

16410

∫ ∫

∫

= =

+ = +

= −( )

+

−( ) = −( )

= −( )

− −( )

= − −( )

= −

=

du

2

x x dx2 7

1−( )∫ .

x x dx2 7

0

2

1−( )∫

Evaluate:

Rewrite:

1

4

1

4

1

4

x

xdx

x

x

dx x x dx

x x xx

3

35

21

2

72

12

1

4 72 1

2

1

4

72

12

72

12

8

88

7 2

8

1 2

2

716

2 4

716 4

2 1

716 1

142

7

−

−= −( )

= −

= − ]

= ( )

− ( )

− ( )

− ( )

=

∫

∫ ∫

−

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Definite Integrals Involving Absolute Value

Example 1Evaluate:

Set 3x − 6 = 0; x = 2; Thus

Rewrite Integral:


Example 2

Evaluate

Set x2 − 4 = 0; x = ± 2

Thus


Thus x dx x dx x dx

x x x x

2

0

42

0

22

2

4

3

0

2

3

2

4

3 3 3

4 4 4

34

34

2

34 2 0

4

34 4

2

34 2

8

38

− = − −( ) + −( )

= − +

+ −

= −

+ ( )

− ( ) + − ( )

− − ( )

= −

+

∫ ∫ ∫

++ −

− −

=

64

316

8

38 16

xx x x

x x

2

2

24

4 2 2

4 2 2− =

− ≥ ≤ −

− −( ) − < <

if or

if

x dx2

0

4

4−∫

3 6 3 6 3 6

3

26

3

26

3 2

26 2

3 1

26 1

3 4

26 4

1

4

1

2

2

4

2

1

2 2

2

4

2 2

2

x dx x dx x dx

xx

xx

− = − −( ) + −( )

= −

+

+ −

= − ( )

+ ( )

−

− ( )+ ( )

+ ( )

− ( )

∫ ∫ ∫

−

( )− ( )

= − +( ) − − +

+ −( ) − −( )

= − + + =

3 2

26 2

6 123

26 24 24 6 12

6 41

20 6

15

2

2

3 63 6 2

3 6 2x

x x

x x− =

− ≥

− −( ) <

if

if

3 61

4

x dx−∫

• You may bring up to 2 (but no more than 2) approved graphing calculators tothe exam.

• You are not required to clear the memories in your calculator for the exam.

7/23/2019 007137714 x



Definite Integrals Involving Trigonometric, Logarithmic, andExponential Functions

Example 1


Example 2


Example 3


Example 4

Evaluate: xe dxx2 1

1

2 +( )−∫

Evaluate

Let

Rewrite

:ln

ln ,

:ln ln

ln ln ln ln

t

t dt

u t dut dt

t

t

dt u du u

c t

c

t

t dt

t e

e

e

e

1

22

1

2

1

2 2

1

2 2

2 2

1

2

1

20

1

2

∫

∫ ∫

∫

= =

= = + = ( )

+

= ( )

= ( )

− ( )

= − =

Rewrite the indefinite integral csc

csc

2

2

= = − +

= − ( ) +

( ) = − ( )]

= −

−

= − − −( )[ ] = −

∫

∫

u du

u c

t c

t dt t

3

1

3

1

33

31

33

1

3

3

2

3

4

1

3 0 1

1

3

4

2

4

2

cot

cot

cot

cot cot

π

π

π

π

π π

Evaluate csc

Let or

2

π

π

4

2 3

3 33

∫ ( )

= = =

t dt

u t du dt du

dt ;

Evaluate

Rewrite:

x x dx

x x dx x x

x

x

+( )

+( ) = − = − − −( )

= + + = +

∫

∫

sin

sin cos cos cos

0

0

2

0

2

2 2

2 20 0

21 1

22

π

π π

π π

7/23/2019 007137714 x




Definite Integrals Involving Odd and Even Functions

If f is an even function, that is, f (−x) = f (x), and is continuous on [−a, a], then

If f is an odd function, that is, F (x) = −f (−x), and is continuous on [−a, a] then

Example 1

Evaluate:

Since f (x) = cos x is an even function,

Verify your result a calculator.

Example 2

Evaluate:

Since f (x) = x4 − x2 is an even function, i.e., f (−x) = f (x), thus


Example 3

Evaluate:

Since f (x) = sin x is an odd function, i.e., f (−x) = −f (x), thus

sin x dx−∫ π

π

x x dx x x dx x x4 2

3

34 2

0

35 3

0

3

5 3

2 25 3

23

5

3

30

396

5

−( )

= −( )

= −

= −

−

=

−∫ ∫

x x dx4 2

3

3

−( )−∫

cos cos sin sin sinx dx x dx x−∫ ∫

= = [ ] =

− ( )

= −( ) =

π

π π π π2

2

0

2

0

22 2 2

2

0

2 1 0 2

cos x dx−∫ π

π

2

2

f x dxa

a

( ) =−∫ 0

f x dx f x dxa

a a

( ) = ( )−∫ ∫ 2

0

Let or

Rewrite

;

:

u x du x dx du

xdx

xe dx e du

e c e c

xe dx e e e

e e

x u u x

x x

2 2

2 2

2

1 1

1

1

2 1

1

2

5 2

2 3

1 22

2

1

2

1

2

1

2

1

2

1

2

12

1

+( ) +( )

+( )−

+( )

−

∫ ∫

∫

= + = =

= = + = +

=

= −

= −(( )

7/23/2019 007137714 x



Verify your result algebraically.

You can also verify the result with a calculator.

Example 4

If for all values of k, then which of the following could be the

graph of f ? See Figure 8.3-1.

f x dx f x dxk

k

k

( ) = ( )∫ ∫ − 20

sin cos cos cosx dx x− −∫ = − ] = −( ) − − −( )[ ]

= − −( )[ ] − − −( )[ ] = ( ) − ( ) =

π

π

π

ππ π

1 1 1 1 0

sin x dx−∫ =

π

π0

0

y

x

(A)

0

y

x

(B)

0

y

x

(D)

0

y

x

(E)

0

y

x

(C)

Figure 8.3-1

Thus f is an even function. Choice (c).


f x dx f x dx f x dx f x dx

kk

k k

k

k

k k

k

( ) = ( ) + ( )

( ) = ( ) ( ) = ( )

−−

− −

∫ ∫ ∫

∫ ∫ ∫ ∫

0

0

0 0

0

2Since then,

8.4 RAPID REVIEW

1. Evaluate

Answer: sin sin sin sin .t x xx

x

] = − ( ) = −2 2

1π

cos .t dt x

π2

∫

7/23/2019 007137714 x



2. Evaluate

Answer:

3. If find G′(4).

Answer:

4. If find k.

Answer:

5. If G(x) is a antiderivative of (e x + 1) and G(0) = 0, find G(1).

Answer: G(x) = ex + x + c

G(0) = e0 + 0 + c = 0 ⇒ c = −1.

G(1) = e1 + 1 − 1 = e.

6. If G′(x) = g (x), express in terms of G(x).

Answer: Let u = 4x;

g u du

G u Thus x dx G x G G( ) = ( ) ( ) = ( )

= ( ) − ( )[ ]∫ ∫ 4

1

44

1

44

1

48 0

0

2

0

2

. .

dudx

4= .

g x dx40

2

( )∫

x k kk2

1

28 1 8 3] = ⇒ − = ⇒ = ± .

2 81

x dxk

∫ = ,

′( ) = +( ) ′( ) = =G x x G2 1 4 9 273

23

2and .

G x t dt x

( ) = +( )∫ 2 13

2

0,

ln ln ln ln .x +( )] = − =1 2 1 20

1

1

10

1

x dx

+∫



Evaluate the following definite integrals.

1.

2.

3.

4.

5. If find k.

6.

7. If f ′(x) = g (x) and g is a continuous function for

all real values of x, express in terms of f.

8. 102

3

e dxx

ln

ln

∫

g x41

2

( )∫

sin

cos

x

xdx

10 +∫

π

6 1 40

x dxk

−( ) =∫ ,

x dx−∫ 30

6

t

t dt

+∫ 11

3

x dx−( )∫ 21

2

6

11

1 3

1

0

+ −( )−∫ x x dx

9.

10.

11.

12.


13. Find k if

14. Evaluate to the nearest 100th.

15. If

16. Use a midpoint Riemann sum with four subdivi-sions of equal length to find the approximate

value of x dx3

0

8

1+( )∫ .

y t dt dy

dx

x

= +∫ 2

11

3

, .find

21 2

3 1

θ θ θcos.

.

d −∫

x k dx3

0

2

10+( ) =∫

cos x x dx−( )−∫

2

π

π

42

1

1

xe dxx

−∫

If findf x t dt f x

( ) = ( ) ′

−∫ tan , .2

4 6

ππ

1

3

2

t dt e

e

+∫

7/23/2019 007137714 x



17. Given

find (a)

(b)

(c)

(d)

18. Evaluate:dx

x1 20

12

−∫

22

2

g x dx( )−

−

∫

50

2

g x dx( )−

∫

g x dx( )−

∫ 22

g x dx( )−∫ 20

g x dx g x dx( ) = ( ) =−∫ ∫ 2

2

0

2

8 3and

P i p e

6 ft.

10 ft.

Figure 8.6-2



21. Evaluate

22. Find

23. The graph of f ′, the derivative of f, −6 ≤ x ≤ 8 isshown in Figure 8.6-1.

dy

dx x y xat if= = −3 42ln .

limx

x

x→−∞

−−

2 4

3 9

(c) Find all values of x such that f has a change

of concavity.

24. (Calculator) Given the equation 9x2 + 4y2 −18x + 16y = 11, find the points on the graphwhere the equation has a vertical or horizontaltangent.

25. (Calculator) Two corridors, one 6 feet wideand another 10 feet wide meet at a corner. SeeFigure 8.6-2. What is the maximum length of apipe of negligible thickness that can be carried

horizontally around the corner?

x 0 5 10 15 20 25 30

f (x) 1.4 2.6 3.4 4.1 4.7 5.2 5.7

x 0 1 2 3 4 5 6–6 –5 –4 –3 –2 –1

–1

1

2

3

–2

–3

7 8

f

Figure 8.6-1

19. Find

20. Let f be a continuous function defined on [0, 35]with selected values as shown below:

dy

dx y t dt

x

x

if = +( )∫ 2 1cos

sin

Use a midpoint Riemann sum with three sub-divisions of equal length to find the approximate

value of f x dx( )∫ 030

.

(a) Find all values of x such that f attains a rela-tive maximum or a relative minimum.

(b) Find all values of x such that f is concaveupward.


1. 12 4

3

1

02 4

1

0

+ −( ) = + − −

−∫ x x dx x

x x

0 11

2

1

4

3

4

2 4

= − −( ) + −( )

− −( )

=

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2. Let u = x − 2 du = dx

3. Let u = t + 1; du = dt and t = u − 1.

Rewrite:

4. Set x − 3 = 0; x = 3.

x dx x dx x dx

xx

xx

− = − −( ) + −( )

= −

+

+ −

= − ( )

+ ( )

−

+ − ( )

− − ( )

= + =

∫ ∫ ∫ 3 3 3

23

23

3

23 3 0

6

23 6

3

23 3

9

2

9

29

0

6

0

3

3

6

2

0

3 2

3

6

2

2 2

xx x

x x− =

−( ) ≥

− −( ) <

3

3 3

3 3

if

if

t

t dt

u

u du

u du

u u c

t t c

t

t dt t t

+ =

−

= −

= − +

= + − + +

+ = + − +[ ]

= ( ) + − +[ ]

− ( ) + − +( )

= − − +

= − += − ( )

∫ ∫

∫

∫

1

1

11

1 1

11 1

3 1 3 1

1 1 1 1

4 4 2 2

2 4 22 2

1

3

1

3

ln

ln

ln

ln

ln

ln ln

ln lnln

22

2

2 2 2 2

2 2

+

= − +

= −

ln

ln ln

ln

x dx u du u

c

x c

x dx x

−( ) = = +

= −( ) +

−( ) = −( ) ]

= −( ) − −( )[ ]

= −( ) =

∫ ∫

∫

22

32

32

22

32

2

311 2 6 2

2

327 8

38

3

12

12

32

32

12

11 32

6

11

32

32

Thus6

5.

Verify your results by evaluating

6. Let u = 1 + cos x; du = −sin x dx or−du = sin x dx.

7. Let u = 4x; du = 4 dx or = dx

8. 10 10 10

10 3 2 10

2

3

2

33 2e dx e e ex x= ] = ( ) − ( )[ ]

= −( ) =

∫ lnln

ln

lnln ln

.

g x dx g u du

g u du

f u c

f x c

g x dx f x

f f

f f

44

1

41

4

1

44

4 14

4

1

44 2

1

44 1

1

48

1

44

1

2

1

2

( ) = ( ) = ( )

= ( ) +

= ( ) +

( ) = ( )]

= ( )( ) − ( )( )

= ( ) − ( )

∫ ∫ ∫

∫

.

du

4

sin

cos

cos

sin

coscos

cos

cos

x

xdx

udu

udu

u du u

c

u c

x c

x

xdx x

1

1 1

1

2

2

2 1

12 1

2 1

1 0

2 0 2

12

12

12

12

12

12

12

12

12

0 0

+=

−( ) = −

= − = − +

= − +

= − +( ) +

+= − +( ) ]

= − +( )[

− +( ) ]

= − −

∫ ∫ ∫

∫

∫

−

π π

π

[[ ] = 2 2

6 1 6 10

43

1

x dx x dx−( ) −( )∫ ∫ −

and0

.

6 1 3 3

3 4 3 4 0

3 4 1 0

4

31

0

2

0

2

2 2

x dx x x k k

k k k k

k k

k k

k k

−( ) = − ] = −

− = ⇒ − − =

⇒ −( ) +( ) =

⇒ = = −

∫ Set

or

7/23/2019 007137714 x



9. Let u = t + 3; du = dt.

10.

11. Let u = x2; du = 2x dx or = x dx

Note that f (x) = 4xex2

is an odd function. Thus

12.

Note that f (x) = cos x − x2 is an even function.

Thus you could have written

and

obtain the same result.


13.

Set 4 + 2k = 10 and thus k = 3.

x k dx x

kx k

k

3

0

24

0

24

4

2

42 0

4 2

+( ) = + ] = + ( )

−

= +

∫

cos cosx x dx x x dx−( ) = −( )∫ ∫ −2 2

02

π

π

π

cos sin

sin

sin

x x dx x x

−( ) = −

= −

− −( ) − −( )

= − − − −

= −

−−

∫ 2

3

3

3

3 3

3

3

3

3

30

3

2

3

π

π

π

π

π π

π π

π π

π

f x dxa

a

( ) =−∫ 0.

4 42

2 2 2

4 2

2 2 0

2

2

2 2

2 2

1

1

1

1

1 1

xe dx e du

e du e c e c

xe dx e

e e e e

x u

u u x

x x

∫ ∫

∫ ∫

=

= = + = +

= ]

= −[ ] = −( ) =

− −

( ) −( )

du

2

′( ) = ′

=

=

=f x x f tan ; tan .2 2

2

6 6

1

3

1

3

π π

1

3

13

1

33

3 3

33

2 2

2

2

t dt

u du u c t c

t dt t

e e

ee

e

e

e

e

+ = = + = + +

+ = + ]

= +( ) − +( )

= ++

∫ ∫

∫

ln ln

ln

ln ln

ln

14. Enter ∫ (2x cos(x), x, −1.2, 3.1) and obtain−4.70208 ≈ −4.702.

15.

16. Midpoints are x = 1, 3, 5 and 7.

17.

18.

19. 2 1 2 1

2 1

2 1 2 1

2 1

2 1

2 1

2

0

0

t dt t dt

t dt

dy

dx

d

dx t x

d

dx

x x d

dx x

x x

x x

x x x

x

x

x

x

x

x

+( ) = +( )

− +( )

= +( ) = +( )

− +( ) ( )

= +( )

− +( ) −( )

= + +

∫ ∫

∫ ∫

sin

cos

sin

cos

cos

sin

sin

sin cos cos

sin cos

cos sin

sin cos cos 22

4

sin cos

sin sin cos cos sin .

x x

x x x x x+ = + +

dx

xx

1

1

20

60

6

20

1

0

1 1

12

12

−= ( )]

=

− ( )

= − =

∫ −

− −

sin

sin sin

π π

(d) 2 2 2 8 162

2

2

2

g x dx g x dx( ) = ( ) = ( ) =− −∫ ∫

(c) 5 5

5

5 5 25

0

2

0

2

2

0

g x dx g x dx

g x dx

( ) = ( )

= − ( )( )= −( ) = −

− −

−

∫ ∫

∫

(b) g x dx g x dx( ) ) = − ( ) = −−

−∫ ∫ 2

2

2

2

8

(a)

Thus

g x dx g x dx g x dx

g x dx g x dx

( ) + ( ) = ( )

( ) + = ( ) =

−−

−−

∫ ∫ ∫

∫ ∫

2

2

0

2

2

0

2

0

2

0

3 8 5.

x dx3

0

123 3

3 3

1 1 1 2 3 1 2

5 1 2 7 1 2

2 2 28 2 126 1

344 2 874

+( ) ≈ +( )( ) + +( )( )

+ +( )( ) + +( )( )

≈ ( )( ) + ( )( ) + ( )( )

+ ( )( ) =

∫

∆x = −

=8 0

4 2

d

dx t dt x

d

dx x

x x

x2

1

3 2 3

2 6

1 1

3 1

3

+

= ( ) + ( )

= +

∫

.

7/23/2019 007137714 x



20.

Midpoints are x = 5, 15 and 25.

∆x = −

=30 0

310 f x dx f f f ( ) ≈ ( )[ ] + ( )[ ] + ( )[ ]

≈ ( )( ) + ( )( ) + ( )( )

≈

∫ 030

5 10 15 10 25 10

2 6 10 4 1 10 5 2 10

119

. . .

.


21.

22.

23. (a) See Figure 8.8-1.

y x dy

dx x x

dy

dxx

= − =−( )

( )

= ( )

−( ) =

=

ln ,2

2

3

2

41

42

2 3

3 4

6

5

As x x x

x

x

x x

x x

x x

x

x

x

x x

x

x

→ −∞ = −

−−

= − −

−( )

= − −( )

− ( )

= − − ( )

−

= − −−

= −

→−∞ →−∞

→−∞

→−∞

,

lim lim

lim

lim

.

2

2 2 2

2 2

2

4

3 9

4

3 9

4

3 9

1 4

3 9

1 03 0

13

The function f is concave upward on intervals(−6, −3) and (1, 5).

(c) A change of concavity occurs at x = −3,x = 1 and x = 5.

24. (Calculator) Differentiate both sides of 9x2 + 4y2

− 18x + 16y = 11.

Horizontal tangent

Using a calculator, enter solve (4y^2 + 16y −20 = 0, y); obtaining y = −5 or y = 1.

Thus each of the points at (1, 1) and (1, −5)the graph has a horizontal tangent at eachpoint.

Vertical tangent ⇒ is undefined.

Set 8y + 16 = 0 ⇒ y = −2.

At y = −2, 9x2 + 16 − 18x − 32 = 11

9x2 − 18x − 27 = 0

Enter solve (9x2 − 8x − 27 = 0, x) and obtainx = 3 or x = −1.

Thus at each of the points (3, −2) and(−1, −2), the graph has a vertical tangent.(See Figure 8.8-3.)

dy

dx

Set or

At

dy

dx

x x

x y y

y y

= ⇒ − + = =

= + − + =

+ − =

0 18 18 0 1

1 9 4 18 16 11

4 16 20 0

2

2

,

⇒ =dy

dx0

18 8 18 16 0

8 16 18 18

8 16 18 18

18 18

8 16

x y dy

dx

dy

dx

y dydx

dydx

x

dy

dx y x

dy

dx

x

y

+ − + =

+ = − +

+( ) = − +

= − +

+

x

decr. incr. decr. decr.incr.

–6 8–5 –1 3 7

rel.min.

rel.max.

rel.min.

rel.max.

– + – + –

Figure 8.8-1

x

incr.

–6

+ – + –

concavedownward

concaveupward

concavedownward

8–3 1 5

decr. decr.incr.

concaveupward

Figure 8.8-2

The function f has a relative minimum at x = −5and x = 3, and f has a relative maximum atx = −1 and x = 7.

(b) See Figure 8.8-2.

7/23/2019 007137714 x



25. (Calculator)

Step 1. See Figure 8.8-4. Let P = x + y where Pis the length of the pipe and x and y areas shown. The minimum value of P isthe maximum length of the pipe to beable to turn in the corner. By similar

triangles, and thus

Step 2. Find the minimum value of P.

Use the Minimum function of the calcu-lator and obtain the minimum point(9.306, 22.388).

Enter y x x x1 10 2 36= + −( )( ) ^ .

P x y x x

x= + = +

−

10

362

y x

xx=

−>

10

366

2,

y x

x10 362=

−

–1 1 2 30

2

1

–1

–2

–3

–4

–5

x = –3 x = –1

y = –5

y = 1

x

y

Figure 8.8-3

x

y

10

6

x 2 – 36

Figure 8.8-4

9.306

rel. min.

incr.decr.

+–

y1 = f

y2 = f

Figure 8.8-5

Step 4. Check endpoints.

The domain of x is (6, ∞)

Since at x = 9.306 is the only relativeextremum, it is the absolute minimum.Thus the maximum length of the pipe is22.388 feet.

Step 3. Verify with the First Derivative Test.

Enter y2 = (y1(x), x) and observe. (SeeFigure 8.8-5.)

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9.1 THE FUNCTION

The Second Fundamental Theorem of Calculus defines

and states that if f is continuous on [a,b], then F ′(x) = f (x) for every point x in [a,b].If f ≥ 0, then F ≥ 0. F (x) can be interpreted geometrically as the area under the curve of f from t = a to t = x. (See Figure 9.1-1.)

F x f t dt a

x

( ) = ( )∫

F x f t dt ( ) = ( )∫ ax

Chapter 9

Areas and Volumes

y f (t )

t x a0

Figure 9.1-1

If f < 0, F < 0. F (x) can be treated as the negative value of the area between the curve of f and the t-axis from t = a to t = x. (See Figure 9.1-2.)

Example 1

If for 0 ≤ x ≤ π, find the value(s) of x where f has a local minimum.f x t dt x

( ) = ∫ 20cos


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Method 1:

Thus at f has a local minimum.

Method 2:

You can solve this problem geometrically by using area. See Figure 9.1-3.

x =3

2

π,

Since cos

Set or

and and

f x t dt f x x

f x x x

f x x f f

x

( ) = ′( ) =

′( ) = = =

′′( ) = − ′′

= − ′′

=

∫ 2 2

0 2 02

32

22

23

22

0, cos .

