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CalculusPART TWO

BARNMC03_0132328186.QXD 2/21/07 1:15 PM 27

128

I N T R O D U C T I O N

How do algebra and calculus differ? The two words static and dynamic probably

come as close as any to expressing the difference between the two disciplines. In al-

gebra, we solve equations for a particular value of a variable—a static notion. In cal-

culus, we are interested in how a change in one variable affects another variable—a

dynamic notion.

Parts (A)–(C) of the figure at the top of the next page illustrate three basic prob-

lems in calculus. It may surprise you to learn that all three problems—as different as

they appear—are related mathematically. The solutions to these problems and the dis-

covery of their relationship required the creation of a new kind of mathematics. Isaac

Newton (1642–1727) of England and Gottfried Wilhelm von Leibniz (1646–1716) of

Germany developed this new mathematics, called the calculus, simultaneously and

independently. It was an idea whose time had come.

In addition to solving the problems described in the figure, calculus will enable us

to solve many other important problems. Until fairly recently, calculus was used pri-

marily in the physical sciences, but now people in many other disciplines are finding

it a useful tool as well.

3-1 Introduction to Limits

3-2 Continuity

3-3 Infinite Limits and Limits at Infinity

3-4 The Derivative

3-5 Basic DifferentiationProperties

3-6 Differentials

3-7 Marginal Analysis inBusiness and Economics

Chapter 3 Review

Review Exercise

Limits and the Derivative

3CHAPTER

BARNMC03_0132328186.QXD 2/21/07 1:15 PM Page 128

S e c t i o n 3 - 1 Introduction to Limits 129

x

y

Tangent line

(A) Find the equation of thetangent line at (x1, y1),given y � f (x)

(x1, y1)y � f (x) Velocity

here

(B) Find the instantaneousvelocity of a fallingobject

x

y

(C) Find the indicated areabounded by y � g(x), x � a,x � b, and the x axis

y � g(x)

a b

Area

g(x)

x

5

�3 5

y � g(x)

Section 3-1 INTRODUCTION TO LIMITS� Functions and Graphs: Brief Review� Limits: A Graphical Approach� Limits: An Algebraic Approach� Limits of Difference Quotients

Basic to the study of calculus is the concept of a limit. This concept helps us todescribe, in a precise way, the behavior of f(x) when x is close, but not equal, to aparticular value c. In this section, we develop an intuitive and informal approach toevaluating limits. Our discussion concentrates on developing and understandingconcepts rather than on presenting formal mathematical details.

� Functions and Graphs: Brief ReviewThe graph of the function is the graph of the set of all ordered pairs(x, f(x)). For example, if then and is a point on thegraph of f. Figure 1 shows and (2, f(2)) plotted on the graphof f. Notice that the domain values and 2 are associated with the x axis and therange values and are associated with the y axis.

Given x, it is sometimes useful to be able to read f(x) directly from the graph of f.Example 1 reviews this process.

f122 = 4f1-12 = 1, f112 = 3,-1, 1,

1-1, f1-122, 11, f1122, 12, f1222 = 12, 42f122 = 4x = 2,y = f1x2 = x + 2

f (x)

x1�4 �1

2

f (1) � 3

f (�1) � 1

f (2) � 4

5

6

2 3 4�1

�2

f (x) � x � 2

(2, f (2)) � (2, 4)

FIGURE 1

x g(x)

1

3

4

- 2

E X A M P L E 1 Finding Values of a Function from Its Graph Complete the following table,using the given graph of the function g:


SOLUTION To determine g(x), proceed vertically from the x value on the x axis to the graph of gand then horizontally to the corresponding y value g(x) on the y axis (as indicated bythe dashed lines):

130 C H A P T E R 3 Limits and the Derivative

x g(x)

4.0

1 2.5

3 1.5

4 1.0

- 2

x h(x)

�2

�1

0

1

2

3

4

g(x)

x

5

�3 5

y � g(x)

h(x)

x

5

�4 5

y � h(x)

20

4

x x

f (x)

f (x) f (x) � x � 2

f (x)

x

FIGURE 2

MATCHED PROBLEM 1 Complete the following table, using the given graph of the function h:

� Limits: A Graphical ApproachWe introduce the important concept of a limit through an example, which will leadto an intuitive definition of the concept.

E X A M P L E 2 Analyzing a Limit Let Discuss the behavior of the values of f(x)when x is close to 2.

SOLUTION We begin by drawing a graph of f that includes the domain value (Fig. 2).x = 2

f1x2 = x + 2.

In Figure 2, we are using a static drawing to describe a dynamic process. This requirescareful interpretation. The thin vertical lines in Figure 2 represent values of x that areclose to 2. The corresponding horizontal lines identify the value of f(x) associated with



each value of x. [Example 1 dealt with the relationship between x and f(x) on a graph.]The graph in Figure 2 indicates that as the values of x get closer and closer to 2 oneither side of 2, the corresponding values of f(x) get closer and closer to 4. Symbol-ically, we write

This equation is read as “The limit of f(x) as x approaches 2 is 4.” Note that That is, the value of the function at 2 and the limit of the function as x approaches 2are the same. This relationship can be expressed as

Graphically, this means that there is no hole or break in the graph of f at x = 2.

limx:2

f1x2 = f122 = 4

f122 = 4.

limx:2

f1x2 = 4

f (x)

x

5

�5 5

FIGURE 3 f1x2 = ƒ x ƒ

MATCHED PROBLEM 2 Let Discuss the behavior of the values of f(x) when x is close to 1.f1x2 = x + 1.

We now present an informal definition of the important concept of a limit. Aprecise definition is not needed for our discussion, but one is given in a footnote.*

DEFINITION Limit

We write

if the functional value f(x) is close to the single real number L whenever x is close,but not equal, to c (on either side of c).

Note: The existence of a limit at c has nothing to do with the value of the functionat c. In fact, c may not even be in the domain of f. However, the function must be de-fined on both sides of c.

limx:c

f1x2 = L or f1x2: L as x : c

The next example involves the absolute value function:

The graph of f is shown in Figure 3.

f132 = ƒ 3 ƒ = 3 f1-22 = ƒ -2 ƒ = -1-22 = 2

f1x2 = ƒ x ƒ = e -x if x 6 0x if x Ú 0

* To make the informal definition of limit precise, the use of the word close must be made more precise.This is done as follows: We write if, for each there exists a such that

whenever This definition is used to establish particular limits and toprove many useful properties of limits that will be helpful to us in finding particular limits. [Even thoughintuitive notions of a limit existed for a long time, it was not until the nineteenth century that a precisedefinition was given by the German mathematician Karl Weierstrass (1815–1897).]

0 6 ƒ x - c ƒ 6 d.ƒ f1x2 - L ƒ 6 ed 7 0e 7 0,limx:c f1x2 = L



E X A M P L E 3 Analyzing a Limit Let Explore the behavior of h(x) for x near, butnot equal, to 0. Find if it exists.

SOLUTION The function h is defined for all real numbers except 0. For example,

Not defined

In general, h(x) is for all negative x and 1 for all positive x. Figure 4 illustrates thebehavior of h(x) for x near 0. Note that the absence of a solid dot on the vertical axisindicates that h is not defined when

When x is near 0 (on either side of 0), is h(x) near one specific number? The an-swer is “No,” because h(x) is for and 1 for Consequently, we saythat

Thus, neither h(x) nor the limit of h(x) exists at However, the limit from theleft and the limit from the right both exist at 0, but they are not equal.

x = 0.

limx:0

ƒ x ƒ

x does not exist

x 7 0.x 6 0-1

x = 0.

-1

h122 =

ƒ 2 ƒ

2=

22

= 1

h102 =

ƒ 0 ƒ

0=

00

h1-22 =

ƒ -2 ƒ

-2=

2-2

= -1

limx:0 h1x2h1x2 = ƒ x ƒ>x.

1 2

�1

�2

�2

11

2

0

0x

x

h(x)

h(x)

x

h(x)

FIGURE 4

MATCHED PROBLEM 3 Graph

and find if it exists.limx:2 h1x2h1x2 =

x - 2ƒ x - 2 ƒ

In Example 3, we saw that the values of the function h(x) approached two differ-ent numbers, depending on the direction of approach, and it was natural to refer tothese values as “the limit from the left” and “the limit from the right.” These expe-riences suggest that the notion of one-sided limits will be very useful in discussing basiclimit concepts.

DEFINITION One-Sided Limits

We write

is read “x approaches c from the left” and means and

and call K the limit from the left or the left-hand limit if f(x) is close to K wheneverx is close to, but to the left of, c on the real number line. We write

is read “x approaches c from the right” and means and

and call L the limit from the right or the right-hand limit if f(x) is close to L when-ever x is close to, but to the right of, c on the real number line.

x 7 c.x :cx : c+lim

x:c +

f1x2 = L

x 6 c.x :cx : c-lim

x:c-

f1x2 = K

If no direction is specified in a limit statement, we will always assume that the limitis two sided or unrestricted. Theorem 1 states an important relationship betweenone-sided limits and unrestricted limits.



1 2

f (x)

x�1�2�3 3

3

2

1

4

5

FIGURE 51 2x x

f (x)

x�1�3 3

3

2

4

5

1

f (x)

f (x)

f1-12 = 1

limx: -1

f1x2 = 1

limx: -1+

f1x2 = 1

limx: -1-

f1x2 = 1

1x x

f (x)

x�1�2�3 3

3

4

1

5

f (x)

f(1) not defined

limx:1

f1x2 = 3

limx:1+

f1x2 = 3

limx:1-

f1x2 = 3

THEOREM 1 ON THE EXISTENCE OF A LIMITFor a (two-sided) limit to exist, the limit from the left and the limit from the right mustexist and be equal. That is,

In Example 3,

Since the left- and right-hand limits are not the same,

limx:0

ƒ x ƒ

x does not exist

limx:0-

ƒ x ƒ

x= -1 and lim

x:0+

ƒ x ƒ

x= 1

limx:c

f1x2 = L if and only if limx:c-

f1x2 = limx:c+

f1x2 = L

E X A M P L E 4 Analyzing Limits Graphically Given the graph of the function f shown in Figure 5, discuss the behavior of f(x) for x near (A) (B) 1, and (C) 2.

SOLUTION (A) Since we have only a graph to work with, we use vertical and horizontal lines torelate the values of x and the corresponding values of f(x). For any x near on either side of we see that the corresponding value of f(x), determined bya horizontal line, is close to 1.

-1,-1

-1,

(B) Again, for any x near, but not equal to, 1, the vertical and horizontal lines indicatethat the corresponding value of f(x) is close to 3. The open dot at (1, 3), togetherwith the absence of a solid dot anywhere on the vertical line through indicates that f(1) is not defined.

x = 1,


(C) The abrupt break in the graph at indicates that the behavior of the graphnear is more complicated than in the two preceding cases. If x is close to2 on the left side of 2, the corresponding horizontal line intersects the y axis ata point close to 2. If x is close to 2 on the right side of 2, the correspondinghorizontal line intersects the y axis at a point close to 5. Thus, this is a case wherethe one-sided limits are different.

x = 2x = 2


f (x)

x�1�2�3

2

5

1

2x x

f (x)

f (x)

f122 = 2

limx:2

f1x2 does not exist

limx:2+

f1x2 = 5

limx:2-

f1x2 = 2

f (x)

x1 2 3 4 5 6�2 �1

2

3

4

FIGURE 6

MATCHED PROBLEM 4 Given the graph of the function f shown in Figure 6, discuss the following, as we didin Example 4:

(A) Behavior of f(x) for x near 0(B) Behavior of f(x) for x near 1(C) Behavior of f(x) for x near 3

In Example 4B, note that exists even though f is not defined at and thegraph has a hole at In general, the value of a function at has no effect on thelimit of the function as x approaches c.

x = cx = 1.x = 1lim

x:1 f1x2

I N S I G H T

� Limits: An Algebraic ApproachGraphs are very useful tools for investigating limits, especially if something unusualhappens at the point in question. However, many of the limits encountered in calcu-lus are routine and can be evaluated quickly with a little algebraic simplification,some intuition, and basic properties of limits. The following list of properties of lim-its forms the basis for this approach:



Each property in Theorem 2 is also valid if is replaced everywhere by or replaced everywhere by x : c+.

x : c-x : c

THEOREM 2 PROPERTIES OF LIMITSLet f and g be two functions, and assume that

where L and M are real numbers (both limits exist). Then

1.

2.

3.

4.

5. for any constant k

6.

7.

8. for n evenlimx:c2n f1x2 = 2n lim

x:c f1x2 = 2n L L 7 0

limx:c

f1x2g1x2 =

limx:c

f1x2limx:c

g1x2 =

L

M if M Z 0

limx:c3f1x2 � g1x24 = 3 lim

x:c f1x243 lim

x:c g1x24 = LM

limx:c

kf1x2 = k limx:c

f1x2 = kL

limx:c3f1x2 - g1x24 = lim

x:c f1x2 - lim

x:c g1x2 = L - M

limx:c3f1x2 + g1x24 = lim

x:c f1x2 + lim

x:c g1x2 = L + M

limx:c

x = c

limx:c

k = k for any constant k

limx:c

f1x2 = L limx:c

g1x2 = M

Explore & Discuss 1 The properties listed in Theorem 2 can be paraphrased in brief verbal statements.For example, property 3 simply states that the limit of a sum is equal to the sumof the limits. Write brief verbal statements for the remaining properties in Theorem 2.

E X A M P L E 5 Using Limit Properties Find

SOLUTION

Property 4

Properties 5 and 6

Property 2With a little practice, you will soon be able to omit the steps in the dashed boxes andsimply write

limx:31x2

- 4x2 = 3 � 3 - 4 � 3 = -3

= 3 � 3 - 4 � 3 = -3

= a limx:3

xb � a limx:3

xb - 4 limx:3

x

limx:31x2

- 4x2 = limx:3

x2

- limx:3

4x

limx:31x2

- 4x2.

MATCHED PROBLEM 5 Find limx: -2

1x2+ 5x2.

What happens if we try to evaluate a limit like the one in Example 5, but with xapproaching an unspecified number, such as c? Proceeding as we did in Example 5,we have

limx:c1x2

- 4x2 = c � c - 4 � c = c2- 4c


If we let we have

That is, this limit can be evaluated simply by evaluating the function f at c. It wouldcertainly simplify the process of evaluating limits if we could identify the functionsfor which

(1)

since we could use this fact to evaluate the limit. It turns out that there are manyfunctions that satisfy equation (1). We postpone a detailed discussion of these func-tions until the next section. For now, we note that if

is a polynomial function, then, by the properties in Theorem 1,

and if

is a rational function, where n(x) and d(x) are polynomials with then, byproperty 7 and the fact that polynomials n(x) and d(x) satisfy equation (1),

These results are summarized in Theorem 3.

limx:c

r1x2 = limx:c

n1x2d1x2 =

limx:c

n1x2limx:c

d1x2 =

n1c2d1c2 = r1c2

d1c2 Z 0,

r1x2 =

n1x2d1x2

= ancn+ an - 1c

n - 1+

Á+ a0 = f1c2

limx:c

f1x2 = limx:c1anxn

+ an - 1xn - 1

+Á

+ a02

f1x2 = anxn+ an - 1x

n - 1+

Á+ a0

limx:c

f1x2 = f1c2

limx:c

f1x2 = limx:c1x2

- 4x2 = c2- 4c = f1c2

f1x2 = x2- 4x,


THEOREM 3 LIMITS OF POLYNOMIAL AND RATIONAL FUNCTIONS1. f any polynomial function

2. r any rational function with a nonzero denominator at x = climx:c

r1x2 = r1c2limx:c

f1x2 = f1c2

E X A M P L E 6 Evaluating Limits Find each limit:

(A) (B) (C)

SOLUTION (A) Theorem 3

(B) Property 8

Theorem 3

(C) Theorem 3

=

813

limx:4

2x

3x + 1=

2 � 43 � 4 + 1

= 25

= 221-122 + 3

limx: -1

22x2+ 3 = 2 lim

x: -112x2

+ 32limx:21x3

- 5x - 12 = 23- 5 � 2 - 1 = -3

limx:4

2x

3x + 1lim

x: -122x2

+ 3limx:21x3

- 5x - 12

MATCHED PROBLEM 6 Find each limit:

(A) (B) (C) limx: -2

x2

x2+ 1

limx:223x2

- 6limx: -1

1x4- 2x + 32



E X A M P L E 7 Evaluating Limits Let

Find

(A) (B) (C) (D) f(2)

SOLUTION (A) If

(B) If

(C) Since the one-sided limits are not equal, does not exist.

(D) Because the definition of f does not assign a value to f for only for and does not exist.x 7 2, f122 x 6 2x = 2,

limx:2

f1x2 = 2 - 1 = 1

x 7 2, f1x2 = x - 1. limx:2+

f1x2 = limx:2+

1x - 12 = 22

+ 1 = 5

x 6 2, f1x2 = x2+ 1. lim

x:2-

f1x2 = limx:2-

1x2+ 12

limx:2

f1x2limx:2+

f1x2limx:2-

f1x2

f1x2 = ex2+ 1 if x 6 2

x - 1 if x 7 2

MATCHED PROBLEM 7 Let

Find each limit:

(A) (B) (C) (D) f(5)limx:5

f1x2limx:5+

f1x2limx:5-

f1x2

f1x2 = e 2x + 3 if x 6 5-x + 12 if x 7 5

It is important to note that there are restrictions on some of the limit properties.In particular, if

may present some difficulties, since limit property 7 (the limit of a quotient) does notapply when The next example illustrates some techniques that can beuseful in this situation.

limx:c

g1x2 = 0.

limx:c

f1x2 = 0 and limx:c

g1x2 = 0, then finding limx:c

f1x2g1x2

E X A M P L E 8 Evaluating Limits Find each limit:

(A)

(B)

SOLUTION (A) Algebraic simplification is often useful when the numerator and denominatorare both approaching 0.

(B) One-sided limits are helpful for limits involving the absolute value function.

If then

If then ƒx + 1 ƒ

x + 1= - 1.x 6 - 1, lim

x: -1-

x ƒ x + 1 ƒ

x + 1= lim

x: -1-

1-x2 = 1

ƒ x + 1 ƒ

x + 1= 1.x 7 - 1, lim

x: -1+

x ƒ x + 1 ƒ

x + 1= lim

x: -1+

1x2 = -1

limx:2

x2

- 4x - 2

= limx:2

1x - 221x + 22x - 2

= limx:21x + 22 = 4

limx: -1

x ƒ x + 1 ƒ

x + 1

limx:2

x2

- 4x - 2


Since the limit from the left and the limit from the right are not the same, weconclude that

limx: -1

x ƒ x + 1 ƒ

x + 1 does not exist


MATCHED PROBLEM 8 Find each limit:

(A) (B) limx:4

x2

- 16

ƒ x - 4 ƒ

limx: -3

x2

+ 4x + 3x + 3

In the solution to part A of Example 8, we used the following algebraic identity:

The restriction is necessary here because the first two expressions are not definedat Why didn’t we include this restriction in the solution in part A? When xapproaches 2 in a limit problem, it is assumed that x is close, but not equal, to 2. It isimportant that you understand that both of the following statements are valid:

limx:2

x2

- 4x - 2

= limx:21x + 22

x2- 4

x - 2= x + 2, x Z 2

x = 2.x Z 2

x2- 4

x - 2=

1x - 221x + 22x - 2

= x + 2 x Z 2

I N S I G H T

Limits like those in Example 8 occur so frequently in calculus that they are given aspecial name.

DEFINITION Indeterminate Form

If and then is said to be indeterminate, or, more

specifically, a 0�0 indeterminate form.

limx:c

f1x2g1x2lim

x:c g1x2 = 0,lim

x:c f1x2 = 0

The term indeterminate is used because the limit of an indeterminate form may ormay not exist (see parts A and B of Example 8).

CAUTION The expression does not represent a real number and should never beused as the value of a limit. If a limit is a indeterminate form, further investiga-tion is always required to determine whether the limit exists and to find its value if itdoes exist.

If the denominator of a quotient approaches 0 and the numerator approaches anonzero number, the limit of the quotient is not an indeterminate form. In fact, alimit of this form never exists, as Theorem 4 states.

0>00>0

THEOREM 4 LIMIT OF A QUOTIENTIf and

then

limx:c

f1x2g1x2 does not exist.

limx:c

g1x2 = 0,limx:c

f1x2 = L, L Z 0,



Explore & Discuss 2 Use algebraic and/or graphical techniques to analyze each of the followingindeterminate forms:

(A) (B) (C)

� Limits of Difference QuotientsLet the function f be defined in an open interval containing the number a. One of themost important limits in calculus is the limit of the difference quotient:

(3)If

as it often does, then limit (3) is an indeterminate form. The examples that followillustrate some useful techniques for evaluating limits of difference quotients.

limh:0

3f1a + h2 - f1a24 = 0

limh:0

f1a + h2 - f1a2h

limx:1

x2

- 1

1x - 122limx:1

1x - 122x2

- 1limx:1

x - 1

x2- 1

E X A M P L E 9 Limit of a Difference Quotient Find the following limit for

SOLUTION

= limh:0

4h

h= lim

h:0 4 = 4

= limh:0

12 + 4h - 5 - 12 + 5

h

limh:0

f13 + h2 - f132h

= limh:0

3413 � h2 - 54 - 34132 - 54h

limh:0

f13 + h2 - f132h

f1x2 = 4x - 5:

Since this is a 0/0indeterminate formand property 7 inTheorem 2 does not apply, we pro-ceed with algebraicsimplification.

Explore & Discuss 3 The following is an incorrect solution to Example 9, with the invalid statementsindicated by

Explain why each is used.Z

Z limh:0

-10 + 4

1= -6

= limh:0

-10 + 4h

h

limh:0

f13 + h2 - f132h

Z limh:0

413 + h2 - 5 - 4132 - 5

h

Z :

MATCHED PROBLEM 9 Find the following limit for f1x2 = 7 - 2x: limh:0

f14 + h2 - f142h

.

E X A M P L E 1 0 Limit of a Difference Quotient Find the following limit for

limh:0

f1-5 + h2 - f1-52h

ƒ x + 5 ƒ :f1x2 =

BARNMC03_0132328186.QXD 2/21/07 1:15 PM 9

SOLUTION

= limh:0

ƒ h ƒ

h does not exist

limh:0

f1-5 + h2 - f1-52h

= limh:0

ƒ1�5 � h2 + 5 ƒ - ƒ �5 + 5 ƒ

h


Since this is a 0/0 indetermi-nate form and property 7 inTheorem 2 does not apply, weproceed with algebraic simplification.

MATCHED PROBLEM 10 Find the following limit for f1x2 = ƒ x - 1 ƒ : limh:0

f11 + h2 - f112h

.

2. 3. does not existlimx:2

x - 2

ƒ x - 2 ƒ

limx:1

f1x2 = 2

1

2

3

�2

�3

�1 1 2 3 4 5

h(x)

x

h(x) � �x � 2�x � 2

x 0 1 2 3 4

h(x) 1.0 1.5 2.0 2.5 3.0 3.5 4.0

- 1- 2

E X A M P L E 1 1 Limit of a Difference Quotient Find the following limit for

SOLUTION

=

122 + 22=

1

222

= limh:0

122 + h + 22

= limh:0

2 + h - 2

h122 + h + 222

= limh:0

22 + h - 22

h � 22 + h + 2222 + h + 22

limh:0

f12 + h2 - f122h

= limh:0

22 � h - 22

h

limh:0

f12 + h2 - f122h

f1x2 = 2x:

This is a 0�0 indeterminate form, so property 7 in Theorem 2does not apply. Rationalizing the numerator will be of help.

1A - B21A + B2 = A2- B2

MATCHED PROBLEM 11 Find the following limit for f1x2 = 2x: limh:0

f13 + h2 - f132h

.

Answers to Matched Problem 1.

4. (A)

f102 = 0

limx:0

f1x2 = 0

limx:0+

f1x2 = 0

limx:0-

f1x2 = 0 (B)

does not exist

f(1) not defined

limx:1

f1x2 limx:1+

f1x2 = 2

limx:1-

f1x2 = 1

BARNMC03_0132328186.QXD 2/21/07 1:15 PM 0


x

�5

5�5

5

f (x)

A In Problems 1–4, use the graph of the function f shown toestimate the indicated limits and function values.

5. (A) (B)

(C) (D) g(1)

(E) Is it possible to define g(1) so thatExplain.

6. (A) (B)

(C) (D) g(2)

7. (A) (B)

(C) (D) g(3)

(E) Is it possible to redefine g(3) so thatExplain.

8. (A) (B)

(C) (D) g(4)

In Problems 9–12, use the graph of the function f shown toestimate the indicated limits and function values.

limx:4

g1x2lim

x:4+

g1x2limx:4-

g1x2limx:3

g1x2 = g132?

limx:3

g1x2lim

x:3+

g1x2limx:3-

g1x2limx:2

g1x2lim

x:2+

g1x2limx:2-

g1x2limx:1

g1x2 = g112?

limx:1

g1x2lim

x:1+

g1x2limx:1-

g1x2

f (x)

x

4

�2

�2 2 4

1. (A) (B)

(C) (D) f(0)

2. (A) (B)

(C) (D) f(1)

3. (A) (B)

(C) (D) f(2)

(E) Is it possible to redefine f(2) so thatExplain.

4. (A) (B)

(C) (D) f(4)

(E) Is it possible to define f(4) so thatExplain.

In Problems 5–8, use the graph of the function g shown toestimate the indicated limits and function values.

limx:4

f1x2 = f142?

limx:4

f1x2lim

x:4+

f1x2limx:4-

f1x2limx:2

f1x2 = f122?

limx:2

f1x2lim

x:2+

f1x2limx:2-

f1x2limx:1

f1x2lim

x:1+

f1x2limx:1-

f1x2limx:0

f1x2lim

x:0+

f1x2limx:0-

f1x2

g(x)

x

4

2

�2

�2 2 4

(C)

f(3) not defined

limx:3

f1x2 = 3

limx:3+

f1x2 = 3

limx:3-

f1x2 = 3 5.

6. (A) 6

(B)

(C) 45

26

-6

7. (A) 13

(B) 7

(C) Does not exist

(D) Not defined

8. (A)

(B) Does not exist

-2

9. 10. Does not exist 11. 1>12232-2

Exercise 3-1

9. (A) (B)

(C) (D)

(E) Is it possible to redefine so thatExplain.

10. (A) (B)

(C) (D)

(E) Is it possible to define so thatExplain.lim

x: -2 f1x2 = f1-22?

f1-22f1-22lim

x: -2 f1x2

limx: -2-

f1x2limx: -2+

f1x2lim

x: -3 f1x2 = f1-32?

f1-32f1-32lim

x: -3 f1x2

limx: -3-

f1x2limx: -3+

f1x2


11. (A) (B)

(C) (D) f(0)


12. (A) (B)

(C) (D) f(2)


In Problems 13–22, find each limit if it exists.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

Given that and find theindicated limits in Problems 23–34.

23. 24.

25. 26.

27. 28.

29. 30.

31. 32.

33. 34.

In Problems 35–38, sketch a possible graph of a function that satisfies the given conditions.

35.

36.

37.

38.

B In Problems 39–54, find each indicated quantity if it exists.

