mth-105-02 algebra and trigonometry spring 2011 …

MTH-105-02

ALGEBRA and TRIGONOMETRY

SPRING 2011

SCIENCE 318

MTWF 11:30 – 12:30

CONTENTS:

1. Syllabus

2. Reviews for tests 1 – 7

3. Review for the final exam

SYLLABUS

Math 105: Algebra and Trigonometry

Instructor: Dr. Arkady Kitover

Office: Science Hall 343

Email: (the best way to contact me) [email protected]

Alternative emails [email protected] or [email protected]

Web page: http://faculty.ccp.edu/FACULTY/akitover (syllabus and all the reviews with solutions are available on the web page)

Telephone: (215) 751-8723

Office Hours: Tuesday, Friday, 10:20 – 11:20, or by appointment.

(Help is also available in the Math Skills Lab, located in Room 23 of the Annex.)

Textbook: Fundamentals of Algebra and Trigonometry, (9th Edition) by Swokowski and Cole.

Technology: you will need a scientific calculator for this course.

Class meets: Monday, Tuesday, Wednesday, and Friday, 11:30 - 12:30.

COURSE OUTLINE

Part 1 Linear and quadratic functions. Sections 3.3 and 3.6

(a) Linear functions y ax b . The meaning of the coefficients a(the slope) and b (the y -intercept). Graphing linear functions.

(b) Compositions and inverses of linear functions. Symmetry between the graphs of a linear function and its inverse.

(c) Quadratic functions. The standard form. The vertex. The range. The quadratic formula.

(d) Graphs of quadratic functions

(e) The inverse of a quadratic function with restricted domain. Graphs of inverse functions.

(f) Applications of quadratic functions.

Review 1

Test 1 (50 points)

Part 2 Equations. Sections 2.3 – 2.5, 4.2 – 4.4

(a) Complex numbers

(b) The main theorem of algebra. The synthetic division. The factor and remainder theorems.

(c) Quadratic equations and equations reducible to them.

(d) Rational roots test.

(e) Approximating solutions of polynomial equations by the method of halving intervals.

Review 2

Test 2 (50 points)

Part 3 General relations and functions. Sections 3.4, 3.7, 3.8

(a) Relations. The domain and the range of a relation. Inverse relations.

(b) When a relation is a function. Vertical line test.

(c) Compositions of functions. Inverse functions.

(d) Radical functions n x as inverses to power functionsnx .

Review 3

Test 3 (50 points)

Part 4 Polynomial and rational functions. Sections 4.1 and 4.5

(a) Graphs of completely factored polynomials.

(b) Polynomials of quadratic type2( ) n nP x ax bx c , their

range, critical points, and graphs.

(c) Linear fractions ( )ax b

R xcx d

, horizontal and vertical

asymptotes, graphs. The inverse of a linear fraction.

(d) Rational functions of the form

2

( )ax bx c

R xdx e

, slant

asymptotes, range, critical points, and graphs.

(e) Completely factored rational functions

Review 4

Test 4 (50 points)

Part 5 Exponential and logarithmic functions. Chapter 5

(a) Exponential functions and their graphs. Numbere . Logistic curve. Hyperbolic functions.

(b) Logarithmic functions. Properties of logarithms. Graphs of logarithmic functions.

(c) Exponential and logarithmic equations. Inverses of logistic and hyperbolic functions.

(d) Exponential models of growth and decay.

Review 5

Test 5 (50 points)

Part 6 Basic trigonometric functions and their properties. Sections 6.1 – 6.7 and 7.6

(a) Angles. The degree measure. The radian measure. Complementary and supplementary angles.

(b) The definition of six basic trigonometric functions; their graphs.

(c) Graphs of functions sin( )a bx c and cos( )a bx c . Amplitude, period, and shift.

(d) Graphs of functions tan( ), cot( ), sec( )a bx c a bx c a bx c and csc( )a bx c .

(e) Inverse trigonometric functions and their graphs.

Review 6

Test 6 (50 points)

Part 7 Analytic Trigonometry Sections 7.1 – 7.5.

(a) Pythagorean identities; complementary angle identities; even and odd trigonometric functions.

(b) Verifying trigonometric identities.

(c) Addition and subtraction formulas, multiple angle formulas, product-to-sum and sum-to-product formulas.

(d) Basic trigonometric equationssin ,cosx a x a , and

tan x a ; general solutions; solutions in the interval[0,2 ) .

(e) Solving trigonometric equations.

Review 7

Test 7 (50 points)

Review for the final exam

Cumulative final exam (100 points)

Grading

I will drop the lowest of the seven one-hour test grades. If you miss a test without a valid excuse I will consider it as the lowest grade test.

Grade for the final will not be dropped.

The homework will consist of seven parts. Each part will be due on the day of the corresponding test. The first six parts will be 14 points each and the last part will be 16 points.

The total weight of the homework will be 100 points.