; cos , .

sin

π π

π π

y

f (t )

t x a

0

Figure 9.1-2

[0,2π] by [−3,3]

Figure 9.1-3

The area “under the curve” is above the t-axis on and below the x-axis on

Thus the local minimum occurs at

Example 2

Let and the graph of f is shown in Figure 9.1-4.

(a) Evaluate: p(0), p(1), p(4)(b) Evaluate: p(5), p(7), p(8)(c) At what value of t does p have a maximum value?

p x f t dt x

( ) = ( )∫ 0

32

π .π π

23

2, .

02

, π

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(d) On what interval(s) is p decreasing?(e) Draw a sketch of the graph of p.

Solution:

(a)

(Note: f (t ) forms a trapezoid from t = 0 to t = 4.)

(b)

(c) Since f ≥ 0 on the interval [0,4], p attains a maximum at t = 4.

(d) Since f (t ) is below the x-axis from t = 4 to t = 8, if x > 4,

Thus p is decreasing on the interval (4, 8).

(e) See Figure 9.1-5 for a sketch. p x f t dt x

( ) = ( )∫ 0 .

f t dt f t dt f t dt f t dt x x x

( ) = ( ) + ( ) ( ) <∫ ∫ ∫ ∫ 0

4

0 4 40where .

p f t dt f t dt f t dt

p f t dt f t dt f t dt f t dt

p f t dt f t dt

5

121 4

210

7

12 2 2 4 2

8

0

5

4

5

0

4

0

7

4

5

0

4

5

7

0

8

( ) = ( ) = ( ) + ( )

= − ( )( )

=

( ) = ( ) = ( ) + ( ) + ( )

= − − ( )( ) =

( ) = ( ) = ( ) +

∫ ∫ ∫

∫ ∫ ∫ ∫

∫ f f t dt ( )

= − =

∫ ∫ 4

8

0

4

12 12 0

p f t dt

p f t dt

p f t dt

0 0

11 4

22

41

22 4 4 12

0

0

0

1

0

4

( ) = ( ) =

( ) = ( ) = ( )( )

=

( ) = ( ) = +( )( ) =

∫

∫

∫

x 0 1 2 3 4 5 6 7 8

p(x) 0 2 6 10 12 10 6 2 0

Areas and Volumes • 239

t

y

f (t )

1 2 3 4 5 6 7 8

0

–4

4

Figure 9.1-4

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(b)

(c) a(t ) = v′(t ). Since v′(t ) = f ′(t ), v′(t ) = 0 at t = 4. Thus a(4) = 0.

(d) The particle is moving to the right when v(t ) > 0. Thus the particle is moving to theright on intervals (0, 2) and (7, 8)

(e) The area of f below the x-axis from x = 2 to x = 7 is larger then the area of f above the

x-axis from x = 0 to x = 2 and x = 7 to x = 8. Thus and the particle ison the left side of the origin.

f x dx( ) <∫ 00

8

s f x dx f x dx f x dx31

210 2

1

21 5

15

22

3

0

2

0

3

( ) = ( ) = ( ) + ( ) = ( )( ) − ( )( ) =∫ ∫ ∫


9.2 APPROXIMATING THE AREA UNDER A CURVE

Main Concepts: Rectangular Approximations, Trapezoidal Approximations

Rectangular Approximations

If f ≥ 0, the area under the curve of f can be approximated using three common types of rectangles: left-endpoint rectangles, right-endpoint rectangles, or midpoint rectangles.(See Figure 9.2-1.)

• Don’t forget that (fg )′ = f ′ g + g ′f and not f ′ g ′. However, lim(fg ) = (lim f ) (lim g )

a0

y f ( x )

x b x 1 x 2 x 3

left-end point

a0

y f ( x )

x b x 1 x 2 x 3

right-end point

a0

y f ( x )

x b x 1 x 2 x 3

midpoint

Figure 9.2-1

The area under the curve using n rectangles of equal length is approximately:

area of rectangle

left-endpoint rectangles

right-endpoint rectangles

midpoint rectangles

where and 1 2

i

i

i

n

i

i

n

i i

i

n

i

n

n

f x x

f x x

f x x

x

x b an a x x x x b

( ) =

( )

( )

+

= − = < < < < =

−=

=

−

=

=

∑

∑

∑

∑

1

1

1

1

1

1

0

2

∆

∆

∆

∆ . . .

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If f is increasing on [a,b], then left-endpoint rectangles are inscribed rectangles andthe right-endpoint rectangles are circumscribed rectangles. If f is decreasing on [a,b], thenleft-endpoint rectangles are circumscribed rectangles and the right-endpoint rectangles areinscribed. Furthermore,

inscribed rectanglei ≤ area under the curve ≤ circumscribed rectanglei.

Example 1Find the approximate area under the curve of f (x) = x2 + 1 from x = 0 to x = 2. Using4 left-endpoint rectangles of equal length. (See Figure 9.2-2.)

i

n

=∑

1i

n

=∑

1

Let ∆xi be the length of i th rectangle. The length

Area under the curve ≈

Enter and obtain 3.75.

Or, finding the area of each rectangle:

Area of (RectI + RectII + RectIII + RectIV) = 3.75Thus the approximate area under the curve of f (x) is 3.75.

Example 2

Find the approximate area under the curve of from x = 4 to x = 9 using 5 right-

endpoint rectangles. (See Figure 9.2-3.)

f x x( ) =

Area of Rect

Area of Rect

Area of Rect

Area of Rect

I

II

III

IV

= ( )( ) = ( )

=

= ( ) = ( ) +( )

=

= ( ) = +( ) =

= ( ) = +( )

f x

f x

f x

f x

0 11

2

1

2

0 5 0 5 11

20 625

1 1 11

2 1

1 5 1 5 11

2

1

2

2

3

2

4

2

∆

∆

∆

∆

. . .

. .

= 1 625.

. * . , , ,5 1 1 5 1 42

x x−( )( ) +( )( )∑

f x x iii

ii

−= =( ) = −( )

+

∑ ∑

11

42

1

4 1

2 1 1

1

2∆

∆x x ii i= −

= = −( )−

2 0

4

1

2

1

211;


I II

IIIIV

0.5 10 21.5

y(2,5)

f ( x )

x

Figure 9.2-2

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Let ∆xi be the length of i th rectangle. The length xi = 4 + (1)i = 4 + i

Or, using notation:

Enter and obtain 13.16

Thus the area under the curve is approximately 13.16.

Example 3

The function f is continuous on [1,9] and f > 0. Selected values of f are given below:

4 1 5+( )( )∑ x x, , ,

f x x f ii i

iii

( ) = +( )( ) = +===∑∑∑ ∆ 4 1 4 1

1

5

1

5

1

5

∑

Area of Recti

i

( ) = + + + + ≈=∑ 5 6 7 8 3 13 16

1

5

.

Area of Rect

Area of Rect

Area of Rect

Area of Rect

Area of Rect

I

II

III

IV

V

= ( ) = ( )( ) =

= ( ) = ( )( ) =

= ( ) = ( )( ) =

= ( ) = ( )( ) =

= ( ) = ( )( ) = =

f x x f

f x x f

f x x f

f x x f

f x x f

1 1

2 2

3 3

4 4

5 5

5 1 5

6 1 6

7 1 7

8 1 8

9 1 9 3

∆

∆

∆

∆

∆

∆xi = −

=9 4

51;


x 1 2 3 4 5 6 7 8 9

f (x) 1 1.41 1.73 2 2.37 2.45 2.65 2.83 3

0 4 5 6 7 8 9

I II IV VIII

x

f ( x ) = √ x y

Figure 9.2-3

Using 4 midpoint rectangles, approximate the area under the curve of f for x = 1 to x = 9.(See Figure 9.2-4.)

Let ∆xi be the length of i th rectangle. The length

Area of RectI = f (2)(2) = (1.41)2 = 2.82

Area of RectII = f (4)(2) = (2)2 = 4

Area of RectIII = f (6)(2) = (2.45)2 = 4.90

Area of RectIV = f (8)(2) = (2.83)2 = 5.66

∆xi = −

=9 1

42.

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Area of (RectI + RectII + RectIII + RectIV) = 2.82 + 4 + 4.90 + 5.66 = 17.38.Thus the area under the curve is approximately 17.38.

Trapezoidal Approximations

Another method of approximating the area under a curve is to use trapezoids. SeeFigure 9.2-5.


I II

IIIIV

10

1

2

3

2 3 4 5 6 7 8 9

y

x

f

Figure 9.2-4

a = x 0 x 1 x 2 b = x 3

y

0 x

f ( x )

Figure 9.2-5

Formula for Trapezoidal Approximation:

If f is continuous, the area under the curve of f from x = a to x = b is:

Example 1

Find the approximate area under the curve of from x = 0 to x = π, using

4 trapezoids. (See Figure 9.2-6.)

f x x

( ) =

cos

2

Area ≈ −

( ) + ( ) + ( ) + ( ) + ( )[ ]−

b a

n f x f x f x f x f xn n

22 2 20 1 2 1

K

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If x = g (y) is continuous and non-negative on [c, d ], then the area under the curve of g from c to d is:

See Figure 9.3-2.Area = ( )∫ g y dyc

d

.


y f

x 0 a b

(+)

y

x 0

a b

( –

)

f

Figure 9.3-1

d

y

x 0

c

g( y)

Figure 9.3-2

Example 1

Find the area under the curve of f (x) = (x − 1)3 from x = 0 to x = 2.

Step 1. Sketch the graph of f (x). See Figure 9.3-3.

10

–1

2

y

x

f ( x )

Figure 9.3-3

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Step 2. Set up integrals:

Step 3. Evaluate integrals:

Thus the total area is

Another solution is to find the area using a calculator.

Example 2Find the area of the region bounded by the graph of f (x) = x2 − 1, the lines x = −2 andx = 2 and the x-axis.

Step 1. Sketch the graph of f (x). See Figure 9.3-4.

Enter and obtainabs x x−( )( )( )∫ 1 3 0 21

2^ , , , .

1

4

1

4

1

2+ = .

x dx x

x dx x

−( ) = −( )

= − =

−( ) = −( )

=

∫

∫

11

4

1

4

1

4

11

4

1

4

3

0

1

4

0

1

3

1

2

4

1

2

Area = ( ) + ( )∫ ∫ f x dx f x dx0

1

1

2

.


(+) (+)

–2 –1 10 2(–)

y

x

f ( x )

Figure 9.3-4

Step 2. Set up integrals.

Step 3. Evaluate the integrals:

x dx x

x

x dx x

x

x dx x

x

2

2

13

2

1

2

1

13

1

1

2

1

23

1

2

13

2

3

2

3

4

3

13

2

3

2

3

4

3

4

3

13

2

3

2

3

−( ) = −

= − −

=

−( ) = −

= − −

= − =

−( ) = −

= − −

−

−

−

−

−−

∫

∫

∫ =4

3

Area = ( ) + ( ) + ( )−

−

−∫ ∫ ∫ f x dx f x dx f x dx2

1

1

1

1

2

.

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Thus the total area =

Note: Since f (x) = x2 − 1 is an even function, you can use the symmetry of

the graph and set

An alternate solution is to find the area using a calculator.Enter and obtain 4.

Example 3

Find the area of the region bounded by x = y2, y = −1 and y = 3. See Figure 9.3-5.

abs x x^ , , ,2 1 2 2−( ) −( )

area = ( ) + ( )( )∫ ∫ 20

1

1

2

f x dx f x dx .

total area = + + =4

3

4

3

4

34.


3

0

–1 x = y2

x

y

Figure 9.3-5

Example 4Using a calculator, find the area bounded by f (x) = x3 + x2 − 6x and the x-axis. SeeFigure 9.3-6.

Area = =

= − −( )

=−

−∫ y dy

y2

1

33

1

3 33

3

3

3

1

3

28

3.

[−4,3] by [−6,10]

Figure 9.3-6

Step 1. Enter y1 = x ^ 3 + x ^ 2 − 6x

Step 2. Enter and obtain 21.083.

Example 5

The area under the curve y = ex from x = 0 to x = k is 1. Find the value of k.

Take ln of both sides:

ln(ek) = ln 2; k = ln 2.

Area = = ] − − = − ⇒ =∫ e dx e e e e exk

k k k k k

0 0

0 1 2.

abs x x x x^ ^ , , ,3 2 6 3 2+ − ∗ ) − )

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Example 6

The region bounded by the x-axis, and the graph of y = sin x between x = 0 and x = πis divided into 2 regions by the line x = k. The area of the region 0 ≤ x ≤ k is twice thearea of the region k ≤ x ≤ π, find k. (See Figure 9.3-7.)


1

0 k π

y = sin x

Figure 9.3-7

Area between Two Curves

Area Bounded by Two Curves: See Figure 9.3-8.

sin sin

cos cos

cos cos cos cos

cos cos

cos cos

cos

cos

cos . .

x dx x dx

x x

k k

k k

k k

k

k

k

k

k

k

k

0

0

2

2

0 2

1 2 1

1 2 2

3 1

1

3

1

31 91063

∫ ∫ =

− ] = −[ ]

− − − ( )( ) = − − −( )( )

− + = +( )

− + = +

− =

= −

= −

=

π

π

π

arc

a 0 b c d

f

y

g

x

Figure 9.3-8

Area

Area upper curve lower curve

= ( ) − ( )[ ] + ( ) − ( )[ ]

= −( )

∫ ∫

∫

f x g x dx g x f x dx

dx

a

c

c

d

a

d

Note:

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Step 1. Sketch the graphs of f (x) and g (x).

Step 2. Find the points of intersection.

Set f (x) = g (x)

x3 = x

x(x2 − 1) = 0

x(x − 1)(x − 1) = 0

x = 0, 1, −1

Step 3. Set up integrals.

Note: You can use the symmetry of the graphs and let An alter-nate solution is to find the area using a calculator.

Example 2

Find the area of the region bounded by the curve y = ex, the y-axis and the line y = e2.

Step 1. Sketch a graph. See Figure 9.3-10.

Enter and obtainabs x x x^ , , , .3 1 11

2−( ) −( )∫

area = −( )∫ 2 3

0

1

x x dx.

Area = ( ) − ( )( ) + ( ) − ( )( )

= −( ) + −( )

= −

+ −

= − −( )

− −( )

+ −

−

−

−

∫ ∫

∫ ∫

f x g x dx g x f x dx

x x dx x x dx

x x x x

1

0

0

1

3

1

03

0

1

4 2

1

0 2 4

0

1

4 22

4 2 2 4

01

4

1

2

1

2

11

40

1

4

1

4

1

2

4

−

= − −

+ = .


–1 10

(–1,1)

(1,1)

y

x

g( x )

f ( x )

Figure 9.3-9

Example 1

Find the area of the region bounded by the graphs of f (x) = x3 and g (x) = x. (SeeFigure 9.3-9.)

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Step 2. Find the point of intersection. Set e2 = ex ⇒ x = 2.

Step 3. Set up an integral:

Or using a calculator, enter

Example 3

Using a calculator, find the area of the region bounded by y = sin x and between0 ≤ x ≤ π.


y x=

2

e e x x e^ ^ , , , .2 0 2 1( ) − ( )( )( ) +( )∫ and obtain 2

Area = −( ) = ( ) − ]= −( ) − −( )

= +

∫ e e dx e x e

e e e

e

x x2

0

22

0

2

2 2 0

2

2 0

1.


y = e x

y = e2

x

y

1

1 20

Figure 9.3-10

Step 2. Find the points of intersection.

Using the Intersection function of the calculator, the intersection pointsare x = 0 and x = 1.89549.

Step 3. Enter nInt(sin(x) − .5x, x, 0, 1.89549) and obtain 0.420798 ≈ 0.421.

(Note: You could have also used the ∫ function on your calculator and getthe same result.)

Example 4

Find the area of the region bounded by the curve xy = 1 and the lines y = −5, x = e andx = e3.

[−π,π] by [−1.5,1.5]

Figure 9.3-11

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0

–5

e e3

y x = e x = e3

x

xy = 1

y = –5

Figure 9.3-12

Step 2. Set up an integral:

Step 3. Evaluate the integral:

Area = − −( )

= +

= + ] = ( ) + ( )[ ] − ( ) + ( )[ ]

= + − − = − +

∫ ∫ 1

51

5

5 5 5

3 5 1 5 2 5 5

3 3

33 3

3 3

x dx

x dx

x x e e e e

e e e e

e

e

e

e

e

e

ln ln ln

.

Area = − −( )

∫

15

3

x dx

e

e

• Remember: if f ′ > 0, then f is increasing and if f ″ > 0 then the graph of f is con-cave upward.

9.4 VOLUMES AND DEFINITE INTEGRALS

Main Concepts: Solids with Known Cross Sections, The Disc Method,

The Washer Method

Solids with Known Cross Sections

If A(x) is the area of a cross section of a solid and A(x) is continuous on [a,b], then thevolume of the solid from x = a to x = b is:

See Figure 9.4-1.

V A x dxa

b

= ( )∫

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Note: A cross section of a solid is perpendicular to the height of the solid.

Example 1

The base of a solid is the region enclosed by the ellipse The cross sections

are perpendicular to the x-axis and are isosceles right triangles whose hypotenuse are onthe ellipse. Find the volume of the solid. See Figure 9.4-2.

x y2 2

4 25

1+ = .


a b

y

x 0

Figure 9.4-1

y

x

– 2

2

0

5

– 5

a

a

y

x 2 y2

4 25= 1

Figure 9.4-2

Step 1. Find the area of a cross section A(x).

Pythagorean Theorem: a2 + a2 = (2y)2

2a2 = 4y2

Step 2. Set up an integral.

V x

dx= −

−∫ 2525

4

2

2

2

A x x

( ) = −2525

4

2

Since orx y y x

y x2 2 2 2

2

2

4 251

251

425

25

4+ = = − = −,

A x a y y( ) = = ( ) =1

2

1

222

22

a y a= >2 0,

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Step 3. Evaluate the integral.

The volume of the solid is

Verify your result with a graphing calculator.

Example 2

Find the volume of a pyramid whose base is a square with a side of 6 feet long , and aheight of 10 feet . See Figure 9.4-3.

200

3.

V x

dx x x= −

= −

= ( ) − ( )

− −( ) − −( )

= − −

=

−−

∫ 2525

425

25

12

25 225

122 25 2

25

122

100

3

100

3

200

3

2

2

23

2

2

3 3

.


x

y

0

10

6

6

s

x

3

10

Figure 9.4-3

Step 1. Find the area of a cross section A(x). Note each cross section is a squareof side 2s.

Similar triangles:


V x

dx= ∫ 9

25

2

0

10

A x s s x x

( ) = ( ) = =

=2 4 4

3

10

9

25

2 2

2 2

x

s s

x= ⇒ =

10

3

3

10

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The volume of the pyramid is 120 ft 3.

Example 3

The base of a solid is the region enclosed by a triangle whose vertices are (0,0), (4,0)and (0,2). The cross sections are semicircles perpendicular to the x-axis. Using a calcu-lator, find the volume of the solid. (See Figure 9.4-4.)

V x

dx x

= =

= ( )

− =∫ 9

25

3

25

3 10

250 120

2

0

103

0

10 3

.


2

0

4

y

x

Figure 9.4-4

Step 1. Find the area of a cross section. Equation of the line passing through




Thus the volume of the solid is 2.0944.

π2

25 1 2 0 4

∗ − +( )

∫ . ^ , , ,x x

V A x dx x= ( ) = − +

∫ ∫ 0

42

0

4

2

1

4

1π

.

A x y

x( ) =

= − +

1

2 2 2

1

41

2 2

π π

.

Area of semicircle = = = − +

= − +

1

2

1

2

1

2

1

22

1

412πr r y x x;

( , ) and ( , ): ;0 2 4 00 2

4 0

1

22

12

2

y mx b m b

y x

= + = −

− = − =

= − +

;

• Remember: if f ′ < 0, then f is decreasing and if f ″ < 0 then the graph of f is con-cave downward.

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Step 1. Draw a sketch. See Figure 9.4-7.


y

x

f ( x )

0

k

a b

y = k

h 0

g( y)

y

x

x = h

d

c

Figure 9.4-6

y

x

y = √ x – 1

1 50

Figure 9.4-7

Step 2. Determine the radius of a disc from a cross section.

r f x x= ( ) = − 1

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Example 2

Find the volume of the solid generated by revolving the region bounded by the graph of

, and the x-axis about the x-axis.


y x x= ≤ ≤cos ,where 02

π

V x dx x x

x= −( ) = −( )[ ] = −

= −

− −

=

∫ π π π

π π

1 12

52

5 12

1 8

2

1

5

1

52

1

5

2 2

V f x dx x dx= ( )( ) = −( )∫ ∫ π π2

1

5 2

1

5

1


y

x

y = √cos x

0

1

π2

Figure 9.4-8

Step 2. Determine the radius from a cross section.



Thus the volume of the solid is π.


Example 3

Find the volume of the solid generated by revolving the region bounded by the graphof y = x2, the y-axis, and the line y = 6 about the y-axis.


V x dx x= = [ ] =

−

=∫ π π π π ππ π

cos sin sin sin .0

2

0

2

20

V x dx x dx= ( ) =∫ ∫ π ππ π

cos cos .2

0

2

0

2

r f x x= ( ) = cos

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The volume of the solid is 18π.


Example 4

Using a calculator, find the volume of the solid generated by revolving the regionbounded by the graph of y = x2 + 4, the line y = 8 about the line y = 8.


V y dy y

= =

=∫ π π π0

62

0

6

218 .

V x dy y dy y dy= = ( ) =∫ ∫ ∫ π π π2

0

6 2

0

6

0

6

.

y x x y

x y

r x y

= ⇒ = ±

=

= =

2

is the part of the curve involved in the region.


6

0

y

y = 6

x = √ y

x

Figure 9.4-9

–2 2

4

8 y = 8

y = x 2 + 4 y

x

Figure 9.4-10

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r = 8 − y = 8 − (x2 + 4) = 4 − x2


To find the intersection points, set 8 = x2 + 4 ⇒ x = ±2.


Enter and obtain

Thus the volume of the solid is


Example 5

Using a calculator, find the volume of the solid generated by revolving the region boundedby the graph of y = e x, the y-axis, the lines x = ln 2 and y = −3 about the line y = −3.


512

15π.

512

15π.π 4 2 2 2 2−( ) −( )∫ x x^ ^ , , ,

V x dx= −( )−∫ π 4 2 2

2

2


x

y

y = –3

y = e x

0

–3

ln 2

Figure 9.4-11


r = y − (−3) = y + 3 = ex + 3.



Enter and obtain ≈ 13.7383π

The volume of the solid is approximately 13.7383π.

π 9 215

2ln +

π e x x^ ^ , , ln( ) +( ) ( )( )∫ 3 2 0 2

V e dxx= +( )∫ π 32

0

2ln

.