39. Let Find

(A) (B)

(C) (D) f(0)

40. Let Find

(A) (B)

(C) (D) f(0)limx:0

f1x2lim

x:0-

f1x2limx:0+

f1x2f1x2 = e2 + x if x … 0

2 - x if x 7 0.

limx:0

f1x2lim

x:0-

f1x2limx:0+

f1x2f1x2 = e1 - x2 if x … 0

1 + x2 if x 7 0.

f102 = -1; limx:0-

f1x2 = 2; limx:0+

f1x2 = 2

f1-22 = 2; limx: -2-

f1x2 = 1; limx: -2+

f1x2 = 1

f112 = -2; limx:1-

f1x2 = 2; limx:1+

f1x2 = -2

f102 = 1; limx:0-

f1x2 = 3; limx:0+

f1x2 = 1

limx:132 - g1x243lim

x:13f1x2 + 142

limx:1

322x + 2f1x2limx:12g1x2 - f1x2

limx:13f1x2 - 7x4g1x2lim

x:1 f1x232 - g1x24

limx:1

3 - f1x21 - 4g1x2lim

x:1

2 - f1x2x + g1x2

limx:13g1x2 - 3f1x24lim

x:132f1x2 + g1x24

limx:1

2g1x2limx:11-32f1x2

limx:1 g1x2 = 4,limx:1 f1x2 = -5

limx:0216 - 7xlim

x:125x + 4

limx:4

x - 2

xlim

x: -3

x

x + 5

limx: -1

x1x + 32limx:2

x1x - 42limx:51x - 32lim

x: -41x + 52

limx: -2

3xlimx:3

4x

limx:2

f1x2 = f122?

limx:2

f1x2lim

x:2-

f1x2limx:2+

f1x2limx:0

f1x2 = f102?

limx:0

f1x2lim

x:0-

f1x2limx:0+

f1x241. Let Find

(A) (B)

(C) (D) f(1)

42. Let Find

(A) (B)

(C) (D)

43. Let Find

(A) (B)

(C)

44. Let Find

(A) (B)

(C)

45. Let Find

(A) (B)

(C) (D) f(1)

46. Let Find

(A) (B)

(C) (D) f(3)

47. Let Find

(A) (B)

(C)

48. Let Find

(A) (B)

(C)

49. Let Find

(A) (B)

(C)

50. Let Find

(A) (B)

(C) limx:2

f1x2limx:0

f1x2limx: -3

f1x2f1x2 =

x2+ x - 6x + 3

.

limx:3

f1x2limx:0

f1x2limx: -2

f1x2f1x2 =

x2- x - 6x + 2

.

limx:3

f1x2limx:0

f1x2limx: -3

f1x2f1x2 =

x + 3

x2+ 3x

.

limx:4

f1x2limx:2

f1x2limx:0

f1x2f1x2 =

x - 2

x2- 2x

.

limx:3

f1x2lim

x:3-

f1x2limx:3+

f1x2f1x2 =

x - 3

ƒ x - 3 ƒ

.

limx:1

f1x2lim

x:1-

f1x2limx:1+

f1x2f1x2 =

ƒ x - 1 ƒ

x - 1.

limx:3

f1x2limx:0

f1x2limx: -3

f1x2

f1x2 = d x

x + 3if x 6 0

x

x - 3if x 7 0

.

limx:3

f1x2limx:0

f1x2limx: -3

f1x2

f1x2 = d x2- 9

x + 3if x 6 0

x2- 9

x - 3if x 7 0

.

f1-22limx: -2

f1x2lim

x: -2-

f1x2limx: -2+

f1x2f1x2 = e x + 3 if x 6 -22x + 2 if x 7 -2

.

limx:1

f1x2lim

x:1-

f1x2limx:1+

f1x2f1x2 = e x2 if x 6 1

2x if x 7 1.




51. Let Find

(A) (B)

(C)

52. Let Find

(A) (B)

(C)

53. Let Find

(A) (B)

(C)

54. Let Find

(A) (B)

(C)

Compute the following limit for each function inProblems 55–64:

55. 56.

57. 58.

59. 60.

61. 62.

63. 64.

C65. Let f be defined by

where m is a constant.

(A) Graph f for and find

limx:1-

f1x2 and limx:1+

f1x2m = 1,

f1x2 = e1 + mx if x … 14 - mx if x 7 1

f1x2 =

1x + 2

f1x2 =

2x - 1

f1x2 = 2 + ƒ x - 2 ƒf1x2 = ƒ x - 2 ƒ - 3

f1x2 = 1 + 2xf1x2 = 2x - 2

f1x2 = x2- 2f1x2 = x2

+ 1

f1x2 = 5x - 1f1x2 = 3x + 1

limh:0

f12 + h2 - f122h

limx:2

f1x2lim

x: -1 f1x2lim

x: -3 f1x2

f1x2 =

3x2+ 2x - 1

x2+ 3x + 2

.

limx:1

f1x2limx:0

f1x2limx:2

f1x2f1x2 =

2x2- 3x - 2

x2+ x - 6

.

limx:1

f1x2limx:0

f1x2limx: -1

f1x2f1x2 =

x2- 1

1x + 122.

limx:2

f1x2limx:0

f1x2limx: -2

f1x2f1x2 =

1x + 222x2

- 4.

(B) Graph f for and find

(C) Find m so that

and graph f for this value of m.

(D) Write a brief verbal description of each graph. Howdoes the graph in part (C) differ from the graphs inparts (A) and (B)?

66. Let f be defined by




(C) Find m so that


(D) Write a brief verbal description of each graph. Howdoes the graph in part (C) differ from the graphs inparts (A) and (B)?

Find each limit in Problems 67–70, where a is a real constant.

67.

68.

69.

70. limh:0

1a + h

-

1a

h, a Z 0

limh:0

2a + h - 2a

h, a 7 0

limh:0

331a + h2 - 24 - 13a - 22h

limh:0

1a + h22 - a2

h

limx:2-

f1x2 = limx:2+

f1x2

limx:2-

f1x2 and limx:2+

f1x2m = 1,

limx:2-

f1x2 and limx:2+

f1x2m = 0,

f1x2 = e -3m + 0.5x if x … 23m - x if x 7 2

limx:1-

f1x2 = limx:1+

f1x2

limx:1-

f1x2 and limx:1+

f1x2m = 2,

71. Telephone rates. A long-distance telephone service charges$0.99 for the first 20 minutes or less of a call and $0.07 perminute for each additional minute or fraction thereof.

(A) Write a piecewise definition of the charge F(x) for a long-distance call lasting x minutes.

(B) Graph F(x) for

(C) Find and

whichever exist.

limx:20

F1x2,limx:20-

F1x2, limx:20+

F1x2,0 6 x … 40.

72. Telephone rates. A second long-distance telephoneservice charges $0.09 per minute or fraction thereof forcalls lasting 10 minutes or more and $0.18 per minute orfraction thereof for calls lasting less than 10 minutes.(A) Write a piecewise definition of the charge G(x) for a

long-distance call lasting x minutes.(B) Graph G(x) for (C) Find and

whichever exist.

limx:10

G1x2,limx:10-

G1x2, limx:10+

G1x2,0 6 x … 40.

Applications


73. Telephone rates. Refer to Problems 71 and 72. Write abrief verbal comparison of the two services described forcalls lasting 20 minutes or less.

74. Telephone rates. Refer to Problems 71 and 72. Write abrief verbal comparison of the two services described forcalls lasting more than 20 minutes.

A company sells custom embroidered apparel and promotionalproducts. The table that follows shows the volume discountsoffered by the company, where x is the volume of a purchase in dollars. Problems 75 and 76 deal with two different interpre-tations of this discount method.

interval in the table. That is, the discounted price for a$4,000 purchase would be computed as follows:

(A) If x is the volume of a purchase before the discountis applied, write a piecewise definition for the dis-counted price P(x) of this purchase.

(B) Use one-sided limits to investigate the limit of P(x)as x approaches $1,000; as x approaches $3,000.

(C) Compare this discount method with the one in Prob-lem 75. Does one always produce a lower price thanthe other? Discuss.

77. Pollution. A state charges polluters an annual fee of $20per ton for each ton of pollutant emitted into the atmos-phere, up to a maximum of 4,000 tons. No fees are chargedfor emissions beyond the 4,000-ton limit. Write a piecewisedefinition of the fees F(x) charged for the emission of xtons of pollutant in a year. What is the limit of F(x) as xapproaches 4,000 tons? As x approaches 8,000 tons?

78. Pollution. Refer to Problem 77. The fee per ton of pollu-tion is given by Write a piecewise defini-tion of A(x). What is the limit of A(x) as x approaches4,000 tons? As x approaches 8,000 tons?

79. Voter turnout. Statisticians often use piecewise-definedfunctions to predict outcomes of elections. For the follow-ing functions f and g, find the limit of each function as xapproaches 5 and as x approaches 10:

g1x2 = L 0 if x … 50.8x - 0.04x2

- 3 if 5 6 x 6 101 if 10 … x

f1x2 = L 00.8 - 0.08x

0

if x … 5if 5 6 x 6 10if 10 … x

A1x2 = F1x2>x.

300 + 0.9717002 + 0.9512,0002 + 0.9311,0002 = 3,809


Volume Discount(Excluding Tax)

Volume ($x) Discount Amount

3%

5%

7%

10% $5,000 … x

$3,000 … x 6 $5,000

$1,000 … x 6 $3,000

$300 … x 6 $1,000

75. Volume discount. Assume that the volume discounts in the table apply to the entire purchase. That is, if thevolume x satisfies the entire purchase is discounted 3%. If the volume x satisfies

the entire purchase is discounted5%, and so on.

(A) If x is the volume of a purchase before the discountis applied, write a piecewise definition for thediscounted price D(x) of this purchase.

(B) Use one-sided limits to investigate the limit of D(x)as x approaches $1,000; as x approaches $3,000.

76. Volume discount. Assume that the volume discounts inthe table apply only to that portion of the volume in each

$1,000 … x 6 $3,000,

$300 … x 6 $1,000,

Section 3-2 CONTINUITY� Continuity� Continuity Properties� Solving Inequalities by Using Continuity Properties

Theorem 3 in Section 3-1 states that if f is a polynomial function or a rational func-tion with a nonzero denominator at then

(1)

Functions that satisfy equation (1) are said to be continuous at A firm under-standing of continuous functions is essential for sketching and analyzing graphs. Wewill also see that continuity properties provide a simple and efficient method for solv-ing inequalities—a tool that we will use extensively in later sections.

� ContinuityCompare the graphs shown in Figure 1. Notice that two of the graphs are broken; thatis, they cannot be drawn without lifting a pen off the paper. Informally, a function iscontinuous over an interval if its graph over the interval can be drawn withoutremoving a pen from the paper. A function whose graph is broken (disconnected) at

x = c.

limx:c

f1x2 = f1c2x = c,


S e c t i o n 3 - 2 Continuity 145

is said to be discontinuous at Function f (Fig. 1A) is continuous for all x. Function g (Fig. 1B) is discontinuous at but is continuous over anyinterval that does not include 2. Function h (Fig. 1C) is discontinuous at butis continuous over any interval that does not include 0.

Most graphs of natural phenomena are continuous, whereas many graphs inbusiness and economics applications have discontinuities. Figure 2A illustrates tem-perature variation over a 24-hour period—a continuous phenomenon. Figure 2B il-lustrates warehouse inventory over a 1-week period—a discontinuous phenomenon.

x = 0,x = 2,

x = c.x = c

Num

ber

of it

ems

Days after the first of the week

t

T

3 6 9 12 15 18 21 24

70

40

50

60

Tem

pera

ture

(�F

)

Hours after midnight

d

Q

10 2 3 4 5

100

200

300

400

500

(A) Temperature for a 24- hour period

(B) Inventory in a warehouse during 1 week

0

FIGURE 2

f (x)

x

�5

5�5

f (x) � x � 2x2 � 4x � 2

g(x) �

�x�xh(x) �

(A) lim f (x) � 4

f (2) � 4x�2

(B) lim g(x) � 4x�2

g(2) is not defined

(C) lim h(x) does not exist

h(0) is not definedx�0

4

0 2

g(x)

x

�5

5�5

4

2

h(x)

x

�12�2 �1

1

10 0

FIGURE 1

Explore & Discuss 1 (A) Write a brief verbal description of the temperature variation illustrated inFigure 2A, including estimates of the high and low temperatures during theperiod shown and the times at which they occurred.

(B) Write a brief verbal description of the changes in inventory illustrated inFigure 2B, including estimates of the changes in inventory and the times atwhich those changes occurred.

The preceding discussion leads to the following formal definition of continuity:

DEFINITION Continuity

A function f is continuous at the point x � c if

1. exists 2. f(c) exists 3.

A function is continuous on the open interval* (a, b) if it is continuous at each pointon the interval.

limx:c

f1x2 = f1c2limx:c

f1x2

* See Section 1-1 for a review of interval notation.

If one or more of the three conditions in the definition fails, the function isdiscontinuous at x = c.


Explore & Discuss 2 Sketch a graph of a function that is discontinuous at a point because it fails tosatisfy condition 1 in the definition of continuity. Repeat for conditions 2 and 3.


E X A M P L E 1 Continuity of a Function Defined by a Graph Use the definition of continu-ity to discuss the continuity of the function whose graph is shown in Figure 3.

f (x)

x

�2

5�5

5

FIGURE 3

SOLUTION We begin by identifying the points of discontinuity. Examining the graph, we see breaks or holes at and 3. Now we must determine which conditions in the definition of continuity are not satisfied at each of these points. Ineach case, we find the value of the function and the limit of the function at the pointin question.

f1-42 = 3

limx: -4

f1x2 does not exist

limx: -4+

f1x2 = 3

limx: -4-

f1x2 = 2Discontinuity at x = -4:

x = -4, -2, 1,

Since the one-sided limits are different,the limit does not exist (Section 3-1).

Thus, f is not continuous at because condition 1 is not satisfied.

f1-22 does not exist

limx: -2

f1x2 = 5

limx: -2+

f1x2 = 5

limx: -2-

f1x2 = 5Discontinuity at x = -2:

x = -4

The hole at indicates that 5 isnot the value of f at Since there isno solid dot elsewhere on the verticalline is not defined.x = -2, f1-22

-2.1-2, 52

Thus, f is not continuous at because condition 2 is not satisfied.

f(1) does not exist

This time, f is not continuous at because neither of conditions 1 and 2 is satisfied.

f132 = 1

limx:3

f1x2 = 3

limx:3+

f1x2 = 3

limx:3-

f1x2 = 3Discontinuity at x = 3:

x = 1

limx:1

f1x2 does not exist

limx:1+

f1x2 = 1

limx:1-

f1x2 = 4Discontinuity at x = 1:

x = -2

The solid dot at (3, 1) indicates that f132 = 1.

Conditions 1 and 2 are satisfied, but f is not continuous at because condition3 is not satisfied.

Having identified and discussed all points of discontinuity, we can now concludethat f is continuous except at and 3.x = -4, -2, 1,

x = 3



For functions defined by equations, it is important to be able to locate points of dis-continuity by examining the equation.

f (x)

x

�2

5�5

5

FIGURE 4

Rather than list the points where a function is discontinuous, sometimes it is useful tostate the intervals on which the function is continuous. Using the set operation union, de-noted by we can express the set of points where the function in Example 1 is continu-ous as follows:

1- q , -42 ´ 1-4, -22 ´ 1-2, 12 ´ 11, 32 ´ 13, q2

´,

I N S I G H T

MATCHED PROBLEM 1 Use the definition of continuity to discuss the continuity of the function whose graphis shown in Figure 4.

E X A M P L E 2 Continuity of Functions Defined by Equations Using the definition of con-tinuity, discuss the continuity of each function at the indicated point(s).

(A) at (B) at

(C) at and at

SOLUTION (A) f is continuous at since

See Figure 1A.

(B) g is not continuous at since is not defined (see Fig. 1B).

(C) h is not continuous at since is not defined; also, does not exist.

h is continuous at since

See Figure 1C.limx:1

ƒ x ƒ

x= 1 = h112

x = 1,

limx:0 h1x2h102 = ƒ 0 ƒ>0x = 0,

g122 = 0>0x = 2,

limx:2

f1x2 = 4 = f122x = 2,

x = 1x = 0h1x2 =

ƒ x ƒ

x

x = 2g1x2 =

x2- 4

x - 2x = 2f1x2 = x + 2

MATCHED PROBLEM 2 Using the definition of continuity, discuss the continuity of each function at theindicated point(s).

(A) at (B) at

(C) at and at x = 0x = 2h1x2 =

x - 2

ƒ x - 2 ƒ

x = 1g1x2 =

x2- 1

x - 1x = 1f1x2 = x + 1

We can also talk about one-sided continuity, just as we talked about one-sidedlimits. For example, a function is said to be continuous on the right at if

and continuous on the left at if Afunction is continuous on the closed interval [a, b] if it is continuous on the open in-terval (a, b) and is continuous both on the right at a and on the left at b.

Figure 5A illustrates a function that is continuous on the closed interval Figure 5B illustrates a function that is continuous on the half-closed interval 30, q2.3-1, 14.

limx:c- f1x2 = f1c2.x = climx:c+ f1x2 = f1c2 x = c



xx

g(x)f (x)

1 100 2 3 4 5�1

1

1

2

3f (x) � �1 � x2g(x) � �x

(A) f is continuous on the closed interval [�1, 1]

(B) g is continuous on the half-closed interval [0, �)

FIGURE 5 Continuity on closed and half-closed intervals

� Continuity PropertiesFunctions have some useful general continuity properties:

If two functions are continuous on the same interval, then their sum, difference,product, and quotient are continuous on the same interval, except for values of xthat make a denominator 0.

These properties, along with Theorem 1 below, enable us to determine intervals ofcontinuity for some important classes of functions without having to look at theirgraphs or use the three conditions in the definition.

THEOREM 1 CONTINUITY PROPERTIES OF SOME SPECIFIC FUNCTIONS(A) A constant function where k is a constant, is continuous for all x.

is continuous for all x.

(B) For n a positive integer, is continuous for all x.


(C) A polynomial function is continuous for all x.


(D) A rational function is continuous for all x except those values that make adenominator 0.

is continuous for all x except a value that makes the denominator 0.

(E) For n an odd positive integer greater than 1, is continuous wherever f(x)is continuous.


(F) For n an even positive integer, is continuous wherever f(x) is continuousand nonnegative.

is continuous on the interval 30, q2.41x

2n f1x232x2

2n f1x2x = 1,

x2+ 1

x - 1

2x3- 3x2

+ x - 5

f1x2 = x5

f1x2 = xn

f1x2 = 7

f1x2 = k,

Parts C and D of Theorem 1 are the same as Theorem 3 in Section 3-1. They arerepeated here to emphasize their importance.

E X A M P L E 3 Using Continuity Properties Using Theorem 1 and the general properties ofcontinuity, determine where each function is continuous.

(A) (B)

(C) (D) f1x2 = 2x - 2f1x2 =

32x2- 4

f1x2 =

x

1x + 221x - 32f1x2 = x2- 2x + 1



SOLUTION (A) Since f is a polynomial function, f is continuous for all x.

(B) Since f is a rational function, f is continuous for all x except and 3 (values thatmake the denominator 0).

(C) The polynomial function is continuous for all x. Since is odd, f iscontinuous for all x.

(D) The polynomial function is continuous for all x and nonnegative for Since is even, f is continuous for or on the interval 32, q2.x Ú 2,n = 2

x Ú 2.x - 2

n = 3x2- 4

-2

x

f (x)

5

�5

50

(7, �3)

(2, 5)

FIGURE 6

MATCHED PROBLEM 3 Using Theorem 1 and the general properties of continuity, determine where eachfunction is continuous.

(A) (B)

(C) (D) f1x2 = 32x3+ 1f1x2 = 2x - 4

f1x2 =

x2

1x + 121x - 42f1x2 = x4+ 2x2

+ 1

� Solving Inequalities by Using Continuity PropertiesOne of the basic tools for analyzing graphs in calculus is a special line graph called asign chart. We will make extensive use of this type of chart in later sections. In thediscussion that follows, we use continuity properties to develop a simple and effi-cient procedure for constructing sign charts.

Suppose that a function f is continuous over the interval (1, 8) and forany x in (1, 8). Suppose also that a positive number. Is it possible for f(x)to be negative for any x in the interval (1, 8)? The answer is “no.” If f(7) were for example, as shown in Figure 6, how would it be possible to join the points (2, 5)and with the graph of a continuous function without crossing the x axis be-tween 1 and 8 at least once? [Crossing the x axis would violate our assumption that

for any x in (1, 8).] Thus, we conclude that f(x) must be positive for all xin (1, 8). If f(2) were negative, then, using the same type of reasoning, f(x) would haveto be negative over the entire interval (1, 8).

In general, if f is continuous and f(x) � 0 on the interval (a, b), then f(x) cannotchange sign on (a, b). This is the essence of Theorem 2.

f1x2 Z 0

17, -32-3,

f122 = 5,f1x2 Z 0

THEOREM 2 SIGN PROPERTIES ON AN INTERVAL (a, b)If f is continuous on (a, b) and for all x in (a, b), then either forall x in (a, b) or for all x in (a, b).f1x2 6 0

f1x2 7 0f1x2 Z 0

Theorem 2 provides the basis for an effective method of solving many types of inequalities. Example 4 illustrates the process.

E X A M P L E 4 Solving an Inequality Solve

SOLUTION We start by using the left side of the inequality to form the function f:

The rational function f is discontinuous at and for (a frac-tion is 0 when the numerator is 0 and the denominator is not 0). We plot and

which we call partition numbers, on a real-number line (Fig. 7). (Note thatthe dot at 2 is open, because the function is not defined at ) The partition num-bers 2 and determine three open intervals: and Thefunction f is continuous and nonzero on each of these intervals. From Theorem 2, we know that f(x) does not change sign on any of these intervals. Thus, we can find

12, q2.1- q , -12, 1-1, 22,-1x = 2.

x = -1,x = 2

x = -1f1x2 = 0x = 2,

f1x2 =

x + 1x - 2

x + 1x - 2

7 0.

�1�2 2 30 1x

FIGURE 7


the sign of f(x) on each of the intervals by selecting a test number in each interval andevaluating f(x) at that number. Since any number in each subinterval will do, wechoose test numbers that are easy to evaluate: and 3. The table in the marginshows the results.

The sign of f(x) at each test number is the same as the sign of f(x) over the interval containing that test number. Using this information, we construct a sign chartfor f(x):

-2, 0,


� � � � � � � � � � � � � � �

Test numbers

(2, �)(��, �1) (�1, 2)

�1�2 2 30 1x

f (x)

From the sign chart, we can easily write the solution of the given nonlinear inequality:

Inequality notationInterval notation

Most of the inequalities we encounter will involve strict inequalities . Ifit is necessary to solve inequalities of the form or we simply include the end-point x of any interval if f is defined at x and f(x) satisfies the given inequality. Forexample, from the sign chart in Example 4, the solution of the inequality

Inequality notationInterval notation

In general, given a function f, a partition number is a value of x such that f is dis-continuous at x or Partition numbers determine open intervals in which f(x)does not change sign. By using a test number from each interval, we can construct asign chart for f(x) on the real-number line. It is then an easy matter to determinewhere or that is, it is easy to solve the inequality or

We summarize the procedure for constructing sign charts in the following box:f1x2 7 0.

f1x2 6 0f1x2 7 0;f1x2 6 0

f1x2 = 0.

x + 1x - 2

Ú 0 is x … -1 or x 7 21- q , -14 ´ 12, q2

… ,Ú

17 or 6 2f1x2 7 0 for

x 6 -1 or x 7 21- q , -12 ´ 12, q2

PROCEDURE Constructing Sign ChartsGiven a function f,Step 1. Find all partition numbers. That is,

(A) Find all numbers such that f is discontinuous. (Rational functions are discon-tinuous for values of x that make a denominator 0.)

(B) Find all numbers such that (For a rational function, this occurs wherethe numerator is 0 and the denominator is not 0.)

Step 2. Plot the numbers found in step 1 on a real-number line, dividing the numberline into intervals.

Step 3. Select a test number in each open interval determined in step 2, and evalu-ate f(x) at each test number to determine whether f(x) is positive ornegative in each interval.

Step 4. Construct a sign chart, using the real-number line in step 2. This will showthe sign of f(x) on each open interval.

Note: From the sign chart, it is easy to find the solution of the inequality or f1x2 7 0.

f1x2 6 0

1-2 1+2

f1x2 = 0.

MATCHED PROBLEM 4 Solve x2

- 1x - 3

6 0.

Test Numbers

x f(x)

0

3 4 1�2 -1

2 1�2 14 1�2- 2



Answers to Matched Problems 1. f is not continuous at and 4.

2. (A) f is continuous at since

(B) g is not continuous at since g(1) is not defined.(C) h is not continuous at for two reasons: h(2) does not exist and does

not exist.h is continuous at since

3. (A) Since f is a polynomial function, f is continuous for all x.

(B) Since f is a rational function, f is continuous for all x except and 4 (values that makethe denominator 0).

(C) The polynomial function is continuous for all x and nonnegative for Sinceis even, f is continuous for or on the interval

(D) The polynomial function is continuous for all x. Since is odd, f is continuousfor all x.

4. or 1 6 x 6 3; 1- q , -12 ´ 11, 32- q 6 x 6 -1

n = 3x3+ 1

34, q2.x Ú 4,n = 2x Ú 4.x - 4

-1

limx:0

h1x2 = -1 = h102.x = 0,

limx:2

h1x2x = 2x = 1,

limx:1

f1x2 = 2 = f112.x = 1,

x = 4: limx:4

f1x2 does not exist, and f142 does not exist

x = 2: limx:2

f1x2 = 5, but f122 = 3

x = -1: f1-12 = 1, but limx: -1

f1x2 does not exist

x = -3: limx: -3

f1x2 = 3, but f1-32 does not exist

x = -3, -1, 2,

9. (A) (B)

(C) (D)

(E) Is f continuous at Explain.

10. (A) (B)

(C) (D)


Problems 11–14 refer to the function g shown in the figure. Usethe graph to estimate the indicated quantities to the nearest integer.

x = -1?

f1-12limx: -1

f1x2lim

x: -1+

f1x2limx: -1-

f1x2x = -2?

f1-22limx: -2

f1x2lim

x: -2+

f1x2limx: -2-

f1x2Exercise 3-2

A In Problems 1–6, sketch a possible graph of a function that sat-isfies the given conditions at and discuss the continuityof f at

1. and

2. and

3. and

4. and

5. and

6. and

Problems 7–10 refer to the function f shown in the figure. Use thegraph to estimate the indicated quantities to the nearest integer.

limx:1+

f1x2 = -2f112 = 2, limx:1-

f1x2 = 2,

limx:1+

f1x2 = -2f112 = -2, limx:1-

f1x2 = 2,

limx:1

f1x2 = -2f112 = -2

limx:1

f1x2 = -2f112 = 2

limx:1

f1x2 = 2f112 = -2

limx:1

f1x2 = 2f112 = 2

x = 1.x = 1,

f (x)

x5�5

5x

�5

5

5

g(x)

7. (A) (B)

(C) (D) f(1)


8. (A) (B)

(C) (D) f(2)

(E) Is f continuous at Explain.x = 2?

limx:2

f1x2lim

x:2+

f1x2limx:2-

f1x2x = 1?

limx:1

f1x2lim

x:1+

f1x2limx:1-

f1x2 11. (A) (B)

(C) (D)

(E) Is g continuous at Explain.

12. (A) (B)

(C) (D)

(E) Is g continuous at Explain.x = -2?

g1-22limx: -2

g1x2lim

x: -2+

g1x2limx: -2-

g1x2x = -3?

g1-32limx: -3

g1x2lim

x: -3+

g1x2limx: -3-

g1x2


13. (A) (B)

(C) (D) g(2)


14. (A) (B)

(C) (D) g(4)


Use Theorem 1 to determine where each function in Problems 15–24 is continuous.

15. 16.

17. 18.

19.

20.

21. 22.

23. 24.

B25. Given the function

(A) Graph f.(B)

(C)(D) Is f continuous at (E) Where is f discontinuous?

26. Given the function

(A) Graph g.

(B)

(C)(D) Is g continuous at (E) Where is g discontinuous?

In Problems 27–34, use a sign chart to solve each inequality.Express answers in inequality and interval notation.

27. 28.

29. 30.

31. 32.

33. 34.

