Smith R., Minton R. Calculus
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CONFIRMING PAGES
6-19 SECTION 6.3
..
The Exponential Function 393
PROOF
These laws are all obvious when the exponents are rational. If the exponent is irrational
though,weonlyknowthevalueoftheseexponentialsindirectly,throughtheinversefunction
relationship with ln x, given in Definition 3.2.
(i) Note that using the rules of logarithms, we have
ln(e
r
e
s
) = ln(e
r
) + ln(e
s
) = r + s = ln(e
r+s
).
Since ln x is one-to-one, it must follow that
e
r
e
s
= e
r+s
.
The proofs of (ii) and (iii) are similar and are left as exercises. (See exercise 59.)
Derivative of the Exponential Function
Your initial thought might be to find the derivative of e
x
using the usual limit definition of
derivative. However, for f (x) = e
x
, we have that
d
dx
e
x
= f
(x) = lim
h→0
f (x +h) − f (x)
h
= lim
h→0
e
x+h
− e
x
h
= lim
h→0
e
x
e
h
− e
x
h
= e
x
lim
h→0
e
h
− 1
h
. (3.3)
While we do not know how to compute this limit exactly, it is an easy exercise to show
its value is approximately 1. We revisit this limit in exercises 57 and 58, at the end of this
section. We now present an alternative derivation, based on Definition 3.2. We have that
y = e
x
if and only if ln y = x.
Differentiating this last equation with respect to x gives us
d
dx
ln y =
d
dx
x = 1.
From the chain rule, we now have
1 =
d
dx
ln y =
1
y
dy
dx
. (3.4)
Multiplying both sides of (3.4) by y,wehave
dy
dx
= y = e
x
or
d
dx
e
x
= e
x
. (3.5)
Note that (3.5) is consistent with (3.3) and confirms that the limit in (3.3) is 1, as we had
conjectured. Of course, this also gives us the corresponding integration rule
e
x
dx = e
x
+ c.
We give several examples of the calculus of exponential functions in examples 3.1 and 3.2.
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394 CHAPTER 6
..
Exponentials, Logarithms and Other Transcendental Functions 6-20
EXAMPLE 3.1 Computing the Derivative of Some Exponentials
Compute the derivatives of the functions (a) e
3x
and (b) e
sin x
.
Solution For (a), we have
d
dx
e
3x
= e
3x
d
dx
(3x)
chain rule
= e
3x
(3) = 3e
3x
.
Similarly, for (b), we get
d
dx
e
sin x
= e
sin x
d
dx
(sin x)
chain rule
= e
sin x
(cos x) = cos xe
sin x
.
y
x
2 424
2
4
6
8
10
FIGURE 6.15
y = e
x
y
x
2 424
2
4
6
8
10
FIGURE 6.16
y = e
−x
EXAMPLE 3.2 Evaluating the Integral of Some Exponentials
Evaluate the integrals (a)
e
−5x
dx and
x
3
e
x
4
dx.
Solution We can resolve these with simple substitutions. For (a), we have
e
−5x
dx =−
1
5
e
−5x
e
u
(−5)dx
du
=−
1
5
e
u
+ c =−
1
5
e
−5x
+ c.
For (b), taking u = x
4
,weget
x
3
e
x
4
dx =
1
4
e
x
4
e
u
(4x
3
)dx
du
=
1
4
e
u
+ c =
1
4
e
x
4
+ c.
We now have the tools to produce the graph of f (x) = e
x
. Since e = 2.718... > 1, we
have
lim
x→∞
e
x
=∞ and lim
x→−∞
e
x
= 0.
We also have that
f
(x) = e
x
> 0,
so that f is increasing for all x and
f
(x) = e
x
> 0,
so that the graph is concave up everywhere. You should now readily obtain the graph in
Figure 6.15. (Notice that you can also obtain this graph by reflecting the graph of y = ln x
through the line y = x.)
Similarly, for f (x) = e
−x
,wehave
lim
x→∞
e
−x
= 0 and lim
x→−∞
e
−x
=∞.
