For Use - Westlake City School · PDF filee = 2.718281828459 c. The notation e was proposed...

28 Chapter 1 Functions

1.3 Inverse, Exponential, and Logarithmic Functions

Exponential functions are fundamental to all of mathematics. Many processes in the world

around us are modeled by exponential functions—they appear in finance, medicine, ecol-

ogy, biology, economics, anthropology, and physics (among other disciplines). Every ex-

ponential function has an inverse function, which is a member of the family of logarithmic functions, also discussed in this section.

Exponential Functions

Exponential functions have the form f 1x2 = bx, where the base b ≠ 1 is a positive real

number. An important question arises immediately: For what values of x can bx be evalu-

ated? We certainly know how to compute bx when x is an integer. For example, 23 = 8

and 2-4 = 1>24 = 1>16. When x is rational, the numerator and denominator are inter-

preted as a power and root, respectively:

powerf

163>4 = 163>4 = 124 1623 = 8. froot

But what happens when x is irrational? How should 2p be understood? Your calcula-

tor provides an approximation to 2p, but where does the approximation come from? For

now, we assume that bx can be defined for all real numbers x and that it can be approxi-

mated as closely as desired by using rational numbers as close to x as needed.

➤ Exponent Rules

For any base b 7 0 and real numbers

x and y, the following relations hold:

E1. bxby = bx+y

E2. bx

by = bx-y

awhich includes 1

by = b-ybE3. 1bx2y = bxy

E4. bx 7 0, for all x

90. Analyzing an algebraic function Consider the function

f 1x2 = x2 - x + 1

2x4 + 2.

a. What is the domain of f ?

b. Does the graph have symmetry with respect to the origin or y-axis?

c. Graph the function and experiment with various graphing win-

dows to be sure you have displayed all the interesting features

of the graph.

d. Find the roots (zeros) of f (approximately).

e. Using the graph of part (c), find the y-intercept of the graph of f .

f. Find the points (approximately) at which f has a local maxi-

mum or a local minimum value. Give the function value at

these points (see Exercise 88f for definitions).

g. Imagine moving along the curve in the positive x-direction.

Estimate the coordinates of the point at which the curve is

falling most rapidly. Estimate the coordinates of the point at

which the curve is rising most rapidly.

h. Describe the behavior of f as x becomes increasingly large and

positive. Describe the behavior of f as x becomes increasingly

large in magnitude and negative.

91. Peaks, valleys, and cusps Consider the function

f 1x2 = x2 � x2 - 1 �x4 + 1

.





of the graph.

d. Find the roots (zeros) of f .

e. At what point does the graph cross the y-axis?

T

T

f. A sharp point on a graph is called a cusp. At what points does

the graph of f have a cusp?

g. Approximate the points at which f has a local maximum or a

local minimum value. Give the function value at these points.

Be sure to include cusps (see Exercise 88f for definitions).

h. Describe the behavior of f as x becomes increasingly large and

positive. Describe the behavior of f as x becomes increasingly

large in magnitude and negative.

92. Analyzing an algebraic function Consider the function

f 1x2 = 62x2 - x - 12

x2 + 10.





of the graph.

d. Find the roots (zeros) of f .

e. Approximate the points at which f has a local maximum or a

local minimum value. Give the function value at these points

(see Exercise 88f for definitions).

f. Describe the behavior of f as x becomes increasingly large

and positive. Describe the behavior of f as x becomes

increasingly large in magnitude and negative.

T

QUICK CHECK ANSWERS

1. Yes; no 2. 1-∞ , ∞2, 30, ∞2 3. Domain and range are 1-∞ , ∞2. Domain and range are 30, ∞2. 4. Shift the graph

of f horizontally 4 units to the left.

➤

F 2013 2014 Only or Use in Pilot Program–

Copyright © 2014 Pearson Education, Inc.

1.3 Inverse, Exponential, and Logarithmic Functions 29

Properties of Exponential Functions f 1x 2 = bx

1. Because bx is defined for all real numbers, the domain of f is 5x: -∞ 6 x 6 ∞6 .

Because bx 7 0 for all values of x, the range of f is 5y: 0 6 y 6 ∞6 .

2. For all b 7 0, b0 = 1, and thus f 102 = 1.

3. If b 7 1, then f is an increasing function of x (Figure 1.45). For example, if b = 2,

then 2x 7 2y whenever x 7 y.

4. If 0 6 b 6 1, then f is a decreasing function of x. For example, if b = 12,

f 1x2 = a1

2b x

=1

2x = 2-x,

and because 2x increases with x, 2-x decreases with x (Figure 1.46).

QUICK CHECK 1 Is it possible to raise a positive number b to a power and obtain a nega-

tive number? Is it possible to obtain zero?

➤

20

15

10

5

21

y

x

y � 10x

y � 5x

y � 3x

y � 2x

Larger values of bproduce greater ratesof increase in bx if b � 1.

FIGURE 1.45

20

15

10

5

�2 �1 x

y

y � 0.1x

y � 0.5x

y � 0.9x

Smaller values of b producegreater rates of decreasein bx if 0 � b � 1.

FIGURE 1.46

QUICK CHECK 2 Explain why f 1x2 = 11>32x is a decreasing function.

➤

The Natural Exponential Function One of the bases used for exponential functions

is special. For reasons that will become evident in upcoming chapters, the special base is e,

one of the fundamental constants of mathematics. It is an irrational number with a value of

e = 2.718281828459 c.

➤ The notation e was proposed by the

Swiss mathematician Leonhard Euler

(pronounced oiler) (1707–1783).

The base e gives an exponential function that has the following valuable property.

As shown in Figure 1.47a, the graph of y = ex lies between the graphs of y = 2x and

y = 3x (because 2 6 e 6 3). At every point on the graph of y = ex, it is possible to draw

a tangent line (discussed in Chapters 2 and 3) that touches the graph only at that point.

The natural exponential function is the only exponential function with the property that

the slope of the tangent line at x = 0 is 1 (Figure 1.47b); thus, ex has both value and slope

equal to 1 at x = 0. This property—minor as it may seem—leads to many simplifications

when we do calculus with exponential functions.

DEFINITION The Natural Exponential Function

The natural exponential function is f 1x2 = ex, which has the base

e = 2.718281828459 c.




Inverse Functions

Consider the linear function f 1x2 = 2x, which takes any value of x and doubles it. The

function that reverses this process by taking any value of f 1x2 = 2x and mapping it back

to x is called the inverse function of f, denoted f -1. In this case, the inverse function is

f -11x2 = x>2. The effect of applying these two functions in succession looks like this:

x ¡ 2x ¡ x

We now generalize this idea.

➤ The notation f -1 for the inverse

can be confusing. The inverse is not

the reciprocal; that is, f -11x2 is not

1>f 1x2 = 1 f 1x22-1. We adopt the

common convention of using simply

inverse to mean inverse function.

QUICK CHECK 3 What is the inverse of f 1x2 = 13

x? What is the inverse of f 1x2 = x - 7?

➤

f f �1

f �1( f (x)) � x

x xy

y yx

f ( f �1(y)) � y

f �1 f

f

f �1

y � f (x)x

y is in the domain of f �1 and

y � f (x) is in the range of f.

x is in the domain of f and

x � f �1(y) is in the range of f �1.

FIGURE 1.48

DEFINITION Inverse Function

Given a function f, its inverse (if it exists) is a function f -1 such that whenever

y = f 1x2, then f -11y2 = x (Figure 1.48).

Because the inverse “undoes” the original function, if we start with a value of x, apply

f to it, and then apply f -1 to the result, we recover the original value of x; that is,

Similarly, if we apply f -1 to a value of y and then apply f to the result, we recover

the original value of y; that is,

One-to-One Functions We have defined the inverse of a function, but said nothing

about when it exists. To ensure that f has an inverse on a domain, f must be one-to-one on

that domain. This property means that every output of the function f must correspond to

1

1

y

x

1

1

y

x

y � ex

y � exy � 3x

y � 2x

Tangent line hasslope 1 at (0, 1).

(a) (b)

FIGURE 1.47

f f -1




exactly one input. The one-to-one property is checked graphically by using the horizontal line test.

For example, in Figure 1.50, some horizontal lines intersect the graph of f 1x2 = x2 twice.

Therefore, f does not have an inverse function on the interval 1-∞ , ∞2. However, if f is

restricted to the interval 1-∞ , 04 or 30, ∞2, then it does pass the horizontal line test and it

is one-to-one on these intervals.

DEFINITION One-to-One Functions and the Horizontal Line Test

A function f is one-to-one on a domain D if each value of f 1x2 corresponds to exactly

one value of x in D. More precisely, f is one-to-one on D if f 1x12 ≠ f 1x22 whenever

x1 ≠ x2, for x1 and x2 in D. The horizontal line test says that every horizontal line

intersects the graph of a one-to-one function at most once (Figure 1.49).

y

x

y

y

xx2

x1

x3

x1

One-to-one function:Each value of y in therange corresponds toat most one value of x.

Not one-to-one function:Some values of y correspondto more than one value of x.

y1

FIGURE 1.49

�4 �2 42

8

6

2

x

y

�4 �2 42

8

6

2

x

y

�4 �2 42

8

6

2

x

y

f (x) � x2 is 1–1on [0, �).

f (x) � x2 is 1–1on (��, 0].

f (x) � x2 is not1–1 on (��, �).

