SOLUTIONS MANUAL for Art of problem solving, precalculus, solutions manual -- Satō, Naoki, Rusczyk, by welldoneassistant

PHILLIPS ACADEMY

Solutions Manual Naoki Sato

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K_An.no 1778 •

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# OLIVER • WENDELL • HOLMES

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LIBRARY

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anual

Richard Rusczyk Art of Problem Solving J

Reproduction of any portion of this book without the written permission of AoPS Incorporated is strictly prohib¬ ited, except for "fair use" or other noncommercial uses as defined in Sections 107 and 108 of the U.S. Copyright Act.

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Printed in the United States of America. First Printing 2009.

FOREWORD

Foreword

This book contains the full solution to every Exercise, Review Problem, and Challenge Problem in the text Precalculus. In most problems, the final answer is contained in a box, like this However, we strongly recommend against just looking up the final answer and moving on to the next problem. Instead, even if you got the right answer, read the solution in this book. It might show you a different way of solving the problem that you might not have thought of. If you don't understand a solution, or you think you have a better way of solving the problem, or (gasp!) find an error in one of our solutions, we invite you to come to our message board at www.artofproblemsolving.com and discuss it. Our message board is free to use and includes thousands of the world's most eager mathematical problem-solvers.

CONTENTS

Contents

Foreword

• • •

Functions Review

Introduction to Trigonometric Functions

Trigonometric Identities

Applications to Geometry

ir>

Parameterization and Trigonometric Coordinate Systems

Basics of Complex Numbers

104

Trigonometry and Complex Numbers

123

6 7 IV

CONTENTS

Geometry of Complex Numbers

151

Vectors in Two Dimensions

177

Matrices in Two Dimensions

187

1 1

Vectors and Matrices in Three Dimensions, Part 1

207

19 1 mm

Vectors and Matrices in Three Dimensions, Part 2

221

IQ 1 w

Vector Geometry

241

—

■

CHAPTER

1 Functions Review

Exercises for Section 1.1 1.1.1

(a)

/(2) = 22 - 2 - 6 = 1-4 |.

(b)

Since f(x) = x2 - x - 6 = (x - 3)(x + 2), we have /(x) = 0 when x = 3 and x - -2

(c)

f(x - 1) = (x - l)2 - (x - 1) - 6 = x2 - 2x + 1 - x + 1 - 6 =

(d)

If /(x) = 6, then x2 - x - 6 = 6, so x2 - x - 12 = 0. Factoring gives (x - 4)(x + 3) = 0, which means x = 4 or x = -3. Hence, /(4) = /(-3) = 6, so yes , we can have /(x) = 6.

(e)

3x - 4

1 2 25 Completing the square, we get /(x) = x-x-6=(x--J - —. Hence, the range of / is [-25/4, +oo)

1.1.2 (a)

We can evaluate the absolute value of 1 - x for any real number x, so the domain of / is R . The absolute value of a real number is always nonnegative, and 1 - x can equal any nonnegative real number, so the range of / is [0, +oo) .

(b)

We can only take the s quare root of a nonnegative real number, so 2 - t must be nonnegative. Hence, the domain of / i s (-oo,2 . Also, the square root of a nonnegative real number is always nonnegative, so the range of / is [0, +oo) .

(c)

We can evaluate 1 - x2 for any real number x, so the domain of h is R |. Since x2 is always nonnegative, 1 - x2 is always less than or equal to 1, so the range of h is (-oo, 1] .

(d)

First, we note that we cannot have u = 0, since this makes the denominator of 1/u equal to 0. We also cannot have the denominator of the whole function, 1 + 1/m, equal to 0. If 1 + 1/u = 0, then u = —1, so the domain of g is all real numbers except 0 and -1 To find the range of g, let v = g(u) =

1+ - = M

Taking the reciprocal of both sides and rearranging gives

11 U

1 V

u = 1 - V

Hence, for every value of v, there is a corresponding value of u, except when v = 1. Furthermore, as determined above, u cannot be equal to 0 or -1. There is no real number v that corresponds to u - -1, but

CHAPTER 1. FUNCTIONS REVIEW

the real number v = 0 corresponds to u = 0, so v cannot be equal to 0 either. Therefore, the range of g is all real numbers except 0 and 1 1.1.3 The function (/ • g)(x) = V4 - x • y/2x - 6 is defined only where both the functions f(x) = V4 - x and

g(x) = V2x-6 are defined. The function V4 — x is defined only for x < 4, and the function V2x — 6 is defined [3,4] As with / • g, the function

JIM =

V4-- X V2x-6

is defined only where both the functions f(x) = V4 - x and g(x) = V2x - 6 are defined. However, we must also exclude points where the denominator is 0. In this case, the denominator is 0 when x = 3. Hence, the domain of the function f/g is (3,4] The domains of f ■ g and f/g are different because we must exclude those values where g(x) = 0 from the domain of f/g, but we include these values in the domain of / • g if they are also in the domain of /. 1.1.4 The function