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The Washer Method

The volume of a solid (with a hole in the middle) generated by revolving a region boundedby 2 curves:

About the x-axis:

where f (x) = outer radius & g (x) = inner radius

About the y-axis:

where p(y) = outer radius & q(y) = inner radius

About a line x = h:

where R(x) = outer radius & r(x) = inner radius

About a line y = k:

where R(y) = outer radius & r(y) = inner radius

Example 1

Using the Washer Method, find the volume of the solid generated by revolving theregion bounded by y = x3 and y = x in the first quadrant about the x-axis.


V R y r y dyc

d

= ( )( ) − ( )( )[ ]∫ π2 2

;

V R x r x dxa

b

= ( )( ) − ( )( )[ ]∫ π2 2

;

V p y q y dyc

d

= ( )( ) − ( )( )[ ]∫ π2 2

;

V f x g x dxa

b

= ( )( ) − ( )( )[ ]∫ π2 2

;


• Remember: if f ′ is increasing, then f ″ > 0 and the graph of f is concave upward.

y

x 0

y =

x

y =

x 3

(1,1)

Figure 9.4-12

To find the points of intersection, set x = x3 ⇒ x3 − x = 0 or x(x2 − 1) = 0 or x = −1,0, 1. In the first quadrant x = 0, 1.

Step 2. Determine the outer and inner radii of a washer, whose outer radius = x;and inner radius = x3.

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Example 2

Using the Washer Method and a calculator, find the volume of the solid generated byrevolving the region in Example 1 about the line y = 2.


V x x dx x x

= −( ) = −

= −

=

∫ 2 6

3 7

0

1

0

1

3 7

13

17

421

π

π π

V x x dx= − ( )[ ]∫ 2 3 2

0

1


y = x 3 y

= x

y = 2

x

0

y

Figure 9.4-13

Step 2. Determine the outer & inner radii of a washer.

The outer radius = (2 − x3) and inner radius = (2 − x).



Enter and obtain

The volume of the solid is

Example 3

Using the Washer Method and a calculator, find the volume of the solid generated byrevolving the region bounded by y = x2 and x = y2 about the y-axis.

17

21

π.

17

21

π.π 2 3 2 2 2 0 1−( ) − −( )( )( )∫ x x x^ ^ ^ , , ,

V x x dx= −( ) − −( )[ ]∫ π 2 23 2 2

0

1

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Answer:

2. The function f is continuous on [1,5] and f > 0 and selected values of f are givenbelow.

f g t dt g t dt g t dt 3

0 5 1 5 0 5

0

1

1

3

0

3

( ) = ( ) = ( ) + ( )

= − = −

∫ ∫ ∫ . . . .


x 1 2 3 4 5f(x) 2 4 6 8 10

Using 2 midpoint rectangles, approximate the area under the curve of f for x = 1to x = 5.

Answer: Midpoints are x = 2 and x = 4 and the width of each

Area ≈ Area of Rect1 + Area of Rect2 ≈ 4(2) + 8(2) ≈ 24.

3. Set up an integral to find the area of the regions bounded by the graphs of y = x3

And y = x. Do not evaluate the integral.Answer: Graphs intersect at x = −1 and x = 1. See Figure 9.5-2.

rectangle = −

=5 1

22.

(–1,–1)

0

(1,1) y = x

x

y

y = x 3

Figure 9.5-2

Or, using symmetry,

4. The base of a solid is the region bounded by the lines y = x, x = 1, and the x-axis. Thecross sections are squares perpendicular to the x-axis. Set up an integral to find thevolume of the solid. Do not evaluate the integral.

Answer: Area of cross section = x2

5. Set up an integral to find the volume of a solid generated by revolving about the x-axisthe region bounded by the graph of y = sin x, where 0 ≤ x ≤ π and the x-axis. Do notevaluate the integral.

Answer: Volume = ( )∫ π π

sin .x dx2

0

Volume of solid = ∫ x dx2

0

1

.

Area = −( )∫ 2 3

0

1

x x dx.

Area = −( ) + −( )−∫ ∫ x x dx x x dx3

1

03

0

1

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6. The area under the curve of fromx = a to x = 5 is approximately 0.916 where

1 ≤ a < 5. Using your calculator, find a.

Answer:

ln a = ln 5 − 0.916 ≈ .693

a ≈ e0.693

≈ 2.

15 0 916

5 5

x dx x a

a a∫ = ] = − =ln ln ln .

yx

=1




1. Let where the graph of f is

given in Figure 9.6-1.

F x f t dt x

( ) = ( )∫ 0

5. Find the approximate area under the curve y = x2

+ 1 from x = 0 to x = 3, using the TrapezoidalRule with n = 3.

6. Find the area of the region bounded by the

graphs , y = −x and x = 4.

7. Find the area of the region bounded by thecurves x = y2 and x = 4.

8. Find the area of the region bounded by thegraphs of all four equations:

x-axis; and the lines,

and x = π.

9. Find the volume of the solid obtained by revolv-ing about the x-axis, the region bounded by thegraph of y = x2 + 4, the x-axis, the y-axis, andthe line x = 3.

10. The area under the curve from x = 1 to

x = k is 1. Find the value of k.

11. Find the volume of the solid obtained by revolv-ing about the y-axis the region bounded byx = y2 + 1, x = 0, y = −1 and y = 1.

12. Let R be the region enclosed by the graph y = 3x,the x-axis and the line x = 4. The line x = adivides region R into two regions such that whenthe regions are revolved about the x-axis, theresulting solids have equal volume. Find a.


13. Find the volume of the solid obtained by revolv-ing about the x-axis the region bounded by thegraphs of f (x) = x3 and g (x) = x2.

yx

=1

x = π2

f x x

( ) =

sin ;

2

y x=

0 1 2 3 4

4

–4

5

f

y

x

Figure 9.6-1

a. Evaluate F (0), F (3), and F (5).

b. On what interval(s) is F decreasing?

c. At what value of t does F have a maximumvalue?

d. On what interval is F concave up?

2. Find the area of the region(s) enclosed by thecurve f (x) = x3, the x-axis, and the lines x = −1and x = 2.

3. Find the area of the region(s) enclosed by thecurve y = 2x − 6 , the x-axis, and the lines x = 0and x = 4.

4. Find the approximate area under the curve

from x = 1 to x = 5, using four

right-endpoint rectangles of equal lengths.

f xx

( ) =1

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1. (a)

(b) Since F is decreasing on the

interval (3,5].

(c) At t = 3, F has a maximum value.

(d) F ′(x) = f (x), F ′(x) is increasing on (4,5)which implies F ≤ (x) > 0. Thus F is concaveupwards on (4,5).


f t dt b

( ) ≤∫ 03

,

F f t dt f t dt f t dt 5

10 4 6

0

5

0

3

3

5

( ) = ( ) = ( ) + ( )

= + −( ) =

∫ ∫ ∫ .

F f t dt 31

23 2 4 10

0

3

( ) = ( ) = +( )( ) =∫ .

F f t dt 0 00

0

( ) = ( ) =∫ .


A x dx x dx

x x x x

= − −( ) + −( )= − +[ ] + −[ ] = −( ) + ( )[ ]

− + − ( )[ ] − − ( )[ ]

= + =

∫ ∫ 2 6 2 6

6 6 3 6 3

0 4 6 4 3 6 3

9 1 10

0

3

3

4

2

0

3 2

3

4 2

2 2

.


y

x

x = –1 x = 2

y = x 3

–1 20

Figure 9.8-1


Set 2x − 6 = 0; x = 3 and

f xx x

x x( ) =

− ≥

− −( ) <

2 6 3

2 6 3

if

if

A x dx x dx

x x

= +

=

+

= −

−( )+ −

= + =

−

−

∫ ∫ 3

1

03

0

2

4

1

0 4

0

2 44

4 4 0

1

4

2

4 0

1

44

17

4

0 43

y

x

y = 2 x –6

x = 0 x = 4

Figure 9.8-2

I IIIII IV

10 2 3 4 5

y f ( x ) =

1

x

Not to Scale

Figure 9.8-3

Length of

Area of Rect III = ( ) = ( ) =f x41

41

1

43∆

Area of Rect II = ( ) = ( ) =f x31

31

1

32∆

Area of Rect I = ( ) = ( ) =f x21

21

1

21∆

∆xt = −

=5 1

41

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A x

dx x

=

= −

= −

−

= −

−

= − −

=

∫ sin cos

cos cos

cos cos

22

2

22

22

22 4

2 02

22

22

π

π

π

π

π π

π π

sin sin sin

cos cos

xdx u du u du

u c x c

22 2

2 22

= ( ) =

= − + = − +

∫ ∫ ∫

Disc Method:

10.

Set ln k = 1. Thus eln k = e1 or k = e.


Area =1

x dx x k k

k k

1 11∫ = ] = − =ln ln ln ln

V x dx x x dx

x xx

= +( ) = + +( )

= + +

= + ( )

+ ( )

− =

∫ ∫ π π

π

π π

2 2

0

34 2

0

3

5 3

0

3

53

4 8 16

5

8

316

3

5

8 3

316 3 0

843

5.


0 x

y

π 2ππ

2

x

2

π2

x = x = π

f ( x ) = sin ( (

Figure 9.8-7

0 3

4

y y = x 2 + 4

Not to Scale

Figure 9.8-8

1

0

–1

y = 1

y = –1

x

x = y2 + 1

y

Figure 9.8-9

Disc Method:

Note: You can use the symmetry of the regionand find the volume by

2 12 2

0

1

π y dy+( )∫ .

V y dy y y dy

y yy

= +( ) = + +( )

= + +

= + ( )

+

− −( )

+ −( )

+ −( )

= +

− −

−

∫ ∫ π π

π

π

π

2 2

1

14 2

1

1

5 3

1

1

53

5 3

1 2 1

5

2

3

1

5

2 1

3

1

1

5

2 1

31

28

15

28

15

=

56

15

π.

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12. Volume of solid by revolving R:

You can verify your result by evaluating

The result is 96π.



π 32

0

2 22

3

x dx( )( )

∫ .

Set π π

π π

3192

2

3 96

32

32 2 2

2

0

3

3

13

23

x dx

a

a

a

a

( ) =

⇒ =

=

= ( ) = ( )

∫

.

V x dx x dx xR = ( ) = = [ ]

=

∫ ∫ π π π

π

3 9 3

192

2

0

42

0

43

0

4

.

Let s = a side of an equilateral triangle

Step 2. Area of a cross section:

Step 3.

Step 4.


Step 1. Washer Method

Enter

and obtain

3 4 2 2 2

32 3

3

( ) ∗ −( ) −( )∫ x x^ , , ,

.

V x dx

x dx

= −( )

= −( )

−

−

∫

∫

2 43

4

3 4

22

2

2

2

2

2

A x

s x

( ) = =

−( )22

2

3

4

2 4 3

4 .

s x= −2 4 2 .


y y = x 3

y = x 2

x 0 1

Figure 9.8-10


Points of intersection: Set x3 = x2 ⇒ x3 −x2 = 0 ⇒ x2 (x − 1) = 0 or x = 0 or x = 1.Outer radius = x2; Inner radius = x3.

Step 2.

Step 3. Enter and

obtain


Step 1. x2 + y2 = 4 ⇒ y2 = 4 − x2 ⇒

y x= ± −4 2

2

35

π.

π x x x^ ^ , , ,4 6 0 1−( )( )∫

V x x dx

x x dx

= ( ) − ( )( )= −( )

∫ ∫

ππ

2 2 3 2

0

1

4 6

0

1

y

x

0

2

2

–2

–2

Figure 9.8-11

y y = x – 2

x = y2

(4,2)

(1,–1)

x

–2

0

–1

Figure 9.8-12

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Points of Intersection: y = x − 2 ⇒x = y + 2

Set y2 = y + 2 ⇒ y2 − y − 2 = 0 ⇒ (y − 2)(y + 1) = 0 or y = −1 or y = 2.

Outer radius = y + 2 and innerradius = y2.

Step 2.

Step 3.


Enter

and obtain

π

π

y y y+( ) − −( )∫ 2 2 4 1 2

72

5

^ ^ , , ,

.

V y y dy= +( ) − ( )

( )−∫

π 22 2 2

1

2

.


0

(0,8) C

y

x

R1

2

Figure 9.8-13


y = 8, y = x3

Outer radius = 8; Inner radius = x3

Step 2.


Using the Washer Method:

Outer radius: x = 2 and Inner radius:

V y dy= − ( )

∫ π 22

13

2

0

8

x y=1

3

Enter

and obtain

π

π

8 2 6 0 2

7687

^ ^ , , ,

.

−( )∫ x x

V x dx= − ( )( )∫ π 82 3 2

0

2

Using your calculator, you obtain V =64

5

π.

8

y

B (2,8)

R2

0 A (2,0) x

Figure 9.8-14

R1

0

y

x 2

(0,8) C B (2,8)

Figure 9.8-15

Step 1. Disc Method:

Step 2.


Using the Disc Method:

Radius = − = −( )= −( )∫

2 2

2

13

13

2

0

8

x y

V y dyπ

Enter

and obtain

ππ

∗ −( )( )∫ 8 3 2 0 2

576

7

x x^ ^ , , ,

.

Radius = −( )

= −( )∫

8

8

3

3 2

0

2

x

V x dxπ

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y

x 0

B (2,8)

A (2,0)

R2

Figure 9.8-16

[−3,3] by [−1,7]

Figure 9.9-1

9

x

y

θ

Figure 9.9-2

Since is an even function, thusex2


Differentiate both sides:

when x = 15, 92 + y2 = 152 ⇒ y = 12

cos θ θd

dt x

dx

dt = ( ) −( )−9 2

sin θ =9

x

k e dx e dx kx

ax

a

= =∫ ∫ 22

2 2

0 0 and .

e dx e dxx

a

xa2 20

0−∫ ∫ = .

Using you calculator, you obtain

20.

xi = midpoint of the i th interval.

Length of

Area of RectI = f (2)∆x1 = (2.24)(4) = 8.96

Area of RectII = f (6)∆x2 = (3.16)(4) = 14.44

Area of RectIII = f (10)∆x3 = (4.58)(4) = 18.32

Total Area = 8.96 + 14.44 + 18.32 = 41.72

The area under the curve is approximately 41.72.

∆xi = −

=12 0

34

Area ≈ ( )=∑ f x xi i

i

∆1

3

V =16

5

π.

Thus


4

59

1

152

18

15

5

4

1

10

2

2

d

dt

d

dt

θ

θ

= −

−( )

= = = radian/sec.

cos ;θ = = = −12

15

4

52 ft/sec.

dx

dt


21. (See Figure 9.9-1.) e dx e dx e dxx

a

a x

a

xa2 2 20

0− −∫ ∫ ∫ = +

[−2,5] by [−2,6]

Figure 9.9-3

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[−3,3] by [−15,15]

Figure 9.9-4

Step 1. Distance Formula

where the domain of x is all real numbers.

Step 2. Enter

Enter

Step 3. Use the Zero Function and obtain x = 2for y2.

Step 4. Use the First Derivative Test. (SeeFigures 9.9-4 and 9.9-5.)

y d y x x2 1= ( )( ),

y x x1 4 2 5 2 1 2= −( ) + −( )( )^ . ^ ^

L x y

x x

= −( ) + −( )

= −( ) + −

4 1

42

1

2 2

22 2

At x = 2, L has a relative mimimum.

Since at x = 2, L has the only relativeextremum, it is an absolute minimum.

Step 5. At x = 2,

Thus the point on closest to

the point (4,1) is the point (2,2).

y x= ( )1

22

y x= ( ) = ( ) =1

2

1

22 22 2

24. (a) s(0) = 0 and

Thus

(b) v(2) = 2 cos (22 + 1) = 2 cos(5) = 0.567324

Since v(2) > 0, the particle is moving to theright at t = 2.

(c) a(t ) = v ′(t )

Enter d (x cos(x ^ 2 + 1), x) x = 2 andobtain 7.95506.

Thus, the velocity of the particle is increas-ing at t = 2, since a(2) > 0.


s t t

s

( ) = +( ) −

( ) = +( )

−

= − ≈ −

sin .

sin.

. .

2

2

12

0 420735

22 1

20 420735

0 900197 0 900

Since s c

c

c

0 00 1

20

841471

20

0 420735 0 421

2

( ) = ⇒ +( )

+ =

⇒ + =

⇒ = − ≈ −

sin

.

. .

s t t

c( ) = +( )

+sin

.2 1

2

Enter

and obtain

x x x

x

∗ +( )( )

+( )

∫ cos ^ ,

sin.

2 1

1

2

2

s t v t dt t t dt ( ) = ( ) = +( )∫ ∫ cos .2 1

– 0 +

decr. incr.2

rel. min.

y2 = ( (d L

dx

L

Figure 9.9-5

[−π,π] by [−1,2]

Figure 9.9-6

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(a) Point of Intersection: Use the IntersectionFunction of the calculation and obtain(0.517757, 0.868931)

Enter

and obtain .304261.

The area of the region is approximately0.304.

cos ^ , , , .x x e x x( ) − ∗ ( )( )∫ 0 51775

Area = −( )∫ cos.

x xe dxx

0

0 51775

(b) Step 1. Washer Method:

Outer radius = cos x and Inner radius = x ex

Step 2. Enter π cos ^x( )( ) −((∫ 2

V x xe dxx= ( ) − ( )[ ]∫ π cos. 2 2

0

0 51775

and

obtain 1.16678.

The volume of the solid is approximately1.167.

x e x x∗ ( )( ) ) )^ ^ , , .2 0 51775

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10.1 AVERAGE VALUE OF A FUNCTION

Main Concepts: Mean Value Theorem for Integrals, Average Value of a Function on [a,b]

Mean Value Theorem for Integrals

Mean Value Theorem for Integrals:

If f is continuous on [a,b], then there exists a number c in [a,b] such that

See Figure 10.1-1.f x dx f c b aa

b

( ) = ( ) −( )∫ .

Chapter 10

More Applications of Definite Integrals

275

a c b x 0

y

(c, f (c))

f ( x )

Figure 10.1-1


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Example 1

Given verify the hypotheses of the Mean Value Theorem for Integralsfor f on [1, 10] and find the value of c as indicated in the theorem.

The function f is continuous for x ≥ 1, thus:

Example 2

Given f (x) = x2, verify the hypotheses of the Mean Value Theorem for Integrals for f on[0, 6] and find the value of c as indicated in the theorem.

Since f is a polynomial, it is continuous everywhere, thus:

Since only is in the interval [0, 6], c = .2 32 3

x dx f c

xf c

f c f c c

c

2

0

6

3

0

6

2

6 0

36

72 6 12 12

12 2 3 2 3 3 4641

∫ = ( ) −( )

= ( )

= ( ) = ( ) =

= ± = ± ± ≈ ±( )

; ;

.

x dx f c

xf c

f c

f c f c c c

c

− = ( ) −( )

−( )

= ( )

−( ) −[ ] = ( )

= ( ) = ( ) = − = −

=

∫ 1 10 1

2 1

3 9

2

310 1 0 9

18 9 2 2 1 4 1

5

1

10

1

1032

32

; ; ;

f x x( ) = − 1,

• Remember: if f ′ is decreasing, then f ″ < 0 and the graph of f is concavedownward.

Average Value of a Function on [a,b ]

Average Value of a Function on an Interval:

If f is a continuous function on [a,b], then the Average Value of f on [a,b]

Example 1

Find the average value of y = sin x between x = 0 to x = π.

Average value =−

= −[ ] = − − − ( )( )[ ]

= +[ ] =

∫ 1

0

1 10

11 1

2

0

0

π

π π π

π π

π

π

sin

cos cos cos

.

xdx

x

=

−

( )∫ 1

b a

f x dxa

b

.

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Example 1

See Figure 10.2-1.

Displacement from to

Total Distance Traveled from to

1 2

1 2

t t v t dt s t s t

t t v t dt

t

t

t

t

= ( ) = ( ) − ( )

= ( )

∫

∫

2 11

2

1

2


0

10

–10

20

2 4 6 8 10 12

V (t )

V (t )

t

(seconds)

( f e e t / s e c )

Figure 10.2-1

The graph of the velocity function of a moving particle is shown in Figure 10.2-1.What is the total distance traveled by the particle during 0 ≤ t ≤ 12?

Example 2

The velocity function of a moving particle on a coordinate line is v(t ) = t 2 + 3t − 10 for0 ≤ t ≤ 6. Find (a) the displacement by the particle during 0 ≤ t ≤ 6, and (b) the total dis-tance traveled during 0 ≤ t ≤ 6.

Let or

if

if

t t t t t t

t t t t t

t t t

2

2

2

2

3 10 0 5 2 0 5 2

3 103 10 0 2

3 10 2

+ − = ⇒ +( ) −( ) = ⇒ = − =

+ − = − + −( ) ≤ ≤

+ − >

( )b Total Distance Traveled = ( )

= + −

∫

∫

v t dt

t t dt

t

t

1

2

2

0

6

3 10

( )

.

a Displacement = ( )

= + −( ) = + −

=

∫

∫

v t dt

t t dt t t

t

t

t

1

2

2

3 2

0

6

0

6

3 103

3

210

66

Total Distance Traveled = ( ) + ( )

= ( )( ) + ( )( ) = + =

∫ ∫ v t dt v t dt

feet

0

4

4

12

1

24 10

1

28 20 20 80 100 .

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Let −4t + 12 = 0 ⇒ t = 3.

Total distance traveled by the particle is 68.

Example 5

The velocity function of a moving particle on a coordinate line is v(t ) = 3 cos(2t ) for0 ≤ t ≤ 2π. Using a calculator.

(a) Determine when the particle is moving to the right.

(b) Determine when the particle stops.(c) The total distance traveled by the particle during 0 ≤ t ≤ 2π.

Solution:

(a) The particle is moving to the right when v(t ) > 0.

Enter y1 = 3 cos(2x). Obtain y1 = 0 when

The particle is moving to the right when:

(b) The particle stops when v(t ) = 0.

Thus the particle stops at

(c) Total distance traveled =

Enter and obtain 12.

The total distance traveled by the particle is 12.