Further, from the chain rule,
f
(x) =−e
−x
< 0,
so that f is decreasing for all x. We also have
f
(x) = e
−x
> 0,
so that the graph is concave up everywhere. You should easily obtain the graph in
Figure 6.16.
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6-21 SECTION 6.3
..
The Exponential Function 395
More general exponential functions, such as f (x) = b
x
, for any base b > 0, are easy to
express in terms of the natural exponential function, as follows. Notice that for any constant
b > 0, we have by the usual rules of logs and exponentials that
b
x
= e
ln(b
x
)
= e
x lnb
.
In particular, observe that this says that
d
dx
b
x
=
d
dx
e
x lnb
= e
x lnb
d
dx
(x lnb)
= e
x lnb
(lnb) = b
x
(lnb).
NOTES
We will occasionally write
e
x
= exp(x). This is particularly
helpful when the exponent is a
complicated expression. For
example,
exp(x
3
− 5x
2
+ 2x + 7)
= e
x
3
−5x
2
+2x+7
,
where the former is more easily
read than the latter.
Similarly, for b > 0(b = 1), we have
b
x
dx =
e
x lnb
dx =
1
lnb
e
x lnb
u
(lnb)dx
du
=
1
lnb
e
x lnb
+ c =
1
lnb
b
x
+ c.
You can now see that the general exponential functions are easily dealt with in terms of
the natural exponential. In fact, you should not bother to memorize the formulas for the
derivatives and integrals of general exponentials. Rather, each time you run across the
exponential function f (x) = b
x
, simply rewriteit as f (x) = e
x lnb
and then use the familiar
rules for the derivative and integral of the natural exponential and the chain rule.
EXAMPLE 3.3 Differentiating an Exponential Function
Find the derivative of f (x) = 2
x
2
.
Solution We first rewrite the function as
f (x) = 2
x
2
= e
ln2
x
2
= e
x
2
ln2
.
From the chain rule, we now have
f
(x) = e
x
2
ln2
(2x ln2) = (2ln2)x2
x
2
.
In a similar way, we can use our knowledge of the natural logarithm to discuss more
general logarithms. First, recall that for any base a > 0(a = 1) and any x > 0, y = log
a
x
if and only if x = a
y
. Taking the natural logarithm of both sides of this equation, we have
ln x = ln(a
y
) = y ln a.
Solving for y gives us
y =
ln x
lna
.
This proves the following result.
THEOREM 3.2
For any base a > 0(a = 1) and any x > 0, log
a
x =
ln x
lna
.
Among other things, Theorem 3.2 enables us to use a calculator to evaluate logarithms
with any base. Calculators typically do not have built-in functions for evaluation of general
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396 CHAPTER 6
..
Exponentials, Logarithms and Other Transcendental Functions 6-22
logarithms, opting instead for only ln x and log
10
x keys. Notice that evaluating general
logarithms is now easy. For instance, we have
log
7
3 =
ln3
ln7
≈ 0.564575.
More importantly, observe that we can use Theorem 3.2 to express derivatives of general
logarithms in terms of the familiar derivative of the natural logarithm. In particular, for any
base a > 0(a = 1), we have
d
dx
log
a
x =
d
dx
ln x
lna
=
1
lna
d
dx
(ln x)
=
1
lna
1
x
=
1
x lna
.
As with the derivative formula for general exponentials, there is little point in learning this
as a new differentiation rule. Rather, when you need to differentiate the general logarithmic
function f (x) = log
a
x, simplyrewriteit first as f (x) =
ln x
lna
anduse the familiar derivative
of the natural logarithm.
Applications of exponential functions are found everywhere, making e
x
one of the
most important functions you will study. In example 3.4, we analyze the concentration of
a chemical in a reaction.
EXAMPLE 3.4 Analyzing the Concentration of a Chemical
The concentration c of a certain chemical after t seconds of an autocatalytic reaction is
given by c(t) =
10
9e
−20t
+ 1
. Show that c
(t) > 0 and use this information to determine
that the concentration of the chemical never exceeds 10.