35. Use the graph of f to determine where

(A) (B) f1x2 6 0f1x2 7 0

x - 4

x2+ 2x

6 0x2

+ 5x

x - 37 0

x4- 9x2

7 0x36 4x

x2+ 7x 7 -10x2

+ 21 7 10x

x2- 2x - 8 6 0x2

- x - 12 6 0

x = 1?g112 = ?

limx:1

g1x2 = ?

g1x2 = e -1 if x is an even integer1 if x is not an even integer

x = 2?f122 = ?

limx:2

f1x2 = ?

f1x2 = e2 if x is an integer1 if x is not an integer

N1x2 =

x2+ 4

4 - 25x2M1x2 =

x - 1

4x2- 9

G1x2 =

1 - x2

x2+ 1

F1x2 =

2x

x2+ 9

n1x2 =

x - 21x - 321x + 12

m1x2 =

x + 11x - 121x + 42

k1x2 =

2x

x - 4g1x2 =

3x

x + 2

h1x2 = 4 - 2xf1x2 = 3x - 4

x = 4?

limx:4

g1x2lim

x:4+

g1x2limx:4-

g1x2x = 2?

limx:2

g1x2lim

x:2+

g1x2limx:2-

g1x2152 C H A P T E R 3 Limits and the Derivative

Express answers in interval notation.

x

�150

5

150

f (x)

�5

36. Use the graph of g to determine where

(A) (B)


g1x2 6 0g1x2 7 0

x

�150

5

150

g(x)

�5

In Problems 37–40, use a graphing calculator to approximatethe partition numbers of each function f(x) to four decimalplaces. Then solve the following inequalities:

(A) (B)


37.

38.

39.

40.

Use Theorem 1 to determine where each function in Problems 41–48 is continuous. Express the answer in interval notation.

41. 42.

43. 44.

45. 46.

47. 48.

In Problems 49–54, graph f, locate all points of discontinuity,and discuss the behavior of f at these points.

49.

50. f1x2 = e x2 if x … 12x if x 7 1

f1x2 = e1 + x if x 6 15 - x if x Ú 1

32x2

+ 22x2+ 1

24 - x22x2- 9

32x - 8

325 - x

27 - x2x - 6

f1x2 =

x3- 5x + 1

x2- 1

f1x2 =

3 + 6x - x3

x2- 1

f1x2 = x4- 4x2

- 2x + 2

f1x2 = x4- 6x2

+ 3x + 5

f1x2 6 0f1x2 7 0



51.

52.

53.

54.

In Problems 55–58, use a graphing calculator to locate allpoints of discontinuity of f, and discuss the behavior of f atthese points. [Hint: Select Xmin and Xmax so that the suspectedpoint of discontinuity is the midpoint of the graphing interval (Xmin, Xmax).]

55. 56.

57. 58.

C59. Use the graph of the function g to answer the following

questions:

(A) Is g continuous on the open interval

(B) Is g continuous from the right at That is,does

(C) Is g continuous from the left at That is, does

(D) Is g continuous on the closed interval 3-1, 24?limx:2- g1x2 = g122?

x = 2?

limx:-1+ g1x2 = g1-12?x = -1?

1-1, 22?

f1x2 =

x3- 8

ƒ x ƒ - 2f1x2 =

x2- 1

ƒ x ƒ - 1

f1x2 = x +

ƒ 3x + 9 ƒ

x + 3f1x2 = x +

ƒ 2x - 4 ƒ

x - 2

f1x2 = L 1 if x 6 00 if x = 01 + x if x 7 0

f1x2 = L -x if x 6 01 if x = 0x if x 7 0

f1x2 = e x2 if x … 22x if x 7 2

f1x2 = e1 + x if x … 25 - x if x 7 2

Problems 61 and 62 refer to the greatest integer function,which is denoted by and is defined as

For example,

The graph of is shown. There, we can see that

and so on.

Œx œ = 2 for 2 … x 6 3 Œx œ = 1 for 1 … x 6 2 Œx œ = 0 for 0 … x 6 1 Œx œ = -1 for - 1 … x 6 0 Œx œ = -2 for - 2 … x 6 -1

f1x2 = Œx œ Œ2.5 œ = greatest integer … 2.5 = 2

Œ2 œ = greatest integer … 2 = 2 Œ -3.6 œ = greatest integer … -3.6 = -4

Œx œ = greatest integer … x

Œx œ

x

g(x)

420�2

4

2

g(x) � �x2 � 2x � 2

Figure for 59

x

f (x)

�5

5�5

5 f (x) � �x�

Figure for 61 and 62

60. Use the graph of the function f to answer the followingquestions:

(A) Is f continuous on the open interval (0, 3)?

(B) Is f continuous from the right at That is, does

(C) Is f continuous from the left at That is, does

(D) Is f continuous on the closed interval [0, 3]?

limx:3- f1x2 = f132?x = 3?

limx:0+ f1x2 = f102?x = 0?

x

f (x)

50

5

f (x) � x2 � 2x � 2

Figure for 60

61. (A) Is f continuous from the right at (B) Is f continuous from the left at (C) Is f continuous on the open interval (0, 1)?(D) Is f continuous on the closed interval [0, 1]?(E) Is f continuous on the half-closed interval [0, 1)?

62. (A) Is f continuous from the right at (B) Is f continuous from the left at (C) Is f continuous on the open interval (1, 2)?(D) Is f continuous on the closed interval [1, 2]?(E) Is f continuous on the half-closed interval [1, 2)?

In Problems 63–66, sketch a possible graph of a function f thatis continuous for all real numbers and satisfies the givenconditions. Find the x intercepts of f.

63. on and on 1-5, 2212, q2; f1x2 7 01- q , -52f1x2 6 0

x = 2?x = 2?

x = 0?x = 0?


64. on and on

65. on and onand

66. on and (2, 7); on and

67. The function satisfies and Is f equal to 0 anywhere on the interval

Does this contradict Theorem 2? Explain.

68. The function satisfies andIs f equal to 0 anywhere on the interval (0, 9)?

Does this contradict Theorem 2? Explain.f172 = 2.

f122 = -3f1x2 = 6>1x - 421-1, 32?

f122 = -2.f102 = 2f1x2 = 2>11 - x2

17, q21-3, 22f1x2 6 01- q , -32f1x2 7 0

14, q21-6, -121-1, 42; f1x2 7 01- q , -62f1x2 6 0

1-4, 3213, q2; f1x2 6 01- q , -42f1x2 7 0 69. The function f is continuous and never 0 on the interval(0, 4) and is continuous and never 0 on the interval (4, 8).Also, and Discuss the validity of thefollowing statement, and illustrate your conclusions withgraphs: Either or f is discontinuous at

70. The function f is continuous and never 0 on the intervaland is continuous and never 0 on the interval

(1, 4). Also, and Discuss thevalidity of the following statement and illustrate yourconclusions with graphs: Either or f isdiscontinuous at x = 1.

f112 = 0

f132 = 4.f1-22 = -31-3, 12

x = 4.f142 = 0

f162 = -3.f122 = 3


Applications71. Postal rates. First-class postage in 2006 was $0.39 for the

first ounce (or any fraction thereof) and $0.24 for eachadditional ounce (or fraction thereof).

(A) Write a piecewise definition of the first-class postageP(x) for a letter weighing x ounces.

(B) Graph P(x) for (C) Is P(x) continuous at At Explain.x = 4?x = 4.5?

0 6 x … 5.

76. Natural-gas rates. Table 2 shows the rates for natural gascharged by the Middle Tennessee Natural Gas UtilityDistrict during the winter months. The customer charge isa fixed monthly charge, independent of the amount of gasused during the month.

(A) Write a piecewise definition of the monthly chargeS(x) for a customer who uses x therms in a wintermonth.

(B) Graph S(x).(C) Is S(x) continuous at At Explain.x = 50?x = 5?

72. Telephone rates. A long-distance telephone servicecharges $0.07 for the first minute (or any fraction thereof)and $0.05 for each additional minute (or fraction thereof).

(A) Write a piecewise definition of the charge R(x) for along-distance call lasting x minutes.

(B) Graph R(x) for (C) Is R(x) continuous at At Explain.

73. Postal rates. Discuss the differences between the functionand the function P(x) defined in

Problem 71.

74. Telephone rates. Discuss the differences between thefunction and the function R(x)defined in Problem 72.

75. Natural-gas rates. Table 1 shows the rates for natural gascharged by the Middle Tennessee Natural Gas UtilityDistrict during the summer months. The customer chargeis a fixed monthly charge, independent of the amount ofgas used during the month.

(A) Write a piecewise definition of the monthly chargeS(x) for a customer who uses x therms* in a summermonth.

S1x2 = 0.07 + 0.05 Œx œ

Q1x2 = 0.39 + 0.24 Œx œx = 3?x = 3.5?

0 6 x … 6.

TABLE 1 Summer (May–September)

Base charge $5.00

First 50 therms 0.63 per therm

Over 50 therms 0.45 per therm

*A British thermal unit (Btu) is the amount of heat required to raise the tem-perature of 1 pound of water 1 degree Fahrenheit, and a therm is 100,000 Btu.

TABLE 2 Winter (October– April)

Base charge $5.00

First 5 therms 0.69 per therm

Next 45 therms 0.65 per therm

Over 50 therms 0.63 per therm

77. Income. A personal-computer salesperson receives a basesalary of $1,000 per month and a commission of 5% of allsales over $10,000 during the month. If the monthly salesare $20,000 or more, the salesperson is given an additional$500 bonus. Let E(s) represent the person’s earnings dur-ing the month as a function of the monthly sales s.

(A) Graph E(s) for

(B) Find and E(10,000).

(C) Find and E(20,000).

(D) Is E continuous at At

78. Equipment rental. An office equipment rental andleasing company rents copiers for $10 per day (and any

s = 20,000?s = 10,000?

lims:20,000 E1s2lims:10,000 E1s2

0 … s … 30,000.

(B) Graph S(x).(C) Is S(x) continuous at Explain.x = 50?


S e c t i o n 3 - 3 Infinite Limits and Limits at Infinity 155

fraction thereof) or for $50 per 7-day week. Let C(x) bethe cost of renting a copier for x days.

(A) Graph C(x) for (B) Find and C(4.5).(C) Find and C(8).(D) Is C continuous at At

79. Animal supply. A medical laboratory raises its own rab-bits. The number of rabbits N(t) available at any time tdepends on the number of births and deaths. When abirth or death occurs, the function N generally has adiscontinuity, as shown in the figure.

x = 8?x = 4.5?limx:8 C1x2limx:4.5 C1x2

0 … x … 10.

80. Learning. The graph shown might represent the historyof a particular person learning the material on limits andcontinuity in this book. At time the student’s mindgoes blank during a quiz. At time the instructorexplains a concept particularly well, and suddenly a bigjump in understanding takes place.

(A) Where is the function p discontinuous?(B)(C)(D) limt:t4

p1t2 = ?; p1t42 = ?limt:t2

p1t2 = ?; p1t22 = ?limt:t1

p1t2 = ?; p1t12 = ?

t4,t2,

tt1 t2 t3 t4 t5 t6 t7

N(t)

0

5

10

Time

Num

ber

of r

abbi

ts

Figure for 79

(A) Where is the function N discontinuous?(B)(C) limt:t3

N1t2 = ?; N1t32 = ?limt:t5

N1t2 = ?; N1t52 = ?

tt1 t2 t3 t4

p(t)

0

100

50

Time

Perc

enta

ge o

f m

aste

ry

Figure for 80

Section 3-3 INFINITE LIMITS AND LIMITS AT INFINITY� Infinite Limits� Locating Vertical Asymptotes� Limits at Infinity� Finding Horizontal Asymptotes

In this section, we consider two new types of limits: infinite limits and limits at infinity. Infinite limits and vertical asymptotes are used to analyze the behavior of afunction that is unbounded near . Limits at infinity and horizontal asymptotesare used to describe the behavior of a function as x assumes arbitrarily large posi-tive values or arbitrarily large negative values. Although we will include graphs toillustrate basic concepts, we postpone a discussion of graphing techniques until Chapter 5.

� Infinite LimitsA function is discontinuous at any point a where does not exist. For ex-ample, if the one-sided limits are different at , then the limit does not existand the function is discontinuous at . Another situation in which a limit mayfail to exist involves functions whose values become very large as x approaches a. The special symbol is often used to describe this type of behavior. To illustrate,consider the function , which is discontinuous only at . As xapproaches 1 from the right, the values of are positive and become larger f(x)

x = 1f(x) = 1>(x - 1)q

x = ax = a

limx:af(x)

x = a


and larger; that is, increases without bound (Table 1). We express this behaviorsymbolically as

(1)

Since is a not a real number, the limit in (1) does not exist. We are using the symbol to describe the manner in which the limit fails to exist, and we call thissituation an infinite limit. If, now, x approaches 1 from the left, the values of arenegative and become larger and larger in absolute value; that is, decreasesthrough negative values without bound (see Table 2). We express this behavior sym-bolically as

(2)

From (1) and (2), we have

limx:1+

1

x - 1= q and lim

x:1-

1

x - 1= - q

f(x) =

1x - 1

: - q as x : 1-

f(x)f(x)

q

q

f(x) =

1x - 1

: q as x : 1+

f(x)

The one-sided limits in (1) and (2) also illustrate the behavior of the graph near(Fig. 1). Does the unrestricted limit of as exist? No, because neither

of the one-sided limits exist. Also, there is no reasonable way to use the symbol to describe the behavior of as on both sides of 1. Thus, we say simply that

Explore & Discuss 1 Let

Construct tables for as and as . Use these tables and infinitelimits to discuss the behavior of near .

We used the dashed vertical line in Figure 1 to illustrate the infinite limits as xapproaches 1 from the right and from the left. We call this line a vertical asymptote.

x = 1

x = 1g(x)x : 1-x : 1+g(x)

g(x) =

1

(x - 1)2

limx:1

1

x - 1 does not exist

x : 1f(x)q

x = 1f(x)x : 1

x

f (x)

�10

�5

10

5

21

1x � 1

f (x) �

FIGURE 1


DEFINITION Infinite Limits and Vertical AsymptotesThe vertical line is a vertical asymptote for the graph of if

[that is, if either increases or decreases without bound as x approaches a fromthe right or from the left].

f(x)

f(x) : q or f(x) : - q as x : a+ or x : a-

y = f(x)x = a

TABLE 1

x

1.1 10

1.01 100

1.001 1,000

1.0001 10,000

1.00001 100,000

1.000001 1,000,000

f(x) �1

x � 1

TABLE 2

x

0.9 �10

0.99 �100

0.999 �1,000

0.9999 �10,000

0.99999 �100,000

0.999999 �1,000,000

f(x) �1

x � 1



� Locating Vertical AsymptotesHow do we locate vertical asymptotes? If a function f is continuous at , then

(3)

Since all of the limits in (3) exist and are finite, f cannot have a vertical asymptote at. In order for f to have a vertical asymptote at , at least one of the limits

in (3) must be an infinite limit and f must be discontinuous at . We know thatpolynomial functions are continuous for all real numbers, so a polynomial has novertical asymptotes. Since a rational function is discontinuous at the zeros of its de-nominator, a vertical asymptote of a rational function can occur only at a zero of itsdenominator. Theorem 1 provides a simple procedure for locating the vertical asymp-totes of a rational function.

x = ax = ax = a

limx:a

f(x) = limx:a+

f(x) = limx:a-

f(x) = f(a)

x = a

THEOREM 1 LOCATING VERTICAL ASYMPTOTES OF RATIONAL FUNCTIONSIf is a rational function, and , then the line is a vertical asymptote of the graph of f.

x = cn(c) Z 0d(c) = 0f(x) = n(x)>d(x)

If and both and then the limit of as xapproaches c involves an indeterminate form and Theorem 1 does not apply:

Algebraic simplification is often useful in this situation.

limx:c

f(x) = limx:c

n(x)

d(x)

00

indeterminate form

f(x)d1c2 = 0,n(c) = 0f(x) = n(x)>d(x)

E X A M P L E 1 Locating Vertical Asymptotes Let

Describe the behavior of f at each point of discontinuity. Use and when ap-propriate. Identify all vertical asymptotes.

SOLUTION Let and . Factoring the denominator, we see that

has two zeros: and . These are the points of discontinuity of f. First we consider . Since and , Theorem 1 tells

us that the line is a vertical asymptote. So at least one of the one-sided limitsat must be either or . Examining tables of values of f for x near �1or a graph on a graphing calculator will show which is the case. From Tables 3 and 4,we see that

- qqx = -1x = -1

n(-1) = -2 Z 0d(-1) = 0x = -1x = 1x = -1

d(x) = x2- 1 = (x - 1)(x + 1)

d(x) = x2- 1n(x) = x2

+ x - 2

- qq

f(x) =

x2+ x - 2

x2- 1

limx: -1-

x2+ x - 2

x2- 1

= -q and limx: -1+

x2+ x - 2

x2- 1

= q

TABLE 3

x

�1.1 �9

�1.01 �99

�1.001 �999

�1.0001 �9,999

�1.00001 �99,999

f(x) �x2 � x � 2

x2 � 1

TABLE 4

x

�0.9 11

�0.99 101

�0.999 1,001

�0.9999 10,001

�0.99999 100,001

f(x) �x2 � x � 2

x2 � 1


Now we consider the other zero of , . This time and Theorem 1does not apply. We use algebraic simplification to investigate the behavior of thefunction at :

Reduced to lowest terms (see Appendix A, Section A-4)

Since the limit exists as x approaches 1, f does not have a vertical asymptote at The graph of f (Fig. 2) shows the behavior at the vertical asymptote and alsoat .x = 1

x = -1x = 1.

=

32

= limx:1

x + 2x + 1

= limx:1

(x - 1)(x + 2)

(x - 1)(x + 1)

00

indeterminate form limx:1

f(x) = limx:1

x2+ x - 2

x2- 1

x = 1

n(1) = 0x = 1d(x)

TABLE 6

x

1.9 472.2

1.99 47,920

1.999 4,799,200.2

f(x) �x2 � 20

5(x � 2)2

TABLE 5

x

2.1 488.2

2.01 48,080

2.001 4,800,800.2

f(x) �x2 � 20

5(x � 2)2

x

f (x)

�5 5

�5

5

FIGURE 2 f1x2 =

x2+ x - 2

x2- 1

x

f (x)

20

5

10

15

321 4

x2 � 205(x � 2)2

f (x) �

FIGURE 3


MATCHED PROBLEM 1 Let f(x) =

x - 3

x2- 4x + 3

Describe the behavior of f at each discontinuity. Use and when appropriate.Identify all vertical asymptotes.

- qq

E X A M P L E 2 Locating Vertical Asymptotes Let


SOLUTION Let and . Since only at , f is discontin-uous only at 2. Since , f has a vertical asymptote at (Theorem 1).Tables 5 and 6 show that as from either side, and we have

limx:2+

x2

+ 20

5(x - 2)2 = q and limx:2-

x2

+ 20

5(x - 2)2 = q

x : 2f(x) : q

x = 2n(2) = 24 Z 0x = 2d = 0d(x) = 5(x - 2)2n(x) = x2

+ 20

- qq

f(x) =

x2+ 20

5(x - 2)2

The denominator d has no other zeros, so f does not have any other vertical asymp-totes. The graph of f (Fig. 3) shows the behavior at the vertical asymptote .With the left- and right-hand limits both infinite, we have

limx: q

x2

+ 20

5(x - 2)2 = q

x = 2



� Limits at InfinityThe symbol can also be used to indicate that an independent variable is increas-ing or decreasing without bound. We will write to indicate that x is increas-ing without bound through positive values and to indicate that x isdecreasing without bound through negative values. We begin by considering powerfunctions of the form xp and where p is a positive real number.

Explore & Discuss 2 (A) Complete the following table:

1>xp

x : - q

x : q

q

When is it correct to say that a limit does not exist, and when is it correct to use ? Itdepends on the situation. Table 7 lists the infinite limits we discussed in Examples 1 and 2.

; q

I N S I G H T

x 100 1,000 10,000 100,000 1,000,000

1/x2

x2

(B) Describe verbally the behavior of as x increases without bound. Thenuse limit notation to describe this behavior.

(C) Repeat part B for .1>x2

x2

MATCHED PROBLEM 2 Let


- qq

f(x) =

x - 1

(x + 3)2

TABLE 7

Right-hand limit Left-hand limit Unrestricted limit

limx:2

x2

+ 20

5(x - 2)2= qlim

x:2-

x2+ 20

5(x - 2)2= qlim

x: -2+

x2

+ 20

5(x - 2)2= q

limx: -1

x2+ x - 2

x2- 1

does not existlimx: -1-

x2+ x - 2

x2- 1

= - qlimx: -1+

x2+ x - 2

x2- 1

= q

The instructions in Examples 1 and 2 said that we should use infinite limits to describe thebehavior at vertical asymptotes. If we had been asked to evaluate the limits, with no men-tion of or asymptotes, then the correct answer would be that all of these limits do notexist. Remember, is a symbol used to describe the behavior of limits that do not exist. q

q

If p is a positive real number, then increases as x increases and it can be shownthat there is no upper bound on the values of . We indicate this behavior by writing

as or

Since the reciprocals of very large numbers are very small numbers, it follows thatapproaches 0 as x increases without bound. We indicate this behavior by writing

as or limx: q

1xp = 0x : q

1xp : 0

1>xp

limx: q

xp= qx : qxp : q

xpxp


Figure 4 illustrates the preceding behavior for and , and wesay that

and

Limits of power forms as x decreases without bound behave in a similar manner,with two important differences. First, if x is negative, then is not defined for all val-ues of p. For example, is not defined for negative values of x. Second, if

is defined, then it may approach or , depending on the value of p. Forexample,

but

For the function g in Figure 4, the line (the x axis) is called a horizontal as-ymptote. In general, a line is a horizontal asymptote of the graph of if approaches b as either x increases without bound or x decreases without bound.Symbolically, is a horizontal asymptote if either

or

In the first case, the graph of f will be close to the horizontal line for large (inabsolute value) negative x. In the second case, the graph will be close to the hori-zontal line for large positive x. Figure 5 shows the graph of a function withtwo horizontal asymptotes: and .

Theorem 2 summarizes the various possibilities for limits of power functions as xincreases or decreases without bound.

y = -1y = 1y = b

y = b

limx: q

f(x) = blimx: -q

f(x) = b

y = bf(x)

y = f(x)y = by = 0

limx: -q

x3= - qlim

x: q

x2= q

- qqxpx1>2

= 2xxp

limx: q

g(x) = 0limx: q

f(x) = q

g(x) = 1>x2f(x) = x2


x

y

5

4

3

2

1

21

1x2g(x) �

f (x) � x2

FIGURE 4

THEOREM 2 LIMITS OF POWER FUNCTIONS AT INFINITYIf p is a positive real number and k is any real constant, then

1. 2.

3. 4.

provided that is a real number for negative values of x. The limits in 3 and 4 willbe either or , depending on k and p.q- q

xp

limx: q

kxp= ; qlim

x: -q

kxp= ; q

limx: q

k

xp = 0limx: -q

k

xp = 0

How can we use Theorem 2 to evaluate limits at infinity? It turns out that the limitproperties listed in Theorem 2, Section 3-1, are also valid if we replace the statement

with or .x : - qx : qx : c

E X A M P L E 3 Limit of a Polynomial Function at Infinity Let .Find the limit of as x approaches and as x approaches .

SOLUTION Since limits of power functions of the form approach 0 as x approaches or, it is convenient to work with these reciprocal forms whenever possible. If we

factor out the term involving the highest power of x, we can write as

Using Theorem 2 in Section 3-3 and Theorem 2 in Section 3-1, we can write

= 1 - 0 - 0 + 0 = 1limx: q

a1 -

12x

-

7

2x2 +

3

2x3 b

p(x) = 2x3a1 -

12x

-

7

2x2 +

3

2x3 bp(x)

- q

q1>xp

- qqp(x)p(x) = 2x3

- x2- 7x + 3

FIGURE 5

f (x)

�1

1

�5 5x

lim f (x)=1xB�

lim f (x)=–1xB–�



For large values of x,

and

Since as , it follows that

Similarly, as implies

So the behavior of for large values is the same as the behavior of the highest-degree term, .2x3

p(x)

limx: -q

p(x) = limx: -q

2x3= - q

x : - q2x3 : - q

limx: q

p(x) = limx: q

2x3= q

x : q2x3 : q

p(x) = 2x3a1 -

12x

-

7

2x2 +

3

2x3 b L 2x3

a1 -

12x

-

7

2x2 +

3

2x3 b L 1

MATCHED PROBLEM 3 Let . Find the limit of as x approaches and as xapproaches .- q

qp(x)p(x) = -4x4+ 2x3

+ 3x

The term with highest degree in a polynomial is called the leading term. In the so-lution in Example 3, the limits at infinity of were the sameas the limits of the leading term . Theorem 3 states that this is true for any poly-nomial of degree greater than or equal to 1.

2x3p(x) = 2x3

- x2- 7x + 3

THEOREM 3 LIMITS OF POLYNOMIAL FUNCTIONS AT INFINITYIf

then

and

Each limit will be either or , depending on and n.anq- q

limx: -q

p(x) = limx: -q

anxn= ; q

limx: q

p(x) = limx: q

anxn= ; q

p(x) = anxn+ an - 1 x

n - 1+

Á+ a1x + a0, an Z 0, n Ú 1

A polynomial of degree 0 is a constant function , and its limit as xapproaches or is the number . For any polynomial of degree 1 or greater,Theorem 3 states that the limit as x approaches or cannot be equal to anumber. This means that polynomials of degree 1 or greater never have horizontalasymptotes.

A description of the behavior of the limits of any function f at infinity is calledthe end behavior of f. determines the right end behavior and

determines the left end behavior. According to Theorem 3, the rightand left end behavior of a nonconstant polynomial is always an infinite limit. p(x)limx : -q f(x)

Limx : q f(x)

- qq

a0- qq

p(x) = a0

E X A M P L E 4 End Behavior of a Polynomial Describe the end behavior of each polynomial:

(A) (B) p(x) = 3x3- 500x4p(x) = 3x3

- 500x2


SOLUTION (A) According to Theorem 3, as x increases without bound to the right, the rightend behavior is

and as x increases without bound to the left, the end behavior is

(B) As x increases without bound to the right, the end behavior is

and as x increases without bound to the left, the end behavior is

limx: -q

(3x3- 500x4) = lim

x: -q

(-500x4) = - q

limx: q

(3x3- 500x4) = lim

x: q

(-500x4) = - q

limx: -q

(3x3- 500x2) = lim

x: -q

3x3= - q

limx: q

(3x3- 500x2) = lim

x: q

3x3= q

� Finding Horizontal AsymptotesSince a rational function is the ratio of two polynomials, it is not surprising that re-ciprocals of powers of x can be used to analyze limits of rational functions at infini-ty. For example, consider the rational function

Factoring the highest-degree term out of the numerator and the denominator, wecan write

Thus, the behavior of this rational function as x approaches infinity is determined bythe ratio of the highest-degree term in the numerator (3x2) to the highest-degreeterm in the denominator (2x2). Theorem 4 generalizes this result to any rational func-tion and lists the three possible outcomes.

limx: q

f(x) = limx: q

3x2

2x2 # limx: q

1 -

53x

+

3

x2

1 +

7

2x2

=

32

#1 - 0 + 0

1 + 0 =

32

f(x) =

3x2

2x2 # 1 -

53x

+

3

x2

1 +

7

2x2

f(x) =

3x2- 5x + 9

2x2+ 7


MATCHED PROBLEM 4 Describe the end behavior of each polynomial:

(A) (B) p(x) = 300x6- 4x5p(x) = 300x2

- 4x5

Explore & Discuss 3 Let be a polynomial of degree with . Complete the followingtable:

an Z 0n > 0p(x)

n Left End Behavior Right End Behavior

Even Positive

Even Negative

Odd Positive

Odd Negative

p(x) : qp(x) : q

an



Notice in cases 1 and 2 of Theorem 4 that the limit is the same if x approaches or . Thus, a rational function can have at most one horizontal asymptote(see Fig. 6).

- q

q

The graph of f in Figure 6 dispels the misconception that the graph of a function cannotcross a horizontal asymptote. Horizontal asymptotes give us information about the graphof a function only as and , not at any specific value of x.x : - qx : q

I N S I G H T

THEOREM 4 LIMITS OF RATIONAL FUNCTIONS AT INFINITY AND HORIZONTALASYMPTOTES OF RATIONAL FUNCTIONS

(A)

(B) There are three possible cases for these limits:

1. If , then , and the line (the x axis) is a

horizontal asymptote of .

2. If , then , and the line is a horizon-

tal asymptote of .

