2 LIMITS
2.1 The Limit Idea: Instantaneous Velocity and Tangent Lines
Preliminary Questions
1. Average velocity is equal to the slope of a secant line through two points on a graph. Which graph?
Average velocity is the slope of a secant line through two points on the graph of position as a function of
SOLUTION
time.
2. Can instantaneous velocity be defined as a ratio? If not, how is instantaneous velocity computed?
Instantaneous velocity cannot be defined as a ratio. It is defined as the limit of average velocity as time
elapsed shrinks to zero.
SOLUTION
= 0 Dale entered Highway 1. At t = 2 he was 126 miles down the highway, on the side of the road
with a flat tire. At t = 3 he was still on the side of the road, waiting for road assistance. What was Dale's average velocity
over each of the time intervals:
3. With tin hours, at t
= 0 to t = 2
(a) From t
(b) From t = 0 tot= 3
(c) From t
= 2 tot= 3
SOLUTION
(a) Over the time interval from t = 0 tot= 2, Dale traveled 126 miles. His average velocity was therefore
126
- = 63 miles/hour
2-0
(b) Over the time interval from t
= 0 tot= 3, Dale traveled 126 miles. His average velocity was therefore
126
_
3 0
= 42 miles/hour
(c) Over the time interval from t = 2 tot= 3, Dale traveled 0 mile~. His average velocity was therefore
0
-3 -2
.
= 0 rmles/hour
4. What is the graphical interpretation of instantaneous velocity at a specific time t
SOLUTION
time at t =
= t0 ?
Instantaneous velocity at time t = to is the slope of the line tangent to the graph of position as a function of
t0 .
Exercises
1. A ball dropped from a state of rest at time t = 0 travels a distance s(t)
(a) How far does the ball travel during the time interval [2, 2.5]?
= 4.9t2 min t seconds.
(b) Compute the average velocity over [2, 2.5].
(c) Compute the average velocity for the time intervals in the table and estimate the ball's instantaneous velocity at t = 2.
Interval
[2, 2.01]
[2, 2.005]
[2, 2.001]
[2, 2.00001]
Average
velocity
SOLUTION
(a) Given s(t) = 4.9t2, the ball travels tu = s(2.5) - s(2) = 4.9(2.5)2
-
4.9(2)2 = 11.025 m during the time interval
[2, 2.5].
(b) The average velocity over [2, 2.5] is
tis s(2.5) - s(2)
tit = 2 .5 _ 2
11.025
= -0.S = 22.05 m/s
49
50
CHAPTER 2
I
LIMITS
(c)
time interval
[2, 2.01]
[2, 2.005]
[2,2.001]
[2, 2.00001]
average velocity
19.649
19.6245
19.6049
19.600049
The instantaneous velocity at t = 2 is approximately 19.6 m/s.
3. On her bicycle ride Fabiana's position (in km) as a function of time (in hours) is s(t)
average velocity between t = 2 and t = 3? What was her instantaneous velocity at t = 2.5?
SOLUTION
= 22t + 17. What was her
Fabiana's average velocity between t = 2 and t = 3 was
s(3) - s(2)
3-2
= 83 -
61
1
= 22 km/hour
To estimate the instantaneous velocity, we compute the average velocities:
time interval
[2.5, 2.51]
[2.5, 2.501]
[2.5, 2.5001]
[2.5, 2.50001]
average velocity
22
22
22
22
The instantaneous velocity is 22 km/hour.
In Exercises 5-6, a ball is dropped on Mars where the distance traveled is s(t) = l.9t2 meters int seconds.
S. Compute the ball's average velocity over the time interval [3, 6] and estimate the instantaneous velocity at t = 3.
SOLUTION
The ball's average velocity over the time interval [3, 6] is
s(6) - s(3) = 68.4 - 17.1 =
_ m/s
17 1
6-3
3
To estimate the instantaneous velocity, we compute the average velocities:
time interval
[3,3.1]
[3, 3.01]
[3, 3.001]
[3, 3.0001]
average velocity
11.59
11.419
11.4019
11.40019
The instantaneous velocity is approximately 11.4 m/ s.
In Exercises 7-8, a stone is tossed vertically into the air from ground level with an initial velocity of 15 mis. Its height at
time tis h(t) = l5t - 4.9t2 m.
7. Compute the stone's average velocity over the time interval [0.5, 2.5] and indicate the corresponding secant line on
a sketch of the graph of h.
SOLUTION
The average velocity is equal to
h(2.5) - h(0.5) =
2
I
0 .3 ms
The secant line is plotted with h(t) below.
h
10
8
6
4
2
0.511.522.53
9. The position of a particle at time t is s(t)
estimate the instantaneous velocity at t = 2.
SOLUTION
= 2t3.
Compute the average velocity over the time interval [2, 4] and
The average velocity over the time interval [2, 4] is
s(4) - s(2) _ 128 - 16 _
4-2
2
- 56
To estimate the instantaneous velocity at t
= 2, we examine the following table.
time interval
[2, 2.01]
[2,2.001]
[2, 2.0001]
[1.99, 2]
[l.999, 2]
[1.9999, 2]
average velocity
24.1202
24.012
24.0012
23.8802
23.988
23.9988
The instantaneous velocity at t = 2 is approximately 24.0.
s E c T 1o N
2.1
I
The Limit Idea: Instantaneous Velocity and Tangent Lines
51
In Exercises 11-20, estimate the slope of the tangent line at the point indicated.
x2 + x;
11. f(x) =
x =0
SOLUTION
x interval
[0,0.01]
[0,0.001]
[0,0.0001]
[-0.01,0]
[-0.001,0]
[-0.0001, O]
slope of secant
1.01
1.001
1.0001
0.99
0.999
0.9999
The slope of the tangent line at x = 0 is approximately 1.0.
13. f(t)
= 12t -
7;
= -4
t
SOLUTION
t interval
[-4, -3.99]
[-4,-3.999]
[-4, -3.9999]
slope of secant
12
12
12
t interval
[-4.01,-4]
[-4.001,-4]
[-4.0001,-4]
slope of secant
12
12
12
= -4 is 12, coinciding with the graph ofy = f(t).
The slope of the tangent line at t
1s. y(t) = v3t+ l;
t=1
SOLUTION
t interval
[l, 1.01]
[l, 1.001]
[l, 1.0001]
[0.99, l]
[0.999, l]
[0.9999, l]
slope of secant
.7486
.7499
.7500
.7514
.7501
.7500
= 1 is approximately 0. 75.
The slope of the tangent line at t
17. f(x) = ex;
x = e
SOLUTION
[e - 0.01, e] [e - 0.001, e] [e - 0.0001, e] [e, e + 0.01] [e, e + 0.001] [e, e + 0.0001]
x interval
15.0787
slope of secant
15.1467
15.1535
15.2303
15.1550
15.1618
The slope of the tangent line at x = e is approximately 15.15.
19. f (x) = tan x;
7r
x= -
4
SOLUTION
x interval
slope of secant
[~
- 0.01,
~]
1.98026
[~
- 0.001,
~]
[~
1.99800
The slope of the tangent line at x
- 0.0001,
~]
1.99980
[~, ~
+ 0.01]
[~, ~
2.02027
[~, ~
+ 0.001]
+ 0.0001]
2.00020
2.00200
= ~ is approximately 2.00.
21. The height (in centimeters) at time t (in seconds) of a small mass oscillating at the end of a spring is h(t) = 3 sin(27rt).
Estimate its instantaneous velocity at t = 4.
SOLUTION
To estimate the instantaneous velocity at t
= 4, we examine the following table.
time interval
[4, 4.01]
[4,4.001]
[4,4.0001]
[3.99,4]
[3.999,4]
[3.9999, 4]
average velocity
18.8732
18.8494
18.8496
18.8732
18.8494
18.8496
The instantaneous velocity at t;;;:; 4 is approximately 18.85 cm/s.
23. Consider the function f(x) = yx.
(a) Compute the slope of the secant lines from (0, 0) to (x,f(x)) for x = 1, 0.1, 0.01, 0.001, 0.0001.
(b) Discuss what the secant-line slopes in (a) suggest happens to the tangent line at 0.
(c)
iGU[I
Plot the graph off near x = 0 and verify your observation from (b).
52
CHAPTER 2
I
LIMITS
SOLUTION
(a) The slope of the secant line from (0, 0) to (x, f(x)) for the function f(x)
f(x) -f(O)
Yx
x-0
x
= Yx is
1
-vx·
Thus, for x = 1, 0.1, 0.01, 0.001, 0.0001, the slope of the secant line is
x
1
0.1
0.01
0.001
0.0001
10
31.62278
100
slope of secant 1 3.16228
(b) The secant line slopes from part (a) suggest that the slope of the tangent line at x = 0 grows without bound; that is,
the tangent line at x = 0 is a vertical line.
(c) The graph off shown below confirms that at x = 0 the tangent line is vertical line.
0.01
0.00
0.005
0.010
0.015
0.020
25. ~ If an object in linear motion (but with changing velocity) covers tis meters in M seconds, then its average
velocity is v0 = tis/ M mis. Show that it would cover the same distance if it traveled at constant velocity v0 over the same
time interval. This justifies our calling tis IM the average velocity.
SOLUTION At constant velocity, the distance traveled is equal to velocity times time, so an object moving at constant
velocity v0 for flt seconds travels vo& meters. Since Vo= tis/tit, we find
distance traveled = v06t = (
~;) M = tis
So the object covers the same distance tis by traveling at constant velocity v0 •
27. ~ Which graph in Figure 5 has the following property: For all x, the slope of the secant line over [0,x] is
greater than the slope of the tangent line at x. Explain.
(A)
(B)
FIGURE 5
SOLUTION The graph in (B) bends downward, so the slope of the secant line through (0, 0) and (x, f(x)) is larger than
the slope of the tangent line at (x,f(x)). On the other hand, the graph in (A) bends upward, so the slope of the tangent line
at (x, f(x)) is larger than the slope of the secant line through (0, 0) and (x, f(x)). Thus, the graph in (B) has the desired
property.
29. Let Q(t) = t2. Find a formula for the slope of the secant line over the interval [1, t] and use it to estimate the slope of
the tangent line at t = 1. Repeat for the interval [2, t] and for the slope of the tangent line at t = 2.
SOLUTION
Let Q(t)
= t2. The slope of the secant line over the interval [1, t] is
Q(t) - Q(l)
t2 - 1 (t - l)(t + 1)
----=--=
=t+l
t-1
t-1
t-1
provided t if:. 1. To estimate the slope of the tangent line at t
= 1, examine the values in the table below.
sEc T Io N
2.1
I
The Limit Idea: Instantaneous Velocity and Tangent Lines
53
0.99 0.999 0.9999 1.01 1.001 1.0001
t
slope of secant 1.99 1.999 1.9999 2.01 2.001 2.0001
The slope of the tangent line at t = 1 is approximately 2.0.
The slope of the secant line over the interval [2, t] is
Q(t) - Q(2) =
provided t
t2 - 4 = (t - 2)(t + 2) = t + 2,
t-2
t-2
t-2
* 2. To estimate the slope of the tangent line at t = 2, examine the values in the table below.
t
1.99 1.999 1.9999 2.01 2.001 2.0001
slope of secant 3.99 3.999 3.9999 4.01 4.001 4.0001
The slope of the tangent line at t = 2 is approximately 4.0.
31. For f(x) = x3, show that the slope of the secant line over [-3, x] is :i2 - 3x + 9, and use this to estimate the slope of
the tangent line at x = -3.
SOLUTION
Let f(x)
=.x3. The slope of the secant line over the interval [-3, x] is
f(x)-/(-3) = .x3+27 = (x+3)(i2-3x+9)
x-(-3)
x+3
x+3
provided x
=x2- 3x+ 9
* -3. To estimate the slope of the tangent line at x = -3, examine the values in the table below.
x
-3.01
-3.001
-3.0001
-2.99
-2.999
-2.9999
slope of secant 27.0901 27.009001 27.000900 26.9101 26.991001 26.999100
The slope of the tangent line at x = -3 is approximately 27.0.
Further Insights and Challenges
The next two exercises involve limit estimates related to the definite integral, an important topic introduced in Chapter 5.
33. Let A represent the area under the graph of y = x3 between x = 0 amd x = 1. In this problem, we will follow the
process in Exercise 32 to approximate A.
(a) As in (a)-(d) in Exercise 32, separately divide [O, l] into 2, 3, 5, and 10 equal-width subintervals, and in each case
compute an overestimate of A using rectangles on each subinterval whose height is the value of .x3 at the right end of the
subinterval.
In this case, it can be shown that if we use n equal-width subintervals, then the total area A(n) of then rectangles is
A(n) = (n
+ 1)2
4n2
(b) Compute A(n) for n
= 2, 3, 5, 10 to verify your results from (a).
(c) Compute A(n) for n = 100, 1000, and 10,000. Use your results to conjecture what the area A equals.
SOLUTION
(a)
• Dividing [O, 1] into 2 equal-width subintervals produces two rectangles with width 1/2 and heights 1/8 and 1.
The combined area of the two rectangles is then
1 1 1
9
-·-+-·l=2 8 2
16
• Dividing [O, 1] into 3 equal-width subintervals produces three rectangles with width 1/3 and heights 1/27, 8/27 and
1. The combined area of the three rectangles is then
1 1
I 8
1
4
-·-+-·-+-·1=3 27 3 27 3
9
• Dividing [0, 1] into 5 equal-width subintervals produces five rectangles with width 1/5 and heights
54
CHAPTER 2
I
LIMITS
respectively. The combined area of the five rectangles is then
(~)3 + !5 (~)3
+ ! (~)3 + ! . 13 = _2._
5
5 5
5
25
! (!)3 + !
5 5
5 5
• Dividing [O, 1] into 10 equal-width subintervals produces 10 rectangles with width 1/10 and heights
3
3
3
3
3
3
3
3
3
1 ) ( 1 ) ( 3 ) ( 2 ) ( 1 ) ( 3 ) ( 7 ) (4 ) ( 9 )
3
( 10
' S ' 10 ' S ' 2 ' S ' 10 ' S ' 10 ' and l
The combined area of the five rectangles is then
(b) Let
= (n;;>2
A(n)
Then,
9
9
16
4
A( 2)
= 4(4) = 16
A( 3 )
= 4(9) = 9
A(5 )
= 4(25) = 25'
36
9
121)
A(lO)
and
121
= 4(100) = 400
confirming the results from (a).
(c) We find
A(lOO) =
A(lOOO)
4
101 2
(100)2 = 0.255025
1001 2
= 4 (1000)2 = 0.25050025,
A(lO, 000) =
4
and
10,001 2
(l0, 000) 2 = 0.2500500025
Based on these results, we conjecture that the area A equals ~.
2.2 Investigating Limits
Preliminary Questions
1. What is the limit of f(x) = 1 as x
SOLUTION
limx~"
n?
1 = 1.
2. What is the limit of g(t)
SOLUTION
~
lim14,, t =
= t as t ~ n?
7r.
3. Is lim 20 equal to 10 or 20?
x~lO
SOLUTION
limx~w20
= 20.
4. Can f(x) approach a limit as x
~
c if /(c) is undefined? If so, give an example.
Yes. The limit of a function fas x ~ c does not depend on what happens at x
c. As an example, consider the function
SOLUTION
f
as x
~
f(x)
= x2 -1
x-1
= c, only on the behavior of
s Ec T Io N
The function is clearly not defined at x
I
2.2
Investigating Limits
55
= 1 but
limf(x) = lim i2- X--+1
x--+l
1
x- I
= lim(x +I)= 2
x--+1
5. What does the following table suggest about lim f(x) and lim f(x)?
x--+I-
x
f(x)
SOLUTION
x--+t+
0.9
0.99
0.999
1.001
1.01
1.1
7
25
4317
3.00011
3.0047
3.0126
The values in the table suggest that limx_, 1- f(x)
= oo and limx_, 1+ f(x) = 3.
6. Can you tell whether limf(x) exists from a plot off for x > 5? Explain.
x->5
No. By examining values of f(x) for x close to but greater than 5, we can determine whether the one-sided
limit limx_, 5+ f(x) exists. To determine whether limx--.5 f(x) exists, we must examine value of f(x) on both sides of x = 5.
SOLUTION
7. If you know in advance that limf(x) exists, can you determine its value from a plot off for all x > 5?
x->5
SOLUTION
Yes. If limx....5 f(x) exists, then both one-sided limits must exist and be equal.
Exercises
In Exercises 1-5,fill in the table and guess the value of the limit .
= -.x3
X2 -
1. limf(x), where f(x)
x--+1
1
1
x
f(x)
x
f(x)
1.002
0.998
1.001
0.999
1.0005
0.9995
1.00001
0.99999
SOLUTION
x
0.998
0.999
0.9995
0.99999
1.00001
1.0005
1.001
1.002
f(x)
1.498501
1.499250
1.499625
1.499993
1.500008
1.500375
1.500750
1.501500
The limit as x
--t
I is ~.
3. limf(y), where f(y)
y->2
=~ - y -
2
y + y- 6
y
y
f(y)
f(y)
2.002
1.998
2.001
1.999
2.0001
1.9999
SOLUTION
The limit as y
--t
y
1.998
1.999
1.9999
2.0001
2.001
2.002
f(y)
0.59984
0.59992
0.599992
0.600008
0.60008
0.60016
2 is ~.
56
CHAPTER 2
I
LIMITS
5. limf(t), where f(t) =
l -cos2t
t
t-->0
f(t)
t
t
0.002
-0.002
0.001
-0.001
0.0005
-0.0005
0.00001
-0.00001
f(t)
SOLUTION
f(t)
t
t
f(t)
0.002
0.004
-0.002
-0.004
0.001
0.002
-0.001
-0.002
0.0005
0.001
-0.0005
-0.001
0.00001
0.00002
-0.00001
-0.00002
The limit as t ~ 0 is 0.
7. Determine lim f(x) for fas in Figure 10.
x--+0.5
y
FIGURE 10
SOLUTION
The graph suggests that f(x)
~
1.5 as x
~
0.5.
In Exercises 9-IO, evaluate the limit.
9. lim x
x--+21
As x ~ 21, f(x) = x ~ 21. You can see this, for example, on the graph of f(x) = x.
11. Show, via illustration, that the limits limx and lima are equal but the functions in each limit are different.
SOLUTION
X--+a
x--+a
SOLUTION
The figure above displays the graphs of f(x) = x and g(x) = a. Clearly, the two functions are different. It is also clear
that as x approaches a, both graphs approach the point (a, a); that is,
limx = lima =a
x--+a
x--+a
In Exercises I3-20, verify each limit using the limit definition. For example, in Exercise 13, show that l3x - 121 can be
made as small as desired by taking x close to 4.
13. lim3x = 12
x~4
SOLUTION
Ix - 41 small.
13x - 121 = 31x - 41. 13x - 121 can be made arbitrarily small by making x close enough to 4, thus making
S E CT I o N 2.2
15. lim(5x + 2)
x .... 3
I
Investigating Limits
57
= 17
1(5x + 2) - 171 = 15x - 151
1(5x + 2) - 171 as small as desired.
SOLUTION
51x - 31. Therefore, if you make Ix - 31 small enough, you can make
17. limx2 = 0
x ....o
As x - t 0, we have lx2 - 01 = Ix+ Ollx - OI. To simplify things, suppose that lxl < 1, so that Ix+ Ollx - 01 =
lxllxl < lxl. By making lxl sufficiently small, so that Ix+ Ollx - OI = x2 is even smaller, you can make lx2 - 01 as small as
desired.
SOLUTION
19. lim(4x2 + 2x + 5)
x->0
=5
As x - t 0, we have 14x2 + 2x + 5 - 51 = 14x2 + 2xl = lxll4x + 21. If lxl < 1, 14x + 21 can be no bigger than 6,
so lxll4x + 21 < 61xl. Therefore, by making Ix - 01 = lxl sufficiently small, you can make 14x2 + 2x + 5 - 51 = lxll4x + 21 as
small as desired.
SOLUTION
In Exercises 21-38, estimate the limit numerically or state that the limit does not exist. If infinite, state whether the
one-sided limits are oo or -oo.
. yx-1
21. hm--x~i
X- 1
SOLUTION
The limit as x
23. lim
-t
1 is
x
0.9995
0.99999
1.00001
1.0005
f(x)
0.500063
0.500001
0.49999
0.499938
x
1.999
1.99999
2.00001
2.001
f(x)
1.666889
1.666669
1.666664
1.666445
x
-0.Ql
-0.005
0.005
0.01
f(x)
1.999867
1.999967
1.999967
1.999867
x
-0.01
-0.001
0.001
0.01
f(x)
0.999850
0.999999
0.999999
0.999850
x
-0.05
-0.001
0.001
0.05
f (x)
0.0249948
0.0005
-0.0005
-0.0249948
t.
x2+x-6
-x- 2
x->2 X 2
SOLUTION
The limit as x--+ 2 is ~-
25. lim sin 2x
x~o
x
SOLUTION
The limit as x --+ 0 is 2.
27• lim sin3x
x->0
3x
SOLUTION
The limit as x --+ 0 is 1.
29 . Jim cosB-1
o-.o
B
SOLUTION
The limit as x --+ 0 is 0.
58
CHAPTER 2
I
LIMITS
.
1
31. h
)
m-(
x-+4 X- 4 3
SOLUTION
The limit does not exist. As x
x
3.9
3.99
3.999
4.001
4.01
4.1
f (x)
-1000
-106
-109
109
106
1000
4+, f(x)
~
~
~
4-, f(x)
-oo; similarly, as x
~
oo.
x+3
33. l i m
--x--+-3 x 2 + x-6
SOLUTION
x
-3.1
-3.01
-3.001
-2.999
-2.99
-2.9
f(x)
-0.196078
-0.199601
-0.199960
-0.200040
-0.200401
-0.204082
The limit as x ~ -3 is
-k-
x- 4
35. Jim ---zx-+3+ X - 9
SOLUTION
The limit does not exist. As x
~
x
3.001
3.01
3.1
f(x)
-166.472
-16.473
-1.475
3+, f(x)
~
-oo.
1
37. limsinhcos -h
h--+0
SOLUTION
~
The limit as x
h
-0.01
-0.001
-0.0001
0.0001
0.001
0.01
f(h)
-0.008623
-0.000562
0.000095
-0.000095
0.000562
0.008623
0 is 0.
39. limlxl-"
x--+0
SOLUTION
The limit as x
~
x
-0.05
-0.001
-0.00001
0.00001
0.001
0.05
f(x)
1.161586
1.006932
1.000115
0.999885
0.993116
0.860892
0 is 1.
t-e
41. l i m - He lnt-1
SOLUTION
The limit as t
~
t
e - 0.01
e - 0.001
e -0.0001
e + 0.0001
e + 0.001
e+O.Ql
f(t)
2.713279
2.717782
2.718232
2.718332
2.718782
2.723279
e is approximately 2.718. (The exact answer is e.)
1
. tan- x
43. h m - x--+1-
cos-IX
SOLUTION
x
0.9
0.99
0.999
0.9999
f(x)
1.624771
5.513466
17.549388
55.532038
The limit does not exist. As x--> 1-, f(x)
~
oo.
s Ec T I o N
2.2
(b) lim LxJ
x-+c-
SOLUTION
Investigating Limits
59
= n, where n is the unique integer
45. The greatest integer function, also known as the floor function, is defined by LxJ
such that n ::; x < n + 1. Sketch the graph of y = LxJ. Calculate for c an integer:
(a) lim LxJ
I
(c) lim LxJ
x-+c+
x-+2.6
The graph ofy = LxJ is shown below.
y
3
2
---0
---0
--1--+---+---+---o---+---<f-----+--X
-3
-2
-]
3
4
2
-2
---0
-3
---0
(a) From the graph of the greatest integer function, we see that lim LxJ
x-+c-
=c -
l, where c is an integer.
(b) Again from the graph of the greatest integer function, we see that lim LxJ
x-+c+
= c, where c is an integer.
(c) Examining the graph in part (a), we see that lim LxJ = 2.
x-+2.6
In Exercises 47-54, determine the one-sided limits numerically or graphically. If infinite, state whether the one-sided
limits are oo or -oo, and describe the corresponding vertical asymptote. In Exercise 54, f(x) = LxJ is the greatest integer
function defined in Exercise 45.
.
sinx
47. hm x-+O±
1X 1
SOLUTION
x
-0.2
-0.02
0.02
0.2
f(x)
-0.993347
-0.999933
0.999933
0.993347
= 1.
The left-hand limit is lim f(x) = -1, whereas the right-hand limit is lim f(x)
x-+o-
x-+O+
. x_-_s_in_lx_I
49• 1Im
x--+O±
x3
SOLUTION
x
-0.1
-0.Ql
O.Ql
0.1
f(x)
199.853
19999.8
0.166666
0.166583
The left-hand limit is lim f(x) = oo, whereas the right-hand limit is lim f(x) =
x-+O-
x--+O+
.!..
Because lim f(x) = oo, the line x =0
6
x-+O'
is a vertical asymptote.
4.x2 + 7
51. lim - 3 - x-+-2± X + 8
SOLUTION
x
-2.1
-2.01
-1.99
-1.9
f(x)
-19.540048
-192.041525
191.291530
18.790535
The left-hand limit is lim_ f(x)
X-+-2
infinite, the line x
= -oo, whereas the right-hand limit is X-+-2+
lim f(x) = oo. Because the one-sided limits are
= -2 is a vertical asymptote.
53. lim x5 + x - 2
x-->I°' if + X - 2
SOLUTION
x
0.99
0.999
1.001
1.01
f(x)
1.973577
1.997336
2.002669
2.026912
60
CHAPTER 2
I
LIMITS
= 2, whereas the right-hand limit is lim f(x) = 2.
Determine the one-sided limits at c = 2 and c = 4 of the function f in Figure 13. What are the vertical asymptotes
The left-hand limit is lim f(x)
x-+1-
x-+t+
55.
off?
y
''
'
15
~
IO
5
x
-5
FIGURE 13
SOLUTION
= 2, we have x-+2lim f(x) = oo and lim f(x) = oo.
x--+2+
For c = 4, we have lim f(x) = -oo and lim f(x) = 10.
x--+4x-+4+
• For c
•
The vertical asymptotes are the vertical lines x = 2 and x = 4.
In Exercises 57-60, sketch the graph of a function with the given limits.
57. limf(x)
x--+ I
= 2,
Jim f(x)
x-+3-
= 0,
Jim f(x)
x--+3+
=4
SOLUTION
y
6
4
2
2
59. lim f(x)
x--+2+
= /(2) = 3,
= -1,
lim f(x)
x--+2-
limf(x)
x-+4
3
4
= 2 * /(4)
SOLUTION
y
-1
2
4
5
61. Determine the one-sided limits of the function fin Figure 15, at the points c
y
5-k::'.....................................................................,
4
3
2
I
-1
-2
-3
--4
FIGURE 15 Graph off.
= 1, 3, 5, 6.
S E CT I O N 2.2
SOLUTION
Investigating Limits
61
Based on the graph of the function fin Figure 15,
lim f(x)
x--+1-
lim f(x)
x--+3-
= 3,
lim f(x) = 3,
x--+I+
= -oo,
lim f(x) = 4,
x--+3+
lim f(x) = -3,
lim f(x) = 2,
x--+5+
x--+5-
lim f(x)
x--+6-
jGq
I
= oo, and x--+6+
lim f(x) = oo.
In Exercises 63--68, plot the function and use the graph to estimate the value of the limit.
sin5B
63. lim -:--8-+0 sm 28
SOLUTION
y
2.50
The limit as B-+ 0 is~·
. 2x -COSX
65. h m - - - x--+O
X
SOLUTION
y
0.6940
0.6920
The limit as x-+ 0 is approximately 0.693. (The exact answer is In 2.)
67
• lim cos7B-cos58
8-+0
82
SOLUTION
y
The limit as B -+ 0 is -12.
69. Let n be a positive integer. For which n are the two infinite one-sided limits lim 1/x!' equal?
x-+()±
SOLUTION
If x > 0, then x!' > 0 for any positive integer n. Moreover, as x -+
o+, x!'
-+
o+, so
.
1
1Im - =oo
x-+O+
x"
for any positive integer n. On the other hand, if x < 0, then x!' < 0 when n is an odd positive integer and x!' > 0 when n
is an even positive integer. Accordingly,
. -1 = { -oo,
IIm
x--+O-
xn
oo,
n is an odd positive integer
n is an even positive integer.
Thus, the two infinite one-sided limits lim 1/ x!' are equal when n is an even positive integer.
x--+O±
62
CHAPTER 2
I
LIMITS
71. !Gull
In some cases, numerical investigations can be misleading. Plot f(x) =cos~­
(a) Does Iimf(x) exist?
x-+0
(b) Show, by evaluating f(x) at x = ±~, ±~, ±k, ... , that you might be able to trick your friends into believing that the
limit exists and is equal to L = 1.
(c) Which sequence of evaluations might trick them into believing that the limit is L = -1?
SOLUTION
A graph off is shown below:
y
(a) Based on the graph off, it appears that the function values oscillate more and more rapidly between +1 and -1 as
x-+ 0. Accordingly, it appears that limf(x) does not exist.
x-+0
(b) Evaluating f(x) at x = ±~, ±~, ±k, ... , we find
t(±~) = cos(±2:ir) = 1
t(±~) = cos(±4:ir) = 1
t(±~) = cos(±6:ir) = 1
and so on.
(c) To trick your friends into believing that L = -1, evaluate f(x) at x = ±1, ±i, ±~, ....
Further Insights and Challenges
.
1·im sin nB numenc
. all y "ior severaI pos1t:J.ve
. . mteger
.
. general.
vaI ues of n. Then guess the value m
73• Invest:J.gate
9... 0
e
SOLUTION
• For n = 3, we have
e
-0.l
-0.Ql
-0.001
0.001
0.01
0.1
sinnB
--
2.955202
2.999550
2.999996
2.999996
2.999550
2.955202
e
-0.1
-0.01
-0.001
0.001
O.Ql
0.1
sinnB
--
4.794255
4.997917
4.999979
4.999979
4.997917
4.794255
e
The limit as e-+ 0 is 3.
• For n = 5, we have
e
The limit as B -+ 0 is 5.
,.,
.
• ne sunnise
th
.
. sinne
at, m general, hm - 9... 0
e = n.
x" - 1
75. Investigate Iim - - for (m,n) equal to (2, 1), (1, 2), (2,3), and (3,2). Then guess the value of the limit in general
x-+1 x"' - 1
and check your guess for two additional pairs.
SOLUTION
The limit as x-+ 1 is!-
x
0.99
0.9999
1.0001
1.01
x-l
x 2 -1
0.502513
0.500025
0.499975
0.497512
s EC T I O N
x
x2- l
x-1
0.99
0.9999
1.0001
1.01
1.99
1.9999
2.0001
2.01
2.2
I
The limit as x --) 1 is 2.
x
0.99
0.9999
1.0001
1.01
x2- l
x 3 -1
0.670011
0.666700
0.666633
0.663344
x
0.99
0.9999
1.0001
1.01
1.492513
1.499925
1.500075
1.507512
The limit as x --) 1 is ~.
x3- l
x 2 -1
The limit as x --) 1 is ~.
x"-1
• For general m and n, we have lim - x .... 1 :!!"- 1
= -mn .
x
0.99
0.9999
1.0001
1.01
x-1
x 3 -1
0.336689
0.333367
0.333300
0.330022
The limit as x--) 1 is ~x
x3- l
x-1
0.99
0.9999
1.0001
1.01
2.9701
2.9997
3.0003
3.0301
The limit as x --) 1 is 3.
x
x3- l
x1 -1
0.99
0.9999
1.0001
1.01
0.437200
0.428657
0.428486
0.420058
The limit as x--) 1 is ~ ~ 0.428571.
2x 8
Plot the graph of f(x) = x ~ .
3
(a) Zoom in on the graph to estimate L = Iimf(x).
77.
~
!Gq
~
x .... 3
(b) Explain why
f(2.99999) :s; L :s; f(3.00001)
Use this to determine L to three decimal places.
SOLUTION
(a)
Y
5.525
x=3
Investigating Limits
63
64
CHAPTER 2
I
LIMITS
(b) It is clear that the graph off rises as we move to the right. Mathematically, we may express this observation as:
whenever u < v, f(u) < f(v). Because
2.99999 < 3 == limx < 3.00001
x->3
it follows that
f(2.99999) < L == Iimf(x) < f(3.00001)
x->3
With f(2.99999) "' 5.54516 and f(3.00001) "' 5.545195, the above inequality becomes 5.54516 < L < 5.545195; hence,
to three decimal places, L == 5.545.
2.3 Basic Limit Laws
Preliminary Questions
1. State the Sum Law and Quotient Law.
SOLUTION
Suppose limx_,c f(x) and limx->c g(x) both exist. The Sum Law states that
lim(f(x) + g(x)) == Iimf(x) + Iimg(x)
X->C
Provided limx_,c g(x)
X->C
X->C
* 0, the Quotient Law states that
limf(x)
f( X )
X->C
Iim - - = = - - x->c g(x)
limg(x)
X->C
2. Which of the following is a verbal version of the Product Law (assuming the limits exist)?
(a) The product of two functions has a limit.
(b) The limit of the product is the product of the limits.
(c) The product of a limit is a product of functions.
(d) A limit produces a product of functions.
SOLUTION
The verbal version of the Product Law is (b): The limit of the product is the product of the limits.
3. Which statement is correct? The Quotient Law does not hold if
(a) The limit of the denominator is zero
(b) The limit of the numerator is zero
SOLUTION
Statement (a) is correct.
Exercises
In Exercises 1-26, evaluate the limit using the Basic Limit Laws and the limits Iim xPfq == cpfq and lim k == k.
X->C
1. limx
x->9
SOLUTION
limx == 9
x->9
3. lim x
4
X-7~
SOLUTION
!~11 x
4
==
Gr
16
5. limr 1
1-->2
SOLUTION
We apply the definition of r
1
,
and then the Quotient Law:
lim 1
. -I
. 1 t->2
1
1Im
t == 1Im - == - - == 1-->2
1-->2
t
lim t
t->2
7. Jim (3x + 4)
X->0.2
2
X->C
sEc T Io N
SOLUTION
2.3
We apply the Laws for Sums and Constant multiples:
lim(3x+4)
x--+0.2
= x--+0.2
lim 3x+ lim 4
x--+0.2
= 3 x--+0.2
lim x + lim 4 = 3(0.2) + 4 = 4.6
x--+0.2
9. lim (3x4
-
2.x3 + 4x)
x--+-1
SOLUTION
We apply the Laws for Sums, Constant multiples, and Powers:
lim (3x4
x--+-1
-
2.x3 + 4x)
= x--+-1
lim 3x4 -
lim 2.x3 + lim
x--+-1
= 3 x--+-I
lim x4 = 3(-1
4
) -
4x
x--+-1
2 lim .x3 + 4 lim x
x--+-1
x--+-l
2(-1
3
) -
4
=3 + 2 -
4
=1
11. lim(x + 1)(3x2 - 9)
x--+2
SOLUTION
We apply the Laws for Products, Sums, Constant multiples and Powers:
lim (x + 1) (3x2 x--+2
9) = {limx + lim 1) {3 limx2 - lim 9)
x--+2
x--+2
x--+2
x--+2
= (2 + 1) ( 3(2)2 - 9) = 3 . 3 = 9
.
1
13. h m t--+4 t + 4
SOLUTION
We apply the Laws for Quotients and Sums:
lim 1
lim-1H4f+4
= ~ = _1_ = !
limt+4
4+4
t--+4
8
3t- 14
15. lim--t--+4
t+1
SOLUTION
We apply the Laws for Quotients, Sums, and Constant multiples:
3t - _
14
lim _
1--+4 t + 1
=
3limt-lim14
t--+4
3. 4- 14
t--+4
4+1
limt +Jim 1
t--+4
t--+4
2
=--
5
17. lim(l6y + 1)(2y112 + 1)
y--+i
SOLUTION
We apply the Laws for Products, Sums, Constant multiples, and Roots:
li~(l6y + 1)(2y
112
+ 1) = (16
y--+4
19. Jim
y--+ 4
li~ y + Jim 1) · (2 lim y 1
12
y--+4
y--+!
y--+!
+Jim 1)
y--+!
l
..j6y + 1
SOLUTION
We apply the Laws for Quotients, Roots, Sums, and Constant multiples:
.
11m
y--+ 4
1
..j6y + 1
liml
y--+4
61imy +Jim 1
y--+4
21. Jim __
x_
x--+-1x3 +4x
y--+4
1
y6(4) + 1
1
1
= .../25 = 5
I
Basic Limit Laws
65
66
CHAPTER 2
I
LIMITS
We apply the Laws for Quotients, Sums, Powers, and Constant multiples:
SOLUTION
lim x
lim __x_ =
x-+-1
x3
-1
(-1)3 +4(-1)
x-+-1
+ 4x
lim
X-+-1
x3 + 4
lim x
-1
1
= -1-4 = 5
X-+-1
3 Vi- !t
23. t-+25
lim (t ~2
20
We apply the Laws for Quotients, Sums, Constant multiples, and Powers and Roots:
SOLUTION
1
3limVt-!limt
3 -r.
.
yt- 5f
H25
5 H25
hm
=------""'""
H25 (f - 20)2
(
)2
lim t - lim 20
f-425
3 VE - k<25)
(25 - 20)2
10
2
= 25 = 5
t--+25
25. lim(4t2 + 8t - 5) 312
t--+~
We apply the Laws for Powers, Sums, and Constant multiples:
SOLUTION
lim(4t2 + 8t - 5)3'
2
t-+~
2
= (4 (~) + 8 · ~
2
2
)3/2
- 5
= 163' 2 = 64
27. Use the Quotient Law to prove that if limf(x) exists and is nonzero, then
x-+c
.
1
1
hm--=--x-+c f(x)
limf(x)
X->C
Since limf(x) is nonzero, we can apply the Quotient Law:
SOLUTION
X->C
r
x~
1 )
(
f(x)
(lim
1)
X->C
1
= ( ~f(x)) = 1~f(x)
1
In Exercises 29-32, evaluate the limit assuming that lim f(x)
X-+-4
= 3 and X--+-4
lim g(x) = 1
29. lim f (x)g(x)
x--+-4
lim f(x)g(x)
SOLUTION
X-+-4
= x-+-4
lim f (x)
lim g(x)
X-+-4
=3 · 1 =3
31. lim g(x)
x-+-4 x2
Since lim x2
SOLUTION
X-+-4
* 0, we may apply the Quotient Law, followed by the Law for Powers:
lim
( ) =
~
x-+-4 x 2
lim g(x)
_x->_-_4 _ _
lim x 2
= (-4)2 = 16
x->-4
33. Can the Quotient Law be applied to evaluate lim sin x? Explain.
x--+0
X
SOLUTION The limit Quotient Law cannot be applied to evaluate lim sinx since limx = 0. This violates a condition
x-+0
X
x--+0
of the Quotient Law. Accordingly, the rule cannot be employed.
35. Assume that iflimf(x)
x--+a
not exist.
(a) lim sin ( ~ 1 )
x--+0
(b) lim
X-+7r/2
x
sinx
X
(c) lim
. (3x l
x--+1 SID 1-x
(d)
limx2 sin(JTx2)
x-+I
= L, then limsinf(x)
= sinL. In each case evaluate the limit or indicate that the limit does
x-+a
s E c T 1o N 2.3 I Basic Limit Laws 67
SOLUTION
(a) Because limo
x=I
X->
= O~I = 0, it follows that
limsin(-x-)
x->O
x-1
= sinO = 0
(b) By the Quotient Law,
sinx
lim
x-m/2
X
. 7r
sm 2
= ---yr-
2
= :;;:
~'
2
(c) Because lim(l - x)
x->I
= 1 - 1 = 0, it follows that
= sinO = 0
limsin(l-x)
x->0
Now, lim 3x = 3(1)
x->1
= 3 * 0, so
.
hm
x->I
3x
d
.
oes not exist
sin(l - x)
(d) Because lim7rx2 = 7r(l) 2 = 7r, it follows that
x->I
limsin(7rx2)
x->1
= sin7r = 0
Then, by the Product Law,
limx2 sin(7rx2) = limx2 · limsin(nx2) = 12 • 0 = 0
x--+1
x--+1
x--+1
37. Give an example where lim(f(x) + g(x)) exists but neither limf(x) nor limg(x) exists.
x-+0
SOLUTION
limg(x)
x-+0
x--+0
x-+0
= lim-1/x
do not exist.
x-+0
39. Give an example where Jim
x--+0
x-+0
x-+O
x-ioO
x--+0
Let
f(x)
Then lim f(x)
x--+0
exists but neither limf(x) nor limg(x) exists.
f((x))
x--+0 g x
SOLUTION
x-+0
Let f(x) = 1/x and g(x) = -1/x. Then lim(f(x) + g(x)) = limO = 0 However, limf(x) = lim 1/x and
g(x)
={
x <0
..,.1,
1,
x
~
1
and
0
g(x) = { -'1,
x<0
x ~0
= lim(-1)
= -1; however, neither limf(x)
nor limg(x) exists.
x--+0
x--+0
x-+0
Further Insights and Challenges
41. Suppose that limtg(t)
,..... 3
equals 4.
SOLUTION
= 12. Show that limg(t)
exists and
3
,.....
We are given that lim tg(t)
t-+3
= 12. Since lim t = 3 * 0, we may apply the Quotient Law:
t--+3
limtg(t)
limg(t)
= lim tg(t) = ~ = 12 = 4
t->3
t->3
t
hmt
3
t->3
f(x)
43.~ Assuming that lim
x--+0 x
(a) /(0)
= 1, which of the following statements is necessarily true? Why?
=0
(b) limf(x)
x->0
=0
SOLUTION
(a) Given that lim f(x) = 1, it is not necessarily true that f(O) = 0. A counterexample is provided by f(x) = {x,
x->O
(b) Given that lim f(x)
x-+0
5,
X
X
= l, it is necessarily true that limf(x) = 0. For note that limx = 0, whence
x-+0
lim f(x)
x--+0
= lim
x
x-+0
f (x)
X
x--+0
= (lim
x) (lim f(x)) = 0 · 1 = O
x--+0
x--+0
X
68
CHAPTER 2
I
LIMITS
45. ~ Suppose that limg(h) = L.
h-->O
= L for any constant a* 0.
Ifwe assume instead that limg(h) = L, is it still necessarily true that Iimg(ah) = L?
(a) Explain why Iimg(ah)
h-->O
(h)
~·
(c) Illustrate (a) and (b) with the function f(x)
~·
= x2.