10.3 DEFINITE INTEGRAL AS ACCUMULATED CHANGEMain Concepts: Business Problems, Temperature Problems, Leakage Problems,

Growth Problems

Business Problems

P(x) = R(x) − C(x) Profit = Revenue − Cost

R(x) = px Revenue = (price)(items sold)

abs x x3 2 0 2cos , , ,( )( )( )∫ π

3 20

2

cos t dt ( )∫ π

t = π π π π4

3

4

5

4

7

4, , .and

04

3

4

5

4

7

42< < < < < <t t t

π π π ππ, , .

t = π π π π4

3

4

5

4

7

4, , .and

− + = − + ≤ ≤

− − +( ) >

− + = − +( ) + − − +( )

= − +[ ] + −[ ]

= + =

∫ ∫ ∫

4 124 12 0 3

4 12 3

4 12 4 12 4 12

12 12 2 12

18 50 68

0

8

3

8

0

3

2

0

3 2

3

8

t t t

t t

t dt t dt t dt

t t t t

if

if

Total Distance Traveled = − +∫ 4 120

8

t dt


7/23/2019 007137714 x


P′(x) Marginal Profit

R′(x) Marginal Revenue

C′(x) Marginal Cost

P′(x), R′(x), C′(x) are the instantaneous rates of change of profit, revenue and costrespectively.

Example 1

The marginal profit of manufacturing and selling a certain drug is P′(x) = 100 − 0.005x.How much profit should the company expect if it sells 10,000 units of this drug?

P t P x dx

x dx

x x

t

( ) = ′( )

= −( )

= −

= ( ) − ( )

=

∫

∫

0

0

10 000

2

0

10 000

2

100 0 005

1000 005

2

100 10 0000 005

2

10 000

750 000

.

.

,.

,

, .

,

,

More Applications of Definite Integrals • 281

• If f ″ (a) = 0, f may or may not have a point of inflection at x = a, e.g., as in thefunction f (x) = x4, f ″ (0) = 0, but at x = 0, f has an absolute minimum.

Example 2

If the marginal cost of producing x units of a commodity is C′(x) = 5 + 0.4x.

Find (a) the marginal cost when x = 50.

(b) the cost of producing the first 100 units.

Solution:

(a) marginal cost at x = 50:

C′(50) = 5 + 0.4(50) = 5 + 20 = 25.

(b) cost of producing 100 units:

Temperature Problems

Example

On a certain day, the changes in the temperature in a greenhouse beginning at 12 noon

is represented by degree Fahrenheit, where t is the number of hoursf t t

( ) =

sin

2

C t C x dx

x dx

x x

t

( ) = ′( )

= +( )

= − ]= ( ) + ( )( ) − =

∫

∫

0

0

100

2

0

100

2

5 0 4

5 0 2

5 100 0 2 100 0 2500

.

.

. .

7/23/2019 007137714 x


elapsed after 12 noon. If at 12 noon, the temperature is 95°F, find the temperature inthe greenhouse at 5 p.m.

Let F (t ) represent the temperature of the greenhouse.

The temperature in the greenhouse at 5 p.m. is 98.602°F.

Leakage Problems

Example

Water is leaking from a faucet at the rate of l (t ) = 10e−0.5t gallons/per hour, where t ismeasured in hours. How many gallons of water will have leaked from the faucet aftera 24 hour period?

Let L(x) represent the number of gallons that have leaked after x hours.

Using your calculator, enter and obtain 19.9999. Thus, the

number of gallons of water that have leaked after x hours is approximately 20 gallons.

10 0 5 0 24e x x^ . , , ,−( )( )∫

L x l t dt e dt t x

( ) = ( ) = −∫ ∫ 10 0 5

0

2

0

. .

F

F t f x dx

F x dx

x

t

0 95

95

5 952

95 22

95 25

22 0

95 3 602 98 602

0

0

5

0

5

( ) = °

( ) = + ( )

( ) = +

= + −

= + −

− − ( )( )

= + =

∫

∫

F

sin

cos cos cos

. .


• You are permitted to use the following 4 built-in capabilities of your calculatorto obtain an answer: plotting the graph of a function, finding the zeros of a func-tion, finding the numerical derivative of a function, and evaluating a definiteintegral. All other capabilities of your calculator can only be used to check youranswer. For example, you may not use the built-in Inflection function of yourcalculator to find points of inflection. You must use calculus using derivativesand showing change of concavity.

Growth Problems

Example

In a farm, the animal population is increasing at a rate which can be approximately rep-resented by g (t ) = 20 + 50 ln(2 + t ), where t is measured in years. How much will theanimal population increase to the nearest tens between the 3rd and 5th year?

Let G(x) be the increase in animal population after x years.

G x g t dt x

( ) = ( )∫ 0

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Thus, the population increase between the 3rd and 5th years


Thus the animal population will increase by approximately 220 between the 3rd and5th years.

10.4 DIFFERENTIAL EQUATIONS

Main Concepts: Exponential Growth/Decay Problems, Separable Differential Equations

Exponential Growth/Decay Problems

1. If then the rate of change of y is proportional to y.

2. If y is a differentiable function of t with y > 0 and then y(t ) = y0ekt ; where

y0 is the initial value of y and k is constant. It k > 0, then k is a growth constant andif k < 0, then k is the decay constant.

Example 1—Population Growth

If the amount of bacteria in a culture at any time increases at a rate proportional to theamount of bacteria present and there are 500 bacteria after one day and 800 bacteriaafter the third day:

(a) approximately how many bacteria are there initially, and

(b) approximately how many bacteria are there after 4 days?

Solution:

(a) Since the rate of increase is proportional to the amount of bacteria present,

Then:

where y is the amount of bacteria at any time

Therefore this is an exponential growth/decay model: y(t ) = y0ekt

Step 1. y(1) = 500 and y(3) = 800

500 = y0e1k and 800 = y0e

3k

Step 2. 500 = y0ek ⇒ y0 = = 500 e−k500

ek

dy

dt ky=

dy

dt ky= ,

dy

dt ky= ,

20 50 2 3 5+ +( )( )∫ ln , , ,x x

= ( ) − ( )

= + +( ) − + +( )( )( )= + +( )[ ]

∫ ∫

∫

G G

t dt t dt

t dt

5 3

20 50 2 20 50 2

20 50 2

0

3

0

5

3

5

ln ln

ln


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Example 2

Find a solution of the differentiation equation y(0) = −1.

Step 1. Separate variables: dy = x sin(x2) dx

Step 2. Integrate both sides:

Let u = x2; du = 2x dx or

Thus:

Step 3. Substitute given condition:

Thus,

Step 4. Verify result by differentiating:

Example 3

If and at x = 0, y′ = −1 and y = 3, find a solution of the differential equation.

Step 1. Rewrite:

Step 2. Separate variables: dy′ = (2x + 1) dx.


Step 4. Substitute given condition: At x = 0, y′ = −1; −1 = 0 + 0 + c1 ⇒ c1 = −1.

Thus y′ = x2 + x − 1.

Step 5. Rewrite:

Step 6. Separate variables: dy = (x2 + x − 1) dx.


y x x

x c= + − +3 2

23 2

.

dy x x dx= + −( )∫ ∫ 2 1

′ = = + −y dy

dx

dy

dx x x; .2 1

dy x dx y x x c′ = +( ) ′ = + +∫ ∫ 2 1 21; .

d y

dx

dy

dx

dy

dx x

2

22 1as

′ ′= +; .

d y

dx x

2

22 1= +

dy

dx x x x x= ( )[ ] ( ) = ( )

1

222 2sin sin .

y x= − ( ) −1

21

22cos .

y c c c0 1 11

20 1

1

2

1

2( ) = − − = − ( ) + − =

−+ − =; cos ; ;

y x c= − ( ) +1

22

cos .

x x dx u du

u du u c

x c

sin sin sin cos

cos

2

2

2

1

2

1

2

1

2

( ) =

= = − +

= − ( ) +

∫ ∫ ∫

dux dx

2 =

dy x x dx dy y∫ ∫ ∫ = ( ) =sin ;2

dy

dx x x= ( )sin ;2


• To get your AP grade, you can call 888-308-0013 in July. There is a charge of about $13.

7/23/2019 007137714 x


7/23/2019 007137714 x


10.5 RAPID REVIEW

1. Find the average value of y = sin x on [0, π].

Answer:

2. Find the total distance traveled by a particle during 0 ≤ t ≤ 3 whose velocity func-tion is shown in Figure 10.5-1.

Average value =−

= −[ ] =

∫ 1

0

1 2

0

0

π

π π

π

π

sin

cos

x dx

x


V

t

V (t )

321

–1

1

2

Figure 10.5-1

Answer:

3. If oil is leaking from a tank at the rate of f (t ) = 5e−0.1t gallons/hour, where t is mea-sured in hours. Write an integral to find the total number of gallons of oil that willhave leaked from the tank after 10 hours. Do not evaluate the integral.

Answer: Total number of gallons leaked =

4. How much money should Mary invest at 7.5% interest a year compounded contin-uously, so that Mary will have $100,000 after 20 years.

Answer: y(t ) = y0ekt , k = 0.075 and t = 20. y(20) = 100,000 = y0e

(0.075)(20). Thus youobtain y0 ≈ 22313, using a calculator.

5. Given and y(1) = 0, solve the differential equation.

Answer:

Thusy x

y x2 2

2 2

2 2

1

21= − = −or .

ydy xdx ydy y x

c c c= ⇒ ⇒ = + ⇒ = + ⇒ = −∫ 2 2

2 20

1

2

1

2.

dy

dx

x

y=

50 1

0

10

e dt t −

∫ .

The Total Distance Traveled = ( ) + ( )

= + =

∫ ∫ v t dt v t dt 0

2

2

3

2 0 5 2 5. .


1. Find the value of c as stated in the Mean Value

Theorem for Integrals for f (x) = x3

on [2,4].


2. The graph of f is shown in Figure 10.6-1. Findthe average value of f on [0,8].

3. The position function of a particle moving on a

coordinate line is given as s(t ) = t 2

− 6t − 7, 0 ≤ t

7/23/2019 007137714 x


≤ 10. Find the displacement and total distancetraveled by the particle from 1 ≤ t ≤ 4.

4. The velocity function of a moving particle on

a coordinate line is v(t ) = 2t + 1 for 0 ≤ t ≤ 8.At t = 1, its position is −4. Find the position of the particle at t = 5.

5. The rate of depreciation for a new piece of equip-ment at a factory is given as p(t ) = 50t − 600 for0 ≤ t ≤ 10, where t is measured in years. Find thetotal loss of value of the equipment over the first5 years.

6. If the acceleration of a moving particle on acoordinate line is a(t ) = −2 for 0 ≤ t ≤ 4, and the

initial velocity v0 = 10. Find the total distancetraveled by the particle during 0 ≤ t ≤ 4.

7. The graph of the velocity function of a movingparticle is shown in Figure 10.6-2. What is thetotal distance traveled by the particle during0 ≤ t ≤ 12?

9. The change of temperature of a cup of coffeemeasured in degrees Fahrenheit in a certainroom is represented by the function

for 0 ≤ t ≤ 5, where t is

measured in minutes. If the temperature of thecoffee is initially 92°F , find its temperature afterthe first 5 minutes.

10. If the half-life of a radioactive element is4500 years, and initially there are 100 grams of this element, approximately how many gramsare left after 5000 year?

11. Find a solution of the differential equation:

12. If and at x = 0, y ′ = −2 and y = 1,

find a solution of the differential equation.


13. Find the average value of y = tan x from

14. The acceleration function of a moving particleon a straight line is given by a(t ) = 3e2t , t is mea-

sured in seconds, and the initial velocity is .Find the displacement and total distance traveledby the particle in the first 3 seconds.

15. The sales of an item in a company follows anexponential growth/decay model, where t ismeasured in months. If the sales drops from5000 units in the first months of 4000 units inthe third month, how many units should thecompany expect to sell during the seventhmonth?

16. Find an equation of the curve that has a slopefor at the point (x, y) and passes through

the point (0,4).

17. The population in a city is approximately750,000 in 1980, and growing at a rate of 3%per year. If the population growth follows anexponential growth model, find the city’s popu-lation in the year 2002.

18. Find a solution of the differential equation

4ey

= y′ −3xey

and y(0) = 0.

2

1

y

x +

1

2

x x= =π π4 3

to .

d y

dx

x2

25= −

dy

dx x x y= ( ) ( ) =cos ;2 0 π

f t t

( ) = −

cos

4


(4,4)

1

0 1 2 3 4 5 6 7 8

2

3

4

x

y

f

Figure 10.6-1

V (t )

V

t

20

10

–10

0 1 2 3 4 5 6 7 8 9 10 11 12 (meters)

Figure 10.6-2

8. If oil is leaking from a tanker at the rate off (t ) = 10e0.2t dt gallons per hour where t ismeasured in hours, how many gallons of oilwill have leaked from the tanker after the first

3 hours?

7/23/2019 007137714 x


19. How much money should a person invest at6.25% interest compounded continuouslyso that the person will have $50,000 after10 years?

20. The velocity function of a moving particle isgiven as v(t ) = 2 − 6e−t , t ≥ 0 and t is measured inseconds. Find the total distance traveled by theparticle during the first 10 seconds.



(“Calculator” indicates that calculators are permitted)

21. If 3ey = x2y, find

22. Evaluate

23. The graph of a continuous function f whichconsists of three line segments on [−2, 4] is

shown in Figure 10.7-1. If for−2 ≤ x ≤ 4,

F x f t dt

x

( ) = ( )−∫ 2

x

x dx

2

30

1

1+∫ .

dy

dx.

(a) find F (−2) and F (0).

(b) find F ′(0) and F ′(2).

(c) find the value of x such that F has a maxi-mum on [−2,4].

(d) on which interval is the graph of F concaveupward.

24. (Calculator) The slope of a function y = f (x) at

any point (x, y) is and f (0) = 2.

(a) Write an equation of the line tangent to thegraph of f at x = 0.

(b) Use the tangent in part (a) to find theapproximate value of f (0.1).

(c) Find a solution y = f (x) for the differentialequation.

(d) Use the result in part (c), find f (0.1).

25. (Calculator) Let R be the region in the first quad-

rant bounded by f (x) = ex

− 1 and g (x) = 3 sin x.(a) Find the area of region R.

(b) Find the volume of the solid obtained byrevolving R about the x-axis.

(c) Find the volume of the solid having R as itsbase and its cross sections are semicirclesperpendicular to the x-axis.

y

x2 1+

7

6

5

4

3

2

1

0–1 1 2 3 4 5–2

y

f

t

Figure 10.7-1



1. x dx f c

x dx x

f c f c

c c

3

2

4

34

2

4 2 4

2

4

3

4 2

4

4

4

2

460

2 60 30

30 301

3

= ( ) −( )

=

=

−

=

( ) = ⇒ ( ) =

= ⇒ = ( )

∫

∫

.

2.

3. Displacement = s(4) − s(1) = −15 − (−12) = −3.

v(t ) = s′(t ) = 2t − 6

Distance Traveled = ( )∫ v t dt 1

4

Average Value =−

( )

= ( )( )

=

∫ 1

8 0

1

8

1

28 4 2

0

8

f x dx

.

7/23/2019 007137714 x


Set 2t − 6 = 0 ⇒ t = 3

4.

5.

6. v(t ) = ∫ a(t ) dt = ∫ − 2 dt = −2t + c

v0 = 10 ⇒ −2(0) + c = 10 or c = 10

v(t ) = −2t + 10.

Set v(t ) = 0 ⇒ −2t + 10 = 0 or t = 5.

−2t + 10 = −2t + 10 if 0 ≤ t < 5

7.

8. Total Leakage

gallons

= = ]

= −

= ≈

∫ 10 50

91 1059 50

41 1059 41

0 2 0 2

0

3

0

3

e dt et t . .

.

.

Total Distance Traveled

meters

= ( ) + ( )

= ( )( ) + ( )( )

=

∫ ∫ v t dt v t 0

8

8

12

1

28 10

1

24 10

60 .

v t dt t dt

t t

( ) = − +( )

= − + ] =

∫ ∫ 0

4

0

4

2

0

4

2 10

10 24.

Distance Traveled = ( )∫ v t dt .0

4

Total Loss = ( ) = −( )

= − ] =

∫ ∫ p t dt t dt

t t

50 600

25 600 2375

0

5

0

5

2

0

5

.

Position Function

or

s t v t dt

t dt

t t c

s c

c

s t t t

s

( ) = ( )

= +( )

= + +

( ) = − ⇒ ( ) + +

= − = −

( ) = + −

( ) = + − =

∫ ∫ 2 1

1 4 1 1

4 6

6

5 5 5 6 24

2

2

2

2

.

.

v t dt t dt t dt

t t t t

( ) = − −( ) + −( )

= − +[ ] + −[ ]

= + =

∫ ∫ ∫ 1

4

3

4

1

3

2

1

3 2

3

4

2 6 2 6

6 6

4 1 5.

2 62 6 0 3

2 6 3 10t

t t

t t − =

− −( ) ≤ <

− ≤ ≤

if

if


9. Total change in temperature

Thus the temperature of coffee after 5 minutes is(92 − 3.79594) ≈ 88.204°F

10. y(t ) = y0ekt

Take ln of both sides:

There are approximately 46.29 grams left.

11. Step 1. Separate variables: dy = x cos(x2) dx

Step 2. Integrate both sides: ∫ dy = ∫ x cos(x2) dx.

Step 3. Substitute given values.

Step 4. Verify result by differentiating

dy

dx

x xx x=

( )( )= ( )

coscos .

2

22

2

y c c

y x

00

2

2

2

( ) = ( )

+ = ⇒ =

= ( )

+

sin

sin.

π π

π

dy y

x x dx u x

du x dx du

xdx

x x dx u du

uc

xc

y x

c

=

( ) = =

= =

( ) =

= + = ( )

+

= ( )

+

∫

∫ ∫

cos ;

,

cos cos

sin sin.

sin

2 2

2

2

2

22

2

2 2

2

Let

Thus

y t e( ) = = ( ) ≈−

( )100 25 2 46 29

2

45005000 8

9

ln

. .

ln ln

ln

ln

1

2

2 4500

2

4500

4500

= ⇒

− =

= −

e

k

k

k

or

Half-life years= ⇒ =45001

24500e k

= −

= −

= − −

= − °

∫ cos

sin

.

.

t dt

t

4

44

3 79594 0

3 79594

0

5

0

5

F.

7/23/2019 007137714 x


12. Step 1. Rewrite

Step 2. Separate variables: dy′ = (x − 5) dx

Step 3. Integrate both sides: ∫ dy′ = ∫ (x − 5) dx

Step 4. Substitute given values:

Step 5. Rewrite:

Step 6. Separate variables:


Step 8. Substitute given values:

At x = 0, y = 0 − 0 − 0 + c2 = 1 ⇒ c2 = 1



13.

Enter

and obtain

=−

( )

( )≈

∫ 1

3 44 3

6 21 32381

π π π π

π

tan , ,

ln. .

x x

Average Value =−

∫ 1

3 44

3

π π π

π

tan x dx

dy

dx

xx

d y

dx x

= − −

= −

2

2

2

25 2

5.

y x x

x= − − +3 2

6

5

22 1.

y x x

x c= − − +3 2

26

5

22 .

dy x

x dx= − −

∫ ∫

2

25 2

dy x

x dx= − −

2

25 2 .

′ = = − −y

dy

dx

dy

dx

x

x; .

2

2 5 2

At x y

c c

y x

x

= ′ = − ( )

+ = − ⇒ = −

′ = − −

00

25 0

2 2

25 2

1 1

2

,

′ = − +y x x c2

12

5

dy

dx x

′= − 5

d y

dx

dy

dx

2

2as

′


14.

15. Step 1. y(t ) = y0ekt

y(1) = 5000 ⇒ 5000 = y0ek ⇒ y0

= 5000e−k

y(3) = 4000 ⇒ 4000 = y0e3k

Substituting y0 = 5000e−k, 4000= (5000e−k)e3k

Step 2. 5000 = y0e−0.111572

y0 = (5000)e0.111572 ≈ 5590.17

y(t ) = (5590.17)e−0.111572t

Step 3. y(7) = (5590.17)e−0.111572(7) ≈ 2560

Thus sales for the 7th month is approxi-mately 2560 units.

4000 5000

4

5

4

52

1

2

4

50 111572

2

2

2

=

=

= ( ) =

=

≈ −

e

e

e k

k

k

k

kln ln

ln . .

Since for

0

3

3

21 0 0

3

21 298 822

2

2

0

3

e t

v t dt e dt

t

t

− > ≥

( ) = −

=∫ ∫

,

. .

Displacement

Enter

and obtain

Distance Traveled

= −

( ) −( )

= ( )

∫

∫

∫

3

21

32

2 1 0 3

298 822

2

0

3

0

3

e dt

e x x

v t dt

t

^ , , ,

. .

v t a t dt

e e c

v e c

c c

v t e

t t

t

( ) = ( )

= = +

( ) = + =

⇒ + = = −

( ) = −

∫ 33

2

03

2

1

2

32

12

1

3

21

2 2

0

2

or

7/23/2019 007137714 x


16. Step 1. Separate variables:


Step 3. Substitute given value (0,4):

Step 4. Verify result by differentiating:

17. y(t ) = y0ekt

y0 = 750,000

y e

E

22 750 000

1 45109 6 1 451 090 89

1 451 094

0 03 22

( ) = ( )

≈ ≈

( )( ),

. , ,

, ,

.

using an TI-

using an TI-85.

dy

dx x x

dy

dx

y

x

x

x

x

= ( ) +( ) = +( )

=+

=

+( )( )+( )

= +( )

4 2 1 8 1

2

1

2 4 1

1

8 1

2

Compare with

.

1

24 1

2

1

21 2

12

12

2 1

2 1

4 1

12

12

12

12

1 2

2 2

2

ln ln

ln

ln ln ln

ln ln

.

lnln

( ) = ( ) +

=

− + =

+ =

=

+ =

= +( )

= ( ) +( )

= +( )

+

c

c

y x

y

x

e e

y

x

y x

y x

y x

y

x

dy

y

dx

x

y x c

2 1

1

21

=+

= + +

∫ ∫

ln ln .

dy

dx

y

x

dy

y

dx

x

=+

=+

2

1

2 1


18. Step 1. Separate variables:


Step 3. Substitute given value: y(0) = 0 ⇒ e0 = 0− 0 + c ⇒ c = 1

Step 4. Take ln of both sides:

Step 5. Verify result by differentiating:Enter d (−ln(1 − 4x − 3(x^2)/2), x) and

obtain

Which is equivalent to ey(4 + 3x).

19. y(t ) = y0ekt

k = 0.0625, y(10) = 50,000

50,000 = y0e10(0.0625)

20. Set v(t ) = 2 − 6e−t = 0. Using the Zero Functionon your calculator, compute t = 1.09861.

2 62 6 0 1 09861

2 6 1 09861− =

− −( ) ≤ <

− ≥

−

−

−e

e t

e t

t

t

t

if

if

.

.

Distance Traveled = ( ) =∫ v t dt x

100

ye

0 0 625

50 000

26763 1

26763 071426 26763 07= ≈ ≈

,

$ .

$ . $ ..

using an TI-89

using an TI-85.