3. If , then each limit will be or , depending on m, n, am, and bn, anddoes not have a horizontal asymptote.f(x)

- qqm > n

f(x)

y =

am

bnlim

x: q

f(x) = limx: -q

f(x) =

am

bnm = n

f(x)

y = 0limx: q

f(x) = limx: -q

f(x) = 0m < n

then limx: q

f(x) = limx: q

amxm

bnxn and limx: -q

f(x) = limx: -q

amxm

bnxn

If f(x) =

amxm+ am - 1x

m - 1+

Á+ a1x + a0

bnxn+ bn - 1x

n - 1+

Á+ b1x + b0

, am Z 0, bn = 0

x

y

1

3

�10�20 10 20

y � f (x)

y � 1.5

FIGURE 6 f1x2 =

3x2- 5x + 9

2x2+ 7

E X A M P L E 5 Finding Horizontal Asymptotes Find all horizontal asymptotes, if any, of eachfunction.

(A) (B)

(C)

SOLUTION We will make use of part A of Theorem 3:

(A)

The line is a horizontal asymptote of .

(B)

The line (the x axis) is a horizontal asymptote of .

(C)

The function has no horizontal asymptotes.f(x)

limx: q

f(x) = limx: q

2x5- x3

- 1

6x3+ 2x2

- 7= lim

x: q

2x5

6x3 = limx: q

x2

3= q

f(x)y = 0

limx: q

f(x) = limx: q

3x4- x2

+ 1

8x6- 10

= limx: q

3x4

8x6 = limx: q

3

8x2 = 0

f(x)y = 5>3lim

x: q

f(x) = limx: q

5x3- 2x2

+ 1

4x3+ 2x - 7

= limx: q

5x3

4x3 =

53

f(x) =

2x5- x3

- 1

6x3+ 2x2

- 7

f(x) =

3x4- x2

+ 1

8x6- 10

f(x) =

5x3- 2x2

+ 1

4x3+ 2x - 7


An accurate sketch of the graph of a rational function requires knowledge of bothvertical and horizontal asymptotes. As we mentioned earlier, we are postponing adetailed discussion of graphing techniques until Section 5-4.


MATCHED PROBLEM 5 Find the horizontal asymptotes of each function.

(A) (B)

(C) f(x) =

2x3- x + 7

4x3+ 3x2

- 100

f(x) =

5x6+ 3x

2x5- x - 5

f(x) =

4x3- 5x + 8

2x4- 7

E X A M P L E 6 Finding the Asymptotes of a Rational Function Find all the asymptotes ofthe function

SOLUTION Let and . Since only atand , the only vertical asymptote of f is the line (Theorem 1).

Since

the horizontal asymptote is the line (Theorem 3).y = 2

limx: q

f(x) = limx: q

2x2- 5

x2+ 4x + 4

= limx: q

2x2

x2 = 2

x = 2n(2) = 3x = 2d(x) = 0d(x) = x2

+ 4x + 4 = (x + 2)2n(x) = 2x2- 5

f(x) =

2x2- 5

x2+ 4x + 4

MATCHED PROBLEM 6 Find all the asymptotes of the function f(x) =

x2- 9

x2- 4

Answers to Matched Problems 1. Vertical asymptote: ; ,

2. Vertical asymptote: ;

3.

4. (A) ,

(B) ,

5. (A) (B) (C) No horizontal asymptotes

6. Vertical asymptotes: , ; horizontal asymptote: y = 1x = 4x = -2

y = 0y =

54

limx: -q

p(x) = - qlimx: q

p(x) = - q

limx: -q

p(x) = - qlimx: q

p(x) = q

limx: q

p(x) = limx: -q

p(x) = - q

limx: -3+

f(x) = limx: -3-

f(x) = qx = -3

limx:1-

f(x) = - qlimx:1+

f(x) = qx = 1



A Problems 1–8 refer to the following graph of :y = f(x)15.

(A) (B) (C)

16.

(A) (B) (C)

In Problems 17–20, find the limit of each polynomial p(x) (A) as x approaches (B) as x approaches

17.

18.

19.

20.

B In Problems 21–30, use or where appropriate to describe the behavior at each discontinuity and identify all vertical asymptotes.

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

In Problems 31–36, evaluate the indicated limit. Use or where appropriate.

31.

32. limx: q

2 - 3x3

7 + 4x3

limx: q

4x + 75x - 9

q

- q

S(x) =

6x + 9

x4+ 6x3

+ 9x2

T(x) =

8x - 16

x4- 8x3

+ 16x2

K(x) =

x2+ 2x - 3

x2- 4x + 3

H(x) =

x2- 2x - 3

x2- 4x + 3

G(x) =

x2+ 9

9 - x2

F(x) =

x2- 4

x2+ 4

k(x) =

x2- 9

x2+ 9

h(x) =

x2+ 4

x2- 4

g(x) =

x

4 - x

f(x) =

1x + 3

q- q

p(x) = 2x4- 2x3

+ 9x

p(x) = 2x5- 2x6

- 11

p(x) = 4x3- 3x4

+ x2

p(x) = 4x5- 3x4

+ 1

- qq

limx: - 2

f(x)limx: - 2+

f(x)limx: - 2-

f(x)

f(x) =

x2+ x - 2

(x + 2)

limx: - 2

f(x)limx: - 2+

f(x)limx: - 2-

f(x)

f(x) =

x2- 3x + 2

(x + 2)

Exercise 3-3

1. 2.

3. 4.

5. 6.

7. 8.

In Problems 9–16, find each limit. Use and whenappropriate.

9.

(A) (B) (C)

10.

(A) (B) (C)

11.

(A) (B) (C)

12.

(A) (B) (C)

13.

(A) (B) (C)

14.

(A) (B) (C) limx:1

f(x)limx:1+

f(x)limx:1-

f(x)

f(x) =

x2+ x + 2

(x - 1)

limx:1

f(x)limx:1+

f(x)limx:1-

f(x)

f(x) =

x2+ x - 2

(x - 1)

limx:2

f(x)limx:2+

f(x)limx:2-

f(x)

f(x) =

2x + 2

(x + 2)2

limx:4

f(x)limx:4+

f(x)limx:4-

f(x)

f(x) =

2x - 4

(x - 4)2

limx: -3

f(x)limx: -3+

f(x)limx: -3-

f(x)

f(x) =

x2

x + 3

limx:5

f(x)limx:5+

f(x)limx:5-

f(x)

f(x) =

x

x - 5

q- q

limx:2

f(x) = ?limx:2-

f(x) = ?

limx:2+

f(x) = ?limx: -2

f(x) = ?

limx: -2-

f(x) = ?limx: -2+

f(x) = ?

limx: -q

f(x) = ?limx: q

f(x) = ?

y

�3

�2

�1

3

2

1

�2�4 2 4x


33.

34.

35.

36.

In Problems 37–50, find all horizontal and vertical asymptotes.

37. 38.

39. 40.

41. 42.

43. 44.

45. 46.

47. 48.

49. 50.

C In Problems 51–54, determine a so that f is continuouson ( , ).

51. f(x) = c x2+ 2x - 3x - 1

if x Z 1

a if x = 1

q- q

f(x) =

x2- x - 12

2x2+ 5x - 12

f(x) =

2x2- 5x + 2

x2- x - 2

f(x) =

2x2+ 7x + 12

2x2+ 5x - 12

f(x) =

2x2+ 3x - 2

x2- x - 2

f(x) =

x + 5

x2f(x) =

x2

x - 3

f(x) =

x2+ 9x

f(x) =

x

x2+ 4

f(x) =

x

x2- 4

f(x) =

x3

x2+ 6

f(x) =

x2- 1

x2+ 2

f(x) =

x2+ 1

x2- 1

f(x) =

3x + 2x - 4

f(x) =

2x

x + 2

limx: q

4x7- 8x

6x4+ 9x2

limx: q

7x4- 14x2

6x5+ 3

limx: q

5x + 11

7x3- 2

limx: q

5x2+ 11

7x - 2 52.

53.

54.

55. Theorem 3 states that

What

conditions must n and satisfy for the limit to be ?For the limit to be ?

56. Theorem 3 also states that

. Whatconditions must n and satisfy for the limit to be ?For the limit to be ?

Describe the end behavior of each function in Problems 57–64.

57.

58.

59.

60.

61.

62.

63.

64. f(x) =

-3x6+ 4x4

+ 2

2x2- 2x - 3

f(x) =

5x5+ 7x4

- 10

-x3+ 6x2

+ 3

f(x) =

5x4+ 7x7

- 10

x2+ 6x4

+ 3

f(x) =

x2- 5x - 7x + 11

f(x) = 4x3- 5x2

- 6x6

f(x) = 7x2+ 9x3

+ 5x

f(x) = 3x - 7x5- 2

f(x) = 2x4- 5x + 11

- q

qan

(anxn+ an - 1x

n - 1+

Á+ a0) = ; qlim

x: - q

- q

qan

= ; q .(anxn+ an - 1x

n - 1+

Á+ a02lim

x: q

f(x) = c x2- 5x + 6x - 2

if x Z 2

a if x = 2

f(x) = c x2+ 4x + 3x + 3

if x Z -3

a if x = -3

f(x) = c x2+ 3x + 2x + 2

if x Z -2

a if x = -2

Applications65. Average cost. A company manufacturing snowboards

has fixed costs of $200 per day and total costs of $3,800per day at a daily output of 20 boards.

(A) Assuming that the total cost per day C(x) is linearlyrelated to the total output per day x, write an equa-tion for the cost function.

(B) The average cost per board for an output of xboards is given by . Find the averagecost function.

(C) Sketch a graph of the average cost function, includ-ing any asymptotes, for .

(D) What does the average cost per board tend to asproduction increases?

66. Average cost. A company manufacturing surfboards hasfixed costs of $300 per day and total costs of $5,100 perday at a daily output of 20 boards.

1 … x … 30

C(x) = C(x)>x

(A) Assuming that the total cost per day C(x) is linearlyrelated to the total output per day x, write an equa-tion for the cost function.

(B) The average cost per board for an output of xboards is given by . Find the averagecost function.

(C) Sketch a graph of the average cost function, includ-ing any asymptotes, for .

(D) What does the average cost per board tend to asproduction increases?

67. Energy costs. Most appliance manufacturers produceboth conventional models and energy-efficient models.The energy-efficient models are more expensive tomake, but cheaper to operate. The costs of purchasingand operating a 23-cubic-foot refrigerator of each typeare given in Table 8. These costs do not include mainte-nance charges or changes in the price of electricity.

1 … x … 30

C(x) = C(x)>x




(A) Express the total cost and the average costof purchasing and operating an

energy-efficient refrigerator for x years.(B) Express the total cost and the average cost

of purchasing and operating a con-ventional refrigerator for x years.

(C) Are the total cost for an energy-efficient model andthe total cost for a conventional model ever thesame? If so, when?

(D) Are the average cost for an energy-efficient modeland the average cost for a conventional model everthe same? If so, when?

(E) Find the limit of each average cost function as, and discuss the implications of the results.

68. Energy costs. Most appliance manufacturers produceboth conventional models and energy-efficient models.The energy-efficient models are more expensive to make,but cheaper to operate. The costs of purchasing and op-erating a 36,000-Btu central air-conditioner of each typeare given in Table 9. These costs do not include mainte-nance charges or changes in the price of electricity.

x : q

Cc(x) = Cc(x)>xCc(x)

Ce(x) = Ce(x)>xCe(x)

70. Drug concentration. A drug is administered to a patientthrough an IV drip. The concentration of the drug (inmilligrams/milliliter) in the bloodstream t hours after the

drip was started is given by . Find andinterpret .

71. Pollution. In Silicon Valley (California), a number ofcomputer-related manufacturing firms were found to becontaminating underground water supplies with toxicchemicals stored in leaking underground containers. Awater quality control agency ordered the companies totake immediate corrective action and to contribute to amonetary pool for testing and cleanup of the under-ground contamination. Suppose that the required mone-tary pool (in millions of dollars) for the testing andcleanup is estimated to be given by

where x is the percentage (expressed as a decimalfraction) of the total contaminant removed.

(A) How much must be in the pool to remove 90% ofthe contaminant?

(B) How much must be in the pool to remove 95% ofthe contaminant?

(C) Find and discuss the implications of this limit.

72. Employee training. A company producing computercomponents has established that, on the average, a newemployee can assemble components per day after tdays of on-the-job training, as given by

(A) How many components per day can a new employeeassemble after 6 days of on-the-job training?

(B) How many days of on-the-job training will a newemployee need to reach the level of 70 componentsper day?

(C) Find and discuss the implications of this limit.

73. Biochemistry. In 1913, biochemists Leonor Michaelis(1875–1949) and Maude Menten (1879–1960) proposedthe rational function model (see figure)

for the velocity of the enzymatic reaction , where s isthe substrate concentration. The constants and are determined from experimental data.

KMVmax

v

v(s) =

Vmax s

KM + s

limt: q

N(t)

N(t) =

100t

t + 9 t Ú 0

N(t)

limx:1

P(x)

P(x) =

2x

1 - x 0 … x 6 1

limt: q

C(t)C(t) =

5t(t + 50)

t3+ 100

TABLE 8 Refrigerators

Energy-Efficient Model Conventional Model

Initial cost $950 $900Total volume 23 ft3 23 ft3

Annual cost of electricity $56 $66

TABLE 9 Central Air-Conditioner

Energy Efficient Model Conventional Model

Initial cost $4,000 $2,700Total capacity 36,000 Btu 36,000 BtuAnnual cost of electricity $932 $1,332

(A) Express the total cost and the average costof purchasing and operating an

energy-efficient central air-conditioner for x years.(B) Express the total cost and the average cost

of purchasing and operating a con-ventional central air-conditioner for x years.

(C) Are the total cost for an energy-efficient model andthe total cost for a conventional model ever thesame? If so, when?

(D) Are the average cost for an energy-efficient modeland the average cost for a conventional model everthe same? If so, when?

(E) Find the limit of each average cost function as, and discuss the implications of the results.

69. Drug concentration. A drug is administered to a patientthrough an injection. The concentration of the drug (inmilligrams/milliliter) in the bloodstream t hours after the

injection is given by . Find and inter-pret .lim

t: q

C(t)C(t) =

5t2(t + 50)

t3+ 100

x : q

Cc(x) = Cc(x)>xCc(x)

Ce(x) = Ce(x)>xCe(x)

s

v

2

Michaelis–Menten function

Vmax s

Vmax

KM

Vmax

KM � sv(s) �

Figure for 73


(A) Show that .

(B) Show that .

(C) Table 10* lists data for an enzyme treated with thesubstrate saccharose.

v(KM) =

Vmax

2

lims: q

v(s) = Vmax

Plot the preceding points on graph paper and estimateto the nearest integer. To estimate , add

the horizontal line to your graph, connect

successive points on the graph with straight-line seg-ments, and estimate the value of s (to the nearest

multiple of 10) that satisfies .

(D) Use the constants and from part C to form a Michaelis–Menten function for the data inTable 10.

(E) Use the function from part D to estimate the velocity of the enzyme reaction when the saccharoseis 15 and to estimate the saccharose when the velocity is 3.

74. Biochemistry. Table 11† lists data for the enzyme inver-tase treated with the substrate sucrose. We want to modelthese data with a Michaelis–Menten function.

KMVmax

v(s) =

Vmax

2

v =

Vmax

2

KMVmax

connect successive points on the graph with straight-line segments, and estimate the value of s

(to the nearest integer) that satisfies .

(B) Use the constants and from part A to forma Michaelis–Menten function for the data in Table 11.

(C) Use the function from part B to estimate the velocityof the enzyme reaction when the sucrose is 9 and toestimate the sucrose when the velocity is 32.

75. Physics: Thermal expansion. The coefficient of thermalexpansion (CTE) is a measure of the expansion of an ob-ject subjected to extreme temperatures. We want to use aMichaelis–Menten function of the form

(Problem 73)

where , T is temperature in (degrees Kelvin),and and M are constants, to model this coefficient.Table 12* lists the coefficients of thermal expansion fornickel and for copper at various temperatures.

Cmax

KC = CTE

C(T) =

C max T

M + T

KMVmax

v(s) =

Vmax

2

TABLE 10

s

5.2 0.86610.4 1.46620.8 2.11441.6 2.66683.3 3.236

167 3.636333 3.636

v

TABLE 11

s

2.92 18.25.84 26.58.76 31.1

11.7 3314.6 34.917.5 37.223.4 37.1

v

TABLE 12 Coefficients of thermal expansion

T (K) Nickel Copper

100 6.6 10.3200 11.3 15.2293 13.4 16.5500 15.3 18.3800 16.8 20.3

1,100 17.8 23.7

*Michaelis and Menten (1913) Biochem. Z. 49, 333–369.†Institute of Chemistry, Macedonia.

*National Physical Laboratory


(A) Plot the points in Table 11 on graph paper and esti-mate to the nearest integer. To estimate ,

add the horizontal line to your graph, v =

Vmax

2

KMVmax

(A) Plot the points in columns 1 and 2 of Table 12 ongraph paper and estimate to the nearest integer. To estimate M, add the horizontal line

to your graph, connect successive

points on the graph with straight-line segments, and estimate the value of T (to the nearest multiple

of fifty) that satisfies .

(B) Use the constants and M from part A to form

a Michaelis–Menten function for the CTE of nickel. (C) Use the function from part B to estimate the CTE of

nickel at 600 K and to estimate the temperaturewhen the CTE of nickel is 12.

76. Physics: Thermal expansion. Repeat Problem 75 for theCTE of copper (column 3 of Table 12).

Cmax

2

C(T) =

Cmax

2

CTE =

Cmax

2

Cmax


S e c t i o n 3 - 4 The Derivative 169

Section 3-4 THE DERIVATIVE� Rate of Change� Slope of the Tangent Line� The Derivative� Nonexistence of the Derivative

We will now make use of the limit concepts developed in Sections 3-1, 3-2, and 3-3to solve the two important problems illustrated in Figure 1. The solution of each ofthese apparently unrelated problems involves a common concept called the derivative.

x

y

Tangent line

(A) Find the equation of thetangent line at (x1, y1)given y � f (x)

(x1, y1)y � f (x) Velocity

here

(B) Find the instantaneousvelocity of a fallingobject

FIGURE 1 Two basic problems of calculus

x500 1,0000

1,000

2,000

3,000

4,000

5,000

R(x)

3,000

300

(400, 4,800)

(100, 1,800)

FIGURE 2 R1x2 = 20x - 0.02x2

� Rate of ChangeLet us start by considering a simple example.

E X A M P L E 1 Revenue Analysis The revenue (in dollars) from the sale of x plastic planter boxesis given by

which is graphed in Figure 2.

R1x2 = 20x - 0.02x2 0 … x … 1,000

(A) What is the change in revenue if production is changed from 100 planters to400 planters?

(B) What is the average change in revenue for this change in production?


SOLUTION (A) The change in revenue is given by

Thus, increasing production from 100 planters to 400 planters will increase rev-enue by $3,000.

(B) To find the average change in revenue, we divide the change in revenue by thechange in production:

Thus, the average change in revenue is $10 per planter when production isincreased from 100 to 400 planters.

R14002 - R11002400 - 100

=

3,000300

= $10

= 4,800 - 1,800 = $3,000 R14002 - R11002 = 2014002 - 0.02140022 - 32011002 - 0.021100224

MATCHED PROBLEM 1 Refer to the revenue function in Example 1.

(A) What is the change in revenue if production is changed from 600 planters to800 planters?

(B) What is the average change in revenue for this change in production?

In general, if we are given a function and if x is changed from a to then y will change from f(a) to The average rate of change is the ratio of thechange in y to the change in x.

f1a + h2. a + h,y = f1x2

DEFINITION Average Rate of Change

For the average rate of change from x � a to x � a � h is

(1)f1a + h2 - f1a21a + h2 - a

=

f1a + h2 - f1a2h

h Z 0

y = f1x2,

As we noted in Section 3-1, the mathematical expression (1) is called the differencequotient. The preceding discussion shows that the difference quotient can be inter-preted as an average rate of change. The next example illustrates another interpre-tation of this quotient: the velocity of a moving object.

E X A M P L E 2 Velocity A small steel ball dropped from a tower will fall a distance of y feet in xseconds, as given approximately by the physics formula

Figure 3 shows the position of the ball on a coordinate line (positive direction down)at the end of 0, 1, 2, and 3 seconds.

y = f1x2 = 16x2

y

y

Ground

Position at x � 3 seconds [y � 16(32) � 144 feet]

Position at x � 2 seconds [y � 16(22) � 64 feet]

Position at x � 1 second [y � 16(12) � 16 feet]Position at start (x � 0 seconds)0

16

64

144

FIGURE 3 Note: Positive y direction is down.




(A) Find the average velocity from seconds to seconds.

(B) Find and simplify the average velocity from seconds to seconds,

(C) Find the limit of the expression from part B as if that limit exists.

(D) Discuss possible interpretations of the limit from part C.

SOLUTION (A) Recall the formula which can be written in the form

For example, if a person drives from San Francisco to Los Angeles (a distanceof about 420 miles) in 7 hours, then the average velocity is

Sometimes the person will be traveling faster and sometimes slower, but theaverage velocity is 60 miles per hour. In our present problem, the average velocityof the steel ball from seconds to seconds is

Thus, we see that if is the position of the falling ball, then the averagevelocity is simply the average rate of change of f(x) with respect to time x, andwe have another interpretation of the difference quotient (1).

(B) Proceeding as in part A, we have

=

h164 + 16h2h

= 64 + 16h h Z 0

=

64 + 64h + 16h2- 64

h

=

1612 + h22 - 161222h

=

f12 + h2 - f122h

Average velocity =

Distance coveredElapsed time

y = f1x2 =

161322 - 1612221

= 80 feet per second

=

f132 - f1223 - 2

Average velocity =


x = 3x = 2

r =

d

t=

4207

= 60 miles per hour

r =

d

t=


= Average velocity

d = rt,

h : 0

h Z 0.x = 2 + hx = 2

x = 3x = 2

Notice that if the average velocity is 80 feet per second, our result inpart A.

(C) The limit of the average velocity expression from part B as is

(D) The average velocity over smaller and smaller time intervals approaches 64 feetper second. This limit can be interpreted as the velocity of the ball at the instantthe ball has been falling for exactly 2 seconds. Thus, 64 feet per second is referredto as the instantaneous velocity at seconds, and we have solved the oneof the basic problems of calculus (see Fig. 1B).

x = 2

= 64 feet per second

limh:0

f12 + h2 - f122h

= limh:0164 + 16h2

h : 0

h = 1,

Difference quotient

Simplify this O�O indeterminate form.


The adjective instantaneous is often omitted with the understanding that the phraserate of change always refers to the instantaneous rate of change and not the averagerate of change. Similarly, velocity always refers to the instantaneous rate of changeof distance with respect to time.

� Slope of the Tangent LineSo far, our interpretations of the difference quotient have been numerical in nature.Now we want to consider a geometric interpretation. A line through two points onthe graph of a function is called a secant line. If (a, f(a)) and aretwo points on the graph of we can use the slope formula from Section 1-2to find the slope of the secant line through these points (see Fig. 4).

Difference quotient

Thus, the difference quotient can be interpreted as both the average rate of changeand the slope of the secant line.

=

f1a + h2 - f1a2h

Slope of secant line =

f1a + h2 - f1a21a + h2 - a

y = f1x2, 1a + h, f1a + h22


MATCHED PROBLEM 2 For the falling steel ball in Example 2, find

(A) The average velocity from second to seconds(B) The average velocity (in simplified form) from second to

seconds,(C) The instantaneous velocity at second

Explore & Discuss 1 Recall the revenue function in Example 1: Find

Discuss possible interpretations of this limit.

The ideas introduced in Example 2 are not confined to average velocity, but canbe applied to the average rate of change of any function.

limh:0

R1100 + h2 - R11002h

R1x2 = 20x - 0.02x2.

x = 1h Z 0

x = 1 + hx = 1x = 2x = 1

DEFINITION Instantaneous Rate of Change

For the instantaneous rate of change at x � a is

(2)

if the limit exists.

limh:0

f1a + h2 - f1a2h

y = f1x2,

x

f (x)y � f (x)

(a � h, f (a � h))

Secant line

(a, f (a))

a a � h

h

f (a � h) � f (a)

FIGURE 4 Secant line

If and are two points in the plane with and L is the line passingthrough these two points (see Section 1-2), then

Slope of L Point–slope form for L

These formulas will be used extensively in the remainder of this chapter.

m =

y2 - y1

x2 - x1 y - y1 = m1x - x12

x1 Z x21x2, y221x1, y12I N S I G H T



(C) The limit of the secant line slope from part B is

(D) In part C, we saw that the limit of the slopes of the secant lines through the point(1, f(1)) is 2. If we graph the line through (1, f(1)) with slope 2 (Fig. 6), this lineappears to be the limit of the secant lines. The slope obtained from the limit ofslopes of secant lines is called the slope of the graph at the line throughthe point (1, f(1)) with this slope is called the tangent line, and we have solvedanother basic problem of calculus (see Fig. 1A).

x = 1,

= 2

limh:0

f11 + h2 - f112h

= limh:012 + h2

x

f (x)

�1 1

5

10

�2�3�4

f (x) � x2

Secant line:slope � 4


2 3 4 5

FIGURE 5 Secant lines

x�1 1

5

10

�2�3�4

f (x) � x2



Tangent line:slope � 2

2 3 4 5

f (x)

FIGURE 6 Tangent line

E X A M P L E 3 Slope of a Secant Line Given

(A) Find the slope of the secant line for and and 1, respectively. Graphand the two secant lines.

(B) Find and simplify the slope of the secant line for and h any nonzeronumber.

(C) Find the limit of the expression in part B.

(D) Discuss possible interpretations of the limit in part C.

SOLUTION (A) For and the secant line goes through andand its slope is

For and the secant line goes through andand its slope is

The graphs of and the two secant lines are shown in Figure 5.

(B) For and h any nonzero number, the secant line goes throughand and its slope is

= 2 + h h Z 0

=

h12 + h2h

=

1 + 2h + h2- 1

h

f11 + h2 - f112

h=

11 + h22 - 12

h

11 + h, f11 + h22 = 11 + h, 11 + h222,11, f1122 = 11, 12a = 1

y = f1x2

f11 + 12 - f1121

=

22- 12

1= 3

12, f1222 = 12, 42, 11, f1122 = 11, 12h = 1,a = 1

f11 + 22 - f1122

=

32- 12

2= 4

13, f1322 = 13, 92, 11, f1122 = 11, 12h = 2,a = 1

a = 1

y = f1x2 h = 2a = 1

f1x2 = x2,

Square the binomial.

Combine like terms andfactor the numerator.

Cancel.

MATCHED PROBLEM 3 Given

(A) Find the slope of the secant line for and and 1, respectively.

(B) Find and simplify the slope of the secant line for and h any nonzeronumber.

(C) Find the limit of the expression in part B.

(D) Find the slope of the graph and the slope of the tangent line at a = 2.

a = 2

h = 2a = 2

f1x2 = x2,


From plane geometry, we know that a line tangent to a circle is a line that passesthrough one and only one point of the circle (see Fig. 7A). Although this definitioncannot be extended to graphs of functions in general, the visual relationship betweengraphs of functions and their tangent lines (see Fig. 7B) is similar to the circle case.Limit (3) provides both a mathematically sound definition of a tangent line and amethod for approximating the slope of the tangent line.


The ideas introduced in the preceding example are summarized next:

DEFINITION Slope of a Graph

Given the slope of the graph at the point (a, f(a)) is given by

(3)

provided that the limit exists. The slope of the graph is also the slope of the tangentline at the point (a, f(a)).

limh:0

f1a + h2 - f1a2h

y = f1x2,

If the function f is continuous at a, then

and limit (3) will be a 0�0 indeterminate form. As we saw in Examples 2 and 3, evalua-tion of this type of limit typically involves algebraic simplification.

limh:0

f1a + h2 = f1a2

I N S I G H T

x

y

Circle

Tangent line

(A) Tangent line for a circle (B) Tangent lines for the graph of a function

x

f (x)

y � f (x)

FIGURE 7

� The DerivativeWe have seen that the limit of a difference quotient can be interpreted as a rate ofchange, as a velocity, or as the slope of a tangent line. In addition, this limit providessolutions to two of the three basic problems stated at the beginning of the chapter.We are now ready to introduce some terms that are used to refer to that limit. Tofollow customary practice, we use x in place of a and think of the difference quotient

as a function of h, with x held fixed as h tends to 0.

f1x + h2 - f1x2h

DEFINITION The Derivative

For we define the derivative of f at x, denoted by f�(x), to be

If exists for each x in the open interval (a, b), then f is said to be differentiableover (a, b).

f¿1x2f¿1x2 = lim

h:0

f1x + h2 - f1x2h

if the limit exists

y = f1x2,



(Differentiability from the left or from the right is defined by using or respectively, in place of in the preceding definition.)