SOLUTION
(a) As h ---> 0, ah ---> 0 as well; hence, if we make the change of variable w
Iimg(ah)
h-tO
= ah, then
= w-+0
Iimg(w) = L
(b) No. Ash---> 1, ah---> a, so we should not expect limg(ah) = limg(h).
h-->I
(c) Letg(x)
h-->I
= x2. Then
limg(h) = 0
and
h-->O
limg(ah) = lim(ah) 2 = 0
h-->0
h-->0
On the other hand,
Iimg(h) = 1
h-->I
Iimg(ah) = Iim(ah) 2 = a 2
while
which is equal to the previous limit if and only if a
h-->I
h-->I
= ± 1.
2.4 Limits and Continuity
Preliminary Questions
1. Which property of f(x)
lim.x3 = 8?
=x3 allows us to conclude that
x-->2
SOLUTION
We can conclude that lim x3
x-->2
= 8 because the function x3 is continuous at x = 2.
2. What can be said about f (3) if f is continuous and
Iimf(x) =
x-->3
SOLUTION
!?
Iffis continuous and limf(x)
x-->3
1
= -2 , then f(3)
=
!·
3. Suppose that f(x) < 0 if xis positive and f(x) > I if xis negative. Can f be continuous at x = O?
SOLUTION
Since f(x) < 0 when xis positive and f(x) > I when xis negative, it follows that
lim f(x)
x-+O+
:<::;
0
lim f(x) 2: I
and
x-+O-
Thus, Iimf(x) does not exist, so f cannot be continuous at x = 0.
x-->O
4. Is it possible to determine f(7) if f(x)
= 3 for all x < 7 and f
is right-continuous at x
= 7? What if f
is left-continuous?
SOLUTION No. To determine f(7), we need to combine either knowledge of the values of f(x) for x < 7 with leftcontinuity or knowledge of the values of f(x) for x > 7 with right-continuity.
5. Are the following true or false? If false, then draw or give a counterexample, and state a correct version.
(a) f is continuous at x = a if the left- and right-hand limits of f(x) as x ---> a exist and are equal.
(b) f is continuous at x =a if the left- and right-hand limits of f(x) as x ---> a exist and equal f(a).
(c) If the left- and right-hand limits of f(x) as x---> a exist, then f has a removable discontinuity at x =a.
(d) If f and g are continuous at x =a, then f + g is continuous at x =a.
(e) If f and g are continuous at x = a, then f / g is continuous at x = a.
SOLUTION
(a) False. The correct statement is "f(x) is continuous at x = a if the left- and right-hand limits of f(x) as x ---> a exist
and equal f(a)."
(h) True.
(c) False. The correct statement is "If the left- and right-hand limits of f(x) as x ---> a are equal but not equal to f(a),
thenf has a removable discontinuity at x = a."
(d) True.
(e) False. The correct statement is "If f(x) and g(x) are continuous at x = a and g(a) * 0, then f(x)/ g(x) is continuous
at x =a."
sEcT IoN
2.4
I
Limits and Continuity
69
Exercises
1. Referring to Figure 15, state whether f is left- or right-continuous (or neither) at each point of discontinuity. Does f
have any removable discontinuities?
y
5
4
3
2
•
2
3
4
5
FIGURE 15 Graph ofy
6
= f(x).
SOLUTION
• The function f is discontinuous at x
• The function f is discontinuous at x
• The function f is discontinuous at x
= 1; it is left-continuous there.
= 3; it is neither left-continuous nor right-continuous there.
= 5; it is left-continuous there.
However, these discontinuities are not removable.
Exercises 2-4 refer to the function g whose graph appears in Figure 16.
y
5
4
2
'---+---+~+---+---+~+--~x
2
4
5
FIGURE 16 Graph ofy
x
6
=g(x).
3. At which point c does g have a removable discontinuity? How should g(c) be redefined to make g continuous at
= c?
SOLUTION Because limg(x) exists, the function g has a removable discontinuity at x
x-+3
g continuous at x = 3.
= 3. Assigning g(3) = 4 makes
5. In Figure 17, determine the one-sided limits at the points of discontinuity. Which discontinuity is removable and
how should f be redefined to make it continuous at this point?
y
-2
2
4
FIGURE 17
SOLUTION
= 0, at which x-+olim f(x) = oo and lim f(x) = 2. The function f is also
x-+O+
= 6 and x--+2+
lim f(x) = 6. The discontinuity at x = 2 is removable. Assigning
The function f is discontinuous at x
discontinuous at x = 2, at which x--+2lim f(x)
/(2) = 6 makes f continuous at x = 2.
In Exercises 7-16, use Theorems 1-5 to show that the function is continuous.
7. f(x) = x + sinx
SOLUTION The polynomial function xis continuous by Theorem 2, and the trigonometric function sinx is continuous
by Theorem 3. Therefore, x + sinx is continuous by Theorem l(i).
9. f(x) = 3x + 4sinx
70
CHAPTER 2
I
LIMITS
SOLUTION The polynomial function 3x is continuous by Theorem 2, and the trigonometric function sin xis continuous
by Theorem 3. Moreover, 4 sinx is continuous by Theorem l(ii). Therefore, 3x + 4 sinx is continuous by Theorem l(i).
11. f(x) = - 2-
1
x +1
1
The function f (x) = - 2- -
SOLUTION
f is continuous by Theorem 2.
x +1
is a rational function whose denominator, x2 + 1, is never equal to 0. Therefore,
13. f(x) = cos(x2)
SOLUTION The polynomial function x2 is continuous by Theorem 2, and the trigonometric function cos xis continuous
by Theorem 3. The function f is the composition of these two continuous functions, so f is continuous by Theorem 5.
15. f(x) = 3x cos 3x
SOLUTION The exponential function 3x is continuous by Theorem 3. Moreover, the polynomial function 3x is continuous by Theorem 2 and the trigonometric function cos x is continuous by Theorem 3, so the composite function cos 3x
is continuous by Theorem 5. Therefore, 3x cos 3x is continuous by Theorem l(iii).
In Exercises 17-38, determine the points of discontinuity. State the type of discontinuity (removable, jump, infinite, or
none of these) and whether the function is left- or right-continuous.
1
17. f(x) = x
SOLUTION The function 1/x is discontinuous at x
neither left-continuous nor right-continuous at x = 0.
0, at which there is an infinite discontinuity. The function is
x-2
19. f(x) = Ix - II
x-2
The function Ix_ ll is discontinuous at x = 1, at which there is an infinite discontinuity. The function is
SOLUTION
neither left-continuous nor right-continuous at x = 1.
l~J
21. f(x) =
The function
SOLUTION
l~ J
is discontinuous at even integers, at which there are jump discontinuities. For every
integer n,
lim
x-}2n+
so that
l_:J2 = n = f(2n)
l~ J
is right-continuous at x = 2n. On the other hand,
l- =
lim_ xj
x--+2n
. so that
2
n- I
=fa
f(2n)
l~J is not left-continuous at x = 2n.
1
23. h(x) = 2 - Ix/
The function h(x)
1
_ lxl is discontinuous at x
2
function is neither left-continuous nor right-continuous at x = 2.
SOLUTION
x+I
x _
4
2
25. f(x) =
.
x+l
The functmn f(x) = - - is discontinuous at x
4x-2
function is neither left-continuous nor right-continuous at x = ~·
SOLUTION
27. f(x)
=3x213 -
SOLUTION
=
Ix { -1
21
~, at which there is an infinite discontinuity. The
9x3
The function f(x) = 3x21 3
x-2
29. f(x)
2, at which there is an infinite discontinuity. The
x=F2
x=2
-
9x3 defined and continuous for all x.
sEcT IoN
x- 2
For x > 2, f(x) = - - = 1. For x < 2, f(x)
SOLUTION
~-~
2.4
I
Limits and Continuity
71
(x - 2)
. h
.
d'
. .
= --= -1. The f unction
as a Jump iscontmmty at
0-zj
x = 2. Because
lim f (x)
x--+2-
= -1 = f (2)
the function is left-continuous at x = 2.
31. f(x) =
2x;;;o
= 2~;;0 is discontinuous at x = -5. Now,
The function f(x)
SOLUTION
. 2(x - 5)(x + 5)
. 2x2 - 50
hm
= 1im
X +5
x-+-5
X +5
x-+-5
.
(
)
1im 2 x- 5 = - 20
= x-+-5
Because lim f(x) exists but f(-5) is not defined, f has a removable discontinuity at x
x--+-5
left-continuous nor right-continuous at x
33. g(t)
= -5. The function is neither
= -5.
= tan 2t
SOLUTION
The function g(t)
sin2t is
. d'iscontmuous
.
.
= tan 2t = w henever cos 2 t = 0 ; 1.e.,
whenever
cos2t
(2n+ l)7r
2t = ,;___--'2
or
t=
(2n + l)7r
4
where n is an integer. At every such value oft there is an infinite discontinuity. The function is neither left-continuous
nor right-continuous at any of these locations.
35. f(x)
= tan(sin x)
SOLUTION
The function f (x) = tan(sin x) is continuous everywhere. Reason: sin xis continuous everywhere and tan u
is continuous on (- ~, ~)-and in particular on -1
composite functions.
:<::;
u
= sin x :<::; 1. Continuity of tan(sin x) follows by the continuity of
1
37. f(x)= - ex -e-X
SOLUTION
The function f(x)
=-1eX -e-X
is discontinuous at x
function is neither left-continuous nor right-continuous at x
= 0, at which there is an infinite discontinuity. The
= 0.
In Exercises 39-52, determine the domain of the function and prove that it is continuous on its domain using the Laws of
Continuity and the facts quoted in this section.
39. f(x)
= 2sinx+ 3cosx
SOLUTION The domain of 2 sin x + 3 cos xis all real numbers. Because the trigonometric functions sin x and cos x are
both continuous by Theorem 3, so are the functions 2 sin x and 3 cos x by Theorem l(ii) and the function 2 sin x + 3 cos x
by Theorem l(i).
41. f(x)
= vxsinx
SOLUTION The domain of vxsinx is {xix~ 0). Because
by Theorem 3, so is Vi sin x by Theorem l(iii).
43. f(x)
Vx and the trigonometric function sinx are both continuous
= x2132x
The domain of x21 3 2x is all real numbers as the denominator of the rational exponent is odd. Both x21 3 and
2x are continuous on this domain by Theorem 3, so x2' 32x is continuous by Theorem l(iii).
SOLUTION
45.
f (x)
= x-4 / 3
SOLUTION
zero for x
The domain of x-4 13 is {xix* 0). Because the function x4 ' 3 is continuous by Theorem 3 and not equal to
* 0, it follows that
x
is continuous for x
47. f(x)
= tan
2
-4/3
1
= x413
* 0 by Theorem l(iv).
x
The domain oftan2 xis all {xix* ±(2n - l)7r/2} where n is a positive integer. Because tan xis continuous on
this domain by Theorem 3 and Theorem 1(iv), it follows from Theorem 1(iii) that tan2 x = tan x · tan xis also continuous
on this domain.
SOLUTION
49.
f (x) =
(x4 + 1)3/2
72
CHAPTER 2
I
LIMITS
The domain of (x4 + 1)312 is all real numbers as x4 + 1 > 0 for all x. Because x31 2 is continuous by Theorem
3 and the polynomial function x4 + 1 is continuous by Theorem 2, so is the composite function (x4 + 1)312 by Theorem 5.
SOLUTION
51. f(x)
= -cos(x2)
x2 - 1
The domain for this function is all {xix* ±1}. Because the trigonometric function cos x and the polynomial
function x2 are continuous on this domain by Theorems 3 and 2, respectively, so is the composite function cos(x2) by
Theorem 5. Finally, because the polynomial function x2 - 1 is continuous by Theorem 2 and not equal to zero for x ±1,
.
by Theorem 1("iv ) .
. contmuous
the function cos(x2)
x2 _
is
1
SOLUTION
*
53. The graph of the following function is shown in Figure 18.
f (x)
Show that f is continuous for x
=
x2+3
10 - x
{
6x-x2
forx<l
for 1 :5 x :5 2
forx>2
* 1, 2. Then compute the right- and left-hand limits at x = 1, 2, and determine whether
f is left-continuous, right-continuous, or continuous at these points.
y
2
6
FIGURE 18
SOLUTION
Let's start with x
* 1, 2.
• The polynomial function x2 + 3 is continuous by Theorem 2; therefore, f(x) is continuous for x < 1.
• The polynomial function 10 - xis continuous by Theorem 2; therefore, f (x) is continuous for 1 < x < 2.
• The polynomial function 6x - x2 is continuous by Theorem 2; therefore, f (x) is continuous for x > 2.
At x
= 1, f(x) has a jump discontinuity because the one-sided limits exist but are not equal:
Jim f(x)
x--+1-
= x-+Ilim (x2 + 3) = 4,
lim f(x)
x--+I+
= x-+I+
lim (10 -
x)
=9
Furthermore, the right-hand limit equals the function value f(l) = 9, so f(x) is right-continuous at x = 1. At x = 2,
lim f(x)
x--+2-
= x--+2lim (10 -
x) = 8,
lim f(x) = Jim (6x X--+2+
X--+2+
The left- and right-hand limits exist and are equal to f(2), so f(x) is continuous at x
x2) = 8
= 2.
In Exercises 55-56, fxl refers to the least integer function. It is defined by fxl = n, where n is the unique integer such
that n - 1 < x :5 n. In each case, provide the graph off, indicate the points of discontinuity and type of each (removable,
jump, infinite, or none of these), and indicate whether f is left- or right-continuous.
55. f(x) = fxl
The graph off(x) = fx l is shown in the figure below. From the graph we see that f
= n for all integers n. Because
SOLUTION
at x
lim f(x)
x-+n-
= n = f(n)
but
f is left-continuous at each discontinuity.
-2
0---.
0---.
-2
lim f(x)
x--+n+
=n + 1 *
f(n)
has a jump discontinuity
sEc T Io N
2.4
I
Limits and Continuity
73
In Exercises 57-60, sketch the graph off. At each point of discontinuity, state whether f is left- or right-continuous.
57. f(x)
r
= {2-x
for x ~ 1
forx> 1
SOLUTION
y
The function f is continuous everywhere.
r-3x+2
59. f(x) =
{0
Ix - 21
SOLUTION
y
The function is neither left-continuous nor right-continuous at x
= 2.
61. Show that the function
r-16
f(x) =
x-4
{ 10
x*4
x= 4
has a removable discontinuity at x = 4.
SOLUTION
Note that
.
hmf(x)
x-+4
. r - 16
. (x + 4)(x = hm
- - = hm
x-+4 X - 4
x-+4
X - 4
Because limf(x) exists but is not equal to f(4)
x->4
= 10, the function f
4)
.
= hm(x
+ 4) = 8
x-+4
has a removable discontinuity at x
= 4.
In Exercises 63-64, His the Heaviside function, defined by
H(x) =
{O1
when x <
whenx~
00
63. In each case, sketch the graph off, indicate whether or not f is continuous, and-if f is not continuous-identify
the points of discontinuity.
(a) f(x) = H(x)(r + 1)
(b) f(x) = H(x)x
(c) f(x) = H(x - 2) Vi
(d) f(x) = H(l + x)H(l - x)(l - r)
SOLUTION
(a) The graph of f(x)
=H(x)(r + 1) is shown below. From the graph, we see that f
-3
-2
has a jump discontinuity at x
= 0.
74
CHAPTER 2
I
I
LIMITS
(b) The graph of f(x) = H(x)x is shown below. From the graph, we see that f is continuous.
3.0
2.5
2.0
1.0
0.5
_,
-2
(c) The graph of f(x)
= H(x -
2)
vx is shown below. From the graph, we see that f has a jump discontinuity at x = 2.
3.0
2.5
20
1.5
1.0
0.5
-2
(d) The graph of f(x)
= H(l + x)H(l -
x)(l - .x2) is shown below. From the graph, we see that f is continuous.
1.5
-3
-2
In Exercises 65-67, find the value of the constant (a, b, or c) that makes the function continuous.
65. f(x) =
SOLUTION
L
.x2-c
{ 4x + 2c
forx<5
for x ?: 5
As x ~ 5-, we have .x2 - c ~ 25 - c
= R or 25 - c = 20 + 2c implies c = ~.
= L. As x ~ 5+, we have 4x + 2c ~ 20 + 2c = R. Match the limits:
x- 1
for x < -1
67. f(x) = ax+ b for - 1 ~ x ~ ~
{ 1
xfor x > ~
r,
As x ~ -1-, we have x- 1 ~ -1, while as x ~ -1+, we have ax+ b ~-a+ b. Additionally, as x ~
we have ax+ b ~ ~a+ b, while as x ~ ~+,we have x- 1 ~ 2. In order for f to be continuous for all x, a and b must
satisfy the system of equations
SOLUTION
-1 = -a+b
The solution of this system of equations is a
and
= 2 and b = 1.
1
69. Define g(t) = tan- 1 ( --) fort -:P l. Answer the following questions, using a plot if necessary.
t- 1
(a) Can g(l) be defined so that g is continuous at t = 1?
(b) How should g( 1) be defined so that g is left-continuous att = 1?
SOLUTION
To answer these questions, consider the graph of
,2 1 shown here:
y
100
50
0.5
-50
1.5
s E c TIO N
2.4
I
Limits and Continuity
75
(a) From the graph shown above, we see that
.
1
1Im - - =-oo
1
and
lim g(t) = -~
2
while
HI- ( -
.
1
11m--=oo
t-1
HI+
It follows that
HI-
lim g(t) = ~
2
,_,1+
Because the two one-sided limits of gas t -t 1 are not equal, the discontinuity at t = 1 is not removable; that is, g(l)
cannot be defined so that g is continuous at t = 1.
(b) In order for g to be left-continuous at t = 1. we must have
2
g(l} = lim g(t) = -~
t-+1-
In Exercises 71-74, draw the graph of a function on [O, 5] with the given properties.
71.
f
is not continuous at x = 1, but lim f(x) and lim f(x) exist and are equal.
x-+l+
x-+1-
SOLUTION
y
2
4
5
73. f has a removable discontinuity at x = 1, a jump discontinuity at x = 2, and
lim f(x) = -oo,
lim f(x) = 2
x-+3-
x-+3+
SOLUTION
y
4
3
2
/
3
In Exercises 75-88, evaluate using substitution.
75. lim(2.x3 -4)
x-+-1
SOLUTION
Jim (2.x3 - 4) = 2(-1) 3 - 4 = -6
x-+-1
77. lim x+ 2
x->3 x2 + 2x
.
SOLUTION
x+2
3+2
~~ -x2_+_2_x :;; 32 + 2. 3
5
1
= 15 = 3
79. lim tan(3x)
x-+~
SOLUTION
lim tan(3x) = tan(3. !!4) =tan(~)= -1
-i
4
4
5
76
CH APT E R 2
I
LIMITS
81. limx- 512
x->4
SOLUTION
limx-5/2
= 4-5/2 = _.!__
x->4
32
83. lim (1 - 8x3)312
x-+-1
SOLUTION
lim (1 - 8x3)312 = (1 - 8(-1) 3 ) 3 /~
x--+-1
= 27
85. lim 10x2-2x
x->3
SOLUTION
87.
lim 10x2-2x
x-->3
= 1032 - 2<3l = 1000
}~1!1- sin- 1 {~)
SOLUTION
~~sin- 1 (~) = sin- (~~ ~) = sin- (~) = ~
1
1
89. Suppose that f and g are discontinuous at x = c. Does it follow that f + g is discontinuous at x = c? If not, give a
counterexample. Does this contradict Theorem l(i)?
SOLUTION Even if f and g are discontinuous at x = c, it is not necessarily true that f + g is discontinuous at x = c.
For example, suppose f(x) = -x- 1 and g(x) = x- 1 • Both f and g are discontinuous at x = O; however, the function
f(x) + g(x) = O is continuous everywhere, including x = 0. This does not contradict Theorem 1 (i), which deals only
with continuous functions.
91. Use the result of Exercise 90 to prove that if g is continuous, then f(x) = lg(x)I is also continuous.
SOLUTION Let c be an arbitrary real number at which g is continuous. Following the logic of Exercise 90 depending
upon whether g(c) is positive, negative, or zero, we find
Iimf(x)
x-+c
= lim
lg(x)I = lg(c)I = f(c)
x--+c
Thus, if g is continuous, then f(x) = lg(x)I is continuous also.
93.
LSJ
In 2017, the federal income tax Ton income of x dollars (up to $91,900) was determined by the formula
0.15x - 466.25
for 0 ~ x < 9325
for 9325 ~ x < 37 ,950
0.25x - 4261.25
for 37,950
O.lOx
T(x)
=
{
~
x
~
91,900
Sketch the graph of T. Does T have any discontinuities? Explain why, if Thad a jump discontinuity, it might be advantageous in some situations to earn less money.
SOLUTION
Here is a graph of T(x) for 2017:
20000
10000
5000
100000
Note that the graph of T has no discontinuities. If T(x) had a jump discontinuity (say at x = c ), it might be advantageous
to earn slightly less income than c (say c - E) and be taxed at a lower rate than to earn c or more and be taxed at a higher
rate. Your net earnings may actually be more in the former case than in the latter one.
Further Insights and Challenges
95. Give an example of functions f and g such that /(g(x)) is continuous but g has at least one discontinuity.
SOLUTION Answers may vary. The simplest examples are the functions f(g(x)) where f(x) = C is a constant function,
and g(x) is defined for all x. In these cases, f(g(x)) = C. For example, if f(x) = 3 and g(x) = [x], g is discontinuous at
all integer values x = n, but f(g(x)) = 3 is continuous.
SE c TI o N 2.5
I
Indeterminate Forms
77
97. Show that f is a discontinuous function for all x, where f(x) is defined as follows:
1
f(x) = { -1
for x rational
for x irrational
Show that f 2 is continuous for all x.
SOLUTION
limf(x) does not exist for any c. If c is irrational, then there is always a rational number r arbitrarily close
to c so that If(~)"._ f(r)I = 2. If, on the other hand, c is rational, there is always an irrational number z arbitrarily close to
c so that lf(c) - f(z)I = 2.
On the other hand, f(x) 2 is a constant function that always has value 1, which is obviously continuous.
2.5 Indeterminate Forms
Preliminary Questions
1. Which of the following is indeterminate at x = 1?
x2 + 1
SOLUTION
At x = 1,
x2- l
x2-1
x+2'
x-1'
x2+1
.Yx + 3 -
2
1 is of the form -°; hence, this function is indeterminate. None of the remaining functions
_t.;
0
vx+3-2
1
is indeterminate at x = 1: x2+
and
x- 1
_gt
-vx+3-2
are undefined because the denominator is zero but the numerator is not, while
~;i is equal to 0.
2. Give counterexamples to show that these statements are false:
(a) If f(c) is indeterminate, then the right- and left-hand limits as x
(b) Iflimf(x) exists, then f(c) is not indeterminate.
~
care not equal.
x-+c
(c) If f(x) is undefined at x = c, then f(x) has an indeterminate form at x = c.
SOLUTION
(a) Let f(x) =
~~/.At x = 1, f is indeterminate of the form
§ but
Jim x2- l = lim(x+ 1) = 2=lim(x+1) = lim x2- l
x-+I-
X -
1
x-+1+
x-+1-
x-+t+
X -
1
(b) Again, let f(x) = ~:}. Then
limf(x) = lim
x-+1
x2 -
x-+1 X -
1
= lim(x + 1) = 2
1
x-+l
but f (l) is indeterminate of the form §.
(c) Let f(x) = ~. Then f is undefined at x = 0 but does not have an indeterminate form at x = 0.
3. The method for evaluating limits discussed in this section is sometimes called simplify and plug in. Explain how it
actually relies on the property of continuity.
SOLUTION If f is continuous at x = c, then, by definition, limx-.c f(x) = f(c); in other words, the limit of a continuous
function at x = c is the value of the function at x = c. The "simplify and plug-in" strategy is based on simplifying
a function which is indeterminate to a continuous function. Once the simplification has been made, the limit of the
remaining continuous function is obtained by evaluation.
Exercises
In Exercises 1-4, show that the limit leads to an indeterminate form. Then carry out the two-step procedure: Transform
the function algebraically and evaluate using continuity.
x2-36
1. lim--x->6 X - 6
When we substitute x = 6 into ~=~6 , we obtain the indeterminate form
and denominator and then simplifying, we find
SOLUTION
. x2 - 36 . (x - 6)(x + 6) .
h m - - =hm
=lim(x+6)= 12
X - 6
x-+6
X - 6
x-+6
x-+6
§. Upon factoring the numerator
78
c H APTER
2
I
LIMITS
x2+2x+l
3. lim - - - x--+-1
X+ 1
SOLUTION When we substitute x = -1 into x2::~+ 1 , we obtain the indeterminate form
and simplifying, we find
g. Upon factoring the numerator
2
. x2 + x + 1
lIm
X--+-1
X+ 1
. (x + 1)
li (
1Im - - - = m x + 1) = 0
= X--+-1
X+ 1
X--+-1
In Exercises 5-34, evaluate the limit, if it exists. If not, determine whether the one-sided limits exist (finite or infinite).
5.
x-7
lim~
X--41
49
X -
.
x- 7
l Im -2 - x--+1 x - 49
SOLUTION
x- 7
l"
1
l
= l"x--+7
Im
= x--+7
Im - - = (x - 7)(x + 7)
x+7
14
x2 + 3x+2
7. lim - - - x--+-2
X+ 2
. x2 + 3x + 2
l Im
X+2
SOLUTION
x--+-2
=
li
m
(x + 2)(x + 1)
X+ 2
x--+-2
.
= 1Im(x+ 1) = -1
x--+-2
2x2-9x-5
9. lim
X-45
X
2
25
-
lim 2.x2 - 9x - 5
x2 - 25
SOLUTION
= lim (2x +
x--+5
l)(x - 5)
= lim 2x + 1 = .!..!..
x--+5 (x - 5)(x + 5)
10
x--+5 x + 5
2x+ 1
11. lim - 2- - - x--+-t 2x + 3x + 1
SOLUTION
13. lim
X-42
2x + 1
2
lim
x--+-i 2x + 3x + 1
=
2x + 1
.
lim
=
x--+-t (2x + l)(x + 1)
.
1
hm - -
x--+-~ x + 1
=-
1
1/2
=2
3.x2-4x-4
2 X2 - 8
SOLUTION
. 3.x2 - 4x - 4
1Im
x--+2 2x2 - 8
(3x + 2)(x - 2)
.
3x + 2
8
= Iix--+2
m
= hm
--- = - = 1
2(x - 2)(x + 2)
x--+2 2(x + 2)
8
15. lim 42r - 1
t--+O 4' - 1
SOLUTION
4 2' - 1
lim - 1--+04'-1
17. lim ../X-
= lim
t--+O
(4' - 1)(4' + 1)
4'-1
= lim(4'
+ 1) = 4° + 1 = 2
t--+O
4
x--+16 x-16
SOLUTION
.
..(X-4
hm - - -
x--+16 x-16
19 r (h + 2)2
• h1:!!J
h
.
= hm
../X-4
x--+16(..fX+4)(..fX-4)
.
= hm
1
---
x--+16 ..(X+4
= -1
8
4
SOLUTION
1
1
lim (h+2)2 - 4
h--+0
h
4-(h+2J2
li
4(h+2)2
= h!!J --h-
= lim
h
h--+0
21. lim
h--+O
-h-4
4(h+2i2
h
4-(h2+4h+4)
-h2 -4h
4(h+2)2
. 4(h+2)2
1
=
I
m
-= h1:!!J
h
h--+O h
r
= lim
-h - 4
h--+0 4(h + 2) 2
= -4 = _.!.
16
4
Yif+h - 2
SOLUTION
h
Observe that ash ~ 0, V2 + h - 2 ~
V2 - 2 * 0 and h -4 0. Accordingly,
. v2+h-2
hm
h
does not exist.
h--+0
S E c TI 0 N 2.5
As for the one-sided limits, Y2 + h - 2 ---)
I
Indeterminate Forms
79
.../2 - 2 < 0 and h ---) o- as h ---) o-; therefore
lim
h->O-
.../2 + h-2
= oo
h
.../2 - 2 < 0 and h---) o+ ash---) o+; therefore
On the other hand, .../2 + h - 2---)
lim
h->O+
.../2+h-2
= -oo
h
x-4
23. lim - - - = =
x->4 yx - y8 - X
SOLUTION
lim
x->4
x-4
. (
Vx _ y8 _ X
= hm
x->4
x-4
. ..:....__.;,..;....;.._-=---(x-4)(yx+ .../8-x)
· vx+ .../8-x) = hm
.../8 _ x
x->4
2x - 8
Vx _ y8 _ x yx +
=limvx+~=
2
x->4
V4+V4=2
2
1 - - -4- )
25. lim ( - x->4
yx-2 x-4
. ( 1
4 )
.
lim - - - - - - = hm
SOLUTION
x->4
yx- 2
X -
4
x->4 (
yx+2-4
yx- 2)( yx + 2)
.
= hm
x->4 (
yx-2
yx- 2)( yx + 2)
1
= -
4
cotx
27. lim-x->O CSC X
cotx
COSX
•
Jim--= Jim-.-· smx = cosO = 1
SOLUTION
x->O CSC X
x->0 Slil X
29. Jim ( -1- - -2-2 )
x->I 1 - x
1- x
x-1
x-1
-1
1
1
2 )
Jim ( - - - - -2 =Jim - -2 =Jim
=Jim - - = -HI 1 - x
1- x
HI 1 - x
x->I (1 - x)(l + x)
x->I 1 + x
2
SOLUTION
221+21 -20
31. l i m -1- - - H2
2 -4
221 + 21 - 20
(21 + 5)(21 - 4)
SOLUTION lim
= lim
4
= lim(21 + 5) = 9
1
1->2
2 - 4
1->2
21 1->2
2
1
- ~--)
tanB-1 tan2 8-1
33. lim (
O->i
lim (
SOLUTION
0->i
1
2
)
. tanB-1
tanB-1
1
= hm
= Jim
= Jim - - 2
tanB-1 tan2 8-1
O->i tan 8-1
o->~ (tanB- l)(tanB+ 1)
O->~ tanB+ 1
1
1+1
2
35. The following limits all have the indeterminate form 0/0. One of the limits does not exist, one is equal to 0, and one
is a nonzero limit. Evaluate each limit algebraically if you can or investigate it numerically if you cannot.
. x2 + 3x+ 2
• hm---x->-2
X +2
l -x- 1
• lim---x->I X - 2 + x-1
.
x2
•hm-x->O
1-
ex
SOLUTION
.
• l1m
X->-2
}"
x2+3x+2
. (x+2)(x+l)
.
= 11m
= 11m(x+l)=-l
X + 2
X->-2
X + 2
X->-2
1-
X-I
Ji
X -
1
}"
X -
1
}"
1
hi h d
• ).!!} x _ 2 + x-I = x.!!t x2 _ 2x + 1 = x1!!1 (x _ l) 2 = x1!!1 x _ 1 , w c
proaches 1
* 0 while the denominator approaches 0.
. b
h
oes not exist ecause t e numerator ap-
• We will investigate this limit numerically. From the table below, it appears that lim _!____ = O.
x->O
1-
ex
80
c HA p T E R
2
I
LIMITS
x
-0.01
-0.001
-0.0001
0.0001
0.001
0.01
X1
0.0100501
0.0010005
0.0001000
-0.000099995
-0.0009995
-0.00995
1-eX
In Exercises 37 and 38, a projectile is launched straight up in the air and is acted on by air resistance and gravity as in
Example 7. The function H gives the maximum height that the ball attains as a function of the air-resistance parameter
.
~
37. If tne mass of the ball is one kilogram and it is launched upward with an initial velocity of 60 m/sec, then
=
H(k)
60k - 9.8 ln ( JOOk +
1)
49
k2
Estimate the maximum height without air resistance by investigating lim H(k) numerically.
k->0
Based on the values in the table below, we estimate that the maximum height with no air resistance is
183.67 meters.
SOLUTION
39. !GUii
k
0.001
0.0001
0.00001
0.000001
H(k)
182.9272
183.5985
183.6660
183.6729
Use a plot of f(x) =
x
Vi-
~
8-x
to estimate limf(x) to two decimal places. Compare with the answer
x->4
obtained algebraically in Exercise 23.
SOLUTION Let f(x) = vx~~- From the plot of f(x) shown below, we estimate ~~f(x) "" 2.00; to two decimal
places, this matches the value of2 obtained in Exercise 23.
y
3.6
3.8
1
41. Show numerically that for b = 3 and b = 5, lim bx x--+0
SOLUTION
4.0
4.2
4.4
appears to equal ln 3 and ln 5, respectively.
X
Based on the values in the table below,
Y-1
lim - - "" 1.0986
x--+0
X
To four decimal places, ln 3 = 1.0986, so it appears that lim
x--+O
x
3x_l
3x-i
x
= ln 3.
x
x
JX-1
x
0.01
1.104669
-0.01
1.092600
0.001
1.099216
-0.001
1.098009
0.0001
1.098673
-0.0001
1.098552
0.00001
1.098618
-0.00001
1.098606
Based on the values in the table below,
Y-1
lim - - "" 1.6094
x--+0
To four decimal places, In 5
X
= 1.6094, so it appears that Jim
sx- 1 = ln 5.
x--+O x
x
5x_I
x
x
5x_l
x
0.001
1.610734
-0.001
1.608143
0.0001
1.609567
-0.0001
1.609308
0.00001
1.609451
-0.00001
1.609425
0.00,0001
1.609439
-0.000001
1.609437
s E c TI o N 2.5 I Indeterminate Forms
81
In Exercises 43-48, evaluate using the identity
a 3 -b3 = (a-b)(a2 +ab+ b2 )
x3 - 8
43. lim-x-+2 x- 2
SOLUTION
45.
x3 - 8
(x - 2)(x2 + 2x + 4)
.
2
lim - - = lim
= hm(r + 2x + 4) = 12
x->2 X - 2
x->2
X - 2
x->2
x2-5x+4
lim-~-­
x3 - 1
x->I
SOLUTION
. x2-5x+4
hm
x->1
x3 - 1
(x- l)(x-4)
x-4
__ -3 ___
1
= lim - - - l)(x2 + x + 1) x->l x 2 + x + 1
3
.
=1x->l
lffi
(x -
x4 -1
47. lim - 3- x->I X - 1
SOLUTION
x"-1
.
(x2- l)(x2+1)
. (x- l)(x+ l)(x2 +1)
. (x+ l)(x2+1)
4
lim-- =hm
=hm
=hm
= x->IX3-1
x->I(x-l)(x2+x+·l) x->I (x-l)(x2 +x+l)
x->I
x 2 +x+l
3
In Exercises 49-56, evaluate in terms of the constant a.
49. lim(2a + x)
x->0
SOLUTION
lim(2a + x)
x->0
= 2a
51. lim(4t - 2at + 3a)
t-+-1
SOLUTION
lim(4t- 2at + 3a) = -4 + 5a
t-+-l
53. lim
x-+a
Yx - Va
x- a
SOLUTION
lim
..rx-va = hm.
X -
x->a
a
X->a (
..rx-va
Yx - va)( Yx + va)
.
= hm
x->a
1
Yx + Va
1
= -2 Va
(x+a) 3 -a3
55. l i m - - - x--+O
SOLUTION
X
lim
(x + a)3 - a 3
x--+0
X
= lim
x3 + 3x2a + 3xa2 + a 3 - a 3
X
X-i-0
= lim(x2 + 3xa + 3a2) = 3a2
x-+0
. of x, and rewnte
. as a l"1rmt
. as x
. {'l + h - l . n·mt: Set x = -·~1
57. Evaluate I1m
v1 + h
n, express h as a fu nctlon
h->0
h
SOLUTION
Let x = {'1 + h. Then h = x4
-
-7
1.
l, and
. {'1 + h - 1 l" x - 1
l"
x- 1
l"
1
= lffi - - = lffi
= lffi - - - : : - - - : - : Ilffi
h->0
h
x->1 x4 - 1
x->l (X - l)(X + l)(x2 + 1) x->I (X + l)(x2 + 1)
4
Further Insights and Challenges
In Exercises 59-62,find all values of c such that the limit exists.
59. lim x2 - Sx - 6
x--+c
SOLUTION
X- C
lim
x2 -5x-6
X- C
x--+c
will exist provided that x - c is a factor of the numerator. (Otherwise there will be an
infinite discontinuity at x = c.) Since x2 - 5x - 6 = (x + l)(x - 6), this occurs for c = -1 and c = 6.
c- )
61. lim ( -l - - x->I x - 1
x3 - 1
SOLUTION
Because x3
- 1 = (x- l)(x2 + x + 1),
c
x - 1
---3 x- 1
1
x2+x+l-c
(x - l)(x2 + x + 1)
Therefore, lim (~ - -/--- ) exists as long as x- 1 is a factor of x2 + x + 1 - c. Now, if x2 + x + 1 - c = (x - l)(x + q),
x-+I XX - 1
then q - 1 = 1 and -q = 1 - c. Hence, q = 2 and c = 3.
s E CT I o N
SOLUTION
2.6
I
The Squeeze Theorem and Trigonometric Limits
The sine function takes on values between -1 and l; therefore, \sin x~i I:::; 1 for all x
83
* 1. Multiplying by
Ix - 11 yields
l(x- l)sin~I:::; lx-11
or
x-l
-lx-11:::; (x-
l)sin~ :::; lx-11
x-
1
Now lim( -Ix - 11) = lim Ix - 11 = 0, so we can apply the Squeeze Theorem to conclude that
x-+1
X-+1
lim(x- l)sin~ =0
x-->I
X- 1
1
5. lim(2' - 1) cos t-->O
t
SOLUTION The cosine function takes on values between -1 and l; therefore, \cos~\ :::; 1 for all t
12' - 11 yields
1<2' - l)cos
~I:::; 12' -
or
11
* 0. Multiplying by
- 12' - 1I :::; (2' - 1) cos ! :::; 12' - 11
t
Now lim(-12' - 11) = liml2' - 11=0, so we can apply the Squeeze Theorem to conclude that
t-+O
t-+0
lim(2' - 1) cos! = 0
t-->O
t
7. lim(t2 - 4) cos __!_
1-->2
SOLUTION
2
The cosine function takes on values between -1 and 1; therefore, \cos 1 ~ 2 \
(-
:::;
1 for all t
* 2. Multiplying by
1t2 - 41 yields
1
(t2 -4)cos - -1:::; 1t2 -41
t-2
l
or
- lt2 - 41 :::; (t2
-
4) cos __!_2 :::; 1t2 - 41
t-
Now lim(-lt2 - 41) = lim 1t2 - 41 = 0, so we can apply the Squeeze Theorem to conclude that
t-+2
t-+2
1
lim(t2 - 4) cos - - = O
t-->2
t-2
9. limcosBcos(tanB)
9_,.~
The cosine function takes on values between -1 and 1; therefore, lcos (tan B)I :::; 1 for all B near
plying by Icos Bl yields
SOLUTION
or
lcos Bcos (tan B)I :::; Icos Bl
~.
Multi-
- Icos Bl :::; cos Bcos (tan B) :::; Icos Bl
Now lim(-1 cos Bl) = lim I cos Bl = 0, so we can apply the Squeeze Theorem to conclude that
0--+~
8-+~
lim cos Bcos (tan B) = 0
8--+~
11. State precisely the hypothesis and conclusions of the Squeeze Theorem for the situation in Figure 6.
y
y=u(x)
2
1
FIGURE 6
SOLUTION
Because there is an open interval containing x
= 1 on which l(x) :::; f(x)
it follows that limf(x) exists and
x-->I
limf(x)
x-->1
2
=2
:::; u(x) and lim l(x)
X-+l
= lim u(x) = 2,
X-+]
84
C H A PT E R 2
I
LIMITS
13. What does the Squeeze Theorem say about limf(x) ifthe limits
x-->1
lim l(x) = lim u(x) = 6 and f, u, and l are related as in Figure 8? The inequality f(x) :5 u(x) is not satisfied for all x. Does
X_.7
X-i-7
this affect the validity of your conclusion?
6
7
FIGURE 8
SOLUTION The Squeeze Theorem does not require that the inequalities l(x) :5 f(x) :5 u(x) hold for all x, only that the
inequalities hold on some open interval containing x = c. In Figure 8, it is clear that l(x) :5 f(x) :5 u(x) on some open
interval containing x = 7. Because lim u(x) = lim l(x) = 6, the Squeeze Theorem guarantees that Iimf(x) = 6.
x-+7
x-t?
x--+1
15. State whether the inequality provides sufficient information to determine lim f(x), and if so, find the limit.
X-->1
(a) 4x - 5 :5 f (x) :5
(b)
(c)
x2
2x - 1 :5 f(x) :5 x2
4x - x2 :5 f(x) :5 x 2 + 2
SOLUTION
(a) Because lim(4x - 5) = -1
x--+1
* 1 = lim x2, the given inequality does not provide sufficient information to determine
x--+1
limf(x).
X-->1
(b) Because lim(2x - 1) = 1 = limx2, it follows from the Squeeze Theorem that limf(x) = 1.
x--+1
x--+1
x-+I
(c) Because lim(4x - x2) = 3 = lim(x2 + 2), it follows from the Squeeze Theorem that limf(x) = 3.
x-+1
x-+1
x-+1
In Exercises 17-26, evaluate using Theorem 2 as necessary.
. tanx
17• hm-x-->O
X
. tanx
. sinx 1
sinx
1
hm - - = hm - - - - = lim - - . lim - - = 1 . 1 = 1
SOLUTION
x-->O
X
x-->O
X
COS X
x-->O
X
x-->O COS X
. Yt3 + 9 sin t
19. h m - - - - t-->O
t
.
hm
SOLUTION
-../r3 +9sint
t
t-->O
. .~sint
-~
sint
= lim vt3 + 9 - = lim vt3 + 9 · lim =
t-->O
t
t-->O
t-->O
t
V9. 1 = 3
21. lim_±_
2
x-->O sin X
x2
Iim-.--.1
l"
1 l"Im-.-=
1
.
1Im--=
=Im-.-.
-1 . -1 = 1
SOLUTION
X-i-0 Sifl2 X
X--+0 ~ ~
x
x
X-+0
SlDX
x
X--+0 ~
x
1 1
23. lim sec () - 1
8-->0
()
SOLUTION
. sec e - I _ . sec e - I cos e _ . I - cos e I
. I - cos() .
I
1
lim
1
1
8-->o
e -Im
8-->o
e cos e -Im
8-->o
e --=hm
cos e 8-->o
e ·hm--=0·-=0
8-->o cos e
I
. sint
25. h m r-~
t
SOLUTION
~nt
-
t
.
n
is continuous at t = - . Hence, by substitution
4
27. Evaluate lim sin llx using a substitution() = llx.
x-i-0
x
s E c T 1o N
SOLUTION
2.6
Let () = 1lx. Then () -+ 0 as x -+ 0, and x =
I
The Squeeze Theorem and Trigonometric Limits
()I 11. Hence,
. sin llx
. sin()
. sin()
h m - - = lim-- = llhm- = 11(1) = 11
x--+0
x
0--+0()111
0--+0
0
In Exercises 29-48, evaluate the limit.