− +( )+ −

2 3 4

3 8 22

x

x x,

e x

x

e x

x

y x x

y

y

−

−

= − − +

( ) = − − +

= − − −

3

24 1

3

24 1

1 43

2

2

2

2

ln ln

ln .

4 3

43

2

3

24

2

2

+( ) =

+ = − +

= − − +

−

−

−

x dx e dy

x x

e c

e x

x c

y

y

yswitch sides:

4 3

4 3

4 3

4 3

e dy

dx xe

e xe dy

dx

e x dy

dx

x dx dy

e e dy

y y

y y

y

y

y

= −

+ =

+( ) =

+( ) = = −

7/23/2019 007137714 x


2 6 2 6

2 6

1 80278 15 803

0

10

0

1 09861

1 09861

10

− = − −( )

+ −( )

= +

− −

−

∫ ∫

∫

e dt e dt

e dt

t t

t

.

.

. .


Alternatively, use the nInt Function on thecalculator.

Enter nInt(abs(2 − 6e^(−x)), x, 0, 10) andobtain the same result.


21. 3ey = x2y

22. Let u = x3 + 1; du = 3x2 dx or

23. (a)

(b) F ′(x) = f (x); F ′(0) = 2 and F ′(2) = 4

(c) Since f > 0 on [−2,4], F has a maximum

value at x = 4.

(d) The function f is increasing on (1,3) whichimplies that f ′ > 0 on (1,3).

Thus, F is concave upward on (1,3). (Note:f ′ is equivalent to the 2nd derivative of F.)

24. (a)

dy

dx m x

x =

=( ) +

= ⇒ = =0

2

2 0 12 2 0at

dy

dx

y

x f =

+ ( ) =

2 10 2;

F f t dt

F f t dt

−( ) = ( ) =

( ) = ( ) = +( ) =

−

−

−

∫

∫

2 0

01

24 2 2 6

2

2

2

0

x

x dx

u

duu

c x c

x

x dx x

2

3

3

2

30

13

0

1

1

1

3

1

31

31

1

1

31

1

32 1

2

3

+ = =

+ = + +

+ = +

= −( ) =

∫ ∫

∫

ln

ln

ln

ln lnln

.

dux dx

32−

3 2

3 2

3 2

23

2

2

2

2

e dy

dx xy

dy

dx x

e dy

dx

dy

dx x xy

dy

dx e x xy

dydx xye x

y

y

y

y

= + ( )

− =

−( ) =

= −

y − y1 = m(x − x1)

y − 2 = 2(x − 0) ⇒ y = 2x + 2

The equation of the tangent to f at x = 0 isy = 2x + 2.

(b) f (0.1) = 2(0.1) + 2 = 2.2

(c) Solve the differential equation:

Step 1. Separate variables

Step 2. Integrate both sides

Step 3. Substitute given values (0,2)


y x= +( )2 2 11

2

ln ln ln

ln ln

ln ln

ln ln

.

ln

ln

21

21 2

1

22 1 2

1

22 1 2

2 12

2 12

2 2 1

12

12

12

12

2 1 2

= + ⇒ =

= + +

− + =

+( )=

=

+( )=

= +( )

+( )

c c

y x

y x

y

x

e e

y

x

y x

y

x

dy

y

dx

x

y x c

∫ ∫ =+

= + +

2 1

1

22 1ln ln

dy

y

dx

x=

+2 1

dy

dx

y

x=

+2 1

7/23/2019 007137714 x


Compare this with:

Thus the function is

(d)


f x x

f

( ) = +( )

( ) = ( ) +( ) = ( ) ≈

2 2 1

0 1 2 2 0 1 1 2 1 2 2 19089

12

12

12. . . .

y f x x= ( ) = +( )2 2 11

2 .

dydx

yx

xx

x

= + = +( )+

=+

2 12 2 1

2 1

2

2 1

12

.

dy

dx x

x

=

+( ) ( )

=+

−2

1

22 1 2

2

2 1

12

.


(a) Intersection points: Using the IntersectionFunction on the calculator, you have x = 0and x = 1.37131.

Area of R

Enter ∫ (3 sin (x)) − (e^(x) − 1), x, 0,1.37131) and obtain 0.836303.

The area of region R is approximately 0.836.(b) Using the Washer Method, volume of

R =

Enter π ∫ ((3 sin(x))^2 − (e^(x) − 1)^2, x, 0,1.37131) and obtain 2.54273π or 7.98824.

The volume of region R is 7.988.

(c) Volume of Solid = (Area of CrossSection) dx

Area of Cross Section =

Enter ∫ ((3 sin(x) − (e^(x) − 1))^2,

x, 0, 1.37131) and obtain 0.077184π or0.24248.

The volume of the solid is approximately0.077184π or 0.242.

π2

1

4

= − −( )( )

1

2

1

23 1

2

π sin x ex

1

22

πr

π0

1 37131.

∫

π 3 12 2

0

1 37131

sin.

x e dxx( ) − −( )[ ]∫

= − −( )[ ]∫ 3 10

1 37131

sin.

x e dxx

Figure 10.9-1

[−π,π] by [−4,4]

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7/23/2019 007137714 x


PRACTICE MAKES PERFECT

PART IV


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7/23/2019 007137714 x


PRACTICE EXAM 1

Answer Sheet for Practice Exam 1—Section I

Practice Exam 1 • 299

Part A

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

Part B

76.

77.

78.

79.

80.

81.

82.

83.

84.

85.

86.

87.

88.

89.

90.

91.

92.


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7/23/2019 007137714 x


Section I—Part A


Directions:

Use the answer sheet provided in the previous page. All questions are given equal weight. There is no penaltyfor unanswered questions. However, 1 ⁄ 4 of the number of the incorrect answers will be subtracted fromthe number of correct answers. Unless otherwise indicated, the domain of a function f is the set of all

real numbers. The use of a calculator is not permitted in this part of the exam.

1. The is

(A) −1 (B) 0 (C) 1(D) 2 (E) nonexistent

2.

(A) cos x (B) −sin x (C) sin x − 1(D) sin x + 1 (E) −sin x + 1

3. The radius of a sphere is increasing at a constantof 2 cm/sec. At the instant when the volume of

the sphere is increasing at 32π cm3 /sec, thesurface area of the sphere is

(A) 8π (B) (C) 16π

(D) 64π (E)

4. Given the equation , what is

the instantaneous rate of change of A withrespect to s at s = 1?

(A) (B) (C)

(D) (E)

5. What is the g (x), if g (x)

=e x

e x

x

x

if 2

if

>

− ≤

ln

ln?

4 2

limlnx→ 2

10 34 3

5

232 32 3 5+

A s= −( )3

45 1

2

256

3

π

32

3

π

cos t dt x

π2

∫

limx

x

x→−∞

−+

2 1

1 2

(A) −2 (B) ln 2 (C) e2

(D) 2 (E) nonexistent

6. The graph of f ′ is shown in Figure 1T-1.


28 55 Minutes No

x

f '

a b0

Figure 1T-1

A possible graph of f is (see Figure 1T-2):

7. If g (x) = −2 x + 3 , what is the

(A) −6 (B) −2 (C) 2(D) 6 (E) nonexistent

8. What is

(A) (B) 0 (C)

(D) (E) nonexistent3

2

1

2−

1

2

lim

sin sin

?∆

∆

∆x

x

x→

−

0

3 3

π π

lim ?x

g x→− −

′( )3

7/23/2019 007137714 x


I. f ′(0) = 0II. f has an absolute maximum value on [a, b]III. f ″ < 0 on (0, b)

(A) III only(B) I and II only(C) II and III only(D) I and II only

(E) I, II and III

12.

(A)

(B)

(C)

(D)

(E) 0

13. The graph of f is shown in Figure 1T-4 and f is twice differentiable. Which of the followinghas the smallest value?

I. f (−1)II. f ′(−1)III. f ″ (−1)

(A) I (B) II (C) III(D) I and II (E) II and III

x x c+ +2

3

32

22

3

32x x c+ +

xx c

2

3

2

32+ +

22

2

x x

c+ +

1 +=∫

x

xdx

302 • Practice Makes Perfect

9. If f (x) is an antiderivative of xe−x 2and f (0) = 1,

then f (1) =

(A) (B) (C)1

2

1

2e −

1

2

3

2e −

1

e

a b

y

x

(A)

a 0 b

y

x

(B)

a 0 b

y

x

(C)

Figure 1T-2

y

x a 0 b

f

a b0

y

x

(D)

a 0 b

y

x

(E)

Figure 1T-3

(D) (E)

10. If g (x) = 3 tan2 (2x), then is

(A) 6 (B) (C) 12

(D) (E) 24

11. The graph of the function f is shown in Figure 1T-3,which of the following statements are true?

12 2

6 2

′

g

π8

− +1

2

1

2e− +

1

2

3

2e

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14. If and at x = 0, a solution

to the differential equation is

(A) (B) (C)

(D) (E)

15. The graph of the velocity function of a movingparticle is shown in Figure 1T-5 What is the totaldisplacement of the particle during 0 ≤ t ≤ 20?

3

252e x +

3

222e x +

32

12e x +3 12

2e x +3 12

2e x −

y =5

2,

dy

dx e x= 3 2 ,

x −1 0 1

f (x) 2 b −2

x

f

0–1

Figure 1T-4

40

20

–20

0 5 10 15 20

–40

t (seconds)

V (t )

V (t )

m e t e r s / s e c .

Figure 1T-5

(A) 20m (B) 50m (C) 100m(D) 250m (E) 500m

16. The position function of a moving particle is

for 0 ≤ t ≤ 4. What is

the maximum velocity of the particle on theinterval 0 ≤ t ≤ 4?

(A) (B) 1 (C) (D) 4 (E) 5

17. If and k > 0, the value(s) of k are2 18x dxk

k

−∫ =

14

6

1

2

s t t t

t ( ) = − + −3 2

6 23

(A) −3 (B) (C) 3

(D) (E) 9

18. A function f is continuous on [−1,1] and someof the values of f are shown below:

3 2

−3 2

If f (x) = 0 has only one solution, r, and r < 0,then a possible value of b is

(A) 3 (B) 2 (C) 1 (D) 0 (E) −1

19.

(A) (B) 3 (C) 4

(D) (E) 2e2 − 1

20. The area of the region enclosed by the graph of

and the x-axis is

(A) 36 (B) (C) 9π

(D) 18π (E) 36π

21. If a function f is continuous for all values of x,and a > 0 and b > 0, which of the followingintegrals always have the same value?

I.

II.

III.

(A) I and II only(B) I and III only(C) II and III only

(D) I, II, and III(E) none

22. What is the average value of the function

y = 2 sin(2x) on the interval

(A) (B) (C) (D) (E) 6π

23. Given the equation what is an

equation of the tangent line to the graph at x = π?

y x

=

3

22sin ,

3

2π3

π1

2−

3

π

06

, ?π

f x b dxb

a b

+( )+

∫

f x b dxb

a b

−( )+

∫

f x dxa

( )∫ 0

9

2

π

y x= −9 2

e2 1

2−

3

2

e dxx2

0

2

=∫ ln

7/23/2019 007137714 x



(A) y = 3(B) y = π(C) y = π + 3(D) y = x − π + 3(E) y = 3 (x − π) + 3

24. The position function of a moving particle onthe x-axis is given as s(t ) = t 3 + t 2 − 8t for

0 ≤ t ≤ 10. For what values of t is the particlemoving to the right?

(A) t < −2 (B) t > 0 (C) t <

(D) 0 < t < (E) t >

25. See Figure 1T-6.

4

3

4

3

4

3

y

x 0

(A) y

x 0

(B)

y

x 0

(D) y

x 0

(E)

y

x 0

(C)

2

1

0–1 1 2 3

–1

–2

f

y

x

Figure 1T-6

The graph of f consist of two semicircles, for−1 ≤ x ≤ 3 as shown in Figure 1T-6. What is the

value of

(A) 0 (B) π (C) 2π (D) 4π (E) 8π

26. If then f ′(2) is

(A) (B) 54 − (C) 54

(D) (E) 135

27. If for all positive

values of k, then which of the following couldbe the graph of f? (See Figure 1T-7.)

28. If h′(x) = k(x) and k is a continuous function for

all real values of x, then is

(A) h(5) − h(−5)

(B) 5h(5) + 5h(−5)

(C) 5h(5) − 5h(−5)

(D)

(E) 1

55

1

55h h( ) − −( )

1

55

1

55h h( ) + −( )

k x dx51

( )−∫ 1

f x dx f x dxk

k

k( ) = ( )

− −∫ ∫ 20

13513 2

2−

23

223

2

f x t t dt x

( ) = +( )∫ 3

11

32 ,

f x dx( )−∫ 13

?

Figure 1T-7

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17 50 Minutes Yes

Section I—Part B

76. If on [0, 2π], then f has a

local maximum at x =

(A) 0 (B) (C) π

(D) (E) 2π

77. The equation of the normal line to the graph

y = e2x at the point where is

78. The graph of f ′, the derivative of f, is shown inFigure 1T-8. At which value of x does the graphof f have a point of inflection?

(A) 0 (B) x1 (C) x2 (D) x3 (E) x4

79. The temperature of a metal is dropping at therate of g (t ) = 10e−0.1t for 0 ≤ t ≤ 10 where g ismeasured in degrees in Fahrenheit and t in

(A)

(B)

(C)

(D)

(E)

y x

y x

y x

y x

y x

= − −

= − +

= +

= − −

+

= −

+

1

21

1

21

2 1

1

2

2

22

22

22

ln

ln

dy

dx = 2

3

2

π

π2

f x t dt x

( ) = −∫ cos0

minutes. If the metal is initially 100°F , what isthe temperature to the nearest degree Fahrenheitafter 6 minutes?

(A) 37 (B) 45 (C) 55 (D) 63 (E) 82

80. What is the approximate volume of the solidobtained by revolving about the x-axis theregion in the first quadrant enclosed by thecurves y = x3 and y = sin x?

(A) 0.061 (B) 0.139 (C) 0.215

(D) 0.225 (E) 0.278

81. Let f be a differentiable function on (a,b).If f has a point of inflection on (a,b), which of the following could be the graph of f ″ on (a,b)?See Figure 1T-9.

Directions:

Use the same answer sheet for Part A. Please note that the questions begin with number 76.This is not an error. It is done to be consistent with the numbering system of the actual AP Calculus AB Exam.

All questions are given equal weight. There is no penalty for unanswered questions. However, 1 ⁄ 4 of thenumber of incorrect answers will be subtracted from the number of correct answers. Unless otherwise

indicated, the domain of a function f is the set of all real numbers. If the exact numerical valuedoes not appear among the given choices, select the best approximate value. The use of a calculator

is permitted in this part of the exam.

x 10 x 2 x 3 x 4

f

x

y

Figure 1T-8

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ba 0

y(A)

x ba 0

y(B)

x

ba 0

y(C)

x ba 0

y(D)

x

Figure 1T-9

(A) A (B) B (C) C (D) D (E) None

82. The base of a solid is a region bounded bythe lines y = x, y = −x and x = 4 as shown inFigure 1T-10. What is the volume of the solid if the cross sections perpendicular to the x-axis areequilateral triangles?

83. Let f be a continuous function on [0, 6] and hasselected values as shown below.

Not to Scale

y

x

x = 4 y = – x

y = x

0

Figure 1T-10

(A) (B) (C)

(D) (E)3072

5

π256

3

π

64 3

3

32 3

3

16 3

3

If you use the subintervals [0,2], [2,4] and[4,6], what is the trapezoidal approximation of

(A) 9.5 (B) 12.75 (C) 19(D) 25.5 (E) 38.25

84. The amount of a certain bacteria y in apetri dish grows according to the equation

where k is a constant and t is

measured in hours.

If the amount of bacteria triples in 10 hours,then k ≈

(A) −1.204 (B) −0.110 (C) 0.110(D) 1.204 (E) 0.3

85. The volume of the solid generated by revolvingabout the y-axis the region bounded by the

graphs of and y = x isy x=

dy

dt ky=

,

f x dx( )∫ 06

?

x 0 2 4 6

f (x) 0 1 2.25 6.25

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(A) (B) (C)

(D) (E)

86. How many points of inflection does the graph of

have on the interval (−π, π)?

(A) 0 (B) 1 (C) 2 (D) 3 (E) 4

87. Given f (x) = x2 ex, what is an approximate valueof f (1.1), if you use a tangent line to the graphof f at x = 1.

(A) 3.534 (B) 3.635 (C) 7.055(D) 8.155 (E) 10.244

88. The area under the curve y = sin x from x = b tox = π is 0.2. If 0 ≤ b < π, then b =

(A) −0.927 (B) −0.201 (C) 0.644(D) 1.369 (E) 2.498

89. At what value(s) of x do the graphs of y = x2

and have perpendicular tangent lines?

(A) −1 (B) 0 (C)

(D) 1 (E) none

90. What is the approximate slope of the tangent tothe curve x3 + y3 = xy at x = 1?

1

4

y x= −

y x

x=

sin

56

15

π16

15

π

2

3

ππ6

2

15

π (A) −2.420 (B) −1.325 (C) −1.014(D) −0.698 (E) 0.267

0a b

f

y

x

Figure 1T-11

a 0 b

(A) y

x a 0 b

(B) y

x

a 0 b

(D) y

x a 0 b

(E) y

x

a

0

b

(C) y

x

Figure 1T-12

91. The graph of f is shown in Figure 1T-11, and

Which of the following

is a possible graph of g ? See Figure 1T-12.

92. If g (x) = xex, which of the followingstatements about g are true?

I. g has a relative minimum at x = 0.II. g changes concavity at x = 0.III. g is differentiable at x = 0.

(A) I only(B) II only(C) III only(D) I and II only(E) I and III only

g x f t dt x aa

x

( ) = ( ) >∫ , .

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Section II—Part A


3 45 Minutes Yes

Directions:

Show all work. You may not receive any credit for correct answers without supporting work. You may use anapproved calculator to help solve a problem. However, you must clearly indicate the setup of your solution

using mathematical notations and not calculator syntax. Calculators may be used to find the derivativeof a function at a point, compute the numerical value of a definite integral, or solve an equation.

Unless otherwise indicated, you may assume the following: (a) the numeric or algebraic answers need not besimplified, (b) your answer, if expressed in approximation, should be correct to 3 places after the

decimal point, and (c) the domain of a function f is the set of all real numbers.

1. The slope of a function at any point (x, y) is

The point (0, 2 ln 2) is on the graph of f.

(a) Write an equation of the tangent line to thegraph of f at x = 0.

(b) Use the tangent line in part (a) to approximatef (0.1) to the nearest thousandth.

(c) Solve the differential equation

with the initial condition f (0) = 2 ln2.

(d) Use the solution in part (c) and find f (0.1) tothe nearest thousandth.

2. The temperature in a greenhouse from 7:00 p.m.

to 7:00 a.m. is given by

where f (t ) is measured in Fahrenheit and t ismeasured in hours.

(a) What is the temperature of the greenhouse at1:00 a.m. to the nearest degree Fahrenheit?

(b) Find the average temperature between

7:00 p.m. and 7:00 a.m. to the nearest tenthof a degree Fahrenheit.

(c) When the temperature of the greenhousedrops below 80°F, a heating system willautomatically be turned on to maintain the

f t t

( ) = −

96 20

4sin ,

dy

dx

e

e

x

x=

+ 1

e

e

x

x + 1.

temperature at a minimum of 80°F. At whatvalue of t to the nearest tenth is the heatingsystem turned on.

(d) The cost of heating the greenhouse is$0.25 per hour for each degree. What is thetotal cost to the nearest dollar to heat thegreenhouse from 7:00 p.m. and 7:00 a.m.

3. A particle is moving on a straight line. Thevelocity of the particle for 0 ≤ t ≤ 30 is shown inthe table below for selected values of t.

0 3 6 9 12 15 18 21 24 27 30

0 7.5 10.1 12 13 13.5 14.1 14 13.9 13 12.2

t (sec)

v(t)(m/sec)

(a) Using the midpoints of five subintervals of equal length, find the approximate value of

(b) Using the result in part (a), find the averagevelocity over the interval 0 ≤ t ≤ 30.

(c) Find the average acceleration over theinterval 0 ≤ t ≤ 30.

(d) Find the approximate acceleration at t = 6.(e) During what intervals of time is the

acceleration negative?

v t dt ( )∫ 030

.

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(d) If f (−5) = 2, draw a possible sketch of f on

−6 < x < 3.

5. Given the equation y2 − x + 2y − 3 = 0:

(a) Find

(b) Write an equation of the line tangent to thegraph of the equation at the point (0, −3).

(c) Write an equation of the line normal to thegraph of the equation at the point (0, −3).

(d) The line is tangent to the graph

at point P. Find the coordinates of point P.

6. Let R be the region enclosed by the graph ofy = x2 and the line y = 4.

(a) Find the area of region R.(b) If the line x = a divides region R into two

regions of equal area, find a.(c) If the line y = b divides the region R into two

regions of equal area, find b.(d) If region R is revolved about the x-axis, find

the volume of the resulting solid.

y x= +1

43

dy

dx.


4. See Figure 1T-13.

Section II—Part B


3 45 Minutes No

Directions:

The use of a calculator is not permitted in this part of the exam. When you have finished this part of the exam,you may return to the problems in Part A of Section II and continue to work on them. However, you may

not use a calculator. You should show all work. You may not receive any credit for correct answers withoutsupporting work. Unless otherwise indicated, the numeric or algebraic answers need not be simplified,

and the domain of a function f is the set of all real numbers.

–6 –5 –4 –3 –2 –1 1 2 30

–1

1

2

3

–2

–3

y = f ( x )

y

x

Figure 1T-13

The graph of f ′, the derivative of a function f, for−6 ≤ x ≤ 3 is shown in Figure 1T-13.

(a) At what value(s) of x does f have a relativemaximum value? Justify your answer.

(b) At what value(s) of x does f have a relativeminimum value? Justify your answer.

(c) At what value(s) of x does the function havea point of inflection? Justify your answer.

7/23/2019 007137714 x



Answers to Practice Exam 1—Section I

Part A

1. C

2. C

3. C

4. E

5. D

6. A

7. C

8. C

9. D

10. E

11. C

12. C

13. A

14. C

15. B

16. E

17. C

18. E

19. A

20. B

21. A

22. C

23. A

24. E

25. A

26. C

27. B

28. E

Part B

76. D

77. B

78. C

79. C

80. B

81. A

82. C

83. B

84. C

85. A

86. C

87. A

88. E

89. D

90. C

91. E

92. D

7/23/2019 007137714 x


Part B

4. (a) x = 2 (2 pts.)(b) x = −5 (2 pts.)

(c) x = −4, x = −2 and x = 0 (2 pts.)(d) See solution. (3 pts.)

5. (a) (3 pts.)

(b) (2 pts.)