Solution Before computing the derivative, look carefully at the function c(t). The
independent variable is t and the only term involving t is in the denominator. So, we
don’t need to use the quotient rule. Instead, first rewrite the function as
c(t) = 10(9e
−20t
+ 1)
−1
and use the chain rule. We get
c
(t) =−10(9e
−20t
+ 1)
−2
d
dt
(9e
−20t
+ 1)
=−10(9e
−20t
+ 1)
−2
(−180e
−20t
)
= 1800e
−20t
(9e
−20t
+ 1)
−2
=
1800e
−20t
(9e
−20t
+ 1)
2
> 0.
Since all of the tangent lines have positive slope, the graph of y = c(t) rises from left to
right, as shown in Figure 6.17.
c
t
2
4
6
8
10
0.2 1.00.80.60.4
FIGURE 6.17
Chemical concentration
Since the concentration increases for all time, the concentration is always less than
the limiting value lim
t→∞
c(t), which is easily computed to be
lim
t→∞
10
9e
−20t
+ 1
=
10
0 + 1
= 10.
EXERCISES 6.3
WRITING EXERCISES
1. Thinking of e as a number (larger than 1), explain why
lim
x→∞
e
x
=∞and lim
x→−∞
e
x
= 0.
2. Explain why the graph of y = e
−x
in Figure 6.16 is the mirror
image (about the y-axis) of the graph of y = e
x
in Figure 6.15.
3. The graph of f (x) = e
x
curves upward in the interval from
x =−1tox = 1. Interpreting f
(x) = e
x
as the slopes of
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CONFIRMING PAGES
6-23 SECTION 6.3
..
The Exponential Function 397
tangent lines and noting that the larger x is, the larger e
x
is,
explain why the graph curves upward. For larger values of x,
the graph of f (x) = e
x
appears to shoot straight up with no
curve.Using the tangent line,determine whether this is correct
or just an optical illusion.
4. There are a variety of equivalent ways of defining the num-
ber e. Discuss the definition given in this section versus the
alternative definition e = lim
n→∞
1 +
1
n
n
. Explain why the limit
definition is more convenient in a precalculus class, whereas
the inverse function definition is more convenient now.
In exercises 1–8, find the derivative of the given function.
1. 4e
3x
2. 3xe
−2x
3.
e
4x
x
4.
x
2
e
6x
5. 2e
x
3
−3x
6. e
2x−x
2
7.
√
e
2x
2
8. (e
3x
+ 1)
2
............................................................
In exercises 9–24, evaluate the given integral.
9.
e
3x
dx 10.
e
2−3x
dx
11.
xe
x
2
dx 12.
2x
2
e
−x
3
dx
13.
sin xe
cos x
dx 14.
e
x
cos(e
x
)dx
15.
e
1/x
x
2
dx 16.
e
x
1 + e
x
dx
17.
(1 + e
x
)
2
dx 18.
2
√
xe
√
x
dx
19.
lne
x
2
dx 20.
e
−lnx
dx
21.
1
0
e
3x
dx 22.
2
0
e
−2x
dx
23.
2
−2
xe
−x
2
dx 24.
1
0
e
x
− 1
e
2x
dx
............................................................
In exercises 25–30, graph the indicated function.
25. 3e
2x
26. 3e
−2x
27. 3xe
−2x
28. 2xe
−3x
29. e
1/x
30. e
−2/(x
3
−x)
............................................................
In exercises 31–34, find all extrema and inflection points.
31. xe
−2x
32. xe
−4x
33. x
2
e
−2x
34.
e
x
x
............................................................
35. The concentration of a certain chemical after t seconds of an
autocatalytic reaction is givenby x(t) =
6
2e
−8t
+ 1
. Show that
x
(t) > 0 and use this information to determine that the con-
centration of the chemical never exceeds 6.