Domain: [0, �)Domain: (��, 0]

f fails thehorizontal line test.

Domain: (��, �)

FIGURE 1.50

EXAMPLE 1 One-to-one functions Determine the (largest possible) intervals on

which the function f 1x2 = 2x2 - x4 (Figure 1.51) is one-to-one.

SOLUTION The function is not one-to-one on the entire real line because it fails the hori-

zontal line test. However, on the intervals 1-∞ , -14 , 3-1, 04 , 30, 14 , and 31, ∞2, f

is one-to-one. The function is also one-to-one on any subinterval of these four intervals.

Related Exercises 11–14

➤

2

�1 1 x

y

y � 2x2 � x4

FIGURE 1.51




Existence of Inverse Functions Figure 1.52a illustrates the actions of a one-to-one

function f and its inverse f -1. We see that f maps a value of x to a unique value of y. In

turn, f -1 maps that value of y back to the original value of x. When f is not one-to-one,

this procedure cannot be carried out (Figure 1.52b).

(a) (b)

??

y

O

f maps x to y

y � f (x)(x, y)

(x, y)

x � f �1(y)

x � f �1(y)f �1 maps y to x

y � f (x)

x

Two values of xcorrespond to y.

x

y

x

FIGURE 1.52

THEOREM 1.1 Existence of Inverse FunctionsLet f be a one-to-one function on a domain D with a range R. Then f has a

unique inverse f -1 with domain R and range D such that

f -11 f 1x22 = x and f 1 f -11y22 = y,

where x is in D and y is in R.

EXAMPLE 2 Does an inverse exist? Determine intervals on which f 1x2 = x2 - 1

has an inverse function.

SOLUTION On the interval 1-∞ , ∞2 the function does not pass the horizontal line test

and is not one-to-one (Figure 1.53). However, if f is restricted to the intervals 1-∞ , 04 or 30, ∞2, it is one-to-one and an inverse exists.


➤

Finding Inverse Functions The crux of finding an inverse for a function f is

solving the equation y = f 1x2 for x in terms of y. If it is possible to do so, then we have

found a relationship of the form x = f -11y2. Interchanging x and y in x = f -11y2 so

that x is the independent variable (which is the customary role for x), the inverse has the

form y = f -11x2. Notice that if f is not one-to-one, this process leads to more than one

inverse function.

➤ The statement that a one-to-one function

has an inverse is plausible based on

its graph. However, the proof of this

theorem is fairly technical and is omitted.

➤ Once you find a formula for f -1, you

can check your work by verifying that

f -11 f 1x22 = x and f 1 f -11x22 = x.

QUICK CHECK 4 The function that gives degrees Fahrenheit in terms of degrees Celsius is

F = 9C>5 + 32. Explain why this function has an inverse.

➤

1

21 x

y

y � x2 � 1

f is one-to-oneand has an inverseon (��, 0].

f is one-to-oneand has an inverseon [0, �).

FIGURE 1.53

PROCEDURE Finding an Inverse Function

Suppose f is one-to-one on an interval I. To find f -1:

1. Solve y = f 1x2 for x. If necessary, choose the function that corresponds to I.

2. Interchange x and y and write y = f -11x2.




EXAMPLE 3 Finding inverse functions Find the inverse(s) of the following func-

tions. Restrict the domain of f if necessary.

a. f 1x2 = 2x + 6 b. f 1x2 = x2 - 1

SOLUTION

a. Linear functions (except constant linear functions) are one-to-one on the entire real

line. Therefore, an inverse function for f exists for all values of x.

Step 1: Solve y = f 1x2 for x: We see that y = 2x + 6 implies that 2x = y - 6, or

x = 1y - 62>2.

Step 2: Interchange x and y and write y = f -11x2:y = f -11x2 = x - 6

2.

It is instructive to verify that the inverse relations f 1 f -11x22 = x and

f -11 f 1x22 = x are satisfied:

f 1 f -11x22 = f ax - 6

2b = 2 ax - 6

2b + 6 = x - 6 + 6 = x,

u

f 1x2 = 2x + 6

f -11 f 1x22 = f -112x + 62 = 12x + 62 - 6

2= x.

u

f -11x2 = 1x - 62>2b. As shown in Example 2, the function f 1x2 = x2 - 1 is not one-to-one on the entire

real line; however, it is one-to-one on 1-∞ , 04 and on 30, ∞2. If we restrict our atten-

tion to either of these intervals, then an inverse function can be found.

Step 1: Solve y = f 1x2 for x:

y = x2 - 1

x2 = y + 1

x = e 1y + 1

-1y + 1.

Each branch of the square root corresponds to an inverse function.

Step 2: Interchange x and y and write y = f -11x2:y = f -11x2 = 1x + 1 or y = f -11x2 = -1x + 1.

The interpretation of this result is important. Taking the positive branch

of the square root, the inverse function y = f -11x2 = 1x + 1 gives

positive values of y; it corresponds to the branch of f 1x2 = x2 - 1 on

the interval 30, ∞2 (Figure 1.54). The negative branch of the square root,

y = f -11x2 = -1x + 1, is another inverse function that gives negative

values of y; it corresponds to the branch of f 1x2 = x2 - 1 on the interval 1-∞ , 04 . Related Exercises 21–30

➤

QUICK CHECK 5 On what interval(s) does the function f 1x2 = x3 have an inverse?

➤

➤ A constant function (whose graph is a

horizontal line) fails the horizontal line

test and does not have an inverse.

1

1

y

x

y � f �1(x) � x � 1

y � f �1(x) � � x � 1

Inverse functionsfor f (x) � x2 �1

FIGURE 1.54




Graphing Inverse Functions

The graphs of a function and its inverse have a special relationship, which is illustrated in

the following example.

EXAMPLE 4 Graphing inverse functions Plot f and f -1 on the same coordinate axes.

a. f 1x2 = 2x + 6 b. f 1x2 = 1x - 1

SOLUTION

a. The inverse of f 1x2 = 2x + 6, found in Example 3, is

y = f -11x2 = x - 6

2=

x

2- 3.

The graphs of f and f -1 are shown in Figure 1.55. Notice that both f and f -1 are in-

creasing linear functions and they intersect at 1-6, -62.b. The domain of f 1x2 = 1x - 1 is 31, ∞2 and its range is 30, ∞2. On this domain f is

one-to-one and has an inverse. It can be found in two steps:

Step 1: Solve y = 1x - 1 for x:

y2 = x - 1 or x = y2 + 1.

Step 2: Interchange x and y and write y = f -11x2:y = f -11x2 = x2 + 1.

The graphs of f and f -1 are shown in Figure 1.56; notice that the domain of f -1 1x Ú 02 corresponds to the range of f 1y Ú 02. Related Exercises 31–40

➤Looking closely at the graphs in Figure 1.55 and Figure 1.56, you see a symmetry that

always occurs when a function and its inverse are plotted on the same set of axes. In each

figure, one curve is the reflection of the other curve across the line y = x. These curves

have symmetry about the line y = x, which means that the point 1a, b2 is on one curve

whenever the point 1b, a2 is on the other curve (Figure 1.57).

The explanation for the symmetry comes directly from the definition of the inverse.

Suppose that the point 1a, b2 is on the graph of y = f 1x2, which means that b = f 1a2. By

the definition of the inverse function, we know that a = f -11b2, which means that the

point 1b, a2 is on the graph of y = f -11x2. This argument applies to all relevant points

1a, b2, so whenever 1a, b2 is on the graph of f, 1b, a2 is on the graph of f -1. As a conse-

quence, the graphs are symmetric about the line y = x.

Logarithmic Functions

Everything we learned about inverse functions is now applied to the exponential func-

tion f 1x2 = bx. For any b 7 0, with b ≠ 1, this function is one-to-one on the interval 1-∞ , ∞2. Therefore, it has an inverse.

2

2

y

x

2x

The function f(x) � 2x � 6 and its

� 3 areinverse

f (x) � 2x � 6

y � x

x2

� 3

symmetric about the line y � x.

f �1(x) �

f �1(x) �

FIGURE 1.55

1

1

y

x

f �1(x) � x2 � 1 (x � 0)

y � x

The function f(x) � x � 1 (x � 1)

and its inverse f �1(x) � x2 � 1 (x � 0)

are symmetric about y � x.

f (x) � x � 1

FIGURE 1.56

y

x

Symmetry about y � x means…

y � x

(a, b)

(b, a)

y � f (x)

y � f �1(x)

…then (b, a) ison the graph of f �1.

if (a, b) is on thegraph of f, …

FIGURE 1.57

DEFINITION Logarithmic Function Base b

For any base b 7 0, with b ≠ 1, the logarithmic function base b, denoted y = logb x,

is the inverse of the exponential function y = bx. The inverse of the natural exponential

function with base b = e is the natural logarithm function, denoted y = ln x.

The inverse relationship between logarithmic and exponential functions may be stated

concisely in several ways. First, we have

y = logb x if and only if by = x.




Combining these two conditions results in two important relations.➤ Logarithms were invented around 1600

for calculating purposes by the Scotsman

John Napier and the Englishman

Henry Briggs. Unfortunately, the word

logarithm, derived from the Greek

for reasoning (logos) with numbers

(arithmos), doesn’t help with the

meaning of the word. When you see

logarithm, you should think exponent.