(c) y = 4x − 3 (2 pts.)(d) (0,1) (2 pts.)

6. (a) (3 pts.)

(b) a = 0 (1 pt.)

(c) (2 pts.)

(d) (3 pts.)256

5

πb = 4

23

32

3

y x= − −1

43

dy

dx y=

+1

2 2


Part A

1. (a) (3 pts.)

(b) 1.436 (1 pt.)

(c) y = ln(ee + 1) + ln 2 (4 pts.)

(d) 1.438 (1 pt.)

2. (a) 76° (2 pts.)

(b) 82.7° (2 pts.)

(c) 3.7 ≤ t ≤ 8.9 (2 pts.)

(d) $3 (3 pts.)

3. (a) 360 (3 pts.)(b) 12 m/sec (1 pt.)

(c) 0.407 m/sec2 (2 pts.)

(d) 0.75 m/sec2 (1 pt.)

(e) 18 < t < 30 (2 pts.)

y x= +1

2

2 2ln

Answers to Practice Exam 1—Section II

7/23/2019 007137714 x


Thus

8. The correct answer is (C).

9. The correct answer is (D).

Thus f x e du

e c

e c

u u

x

( ) = −

= − +

= − +

∫

−

2

1

2

1

2

2

Since let

or

f x xe dx u x

du x dx du

x dx

x( ) = = −

= − −

=

−∫ 2 2

22

, ,

.

Thus limsin sin

sin

cos .

∆

∆

∆x

x

x

x

d x

dx

→

=

+( ) − ( )

= ( )

= ( ) =

0

3

3 3

3

1

2

π π

π

π

The definition of is′( ) ′( )

= +( ) − ( )

→

f x f x

f x x f x

xxlim .∆

∆∆0

lim .x

g x→− −

′( ) =3

2

= − − ≥ −

+ < −

′( ) = − > −

< −

2 6 3

2 6 3

2 3

2 3

x x

x x

g xx

x

if

if

if

if






4. The correct answer is (E).


Since the two, one-sided limits are the same,

6. The correct answer is (A).See Figure 1TS-1.The only graph that satisfies the behavior of f is (A).


g xx x

x x( ) =

− +( ) ≥ −

−( ) − +( )[ ] < −

2 3 3

2 3 3

if

if

lim .lnx

g x→

) =2

2

lim limln

ln

ln

x

x

x

xe e e

e

→( ) →( )+ −( ) = = −( )

= − = − =

ln

and2

2

2

2

2 4

4 4 2 2

A s dA

ds s

s

dA

dss

= −( ) = ( )

−( )( )

= −( )

= ( ) ==

3

45 1 2

3

45 1 5

5 3

25 1

5 3

24 10 3

2

1

,

.

V r dV dt

r r

dV

dt cm r r

r

= = ( ) =

= = ⇒ =

= = ( ) =

43

4 2 8

32 8 32 2

4 4 2 16

3 2 2

3 2

2 2

π π π

π π π

π π π

and

Surface Area

sec;

.

cos sin sin sin

sin .

t dt t x

x

x x

= ] = − ( )= −

∫ π ππ

2 2 2

1

lim lim .x x

x

xx

x→−∞ →−∞

−+

= −

+=

2 1

1 2

2 1

1 21

Solutions to Practice Exam 1—Section I

incr. decr. incr.

+ +–0 0

a b

decr. incr. f

0

concave

downward

concave

upward

f

Figure 1TS-1

7/23/2019 007137714 x


15. The correct answer is (B).


− t + 1 and a(t ) = t − 1 and a′(t ) = 1.

Set a(t ) = 0 ⇒ t = 1. Thus, v(t ) has a relative

minimum at t = 1 and Since it is the

only relative extremum, it is an absolute

minimum. And since v(t ) is continuous on theclosed interval [0,4], thus v(t ) has an absolutemaximum at the endpoints.

v(0) = 1 and v(4) = 8 − 4 + 1 = 5.

Therefore, the maximum velocity of the particleon [1,4] is 5.


Since y = 2x is symmetrical with respect to they-axis,

Set 2k2 = 18 ⇒ k2 = 9 ⇒ k = ±3. Since k > 0, k = 3.

18. The correct answer is (E).See Figure 1TS-2.

2 2 2 2 20

2

0

2x dx x dx x kk

k k k

−∫ ∫ = = [ ] =

v 11

2( ) = .

v t s t t

( ) = ′( ) =2

2

v t dt ( ) = ( )( ) + ( ) −( ) + ( )( )=

∫ 1

2 40 5

1

2 10 20

1

2 5 20

50

0

20

.

Therefore, y e x

= +3

21

2

.

At 0,5

2x

ec c c= =

( )+ ⇒ = + ⇒ =

3

2

5

2

3

21

0



11. The correct answer is (C).(I) f ′(0) ≠ 0 since the tangent to f (x) at x = 0 is

not parallel to the x-axis.(II) f has an absolute maximum at x = a.(III) f ″ is less than 0 on (0,b) since f is concave

downward.

Thus the only statement II and III are true.


13. The correct answer is (A).

(I) f (−1) = 0;(II) Since f is increasing, f ′(−1) > 0(III) Since f is concave upward, f ″ (−1) > 0

Thus f (−1) has the smallest value.


Since dy e dx dy e dx

y e

c

x x

x

= ⇒ ⇒ ⇒

= +

∫ ∫ 3 1 3

3

2

2 2

2

.

1 1

12

32

2 23

2 23

12

12

12

32

12

32

32

+= +

= +( )

= + +

= + +

= + +

∫ ∫

∫ −

x

xdx

x

x

xdx

x x dx

x xc

x x c

x x c.

′( ) = ( ) ( )

= ( ) ( ) =

g π π π8

124 4

12 1 2 24

2

2 2

tan sec

.

g x x g x x x

x x

( ) = ( )[ ] ′( ) = ( )[ ] ( )

= ( ) ( )

3 2 6 2 2 2

12 2 2

2 2

2

tan ; tan sec

tan sec

Therefore, andf x e

f e e

x( ) = − +

( ) = − + = − +

−

−

1

2

3

2

11

2

3

2

1

2

3

2

2

1

And f e c c

c

0 11

21

1

2

13

2

0( ) = ⇒ − ( ) + = ⇒ − +

= ⇒ =

A possible graph of

x

y

(1,–2)

(–1,2)

–1 10

–1

1

2

3

–2

Figure 1TS-2

7/23/2019 007137714 x



Equation of tangent at x = π: y = 3.


s(t ) = t 3 + t 2 − 8t; v(t ) = 3t 2 + 2t − 8

Since 0 ≤ t ≤ 10, thus t = −2 is not in the domain.

the particle is moving to

the right.



′( ) = +( ) ′( ) = +( )

= ( ) =

f x x x f 33

2 33

2

32

1 2 2 2 1

2 9 54

;

.

f x dx f x dx f x dx( ) = ( ) + ( )

= π( ) − π( ) =

− −∫ ∫ ∫ 1

3

1

3

1

1

2 21

21

1

21 0.

If t v t > ( ) > ⇒4

30,

Set

or or

v t t t

t t

t t

( ) = ⇒ + −

= ⇒ −( ) +( )

= = = −

0 3 2 8

0 3 4 2

04

32

2

At x y= π = π

= ( ) = π( ), sin ; ,32

3 1 3 32 2

dy

dx x =π =

π

π

= ( )( ) =3 2 2 3 1 0 0sin cos .

y x dy

dx

x x

x x

=

=

=

32

62 2

1

2

32 2

2sin ; sin cos

sin cos


y

x a0

y

x b a + b0

y

x b –b a + b0

Figure 1TS-3

If b = 0, then 0 is a root and thus r = 0.

If b = 1, 2, or 3, then the graph of f must crossthe x-axis which implies there is another root.

Thus, b = −1.



The graph of is a semicircle above

the x-axis and whose endpoints are (−3,0)and (3.0). Thus the radius of the circle is r = 3.

21. The correct answer is (A).See Figure 1TS-3.The graphs f (x − b) and f (x + b) are the same asthe graph of f (x) shifted b units to the right andleft, respectively. Looking at Figure 1TS-3, onlyI and II have the same value.


Average value =−

( )

=π

− ( )[ ]

=π

− π

− −( )

=π

− +

=

π

∫

π

1

6

02 2

62

6

30

6 1

21

3

0

6

0

6

π

π

sin

cos

cos cos

.

x dx

x

Area = =1

2

9

2

2π π

r .

y x= −9 2

e dx e e e

e e

x

x

22

0

2 2 2 2 0

2 20

2

2 2 2

2 2

2

2

1

2

3

2

=

= −

= ( )

− = ( )

− =

∫

( ) ( )ln2

0

ln ln

ln

.

7/23/2019 007137714 x


The graph of f has a point of inflection at x = x2.

79. The correct answer is (C).Temperature of metal =


Temperature of metal = 100 − 45.1188= 54.8812 ≈ 55°F.

80. The correct answer is (B).See Figure 1TS-6.

100 10 0 1

0

6

− −∫ e dt t .



f (x) is an even

function, i.e., f (x) = f (−x).

The graph in (B) is the only even function.




See Figure 1TS-4.

f x t dt f x xx

( ) = − ′( ) = −∫ cos ; cos0

Let or

1

u x du dx du dx

k x dx k u du h u c

h x c

k x dx h x h h

= = =

( ) = ( ) = ( ) +

= ( ) +

( ) = ( )

= ( ) − −( )

∫ ∫

∫ −−

5 55

51

5

1

5

1

55

51

55

1

55

1

55

11

1

;

.

f x dx f x dxkk

k

( ) = ( ) ⇒−− ∫ ∫ 20

[ ]

– + –

decr.

π2

0 2π3π2

rel. min. rel. max.

incr. decr.

x

Figure 1TS-4

Thus f has a local maximum at


y e dy

dx e e

dy

dx e e e

x x

x y e e

y x

x x x

x x x

x

= = ( ) =

= ⇒ = ⇒ = ⇒ ( )

= ⇒ = =

= = =

= ( ) = − +

( )

2 2 2

2 2 2

2 2 0

2 2

2 2 2 1

1 2 0 0

0

1 0 11

21

;

ln

ln .

,

; , .

Set

or

At

or

x = π3

2.

decr. incr.

x 2

0– +

concave

downward

point of inflation

concave

upward

Figure 1TS-5

[π,π]by [2,2]

Figure 1TS-6

Using the Intersection function on your

calculator, you obtain the points of intersection:(0, 0) and (0.929, 0.801).

81. The correct answer is (A).A point of inflection ⇒ the graph of f changesits concavity ⇒ f ″ changes signs. Thus, thegraph in (A) is the only one that goes frombelow the x-axis (negative) to above the x-axis(positive).

v x x dx= π ( ) − ( )( ) =∫ sin . .. 2 3 2

0

0 929

0 139

78. The correct answer is (C).See Figure 1TS-5.

7/23/2019 007137714 x


Using your calculator, you obtain f (1) ≈ 2.7183and f ′(1) ≈ 8.15485.

Equation of tangent line at x = 1:y − 2.7183 = 8.15485(x − 1)y = 8.15485(x − 1) + 2.7183f (.01) ≈ 8.15485(1.1 − 1) + 2.7183 ≈ 3.534.


See Figure 1TS-9.





Triple in 10 hours ⇒ y = 3y0 at t = 10.

85. The correct answer is (A).See Figure 1TS-7.

3 3 3

3 103

10

0 109861 0 110

0 0 10 10 10y y e e e

k k

k k k= ⇒ = ⇒ = ( )

⇒ = =

≈ ≈

ln ln

lnln

. . .

or

dy

dx ky y y ekt = ⇒ = 0

f x dx( ) ≈ −

( ) + ( ) + ( ) +[ ]

≈

∫ 06 6 0

2 30 2 1 2 2 25 6 25

12 75

. .

. .

Using your calculator, you have:

Volume of solid = ( ) =∫ 364 3

32

0

4

x dx .

Area of a cross section = ( ) =3

42 3

2 2x x

[0.5,1.5] by [1,2]

Figure 1TS-7

Points of intersection: (0, 0) and (1, 1)

Volume of solid =


Volume of solid =

86. The correct answer is (C).See Figure 1TS-8.Using the Inflection function on your calculator,you obtain x = −2.08 and x = 2.08. Thus, thereare two points of inflection on (−π, π).


f (x) = x2 ex

2

15

π.

π − ( )( )∫ y y dy2 2 2

0

1

[1.5π,1.5π]by [1,2]

Figure 1TS-8

x b π0

1 y = sin x

Figure 1TS-9

Set 1 + cos b = 0.2 ⇒ cos b = −0.8 ⇒b = cos−1 (−0.8) ⇒ b ≈ 2.498.


Perpendicular tangent lines ⇒ slopes arenegative reciprocals.

y x dy

dx x

y x x dy

dx x

x

= =

= − = − = − = −−

2

12

12

2

1

2

1

2

;

;

Area = = − ]

= − − −( )

= − −( ) + = +

π π

∫ sin cos

cos cos

cos cos

x dx x

b

b b

b b

π

1 1

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7/23/2019 007137714 x


(c) Let y1 = f (x) = 96 − 20 sin and y2 = 80

Using the Intersection function of your

calculator, you obtain

x = 3.70 ≈ 3.7 or x = 8.85 ≈ 8.9

Thus, heating system is turned on when3.7 ≤ t ≤ 8.9.

(See Figure 1TS-12.)

x

4


Part A—Calculators are permitted.

1. (a)

At x = 0,

Equation of tangent line at x = 0:

(b) f (0.1) ≈ (0.1) + 2 ln 2 ≈ 1.43629 ≈ 1.436.

(c)

Let u = ex + 1, du = ex dx

y = ln (ex + 1) + cThe point (0, 2 ln 2) is on the graph of f.2 ln 2 = ln (eo + 1) + c2 ln 2 = ln 2 + c ⇒ c = ln 2y = ln (ex + 1) + ln 2.

(d) f (0.1) = ln (e0.1 + 1) + ln 2 ≈ 1.43754 ≈ 1.438

2. (a) At 1:00 am, t = 6.

(b) Average temperature


Average temperature = ( )

= ≈

1

12992 80

82 73 82 7

.

. . .

= −

∫ 1

1296 20

40

12

sin t

dt

f 6 96 20 4

76 05 76

( ) = −

6

= ° ≈ °

sin

. Fahrenheit.

e

e dx

u du u c

e c

x

x

x

+ = = +

= + +

∫ ∫ 1

1

1

ln

ln

dy

dx

e

e

dy e

e dx

dy e

e dx

x

x

x

x

x

x

=+

=+

=+∫ ∫

1

1

1

1

2

y x

y x y x

− = −( )

− = = +

2 21

20

2 21

2

1

22 2

ln

ln ln .or

dy

dx

e

e

o

o=

+ =1

1

2

dy

dx

e

e

x

x=

+ 1

Solutions—Section II

[2,10] by[10,100]

Figure 1TS-12

(d) Total cost


= (0.25)(13.629) = 3.407

≈ 3 dollars.

3. (a) Midpoints of 5 subintervals of equal lengthare t = 3, 9, 15, 21, and 27.

The length of each subinterval is

(b) Average velocity1

30 0=

− ( )∫ v t dt 0

30

Thus30

v t dt v v v

v v

( ) = ( ) + ( ) + ( )[

+ ( ) + ( )]

= + + +[

+ ] = [ ] =

∫ 1 6 3 9 15

21 27

6 7 5 12 13 5 14

13 6 60 360

. .

.

30 0

56

−= .

Total cost = ( ) − ( )( )

= ( )

−

−( −

= ( ) − +

∫

∫

∫

0 25 80

0 25 80

96 204

0 25 16 204

3 7

8 9

3 7

8 9

3 7

8 9

.

.

sin

. sin .

.

.

.

.

.

.

f t dt

t dt

t dt

7/23/2019 007137714 x


5. (a)

(b)

y − y1 = m(x − x1)

(c)

y − (−3) = 4 (x − 0)

y + 3 = 4x or y = 4x − 3.

(d)

y2 − x + 2y − 3 = 0.

At y = 1, 12 − x + 2(1) − 3 = 0 ⇒ x = 0.

Thus, point P is (0,1).

Setdy

dx y y=

+ = ⇒ =

1

2 2

1

41

y x m= + ⇒ =1

43

1

4

At normal0 31

14

4, , .−( ) = −

−=m

mm

normal

tangent

= −1

y x y x+ = − = − −31

4

1

43or .

y x− −( ) = − −( )3 14

0

At 0 31

2 3 2

1

4, −( ) =

−( ) + = −

dy

dx

dy

dx y

dy

dx y2 2 1

1

2 2+( ) = ⇒ =

+.

Differentiating: 2 1 2 0y dy

dx

dy

dx− + =


(c)

(d) Approximate acceleration at t = 6

(e) Looking at the velocity in the table, you seethat the velocity decreases beginning at t = 18to t = 30. Thus the acceleration is negative for18 < t < 30.

Part B—No calculators.

4. (a) See Figure 1TS-13.

= ( ) − ( )

− =

−=

v v9 3

9 3

12 7 5

67 5 2.

. sec .m

Average acceleration m

m

= −

−

=

12 2 0

30 0

0 407

2

2

.sec

. sec .

≈ ( )

≈

1

30360

12 m sec.

[ ]

– 0 + 0 + 0 –

decr.

–6 32–2–5

rel. min. rel. max.

incr. decr.incr.

x

f

f

Figure 1TS-13

Since f increases on (−5, 2) and decreases on(2, 3), it has a relative maximum at x = 2.

(b) Since f decreases on (−6, −5) and increases on(−5, 2), it has a relative minimum at x = −5.

(c) See Figure 1TS-14.

[ ]

incr. incr.decr. decr.

concaveupward

concaveupward

concavedownward

points of inflection

concavedownward

+ +– ––6 30–2–4

x

Figure 1TS-14

A change of concavity occurs at x = −4, −2,and 0, and since f ′ exists at these x-values,therefore f has a point of inflection at x = −4,x = −2, and x = 0.

–6 –5

(–5,2)

–4 –3 –2 –1 0 1 2 3

f

y

x

Figure 1TS-15

(d) See Figure 1TS-15.

7/23/2019 007137714 x


(d) Washer Method

V x dx x dx

x x

= π − ( )( ) = π −( )

= π −

= π

−−

−

∫ ∫ 4 16

165

256

5

2 2 2 4

2

2

2

2

5

2

2

.

Set4

orb

b b

32 3

22

3

3

6

34 4=

1⇒ = = .

= −

=

( ) −

= −

=

=

23

23

23

22

3

4

3

3

0

3

32

32

32

32

bx x

b bb

b b b b

b


(a) Set x2 = 4 ⇒ x = ±2

(b) Since y = x2 is an even function, x = 0 dividesR into two regions of equal area. Thus a = 0.

(c) See Figure 1TS-17.

Area R b x dx

b x dx

b

b

b

22

2

02

= −( )

= −( )

−∫ ∫

Area AreaR R1 2

16

3= =

Area of R x dx x x

= −( ) = −

= ( ) −

− −( ) −

−( )

= − −

=

−−

∫ 4 43

4 22

3

4 22

3

16

3

16

3

32

3

2

3

2

2

2

2

33

.

[3,3] by [1,5]

Figure 1TS-16

y

x

y = b

y = 4

y = x 2

0–√b √b

Figure 1TS-17

6. See Figure 1TS-16.

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PRACTICE EXAM 2

Answer Sheet for Practice Exam 2—Section I


Part A

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

22.

23.

24.

25.

26.

27.

28.

Part B

76.

77.

78.

79.

80.

81.

82.

83.

84.

85.

86.

87.

88.

89.

90.

91.

92.


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Section I—Part A



28 55 Minutes No

Directions:

Use the answer sheet provided in the previous page. All questions are given equal weight. There is no penaltyfor unanswered questions. However, 1 ⁄ 4 of the number of the incorrect answers will be subtracted from the

number of correct answers. Unless otherwise indicated, the domain of a function f is the set of all realnumbers. The use of a calculator is not permitted in this part of the exam.

5.

(A) 2 cos x (B) −2 cos x

(C) 2 sin x (D) −2 sin x + 2

(E) 2 sin x − 2

6. Given the equation y = 3e−2x, what is an equationof the normal line to the graph at x = ln 2?

7. What is ?

(A) (B) (C) 0

(D) (E) undefined

8. If f (x) is an antiderivative of andf (2) = 0, then f (0) =

(A) −6 (B) 6 (C)

(D) (E)56

9

−52

9

2

9

x x2 3 1+

−2

2

− 22

limcsc csc

h

h

h→

+( ) − ( )0

4 4π π

(A)

(B)

(C)

(D)

(E)

y x

y x

y x

y x

y x

= −( ) +

= +( ) −

= − −( ) +

= − −( ) −

= −( ) +

2

32

3

4

2

32

3

4

32

2 34

3

22

3

4

24 2 12

ln

ln

ln

ln

ln

22

cos t dt x

x

=∫

a b0

y

x

f '

Figure 2T-1

1.

(A) (B) (C)

(D) (E)

2. The is

(A) 0 (B) (C) 5

(D) −∞ (E) ∞

3. What is the if

?

(A) −3 (B) 1 (C) 3(D) 11 (E) nonexistent

4. The graph of f ′ is shown in Figure 2T-1.

Which of the graphs in Figure 2T-2 on page 326is a possible graph of f ?

f x x x

x x( ) = − > −

+ ≤ −

1 2

2 7 2

if

if

lim ,x

f x→−

( )2

1

3

limx

x x

x→−∞

+ −−

2

3

4 5

1

−96

5−

1

3

4

3

96

5

1

3

x dx2

3

0

8

∫

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(A) f (0)(B) f ′(0)(C) f ″ (0)(D) f (0) and f ′(0)(E) f ′(0) and f ″ (0)

11.

12. If p′(x) = q(x) and q is a continuous function for

all values of x, then is

(A)

(B) 4

(C)1

4

(D)1

4

(E)

p p

p p

p p

p p

p p

0 4

0 4 4

01

44

01

44

0 4

( ) − −( )

( ) − −( )

( ) − −( )

( ) + −( )

( ) + −( )

q x dx41

0

( )−∫

(A)

(B)

(C)

(D)

(E)

xx c

xx c

x

x c

x

x c

x

x c

3

3

3

3

3

3

3

3

3

3

3

1

3

1

+ +

− +

+ +

+ +

− +

x

x dx

4

2

1−=∫


a b x

y

0 a b a b x x

y y

0 0

(A)

a b x

y

0

(D)

(B) (C)

a b x

y

0

(E)

Figure 2T-2

f

x

y

0

Figure 2T-3

9. If a function f is continuous for all values of x,which of the following statements is always true?

I.

II.

III.