36. The concentration of a certain chemical after t seconds of an
autocatalyticreactionisgivenby x(t) =
10
9e
−10t
+ 2
.Showthat
x
(t) > 0 and use this information to determine that the con-
centration of the chemical never exceeds 5.
............................................................
In exercises 37–42, find the derivative of the given function.
37. 3
2x
38. 5
−2x
39. 3
x
2
40. 4
−x
41. log
4
x
2
42. log
6
(x
2
+ 5)
............................................................
In exercises 43–46, evaluate each integral.
43.
2
x
dx 44.
4
3x
dx
45.
x2
x
2
dx 46.
2x3
−x
2
dx
............................................................
In exercises 47–50, graph each function.
47. 3
x
48.
1
2
x
49. 3
−x
50.
1
2
−x
............................................................
51. Based on exercises 47 and 48,describe the graph of y = b
x
for
b > 1;0 < b < 1.
52. Based on exercises 49 and 50, describe the graph of y = b
−x
for b > 1;0 < b < 1.
............................................................
In exercises 53–56, use a CAS or graphing calculator.
53. Find the derivative of f (x) = e
ln x
2
on your CAS. Compare its
answer to 2x. Explain how to get this answer and your CAS’s
answer, if it differs.
54. Findthederivativeof f (x) = e
ln(−x
2
)
onyourCAS.Thecorrect
answer is that it does not exist. Explain how to get this answer
and your CAS’s answer, if it differs.
55. Findthederivative of f (x) = ln
√
4e
3x
onyourCAS. Compare
its answer to
3
2
. Explain how to get this answer and your CAS’s
answer, if it differs.
56. Find the derivative of f (x) = ln
e
4x
x
2
on your CAS. Com-
pare its answer to 4 −2/x. Explain how to get this answer and
your CAS’s answer, if it differs.
............................................................
57. In the text, we referred the proof of lim
h→0
e
h
− 1
h
= 1 to the ex-
ercises. In this exercise, we guide you through one possible
proof. (Another proof is given in exercise 58.) Starting with
h > 0, write h = ln e
h
=
e
h
1
1
x
dx. Use the Integral Mean
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CONFIRMING PAGES
398 CHAPTER 6
..
Exponentials, Logarithms and Other Transcendental Functions 6-24
Value Theorem to write
e
h
1
1
x
dx =
e
h
− 1
¯
x
for some number
¯
x between 1 and e
h
. This gives you
e
h
− 1
h
=
¯
x. Now, take
the limit as h → 0
+
.Forh < 0, repeat this argument, with h
replaced with −h.
58. In this exercise, we guide you through a different proof
of lim
h→0
e
h
− 1
h
= 1. Start with f (x) = ln x and the fact that
f
(1) = 1. Using the alternative definition of derivative, we
write this as f
(1) = lim
x→1
ln x − ln1
x − 1
= 1. Explain why this
implies that lim
x→1
x − 1
ln x
= 1. Finally, substitute x = e
h
.
59. Prove parts (ii) and (iii) of Theorem 3.1.
60. The derivative of e
x
is derived in the text from (3.4) and (3.5).
As an alternative, start with f (x) = e
x
and apply Theorem 2.3
from section 6.2, to obtain the same derivative formula.
61. In statistics, the function f (x) = e
−x
2
/2
is used to analyze ran-
dom quantities that have a bell-shaped distribution. Solutions
of the equation f
(x) = 0 give statisticians a measure of the
variability of the random variable. Find all solutions.
Repeat for the function g(x) = e
−x
2
/8
. Comparing the
graphs of the two functions, explain why you would say that g
is more spread out than f .
62. Apply Newton’s method to the function f (x) = ln x − 1to
find an iterative scheme for approximating e. Discover how
many steps are needed to start at x
0
= 3 and obtain five digits
of accuracy.
APPLICATIONS
63. A water wave of length L meters in water of depth d meters
has velocity v satisfying the equation
v
2
=
4.9L
π
e
2πd/L
− e
−2πd/L
e
2πd/L
+ e
−2πd/L
.