. sin9h
29. hm-hh--+O
SOLUTION
sin 9h
.
sin 9h
lim - - = hm 9 -9h = 9.
h--+0
h
h--+O
sink
5
31. limh
h--+0
SOLUTION
. sink
1
. 1 sink
hm - - = 1im - - - = 5h
h--+0 5
h
5
h--+0
sin 1()
33. lim -.- 0--+0 sm 30
SOLUTION
We have
sin 7e = 2 (sin 70) (~)
sin 30 3 1()
sin 30
Therefore,
lim sin 1()
O--+O
3()
= 2(um sin 7()) (1im ~) = 2(1)
3
O--+O
1()
O--+O
sin 3()
3
1
( )
=2
3
35. lim x csc 25x
x--+0
SOLUTION
37
r
• h1!!/i
.
.
x
1 .
25x
1
hmxcsc25x = hm-.-- = - h m - - - = x--+0
x--+0 sm 25x
25 x--+0 sin 25x 25
sin 2h sin 3h
h2
SOLUTION
Iim
h--+O
sin 2h sin 3h
h2
sin 2h sin 3h
= Iih--+0
m
h·h
. sin 2h sin 3h
=lim-- - h--+O
h
h
sin 2h sin3h
.
sin 2h .
sin3h
.
= 1im 2 - - 3 - - = 1im 2 - - hm3-- =2·3 =6
2h
h--+0
39
r
• /-To
3h
h--+0
2h
h--+0
sin(-30)
sin40
SOLUTION
lim sin(- 30)
O--+O sin(40)
= lim -sin(3(J) · ~. ~
4
30
O--+O
sin(40)
= -~
4
41. lim csc 8t
HO csc4t
SOLUTION
lim csc 8t = lim sin4t . ~ . ~ = ~
HO csc4t
HO sin8t 4t
2
2
. sin 3x sin 2x
43. 1i m - - - x--+O
xsin5x
SOLUTION
45
lim sin 3~ sin 2x
x--+o
x sm 5x
= lim (3 sin 3x . ~
x--+o
3x
(sin 2x) I (2x)) = ~
5 (sin 5x) / (5x)
s
. lim sin(2h)(l - cos h)
h--+0
h2
SOLUTION
. sin(2h)(l - cos h) _ .
sin(2h) . 1 - cos h
1lffi
h2
- 1lffi 2 - - - 1lffi
=2·0=0
h--+0
. cos 2() - cos ()
47. 1i m - - - - O--+O
()
h--+0
2h
h--+O
h
3h
85
86
CHAPTER 2
I
LIMITS
SOLUTION
lim
9_,o
cos 2B - cos e
(cos 2B - 1) + (1 - cos e)
cos 2B - 1
1 - cos e
= lim
= lim
+ lim - - e
9_,o
e
9_,o
e
8_,o
e
. 1 - cos e
. 1 - cos 2B
O O O
2 ·+=
1
=-Im
21
+Im
=9_,o
2e
8_,o
e
. sin 2B - 2 sine
49. Use the identity sin 2B = 2 sine cos e to evaluate hm
ez
8--'>0
SOLUTION
Using the identity sin 2B = 2 sine cos e,
sinW - 2 sine= 2 sin ecos e - 2 sine= 2 sin e(cos e - 1) = -2 sine(l - cos e)
Then
. sin2B-2sine_
. sine(cose-1)_
. sine 1-cose
(l)(O)-O
hm
- - 211m
- - 21 Im - - ·
=-2
8--'>0
ez
8--'>0
e2
8--'>0
e
e
51. Explain why lim(csc e- cot 8) involves an indeterminate form, and then prove that the limit equals 0.
8_,o
SOLUTION Ase approaches 0 from the right, csce ~ oo and cote ~ oo, and as e approaches 0 from the left,
csc e ~ -oo and cote ~ -oo. Thus, csc e - cote has the indeterminate form oo - oo as e ~ 0.
Now,
1
cose
1-cose
csce-cote= - - - - - = - - sine sine
sine
so
.
. 1-cose
. 1-cose e
hm( csc e - cote) = hm
.
= lim
-.- = 0 · 1 = 0
8--'>0
8--'>0
sm e
8--'>0
e
sm e
imTii
.
.
53. ~ . Investigate l~
1 - cos 2h
.
h2
numerically or graphically. Then evaluate the limit using the double angle formula
cos2h = 1 - 2sin2 h.
SOLUTION
h
-0.1
-0.01
O.Ql
0.1
1.993342
1.999933
1.999933
1.993342
l-cos2h
h2
1.5
Both the numerical estimates and the graph suggest that the value of the limit is 2.
• Using the double angle formula cos 2h = 1 - 2 sin2 h,
l -cos2h = 1- (l -2sin2 h) = 2sin2 h
Then
. 1 - cos 2h
2 sin2 h
sin h sin h
hm
h2
= lim-- = 2lim-- · - - = 2(1)(1) = 2
h--'>0
h--'>0
h2
h--'>0
h
h
In Exercises 55-57, evaluate using the result of Exercise 54.
. cos3h- 1
55. hm
h2
h--'>0
S E CT I 0 N 2.6
SOLUTION
I
The Squeeze Theorem and Trigonometric Limits
We make the substitution B = 3h. Then h = 0/3, and
. cos 3h - i
hm
h->O
h1
cos e - i
= Ii8->0
m
<e/3)2
1 - cos e
.
9
2
=-911 m - - -
ei
8->o
.Yl -cost
57. l i m - - - ,_,o
t
SOLUTION
lim
.Yl - cost
=
t
t->0
Further Insights and Challenges
59. Use the result of Exercise 54 to prove that form
* 0,
. cosmx- 1
m2
=
11m
1
x->O
SOLUTION
x
. x
.
.
cos mx - 1 • u'
vve obtam
Sub stltute u = mx mto
X2
. cos mx - 1
11m
x1
x->O
2
= -mu . As x --+ 0 , u --+ 0 ; therefiore,
. cos u - 1
.
cos u = hm
= 1u->O
1mm2
u->O (ujm)1
u1
1
1
m
= m2 ( --21 ) = -2
.
. sinx-sine
.
61. (a) Investigate hm
numencally for the five values e = 0, ~, ~, } , ~.
x-+c
X -
C
(b) Can you guess the answer for general c?
(c) Check numerically that your answer to (b) works for two other values of e.
SOLUTION
(a) Here e
= 0 and cose = 1.
Here e = ~ and cos e =
Here e
e -0.01
e -0.001
e + 0.001
c + 0.01
sinx - sine
x-e
0.999983
0.99999983
0.99999983
0.999983
f , , .866025.
x
c -0.01
e -0.001
e +0.001
e + 0.01
sinx- sine
x-e
0.868511
0.866275
0.865775
0.863511
x
e - 0.01
e - 0.001
e + 0.001
e + 0.01
sinx - sine
x-e
0.504322
0.500433
0.499567
0.495662
x
c -0.01
e-0.001
c + 0.001
c + 0.01
sinx - sine
x-e
0.710631
0.707460
0.706753
0.703559
= } and cose = ~·
Here e = ~ and cos e =
Here e
x
Y{ ,,, 0.707107.
= ~ and cos e = 0.
. sinx-sine
(b)l1m
x-+c
X -
C
x
e- 0.01
e -0.001
e + 0.001
e + 0.01
sinx-sinc
x-e
0.005000
0.000500
-0.000500
-0.005000
=cose.
87
88
cHAp TER
2
I
LIMITS
(c) Here c = 2 and cos c =cos 2"" -.416147.
x
c -0.01
c- 0.001
c+0.001
c + 0.01
sinx - sine
x-c
-0.411593
-0.415692
-0.416601
-0.420686
Here c = -~ andcosc = "{ "".866025.
x
c-0.01
c -0.001
c+0.001
c + 0.01
sinx - sin c
x-c
0.863511
0.865775
0.866275
0.868511
2. 7 Limits at Infinity
Preliminary Questions
1. Assume that
Jim f(x) = L
and
x--+oo
limg(x) = oo
x--+L
Which of the following statements are correct?
(a) x =Lis a vertical asymptote of g.
(b) y =Lis a horizontal asymptote of g.
(c) x =Lis a vertical asymptote off.
(d) y =Lis a horizontal asymptote off.
SOLUTION
(a) Because lim g(x) = oo, x = L is a vertical asymptote of g(x). This statement is correct.
x->L
(b) This statement is not correct.
(c) This statement is not correct.
(d) Because Jim f(x) = L, y =Lis a horizontal asymptote of f(x). This statement is correct.
x--+oo
2. What are the following limits?
(a) lim x3
X--+OO
(b) Jim
x3
(c) Jim
X--+-DO
x--+-oo
.x4
SOLUTION
(a) limx--+oo
x3 = oo
(b) limx--+-oo x3 = -oo
(c) limx--+-oo x 4 = oo
3. Sketch the graph of a function that approaches a limit as x ~ oo but does not approach a limit (either finite or infinite)
as x ~ -oo.
SOLUTION
y
4. What is the sign of a if f(x) = ax3 + x + 1 satisfies
lim f(x) = oo?
X--+-00
SOLUTION
Because lim
x--+-oo
x3 = -oo, a must be negative to have x--+-oo
Jim f(x) == oo.
5. What is the sign of the coefficient multiplying x7 if f is a polynomial of degree 7 such that lim f(x) "" oo?
x--+-oo
s Ec T Io N
2. 7
I
Limits at Infinity
89
The behavior of f(x) as x ~ -oo is controlled by the leading term; that is, limx--+-oo f(x) = limx--+-oo a1x1.
Because x 1 ~ -oo as x ~ -oo, d7 must be negative to have limx--+-oo f(x) = oo.
SOLUTION
6. Explain why lim sin l exists but lim sin l does not exist. What is lim sin l?
x--+oo
SOLUTION
x--+0
x
x
x--+oo
x
As x ~ oo, ; ~ 0, so
. sm. 1
lim
. 0 =0
= sm
X
X-HX)
On the other hand, ; ~ ±oo as x ~ 0, and as ; ~ ±oo, sin ; oscillates infinitely often. Thus
1
limsin X
X--+0
does not exist.
Exercises
1. What are the horizontal asymptotes of the function in Figure 7?
y
2
-20
20
40
60
80
FIGURE 7
SOLUTION
Because
lim f(x)
X--+-oo
the function f has horizontal asymptotes of y
=1
and
lim f(x)
X--+OO
=2
= 1 and y = 2.
3. Sketch the graph of a function f with a single horizontal asymptote y = 3.
SOLUTION
y
---4-r--_+z--~l..-.......-,__x
-5
-9
-13
5. !GUii Investigate the asymptotic behavior off(x) = _£___
numerically and graphically:
x2 + 1
(a) Make a table of values of f(x) for x = ±50, ±100, ±500, ±1000.
(b) Plot the graph off.
(c) What are the horizontal asymptotes off?
SOLUTION
(a) From the table below, it appears that
.
x2
1im - - = 1
2
x--+±00
x
! <x)
I
±50
I o.999600
(b) From the graph below, it also appears that
x +1
±100
±500
±1000
0.999900
0.999996
0.999999
90
CHAPTER 2
I
LIMITS
y
-5
5
(c) The horizontal asymptote off is y = 1.
In Exercises 7-16, evaluate the limit.
7. lim -2_
x-+oo x+ 9
SOLUTION
x
x- 1(x)
.
1
1
lim--=lim 1
=hm--=--=1
X->00 x + 9
X->OO x- (x + 9)
X->OO 1 + ~
1+0
3.x2 + 20x
9. lim ---,,..-x->oo 2x4 + 3x3 - 29
SOLUTION
.
hm
x->oo
3.x2 + 20x
.
x-4 (3.x2 + 20x)
~ + ~ ___
o __ 0
.
=
hm
= 1Im - - - 2x4 + 3x3 - 29 x->oo x- 4 (2x4 + 3x3 - 29) x->oo 2 + ~ - ~
2
7x-9
11. l i m - x->oo 4x + 3
SOLUTION
7
. 7x-9
. x- 1(7x-9)
. 7-~
1I m - - = 1Im 1
=Im--=1
x->oo 4x + 3
X->OO x- (4x + 3)
X->OO 4 + ~
4
7.x2-9
4X + 3
13. lim x-+-oo
SOLUTION
.
.
7.x2-9
x- 1(7x2-9)
7x-~
.
1Im - - - = 1Im
= 1Im - - = -oo
x->-oo 4x + 3
x->-oo x- 1(4x + 3)
x->-oo 4 + ~
15. lim
3x3 - 10
x-+-oo
X
+4
SOLUTION
lim
x->-oo
3x3 - 10
x +4
= lim
x- 1(3x3 - 10)
x->-oo
x- 1(x + 4)
= lim
x->-oo
3.x2 -
x
In Exercises 17-24,find the horizontal asymptotes.
2.x2 - 3x
17. f(x) = 8x2 + 8
SOLUTION
First calculate the limits as x
~
lim
x->oo
±oo. For x
2.x2 - 3x
8x2 + 8
~
oo,
2 - :!
2
1
=lim--x-=-=x->oo 8 +
8
4
!x
Similarly,
r
2x2
-
3x
x-l~oo 8x2 + 8
Thus, the horizontal asymptote off is y =
~.
2
8
lQ
x
1 + '.!
4
= oo
s E c TI o N 2.7 I Limits at Infinity 91
19.
v36x2 +1
x+4
9
f (x) =
For x > 0, x- 1
SOLUTION
= lx- 11= R, so
V36
. v36x2 + 7
l"
hm
= im
X-->OO
9x + 4
X-->OO
On the other hand, for x < 0, x- 1
v36x2 + 7
- 36 + :;,
- V36
2
= lim
= - - = -9x + 4
x-->-oo
9 + ix
9
3
~
x-->-oo
Thus, the horizontal asymptotes off are y
f (t) =
1
SOLUTION
2
3
= -lx- 11 = - R, so
lim
21.
9
= ~ and y = - ~.
e'
+ e-i
With
e1
l i m - -1 = oo
1 + e-
1-->oo
and
e'
lim - -1 =0
I + e-
1-->-oo
the function f has one horizontal asymptote, y = 0.
23. g(t) = 1:~-1
Because
SOLUTION
lim
t--+-oo
r' =
and
00
lim 3-I
t--+oo
=0
it follows that
.
10
hm - -1 =0
1-->-00
Thus, the horizontal asymptotes of g are y
1 + 3-
and
.
10
hm - -1
1 + 3-
1-->-00
= 10
= 0 and y = 10.
The following statement is incorrect: "If f has a horizontal asymptote y = Lat oo, then the graph off approaches the
line y = Las x gets greater and greater, but never touches it." In Exercises 25 and 26, determine lim f(x) and indicate
x-->oo
how f demonstrates that the statement is incorrect.
25• f(x) =
SOLUTION
2x;lxl
For x > 0, lxl = x and
2x+x
x
f(x)= - - =3
Thus, lim f (x) = 3, so f has a horizontal asymptote of y = 3. The statement that the graph of f never touches this
X-'>00
horizontal asymptote is incorrect because, for all x > 0, the graph off coincides with the horizontal asymptote y = 3.
In Exercises 27-34, evaluate the limit.
. v9.x4 + 3x + 2
27. hm
x--+oo
4 X3 + 1
SOLUTION
For x > 0, x- 3 = ix- 31 =
-..!r6, so
lim
x-->oo
8x2 + 7x1t3
29. lim --;::===
x+oo v16.x4 + 6
v9.x4 + 3x + 2
= lim
4x3 + 1
x--.oo
~ !r + !s + ~
4 + J..
x3
=0
92
CH APT E R 2
I
LIMITS
For x < 0, x- 2 =
SOLUTION
IX-21= V?, so
8.x2 + 7x 1i 3
8
8 + )/3
=lim
=--=2
lim
x+oo Yl6x4 + 6 x+oo ~16 + -3i
-{[6
t4f3
+ tlf3
31. Hoo
lim (4 f 2/3 + 1)2
t4f3
+ tlf3
l~1!! (4t2/3 + 1)2
SOLUTION
l +l
=
1
16
l.To! (4 + ,2;3 )2
. lxl +x
33. hm - x~-oo X + 1
For x < 0, lxl = -x. Therefore, for all x < 0,
SOLUTION
lxl + x _ -x + x _
x+l -7+1- 0
consequently,
lim lxl + x = 0
X->-oo X + 1
35. ~
SOLUTION
Determine lim tan- 1 x. Explain geometrically.
X->OO
As an angle(} increases from 0 to
~,
its tangent x = tan(} approaches oo. Therefore,
lim tan- 1 x = ~
X->OO
2
Geometrically, this means that the graph of y = tan- x has a horizontal asymptote at y = ~.
1
37. In 2009, 2012, and 2015, the number (in millions) of smart phones sold in the world was 172.4, 680.1, and 1423.9,
respectively.
(a) CR 5 Lett represent time in years since 2009, and let S represent the number of smart phones sold in millions.
Determine M, A, and k for a logistic model, S (t) = i+:!-k,, that fits the given data points.
(b) What is the long-term expected maximum number of smart phones sold annually? That is, what is lim S (t)?
t->oo
(c) In what year does the model predict that smart-phone sales will reach 98% of the expected maximum?
SOLUTION
(a) With t representing time in years since 2009, 2009 corresponds to t = 0, 2012 corresponds to t = 3, and 2015
corresponds tot = 6. To fit the given data to the logistic model S (t) = i+:!-k,, we must solve the system of equations
M
172.4 = l +A'
M
680.1 = 1 + Ae-3k,
and
1423.9 =
M
6k
1 +Ae-
for M, A, and k. Using a computer algebra system to solve this system yields M"" 1863.3, A"" 9.808, and k "" 0.576.
Hence, the model for the number of smart phones sold, in millions, is
1863.3
S (t) = 1 + 9.808e-o.s161
(b) Because lim e-0576' = 0, it follows that
1-->00
.
1863.3
1863.3
11m S(t) = lim
= - - = 1863.3
Hoo 1 + 9.808e-0.576t
1+ 0
t tooo
The long-term expected maximum number of smart phones sold annually is approximately 1863.3 million.
(c) To determine the year in which the model predicts that smart-phone sales will reach 98% of the expected maximum,
we must solve the equation
1863.3
(0.98)1863.3 = 1 + 9.808e-0.576r
for t. This equation is equivalent to
1 + 9.808e-0.576r =
1863.3
(0.98)1863.3
0.98
and then
1
0 02
9.808e-0·576' = - - - 1 = · = ..!.._
0.98
0.98
49
s Ec T Io N
2. 7
I
Limits at Infinity
93
Finally,
1
1
t = -0.576 ln 49(9.808) ""
10 12
·
Thus, the year in which the model predicts that smart-phone sales will reach 98% of the expected maximum is approximately 10.72 years after 2009, or sometime in 2019.
In Exercises 39-46, calculate the limit.
39. lim( ../4x4 + 9x - 2x2)
X-+OO
Write
SOLUTION
4
../4x4 + 9x - 2x2 = ( ../4x4 + 9x - 2x2) ../ x4 + 9x + 2x2
../4x4 + 9x + 2x2
(4x4 + 9x) - 4x4
9x
../4x4 + 9x + 2x
V4x4 + 9x + 2xz
2
Thus,
9
x
lim( ../4x4 + 9x - 2x2) = lim
x-+oo
../4x4 + 9x + 2x2
x-+oo
=
41. lim(2..;i- .Yx+ 2)
X-+OO
Write
SOLUTION
_r::. _r--;;
_r::. _r--;; 2..;x+ vx+2
2v .... - vx+2=(2vx- v x + 2 ) - - - - 2..;i+ ../x+2
4x- (x+ 2)
3x-2
2..;i+.Yx+2
Thus,
lim(2../i- ../x+2) = lim
x-+oo
x-+oo
2
3
x-
2 ..jX + .../X + 2
43. lim (ln(3x + 1) - ln(2x + 1))
x-+oo
Because
SOLUTION
3x+ 1
ln(3x + 1) - ln(2x + 1) = ln - 2x + 1
and
lim 3x+ 1
2X + 1
=~
2
X-+OO
it follows that
lim (ln(3x + 1) - ln(2x + 1))
=
X-+OO
ln ~
2
9)
45. lim tan- 1 (x2+
x-+oo
9- X
SOLUTION
Because
. x2+9
. x+~
oo
hm - - = hm - - = - = -oo
X-+oo
9-
X
x-+oo
~
-
}
-1
it follows that
. tan- l(x2+9)
hm
- - = - -1f
9- X
2
x-+oo
=
00
o
94
CHAPTER 2
I
LIMITS
47.
($J
Let P(n) be the perimeter of an n-gon inscribed in a unit circle (Figure 8).
(a) Explain, intuitively, why P(n) approaches 2n as n ~ oo.
(b) Show that P(n) = 2n sin(~).
(c) Combine (a) and (b) to conclude that lim
!!
n--+oo Tr
sin(!!)=
1.
n
sinB
(d) Use this to give another argument that lim -() = 1.
·
.
"
·
.
·
.
·
.
00.
e~o
''
'
~'
'
"
"
'
<
,'
-
·,
'
n=6
'
n= 12
n=9
FIGURE 8
SOLUTION
(a) As n ~ oo, the n-gon approaches a circle of radius 1. Therefore, the perimeter of the n-gon approaches the circumference of the unit circle as n ~ oo. That is, P(n) ~ 2n as n ~ oo.
(b) Each side of the n-gon is the third side of an isosceles triangle with equal length sides of length 1 and angle () = ~
between the equal length sides. The length of each side of the n-gon is therefore
2
/l2+l2-2cos 7f =
-V
n
2
/2(1-cos 7f)=
-V
n
~4sin2 '.!'.=2sin'.!'.
n
n
Finally,
. 7f
n
P(n) = 2nsm -
(c) Combining parts (a) and (b),
7f
lim P(n) = lim 2n sin - = 2n
n-HlO
n
n--+OO
Dividing both sides of this last expression by 2n yields
lim~sin'.!'.=1
7f
n
n--+oo
(d) Let()= ~-Then()~ 0 as n ~ oo,
n . 7f
I .
sin()
-sm- = -smB= - 7f
nB
()
and
n.n
sin()
lim -sm- =lim-- =I
7f
n e~o ()
n~oo
49. According to the Michaelis-Menten equation, when an enzyme is combined with a substrate of concentration s (in
millimolars), the reaction rate (in micromolars/min) is
As
R(s)= - K
(A, K constants)
+s
(a) Show, by computing!~~ R(s), that A is the limiting reaction rate as the concentrations approaches oo.
(b) Show that the reaction rate R(s) attains one-half of the limiting value A when s = K.
(c) For a certain reaction, K = 1.25 mM and A = 0.1. For which concentrations is R(s) equal to 75% of its limiting
value?
SOLUTION
As
.
.
.
(a) limR(s) = 11m - - = lim
s-Hio
s--+oo
K+
S
s--+oo
A
--K
1+ -
=A
s
(b) Observe that
R(K) =
AK
K+K
half of the limiting value.
= AK = ~
2K
2
S EC TI o N 2.B
I
The Intermediate Value Theorem
95
(c) By part (a), the limiting value is 0.1, so we need to determine the value of s that satisfies
R(s) = 1.
O.ls
+ s = 0.075
25
Solving this equation for s yields
s
= (1.25)(0.075) = 3.75 mM
0.025
Further Insights and Challenges
51. Rewrite the following as one-sided limits as in Exercise 50 and evaluate.
3 - 12.x3
(b) lim e 11 x
(a) l~~ -4x_3_+_3_x_+_l
x->oo
1
(c) lim xsin x-+oo
1)
x +(d) lim In ( X-+CO
X - 1
X
SOLUTION
(a) Lett= x- 1• Then x = i- 1, t--+ O+ as x--+ oo, and
3 - 12i-3
4r 3 + 3r 1 + I
3 - 12.x3
4x3 + 3x + 1
3t3 - 12
4 + 3t2 + t3
Thus,
r
3- I2.x3
= lim
x~~ 4x3 + 3x + 1
t->O+
3t3 -12 = -12 = _ 3
4 + 3t2 + t3
4
(b) Lett = x- 1 • Then x = t- 1 , t--+ O+ as x --+ oo, and e'fx = e'. Thus,
lim e 11 x = lim e' = e0 = 1
t-+0+
X-+OO
1
1
(c) Lett = x- • Then x = t- , t --+ O+ as x --+ oo, and
. 1
1 .
sin t
= - Sill t = x
t
t
X Sill -
Thus,
. 1
. sin t
limxsm- = 11m = 1
X
x-+oo
f
t-+0+
(d) Lett = x- 1 • Then x = r', t --+ O+ as x --+ oo, and
x+ 1
r' + 1
x- I
r
1 -
1+ t
1- t
1
Thus,
x+l)
.
(l+t)
limln ( - = hmln
- - =lnl =0
X->OO
X -
1
1->0+
1- (
2.8 The Intermediate Value Theorem
Preliminary Questions
x2 takes on the value 0.5 in the interval [O, l].
Observe that f(x) = x2 is continuous on [O, 1] with f(O) = 0 and f(l) = 1. Because f(O) < 0.5 < f(l), the
1. Prove that f(x) =
SOLUTION
Intermediate Value Theorem guarantees there is a c
E
[0, 1] such that f(c) = 0.5.
2. The temperature in Vancouver was 8°C at 6 AM and rose to 20°C at noon. Which assumption about temperature allows
us to conclude that the temperature was 15°C at some moment of time between 6 AM and noon?
SOLUTION
We must assume that temperature is a continuous function of time.
3. What is the graphical interpretation of the IVT?
SOLUTION Iff is continuous on [a, b], then the horizontal line y = k for every k between f(a) and f(b) intersects the
graph of y = f(x) at least once.
96
CHAPTER 2
I
LIMITS
4. Show that the following statement is false by drawing a graph that provides a counterexample:
If f is continuous and has a root in [a, b], then f(a) and f(b) have opposite signs.
SOLUTION
y
5. Assume that f is continuous on [1,5] and that f(l)
statements is always true, never true, or sometimes true.
(a) f(c) = 3 has a solution with c E [l, 5].
(b) f(c) = 75 has a solution with c E [1, 5].
= 20, f(5) =
100. Determine whether each of the following
(c) f(c) = 50 has no solution with c E [l, 5].
(d) f(c) = 30 has exactly one solution with c E [1, 5].
SOLUTION
(a) This statement is sometimes true. Because 3 does not lie between 20 and 100, the IVT cannot be used to guarantee
that the function takes on the value 3 but it may still do so.
(b) This statement is always true. Because f is continuous on [l, 5] and 20 = f(l) < 75 < f(5) = 100, the IVT
guarantees there exists a c E [1, 5] such that f(c) = 75.
(c) This statement is never true. Because f is continuous on [1, 5] and 20 = f(l) < 50 < f(5) = 100, the IVT guarantees
there exists a c E [l, 5] such that f(c) = 50.
(d) This statement is sometimes true. Because f is continuous on [1, 5] and 20 = f(l) < 30 < f(5) = 100, the IVT
guarantees there exists a c E [l, 5] such that f(c) = 30 but there may be more than one such value for c.
Exercises
1. Use the IVT to show that f(x)
= x3 + x takes on the value 9 for some x in [1, 2].
Observe that f(l) = 2 and f(2) = 10. Since f is a polynomial, it is continuous everywhere; in particular
on [1,2]. Therefore, by the IVTthereis ac E [1,2] such thatf(c) = 9.
SOLUTION
3. Show that g(t)
SOLUTION
there is a c
E
= r tan t takes on the value t for some tin [O, H
g(O) = 0 and g(~) = ~- g(t) is continuous for all t between 0 and~. and O <
[O, ~]such that g(c) =
t-
t < ~;therefore, by the IVT,
= x has a solution in the interval [0, l]. Hint: Show that f(x) = x - cos x has a zero in [O, l].
SOLUTION Let f(x) = x - cos x. Observe that f is continuous with f(O) = -1 and f(l) = 1 - cos 1 ~ .46. Therefore,
by the IVT there is a c E [0, l] such that f(c) = c - cos c = 0. Thus c = cos c and hence the equation cos x = x has a
5. Show that cos x
solution c in [O, 1].
In Exercises 7-16, prove using the /VT.
.ye+ -Ve+ 2 = 3 has a solution.
SOLUTION Let f(x) = Vx + -Vx + 2 - 3. Note that f is continuous on [0,2] with f(O) = VO+ -.../2- 3 ~ -1.59 and
f(2) = V2 + '14- 3 ~ 0.41. Therefore, by the IVT there is a c E [0,2] such that f(c) = .ye+ Ve+ 2 - 3 = 0. Thus
.ye+ Ve+ 2 = 3, and the equation .ye+ vc + 2 = 3 has a solution c in [O, 2].
9. V2 exists. Hint: Consider f(x) = x2.
7.
Let f(x) = x2. Observe that f is continuous with f(l) = 1 and f(2) = 4. Therefore, by the IVT there is a
[l, 2] such that f(c) = c2 = 2. This proves the existence of -.../2, a number whose square is 2.
SOLUTION
c
E
11. For all positive integers k, cos x
SOLUTION
= :x!' has a solution.
For each positive integer k, let f (x)
= :x!' -
cos x. Observe that f is continuous on [O, ~] with f(O)
and f(~) = (~t > 0. Therefore, by the IVT there is a c E
~]such that f(c)
hence the equation cos x = :x!' has a solution c in the interval [0, ~].
[o,
13. 2x + 3x
= 4x has a solution.
= c" -
cos(c)
= -1
= O. Thus cosc = c" and
s E c TI o N
2.8
I
The Intermediate Value Theorem
97
Let f(x) = 2x + Y - 4x_ Observe that f is continuous on [0, 2] with f(O) = 2° + 3° - 4° = 1 + 1 - 1 = 1
and f(2) = 22 + 32 - 42 = 4 + 9 - 16 = -3. Therefore, by the IVT, there is a c E [O, 2] such that f(c) = 2c + 3c - 4c = 0.
Hence, the equation 2x + 3x = 4x has a solution.
15. ex + In x = 0 has a solution.
2
SOLUTION Let f(x) =ex+ Inx. Observe that f is continuous on [e- 2, 1] with f(e- 2 ) = e'- - 2 < 0 and f(l) = e > 0.
2
Therefore, by thelVT, there is ac E [e- , 1] such thatf(c) = ec + lnc = 0.
17. Use the Intermediate Value Theorem to show that the equation x 6 - 8.x4 + 10x2 - 1 = 0 has at least six distinct
SOLUTION
solutions.
Let f(x) = x 6 - 8x4 + IOx2 - 1. Then f(O) = -1,f(±l) = 2,f(±2) = -25 and f(±3) = 170. Hence as we
move along the number line from left to right through the points -3, -2, -1, 0, 1, 2, 3, the function changes sign at least
6 times. Hence there must be a zero of the function between any two of these integers, and therefore, there must be at
least six distinct solutions to the equation x 6 - 8x4 + IOx2 - 1 = 0.
SOLUTION
In Exercises 18-20, determine whether or not the /VT applies to show that the given function takes on all values between
f(a) and f(b) for x E (a, b). If it does not apply, determine any values between f(a) and f(b) that the function does not
take on for x E (a,b).
19.
-x
f(x)
= { x3 + 1
for x < 0
for x ~ 0
for the interval [-1, 1].
SOLUTION The graph off over the interval [-1, 1] is shown below. From the graph, we see that f is not continuous
on [-1, 1] because of a jump discontinuity at x = 0. Therefore, the IVT does not apply. However, from the graph, we see
that f does take on every value between f(-1) = 1 and f(l) = 2 for x E (-1, 1).
y
-1.0
-0.5
0.5
1.0
Carry out three steps of the Bisection Method for f(x) = 2x -x3 as follows:
Show that f has a zero in [l, 1.5].
Show that f has a zero in [1.25, 1.5].
Determine whether [1.25, 1.375] or [1.375, 1.5] contains a zero.
SOLUTION Note that f(x) is continuous for all x.
(a) f(l) = 1, f(l.5) = 21. 5 - (1.5) 3 < 3 - 3.375 < 0. Hence, f(x) = 0 for some x between 1 and 1.5.
(b) f(l.25) "'0.4253 > 0 and f(l.5) < 0. Hence, f(x) = 0 for some x between 1.25 and 1.5.
(c) f(l.375)"' -0.0059. Hence, f(x) = 0 for some x between 1.25 and 1.375.
21.
(a)
(b)
(c)
23. Find an interval of length ~ in [1, 2] containing a root of the equation x7 + 3x - 10 = 0.
Let f(x) = x 1 + 3x - 10. Observe that f is continuous on [1, 2] with f(l) = -6 and f(2) = 124, so the IVT
guarantees that the equation x7 + 3x - 10 = 0 has a root on the interval [l, 2]. The midpoint of the interval [1, 2] is 1.5
and f(l.5) = 11.585938 > 0, so we can conclude that the equation x7 + 3x - 10 = O has a root on the interval [1, 1.5].
Finally, the midpoint of the interval [1, 1.5] is 1.25 and f(l.25) = -1.481628 < 0, so we can conclude that the equation
x1 + 3x - 10 = 0 has a root on the interval [l.25, 1.5].
SOLUTION
In Exercises 25-28, draw the graph of a function f on [O, 4] with the given property.
25. Jump discontinuity at x = 2 and does not satisfy the conclusion of the IVT
SOLUTION The function graphed below has a jump discontinuity at x = 2. Note that while f(O) = 2 and f(4) = 4,
there is no point c in the interval [O, 4] such that f(c) = 3. Accordingly, the conclusion of the IVT is not satisfied.
y
4
-
2
0
-+-+--+---+-+-+-x
2
4
98
CH APT E R 2
I
LIMITS
= 2 and does not satisfy the conclusion of the IVT
The function graphed below has infinite one-sided limits at x = 2. Note that while f(O) = 2 and f(4) = 4,
27. Infinite one-sided limits at x
SOLUTION
there is no point c in the interval [0, 4] such that f(c) = 3. Accordingly, the conclusion of the IVT is not satisfied.
y
6
-~
5
4
2
x
2
4
-1
29. ~
Can Corollary 2 be applied to f(x) = x- 1 on (-1, 1]? Does f have any roots?
SOLUTION Although f(-1) = -1 < 0 and f(l) = 1 > 0 are of opposite sign, Corollary 2 cannot be applied because f
is not continuous on the interval [-1, 1]. This function does not have any roots.
31. At 1:00 PM. Jacqueline began to climb up Waterpail Hill from the bottom. At the same time Giles began to climb
down from the top. Giles reached the bottom at 2:20 PM, when Jacqueline was 85% of the way up. Jacqueline reached
the top at 2:50. Use the result in Exercise 30 to prove that there was a time when they were at the same elevation on the
hill.
SOLUTION Lett represent time in minutes since 1:00 PM, and let J(t) and G(t) represent the elevation in percentage of
the way up the hill at time t of Jacqueline and Giles, respectively. Then J(O) = 0 < 1 = G(O) and G(80) = 0 < 0.85 =
J(80). Assuming that J and G are continuous on (0, 80], the result in Exercise 30 implies there exists c E (0, 80] such
that J(c) = G(c). At time c, Jacqueline and Giles were at the same elevation on the hill.
Further Insights and Challenges
Exercises 33 and 34 address the I-Dimensional Brouwer Fixed Point Theorem. It indicates that every continuous function
f mapping the closed interval [0, 1] to itself must have a fixed point; that is, a point c such that f(c) = c.
33. ~
Show that if f is continuous and 0
~ f(x) ~ 1 for 0 ~ x ~ 1, then f(c)
y
1 ___________/
= c for some c in (0, 1] (Figure 7).
Y=X
/I
/
/
/
/I
Y=f(x)
I
I
I
I
I
I
I
I
c
FIGURE 7 A function satisfying 0 ~ f(x) ~ 1for0 ~ x ~ 1.
If f(O) = 0, the proof is done with c = 0. We may assume that f(O) > 0. Let g(x) = f(x) - x. g(O) =
f (0) - 0 = f (0) > 0. Since f (x) is continuous, the Rule of Differences dictates that g(x) is continuous. We need to prove
that g(c) = 0 for some c E [0, I]. Since f(l) ~ 1, g(l) = f(l) - 1 ~ 0. If g(l) = 0, the proof is done with c = 1, so let's
assume that g(l) < 0.
We now have a continuous function g(x) on the interval [O, l] such that g(O) > 0 and g(l) < 0. From the IVT, there
must be some c E [O, 1] so that g(c) = 0, so f(c) - c = 0 and so f(c) = c.
SOLUTION
35. Use the IVT to show that if f is continuous and one-to-one on an interval [a, b], then f is either an increasing or a
decreasing function.
Let f(x) be a continuous, one-to-one function on the interval [a, b]. Suppose for sake of contradiction that
f(x) is neither increasing nor decreasing on [a, b]. Now, f(x) cannot be constant, for that would contradict the condition
that f(x) is one-to-one. It follows that somewhere on [a, b], f(x) must transition from increasing to decreasing or from
decreasing to increasing. To be specific, suppose f(x) is increasing for x 1 < x < x 2 and decreasing for x 2 < x < x 3 •
Let k be any number between max{f(x1), f(x 3 )) and f(x 2 ). Because f(x) is continuous, the IVT guarantees there exists a
c1 E (x1, X2) such that f(c1) = k; moreover, there exists a c2 E (x2 , X3) such that f(c 2 ) = k. However, this contradicts the
condition that f(x) is one-to-one. A similar analysis for the case when f(x) is decreasing for x 1 < x < x 2 and increasing
for Xz < x < X3 again leads to a contradiction. Therefore, f(x) must either be increasing or decreasing on [a, b ].
SOLUTION
SECT Io N 2.9
I
The Formal Definition of a Limit
99
37. ($1 Figure 8(B) shows a slice of ham on a piece of bread. Prove that it is possible to slice this open-faced
sandwich so that each part has equal amounts of ham and bread. Hint: By Exercise 36, for all 0 :5 (} :5 n there is a line
L((}) of incline(} (which we assume is unique) that divides the ham into two equal pieces. Let B((}) denote the amount of
bread to the left of (or above) L((}) minus the amount to the right (or below). Notice that L(n) and L(O) are the same line,
but B(n) = -B(O) since left and right get interchanged as the angle moves from 0 ton. Assume that Bis continuous and
apply the IVT. (By a further extension of this argument, one can prove the full Ham Sandwich Theorem, which states
that if you allow the knife to cut at a slant, then it is possible to cut a sandwich consisting of a slice of ham and two slices
of bread so that all three layers are divided in half.)
y
L(8)
L(O) = L(lf)
(B) A slice of ham on top
(A) Cutting a slice of ham
at an angle (J
of a slice of bread
FIGURE 8
For each angle(}, 0 :5 (} < rr, let L((}) be the line at angle(} to the x-axis that slices the ham exactly in half,
as shown in Figure 8. Let L(O) = L(rr) be the horizontal line cutting the ham in half, also as shown. For(} and L((}) thus
defined, let B((}) = the amount of bread to the left of L((}) minus that to the right of L((}).
To understand this argument, one must understand what we mean by "to the left" or "to the right". Here, we mean to
the left or right of the line as viewed in the direction e. Imagine you are walking along the line in direction(} (directly
right if(} = 0, directly left if(} = rr, etc).
We will further accept the fact that B is continuous as a function of(}, which seems intuitively obvious. We need to
prove that B(c) = 0 for some angle c.
Since L(O) and L(rr) are drawn in opposite direction, B(O) = -B(rr). If B(O) > 0, we apply the IVT on [0, rr] with
B(O) > 0, B(rr) < 0, and B continuous on [O, 7r]; by IVT, B(c) = 0 for some c E [0, rr]. On the other hand, if B(O) < 0, then
we apply the IVT with B(O) < 0 and B(rr) > 0. If B(O) = 0, we are also done; L(O) is the appropriate line.
SOLUTION
2.9 The Formal Definition of a Limit
Preliminary Questions
1. Given that lim cos x = 1, which of the following statements is true?
x->O
(a) If lcos x - 11 is very small, then xis close to 0.
(b) There is an E > 0 such that if if 0 < lcos x - 11 < E, then lxl < 10-5 •
(c) There is a 8 > 0 such that ifO < lxl < 8, then lcos x - 11 < 10-5 •
(d) There is a 8 > 0 such that if 0 < Ix - 11 < 8, then lcos xi < 10-5 •
SOLUTION
The true statement is (c): There is a 8 > 0 such that if O < lxl < 8, then lcosx - 11 < 10-5.
2. Suppose it is known that for a given E and 8, if 0 < Ix - 31 < 8, then lf(x) - 21 < E. Which of the following statements
must also be true?
(a) IfO <Ix - 31<28, then lf(x)- 21 < E.
(b) If 0 < Ix - 31 < 8, thenlf(x) - 21 < 2E.
0
E
(c) lfO <Ix- 31 <
2' then lf(x)- 21 < 2·
(d) IfO <Ix- 31 <
2' then lf(x)-21 < E.
SOLUTION
8
Statements (b) and (d) are true.
Exercises
1. Based on the information conveyed in Figure S(A), find values of L,
holds: If lxl < 8, then lf(x) - LI < E.
E,
and o > O such that the following statement
We see -0.1 < x < 0.1 forces 3.5 < f(x) < 4.8. Rewritten, this means that lxl < 0.1 implies that
lf(x) - 41 < 0.8. Looking at the limit definition lxl < implies If(x) - LI < E, we can replace so that L = 4, E = 0.8, and
8 = 0.1.
SOLUTION
o
100
C HA PT ER 2
I
LIMITS
3. Make a sketch illustrating the following statement: To prove lim x =a, given c > 0, we can take 8
X->a
= c to have the
gap be small enough.
SOLUTION
See the figure below. With 8 = c, the gap is within c of a.
f(x)=x
5. Consider limf(x), where f(x) = 8x + 3.
x->4
(a) Show that lf(x) - 3SI = 81x - 41.
(b) Show that for any E > 0, if 0 < Ix - 41 < 8, then If(x) - 3SI < c, where 8
that limf(x) = 3S.
= ~. Explain how this proves rigorously
x->4
SOLUTION
(a) lf(x) - 3SI = 18x + 3 - 3SI = 18x - 321 = 18(x - 4)1 = 8 Ix - 41. (Remember that the last step is justified because 8 >
0.)
(b) Let c > 0. Let 8 = c/8 and suppose Ix - 41 < 8. By part (a), lf(x) - 3SI = 81x - 41 < 88. Substituting 8 = c/8, we see
lf(x) - 3SI < 8c/8 = c. We see that, for any c > 0, we found an appropriate 8 so that Ix - 41 < 8 implies lf(x) - 3SI < c.