(A) I only(B) I and II only(C) II only(D) II and III only(E) I, II, and III

10. The graph of f is shown in Figure 2T-3 and f istwice differentiable. Which of the following hasthe largest value: f (0), f ′(0), f ″ (0)?

f x dx f x dxa

b

a

b

( ) = ( )∫ ∫

f x dx f x dxb

a

a

b

( ) = − ( )∫ ∫

22

2

f x dx f x dxa

b

a

b

( ) = ( )∫ ∫

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15. , then g ′(2) is

(A) 0 (B) (C)

(D) (E)

16. If a possible value of k is

(A) −2 (B) 0 (C) 1(D) 2 (E) 3

17. If for all positive values

of a, then which of the following could be thegraph of f? See Figure 2T-5.

18. A function f is continuous on [1, 5] and some of the values of f are shown below:

If f has only one root, r, on the closed interval[1, 5], and r ≠ 3, then a possible value of b is

(A) −1 (B) 0 (C) 1(D) 3 (E) 5

19. Given the equation , what is

the instantaneous rate of change of V withrespect to r at r = 5?

V r r= −( )1

3

52π

f x dx f x dxa

a

( ) = − ( )−∫ ∫ 0

0

2 2 32

x dxk

−( ) = −∫ ,

5

6−

5

6

2

3−

2

3

g x t

t dt

x

( ) =+∫

3

131


13. Water is leaking from a tank at a rate representedby f (t ) whose graph is shown in Figure 2T-4.Which of the following is best approximationof the total amount of water leaked from thetank for 1 ≤ t ≤ 3 ?

y

x 0

y

x 0

(A)

y

x 0

(D)

(B) y

x 0

(C)

y

x 0

(E)

Figure 2T-5

1 2 3

50

100

150

200

f (t )

g a l l o n s / h o u r

t

(hours)

Figure 2T-4

(A) gallons

(B) 5 gallons(C) 175 gallons(D) 350 gallons(E) 450 gallons

14. If f (x) = 5 cos2 (π − x), then is

(A) 0 (B) −5 (C) 5(D) −10 (E) 10

′

f

π2

9

2

x 1 3 5

f (x) −2 b −1

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23. The graph of f consists of two line segments and asemicircle for −2 ≤ x ≤ 2 as shown in Figure 2T-7.

What is the value of ?f x dx( )−∫ 22

(D) 0 < and only

(E) < < 5 only

3

25

3

2

< >t t

t


(A) (B) (C)

(D) 25π (E)

20. What is the slope of the tangent to the curvex3 − y2 = 1 at x = 1?

(A) (B) 0 (C)

(D) (E) undefined

21. The graph of function f is shown in Figure 2T-6.Which of the following is true for f on the inter-val (a,b)?

3

2

3

2 2− 3

2

125

3

π

50

3

π25

3

π−25

3

π

y

x a b0

f

Figure 2T-6

I. The function f is differentiable on (a,b).II. There exists a number k on (a,b) such that

f ′(k) = 0.III. f ″ > 0 on (a,b)

(A) I only(B) II only(C) I and II only(D) II and III only(E) I, II and III

22. The velocity function of a moving particle onthe x-axis is given as v(t ) = t 2 − 3t − 10. Forwhat positive values of t is the particle’sspeed increasing?

(A) only

(B) only

(C) only

03

2

3

2

5

< <

>

>

t

t

t

(A) −4 −4π (B) −4 −π (C)

(D) (E) −2 −π

24. What is the average value of the function

y = 3 cos (2x) on the interval

(A) −2 (B) (C) 0

(D) (E)

25. If , what is the value of

(A) −3 (B) 0 (C) 1

(D) 3 (E) undefined

26. A spherical balloon is being inflated. Atthe instant when the rate of increase of thevolume of the sphere is four times the rateof increase of the radius, the radius of thesphere is

(A) (B) (C)1

π1

π

1

4 π

lim ?x

f x→−

′( )1

f x x( ) = 3

3

2π1

π

−2

π

−

π π2 2

, ?

22

+ π

− −22

π

y

x

f

1–1–2 2

1

0–1

–2

–3

2

Figure 2T-7

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79. Let f be a continuous function on [4,10] and has

selected values as shown below:

76. The graph of f ′, the derivative of f, is shown in

Figure 2T-8. At which value of x does thegraph f has a horizontal tangent?

Section I—Part B


17 50 Minutes Yes

Directions:

Use the same answer sheet for Part A. Please note that the questions begin with number 76.This is not anerror. It is done to be consistent with the numbering system of the actual AP Calculus AB Exam. All questionsare given equal weight. There is no penalty for unanswered questions. However, 1 ⁄ 4 of the number of incorrectanswers will be subtracted from the number of correct answers. Unless otherwise indicated, the domain of a

function f is the set of all real numbers. If the exact numerical values does not appear among the given choices,select the best approximate value. The use of a calculator is permitted in this part of the exam.

(A) x1 (B) 0 (C) x2

(D) x1 and x2 (E) x3

77. The position function of a moving particle iss(t ) = 5 + 4t − t 2 for 0 ≤ t ≤ 10 where s is in

meters and t is measured in seconds. What is themaximum speed in m/sec of the particle on theinterval 0 ≤ t ≤ 10?

(A) −16 (B) 0 (C) 2

(D) 4 (E) 16

78. How many points of inflection does the graph of y = cos(x2) have on the interval (0, π)?

(A) 0 (B) 1 (C) 2

(D) 3 (E) 4

x 1 x 2 x 30

y

x

f '

Figure 2T-8

Using three right endpoint rectangles of equallength, what is the approximate value of

? f (x) dx?

(A) 8.4 (B) 9.6 (C) 14.4

(D) 16.8 (E) 20.8

80. Given a differentiable function f with f (−1) = 2

and Using a tangent line to the

graph of f at x = −1, what is an approximatevalue of f (−1.1)?

(A) −3.05 (B) −1.95 (C) 0.95

(D) 1.95 (E) 3.05

81. The area under the curve of from x = 1

to x = b, where b > 1 is 0.66. Then the value of b is approximately,

(A) 1.93 (B) 2.25 (C) 3.15

(D) 3.74 (E) 5.71

82. The base of a solid is a region enclosed by thecircle x2 + y2 = 4. What is the approximate vol-ume of the solid if the cross sections of the solidperpendicular to the x-axis are semicircles?

y xx= ln

′ −( ) =f 11

2.

f x dx( )∫ 410

x 4 6 8 10

f (x) 2 2.4 2.8 3.2

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(A) 36π (B) (C)

(D) (E)

86. At what value(s) of x do the graphs of y = ee andy = x2 + 5x have parallel tangent lines?

(A) −2.5 (B) 0

(C) 0 and 5 (D) −5 and 0.24

(E) −2.45 and 2.25

87. Let y represent the population in a town. If y

decreases according to the equation

and the population decreases by 25% in 6 years ,then k =

(A) −

8.318 (B) −

1.726 (C) −

0.231

(D) −0.120 (E) −0.048

88. If on [4, 8], then h has a

local minimum at x =

(A) 4 (B) 5 (C) 6

(D) 7 (E) 8

h x t dt x

( ) = −( )∫ 54

3

dy

dx ky=

1994

5

π1994

5

486

5

π81

2

π


(A) 8π (B) (C)

(D) (E)

83. The temperature of a cup of coffee is dropping at

the rate of degrees for 0 ≤ t ≤ 5,

where f is measured in Fahrenheit and t in min-utes. If initially, the coffee is 95°F , find its tem-perature to the nearest degree Fahrenheit5 minutes later.

(A) 84 (B) 85 (C) 91

(D) 92 (E) 94

84. The graphs of f ′, g ′, p′, and q′ are shown inFigure 2T-9. Which of the functions f, g, p,or q have a relative minimum on (a, b)?

(A) f only (B) g only

(C) p only (D) q only

(E) q and p only

85. What is the volume of the solid obtained by re-volving about the y-axis the region enclosedby the graphs of x = y2 and x = 9?

f t

t

( ) =

4 4sin

512

15

π64

3

π

32

3

π16

3

π

0 x

y

f '

a b 0 x

y

g'

a b

0 x

y

q'

a b

0 x

y

p'

a b

Figure 2T-9

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(d) If R is the base of a solid whose cross sec-tions perpendicular to the x-axis are semi-circles, find the volume of the solid.

3. The temperature of a liquid at a chemical plantduring a 20-minute period is given as g (t ) =

, where g (t ) is measured in

degrees in Fahrenheit and 0 ≤ t ≤ 20, t ismeasured in minutes.

(a) Sketch the graph of g on the provided grid.

What is the temperature of the liquid to thenearest hundredth of a degree Fahrenheitwhen t = 10? See Figure 2T-13.

90 420

−

tan

t

1. A particle is moving on a coordinate line. Thegraph of its velocity function v(t ) for 0 ≤ t ≤ 24 seconds is shown in Figure 2T-12.

Section II—Part A


3 45 Minutes Yes

Directions:

Show all work. You may not receive any credit for correct answers without supporting work. You may use anapproved calculator to help solve a problem. However, you must clearly indicate the setup of your solutionusing mathematical notations and not calculator syntax. Calculators may be used to find the derivative of a

function at a point, compute the numerical value of a definite integral, or solve an equation. Unless otherwiseindicated, you may assume the following: (a) the numeric or algebraic answers need not be simplified,

(b) your answer, if expressed in approximation, should be correct to 3 places after the decimal point, and(c) the domain of a function f is the set of all real numbers.

(a) Using midpoints of three subintervals of equal length, find the approximate value

of .

(b) Using the result in part (a), find the averagevelocity over the interval 0 ≤ t ≤ 24 seconds.

(c) Find the average acceleration over the inter-

val 0 ≤ t ≤ 24 seconds.(d) When is the acceleration of the particle

equal to zero?(e) Find the approximate acceleration at

t = 20 seconds.

2. Let R be the region in the first quadrant enclosedby the graph of y = 2 cos x, the x-axis and they-axis.

(a) Find the area of the region R.(b) If the line x = a divides the region R into

two regions of equal area, find a.

(c) Find the volume of the solid obtained byrevolving region R about the x-axis.

v t dt ( )∫ 024

90

88

86

84

82

800 5 10 15 20

t (minutes)

d e g r e e s ( F a r e n h e i t )

Figure 2T-13

V (t )

V (t )

20

5

10

15

20

25

30

4 6 8 10 12 14 16 18 20 22 24

(seconds)

( f e e t / s e c )

t

Figure 2T-12

(b) What is the average temperature of theliquid to the nearest hundredth of a degreeFahrenheit during the 20-minute period?

(c) At what values of t is the temperature of theliquid below 86°F .

(d) During the time within the 20-minute periodwhen the temperature is below 86°F, what is

the average temperature to the nearest hun-dredth of a degree Fahrenheit?

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Answers to Practice Exam 2—Section I

Part A

1. B

2. A

3. C

4. D

5. E

6. A

7. B

8. D

9. C

10. A

11. D

12. C

13. C

14. A

15. C

16. E

17. A

18. A

19. A

20. E

21. C

22. D

23. C

24. C

25. A

26. B

27. E

28. D

Part B

76. E

77. E

78. D

79. D

80. D

81. C

82. B

83. A

84. A

85. E

86. E

87. E

88. B

89. E

90. C

91. E

92. B

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Let u = x3 + 1, du = 3x2 dx or

f x x x dx u du

u du

uc x c

( ) = + = =

= + = +( ) +

∫ ∫ ∫ 2 3

3

13

1

3

1

3 32

2

91

12

32 3

2

dux dx

32=

′( ) = +( ) − ( )

+

−

= ( )

= −

= − ( )

= −

→

→=

f x f x h f x

h

h

h

d x

dx

h

h

x

lim

lim

csc csccsc

csc cot

.

0

0

4

4 4

4 4

2 1 2

Thus

π π

π π

π

y x y x− = −( ) = −( ) +3

4

2

32

2

32

3

4ln ln .or

At pointx y e= = =

−ln , ; ln ,ln2 3

3

42

3

42 2

y e dy

dx e e

dy

dx e e

x

x x x

x

= = −( ) = −

= − = − ( ) = − ( )

= −

= −

=

− − −

=

− − −

3 3 2 6

6 6 6 2

61

4

3

2

2

2 2 2

2

2 2 2 2 2

;

ln .

ln

ln ln

Slope of normal line at is2

3

2 2 2 2 1

2 2

22

cos sin sin

sin .

t dt t x

x

x x

xπ∫ = ] = − ( )

= −






See Figure 2TS-1.

lim

lim

lim .

x

x

x

x

x

f x

→−

→−

→−

+

−

− = − − =

+( ) = −( ) + =

( ) =

2

2

2

1 2 1 3

2 7 2 2 7 3

3Thus

lim lim

lim .

x x

x

x x

x

xx

xx x

x x x

x x x

x

→−∞ →−∞

→−∞

+ −

−

=− −

−

=− −

−=

2

3

2

3 3 3

3

3 3

2 3

3

4 5

1

4 5

1

1 4 5

1 10

x dx x x2

3

53

53

53

0

8

0

8

0

8

53

3

5

3 8

50

3 32

5

96

5

∫ =

=

= ( )

− = ( )

= .

Solutions to Practice Exam 2—Section I

decr decr

decr

incr

incr

a b

0

– '

f '

f

f"

+ –

concave

upward

concave

downward

+ –

Figure 2TS-1

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Statement I is not true, e.g.

Statement II is always true since

by the properties of definite integrals.

Statement III is not true, e.g.


f (0) = 0; f ′(0) ≤ 0 since f is decreasing and f ″ (0)≤ 0 since f is concave downward. Thus f (0) hasthe largest value.




The total amount of water leaked from the tank for

1 3

100 25 50 175

1

3

≤ ≤ = ( )

≈ + + ≈

∫ t f t dt

gallons.

Let or

Thus

u x du dx du

dx

q x dx q u du

p u

c p x c

q x dx p x

p p

= = =

( ) = ( ) = ( )

+ = ( ) +

( ) = ( )

= ( ) − −( )

∫ ∫

∫ − −

4 44

44

1

4

1

44

41

4

4

1

40

1

44

1

0

1

0

;

.

x

x dx x

x dx x x dx

x xc

x

x c

4

2

2

2

2 2

3 1 3

1 1

3 1 3

1

−= −

= −( )

= −−

+ = + +

∫ ∫ ∫ −

−

x dx x dx− −∫ ∫ ≠2

2

2

2

.

= − ( )∫ f x dxb

a

f x dxa

b

( )∫

2 20

8

0

4

x dx x dx≠ ∫ ∫ .

f c

c c c

f x x

f

( ) = ⇒ +( ) + = ⇒ ( )

+ = ⇒ + = = −

( ) = +( ) −

( ) =

3

3

2 02

92 1 0

2

99

0 6 0 6

2

91 6

02

9

32

32

32

or

−− = −

652

9

.


0 1 3 5

(3,5)

(3,3)

(3,1)

(3,–1)(5,–1)

(1,–2)

y

x

A possible

graph of f

Figure 2TS-2





f (x) is an odd function.

The function whose graph is shown in (A) is theonly odd function.


See Figure 2TS-2.

If b = 0, then r = 3, but r cannot be 3.

If b = 1, 3, or 5, f would have more than one root.

Thus, of all the choices, the only possible valuefor b is −1.

f x dx f x dxa

a

( ) = − ( ) ⇒−∫ ∫ 0

0

2 2 2 2 2 2 2

0 2 2

2 3 0 2 3

0 3 1 3 1

22 2 2 2

2 2

2 2

x dx x x k k

k k k k

k k k k

k k k k

k k−( ) = − ] − ( )( ) − −( )

= − −( ) = − +

− + = − ⇒ = − −= −( ) +( ) ⇒ = = −

∫

Set

or

′( ) =+

′( ) = ( )

+ = = g x

x

x g

3

12

3 2

2 1

6

9

2

33 3; .

′( ) = −( )[ ] − −( )[ ] −( )

= −( ) −( )

′

= −

−

= −

= ( )( ) =

f x x x

x x

f

10 1

10

210

2 2

10 2 2 10 0 1 0

cos sin

cos sin

cos sin

cos sin .

π π

π π

ππ

ππ

π

π π

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Thus a solution is


See Figure 2TS-4.

y x2 3

2 38= + .

dy

dx

x

y y dy x dx

ydy x dx

y x c

c c

= =

=

= + ( )

= + ⇒ =

∫ ∫

2

2

2

2 3

2

2 3 0 4

4

20 8

;

. ,Substituting

V r dV

dt r

dr

dt

dV dt drdt r

r r

= =

= ⇒ =

= =

4

34

4 4 4

1 1

3 2

2

2

π π

π

π π

;

.

Since or

or

f x x x xx x

f x x x

x x

f x xx x

( ) = = ≥− <

′( ) = ≥− <

′( ) = −( ) = −→− →−

3 3

3

2

2

1 1

2

00

3 03 0

3 3

ifif

ifif

lim lim .

Average value =− −( )

( )

= ( )

= − −[ ]( )[ ] =

−

−

∫ 1

2 2

3 2

1 3 2

2

3

20

2

2

2

2

π π

π

π π π

π

π

π

π

cos

sin

sin sin .

x dx

x

f x dx f x dx f x dx( ) = ( ) + ( )

= ( ) −( ) + −

( )

= − −

− −∫ ∫ ∫ 2

2

2

0

0

2

21

22 2

1

21 2

2π

π.




21. The correct answer (C).

I. f is differentiable on (a,b) since the graph isa smooth curve.

II. There exists a horizontal tangent to the graphon (a,b); thus f ′(k) = 0 for some k on (a,b).

III. The graph is concave downward; thus f ″ < 0.


See Figure 2TS-3.

or t =3

2.

v t t t v t t t

t t

a t t a t t

( ) = − − ( ) = ⇒ −( ) +( )

= ⇒ = = −

( ) = − ( ) = ⇒ − =

2 3 10 0 5 2

0 5 2

2 3 0 2 3 0

; set

or

Set

x y x y dy

dx

dy

dx

x

y

x y y

dy

dx x

3 2 2

2

3 2

1

2

1 3 2 03

2

1 1 1 0 1 0

3 1

2 0

− = − = ⇒ =

= − = ⇒ = ⇒ ( )

= ( )

( )=

;

, ,At

underfined.

V r r r

r r

dV

dr

r r

dV

drr

= ( ) − ( )

= −

= −

= ( ) − ( ) = −

=

1

35

1

3

5

3

1

3

10

310

35 25

25

3

2 2

2 3

2

5

π π

π π

π π

π π π

.

V (t )

a(t )

[ [

– – – – – – – – – – – – – – – – – – – + + + +

– – – – – – – + + + + + + + + + + + + + + + +

t 3

20 5

Figure 2TS-3

Since v(t ) and a(t ) are both negative on

and are both positive on (5, ∞), the particle’sspeed is increasing on these intervals.

0 32

,( )

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Equation of tangent at x = −1 is y − 2 =

or

Thus


Let u = ln x;

Let


See Figure 2TS-7.

ln

. .

ln .

b

e e

b

b

=

=

≈

1.32

1 32

3 15

ln. ln .

bb

( )= ( ) =

2

2

20 66 1 32

ln ln

ln ln ln ln

ln

x

x dx udu

uc

xc

x

x dx

x b

b

b

b

∫ ∫

∫

= = + = ( )

+

= ( )

= ( )

− ( )

= ( )

22

1

2

1

2 2

2

2 2

2 2

1

2

2

dux

dx=1

Area = =∫ ln

.

x

x dx

b

1 0 66

f −( ) ≈ − +( ) + ≈1 11

21 1 1 2 1 95. . . .

y x= +( ) +1

21 2.

1

21x +( )

f

f x

−( ) = ⇒ −( )

′ −( ) = ⇒ = −

1 2 1 2

1 12

1

a point

the slope at is 12

,

.

f x dx f f f ( ) ≈ ( ) + ( ) + ( )( )

≈ + +( ) ≈

∫ 2 6 8 10

2 2 4 2 8 3 2 16 8

4

10

. . . . .


Since v(t ) is a straight line with a negative slope,the maximum speed for 0 ≤ t ≤ 10 occurs att = 10 where v(t ) = 4 − 2 (10) = −16. Thusmaximum speed = 16.


See Figure 2TS-6.

Using the Inflection function of your calculator,you will find three points of inflection. Theyoccur at x = 1.355, 2.195, and 2.8.



At x = x3, f ′ = 0. Thus the tangent to the graphof f at x = x3 is horizontal.

77. The correct answer is (E).s(t ) = 5 + 4t − t 2; v(t ) = s′(t ) = 4 − 2t.

See Figure 2TS-5.

A y dy y

y= −( ) = −

= −

− − − −( )

=

−−

∫ 2

1

13

1

1

13

1

3

11

3

14

3

.

x = y2 – 1

x

y

–1 0

–1

1

Figure 2TS-4

Figure 2TS-5

[−2,12] by[−30,5]

Figure 2TS-6

[0,π]by [−2,2]

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If the graphs have parallel tangents at a point,then the slopes of the tangents are equal. Entery1 = ex and y2 = 2x + 5. Using the Intersectionfunction on your calculator, you obtain x = −2.45and x = 2.25. (See Figure 2TS-10.)

y e dydx

e

y x x dy

dx x

x x= =

= + = +

;

;2 5 2 5



See Figure 2TS-8.

Temperature of coffee 950

5

= −

≈ − ≈

∫ 4 4

95 10 9548 84

sin

. .

t dt

Area of a cross section1

2

Volume of the solid

Using your calculator, you obtain

2

2

= −( )

= −( )

= −( )

=

−∫

π

π

π

π

4

12

4

1

24

16

3

22

2

2

x

x

x dx

V .


0– +

0

decr. f

'

incr.

rel. min.

Figure 2TS-8

0

2

–2

2–2

y = –√4 – x 2

y = √4 – x 2

y

x

Figure 2TS-7

Only f has a relative minimum on (a,b).


See Figure 2TS-9.

Volume = 92

-3

3

π π

− ( )( ) =∫ y dy2 2 1944

5.

3

0 9

–3

x = 9

x = y2

y

x

Figure 2TS-9


88. The correct answer is (B).h′(x) = (x − 5)3

See Figure 2TS-11.

Since

or

dy

dx ky y y e

y y e e e

k k

kt

k k k

= ⇒ =

= ⇒ ⇒

= ( )

⇒ = = ( )

≈ −

( )

0

0 0

6 6 63

4

3

4

3

4

3

46

34

60 048

ln ln

lnln

. .

Figure 2TS-10

[−4,3] by[−5,12]

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7/23/2019 007137714 x


3. (a) See Figure 2TS-16.

g 10 90 410

2090 4

1

2

90 4 5463 90 2 1852

87 815 87 82

( ) = −

= −

≈ − ( ) ≈ −

≈ ≈ °

tan tan

. .

. . Fahrenheit.