Treating L as a constant and thinking of v
2
as a function f (d),
use a linear approximation to show that f (d) ≈ 9.8d for small
values of d. That is, for small depths, the velocity of the
wave is approximately
√
9.8d and is independent of the wave-
length L.
64. Planck’s law states that the energy density of blackbody radi-
ation of wavelength x is given by
f (x) =
8πhcx
−5
e
hc/(kTx)
− 1
.
Use the linear approximation e
x
≈ 1 + x to show that
f (x) ≈ 8π kT/x
4
, which is known as the Rayleigh-Jeans
law.
65. If two soccer teams each score goals at a rate of r goals per
minute, the probability that n goals will be scored in t minutes
is P =
(rt)
n
n!
e
−rt
. Take r =
1
25
. Show that for a 90-minute
game, P is maximized with n = 3. Briefly explain why this
makes sense. Find t to maximize the probability that exactly 1
goal has been scored. Briefly explain why your answer makes
sense.
66. The atmospheric pressure at height h feet above sea level is
approximately p = 2116e
−0.0000318h
. If a balloon is at height
1000 feet and rising at the rate of 160 ft/s, at what rate is the
atmospheric pressure changing?
67. The function f (t) = a/(1 +3e
−bt
) has been used to modelthe
spread of a rumor. Suppose that a = 70 and b = 0.2. Compute
f (2), the percentage of the population that has heard the rumor
after 2 hours. Compute f
(2) and describe what it represents.
Compute lim
t→∞
f (t) and describe what it represents.
68. After an injection, the concentration of drug in a muscle is
given by a function of time, f (t). Suppose that t is measured
in hours and f (t) = e
−0.02t
− e
−0.42t
. Determine the time when
the maximum concentration of drug occurs.
69. The sigmoid function f (x) =
1
1 + e
−x
is used to model situ-
ations with a threshold. For example, in the brain each neuron
receives inputs from numerous other neurons and fires only
after its total input crosses some threshold. Graph y = f (x)
and find lim
x→∞
f (x) and lim
x→−∞
f (x). Define the function g(x)to
be the value of f (x) rounded off to the nearest integer. What
valueof x is the threshold for this function to switch from “off”
(0) to “on” (1)? How could you modify the function to move
the threshold to x = 4 instead?
70. A human being starts with a single fertilized egg cell, which
divides into 2 cells, which then divide into 4 cells and so on,
until the newborn infant has about 1 quadrillion (10
15
) cells.
Without doing any calculations, guess how many divisions are
required to reach 10
15
. Then, determine the valueof n such that
2
n
≈ 10
15
. Are you surprised?
71. Suppose a certain type of cell grows for three days and
then divides into two cells. The distribution of ages of cells
will have a probability distribution function (pdf) of the
form f (x) = 2ke
−kx
for 0 ≤ x ≤ 3. Find the value of k such
that f (x) is a pdf; that is,
3
0
f (x) dx = 1. Then find the
probability that a given cell is between one and two days
old, given by
2
1
f (x) dx.
72. Suppose you have a 1-in-10 chance of winning a prize with
some purchase (like a lottery). If you make 10 purchases (i.e.,
you get 10 tries), the probability of winning at least one prize
is 1 −(9/10)
10
. If the prize had probability 1-in-20 and you
tried 20 times, would the probability of winning at least once
be higher or lower? Compare 1 −(9/10)
10
and 1 −(19/20)
20
to find out. To see what happens for larger and larger odds,
compute lim
n→∞
{1 − [(n − 1)/n]
n
}.
EXPLORATORY EXERCISES
1. Find the number of intersections of y = x
a
and y = a
x
for
a = 3, a = e and a = 2. The number of intersections changes
betweena = 2anda = 3.Repeatthisfora = 2.1, a = 2.2and
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CONFIRMING PAGES
6-25 SECTION 6.4
..
The Inverse Trigonometric Functions 399
so on, to explore whether other a-values have different num-
bers of intersections. Prove your conjecture. You will want to
use the following fact: if f (c) = g(c), f
(x) > g
(x) for x > c
and f
(x) < g
(x)for x < c,then x = c is the only intersection
of f and g.