Hence limf(x) = 3S.
x->4
7. Consider lim x2 = 4 (refer to Example 2).
x->2
(a) Show that if 0 < Ix - 21 < 0.01, then lx2 - 41 < O.OS.
(b) Show that if 0 < Ix - 21 < 0.0002, then lx2 - 41 < 0.0009.
(c) Find a value of 8 such that ifO <Ix - 21<8, then lx2 - 41 is less than 10-4.
SOLUTION
(a) If 0 < Ix - 21 < 8 = .01, then lxl < 3 and lx2 - 41 = Ix - 211x + 21 ~Ix - 21(lxl+2) <Six- 21 < .OS.
(b) If 0 < Ix - 21 < t5 = .0002, then lxl < 2.0002 and
lx2 - 41
= Ix -
211x + 21 ~ Ix - 21 (lxl + 2) < 4.00021x - 21 < .00080004 < .0009
(c) Note that lx2 - 41 = l(x + 2)(x - 2)1 ~ Ix+ 21 Ix - 21. Since Ix - 21 can get arbitrarily small, we can require Ix - 21 < 1
so that 1 < x < 3. This ensures that Ix+ 21 is at most S. Now we know that lx2 - 4j ~ Six - 21. Let 8 = 10-5 • Then, if
Ix - 21 < 8, we get lx2 - 41 ~ Six - 21 < S x 10-5 < 10-4 as desired.
9. Refer to Example 3 to find a value of t5 > 0 such that
SOLUTION
.
If 0 <Ix - 31<8,
then
The Example shows that for any c >Owe have
.
1 11
1~ - 3 ~ E 1f Ix -
i! -!i
x
3
< 10-4
31 < 8
where t5 is the smaller of the numbers 6c and 1. In our case, we may take t5 = 6 x 1o-4.
11. ~GU!J Plot f(x) = -../2x - 1 together with the horizontal lines y
t5 > 0 such that ifO <Ix - SI< 8, then I V2x - 1 - 31<0.1.
SOLUTION
= 2.9 and y = 3.1. Use this plot to find a value of
From the plot below, we see that 8 = 0.2S will guarantee that I -../2x - 1 - 31 < 0.1 whenever Ix - SI ~ 8.
y
3.1
2.9
11 J_ l
sEcT IoN
13.
!GUJI
2.9
I
The Formal Definition of a Limit
101
ex-1
ex-1
The number e has the following property: lim - - = 1. Use a plot of f(x) = - - to find a value of 8 > 0
~a
x
x
such that if 0 < lxl < 8, then 1/(x) - 11 < 0.01.
SOLUTION
From the plot below, we see that 8 = 0.02 will guarantee that
e-"-1
I
l- x - -1 <0.01
whenever lxl < 8.
y
1.02
1.01
1.00
0.99
~::~
!! x
-0.04 -0.02
0
0.02 0.04
1
15. Consider lim - .
x-+2 X
(a) Show that if Ix - 21 < 1, then
!_ !I< !1x-21
2
Ix 2
(b) Find a 8 > 0 such that if 0 <Ix - 21<8, then I~ - ~I< 0.01.
(c) Let 8 be the smaller of 1 and 2E. Prove the following:
If
o < Ix -
21 < 8,
1
Then explain why this proves that lim ! = - .
x-+2 X
2
SOLUTION
1
(a) Since Ix - 21 < 1, it follows that 1 < x < 3, in particular that x > 1. Because x > 1, then - < 1 and
x
1
~
1
11
12 - x I
Ix - 21
1
- 2 = ~ = ~ < 2lx- 21
1 = O.Ql by part (a).
(b) Choose 8 = 0.02. Then -1 - -11 < -8
2
2
1x
(c) Let 8 = min{ 1, 2i:) and suppose that Ix - 21 < 8. Then by part (a) we have
! - !I < !1x - 21 < !8 < ! · 2i: = i:
2
2
2
Ix 2
Let E > 0 be given. Then whenever 0 <Ix- 21 < 8 =min {1, 2i:), we have
!_ !I< !8 < i:
2 Ix 2
Since E was arbitrary, we conclude that lim ! =
x-->2 X
17. ~
!2 .
Let f (x) = sin x. Using a calculator, we find
t(i- 0.1) ""0.633, t(i) ""0.707, t(i + 0.1)"" 0.774
Use these values and the fact that f is increasing on
If
[o, ~]to justify the statement
o <Ix - ~I< 0.1.
then
lt(x)- f (~)I< 0.08
Then draw a figure like Figure 3 to illustrate this statement.
Since f(x) is increasing on the interval, the three f(x) values tell us that .633 ~ f(x) ~ .774 for all x
- .1 and~+ .1. We may subtract f(~) from the inequality for f(x). This show that, for~ - .1 < x < ~ + .1,
.633 - f(D ~ f(x) - /(~) ~ .774 - /(~).This means that, if Ix - ~I< .1, then .633 - .707 ~ f(x)- /(~) ~ .774 - .707,
so -0.074 ~ f(x) - f(~) ~ 0.067. Then -0.08 < f(x) - /CD < 0.08 follows from this, so Ix - ~I < 0.1 implies
lf(x) - JCDI < .08. The figure below illustrates this.
SOLUTION
between~
102
CHAPTER 2
I
LIMITS
y
0.8
0.6~~~~~~~
0.4
0.2
0.25 0.5 0.75
I
1.25 1.5
19. Adapt the argument in Example 2 to prove rigorously that
lim.x2 = c2 for all c.
X-->C
SOLUTION
To relate the gap to Ix - cl, we take
J.x2 -c2J = l(x+c)(x-c)I =Ix+ cl Ix-cl
We choose oin two steps. First, since we are requiring Ix - cl to be small, we require o < 1 + lcl, where we have added
1 to avoid complications associated with c = 0. Then lxl < 21cl + 1 and Ix+ cl < 31cl + 1, so Ix - cllx +cl < (31cl + l)o.
Next, we require that o <
3
lcr+ l' so
E
Ix - cllx +cl <
3
lcl + (3lcl + 1) =
1
E
and we are done.
Therefore, given E > 0, we let
o= min { 1 + lei, 31c1E+ 1 }
Then, for Ix - cl <
o, we have
lx2 - c2 1= Ix - cl Ix+ cl < (31cl + l)o < (31cl + 1 )
E - =
3lcl + 1
E.
In Exercises 21-26, use the formal definition of the limit to prove the statement rigorously.
21. lim Vx = 2
x-->4
SOLUTION
Let E > 0 be given. We bound I yx - 21 by multiplying
~ + 2.
vx+2
2
4
1
1...rx-21 = <vx-2>- - = yx+
x-- 1 = lx-41 - vx+2
vx+2
vx+2
1I
I
We can assume o < 1, so that Ix- 41 < I, and hence
I
yx + 2 > V3 + 2 > 3. This gives us
1
lvx-21 = lx-41J-- J <
yx+2
Leto= min(l, 3E). If Ix- 41 <
I
lx-41~.
3
o,
I
IVx-21 = lx-41 -I- < lx-41-1 < o-1 < 3E-1 =
yx+2
3
3
3
I
E
thus proving the limit rigorously.
23. limx3 = I
X-->I
SOLUTION
Let E > 0 be given. We bound Jx3 - 1J by factoring the difference of cubes:
Jx3 - IJ = J<x2 + x + I)(x- I)j =Ix - 11J.x2+x+1J
Leto= rnin(l, ~),and assume Ix- 11 <
x2 + x + 1 < 7 for 0 < x < 2, and so
o. Since o <
1, 0 < x < 2. Since x2 + x +I increases as x increases for x > O,
Jx3 - lJ =Ix- 11 Jx2 + x + lJ < 71x- 11<1; =
and the limit is rigorously proven.
E
s Ec T Io N
I
2.9
The Formal Definition of a Limit
103
1
25. limx-2 = x-+2
4
SOLUTION
Let€> 0 be given. First, we bound lx-2 - ~I:
lx-2 - !4 I= 144x- x2 I= 12 - xi 124x+ x I
2
2
Let 8 = min(l, ~€),and suppose Ix - 21 < 8. Since 8 < l, Ix - 21 < 1, so 1 < x < 3. This means that 4x2 > 4 and
2+x
12 +xi< 5, so that x2 < ~·We get
4
-2
1x
I
I
I
2 +x <-Ixs 21 < -s · -€
4 =
- -1 = 12-xl 4
4x2
4
4 5
€
and the limit is rigorously proven.
1
27. Let f(x) = ~ . Prove rigorously that limf(x) does not exist. Hint: Show that for any L, there always exists some x
Ix
x-+0
such that lxl < 8 but lf(x) - LI~ no matter how small 8 is taken.
!,
SOLUTION Let L be any real number. Let 8 > 0 be any small positive number. Let x
f(x) = 1. We consider two cases:
= ~.which satisfies lxl < 8, and
!) : we are done.
!): This means ! < L < ~·In this case, let x = -~. f(x) = -1, and so ~ < L In either case, there exists an x such that lxl < ~, but If(x) - LI ~ ! .
• (lf(x) - LI~
• (lf(x) - LI<
f(x).
29. Let f(x) = min(x, x2), where min(a, b) is the minimum of a and b. Prove rigorously that lim f (x) = 1.
x-+l
Let E > 0 be given, and take 8 = min(l, D· Suppose 0 < Ix - 11 < 8. Then either 1 - 8 < x < 1 or
1 < x < 1 + 8. If 1 < x < 1 + 8, then
SOLUTION
lf(x) - 11 = Ix - 11 < 8 <
€
2<€
as required. On the other hand, if 1 - 8 < x < 1, then 11 +xi< 2 and
lf(x) - 11 = lx2 - 11 =Ix - lllx + 11 < 21x - 11 < 28 <
Thus, limf(x) = 1.
x-+1
31. Use the identity
sinx + siny = 2sin{
x; Y)cosC; Y)
to prove that
h)
.
.
sin(h/2)
( a+ = h---cos
sm(a
+ h)-sma
h/2
2
I I
Then use the inequality si; x :$ 1 for x
* 0 to show that
lsin(a + h) - sin al< lhl for all a. Finally, prove rigorously that
lim sin x = sin a.
X-+a
SOLUTION
We first write
sin(a+ h) - sin a = sin(a+ h) + sin(-a)
Applying the identity with x = a + h, y = -a, yields:
· (
h) -sma=sma+
·
· (
h) +sm(-a)=2sm
.
. (a+h-a) cos (2a+h)
sma+
- 2
2
Therefore,
lsin(a + h) - sin al = lhl
Isi~;~ 2) I!cos (a+ ~)I
€
104
CHAPTER 2
I
LIMITS
Using the fact that
Isi~ 8 I < 1 and that Icos 81 ~ 1, and making the substitution h = x - a, we see that this last relation is
equivalent to
lsinx - sin al< Ix - al
Now, to prove the desired limit, let E > 0, and take 6 =
If Ix - al < 6, then
E.
lsin x - sin al < Ix - al < 6 =
Therefore, a 6 was found for arbitrary
E,
E
and the proof is complete.
Further Insights and Challenges
In Exercises 33-35, prove the statement using the formal limit definition.
33. The Constant Multiple Law [Theorem 1, part (ii) in Section 2.3]
SOLUTION
Let
E
Suppose that limf(x)
= L. We wish to prove that limaf(x) = aL.
> 0 be given. c/la(~calso a positive number. Since limx.f(".x)
X-->C
forces lf(x) - LI < c/lal. Suppose Ix - cl < 6. laf(x) - aLI
= lallf(x) -
= L, we know there is a 6 > 0 such that Ix aLI < lal(c/lal)
cl < 6
= E, so the rule is proven.
35. The Product Law [Theorem 1, part (iii) in Section 2.3]. Hint: Use the identity.
f(x)g(x) - LM
= (f(x) -
L) g(x) + L(g(x) - M)
Before we can prove the Product Law, we need to establish one preliminary result. We are given that
Consequently, if we set E = 1, then the definition of a limit guarantees the existence of a 6 1 > 0
such that whenever 0 < Ix - cl < 61, lg(x) - Ml < 1. Applying the inequality lg(x)l - IMI ~ lg(x) - Ml, it follows that
lg(x)I < 1 + IMI. In other words, because limx-->c g(x) = M, there exists a 61 > 0 such that lg(x)I < 1 + IMI whenever
0 < Ix - cl < 61.
We can now prove the Product Law. Let E > 0. As proven above, because limx-->c g(x) = M, there exists a 6 1 > 0 such
that lg(x)I < 1 + IMI whenever 0 < Ix - cl < 61. Furthermore, by the definition of a limit, limx--+c g(x) = M implies there
exists a 62 > 0 such that lg(x) - Ml < Z(l:ILll whenever 0 < Ix - cl < 62. We have included the "1 +" in the denominator
to avoid division by zero in case L = 0. The reason for including the factor of 2 in the denominator will become clear
shortly. Finally, because limx--+c f(x) = L, there exists a 03 > 0 such that lf(x) - LI < Z(l:IMll whenever 0 < Ix - cl < 03.
Now, let 6 = min(6 1, 62, 63 ). Then, for all x satisfying 0 < Ix - cl < 6, we have
SOLUTION
limx-->c g(x)
= M.
lf(x)g(x) - LMI
= l(f(x) ~
L)g(x) + L(g(x) - M)I
lf(x) - LI lg(x)I + ILi lg(x) - Ml
€
€
< 2(1 + IMD (l + IMI) + ILi 2(1 + ILD
Hence,
limf(x)g(x)
x-+c
37.
~
= LM = limf(x)
· limg(x)
x-+c
x-+c
Here is a function with strange continuity properties:
f(x) =
(
~
q
if xis the rational number p/q in
lowest terms
0
if x is an irrational number
(a) Show that f is discontinuous at c if c is rational. Hint: There exist irrational numbers arbitrarily close to c.
(b) Show that f is continuous at c if c is irrational. Hint: Let I be the interval {x: Ix - cl < 1). Show that for any Q > O, I
contains at most finitely many fractions p/q with q < Q. Conclude that there is a6 such that all fractions in {x: Ix- cl < 6)
have a denominator larger than Q.
SOLUTION
(a) Let c be any rational number and suppose that, in lowest terms, c = p/q, where p and q are integers. To prove the
discontinuity off at c, we must show there is an E > 0 such that for any 6 > 0 there is an x for which Ix - cl < 6, but
that lf(x) - f(c)I > E. Let E = -dq and 6 > 0. Since there is at least one irrational number between any two distinct real
numbers, there is some irrational x between c and c + 6. Hence, Ix - cl < 6, but lf(x) - f(c)I = 10 - 11 = l > 1._ = E
q
q
2q
•
Chapter Review Exercises
105
(b) Let c be irrational, let E > 0 be given, and let N > 0 be a prime integer sufficiently large so that~ < E. Let~, ... ,::
be all rational numbers 2q in lowest terms such that 12q - cl < 1 and q < N. Since N is finite, this is a finite list; hence, one
number !!1. in the list must be closest to c. Let 6 = -21 11!.i. - cl. By construction, 11!.i. - cl > 6 for all i = 1 ... m. Therefore, for
.
~
.
any rational number 2 such that 12 - cl< 6, q > N, so! < ~ < E.
Therefore, for an; rational n~mber x such that Ix '!_ cl < 6, lf(x) - f(c)I <
number x, so Ix - cl < 6 implies that lf(x) - f(c)I < E for any number x.
39.
E.
lf(x) - f(c)I
= 0 for any irrational
~ Write a formal definition of the following:
limf(x) =
oo
x-+a
SOLUTION
= oo if, for any M
limf(x)
x-+a
> 0, there exists an 6 > 0 such that f(x) > M whenever 0 < Ix - al < 6.
CHAPTER REVIEW EXERCISES
1. The position of a particle at time t (s) is s(t) = Vt2 + 1 m. Compute its average velocity over [2, 5] and estimate its
instantaneous velocity at t = 2.
SOLUTION
Let s(t)
= ..fi2+1. The average velocity over [2, 5] is
s(5~ =;<2) =
.../26; VS ,,, 0.954 mis
From the data in the table below, we estimate that the instantaneous velocity at t
[1.9, 2]
[l.99, 2]
[1.999, 2]
[2,2.001]
[2, 2.01]
[2, 2.1]
0.889769
0.893978
0.894382
0.894472
0.894873
0.898727
interval
average ROC
= 2 is approximately 0.894 mis.
3. For f(x) = -..ffX compute the slopes of the secant lines from 16 to each of 16 ± 0.01, 16 ± 0.001, 16 ± 0.0001 and
use those values to estimate the slope of the tangent line at x = 16.
SOLUTION
x interval
[15.99, 16]
[15.999, 16]
[15.9999, 16]
[16, 16.0001]
[16, 16.001]
[16, 16.01]
slope of secant
0.176804
0.176779
0.176777
0.176776
0.176774
0.176749
The slope of the tangent line at t
= 16 is approximately 0.1768.
In Exercises 5-10, estimate the limit numerically to two decimal places or state that the limit does not exist.
5
r
•
x~
1 - cos 3 (x)
x2
SOLUTION
Let f(x) = 1-c_:zs
3
x.
The data in the table below suggests that
. 1-cos3 x
hm
2
x--+0
<><
X
1.50
In constructing the table, we take advantage of the fact that f is an even function.
x
±0.001
±0.01
±0.1
f(x)
1.500000
1.499912
1.491275
(The exact value is~-)
r-4
7. lim-2- x-+2 X - 4
SOLUTION
Let f(x)
= $=:.The data in the table below suggests that
r-4
lim -i---x-+2 X - 4
<><
1.69
x
1.9
1.99
1.999
2.001
2.01
2.1
f(x)
1.575461
1.680633
1.691888
1.694408
1.705836
1.828386
106
C H A PT E R 2
I
LIMITS
(The exact value is 1 +In 2.)
7- --X
313)
9. lim ( -71
x-->1
X
-
Let f(x) =
SOLUTION
2x1 - 1!x3. The data in the table below suggests that
1
. ( -7- - -3- ) ,,,,2.00
hm
x-->I l -x7
l -x3
x
0.9
0.99
0.999
1.001
1.01
1.1
f(x)
2.347483
2.033498
2.003335
1.996668
1.966835
1.685059
(The exact value is 2.)
In Exercises I I-50, evaluate the limit if it exists. If not, determine whether the one-sided limits exist. For limits that don't
exist indicate whether they can be expressed as "= -oo" or "= oo ".
11. lim(3 + x 112 )
x-->4
lim(3 + x 112 ) = 3 +
SOLUTION
x-->4
.../4 = 5
4
13. lim 3
X--+-2 X
4
4
1
lim - = - - = --
SOLUTION
( -2)3
x-->-2 x3
2
. Yi-3
15. hm--1--.9 t - 9
Yt-3
Yt-3
1
1
1
lim - - = lim
= lim - - = - - - = H9 t - 9
H9 (Yi- 3)( Yi+ 3)
H9 Yi+ 3
-../§ + 3 6
SOLUTION
x3 -x
17. lim-x--+1 X- 1
x3 -
x
lim - - = lim
x-->1 X - 1
x-->I
t-6
19. lim~
H9 yt - 3
SOLUTION
As
SOLUTION
t-+
x(x - l)(x + 1)
X -
9, the numerator t
1
-
.
= hmx(x + 1) = 1(1+1) = 2
x-->I
6 -+ 3
* 0 while the denominator Yi - 3 -+ 0. Accordingly,
t- 6
lim - H9 Yt-3
does not exist.
Similarly, the one-sided limits as t -+ 9- and as t -+ 9+ also do not exist. Let's take a closer look at the limit as
t -+ 9-. The numerator approaches a positive number while the denominator Yi - 3 -+ o-. We may therefore express
this one-sided limit as
t-6
lim - - =-oo
Yt-3
H9-
On the other hand, as t -+ 9+, the numerator approaches a positive number while the denominator
can express this one-sided limit as
Yi - 3 -+ o+, so we
t-6
l i m - - =oo
Yt-3
H9+
Because one of the one-sided limits approaches -oo and the other approaches oo, the two-sided limit can be expressed
neither as"= -oo" nor as"= oo".
.
1
21. hm - x--.-1+ X + 1
SOLUTION
As x -+ -1 +, the numerator remains constant at 1
.
1
hm - x--+-I + X
+}
* 0 while the denominator x + 1 -+ 0. Accordingly,
does not exist.
Taking a closer look at the denominator, we see that x + 1 -+ o+ as x -+ -1 +. Because the numerator is also approaching
a positive number, we may express this limit as
Jim - - = oo
x+I
x--.-1•
Chapter Review Exercises
x3-2x
23. lim--x-+I X - 1
SOLUTION
As x---+ 1, the numerator x3 - 2x---+ -1
107
* 0 while the denominator x - 1 ---+ 0. Accordingly,
x3-2x
lim--x.-.1 x-1
does not exist.
Similarly, the one-sided limits as x ---+ 1- and as x ---+ 1+ also do not exist. Let's take a closer look at the limit as
x ---+ i - . The numerator approaches a negative number while the denominator x - 1 ---+ o-. We may therefore express this
one-sided limit as
. x3 - 2x
11m
x.-.J-
---=oo
x- l
On the other hand, as x ---+ 1+, the numerator approaches a negative number while the denominator x - 1 ---+ o+, so we
can express this one-sided limit as
x3 -2x
lim - - x-1
x.-.1+
Because one of the one-sided limits approaches
neither as "= -oo" nor as "= oo".
=-oo
-oo and the other approaches oo, the two-sided limit can be expressed
e3x - ex
25. lim--x-+O ex - 1
SOLUTION
27. x.-.1.5
lim
l!J
X
SOLUTION
29. lim
z->-3
lim
x-.1.5
l!J =l-J=l~J =
X
1
1.5
3
0
z+3
z2 + 4z + 3
SOLUTION
Jim
z->-3
z + 3 = lim
z+ 3
= lim _l_
z2 + 4z + 3 z->-3 (z + 3)(z + 1) z->-3 z + 1
=_!
2
x3 -b3
31. lim--bx-+b
X -
SOLUTION
33. lim
x->O
. x3-b 3
(x-b)(x2+xb+b 2 )
.
hm - - = hm
= lim(x2 + xb + b1 ) = b1 + b(b) + b 2 = 3b1
x-.b X - b
x->b
X - b
x->b
(J_ --
1
-)
x(x + 3)
3x
SOLUTION
lim
x-.O
35. lim
x-+O-
(J_ ---1+- ) = x-.o
lim
3x
x(x
3)
3
(x + ) - 3 = lim __l _ = __l _ 3x(x + 3)
x-.o 3(x + 3)
!
3(0 + 3) - 9
LxJ
X
SOLUTION For x sufficiently close to zero but negative, LxJ = -1. Therefore, as x ---+ o-, the numerator LxJ ---+ -1 if= O
while the denominator x ---+ 0. Accordingly,
. LxJ
11m-
x-+o-
x
Taking a closer look at the denominator, we see that x ---+
negative number, we may express this limit as
lim
x-+O-
37. lim esec e
8-+~
does not exist.
o- as x ---+ o-. Because the numerator is also approaching a
LxJ = oo
X
108
CHAPTER 2
I
LIMITS
SOLUTION
First note that
(}
(}sec (} = - cos 0
As (} ->
~,
the numerator (} ->
~
* 0 while the denominator cos (} -> 0. Accordingly,
. (}sec (} = l"1m -(}11m
O-> ~
does not exist.
0-->~ COS(}
Similarly, the one-sided limits as (} -> ~- and as (} -> ~+ also do not exist. Let's take a closer look at the limit as
(} -> ~-.The numerator approaches a positive number while the denominator cos(} -> o+. We may therefore express this
one-sided limit as
On the other hand, as (} -> ~ +, the numerator approaches a positive number while the denominator cos (} ->
can express this one-sided limit as
. OsecO
1im
B-41+
=
l"im -(}-
8-4~+ COS 8
o-, so we
= -oo
Because one of the one-sided limits approaches -oo and the other approaches oo, the two-sided limit can be expressed
neither as "= -oo" nor as "= oo".
. cos(} - 2
39• 1
o~
(}
SOLUTION
As(}-> 0, the numerator cos(} - 2 -> -1
* 0 while the denominator(} -> O. Accordingly,
. cos(}- 2
1i m - - - o_,o
does not exist.
(}
Similarly, the one-sided limits as (} -> o- and as(}-> o+ also do not exist. Let's take a closer look at the limit as(} -> o-.
The numerator approaches a negative number while the denominator (} -> o-. We may therefore express this one-sided
limit as
. cos(} - 2
1tm
0-->0-
(}
= 00
On the other hand, as (} -> o+, the numerator approaches a negative number while the denominator e ->
express this one-sided limit as
. cose- 2
1tm
0-->0+
(}
o+, so we can
= -oo
Because one of the one-sided limits approaches -oo and the other approaches oo, the two-sided limit can be expressed
neither as"= -oo" nor as"= oo".
41. lim x- 3
x->2- X- 2
SOLUTION
As x -> 2-, the numerator x - 3 -> -1
* 0 while the denominator x - 2 -> O. Accordingly,
. x-3
1I m - 2
does not exist.
x->2+ X-
Taking a closer look at the denominator, we see that x - 2 ->
a negative number, we may express this limit as
o- as x -> 2-. Because the numerator is also approaching
x-3
lim - - = oo
2
x->2- X -
1
43. lim ( -x->J+
~
SOLUTION
1
- - -)
VX2 - 1
First note that
1
vx-1- vx2 -1
1
= vx-l
0
.y;:+f
l
vx+l - vx2 -l
=
.y;:+f - 1
vxz-1
Chapter Review Exercises
As x ~ 1+, the numerator
-.fX+1 -
1 ~ ..(2, - 1 ::f:. 0 while the denominator
1
lim ( - - ~
x--+i+
1
-)
Vx2 -
1~
1 ~ 0. Accordingly,
does not exist.
vxz-1
Taking a closer look at the denominator, we see that
a positive number, we may express this limit as
Vx 2 -
109
o+ as x ~ 1+. Because the numerator is also approaching
1 - - -1 - ) = oo
lim ( - ~
x--+1+
vxz-1
45. lim tanx
x-+1
SOLUTION
First note that
sinx
tanx = - cosx
As x
~ ~,
the numerator sin x
~
1 ::f:. 0 while the denominator cos x
.
. sinx
1im tanx = 1im - -
x-+1
x-+!
~
0. Accordingly,
does not exist.
COS X
Similarly, the one-sided limits as x ~ ~- and as x ~ f also do not exist. Let's take a closer look at the limit as
x ~ ~ - . The numerator approaches a positive number while the denominator cos x ~ o+. We may therefore express this
one-sided limit as
.
. sinx
11m tanx = 11m - - =
x-+~-
oo
x-i-l- COS X
On the other hand, as x ~ ~ +, the numerator approaches a positive number while the denominator cos x ~
can express this one-sided limit as
.
1im tanx
x-+1+
=
Ii
sinx
m --
x-+~+ COS X
o-, so we
= -oo
Because one of the one-sided limits approaches -oo and the other approaches oo, the two-sided limit can be expressed
neither as "= -oo" nor as "= oo".
SOLUTION
Fort > 0,
so
Because
lim-Vt= lim
t-+O+
t-+O+
Vt=O
it follows from the Squeeze Theorem that
lim
1--.0+
Vt cos(~)=
0
t
. cosx- 1
49 • 1lffi
.
x->O
smx
SOLUTION
Ii
x-To
cosx-1 r cosx-1
sin x = x~ sin x
cosx+ 1
.
-sin2 x
.
sinx
O
---= 1im
= - hm
= --=0
cos x + 1 x--.o sin x( cos x + 1)
x--+O cos x + 1
1+ 1
110
c HAPTE R
2
I
LIMITS
51. Find the left- and right-hand limits of the function
continuous (or both) at these points.
f in Figure 1 at x = 0, 2, 4. State whether f is left- or right-
y
FIGURE 1
SOLUTION
According to the graph of f(x),
lim f(x) = lim f(x) = l
x--+O-
x--+O+
lim f(x) = lim f(x) = oo
x-.2-
x--+2+
lim f(x) = -oo
x--+4-
lim f(x) = oo
x--+4+
The function is both left- and right-continuous at x = 0 and neither left- nor right-continuous at x = 2 and x = 4.
53. Graph h and describe the discontinuity:
h(x) =
ex
for x ~ 0
{ lnx forx>O
Is h left- or right-continuous?
SOLUTION
The graph of h(x) is shown below. At x = 0, the function has an infinite discontinuity but is left-continuous.
y
55. Find the points of discontinuity of
g(x) =
cos (nx)
2
{
Ix - 11
for lxl < 1
for lxl :::: 1
Determine the type of discontinuity and whether g is left- or right-continuous.
First note that cos ( ¥) is continuous for -1 < x < 1 and that Ix - 11 is continuous for x ~ -1 and for x :::: 1.
Thus, the only points at which g(x) might be discontinuous are x = ±1. At x = 1, we have
SOLUTION
lim g(x) = lim cos(nx) =cos(:!.)= 0
x--.12
2
x--.1-
and
lim g(x) = lim lx-11=11-11=0
X--+I+
x--+l+
so g(x) is continuous at x = 1. On the other hand, at x = -1,
lim g(x) = Jim cos (nx)
X->-1+
2
X->-[+
=cos(-~)=
O
2
and
lim g(x) = lim Ix- 11 =I - 1 - 11=2
X--+-1-
X--+-}-
so g(x) has a jump discontinuity at x = -1. Since g(-1) = 2, g(x) is left-continuous at x = -1.
Chapter Review Exercises
111
57. Find a constant b such that his continuous at x = 2, where
x+1
h(x)
= { b- .x-~'
for lxl < 2
for lxl ~ 2
With this choice of b, find all points of discontinuity.
To make h(x) continuous at x = 2, we must have the two one-sided limits as x approaches 2 be equal. With
SOLUTION
lim h(x) = lim (x + 1) = 2 + 1 = 3
X-+2-
X-+2-
and
lim h(x) = lim(b-.x2) = b-4
x-+2+
x-+2+
it follows that we must choose b = 7. Because x + 1 is continuous for -2 < x < 2 and 7 - x2 is continuous for x :s; -2
and for x ~ 2, the only possible point of discontinuity is x = -2. At x = -2,
limh(x)= lim(x+l)=-2+1=-1
X-+-2+
X-+-2+
and
lim h(x) = lim (7 -.x2) = 7 - (-2)2 = 3
X-+-2-
X-+-2-
so h(x) has a jump discontinuity at x = -2.
In Exercises 58-65,find the horizontal asymptotes of the function by computing the limits at infinity.
x2- 3x4
59. f(x) = - -x- 1
SOLUTION Because
.x2 - 3x4
x - 3.x3
lim - - - = lim - - - =
X-+00
X - 1
X->OO 1 - 1/x
-oo
and
x2 - 3x4
x - 3.x3
lim - - - = lim - - - =
X - 1
x-+-oo 1 - 1/x
oo
x-+-oo
it follows that the graph of y
61. f(u)
=
SOLUTION
x2 - 3x4
= -x-1
- - does not have any horizontal asymptotes.
2u 2 -1
-~
v6 + u4
Because
lim 2u2 - 1
= lim
V6 + u4
U->OO
U->OO
2- l/u2 =
2_ = 2
--/6Ju4 + 1
VI
and
.
.
2u2 -1
lI m - - - = 1Im
u+oo
V6 + u4
2-1/u2
--/6Ju4 + 1
u+oo
2
=-=2
VI
2u 2 -1
it follows that the graph of y = - - - has a horizontal asymptote of y = 2.
v6+u4
t1;3 _ rl/3
63. f(t)
= (t -
SOLUTION
rI )1/3
Because
1 _ r213
tlf3 _ rl/3
lim
t tooo
(t - r l )1/3
=lim
Hoo
(1 - r2)1/3
1
=-=1
11/3
and
tlf3 _ rl/3
lim
t to-oo
.
It follows that the graph of y =
(t- r l )1/3
= lim
H-oo
1_
r213
(1 - r2)1/3
1
= ]173 =
t1;3 _ r1t3
(t - rl)l/3
has a horizontal asymptote of y = 1
.
1
112
C HA PT E R 2
I
LIMITS
65. g(x) =
7r
+ 2 tan- 1 x
Because
SOLUTION
lim (7r + 2 tan- 1 x) =
7r
x---+-oo
it follows that the graph of y =
7r
+ 2 (-~) = 0 and
lim(7r + 2 tan- 1 x) =
2
7r
x---+oo
+ 2 (~) =
2
27r
+ 2 tan- 1 x has horizontal asymptotes of y = 0 and y = 27r.
67. CR S Determine M, A, and k for a logistic function p(t) = i+.7.-ki satisfying p(O) = 10, p(4) = 35, and p(lO) = 60.
What are the horizontal asymptotes of p?
To fit a logistic function p(t) = i+:.-k, to the data p(O) = 10, p(4) = 35, and p(lO) = 60, we must solve the
system of equations
SOLUTION
10 =
__!!__
1 +A'
35
M
= ----::1 + Ae-4k'
and
M
60= ---...,.,,..,.
1 + Ae-IOk
for M, A, and k. Using a computer algebra system to solve this system yields M,,, 62.777, A ,,, 5.278, and k,,, 0.474.
Thus,
62.777
p(t) = 1 + 5.278e-0 .474'
Because lim e- 0.474' =
oo
and lim e- 0.474' = 0, it follows that
t---+-oo
t---+oo
lim
1->-00
62.777
1 + 5.278e-0.474t
=0
and
.
1Im
1->oo
62.777
62.777
= - - = 62.777
1 + 5.278e-0·4741
1+ 0
The horizontal asymptotes of p are y = 0 and y = 62. 777.
69. Assume that the following limits exist:
A= lim/(x),
B = limg(x),
x->a
X->a
L = lim f(x)
x->a
g(x)
Prove that if L = 1, then A= B. Hint: You cannot use the Quotient Law if B = 0, so apply the Product Law to Land B
instead.
SOLUTION
Suppose the limits A, B, and Lall exist and L = 1. Then
B = B · 1 = B · L = limg(x) · lim f(x) = limg(x/(x) = lim/(x) =A
x->a
71.
lSJ
SOLUTION
x->a
g(x)
x->a
g(x)
x->a
In the notation of Exercise 69, give an example where L exists but neither A nor B exists.
Suppose
1
f(x)=--
(x - a) 3
I
g(x)=--
and
(x - a) 5
Then, neither A nor B exists, but
.
L = hm
x---+a
(x - a)- 3
(x - a)-5
.
= hm(x - a) 2 = O
x--+a
Chapter Review Exercises
73.
L$J
Let f(x) =
113
lH where LxJ is the greatest integer function. Show that for x * 0,
Then use the Squeeze Theorem to prove that
limxl!J=l
x-+0
X
Hint: Treat the one-sided limits separately.
SOLUTION Lety be any real number. From the definition of the greatest integer function, it follows thaty- 1 < LYJ ~ y,
with equality holding if and only if y is an integer. If x 0, then ~ is a real number, so
*
Upon multiplying this inequality through by x, we find
l-x<xl~J~1
Because
lim(l - x) = lim 1 = 1
x....,.Q
X-40
it follows from the Squeeze Theorem that
limxl!J=l
x-+0
X
75. Use the IVT to prove that the curves y =
SOLUTION
Let f(x) =
x2 -
x2 and y = cos x intersect.
cos x. Note that any root of f(x) corresponds to a point of intersection between the curves
y = x2 and y = cosx. Now, f(x) is continuous over the interval [0, ~], f(O) = -1 < 0 and f(D = ~ > 0. Therefore,
by the Intermediate Value Theorem, there exists a c E (0, ~) such that f(c) = O; consequently, the curves y = x2 and
y = cos x intersect.
77. Use the IVT to show that e-x2 = x has a solution on (0, 1).
SOLUTION
Let f(x) = e-x2 - x. Observe that f is continuous on [0, 1] with f(O) = e 0 - 0 = 1 > 0 and f(l) = e- 1 - 1 <
0. Therefore, the IVT guarantees there exists a c
E
(0, 1) such that f(c) = e-c2 - c = 0.
79. ~ Give an example of a (discontinuous) function that does not satisfy the conclusion of the IVT on [-1, 1].
Then show that the function
f(x)
=
.
1
1
sm0 x
satisfies the conclusion of the IVT on every interval [-a, a].
SOLUTION
c
Let g(x) = LxJ. This function is discontinuous on [-1, 1] with g(-1) = -1 and g(l) = 1. For all c E (-1, 1),
* 0, there is no x such that g(x) = c; thus, g(x) does not satisfy the conclusion of the Intermediate Value Theorem on
[-1, l].
Now, let
1
f(x)
= { 0sin ( x )
for x
for x
*0
=0
and let a > 0. On the interval
xE
[--a-,'.:]
c [-a,a]
2 + 2rra 2
~ runs from ~ to ~ + 27f, so the sine function covers one full period and clearly takes on every value from - sin a through
sin a.
81.
~Guij
0.05.
Plot the function f(x) = x 113 • Use the zoom feature to find a
o > 0 such that if Ix -
81
<
o, then lx113 -
21
<
114
CHAPTER 2
I
LIMITS
The graphs of y = f (x) = x 1' 3 and the horizontal lines y = 1.95 and y = 2.05 are shown below. From this
plot, we see that 6 = 0.55 guarantees that whenever Ix - 81 < 6, then lx 1' 3 - 21 < 0.05.
SOLUTION
y
2.05
:: Jll i il[j lliJ 111
83. Prove rigorously that lim (4 + 8x)
x~-1
SOLUTION
= -4.
Let E > 0 and take 6 = c/8. Then, whenever Ix - (-1)1 = Ix+ 11 < 6,
lf(x)-(-4)1=14+8x+41=81x+ 11<86=
E
3 DIFFERENTIATION
3.1 Definition of the Derivative
Preliminary Questions
1. Which of the lines in Figure 12 are tangent to the curve?
FIGURE 12
SOLUTION
Lines B and D are tangent to the curve.
2. What are the two ways of writing the difference quotient?
SOLUTION
The difference quotient may be written either as
f(x)- f(a)
x-a
or as
f(a + h) - f(a)
h
.
f(a + h) - f(a) .
.
3. Fmd a and h such that
h
is equal to the slope of the secant !me between (3, /(3)) and (5, /(5)).
SOLUTION
With a
= 3 and h = 2,
f(a +
k-
f(a) is equal to the slope of the secant line between the points (3, /(3))
and (5, /(5)) on the graph of f(x).
4. Which derivative is approximated by
SOLUTION
tan(!!+
0.0001) - I
4
O.OOOI
?
tan(!!+
4 0.0001) - I
O.OOOI
is a good approximation to the derivative of the function f(x)
= tanx at x = ~·
5. wliat do the following quantities represent in terms of the graph of f(x) = sin x?
(a) sin 1.3 - sin 0.9
SOLUTION
sin 1.3 - sin 0.9
0.4
(b)----
(c) f'(0.9)
Consider the graph ofy = sinx.
(a) The quantity sin 1.3 - sin 0.9 represents the difference in height between the points (0.9, sin 0.9) and (1.3, sin 1.3).
. sin 1.3 - sin 0.9
(b) The quantity
0.4
represents the slope of the secant line between the points (0.9, sin 0.9) and (1.3, sin 1.3)
on the graph.
(c) The quantity f'(0.9) represents the slope of the tangent line to the graph at x
= 0.9.
6. Choose (a) or (b). The derivative at a point is zero ifthe tangent line at that point is (a) horizontal (b) vertical.
SOLUTION
The derivative at a point is zero if the tangent line at that point is (a) horizontal.
7. Choose (a) or (b). The derivative at a point does not exist ifthe tangent line at that point is (a) horizontal (b) vertical.
SOLUTION
The derivative at a point does not exist if the tangent line at that point is (b) vertical.
111';
116
CHAPTER 3
I
DIFFERENTIATION
Exercises
1. Let f(x)
= 5x2. Show that f(3 + h) = 5h 2 + 30h + 4S. Then show that
f(3 + h) - f(3)
h
and compute f' (3) by taking the limit as h
SOLUTION
With f(x)
= Sh+ 30
0.
-7
= 5x2' it follows that
f(3 + h)
= S(3 + h)2 = S(9 + 6h + h 2) = 4S + 30h + Sh2
Using this result, we find
h
4S + 30h + 5h2 - 5 . 9
h
30, so /'(3)
= 30.
f(3 + h) - f(3)
Ash
-7
0, 30 +Sh
-7
4S + 30h + Sh 2 - 4S
h
30h + 5h2
h
= 30 + 5h
In Exercises 3--8, compute f'(a) in two ways, using Eq. (I) and Eq. (2).
3. f(x)
= x2 + 9x,
SOLUTION
a=0
Let f(x)
f'(O)
= x2 + 9x. Then
= lim f(O + h) -
f(O)
= lim (0 + h)2 + 9(0 + h) -
h
h->O
0
= lim 9h + h2 = lim(9 + h) = 9
h
h->O
h->0
h
h->0
Alternately,
f'(O) = lim f(x)- f(O) = lim x2 + 9 x - O = lim(x + 9) = 9
x->0
X - 0
x->O
X
x->0
5. f(x)
= 3x2 + 4x + 2,
SOLUTION
a
= -1
Let f(x) = 3.x2 + 4x + 2. Then
f' (_1) = lim f (-1 + h) - f (-1) = lim 3(-1 + h)2 + 4(-1
h
h->0
.
= !rm
h-+0
3h2 -2h
h
= h-+0
Iim(3h -
+ h) + 2 - 1
h
h->0
2) = -2
Alternately,
f'(-l)= lim f(x)-f(-1) = lim 3x2+4x+2-I
x->-1
x-(-1)
X->-1
x+ I
. (3x+ I)(x+ I)
.
= hm
= hm(3x+ I)= -2
x->-1
X + 1
x->-1
7. f(x) = x3 + 2x,
SOLUTION
Let f(x)
a= 1
= x3 + 2x. Then
f'(l)
3
= lim f(l
+ h) - f(l) = lim (I + h) + 2(1 + h) - 3
h->O
h
h->0
h
2
3
. I + 3h + 3h + h + 2 + 2h - 3
= Iim
h
= lim(S + 3h + h 2 ) = S
h->0
h->O
Alternately,
f'(I)
= lim f(x) x->I
.
= hm
X-+l
X -
f(I) = lim
I
x->1
(x - I)(x2
X-
+ x + 3)
1
x3 + 2x - 3
X -
I
= lim(x2 + x + 3) = 5
X-+1
sEcT IoN
3.1
I
Definition of the Derivative
117
In Exercises 9-I 2, refer to Figure I 3.
2.5 +···"·······+-············-;···············i··
2.0k········•···············,.········•·····•··
1.5.j....... +~~=1"
1.0+············;·-,,-... ,. ............. , ..
0.5 ..._.r....... ; ...............;................. .