A: 475 – 500 points

A-: 450 – 474 points

B+: 425 – 449 points

B: 400 – 424 points

B-: 375 – 399 points

C+: 350 – 374 points

C: 325 – 349 points

C-: 300 – 324 points

D: 275 – 299 points

F: 0 – 274 points

Attendance:

I will take attendance every day, since I view attendance as important. For every 3 classes missed for an unexcused absence, I will decrease your course grade by 1/3 of a letter grade.

Class Rules:

(a) No food in the class room.

(b) Cell phones must be put in vibration mode.

(c) Use of cell phones and laptops is not allowed during a test.

(d) Lateness is strictly discouraged. If you are late 10 minutes or more I will consider it as an absence.

(e) Please refrain from going out of the class room during the class time. It distracts other students. If it is necessary for you to leave early let me know before the class starts.

HOMEWORK

Part Page Problems

1 159 62

198 – 199 11, 21, 39, 40

2 92 8, 18, 24

99 35, 43

252 10, 14, 27

272 16, 20

3 172 – 173 25, 29, 39, 45, 51

211 – 212 23, 27, 31,59

220 – 221 7, 15, 25

4 244 11, 19, 21, 23, 25

287 6, 15, 25,27

5 335 – 336 5, 11, 15, 19

346 5, 7, 25

360 13, 19, 35

369 – 370 3, 11, 21, 27

379 – 380 13, 19, 23, 29, 33, 51, 55, 57

6 395 – 397 3, 5, 7, 11, 15, 37, 41

409 – 410 7, 17, 29, 37, 51, 59

431 – 432 13, 27, 33

439 11, 15, 17

452 – 453 7, 19, 49

462 – 463 27, 37, 47

541 – 542 5, 9, 15, 21, 25, 29

7 489 - 490 7, 17, 33, 35

500 – 501 13, 17, 25, 47, 55, 65

511 – 513 5, 9, 19, 37, 43, 55, 61

520 – 521 5, 19, 23, 35, 39

528 5, 11, 19, 23, 29, 31, 33

105. Algebra and Trigonometry Review 1

Problem 1 Let1 1

( )3 2

y x x . Find the inverse function 1( )y x and graph

both functions in the same coordinate system.

Solution First we solve the equation 1 1

3 2y x for x . Multiplying both parts

of it by 3 we get 3

32

y x whence3

32

x y . Interchanging the input x

and the output y in the last formula we get

1 3( ) 3

2y x x .

The graphs of the functions ( )y x and 1( )y x are shown below.

In problems 2 – 4 consider the following quadratic function

23 7 1y x x .

Problem 2 Write the function in the standard form

2( )v vy a x x y.

Find the coordinates of the vertex. Find the x -intercepts of the graph of y (if

any). Graph the function.

Solution We first factor out the coefficient by2x ,

2 7 1

33 3

y x x .

Next we complete the square inside the parentheses using the formula2 2

2

2 2

b bx bx x .

In our case we have

2

2 1 7 49

3 6 36x x x . After we plug in this

expression into the formula for y we get

27 61

36 12

y x

This is the standard form of the function23 7 1y x x .

From the standard form we see that the coordinates of the vertex are

7 61

,6 12

v vx y .

The x -intercepts can be found by quadratic formula

22 7 7 4 ( 3) 14 7 61

2 2 ( 3) 6

b b acx

a

The x -intercepts are located approximately at -0.13 and 2.47.

The graph of the function is shown below.

Problem 3 Consider function y over the restricted domain[ , )vx . Find the

inverse function for this restriction of y . State the domain and the range of the

inverse function.

Solution The restricted domain of y is the interval7

,6

. To find an

expression for the inverse function we have to solve the equation 23 7 1y x x for x . Writing it as

23 7 1 0x x y and applying

the quadratic formula we get

7 49 4 3 ( 1) 7 61 12

6 6

y yx

Because according to our choice of the restricted domain 7

6x the correct sign

in the above formula is +. Interchanging the input x and the output y we get

the inverse function in the form

7 61 12

6

xy

Notice now that that the original function has the domain 7

,6

and the

range61

,12

. Because for the inverse function we have to interchange the

domain and the range we see that the domain of the inverse function is

61,12

and its range is7

,6

.

Problem 4. Graph the function y with the restricted domain and its inverse in

the same coordinate system.

Solution Notice that the graph of y is the part of the graph from problem 2to

the right of the vertex. The graph of the inverse function can be obtained by

interchanging the x and y coordinates, in other words by reflecting the graph of

y about the line y x . The graphs are shown below.

Problem 5 A penny is thrown up from the observation deck of the Empire State

building (1050ft) with the velocity 50 ft/sec. Answer the following questions

(a) At what moment will the penny reach its maximum height?

(b) What is the maximum height?

(c) At what moment will the penny hit the ground?

Solution. In a problem like this the height of the object aftert seconds is given

by the formula

2

0 0( ) 16h t t v t h

where 0v is the initial velocity of the object in ft/sec and 0h is its initial height. In

our case 0 50v and 0 1050h whence

2( ) 16 50 1050h t t t

Because the leading coefficient of the quadratic function h is negative the

function takes it greatest value at the vertex.