The process of finding the derivative of a function is called differentiation. Thatis, the derivative of a function is obtained by differentiating the function.

h : 0h : 0+,h : 0-

SUMMARY Interpretations of the DerivativeThe derivative of a function f is a new function The domain of is a subset of thedomain of f. The derivative has various applications and interpretations, includingthe following:

1. Slope of the tangent line. For each x in the domain of is the slope of theline tangent to the graph of f at the point (x, f(x)).

2. Instantaneous rate of change. For each x in the domain of is the instan-taneous rate of change of with respect to x.

3. Velocity. If f(x) is the position of a moving object at time x, then isthe velocity of the object at that time.

Example 4 illustrates the four-step process that we use to find derivatives in thissection. In subsequent sections, we develop rules for finding derivatives that do notinvolve limits. However, it is important that you master the limit process in order tofully comprehend and appreciate the various applications we will consider.

v = f¿1x2y = f1x2 f¿, f¿1x2

f¿, f¿1x2

f¿f¿.

E X A M P L E 4 Finding a Derivative Find the derivative of f at x, for

SOLUTION To find we use a four-step process.

Step 1. Find

Step 2. Find

Step 3. Find

Step 4. Find

Thus, if then The function is a new functionderived from the function f.

f¿f¿1x2 = 4 - 2x.f1x2 = 4x - x2,

f¿1x2 = limh:0

f1x + h2 - f1x2h

= limh:014 - 2x - h2 = 4 - 2x

f¿1x2 = limh:0

f1x + h2 - f1x2h

.

= 4 - 2x - h, h Z 0

f1x + h2 - f1x2

h=

4h - 2xh - h2

h=

h14 - 2x - h2h

f1x + h2 - f1x2h

.

= 4h - 2xh - h2

f1x + h2 - f1x2 = 4x + 4h - x2- 2xh - h2

- 14x - x22f1x + h2 - f1x2.

= 4x + 4h - x2- 2xh - h2

f1x + h2 = 41x + h2 - 1x + h22f1x + h2.

f¿1x2,f1x2 = 4x - x2.f¿1x2,

MATCHED PROBLEM 4 Find the derivative of f at x, for f1x2 = 8x - 2x2.f¿1x2,


The four-step process used in Example 4 is summarized as follows for easyreference:


PROCEDURE The four-step process for finding the derivative of a function f:

Step 1. Find

Step 2. Find

Step 3. Find

Step 4. Find limh:0

f1x + h2 - f1x2h

.

f1x + h2 - f1x2h

.

f1x + h2 - f1x2.f1x + h2.

E X A M P L E 5 Finding Tangent Line Slopes In Example 4, we started with the function speci-fied by and found the derivative of f at x to be Thus,the slope of a line tangent to the graph of f at any point (x, f(x)) on the graph is

(A) Find the slope of the graph of f at and

(B) Graph and use the slopes found in part (A) to make arough sketch of the lines tangent to the graph at and

SOLUTION (A) Using we have

Slope at Slope at Slope at

(B)

x = 3 f¿132 = 4 - 2132 = -2

x = 2 f¿122 = 4 - 2122 = 0

x = 0 f¿102 = 4 - 2102 = 4

f¿1x2 = 4 - 2x,

x = 3.x = 0, x = 2,y = f1x2 = 4x - x2,

x = 3.x = 0, x = 2,

m = f¿1x2 = 4 - 2x

f¿1x2 = 4 - 2x.f1x2 = 4x - x2

y

f (x) � 4x � x2

x5�3

5Slope � 0

Slope � 4

Slope � �2

(2, 4)

(3, 3)

(0, 0)

MATCHED PROBLEM 5 In Matched Problem 4, we started with the function specified by Using the derivative found there,

(A) Find the slope of the graph of f at and

(B) Graph and use the slopes from part (A) to make a roughsketch of the lines tangent to the graph at and

Explore & Discuss 2 In Example 4 we found that the derivative of isand in Example 5 we graphed f(x) and several tangent lines.

(A) Graph f and on the same set of axes.

(B) The graph of is a straight line. Is it a tangent line for the graph of f ?Explain.

f¿

f¿

f¿1x2 = 4 - 2x,f1x2 = 4x - x2

x = 4.x = 1, x = 2,y = f1x2 = 8x - 2x2,

x = 4.x = 1, x = 2,

f1x2 = 8x - 2x2.



(C) Find the x intercept for the graph of What is the slope of the linetangent to the graph of f for this value of x? Write a verbal description of the relationship between the slopes of the tangent lines of a functionand the x intercepts of the derivative of the function.

f¿.

E X A M P L E 6 Finding a Derivative Find the derivative of f at x, for

SOLUTION We use the four-step process to find

Step 1. Find

Step 2. Find

Combine like terms.

Step 3. Find

Step 4. Find

Thus, the derivative of is a new function.Thedomain of f is Since is not defined, the domain of is a subsetof the domain of f.

10, q2,f¿f¿10230, q2. f¿1x2 = 1>122x2,f1x2 = 2x + 2

=

12x + 2x=

1

22x x 7 0

limh:0

f1x + h2 - f1x2h

= limh:0

12x + h + 2x

f¿1x2 = limh:0

f1x + h2 - f1x2h

.

=

12x + h + 2x h Z 0

=

h

h12x + h + 2x2

=

x + h - x

h12x + h + 2x2

=

2x + h - 2x

h � 2x + h + 2x2x + h + 2x

f1x + h2 - f1x2

h=

2x + h - 2x

h

f1x + h2 - f1x2h

.

= 2x + h - 2x

f1x + h2 - f1x2 = 2x + h + 2 - 12x + 22f1x + h2 - f1x2.

f1x + h2 = 2x + h + 2

f1x + h2.f¿1x2.

f1x2 = 2x + 2.f¿1x2,

We rationalizethe numerator(Appendix A,Section A-6) tochange the formof this fraction.Combine liketerms.

Cancel.

MATCHED PROBLEM 6 Find for f1x2 = 2x + 4.f ¿1x2

E X A M P L E 7 Sales Analysis The total sales of a company (in millions of dollars) t months fromnow are given by Find and interpret S(25) and Use theseresults to estimate the total sales after 26 months and after 27 months.

S¿1252.S1t2 = 2t + 2.



SOLUTION The total sales function S has the same form as the function f in Example 6; only theletters used to represent the function and the independent variable have beenchanged. It follows that and also have the same form:

Evaluating S and at we have

Thus, 25 months from now the total sales will be $7 million and will be increasingat the rate of $0.1 million ($100,000) per month. If this instantaneous rate of changeof sales remained constant, the sales would grow to $7.1 million after 26 months,$7.2 million after 27 months, and so on. Even though is not a constant functionin this case, these values provide useful estimates of the total sales.

S¿1t2

S1252 = 225 + 2 = 7 S¿1252 =

1

2225= 0.1

t = 25,S¿

S¿1t2 =

1

22t f¿1x2 =

1

22x

S1t2 = 2t + 2 f1x2 = 2x + 2

f¿S¿

MATCHED PROBLEM 7 The total sales of a company (in millions of dollars) t months from now are given byFind and interpret S(12) and . Use these results to estimate the

total sales after 13 months and after 14 months. (Use the derivative found in MatchedProblem 6.)

In Example 7, it is instructive to compare the estimates of total sales obtained byusing the derivative with the corresponding exact values of S(t):

Exact values Estimated values

For this function, the estimated values provide very good approximations to the exactvalues of S(t). For other functions, the approximation might not be as accurate.

Using the instantaneous rate of change of a function at a point to estimate valuesof the function at nearby points is a simple, but important, application of the derivative.

� Nonexistence of the DerivativeThe existence of a derivative at depends on the existence of a limit at

, that is, on the existence of

(4)

If the limit does not exist at we say that the function f is nondifferentiable atx � a, or does not exist.


(A) Graph f.

(B) Complete the following table:

f1x2 = ƒ x - 1 ƒ .

f œ1a2 x = a,

f¿1a2 = limh:0

f1a + h2 - f1a2h

x = ax = a

S1272 = 227 + 2 = 7.196 Á L 7.2

S1262 = 226 + 2 = 7.099 Á L 7.1

S¿1122S1t2 = 2t + 4.

h 0.001 0.01 0.1

? ? ? ? ? ?: ? ;f(1 � h) � f(1)

h

-0.001 : 0 ;-0.01-0.1

(C) Find the following limit if it exists:

limh:0

f11 + h2 - f112h



(D) Use the results of parts (A)–(C) to discuss the existence of

Repeat parts (A)–(D) for

How can we recognize the points on the graph of f where does not exist? Itis impossible to describe all the ways that the limit of a difference quotient can failto exist. However, we can illustrate some common situations where does fail toexist (see Fig. 8):

f¿1a2f¿1a2

g1x2 = 23 x - 1.

f¿112.

xa a a a

f (x)

x

f (x)

x

f (x)

x

f (x)

(A) Not continuous atx � a

(B) Graph has sharpcorner at x � a

(C) Vertical tangentat x � a

(D) Vertical tangentat x � a

FIGURE 8 The function f is nondifferentiable at x = a.

1. If the graph of f has a hole or a break at then does not exist (Fig. 8A).

2. If the graph of f has a sharp corner at then does not exist and thegraph has no tangent line at (Fig. 8B). (In Fig. 8B, the left- and right-handderivatives exist, but are not equal.)

3. If the graph of f has a vertical tangent line at then does not exist(Fig. 8C and D).

Answers to Matched Problems 1. (A) (B)

2. (A) 48 ft/s (B) (C) 32 ft/s

3. (A) 6, 5 (B)(C) 4 (D) Both are 4

4.

5. (A)(B)

f¿112 = 4, f¿122 = 0, f¿142 = -8

f¿1x2 = 8 - 4x

4 + h

32 + 16h

-$8 per planter-$1,600

f¿1a2x = a,

x = af¿1a2x = a,

f¿1a2x = a,

f (x)

x21

10

5

3 5

6.

7. 12 months from now, the total sales will be $4 million and willbe increasing at the rate of $0.125 million ($125,000) per month. The estimated total salesare $4.125 million after 13 months and $4.25 million after 14 months.

S1122 = 4, S¿1122 = 0.125;

f¿1x2 = 1>122x + 42



A In Problems 1 and 2, find the indicated quantity forand interpret that quantity in terms of the

following graph.y = f1x2 = 5 - x2,

11. 12.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

23. 24.

25. 26.

B Problems 27 and 28 refer to the graph of shown.

y = f1x2 = x2+ x

f1x2 =

5x

3 + xf1x2 =

3x

x + 2

f1x2 = 162x + 9f1x2 = 102x + 5

f1x2 = 3 - 72xf1x2 = 5 + 32x

f1x2 =

6x

- 2f1x2 = 4 +

4x

f1x2 = -2x3+ 5f1x2 = 2x3

+ 1

f1x2 = -x2+ 9x - 2f1x2 = -x2

+ 4x - 9

f1x2 = 2x2+ 5x + 1f1x2 = 2x2

- 7x + 3

f1x2 = x2+ 4x + 7f1x2 = x2

+ 6x - 10

Exercise 3-4

f (x)

x2

10

8

6

4

2

�2�2

f (x) � 5 � x2

f (x)f (x) � x2 � x

x5�5

5

10

15

20

1. (A) (B)

(C)

2. (A) (B)

(C)

3. Find the indicated quantities for

(A) The average rate of change of f(x) if x changes from 1 to 4.

(B) The slope of the secant line through the points (1, f(1))and (4, f(4)) on the graph of

(C) The slope of the secant line through the points (1, f(1))and Simplify your answer.

(D) The slope of the graph at (1, f(1)).(E) The instantaneous rate of change of with

respect to x at (F) The slope of the tangent line at (1, f(1)).(G) The equation of the tangent line at (1, f(1)).

4. Find the indicated quantities for

(A) The average rate of change of f(x) if x changes from2 to 5.

(B) The slope of the secant line through the points (2, f(2))and (5, f(5)) on the graph of

(C) The slope of the secant line through the points (2, f(2))and Simplify your answer.

(D) The slope of the graph at (2, f(2)).(E) The instantaneous rate of change of with

respect to x at (F) The slope of the tangent line at (2, f(2)).(G) The equation of the tangent line at (2, f(2)).

In Problems 5–26, use the four-step process to find andthen find and

5. 6.

7. 8.

9. 10. f1x2 = 2x2+ 8f1x2 = 2 - 3x2

f1x2 = 4 - 6xf1x2 = 3x - 7

f1x2 = 9f1x2 = -5

f¿132.f¿112, f¿122,f¿1x2

x = 2.y = f1x2

12 + h, f12 + h22, h Z 0.

y = f1x2.

f1x2 = 3x2.

x = 1.y = f1x2

11 + h, f11 + h22, h Z 0.

y = f1x2.

f1x2 = 3x2.

limh:0

f1-2 + h2 - f1-22h

f1-2 + h2 - f1-22h

f1-12 - f1-22-1 - 1-22

limh:0

f11 + h2 - f112h

f11 + h2 - f112h

f122 - f1122 - 1

27. (A) Find the slope of the secant line joining (1, f(1)) and(3, f(3)).

(B) Find the slope of the secant line joining (1, f(1)) and

(C) Find the slope of the tangent line at (1, f(1)).(D) Find the equation of the tangent line at (1, f(1)).

28. (A) Find the slope of the secant line joining (2, f(2)) and(4, f(4)).

(B) Find the slope of the secant line joining (2, f(2)) and

(C) Find the slope of the tangent line at (2, f(2)).(D) Find the equation of the tangent line at (2, f(2)).

In Problems 29 and 30, suppose an object moves along the yaxis so that its location is at time x (y is inmeters and x is in seconds). Find

29. (A) The average velocity (the average rate of change of ywith respect to x) for x changing from 1 to 3 seconds

(B) The average velocity for x changing from 1 to seconds

(C) The instantaneous velocity at second

30. (A) The average velocity (the average rate of change of ywith respect to x) for x changing from 2 to 4 seconds

(B) The average velocity for x changing from 2 to seconds

(C) The instantaneous velocity at secondsx = 2

2 + h

x = 1

1 + h

y = f1x2 = x2+ x

12 + h, f12 + h22.

11 + h, f11 + h22.



Problems 31–38 refer to the function F in the graph shown. Usethe graph to determine whether exists at each indicatedvalue of x.

F¿1x2 44. Let and

(A) How are the graphs of these functions related? Howwould you expect the derivatives of these functions tobe related?

(B) Use the four-step process to find the derivative ofwhere C is any real constant.

45. (A) Give a geometric explanation of the following state-ment: If is a constant function, then

(B) Use the four-step process to verify the statement inpart (A).

46. (A) Give a geometric explanation of the following statement: If is a linear function,then

(B) Use the four-step process to verify the statement in part (A).

C In Problems 47–50, sketch the graph of f and determine where fis nondifferentiable.

47.

48.

49.

50.

In Problems 51–56, determine whether f is differentiable atby considering

51. 52.

53. 54.

55. 56.

57. A ball dropped from a balloon falls feet in x seconds. If the balloon is 576 feet above the groundwhen the ball is dropped, when does the ball hit theground? What is the velocity of the ball at the instant ithits the ground?

58. Repeat Problem 57 if the balloon is 1,024 feet above theground when the ball is dropped.

y = 16x2

f1x2 = 21 + x2f1x2 = 21 - x2

f1x2 = x2>3f1x2 = x1>3f1x2 = 1 - ƒ x ƒf1x2 = ƒ x ƒ

limh:0

f10 + h2 - f102h

x = 0

f1x2 = e2 - x2 if x … 02 if x 7 0

f1x2 = ex2+ 1 if x 6 0

1 if x Ú 0

f1x2 = e2x if x 6 26 - x if x Ú 2

f1x2 = e2x if x 6 12 if x Ú 1

f¿1x2 = m.f1x2 = mx + b

f¿1x2 = 0.f1x2 = C

m1x2 = -x2+ C,

-x2+ 2.h1x2 =f1x2 = -x2, g1x2 = -x2

- 1,

xa b c d e f g h

F(x)

31. 32.

33. 34.

35. 36.

37. 38.

39. Given

(A) Find (B) Find the slopes of the lines tangent to the graph of f at

2, and 4.(C) Graph f, and sketch in the tangent lines at 2,

and 4.

40. Given

(A) Find (B) Find the slopes of the lines tangent to the graph of f at

and 1.(C) Graph f, and sketch in the tangent lines at

and 1.

41. If an object moves along a line so that it is atat time x (in seconds), find the

instantaneous velocity function and find thevelocity at times and 5 seconds (y is measuredin feet).

42. Repeat Problem 41 with

43. Let and

(A) How are the graphs of these functions related? Howwould you expect the derivatives of these functions tobe related?

(B) Use the four-step process to find the derivative ofwhere C is any real constant.m1x2 = x2

+ C,

h1x2 = x2+ 2.f1x2 = x2, g1x2 = x2

- 1,

f1x2 = 8x2- 4x.

x = 1, 3,v = f¿1x2,

y = f1x2 = 4x2- 2x

x = -2, -1,x = -2, -1,

f¿1x2.f1x2 = x2

+ 2x,

x = 0,x = 0,

f¿1x2.f1x2 = x2

- 4x,

x = hx = g

x = fx = e

x = dx = c

x = bx = a

59. Revenue. The revenue (in dollars) from the sale of x carseats for infants is given by

(A) Find the average change in revenue if production ischanged from 1,000 car seats to 1,050 car seats.

(B) Use the four-step process to find (C) Find the revenue and the instantaneous rate of change

of revenue at a production level of 1,000 car seats, andwrite a brief verbal interpretation of these results.

R¿1x2.

R1x2 = 60x - 0.025x2 0 … x … 2,400

60. Profit. The profit (in dollars) from the sale of x car seatsfor infants is given by

(A) Find the average change in profit if production ischanged from 800 car seats to 850 car seats.

(B) Use the four-step process to find (C) Find the profit and the instantaneous rate of change

of profit at a production level of 800 car seats, andwrite a brief verbal interpretation of these results.

P¿1x2.

P1x2 = 45x - 0.025x2- 5,000 0 … x … 2,400

Applications



61. Sales analysis. The total sales of a company (in millionsof dollars) t months from now are given by

(A) Use the four-step process to find (B) Find S(15) and Write a brief verbal interpre-

tation of these results.(C) Use the results in part (B) to estimate the total sales

after 16 months and after 17 months.


(A) Use the four-step process to find (B) Find S(10) and Write a brief verbal interpre-

tation of these results.(C) Use the results in part (B) to estimate the total sales

after 11 months and after 12 months.

63. Mineral production. The U.S. production of zinc (in thou-sands of metric tons) is given approximately by

where t is time in years and corresponds to 1995.

(A) Use the four-step process to find (B) Find the annual production in 2010 and the

instantaneous rate of change of production in2010, and write a brief verbal interpretation of theseresults.

64. Mineral consumption. The U.S. consumption of copper(in thousands of metric tons) is given approximately by


(A) Use the four-step process to find (B) Find the annual production in 2010 and the

instantaneous rate of change of production in2010, and write a brief verbal interpretation of theseresults.

65. Electricity consumption. Table 1 gives the retail sales of electricity (in billions of kilowatt-hours) for theresidential and commercial sectors in the UnitedStates during the 1990s. (Source: Energy InformationAdministration)

p¿1t2.t = 0

p1t2 = 27t2- 75t + 6,015

p¿1t2.t = 0

p1t2 = 14t2- 6.6t + 602.4

S¿1102.S¿1t2.

S1t2 = 22t + 6

S¿1152.S¿1t2.

S1t2 = 22t + 10

(A) Let x represent time (in years) with corresponding to 1996, and let y represent thecorresponding residential sales. Enter theappropriate data set in a graphing calculator andfind a quadratic regression equation for the data.

(B) If denotes the regression equation foundin part (A), find R(20) and and write a briefverbal interpretation of these results. Roundanswers to the nearest tenth of a billion.

66. Electricity consumption. Refer to the data in Table 1.

(A) Let x represent time (in years) with corresponding to 1996, and let y represent the corre-sponding commercial sales. Enter the appropriatedata set in a graphing calculator and find a quadraticregression equation for the data.

(B) If denotes the regression equation foundin part (A), find C(20) and and write a briefverbal interpretation of these results. Roundanswers to the nearest tenth of a billion.

67. Air pollution. The ozone level (in parts per billion) on asummer day in a metropolitan area is given by

where t is time in hours and corresponds to 9 A.M.

(A) Use the four-step process to find (B) Find P(3) and Write a brief verbal interpreta-

tion of these results.

68. Medicine. The body temperature (in degrees Fahrenheit)of a patient t hours after being given a fever-reducingdrug is given by

(A) Use the four-step process to find (B) Find F(3) and Write a brief verbal

interpretation of these results.

69. Infant mortality. The number of infant deaths per 100,000births for males in the United States is given approxi-mately by


(A) Use the four-step process to find (B) Find the number of male infant deaths in 2010 and

the instantaneous rate of change of the number ofmale infant deaths in 2010, and write a brief verbalinterpretation of these results.

70. Infant mortality. The number of infant deaths per 100,000births for females in the United States is given approxi-mately by


(A) Use the four-step process to find (B) Find the number of female infant deaths in 2010 and

the instantaneous rate of change of the number offemale infant deaths in 2010, and write a brief verbalinterpretation of these results.

f¿1t2.t = 0

f1t2 = 0.007t2- 0.4t + 11.6

f¿1t2.t = 0

f1t2 = 0.008t2- 0.5t + 14.4

F¿132.F¿1t2.

F1t2 = 98 +

4t + 1

P¿132.P¿1t2.

t = 0

P1t2 = 80 + 12t - t2

C¿1202,y = C1x2

x = 0

R¿1202,y = R1x2

x = 0

Year Residential Commercial

1996 1,083 887

1997 1,076 926

1998 1,130 979

1999 1,145 1,002

2000 1,192 1,055

2001 1,202 1,083

2002 1,265 1,104

2003 1,276 1,199

2004 1,292 1,230

2005 1,359 1,275

TABLE 1 Electricity Sales


S e c t i o n 3 - 5 Basic Differentiation Properties 183

Section 3-5 BASIC DIFFERENTIATION PROPERTIES� Constant Function Rule� Power Rule� Constant Multiple Property� Sum and Difference Properties� Applications

In the preceding section, we defined the derivative of f at x as

if the limit exists, and we used this definition and a four-step process to find the de-rivatives of several functions. Now we want to develop some rules of differentiation.These rules will enable us to find the derivative of a great many functions withoutusing the four-step process.

Before starting on these rules, we list some symbols that are widely used torepresent derivatives.

f¿1x2 = limh:0

f1x + h2 - f1x2h

NOTATION The Derivative

If then

all represent the derivative of f at x.

f¿1x2 y¿ dy

dx

y = f1x2,

Each of these symbols for the derivative has its particular advantage in certain situ-ations. All of them will become familiar to you after a little experience.

� Constant Function RuleIf is a constant function, then the four-step process can be used to show that

(see Problem 45 in Exercise 3-4). Thus,

The derivative of any constant function is 0.

f¿1x2 = 0f1x2 = C

THEOREM 1 CONSTANT FUNCTION RULEIf then

Also, and

Note: When we write or we mean that when y = C.y¿ =

dy

dx= 0

d

dx C = 0,C¿ = 0

dy>dx = 0.y¿ = 0f¿1x2 = 0

y = f1x2 = C,

The graph of is a horizontal line with slope 0 (see Fig. 1), so we would expectthat f¿1x2 = 0.

f1x2 = C

I N S I G H T

x

f (x)

f (x) � C

0

C

Slope � 0

FIGURE 1



E X A M P L E 1 Differentiating Constant Functions

(A) If then (B) If then

(C) If then (D)d

dx 23 = 0

dy

dx= 0.y = p,

y¿ = 0.y = -1.4,f¿1x2 = 0.f1x2 = 3,

MATCHED PROBLEM 1 Find

(A) for (B) for

(C) for (D)

� Power RuleA function of the form where k is a real number, is called a power function.The following elementary functions (see the inside front cover) are examples of powerfunctions:

(1)

Explore & Discuss 1 (A) It is clear that the functions f, h, and m in (1) are power functions. Explainwhy the functions n and p are also power functions.

(B) The domain of a power function depends on the power. Discuss the domainof each of the following power functions:

The definition of the derivative and the four-step process introduced in thepreceding section can be used to find the derivatives of many power functions. Forexample, it can be shown that

Notice the pattern in these derivatives. In each case, the power in f becomes thecoefficient in and the power in is 1 less than the power in f. In general, for anypositive integer n,

(2)

In fact, more advanced techniques can be used to show that (2) holds for any realnumber n. We will assume this general result for the remainder of the book.

If f1x2 = xn, then f ¿1x2 = nxn - 1.

f¿f¿

If f1x2 = x5, then f¿1x2 = 5x4.

If f1x2 = x4, then f¿1x2 = 4x3.

If f1x2 = x3, then f¿1x2 = 3x2.

If f1x2 = x2, then f¿1x2 = 2x.

u1x2 = x-1>4 v1x2 = x1>5 w1x2 = x-1>5 r1x2 = x4 s1x2 = x-4 t1x2 = x1>4

n1x2 = 2x p1x2 = 23 x

f1x2 = x h1x2 = x2 m1x2 = x3

f1x2 = xk,

d

dx 1-p2y = -27

dy

dx

y = 12y¿f1x2 = -24f¿1x2

THEOREM 2 POWER RULEIf where n is a real number, then

Also, and dy>dx = nxn - 1.y¿ = nxn - 1

f¿1x2 = nxn - 1

y = f1x2 = xn,

Explore & Discuss 2 (A) Write a verbal description of the power rule.

(B) If what is Discuss how this derivative can be obtainedfrom the power rule.

f¿1x2?f1x2 = x,




(A) for (B) for

(C) for (D)

In some cases, properties of exponents must be used to rewrite an expressionbefore the power rule is applied.

d

dx x3>2y = t-2dy

dt

y = x30y¿f1x2 = x6f¿1x2

E X A M P L E 2 Differentiating Power Functions

(A) If then

(B) If then

(C) If then

(D)d

dx x5>3

=

53

x15>32- 1=

53

x2>3.

dy

dt= -3t-3 - 1

= -3t-4= -

3

t4.y = t-3,

y¿ = 25x25 - 1= 25x24.y = x25,

f¿1x2 = 5x5 - 1= 5x4.f1x2 = x5,

E X A M P L E 3 Differentiating Power Functions

(A) If we can write and

(B) If we can write and

(C) or-1

323 x4

d

dx

123 x=

d

dx x-1>3

= -

13

x1-1>32- 1= -

13

x-4>3,

y¿ =

12

u11>22- 1=

12

u-1>2, or 1

22u

y = u1>2y = 2u,

f¿1x2 = -4x-4 - 1= -4x-5, or

-4

x5

f1x2 = x-4f1x2 = 1>x4,


(A) for (B) for (C)

� Constant Multiple PropertyLet where k is a constant and u is differentiable at x. Then, using thefour-step process, we have the following:

Step 1.

Step 2.

Step 3.

Step 4.

Definition of

= ku¿1x2u¿1x2 = k lim

h:0cu1x + h2 - u1x2

hd

limx:c

kg1x2 = k limx:c

g1x2 = limh:0

k cu1x + h2 - u1x2h

d f¿1x2 = lim

h:0

f1x + h2 - f1x2h

f1x + h2 - f1x2h

=

k3u1x + h2 - u1x24h

= k cu1x + h2 - u1x2h

df1x + h2 - f1x2 = ku1x + h2 - ku1x2 = k3u1x + h2 - u1x24f1x + h2 = ku1x + h2

f1x2 = ku1x2,

d

dx

12xy = 23 u2y¿f1x2 =

1x

f¿1x2


Thus,

The derivative of a constant times a differentiable function is the constant timesthe derivative of the function.