➤ Logarithm Rules

For any base b 7 0 1b ≠ 12, positive

real numbers x and y, and real numbers z, the following relations hold:

L1. logb 1xy2 = logb x + logb y

L2. logb ax

yb = logb x - logb y

aincludes logb 1

y= - logb yb

L3. logb 1xz2 = z logb x

L4. logb b = 1

Inverse Relations for Exponential and Logarithmic Functions

For any base b 7 0, with b ≠ 1, the following inverse relations hold.

I1. blogb x = x, for x 7 0

I2. logb bx = x, for real values of x

Properties of Logarithmic Functions The graph of the logarithmic function is gen-

erated using the symmetry of the graphs of a function and its inverse. Figure 1.58 shows

how the graph of y = bx, for b 7 1, is reflected across the line y = x to obtain the graph

of y = logb x.

The graphs of y = logb x are shown (Figure 1.59) for several bases b 7 1. Logarithms

with bases 0 6 b 6 1, although well defined, are generally not used (and they can be ex-

pressed in terms of bases with b 7 1).

y

x

Graphs of bx and logb x aresymmetric about y � x.

(0, 1)

(1, 0)

y � bx, b � 1

y � x

y � logb x

FIGURE 1.58

1

10

y

x

logb 1 � 0 for anybase b � 0, b � 1.

y � log2

x

y � log5

x

y � log10

x

logb x increases on theinterval x � 0 when b � 1.

y � ln x

FIGURE 1.59

Logarithmic functions with base b 7 0 satisfy properties that parallel the properties

of the exponential functions given earlier.

1. Because the range of bx is 5y: 0 6 y 6 ∞6 , the domain of logb x is 5x: 0 6 x 6 ∞6 .

2. The domain of bx is all real numbers, which implies that the range of logb x is all real

numbers.

3. Because b0 = 1, it follows that logb 1 = 0.

4. If b 7 1, then logb x is an increasing function of x. For example, if b = e, then

ln x 7 ln y whenever x 7 y (Figure 1.59).

EXAMPLE 5 Using inverse relations One thousand grams of a particular radioactive

substance decays according to the function m1t2 = 1000e-t>850, where t Ú 0 measures

time in years. When does the mass of the substance reach the safe level deemed to be 1 g?

SOLUTION Setting m1t2 = 1, we solve 1000e-t>850 = 1 by dividing both sides by 1000

and taking the natural logarithm of both sides:

ln 1e-t>8502 = ln a 1

1000b .

➤ Provided the arguments are positive,

we can take the logb of both sides of

an equation and produce an equivalent

equation.

QUICK CHECK 6 What is the domain of

f 1x2 = logb 1x22? What is the range

of f 1x2 = logb 1x22? ➤




This equation is simplified by calculating ln 11>10002 ≈ -6.908 and observing that

ln 1e-t>8502 = -t

850 (inverse property I2). Therefore,

-t

850≈ -6.908.

Solving for t, we find that t ≈ 1-85021-6.9082 ≈ 5872 years.


➤

Change of Base

When working with logarithms and exponentials, it doesn’t matter in principle which base

is used. However, there are practical reasons for switching between bases. For example,

most calculators have built-in logarithmic functions in just one or two bases. If you need

to use a different base, then the change-of-base rules are essential.

Consider changing bases with exponential functions. Specifically, suppose you wish

to express bx (base b) in the form ey (base e), where y must be determined. Taking the

natural logarithm of both sides of ey = bx, we have

ln ey = ln bx which implies that y = x ln b. e e

y x ln b

It follows that bx = ey = ex ln b. For example, 4x = ex ln 4.

The formula for changing from logb x to ln x is derived in a similar way. We let

y = logb x, which implies that x = by. Taking the natural logarithm of both sides of

x = by gives ln x = ln by = y ln b. Solving for y = logb x gives the required formula:

y = logb x =ln x

ln b.

➤ A similar argument is used to derive

more general formulas for changing from

base b to any other positive base c.

Change-of-Base Rules

Let b be a positive real number with b ≠ 1. Then

bx = ex ln b, for all x and logb x =ln x

ln b, for x 7 0.

More generally, if c is a positive real number with c ≠ 1, then

bx = cx logc b, for all x and logb x =logc x

logc b, for x 7 0.

EXAMPLE 6 Changing bases

a. Express 2x+4 as an exponential with base e.

b. Express log2 x using base e and base 32.

SOLUTION

a. Using the change-of-base rule for exponentials, we have

2x+4 = e1x+42ln 2.

b. Using the change-of-base rule for logarithms, we have

log2 x =ln x

ln 2≈ 1.44 ln x.




To change from base 2 to base 32, we use the general change-of-base formula:

log2 x =log32 x

log32 2=

log32 x

1>5 = 5 log32 x.

The middle step follows from the fact that 2 = 321>5, so log32 2 = 15.


➤

1. For b 7 0, what are the domain and range of f 1x2 = bx?

2. Give an example of a function that is one-to-one on the entire real

number line.

3. Explain why a function that is not one-to-one on an interval I cannot have an inverse function on I.

4. Explain with pictures why 1a, b2 is on the graph of f whenever

1b, a2 is on the graph of f -1.

5. Sketch a function that is one-to-one and positive for x Ú 0. Make

a rough sketch of its inverse.

6. Express the inverse of f 1x2 = 3x - 4 in the form y = f -11x2.7. Explain the meaning of logb x.

8. How is the property bx+y = bxby related to the property

logb 1xy2 = logb x + logb y?

9. For b 7 0 with b ≠ 1, what are the domain and range of

f 1x2 = logb x and why?

10. Express 25 using base e.

Basic Skills11–14. One-to-one functions

11. Find three intervals on which f is one-to-one, making each inter-

val as large as possible.

�3 �2 �1 321

y

x

y � f (x)

SECTION 1.3 EXERCISESReview Questions

12. Find four intervals on which f is one-to-one, making each interval

as large as possible.

�4 �2 42

y

x

y � f (x)

13. Sketch a graph of a function that is one-to-one on the interval

1-∞ , 02, but is not one-to-one on 1-∞ , ∞2. 14. Sketch a graph of a function that is one-to-one on the intervals

1-∞ , -22, and 1-2, ∞2 but is not one-to-one on 1-∞ , ∞2. 15–20. Where do inverses exist? Use analytical and/or graphical methods to determine the largest possible set of points on which the following functions have an inverse.

15. f 1x2 = 3x + 4

16. f 1x2 = �2x + 1 �

17. f 1x2 = 1>1x - 52 18. f 1x2 = -16 - x2219. f 1x2 = 1>x2

20. f 1x2 = x2 - 2x + 8 (Hint: Complete the square.)

21–28. Finding inverse functions

a. Find the inverse of each function (on the given interval, if specified) and write it in the form y = f -11x2.

b. Verify the relationships f 1 f -11x22 = x and f -11 f 1x22 = x.

21. f 1x2 = 2x

22. f 1x2 = x>4 + 1




23. f 1x2 = 6 - 4x

24. f 1x2 = 3x3

25. f 1x2 = 3x + 5

26. f 1x2 = x2 + 4, for x Ú 0

27. f 1x2 = 1x + 2, for x Ú -2

28. f 1x2 = 2>1x2 + 12, for x Ú 0

29. Splitting up curves The unit circle x2 + y2 = 1 consists of four

one-to-one functions, f11x2, f21x2, f31x2, and f41x2 (see figure).

a. Find the domain and a formula for each function.

b. Find the inverse of each function and write it as y = f -11x2.

1

�1

�1 1

y

x

y � f1(x)

y � f4(x)

y � f2(x)

y � f3(x)

30. Splitting up curves The equation y4 = 4x2 is associated with four

one-to-one functions f11x2, f21x2, f31x2, and f41x2 (see figure).

a. Find the domain and a formula for each function.

b. Find the inverse of each function and write it as y = f -11x2.

1

�1

�1 1

y

x

y � f1(x)

y � f4(x)

y � f2(x)

y � f3(x)

31–38. Graphing inverse functions Find the inverse function (on the given interval, if specified) and graph both f and f -1 on the same set of axes. Check your work by looking for the required symmetry in the graphs.

31. f 1x2 = 8 - 4x

32. f 1x2 = 4x - 12

33. f 1x2 = 1x, for x Ú 0

34. f 1x2 = 13 - x, for x … 3

35. f 1x2 = x4 + 4, for x Ú 0

36. f 1x2 = 6>1x2 - 92, for x 7 3

37. f 1x2 = x2 - 2x + 6, for x Ú 1 (Hint: Complete the square.)

38. f 1x2 = -x2 - 4x - 3, for x … -2 (Hint: Complete the square.)

39–40. Graphs of inverses Sketch the graph of the inverse function.

39.

1

1

y

x

y � x

y � f (x)

40.

1

�1

�1 1

y

x

y � x

y � f (x)

41–46. Solving logarithmic equations Solve the following equations.

41. log10 x = 3 42. log5 x = -1

43. log8 x = 13 44. logb 125 = 3

45. ln x = -1 46. ln y = 3

47–52. Properties of logarithms Assume logb x = 0.36, logb y = 0.56, and logb z = 0.83. Evaluate the following expressions.

47. logb x

y 48. logb x2

49. logb xz 50. logb 1xy

z

51. logb 1x

23 z 52. logb

b2x5>21y

53–56. Solving equations Solve the following equations.