2x - 5 x-8

is defined only where (2x - 5)/(x - 8) > 0. For this inequality to hold, either both 2x - 5 and x - 8 must be positive, both of them must be negative, or 2x - 5 = 0. If both 2x - 5 and x - 8 are positive, then x > 5/2 and x > 8, so in this case, we get x > 8. If both 2x - 5 and x - 8 are negative, then x < 5/2 and x < 8, so in this case, we get x < 5/2. Finally, if 2x - 5 = 0, then x = 5/2. Therefore, the set of x such that (2x - 5)/(x - 8) > 0 is (—oo, 5/2] U (8, +oo), which is the domain of /. On the other hand, the function

V2x^5 g(x) =

Vx - 8

is defined only where both the functions V2x - 5 and Vx - 8 are defined, and the denominator is nonzero. The function V2x - 5 is defined for x > 5/2, and the function Vx - 8 is defined for x > 8. The denominator is zero when x = 8, so the domain of g is (8, +oo). Thus, the functions / and g have different domains

because those values of x for which 2x - 5 and x - 8 are

negative are in the domain of /, but not in the domain o g1.1.5 /(x) • /(-x) =

x+1 x-1

—X + 1

X + 1

-x - 1

x-1 - =yj,for |x| * 1. X +

1.1.6 We break up the interval [-3,4] into the intervals [-3, -2] and [-2,4]. If -3 < x < -2, then -1 < x + 2 < 0, so 0 < (x + 2)2 < 1. If -2 < x < 4, then 0 < x + 2 < 6, so 0 < (x + 2)2 < 36. Hence, every value in the range of g is in the interval [0,36]. Intuitively, it seems clear that every value in this interval is in the range. To prove it we let y = (x + 2)2 and solve for x. Taking the square root of both sides gives x + 2 = ± sfy, so x = ± yjy - 2. Consider the solution x = yjy - 2. For any y such that 0 < y < 36, we have 0 < yfy ^ 6, so -2 < Vy - 2 < 4. Therefore, the number yjy - 2 is in the domain of g if 0 < y < 36. So, if we let x = y/y - 2 for any y such that 0 < y < 36, we have /(x) = y. Therefore, every value in [0,36] is in the range, and our range is [0,36] 1.1.7 (a) 2

We have T( 2,3, -5) = 3 • 23 - (-5) = 3-8 + 5 = [~29

Section 1.2

(b)

Since T(x, 2,6) - 3x2 — 6, we have 3x2 — 6 = 21. Isolating x2 gives x2 = 9, from which we find x = ±3

Exercises for Section 1.2 1.2.1 (a) If 3x - 7 = 0, then x = 7/3, so the ^-intercept is (7/3,0) I. Since /(0) = -7, the y-intercept is (0,-7)

f(x) = 3x - 7 (b) If 2 - \x\ - 0, then |x| = 2, so x = ±2. Hence, the x-intercepts are (2,0) and (-2,0)

Since /(0) = 2, the y-intercept is (0,2)

/(x) = 2 - |x|

(c)

Note that /(x) = x2 - 5x + 6 factors as (x - 2)(x - 3). If (x - 2)(x - 3) = 0, then

x = 2 or x = 3, so the x-intercepts are (2,0) and (3,0) . Since /(0) = 6, the

y-intercept is (0,6) t

j r—

\ //

.1.L. _l. .._..L ,

1.2.2

. J j

/(x) = x2 — 5x + 6

(a)

Every vertical line intersects the graph at most once, so the graph represents a function.

(b)

Every vertical line intersects the graph at most once, so the graph represents a function. Note that every vertical line does not need to intersect the graph, such as the line x = 0.

(c)

There are many vertical lines that intersect the graph more than once, such as the line x = 1, so the graph does not represent a function.

1.2.3 (a)

All we know about the function / is that /(9) = 2. Hence, if we set x - 3 = 9, then we have x = 12, and y = /(12 - 3) + 5 = /(9) + 5 = 7. Therefore, a point on the graph of y = /(x - 3) + 5 is (12,7) . We can also find this point by noting that the graph of y = /(x - 3) + 5 results from shifting the graph of y = /(x) to the

CHAPTER 1. FUNCTIONS REVIEW

right by 3 and up by 5. So, because (9,2) is on the graph of y = /(x), the point (9 + 3,2 + 5) = (12,7) is on the graph of y = /(x - 3) + 5. (b)

If we set x/4 = 9, then x = 36, and y = 2/(9) = 4. Therefore, a point on the graph of y = 2/(x/4) is (36,4) We can also see this by noting that the graph of y = 2/(x/4) results from scaling the graph of y = /(x vertically by a factor of 2 and horizontally by a factor of 4. So, because (9,2) is on the graph of y = /(x), the point (9 • 4,2 • 2) = (36,4) on the graph of y = 2/(x/4).