(a) The t coordinate of the vertex is given by the formula

50

1.56252 2 ( 16)

v

bt s

a

This answers question (a).

(b) The h coordinate of the vertex is given by

2 2501050 1089.0625

4 4( 16)v

bh c ft

a

This answers question (b).

To answer question (c) we have to find the positive solution of the equation

2( ) 16 50 1050h t t t

216 50 1050 0t t

By the quadratic formula

250 50 4( 16)10509.812

2( 16)t s

105 Algebra and Trigonometry Review 2

Problem 1 Perform computations and present the result as a complex number in

the standard forma bi .

(3 2 )(4 3 )

(6 5 )(5 6 )

i i

i i

Solution

2

2

(3 2 )(4 3 ) 12 9 8 6

(6 5 )(5 6 ) 30 36 25 30

i i i i i

i i i i i.

Recalling that 2 1i we see that the above expression is equal to

18

60 11

i

i.

Finally we perform the division

2 2

18 (18 )(60 11 ) 1091 138 1091 138

60 11 (60 11 )(60 11 ) 60 11 3721 3721

i i i ii

i i i.

Problem 2 Find all the solutions of the equation6 3 6 0x x .

Solution The left part of the equation is a trinomial of quadratic type and it can

be factored as

3 3( 3)( 2)x x

Therefore we have to solve two equations of degree 3 3 3x and

3 2x .

Recall that all the solutions of the equation 3x a where a is a real number

can be obtained by multiplying all the solutions of the equation 3 1x by

3 a .

As we discussed in class the solutions of the last equation are

1 3 1 3

1, ,2 2

i i.

Therefore all six solutions of our original equation can be obtained as

3 3 3

1 2 3

3 3 3

4 5 6

1 3 1 33, 3 , 3 ,

2 2

1 3 1 32, 2 , 2 .

2 2

i ix x x

i ix x x

Problem 3 Solve the radical equation 1 2x x x .

Solution. Squaring both parts of the equation we get

2 ( 1) 1 2x x x x x .

Or equivalently

2 ( 1) 1x x x .

Squaring both parts again we obtain

24 ( 1) 2 1x x x x ,

which is equivalent to the quadratic equation

23 6 1 0x x

By quadratic formula

26 6 4 3 ( 1) 6 48 2 31

2 3 6 3x . Clearly, the

negative value of x is not a solution of the original equation, but the positive

value is (e.g. we can use our calculators to see that 1 (2 / 3) 3 0.1547

and that 0.1547 1.1547 2.1547 ). Finally

2 3

13

x .

Problem 4 Using your calculator and the method of halving the interval locate

the positive solution of the equation 3 2 1 0x x with accuracy 0.01.

Solution Let3 2( ) 1f x x x . Then (0) 1 0f and (1) 1 0f .

Therefore the solution is in the interval[0,1]. The successive steps are shown in

the table below.

Interval Midpoint Sign of the value of f at

the midpoint

[0, 1] 1/2 -

[1/2, 1] 3/4 - [3/4, 1] 7/8 +

[3/4, 7/8] 13/16 + [3/4, 13/16] 25/32 +

[3/4, 25/32] 49/64 + [3/4, 49/64] 97/128 +

[3/4, 97/128]

The length of the last interval is1/128 0.01. We have located the positive

solution of our equation in the interval

3 4,97 128 [0.75,0.7578125]

Problem 5 Using the rational roots test, the factor theorem, and the synthetic

division find all the solutions of the polynomial equation and factor the

polynomial completely.

4 3 26 5 11 6 0x x x x

Solution. By the rational roots test if there are rational solutions of the equation

above then their numerators can be only the numbers 1, 2, 3, 6 , and

their denominators can be only from the list1, 2, 3, 4, 6 . Therefore the list of

all possible rational roots is

1, 2, 3, 6, 1 2, 3 2, 1 3, 2 3, 1 4, 3 4, 1 6.

Plugging in the numbers from the list into the original equation (either by using

the synthetic division or a calculator) we successively eliminate the numbers

1, 2, 3, 6, 1 2, 3 2. But 3 2 is a solution of the original

equation. Indeed the synthetic division goes as follows.

-3/2 6 -1 -5 11 -6

-9 15 -15 6

6 -10 10 -4 0

Whence 3 2 is indeed a root of the original equation and

4 3 2

3 2 3 2

6 5 11 6

( 3 2)(6 10 10 4) (2 3)(3 5 5 2)

x x x x

x x x x x x x x

We have reduced the problem to solving the cubic equation

3 23 5 5 2 0x x x .

The list of possible rational roots for this equation is much shorter:

1 3, 2 3. We easily eliminate 1 3but when we come to 2 3we hit

another root.