0.5
1.0
1.5
2.0
2.5
3.0
FIGURE 13
9. ~
Find the slope of the secant line through (2,f(2)) and (2.5,f(2.5)). Is it largeror smaller than f'(2)? Explain.
From the graph, it appears that f(2.5) = 2.5 and f (2) = 2. Thus, the slope of the secant line through
(2,f(2)) and (2.5,f(2.5)) is
SOLUTION
f(2.5) - f(2)
2.5 -2
= 2.5 -
=I
2
2.5-2
From the graph, it is also clear that the secant line through (2, f(2)) and (2.5, f(2.5)) has a larger slope than the tangent
line at x = 2. In other words, the slope of the secant line through (2, f(2)) and (2.5, f (2.5)) is larger than f' (2).
11. Estimate f' (I) and f' (2).
SOLUTION From the graph, it appears that the tangent line at x = I would be horizontal. Thus, f'(I)"" 0. The tangent
line at x = 2 appears to pass through the points (0.5, 0.8) and (2, 2). Thus
/'(2)"" 2 - 0.8
2-0.5
= 0.8
In Exercises 13-16, refer to Figure I4.
5
4
3
2
I
2
3
4
5
6
7
8
9
FIGURE 14 Graph off.
13. Determine f'(a) for a= I, 2, 4, 7.
Remember that the value of the derivative off at x = a can be interpreted as the slope of the line tangent
to the graph of y = f(x) at x = a. From Figure 14, we see that the graph of y = f(x) is a horizontal line (that is, a line
with zero slope) on the interval 0 ~ x ~ 3. Accordingly, f'(l) = f'(2) = 0. On the interval 3 ~ x ~ 5, the graph of
y = f(x) is a line of slope }; thus, f'(4) = }. Finally, the line tangent to the graph of y = f(x) at x = 7 is horizontal,
so f'(7) = 0.
SOLUTION
15. Which is larger, f'(5.5) or f'(6.5)?
SOLUTION The line tangent to the graph of y = f(x) at x = 5.5 has a larger slope than the line tangent to the graph of
y = f(x) at x = 6.5. Therefore, f'(5.5) is larger than f'(6.5).
In Exercises 17-20, use the limit definition to calculate the derivative of the linear function.
17. f(x)
= 1x-9
SOLUTION
. f(a + h) - f(a)
. 7(a + h) - 9 - (1a - 9)
.
I1m
= 11m
=hm7=7
o
.h
h ..... o
h
h ..... o
h .....
19. g(t) = 8 - 3t
SOLUTION
. g(a + h) - g(a)
. 8 - 3(a. + h) - (8 - 3a)
. -3h
.
I1m
= 11m
=hm-=bm(-3)=-3
o
h
h ..... o
h
h .....o h
h ..... o
h .....
118
CH APT E R 3
I
DIFFERENTIATION
21. Find an equation of the tangent line at x == 3, assuming that f(3) = 5 and f'(3) == 2.
SOLUTION By definition, the equation of the tangent line to the graph of f(x) at x == 3 is y == f(3) + f'(3)(x - 3) =
5 + 2(x - 3) == 2x - I.
23. Describe the tangent line at an arbitrary point on the graph of y == 2x + 8.
SOLUTION
Since y == 2x + 8 represents a straight line, the tangent line at any point is the line itself, y == 2x + 8.
I
I
1
1
.
.
.
25. Let f(x) = -. Does f(-2 + h) equal - - or - + -h? Compute the difference quotient at a == -2 with h = 0.5.
x
-2+h
-2
SOLUTION
Letf(x) ==~-Then
1
f(-2 + h) == - -2 + h
With a== -2 and h == 0.5, the difference quotient is
f(-1.5)- f(-2)
0.5
f(a + h) - f(a)
h
27. Let f(x) == 1/
I
I
=I5 - =z
0.5
3
yx. Compute f' (5) by showing that
f(5 + h)- f(5)
h
SOLUTION
Let f(x) == 1 /
yx. Then
I
I
~--;rs
f(5 + h) - f(5)
h
h
VS+h)
Ys - V5+h ( Ys +
h Ys -../5 + h Ys + -../5 + h
5 - (5 + h)
h Ys -../5
+ h( -./5 + h +
Ys)
Ys -./5 + h(-../5 + h + Ys)
Thus,
f'(5) ==Jim f(5 + h) h->0
1
f(5) ==Jim-
h
h->0
1
Ys -../5 + h( -../5 + h + Ys)
1
Ys Ys(Ys + Ys)
== -
lOYs
In Exercises 29-46, use the limit definition to compute f'(a) and.find an equation of the tangent line.
29. f(x) == 2x2 + lOx,
SOLUTION
a=3
Let f(x) == 3x2 + 2x. Then
f'(2 ) =Jim f(2 + h) ~
h
f(2) =Jim 3(2 + h)2 + 2(2 + h) - 16
~
h
. 12 + l2h + 3h2 + 4 + 2h - 16
= hm
h->0
h
= h->O
Iim(14 + 3h) = 14
At a = 2, the tangent line is
y = f'(2)(x - 2) + f(2) = I4(x - 2) + 16 = l4x - 12.
31. f(t) = t - 2t2,
SOLUTION
a= 3
Let f(t) = t - 2t2. Then
f'( 3 ) =Jim f(3 + h) - f(3) =Jim (3 + h)- 2(3 + h)
h->O
h
h->O
h
2
-
(-15)
. 3 + h - 18 - 12h - 2h 2 + 15
= I1 m - - - - - - - - h->o
h
= h->O
lim(-11 -
2h)
= -11
At a = 3, the tangent line is
y = f'(3)(t -3) + f(3) = -Il(t- 3)- 15 = -Ilt+ 18
I
S E c T I 0 N 3.1
33. f(x)
= x3 + x,
SOLUTION
a
Definition of the Derivative
=0
Let f(x) =
x3 + x. Then
f'(O) = lim f(h) - f(O) = lim h3 + h - 0
h->0
h
h->0
h
= lim(h2 + 1) = I
h->0
At a = 0, the tangent line is
y = f'(O)(x - 0) + /(0) = x
35. f(x) = x- 1 ,
SOLUTION
a
=8
Let f(x) = x- 1• Then
1 (I)
/'(8) = lim /( 8 + h) - /( 8) = lim 8+h - 8 = lim
h->O
h
h->O
h
h->0
The tangent at a
8-8-h
8<8+h)
h
= lim
h->O
-h
= _ _!_
(64 + 8h)h
64
= 8 is
,
I
I
I
I
y = f (8)(x- 8) + /(8) = --(x- 8) + - = - - x + 64
8
64
4
I
37. f(x) = - - , a= -2
x+ 3
SOLUTION
Let f(x) =
x!3 . Then
'( 2) 1· f(-2 +h)-/(-2) =Im
1·
! - =Im
h->O
h
h->O
=z:!w -1
h
t!h
-h
1·Im--=-1
I
=1·I m --1
- = 1·Im---=
h->O
h
h->O h(I + h)
h->O 1 + h
The tangent line at a = - 2 is
y = f'(-2)(x + 2) + /(-2) = -l(x + 2) + 1 = -x- 1
39. f(x) = .Yx+4, a= 1
SOLUTION
Let f(x) = vx + 4. Then
f
'ci) _ . 1c1 + h) - 1c1) _ .
- 1Im
- 11m
h->O
h
h->0
.
=1m
1
h->0
The tangent line at a
h
h( vh + s +
-Yh+5 h
SOLUTION
I
vx'
a
VS
1
h->O
= 1 is
' )
=
Vh + 5 +
h
vs) =hm vh + s + vs -- 2 vs
y= ! (1 (x-1)+/(1)=
41. f(x)
h->0
1
.
vs -_Iim vh + s - vs ·vh
+ s + vs
----
1
_,-;
1 - x +9.k(x-1)+
v5=
2 vs
2-VS
2-VS
=4
Let f(x) =
1
-vx·
Then
I
I
2- -./4+h
2+ -./4+h
/'(4) = lim /(4 + h) - /(4) = lim 74.:'t. - 2 = lim ~ . ~
h->0
h
h->O
h
h->O
h
-1
1
4V4 + h +2(4 + h)
16
= Iim - - - - - h->O
4-4-h
= lim 4 -.f4+h+2<4+hl
h->O
At a = 4 the tangent line is
1
1
I
3
y = f'(4)(x - 4) + /(4) = --(x- 4) + - = - - x + 16
2
16
4
h
119
120
CHAPTER 3
I
DIFFERENTIATION
43.
f (t) = ViZ+T,
SOLUTION
=3
a
Let f(t) =
..fi2+T. Then
j'(3) = lim f(3 + h) - f(3) = lim VIO + 6h + h
h->O
h
h->0
h
.
= hm
VIO + 6h + h2
h->O
v'IO
-
h
The tangent line at a
v'IO)
1
= x2 + 1 ,
SOLUTION
VfO
=lim
6+h
h....0 vIO + 6h + h 2 +
3
vw = vw
= 3 is
3
y=f'(3)(t-3)+f(3)= _r;-n<t-3)+
vIO
45. f(x)
-
-./IO+ 6h + h 2 + v'IO
· -;:===::::;---;::=
-./IO+ 6h + h 2 + VfO
2
=lim
6h+h
h....0 h( vIO + 6h + h 2 +
2
a
Let f (x)
m=
3
1
_r;-nt+ -""
vIO
~IO
=0
1
= x2 + l . Then
- ' - -1
-h2
j'(O) = lim f(O + h) - f(O) = lim (O+h)2+1
= lim h2+1 = lim ~ = 0
h->O
h
h->0
h
h->0 h
h->oh2 +1
The tangent line at a
= 0 is
y = f(O) + f'(O)(x - 0)
47. Show that f is not differentiable and x
= 1 + O(x -
1)
=1
= 1 and has a corner in its graph there.
x~l
x~l
SOLUTION
Note
lim f(l + h) - f(l)
h->Oh
= lim
h->O-
1- 1
h
= 0,
but
Iim
h->O+
f(l + h) - f(l)
h
. (1 + h)
11m
= h->O+
h
2
-
1
.
11m
= h->O+
1 + 2h + h 2 - 1
h
. (Z h)
11m + = 2
= h->O+
Because the two one-sided limits are not equal, the two-sided limit
r
J(l +h)-f(l)
h'!R,
h
does not exist.
Therefore, f is not differentiable at x = 1. Also, because the two one-sided limits exist but are not equal, the graph off
has a corner at x = 1 (see the figure below).
y
4
-+-+-+-+--+-+-++-X
-3
3
S E CT I 0 N 3.1
I
Definition of the Derivative
121
In Exercises 49-5I, sketch a graph off and identify the points c such that f'(c) does not exist. In which cases is there a
corneratc?
49. f(x)
= Ix+ 31
The graph of y
not differentiable at x = -3.
SOLUTION
= Ix+ 31 is shown below. From the graph, we see that there is a corner at x = -3, so f
is
y
-6
-3
51. f(x) = lx2 - 41
The graph of y = lx2 - 41 is shown below. From the graph, we see that there is a corner at x
another at x = 2, so f is not differentiable at either x = -2 or x = 2.
SOLUTION
= -2 and
y
-4
53. !Gq Let f(x)
the slope f'(O).
=
1
:
2
x.
4
Plot f over [-2, 2]. Then zoom in near x
= 0 until the graph appears straight, and estimate
:zx
The figure at the left shows the graph of f(x) = 1
over [-2, 2]. The figure below at the right is a close-up
near x = 0. From the close-up, we see that the graph is nearly straight and passes through the points (-0.22, 2.15) and
(0.22, 1.85). We therefore estimate
SOLUTION
/'(O)"'"
1.85 - 2.15
0.22 - (-0.22)
= -0.3 = _0 _68
y
0.44
y
0.5
-2
-I
2
55. Determine the intervals along the x-axis on which the derivative in Figure 16 is positive.
y
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
0.5
1.0
1.5
2.0 2.5
3.0 3.5 4.0
FIGURE 16
SOLUTION The derivative (that is, the slope of the tangent line) is positive when the height of the graph increases as
we move to the right. From Figure 16, this appears to be true for 1 < x < 2.5 and for x > 3.5.
122
CHAPTER 3
I
DIFFERENTIATION
Jn Exercises 57-62, each limit represents a derivative f'(a). Find f(x) and a.
(5 + h) 3
h
57. lim
h--+0
-
125
.
.
(5 + h)3 - 125
f(a + h) - f(a)
The difference quot:J.ent
h
has the form
h
where f(x)
SOLUTION
sin(~+
59. lim
= x3 and a = 5.
h)-0.5
h
h--+0
The difference quotient
SOLUTION
sin(~+ h) -
h
0.5
has the form
f(a + h) - f(a)
h
where f(x)
= sin x and a = ~.
-25
h
52+h
61. lim
h--+0
SOLUTION
The difference quotient
5(2+h) - 25
f(a + h) - f(a)
h
has the form
h
where f(x)
=Y
and a
= 2.
63. Apply the method of Example 6 to f(x) = sin x to determine f' ( ~) accurately to four decimal places.
SOLUTION
We know that
f'(7r/ 4)
= lim f(7r/4 + h) -
Creating a table of values with h close to zero:
h
sin(~ + h) - ( V2/2)
h
= lim sin(7r/4 + h) -
f(7r/4)
h
-
V2/2
h
-0.001
-0.0001
-0.00001
0.00001
0.0001
0.001
0.7074602
0.7071421
0.7071103
0.7071033
0.7070714
0.7067531
Accurate up to four decimal places, f' ( ~) "" 0. 7071 .
. 65. ~
between x
For each graph in Figure 17, determine whether f' (1) is larger or smaller than the slope of the secant line
= I and x = I + h for h > 0. Explain.
y
(B)
(A)
FIGURE 17
SOLUTION
• On curve (A),f'(I) is larger than
f(l + h) - f(I)
h
the curve is bending downwards, so that the secant line to the right is at a lower angle than the tangent line. We say
such a curve is concave down, and that its derivative is decreasing.
• On curve (B), f'(I) is smaller than
f(I + h) - f(I)
h
the curve is bending upwards, so that the secant line to the right is at a steeper angle than the tangent line. We say
such a curve is concave up, and that its derivative is increasing.
67. jGuij Sketch the graph of f(x) = x51 2 on (0, 6].
(a) Use the sketch to justify the inequalities for h > 0:
/(4) - f(4 - h) < f'(4) < f(4 + h) - /(4)
h
-
-
h
s Ec T Io N
I
3.1
Definition of the Derivative
123
(b) Use (a) to compute f'(4) to four decimal places.
(c) Use a graphing utility to plot y
SOLUTION
= f(x) and the tangent line at x = 4, utilizing your estimate for f'(4).
A sketch of the graph off(x) = x512 on [O, 6] is shown below in the answer to part (c).
(a) The slope of the secant line between points (4,f(4)) and (4 + h,f(4 + h)) is
f(4 + h) - f(4)
h
x512 is a smooth curve increasing at a faster rate as x ~ oo. Therefore, if h > 0, then the slope of the secant line is greater
than the slope of the tangent line at f(4), which happens to be f'(4). Likewise, if h < 0, the slope of the secant line is less
than the slope of the tangent line at f(4), which happens to be f'(4).
(b) We know that
f'(4)
= lim f(4 + h)- f(4) = lim (4 + h)s12 h
h-->0
32
h
h-->0
Creating a table with values of h close to zero:
h
-0.0001
-0.00001
0.00001
0.0001
(4 + h)512 - 32
h
19.999625
19.99999
20.0000
20.0000375
Thus, f'(4)"" 20.0000.
(c) Using the estimate for f'(4) obtained in part (b), the equation of the line tangent to f(x)
y
=f'(4)(x-4) + f(4) =20(x -
= x512 at x = 4 is
4) + 32 = 20x- 48
y
80
60
40
20
-20
-40
--{;0
69. !GU[I
Use a plot of f(x)
= x' to estimate the value c such that f'(c) = 0. Find c to sufficient accuracy so that
lf(c + ~ - f(c)I::; 0.006
SOLUTION
Here is a graph of f(x)
for
h = ±0.001
= x' over the interval [0, 1.5].
y
2
1.5
0.5
0.2 0.4 0.6 0.8
I
1.2 1.4
The graph shows one location with a horizontal tangent line. The figure below at the left shows the graph of f(x) together
with the horizontal lines y = 0.6, y = 0.7 and y = 0.8. The line y = 0.7 is very close to being tangent to the graph of f(x).
The figure below at the right refines this estimate by graphing f (x) and y = 0.69 on the same set of axes. The point of
tangency has an x-coordinate ofroughly 0.37, soc"" 0.37.
y
y
0.2 0.4 0.6 0.8
1 1.2 1.4
0.2 0.4 0.6 0.8
1 1.2 1.4
124
CHAPTER 3
I
DIFFERENTIATION
We note that
lf(0.37 + o~~g~i- f(0.37)' ""o.00491 < o.006
and
f(0.37 - 0.001) - f(0.37)' "" 0.00304 < 0.006
0.001
I
so we have determined c to the desired accuracy.
The vapor pressure of water at temperature T (in kelvins) is the atmospheric pressure Pat which no net evaporation
takes place. In Exercises 71 and 72, use the following table to estimate the indicated derivatives using the difference
quotient approximation.
T (K)
293
303
323
313
333
343
353
P (atm) 0.0278 0.0482 0.0808 0.1311 0.2067 0.3173 0.4754
71. Estimate P' (T) for T = 293, 313, 333. (Include proper units on the derivative.)
SOLUTION
P'
_ P(303) - P(293) _ 0.0482 - 0.0278 _ O OO O
K
303 - 293
10
- . 2 4 atm/
(2 93 ) -
P'( 3 l 3)"" P(323)- P(313)
323 - 313
P'
= 0.1311 -
0.0808
10
= 0.00503
tm/K
a
_ P(343) - P(333) _ 0.3173 - 0.2067 _ O Ol O
K
(3 33 ) 343 - 333
10
- . 1 6 atm/
Let P(t) represent the U.S. ethanol production as shown in Figure 19. In Exercises 73 and 74, estimate the indicated
derivatives using the difference quotient approximation.
P (billions of gallons)
Pl-
~
f'j'b
~
f'j<l)
~
,-SI
-"'
-""'
-"°'
~
,,,..,
_,,.b
_&
_,,.'b
-"C)
~\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~~
~'
~
~"'
~
FIGURE 19 U.S. ethanol production.
73. Estimate P'(t) fort= 1997, 2001, 2005, 2009. (Include proper units on the derivative.)
SOLUTION
,
P(1998) _
1998
,
P(2002) P (2001) ""
_
2002
P (1997) ""
P(1997)
1.40 - 1.30
=
= 0.10 billions of gallons/year
1997
1
P(2001)
2.12 - 1.77
.
=
= 0.35 billions of gallons/year
2001
1
P'( 2005 )"" P(2006) -_ P(2005)
= 4.89 -
P'(2009) "" P(2010) - P(2009)
= 13.30 -l
2006
2010
2005
_
2009
4.00
1
= O.89 b"Ir
f II
I
1 10ns o ga ons year
10.94
= 2.36 b"lli
I
1 ons of gallons year
In the remaining exercises, SDQ refers to the symmetric difference quotient derivative approximation that is based on the
slope of the secant line between (x - h, f(x - h)) and (x + h, f(x + h)):
' ) _ f(x + h) - f(x - h)
f (x 2h
s ECTI0
I
N 3.1
Definition of the Derivative
125
75. With P(T) as in Exercises 71 and 72, estimate P' (T) for T = 303, 313, 333, 343, now using the SDQ.
SOLUTION
P' 303
(
m/K
),,, P(313) - P(293) = 0.0808 - 0.0278 = 0.00
265
313 - 293
20
at
=
P(3:~; :6~03) = 0.1311;O0.0482 = 0.004145 atm/K
P'(313),,,
P' <333
) ,,, P(343) - P(323) = 0.3173 - 0.1311 = 0.00
/K
93 1 atm
343 - 323
20
P' 43 ) ,,, P(353) - P(333)
<3
353 - 333
= 0.4754 -
= O.Ol 3435
0.2067
20
/K
atm
In Exercises 77 and 78, traffic speed S along Kat/nan Road (in kilometers per hour) varies as a function of traffic density
q (number of cars per kilometer of road) according to the data:
q (density)
60
70
80
90
100
S (speed)
72.5
67.5
63.5
60
56
77. Estimate S'(80) using the SDQ. (Include proper units on the derivative.)
SOLUTION
Let S(q) be the function determining S given q. With h = 10 in the SDQ,
= 60- 67.5 = _0375
S'( 0),,, S(90)- S(70)
8
20
20
km/h
cars/km
Exercises 79-81: The current (in amperes) at time t (in seconds)flowing in the circuit in Figure 20 is given by Kirchhoff's
Law:
i(t)
= Cv' (t) + R- 1v(t)
where v(t) is the voltage (in volts, V), C the capacitance (in farads, F), and R the resistance (in ohms, 0).
-
i
R
C
FIGURE 20
79. Calculate the current at t = 3 if
v(t)
where C
= 0.5t + 4 V
= 0.01 F and R = 100 0.
SOLUTION Since v(t) is a line with slope 0.5, v'(t) = 0.5 volts/s for all t. From the formula, i(3)
0.01(0.5) + (1/100)(5.5) = 0.005 + 0.055 = 0.06 amperes.
= Cv'(3) + (1 / R)v(3) =
81. Assume that R = 200 .Q but C is unknown. Use the following data and the SDQ to estimate v'(4), and deduce an
approximate value for the capacitance C.
t
SOLUTION
3.8
3.9
4
4.1
4.2
v(t)
388.8
404.2
420
436.2
452.8
i(t)
32.34
33.22
34.1
34.98
35.86
Solving i(4) = Cv'(4) + (1/R)v(4) for c yields
C
= i(4) -
(l/R)v(4)
v'(4)
= 34.1 -
~
v'(4)
126
CH APTER 3
I
DIFFERENTIATION
To compute C, we first approximate v'(4). Taking h
,
V
(4) ""
v(4.l)-v(3.9)
0.2
=
0.1, we find
=
436.2-404.2
0.2
= 160 VO1ts
I
S
Plugging this in to the equation above yields
C""
34.1 - 2.1
160
= 0.2 farads
Further Insights and Challenges
83. (a) Show that the symmetric difference quotient f(a+h>;t<a-h) is the slope of the secant line to the graph off from
x =a - h to x =a+ h. (Include an illustration.)
(b) Prove that the symmetric difference quotient is the a~erage of the slopes of the secant lines from x to x + h and from
x-h tox.
SOLUTION
(a) The secant line to the graph off from x = a - h to x = a + h passes through the points (a - h,f(a - h)) and
(a+ h, f(a + h)). See the figure below. The slope of this line is
f(a + h) - f(a - h)
(a+ h)-(a-h)
f(a + h) - f(a - h)
2h
which is equal to the symmetric difference quotient.
y
a-h
a+h
(b) The secant line to the graph off from x to x + h passes through the points (x, f(x)) and (x + h, f(x + h)), so the slope
of this secant line is
f(x + h) - f(x)
(x+h)-x
f(x + h) - f(x)
h
On the other hand, the secant line to the graph off from x - h to x passes through the points (x - h,f(x - h)) and
(x, f(x)), so the slope of this secant line is
f(x) - f(x - h)
x-(x-h)
f(x) - f(x - h)
h
The average of these two slopes is
! (f(x + h) 2
h
f(x) + f(x) - f(x - h)) = ! f(x + h) - f(x) + f(x) - f(x - h) = f(x + h) - f(x - h)
h
2
h
2h
which is equal to the symmetric difference quotient.
85. ~ Show that if f is a quadratic polynomial, then the SDQ at x =a (for any h
the graphical meaning of this result.
SOLUTION
* 0) is equal to f'(a). Explain
Let f (x) = px2 + qx + r be a quadratic polynomial. We compute the SDQ at x = a.
f(a + h) - f(a - h) _ p(a + h) 2 + q(a + h) + r - (p(a - h)2 + q(a - h) + r)
2h
2h
= pa2 + 2pah + ph2 + qa + qh + r - pa 2 + 2pah - ph2
2h
4pah + 2qh
2h(2pa + q)
=
=
= 2pa+q
2h
2h
-
qa + qh - r
Since this doesn't depend on h, the limit, which is equal to f'(a), is also 2pa + q. Graphically, this result tells us that the
secant line to a parabola passing through points chosen symmetrically about x = a is always parallel to the tangent line
atx=a.
sECT Io N
3.2
I
The Derivative as a Function
127
3.2 The Derivative as a Function
Preliminary Questions
1. What is the slope of the tangent line through the point (2, f (2)) if f' (x) = i3?
The slope of the tangent line through the point (2,f(2)) is given by f'(2). Since f'(x) = i3, it follows that
!'(2) = 23 = 8.
SOLUTION
2. Evaluate (f - g)'(l) and (3f + 2g)'(l), assuming that f'(l) = 3 and g'(l) = 5.
SOLUTION
(f - g)'(l) = f'(l) - g'(l) = 3 - 5 = -2 and (3f + 2g)'(l) = 3f'(l) + 2g'(l) = 3(3) + 2(5) = 19.
3. To which of the following does the Power Rule apply?
= x2
(a) f(x)
(b) f(x)
= x'
f(x) = r
(c) f(x)
(e)
(d) f(x)
(f) f(x)
= 2'
= ex
= x-4!5
SOLUTION
(a) Yes. x2 is a power function, so the Power Rule can be applied.
(b) Yes. 2' is a constant function, so the Power Rule can be applied.
(c) Yes. x' is a power function, so the Power Rule can be applied.
(d) No.~ is an exponential function (the base is constant while the exponent is a variable), so the Power Rule does not
apply.
r is not a power function because both the base and the exponent are variable, so the Power Rule does not apply.
Yes. x-415 is a power function, so the Power Rule can be applied.
(e) No.
(t)
4. State whether each claim is true or false. If false, give an example demonstrating that it is false.
(a) If f is continuous at a, then f is differentiable at a.
(b) If f is differentiable at a, then f is continuous at a.
SOLUTION
(a) This statement is false. For example, the function f(x) = lxl is continuous at x = O but is not differentiable at x = 0.
(b) This statement is true.
Exercises
In Exercises 1-6, compute f'(x) using the limit definition.
1. f(x)
= 3x-7
Let f(x) = 3x - 7. Then,
SOLUTION
. f(x + h) - f(x)
'( )
. 3h
. 3(x + h) - 7 - (3x - 7)
f x = 11m
= 11m
= hm - = 3
h-->O
h
h-->0
h
h-->0 h
3. f(x)
= i3
SOLUTION
Let f(x) = i3. Then,
f
'( )
X
. f(x + h) - f(x)
. (x + h) 3
= 1!ID
= 1!ID
h
h-->0
2
-
h
h-->0
x3
. i3 + 3x2h + 3xh2 + h 3 - i3
= 1! I D - - - - - - - - -
h
h-->0
. 3x2h + 3xh + h
= hm
h
= lim(3x2 + 3xh + h 2 ) = 3x2
3
h-->O
S. f(x) = x SOLUTION
!
h-->0
Yx
Let f(x) = x - yx. Then,
'(x-1m
) _ 1. f(x + h) h
h-->0
.
= 1 - hm
h-->O
f(x) _ . x + h - Vx + h - (x 1
-1m
h-->0
h
~+~-x
h( Vx + h +
yx)
.
= 1 - hm
h-->O
1
yx)
.
=1-hm
h-->O
1
= 1 - -VX + h + yx
2 Yx
vx + h - yx ( vx + h + yx)
.
h
Vx+h+yx
128
CH APT E R 3
I
DIFFERENTIATION
In Exercises 7-14, use the Power Rule to compute the derivative.
7 . .!!:_ x4 \
dx
x=-2
.!!:_ (x4 ) = 4.x3
SOLUTION
dx
= 4(-2)3 = -32
!!:_x4 \
so
dx
=-2
9. !!:_ t213\
dt
t=S
!!:_ (t2t3)
dt
SOLUTION
= ~r113 so
3
!!:_t2t3\
dt
t=S
= ~( 8 )-113 = !
3
3
d
11. dx X0.35
SOLUTION
13.
~ t-m
SOLUTION
In Exercises 15-18, compute f'(x) and find an equation of the tangent line to the graph at x =a.
15. f(x)
= x4, a = 2
Let f(x)
SOLUTION
at x
= x4. Then, by the Power Rule, f'(x) = 4.x3. The equation of the tangent line to the graph of f(x)
= 2 is
y
17. f(x) = 5x - 32 yx,
Let f(x)
SOLUTION
= f'(2)(x -
2) + f(2)
= 32(x -
2) + 16
= 32x -
48
a= 4
= 5x- 32x112 . Then f'(x) = 5 y
16x- 1t 2 . In particular, f'(4)
= f'(4)(x-4) + f(4) = -3(x -
4) - 44
= -3x -
= -3. The tangent line at x = 4 is
32
19. Calculate:
d
(a) -12ex
d
t
d r-3
(c) dte
(b) dt(25t - 8e)
dx
Hint for (c): Write e'- 3 as e- 3e'.
SOLUTION
d
d
(a) -9ex = 9-ex = 9ex
dx
dx
d
t
d
d t
t
(b) d/3t-4e) = 3 d/- 4";]/ = 3 - 4e
d
(c) -e'-3
dt
d
= e-3-e'
= e-3 . e' = e'-3
dt
In Exercises 21-32, calculate the derivative.
21. f(x) = 2x3 - 3x2 + 5
SOLUTION
23. f(x) =
SOLUTION
25. g(z)
d
dx (2.x3 - 3x2 +
5) = 6x2 -
6x
4x5 13 - 3x-2 - 12
d (
20
dx 4x513 - 3x-2 - 12) = 3x213 + 6x- 3
= 1z-5t 14 + z-5 + 9
SOLUTION
d
(1z- 5/1 4 + z-5 + 9) =
dz
-
5
2
--z- 19114 - 5z- 6
fS + Vs
SOLUTION f(s) = Vs+ Vs= s 114 + s 113 . In this form, we can apply the Sum and Power Rules .
27. f(s) =
.!!:_ (s1f4 + s113)
ds
= !cs<J/4)-1) + !(il/3)-1) = !s-3/4
4
3
4
+ ! _213
3s
s E CT I O N
29. g(x)
3.2
I
The Derivative as a Function
= e2
.
d
Because e2 is a constant, -e2 =0.
dx
SOLUTION
31. h(t) = Se'- 3
!!:._5e1-3 = 5e- 3 !!:._e1=5e- 3 e1 = Se'- 3
dt
dt
SOLUTION
In Exercises 33-38, calculate the derivative by expanding or simplifying the function.
33. P(s)
= (4s -
3)2
P(s)
SOLUTION
= (4s -
3)2 = 16s2 - 24s + 9. Thus,
dP
= 32s-24
ds
-
35. f(x)
= (2 -
x)(2 + x)
f(x)
SOLUTION
= (2 -
x)(2 + x)
= 4 - x2. Thus,
df
- =-2x
dx
37. g(x) =
x2 + 4x112
x2
g(x) =
SOLUTION
x2 + 4xl/2
x
2
= 1 + 4x-3 12 • Thus,
dg =
dx
-6x-5/2
In Exercises 39-44, calculate the derivative indicated.
39. dT'
dC c;s
SOLUTION
,
T
= 3C2i 3
With T(C) = 3C2i3, we have dT = 2c- 113 _ Therefore,
•
dC
dT
dC
I
= 2(8)-1/3 = 1
C;8
41. ds' ,
dz z;2
s=4z-16z2
SOLUTION
Withs= 4z - 16z2 , we have ds = 4- 32z. Therefore,
dz
-dsl
= 4-32(2) = -60
dz z;z
43.
drl ,
dt f;4
SOLUTION
r=t-e'
With r = t- e', we have dr = I - e'. Therefore
dt
,
-drl
dt
=1-e4
f;4
129
130
CH APTER 3
I
DIFFERENTIATION
45. Match the functions in graphs (A)-(D) with their derivatives (1)-(III) in Figure 11. Note that two of the functions
have the same derivative. Explain why.
(I)
(D)
(C)
(B)
(A)
(Ill)
(II)
FIGURE 11
SOLUTION
• Consider the graph in (A). On the left side of the graph, the slope of the tangent line is positive but on the right
side, the slope of the tangent line is negative. Thus, the derivative should transition from positive to negative with
increasing x. This matches the graph in (III) .
• Consider the graph in (B). This is a linear function, so its slope is constant. Thus the derivative is constant, which
matches the graph in (I).
• Consider the graph in (C). Moving from left to right, the slope of the tangent line transitions from positive to
negative then back to positive. The derivative should therefore be negative in the middle and positive to either side.
This matches the graph in (II).
• Consider the graph in (D). On the left side of the graph, the slope of the tangent line is positive but on the right
side, the slope of the tangent line is negative. Thus, the derivative should transition from positive to negative with
increasing x. This matches the graph in (III).
Note that the functions whose graphs are shown in (A) and (D) have the same derivative. This happens because the
graph in (D) is just a vertical translation of the graph in (A), which means the two functions differ by a constant. The
derivative of a constant is zero, so the two functions end up with the same derivative.
47. Assign the labels y
g'(x)
= f(x), y = g(x), and y = h(x) to the graphs in Figure 13 in such a way that f'(x) = g(x) and
= h(x).
(A)
(B)
(C)
FIGURE 13
Consider the graph in (A). Moving from left to right, the slope of the tangent line is positive over the first
quarter of the graph, negative in the middle half, and positive again over the final quarter. The derivative of this function
must therefore be negative in the middle and positive on either side. This matches the graph in (C).
Now focus on the graph in (C). The slope of the tangent line is negative over the left half and positive on the right
half. The derivative of this function therefore needs to be negative on the left and positive on the right. This description
matches the graph in (B).
We should therefore label the graph in (A) as f(x), the graph in (B) as h(x), and the graph in (C) as g(x). Then
f'(x) = g(x) and g'(x) = h(x).
SOLUTION
49. Use the rules in Exercise 48 and the Linearity Rules to prove the first part of Theorem 1 in Section 3.1.
SOLUTION
Let f(x) = mx + b. Then
f'(x) = (mx + b)'
+ b'
Sum Rule
= m(x') + 0
Constant multiple and Constant Rules
= (mx)'
= m(l)
=m
First-Power Rule
S E CT I 0 N 3.2
51.
($.1
I
The Derivative as a Function
131
Use the table of values off to determine which of (A) or (B) in Figure 15 is the graph off'. Explain.
x
0
0.5
1
1.5
2
2.5
3
3.5
4
f(x)
10
55
98
139
177
210
237
257
268
(B)
(A)
FIGURE 15 Which is the graph off'?
The increment between successive x values in the table is a constant 0.5, but the increment between successive f(x) values decreases from 45 to 43 to 41 to 38 and so on. Thus the difference quotients decrease with increasing
x, suggesting that f'(x) decreases as a function of x. Because the graph in (B) depicts a decreasing function, (B) might
be the graph of the derivative off(x).
SOLUTION
53. Compute the derivatives, where c is a constant.
(a)
!!.. ci3
(c)
.!!._ (9c2y3 - 24c)
(b) dd
z
dt
(5z + 4cz2)
dy
SOLUTION
(a)
!!..ci3 = 3ct2
(b)
-(5z + 4cz2) = 5 + 8cz
dt
d
dz
d
(c) -(9c2 y 3
dy
-
24c)
= 27c2y2
55. Find the points on the graph of y
= x2 + 3x -
7 at which the slope of the tangent line is equal to 4.
= x2 + 3x- 7. Solving dy/dx = 2x + 3 = 4 yields x = ~·
57. Determine a and b such that p(x) = x2 +ax+ b satisfies p(l) = 0 and p'(l) = 4.
SOLUTION Let p(x) = x2 +ax+ b satisfy p(l) = 0 and p'(l) = 4. Now, p'(x) = 2x +a. Therefore 0 = p(l) = 1 +a+ b
and4 = p'(l) = 2 +a; i.e., a= 2 andb = -3.
SOLUTION
Lety
59. Let f(x) = x3 - 3x + 1. Show that f'(x) ~ -3 for all x and that, for every m > -3, there are precisely two points
where f' (x) = m. Indicate the position of these points and the corresponding tangent lines for one value of m in a sketch
of the graph off.
SOLUTION
Let P = (a, b) be a point on the graph off(x)
= x3 -
3x + 1.
• The derivative satisfies f'(x) = 3x2 - 3 ~ -3 since 3x2 is nonnegative.
• Suppose the slope m of the tangent line is greater than -3. Then f'(a) = 3a2
a
2
m+3
=->0
3
and thus
a
-
3
= m, whence
= ± ~+3
-3
• The two parallel tangent lines with slope 2 are shown with the graph of f(x) here.
y
61. Compute the derivative of f(x)
= x31 2 using the limit definition. Hint: Show that
f(x + h) - f(x) = (x + h) 3 - x3 (
h
h
1
-V<x + h)3 + W
)
132
CHAPTER 3
I
DIFFERENTIATION
Once we have the difference of square roots, we multiply by the conjugate to solve the problem.
SOLUTION
.
f' (x) = lim
(x + h)3/2 - x3/2
h
h-->O
3
.
(x + h)
h
-
x3 (
=hm---h-->O
w (.,f(X+ii)3 + w)
.
.,,/(x + h)3 = hm - ' - - - - h-->0
h
1
.,,/(x + h)3 +
W
)
.,,/(x + h)3 +
W
The first factor of the expression in the last line is clearly the limit definition of the derivative of x3, which is 3x2. The
second factor can be evaluated, so
.!!_x312
dx
= 3x2_1_ = ~x112
2w
2
63. In each case use the limit definition to compute f' (0), and then find the equation of the tangent line at x = 0
(a)
f (x) =
(b) f(x)
xex
= x2ex
SOLUTION
(a) Let f(x) = xex. Then f(O) = 0, and
f'(O) = lim f(O + h)- f(O) = lim heh - 0 = Iimeh = 1
h-->0
h
h-->0
h
h-->0
The equation of the tangent line is
y = f' (O)(x - 0) + f(O) = 1(x - 0)
+0 = x
(b) Letf(x) = x2ex. Then f(O) = 0, and
J'(O) = lim f(O + h)- f(O) = lim h2eh - 0 = limheh = 0
h-->0
h
h-->0
h
h-->0
The equation of the tangent line is
y=
f' (O)(x - 0) + f (0) = O(x - 0) + 0 = 0
65. The brightness b of the sun (in watts per square meter) at a distance of d meters from the sun is expressed as an
inverse-square law in the form b = 4
where Lis the luminosity of the sun and equals 3.9 x 1026 watts. What is the
derivative of b with respect to d at the earth's distance from the sun (1.5 x 1011 m)?
/:az,
SOLUTION
Let b = 41r1;p = f,;d- 2• Then
,
L
_
L
b =-(-2d 3) = - 411'
2nd3
and
b'(l.5 x 10")
=
3.9 x 1026
Zn(l. 5 x 1011 )3
"" -
18
-8
3
watts/m
• 4 x 10
67. The Clausius-Clapeyron Law relates the vapor pressure of water P (in atmospheres) to the temperature T (in
kelvins):
dP =k.!_
dT
T2
where k is a constant. Estimate dP/dT for T = 303, 313, 323, 333, 343 using the data and the symmetric difference
approximation
dP
P(T + 10) - P(T - 10)
dT""
20
T (K)
293
303
313
323
P(atm) 0.0278 0.0482 0.0808 0.1311
333
343
0.2067 0.3173
353
0.4754
Do your estimates seem to confirm the Clausius-Clapeyron Law? What is the approximate value of k?
s E c T to N
SOLUTION
3.2
I
The Derivative as a Function
133
Using the indicated approximation to the first derivative, we calculate
P'(303),,.,
P(313) - P(293)
20
= 0.0808 -
0.0278
= O.00265 atm /K
P'(313),,.,
P(323)- P(303)
20
= 0.1311 -
0.0482
= O.004145 atm /K
P'(323),,.,
P(333)- P(313)
20
- 0.0808 O
= 0.2067 20
= .006295 atm /K
20
20
P(343)-P(323)
0.3173-0.1311
/K
,
P (333) ,,.,
=
= 000931
.
atm
20
20
P'(
343
),,., P(353)- P(333)
20
= 0.4754 -
0.2067
20
= 0 _013435 atm/K
2
If the Clausius-Clapeyron law is valid, then T dP should remain constant as T varies. Using the data for vapor
P dT
pressure and temperature and the approximate derivative values calculated above, we find
T (K)
303
313
323
333
343
T 2 dP
P dT
5047.59
5025.76
5009.54
4994.57
4981.45
These values are roughly constant, suggesting that the Clausius-Clapeyron law is valid, and that k ,,., 5000.
69. In the setting of Exercise 68, show that the point of tangency is the midpoint of the segment of L lying in the first
quadrant.
SOLUTION In the previous exercise, we saw that the tangent line to the hyperbola xy = 1 or y = ~ at x = a
has y-intercept P = (0, ~) and x-intercept Q = (2a, 0). The midpoint of the line segment connecting P and Q
is thus
(0+2 2a ' ~ 2+ 0)
= (
!)
a, a
which is the point of tangency.
71. Make a rough sketch of the graph of the derivative of the function in Figure l 7(A).
The graph has a tangent line with negative slope approximately on the interval (1,3.6) and has a tangent
line with a positive slope elsewhere. This implies that the derivative must be negative on the interval (1, 3.6) and positive
elsewhere. The graph may therefore look like this:
SOLUTION
y
73. Sketch the graph of f(x) = x Ix!. Then show that f' (0) exists.
SOLUTION
For x < 0, f(x)
= -x2, and f'(x) = -2x. For x > 0, f(x) = x2, and f'(x) = 2x. At x = 0, we find
lim f(O + h) - f(O)
h->O+
h
= lim
h->O+
hz
h
=0
and
2
Jim f(O + h) - f(O) = Jim -h =
0
h
h->0- h
h->0-
134
CHAPTER 3
I
DIFFERENTIATION
Because the two one-sided limits exist and are equal, it follows that f' (0) exists and is equal to zero. Here is the graph
of f(x) = xlxl.
!GU[i
In Exercises 75-80, zoom in on a plot off at the point (a,f(a)) and state whether or not f appears to be
differentiable at x =a. If it is nondifferentiable, state whether the tangent line appears to be vertical or does not exist.
75. f(x) = (x - l)lxl,
a= 0
The graph of f(x) = (x - l)lxl for x near 0 is shown below. Because the graph has a sharp corner at x
it appears that f is not differentiable at x = 0. Moreover, the tangent line does not exist at this point.
SOLUTION
= 0,
y
-0.2
-0.3
77. f(x)
= (x -
3) 113 ,
a
=3
The graph of f(x) = (x - 3) 113 for x near 3 is shown below. From this graph, it appears that
differentiable at x = 3. Moreover, the tangent line appears to be vertical.
f
is not
SOLUTION The graph of f(x) = I sin xi for x near 0 is shown below. Because the graph has a sharp corner at x
appears that f is not differentiable at x = 0. Moreover, the tangent line does not exist at this point.