THEOREM 3 CONSTANT MULTIPLE PROPERTYIf then

Also,

y¿ = ku¿ dy

dx= k

du

dx

f¿1x2 = ku¿1x2y = f1x2 = ku1x2,

E X A M P L E 4 Differentiating a Constant Times a Function

(A) If then

(B) If then

(C) If then , or

(D)

or -

0.62x5= -0.6x-5>2,

d

dx

0.42x3=

d

dx 0.4

x3>2 =

d

dx 0.4x-3>2 = 0.4 c - 3

2 x1-3>22- 1 d

-2

x5 .y¿ =

12

1-4x-4 - 12 = -2x-5y =

1

2x4 =

12

x-4,

dy

dt =

16

# 3t3 - 1=

12

t2.y =

t3

6=

16

t3,

f¿1x2 = 3 # 2x2 - 1= 6x.f1x2 = 3x2,


(A) for (B) for

(C) for (D)

� Sum and Difference PropertiesLet where and exist. Then, using the four-stepprocess, we have the following:

Step 1.Step 2.

Step 3.

Step 4.

= u¿1x2 + v¿1x2 = lim

h:0

u1x + h2 - u1x2h

+ limh:0

v1x + h2 - v1x2h

limx:c3g1x2 + h1x24 = lim

x:c g1x2 + lim

x:c h1x2

= limh:0cu1x + h2 - u1x2

h+

v1x + h2 - v1x2h

d f¿1x2 = lim

h:0

f1x + h2 - f1x2h

=

u1x + h2 - u1x2h

+

v1x + h2 - v1x2h

f1x + h2 - f1x2

h=

u1x + h2 - u1x2 + v1x + h2 - v1x2h

= u1x + h2 - u1x2 + v1x + h2 - v1x2 f1x + h2 - f1x2 = u1x + h2 + v1x + h2 - 3u1x2 + v1x24f1x + h2 = u1x + h2 + v1x + h2

v¿1x2u¿1x2f1x2 = u1x2 + v1x2,

d

dx 0.923 x

y =

1

3x3y¿

y =

t4

12

dy

dtf1x2 = 4x5f¿1x2



Thus,

The derivative of the sum of two differentiable functions is the sum of thederivatives of the functions.

Similarly, we can show that

The derivative of the difference of two differentiable functions is the differenceof the derivatives of the functions.

Together, we then have the sum and difference property for differentiation:

THEOREM 4 SUM AND DIFFERENCE PROPERTYIf then

Also,

Note: This rule generalizes to the sum and difference of any given number offunctions.

With the preceding rule and the other rules stated previously, we will be able tocompute the derivatives of all polynomials and a variety of other functions.

y¿ = u¿ ; v¿ dy

dx=

du

dx;

dv

dx

f¿1x2 = u¿1x2 ; v¿1x2y = f1x2 = u1x2 ; v1x2,

E X A M P L E 5 Differentiating Sums and Differences

(A) If then

(B) If then

(C) If then

(D)

= -

10

3x3 +

8

x5 +

13

x2

=

53

1-22x-3- 21-42x-5

+

19

# 3x2

d

dx a 5

3x2 -

2

x4 +

x3

9b =

d

dx 53

x-2-

d

dx 2x-4

+

d

dx 19

x3

dy

dw=

d

dw w1>3

-

d

dw 3w =

1

3 w-2>3

- 3 =

1

3w2>3 - 3

y = 23 w - 3w,

y¿ = 142¿ + 12x32¿ - 13x-12¿ = 0 + 213x22 - 31-12x-2 = 6x2+ 3x-2

y = 4 + 2x3- 3x-1,

f¿1x2 = 13x22¿ + 12x2¿ = 312x2 + 2112 = 6x + 2

f1x2 = 3x2+ 2x,


(A) for

(B) for

(C) for

(D)d

dx a -

34x

+

4

x3 -

x4

8b

y = 5v3- 24 v

dy

dv

y = 3 - 7x-2y¿

f1x2 = 3x4- 2x3

+ x2- 5x + 7f¿1x2



APPLICATIONS

E X A M P L E 6 Instantaneous Velocity An object moves along the y axis (marked in feet) sothat its position at time x (in seconds) is

(A) Find the instantaneous velocity function v.

(B) Find the velocity at and seconds.

(C) Find the time(s) when the velocity is 0.

SOLUTION (A)

(B)

(C) Factor 3 out of each term.Factor the quadratic term.Use the zero properly.

Thus, at and seconds.x = 3x = 1v = 0

x = 1, 3 31x - 121x - 32 = 0

31x2- 4x + 32 = 0

v = f¿1x2 = 3x2- 12x + 9 = 0

f¿152 = 31522 - 12152 + 9 = 24 feet per second

f¿122 = 31222 - 12122 + 9 = -3 feet per second

v = f¿1x2 = 1x32¿ - 16x22¿ + 19x2¿ = 3x2- 12x + 9

x = 5x = 2

f1x2 = x3- 6x2

+ 9x

MATCHED PROBLEM 6 Repeat Example 6 for f1x2 = x3- 15x2

+ 72x.

E X A M P L E 7 Tangents Let

(A) Find

(B) Find the equation of the tangent line at

(C) Find the values of x where the tangent line is horizontal.

SOLUTION (A)

(B)

Tangent line at (C) Since a horizontal line has 0 slope, we must solve for x:

Factor 4x out of each term.Factor the difference of two squares.

Use the zero properly. x = 0, -23, 23

4x1x + 2321x - 232 = 0

4x1x2- 32 = 0

f¿1x2 = 4x3- 12x = 0

f¿1x2 = 0

x = 1 y = -8x + 13

m = f ¿1x12 = f ¿112 = 41123 - 12112 = -8 y - 5 = -81x - 12y1 = f1x12 = f112 = 1124 - 61122 + 10 = 5 y - y1 = m1x - x12

= 4x3- 12x

f¿1x2 = 1x42¿ - 16x22¿ + 1102¿

x = 1.

f¿1x2.f1x2 = x4

- 6x2+ 10.

MATCHED PROBLEM 7 Repeat Example 7 for

Answers to Matched Problems 1. All are 0.

2. (A) (B) (C) (D)

3. (A) (B) (C)

4. (A) (B) (C) (D)

5. (A) (B)

(C) (D) 3>14x22 - 112>x42 - 1x3>2215v2-

14 v-3>4, or 15v2

- 1>14v3>4214x-3, or 14>x312x3

- 6x2+ 2x - 5

-0.3x-4>3, or -0.3>23 x4-x-4, or -1>x4t3>320x4

-12 x-3>2, or -1>122x322

3 u-1>3, or 2>1323 u2-x-2, or -1>x2

32 x1>2

-2t-3= -2>t330x296x5

f1x2 = x4- 8x3

+ 7.



6. (A)

(B) ft�s; ft�s

(C)

7. (A)

(B)

(C) x = 0 and x = 6

y = -20x + 20

f¿1x2 = 4x3- 24x2

x = 4 and x = 6 seconds

f¿152 = -3f¿122 = 24

v = 3x2- 30x + 72

A Find the indicated derivatives in Problems 1–18.

1. 2.

3. 4.

5. 6.

7. 8.

9.

10.

11. 12.

13.

14.

15. 16.

17. 18.

Problems 19–24 refer to functions f and g that satisfy and In each problem, find for the indicatedfunction h.

19.

20.

21.

22.

23.

24.

B Find the indicated derivatives in Problems 25–48.

25.

26.

27.

28.

29.

30.

31.d

du 15u0.3

- 4u2.22g¿1x2 if g1x2 = 5x-7

- 2x-4

y¿ for y = 5x-2+ 9x-1

dy

dt if y = 2 + 5t - 8t3

f¿1t2 if f1t2 = 2t2- 3t + 1

d

dx 1-4x + 92

d

dx 12x - 52

h1x2 = -4f1x2 + 5g1x2 - 9

h1x2 = 2f1x2 - 3g1x2 + 7

h1x2 = g1x2 - f1x2h1x2 = f1x2 + g1x2h1x2 = 5g1x2h1x2 = 4f1x2

h¿122g¿122 = -1.f¿122 = 3

dy

dx for y =

x5

25d

dx a x3

18b

f¿1x2 for f1x2 = 0.8x4y¿ for y = 0.4x7

d

dx 1-2x32

f¿1x2 for f1x2 = 5x2

y¿ for y =

1

x12

dy

dx for y =

1

x10

f¿1x2 for f1x2 = x9>2g¿1x2 for g1x2 = x8>3

dy

dx for y = x-8y¿ for y = x-4

g¿1x2 for g1x2 = x5d

dx x3

y¿ for y = x6dy

dx for y = x9

d

dx 3f¿1x2 for f1x2 = 7

32.

33.

34.

35.

36.

37.

38.

39.

40.

41.

42.

43.

44.

45.

46.

47.

48.

For Problems 49–52, find

(A)

(B) The slope of the graph of f at and

(C) The equations of the tangent lines at and

(D) The value(s) of x where the tangent line is horizontal

49.

50.

51.

52. f1x2 = x4- 32x2

+ 10

f1x2 = 3x4- 6x2

- 7

f1x2 = 2x2+ 8x

f1x2 = 6x - x2

x = 4x = 2x = 4x = 2

f¿1x2

d

dx a2.8x-3

-

0.623 x2+ 7b

d

dx a 1.22x

- 3.2x-2+ xb

w¿ if w =

1025 u

y¿ if y =

123 x

F¿1t2 if F1t2 =

5

t1>5 -

8

t3>2

h¿1t2 if h1t2 =

3

t3>5 -

6

t1>2

d

du 18u3>4

+ 4u-1>42

d

du 13u2>3

- 5u1>32H¿1w2 if H1w2 =

5

w6 - 22w

G¿1w2 if G1w2 =

5

9w4 + 523 w

d

dx a5x3

4-

2

5x3 b

d

dx a3x2

2-

7

5x2 bw¿ if w =

7

5u2

y¿ if y =

2

5x4

F¿1t2 if F1t2 = 0.2t3- 3.1t + 13.2

h¿1t2 if h1t2 = 2.1 + 0.5t - 1.1t3

d

du 12u4.5

- 3.1u + 13.22

Exercise 3-5



If an object moves along the y axis (marked in feet) so that its position at time x (in seconds) is given by the indicatedfunctions in Problems 53–56, find

(A) The instantaneous velocity function

(B) The velocity when and seconds

(C) The time(s) when

53.

54.

55.

56.

Problems 57–64 require the use of a graphing calculator. For eachproblem, find and approximate (to four decimal places) thevalue(s) of x where the graph of f has a horizontal tangent line.

57.

58.

59.

60.

61.

62.

63.

64.

65. Let Recall that the graph ofis a parabola. Use the derivative to derive a

formula for the x coordinate of the vertex of this parabola.

66. Now that you know how to find derivatives, explain whyit is no longer necessary for you to memorize the formulafor the x coordinate of the vertex of a parabola.

67. Give an example of a cubic polynomial function that has

(A) No horizontal tangents(B) One horizontal tangent(C) Two horizontal tangents

f¿1x2y = f1x2f1x2 = ax2

+ bx + c, a Z 0.

f1x2 = 0.25x4- 2.6x3

+ 8.1x2- 10x + 9

f1x2 = 0.2x4- 3.12x3

+ 16.25x2- 28.25x + 7.5

f1x2 = 0.02x4- 0.06x3

- 0.78x2+ 0.94x + 2.2

f1x2 = 0.05x4+ 0.1x3

- 1.5x2- 1.6x + 3

f1x2 = 323 x4- 2x2

+ 4x

f1x2 = 323 x4- 1.5x2

- 3x

f1x2 = x2+ x - 102x

f1x2 = x2- 3x - 42x

f¿1x2

f1x2 = x3- 9x2

+ 24x

f1x2 = x3- 9x2

+ 15x

f1x2 = 80x - 10x2

f1x2 = 176x - 16x2

v = 0

x = 3x = 0v = f¿1x2

68. Can a cubic polynomial function have more than twohorizontal tangents? Explain.

C Find the indicated derivatives in Problems 69–76.

69.

70.

71.

72.

73.

74.

75.

76.

In Problems 77 and 78, use the definition of the derivative andthe four-step process to verify each statement.

77. 78.

79. The domain of the power function is the setof all real numbers. Find the domain of the derivative

Discuss the nature of the graph of for anyx values excluded from the domain of

80. The domain of the power function is the setof all real numbers. Find the domain of the derivative

Discuss the nature of the graph of for anyx values excluded from the domain of f¿1x2.

y = f1x2f¿1x2.f1x2 = x2>3

f¿1x2.y = f1x2f¿1x2.

f1x2 = x1>3

d

dx x4

= 4x3d

dx x3

= 3x2

f¿1x2 if f1x2 =

2x5- 4x3

+ 2x

x3

y¿ if y =

x4- 3x3

+ 5

x2

d

dx 5x - 3

15x6

dy

dx if y =

3x - 4

12x2

dy

dx if y =

x2+ 25

x2

d

dx 10x + 20

x

y¿ if y = 12x - 522f¿1x2 if f1x2 = 12x - 122


(A) Find

(B) Find S(5) and (to two decimal places). Write abrief verbal interpretation of these results.

(C) Find S(10) and (to two decimal places). Writea brief verbal interpretation of these results.


(A) Find

(B) Find S(4) and (to two decimal places). Write abrief verbal interpretation of these results.

(C) Find S(8) and (to two decimal places). Write abrief verbal interpretation of these results.

S¿182S¿142

S¿1t2.S1t2 = 0.015t4

+ 0.4t3+ 3.4t2

+ 10t - 3

S¿1102S¿152

S¿1t2.S1t2 = 0.03t3

+ 0.5t2+ 2t + 3

83. Advertising. From past records, it is estimated that amarine manufacturer will sell N(x) power boats afterspending $x thousand on advertising, as given by

(see the figure).

N1x2 = 1,000 -

3,780x

5 … x … 30

Applications

N(x)

x10 20 30

Thousands of advertising dollars

Boa

ts s

old

500

1,000

Figure for 83



(A) Find (B) Find and Write a brief verbal inter-

pretation of these results.

84. Price–demand equation. Suppose that, in a givengourmet food store, people are willing to buy x pounds ofchocolate candy per day at $p per quarter pound, as givenby the price–demand equation

This function is graphed in the accompanying figure. Findthe demand and the instantaneous rate of change ofdemand with respect to price when the price is $5. Writea brief verbal interpretation of these results.

x = 10 +

180p

2 … p … 10

N¿1202.N¿1102N¿1x2. (A) Let x represent time (in years) since 1980, and let y

represent the corresponding number of registered lim-ousine operators in the United States. Enter the datain a graphing calculator and find a quadratic regres-sion equation for the data.

(B) If denotes the regression equation foundin part (A), find L(18) and (to the nearest hun-dred), and write a brief verbal interpretation of theseresults.

87. Medicine. A person x inches tall has a pulse rate of ybeats per minute, as given approximately by

What is the instantaneous rate of change of pulse rate atthe

(A) 36-inch level?

(B) 64-inch level?

88. Ecology. A coal-burning electrical generating plant emitssulfur dioxide into the surrounding air. The concentrationC(x), in parts per million, is given approximately by

where x is the distance from the plant in miles. Find theinstantaneous rate of change of concentration at

(A)

(B)

89. Learning. Suppose that a person learns y items in xhours, as given by

(see the figure). Find the rate of learning at the end of

(A) 1 hour (B) 9 hours

y = 502x 0 … x … 9

x = 2 miles

x = 1 mile

C1x2 =

0.1

x2

y = 590x-1>2 30 … x … 75

L¿1182y = L1x2

x

p5 10

Price (dollars)

Dem

and

(pou

nds)

50

100

Figure for 84

Year Limousine Operators

1985 4,000

1990 7,000

1995 8,900

2000 11,000

TABLE 2

Year Limousine Production

1980 2,500

1985 6,500

1990 4,400

1995 3,400

2000 5,700

TABLE 1

85. Motor vehicle production. Annual limousine productionin the United States for selected years is given in Table 1.

(A) Let x represent time (in years) since 1980, and let yrepresent the corresponding U.S. production of lim-ousines. Enter the data in a graphing calculator andfind a cubic regression equation for the data.

(B) If denotes the regression equation found inpart (A), find L(12) and (to the nearest hun-dred), and write a brief verbal interpretation of theseresults.

(C) Repeat part (B) for L(18) and

86. Motor vehicle operation. The total number of registeredlimousine operators in the United States for selectedyears is given in Table 2.

L¿1182.

L¿1122y = L1x2

y

x5 10

Time (hours)

Item

s le

arne

d

50

100

150

Figure for 89

90. Learning. If a person learns y items in x hours, as given by

find the rate of learning at the end of


y = 2123 x2 0 … x … 8


Section 3-6 DIFFERENTIALS� Increments� Differentials� Approximations Using Differentials

In this section, we introduce increments and differentials. Increments are useful intheir own right, and they also provide an alternative notation for defining the deriv-ative. Differentials are often easier to compute than increments and can be used toapproximate increments.

� IncrementsIn Section 3-4, we defined the derivative of f at x as the limit of the difference quotient

We considered various interpretations of this limit, including slope, velocity, andinstantaneous rate of change. Increment notation enables us to interpret the numer-ator and the denominator of the difference quotient separately.

Given if x changes from 2 to 2.1, then y will change fromto The change in x is called the

increment in x and is denoted by (read “delta x”), since is the Greek letter delta.Similarly, the change in y is called the increment in y and is denoted by . In termsof the given example, we write

Change in xf(x) � x3

Use a calculator.

Corresponding change in y = 1.261

= 9.261 - 8

= 2.13- 23

¢y = f 12.12 - f 122 ¢x = 2.1 - 2 = 0.1

¢y¢¢x

y = f12.12 = 2.13= 9.261.y = f122 = 23

= 8y = f1x2 = x3,

f ¿1x2 = limh:0

f1x + h2 - f1x2h


The symbol does not represent the product of and x, but is the symbol for a singlequantity: the change in x. Likewise, the symbol represents a single quantity: the changein y.

¢y¢¢x

I N S I G H T

The definition of these quantities is given next.

f (x)

x1 x2x

(x2, f (x2))

(x1, f (x1))

y � f (x)

�y � f (x2) � f (x1)� f (x1 � �x) � f (x1)

x1 � �x

�x

DEFINITION IncrementsFor

represents the changein y corresponding to achange in x.

can be either positiveor negative.

[Note: depends on the function f, the input x1, and the increment .]¢x¢y

¢x

¢x

¢y

- f 1x12= f 1x1 + ¢x2= f 1x22 - f 1x12

¢y = y2 - y1

x2 = x1 + ¢x, and¢x = x2 - x1, soy = f 1x2,


S e c t i o n 3 - 6 Differentials 193

In Example 1, we observe another notation for the familiar difference quotient

(1)

It is common to refer to h, the change in x, as Then the difference quotient (1)takes on the form

and the derivative is defined by

or

(2)

if the limit exists.

Explore & Discuss 1 Suppose that defines a function whose domain is the set of all realnumbers. If every increment is equal to 0, what is the range of f?

� DifferentialsAssume that the limit in equation (2) exists. Then, for small the difference quo-tient provides a good approximation for Also, provides a goodapproximation for We indicate the latter by writing

is small, but (3)Z 0¢x¢y

¢xL f¿1x2

¢y>¢x.f¿1x2f¿1x2.¢y>¢x

¢x,

¢yy = f 1x2

f¿1x2 = lim¢x:0

¢y

¢x

f¿1x2 = lim¢x:0

f 1x + ¢x2 - f 1x2¢x

¢y = f 1x + ¢x2 - f 1x2f 1x + ¢x2 - f 1x2¢x

or ¢y

¢x

¢x.

f 1x + h2 - f 1x2h

E X A M P L E 1 Increments Given the function

(A) Find and for and

(B) Find for and

SOLUTION (A)

(B)

=

f 132 - f 1122

=

92 -

12

2=

42

= 2

f 1x1 + ¢x2 - f 1x12

¢x=

f 11 + 22 - f 1122

¢y

¢x=

f 1x22 - f 1x12x2 - x1

=

32

1=

32

= f 122 - f 112 =

42

-

12

=

32

¢y = f 1x22 - f 1x12 ¢x = x2 - x1 = 2 - 1 = 1

¢x = 2.x1 = 1f 1x1 + ¢x2 - f 1x12

¢x

x2 = 2.x1 = 1¢y>¢x¢x, ¢y,

y = f 1x2 =

x2

2,

MATCHED PROBLEM 1 Given the function

(A) Find and for and

(B) Find for and ¢x = 2.x1 = 1f 1x1 + ¢x2 - f 1x12

¢x

x2 = 3.x1 = 2¢y>¢x¢x, ¢y,

y = f 1x2 = x2+ 1,


Multiplying both sides of (3) by gives us

is small, but (4)

From equation (4), we see that provides a good approximation for whenis small.

Because of the practical and theoretical importance of we give it thespecial name differential and represent it by the special symbol dy or df :

For example,

In the second example, we usually drop the parentheses in and simply write

In summary, we have the following:

dx � ¢x

d1x2 d1x2 = 1x2¿ ¢x = 1 ¢x = ¢x

d12x32 = 12x32¿ ¢x = 6x2 ¢x

dy � f ¿1x2¢x or df � f ¿1x2¢x

f¿1x2 ¢x,¢x

¢yf¿1x2¢x

Z 0¢x¢y L f¿1x2 ¢x

¢x


DEFINITION DifferentialsIf defines a differential function, then the differential dy, or df, is definedas the product of and where Symbolically,

where

[Note: The differential or is actually a function involving two independentvariables: x and ; a change in either one or both will affect or ]df2.1dydx

df21dy

dx = ¢x

dy = f¿1x2 dx, or df = f¿1x2 dx

dx = ¢x.dx,f¿1x2y = f1x2

E X A M P L E 2 Differentials Find for Evaluate for

(A) and (B) and

(C) and

SOLUTION

(A) When and (B) When and

(C) When and

dy = 32112 + 340.02 = 0.1

dx = 0.02,x = 1

dy = 32132 + 340.1 = 0.9dy = 32122 + 340.1 = 0.7

dx = 0.1,x = 3dx = 0.1,x = 2

= 12x + 32 dx

dy = f¿1x2 dx

dx = 0.02x = 1

dx = 0.1x = 3dx = 0.1x = 2

dyf 1x2 = x2+ 3x.dy

MATCHED PROBLEM 2 Find for Evaluate for

(A) and (B) and (C) and dx = 0.01x = 1

dx = 0.12x = 9dx = 0.1x = 4

dyf 1x2 = 1x + 3.dy

We now have two interpretations of the symbol Referring to the functionin Example 2 with and we have

Derivativedy

dx= f¿122 = 7

dx = 0.1,x = 2y = f 1x2 = x2+ 3x

dy>dx.



and

Ratio of differentialsdy

dx=

0.70.1

= 7

Finding a differential is not a new operation; it is new notation and a new interpretationof the familiar operation of differentiation.

I N S I G H T

� Approximations Using DifferentialsEarlier, we noted that for small

Also, since

it follows that

and can be used to approximate To interpret this result geometrically, we need to recall a basic property of the

slope. The vertical change in a line is equal to the product of the slope and the hori-zontal change, as shown in Figure 1.

Now consider the line tangent to the graph of as shown in Figure 2. Sinceis the slope of the tangent line and is the horizontal change in the tangent

line, it follows that the vertical change in the tangent line is given by asindicated in Figure 2.

dy = f¿1x2 dx,dxf¿1x2 y = f 1x2,

¢y.dy

¢y L dy

dy = f¿1x2 dx

¢y

¢xL f¿1x2 and ¢y L f¿1x2¢x

¢x,

Slope � m

a

b

m �ba or b � ma

FIGURE 1

f (x)

x x � �xx

(x, f (x))

y � f (x)

f (x)Tangent lineat x

(x � �x, f (x � �x))

�x � dx

dy

�ySecant line

FIGURE 2

Explore & Discuss 2 Suppose that defines a differentiable function. Explain why the graphof the function (of the independent variable ) is a straight linethrough the origin for a fixed value of x.

¢x = dxdyy = f 1x2

E X A M P L E 3 Comparing Increments and Differentials Let .

(A) Find and when

(B) Graph and from part A for

(C) Compare and from part A for , and 0.3.¢x = 0.1, 0.2dy¢y

-1 … ¢x … 1.dy¢y

x = 2.dy¢y

y = f1x2 = 6x - x2


SOLUTION (A)

Remove parentheses.

Collect like terms.

Since and

(B) The graphs of and are shown in Figure 3. Exam-ining the graphs, we conclude that the differential provides a linear approxima-tion of the increments and that the approximation is better for values of close to 0.

(C) The table on the left compares the values of and for the indicated valuesof ¢x.

dy¢x

¢x = dx

dy = 2¢x¢y = 2¢x - ¢x2

dy = f¿122 dx = 2¢xdx = ¢x,f¿1x2 = 6 - 2x, f¿122 = 2,

= 2¢x - ¢x2 = 12 + 6¢x - 4 - 4¢x - ¢x2

- 12 + 4

= 612 + ¢x2 - 12 + ¢x22 - 16 # 2 - 222 ¢y = f12 + ¢x2 - f122


dx � �x1

�1

1

�1

dy

�y

FIGURE 3¢y = 2¢x - ¢x2, dy = 2¢x

0.1 0.19 0.2

0.2 0.36 0.4

0.3 0.51 0.6

dy¢y¢x

MATCHED PROBLEM 3 Repeat Example 3 for and and .-0.3¢x = dx = -0.1, -0.2,x = 4

The error in the approximation is usually small when is small (seeExample 3C), but can be quite substantial in certain cases.

¢x = dx¢y L dy

I N S I G H T

E X A M P L E 4 Cost–Revenue A company manufactures and sells x transistor radios per week. Ifthe weekly cost and revenue equations are

find the approximate changes in revenue and profit if production is increased from2,000 to 2,010 units per week.

SOLUTION We will approximate and with and respectively, using and

= $40 per week

= a8 -

2,000500

b 10= $60 per week

= a8 -

x

500b dx= a10 -

2,000500

b10

dP = P ¿1x2 dx= a10 -

x

500b dx

= 8x -

x2

1,000- 5,000 dR = R¿1x2 dx

R1x2 = 10x -

x2

1,000 P1x2 = R1x2 - C1x2 = 10x -

x2

1,000- 5,000 - 2x

dx = 2,010 - 2,000 = 10.x = 2,000dP,dR¢P¢R

C1x2 = 5,000 + 2x R1x2 = 10x -

x2

1,000 0 … x … 8,000

MATCHED PROBLEM 4 Repeat Example 4 with production increasing from 6,000 to 6,010.

Comparing the results in Example 4 and Matched Problem 4, we see that an increase in production results in a revenue and profit increase at the 2,000 productionlevel, but a revenue and profit loss at the 6,000 production level.

Answers to Matched Problems 1. (A) (B) 4

2. (A) 0.025 (B) 0.02 (C) 0.005

3. (A) ¢y = -2¢x - ¢x2; dy = -2¢x

dy =

1

21x dx

¢x = 1, ¢y = 5, ¢y>¢x = 5



(B) (C)

dx � �x1

�1

1

�1

dy

�y�0.1 0.19 0.2

�0.2 0.36 0.4

�0.3 0.51 0.6

dy¢y¢x

4. dR = -$20>wk; dP = -$40>wk

A In Problems 1–6, find the indicated quantities for

1. and given and

2. and given and

3. given and

4. given and

5. given and

6. given and

In Problems 7–12, find dy for each function.

7. 8.

9. 10.

11. 12.

B In Problems 13 and 14, find the indicated quantities for

13. (A) (simplify)

(B) What does the quantity in part (A) approach as approaches 0?

14. (A) (simplify)

(B) What does the quantity in part (A) approach as approaches 0?

In Problems 15–18, find dy for each function.

15.

16.

17.