2/3 3 -5 5 -2

2 -2 2 3 -3 3 0

Therefore

3 2 2

2

3 5 5 2 ( 2 3)(3 3 3)

(3 2)( 1).

x x x x x x

x x x

It remains to solve the quadratic equation2 1 0x x . By the quadratic

formula

2( 1) ( 1) 4 1 1 1 3

2 1 2

ix . Finally, the complete list

of solutions of the original equation is

3 2 1 3 1 3

, , ,2 3 2 2

i i

The polynomial can be completely factored as

4 3 2 1 3 1 36 5 11 6 (3 2)(2 3)

2 2

i ix x x x x x x x


Problem 1

(a) A relation is given as the set of pairs

{(1,0),(2, 1),(3,1),(1,2)}R

Find the domain and the range of this relation, graph the relation, and decide

whether this relation is a function.

Solution. The domain is the set of inputs{1,2,3}. The range is the set of

outputs { 1,0,1,2}. The graph of the relation is shown below.

The relation is not a function because to the same input 1 correspond two

different outputs 0 and 2. Also we see that the graph does not pass the vertical

line test.

(b) Describe the relation inverse to the relation from Problem 1. Is this

relation a function? Graph it.

Solution We get the inverse relation if we interchange the input and the output

in each pair.

1 {(0,1),( 1,2),(1,3),(2,1)}R .

This relation is a function: to each input corresponds only one output.

The graph below passes the vertical line test.

Problem 2 Consider the following two functions

2( ) , ( ) 2f x x g x x x

Describe the composition f g . Find the domain and the range of this

composition. Graph the composition.

Solution 2( ( )) ( ) 2f g x g x x x . To find the domain of this

function notice that the expression under the square root cannot be negative;

thus we have to solve the inequality2 2 0x x . To solve it we factor the

left part and write the equation as( 1)( 2) 0x x . The left part will be

nonnegative either on the interval ( , 2] or on the interval[1, ) .

Therefore the domain is( , 2] [1, ). On this domain the quadratic

function g takes all nonnegative values whence the composition f g g

also takes all nonnegative values and its range is[0, ) .

Problem 3 Consider the composition g f where g and f are functions from

Problem 2. Describe the composition; find its domain and its range; graph the

composition function.

Solution 2( ( ) ( ) 2 2g f x x x x x . Clearly this expression

is defined if and only if 0x and therefore the domain of the composition is

[0, ). The question about the range can be reduced to the following: what

values does the quadratic trinomial 2 1u u take if 0u ? To answer this

question notice that the vertex of this quadratic trinomial is at

1

2 2v

bu

a. To the right of the vertex the quadratic function is

increasing because 0a and therefore the range of the composition is

[ ( (0), ) [ 2, )g f .

The graph of the composition is shown below

Problem 4 Consider the function3( ) 1f x x . Find its domain and its

range. Prove that the function f is one-to-one and find its inverse1f .

Solution The function f is defined if3 1 0x . Factoring the left part of this

inequality we write it as2( 1)( 1) 0x x x . Because

2

2 1 31 0

2 4x x x the last inequality is equivalent to 1 0x

or 1x . Thus the domain of f is[ 1, ) . The range of f is clearly[0, ) .

To prove that f is one-to-one and to find its inverse let3 1y x . Notice

that 0y . By squaring both parts we obtain2 3 1y x , and

23 1x y .

Thus, if the output y is given the input x is defined in the unique way which

proves that f is one-to-one. Finally we get an expression for 1f by

interchanging the input and the output.

31 2( ) 1, 0f x x x .

Problem 5 Graph the functions f and1f from Problem 4 in the same

coordinate system.

Solution


Problem 1 Graph the polynomial function2( ) ( 2) ( 1)P x x x x .

Solution The polynomial is of degree 4 and therefore it is positive to the left of

its smallest real root and to the right of its largest real root. The roots of the

polynomial are 2,0, and 1. The sign of the polynomial changes at 2and at

1, because they are simple roots, and does not change at0 because 0 is a root

of multiplicity two. The sign of the polynomial is shown in the table below

Interval ( , 2) ( 2,0) (0,1) (1, )

Sign + - - +

The graph of the polynomial is shown below.

Problem 2 Consider the polynomial4 2( ) 3 1P x x x .

(a) Find the x -intercepts.

(b) Find the critical points and the range.

(c) Graph the polynomial

Solution (a) to find the x -intercepts we have to find the real solutions of the

equation4 23 1 0x x . This equation is of quadratic type and applying the

quadratic formula we get

2

2 ( 3) ( 3) 4 1 ( 1) 3 13

2 2x

a

Sign minus does not provide any real solutions and taking sign plus we obtain

3 131.82

2x

(b) To find the range and the position of critical points we have to solve the

equation 4 23 1y x x for x . Applying again the quadratic formula we see

that

2 3 9 4( 1)

2

yx

The expression under square root cannot be negative whence4 13 0y .

Therefore 13

4y and the range of P is[ 13 / 4, ) .

There are two critical points corresponding to the value 13 4y . For this

value of y we obtain2 3

2x and 3 2 1.22x . Because polynomial

P is an even function there is one more critical point at(0, 1) . The list of

critical points is

( 3 2, 13 4),(0, 1),( 3 2, 13 4)

(c) The graph of P is shown below.