53. 7x = 21 54. 2x = 55

55. 33x-4 = 15 56. 53x = 29

T




57. Using inverse relations One hundred grams of a particular radioac-

tive substance decays according to the function m1t2 = 100 e-t>650,

where t 7 0 measures time in years. When does the mass reach

50 grams?

58. Using inverse relations The population P of a small town is

growing according to the function P1t2 = 100 et>50, where t mea-

sures the number of years after 2010. How long does it take the

population to double?

59–62. Calculator base change Write the following logarithms in terms of the natural logarithm. Then use a calculator to find the value of the logarithm, rounding your result to three decimal places.

59. log2 15 60. log3 30 61. log4 40 62. log6 60

63–68. Changing bases Convert the following expressions to the indi-cated base.

63. 2x using base e

64. 3sin x using base e

65. ln � x � using base 5

66. log2 1x2 + 12 using base e

67. a1>ln a using base e, for a 7 0 and a ≠ 1

68. a1>log a using base 10, for a 7 0 and a ≠ 1

Further Explorations69. Explain why or why not Determine whether the following state-

ments are true and give an explanation or counterexample.

a. If y = 3x, then x = 13 y.

b. logb x

logb y= logb x - logb y

c. log5 46 = 4 log5 6

d. 2 = 10log10 2

e. 2 = ln 2e

f. If f 1x2 = x2 + 1, then f -11x2 = 1>1x2 + 12. g. If f 1x2 = 1>x, then f -11x2 = 1>x.

70. Graphs of exponential functions The following figure shows

the graphs of y = 2x, y = 3x, y = 2-x, and y = 3-x. Match each

curve with the correct function.

1

�1 1

A B C D

y

x

T

T

T

71. Graphs of logarithmic functions The following figure shows the

graphs of y = log2 x, y = log4 x, and y = log10 x. Match each

curve with the correct function.

1

�1

1

y

x

A

B

C

72. Graphs of modified exponential functions Without using a

graphing utility, sketch the graph of y = 2x. Then, on the same set

of axes, sketch the graphs of y = 2-x, y = 2x-1, y = 2x + 1, and

y = 22x.

73. Graphs of modified logarithmic functions Without using a

graphing utility, sketch the graph of y = log2 x. Then, on the same

set of axes, sketch the graphs of y = log2 1x - 12, y = log2 x2,

y = 1log2 x22, and y = log2 x + 1.

74. Large intersection point Use any means to approximate the

intersection point(s) of the graphs of f 1x2 = ex and g1x2 = x123.

(Hint: Consider using logarithms.)

75–78. Finding all inverses Find all the inverses associated with the following functions and state their domains.

75. f 1x2 = 1x + 123 76. f 1x2 = 1x - 422

77. f 1x2 = 2>1x2 + 22 78. f 1x2 = 2x>1x + 22 Applications79. Population model A culture of bacteria has a population of

150 cells when it is first observed. The population doubles every

12 hr, which means its population is governed by the function

p1t2 = 150 * 2t>12, where t is the number of hours after the first

observation.

a. Verify that p102 = 150, as claimed.

b. Show that the population doubles every 12 hr, as claimed.

c. What is the population 4 days after the first observation?

d. How long does it take the population to triple in size?

e. How long does it take the population to reach 10,000?

80. Charging a capacitor A capacitor is a device that stores electrical

charge. The charge on a capacitor accumulates according to the

function Q1t2 = a11 - e-t>c2, where t is measured in seconds,

and a and c 7 0 are physical constants. The steady-state charge

is the value that Q1t2 approaches as t becomes large.

a. Graph the charge function for t Ú 0 using a = 1 and c = 10.

Find a graphing window that shows the full range of the func-

tion.

b. Vary the value of a holding c fixed. Describe the effect on the

curve. How does the steady-state charge vary with a?

T

T




c. Vary the value of c holding a fixed. Describe the effect on the

curve. How does the steady-state charge vary with c?

d. Find a formula that gives the steady-state charge in terms of a

and c.

81. Height and time The height of a baseball hit straight up from the

ground with an initial velocity of 64 ft>s is given by h = f 1t2 =64t - 16t2, where t is measured in seconds after the hit.

a. Is this function one-to-one on the interval 0 … t … 4?

b. Find the inverse function that gives the time t at which the ball

is at height h as the ball travels upward. Express your answer

in the form t = f -1 1h2. c. Find the inverse function that gives the time t at which the ball

is at height h as the ball travels downward. Express your an-

swer in the form t = f -1 1h2. d. At what time is the ball at a height of 30 ft on the way up?

e. At what time is the ball at a height of 10 ft on the way down?

82. Velocity of a skydiver The velocity of a skydiver 1in m>s2 t sec-

onds after jumping from the plane is v1t2 = 60011 - e-kt>602>k,

where k 7 0 is a constant. The terminal velocity of the skydiver

is the value that v1t2 approaches as t becomes large. Graph v with

k = 11 and estimate the terminal velocity.

Additional Exercises83. Reciprocal bases Assume that b 7 0 and b ≠ 1. Show that

log1>b x = - logb x.

84. Proof of rule L1 Use the following steps to prove that

logb 1xy2 = logb x + logb y.

a. Let x = bp and y = bq. Solve these expressions for p and q,

respectively.

b. Use property E1 for exponents to express xy in terms of b, p,

and q.

c. Compute logb 1xy2 and simplify.

85. Proof of rule L2 Modify Exercise 84 and use property E2 for

exponents to prove that logb 1x>y2 = logb x - logb y.

86. Proof of rule L3 Use the following steps to prove that

logb 1xy2 = y logb x.

a. Let x = bp. Solve this expression for p.

b. Use property E3 for exponents to express xy in terms of b

and p.

c. Compute logb xy and simplify.

87. Inverses of a quartic Consider the quartic polynomial

y = f 1x2 = x4 - x2.

a. Graph f and estimate the largest intervals on which it is one-

to-one. The goal is to find the inverse function on each of these

intervals.

b. Make the substitution u = x2 to solve the equation y = f 1x2 for x

in terms of y. Be sure you have included all possible solutions.

c. Write each inverse function in the form y = f -11x2 for each of

the intervals found in part (a).

88. Inverse of composite functions

a. Let g1x2 = 2x + 3 and h1x2 = x3. Consider the composite

function f 1x2 = g1h1x22. Find f -1 directly and then express it

in terms of g-1 and h-1.

T

T

T

b. Let g1x2 = x2 + 1 and h1x2 = 1x. Consider the composite

function f 1x2 = g1h1x22. Find f -1 directly and then express it

in terms of g-1 and h-1.

c. Explain why if g and h are one-to-one, the inverse of

f 1x2 = g1h1x22 exists.

89–90. Inverses of (some) cubics Finding the inverse of a cubic polynomial is equivalent to solving a cubic equation. A special case that is simpler than the general case is the cubic y = f 1x2 = x3 + ax. Find the inverse of the following cubics using the substitution (known as Vieta’s substitution) x = z - a>13z2. Be sure to determine where the function is one-to-one.

89. f 1x2 = x3 + 2x 90. f 1x2 = x3 - 2x

91. Nice property Prove that 1logb c21logc b2 = 1, for b 7 0,

c 7 0, b ≠ 1, and c ≠ 1.

Technology Exercises92. Finding roots Consider the function f 1x2 = ln x - x2 - 2x + 4.


b. Graph f with a graphing window that shows all the interesting

features of the graph.

c. Use the graph of part (b) to estimate the roots of f .

93. Points of intersection

a. Graph the curves y = ln x and y = 0.3x, and approximate the

coordinates of their intersection points.

b. Graph the curves y = ln x and y = 0.4x and approximate the


c. Estimate the value of a such that the curves y = ln x and

y = ax intersect exactly once. What are the coordinates of the

intersection point?

94. Points of intersection

a. Graph the curves y = ex>2 and y = 2x, and approximate the


b. Graph the curves y = ex>2 and y = x and approximate the


c. Estimate the value of a such that the curves y = ex>2 and

y = ax intersect exactly once. What are the coordinates of the

intersection point?

95. Threshold point Use any means to approximate as closely as

possible the value of A that makes the following statement true.

The curves y = ex and y = ln ax have two intersection points for

a 7 A, and they have no intersection points for a 6 A.

T

T

T

T

T

QUICK CHECK ANSWERS

1. bx is always positive (and never zero) for all x and for positive

bases b. 2. Because 11>32x = 1>3x and 3x increases as x

increases, it follows that 11>32x decreases as x increases.

3. f -11x2 = 3x; f -11x2 = x + 7. 4. For every Fahrenheit

temperature, there is exactly one Celsius temperature, and

vice versa. The given relation is also a linear function. It is

one-to-one, so it has an inverse function. 5. The function

f 1x2 = x3 is one-to-one on 1-∞ , ∞2, so it has an inverse

for all values of x. 6. The domain of logb 1x22 is all real

numbers except zero (because x2 is positive for x ≠ 0). The

range of logb 1x22 is all real numbers.

➤



1.4 Trigonometric Functions and Their Inverses 41

denoted s, divided by the radius of the circle r (Figure 1.60a). Working on a unit circle 1r = 12, the radian measure of an angle is simply the length of the arc associated with u

(Figure 1.60b). For example, the length of a full unit circle is 2p; therefore, an angle with

a radian measure of p corresponds to a half circle 1u = 180°2 and an angle with a radian

measure of p>2 corresponds to a quarter circle 1u = 90°2.