(c)

Ifweset3x-1 = 9, thenx = 10/3,and y = 2/(9)+ 7 = 11. Therefore, a point on the graph of y = 2/(3x-l) + 7 is (10/3,11)

1.2.4 The graph of y = /(2x) is the same as the graph of y = /(x) compressed by a factor of 2 horizontally, and the graph of y = /(x - 6) is the same as the graph of y = /(x) shifted 6 units to the right. Knowing these results, we may think that the graph of y = /(2x - 6) results from compressing the graph of y = /(x) by a factor of 2 horizontally to give the graph of y = /(2x), and then shifting the result 6 units to the right. However, this is not true. The first step is correct; compressing the graph of y = /(x) by a factor of two horizontally gives the graph of y = /(2x). To see how we produce the graph of y = /(2x - 6), we let /z(x) = /(2x). Then, the graph of y = h(x) is the same as the graph of y = /(2x). To get the graph of y = /(2x - 6), we note that h(x - 3) = /(2(x - 3)) = /(2x - 6). The graph of y = h(x - 3), which is the same as the graph of y = /(2x - 6), is the result of shifting the graph of y = h(x), which is the same as the graph of y = /(2x), by 3 units to the right. Therefore, the answer is (c). The graph of y = /(2x - 6) is the same as the graph of y = /(2x) shifted 3 units to the right.

Exercises for Section 1.3 1.3.1

If g(x) = 3x + 7, then g(g(x)) = g(3x + 7) = 3(3x + 7) + 7 = 9x + 28 |. If g(x) = -3x - 14, then g(g(x)) =

g(—3x - 14) = —3(-3x - 14) - 14 = 9x + 28 . Hence, both solutions work.

1.3.2 (a)

First, we note that the domain of g is (-oo, 1], Since the domain of / is all reals, all outputs of g are in the domain of /. For x < 1, we therefore have f(g(x)) = /(Vl - x) = 1 - x - 2 Vl - x, where x < 1

(b)

The domain of f(g(x)) is the set of values of x such that both g(x) is defined, and g(x) lies in the domain of /(x). From g(x) = Vl -x, we see that g(x) is defined only for x < 1. The domain of /(x) consists of all real numbers, so the domain of f(g{x)) is

(c)

(-oo, 1]

First, /(-2) = (-2)2 - 2 • (-2) = 8. Since g(x) is only defined for x < 1, we see that g(f(-2)) is I not defined

1.3.3 We show that /3(x) is the same as /2(/(x)) by expanding both expressions. By definition, /3(x) = /(/(/(x))). Since /2(x) = /(/(x)), we may substitute to get /2(/(x)) = /(/(/(x))), so /2(/(x)) = /3(x). Similarly, /(/2(x)) = /(/(/(x))), so /3(x) and /(/2(x)) are the same, as well. 1.3.4

To get a feel for the problem, we compute /"(x) for the first few values of n:

f\x) = f(x) = ax, f2(x) = /(/(*)) = f(ax) = a2x, f3(x) = f(f2(*)) = f(a2x) = a3x, f4(x) = /(/3(x)) = f(a3x) = a4x, and so on.

Section 1.4

Applying the function / to x multiplies it by a. Therefore, applying the function / to x exactly n times multiplies it by a exactly n times; in other words, it multiplies it by an. Hence, f n(x) = anx 1.3.5 All the statement "(3,0) is the only ^-intercept of the graph of g" tells us is that g(3) = 0. So, we have ^(3) = /(g(3)) = /(0). This tells us that (3, /(0)) is on the graph of y = h(x), but this doesn't tell us anything about the intercepts of h. We cannot conclude anything about the intercepts of h from the given information.

Exercises for Section 1.4 1.4.1

Let y — / 1(3). So, we have /(y) = f(f 1(3)) = 3. This means we have

^=3

y-2

y - 1 = 3(y - 2) = 3y - 6

1.4.2 (a)

Let x = /(y) = 2y - 7. Then y = (x + 7)/2, so / x(x) = me + 7)/2

(b)

Let x = f{y)

(c)

Let x = /(y) = V2 - y. Then x2 = 2 - y, so y = 2 - x2. Therefore, /_1(x) = 2 - x2. Note that since the range of f{x) is the set of all nonnegative real numbers, the domain of /-1 is also all nonnegative real numbers. So,

1 =

2^3.

Then 2y + 3

so y =

- 3)/2 = (1 - 3x)/(2x). Therefore,

/-1

(x) =

we have / 1(x) = 2 - x2, where the domain of / 1 is [0, +00) (d)

Since /(4) = |5 — 4| = 1 and /(6) = |5 — 6[_ *= 1, there are two different values of x that give the same value of

f{x). Therefore, the function fix) does not have an inverse . (e)

Since /(2) = V4 - 2 + V2 - 2 = x/l and /(4) = V4 — 4 + V4 - 2 = V2, there are two different values of x that give the same value of fix). Therefore, the function fix) does not have an inverse

(f)

Completing the square gives fix) = (x - 3)2 - 6. We might at first think that this function does not have an inverse, since /(l) = /(5). However, fix) is only defined for x > 4. By considering the graph of fix), or by noting that the function fix) strictly increases as x increases for x > 4, we see that no two values of x in the domain of / give the same output from /. In particular, /(4) = -5, so the range of / is [-5, +00). So, / does have an inverse. But what is it? To find the inverse of /, we solve x = /(y) = (y - 3)2 - 6 for y in terms of x. Adding 6 to both sides gives (y - 3)2 = x + 6. Since x is in the range of /, we have x > -5, so x + 6 > 1. Therefore, we take the positive square root to find y = Vx + 6 + 3. Hence, / !(x) = Vx + 6 + 3, where the domain of / 1 is [-5, +00) . Note that / *(x) is defined only for x > -5, because this is the range of /.