= 0, it
SOLUTION
79. f(x) =I sin xi,
a= 0
y
0.10
-0.10 -0.05
0.05
0.10
81. Find the coordinates of the point Pin Figure 19 at which the tangent line passes through (5, 0).
y
FIGURE 19
s E c TI 0 N
3.2
I
The Derivative as a Function
135
Let f(x) = 9 - x2, and suppose P has coordinates (a, 9 - a2 ). Because f'(x) = -2x, the slope of the line
tangent to the graph of f(x) at Pis -2a, and the equation of the tangent line is
SOLUTION
y = f' (a)(x - a)+ f(a)
= -2a(x -
a)+ 9 - a 2
= -2ax + 9 + a2
In order for this line to pass through the point (5, 0), we must have
0 = -lOa +9 +a2 = (a-9)(a- I)
Thus, a= 1 or a= 9. We exclude a= 9 because from Figure 19, we are looking for an x-coordinate between 0 and 5.
Thus, the point P has coordinates (1, 8).
Exercises 83-86 refer to Figure 20. Length QR is called the subtangent at P, and length RT is called the subnormal.
y
R
Q
T
FIGURE 20
83. Calculate the subtangent of
f (x) = x2 + 3x
SOLUTION
Let f(x)
at x
=2
= x2 + 3x. Then f'(x) = 2x + 3, and the equation of the tangent line at x = 2 is
y
This line intersects the x-axis at x
= f'(2)(x -
2) + /(2)
= 7(x- 2) + 10 = 7x- 4
= ~·Thus Q has coordinates(~, 0), R has coordinates (2, 0) and the subtangent is
85. Prove in general that the subnormal at Pis lf'(x)f(x)I.
SOLUTION The slope of the tangent line at P is f'(x). The slope of the line normal to the graph at Pis then - I/ f' (x),
and the normal line intersects the x-axis at the point T with coordinates (x + f(x)f'(x), 0). The point R has coordinates
(x, 0), so the subnormal is
Ix+ f(x)f'(x)- xi= lf(x)f'(x)I
87. Prove the following theorem of Apollonius of Perga (the Greek mathematician born in 262 BCE who gave the
parabola, ellipse, and hyperbola their names): The subtangent of the parabolay = x2 at x =a is equal to a/2.
SOLUTION
Let f(x)
= x2. The tangent line to fat x = a is
Y = f'(a)(x- a)+ f(a) = 2a(x - a)+ a 2
The x-intercept of this line (where y
= 0) is ~ as claimed.
y
= 2ax- a 2 •
136
CH APTER 3
I
DIFFERENTIATION
!SJ
= x".
SOLUTION The generalized statement is: The subtangent toy = x" at x = a is equal to ~a.
Proof: Let f(x) = x". Then f'(x) = nx"- 1, and the equation of the tangent line t x = a is
89.
Formulate and prove a generalization of Exercise 88 for y
y
= f'(a)(x- a)+ f(a) = na"-\x -
This line intersects the x-axis at x
subtangent is
= (n -
a)+ a"= na"- 1x - (n - l)a"
I)a/n. Thus, Q has coordinates ((n - l)a/n, 0), R has coordinates (a, 0), and the
n- I
1
a- --a= -a
n
n
Further Insights and Challenges
91. A vase is formed by rotating y = x2 around the y-axis. If we drop in a marble, it will either touch the bottom point
of the vase or be suspended above the bottom by touching the sides (Figure 22). How small must the marble be to touch
the bottom?
FIGURE 22
SOLUTION Suppose a circle is tangent to the parabola y = x2 at the point (t, t2). The slope of the parabola at this point
is 2t, so the slope of the radius of the circle at this point is -f, (since it is perpendicular to the tangent line of the circle).
Thus the center of the circle must be where the line given by y = -f,(x - t) + t2 crosses the y-axis. We can find the
y-coordinate by setting x = 0: We get y = + t2. Thus, the radius extends from (0, + t2) to (t, t2) and
!
!
This radius is greater than ~ whenever t > O; so, if a marble has radius > 1/2 it sits on the edge of the vase, but if it has
radius ::; I /2 it rolls all the way to the bottom.
93. Negative Exponents
Let n be a whole number. Calculate the derivative off(x) = x-n by showing that
f(x
+ h) - f(x)
-1
(x + h)" - x"
x"(x+h)"
h
h
SOLUTION
Let f(x) = x-n where n is a positive integer.
• The difference quotient for f is
f(x + h) - f(x)
h
(x + h)-n - x-n
X'-(x+h)"
x"(x+h)"
h
h
-1
(x + h)" - x"
x"(x + h)"
h
• Therefore,
f'(x) = lim f(x + h)- f(x) = lim
-I
h->O
h
h->O x"(x + h)"
.
-1
= Ih->O
!ID
x"(x + h )"
. (x+h)n
1Im
h->O
h
-~
= -x
_2 d
" - (x")
dx
= -x-2" :x (x") = -x-2" • nx"- 1 = -nx-n- 1 • Since n is a positive integer, k = -n is a
negative integer and we have .!!:.._ (x!') = .!!:.._ (X-") = -nx-n-I = kx!'- 1 ·that is .!!:.... (x!') = kx!'- 1 fornegative integers k
dx
dx
'
'dx
·
• From above, we continue: f' (x)
s E c TI o N
95. Infinitely Rapid Oscillations
SOLUTION
I
Product and Quotient Rules
137
Define
h(x) =
Show that his continuous at x
3.3
l
xsin!
x
0
x;tO
x=O
= 0 but h'(O) does not exist (see Figure 8).
*
0
if x
. As x
if x = 0
Let h(x) = {xo sin 0)
lh(x)- h(o)1 =
--+
0,
ixsin(~)- oi = 1x1 lsinG)i-+ o
since the values of the sine lie between -1and1. Hence, by the Squeeze Theorem, limh(x)
x-->O
= h(O) and thus his contin-
uous atx = 0.
As x --+ 0, the difference quotient at x = 0,
h(x)-h(O) = xsin0)-0 =sin(!)
x-0
x-0
x
does not converge to a limit since it oscillates infinitely through every value between -1and1. Accordingly, h'(O) does
not exist.
97. If lim
f(c + hh- f(c)
and Jim
h-.-,O-
ex'ist but are not equal, then f is not differentiable at c, and the graph off has
f(c + hh- f(c)
h--tO+
a comer at c. Prove that f is continuous at c.
SOLUTION
The function f is continuous at x = c if limf(x) = f(c), or equivalently if lim(f(c + h) - f(c)) = 0. Note
/z-+Q
X--+C
that
. (f( c + h) - f( c))
1im
h-->O-
Now Jim h = 0, and by assumption Jim
h-+O-
. (hf(c + h)- f(c)) = 1Im
. h · 1Im
. f(c + h)- f(c)
1im
= h-->0h
h-->Oh-->Oh
f(c+htf(c)
h-+O-
exists. It follows that Jim (f(c + h) - f(c)) = 0. Similarly, Jim (f(c +
h--+O-
h)- f(c)) = 0. Because the two one-sided limits exist and are equal to 0,
lim(f(c + h) - f(c)) = 0
h-->O
therefore, f is continuous at x = c.
3.3 Product and Quotient Rules
Preliminary Questions
1. Are the following statements true or false? If false, state the correct version.
(a) f g denotes the function whose value at xis f(g(x)).
(b) f /g denotes the function whose value at xis f(x)/g(x).
(c) The derivative of the product is the product of the derivatives.
(d)
(e)
:x
!
(fg)lx=4 = f(4)g'(4)- g(4)J'(4)
(fg)lx=o = f'(O)g(O) + f(O)g'(O)
SOLUTION
(a) False. The notation f g denotes the function whose value at xis f(x)g(x).
(b) True.
(c) False. The derivative of a product f g is f'(x)g(x) + f(x)g'(x).
(d) False. dd (fg)I
= f'(4)g(4)
x
x=4
+ f(4)g'(4).
(e) True.
2. Find (f/g)'(l) if f(l) = f'(l) = g(l) = 2 and g'(l) = 4.
SOLUTION
d
dx (f /g)lx=I = [g(l)J'(l) - f(l)g'(l))/g(1) 2 = [2(2) - 2(4))/22 = -1.
= 0, f'(l) = 2, and (fg)'(l) = 10.
(fg)'(l) = f(l)g'(l) + f'(l)g(l), so 10 = 0. g'(l) + 2g(l) and g(l) = 5.
3. Find g(l) if f(l)
SOLUTION
h-+O+
138
CHAPTER 3
I
DIFFERENTIATION
Exercises
In Exercises 1-6, use the Product Rule to calculate the derivative.
1. f(x)
= .x3(2x2 + 1)
SOLUTION
Let f(x) = .x3(2x2 + 1). Then
f'(x) =
3. f(x)
• ds
(2x2 + 1) +
x3 :x (2x2 + 1) = (3x2)(2x2 + 1) + x3(4x) = 10x4 + 3x2
= x2ex
SOLUTION
5 dhl
(! x3)
s=4'
SOLUTION
Letf(x) = x2ex. Then
h(s) = (s- 112 + 2s)(7 - s- 1)
Let h(s) = (s- 112 + 2s)(7 - s- 1). Then
dh = (.!!:._cs- 112 + 2s)) (7 - s- 1) + (s- 112 + 2s) .!!:._(7 - s- 1)
ds
ds
ds
=
312 + 2) (7 (-~s2
s- 1) + (s- 112 + 2s)(s- 2) =
312 + ~s- 5 12 + 14
-~s2
2
Therefore,
dh
ds
I
= -~(4)-3/2 + ~(4)-5/2 + 14 = 871
2
s=4
2
64
In Exercises 7-12, use the Quotient Rule to calculate the derivative.
x
7. f(x)= x-2
SOLUTION
Let f(x) =
x~ 2 •
Then
f' (x) =
9 dgl
• dt t=-2'
SOLUTION
g(t)
(x-2)..4..x-x..4..(x-2)
dx
dx
(x - 2)
2
=
(x-2)-x
(x - 2)
2
-2
= __
(x - 2)2
t2 + 1
=t2 - 1
Let g(t) =
t2 + 1
Then
t - 1
-2-·
dg - (t2 -
l)fi(t2 + 1) (t2 -
dt -
(t2 +
1)2
l)fi(t2 -
l)
(t2 - 1)(2t) - (t2 + 1)(2t)
(t2 - 1)2
Therefore,
4(-2)
dgl
dt
11. g(x)
t=-2
8
= - ((-2)2 -1)2 = 9
1
=1 + eX
SOLUTION
1
Letg(x) =--.Then
1 + eX
dg
(1 + ex)fx 1 - 1 ;f;O +ex)
dx
(1 + ex)2
(1 + ~)(0) - ex
=
4t
=- (t2 -
1)2
S E CT I 0 N 3.3
I
Product and Quotient Rules
139
In Exercises I3-I8, calculate the derivative in two ways. First use the Product or Quotient Rule; then rewrite thefanction
algebraically and directly calculate the derivative.
x3 x- 3
13. f(x) =
SOLUTION
Let f(x) =
x3 x- 3. By the Product Rule,
J'(x)
= (x3)'x-3 + x3 (x-3)' = 3.x2x-3 + x\-3x-4 ) = 3x- 1 -
Alternatively, simplifying first, we find f(x)
15. f(t)
= (2t + l)(t2 -
SOLUTION
= 1, so f'(x) = 0.
2)
Let f(t) = (2t + l)(t2 - 2). Using the Product Rule,
= (2t + l)'(t2 -
f'(t)
2) + (2t + l)(t2 - 2)'
= 2(t2 -
2) + (2t + 1)(2t)
Alternatively, multiplying out first, we find f(t) = 2t3 + t2 - 4t - 2. Therefore, f'(t)
17. h(t)
3x- 1 = 0
= 6t2 + 2t- 4
= 6t2 + 2t -
4.
t2 - 1
= --1
(-
SOLUTION
Let h(t) = ~~/.Using the Quotient Rule,
,
h (t)
=
(t - l)(t2 - l)' - (t2 - l)(t - 1)'
(t - 1)2
=
(t- 1)(2t) - (t2 - 1)(1)
(t - 1)2
=
t2 - 2t + 1
(t - 1)2
=1
fort* 1. Alternatively, simplifying first, we find fort* 1,
h(t)
Hence h'(t)
= 1 fort*
= (t -
1)(t + 1)
(t- 1)
=t + 1
1.
In Exercises I9-40, calculate the derivative.
19. f(x) = (.x3 + 5)(.x3 + x + 1)
SOLUTION
Let f(x)
= (.x3 + 5)(.x3 + x + 1). Using the Product Rule,
f'(x)
21. dy' ,
dxx=3
1
y=x+lO
SOLUTION
Let y
= (3.x2)(x3 + x + 1) + (x3
+ 5)(3.x2 + 1)
= 6.x5 + 4x3 + 1s.x2 + 5
= xlw. Using the Quotient Rule,
(x + 10)(0) - 1(1)
dy
dx
(x + 10)2
(x + 10)2
Therefore,
169
= ( -Vx + 1)( -Vx- 1)
SOLUTION Let f(x) = ( -Vx + 1)( -Vx- 1). Multiplying through first yields f(x) = x for x ~ 0. If we carry out the Product Rule on f(x) = (x 1i 2 + l)(x 1i 2 - 1), we find
23. f(x)
f'(x)
= (!x-112) (x112 _
l) + (x1;2 + 1) (!cx-112))
2
I,
25. dy
dxx=2
x4 - 4
y=x2 -5
SOLUTION
Let y
2
=!
2
_ !x-112 + ! + !x-112
2
2 2
x4 -4
= -x 2--5
- . Using the Quotient Rule,
dy
dx
4
{x2-5){4.x3)-(x -4)(2x)
=
(x2 -5)2
2.x5-20.x3+8x
= --(x_2___5_)-=-2-
Therefore,
I
dy
dx x=2
= 2(2)5 _
1 for x ~ 0. Therefore, f'(x)
20(2) 3 + 8(2) __
80
(2 2 - 5) 2
=1
=1
140
CHAPTER 3
I
DIFFERENTIATION
27.
dzl
dx
x=I
, z= 3
x +1
Let z = )+ 1 • Using the Quotient Rule,
SOLUTION
dz
(x3 + 1)(0) - 1(3x2)
3.x2
+
(x3 + 1)2
(x3
dx
1)2
Therefore,
3(1)2
(l3 + 1)2
3
4
t
29. h(t) = (t + l)(t2 + 1)
SOLUTION
t
Let h(t) = -(t_+_l-)(...,,t2_+_l_)
'
h (t)
=
t
. Using the Quotient Rule,
t3 +t2 +t+l
(t3 + t2 + t + 1) (1) -
t (3t2 + 2t + 1)
(t3 + t2 + t + l)2
-2t3 - t2 +
1
=- - - - (t3+t2+t+1)2
31. f(x) = x2e2
SOLUTION
Letf(x)
= x2e 2 • Thenf'(x) = 2e2 x.
33. f(x) = (x + 3)(x - l)(x - 5)
SOLUTION
Let f(x) = (x + 3)(x - l)(x - 5). Using the Product Rule inside the Product Rule with a first factor of
(x + 3) and a second factor of (x - l)(x - 5), we find
f'(x) = (l)(x - l)(x - 5) + (x + 3) ((l)(x- 5) + (x- 1)(1)) = 3x2 - 6x - 13
Alternatively,
f(x)
= (x + 3) {x2 -
6x + 5)
= x3 -
3x2 - 13x + 15
Therefore, f'(x) = 3x2 - 6x - 13.
ex
35. f(x) = x+ 1
SOLUTION
Let f(x)
x+l
(zz - 14)(zz + 21)
2
37• g(z) =
ex
= - -. Using the Quotient Rule,
SOLUTION
2
Hint: Simplify first.
Let
z2 -
4) (z- -1) = ((z + 2)(z - 2) (z + l)(z - 1)) = (z - 2)(z + 1)
2
g (z) = ( - -
z-1
for z
-
z+2
z-1
z+2
* -2 and z * 1. Then,
g'(z)
d (xt-4)
39. -d
- t t2 -x
SOLUTION
= (z + 1)(1) + (z -
2)(1)
= 2z -
1
(x constant)
Let f(t)
= :I=!. Using the Quotient Rule,
f'(t)
= (t2 -
x)(x)- (xt - 4)(2t)
(t2 - x)2
= xt2 -x2 - 2xt2 + 8t = -xt2 + 8t-x2
(t2 - x)2
(t2 - x)2
s Ec T Io N
3.3
I
Product and Quotient Rules
Jn Exercises 41-44, calculate f'(x) in terms of P(x), Q(x), and R(x), assuming that P'(x)
R'(x) = P(x).
41. f(x)
Q'(x)
= -R(x), and
xR(x) + Q(x)
=
SOLUTION
Let f(x)
= xR(x) + Q(x). Using the Product Rule on the first term, we find
f'(x)
43. f(x)
= Q(x),
141
= ~i:~ - x
SOLUTION
Let f(x) = ~i~
,
f
= (l)R(x) + xR'(x) + Q'(x) = R(x) + xP(x) -R(x) = xP(x)
- x. Using the Quotient Rule on the first term, we find
_ Q(x)P'(x)- P(x)Q'(x) _ _ Q(x)Q(x) + P(x)R(x) _ _ P(x)R(x)
11 - (Q(x))2
(x) (Q(x))2
(Q(x))2
In Exercises 45-48, calculate the derivative using the values:
/(4)
f'(4)
g(4)
g'(4)
10
-2
5
-1
45. (fg)'(4) and (f /g)'(4)
SOLUTION
Leth= Jg andH =fig. Then h'
h'(4)
= f'g+ Jg' andH' = gf'1g'.
Finally,
g
= f'(4)g(4) + f(4)g'(4) = (-2)(5) + (10)(-1) = -20
and
H'( )
4
= g(4)/'(4)- f(4)g'(4)
= (5)(-2)- (10)(-1)
(g(4))2
=O
(5)2
47. G'(4), where G(x) = (g(x))2
SOLUTION Let G(x) = g(x) 2 = g(x)g(x). Then G'(x) = g'(x)g(x) + g(x)g'(x) = 2g(x)g'(x), and
G'(4)
= 2g(4)g'(4) = 2(5)(-1) = -10
In Exercises 49 and 50, a rectangle's length L(t) and width W(t) (measured in inches) are varying in time (t, in minutes).
Determine A'(t) in each case. ls the area increasing or decreasing at that time?
= 6, L'(3) = -4, and W'(3) = 5.
Because A(t) = L(t)W(t), it follows that A'(t) = L'(t)W(t) + L(t)W'(t). Then
49. At t = 3, we have L(3) = 4, W(3)
SOLUTION
A'(3) = L'(3)W(3) + L(3)W'(3) = (-4)(6) + 4(5) = -4 inches 2/minute
Because A'(3) is negative, the area is decreasing at t = 3.
51. Calculate F'(O), where
F(x)
= x 9 + xs + 4.x5 - 7x
.x4 - 3x2 + 2x + 1
Hint: Do not calculate F'(x). Instead, write F(x) = f(x)/g(x) and express F'(O) directly in terms of f(O),f'(O), g(O), g'(O).
SOLUTION Taking the hint, let
f(x)
= x 9 + xs + 4.x5 - 7x
and let
g(x) = x4
Then F(x)
-
3x2 + 2x + 1
= ~g~. Now,
= 9x8 + 8x7 + 20x4 Moreover, f(O) = 0, f'(O) = -7, g(O) = 1, and g'(O) = 2.
f'(x)
7
and
g'(x)
= 4x3 -
6x + 2
Using the Quotient Rule,
F' (O) = g(O)f' (0) - f (O)g' (0) = - 7 - 0 = _
(g(0))2
1
7
142
CHAPTER 3
I
DIFFERENTIATION
d 2x
53. Use the Product Rule to calculate dx e .
SOLUTION
Note that e2x =ex·~. Therefore
d
d
-e2x
= -(ex.
ex) =ex. ex+ ex. ex= 2e2x
dx
dx
55. ~Gq
Plot f(x) = x/(x2 - 1) (in a suitably bounded viewing box). Use the plot to determine whether f'(x) is positive or negative on its domain {x: xi= ±1). Then compute f'(x) and confirm your conclusion algebraically.
SOLUTION
domain {x: x
Let f(x) = _x_. The graph of f(x) is shown below. From this plot, we see that f(x) is decreasing on its
x 2 -1
±1). Consequently, f'(x) must be negative. Using the quotient rule, we find
*
,
f (x)
=
(x2 - 1)(1) - x(2x)
(x2 _ l)2
=
x2 + 1
(x2 _ 1)2
which is negative for all x i= ± 1.
y
57. Find a> 0 such that the tangent line to the graph of
= x2e-x
f(x)
at x =a
passes through the origin (Figure 5).
y
a
FIGURE 5
SOLUTION
Let f(x) = x2e-x. Then f(a)
= a 2e-a,
f'(x) = 2xe-x +x2(-e-x)
f' (a)
= (2a -
a 2)e-a, and the equation of the tangent line to fat x
y
= f'(a)(x -
= (2x-x2)e-x
= a is
a)+ f(a) = (2a - a 2)e-a(x- a)+ a 2e-a
For this line to pass through the origin, we must have
0
Thus, a = 0 or a
therefore a = 1.
= (2a -
a 2)e-a(-a) + a 2e-a
= 1. The only value a
= e-a (a 2 -
2a 2 + a 3) = a 2e-0 (a - 1)
> 0 such that the tangent line to f(x) = x 2e-x passes through the origin is
59. The revenue per month earned by the Couture clothing chain at time tis R(t) = N(t)S (t), where N(t) is the number of
stores and S (t) is average revenue per store per month. Couture embarks on a two-part campaign: (A) to build new stores
at a rate of five stores per month, and (B) to use advertising to increase average revenue per store at a rate of $10,000 per
month. Assume that N(O) = 50 and S (0) = $150,000.
(a) Show that total revenue will increase at the rate
dR
-
dt
= SS (t) + 10,000N(t)
Note that the two terms in the Product Rule correspond to the separate effects of increasing the riumber of stores on the
one hand and the average revenue per store on the other.
s E c TI o N 3.3 I Product and Quotient Rules
(b) Calculate (idR
143
I.
t t=O
(c) If Couture can implement only one leg (A or B) of its expansion at t = 0, which choice will grow revenue most
rapidly?
SOLUTION
(a) Given R(t) = N(t)S (t), it follows that
dR
- = N'(t)S (t) + N(t)S'(t)
dt
We are told that N' (t)
= 5 stores per month and S '(t) = 10,000 dollars per month. Therefore,
dR
dt = 5S (t) + 10,000N(t)
(b) Using part (a) and the given values of N(O) and S(O), we find
ddRI
t t=O
= 5(150,000) + 10.000(50) = 1,250,000
(c) From part (b), we see that of the two terms contributing to total revenue growth, the term SS (0) is larger than the
term 1O,OOON(O). Thus, if only one leg of the campaign can be implemented, it should be part A: increase the number of
stores by 5 per month.
61. The curve y = l/(.x2 + 1) is called the witch ofAgnesi (Figure 6) afterthe Italian mathematician Maria Agnesi (17181799). This strange name is the result of a mistranslation of the Italian word la versiera, meaning "that which turns."
Find equations of the tangent lines at x = ± 1.
-
~
• x
FIGURE 6 The witch of Agnesi.
SOLUTION
Let f(x) = _I_. Then f'(x) = (x2 + 1)(0)- 1(2x) =
x2 + 1
(x2 + 1)2
2x
(x2 + 1)2.
• At x = -1, the tangent line is
y = f'(-I)(x+ 1) + f(-1) = .!_(x+ I)+.!_= .!_x+ 1
2
2
2
• At x
= 1, the tangent line is
y = f'(l)(x - 1) + f(l) = _.!_(x - 1) + .!_ = _.!_x + 1
2
2
2
63. Use the Product Rule to show that (f2 )' = 2ff'.
SOLUTION
Let g =
J2 =ff. Then g' = (f2)' =(ff)' = f' f
+ff' = 2ff'.
Further Insights and Challenges
65. Let f, g, h be differentiable functions. Show that (fgh)'(x) is equal to
f'(x)g(x)h(x) + f(x)g'(x)h(x) + f(x)g(x)h'(x)
Hint: Write f gh as f(gh).
SOLUTION
Let p
= f gh. Then
p' = (fgh)' =f'gh+ f(g'h+gh') = f'gh+ fg'h+ fgh'
67. Derivative of the Reciprocal Use the limit definition to prove
~(-1 )-_f'(x)
dx f(x) -
J2(x)
Hint: Show that the difference quotient for 1/ f(x) is equal to
f(x) - f(x + h)
hf(x)f(x + h)
144
CH APT E R 3
I
DIFFERENTIATION
SOLUTION
Let g(x) =
f
1
x).
We then compute the derivative of g(x) using the difference quotient:
, x = lim g(x + h)- g(x)
g ( ) h-.o
h
= lim ~
h->O
(--1- __1_)
h f(x + h)
f(x)
= lim ~ (f(x)- f(x + h))
h-.o h
f(x)f(x + h)
--lim(f(x+h)-f(x))(
1
)
h-.O
h
f(x)f(x + h)
We can apply the rule of products for limits. The first parenthetical expression is the difference quotient definition of
f'(x). The second can be evaluated at h = 0 to give (f(~))2 • Hence
,
g (x)
d ( I )
f(x)
= dx
f'(x)
= - f1-(x)
69. Use the limit definition of the derivative to prove the following special case of the Product Rule:
!:_(xf(x))
dx
SOLUTION
= f(x) + xf'(x)
First note that because f(x) is differentiable, it is also continuous. It follows that
limf(x + h) = f(x)
h->0
Now we tackle the derivative:
!:_(xf(x)) = lim (x + h)f(x + h) - xf(x) = Iim(xf(x + h)- f(x) + f(x + h))
dx
h->O
h
h->0
h
.
= x lIm
h->O
f(x + h) - f(x)
h)
. f(
+ 1Im x+
h
h->0
= xf'(x)
+ f(x)
71. The Power Rule Revisited If you are familiar with proof by induction, use induction to prove the Power Rule for
all whole numbers n. Show that the Power Rule holds for n = 1; then write x" as x · x"- 1 and use the Product Rule.
SOLUTION
Let k be a positive integer. If k = I, then X' = x. Note that
d
d
dx (xi)= dx (x) = I = lxo
Hence the Power Rule holds fork= 1. Assume it holds fork= n where n
~
2. Then fork= n + 1, we have
!:..._ (:Y!) = !:..._ (x"+ 1) = !:..._ (x · x") = x!:_ (x") + x" !:..._ (x)
dx
dx
dx
1
= x · nx"- + x" · I
dx
= (n + l)x" = k:Y!-
dx
1
Accordingly, the Power Rule holds for all positive integers by induction.
Exercises 72 and 73: A basic fact of algebra states that c is a root of a polynomial f if and only if f(x) = (x - c)g(x)for
some polynomial g. We say that c is a multiple root if f(x) = (x - c)2h(x), where his a polynomial.
73. Use Exercise 72 to determine whether c
(a) x5 + 2x4 - 4.x3 - 8x2 - x + 2
(b) x4
= -1 is a multiple root.
+x3 -5x2-3x+ 2
SOLUTION
(a) To show that -1 is a multiple root of
f(x)
= x5 + 2x4 - 4x3 -
8.x2 - x + 2
it suffices to check that/( -1) = f' (-1) = 0. We have f (-I) = - I + 2 + 4 - 8 + 1 + 2 = O and
f'(x)
= 5x4 + 8x3 -
12.x2 - 16x- 1
f'(-1) = 5 - 8- 12 + 16- 1=0
(b) Let f(x)
=x
4
+ x3 - 5.x2 - 3x + 2. Then f' (x) = 4.x3 + 3x2 - lOx - 3. Because
f(-l)= 1-1-5+3+2=0
but
f'(-l)
it follows that x
= -4 + 3 + IO - 3 =6 :;e 0
= - l is a root off, but not a multiple root.
s E c TI o N
3.4
I
Rates of Change
145
3.4 Rates of Change
Preliminary Questions
1. Which units might be used for each rate of change?
(a) Pressure (in atmospheres) in a water tank with respect to depth
(b) The rate of a chemical reaction (change in concentration with respect to time with concentration in moles per liter)
SOLUTION
(a) The rate of change of pressure with respect to depth might be measured in atmospheres/meter.
(b) The reaction rate of a chemical reaction might be measured in moles/(liter·hour).
2. Two trains travel from New Orleans to Memphis in 4 h. The first train travels at a constant velocity of 90 mph, but the
velocity of the second train varies. What was the second train's average velocity during the trip?
SOLUTION
Since both trains travel the same distance in the same amount of time, they have the same average velocity:
90mph.
3. Discuss how it is possible to be speeding up with a velocity that is decreasing.
If the velocity of an object in motion is negative and decreasing, then the velocity is becoming more negative.
This means that the magnitude of the velocity is increasing. But the magnitude of the velocity is the speed of the object,
so the speed is increasing. Thus, if the velocity of an object in motion is negative and decreasing, then the object is
speeding up.
SOLUTION
4. Sketch the graph of a function that has an average rate of change equal to zero over the interval [O, l] but has instantaneous rates of change at 0 and 1 that are positive.
Consider the graph of y = f(x) shown in the figure below. Because f(O) = f(l), the average rate of change
off over the interval [0, l] is zero. On the other hand, the tangent lines to the graph off at x = 0 and x = 1 are positive,
so the instantaneous rates of change at 0 and 1 are positive.
SOLUTION
y
Exercises
In Exercises 1-8,.find the rate of change.
1. Area of a square with respect to its side s when s
=5
2
SOLUTION Let the area be A= f(s) = s • Then the rate of change of A with respect to sis d/ds(s 2 ) = 2s. Whens= 5,
the area changes at a rate of 10 square units per unit increase. (Draw a 5 x 5 square on graph paper and trace the area
added by increasing each side length by 1, excluding the corner, to see what this means.)
Vx with respect to x when x = 1, 8, 27
Let f(x) = fi. Writing f(x) = x 1l 3 , we see the rate of change of f(x)
3. Cube root
SOLUTION
f'(x) = ~x- 2 1 3 • The requested rates of change are given in the table that follows;
c
ROC of f(x) with respect to x at x = c.
1
t(l) = t
f'C8) = tc8-213) = KD = -&.
8
27
with respect to x is given by
f'(l) =
f'(27) = 1(27-2/3)
= l( l) = l_
3
3 9
27
5. The diameter of a circle with respect to radius
SOLUTION The relationship between the diameter d of a circle and its radius r is d = 2r. The rate of change of the
diameter with respect to the radius is then d' = 2.
146
CH APT E R 3
I
DIFFERENTIATION
7. Volume V of a cylinder with respect to radius if the height is equal to the radius
SOLUTION
The volume of the cylinder is V
= 7rr2 h = 7rr3 • Thus dV/dr = 37rr2 .
In Exercises 9-11, refer to Figure I l, the graph of distance sfrom the origin as afunction of time for a car trip.
9. Find the average velocity over each interval.
(b) [0.5, l]
(c) [l, 1.5]
(a) [0, 0.5]
(d) [1, 2]
SOLUTION
(a) The average velocity over the interval (0, 0.5] is
50-0
- - = 100 km/hour
0.5-0
(b) The average velocity over the interval [0.5, 1] is
lOO - 50
1-0.5
= 100 km/hour
(c) The average velocity over the interval [l, 1.5] is
1OO - lOO = 0 km/hour
1.5 - 1
(d) The average velocity over the interval [1, 2] is
50 - lOO
2-1
= -50 km/hour
11. Match the descriptions (i)-(iii) with the intervals (a)-(c) in Figure 11.
(i) Velocity increasing
(ii) Velocity decreasing
(iii) Velocity negative
(a) [0, 0.5]
(b) [2.5, 3]
(c) [1.5, 2]
Distance (km)
50
0.5
1.0
l.5
2.0
2.5
3.0
FIGURE 11 Distance from the origin versus time for a car trip.
SOLUTION
(a) (i) The distance curve is increasing, and is also bending upward, so that distance is increasing at an increasing rate.
(b) (ii) Over the interval [2.5, 3], the distance curve is flattening, showing that the car is slowing down; that is, the
velocity is decreasing.
(c) (iii) The distance curve is decreasing, so the tangent line has negative slope; this means the velocity is negative.
Exercises 12 and 13 refer to the data in Example 1. Approximate the derivative with the symmetric difference quotient
.
T(t + 20) - T(t - 20)
(SDQ) approximation: T'(t)""
.
40
13. At what t does the SDQ approximation give the smallest (i.e., closest to 0) rate of change of temperature? What is
the rate of change?
SOLUTION From the values in the table in the solution to Exercise 12, the SDQ approximation gives the smallest (i.e.,
closest to 0) rate of change at t = 180 minutes and at t = 200 minutes. The rate of change at both of these times is
approximately-0.10 °C/min.
Exercises 14-16, refer to the four graphs of s as a function oft in Figure 7.
15. Match each situation with the graph that best represents it.
(a) Rocky slowed down his car as it approached the moose in the road. The distance from the car to the moose is s and
the time since he spotted the moose is t.
S E CT I O N 3.4
I
Rates of Change
147
(b) The rocket's speed increased after liftoff until the fuel was used up. The distance from the rocket to the launchpad is
s and the time since liftoff is t.
(c) The increase in college costs slowed for the fourth year in a row. The cost of college is sand the time since the start
of the 4-year period is t.
SOLUTION
(a) Because the car is getting closer to the moose, the graph of s should be decreasing. Moreover, because the car is
slowing down, the slope of the graph of s should be getting closer to zero. This matches the graph in Figure 7(C).
(b) Because the rocket is getting farther from the launchpad, the graph of s should be increasing. Moreover, because the
speed of the rocket is increasing, the slope of the graph of s should be increasing. This matches the graph in Figure 7(A).
(c) Because college costs are increasing, the graph of s should be increasip.g. Moreover, because the rate of increase in
college costs is slowing down, the slope of the graph of s should be decreasing. This matches the graph in Figure 7(B).
17. Sketch a graph of velocity as a function of time for the shuttle train in Example 6.
SOLUTION Over the interval [0, 2], the velocity is positive and increasing, while over the interval [2, 4], the velocity
is positive and decreasing. Over the interval [4, 6], the velocity is zero. Finally, over the interval [6, 8], the velocity is
negative and decreasing, while over the interval [8, 10], the velocity is negative and increasing. A possible graph of the
velocity as a function of time is shown in the figure below.
v
2
4
6
19. Fred X has to make a book delivery from his warehouse, 15 mi north of the city, to the Amazing Book Store 10 mi
south of the city. Traffic is usually congested within 5 mi of the city. He leaves at noon, traveling due south through the
city, and arrives the store at 12:50. After 15 min at the store, he makes the return trip north to his warehouse, arriving at
2:00. Let s represent the distance from the warehouse in miles and t represent time in minutes since noon. Make sketches
of the graphs of sand s' as functions oft for Fred's trip.
SOLUTION
Possible graphs of sand s' for Fred's trip are shown below.
s'
30
20
120
10
40
40
80
80
120
21. The velocity (in centimeters per second) of blood molecules flowing through a capillary of radius 0.008 cm is
v = 6.4 x 10-s - 0.00lr2 , where r is the distance from the molecule to the center of the capillary. Find the rate of change
of velocity with respect to r when r = 0.004 cm.
SOLUTION The rate of change of the velocity of the blood molecules is v'(r)
is -8 x 10-6 1/s.
= -0.002r. When r = 0.004 cm, this rate
23. Use Figure 14 to estimate dT /dh at h = 30 and 70, where Tis atmospheric temperature (in degrees Celsius) and his
altitude (in kilometers). Where is dT /dh equal to zero?
T(•C)
250
200
150
100
50
-50
-100
'--+10'-+-+-<1-5-t'O-+-+-<l-1+--1100-+-+-+-1-l+-50-+-+-h (km)
FIGURE 14 Atmospheric temperature versus altitude.
148
CH APT E R 3
I
DIFFERENTIATION
At h = 30 km, the graph of atmospheric temperature appears to be linear passing through the points
(23, -50) and (40, 0). The slope of this segment of the graph is then
SOLUTION
0 - (-50)
40-23
= 50 = 2.94
17
so
dT
dh
I
~ 2.94°C/km
h=30
At h = 70 km, the graph of atmospheric temperature appears to be linear passing through the points (58, 0) and (88, -100).
The slope of this segment of the graph is then
_-_IO_O_-_O
88 - 58
= _-_IO_O = _3 33
30
so
dT
dh
I
~ -3.33°C/km
h=?O
!!Ji; = Oat those points where the tangent line on the graph is horizontal. This appears to happen over the interval [13, 23],
and near the points h = 50 and h = 90.
25. Calculate the rate of change of escape velocity Vesc = (2.82 x 107 )r- 112 mis with respect to distance r from the center
of the earth.
SOLUTION
The rate that escape velocity changes is v~sc(r) = -1.41 x I07r- 3i 2 .
= t4 -
18t2 mat time t seconds. At which times does the particle
pass through the origin? At which times is the particle instantaneously motionless (i.e., it has zero velocity)?
27. A particle moving along a line has position s(t)
The particle passes through the origin when s(t) = t4 - 18t2 = t2(t2 - 18) = 0. This happens when t = 0
seconds and when t = 3 '\/2 ~ 4.24 seconds. With s(t) = t4 - 18t2, it follows that v(t) = s'(t) = 4t3 - 36t = 4t(t2 - 9). The
particle is therefore instantaneously motionless when t = 0 seconds and when t = 3 seconds.
SOLUTION
29. A projectile is launched in the air from the ground with an initial velocity v0 = 60 mis. What is the maximum height
that the projectile reaches? (Compare your result with Exercise 37 in Section 2.5, where we considered maximum height
when air resistance is included and we investigated the result of letting the air resistance go to 0.)
With the launch taking place from the ground, so that so
height and velocity of the projectile are given by
SOLUTION
s(t)
= 60t -
4.9t2
and
v(t)
= 0, and an initial velocity of v0 = 60 mis, the
= 60 -
9.8t
Maximum height is attained when the velocity is zero, so we solve
60 - 9.8t = 0
60
9.8
to obtain
t=-
The maximum height reached by the projectile is then
2
s ( 60)
. = 60 (60)
. - 4.9 (60)
.
~ 183.67 m
98
98
98
This value is in agreement with the estimate obtained in Exercise 37 from Section 2.5.
31. A ball tossed in the air vertically from ground level returns to earth 4 s later. Find the initial velocity and maximum
height of the ball.
=
Galileo's formula gives s(t) =so+ Vot- ~gt2 v0t-4.9t2, where the time tis in seconds (s) and the height
sis in meters (m). When the ball hits the ground after 4 seconds its height is 0. Solve O = s(4) = 4v0 - 4.9(4)2 to obtain
v0 = 19.6 m/s. The ball reaches its maximum height when s'(t) = 0, that is, when 19.6 - 9.8t = 0, or t = 2 s. At this
time, t = 2 s,
SOLUTION
s(2)
= 0 + 19.6(2) -
1
2
(9.8)(4)
= 19.6 m
SECTION 3.4
I
RatesofChange
149
33. Show that for an object falling according to Galileo's formula, the average velocity over any time interval [t 1, t2 ] is
equal to the average of the instantaneous velocities at t 1 and t2 •
SOLUTION
over [t 1 , t2 ] is
The simplest way to proceed is to compute both values and show that they are equal. The average velocity
(so+ vot2 - tg~) - (so+ Vot1 -
s(t2) - s(t 1)
~-~
tgFi)
vo(/2 -
~-~
=
Vo(/2 - t1)
12 -11
11)
+ !U2 2 - !1 2)
~-~
g
g
- -(!2 + t1) =Vo - -(!2+11)
2
2
whereas the average of the instantaneous velocities at the beginning and end of [t 1, t2 ] is
The two quantities are the same.
35. By Faraday's Law, if a conducting wire of length t meters moves at velocity v mis perpendicular to a magnetic field
of strength B (in teslas), a voltage of size V = -Btv is induced in the wire. Assume that B = 2 and t = 0.5.
(a) Calculate dV/dv.
(b) Find the rate of change of V with respect to time t if v(t) = 4t + 9.
SOLUTION
(a) Assuming that B
= 2 and l = 0.5, V = -2(0.5)v = -v. Therefore,
dV = _ 1
dv
(b) If v
= 4t + 9, then V = -2(0.5)(4t + 9) = -(4t + 9). Therefore, !lJi- = -4.
37. LS;] Ethan finds that with h hours of tutoring, he is able to answer correctly S (h) percent of the problems on a
math exam. Which would you expect to be larger: S'(3) or S'(30)? Explain.
The derivative S'(h) measures the rate at which the percent of problems Ethan answers correctly changes
with respect to the number of hours of tutoring he receives.
One possible graph of S(h) is shown in the figure below on the left. This graph indicates that in the early hours of
working with the tutor, Ethan makes rapid progress in learning the material but eventually approaches either the limit of
his ability to learn the material or the maximum possible score on the exam. In this scenario, S'(3) would be larger than
S'(30).
An alternative graph of S (h) is shown below on the right. Here, in the early hours of working with the tutor little
progress is made (perhaps the tutor is assessing how much Ethan already knows, his learning style, his personality, etc.).
This is followed by a period of rapid improvement and finally a leveling off as Ethan reaches his maximum score. In this
scenario, S'(3) and S'(30) might be roughly equal.
SOLUTION
I
J
Hours of tutoring
Hours of tutoring
39. To determine drug dosages, doctors estimate a person's body surface area (BSA) (in meters squared) using the
formula BSA = .../hm/60, where h is the height in centimeters and m the mass in kilograms. Calculate the rate of change
of BSA with respect to mass for a person of constant height h = 180. What is this rate at m = 70 and m = 80? Express
your result in the correct units. Does BSA increase more rapidly with respect to mass at lower or higher body mass?
SOLUTION Assuming constant height h = 180 cm, let f(m) = .../hm/60
in terms of weight. The rate of change of BSA with respect to mass is
f'(m)
= 'ft m be the formula for body surface area
= V5 (!m-112) = V5
10
2
20ym
If m = 70 kg, this is
f (70) = -Y5
I
V14 , , 0.0133631m
= -280
20 Y76
kg
2
150
CHAPTER 3
I
DIFFERENTIATION
Ifm
= 80 kg,
,
VS
1
1 m2
1 <8o) = 2ovso = 20Yf6 = 80 kg
Because the rate of change of BSA depends on 1/
41. The tangent lines to the graph of f(x) =
lines increase?
ym, it is clear that BSA increases more rapidly at lower body mass.
x2 grow steeper as x increases. At what rate do the slopes of the tangent
Letf(x) = x2. The slopes s of the tangent lines are given bys= f'(x) = 2x. The rate at which these slopes
are increasing is ds/dx = 2.