Degrees Radians

0 0

30 p>645 p>460 p>390 p>2

120 2p>3135 3p>4150 5p>6180 p

QUICK CHECK 1 What is the radian measure of a 270° angle? What is the degree measure

of a 5p>4-rad angle?

➤

Trigonometric FunctionsFor acute angles, the trigonometric functions are defined as ratios of the sides of a right

triangle (Figure 1.61). To extend these definitions to include all angles, we work in an

xy-coordinate system with a circle of radius r centered at the origin. Suppose that P1x, y2 is a point on the circle. An angle u is in standard position if its initial side is on the

positive x-axis and its terminal side is the line segment OP between the origin and P. An

angle is positive if it is obtained by a counterclockwise rotation from the positive x-axis

(Figure 1.62). When the right-triangle definitions of Figure 1.61 are used with the right

triangle in Figure 1.62, the trigonometric functions may be expressed in terms of x, y, and

the radius of the circle, r = 2x2 + y2.

sr

y

x

On a circle of radius r,

radian measure of is .sr

�

�

(a)

FIGURE 1.60

On a circle of radius 1,

radian measure of is s.

s1

y

x

(b)

�

�

This section is a review of what you need to know in order to study the calculus of trigo-

nometric functions. Once the trigonometric functions are on stage, it makes sense to pre-

sent the inverse trigonometric functions and their basic properties.

Radian MeasureCalculus typically requires that angles be measured in radians (rad). Working with a cir-

cle of radius r, the radian measure of an angle u is the length of the arc associated with u,

Why is it important to use radians? If we divide a full circle into 360 equal parts,

we get degrees. If we divide a full circle into 400 equal parts, we get gradians. In fact,

we could choose any fraction of a full circle as a basis for angle measurement, but the

resulting unit would be arbitrary. Said differently, there is no geometrical reason that 90

(degrees) should be associated with a right angle.

We use the radian because it is the natural unit for measuring angles. It is the ratio of two

lengths that can be measured: the radius of a circle and the length of the arc associated with

the angle. Another less obvious advantage of using radians is that if we work on a unit circle

(radius equal to 1), then the radian measure of the angle equals the length of the correspond-

ing arc. The fact that angle equals arc length on a unit circle is important in upcoming topics.

1.4 Trigonometric Functions and Their Inverses




To find the trigonometric functions of the standard angles (multiples of 30° and 45°), it is helpful to know the radian measure of those angles and the coordinates of the associ-

ated points on the unit circle (Figure 1.63).

➤ When working on a unit circle 1r = 12, these definitions become

sin u = y cos u = x

tan u =y

x cot u =

x

y

sec u =1

x csc u =

1

y

➤ Standard triangles

DEFINITION Trigonometric Functions

Let P1x, y2 be a point on a circle of radius r associated with the angle u. Then

sin u =y

r, cos u =

xr, tan u =

y

x,

cot u =xy, sec u =

rx, csc u =

ry.

45�

1

145�

60�

30�

22

12

22

32

�

90�

� /2( )2 2

,�

180� �

(0, �1)

(0, 1)

(1, 0)(�1, 0)

270�

� 3

/2

0� � 0 radians

360� � 2

( )2 2,

330� � 11 /6

( )2 2

2

1

, �( )2 2,��

225�

� 5

/4210� �

7/6

2 )(� ,

( 2� , )

( 2� , � )

2 )�(� , 2 )�( ,

( 2

2 )( ,

( 2, )

�

�

�

�315� �

7 /4�

300� � 5 /3

�

�

60�

�

/3�45

� �

/4�

30� �

/6�

�

�

150� � 5 /6�

135� � 3 /4

�120� �

2 /3�

240�

� 4

/3

�

12

12

12

12

12

, � )12

12

FIGURE 1.63

�

Hypotenuse (H)

Adjacent side (A)

Oppositeside (O)

sin u =O

H cos u =

A

H

tan u =O

A cot u =

A

O

sec u =H

A csc u =

H

OFIGURE 1.61

y

�

x

P(x, y)

r � x2 � y2

y

x

A positive angle � results froma counterclockwise rotation.

O

FIGURE 1.62




b. The angle u = -11p>3 = -2p - 5p>3 corresponds to a clockwise revolution of

one full circle 12p rad2 plus an additional 5p>3 rad (Figure 1.65). Therefore, this

angle has the same terminal side as the angle p>3. The coordinates of the correspond-

ing point on the unit circle are 11>2, 13>22, so csc 1-11p>32 = 1>y = 2>13.


➤Trigonometric Identities

Reciprocal Identities

tan u =sin u

cos u cot u =

1

tan u=

cos u

sin u

csc u =1

sin u sec u =

1

cos u

Pythagorean Identities

sin2 u + cos2 u = 1 1 + cot2 u = csc2 u tan2 u + 1 = sec2 u

Double- and Half-Angle Formulas

sin 2u = 2 sin u cos u cos 2u = cos2 u - sin2 u

cos2 u =1 + cos 2u

2 sin2 u =

1 - cos 2u

2

Combining the definitions of the trigonometric functions with the coordinates shown in

Figure 1.63, we may evaluate these functions at any standard angle. For example,

sin 2p

3=13

2, cos

5p

6= -

13

2, tan

7p

6=

1

13, tan

3p

2 is undefined,

cot 5p

3= -

1

13, sec

7p

4= 12, csc

3p

2= -1, sec

p

2 is undefined.

EXAMPLE 1 Evaluating trigonometric functions Evaluate the following

expressions.

a. sin 18p>32 b. csc 1-11p>32SOLUTION

a. The angle 8p>3 = 2p + 2p>3 corresponds to a counterclockwise revolution of one

full circle 12p rad2 plus an additional 2p>3 rad (Figure 1.64). Therefore, this angle

has the same terminal side as the angle 2p>3, and the corresponding point on the unit

circle is 1-1>2, 13>22. It follows that sin 18p>32 = y = 13>2.

� �83�

2 )(� ,12

FIGURE 1.64

�2

� �3

311�

12

2 )( ,12

FIGURE 1.65 QUICK CHECK 2 Evaluate cos 111p>62 and sin 15p>42. ➤

Trigonometric IdentitiesTrigonometric functions have a variety of properties, called identities, that are true for all

angles in the domain. Here is a list of some commonly used identities.

QUICK CHECK 3 Use sin2 u + cos2 u = 1 to prove that 1 + cot2 u = csc2 u.

➤

EXAMPLE 2 Solving trigonometric equations Solve the following equations.

a. 12 sin x + 1 = 0 b. cos 2x = sin 2x, where 0 … x 6 2p.




Graphs of the Trigonometric FunctionsTrigonometric functions are examples of periodic functions: Their values repeat over every

interval of some fixed length. A function f is said to be periodic if f 1x + P2 = f 1x2, for all x

in the domain, where the period P is the smallest positive real number that has this property.

Period of Trigonometric Functions

The functions sin u, cos u, sec u, and csc u have a period of 2p:

sin 1u + 2p2 = sin u cos 1u + 2p2 = cos u

sec 1u + 2p2 = sec u csc 1u + 2p2 = csc u,

for all u in the domain.

The functions tan u and cot u have a period of p:

tan 1u + p2 = tan u cot 1u + p2 = cot u,

for all u in the domain.

The graph of y = sin u is shown in Figure 1.66a. Because csc u = 1>sin u, these

two functions have the same sign, but y = csc u is undefined with vertical asymptotes

at u = 0, {p, {2p, c. The functions cos u and sec u have a similar relationship

(Figure 1.66b).

�

�

�

��

3� 4��

1

y

1

�

y

The graphs of y � sin and its reciprocal, y � csc ��The graphs of y � cos and its reciprocal, y � sec

y � sin

y � csc

�

�

y � cos

y � sec

2� 7�

25�2

3�2

2 3� 4��2� 7�

25�2

3�2

2

FIGURE 1.66 (a) (b)

SOLUTION

a. First, we solve for sin x to obtain sin x = -1>12 = -12>2. From the unit circle

(Figure 1.63), we find that sin x = -12>2 if x = 5p>4 or x = 7p>4. Adding integer

multiples of 2p produces additional solutions. Therefore, the set of all solutions is

x =5p

4+ 2np and x =

7p

4+ 2np, for n = 0, {1, {2, {3, c.

b. Dividing both sides of the equation by cos 2x (assuming cos 2x ≠ 0), we obtain

tan 2x = 1. Letting u = 2x gives us the equivalent equation tan u = 1. This equation

is satisfied by

u =p

4,

5p

4,

9p

4,

13p

4,

17p

4, c.

Dividing by two and using the restriction 0 … x 6 2p gives the solutions

x =u

2=p

8,

5p

8,

9p

8, and

13p

8.


➤

➤ By rationalizing the denominator,

observe that 1

12=

1

12# 12

12=12

2.

➤ Notice that the assumption cos 2x ≠ 0 is

valid for these values of x.

The graphs of tan u and cot u are shown in Figure 1.67. Each function has points, sepa-

rated by p units, at which it is undefined.




Transforming GraphsMany physical phenomena, such as the motion of waves or the rising and setting of the

sun, can be modeled using trigonometric functions; the sine and cosine functions are

especially useful. With the transformation methods introduced in Section 1.2, we can

show that the functions

y = A sin 1B1u - C22 + D and y = A cos 1B1u - C22 + D,

when compared to the graphs of y = sin u and y = cos u, have a vertical stretch (or

amplitude) of �A � , a period of 2p> �B � , a horizontal shift (or phase shift) of C, and a

vertical shift of D (Figure 1.68).