Problem 3 Consider the linear fraction2 3

( )3 4

xR x

x.

(a) Find the x and y -intercepts

(b) Find equations of the horizontal and the vertical asymptote

(c) Graph the function together with its horizontal and vertical asymptotes

(d) Find an equation of the inverse function and graph it together with its

horizontal and vertical asymptotes

Solution (a) to find the x -intercept we solve the equation 0y which is

equivalent to 2 3 0x whence 3 2x and the x intercept is

( 3 2,0) . To find the y -intercept we just notice that if 0x then

3 4y and the y -intercept is(0, 3 4) .

(b) Looking at the ratio of leading terms 2 2

3 3

x

xwe see that an equation of the

horizontal asymptote is2

3y .

The function is undefined if 4 3x whence an equation of the vertical

asymptote is 4 3x .

(c) The graph is shown below

(d) Solving the equation 2 3

3 4

xy

xfor x we get 3 4 2 3xy y x

whence (3 2) 4 3y x y and4 3

3 2

yx

y. The inverse function is

1 4 3( )

3 2

xR x

x

The x and y -intercepts are, respectively, ( 3 4,0)and(0, 3 2) . The

horizontal asymptote is 4 3y , the vertical asymptote is 2 3x . The

graph is shown below.

Problem 4 Consider the rational function

2 1( )

1

x xR x

x.

(a) Find the x and the y -intercepts, if any

(b) Find an equation of the vertical and the slant asymptote

(c) Find the critical points, if any, and the range of the function

(d) Graph the function together with its vertical and slant asymptotes

Solution (a) the equation 2 1 0x x has no real solutions whence there are

no x -intercepts. The y -intercept i (0,1) .

(b) The vertical asymptote is clearly 1x . To find an equation of the slant

asymptote we will divide 2 1x x by 1x using e.g. the synthetic division

-1 1 1 1

-1 0 1 0 1

The quotient is x whence an equation of the slant asymptote is

y x

(c)To find the range and the critical points (if they exist) we have to solve the

equation

2 1

1

x xy

xfor x . This equation is equivalent to the following

quadratic equation2 (1 ) (1 ) 0x y x y . The quadratic formula

provides

21 ( 1) 4( 1) 1 ( 1)( 3)

2 2

y y y y y yx

The expression is defined if and only if ( 1)( 3) 0y y which happens

when either 1y or 3y . The range of R is therefore

( , 3] [1, )

To find the critical points we plug in into the expression for x the values

1y and 3y getting 0x and 2x , respectively. The critical points

are located at ( 2, 3)and at(0,1) .

(d) The graph of R and its asymptotes is shown below

Problem 5 Consider the function

2

2

4

1

xR

x

(a) Find the x -intercepts

(b) Find the vertical asymptotes

(c) Find the horizontal or slant asymptote, if any

(d) Find the sign of the function

(e) Graph the function and its asymptotes

Solution (a) the x -intercepts are ( 2,0)and(2,0)

(b) the vertical asymptotes are 1x and 1x

(c) the degree of the numerator equals the degree of the denominator; the ratio

of the leading terms is 1; therefore there is the horizontal asymptote with the

equation 1y

(d) we see from (c) that the sign of R is positive far right and far left. Because

( 2)( 2)( )

( 1)( 1)

x xR x

x xevery root of the numerator and of the denominator

is simple and the sign of the function changes at points 2, 1,1 and 2 .

Interval ( , 2) ( 2, 1) ( 1,1) (1,2) (2, )

Sign of R + - + - +

(f) The graph is shown below


Problem 1 Consider logistic function5

2 3 2 xy .

(a) Describe the domain and the range of y .

(b) Prove that y is one-to-one and find its inverse.

(c) Graph y and its inverse in the same coordinate system.

Solution (a) y is defined for any value of x and therefore its domain is( , ) .

To see what is its range notice that0 3 2 x, whence

2 2 3 2 x. Recalling that if a b then 1 1a bwe get

5 5

02 3 2 2x

Moreover, because the function 3 2 xtakes all values in the interval(0, ) ,

the range of y is exactly the interval(0,5 2) .

(b) Solving the equation5

2 3 2 xy for x we obtain

2 3 2 5,

3 2 5 2 ,

5 22 ,

3

x

x

x

y y

y y

y

y

2

2 2

5 2log ,

3

5 2 3log log .

3 5 2

yx

y

y yx

y y

Therefore the input x is defined in the unique way as soon as we know the

output y , and the inverse function is1

2

3( ) log

5 2

xy x

x.

(c) The graphs of y and1y are shown below.

Problem 2 solve the exponential equation 23 5 2 5 2 0x x

and

approximate its solutions with the accuracy0.0001.

Solution. It is an equation of quadratic type2

3 5 2 5 2 0x x.

Applying the quadratic formula we see that

22 2 4 3( 2) 2 28 1 75

2 6 3

x

a.