1.4.3

To find /-1(x), let x = /(y) = ay + b, so y = (x - b)/a. Then the equation fix) - /-1(x) becomes

ax+ b -

x-b

x a

b a

This holds for all x if and only if a = 1/a and b = -b/a. From the first equation, we have az = 1, so a = 1 or a = -1. If a = 1, then the second equation becomes & = -b, so b = 0. If a = -1, then the second equation becomes b = b, which is true for any real number b. Therefore, the ordered pairs (a,&) that satisfy the problem are (1,0) and (-1, b) for any any real number b

CHAPTER 1. FUNCTIONS REVIEW

We can check these solutions quickly. For (a, b) - (1,0), we have f(x) = x, so /(/(x)) = f(x) - x, which means /'_1 (x) = f(x). If (a, b) = (-1, b), then f(x) = -x + b, so /(/(x)) = f(-x + b) = -(-x + b) + b = x, and again we have

f~\x) = f(x). 1.4.4

First, we find y_1(x)- Let x = g(y) = (ay + b)/(cy + d). Then

x(cy + d) = ay + b

cxy + dx = ay + b

cxy - ay = b - dx

y(cx - a) = b — dx

—^

b - dx cx - a

■

b - dx

Therefore, we have g 2(x) = -——. We see that g !(x) is defined for all x except when cx - a = 0, or x — a/c (Note that we must have ad ± be because if ad = be, then g(x) is a constant for all x such that cx + d means that g does not have an inverse.)

0, which

Review Problems 1.23 (a) Since A(x) = 4x2 + 1 is defined for all real numbers, the domain of A is |R . Since 4x2 can equal any nonnegative real number, but no negative number, the range of A(x) = 4x2 + 1 is [1, +oo) (b) The function o(x) = 3 + y/l6 - (x - 3)2 is defined only when (x - 3)2 < 16, so x - 3 must be between -4 and 4. In other words, -4 < x - 3 < 4, so -1 < x < 7. Therefore, the domain of o is [-1,7] As x varies from -1 to 3, y/l6 - (x - 3)2 varies from 0 to 4, so o(x) = 3 + y/l6 - (x - 3)2 varies from 3 to 7. As x varies from 3 to 7, \Jl6 - (x - 3)2 varies from 4 to 0, so o(x) = 3 + -^16 - (x - 3)2 varies from 7 to 3. Therefore, the range of o is [3,7] (c) The function P(x) = 1/(3 + Vx + 1) is defined if and only if x + 1 > 0, or x > —1. Therefore, the domain of P is [—1, +oo) Since Vx + 1 can take on any nonnegative value, we see that P(x) = 1/(3 + Vx + 1) varies from 1/3 to 0 (without ever reaching 0). Therefore, the range of P is (0,1/3] (d) The function S(x) = (12x - 9)/(6 - 9x) = (4x - 3)/(2 - 3x) is defined for all x except when the denominator 2 - 3x is zero. We have 2 - 3x = 0 when x = 2/3, so the domain of S is all reals except 2/3 To find the range of S, we let y = S(x) - (4x - 3)/(2 - 3x). Then we have y(2 - 3x) = 4x - 3

2y — 3xy = 4x - 3

3xy + 4x = 2y + 3

2y + 3 ~ 3y + 4 ’

Hence, for every value of y except y = -4/3, there is a corresponding value of x such that S(x) = y. Therefore, the range of S is all reals except -4/3 1.24 (a)

The function

V2x-5

is defined only when 2x - 5 > 0, or x > 5/2, and the function V9 - 3x is defined only

when 9 - 3x > 0, or x < 3. Combining these, the domain of / is (5/2,3] (b)

The function | yjx - 2| is defined only when x > 0 and the function | Vx - 2| is defined only when x - 2 > 0, or x > 2. Therefore, the domain of / is [2, +<x>)

Review Problems

(c)

1.25

f{g{x)) = f ( 21 Vx +

1 4)

4 • ggi

2x + 2(x + 4)

4x + 8

x+2

where x £ 4, because -4 is not in the

domain of g. 1.26

The equation /(1) - g(l) + 2 gives us a + b + c = a- b + c +2, so b = 1. Then the equation /(2) = 2 gives