For example, at latitude 40° north (Beijing, Madrid, Philadelphia) there are 12 hours

of daylight on the equinoxes (approximately March 21 and September 21), with a maxi-

mum of 14.8 hours of daylight on the summer solstice (approximately June 21) and a

minimum of 9.2 hours of daylight on the winter solstice (approximately December 21).

Using this information, it can be shown that the function

D1t2 = 2.8 sin a 2p

365 1t - 812b + 12

models the number of daylight hours t days after January 1 (Figure 1.69;

Exercise 100). The graph of this function is obtained from the graph of

y = sin t by (1) a horizontal scaling by a factor of 2p>365, (2) a horizontal

shift of 81, (3) a vertical scaling by a factor of 2.8, and (4) a vertical shift of 12.

Inverse Trigonometric FunctionsThe notion of inverse functions led from exponential functions to loga-

rithmic functions (Section 1.3). We now carry out a similar procedure—

this time with trigonometric functions.

Inverse Sine and Cosine Our goal is to develop the inverses of the

sine and cosine in detail. The inverses of the other four trigonometric functions then follow

in an analogous way. So far, we have asked this question: Given an angle x, what is sin x or

cos x? Now we ask the opposite question: Given a number y, what is the angle x such that

sin x = y? Or, what is the angle x such that cos x = y? These are inverse questions.

There are a few things to notice right away. First, these questions don’t make sense

if � y � 7 1, because -1 … sin x … 1 and -1 … cos x … 1. Next, let’s select an accept-

able value of y, say y = 12, and find the angle x that satisfies sin x = y = 1

2. It is appar-

ent that infinitely many angles satisfy sin x = 12; all angles of the form p>6 { 2np and

5p>6 { 2np, where n is an integer, answer the inverse question (Figure 1.70). A similar

situation occurs with the cosine function.

These inverse questions do not have unique answers because sin x and cos x are not

one-to-one on their domains. To define their inverses, these functions must be restricted

to intervals on which they are one-to-one. For the sine function, the standard choice is

Amplitude |A|

uO

y

y � A sin(B(u � C)) � D

Horizontal shift C

Vertical shift D 2p|B|

D � A

D

D � A

Period

FIGURE 1.68

15

12

9

6

3

356Dec21

265Sep 21

173June 21

81Mar 21Jan 1

y (hours)

0

Daylight function gives lengthof day throughout the year.

y � 12

2p365

(t � 81)) � 12

t (days)

D(t) � 2.8 sin(

FIGURE 1.69

1

5 x

y

Infinitely many values of x satisfy sin x � .

y � sin x

y �

12

12

FIGURE 1.70

�

�

3� 4��

�

2� 7�

25�2

3�2 222

2

1

y

1

3� 4��2� 7�5�3� 2

y

The graph of y � tan has period . � �The graph of y � cot has period .

y � tan �

��

y � cot

FIGURE 1.67 (a) (b)




Any invertible function and its inverse satisfy the properties

f 1 f -11y22 = y and f -11 f 1x22 = x.

These properties apply to the inverse sine and cosine, as long as we observe the restric-

tions on the domains. Here is what we can say:

sin 1sin-1 x2 = x and cos 1cos-1 x2 = x, for -1 … x … 1.

sin-1 1sin y2 = y, for -p>2 … y … p>2.

cos-1 1cos y2 = y, for 0 … y … p.

DEFINITION Inverse Sine and Cosine

y = sin-1 x is the value of y such that x = sin y, where -p>2 … y … p>2.

y = cos-1 x is the value of y such that x = cos y, where 0 … y … p.

The domain of both sin-1 x and cos-1 x is 5x: -1 … x … 16 .

�

�x

1

�1

y

x

1

�1

y

�

Restrict the domain of y � sin x to [� , ]. Restrict the domain of y � cos x to [0, ].

Range � [�1, 1]

Domain � [� , ](a)

Domain � [0, ]

�

(b)

y � cos x

y � sin x

Range � [�1, 1]

2�

2�

2�

2�

2�

2�

2�

FIGURE 1.71

QUICK CHECK 4 Explain why

sin-1 1sin 02 = 0, but

sin-1 1sin 2p2 ≠ 2p.

➤

EXAMPLE 3 Working with inverse sine and cosine Evaluate the following

expressions.

a. sin-1 113>22 b. cos-1 1-13>22 c. cos-1 1cos 3p2 d. sin 1sin-1 122

SOLUTION

a. sin-1 113>22 = p>3 because sin 1p>32 = 13>2 and p>3 is in the interval 3-p>2, p>24 .b. cos-1 1-13>22 = 5p>6 because cos 15p>62 = -13>2 and 5p>6 is in the interval 30, p4 .

3-p>2, p>24 ; for cosine, it is 30, p4 (Figure 1.71). Now when we ask for the angle x on

the interval 3-p>2, p>24 such that sin x = 12, there is one answer: x = p>6. When we ask

for the angle x on the interval 30, p4 such that cos x = -12, there is one answer: x = 2p>3.

We define the inverse sine, or arcsine, denoted y = sin-1 x or y = arcsin x, such that

y is the angle whose sine is x, with the provision that y lies in the interval 3-p>2, p>24 . Similarly, we define the inverse cosine, or arccosine, denoted y = cos-1 x or y = arccos x,

such that y is the angle whose cosine is x, with the provision that y lies in the interval 30, p4 .

➤ The notation for the inverse trigonometric

functions invites confusion: sin-1 x and

cos-1 x do not mean the reciprocals of

sin x and cos x. The expression sin-1 x

should be read “angle whose sine is x,”

and cos-1 x should be read “angle whose cosine is x.” The values of sin-1 and

cos-1 are angles.

c. It’s tempting to conclude that cos-1 1cos 3p2 = 3p, but the result of an inverse

cosine operation must lie in the interval 30, p4 . Because cos 13p2 = -1 and

cos-1 1-12 = p, we have

cos-1 1cos 3p2 = cos-1 1-12 = p. s

-1

d. sin asin-1 1

2b = sin

p

6=

1

2 (++)++*

p>6 Related Exercises 47–56

➤F 2013 2014 Only or Use in Pilot Program–



Graphs and Properties Recall from Section 1.3 that the graph of f -1 is obtained by

reflecting the graph of f about the identity line y = x. This operation produces the graphs

of the inverse sine (Figure 1.72) and inverse cosine (Figure 1.73). The graphs make it easy

to compare the domain and range of each function and its inverse.

�2�

2�

�2�

2�

[� , ]2�

2�

1

�1

�1 1 x

y

[�1, 1]

The graphs of y � sin x and

y � sin�1 x are symmetric

about the line y � x.

Range of sin x� Domain of sin�1 x

Restricted domain of sin x� Range of sin�1 x

y � sin�1 x

y � sin x

y � x

FIGURE 1.72

p

1

�1

�1 1 x

y

p[�1, 1]

[0, p]

The graphs of y � cos x andy � cos�1 x are symmetricabout the line y � x.

Restricted domain of cos x� Range of cos�1 x

Range of cos x� Domain of cos�1 x

y � cos�1 x

y � cos x

y � x

FIGURE 1.73

EXAMPLE 4 Right-triangle relationships

a. Suppose u = sin-1 12>52. Find cos u and tan u.

b. Find an alternative form for cot 1cos-1 1x>422 in terms of x.

SOLUTION

a. Relationships between the trigonometric functions and their inverses can often be

simplified using a right-triangle sketch. The right triangle in Figure 1.74 satisfies the

relationship sin u = 25, or, equivalently, u = sin-1 25. We label the angle u and the

lengths of two sides; then the length of the third side is 121 (by the Pythagorean

theorem). Now it is easy to read directly from the triangle:

cos u =121

5 and tan u =

2

121.

b. We draw a right triangle with an angle u satisfying cos u = x>4, or, equivalently,

u = cos-1 1x>42 (Figure 1.75). The length of the third side of the triangle is

216 - x2. It now follows that

cot acos-1 x

4b =

x

216 - x2.

(+)+*

u Related Exercises 57–62

➤

EXAMPLE 5 A useful identity Use right triangles to explain why

cos-1 x + sin-1 x = p>2.

2

5

�

�

52

sin �

�cos �

�tan �25

FIGURE 1.74

16 � x24

x

u4xcos u �

FIGURE 1.75

SOLUTION We draw a right triangle in a unit circle and label the acute angles u and w

(Figure 1.76). These angles satisfy cos u = x, or u = cos-1 x, and sin w = x, or

w = sin-1 x. Because u and w are complementary angles, we have

p

2= u + w = cos-1 x + sin-1 x.

This result holds for 0 … x … 1. An analogous argument extends the property to

-1 … x … 1. Related Exercises 63–66

➤

y

1

x

(x, y)

u

sin w � x ��w � sin�1 x

x

wcos u � x ��u � cos�1 x

FIGURE 1.76




Other Inverse Trigonometric FunctionsThe procedures that led to the inverse sine and inverse cosine functions can be used to

obtain the other four inverse trigonometric functions. Each of these functions carries a

restriction that must be imposed to ensure that an inverse exists:

The tangent function is one-to-one on 1-p>2, p>22, which becomes the range of

y = tan-1 x.