Because 5xcannot be negative the only possibility is

1 75

3

xwhence

5

1 7log

1 7 3log 0.37313 log5

x .

Problem 3 solve the logarithmic equation ln(3 2) ln(2 3) 5x x and

approximate its solutions with the accuracy0.0001.

Solution By the first law of logarithms the equation can be written as

2ln(6 13 6) 5x x , or2 56 13 6x x e . Applying the quadratic

formula we get

2 513 13 4 6 (6 )

12

ex . It is easy to see that we

cannot take sign minus in the above expression because then the expressions

3 2x and2 3x will be negative and their logarithms – undefined. Therefore

the equation has one solution

513 169 24( 6)3.9076

12

ex

Problem 4 an artifact is discovered at a certain site. If it has 52% of carbon-14 it

originally contained, what is the approximate age of the artifact to the nearest

year? (Carbon-14 decays at the rate 0.0125% annually.)

Solution According to the model of radioactive decay the amount of carbon-14

after t years is given by the formula 0( ) ktA t A e , where 0A is the initial

amount of carbon-14. We can find the coefficient k from the conditions of the

problem in the following way

0 0

0 0

0.000125(1).999875,

ln(.999875) .000125.

k A AAe

A A

k

Now we can reduce the problem to solving fort the equation

0.000125 0.52,te

Whence

.000125 ln(0.52),

ln(0.52)5231

.000125

t

t

Problem 5 A lake is stocked with 438 fish of a new variety. The size of the lake, the availability of food, and the number of other fish restrict growth in the lake to a limiting value of 2737. The population of fish in the lake after timet , in months, is given by the function

0.35

2737P t

1 6.15 te

After how many months will the population be 1323?

Solution all we have to do is to solve the equation0.35

27371321

1 6.15 te.

The solution goes as follows

0.35

0.35

2737 1321 8124.15 ,

2737 13210.1743,

8124.15

0.35 ln(0.1743),

ln(0.1743)5

0.35

t

t

e

e

t

t


Problem 1. Consider the function2

2cos3 3

y x

.

(a) Find the amplitude, the period, and the shift.

(b) Graph the function.

Solution. (a) We identify the coefficients in this expression as

2

2, ,3 3

a b c

.

Next we compute the amplitude| | 2a , the period2

3b

, and the shift

1

2

c

b . The graph of the function is shown below.

Problem 2 (a) Find the standard restricted domain over which the function y

from Problem 1 is one-to-one.

(b) Find the inverse function to this restriction of y .

(c) Graph the function y with the restricted domain and the inverse function in


Solution (a) the standard interval where the function cos x is one-to-one is

[0, ] . Respectively for the function y we have to consider the interval

2

03 3

x

Whence2 2

3 3 3x

, and

1

12

x

(b) We consider the interval above and solve the equation

22cos

3 3y x

for x .

2cos ,

2 3 3

2arccos ,

3 3 2

3 1arccos

2 2 2

yx

yx

yx

Therefore the inverse function is defined as

1 3 1( ) arccos

2 2 2

xy x

.

(C) The graphs of y and1y

are shown below.

Problem 3 Consider the function1

tan3 6

y x

.

(a) Find the period and the shift.

(b) Graph the function.

Solution (a) the period is (1/ 3) 3 , the shift is 6 1 3 2 .

(b) The graph is shown below

Problem 4(a) Find the standard restricted domain over which the function y

from Problem 1 is one-to-one.

(b) Find the inverse function to this restriction of y .

(c) Graph the function y with the restricted domain and the inverse function in


(a) The standard interval where tan x is one-to-one is 2, 2 .

Respectively for y we have1

2 3 6 2x

, whence 2x .

(b) Over this restricted domain we have1

arctan3 6

x y

, whence

3arctan 2x y , and

1( ) 3arctan 2y x x .

(c) The graphs are shown below.

Problem 5 (a) Find the exact value (no calculators!) ofcsc(arctan( 7 5)) .

(b) Rewrite tan(arcsin( ))x as an equivalent algebraic expression.

Solution (a) First notices that because the functions cscandarctan are both

odd functions we have

csc(arctan( 7 5)) csc( arctan(7 5)) csc(arctan(7 5)) .

To find the value of csc(arctan(7 5))consider the right triangle with an acute

angle , opposite side7 , and adjacent side5 . Then

2 2csc(arctan( 7 5)) csc 7 5 7 7 74 , whence the answer

to part (a) is 7 74 .

(b) Consider the right triangle with an acute angle , opposite side x , and

hypotenuse1. Then

2

tan(arcsin( )) tan11

1

x x xx

xxx

.


Problem 1. Verify the identity

2

sin cos 1 2sin cos(?)

sin cos 2sin 1

x x x x

x x x

Solution. To verify that two fractions are identical it is enough to verify that the

cross-products are identical, i.e. that

3 2

2 2

2sin sin 2sin cos cos

sin cos 2sin cos 2sin cos (?)

x x x x x

x x x x x x

After canceling out identical terms in the left and the right parts we get

3 22sin sin sin 2sin cos (?)x x x x x

Dividing both parts by sin x provides

2 22sin 1 1 2cos (?)x x

Finally, moving 22cos x to the left and 1to the right we get

2 22sin 2cos 2x x ,

which is obviously equivalent to the first Pythagorean identity.