SOLUTION
43. The dollar cost of producing x bagels is given by the function C(x) = 300 + 0.25x - 0.5(x/1000) 3 . Determine the
cost of producing 2000 bagels and estimate the cost of the 200lst bagel. Compare your estimate with the actual cost of
the 2001 st bagel.
Expanding the power of 3 yields
SOLUTION
C(x) = 300 + 0.25x - 5 x 10- 10x3
This allows us to get the derivative C'(x) = 0.25 - 1.5 x 10-9 x2. The cost of producing 2000 bagels is
C(2000) = 300 + 0.25(2000)- 0.5(2000/1000)3 = 796
dollars. The cost of the 200lst bagel is, by definition, C(2001) - C(2000). By the derivative estimate, C(2001) C(2000) ,,, C'(2000)(1), so the cost of the 200lst bagel is approximately
C'(2000) = .25 - 1.5 x 10-9 (20002 ) = $0.244
C(2001) = 796.244, so the exact cost of the 2001st bagel is indistinguishable from the estimated cost. The function is
very nearly linear at this point.
45. ~ According to Stevens's Law in psychology, the perceived magnitude of a stimulus is proportional (approximately) to a power of the actual intensity I of the stimulus. Experiments show that the perceived brightness B of a light
satisfies B = kf21 3 , where I is the light intensity, whereas the perceived heaviness Hof a weight W satisfies H = kW312
(k is a constant that is different in the two cases). Compute dB/di and dH/dW and state whether they are increasing or
decreasing functions. Then explain the following statements:
(a) An increase in light intensity is felt more strongly when I is small than when I is large.
(b) An increase in load Wis felt more strongly when Wis large than when Wis small.
SOLUTION
- 2k
(a) dB/di -
31-1/3 --
2k
1'
3 13
As I increases, dB/di shrinks, so that the rate of change of perceived intensity decreases as the actual intensity
increases. Increased light intensity has a diminished return in perceived intensity. A sketch of B against I is shown: See
that the height of the graph increases more slowly as you move to the right.
y
¥
(b) dH I dW = w112 • As W increases, dH I dW increases as well, so that the rate of change of perceived weight increases
as weight increases. A sketch of H against Wis shown: See that the graph becomes steeper as you move to the right.
y
SECTION 3.4
I
RatesofChange
151
Further Insights and Challenges
Exercises 47-49: The Lorenz curve y = F(r) is used by economists to study income distribution in a given country (see
Figure 15). By definition, F(r) is the fraction of the total income that goes to the bottom rth part of the population, where
0 ::;; r ::;; I. For example, if F(0.4) =0.245, then the bottom 40% of households receive 24.5% of the total income. Note
that F(O) =0 and F(I) =I.
F(r)
1.0
0.8
0.6
0.4
0.2
0.2
0.4
0.6
0.8
1.0
(A) Lorenz curve for the United States
in 2010
F(r)
0.2
0.4
0.6
0.8
1.0
(B) Two Lorenz curves: The tangent
lines at P and Q have slope l.
FIGURE 15
47. ~ Our goal is to find an interpretation for F'(r). The average income for a group of households is the total
income going to the group divided by the number of households in the group. The national average income is A = T /N,
where N is the total number of households and T is the total income earned by the entire population.
(a) Show that the average income among households in the bottom rth part is equal to (F(r)/r)A.
(b) Show more generally that the average income of households belonging to an interval [r, r + l'lr] is equal to
( F(r + ~~ -F(r))A
(c) Let 0 ::;; r ::;; I. A household belongs to the I OOrth percentile if its income is greater than or equal to the income of
IOOr % of all households. Pass to the limit as l'lr --t 0 in (b) to derive the following interpretation: A household in the
IOOrth percentile has income F'(r)A. In particular, a household in the IOOrth percentile receives more than the national
average if F'(r) > I and less if F'(r) < 1.
(d) For the Lorenz curves L 1 and Li in Figure 15(B), what percentage of households have above-average income?
SOLUTION
(a) The total income among households in the bottom rth part is F(r)T and there are rN households in this part of the
population. Thus, the average income among households in the bottom rth part is equal to
F(r)T
rN
= F(r) . !.._ = F(r) A
r
N
r
(b) Consider the interval [r,r + l'lr]. The total income among households between the bottom rth part and the bottom
r + l'lrth part is F(r + l'lr)T - F(r)T. Moreover, the number of households covered by this interval is (r + l'lr)N - rN =
l'lrN. Thus, the average income of households belonging to an interval [r, r + l'lr] is equal to
F(r + l'lr)T - F(r)T
MN
=
F(r + l'lr) - F(r) T
F(r + l'lr) - F(r)
l'lr
· N=
l'lr
A
152
CHAPTER 3
I
DIFFERENTIATION
(c) Take the result from part (b) and let flr---) 0. Because
. F(r + flr) - F(r) F'( )
hm
=
r
flr
t.r~o
we find that a household in the lOOrth percentile has income F'(r)A.
(d) The point Pin Figure 15(B) has an r-coordinate of0.6, while the point Q has an r-coordinate of roughly 0.75. Thus,
on curve L 1 , 40% of households have F' (r) > 1 and therefore have above-average income. On curve Lz, roughly 25% of
households have above-average income.
49. Use Exercise 47(c) to prove:
(a) F'(r) is an increasing function of r.
(b) Income is distributed equally (all households have the same income) if and only if F(r)
= r for 0 $
r $ 1.
SOLUTION
(a) Recall from Exercise 47 (c) that F'(r)A is the income of a household in the lOOrth percentile. Suppose 0 $ r1 < rz $
1. Because r 2 > r 1, a household in the 100r2 th percentile must have income at least as large as a household in the 1OOr1 th
percentile. Thus, F'(r2 )A;:::: F'(r 1)A, or F'(r2 );:::: F'(r 1). This implies F'(r) is an increasing function of r.
(b) If F(r) = r for O::;; r::;; 1, then F'(r) = I and households in all percentiles have income equal to the national average;
that is, income is distributed equally. Alternately, if income is distributed equally (all households have the same income),
then F'(r) = I for 0 ::;; r ::;; 1. Thus, F must be a linear function in r with slope l. Moreover, the condition F(O) == 0
requires the F intercept of the line to be 0. Hence, F(r) = I· r + 0 == r.
In Exercises 50 and 51, the average cost per unit at production level xis defined as Cavg(x) = C(x)/ x, where C(x) is the
cost of producing x units. Average cost is a measure of the efficiency of the production process.
51. Show that Cavg(x) is equal to the slope of the line through the origin and the point (x, C(x)) on the graph of y = C(x).
Using this interpretation, determine whether average cost or marginal cost is greater at points A, B, C, D in Figure 17.
c
Production level
FIGURE 17 Graph of y = C(x).
SOLUTION
By definition, the slope of the line through the origin and (x, C(x)), that is, between (0, 0) and (x, C(x)) is
C(x) - 0 == C(x) == Cav
x-0
x
At point A, average cost is greater than marginal cost, as the line from the origin to A is steeper than the curve at this
point (we see this because the line, tracing from the origin, crosses the curve from below). At point B, the average cost is
still greater than the marginal cost. At the point C, the average cost and the marginal cost are nearly the same, since the
tangent line and the line from the origin are nearly the same. The line from the origin to D crosses the cost curve from
above, and so is less steep than the tangent line to the curve at D; the average cost at this point is less than the marginal
cost.
3.5 Higher Derivatives
Preliminary Questions
1. For each headline, rephrase as a statement about first and second derivatives and sketch a possible graph.
• "Stocks Go Higher, Though the Pace of Their Gains Slows"
• "Recent Rains Slow Roland Reservoir Water Level Drop"
• ''Asteroid Approaching Earth at Rapidly Increasing Rate!!"
SECTI
oN
3.5
I
Higher Derivatives
153
SOLUTION
• Because stocks are going higher, stock prices are increasing and the first derivative of stock prices must therefore
be positive. On the other hand, because the pace of gains is slowing, the second derivative of stock prices must be
negative.
Stock price
'--------Time
• Because the water level is dropping, the water level is decreasing and the first derivative of the water level must
therefore be negative. On the other hand, because the drop in water level is being slowed by the recent rains (that
is, the rate of decrease in the water level is getting closer to zero), the second derivative of the water level must be
positive.
Water level
Time
• Because the asteroid is approaching Earth, the distance between the asteroid and Earth is decreasing and the first
derivative of the distance must therefore be negative. On the other hand, because the rate at which the asteroid is
approaching Earth is increasing (that is, the rate of decrease in the distance is becoming more negative), the second
derivative of the water level must also be negative.
Distance
Time
2. Sketch a graph of position as a function of time for an object that is slowing down and has positive acceleration.
Because the object is slowing down but has positive acceleration, the velocity of the object must be negative.
Thus, the graph of position as a function of time should be decreasing and bending upward.
SOLUTION
3. Sketch a graph of position as a function of time for an object that is speeding up and has negative acceleration.
Because the object is speeding up but has negative acceleration, the velocity of the object must be negative.
Thus, the graph of position as a function of time should be decreasing and bending downward.
SOLUTION
154
CH APT E R 3
I
DIFFERENTIATION
4. True or false? The third derivative of position with respect to time is zero for an object falling to Earth under the
influence of gravity. Explain.
This statement is true. The acceleration of an object falling to earth under the influence of gravity is constant;
hence, the second derivative of position with respect to time is constant. Because the third derivative is just the derivative
of the second derivative and the derivative of a constant is zero, it follows that the third derivative is zero.
SOLUTION
5. Which type of polynomial satisfies f'"(x) = 0 for all x?
SOLUTION The third derivative of all quadratic polynomials (polynomials of the form ax2 +bx+ c for some constants
a, band c) is equal to 0 for all x.
6. What is the millionth derivative of f(x) =ex?
Every derivative of f(x) =ex is ex.
SOLUTION
7. What are the seventh and eighth derivatives of f(x) = x 7 ?
Let f(x) = x 7. Then
SOLUTION
f'(x)
= 7x6 ,
J"(x)
= 42x5,
J"'(x)
= 2l0x4, f 4l(x) = 840.x3, f
5l(x)
= 2520x2,
and
f
6
l(x)
= 5040x
Finally,
f
7
l(x)
= 5040
and
f8l(x)
=0
Exercises
In Exercises I-I 6, calculate y" and y"'.
1. y
= 14x2
Let y
SOLUTION
= 14x2. Then y' = 28x, y" = 28, and y"' = 0.
3. y = x 4 - 25x2 + 2x
SOLUTION
5. y =
4
7rr
= x4 -
Let y
= ~JTr
25x2 + 2x. Then y'
= 4.x3 -
50x + 2, y"
= 12x2 -
50, and y"'
= 24x.
3
3
SOLUTION
7. y
Let y
= 20t4/5 -
SOLUTION
3
. Then y'
= 4JTr2 , y" = 8JTr, and y"' = 8JT.
6t213
Let y
= 20t415 -
Let y
= z - 4z- 1. Then y' = I + 4z-2, y" = -sz-3, and y"' = 24z-4.
6t2 13. Then y'
= 16r 115 -
4r 113 , y"
= -lf-r615 + ~r4!3, and y"' = ~runs -11r7f3.
4
9. y=z- -
z
SOLUTION
11. y = 02(28 + 7)
Let y
SOLUTION
= 82 (28 + 7) = 2B3 + 7e2. Then y' = 6e2 + 148, y" = 12e + 14, and y"' = 12.
x-4
13. y= - x
SOLUTION
15. y
Let y
= x~ = 1 4
4x-I. Then y'
= 4x-2 , y" = -Sx-3, and y'" = 24x-4.
= x5ex
SOLUTION
Lety = x5ex. Then
y'
y"
y"'
= 5x4ex + x5 ex = (x5 + 5x4)ex
= (5x4 + 20x3)ex + (x5 + 5x4)ex = (x5 + l0x4 + 20x3)ex
= (5x4 + 40.x3 + 60x2)ex + (x5 + l0x4 + 20.x3V = (x5 + l5x4 + 60x3 + 60x2)ex
In Exercises I7-26, calculate the derivative indicated.
17.
f
4
l(l),
SOLUTION
f(x)
= x4
Letf(x)
= x4. Then f'(x) = 4x3, f"(x) = 12x2, f'"(x) = 24x, and J<4l(x) = 24. Thus J<4l(l) = 24.
S E c T I o N 3.5
19.
d2d ;
t
I
Higher Derivatives
155
I , y = 4,- + 3t2
3
t=l
Let y = 4r3 + 3t2. Then
SOLUTION
~
= -12r4 + 6t and
-d2yl
dt2
21. d4xl
,
dt4 1=16
d4xl
dt4 1=16
23. f"'(-3),
f(x) = 4ex -
25. h"(l),
h(w) =
SOLUTION
= 48(1) -5 + 6 = 54
t=l
2
d x
dt2
= n,-11/4
16
,
3
d x
dt3
= _231,-15/4
and d4x
= 3465,-19/4
Thus
64
,
dt4
256
.
= 3465 16_1914 =
3465
256
134217728
x3
Let f(x) = 4ex -
SOLUTION
= 48r5 + 6. Hence
x = ,-314
Letx(t) = ,-314 • Then~=
_'J.,-7/4
dt
4
,
SOLUTI ON
~~?
x3. Then f'(x)
= 4~ - 3x2, f" (x) = 4~ - 6x, f"' (x) = 4ex - 6, and f"'(-3) = 4e-3 - 6.
-../Wew
Let h(w) = ywew = w 112 ew. Then
and
Thus, h"(l) =
7
4e.
27. Calculate y<kl(O) for 0
SOLUTION
:<::;
k
:<::;
5, where y = x4 + ax3 + bx2 +ex+ d (with a, b, e, d the constants).
Applying the power, constant multiple, and sum rules at each stage, we get (note y<0J is y by convention):
k
y(k)
0
x 4 + ax3 + bx2 + ex+ d
1
4x3 + 3ax2 + 2bx + e
2
12x2 + 6ax + 2b
3
24x+ 6a
4
24
5
0
from which we get y<0l(O) = d, y(ll(O) = e, y<2l(O) = 2b, y<3l(O) = 6a, y<4l(O) = 24, and y(5l(O) = 0.
29. UsetheresultinExample3tofind
~x- 1 •
SOLUTION With the launch taking place from the ground, so that s 0 = 0, and an initial velocity of v0 = 60 mis, the
height and velocity of the projectile are given by
s(t) = 60t - 4.9t2
and
v(t) = 60 - 9.8t.
Maximum height is attained when the velocity is zero, so we solve
60 - 9.8t = 0
60
9.8
to obtain
t=-.
The maximum height reached by the projectile is then
2
s ( 60)
_ = 60 (60)
_ - 4.9 (60)
_ "" 183.67 m.
98
98
98
This value is in agreement with the estimate obtained in Exercise 37 from Section 2.5.
156
CHAPTER 3
I
DIFFERENTIATION
In Exercises 31-36,find a general formula for .fnl(x).
31. f(x)
= x- 2
SOLUTION f'(x) = -2x- 3 , f"(x) = 6x-4, f'"(x) = -24x-5 , j<4l(x) = 5 · 24x-6 , ..•. From this we can conclude that the
nth derivative can be written as J<nl(x) = (-Ir(n + l)tx-<n+2i.
33. f(x)
= x- 112
SOLUTION
f'(x)
= =f x- 312 • We will avoid simplifying numerators and denominators to find the pattern:
f"(x)
-3-1
3Xl
= --x-512
= (-l)2--x-5/2
2
f"'(x)
= -~ 3 x
2 2
2 22
2
1 x_712
=
(-1) 3 5
x 3 x 1 x_712
23
.fnl(x) = (-It (2n - 1) X (2n - 3) X ... X 1 x-<2n+l)/2
2n
35. f(x) = xe-x
SOLUTION
Letf(x) = xe-x. Then
= e-x + x(-e-x) = (1 - x)e-x = -(x - I)e-x
f"(x) = -e-x + (1 - x)(-e-x) = (x - 2)e-x
f"' (x) = e-x + (x - 2)(-e-x) = (3 - x)e-x = -(x - 3)e-x
f'(x)
From this we conclude that the nth derivative can be written as J<nl(x)
37. (a) Find the acceleration at time t
= (-l)n(x -
n)e-x.
= 5 min of a helicopter whose height is s(t) = 300t -
4t3 m.
(b) Plot the accelerations" for 0::;; t::;; 6. How does this graph show that the helicopter is slowing down during this time
interval?
SOLUTION
(a) Let s(t) = 300t - 4t3, with tin minutes ands in meters. The velocity is v(t) = s'(t)
a(t) = s"(t) = -24t. Thus a(5) = -120 m/min2 .
= 300 -
12t2 and acceleration is
(b) The acceleration of the helicopter for 0 ::;; t ::;; 6 is shown in the figure below. As the acceleration of the helicopter is
negative, the velocity of the helicopter must be decreasing. The velocity, v(t) = 300 - 12t2 = 12(25 - t2), is positive for
0::;; t < 5 and negative fort> 5. Thus, the helicopter is slowing down for 0::;; t < 5 and speeding up for 5 ::;; t::;; 6.
y
-20
-40
--00
-80
-100
-120
-140
39. Figure 6 shows f, f', and f". Determine which is which.
(A)
(B)
(C)
FIGURE 6
SOLUTION
(a) f"
(b)
f'
(c)
f
The tangent line to (c) is horizontal at x = 1 and x
x = 2 and x = 0, where (a) has roots.
= 3, where (b) has roots. The tangent line to (b) is horizontal at
s Ec T Io N
3.5
I
Higher Derivatives
157
41. Figure 8 shows the graph of the position s of an object as a function of time t. Determine the intervals on which the
acceleration is positive.
s
20
10
30
Time
40
FIGURE 8
SOLUTION Roughly from time 10 to time 20 and from time 30 to time 40. The acceleration is positive over the same
intervals over which the graph is bending upward.
43. Find all values of n such that y
= x" satisfies
i2y" - 2xy'
SOLUTION
Lety
= x". Theny' = nx"- 1,y" = n(n- l)x"-2 , and
i2y" - 2xy' = n(n - l)x" -
In order for this last expression to be equal to 4y
n(n - 1) - 2n
Thus, y
= 4y
= x" satisfies the equation i2y" -
=4
2xy'
2nx"
= 4x", we must have
or
n2
-
3n - 4
= (n -
4)(n + 1)
=0
= 4y for n = 4 and n = -1.
45. According to one model that takes into account air resistance, the acceleration a(t) (in m/s 2 ) of a skydiver of mass m
in free-fall satisfies
a(t) = -9.8
k
+ -v(t) 2
m
=
=
where v(t) is velocity (negative since the object is falling) and k is a constant. Suppose that m 75 kg and k 0.24 kg/m.
(a) What is the skydiver's velocity when a(t) = -4.9?
(b) What is the skydiver's velocity when a(t) = O? (This velocity is the terminal velocity, the velocity attained when air
resistance balances gravity and the skydiver falls at a constant speed.)
Solving a(t)
skydiver is falling, we find
SOLUTION
= -9.8 + ~v(t)
v(t)
=-
2
m
k(a(t)
for the velocity and taking into account that the velocity is negative since the
+ 9.8) = -
75
_ (a(t) + 9.8) = - ..,/312.5(a(t) + 9.8)
0 24
(a) Substituting a(t) = -4.9 into the above formula for the velocity, we find
v(t)
(b) When a(t)
= -..,/312.5(4.9) = --v'1531.25 ~ -39.13 m/s
= 0,
v(t) = - ..,/312.5(9.8)
= - -v'3062.5 ~ -55.34 m/s
47. In a manufacturing process, a drill press automatically drills a hole into a sheet metal part on a conveyor. In the
drilling operation the drill bit starts at rest directly above the part, descends quickly, drills a hole, and quickly returns to
the start position. The maximum vertical speed of the drill bit is 4 in.ls, and while drilling the hole, it must move no more
than 2.6 in.ls to avoid warping the metal. Let s(t) be the drill bit's height (in inches) above the part as a function of time
tin seconds. Sketch possible graphs of the drill bit's velocity [s'(t)] and acceleration [s"(t)].
SOLUTION
Possible graphs of the drill bit's velocity [s'(t)] and acceleration [s"(t)] are shown below.
s'
4
-4
s"
158
CHAPTER 3
I
DIFFERENTIATION
49.
[AS
2
Let f(x) = x + . Use a computer algebra system to compute the .fk>(x) for 1 :5 k :5 4. Can you find a
x-1
general formula for fk>(x)?
SOLUTION
Let f(x)
2 u·
x+-.
=smg a computer alb
ge ra system,
x-1
f
3
1
_
I
3•l
(x) = - (x-1)2 - (-1) (x-1)1+1
6
3. 2·1
f"(x)= (x-1)3 =(-1)2(x-l)2+1
18
f"'(x)
= - (x-1)4 = (-1)
3
3·3!
(x-1)3+1 and
4
3 . 4!
(x - 1)5 - ( 1) (x - 1)4+1
f 4) (x) --
___}_}:____ -
From the pattern observed above, we conjecture
fk>(x)=(-l)k
3 ·k'
.
(X - l)k+I
Further Insights and Challenges
51. What is p<99>(x) for p(x) as in Exercise 50?
SOLUTION
First note that for any integer n :5 98,
cf9
dx99x"
=0
Now, if we expand p(x), we find
p(x) =
x99 +
terms of degree at most 98
therefore,
cf9
cf9
cf9
99
dx99 p(x) = dx99 (x + terms of degree at most 98) = dx99 x 99
Using logic similar to that used to compute the derivative in Example (3), we compute:
cf9 99
dx99 (x )
= 99 x 98 x ... 1
so that :{;9 p(x) = 99 !.
53. Use the Product Rule to find a formula for (Jg)'" and compare your result with the expansion of (a+ b) 3 • Then try
to guess the general formula for (fg)<nl.
SOLUTION
Continuing from Exercise 52, we have
h"'
= !"' g + !" g' + 2(/" g' + f' g") + f' g" + f g"' = !"' g + 3/" g' + 3f' g" + f g"'
The binomial theorem gives
and more generally
where the binomial coefficients are given by
(n)k = k(k -
1) · · · (k -
n+ 1)
n!
Accordingly, the general formula for Cfgin> is given by
(f gr>=
I
k=O
where p<k> is the kth derivative of p (or p itself when k
= 0).
(n)fn-k>g<k>
k
S E c T I O N 3.6
I
Trigonometric Functions
159
3.6 Trigonometric Functions
Preliminary Questions
1. Determine the sign (+ or - ) that yields the correct formula for the following:
.
()
a -d(.
smx+cosx) = ±smx±cosx
dx
d
(b) -secx=±secxtanx
dx
d
(c) - cotx = ±csc2 x
dx
SOLUTION
The correct formulas are
()
a -d(.
smx+cosx)
dx
.
= -smx+cosx
d
secx = secxtanx
dx
(b) -
d
(c) - cotx = -csc2 x
dx
2. Which of the following functions can be differentiated using the rules we have covered so far?
(a) y = 3 cos x cot x
(b) y = cos(x2)
(c) y = ex sin x
SOLUTION
(a) 3 cos x cot x is a product of functions whose derivatives are known. This function can therefore be differentiated
using the Product Rule.
(b) cos(x2) is a composition of the functions cos x and x2. We have not yet discussed how to differentiate composite
functions.
(c) ex sin xis a product of functions whose derivatives are known. This function can therefore be differentiated using the
Product Rule.
3. For each, give an equation of the tangent line to the graph at x = 0.
(a) y = sinx
(b) y =
COSX
SOLUTION
(a) Let f(x) = sin x. Then f' (x) = cos x, and an equation of the tangent line is
y
= f' (O)(x -
0) + /(0)
= l(x) + 0 = x
(b) Let f(x) =cos x. Then f'(x) = - sinx, and an equation of the tangent line is
0) + f(O) = O(x) + 1 = 1
4. How is the addition formula for sine used in deriving the formula (sin x)' = cos x?
y
= f'(O)(x -
The difference quotient for the function sinx involves the expression sin(x + h). The addition formula for
the sine function is used to expand this expression as sin(x + h) = sin xcos h +sin h cos x.
SOLUTION
Exercises
In Exercises 1-4,find an equation of the tangent line at the point indicated.
1. y = sin x,
SOLUTION
x=
~
Let f(x) = sin x. Then f' (x) = cos x and the equation of the tangent line is
y=f'(~)(x-~)+t(i)= ~(x-~)+ ~ = ~x+ ~(1-~)
3. y = tan x,
SOLUTION
x=~
Let f(x) = tanx. Then f'(x) = sec 2 x and the equation of the tangent line is
160
CHAPTER 3
I
DIFFERENTIATION
In Exercises 5-24, compute the derivative.
5. f(x)
= sinxcosx
SOLUTION
Letf(x)
= sinxcosx. Then
2
f'(x) = cosx(cosx) + sinx(-sinx) = cos 2 x-sin x
7. f(x) = xsinx
SOLUTION
Let f(x) = x sin x. Then
f'(x) = sinx + xcosx
9. H(t)
= sin t sec2 t
SOLUTION
= sint sec2 t. Then
Let H(t)
H' (t) = cost sec2 t +sin t!!._(sec t · sect)
dt
= sect + sin t(sec t tan t sect + sect sect tan t)
= sec t + 2 sin t sec2 ttan t
11. f(B)
= tanBsece
SOLUTION
= tan 8 sec 8. Then
Let f(B)
f'(B)
13. f(x)
= (2x4 -
SOLUTION
= sec2 Bsec e + tanesec Btane = sec3 e +sec 8tan2 B = (tan2 B + sec2 e) sece
4x- 1) sec x
= (2x4 -4x- 1)secx. Then
Letf(x)
f'(x)
15. y
= (8x3 + 4x-2 ) secx + (2x4 -
4[ 1) secxtanx
= sece
e
SOLUTION
sece
-e-· Then
Lety =
,
Y =
Bsec Btane- sec e
e2
17_ R(y) = _3c_o_~.:....y_-_4
smy
SOLUTION
Let R(y) =
R'(y)
3cosy-4
.
. Then
smy
= siny(-3siny)-(3cosy-4)(cosy) = 4cosy-3(sin2 y+cos2 y) = 4cosy-3
sin 2 y
sin2 y
sin 2 y
19• f(x) = 1 + tanx
1 - tanx
SOLUTION
l+tanx h
Letf(x) = 1
. T en
-tanx
(1- tanx) sec2
f'(x)=
x- (1 + tanx) (-sec2 x)
(1 - tan x)
2
2sec2
(1 - tan x)2
21. f(x) = ex sin x
SOLUTION
Let f(x)
x
=---
= ex sin x. Then f'(x) = ex sin x +ex cos x = ex(sin x +cos x).
S E CT I O N 3.6
I
Trigonometric Functions
161
23. f(B) = e8 (5 sin(} - 4 tan B)
SOLUTION
Letf(B) = e8 (5sin8-4tan8). Then
f' (B) = e8 (5 sin(} - 4 tan(})+ e8 (5 cos(} - 4 sec2 B)
= e8 (5 sin(}+ 5 cos(} - 4 tan(} - 4 sec2 B)
In Exercises 25-34,find an equation of the tangent line at the point specified.
x3 + cos x, x = 0
SOLUTION Let f(x) = x3 +cos x. Then f'(x) = 3x2 - sinx and /'(0) = 0. The tangent line at x = 0 is
25. y =
y = f' (O)(x - 0) + f (0) = O(x) + 1 = 1
sint
27. y= - - - ,
1 +cost
SOLUTION
~
t=
Let f(t) = 1 :~'::sr. Then
,
(1 +cos t)(cos t) - sin t(- sin t)
1 +cost
f(t)=
=
=-( 1 + cos t) 2
(1 + cos t) 2
1 + cos t
and
The tangent line at x =
~
is
Y
29. y = 2(sin (} + cos B),
SOLUTION
2
7f
9
!!
(} - 3
Let f(B) = 2(sin (}+cos B). Then f'(B) = 2(cos (} - sin B) and f'<V = 1 -
31. y = ex cos x,
SOLUTION
=f'(~)(x~)+
t{~)
= ~ (x- ~)+ -../3 = ~x+ -../3 3
3
3
3
3
3
3
3
-../3. The tangent line at x = ~is
x=0
Let f(x) = ex cos x. Then
f'(x) =
~ cosx
+ ~(-sinx) =
~(cosx
- sinx)
0
and f'(O) = e (cos 0 - sinO) = 1. Thus, the equation of the tangent line is
y
33. y = e'(l - cost),
SOLUTION
= f' (O)(x -
0) + f(O)
=x + 1
t= ~
Let f(t) = e'(l - cost). Then
f'(t) = e'(l - cost)+ e' sin t = e'(l + sint - cost)
and f'<D = 2e"' 2 • The tangent line at x = ~ is
y
= f' ( 3:) (t - 3:) + f ( 3:) = 2e"'2 (t - 3:) + e"/2
In Exercises 35-37, use Theorem I to verify the formula.
d
35. dx cotx =-csc2 x
SOLUTION
cosx
.
cotx = -.-.
Usmg the quotient rule and the derivative formulas , we compute:
~x
d
d cos x
sin x(- sin x) - cos x(cos x)
-(sin 2 x + cos 2 x)
-1
-cotx=--.-=
·2
=
.
=--=-csc2x
dx
dx sm x
sm x
sm2 x
sin2 x
162
CH APTER 3
I
DIFFERENTIATION
d
37. - cscx = -cscxcotx
dx
Since csc x
SOLUTION
= -.-1-, we can apply the quotient rule and the two known derivatives to get:
smx
d
d 1
-cscx= - - . dx
dx sm x
-cosx
cosx 1
= sinx(O)-2 l(cosx) = --= --.--.= -cotxcscx
2
sin x
sin x
sm x sm x
In Exercises 39-42, calculate the higher derivative.
39. f"(B),
f(8)=8sin8
SOLUTION
Let f(8) =()sin(). Then
f'(B)
= sin()+ ()cos 8
f" (8) = cos()+ cos 8 + 8(- sin 8) = 2 cos 8 41.y",
y"',
SOLUTION
8 sin()
y=tanx
Let y =tan x. Then y' = sec2 x and by the Chain Rule,
d
2
2
y''==-sec x=2(secx)(secxtanx)=2sec xtanx
dx
y"' = 2 sec 2 x(sec2 x) + (2 sec2 xtan x) tanx
= 2 sec4 +4 sec4 xtan2 x
43. Calculate the first five derivatives of f(x) = cos x. Then determine f 8>(x) and f
SOLUTION
37
>(x).
Let f(x) = cos x.
• Then f'(x) = - sin x, f" (x) = - cos x, f"' (x) = sin x, f 4>(x) = cos x, and f 5l(x) = - sin x.
• Accordingly, the successive derivatives off cycle among
{- sinx, -cos x, sinx, cosx)
in that order. Since 8 is a multiple of 4, we have j<8l(x) = j<4 >(x) = cos x.
• Since 36 is a multiple of 4, we have J<36l(x) = j<4>(x) = cos x. Therefore, j<37l(x)
= - sin x.
45. Let f(x) = sin x. We can compute J<n>(x) as follows: First, express n = 4m + r where mis a whole number and r
1, 2, or 3. Then determine J<n>(x) from r. Explain how to do the latter step.
SOLUTION
Let f(x) = sin x. If n = 4m + r where mis a whole number and r
= 0,
= 0, 1, 2, or 3, then J<n>(x) = J<'>(x). In
particular:
•
•
•
•
If r = 0, then j<n>(x)
Ifr = 1, thenJ<nl(x)
If r = 2, then J<n>(x)
If r = 3, then J<n>(x)
=
=
=
=
j<0>(x) = f(x) = sin x.
j<O(x) = cosx.
j<2>(x) = - sinx.
j<3>(x) =-cos x.
47. Let f(x) = sin2 x and g(x) = cos 2 x.
(a) Use an identity and prove f'(x) = -g'(x) without directly computing f'(x) and g'(x).
(b) Now verify the result in (a) by directly computing f'(x) and g'(x).
SOLUTION
2
(a) Using the identity sin x + cos 2 x = 1, we have f(x) + g(x) = 1. Taking the derivative on both sides of this equation
yields f'(x) + g'(x) = 0, or f'(x) = -g'(x).
(b) Working directly, first write f (x) = sin x sin x and g(x) = cos x cos x. Then, by the Product Rule,
f'(x) = cosxsinx + sinxcosx = 2 sinxcos x, and
g'(x) = - sinxcos x +cos x(-sinx) = -2 sinxcos x.
Thus, f'(x) = -g'(x).
49. Find the values of x between 0 and 2lr where the tangent line to the graph of y = sin x cos xis horizontal.
SOLUTION
Lety = sinxcosx. Then
y' = (cosx)(cosx) + (sinx)(-sinx) = cos2 x- sin 2 x
When y' = 0, we have sin x = ±cos x. In the interval [0, 211"], this occurs when x = ~, ~,
¥, 'If.
s EcT Io N
I
3.6
Trigonometric Functions
163
51. !Guij Let g(t) = t - sin t.
(a) Plot the graph of g with a graphing utility for 0 :=:; t :=:; 41f.
(b) Show that the slope of the tangent line is nonnegative. Verify this on your graph.
(c) For which values oft in the given range is the tangent line horizontal?
Let g(t) = t - sin t.
(a) Here is a graph of g over the interval [O, 47r].
SOLUTION
y
2
4
6
8
10 12
(b) Since g'(t) = 1 - cost~ 0 for all t, the slope of the tangent line tog is always nonnegative. We see that the graph is
never decreasing; therefore, the slope of the tangent is never negative.
(c) In the interval [0, 47r], the tangent line is horizontal when t = 0, 27r, 41f.
53.
LSJ
Show that no tangent line to the graph of f(x)
line? Justify by sketching the graph of f'(x) = (tanx)'.
= tanx has zero slope. What is the least slope of a tangent
+
Let f(x) = tanx. Then f'(x) = sec2 x = -cos12-x. Note that f'(x) = cos x has numerator 1; the equation
j'(x) = Otherefore has no solution. Because the maximum value of cos2 xis 1, the minimum value of f'(x) = co:2 xis 1.
Hence, the least slope for a tangent line to tan x is 1. Here is a graph of/'.
SOLUTION
y
14
12
10
8
6
4
-4
x
-2
4
2
55. A projectile is launched from ground level with an initial velocity Vo at an angle e, where 0 :=:; e :=:; 7r/2. Its horizontal
rangeisR = (2vii/g)sinBcose, whereg = 32 ft/s2 with v0 in ft/s, andg = 9.8 mls2 with v0 in mis. Calculate dR/dB. The
maximum range occurs where dR/dB = 0. Show that that occurs ate= 7r/4 and that the maximum range is viif g.
SOLUTION
LetR(B) = (2v~/g)sinBcose. Then
dR
2v2
de =
<cosecose +sin B(- sin B))
-f
2v2
= -f <cos2 e -
sin2 B)
Setting dR/dB = 0 yields the equation cos2 e = sin2 e. With the restriction O :=:; e :=:; 7r/2, this latter equation is equivalent
to cos e = sine, which has e = 1f / 4 as its solution. When e = 1fI 4,
R
(~) =
4
2
v6
g
sin~ cos ~
4
4
=
2
v6 . -./2
g
2
. -./2
2
=
vii
g
Further Insights and Challenges
57. Use the limit definition of the derivative and the addition law for the cosine function to prove that (cos x)' = - sin x.
SOLUTION
Let f(x)
!
= cos x. Then
'()
x
. cos(x+h)-cosx
. cosxcosh-sinxsinh-cosx
= 1h-->O
1m
= h-->O
hm - - - - - - - - - - h
h
. (
.
sinh
cosh-1)
= l~ (-smx) -h- + (cosx)
h
= (-sinx) · 1 + (cosx) · O = -sinx
SECTION 3.7
I
The Chain Rule
165
SOLUTION
(a) The outer function is
.../i, and the inner function is 4x + 9x2.
(b) The outer function is tan x, and the inner function is x2 + 1.
(c) The outer function is x5, and the inner function is sec x.
(d) The outer function is x 4 , and the inner function is 1 + ~.
2. Which of the following can be differentiated without using the Chain Rule?
x
(a) y =tan(7.x2 + 2)
(b) Y = x+ 1
(c) y = -{X · secx
(d) y = x-Vsecx
(e) y = xex
The function x~ 1 can be differentiated using the Quotient Rule, and the functions -{X · sec x and xex can be
differentiated using the Product Rule. The functions tan(7.x2 + 2), x ysec x, and sin( ex) require the Chain Rule.
SOLUTION
3. Which is the derivative of f(5x)?
SOLUTION
(c) f'(5x)
(b) 5f'(5x)
(a) 5f'(x)
The correct answer is (b): 5f'(5x).
4. Suppose that f'(4) = g(4) = g'(4) = l. Do we have enough information to compute F'(4), where F(x) = f(g(x))? If
not, what is missing?
If F(x) = f(g(x)), then F'(x) = f'(g(x))g'(x) and F'(4) = f'(g(4))g'(4). Thus, we do not have enough
information to compute F'(4). We are missing the value of f'(l).
SOLUTION
Exercises
In Exercises I-4,fill in a table of the following type:
f(g(x))
1. f(u)
= u312 ,
g(x)
f'(u)
f'(g(x))
g'(x)
(fa g)'(x)
= x4 + 1
SOLUTION
3. f(u)
= tan u,
g(x)
f(g(x))
f'(u)
f'(g(x))
g'(x)
(f 0 g)'
(x4 + l)3/2
~u112
~(x4
4x3
6x3(x4 + 1)1/2
+ l)112
= x4
SOLUTION
f(g(x))
f'(u)
f'(g(x))
g'(x)
(f
tan(x4)
sec2 u
sec2(x4 )
4x3
4x3 sec2(x4 )
0
g)'
In Exercises 5 and 6, write the function as a composite f(g(x)) and compute the derivative using the Chain Rule.
5. y = (x + sinx)4
SOLUTION
Let f(x) = x 4 , g(x) = x + sinx, andy = f(g(x)) = (x + sinx)4. Then
dy
dx
= f' (g(x))g (x) = 4(x +sin x)3 (1 +cos x)
I
d
7. Calculate dx cos u for the following choices of u(x):
(a) u(x) = 9 -
x2
(b) u(x)
=x- 1
(c) u(x)
=tanx
SOLUTION
(a) cos(u(x))
= cos(9 -
.x2).
d
dx cos(u(x)) = - sin(u(x))u'(x) = - sin(9 - x2)(-2x) = 2x sin(9 - .x2)
166
CHAPTER 3
I
DIFFERENTIATION
(b) cos(u(x)) = cos(x- 1).
d
- cos(u(x))
dx
1
(
.
,
.
1 ) sin(x- )
= - sm(u(x))u
(x) = - sm(x- 1) -2 = --2x
x
(c) cos(u(x)) = cos(tan x) .
.!!_ cos(u(x)) = -sin(u(x))u'(x) = -sin(tanx)(sec2 x) = -sec2 xsin(tanx)
dx
df. df
9. Compute dx if du
du
= 2 and dx = 6.
Assuming f is a function of u, which is in tum a function of x,
SOLUTION
df
dx
= df . du = 2(6) = 12
du
dx
11. Let f(x) = (2x2 - 5) 2 . Compute f'(x) three different ways: 1. Multiplying out and then differentiating, 2. using the
Product Rule, and 3. using the Chain Rule. Show that the results coincide.
Letf(x) = (2x2 - 5)2.
SOLUTION
• Multiplying out yields f(x)
= 4x4 -
20x2 + 25. Then f'(x)
= 16.x3 -
40x
= 8x(2x2 -
5).
• To use the Product Rule, write f (x) = (2x2 - 5)(2x2 - 5). Then,
J'(x)
= 4x(2x2 -
5) + (2x2 - 5)(4x)
= 8x(2x2 -
5)
• Using the Chain Rule,
f'(x)
= 2(2x2 -
d
5)-(2x2 - 5)
dx
= 2(2x2 -
5)(4x)
= 8x(2x2 -
5)
In Exercises 13-24, compute the derivative using derivative rules that have been introduced so far.
13. y = (x4 + 5)3
Using the General Power Rule,
SOLUTION
d
x
4
-d (x + 5)
3
d
4
= 3(x4 + 5)2 -(x
+ 5) = 3(x4 + 5)2 (4.x3) = 12.x3(x4 + 5)2
dx
15. y = 1/7x - 3
Using the General Power Rule
SOLUTION
.!!_ -V7x- 3 = .!!_(7x dx
11. y
dx
3) 1' 2
= ~(7x- 3)- 1' 2 (7) = __7__
2
21/7x-3
= (x2 + 9xr2
SOLUTION
Using the General Power Rule,
d
-d (x2 + 9x)-2
x
d
= -2(x2 + 9x)-3 -(x2
+ 9x) = -2(x2 + 9x)-3(2x + 9)
dx
19. y = cos4 ()
SOLUTION
Using the General Power Rule,
d
4
d() cos ()
d
= 4 cos3 () d()
cos 8 = -4 cos 3 ()sin ()
SECTION 3.7
I
The Chain Rule
21. y = (2 cos 8 + 5 sin 8) 9
SOLUTION
Using the General Power Rule,
!!;_(2cos 8 + 5 sin 8) 9 = 9(2 cos 8 + 5 sin8) 8 dd (2cos 8 + 5 sinB) = 9(2cos B + 5 sin8) 8 (5 cos B - 2 sinB)
dB
B
23. y =
ex-12
SOLUTION
Using the General Exponential Rule,
In Exercises 25-28, compute the derivative off o g.
25. f(u) =sin u,
SOLUTION
g(x) = 2x + 1
Let h(x) = f(g(x)) = sin(2x + 1). Then,
:xf(g(x)) = f'(g(x))g'(x) = cos(2x + 1) · 2 = 2 cos(2x + 1)
27. f(u)
= e",
SOLUTION
g(x)
= x + x- 1
Let h(x) = f(g(x)) = ex+x-'. Then
= f'(g(x))g'(x) = ex+x_, {1 -
:xf(g(x))
x- 2 )
In Exercises 29 and 30,find the derivatives of f(g(x)) and g(j(u)).
29. f(u) = cos u,
g(x) =
x2 + 1
SOLUTION
d
dJ(g(x)) = f'(g(x))g'(x) = -sin(x2+1)(2x) = -2xsin(x2 + 1)
d
dxg(j(x)) = g'(j(x))f'(x) = 2(cosx)(-sinx) = -2sinxcosx
In Exercises 31-44, use the Chain Rule to find the derivative.
31. y = sin(x2)
SOLUTION
33. y
Lety = sin{x2). Then y' =cos (x2) · 2x = 2xcos{x2).