4a + 2fr + c = 2, so 4a + c = 2 - 2b = 0. Therefore, g(2) = 4a - 2b + c = (4a + c) - 2b = 0 - 2 = -2

1.27

For part (a), we shift the graph 3 units to the right. For part (b), we scale the graph vertically away from the x-axis by a factor of 2, then shift the graph 1 unit upwards. For part (c), we reflect the graph over the y-axis. For part (d), we reflect the graph over the y-axis, then shift the graph 2 units to the right. To see why we shift to the right rather than the left, let h(x) = /(—x). Then, the graph of y = /(2 - x) is the same as the graph of y = h(x - 2). The graph of y = h(x - 2) is the result of shifting the graph of y = h(x) to the right 2 units. So, the graph of y = /(2 - x) is the result of shifting the graph of y = /(-x) to the right 2 units. For part (e), we compress the graph horizontally by a factor of 1/2, then shift the graph 1/2 unit to the right. As with part (d), we can see that this shift is 1/2 unit by letting h(x) = /(2x); therefore, h(x — \) = /(2x - 1), so we must shift the graph of h(x), which is the same as the graph of y = /(2x), to the right 1/2 unit to get the graph of y -- /(2x - 1). We then compress the graph of y = /(2x - 1) vertically by a factor of 1/2, then shift the resulting graph 3 units upwards. The results for all five parts are shown below.

y = 2/(x) + 1 1.28

y = /(2 - x)

The functions / and g need not be the same. For example, let f(x) - |x| and g(x) = x. Then f(f(x)) = |/(x)| =

||x|| = |x|, and g(f(x)) = g(|x|) = |x|. So, we have f(f(x)) = g(f(x)), but f(x) * g(x).

CHAPTER 1. FUNCTIONS REVIEW

1.29

(a)

Every vertical line intersects the graph at most once, so the graph can represent a function. We also see that every horizontal line intersects the graph at most once, so the function has an inverse, the graph of which is shown at right. (It is simply

y — —-

the graph in thp prnblpm rpflprtpd ovpr thp linp y — r )

(c)

There are vertical lines that mtersect the graph more than once, so the graph cannot represent a function.

-► v .fC

-f

From f(a, b, c) - 1, we have = 1, so a - c = b - c, which means a = b. Then f(a, c, b) that we know that b ± c because f(a, b, c) is defined, so we know that we don't have divisicm by f(a,c,b).) 1.30

1.31

:4: _

Every vertical line intersects the graph at most once, so the graph can represent a function. However, there are horizontal lines that intersect the graph more than once, so the function does not have an inverse. Oil cries-

(b)

Jote lua ting XI (jva 0

If /(x) = -x2 + bx + c and g(x) = dx + e, then

f(g(x)) = f{dx + e) = -(dx + e)2 + b(dx + e) + c = -d2x2 - 2dex - e2 + bdx + be+ c = -d2x2 + (bd - 2de)x - (e2 - be - c). For f(g(x)) to be equal to x2 for all x, the coefficient of x2 must be 1. However, there is no real number d for which -d2 = 1, so it is not possible for f(g(x)) to be x2. 1.32 To find the inverse of f(x) - ax + b, set x - f(y) - ay + b. Then y = ^ | - -a, so we have / x(x) - \x—\. This is a linear function. We conclude that the inverse of a linear function is always a linear function. 1.33 Consider a point (x,\f(x)\) on the graph of y - |/(x)|. If f(x) > 0, then \f(x)\ = f(x). On the other hand, if f(x) < 0, then \f(x)\ = -f(x), and the point (x, |/(x)|) = (x, —f(x)) is the reflection of the point (x,/(x)) over the x-axis. Thus, the graph of y = \f(x)\ can be constructed from the graph of y = f(x) by reflecting over the x-axis the portion of the graph of y = f(x) that lies below the x-axis. An example is shown below. The dashed lines in the diagram on the right indicates portions of the graph of y - f(x) that are reflected over the x-axis to produce the graph of y = |/(x)|.

V = fix)

V = \f(x)\

1.34 If y =/(x) = x/(l - x), then y(l-x) = x

=»

y - xy - x

=»

xy + x = y

x(y + 1) = y

y + i'

Challenge Problems

We compare this to each of the given possibilities:

-/(y) -f(-y) /(-y) f(y) Therefore, we have x =

y+i 1.35

Vy _ 1-1 /y y-i' _y_ i-y' -y = _y_ l+y

y +1'

-y =_y_ l+y

y + 1'

y i-y'

-/(-y)

c2x We have /(/(x)) =f(C*/(2* + 3) = J J ” J \2x + 3/ 2 • cx/(2x + 3) + 3 2cx + 3(2x + 3)

c2x (2c + 6)x + 9 ’

If /(/(x)) = x for all x except -3/2, then

—-—-- = x (2c + 6)x + 9

c2x = x[(2c + 6)x + 9]

c2x = (2c + 6)x2 + 9x. v ’

This equation holds for all x except -3/2 if and only if 2c + 6 = 0 and c2 = 9. The only value of c that satisfies both equations is c = | —3 |. 1.36 Let g(x) - f(2x). As discussed in the text, we find the inverse of g(x) by solving the equation x = g(y) for y in terms of x. From x - g(y), we have x = f(2y). Because / is invertible, x = f(2y) tells us that 2y = /-1(x), so

y - \f~l(x). Thus, the inverse of f(2x) is |/_1(x) , not /-1(2x). We also could have found the inverse using the fact that the inverse of (f(g(x)) is g l(f l(x)), which we proved in the text. Here, we let g(x) = 2x, so g x(x) = l2x and the inverse of/(2x) is y-1(/-1(x)) = \

(x), as before.