Problem 2. Verify the identity

2 csc cot

cot (?)2 csc cot

u u u

u u

Solution. We will work with the right part of the alleged identity

csc cot 1 sin cos sin

csc cot 1 sin cos sin

u u u u u

u u u u u

Multiplying both the numerator and the denominator of the right part by sinu

we get the expression1 cos

1 cos

u

u. Finally, recalling that by power reduction

formulas21 cos 2cos

2

uu and

21 cos 2sin2

uu we obtain the desired

identity.

In problems 3 – 5

(a) Describe all the solutions of the given trigonometric equation.

(b) List the solutions in the interval [0,2 )(if any) and approximate them with

the accuracy 0.0001.

Problem 3 2sec 5sec 1 0x x .

Solution. (a) We have here an equation of the quadratic type. Applying the

quadratic formula we obtain

5 21

sec2

x

The sign minus provides a positive value smaller than 1which is not in the range

ofsec x ; therefore we only have to consider the possibility

5 21sec

2x .

Or, equivalently

2 5 212cos

5 21 5 21 5 21

2 5 21 5 21.

4 2

x

All the solutions of the last equation can be described as

5 21

arccos 2 ,2

x n n Z

(b) There are two solutions in the interval[0,2 ) :

5 21

arccos 1.36052

x .

And

5 212 arccos 4.923

2x

Problem 42sin 3cos 1x x .

Solution (a) We transform the left part of the equation according to the formula

2 2sin cos sin arctanA x B x A B x B A .

The equation now can be written as

13sin( arctan(3 / 2)) 1x

Or

sin( arctan(3 / 2)) 1 13 13 13x .

From here

arctan(3 / 2) 1 arcsin 13 13 ,n

x n n Z

And finally

1 arcsin 13 13 arctan(3 / 2) ,n

x n n Z

(b) We have two solutions in the interval [0,2 )corresponding to the values

0n and 1n .

arcsin 13 13 arctan(3 2) 1.2638x

And

arcsin 13 13 arctan(3 2) 3.8434x .

Problem 5 cos3 cos7 2sin2x x x .

Solution (a) We will transform the left part according to the sum-to-product

formula

cos cos 2sin sin2 2

A B B AA B

Then the equation becomes

2sin4 sin2 2sin2x x x

Or

sin2 (2sin4 2) 0x x

If sin 2 0x then2 ,x n n Z , or

, (*)2

x n n Z

If sin 4 2 2x then

4 1 arcsin 2 2 1 4 ,n n

x n n n Z ,

Or

1 16 4 , (**)n

x n n Z

Formulas (*) and (**) together define all the solutions of the equation

(b) It is easy to see the solutions provided by (*) in the interval [0,2 )are

0, 2, ,3 2

And the solutions provided by (**) in the same interval are

16,3 16,9 16,11 16,17 16,19 16,25 16,27 16.

105 Algebra and Trigonometry Review for the Final Exam

Problem 1. Consider the quadratic function

2 3 1y x x

(a) Find the coordinates of the vertex. (2 points)

3 3.

2 2 ( 1) 2

2 9 131

4 4 4

bxv a

by cv a

(b) Find the x-intercepts (if any). (2 points)

22

1

2

3 3 4 ( 1) 14.

2 2 ( 1)

3 13.3,

2

3 133.3.

2

b b acx

a

x

x

(c) Find the range of the function. (2 points)

The coefficient a is negative; therefore the range is from the

vertex down: ( ,13 4]

(d) Graph the function. (4 points)

Problem 2. A rectangular plot of land on the edge of a river is to be

enclosed with fence on three sides. Find the dimensions of the

rectangular enclosure of the greatest area if the side that goes

along the river does not require fencing and the total length of the

fence is 200 m. (10 points)

Let x be the side along the river and y be the perpendicular side.

Then

2 200.x y

Whence

200 2 .x y

The area of the rectangle is

2(200 2 ) 2 200 .xy y y y y

The greatest value of this quadratic function will be at the vertex

200

50.2 ( 2)

y

Then 200 2 50 100.x

Problem 3. for the polynomial function

2 2( ) ( 4)P x x x

(a) find the x-intercepts and the sign of the function (3 points)

(b) find the critical points (5 points)

(c) graph the function (5 points)

Solution. (a)The x-intercepts are at 2 ,0 , and 2 . The function is

positive to the left of -2 (because the leading term is4x . It changes sign

at 2 and at2 (because they are simple roots), but does not change

sign at 0 (because it is a root of multiplicity2 ).

(b) Because the function is even there is one critical point at(0,0) . To

find if there are other critical points we will solve the equation 4 24 0x x y for x . The quadratic formula provides

2

4 16 42 4

2

yx y

At each of the remaining critical points we have 4y whence2 2x .