2. Determine the general properties (positive/negative,
increasing/decreasing, concave up/down) of the functions
xe
−bx
and x
2
e
−bx
for a constant b > 0. In a number of ap-
plications, mathematicians need to find a function for which
f (0) = 0 and (for x > 0) f (x) rises to a single maximum at
x = c andgradually drops with lim
x→∞
f (x) = 0. Show that both
families of functions meet these criteria and find the appro-
priate values for b. Taking c = 2, graph xe
−bx
and x
2
e
−bx
.
Compare and contrast the graphs. Explain what an applied
mathematician might use to decide which one is a more real-
istic function for a given application.
3. For n = 1 and n = 2, investigate lim
x→0
e
−1/x
x
n
numerically and
graphically. Conjecture the value of lim
x→0
e
−1/x
x
n
for any posi-
tive integer n and use your conjecture for the remainder of the
exercise. For f (x) =
0ifx ≤ 0
e
−1/x
if x > 0
, show that f is dif-
ferentiable at each x and that f
(x) is continuous for all x.
Then show that f
(0) exists and compare the work needed to
show that f
(x) is continuous at x = 0 and to show that f
(0)
exists.
6.4 THE INVERSE TRIGONOMETRIC FUNCTIONS
In this section, we expand the set of functions available to you by defining inverses to the
trigonometric functions. Notice from the graph in Figure 6.18 that we cannot define an
inverse to f (x) = sin x, since sin x is not one-to-one. We remedy this by restricting the
domain to the interval
−
π
2
,
π
2
. (See Figure 6.19.) The restricted function is one-to-one
and so, has an inverse. We thus define the inverse sine function by
y = sin
−1
x if and only if sin y = xand −
π
2
≤ y ≤
π
2
. (4.1)
It is convenient to think of this definition as follows. If y = sin
−1
x, then y is the angle
(between −
π
2
and
π
2
) for which sin y = x. Note that while we could have selected any
interval on which sin x is one-to-one,
−
π
2
,
π
2
is the most convenient. To see that these are
indeed inverse functions, observe that
sin(sin
−1
x) = x, for all x ∈ [−1, 1]
and sin
−1
(sin x) = x, for all x ∈
−
π
2
,
π
2
. (4.2)
Read equation (4.2) very carefully. It does not say that sin
−1
(sin x) = x for all x, but rather,
only for those in the restricted domain,
−
π
2
,
π
2
. So, while it might be tempting to write
sin
−1
(sinπ ) = π, this is incorrect,as
sin
−1
(sinπ ) = sin
−1
(0) = 0.
REMARK 4.1
Mathematicians often use the notation arcsin x in place of sin
−1
x. People will read
sin
−1
x interchangeably as “inverse sine of x” or “arcsine of x.”
x
p
1
1
y
p
q
q
FIGURE 6.18
y = sin x
y
x
qq
1
1
FIGURE 6.19
y = sin x on
−
π
2
,
π
2
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CONFIRMING PAGES
400 CHAPTER 6
..
Exponentials, Logarithms and Other Transcendental Functions 6-26
EXAMPLE 4.1 Evaluating the Inverse Sine Function
Evaluate (a) sin
−1
√
3
2
and (b) sin
−1
−
1
2
.
Solution (a) We look for the angle θ in the interval
−
π
2
,
π
2
for which sinθ =
√
3
2
.
Note that since sin
π
3
=
√
3
2
and
π
3
∈
−
π
2
,
π
2
, we have that sin
−1
√
3
2
=
π
3
.
(b) In this case, note that sin
−
π
6
=−
1
2
and −
π
6
∈
−
π
2
,
π
2
. Thus,
sin
−1
−
1
2
=−
π
6
.
Judging by example 4.1, you might think that equation (4.1) is a roundabout way of
defininga function.Ifso, you’vegottheidea exactly.Infact,wewanttoemphasizethat what
weknowabout the inversesine function is principally through referencetothesinefunction.