The cotangent function is one-to-one on 10, p2, which becomes the range of y = cot-1 x.

The secant function is one-to-one on 30, p4 , excluding x = p>2; this set becomes the

range of y = sec-1 x.

The cosecant function is one-to-one on 3-p>2, p>24 , excluding x = 0; this set

becomes the range of y = csc-1 x.

The inverse tangent, cotangent, secant, and cosecant are defined as follows.

➤ Tables and books differ on the definition

of the inverse secant and cosecant. In

some books, sec-1 x is defined to lie in

the interval 3 -p, -p>22 when x 6 0.

(� , ).2�

2�

(� , ).2�

2�

2�

�2�

�2�

2�

1

�1

�1 1 x

y

y � tan�1 x

y � tan x

y � x

Range of tan�1 xis

Restricted domain of tan xis

FIGURE 1.77

1

�1

�1 1 x�

�

y

y � cot�1 x

y � cot x

y � x

Range of cot�1 xis (0, �).

Restricted domainof cot x is (0, �).

FIGURE 1.78

The graphs of these inverse functions are obtained by reflecting the graphs of

the original trigonometric functions about the line y = x (Figures 1.77–1.80). The

inverse secant and cosecant are somewhat irregular. The domain of the secant function

( Figure 1.79) is restricted to the set 30, p4 , excluding x = p>2, where the secant has a

vertical asymptote. This asymptote splits the range of the secant into two disjoint intervals 1-∞ , -14 and 31, ∞2, which, in turn, splits the domain of the inverse secant into the

same two intervals. A similar situation occurs with the cosecant.

DEFINITION Other Inverse Trigonometric Functions

y = tan-1 x is the value of y such that x = tan y, where -p>2 6 y 6 p>2.

y = cot-1 x is the value of y such that x = cot y, where 0 6 y 6 p.

The domain of both tan-1 x and cot-1 x is 5x: -∞ 6 x 6 ∞6 .

y = sec-1 x is the value of y such that x = sec y, where 0 … y … p, with y ≠ p>2.

y = csc-1 x is the value of y such that x = csc y, where -p>2 … y … p>2, with

y ≠ 0.

The domain of both sec-1 x and csc-1 x is 5x: � x � Ú16 .




EXAMPLE 6 Working with inverse trigonometric functions Evaluate or simplify

the following expressions.

a. tan-1 1-1>132 b. sec-1 1-22 c. sin 1tan-1 x2SOLUTION

a. The result of an inverse tangent operation must lie in the interval 1-p>2, p>22. Therefore,

tan-1 a- 1

13b = -

p

6 because tan a-p

6b = -

1

13.

b. The result of an inverse secant operation when x … -1 must lie in the interval 1p>2, p4 . Therefore,

sec-1 1-22 = 2p

3 because sec

2p

3= -2.

c. Figure 1.81 shows a right triangle with the relationship x = tan u or u = tan-1 x, in

the case that 0 … u 6 p>2. We see that

sin 1tan-1 x2 = x

21 + x2. s

u

The same result follows if -p>2 6 u 6 0, in which case x 6 0 and sin u 6 0.


➤

�2�

�2�

�2�

2�

�

2�

1

�1

�1 1 x

y

y � sec�1 x

y � sec�1 x

y � sec x

y � sec x

y � x

� �

�

�

��

Range of sec�1 xis [0, ], y � .

2�

�Restricted domain of

sec x is [0, ], x � .

FIGURE 1.79

�2�

2�

�2�

2�

2�

2� 1

�1

�1 1 x

y

y � csc�1 x

y � csc x

y � csc�1 x

y � csc x

y � x

Range of csc�1 xis [� , ], y � 0.

2�

2�

Restricted domain of

csc x is [� , ], x � 0.

FIGURE 1.80

x

1

u

tan u � x ��u � tan�1 x

1 � x2

FIGURE 1.81

QUICK CHECK 5 Evaluate sec-1 1 and tan-1 1.

➤

Review Questions1. Define the six trigonometric functions in terms of the sides of a

right triangle.

2. Explain how a point P1x, y2 on a circle of radius r determines an

angle u and the values of the six trigonometric functions at u.

SECTION 1.4 EXERCISES3. How is the radian measure of an angle determined?

4. Explain what is meant by the period of a trigonometric function.

What are the periods of the six trigonometric functions?

5. What are the three Pythagorean identities for the trigonomet-

ric functions?




6. How are the sine and cosine functions related to the other four

trigonometric functions?

7. Where is the tangent function undefined?

8. What is the domain of the secant function?

9. Explain why the domain of the sine function must be restricted in

order to define its inverse function.

10. Why do the values of cos-1 x lie in the interval 30, p4?

11. Is it true that tan 1tan-1 x2 = x for all x? Is it true that

tan-1 1tan x2 = x for all x?

12. Sketch the graphs of y = cos x and y = cos-1 x on the same set

of axes.

13. The function tan x is undefined at x = {p>2. How does this fact

appear in the graph of y = tan-1 x?

14. State the domain and range of sec-1 x.

Basic Skills15–22. Evaluating trigonometric functions Evaluate the following expressions by drawing the unit circle and the appropriate right triangle. Use a calculator only to check your work. All angles are in radians.

15. cos 12p>32 16. sin 12p>32 17. tan 1-3p>42 18. tan 115p>42 19. cot 1-13p>32 20. sec 17p>6221. cot 1-17p>32 22. sin 116p>32 23–28. Evaluating trigonometric functions Evaluate the following expressions or state that the quantity is undefined. Use a calculator only to check your work.

23. cos 0 24. sin 1-p>22 25. cos 1-p2 26. tan 3p 27. sec 15p>22 28. cot p

29–36. Trigonometric identities

29. Prove that sec u =1

cos u.

30. Prove that tan u =sin u

cos u.

31. Prove that tan2 u + 1 = sec2 u.

32. Prove that sin u

csc u+

cos u

sec u= 1.

33. Prove that sec 1p>2 - u2 = csc u.

34. Prove that sec 1x + p2 = -sec x.

35. Find the exact value of cos 1p>122. 36. Find the exact value of tan 13p>82. 37–46. Solving trigonometric equations Solve the following equations.

37. tan x = 1 38. 2u cos u + u = 0

39. sin2 u = 14, 0 … u 6 2p 40. cos2 u = 1

2, 0 … u 6 2p

41. 12 sin x - 1 = 0 42. sin 3x = 222 , 0 … x 6 2p

43. cos 3x = sin 3x, 0 … x 6 2p

44. sin2 u - 1 = 0

45. sin u cos u = 0, 0 … u 6 2p

46. tan2 2u = 1, 0 … u 6 p

47–56. Inverse sines and cosines Without using a calculator, evaluate, if possible, the following expressions.

47. sin-1 1 48. cos-1 1-12 49. tan-1 1

50. cos-1 1 -222 2 51. sin-1 23

2 52. cos-1 2

53. cos-1 1- 122 54. sin-1 1-12 55. cos 1cos-1 1-122

56. cos-1 1cos 7p>62 57–62. Right-triangle relationships Draw a right triangle to simplify the given expressions.

57. cos 1sin-1 x2 58. cos 1sin-1 1x>322 59. sin 1cos-1 1x>222 60. sin-1 1cos u2, for 0 … u …

p

2

61. sin 12 cos-1 x2 1Hint: Use sin 2u = 2 sin u cos u.2 62. cos 12 sin-1 x2 1Hint: Use cos 2u = cos2 u - sin2 u.2 63–64. Identities Prove the following identities.

63. cos-1 x + cos-1 1-x2 = p 64. sin-1 y + sin-1 1-y2 = 0

65–66. Verifying identities Sketch a graph of the given pair of functions to conjecture a relationship between the two functions. Then verify the conjecture.

65. sin-1 x; p

2- cos-1 x 66. tan-1 x;

p

2- cot-1 x

67–74. Evaluating inverse trigonometric functions Without using a calculator, evaluate or simplify the following expressions.

67. tan-1 13 68. cot-1 1-1>132 69. sec-1 2 70. csc-1 1-12 71. tan-1 1tan p>42 72. tan-1 1tan 3p>42 73. csc-1 1sec 22 74. tan 1tan-1 12 75–80. Right-triangle relationships Draw a right triangle to simplify the given expressions.

75. cos 1tan-1 x2 76. tan 1cos-1 x2 77. cos 1sec-1 x2 78. cot 1tan-1 2x2 79. sin asec-1 a2x2 + 16

4bb 80. cos atan-1 a x

29 - x2bb

81–82. Right-triangle pictures Express u in terms of x using the inverse sine, inverse tangent, and inverse secant functions.

81.

6 x

u

82.

12

x 2x

u

T




Further Explorations83. Explain why or why not Determine whether the following state-

ments are true and give an explanation or counterexample.

a. sin 1a + b2 = sin a + sin b.

b. The equation cos u = 2 has multiple real solutions.

c. The equation sin u = 12 has exactly one solution.

d. The function sin 1px>122 has a period of 12.

e. Of the six basic trigonometric functions, only tangent and

cotangent have a range of 1-∞ , ∞2. f.

sin-1 x

cos-1 x= tan-1 x.

g. cos-1 1cos 115p>1622 = 15p>16.

h. sin-1 x = 1>sin x.

84–87. One function gives all six Given the following information about one trigonometric function, evaluate the other five functions.