The list of critical points is

2, 4 , 0,0 , 2, 4

(c) The graph is shown below

Problem 4. for the function

( ) 1 2f x x

(a) Find the domain. (2 points)

The function is defined when 1 0x whence 1x . The domain is

the interval [1, )

(b) Find the range. (2 points)

Clearly ( ) 2f x and f takes any value in the interval[2, )

whence the range of f is[2, ) (c) Find the inverse function, its domain and range. (5 points)

We will solve the equation 1 2y x for x

2

2

2

2 1,

2 1,

4 4 1,

4 5.

y x

y x

y y x

x y y

The inverse function is

1 2( ) 4 5, [2, )y x x x x

The domain of the inverse function is the range of the original function;

in our case it is the interval[2, ) .

The range of 1( )y x is the domain of the original function, i.e. the

interval[1, ) .

(d) Graph the function and its inverse in the same coordinate system. (5

points)

Problem 5. for the rational function

2 1( )

1

x xf x

x

(a) Find equations of the horizontal and the slant asymptote.

(3 points)

The denominator is equal to 0 if 1x whence the horizontal asymptote

has equation 1x .

To find the slant asymptote we apply the synthetic division

-1 1 -1 -1

-1 2

1 -2 1

The quotient is 2x whence an equation of the slant asymptote is

2y x

(b) Find the x and y intercepts, if any, and the sign of the function

(3 points)

The y -intercept clearly is(0, 1) . We find the x -intercepts by solving

the quadratic equation2 1 0x x . The x -intercepts are

1 5 1 5

,0 , ,02 2

The sign of the function is shown in the table below

Interva

l ( , 1)

1 51,

2

1 5 1 5,

2 2

1 5,

2

Sign minus plus minus plus

(c) Find the critical points, if any, and the range of the function.

We proceed with solving the equation

2 1

1

x xy

x

for x .

2

2

2

1,

(1 ) (1 ) 0,

1 (1 ) 4(1 ),

2

1 ( 1)( 5)

2

yx y x x

x y x y

y y yx

y y yx

The equation ( 5)( 1) 0y y provides two critical points. The first

one is ( 2, 5)and the second( 1,0) . The range of the function is

( , 5] [ 1, )

(d) Graph the function together with its vertical and slant asymptotes.

Problem 6. Consider the following logistic function

10

2 3 xy

e

(a) Find the range of the function.(2points)

For very large negative values of x the expressionxe takes very large

positive values, whence y takes values close to0 . On the other hand, if

x is a large positive number, thenxe is close to 0 and the function takes

values close to5 . The range therefore is(0,5) .

(b) Find the inverse function. (3points)

2 3 10,

10 2,

3

10 2ln ,

3

3ln .

10 2

x

x

y ye

ye

y

yx

y

yx

y

Therefore

1

3( ) ln , 0 5.

10 2

xy x x

x

(c) Graph the function together with its inverse

Problem 7. The half-life of radon222 Rn is 3.82 days. If the initial

amount of radon is 10g after what time in days it will remain 2g of

radon? Round to the nearest hundredth of a day. (5 points)

The equation of radioactive decay is

( ) (0) ktA t A e ,

where the constantk and the half-liveT are connected as ln 2kT . In

our caseln 2

0.18153.82

k . It remains to solve the equation

0.1815

0.1815

10 2,

0.2,

0.1815 ln 0.2,

ln 0.28.87

0.1815

t

t

e

e

t

t

Problem 8 Consider the function 3sin(2 2)y x .

(a) Find the amplitude, the period, and the shift (3 points)

The amplitude is3 .

The period is2

2.

The shift is 2 24

.

(b) Graph the function. (5 points)

(c) Find the standard domain where the function is one-to-one and find

the inverse function. (5points)

The domain is defined by the inequality 22 2 2

x whence

02

x

To find the inverse function we write

sin(2 2) 3,

2 2 arcsin( 3),

1arcsin .

2 3 4

x y

x y

yx

Whence

1

1( ) arcsin .

2 3 4

xy x

(d) Graph the function with restricted domain and the inverse function

in the same coordinate system (5 points)

Problem 9 verify the identity. (5 points)

2

2

1 sin1 cos2

3 cos1 sin

2

x

x

x x

We will work with the left part of the identity and apply the power

reduction formula

2

1 cossin

2 2

x x

Then the left part becomes

1 cos1

2 (1 cos ) 1 cos21 cos 2 (1 cos ) 3 cos

12

xx x

x x x

Problem 10 Find all the solutions of the equation22cos 3cos 1x x . (5points)

Writing the equation as 22cos 3cos 1 0x x and factoring the

left part we get

(2cos 1)(cos 1) 0x x

The problem now is reduced to solving two basic equations

cos 1x

which has solutions

(2 1) ,x n n Z

and

1

cos2

x

with the solutions

1 2

arccos 2 2 , .2 3

x n n n Z

mth-105-02 algebra and trigonometry spring 2011 …

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