Note that substituting f(2x) into the function \

(x) gives x: we have \f~1(f(2x)) = l(2x) = x, as expected.

Also, we have / (2 • |/_1(x)) = f(f~l(x)) = x. 1.37 For the functions / and g to be inverses, we also require that g(f(x)) = x for all x in the domain of /. But

g(f(x)) - Vx2 = |x|, which is not equal to x when x is negative, so / and g are not inverse functions. Since the function f(x) = x2 fails the horizontal line test (for example, /(-1) = /(1) = 1), no function g can be an inverse of /.

Challenge Problems 1.38 If /(x) = /_1 (x), then /(/(x)) = /(/-1 (x)) = x for all x in the domain of /. We find an expression for f(f(x)) as follows: / 2x + a \ _ 2 ■ (2x + a)/(bx -2) + a _ 2(2x + a) + a(bx - 2) /(/(*)) “ J {yx _ 2)~ b ■ (2x + a)/(bx -2)-2~ b(2x + a)~ 2 (bx - 2) 4x + 2a + abx -2a _ (ab + 4)x 2bx + ab - 2bx + 4

ab + 4 9

CHAPTER 1. FUNCTIONS REVIEW

This simplifies to x as long as ab + 4 =£ 0. If ab + 4 = 0, then a = -4/b, and 2x + a _ 2x - 4/b _ 2bx - 4 _ 2(fcx - 2) _ 2 X

bx-2

b(bx- 2)

fe'

which is a constant, so / does not have an inverse if ab = -4. Hence, / x(x) exists and / 1(x) = /(x) for all a and b such that ab £ -4 139 (a) The function f(x + 1) is defined only when -1 < x + 1 < 1, or -2 < x < 0. Therefore, the domain of f(x + 1) is

(-2,0)

(b) The function /(1/x) is defined only when -1 < 1/x < 1. If x is positive, then multiplying all three parts of the inequality chain by x gives —x < 1 < x. Since x is positive, —x is negative, so the inequality -x < 1 is always satisfied. Hence, the solution in this case is x > 1. If x is negative, then multiplying all three parts of the inequality chain by x gives -x > 1 > x. Since x is negative, the inequality 1 > x is always satisfied. Multiplying the inequality —x > 1 by —1, we get x < —1. Hence, the solution in this case is x < -1. Therefore, the domain of /(1/x) is (—oo, -1) u (1, +oo) (c) The function f(xfx) is defined only when both x > 0 and -1 < y/x < 1. Since \[x is always nonnegative, the inequality -1 < y/x is always satisfied. For the same reason, we can square both sides of the inequality y/x < 1 to get x < 1. Therefore, the domain of the function /(y/x) is [0,1) (d) The function /(fiy) is defined only when — 1 < separately.

< 1. We take the cases where x - 1 > 0 and x - 1 < 0

If x - 1 >0, then multiplying all parts of the inequality -1 < |^ < 1 by x - 1, we get -x + l<x+l<x-l. Subtracting x + 1 from all three parts gives —2x < 0 < -2. Therefore, there are no solutions in this case. If x -1 <0, then multiplying all parts of the inequality -l<jzi<lbyx-l (and changing the directions of the inequalities because x - 1 is negative), we get —x +1 >X + 1 > X — 1. Subtracting x + 1 from all three parts gives -2x > 0 > -2, so x < 0. Therefore, the solution is x < 0 in this case. Hence, the domain of /(fzj) is (-oo, 0) 1.40 The function /(x) = V2 - x - x2 is defined only when 2 - x - x2 > 0. Factoring gives (2 + x)(l - x) > 0. Either both factors must be nonnegative, or both factors must be nonpositive. If both factors are nonnegative, then 2 + x > 0, so x > -2, and 1 - x > 0, sox<l. Hence, the solution in this case is -2 < x < 1. If both factors are nonpositive, then 2 + x < 0, so x < -2, and 1 - x < 0, so x > 1. There is no value of x such that x < -2 and x > 1 simultaneously, so there is no solution in this case. Therefore, the domain of /(x) is [-2,1] To find the range of /(x), we complete the square inside the radical:

/(x) = V2 - x - x2 =

Challenge Problems

As x varies from 2 to —1/2, the expression 9/4 — (x +1/2)2 varies from 0 to 9/4, so f(x) varies from 0 to 3/2. Then as x varies from -1/2 to 1, the expression 9/4 — (x + 1/2)2 varies from 9/4 to 0, so f(x) varies from 3/2 to 0. Therefore, the range of f(x) is [0,3/2]

1.41

Let t = x2 + 1, so x2

= t - 1 and x4 = (t - l)2. Hence,

f(t) = f(x2 + l) = x4+5x2 + 3 = (t-l)2 + 5(t-l) + 3 = t2-2t + l+5t-5 + 3 = t2 + 3t-l. Substituting t

x2 - 1 we get f(x2 - 1) = (x2 - l)2 + 3(x2 - 1) - 1

c4 - 2x2 + 1 + 3x2 - 3 - 1 = x4 + x2

1.42 Consider a point (x, /(|x|)) on the graph of y = /(|x|). If x > 0, then |x| = x, so (x,/(|x|)) is the same as (x,/(x)). On the other hand, if x < 0, then |x| = -x, and the point (x, /(|x|)) = (x, /(-x)) is the reflection of the point (—x, /(-x)) over the y-axis. Thus, the graph of y = /(|x|) consists of the portion of the graph of y = /(x) that is on or to the right of the y-axis, together with the curve formed when reflecting this portion of the graph of y = /(x) over the y-axis. An example is shown below.

y(>>

7\ ! / \). \ V 7

7 //

L A\ V—

\\/ / J // \

y = /(M)

V = /(*) 1.43

Let x be a root of /(x) = 0. Applying / to both sides, we get f(f(x)) = /(0) = 2003. But f(f(x)) = x for all

values of x, so x = 2003. Therefore, the only root of the equation f(x) = 0 is x 1.44

To find the inverse of /(x), set x = /(y) = (ay + b)/(cy x(cy + d) = ay + b

cxy + dx = ay + b

Thus, the inverse of f(x), if it exists, must be / 1 (x)

2003

+ d). Then

cxy - ay

b - dx

= b - dx

cx-a'

b - dx cx-a'

To check if this works, we substitute: a ■ (b - dx)/(cx - a) + b _ a(b - dx) + b(cx - a)

(be - ad)x

- a)

be-ad

c- (b - dx)/(cx - a) + d

c(b - dx) + d(cx

This simplifies to x as long as ad - be £ 0. Hence, the function / has an inverse if and only if ad - be ± 0 Note that the function f(x)

is well-defined as long as c and d are not simultaneously 0, and the function

/-V) =

b - dx cx- a'

is well-defined as long as a and c are not simultaneously 0. However, if c = d = 0, then ad-bc = 0, and if a = c = 0, then ad-bc = 0. Hence, the condition ad-bc ± 0 also ensures that both of these functions are well-defined. 11

CHAPTER 1. FUNCTIONS REVIEW

1.45 Let (x, 0) be an x-intercept of the graph of /, so f(x) = 0. Then g(x) = /(x)/(|x|) = 0, so (x, 0) is also an x-intercept of the graph of g. Hence, all x-intercepts of the graph of / must also be x-intercepts of the graph of g. On the other hand, an x-intercept of the graph of g is not necessarily an x-intercept of the graph of /. For example, take /(x) = x - 1, so g(x) = /(x)/(|x|) = (x - l)(|x| - 1). Note that g(-l) = 0, but /(-1) = -2^0. Thus, the point (-1,0) is an x-intercept of the graph of g, but not an x-intercept of the graph of /. 1.46 To construct the graph of y = \f(2x - 2) - 1 from the graph of y = /(x), we describe the construction one step at a time, working from the inside of the function out. First, we look at the expression /(2x - 2) = /(2(x - 1)). As described in the solution to Problem 1.2.4, the graph of y = /(2(x -1)) is the result of scaling the graph of y = f(x) horizontally towards the y-axis by a factor of 1/2 and shifting the ensuing graph to the right 1 unit. This produces the middle graph below. Then, the graph of y = |/(2x - 2) is the result of scaling the graph of /(2x - 2) vertically away from the x-axis by a factor of 3/2, and the graph of y = |/(2x - 2) - 1 results from shifting the graph of y = |/(2x - 2) downward 1 unit. This produces the final graph at right below.

1.47 To find the unique number that is not in the range of /, let y = /(x) = (ax + b)/(cx + d). Then y(cx + d) = ax + b

cxy + dy = ax + b

cxy - ax = b - dy

b - dy cy-a'

Hence, for every value of y, there is a corresponding value of x such that /(x) = y, except when the denominator cy-a is equal to 0. Therefore, the unique number that is not in the range of / is a/c. Now we compute /(/(x)):

/(/(*)) = /

ax + b\ cx + dJ

a - (ax + b)/(cx + d) + b c ■ (ax + b)/(cx + d) + d

a(ax + b) + b(cx + d) c(ax + b) + d(cx + d)

(a2 + bc)x + (ab + bd) (ac + cd)x + (be + d2)'