Wewill not have any other definition of arcsine, nor are there anyalgebraic formulas for this
function. (These things are true of most inverse functions.) Further, you should recall from
our discussion in section 6.2 that we can draw a graph of y = sin
−1
x simply by reflecting
the graph of y = sin x on the interval
−
π
2
,
π
2
(from Figure 6.19) through the line y = x.
(See Figure 6.20.)
y
x
q
q
11
FIGURE 6.20
y = sin
−1
x
y
x
q
p
1
1
FIGURE 6.21
y = cos x on [0,π]
Turning to y = cos x, can you think of how to restrict the domain to make the function
one-to-one? Notice that restricting the domain to the interval
−
π
2
,
π
2
, as we did for the
inverse sine function will not work here. (Why not?) The simplest way to do this is to
restrict its domain to the interval [0,π]. (See Figure 6.21.) Consequently, we define the
inverse cosine function by
y = cos
−1
x if and only if cos y = xand0 ≤ y ≤ π. (4.3)
Note that here, we have
cos(cos
−1
x) = x, for all x ∈ [−1, 1]
and cos
−1
(cos x) = x, for all x ∈ [0,π].
As with the definition of arcsine, it is helpful to think of cos
−1
x as that angle θ in [0,π] for
whichcosθ = x.As with sin
−1
x, itis common to use cos
−1
x and arccosx interchangeably.
EXAMPLE 4.2 Evaluating the Inverse Cosine Function
Evaluate (a) cos
−1
(0) and (b) cos
−1
−
√
2
2
.
Solution (a) You will need to find that angle θ in [0,π] for which cos θ = 0. It’s not
hard to see that cos
−1
(0) =
π
2
. Note that if you calculate this on your calculator and get
90, your calculator is in degrees mode, in which case, you should change it to radians
mode.
(b) Here, look for the angle θ ∈ [0,π] for which cosθ =−
√
2
2
. Notice that
cos
3π
4
=−
√
2
2
and
3π
4
∈ [0,π]. Consequently,
cos
−1
−
√
2
2
=
3π
4
.
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6-27 SECTION 6.4
..
The Inverse Trigonometric Functions 401
Once again, we obtain the graph of this inverse function by reflecting the graph of
y = cos x on the interval [0,π] (seen in Figure 6.21) through the line y = x. (See
Figure 6.22.)
y
x
11
q
p
FIGURE 6.22
y = cos
−1
x
qq
6
4
2
2
4
6
y
x
FIGURE 6.23
y = tan x on
−
π
2
,
π
2
Wecan define inverses for each of the four remainingtrig functions in similar ways. For
y = tan x, we restrict the domain to the interval
−
π
2
,
π
2
. Think about why the endpoints
of this interval are not included. (See Figure 6.23.) Having done this, you should readily
see that we define the inverse tangent function by
y = tan
−1
x if and only if tan y = x and −
π
2
< y <
π
2
. (4.4)
The graph of y = tan
−1
x is then as seen in Figure 6.24, found by reflecting the graph in
Figure 6.23 through the line y = x.
y
x
q
q
4626 4 2
FIGURE 6.24
y = tan
−1
x
EXAMPLE 4.3 Evaluating an Inverse Tangent
Evaluate tan
−1
(1).
Solution You must look for the angle θ on the interval
−
π
2
,
π
2
for which tanθ = 1.
This is easy enough. Since tan
π
4
= 1 and
π
4
∈
−
π
2
,
π
2
, we have that tan
−1
(1) =
π
4
.
y
x
p
10
5
1
5
1
10
q
FIGURE 6.25
y = sec x on [0,π]
y
x
q
10515 110
p
FIGURE 6.26
y = sec
−1
x
We now turn to defining an inverse for sec x. First, we must issue a disclaimer. As
we have indicated, there are any number of ways to suitably restrict the domains of the
trigonometric functions in order to make them one-to-one. With the first three we’ve seen,
there has been an obvious choice of how to do this and there is general agreement among
mathematicianson thechoiceof theseintervals.Inthe caseofsec x,thisis not true. There are
severalreasonablewaysinwhichto suitablyrestrictthedomainand differentauthorsrestrict
these differently. We have (arbitrarily) chosen to restrict the domain to be [0,
π
2
) ∪ (
π
2
,π].