18. y =

x2

1x + 122

y =

x

x2+ 9

y = 13x + 525y = 12x + 123

¢x

f 13 + ¢x2 - f 132¢x

¢x

f 12 + ¢x2 - f 122¢x

y = f 1x2 = 3x2.

y = 521xy =

5901x

y = x3160 - x2y = x2a1 -

x

9b

y = 200x -

x2

30y = 30 + 12x2

- x3

x2 = 3x1 = 2¢y>¢x ;

x2 = 3x1 = 1¢y>¢x ;

¢x = 1x1 = 2f 1x1 + ¢x2 - f 1x12

¢x;

¢x = 2x1 = 1f 1x1 + ¢x2 - f 1x12

¢x;

x2 = 5x1 = 2¢y>¢x ;¢x, ¢y,

x2 = 4x1 = 1¢y>¢x ;¢x, ¢y,

y = f 1x2 = 3x2.In Problems 19–22, evaluate and for each function forthe indicated values.

19.

20.

21.

22.

23. A cube with sides 10 inches long is covered with a coat offiberglass 0.2 inch thick. Use differentials to estimate thevolume of the fiberglass shell.

24. A sphere with a radius of 5 centimeters is coated with ice0.1 centimeter thick. Use differentials to estimate the vol-ume of the ice. [ ]

C In Problems 25–28,

(A) Find and for the function f at the indicated valueof x.

(B) Graph and dy from part A.

(C) Compare the values of and dy from part A at the indi-cated values of .

25.

26.

27.

28.

In Problems 29–32, discuss the validity of each statement.If the statement is always true, explain why. If not, give acounterexample.

29. If the graph of the function is a line, then thefunctions and (of the independent variable

) for at are identical.

30. If the graph of the function is a parabola, thenthe functions and (of the independent variable

) for at are identical.

31. Suppose that defines a differentiable functionwhose domain is the set of all real numbers. If everydifferential at is equal to 0, then is aconstant function.

32. Suppose that defines a function whose domain is the set of all real numbers. If every

y = f 1x2f 1x2x = 2dy

y = f 1x2x = 0f 1x2¢x = dxdy¢y

y = f 1x2x = 3f 1x2¢x = dx

dy¢yy = f 1x2

x = 2, ¢x = dx = - 0.05, - 0.10, - 0.15f 1x2 = x3

- 2x2;

f 1x2 = x3- 2x2; x = 1, ¢x = dx = 0.05, 0.10, 0.15

x = -2, ¢x = dx = -0.1, -0.2, -0.3f 1x2 = x2

+ 2x + 3;

x = -0.5, ¢x = dx = 0.1, 0.2, 0.3f 1x2 = x2+ 2x + 3;

¢x¢y

¢y

dy¢y

Recall that V =43pr 3.

y = f 1x2 = 100ax -

4

x2 b ; x = 2, dx = ¢x = - 0.1

y = f 1x2 = 75a1 -

2xb ; x = 5, dx = ¢x = - 0.5

y = f 1x2 = 30 + 12x2- x3; x = 2, dx = ¢x = 0.1

y = f 1x2 = x2- 3x + 2; x = 5, dx = ¢x = 0.2

¢ydy

Exercise 3-6



increment at is equal to 0, then is a constant function.

33. Find if

34. Find if y = 12x2- 422x + 2.dy

y =

323x2- 2x + 1.dy

f 1x2x = 2 35. Find and for and

36. Find and for and¢x = dx = 1.

y = 590>2x, x = 64,¢ydy

¢x = dx = 0.3.y = 522x, x = 4,¢ydy

Use differential approximations in the following problems.

37. Advertising. From past records, it is estimated that acompany will sell N units of a product after spending $x thousand in advertising, as given by

Approximately what increase in sales will result by in-creasing the advertising budget from $10,000 to $11,000?From $20,000 to $21,000?

38. Price–demand. Suppose, in a grocery chain, the daily demand (in pounds) for chocolate candy at $x per poundis given by

If the price is increased from $3.00 per pound to $3.20 per pound, what is the approximate change in demand?

39. Average cost. For a company that manufactures tennisrackets, the average cost per racket is found to be

where x is the number of rackets produced per hour.What will the approximate change in average cost perracket be if production is increased from 20 per hour to25 per hour? From 40 per hour to 45 per hour?

40. Revenue and profit. A company manufactures and sells x televisions per month. If the cost and revenueequations are

what will the approximate changes in revenue and profitbe if production is increased from 1,500 to 1,510? From4,500 to 4,510?

41. Pulse rate. The average pulse rate y (in beats per minute) of a healthy person x inches tall is givenapproximately by

Approximately how will the pulse rate change for a change in height from 36 to 37 inches? From 64 to 65 inches?

42. Measurement. An egg of a particular bird is very nearly spherical. If the radius to the inside of the shell

y =

5901x 30 … x … 75

0 … x … 6,000R1x2 = 200x -

x2

30

C1x2 = 72,000 + 60x

C =

400x

+ 5 +

12

x x Ú 1

C

D = 1,000 - 40x2 1 … x … 5

N = 60x - x2 5 … x … 30

is 5 millimeters and the radius to the outside of the shell is 5.3 millimeters, approximately what is the volume of the shell? [Remember that ]

43. Medicine. A drug is given to a patient to dilate her arteries. If the radius of an artery is increased from 2 to 2.1 millimeters, approximately how much is the cross-sectional area increased? Assume that the cross section of the artery is circular; that is,

44. Drug sensitivity. One hour after x milligrams of a particular drug are given to a person, the change in body temperature T (in degrees Fahrenheit) is given by

Approximate the changes in body temperature producedby the following changes in drug dosages:(A) From 2 to 2.1 milligrams(B) From 3 to 3.1 milligrams(C) From 4 to 4.1 milligrams

45. Learning. A particular person learning to type has anachievement record given approximately by

where N is the number of words per minute typed after t weeks of practice. What is the approximate improve-ment from 5 to 5.5 weeks of practice?

46. Learning. If a person learns y items in x hours, as givenapproximately by

what is the approximate increase in the number of itemslearned when x changes from 1 to 1.1 hours? From 4 to4.1 hours?

47. Politics. In a newly incorporated city, it is estimated thatthe voting population (in thousands) will increase according to

where t is time in years. Find the approximate change invotes for the following changes in time:(A) From 1 to 1.1 years(B) From 4 to 4.1 years(C) From 7 to 7.1 years

N1t2 = 30 + 12t2- t3 0 … t … 8

y = 521x 0 … x … 9

N = 75a1 -

2tb 3 … t … 20

T = x2a1 -

x

9b 0 … x … 6

A = pr2.43

V =43pr3.

Applications


S e c t i o n 3 - 7 Marginal Analysis in Business and Economics 199

Section 3-7 MARGINAL ANALYSIS IN BUSINESS AND ECONOMICS� Marginal Cost, Revenue, and Profit� Application� Marginal Average Cost, Revenue, and Profit

� Marginal Cost, Revenue, and ProfitOne important use of calculus in business and economics is in marginal analysis. Ineconomics, the word marginal refers to a rate of change—that is, to a derivative. Thus,if C(x) is the total cost of producing x items, then is called the marginal cost andrepresents the instantaneous rate of change of total cost with respect to the numberof items produced. Similarly, the marginal revenue is the derivative of the total rev-enue function and the marginal profit is the derivative of the total profit function.

C¿1x2

DEFINITION Marginal Cost, Revenue, and ProfitIf x is the number of units of a product produced in some time interval, then

Marginal cost (or revenue or profit) is the instantaneous rate of change of cost (or rev-enue or profit) relative to production at a given production level.

= 1marginal revenue2 - 1marginal cost2 marginal profit = P¿1x2 = R¿1x2 - C¿1x2

total profit = P1x2 = R1x2 - C1x2 marginal revenue = R¿1x2

total revenue = R1x2 marginal cost = C¿1x2

total cost = C1x2

To begin our discussion, we consider a cost function C(x). It is important to re-member that C(x) represents the total cost of producing x items, not the cost of pro-ducing a single item. To find the cost of producing a single item, we use the differenceof two successive values of C(x):

Exact cost of producing the 1x + 12 st item = C1x + 12 - C1x2 Total cost of producing x items = C1x2

Total cost of producing x + 1 items = C1x + 12

E X A M P L E 1 Cost Analysis A company manufactures fuel tanks for automobiles. The totalweekly cost (in dollars) of producing x tanks is given by

(A) Find the marginal cost function.

(B) Find the marginal cost at a production level of 500 tanks per week, and interpretthe results.

(C) Find the exact cost of producing the 501st item.

SOLUTION (A)

(B) Marginal cost

At a production level of 500 tanks per week, the total production costs are in-creasing at the rate of $40 tank.

C¿15002 = 90 - 0.115002 = $40

C¿1x2 = 90 - 0.1x

C1x2 = 10,000 + 90x - 0.05x2


(C)

Total cost of producing 501 tanks per week

Total cost of producing 500 tanks per week

Exact cost of producing the 501st tank = $39.95

C15012 - C15002 = 42,539.95 - 42,500.00

= $42,500.00

C15002 = 10,000 + 9015002 - 0.05150022 = $42,539.95

C15012 = 10,000 + 9015012 - 0.05150122

MATCHED PROBLEM 1 A company manufactures automatic transmissions for automobiles.The total weeklycost (in dollars) of producing x transmissions is given by

(A) Find the marginal cost function.

(B) Find the marginal cost at a production level of 200 transmissions per week, andinterpret the results.

(C) Find the exact cost of producing the 201st transmission.

In Example 1, we found that the cost of the 501st tank and the marginal cost at aproduction level of 500 tanks differ by only a nickel. Increments and differentials willhelp us understand the relationship between marginal cost and the cost of a singleitem. If C(x) is any total cost function, then

See Section 3-6

Thus, we see that the marginal cost approximates the exactcost of producing the item. These observations are summarized next andare illustrated in Figure 1.

1x + 12stC1x + 12 - C1x2,C¿1x2

¢x = 1 C¿1x2 L C1x + 12 - C1x2 C¿1x2 L

C1x + ¢x2 - C1x2¢x

C1x2 = 50,000 + 600x - 0.75x2

C

x

y � C(x)

�x � 1 C(x � 1) � C(x)

C�(x)

Tangent line at x

x � 1x

FIGURE 1C1x + 12 - C1x2

C¿1x2 L

C(x)

x

$50,000

$40,000

$30,000

$20,000

$10,000

200 400 600 800

Slope � C �(500)� 40

Slope � C �(200)� 70

FIGURE 2 C1x2 = 10,000 + 90x - 0.05x2

THEOREM 1 MARGINAL COST AND EXACT COSTIf C(x) is the total cost of producing x items, the marginal cost function approximatesthe exact cost of producing the item:

Marginal cost Exact cost

Similar statements can be made for total revenue functions and total profit functions.

C¿1x2 L C1x + 12 - C1x2

1x + 12st

Theorem 1 states that the marginal cost at a given production level x approximates the costof producing the or next, item. In practice, the marginal cost is used more fre-quently than the exact cost. One reason for this is that the marginal cost is easily visual-ized when one is examining the graph of the total cost function. Figure 2 shows the graphof the cost function discussed in Example 1, with tangent lines added at and

The graph clearly shows that as production increases, the slope of the tangentline decreases. Thus, the cost of producing the next tank also decreases, a desirable char-acteristic of a total cost function. We will have much more to say about graphical analy-sis in Chapter 5.

x = 500.x = 200

1x + 12st,

I N S I G H T




APPLICATIONWe next discuss how price, demand, revenue, cost, and profit are tied together in typ-ical applications. Although either price or demand can be used as the independentvariable in a price–demand equation, it is common practice to use demand as the in-dependent variable when marginal revenue, cost, and profit are also involved.

Explore & Discuss 1 The market research department of a company used test marketing to determinethe demand for a new headphone set (Table 1).

(A) Assuming that the relationship between price p and demand x is linear, findthe price–demand equation and write the result in the form Graph the equation and find the domain of f. Discuss the effect of price in-creases and decreases on demand.

(B) Solve the equation found in part (A) for p, obtaining an equation of theform Graph this equation and find the domain of g. Discuss theeffect of price increases and decreases on demand.

p = g1x2.

x = f1p2.Demand Price

x p

3,000 $7

6,000 $4

TABLE 1

E X A M P L E 2 Production Strategy The market research department of a company recommendsthat the company manufacture and market a new headphone set for MP3 players.After suitable test marketing, the research department presents the followingprice–demand equation:

x is demand at price p. (1)

Solving (1) for p gives

(2)

where x is the number of headphones retailers are likely to buy at $p per set.The financial department provides the cost function

(3)

where $7,000 is the estimate of fixed costs (tooling and overhead) and $2 is the esti-mate of variable costs per headphone set (materials, labor, marketing, transporta-tion, storage, etc.).

(A) Find the domain of the function defined by the price–demand equation (2).

(B) Find and interpret the marginal cost function .

(C) Find the revenue function as a function of x, and find its domain.

(D) Find the marginal revenue at 5,000, and 7,000. Interpret these results.

(E) Graph the cost function and the revenue function in the same coordinate system,find the intersection points of these two graphs, and interpret the results.

(F) Find the profit function and its domain, and sketch the graph of the function.

(G) Find the marginal profit at 4,000, and 6,000. Interpret these results.

SOLUTION (A) Since price p and demand x must be nonnegative, we have and

Thus, the permissible values of x are

(B) The marginal cost is Since this is a constant, it costs an additional $2to produce one more headphone set at any production level.

(C) The revenue is the amount of money R received by the company for manufac-turing and selling x headphone sets at $p per set and is given by

R = 1number of headphone sets sold21price per headphone set2 = xp

C¿1x2 = 2.

0 … x … 10,000.

10,000 Ú x 10 Ú 0.001x

p = 10 - 0.001x Ú 0

x Ú 0

x = 1,000,

x = 2,000,

C¿1x2

C1x2 = 7,000 + 2x

p = 10 - 0.001x

x = 10,000 - 1,000p


In general, the revenue R can be expressed as a function of p by using equa-tion (1) or as a function of x by using equation (2). As we mentioned earlier,when using marginal functions, we will always use the number of items x as theindependent variable. Thus, the revenue function is

Using equation (2) (4)

Since equation (2) is defined only for it follows that the domainof the revenue function is

(D) The marginal revenue is

For production levels of and 7,000, we have

This means that at production levels of 2,000, 5,000, and 7,000, the respective ap-proximate changes in revenue per unit change in production are $6, $0, and That is, at the 2,000 output level, revenue increases as production increases; atthe 5,000 output level, revenue does not change with a “small” change in pro-duction; and at the 7,000 output level, revenue decreases with an increase inproduction.

(E) When we graph R(x) and C(x) in the same coordinate system, we obtain Figure 3. The intersection points are called the break-even points, because revenue equals cost at these production levels: The company neither makes nor loses money, but just breaks even. The break-even points are obtained as follows:

Solve by the quadratic formula(see Appendix A-7).

Thus, the break-even points are (1,000, 9,000) and (7,000, 21,000), as shown inFigure 3. Further examination of the figure shows that cost is greater than rev-enue for production levels between 0 and 1,000 and also between 7,000 and10,000. Consequently, the company incurs a loss at these levels. By contrast, forproduction levels between 1,000 and 7,000, revenue is greater than cost and thecompany makes a profit.

C17,0002 = 7,000 + 217,0002 = 21,000

R17,0002 = 1017,0002 - 0.00117,00022 = 21,000

C11,0002 = 7,000 + 211,0002 = 9,000

R11,0002 = 1011,0002 - 0.00111,00022 = 9,000

= 1,000, 7,000

=

8,000 ; 6,0002

=

8,000 ; 236,000,0002

x =

8,000 ; 28,0002- 417,000,0002

2

x2- 8,000x + 7,000,000 = 0

0.001x2- 8x + 7,000 = 0

7,000 + 2x = 10x - 0.001x2

C1x2 = R1x2

-$4.

R¿12,0002 = 6 R¿15,0002 = 0 R¿17,0002 = -4

x = 2,000, 5,000,

R¿1x2 = 10 - 0.002x

0 … x … 10,000.0 … x … 10,000,

= 10x - 0.001x2

R1x2 = xp = x110 - 0.001x2




(F) The profit function is

The domain of the cost function is and the domain of the revenue func-tion is Thus, the domain of the profit function is the set of x val-ues for which both functions are defined—that is, The graph ofthe profit function is shown in Figure 4. Notice that the x coordinates of thebreak-even points in Figure 3 are the x intercepts of the profit function. Fur-thermore, the intervals on which cost is greater than revenue and on which rev-enue is greater than cost correspond, respectively, to the intervals on which profitis negative and the intervals on which profit is positive.

(G) The marginal profit is

For production levels of 1,000, 4,000, and 6,000, we have

This means that at production levels of 1,000, 4,000, and 6,000, the respective ap-proximate changes in profit per unit change in production are $6, $0, and That is, at the 1,000 output level, profit will be increased if production is in-creased; at the 4,000 output level, profit does not change for “small” changes inproduction; and at the 6,000 output level, profits will decrease if production isincreased. It seems that the best production level to produce a maximum profitis 4,000.

Example 2 warrants careful study, since a number of important ideas in eco-nomics and calculus are involved. In the next chapter, we will develop a sys-tematic procedure for finding the production level (and, using the demandequation, the selling price) that will maximize profit.

-$4.

P¿11,0002 = 6 P¿14,0002 = 0 P¿16,0002 = -4

P¿1x2 = -0.002x + 8

0 … x … 10,000.0 … x … 10,000.

x Ú 0

= -0.001x2+ 8x - 7,000

= 110x - 0.001x22 - 17,000 + 2x2P1x2 = R1x2 - C1x2

P

x10,0005,000

$10,000

�$30,000

�$10,000

�$20,000

FIGURE 4

R C

5,000

10,000

15,000

20,000

25,000

30,000

5,000Number of headphone sets produced per week

Rev

enue

and

cos

ts (

dolla

rs)

10,000

Loss

Profit

Break-evenpoints

Revenue function

Loss

Cost function

x

FIGURE 3

MATCHED PROBLEM 2 Refer to the revenue and profit functions in Example 2.

(A) Find and and interpret the results.

(B) Find and and interpret the results.P¿17,0002,P¿12,0002R¿16,0002,R¿13,0002



Then

Explain why is used in the preceding equation. Then find the correct expres-sion for the profit function.

� Marginal Average Cost, Revenue, and ProfitSometimes it is desirable to carry out marginal analysis relative to average cost (costper unit), average revenue (revenue per unit), and average profit (profit per unit).The relevant definitions are summarized as follows:

Z

P1x2 = R1x2 - C1x2 Z 9x - 0.002x2- 12,000 + 5x

C1x2 = 12,000 + 5x and R1x2 = 9x - 0.002x2


DEFINITION Marginal Average Cost, Revenue, and ProfitIf x is the number of units of a product produced in some time interval, then

Cost per unit:

Revenue per unit:

Profit per unit:

marginal average profit = P¿1x2 =

d

dx P1x2

average profit = P1x2 =

P1x2x

marginal average revenue = R¿1x2 =

d

dx R1x2

average revenue = R1x2 =

R1x2x

marginal average cost = C¿1x2 =

d

dx C1x2

average cost = C1x2 =

C1x2x

E X A M P L E 3 Cost Analysis A small machine shop manufactures drill bits used in the petrole-um industry.The shop manager estimates that the total daily cost (in dollars) of pro-ducing x bits is

(A) Find and

(B) Find and and interpret these quantities.

(C) Use the results in part (B) to estimate the average cost per bit at a productionlevel of 11 bits per day.

SOLUTION (A)

Average cost function

Marginal average cost function

(B)

At a production level of 10 bits per day, the average cost of producing a bit is$124, and this cost is decreasing at the rate of $10.10 per bit.

C¿1102 = -

1,000

102 - 0.1 = -$10.10

C1102 =

1,00010

+ 25 - 0.11102 = $124

C¿1x2 =

d

dx C1x2 = -

1,000

x2 - 0.1

=

1,000x

+ 25 - 0.1x

C1x2 =

C1x2x

=

1,000 + 25x - 0.1x2

x

C¿1102,C1102C¿1x2.C1x2

C1x2 = 1,000 + 25x - 0.1x2



(C) If production is increased by 1 bit, then the average cost per bit will decrease byapproximately $10.10. Thus, the average cost per bit at a production level of 11 bitsper day is approximately $124 - $10.10 = $113.90.

MATCHED PROBLEM 3 Consider the cost function for the production of headphone sets from Example 2:

(A) Find and

(B) Find and and interpret these quantities.

(C) Use the results in part (B) to estimate the average cost per headphone set at aproduction level of 101 headphone sets.

C¿11002,C11002C¿1x2.C1x2

C1x2 = 7,000 + 2x

Explore & Discuss 3 A student produced the following solution to Matched Problem 3:

CostMarginal cost

“Average” of the marginal cost

Explain why the last function is not the same as the marginal average costfunction.

CAUTION

1. The marginal average cost function must be computed by first finding the aver-age cost function and then finding its derivative. As Explore & Discuss 3 illus-trates, reversing the order of these two steps produces a different function thatdoes not have any useful economic interpretations.

2. Recall that the marginal cost function has two interpretations: the usual inter-pretation of any derivative as an instantaneous rate of change and the specialinterpretation as an approximation to the exact cost of the item. Thisspecial interpretation does not apply to the marginal average cost function.Referring to Example 3, we would be incorrect to interpretto mean that the average cost of the next bit is approximately In fact,the phrase “average cost of the next bit” does not even make sense. Averagingis a concept applied to a collection of items, not to a single item.

These remarks also apply to revenue and profit functions.

Answers to Matched Problems 1. (A)(B) At a production level of 200 transmissions, total costs are increasing at the

rate of $300 per transmission.(C)

2. (A) At a production level of 3,000, a unit increase in production will increaserevenue by approximately $4.

At a production level of 6,000, a unit increase in production will de-crease revenue by approximately $2.

(B) At a production level of 2,000, a unit increase in production will increaseprofit by approximately $4.

At a production level of 7,000, a unit increase in production will de-crease profit by approximately $6.

3. (A)

(B) At a production level of 100 headphone sets, the av-erage cost per headphone set is $72, and this average cost is decreasing at a rate of $0.70per headphone set.

(C) Approx. $71.30.

C11002 = $72; C¿11002 = -$0.70.

C1x2 =

7,000x

+ 2; C¿1x2 = -

7,000

x2

P¿17,0002 = -6.

P¿12,0002 = 4.

R¿16,0002 = -2.

R¿13,0002 = 4.

C12012 - C12002 = $299.25

C¿12002 = 300.C¿1x2 = 600 - 1.5x

-$10.10.C¿1102 = -$10.10

1x + 12st

C¿1x2

x=

2x

C¿1x2 = 2 C1x2 = 7,000 + 2x



1. Cost analysis. The total cost (in dollars) of producing xfood processors is

(A) Find the exact cost of producing the 21st foodprocessor.

(B) Use the marginal cost to approximate the cost ofproducing the 21st food processor.

2. Cost analysis. The total cost (in dollars) of producing x electric guitars is

(A) Find the exact cost of producing the 51st guitar.

(B) Use the marginal cost to approximate the cost ofproducing the 51st guitar.

C1x2 = 1,000 + 100x - 0.25x2

C1x2 = 2,000 + 50x - 0.5x2

6. Profit analysis. The total profit (in dollars) from the saleof x portable stereos is

(A) Find the exact profit from the sale of the 41st stereo.(B) Use the marginal profit to approximate the profit

from the sale of the 41st stereo.

7. Profit analysis. The total profit (in dollars) from the saleof x videocassettes is

Evaluate the marginal profit at the given values of x, andinterpret the results.

(A) (B)

8. Profit analysis. The total profit (in dollars) from the saleof x cameras is

Evaluate the marginal profit at the given values of x, andinterpret the results.

(A) (B) x = 350x = 200

P1x2 = 12x - 0.02x2- 1,000 0 … x … 600

x = 750x = 450

P1x2 = 5x - 0.005x2- 450 0 … x … 1,000

P1x2 = 22x - 0.2x2- 400 0 … x … 100

Applications

3. Cost analysis. The total cost (in dollars) of manufacturingx auto body frames is

(A) Find the average cost per unit if 500 frames areproduced.

(B) Find the marginal average cost at a production levelof 500 units, and interpret the results.

(C) Use the results from parts (A) and (B) to estimate theaverage cost per frame if 501 frames are produced.

4. Cost analysis. The total cost (in dollars) of printing xdictionaries is

(A) Find the average cost per unit if 1,000 dictionariesare produced.

(B) Find the marginal average cost at a production levelof 1,000 units, and interpret the results.

(C) Use the results from parts (A) and (B) to estimatethe average cost per dictionary if 1,001 dictionariesare produced.

5. Profit analysis. The total profit (in dollars) from the saleof x skateboards is

(A) Find the exact profit from the sale of the 26thskateboard.

(B) Use the marginal profit to approximate the profitfrom the sale of the 26th skateboard.

P1x2 = 30x - 0.3x2- 250 0 … x … 100

C1x2 = 20,000 + 10x

C1x2 = 60,000 + 300x

9. Profit analysis. The total profit (in dollars) from the saleof x lawn mowers is

(A) Find the average profit per mower if 50 mowers areproduced.

(B) Find the marginal average profit at a productionlevel of 50 mowers, and interpret the results.

(C) Use the results from parts (A) and (B) to estimate theaverage profit per mower if 51 mowers are produced.

10. Profit analysis. The total profit (in dollars) from the saleof x charcoal grills is

(A) Find the average profit per grill if 40 grills areproduced.

(B) Find the marginal average profit at a productionlevel of 40 grills, and interpret the results.

(C) Use the results from parts (A) and (B) to estimatethe average profit per grill if 41 grills are produced.

11. Revenue analysis. The price p (in dollars) and thedemand x for a particular clock radio are related by theequation

x = 4,000 - 40p

P1x2 = 20x - 0.02x2- 320 0 … x … 1,000

P1x2 = 30x - 0.03x2- 750 0 … x … 1,000

Exercise 3-7



(A) Express the price p in terms of the demand x,and find the domain of this function.

(B) Find the revenue R(x) from the sale of x clock ra-dios. What is the domain of R?

(C) Find the marginal revenue at a production level of 1,600 clock radios, and interpret the results.

(D) Find the marginal revenue at a production level of 2,500 clock radios, and interpret the results.

12. Revenue analysis. The price p (in dollars) and thedemand x for a particular steam iron are related by theequation

(A) Express the price p in terms of the demand x,and find the domain of this function.

(B) Find the revenue R(x) from the sale of x clock radios. What is the domain of R?

(C) Find the marginal revenue at a production level of 400 steam irons, and interpret the results.

(D) Find the marginal revenue at a production level of 650 steam irons, and interpret the results.

13. Revenue, cost and profit. The price–demand equation andthe cost function for the production of table saws aregiven, respectively, by

where x is the number of saws that can be sold at a priceof $p per saw and C(x) is the total cost (in dollars) ofproducing x saws.

(A) Express the price p as a function of the demand x,and find the domain of this function.

(B) Find the marginal cost.(C) Find the revenue function and state its domain.(D) Find the marginal revenue.(E) Find and and interpret these

quantities.(F) Graph the cost function and the revenue function

on the same coordinate system for Find the break-even points, and indicate regions of loss and profit.

(G) Find the profit function in terms of x.

(H) Find the marginal profit.(I) Find and and interpret these

quantities.

14. Revenue, cost, and profit. The price–demand equationand the cost function for the production of television setsare given, respectively, by

where x is the number of sets that can be sold at a priceof $p per set and C(x) is the total cost (in dollars) ofproducing x sets.

(A) Express the price p as a function of the demand x,and find the domain of this function.

(B) Find the marginal cost.(C) Find the revenue function and state its domain.(D) Find the marginal revenue.

x = 9,000 - 30p and C1x2 = 150,000 + 30x

P¿13,0002,P¿11,5002

0 … x … 6,000.

R¿14,5002,R¿11,5002

x = 6,000 - 30p and C1x2 = 72,000 + 60x

x = 1,000 - 20p

(E) Find and and interpret thesequantities.

(F) Graph the cost function and the revenue function onthe same coordinate system for Findthe break-even points, and indicate regions of lossand profit.