If f(f(x)) = x for all x, then

(a1 + bc)x + (ab + bd) = x, so (a2 + bc)x + (ab + bd) - (ac + cd)x2 + (be + d2)x for all x. (ac + cd)x + (be + d2)

Then the coefficient of x2 must be 0, so ac + cd - 0, which means c(a + d) - 0. Since c is nonzero, we have a + d = 0, so d - -a. Furthermore, if d = -a, then

X 12

(a2 + bc)x + (ab + bd)

(a2 + bc)x + (ab - ab)

(ac + cd)x + (be + d2)

(ac - ac)x + (a2 + be)

(a2 + bc)x a2 + be

Challenge Problems

as desired. Therefore, f(x) =

ax + b

cx + d

cx — a

Since /(19) = 19, we have 19 a + b 19 c - a

192c - 19a = 19 a + b

19 2c = 2-19 a + b.

972c - 97a = 97a + b

972c = 2-9 7a+ b.

Also, /(97) = 97, so we have 9 7a+ b 97c - a

Subtracting the first equation from the second, we get (972 - 192)c = 2(97 - 19)a

(97 + 19)(97 - 19)c = 2(97 - 19)a

97 + 19

Here is a faster way to derive the same equations. Since /(/(x)) = x for all x, we have /(/(0)) = 0. Since /(0) = b/d, we have r/ r/nw ^ mm=nm = a-b/d + b

ab + bd

so /(/(0)) = 0 gives (ab + bd)/(bc + d2) = 0. Therefore, we have ab + bd = 0, so b(a + d) = 0. Since have a + d = 0, so d = -a.

is nonzero, we

Now, we know that x = 19 and j = 97 satisfy the equation /(x) = x. Writing this equation out, we get

ax + b =x cx + d

x(cx + d) = ax + b

cx + (d- a)x-b = 0.

The sum of the roots of this quadratic is -(d-a)/c = la/c. We know that the roots of this quadratic are 19 and 97, since these are the solutions to /(x) = x. Therefore, we have 2a/c = 19 + 97 = 116, which gives us a/c = 116/2 = 58.

CHAPTER 2. INTRODUCTION TO TRIGONOMETRIC FUNCTIONS

CHAPTER

1 Introduction to Trigonometric Functions

Exercises for Section 2.1 2.1.1 (a)

Let P be the terminal point of 120° and let S be the foot of the perpendicular from P to the x-axis. From 30-

60-90 triangle POS, we have PS = V3/2. Since 120° lies in the second quadrant, we have sin 120° = V3/2

Figure 2.1: Diagram for Part (a) (b)

Figure 2.2: Diagram for Part (b)

Let P be the terminal point of 60°, and let S be the foot of the perpendicular from P to the x-axis. From

30-60-90 triangle POS, we have PS = V3/2 and OS = 1/2, so

cot 60° =

(c)

1/2

cos 60°

OS _

sin 60°

PS ~ V3/2

Since cos 45° = V2/2 = 1/ V2, we have sec 45° = cos 45°

1 _ 1/V2

V3 3

Section 2.1 (d)

Let P be the terminal point of 210° and let S be the foot of the altitude from P to the x-axis. From 30-60-90 triangle POS, we have OS = V3/2 and PS = 1/2. Since 210° lies in the third quadrant, we have cos 210° = - V3/2 and sin 210° = -1/2, so tan 210° =

(e)

1/2

sin210c

-V3/2■" V3

cos210c

V3 3

The terminal point of 180° is (-1,0), so cos 180° = -1

(f) anuuuc Hum r iu uit x-dxis. men = ou , so us = i/z ana ri> Since 300° lies in the fourth quadrant, we have sin 300° = — V3/2, so

esc 300° sin 300°

-V3/2

2.1.2 (a)

The terminal point of -90° is (0, -1), so cos(-90°) - 0

(b)

Since 36000° = 100 • 360°, the terminal point of 36000° is the same as the terminal point of 0°. Therefore, tan 36000° = tan0°

(c)

Since 750° = 2 • 360° + 30°, the terminal point of 750° is the same as the terminal point of 30°. Therefore, sin 750° = sin 30° = 1/2

(d)

Since -1200° = -4 • 360° + 240°, the terminal point of -1200° is the same as the terminal point of 240° = 180° + 60°. We let P be this terminal point, and let S be the foot of the altitude from P to the x-axis. Then ZPOS = 60°, so OS = 1/2 and PS = V3/2. Since 240° lies in the third quadrant, we have cos 240° = -1/2, so sec(-1200°) = sec 240° =

cos 240°

1/2

-2

The angle 293° lies in the fourth quadrant, so sin 293° is negative. The angle -68° lies in the fourth quadrant, so cos(-68°) is positive. The angle 206° lies in the third quadrant, so cot 206° = is positive. The angle 90.5° 2.1.3

lies in the second quadrant, so esc 90.5° = sing05<, is positive. Thus, cos(-68°), cot 206°, and esc 90.5° are positive.