You might initially think that this looks strange. Why not use all of [0,π]? You need only
think about the definition of sec x to see why we needed to exclude the point x =
π
2
. See
Figure 6.25 for a graph of sec x on this domain. (Note the vertical asymptote at x =
π
2
.)
Consequently, we define the inverse secant function by
y = sec
−1
x if and only if sec y = x and y ∈
0,
π
2
∪
π
2
,π
. (4.5)
A graph of sec
−1
x is shown in Figure 6.26.
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402 CHAPTER 6
..
Exponentials, Logarithms and Other Transcendental Functions 6-28
EXAMPLE 4.4 Evaluating an Inverse Secant
Evaluate sec
−1
(−
√
2).
Solution You must look for the angle θ with θ ∈
0,
π
2
∪
π
2
,π
, for which
secθ =−
√
2. Notice that if secθ =−
√
2, then cosθ =−
1
√
2
=−
√
2
2
. Recall from
example 4.2 that cos
3π
4
=−
√
2
2
. Further, the angle
3π
4
is in the interval
π
2
,π
and so,
sec
−1
(−
√
2) =
3π
4
.
Calculators do not usually have built-in functions for sec x or sec
−1
x. To compute
values of these, you must convert the desired secant value to a cosine value or use the
inverse cosine function, much as we did in example 4.4.
u cos
1
x
sin u
兹
1 x
2
1
cos u x
FIGURE 6.27
θ = cos
−1
x
REMARK 4.2
We can likewise define inverses for cotx and csc x. As these functions are used only
infrequently, we will omit them here and examine them in the exercises.
Often, as a part of a larger problem (for example, evaluating an integral by some of
the methods discussed in Chapter 7) you will need to recognize some relationship between
the trigonometric functions and their inverses. We present several clues here. When you are
faced with these problems, our best advice is to keep in mind the definitions of the inverse
functions and then draw a picture.
EXAMPLE 4.5 Simplifying Expressions Involving Inverse
Trigonometric Functions
Simplify sin(cos
−1
x) and tan(cos
−1
x).
Solution Do not look for some arcane formula to help you out. Think first: cos
−1
x is
the angle (call it θ) for which x = cosθ. First, consider the case where x > 0. In
Figure 6.27, we have drawn a right triangle, with hypotenuse 1 and adjacent angle θ.
From the definition of the sine and cosine, then, we have that the base of the triangle is
cosθ = x and the altitude is sinθ , which by the Pythagorean Theorem is
sin(cos
−1
x) = sinθ =
1 − x
2
.
There are two subtle points that we should illuminate here. First, what if anything
in Figure 6.27 changes if x < 0? (Think about this one.) Second, since θ = cos
−1
x,we
have that θ ∈ [0,π] and hence, sinθ ≥ 0. Finally, you can also read from Figure 6.27 that
tan(cos
−1
x) = tanθ =
sinθ
cosθ
=
√
1 − x
2
x
.
Note that this last identity is valid,regardless of whether x = cos θ is positive or negative.
5
4 5 sin
u
3 5 cos u
u tan
1
( )
3
4
FIGURE 6.28
θ = tan
−1
4
3
EXAMPLE 4.6 Simplifying an Expression Involving an Inverse Tangent
Simplify sin
tan
−1
4
3
.
Solution Once again, keep in mind that tan
−1
4
3
is the angle θ , in the interval
−
π
2
,
π
2
for which tanθ =
4
3
. As an aid, we visualize the triangle shown in Figure 6.28. Note
that we have made the side opposite the angle θ to be of length 4 and the adjacent side
of length 3, so that the tangent of the angle is
4
3
, as desired. Of course, this makes the