(G) Find the profit function in terms of x.

(H) Find the marginal profit.(I) Find and and interpret these

quantities.

15. Revenue, cost, and profit. A company is planning tomanufacture and market a new two-slice electric toaster.After conducting extensive market surveys, the researchdepartment provides the following estimates: a weeklydemand of 200 toasters at a price of $16 per toaster and aweekly demand of 300 toasters at a price of $14 pertoaster. The financial department estimates that weeklyfixed costs will be $1,400 and variable costs (cost per unit)will be $4.

P¿14,5002,P¿11,5002

0 … x … 9,000.

R¿16,0002,R¿13,0002

(A) Assume that the relationship between the price pand the demand x is linear. Use the research depart-ment’s estimates to express p as a function of x, andfind the domain of this function.

(B) Find the revenue function in terms of x and state itsdomain.

(C) Assume that the cost function is linear. Use the fi-nancial department’s estimates to express the costfunction in terms of x.

(D) Graph the cost function and the revenue function onthe same coordinate system for Findthe break-even points, and indicate regions of lossand profit.

(E) Find the profit function in terms of x.

(F) Evaluate the marginal profit at andand interpret the results.

16. Revenue, cost, and profit. The company in Problem 15 isalso planning to manufacture and market a four-slicetoaster. For this toaster, the research department’s esti-mates are a weekly demand of 300 toasters at a price of$25 per toaster and a weekly demand of 400 toasters at aprice of $20. The financial department’s estimates are fixedweekly costs of $5,000 and variable costs of $5 per toaster.

(A) Assume that the relationship between the price pand the demand x is linear. Use the research depart-ment’s estimates to express p as a function of x, andfind the domain of this function.

(B) Find the revenue function in terms of x and state itsdomain.

x = 475,x = 250

0 … x … 1,000.



(C) Assume that the cost function is linear. Use the financial department’s estimates to express the cost function in terms of x.

(D) Graph the cost function and the revenue function on the same coordinate system for Find the break-even points, and indicate regions of loss and profit.

(E) Find the profit function in terms of x.

(F) Evaluate the marginal profit at andand interpret the results.

17. Revenue, cost, and profit. The total cost and the total rev-enue (in dollars) for the production and sale of x ski jack-ets are given, respectively, by

(A) Find the value of x where the graph of R(x) has ahorizontal tangent line.

(B) Find the profit function P(x).(C) Find the value of x where the graph of P(x) has a

horizontal tangent line.(D) Graph C(x), R(x), and P(x) on the same coordinate

system for Find the break-evenpoints. Find the x intercepts of the graph of P(x).

18. Revenue, cost, and profit. The total cost and the total revenue (in dollars) for the production and sale of x hairdryers are given, respectively, by

(A) Find the value of x where the graph of R(x) has ahorizontal tangent line.

(B) Find the profit function P(x).(C) Find the value of x where the graph of P(x) has a

horizontal tangent line.(D) Graph C(x), R(x), and P(x) on the same coordinate

system for Find the break-even points.Find the x intercepts of the graph of P(x).

19. Break-even analysis. The price–demand equation and thecost function for the production of garden hoses aregiven, respectively, by

where x is the number of garden hoses that can be sold ata price of $p per unit and C(x) is the total cost (in dollars)of producing x garden hoses.

(A) Express the revenue function in terms of x.

(B) Graph the cost function and the revenue function in the same viewing window for Use approximation techniques to find the break-even points correct to the nearest unit.

20. Break-even analysis. The price–demand equation and the cost function for the production of handwovensilk scarves are given, respectively, by

p = 60 - 22x and C1x2 = 3,000 + 5x

0 … x … 400.

p = 20 - 2x and C1x2 = 500 + 2x

0 … x … 400.

0 … x … 400 C1x2 = 5x + 2,340 and R1x2 = 40x - 0.1x2

0 … x … 1,000.

0 … x … 1,000

C1x2 = 24x + 21,900 and R1x2 = 200x - 0.2x2

x = 425,x = 325

0 … x … 800.

where x is the number of scarves that can be sold at aprice of $p per unit and C(x) is the total cost (in dollars)of producing x scarves.

(A) Express the revenue function in terms of x.

(B) Graph the cost function and the revenue function in the same viewing window for Use approximation techniques to find the break-even points correct to the nearest unit.

21. Break-even analysis. Table 2 contains price–demand andtotal cost data for the production of overhead projectors,where p is the wholesale price (in dollars) of a projectorfor an annual demand of x projectors and C is the totalcost (in dollars) of producing x projectors.

0 … x … 900.

x p($) C($)

3,190 581 1,130,000

4,570 405 1,241,000

5,740 181 1,410,000

7,330 85 1,620,000

TABLE 2

x p($) C($)

2,910 1,435 3,650,000

3,415 1,280 3,870,000

4,645 1,125 4,190,000

5,330 910 4,380,000

TABLE 3

(A) Find a quadratic regression equation for the price–demand data, using x as the independent variable.

(B) Find a linear regression equation for the cost data,using x as the independent variable. Use this equa-tion to estimate the fixed costs and the variable costsper projector. Round answers to the nearest dollar.

(C) Find the break-even points. Round answers to thenearest integer.

(D) Find the price range for which the company will makea profit. Round answers to the nearest dollar.

22. Break-even analysis. Table 3 contains price–demand andtotal cost data for the production of treadmills, where p isthe wholesale price (in dollars) of a treadmill for an annualdemand of x treadmills and C is the total cost (in dollars)of producing x treadmills.

(A) Find a linear regression equation for the price–demand data, using x as the independent variable.

(B) Find a linear regression equation for the cost data, using x as the independent variable. Use thisequation to estimate the fixed costs and the variablecosts per treadmill. Round answers to the nearestdollar.

(C) Find the break-even points. Round answers to thenearest integer.

(D) Find the price range for which the company willmake a profit. Round answers to the nearest dollar.


Chapter 3 Review 209

3-1 Introduction to Limits• The graph of the function is the graph of the set of all ordered pairs • The limit of the function as x approaches c is L, written , if the functional value

is close to the single real number L whenever x is close, but not equal, to c (on either side of c).

• The limit of the function as x approaches c from the left is K, written , if isclose to K whenever x is close to, but to the left of, c on the real-number line.

• The limit of the function as x approaches c from the right is L, written , if

is close to L whenever x is close to, but to the right of, c on the real-number line.

• Limit properties are useful for evaluating limits.

• The limit of the difference quotient always results in a indeterminate form.Algebraic simplification is often required to evaluate this type of limit.

3-2 Continuity• Intuitively, the graph of a continuous function can be drawn without lifting a pen off the paper.

Algebraically, a function f is continuous at c if 1. exists 2. f(c) exists 3.

• Continuity properties are useful for determining where a function is continuous and where it is not.

• Continuity properties are also useful for solving inequalities.

3-3 Infinite Limits and Limits at Infinity• If increases or decreases without bound as x approaches a from either side of a, then the line

is a vertical asymptote of the graph of

• If gets close to L as x increases without bound or decreases without bound, then L is called thelimit of f at or .

• The end behavior of a polynomial is described in terms of limits at infinity.

• If approaches L as or as , then the line is a horizontal asymptote of thegraph of Polynomial functions never have horizontal asymptotes. A rational function can have at most one.

3-4 The Derivative• Given a function , the average rate of change is the ratio of the change in y to the change in x.

• The instantaneous rate of change is the limit of the average rate of change as the change in x approaches 0.

• The slope of the secant line through two points on the graph of a function is the ratio of thechange in y to the change in x. The slope of the tangent line at the point is the limit of the slopeof the secant line through the points and as h approaches 0.

• The derivative of at x, denoted , is the limit of the difference quotientas (if the limit exists).

• The four-step method is used to find derivatives.

• If the limit of the difference quotient does not exist at , then f is nondifferentiable at a and does not exist.

3-5 Basic Differentiation Properties• The derivative of a constant function is 0.

• For any real number n, the derivative of is .

• If f is a differential function, then the derivative of

• The derivative of the sum or difference of two differential functions is the sum or difference of thederivatives of the functions.

3-6 Differentials• Given the function , the change in x is also called the increment of x and is denoted by ∆x.

The corresponding change in y is called the increment of y and is given by

• If is differentiable at x, then the differential of x is and the differential of isor In this context, x and dx are both independent variables.df = f¿(x)dx.dy = f¿(x)dx,

y � f (x)dx = ∆xy = f(x)

∆y = f(x + ∆x) - f(x).y = f(x)

kf(x) is kf¿(x).

nxn - 1f(x) = xn

f¿(a)x = a

h : 0[f(x + h) - f(x)]>hf¿(x)y � f (x)

(a + h, f(a + h))(a, f(a))(a, f(a))

y = f(x)

y = f(x)

y = f(x).y = Lx : - qx : qf(x)

- qq

f(x)

y = f(x).x = af(x)

limx:c

f(x) = f(c)limx:c

f(x)

0>0[f(a + h) - f(a)]>h

f(x)

limx:c +

f(x) = Ly = f(x)

f(x)limx:c-

f(x) = Ky = f(x)

f(x)

limx:c

f(x) = Ly = f(x)(x, f(x)).y = f(x)

CHAPTER 3 REVIEWImportant Terms, Symbols, and Concepts

ExamplesEx. 1, p. 129Ex. 2, p. 130Ex. 3, p. 132

Ex. 4, p. 133Ex. 5, p. 135Ex. 6. p. 136Ex. 7, p. 137Ex. 8, p. 137

Ex. 9, p. 139Ex. 10, p. 139Ex. 11, p. 140

Ex. 1, p. 146Ex. 2, p. 147

Ex. 3, p. 148

Ex. 4, p. 149

Ex. 1, p. 157Ex. 2, p. 158

Ex. 3, p. 160

Ex. 4, p. 161

Ex. 5, p. 163

Ex. 6, p. 164

Ex. 1, p. 169

Ex. 2, p. 170

Ex. 3, p. 173

Ex. 4, p. 175

Ex. 5, p. 176

Ex. 6, p. 177

Ex. 1, p. 184

Ex. 2, p. 185

Ex. 3, p. 185

Ex. 4, p. 186Ex. 5, p. 187

Ex. 1, p. 193

Ex. 2, p. 194

Ex. 3, p. 195


Ex. 1, p. 199Ex. 2, p. 201

Ex. 3, p. 204

3-7 Marginal Analysis in Business and Economics• If is the total cost of producing x items, then is the marginal cost and

is the exact cost of producing item . Furthermore, Similar statements can be made regarding total revenue and total profit functions.

• If is the total cost of producing x items, then the average cost, or cost per unit, is

and the marginal average cost is . Similar statements can be made regarding total

revenue and total profit functions.

C¿(x) =

d

dx C(x)

C(x) =

C(x)

xy = C(x)

C¿(x) L C(x + 1) - C(x).x + 1C(x + 1) - C(x)y = C¿(x)y = C(x)

Work through all the problems in this chapter review, and checkyour answers in the back of the book. Answers to all review prob-lems are there, along with section numbers in italics to indicatewhere each type of problem is discussed. Where weaknesses showup, review appropriate sections of the text.

Many of the problems in this exercise set ask you to find aderivative. Most of the answers to these problems contain both anunsimplified form and a simplified form of the derivative. Whenchecking your work, first check that you applied the rules correctly,and then check that you performed the algebraic simplificationcorrectly.

A 1. Find the indicated quantities for

(A) The change in y if x changes from 1 to 3

(B) The average rate of change of y with respect to xif x changes from 1 to 3

(C) The slope of the secant line through the points (1, f(1)) and (3, f (3)) on the graph of

(D) The instantaneous rate of change of y with respectto x at

(E) The slope of the line tangent to the graph ofat

(F)

2. Use the two-step limiting process to find for

3. If and find

(A) (B)

(C) (D)

In Problems 4–6, use the graph of f to estimate the indicatedlimits and function values.

limx:135 + 2x - 3g1x24lim

x:1

g1x2f1x2

limx:13f1x2g1x24lim

x:115f1x2 + 3g1x22

limx:1

g1x2 = 4,limx:1

f1x2 = 2

f1x2 = -3x + 2.f ¿1x2

f¿112x = 1y = f1x2

x = 1

y = f1x2

y = f1x2 = 2x2+ 5:

4. (A)

(B)

(C)

(D) f(1)

5. (A)

(B)

(C)

(D) f(2)

6. (A)

(B)

(C)

(D) f(3)

In Problems 7–9, use the graph of the function f shown in thefigure to answer each question.

limx:3

f1x2lim

x:3+

f1x2lim

x:3-

f1x2

limx:2

f1x2lim

x:2+

f1x2lim

x:2-

f1x2

limx:1

f1x2lim

x:1+

f1x2lim

x:1-

f1x2

x

f (x)

�2

5�2

5

x

f(x)

50

5

7. (A)

(B)(C) Is f continuous at

8. (A)

(B)(C) Is f continuous at

9. (A)

(B)(C) Is f continuous at x = 3?

f132 = ?

limx:3

f1x2 = ?

x = 2?f122 = ?

limx:2

f1x2 = ?

x = 1?f112 = ?

limx:1

f1x2 = ?


REVIEW EXERCISE


Review Exercise 211

27. (Hint: Multiply and thendifferentiate.)

In Problems 28–31, find the indicated quantities for

28.

29.

30.

31.

B Problems 32–34 refer to the function f in the figure.

∆y and dy for x = 1, ∆x = dx = 0.2.

dy for x1 = 1 and x2 = 3.

[f(x1 + ∆x) - f(x1)]>∆x for x1 = 1 and ∆x = 2.

∆x, ∆y, and ∆y>∆x for x1 = 1 and x2 = 3.

y = f(x) = x2+ x.

f(x) = (3x3- 2)(x + 1)

x

f(x)

50 10

5

10

Figure for 32–34:

f1x2 = ex2 if 0 … x 6 28 - x if x Ú 2

32. (A)

(B)

(C)

(D)

(E) Is f continuous at

33. (A)

(B)

(C)

(D)

(E) Is f continuous at

34. Solve each inequality. Express answers in interval notation.

(A)(B)

In Problems 35–37, solve each inequality. Express the answerin interval notation. Use a graphing calculator in Problem 37to approximate partition numbers to four decimal places.

35.

36.

37. x3+ x2

- 4x - 2 7 0

x - 5

x2+ 3x

7 0

x2- x 6 12

f1x2 Ú 0f1x2 6 0

x = 5?

f152 = ?

limx:5

f1x2 = ?

limx:5+

f1x2 = ?

limx:5-

f1x2 = ?

x = 2?

f122 = ?

limx:2

f1x2 = ?

limx:2+

f1x2 = ?

limx:2-

f1x2 = ?

In Problems 10–19, refer to the following graph of

10.

11.

12.

13.

14.

15.

16.

17. Identify any vertical asymptotes.

18. Identify any horizontal asymptotes.

19. Where is discontinuous?

20. Use the four-step limiting process to find

21. If and findfor each of the following functions:

(A)

(B)

(C)

(D)

(E)

In Problems 22–26, find and simplify.

22.

23.

24.

25.

26. f1x2 =

0.5

x4 + 0.25x4

f1x2 =

32x

+

5x3

4

f1x2 = 5

f1x2 = 2x1>2- 3x

f1x2 =

13

x3- 5x2

+ 1

f¿1x2

h(x) = 2f(x) + 3g(x)

h(x) = -g(x) - 1

h(x) = 2f(x) + 5

h(x) = -2g(x)

h(x) = 3f(x)

h¿(5)g¿(5) = -3,f(5) = 4, f¿(5) = -1, g(5) = 2,

f¿(x) for f(x) = 5x2.

y = f(x)

limx:0

f(x) = ?

limx:0+

f(x) = ?

limx:0-

f(x) = ?

limx:2-

f(x) = ?

limx:2+

f(x) = ?

limx: - q

f(x) = ?

limx: q

f(x) = ?

x

y

�10

10

20

2 4 6�2

y = f(x):


38. Let

(A) Find the slope of the secant line through (2, f (2))and (4, f (4)).

(B) Find the slope of the secant line through (2, f (2))and

(C) Find the slope of the tangent line at

In Problems 39–42, find the indicated derivative and simplify.

39. for

40. for

41. for

42. for

43. For find

(A) The slope of the graph at

(B) The equation of the tangent line at in the form

In Problems 44 and 45, find the value(s) of x where the tangentline is horizontal.

44.

45.

In Problems 46 and 47, approximate (to four decimal places) thevalue(s) of x where the graph of f has a horizontal tangent line.

46.

47.

48. If an object moves along the y axis (scale in feet) so that itis at at time x (in seconds), find


(B) The velocity at time seconds

49. An object moves along the y axis (scale in feet) so that attime x (in seconds) it is at Find


(B) The time(s) when the velocity is 0

50. Let and

(A) How are the graphs of f, g, and h related? Illustrateyour conclusion by graphing f, g, and h on the samecoordinate axes.

(B) How would you expect the graphs of the derivatives ofthese functions to be related? Illustrate your conclusionby graphing and on the same coordinate axes.

51. Let f(x) be a differentiable function and let k be a nonze-ro constant. For each function g, write a brief verbal de-scription of the relationship between the graphs of f and g.Do the same for the graphs of and

(A)

(B) g1x2 = f1x2 + k

g1x2 = kf1x), k > 1

g¿.f¿

h¿f¿, g¿,

h1x2 = 1x + 323.f1x2 = x3, g1x2 = 1x - 423,

y = f1x2 = -5x2+ 16x + 3.

x = 3

y = f1x2 = 8x2- 4x + 1

f(x) = x3- 10x3

- 5x + 10

f1x2 = x4- 2x3

- 5x2+ 7x

f1x2 = x3+ 3x2

- 45x - 135

f1x2 = 10x - x2

y = mx + bx = 1

x = 1

y = f1x2 = x2+ 4,

y =

2x3- 3

5x3

dy

dx

g1x2 = 1.832x +

0.931x

g¿1x2

y =

32x

2+

5

32xy¿

y =

13

x-3- 5x-2

+ 1dy

dx

x = 2.

12 + h, f12 + h22, h Z 0.

f1x2 = 0.5x2- 5.


52. Find and for and(correct to four decimal places).

53. If , and use differentials

to approximate (correct to four decimal places). Usea calculator to find (correct to four decimal places)and compare your approximation with the calculator value.

In Problems 54–58, determine where f is continuous. Expressthe answer in interval notation.

54.

55.

56.

57.

58.

In Problems 59–68, evaluate the indicated limits if they exist.

59. Let Find

(A) (B) (C)

60. Let Find

(A) (B) (C)

61. Let Find

(A) (B) (C)

62. Let Find

(A) (B) (C)

63. Let Find

(A) (B) (C)

64. Let . Find

(A)

(B)

(C)

65. Let . Find

(A)

(B)

(C) limx:2

f(x)

limx: - q

f(x)

limx: q

f(x)

f(x) =

2x3

3(x - 2)2

limx:2

f(x)

limx: - q

f(x)

limx: q

f(x)

f(x) =

2x

3x - 6

limx:5

f1x2limx:2

f1x2limx: -1

f1x2f1x2 =

x2- x - 2

x2- 7x + 10

.

limx:0

f1x2limx: -3

f1x2limx:3

f1x2f1x2 =

x - 3

9 - x2.

limx:4

f1x2limx:4+

f1x2limx:4-

f1x2f1x2 =

ƒ x - 4 ƒ

x - 4.

limx:3

f1x2limx: -1

f1x2limx:1

f1x2f1x2 =

x + 1

13 - x22.

limx:0

f1x2limx:3

f1x2limx:1

f1x2f1x2 =

2x

x2- 3x

.

f1x2 = 24 - x2

f1x2 =

324 - x2

f1x2 =

x + 4

x2+ 3x - 4

f1x2 =

x + 1x - 2

f1x2 = x2- 4

318.1,

318.1,

∆x = dx = 0.1,f(x) = x1/3, x = 8

∆x = dx = 0.2,y = 12 1x, x = 3,∆ydy


Applications 213

79.

80.

81. The domain of the power function is the set of all real numbers. Find the domain of the derivative

Discuss the nature of the graph of for anyx values excluded from the domain of

82. Let f be defined by




(C) Find m so that


(D) Write a brief verbal description of each graph. How does the graph in part (C) differ from the graphsin parts (A) and (B)?

83. Let (see the figure).f1x2 = 1 - ƒ x - 1 ƒ , 0 … x … 2

limx:1-

f1x2 = limx:1+

f1x2

limx:1-

f1x2 and limx:1+

f1x2m = 2,

limx:1-

f1x2 and limx:1+

f1x2m = 0,

f1x2 = e x2- m if x … 1

-x2+ m if x 7 1

f¿1x2.y = f1x2f¿1x2.

f1x2 = x1/5

f(x) =

x3- 1

x3- x2

- x + 1

f(x) =

x2- 9

x2+ x - 2

Monthly customer charge $7.47

First 90 therms $0.4000 per therm

All usage over 90 therms $0.2076 per therm

Natural Gas Rates

66. Let . Find

(A)

(B)

(C)

67.

68.

69. Let

Graph f and use zoom and trace to investigate the left- andright-hand limits at the indicated values of c.

(A) (B) (C)

In Problems 70 and 71, use the definition of the derivative andthe four-step process to find

70. 71.

C Problems 72–75 refer to the function f in the figure. Determinewhether f is differentiable at the indicated value of x.

f1x2 = 2x - 3f1x2 = x2- x

f¿1x2.

c = 2c = 0c = -2

f1x2 =

x3- 4x2

- 4x + 16

ƒ x2- 4 ƒ

limh:0

f1x + h2 - f1x2h

for f1x2 =

1x + 2

limh:0

f12 + h2 - f122h

for f1x2 = x2+ 4

limx:2

f(x)

limx: - q

f(x)

limx: q

f(x)

f(x) =

2x

3(x - 2)3

f (x)

x

4

�1�2 5

f (x)

x1 2

1

Figure for 8372. 73. 74. 75.

In Problems 76–80, find all horizontal and vertical asymptotes.

76.

77.

78. f(x) =

x2+ 9

x - 3

f(x) =

-2x + 5

(x - 4)2

f(x) =

5x

x - 7

x = 3x = 2x = 1x = 0

(A)

(B)

(C)

(D) Does exist?f ¿112limh:0

f11 + h2 - f112h

= ?

limh:0+

f11 + h2 - f112h

= ?

limh:0-

f11 + h2 - f112h

= ?

84. Natural-gas rates. The accompanying table shows the win-ter rates for natural gas charged by the Bay State GasCompany. The customer charge is a fixed monthly charge,independent of the amount of gas used during the month.

(A) Write a piecewise definition of the monthly chargeS(x) for a customer who uses x therms in a wintermonth.

(B) Graph S(x).

(C) Is S(x) continuous at Explain.x = 90?

APPLICATIONS


85. Cost analysis. The total cost (in dollars) of producing xtelevision sets is

(A) Find the exact cost of producing the 101st televisionset.

(B) Use the marginal cost to approximate the cost ofproducing the 101st television set.

86. Cost analysis. The total cost (in dollars) of producing xbicycles is

(A) Find the total cost and the marginal cost at aproduction level of 100 bicycles, and interpret theresults.

(B) Find the average cost and the marginal average cost ata production level of 100 bicycles, and interpret theresults.

87. Cost analysis. The total cost (in dollars) of producing x laserprinters per week is shown in the accompanying figure.Which is greater, the approximate cost of producing the201st printer or the approximate cost of producing the 601stprinter? Does this graph represent a manufacturing processthat is becoming more efficient or less efficient as produc-tion levels increase? Explain.

C1x2 = 5,000 + 40x + 0.05x2

C1x2 = 10,000 + 200x - 0.1x2

employee can assemble N(t) components per day after tdays of on-the-job training, as given by

(A) Find the average rate of change of N(t) from 2 daysto 5 days.

(B) Find the instantaneous rate of change of N(t) at 2 days.

90. Sales analysis. Past sales records for a swimming-pool man-ufacturer indicate that the total number of swimming pools,N (in thousands), sold during a year is given by

where t is the number of months since the beginning ofthe year. Find N(9) and and interpret thesequantities.

91. Natural-gas consumption. The data in Table 1 give the U.S.consumption of natural gas in trillions of cubic feet.

N¿192,

N(t) = 2t +

13

t3/2

N1t2 =

40t - 80t

, t Ú 2


x200 400 600 800 1,000

$20,000

$40,000

$60,000

$80,000

$100,000

C(x)

Figure for 87

TABLE 1

88. Cost analysis. Let

be the price–demand equation and the cost function,respectively, for the manufacture of umbrellas.

(A) Find the marginal cost, average cost, and marginalaverage cost functions.

(B) Express the revenue in terms of x, and find the mar-ginal revenue, average revenue, and marginal aver-age revenue functions.

(C) Find the profit, marginal profit, average profit, andmarginal average profit functions.

(D) Find the break-even point(s).(E) Evaluate the marginal profit at and

1,400, and interpret the results.

(F) Graph and on the same coordi-nate system, and locate regions of profit and loss.

89. Employee training. A company producing computercomponents has established that, on the average, a new

C = C1x2R = R1x2x = 1,000, 1,150,

0 … x … 2,500p = 25 - 0.01x and C1x2 = 2x + 9,000

(A) Let x represent time (in years), with corre-sponding to 1960, and let y represent the corre-sponding U.S. consumption of natural gas. Enter thedata set in a graphing calculator and find a cubicregression equation for the data.

(B) If denotes the regression equation found inpart (A), find N(50) and and write a brief ver-bal interpretation of these results.

92. Break-even analysis. Table 2 contains price–demand andtotal cost data from a bakery for the production of kringles(a Danish pastry), where p is the price (in dollars) of akringle for a daily demand of x kringles and C is the totalcost (in dollars) of producing x kringles.

N¿1502,y = N1x2

x = 0

Natural-Gas Year Consumption

1960 12.0

1970 21.1

1980 19.9

1990 18.7

2000 21.9

TABLE 2

x p($) C($)

125 9 740

140 8 785

170 7 850

200 6 900

(A) Find a linear regression equation for the price–demand data, using x as the independent variable.

(B) Find a linear regression equation for the cost data,using x as the independent variable. Use this equa-tion to estimate the fixed costs and the variable costsper kringle.


Applications 215

(C) Find the break-even points.

(D) Find the price range for which the bakery will make a profit.

In all answers, round dollar amounts to the nearest cent.

93. Pollution. A sewage treatment plant disposes of its effluentthrough a pipeline that extends 1 mile toward the center ofa large lake.The concentration of effluent C(x) in parts permillion, x meters from the end of the pipe is givenapproximately by

What is the instantaneous rate of change of concentrationat 10 meters? At 100 meters?

94. Medicine. The body temperature (in degrees Fahrenheit) ofa patient t hours after being given a fever-reducing drug isgiven by

Find F(4) and Write a brief verbal interpretationof these quantities.

95. Learning. If a person learns N items in t hours, as given by

find the rate of learning after


96. Physics:Thermal expansion. The coefficient of thermal ex-pansion (CTE) is a measure of the expansion of an objectsubjected to extreme temperatures. We want to use aMichaelis–Menten function of the form

C(T) =

CmaxT

M + T

N1t2 = 202t

F¿142.F(t) = 0.16 t2

- 1.6 t + 102

C1x2 =

500

x2 , x Ú 1

(A) Plot the points in columns 1 and 2 of Table 3 on graphpaper and estimate to the nearest integer. To

estimate M, add the horizontal line to

your graph, connect successive points on the graphwith straight-line segments, and estimate the value of T (to the nearest multiple of fifty) that satisfies

(B) Use the constants and M from part A to form a

Michaelis–Mentenfunctionfor theCTE of germanium.

(C) Use the function from part B to estimate the CTE ofgermanium at 600 K and to estimate the temperaturewhen the CTE of germanium is 10.

Cmax

2

C(T) =

Cmax

2.

CTE =

Cmax

2

Cmax

T (K) Titanium

100 4.5

200 7.4

293 8.6

500 9.9

800 11.1

1100 11.7

TABLE 3 Coefficients ofThermal Expansion

where C � CTE, T is temperature in K (degrees Kelvin),and and M are constants. Table 3 lists the coeffi-cients of thermal expansion for titanium at varioustemperatures.

Cmax


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