84. sin u = -4

5 and p 6 u 6 3p>2 (Find cos u, tan u, cot u, sec u,

and csc u.2 85. cos u =

5

13 and 0 6 u 6 p>2

86. sec u =5

3 and 3p>2 6 u 6 2p

87. csc u =13

12 and 0 6 u 6 p>2

88–91. Amplitude and period Identify the amplitude and period of the following functions.

88. f 1u2 = 2 sin 2u

89. g1u2 = 3 cos 1u>32 90. p1t2 = 2.5 sin 11

21t - 322 91. q1x2 = 3.6 cos 1px>242 92–95. Graphing sine and cosine functions Beginning with the graphs of y = sin x or y = cos x, use shifting and scaling transforma-tions to sketch the graph of the following functions. Use a graphing utility only to check your work.

92. f 1x2 = 3 sin 2x

93. g1x2 = -2 cos 1x>32 94. p1x2 = 3 sin 12x - p>32 + 1

95. q1x2 = 3.6 cos 1px>242 + 2

96–97. Designer functions Design a sine function with the given properties.

96. It has a period of 12 hr with a minimum value of -4 at t = 0 hr

and a maximum value of 4 at t = 6 hr.

97. It has a period of 24 hr with a minimum value of 10 at t = 3 hr

and a maximum value of 16 at t = 15 hr.

98. Field goal attempt Near the end of the 1950 Rose Bowl football

game between the University of California and Ohio State University,

Ohio State was preparing to attempt a field goal from a distance of

23 yd from the endline at point A on the edge of the kicking region

(see figure). But before the kick, Ohio State committed a penalty and

T

the ball was backed up 5 yd to point B on the edge of the kicking

region. After the game, the Ohio State coach claimed that his team

deliberately committed a penalty to improve the kicking angle. Given

that a successful kick must go between the uprights of the goal posts

G1 and G2, is ∠G1BG2 greater than ∠G1AG2? (In 1950, the uprights

were 23 ft, 4 in apart, equidistant from the origin on the end line. The

boundaries of the kicking region are 53 ft, 4 in apart and are equidis-

tant from the y-axis.) (Source: The College Mathematics Journal, 27, 4 (Sep 1996))

y

xG1 G2

End line

Kicking region

A

B

53�4

23�4

99. A surprising result The Earth is approximately circular in cross

section, with a circumference at the equator of 24,882 miles.

Suppose we use two ropes to create two concentric circles; one

by wrapping a rope around the equator and another using a rope

38 ft longer (see figure). How much space is between the ropes?

100–103. Graphs to trigonometric functions Use shifts and scalings of either y = sin x or y = cos x to find a function that describes each of the following curves. Answers are unique up to trigonometric identities.

100.

��

1

2

�1

�2

�2 2� x

y

�

101.

��

1

2

�1

�2

�2 2� x

y

�

T




102.

��

1

2

�1

�2

�2 2� x

y

�

103.

��

1

2

�1

�2

�2 2� x

y

��23�

23�

�2�

2�

Applications104. Daylight function for 40° N Verify that the function

D1t2 = 2.8 sin a 2p

365 1t - 812b + 12

has the following properties, where t is measured in days and D is

the number of hours between sunrise and sunset.

a. It has a period of 365 days.

b. Its maximum and minimum values are 14.8 and 9.2, respec-

tively, which occur approximately at t = 172 and t = 355,

respectively (corresponding to the solstices).

c. D1812 = 12 and D12642 ≈ 12 (corresponding to the

equinoxes).

105. Block on a spring A light block hangs at rest from the end of a

spring when it is pulled down 10 cm and released. Assume the

block oscillates with an amplitude of 10 cm on either side of its

rest position with a period of 1.5 s. Find a function d1t2 that gives

the displacement of the block t seconds after it is released, where

d1t2 7 0 represents downward displacement.

Rest position, d � 0

d(t) � 0

106. Approaching a lighthouse A boat approaches a 50-ft-high light-

house whose base is at sea level. Let d be the distance between the

boat and the base of the lighthouse. Let L be the distance between

the boat and the top of the lighthouse. Let u be the angle of eleva-

tion between the boat and the top of the lighthouse.

a. Express d as a function of u.

b. Express L as a function of u.

107. Ladders Two ladders of length a lean against opposite walls of

an alley with their feet touching (see figure). One ladder extends h

feet up the wall and makes a 75° angle with the ground. The other

ladder extends k feet up the opposite wall and makes a 45° angle

with the ground. Find the width of the alley in terms of h. Assume

the ground is horizontal and perpendicular to both walls.

h a a k

75� 45�

108. Pole in a corner A pole of length L is carried horizontally around

a corner where a 3-ft-wide hallway meets a 4-ft-wide hallway.

For 0 6 u 6 p>2, find the relationship between L and u at the

moment when the pole simultaneously touches both walls and the

corner P. Estimate u when L = 10 ft.

Pole, length L

3 ft

P

4 ft

u

109. Little-known fact The shortest day of the year occurs on the

winter solstice (near December 21) and the longest day of the year

occurs on the summer solstice (near June 21). However, the latest

sunrise and the earliest sunset do not occur on the winter solstice,

and the earliest sunrise and the latest sunset do not occur on the

summer solstice. At latitude 40° north, the latest sunrise occurs on

January 4 at 7:25 a.m. (14 days after the solstice), and the earliest

sunset occurs on December 7 at 4:37 p.m. (14 days before the sol-

stice). Similarly, the earliest sunrise occurs on July 2 at 4:30 a.m.

(14 days after the solstice) and the latest sunset occurs on June 7

at 7:32 p.m. (14 days before the solstice). Using sine functions,

devise a function s1t2 that gives the time of sunrise t days after

January 1 and a function S1t2 that gives the time of sunset t days

after January 1. Assume that s and S are measured in minutes and

s = 0 and S = 0 correspond to 4:00 a.m. (all times are standard

times). Graph the functions. Then graph the length of the day

function D1t2 = S1t2 - s1t2 and show that the longest and

shortest days occur on the solstices.

T




110. Viewing angles An auditorium with a flat floor has a large flat-

panel television on one wall. The lower edge of the television is

3 ft above the floor, and the upper edge is 10 ft above the floor

(see figure). Express u in terms of x.

10 ft

3 ft

xu

Additional Exercises111. Area of a circular sector Prove that the area of a sector of a

circle of radius r associated with a central angle u (measured in

radians) is A = 12 r2 u.

ru

112. Law of cosines Use the figure to prove the law of cosines

(which is a generalization of the Pythagorean theorem):

c2 = a2 + b2 - 2ab cos u.

x

y

(b cos u, b sin u)

(a, 0)

c

a

bu

113. Law of sines Use the figure to prove the law of sines:

sin A

a=

sin B

b=

sin C

c.

A

B

C

b

h

c a

Technology Exercises114. Intersection points Consider the curves y = cos x and y =

x

a,

where a 7 0 and x Ú 0.

a. At how many points do the curves intersect in the first quad-

rant when a = 2? Find the (approximate) coordinates of the

intersection point(s).

T

b. At how many points do the curves intersect in the first quad-

rant when a = 8? Find the (approximate) coordinates of the

intersection point(s).

c. Estimate the value of a such that y = cos x and y =x

a

intersect exactly twice in the first quadrant. Find the (approxi-

mate) coordinates of the intersection points.

115. The sinc function The sinc (pronounced sink) function

sinc x =sin x

xif x ≠ 0

1if x = 0

is used in applications such as digital signal processing.

a. What is the domain of the function f 1x2 = sin x

x?

b. Graph sinc x on the interval 3 -3p, 3p4 . c. In the definition of sinc x, why must a value be specified

at x = 0? Based on the graph in part (b), why is the value

sinc 0 = 1 given?

d. What is the range of sinc x?

e. Describe the roots of sinc x. f. Describe the behavior of sinc x as � x � increases, with x both

positive and negative.

116. Hidden oscillations Consider the function f 1x2 = 1 - sin3 x

x2 + 1 on

30, 104 .a. Graph the function using a graphing window that reveals all

the interesting features of the graph.

b. Using the graph of part (a), find the range of f .

c. Using the graph of part (a), find the roots of f on 30, 104 . You may need to use different graphing windows

to locate all the roots.

d. Find the points (approximately) at which f has a peak (a local

maximum value) or a valley (a local minimum value). Give the

function values at these points. Again, different graphing win-

dows or a zoom feature may be needed.

117. A Fourier series Working on the interval 3-p, p4 , consider the

sum

1

2+

4

p2 acos x +

cos 3x

9+

cos 5x

25+

cos 7x

49+ gb ,

where the dots mean that the pattern in the sum continues

indefinitely.

a. Graph the first two terms of the sum a1

2+

4

p2 cos xb on the

interval 3-p, p4 . b. Now successively add new terms to the sum and plot the sum

with three, four, and five terms. Comment on how the graph

changes as you add new terms to the sum.

c. Graph the sum with six, seven, and eight terms. Do the graphs

appear to approach a specific function? Describe that function.

T

T

T

QUICK CHECK ANSWERS

1. 3p>2; 225° 2. 13>2 ; -12>2 3. Divide both sides of

sin2 u + cos2 u = 1 by sin2 u. 4. sin-1 1sin 02 = sin-1 0 = 0

and sin-1 1sin 12p22 = sin-1 0 = 0 5. 0, p>4 ➤



For Use - Westlake City School · PDF filee = 2.718281828459 c. The notation e was proposed...

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