V x dx x dx

xdx

x dx

x x

= =

= + ( )

= + ( )( )

= + ( )

= +

−

=

∫ ∫

∫

∫

1

2

1

2

1

2

1 2

2

41 2

4

2

2

4 2 20

8

0

2 2 2

0

2

0

2

0

2

0

2

2

π π

π

π

π

π π

π

π

π

π π π π

cos cos

cos

cos

sin

sin.

(d) Area of cross section =

=

1

2

2

2

1

2

2

2

π

π

cos

cos

x

x

(c) Volume = ( ) =

=

= + ( )

= + ( )[ ]

= + ( )

= +

−

=

∫ ∫

∫

∫

∫

π π

π

π

π

π

π π π

π

ππ

π

π

π

π

2 4

4

41 2

2

2 1 2

22

2

22 2

0

2 2

0

2

0

2

2

0

2

0

2

0

2

0

2

2

cos cos

cos

cos

cos

sin

sin.

x dx xdx

dx

xdx

x dx

x x

(b) cos

sin sin sin sin

sin sin

sin .

x dx

x a a

a a

a

a

a

=

] = − ( ) =

= ⇒ =

=

=

∫

−

1

2 2 2 0 2

2 1 12

1

2 6

0

0

1 π

Part A—Calculators are permitted.

1. (a) The midpoints of 3 subintervals of equallength are:

t = 4, 12, and 20.

(d) a(t ) = 0 at t = 6 and t = 14, since the slope of tangents at t = 6 and t = 14 is 0.

2. See Figure 2TS-15.

(e) a v v

ft

2022 18

22 18

25 15

4

10

4

2 25 2

( ) ≈ ( ) − ( )

− ≈

−≈

≈ . sec .

(c) Average acceleration = ( ) − ( )

− =

=

v v

ft

24 0

24 0

30

24

1 25 2. sec .

(b) Average velocity = ( )

≈ ( ) =

∫ 1

24

124

480 20

0

24

v t dt

ft sec .

The length of each interval is

Thus

24 0

38

8 4 12 20

8 25 15 20

8 60 480

0

24

−=

( ) ≈ ( ) + ( ) + ( )[ ]

≈ + +[ ]= ( ) =

∫

.

.

v t dt v v v


Solutions—Section II

2

02

R

y = 2 cos x

x

y

π

Figure 2TS-15

(a) Area of R x dx x= =

=

− ( ) =

∫ 2 2

22

2 0 2

0

2

0

2

cos sin

sin sin .

π π

π

7/23/2019 007137714 x


(d) See Figure 2TS-17.

(c) lim lim

lim lim .

x

x

x x

x

x

x x

ee

ee

→∞

−

→∞

→−∞

−

→−∞

= =

= =

33

0

33

0

2

2

2

2

2

2

2

2

Setting

or

′′( ) = − + =

⇒ =

⇒ = ⇒ = = ±

− −

− −

f x e x e

x e e

x x x

x x

x x

0 12 48 0

48 12

48 121

4

1

2

2 2 2

2 2 2

2 2

2 2

2 2

,

.

(b) ′′( ) = −( )( ) + −( )( ) −( )

= − +

− −

− −

f x e x e x

e x e

x x

x x

12 12 4

12 48

2 2

2 2 2

2 2

2 2


(b) Average Temperature


(c) Set the temperature of the liquid equalto 86°F . Using your calculator, let

; and y2 = 86.

To find the intersection point of y1 and y2,let y3 = y1 − y2 and find the zeros of y5.

Using the zero function of your calculator,you obtain x = 15.708.

Since y1 < y2 on the interval 15.708 < x ≤ 20,the temperature of the liquid is below 86°Fahrenheit when 15.708 < t ≤ 20. (Note thatthe intersection function of a calculator isnot among the four calculator capabilitiesallowed without supporting explanation.)

(d) Average temperature below 86°


Part B—No calculators.

4. (a) ′( ) = ( ) −( ) = −

′( ) = − = ⇒ =

− −

−

f x e x xe

f x xe x

x x

x

3 4 12

0 12 0 0

2 2

2

2 2

2

Setting , .

Average temperature

Fahrenheit.

= ( )

≈

≈ °

1

4 292364 756

84 9851

84 99

..

.

.

=−

−

−

∫ 1

20 15 70890 4

2015 708

20

.tan .

.

xdx

y x

1 90 420

= −

tan

= ( ) =

≈ °

1

201750 75 87 5375. .

87.54 Fahrenheit.

= −

∫ 1

2090 4

200

20

tan t

dt

080

82

84

86

88

90

5 10 15 20

t (minutes)

d e g r e e s ( F a r e n h e i t )

g(t ) = 90 – 4 tan( (t

20

Figure 2TS-16

incr. decr.

rel. max.

+'

'

–0

x

0

Figure 2TS-17

f (0) = 3, since f has only one critical point (atx = 0), thus at x = 0, f has an absolute maxi-mum. The absolute maximum value is 3.

f ′(x) > 0 if x < 0 and f ′(x) < 0 if x > 0. Thus f has a relative maximum at x = 0 and since it

is the only critical point, f has an absolutemaximum at x = 0. Since f (0) = a, the absolutemaximum for f is a.

5. (a) f g t dt g t dt

g t dt g t dt

−( ) = ( ) = − ( )

= − ( ) − ( )

= − − ( )( )

− ( )( )

= − =

−

−

−

−

−

∫ ∫

∫ ∫

3

1

22 2

1

21 2

2 1 1

0

3

3

0

3

1

1

0

.

(e)

Setting

f x a e a b

f x a e b x a bx e

f x a bx e x

f x a bx

e

bx

bx bx

bx

bx

( ) = > >

′( ) = −( ) = −

′( ) = − = ⇒ =

′( ) = −

−

− −

−

2

2 2

2

2

0 0

2 2

0 2 0 0

2

, ,

,

′( ) = − = −− =

f x xe x

e

x

x12

122

2

2

2

7/23/2019 007137714 x


(b) f 2 51

82 5 2 1 1 0625 1 063. . . .( ) ≈ −( ) + = ≈

6. (a)

Equation of tangent:

or

dy

dx

y

x

dy

dx

y x y x

x y

= ( )

=( )

=

− = −( ) = −( ) +

= =

22 1

1

2 2

1

8

11

82

1

82 1

2

2 1

2

; ,

.

,


(b) The function f increases on (0,1) and de-creases on (1,3). Thus f has a relative maxi-mum at x = 1. And f decreases on (−3,−1)

and increases on (−1,0). Thus f has a relativeminimum at x = −1.

(c) f ′(x) = g (x) and f ″ (x) = g ′(x). See Figure2TS-18.

f g t dt g t dt g t dt 3

1

21 2

1

21 2 1 1 0

0

3

0

1

1

3

( ) = ( ) = ( ) + ( )

= ( )( ) + − ( )( )

= − =

−

∫ ∫ ∫

" ( x ) = g' ( x ) + +–

x

–3 0 2 3

g( x ) incr. decr. incr.

f concave

upward

concav

upwar

concave

downward

change of

concavity

change of

concavity

[ [

Figure 2TS-18

The function f has a change of concavity atx = 0 and x = 2. Thus f has a point of inflec-tion at x = 0 and x = 2.

Thus, m = 0, point (1, 1); the equation of thetangent line to f (x) at x = 1 is y = 1.

(d) f g t dt

f g

1

1

2 1 2 1

1 1 0

0

1

( ) = ( ) = ( )( ) =′( ) = ( ) =

∫

(c)

and

Since

dy

dx

y

x

dy

y

dx

x

dy

y

dx

x

y x dx

x

c x c

e e

y e f

e e

e

y xc

xc

c c

=

= =

= = ( )

− + = − +

=

= ( ) =

= ⇒ =

= − +

∫ ∫

∫ −

−

+

− +

−( )

+ − +

2

2 2

1

2

1

2 1

1

2

2 1

1 1

11

4

2

2 2

2

1

1

2

1

2

1

2 21

4

0

ln

;

,

ln

cc c

y e x

= ⇒ =

= − +

01

41

2

1

4

.

.Thus,

(d) f e e e

or

2 5

1 05127 1 051

1

2 2 5

1

4

1

5

1

41

20.

. . .

.

( ) = = =

≈ ≈( )

−( )

+

− +

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APPENDIXES


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Appendix I

Formulas and Theorems

1. Quadratic Formula:

2. Distance Formula:

3. Equation of a Circle:

x2 + y2 = r2 center at (0,0) and radius = r

4. Equation of an Ellipse:

x

a

y

b

2

2

2

21 0 0+ = ( )center at ,

d x x y y= −( ) + −( )2 1

2

2 1

2

ax b c a

x b b ac

a

2

2

0 0

4

2

+ + = ≠( )

= − ± − 5. Area and Volume Formulas:

x h

a

y k

b h k

−( )+

−( )= ( )

2

2

2

21 center at ,

Figure Area Formula

Trapezoid [base1 + base2] (height)

Parallelogram (base)(height)

Equilateral Triangle

Circle π r2 (circumference = 2πr)

s2 3

4

1

2

Solid Volume Surface Area

Sphere 4π r2

Right Circular Cylinder π r2h 2π rh

Right Circular ConeLateral S.A.:

Total S.A.:

π

π π

r r h

r r r h

2 2

2 2 2

+

+ +

1

32πr h

43

3πr


7/23/2019 007137714 x


350 • Appendixes

7. Double Angles:

• sin 2θ = 2sinθ cosθ• cos 2θ = cos2 θ−sin2 θ or

1 − 2 sin2 θ or 2 cos2 θ − 1

• cos2 θ =

• sin2 θ =

8. Pythagorean Identities:

• sin2 θ + cos2 θ = 1• 1 + tan2 θ = sec2 θ• 1 + cot2 θ = csc2 θ

9. Limits:

10. Rules of Differentiation:

a. Definition of the Derivative of a Function:

b. Power Rule:

c. Sum & Difference Rules:

d

dx u v

du

dx

du

dx±( ) = ±

d

dx x nxn n( ) = −1

′( ) = ′ +( ) − ( )

→f x

f x h f x

hhlim

0

lim limcos

limsin

lim

lim lim

x x

x x

h

h

h

x

x

x

x

x

x

x h e

e

h x e

→∞ →

→ →∞

→ →

= −

=

= +

=

−= +( ) =

10

10

1 11

11 1

0

0

0 0

1

1 2

2

− cos θ

1 22

+ cos θ

d. Product Rule:

e. Quotient Rule:

Summary of Sum, Difference, Product andQuotient Rules:

f. Chain Rule:

11. Inverse Function and Derivatives:

12. Differentiation and Integration Formulas:

Integration Rules

a.

b.

c.

d. f x g x dx f x dx g x dx( ) ± ( )[ ] = ( ) ± ( )∫ ∫ ∫

− ( ) = − ( )f x dx f x dx

af x dx a f x dx( ) = ( )

f x dx F x C F x f x) = ) + ⇔ ′ ) = )

f xf f x

dy

dx dx

dy

−

−( )′ ( ) =′ ( )( )

=1

1

1 1or

d

dx f g x f g x g x

dy

dx

dy

du

du

dx

( )( )[ ] = ′ ( )( ) ′( )

=

or

u v u v uv u v v u

u

v

u v v u

v

±( )′ = ′ ± ′ ( )′ = ′ + ′

′=

′ − ′2

d

dx

u

v

v du

dx u

dv

dxv

v

=

−≠

20,

d

dx uv v

du

dx u

dv

dx( ) = +

Angle

Function0° π

180°2π360°

Sin 0 1 0 −1 0

Cos 1 0 −1 0 1

Tan 0 1 Undefined 0 Undefined 03

33

12

22

32

32

22

12

32

2

π

70°

π2

90°

π3

60°

π4

45°

π6

30°

6. Special Angles:

7/23/2019 007137714 x


g.

h.

i.

j.

k.

l.

m.

n.

o.

More Integration Formulas:

a.

b.

c.

d.

e.

f.

g.

h.

i.

Note: After evaluating an integral, alwayscheck the result by taking the derivative ofthe answer (i.e., taking the derivative of theantiderivative).

sinsin

.

sincos

2

2

2

2

4

1 2

2

xdx x x

c

x x

= − ( )

+

= −

∫

Note:

1 1

1

2 2

1

1

x x adx

a

x

a c

a

a

x

c

−= +

+

∫ −

−

sec

cos

or

1 12 2

1

a x dx

a

x

a c

+ =

+∫ −tan

12 2

1

a xdx

x

a c

−=

+∫ −sin

ln lnxdx x x x c= − +∫

csc ln csc cotxdx x x c= − +

sec lnsec tanxdx x x c= + +∫

cot lnsin ln cscxdx x c x c= + − +or

tan lnsec ln cosxdx x c x c= + − +∫ or

1

12

1

x xdx x c

−= +∫ −sec

1

1 2

1

+ = +∫ −

x dx x ctan

1

1 2

1

−= +−∫

xdx x csin

a dx a

a c

a ax

x

= +> ≠∫ ln ,0 1

e dx e cx x= +

1

x dx x c= +∫ ln

csc cot cscx x dx x c( ) = − +

sec tan secx x dx x c) = +

csc cot2 = − +xdx x c

Formulas and Theorems • 351

Differentiation Formulas:

a.

b.

c.

d.

e.

f.

g.

h.

i.

j.

k.

l.

m.

n.

o.

Integration Formulas:

a.

b.

c.

d.

e.

f. sec tan2 xdx x c= +

cos sinxdx x c= +

sin cosxdx x c= − +

x dx x

n c nn

n

=+

+ ≠ −+

∫ 1

11,

adx ax c= +

1dx x c= +

d

dx x

x xsec−( ) =

−1

2

1

1

d

dx x

xtan−( ) =

+1

2

1

1

d

dx x

xsin−( ) =

−1

2

1

1

d

dx a a ax x( ) = ( )ln

d

dx e ex x( ) =

d

dx x

xln( ) =

1

d

dx x x xcsc csc cot( ) = − ( )

d dx

x x xsec sec tan( ) =

d

dx x xcot csc( ) = − 2

d

dx x xtan sec( ) = 2

d

dx x xsin cos( ) =

d

dx x xcos sin( ) = −

d

dx x nxn n( ) = −1

d

dx ax a( ) =

d

dx x( ) = 1

7/23/2019 007137714 x


21. Washer Method:

where f (x) = outer radius and g (x) = inner radius

22. Distance Traveled Formulas:

• Position Function:

• Velocity:

• Acceleration: a(t ) =

• Speed:

• Displacement from t 1 to

• Total Distance Traveled from t 1 to

23. Business Formulas:

P(x) = R(x) − C(x) Profit = Revenue − Cost

R(x) = px Revenue = (price)(items sold)

P′(x) Marginal Profit

R′(x) Marginal Revenue

C′(x) Marginal Cost

P′(x), R′(x), C′(x) are the instantaneous ratesof change of profit, revenue and costrespectively.

24. Exponential Growth/Decay Formulas:

dy

dt ky y y t y ekt = > ( ) =, 0 0and

t v t dt t

t

21

2

= ( )∫

s t s t 2 1= ( ) − ( )

t v t dt t

t

21

2

= ( )∫

v t ( )

a t (

v t ds

dt v t a t dt ( ) = ( ) = ( )∫ ;

s t s t v t dt ( ) ( ) = ( );

V f x g x dxa

b

= ( )( ) − ( )( )( )∫ π2 2

352 • Appendixes

13. The Fundamental Theorems of Calculus

14. Trapezoidal Approximation:

15. Average Value of a Function:

16. Mean Value Theorem:

For some c in (a,b)

17. Mean Value Theorem for Integrals:

For some c in (a,b)

18. Area Bounded by 2 Curves:

19. Volume of a Solid with Known Cross Section:

where A(x) is the cross section.

20. Disc Method:

where f (x) = radiusV f x dxa

b

= ( )( )∫ π2

,

V A x dxa

b

= ( )∫ ,

Area where= ( ) − ( )( ) ( ) ≥ ( )∫ f x g x dx f x g xx

x

,1

2

f x dx f c b aa

b

( ) = ( ) −( )∫

′( ) = ( ) − ( )−

f c f b f ab a

f cb a

f x dxa

b

( ) =−

( )∫ 1

f x dx b a

n

f x f x f x

f x f xa

b

n n

( ) ≈ − ( ) + ( ) + ( )

+ ( ) + ( )

∫

−2

2 2

20 1 2

1

K

f x dx F b F a F x f x

F x f t dt F x f x

a

b

a

x

( ) = ( ) − ( ) ′( ) = ( )

( ) = ( ) ′( ) = ( )

∫

∫

, .

, .

where

If then

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Appendix II

Special Topic: Slope Fields

Slope field is a relatively new topic in AP Calculus. It has been part of the AP CalculusBC curriculum for the past several years. It will be introduced in the AP Calculus ABcurriculum beginning with the academic year 2003–2004.*

A slope field (or a direction field ) for a first-order differential equation is a graphicrepresentation of the slopes of a family of curves. It consists of a set of short line segmentsdrawn on a pair of axes. These line segments are the tangents to a family of solution curvesfor the differential equation at various points. The tangents show the direction in whichthe solution curves will follow. Slope fields are useful in sketching solution curves with-out having to solve a differential equation algebraically.

Example 1

If = 0.5x, draw a slope field for the given differential equation.

Solution:

Step 1: Set up a table of values for for selected values of x.dy

dx

dy

dx

x −4 −3 −2 −1 0 1 2 3 4

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2dy

dx

Note that since = 0.5x, the numerical value of is independent of the value of y.

For example, at the points (1,−1), (1,0), (1,1), (1,2), (1,3) and at all the points whose

x-coordinates are 1, the numerical value of is 0.5 regardless of their y-coordinates.

Similarly, for all the points whose x-coordinates are 2 (e.g., (2,−1), (2,0), (2,3), etc.),

dy

dx

dy

dx

dy

dx

*This topic will not appear on the AP Calculus AB Exam until, at the earliest, May 2004.


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Example 2

Figure A-2 shows a slope field for one of the differential equations given below. Identifythe equation.

354 • Appendixes

(a) = 2x (b) = −2x (c) = y

(d) = −y (e) = x + y

Solution:

If you look across horizontally at any row of tangents, you’ll notice that the tangentshave the same slope. (Points on the same row have the same y-coordinate but different

x-coordinates.) Therefore, the numerical value of (which represents the slope of the

tangent) depends solely on the y-coordinate of a point and it is independent of thex-coordinate. Thus only choice (c) and choice (d) satisfy this condition. Also notice thatthe tangents have a negative slope when y > 0 and have a positive slope when y < 0.

dy

dx

dydx

dydx

dy

dx

dy

dx

dy

dx

Figure A-1

Figure A-2

= 1. Also, remember that represents the slopes of the tangent lines to the curve

at various points. You are now ready to draw these tangents.

Step 2: Draw short line segments with the given slopes at the various points. The slope

field for the differential equation = 0.5x is shown in Figure A-1.dy

dx

dy

dx

dy

dx

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Solution:

Begin by locating the point (0,−2) as given in the initial condition. Follow the flow of thefield and sketch the graph of the function. Repeat the same procedure with the point(0, 0). See the curves as shown in Figure A-4.

Special Topic: Slope Fields • 355

Example 4

Given the differential equation= −

xy.

(a) draw a slope field for the differential equation at the 15 points indicated on the pro-vided set of axes in Figure A-5.

(b) sketch a possible graph for the particular solution y = f (x) to the differential equa-tion with the initial condition f (0) = 3

(c) find, algebraically, the particular solution y = f (x) to the differential equation withthe initial condition f (0) = 3.

Solution:

(a) Set up a table of values for at the 15 given points.dy

dx

dy

dx

Figure A-3

Figure A-4

Therefore, the correct choice is (c) = −y.

Example 3

A slope field for a differential equation is shown in Figure A-3.Draw a possible graph for the particular solution y = f (x) to the differential equation

function, if (a) the initial condition is f (0) = −2 and (b) the initial condition is f (0) = 0.

dy

dx

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Then sketch the tangents at the various points as shown in Figure A-6.

356 • Appendixes

3

2

1

0–1–2 1 2

y

x

Figure A-5

3

2

1

0–1–2 1 2

y

x

Figure A-6

x = −2 x = −1 x = 0 x = 1 x = 2

y = 1 2 1 0 −1 −2

y = 2 4 2 0 −2 −4

y = 3 6 3 0 −3 −6

(b) Locate the point (0,3) as indicated in the initial condition. Follow the flow of thefield and sketch the curve as shown in Figure A-7.

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Let k = ec and you have

Step 5: Substitute initial condition (0, 3) and obtain k = 3. Thus you have y

ex

=3

2

2

.

y k

ex

=2

2

.

Special Topic: Slope Fields • 357

3

2

1

0–1–2 1 2

y

x

Figure A-7

(c) Step 1: Rewrite = −xy as = −x dx.

Step 2: Integrate both sides and obtain

Step 3: Apply the exponential function to both sides and obtain

Step 4: Simplify the equation and get y e e

e

e

xc

c

x=

( ) =

2

2

2

2.

e ey

xcln =

+2

2

ln .y x

c= − +2

2

dy

y xdx∫ ∫ = −

dy

y

dy

dx

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Appendix III

Bibliography

Advanced Placement Program Course Description. New York: The College Board,2000.

Anton, Howard. Calculus. New York: John Wiley & Sons, 1984.Apostol, Tom M. Calculus. Waltham, MA: Blaisdell Publishing Company, 1967.Berlinski, David. A Tour of the Calculus. New York: Pantheon Books, 1995.Boyer, Carl B. The History of the Calculus and Its Conceptual Development.

New York: Dover, 1959.Finney, R., Demana, F. D., Waits, B. K., Kennedy, D. Calculus Graphical, Numerical,

Algebraic. New York: Scott Foresman Addison Wesley, 1999.Kennedy, Dan. Teacher’s Guide–AP Calculus. New York: The College Board, 1997.

Larson, R. E., Hostetler, R. P., Edwards, B. H. Calculus. New York: HoughtonMifflin Company, 1998.

Leithold, Louis. The Calculus with Analytic Geometry. New York: Harper & Row,1976.

Sawyer, W.W. What Is Calculus About? Washington, DC: Mathematical Associationof America, 1961.

Spivak, Michael. Calculus. New York: W. A. Benjamin, Inc., 1967.Stewart, James. Calculus. New York: Brooks/Cole Publishing Company, 1995.


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Appendix IV

Websites

www.collegeboard.org/ap/calculus

http://www.maa.org/features/mathed_disc.html

http://www.askdrmath.com/

http://www.askdrmath.com/calculus/calculus.html

http://www.askdrmath.com/library/topics/svcalc/

http://www-history.mcs.st-and.ac.uk/history/HistTopics/The_rise_of_calculus.html

http://mathforum.com/epigone/ap-calc/

http://www.sparknotes.com/math/calcab/.dir/


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