= Yt2 + 9
SOLUTION
Lety =
Yt2 + 9 = (t2 + 9) 112 • Then
y'
1
t
= -(t2
+ 9)- 112 (2t) = - 2
2
vr
+9
35. y = (x4 - x3 - 1)2/3
SOLUTION
37. y =
Let y = (x 4
x3 - 1)2' . Then
3
-
(x+
1)4
x-l
4
SOLUTION
Lety =
1
) . Then
(x+
x-1
y'= 4 (x+1)
x-1
3
(x-1)·1-(x+l)·l
(x-1)2
8(x+ 1)3
8(1 + x) 3
(x-1)5
(1- x)5
167
168
CH APTER 3
I
DIFFERENTIATION
1
39. y =sec x
SOLUTION
Let f(x) =sec {x- 1). Then
f'(x) =sec (x
-l)
( _) ( _)
sec(l/x)tan(l/x)
tan x 1 · -x 2 = x2
41. y=tan(8+cos8)
SOLUTION
Let y = tan (8 +cos 8). Then
y' = sec2 (8 +cos 8) · (1 - sin 8) = (1 - sin 8) sec2 (8 +cos 8)
43. y = e1- 9t2
SOLUTION
Let y = e1- 9t2. Then
= e1- 9t2 (- l 8t) = - l 8te2- 9t2
y'
Jn Exercises 45-74,jind the derivative using the appropriate rule or combination of rules.
45. y
= tan(x2 + 4x)
SOLUTION
Lety = tan(x2 + 4x). By the Chain Rule,
y' = sec 2 (x2 + 4x) · (2x + 4) = (2x + 4) sec2 (x2 + 4x)
47. y = xcos(l -3x)
SOLUTION
Let y = xcos (1 - 3x). Applying the Product Rule and then the Chain Rule,
y' = cos(l -3x) +x(-sin(l - 3x)) · (-3) = cos(l -3x) + 3xsin(l -3x)
49. y
= (4t + 9) 112
SOLUTION
Lety = (4t + 9) 112 • By the General Power Rule,
dy
dt
51. y =
= 4(~)
(4t +
2
9f
112
= 2(4t +
9f
112
(x3 + cos x)-4
SOLUTION
Let y = (x3 + cos x)- 4 • By the General Power Rule,
y' = -4(.x3 +cos xf5(3x2 - sin.x) = 4(sinx - 3x2)(.x3 +cos x)-5
53. y = Ysinxcosx
SOLUTION
We start by using a trig identity to rewrite
y = Ysinxcosx =
~~ sin2x = ~ (sin2x) f
12
Then, after two applications of the Chain Rule,
y
I
1 .
1/2
COS 2x
= - 1 · -(sm2xf
·cos2x·2 = - - -
yz
2
Y2sin2x
55. y = (cos6x + sin.x2) 112
SOLUTION
Let y = (cos 6x + sin(x2)) 1f 2. Applying the General Power Rule followed by two applications of the Chain
Rule,
-;:=======-
. 1 -112
•
xcos(x2)-3sin6x
, 1(
2
y = 2 cos6x+sm(x-)) (-sm6x·6+cos(x-)·2x)=
--/cos 6x + sin(x2 )
SECTION 3.7
57. y
I
TheChainRule
169
=tan3 x + tan(x3)
Let y = tan3 x + tan(x3) = (tanx) 3 + tan(x3). Applying the General Power Rule to the first term and the
Chain Rule to the second term,
SOLUTION
y'
= 3(tanx)2 sec2 x + sec2 (x3) · 3x2 = 3(x2 sec2 (x3) + sec2 xtan2 x)
{Z+f
59. y
= " "i=1
SOLUTION
Let y =
z+
(
1/2
1
z-1 )
. Applying the General Power Rule followed by the Quotient Rule,
dy = .!_ (z + 1 )dz
2 z- 1
61
•
y
112
.
(z - 1) · 1 - (z + 1) · 1
(z - 1)2
=
-1
Yz+f. (z -
.
1)312
= cos(l + x)
1 +cosx
SOLUTION
Let
cos(l + x)
y= l+cosx
Then, applying the Quotient Rule and the Chain Rule,
dy
dx
-(1 +cos x) sin(l + x) + cos(l + x) sin x
(l +cos x)2
cos(l + x) sin x - cos x sin(l + x) - sin(l + x)
(1 +cos x)2
sin(-1) - sin(l + x)
(1 + cosx)2
The last line follows from the identity
sin(A- B)
with A
63. y
= sinAcosB-cosAsinB
= x and B = 1 + x.
= cot7 (x5)
SOLUTION
Let y = cot7 ( x5). Applying the General Power Rule followed by the Chain Rule,
:~ = 7 cot6 (x5) · (-csc2 (x5)) · 5x4 = -35x4 cot6 (x5) csc2 (x5)
65. y = (1+cot5<x4 +1))9
SOLUTION
Let y
= (1 + cot5 (x4 + 1))9. Applying the General Power Rule three times, we get
:~ = 9 ( 1 + cot5 (x4 + 1))
4
= -180x3 cot
67. Y
8
4
• 5 cot
(x4 + 1) · (-csc2 (x4 + 1)).4x3
(x4 + 1)csc2 (x4 +1)(1 + cot5 (x4 + 1))8
= (2e3x + 3e-2x)4
SOLUTION Let y = (2e 3x + 3e-2x)4 • Applying the General Power Rule followed by two applications of the General
Exponential Rule, we find
69. y
= e<x2+2x+3)2
SOLUTION
Let y = e<x2+2x+ 3>2. By the General Exponential Rule and the General Power Rule, we obtain
dy
dx
= e<x2+2x+ 3>2 • 2(x2 + 2x + 3)(2x + 2) = 4(x + l)(x2 + 2x + 3)e<x2+2x+ 3>2
170
CH APT E R 3 . I
DIFFERENTIATION
71. y
= ~ 1 + ~ 1 + Vx
SOLUTION
73. y
Let y = ( 1 + ( 1 + x 112 )
= (kx + b)- 113 ;
SOLUTION
1/2)1/2
•
Applying the General Power Rule twice,
k and b any constants
Let y = (kx + b)- 113 , where band k are constants. By the General Power Rule,
y'
= _! (kx + b)-413 · k = -~(kx + b)-413
3
3
In Exercises 75-78, compute the higher derivative.
75.
f2x2 sin(x2)
SOLUTION
Let f(x)
= sin (x2). Then, by the Chain Rule, f'(x) = 2xcos (x2) and, by the Product Rule and the Chain
Rule,
f"(x)
= 2cos (x2) + 2x(- sin (x2) · 2x) = 2 cos (x2)- 4x2 sin (x2)
d3
8
77. -(9-x)
dx3
SOLUTION
Let f(x)
= (9 -
x) 8 . Then, by repeated use of the General Power Rule,
f'(x) = 8(9 -x) 7 · (-1) = -8(9-x) 7
f"(x)
f"'(x)
= -56(9 - x) 6 • (-1) = 56(9 - x) 6
= 336(9- x) 5 · (-1) = -336(9-x)5
79. Assume that the average molecular velocity v of a gas in a particular container is given by v(T) = 29 Vf mis, where
Tis the temperature in kelvins. The temperature is related to the pressure (in atmospheres) by T = 200P. Find dv
dP
SOLUTION
First note that when P
dv
dP
Alternately, substituting T
I
P=I.5
=
I .
P=l.5
= 1.5 atmospheres, T = 200(1.5) = 300K. Thus,
dv
dT
I . dT I
T=300
dP
= ~ . 200 = 290 Y3
2 V3QO
P=l.5
3
m
s ·atmospheres
= 200P into the equation for v gives v = 290 -..fiP. Therefore,
dv
dP
290-../2
290
= 2 VP· = Y2P
so
dv
dP
I
290
P=l.5
=
290.../3
m
= - 3 - s · atmospheres
Y3
81. An expanding sphere has radius r = 0.4t cm at time t (in seconds). Let V be the sphere's volume. Find dV/dt when·
(a) r = 3 and (b) t = 3.
SOLUTION
Let r
= 0.4t, where tis in seconds (s) and r is in centimeters (cm). With V = ~rrr
dV
= 4rrr2
dr
-
Thus
dV
dt
= dV dr = 4irr2. (0.4) = l.6rrr2
dr dt
3
, we have
SECTION 3.7
I
TheChainRule
171
dV
(a) When r = 3, - = l.6n(3)2 ,,, 45.24 emfs.
dt
dV
(b) When t = 3, we haver= 1.2. Hence - = l.6n(l.2) 2 ,,, 7.24 emfs.
dt
83. The function L(t) = 12 + 3.4 sin( 32:s t) models the length of a day from sunrise to sunset in Sapporo, Japan, where tis
the day in the year after the spring equinox on March 21. Determine L'(t), and use it to calculate the rate that the length
of the days are changing on December 1, January 1, and February 1. Discuss what the results say about the changing day
length in the late fall and winter in Sapporo.
SOLUTION Let L(t) = 12 + 3.4 sin( 32:s t) model the length of a day from sunrise to sunset in Sapporo, Japan, where tis
the day in the year after the spring equinox on March 21. Then
,
( 2n ) 2n
6.8n
( 2n )
L (t) = 3.4 cos 365 t . 365 = 365 cos 365 t
December 1 corresponds to t = 255, so the rate of change in the length of the day on December 1 is
L'(255) = 6.811" cos (5101!") ,,, -0.0186 hours/day
365
365
Next, January 1 corresponds to t = 286, so the rate of change in the length of the day on January 1 is
L I (286) = 6.81f cos (5721f) ,,, 0.0123 hours/day
365
Finally, February 1 corresponds to t
365
= 317, so the rate of change in the length of the day on February 1 is
L I (317) = 6.81!" cos ( 6341!") ,,, 0.0387 hours/day
365
365
The length of the day is decreasing in late fall in Sapporo but then is increasing once winter begins. As the winter
progresses, the rate at which the length of the day is increasing is itself increasing.
85. The general shape of the graph of S(t) = Ate-kt1 for A,k > 0 is shown in Figure 2(B). There are two points on the
graph where the tangent line is horizontal and therefore where S'(t) = 0. Determine tin terms of k for these points by
solving S'(t) = 0.
SOLUTION
Let S (t) = Ate-kt1. Then,
Thus, S'(t) = 0 when t =
±Ju·
87. A 1999 study by Starkey and Scamecchia developed the following model for the average weight (in kilograms) at
age t (in years) of channel catfish in the Lower Yellowstone River (Figure 4):
W(t)
= (3.46293 -
Find the rate at which average weight is changing at age t
3.32173e-0·03456')3·4026
= 10.
8
7
6
5
4
3
2
10
15
20
t(year)
FIGURE 4 Average weight of channel catfish at age t.
SOLUTION
Let W(t)
= (3.46293 -
W' (t)
3.32173e-0.03456<)3-4026. Then
=3.4026(3.46293 -
3.32173e-0·03456') 2·4026 (-3.32173)(-0.03456)e-o.034:;6r
= 0.3906(3.46293 _ 3.32173e-0.03456r)2-4026e-o.034561
At age t
= 10,
W'(lO)
= 0.3906(1.1118)2·4026 (0.7078),,, 0.3566 kg/yr
172
CHAPTER 3
I
DIFFERENTIATION
89. Assume that
f(I) = 4, J'(l) = -3, g(2) = 1, g'(2) = 3
Calculate the derivatives of the following functions at x = 2:
(a) f(g(x))
(b) f(x/2)
(c) g(2g(x))
SOLUTION
(a) Let h(x) = f(g(x)). Then
h'(2) = f'(g(2))g'(2) = f'(l)g'(2) = -3(3) = -9
(b) Let h(x) = f(x/2). Then
h'(2) = ! J'(2/2) = ! f'(l) = !(-3) =
2
2
2
-~2
(c) Let h(x) = g(2g(x)). Then
h'(2) = g'(2g(2)). 2g'(2) = g'(2. 1). 2g'(2) = 2g'(2)g'(2) = 2(3) 2 = 18
91. Compute the derivative of h(sinx) at x =~·assuming that h'(0.5) = 10.
Let u =sin x and suppose that h' (0.5)
SOLUTION
=10. Then
d
- (h(u))
dx
dh du
dh
= -du -dx = -du cosx
When x = ~.we have u = 0.5. Accordingly, the derivative of h(sinx) at x = ~is lOcos (~) = 5 --./3.
Jn Exercises 93-96, use the table of values to calculate the derivative of the function at the given point.
93. f(g(x)),
x=6
SOLUTION
!!:.._ f(g(x))'
9s.
g(
vx).
SOLUTION
dx
x = 16
:xg(
x
1
4
6
f(x)
f'(x)
g(x)
g'(x)
4
5
4
5
0
7
6
4
6
3
1
I
2
= f'(g(6))g'(6) = f'(6)g'(6) = 4 X 3 = 12
=6
vx)lx=l6 = g'(4)U)o; v'I6) = (4)(~)(~) =
/6
97. The price (in dollars) of a computer component is P = 2C - l8c- 1, where C is the manufacturer's cost to produce it.
Assume that cost at time t (in years) is C = 9 + 3r 1 . Determine the rate of change of price with respect to time at t = 3.
dC
SOLUTION = -3r2. C(3) = 10 and C'(3) =-~,so we compute:
dt
dP'
,
18
,
dt t=J = 2C (3) + (C(3))2C (3) =
2
-3 +
18 ( 1)
dollars
100 -3 = -0.?27 year
99. According to the U.S. standard atmospheric model, developed by the National Oceanic and Atmospheric Administration for use in aircraft and rocket design, atmospheric temperature T (in degrees Celsius), pressure P (kPa = 1000 pascals), and altitude h (in meters) are related by these formulas (valid in the troposphere h:::; 11,000):
p = lOl. 29 T + 273.1
( 288.08 )
T = 15.04 - 0.00649h,
5.256
Use the Chain Rule to calculate dP /dh.
SOLUTION
4 256
dP
( T + 273 1 ) .
dT = 101.29(5.256) 288.08.
(
I )
288.08 = 6.29537 x 10-11 (T + 273.1)4.256
and~= -0.00649°C/m. !ijf;_ =~~.so
dP
dh = (6.29537 x 10-11 (T + 273.1)4·256 ) (-0.00649) = -4.08569 x 10- 13 (288.14 - 0.00649 h)4·256
SECTION 3.7
I
TheChainRule
173
101. In the setting of Exercise 100, calculate the rate of change of T (in K/yr) if T = 283 Kand R increases at a rate of
0.5 Js- 1m-2 per yr.
SOLUTION
By the Chain Rule,
dR _ dR
dt
dT
dT _
T3 dT
4<T dt
dt
---·--
Assuming T = 283 Kand !!ff = 0.5 Js- 1m- 2 per year, it follows that
3 dT
0.5 = 4<r(283) dt
0.5
= <r(
)
4 283 3
dT
=?
dt
""
0.0973 K/yr
103. Use the Chain Rule to express the second derivative off o gin terms of the first and second derivatives off and g.
SOLUTION
Let h(x) = f (g(x)). Then
h'(x) = f'(g(x))g'(x)
and
h"(x) = J'(g(x))g"(x) + g'(x)f"(g(x))g'(x) = f'(g(x))g"(x) + f"(g(x)) (g'(x))
2
Further Insights and Challenges
105. Show that if f, g, and hare differentiable, then
[f(g(h(x)))]'
SOLUTION
= f' (g(h(x)))g' (h(x))h' (x)
Let f, g, and h be differentiable. Let u = h(x), v = g(u), and w = f(v). Then
dw = df dv = df dg du = J'(g(h(x))g'(h(x))h'(x)
dx
dv dx
dv du dx
.
lSJ
107. (a)
Sketch a graph of any even function f and explain graphically why f' is odd.
(b) Suppose that f' is even. Is f necessarily odd? Hint: Check whether this is true for linear functions.
SOLUTION
(a) The graph of an even function is symmetric with respect to the y-axis. Accordingly, its image in the left half-plane is
a mirror reflection of that in the right half-plane through the y-axis. If at x = a ~ 0, the slope off exists and is equal to
m, then by reflection its slope at x = -a :o; 0 is -m. That is, f'(-a) = -f'(a). Note: This means that if f'(O) exists, then
it equals 0.
y
-2
-I
2
(b) Suppose that f' is even. Then f is not necessarily odd. Let f(x) = 4x + 7. Then f'(x) = 4, an even function. But f is
not odd. For example, f(2) = 15, f(-2) = -1, but f(-2) -f(2).
*
109. Prove that for all whole numbers n
~
1,
d"
dx" sinx=sin(x+
1l1f
-z)
Hint: Use the identity cos x = sin (x + ~ ).
SOLUTION
We will proceed by induction on n. For n = I, we find
1f)
.
. (x + -d smx
= cosx = sm
dx
2
as required. Now, suppose that for some positive integer k,
dk
dxk sin x = sin (x +
k7r
Z)
174
CHAPTER 3
I
DIFFERENTIATION
Then
dk+I
d
k;r
)
d:xf'+I sinx= dxsin(x+
2
( 2hr)
=cos x+
(
(k + -l)n)
=sin x+ 2
111. Chain Rule This exercise proves the Chain Rule without the special assumption made in the text. For any number
b, define a new function
F(u) =
f(u~ = ~(b)
for all u * b
(a) Show that if we define F(b) = f'(b), then Fis continuous at u = b.
(b) Take b = g(a). Show that if x *a, then for all u,
f(u) - f(g(a)) = F(u) u - g(a)
x-a
x-a
Note that both sides are zero if u = g(a).
(c) Substitute u = g(x) in Eq. (1) to obtain
f(g(x)) - f(g(a)) = F(g(x)) g(x) - g(a)
x-a
x-a
Derive the Chain Rule by computing the limit of both sides as x -> a.
SOLUTION
For any differentiable function f and any number b, define
F(u)
= f(u) -
f(b)
u-b
*
for all u b.
(a) Define F(b) = f'(b). Then
limF(u) = lim f(u)- f(b) = f'(b) = F(b)
u->b
u-+b
U - b
i.e., limF(u) = F(b). Therefore, Fis continuous at u = b.
u-+b
(b) Let g be a· differentiable function and take b = g(a). Let x be a number distinct from a. If we substitute u = g(a) into
Eq. (1), both sides evaluate to 0, so equality is satisfied. On the other hand, if u g(a), then
*
f(u) - f(g(a)) = f(u) - f(g(a)) u - g(a) = f(u) - f(b) u - g(a) = F(u) u - g(a)
x-a
u-g(a)
x-a
u-b
x-a
x-a
(c) Hence for all u, we have
f(u) - f(g(a)) = F(u) u - g(a)
x-a
x-a
Substituting u = g(x) in Eq. (1), we have
f(g(x)) - f(g(a)) = F(g(x)) g(x) - g(a)
x-a
x-a
Letting x -> a gives .
~~ f(g(x)~ = ~(g(a)) = ~~(F(g(x)) g(x~ =!(a))= F(g(a))g'(a) = F(b)g'(a) = J'(b)g'(a)
= f'(g(a))g'(a)
Therefore (fag)' (a) = f'(g(a))g'(a), which is the Chain Rule.
s EcT Io N
3.8
I
Implicit Differentiation
175
3.8 Implicit Differentiation
Preliminary Questions
. used to show dx
d sm
. y = cos Y dx
dy?·
. . ru1e is
. h d"""
1• Whic
iuerentlat10n
fx
The chain rule is used to show that siny =cosy~.
2. One of (a)-(c) is incorrect. Find and correct the mistake.
SOLUTION
(a) !!:..._ sin(y2) = 2y cos(y2)
dy
(b) !!:..._ sin(x2) = 2x cos(x2)
dx
(c) !!:..._ sin(y 2 ) = 2y cos(y2 )
dx
SOLUTION
(a) This is correct. Note that the differentiation is with respect to the variable y.
(b) This is correct. Note that the differentiation is with respect to the variable x.
(c) This is incorrect. Because the differentiation is with respect to the variable x, the chain rule is needed to obtain
!!:..._ sin(y2) = 2y cos(y2) dy
dx
dx
3. On an exam, Jason was asked to differentiate the equation
x2+2.xy+y3 =7
Find the errors in Jason's answer: 2x + 2.xy' + 3y2 = 0.
There are two mistakes in Jason's answer. First, Jason should have applied the product rule to the second
term to obtain
SOLUTION
!!:_(2.xy) = 2xddy + 2y
dx
x
Second, he should have applied the general power rule to the third term to obtain
!!:_y3 = 3y2 dy
dx
dx
4. Which of (a) or (b) is equal to
:x (x sin
t)?
dt
(a) (x cost) dx
SOLUTION
(b) (xcos t) dt +sin t
dx
Using the product rule and the chain rule we see that
d ( . )
dt
-d
xsmt = xcost+ sint
x
dx
so the correct answer is (b).
5. Determine which inverse trigonometric function g has the derivative
1
g '( x)= -2-1
x +
SOLUTION
g(x) = tan- 1 x has the derivative
g '(x)= -1x2 + 1
6. What does the following identity tell us about the derivatives of sin- 1 x and cos- 1 x?
sin- 1 x + cos- 1 x = ~
2
SOLUTION
Differentiating both sides of the identity with respect to x yields
d . I
d
I
- sm- x + - cos- x = 0
dx
dx
or
d . -1
d
l
-d sm x= --cos- x
dx
x
In other words, the derivatives of sin- 1 x and cos- 1 x are negatives of each other.
176
CH APT E R 3
I
DIFFERENTIATION
7. Assume that a is a constant and that y is implicitly a function of x. Compute the derivative with respect to x of each of
a 2 , X2, andy2.
SOLUTION
Because a is a constant,
!!:._a2 = 0
dx
On the other hand,
d
dx
-x2
= 2x
Finally, because y is implicitly a function of x,
Exercises
1. Show that if you differentiate both sides of X2 + 2y 3
= 6, the result is 2x + 6y2 ~ = 0. Then solve for dy/dx and
evaluate it at the point (2, l ).
SOLUTION
Let X2 + 2y3 = 6. Then
d
2
-(xdx
2x+
d
dx
3
+ 2y ) = -6
6y2~~ = 0
Solving for dy/dx yields
2dY
dx
= -2x
dy
dx
-2x
6y -
6y2
dy - -4 2
A t (2 , 1), dx
- 6 - -3.
In Exercises 3-10, differentiate the expression with respect to x, assuming that y is implicitly a function of x.
3. :(ly3
SOLUTION
Assuming that y is implicitly a function of x, then
!!:._
dx
(x2y3) = x2 · 3y2y' + y3 · 2x = 3x2y2y' + 2xy3
s. <r + y2)3/2
SOLUTION
Assuming that y is implicitly a function of x, then
d ((x-2+ Y2)3/2)
dx
3(
=2
x2 + y 2)1/2 (2x + 2yy') = 3 (x + yy')
..,/x2 + y2
7. x-f!Y
SOLUTION
Assuming that y is implicitly a function of x, then
(1
d
-xVY
= VY+ x -y-213y'
dx
3
9.
)=
xy'VY+ -
3i{y2
3
+
=~
y
y+l
SOLUTION
Assuming that y is implicitly a function of x, then
(-y-) =
!!:._
dx y+ 1
(y + I)y' - yy'
(y+ 1)2
y'
= (y+ 1)2
3i{y2
I
s EcT I0 N
3.8
I
Implicit Differentiation
In Exercises I 1-28, calculate the derivative with respect to x of the other variable appearing in the equation.
11. 3y3 + x2 = 5
Let 3y3 +
SOLUTION
x2
2x
= 5. Then 9y2y' + 2x = 0, and y' = - 9y2.
13• .x2y + 2x3y = x + y
Let .x2y + 2xy2 = x + y. Then
SOLUTION
.x2y' + 2xy + 2x · 2yy' + 2y2 = 1 + y'
x2y' + 4xyy' - y' = 1 - 2xy - 2y2
' 1- 2xy- 2y2
y = x 2 + 4xy - 1 ·
=1
15. x3R5
SOLUTION
17. ~ + ~
x y
SOLUTION
Let x3R 5 = 1. Then x3 · 5R.4R' + R5 • 3x2 = 0, andR' = -
3x2R5
3R
5x
"" = - - .
5x3.n·
= 2y
Let
y x
-+-=2y
x y
Then
xy' - y + y - xy' = 2y'
x2
y2
( ~x -
.::_ - 2)y'
y2
= 2'._
- ~
x2 y
y2-.x2-2xy2' y2-x2
xy2
y =~
y'
19.
y-2/3
+ x312
SOLUTION
=
=1
Let y-213 + x31 2 = 1. Then
2
3
3
2
--y-5f3y' + -x112 = 0
1
21. y + y
y(y2 - x2)
x(y2 - x2 - 2xy2)
or y'
= -9 xl/2y5/3
4
= x2 + x
SOLUTION
Let y + .!.y =
x2 + x. Then
y ' - -1y' = 2x+ 1
y2
or y'
= 2x+ 1 = (2x+ l)y2
1 -y-2
y2 - 1
23. sin(x + y) = x +cosy
SOLUTION
Let sin(x + y) = x +cosy. Then
(1 + y') cos(x + y) = 1 - y' siny
cos(x + y) + y' cos(x + y) = 1 - y' siny
(cos(x + y) + siny)y' = 1 - cos(x + y)
y
,
1 - cos(x + y)
cos(x + y) + siny
= ------'---
1n
178
CH APTER 3
I
DIFFERENTIATION
25. xeY
= 2xy + y3
SOLUTION
Let xeY
= 2xy + y3. Then xy' eY + eY = 2xy' + 2y + 3y2y', whence
eY-2y
y'=---,--2x+ 3y2 -xeY
27. ex + eY =
X-
SOLUTION
Let ex + eY = x - y. Then
y
ex + eYy' = 1 - y'
(eY + l)y' = 1 - ex
1-~
y'=eY+l
In Exercises 29-32, compute the derivative at the point indicated without using a calculator.
29. y
x=~
= sin- 1 x,
SOLUTION
Lety
= sin- 1 x. Theny' = '11-x_j....,,. and
1
4/5
y' (-3)
5 = yl -19/25
31. y
= sec 1 x,
SOLUTION
5
4
=4
Let y = sec- 1 x. Then y' = Ix! _br--o
and
vx--1
x
= -~·
y'(4)
4 v15
In Exercises 33-46,find the derivative.
33. y = sin- 1(7x)
d
1
SOLUTION dx sin- (7x)
35. y
.../1 - (7x)2
d
7
dx
.../1 - (7x)2
. -7x = ----;:..=="='
= cos- (x2)
d
-
dx
d
-1
cos- 1(x2)
-2x
= - - - . -x2 = - - Vl -.x4 dx
Vl -.x4
= xtan- 1 x
SOLUTION
39. y
1
1
SOLUTION
37. y
=
1
1
:xxtan- x=x(1: x2) + tan- x
= arcsin(ex)
SOLUTION
dd sin- 1(ex) = -~ · dd ex= -~
x
vl-e2x
x
vl-e2x
41. y= ~+sin- 1 t
SOLUTION
!.._ ( ~ + sin- 1 t) = !(1-rt 1' 2(-2t) + _ l _ = __::!_ + __l _
= __!_:!_
~~~~
2
dt
43. y = (tan- 1 x) 3
d
3(tan- 1 x)2
(<tan- 1 x) 3 ) = 3(tan- 1 x)2- tan- 1 x = -~-dx
dx
x2 + 1
1
1
1
45. y = cos- r - sec- t
d
SOLUTION
-
SOLUTION
!....(cos- 1 r
dt
1
-
sec- 1 t)
(-l )..Ji - (1/t) t2 ltl ~
1
-l
2
1
=--2
vt4-t
Alternatively, let t = sec e. Then r 1
1
ltl~
= cos e and cos- 1 r 1 d
-(cos- 1 r
dx
=
1
ltl~
sec- 1 t
1
-
-
1
ltl~
=0
=e - e =0. Consequently,
sec- 1 t)
=0
sEc T Io N
47. Use Figure 7 to prove that (cos- 1 x)'
Let y
vl -x
= siny = '11 d
- cos- 1 x
dx
49. Lety = sec- 1 x. Show that tany
and tany ~ 0 on G,n).
= cos- 1 x.
In general, 1 + tan2 y
dy
1
- = --siny
dx
1
= -------,1
sin(cos- x)
x 2 ; hence,
1
= -.---sm(cos- 1 x)
= .Yx2 -1ifx;?:1 and that tany = -Vx2 -1 if x ~ -1. Hint: tany;?: 0 on {O, D
= sec2 y,
so tany
'1x2 E
179
= cos- 1 x. Then cosy = x and
From Figure 7, we see that sin(cos- 1 x)
tany = ±
y = sec- 1 x
Implicit Differentiation
1
. dy
1 or
-smy=
dx
SOLUTION
I
= - _,,..-----.,,2 .
FIGURE 7 Right triangle withy
SOLUTION
3.8
1. Finally, if x ;?: l then y =
(n/2,n] so tany is negative.
= ± ../sec2 y -
sec- 1 x E
1. Withy
= sec- 1 x, it follows that secy = x, so
[0, n/2) so tany is positive; on the other hand, if x ~ 1 then
51. Show that x + yx- 1 = 1 and y = x - x2 define the same curve [except that (0, 0) is not a solution of the first equation]
and that implicit differentiation yields y' = yx- 1 - x and y' = 1 - 2x. Explain why these formulas produce the same
values for the derivative.
SOLUTION
Multiply the first equation by x and then isolate they term to obtain
x2+y=x
or y=x-x2
Implicit differentiation applied to the first equation yields
l-
y[
2
+ x- 1y' = 0 or y'
= yx- 1 -
x
From the first equation, we find yx- 1 = 1 - x; upon substituting this expression into the previous derivative, we find
y'=l-x-x=l-2x
which is the derivative of the second equation.
In Exercises 53 and 54,find dy/dx at the given point.
53. (x + 2)2 - 6(2y + 3)2
SOLUTION
= 3,
(1, -1)
By the chain rule,
2(x + 2) - 24(2y + 3)y' = 0
If x = 1 andy = -1, then
2(3) - 24(1)y'
so that 24y'
=0
= 6, or y' = ±-
In Exercises 55--62,find an equation of the tangent line at the given point.
55. xy + x2y2 = 6,
SOLUTION
(2, 1)
Taking the derivative of both sides of xy + x2y 2
= 6 yields
xy' + y + 2.xy2 + 2x2yy' = 0
Substituting x = 2,y
= 1, we find
2y' + 1+4 + 8y'
Hence, the equation of the tangent line at (2, 1) is y - l
=0
= -! (x -
or y'
= -~
2) or y =
2
-! x + 2.
180
CHAPTER 3
I
DIFFERENTIATION
57.
x2 + siny = xy2 + 1,
(1,0)
Taking the derivative of both sides of x2 + siny = xy2 + 1 yields
SOLUTION
2x + cos yy' = y2 + 2.xyy'
Substituting x = 1, y = 0, we find
2 + y'
=0
= -2
or y'
Hence, the equation of the tangent line is y - 0 = -2(x - 1) or y = -2x + 2.
59. 2x112 + 4y-112
= xy,
(1,4)
Taking the derivative of both sides of 2x 112 + 4y- 112 = xy yields
SOLUTION
x-112 _ 2y-3f2y' = xy' + y
Substituting x = l,y = 4, we find
1-2Uk=y'+4 or
y'=-~2
-lj-(x -
-lj-x + ?f.
Hence, the equation of the tangent line is y - 4 =
61. e2x-y
= x2 ,
SOLUTION
y
1) or y =
(2, 4)
Taking the derivative of both sides of e2x-y =
x2 yields
y
e2x-y(2 - y') = 2.xy - x2y'
y2
Substituting x = 2, y = 4, we find
16-4y'
e0 (2-y') = - - .
16
I
4
or y = 3
Hence, the equation of the tangent line is y - 4 = ~ (x - 2) or y = ~ x + ~.
63. Find the points on the graph of y2 = x3 - 3x + 1 (Figure 9) where the tangent line is horizontal.
(a) First show that 2yy' = 3x2 - 3, where y' = dy/dx.
(b) Do not solve for y'. Rather, set y' = 0 and solve for x. This yields two values of x where the slope may be zero.
(c) Show that the positive value of x does not correspond to a point on the graph.
(d) The negative value corresponds to the two points on the graph where the tangent line is horizontal. Find their coordinates.
y
2
-2
FIGURE 9 Graph ofy2 =
x3 - 3x + 1.
SOLUTION
(a) Applying implicit differentiation to y2 = x3 - 3x + 1, we have
2/Ydx = 3:x2-3
(b) Setting y' = 0 we have 0 =
3x2 - 3, so x = 1 or x = -1.
(c) If we return to the equation y2 = x3 - 3x + 1 and substitute x = 1, we obtain the equation y2 = -1, which has no real
solutions.
S EC T I 0 N 3.8
(d) Substituting x = -1 into -y2 =
x3 -
I
Implicit Differentiation
181
3x + 1 yields
y2 = (-1) 3 - 3(-1) + 1=-1+3 + 1 = 3
soy =
-../3 or - -../3. The tangent is horizontal at the points (-1, -../3) and (-1, - -../3).
65. Find all points on the graph of 3x2 + 4y2 + 3.xy = 24 where the tangent line is horizontal (Figure 10).
FIGURE 10 Graph of 3.x2 + 4y2 + 3.xy = 24.
SOLUTION
(a) Differentiating the equation 3.x2 + 4y 2 + 3.xy = 24 implicitly yields
6x + Syy' + 3.xy' + 3y = 0
so
= _ 6x+ 3y
y'
Sy+3x
Setting y' = 0 leads to 6x + 3y = 0, or y = -2x.
(b) Substituting y = -2x into the equation 3.x2 + 4y2 + 3.xy = 24 yields
3.x2 + 4(-2x) 2 + 3x(-2x) = 24
or 13.x2 = 24. Thus, x = ±2 -..fiS/13, and the coordinates of the two points on the graph of 3.x2 + 4y2 + 3.xy = 24 where
the tangent line is horizontal are
(
2-..fiS _
13 '
4-..fiS)
and
13
(
_2-..fiS
13 ,
4-..fiS)
13
67. Figure 1 shows the graph ofy4 + xy = x3 - x + 2. Find dy/dx at the two points on the graph with x-coordinate 0 and
find an equation of the tangent line at each of those points.
SOLUTION
Consider the equation y 4 + xy =
x3 - x + 2. Then 4y3y' + xy' + y = 3.x2 - 1, and
,
3x2 -y- 1
y = x+4y3
Substituting x = 0 into y4 + xy =
x3 - x + 2 gives y 4 = 2, which has two real solutions, y = ± {/2. When y = {/2, we have
-Vi- 1
y' = 4 (
so an equation of the tangent line at the point (0,
V2)3
0+ {/2
8
Vi) is
y=
-../2+{/2x+V1
8
When y = - {/2, we have
y'
Vi-1
0-Y2
= 4(-Y2)3 =
8
so an equation of the tangent line at the point (0, -
Vi) is
y=
182
CH APT E R 3
I
DIFFERENTIATION
69. Find a point on the folium x3 + y3
SOLUTION
= 3.xy other than the origin at which the tangent line is horizontal.
Using implicit differentiation, we find
!!_ (x 3 + l) = !!_(3.xy)
dx
dx
3x2 + 3y2y' = 3(.xy' + y)
= 0 in this equation yields 3x2 = 3y or y = x2. If we substitute this expression into the original equation
= 3.xy, we obtain:
Setting y'
x3 + y 3
x3 + x6
= 3x(x2) = 3x3
One solution of this equation is x = 0 and the other is x
the tangent line is horizontal are (0, 0) and (2 113 ' 2 213 ).
or
= 2113 • Thus, the two points on the folium x3 + y 3 = 3.xy at which
71. Find the x-coordinates of the points where the tangent line is horizontal on the trident curve xy = x3 - 5x2 + 2x - 1,
so named by Isaac Newton in his treatise on curves published in 1710 (Figure 12).
Hint: 2x3 - 5x2 + 1 = (2x - l)(x2 - 2x - 1).
y
-20
FIGURE 12 Trident curve: xy = x3 - 5x2
SOLUTION
+ 2x - 1.
Take the derivative of the equation of a trident curve:
xy
= x3 -
sx2 + 2x - 1
to obtain
xy' + y = 3x2 - lOx + 2
Setting y' = 0 in (a) gives y = 3x2 - lOx + 2. Substituting this into the equation of the trident, we have
xy
= x(3x2 -
lOx + 2)
= x3 -
3x3 - 10x2
+ 2x = x3 - 5x2 + 2x - 1
5x2 + 2x - 1
or
Collecting like terms and setting to zero, we have
0
Hence, x =
= 2x3 -
5x2 + 1 = (2x - l)(x2 - 2x - 1)
!, 1 ± .../2.
73. Find the derivative at the points where x
= 1 on the folium (x2 + y2) 2 = ¥.xy2 • See Figure 14.
y
2
-2
FIGURE 14 Foliumcurve: (x2
.
+y2)2
= 25 xy2.
4
S E c T I O N 3.8
SOLUTION
I
Implicit Differentiation
2
First, find the points (l,y) on the curve. Setting x = 1 in the equation (x2 + y2) =
25
(1 + y2)2 = -y2
4
y4 + 2y2 + 1 =
25
4
-y2
4y4 + sy2 + 4 = 25y2
4y4-17y2+4=0
(4y2
l)(f - 4) = 0
-
1
y2 = - ory2 = 4
4
!
Hence y = ± or y = ±2. Taking
fx of both sides of the original equation yields
25
2(x2 + y2)(2x + 2yy')
=4 l
4(x2 + y2)x + 4(x2 + y2)yy'
=4 l
(4(x2 + y2)-
25
25
25
+
2 w'
+
2 w'
25
25
I
2x)yy = 4f - 4(x2 + f)x
lly2_
- 4(x2
+ y2)x
y' = _4,__
_ _ __
2
2
y(4(x + y ) x)
¥
• At(l,2),x2+y2 =5,and
¥22 -4(5)(1)
y = 2(4(5) - ¥<1))
1
I
=3
Hence, at (1, 2), the equation of the tangent line is y- 2 = ~(x - 1) or y = ~x + ~·
• At (1, -2), x2 + y2 = 5 as well, and
y' =
ll(-2)2
-
4(5)(1)
--'4'-------
-2(4(5) -
¥<1))
1
3
Hence, at(l,-2), the equation of the tangent line isy+ 2 = -~(x-1) ory = -~x- ~·
• At (1, !), x2 + y2
= ~,and
y'=?f0)2-4(~)(1)
11
¥<1))
12
!(4(~)-
Hence, at (1, ~),the equation of the tangent line is y- ~ = f!Cx - 1) or y =
f!x -
fi.
• At (1, -~), x2 + y2 = ~.and
Hence, at (1, --21 ), the equation of the tangent line is y + -21
The folium and its tangent lines are plotted below:
y
= -ll(x
12
1) ory
= -llx
12 + i12.
¥xy2 yields
183
184
CH APT E R 3
I
DIFFERENTIATION
75. Calculate dx/ dy for the equation y4 + 1 =I + r and find the points on the graph where the tangent line is vertical.
SOLUTION
Let y4 + 1 = y2 +
r. Differentiating this equation with respect toy yields
dx
4y3 = 2y + 2x dy
so
dx 4y3 - 2y y(2y2 - 1)
=---=
2x
x
dy
Thus,
~; = 0 when y = 0 and when y = ±
x = ±1. Substituting y = ±
-v;, gives r
v;.
Substituting y = 0 into the equation y4 + 1 =
(1,0),(-1,0),(
+ r gives 1 =
r, so
v;.
v;. -v;).(- v;. -v;),( v;,- -v;).(- v;,- -v;)
Thus, there are six points on the graph ofy4 + 1 = y2 + r
= 3/4, so x = ±
where the tangent line is vertical:
I
77. [ R 5 Use a computer algebra system to plot y2 = x3 - 4x for x and y between -4 and 4. Show that if dx/dy = 0,
then y = O. Conclude that the tangent line is vertical at the points where the curve intersects the x-axis. Does your plot
confirm this conclusion?
SOLUTION
A plot of the curve y2
= x3 - 4x is shown below.
y
Differentiating the equation y 2 =
x3 - 4x with respect toy yields
2y = 3x2 dx _ 4 dx
dy
dy
or
dx
dy
2y
3x2 -4
From here, it follows that ~ = 0 when y = 0, so the tangent line to this curve is vertical at the points where the curve
intersects the x-axis. This conclusion is confirmed by the plot of the curve shown above.
In Exercises 79 and 80, first compute y' and y" by implicit differentiation. Then solve the given equation for y, and
compute y' and y" by direct differentiation. Finally, show that the results obtained by each approach are the same.
79. xy =y-2
SOLUTION
Let xy = y - 2. Then
y + xy'
= y'
which yields
y'
= _Y_
l-x
Additionally,
,, (l-x)y' -y(-1)
(1-x)it +y
2y
y =
=
=-2
2
(1 - x)
(1 - x)
Solving the original equation for y, we find
y
2
=- = 2(1 1-x
x)- 1
(1 - x)2
s E CT I o N
3.8
I
Implicit Differentiation
185
Then,
2
y' = -2(1-x)- 2(-1)=2(1 - x)-2 = (1- x)2
If we substitute y =
4
3
3
y11 = -4(1 - x)- (-1) = 4(1 -x)- = (1- x) 3
and
into the expressions for y' and y11 obtained by implicit differentiation, the expressions obtained
by direct differentiation result.
1
: x
In Exercises 81-84, use implicit differentiation to calculate higher derivatives.
81. Consider the equation y3 - ~ x2 = 1.
(a) Show that y' = x/y2 and differentiate again to show that
y2 -2.xyy'
yll=----
y4
(b) Express y 11 in terms of x and y using part (a).
SOLUTION
(a) Lety3 - ~x2 = 1. Then 3y2 y' - 3x = 0, andy' = x/y2 • Therefore,
II
y
=
y 2 ' 1 - X ' 2yy'
y2 - 2.xyy'
y4
y4
(b) Substituting the expression for y' into the result for y11 gives
y2 - 2xy(x!f)
II
y4
y =
y3 - 2x2
=-y-5-
83. Calculate y" at the point (1, 1) on the curve xy2 + y - 2 =0 by the following steps:
(a) Find y' by implicit differentiation and calculate y' at the point (1, 1).
(b) Differentiate the expression for y' found in (a). Then compute y11 at (1, 1) by substituting x = 1, y = 1, and the value
ofy' found in (a).
Let xy2 + y - 2 = 0.
SOLUTION
2
1
(a) Then x · 2yy' + y2 · 1+y'=0, andy' = -Y- . At (x,y) = (1, 1), we have y' = -- .
2 xy + 1
3
(b) Differentiating the expression for y' from (a) yields
y" =
Substituting x
= 1, y = l, and y'
(2xy + 1) (2yy') - y 2 (2.xy' + 2y)
(2xy + 1)2
= - ~ gives
yll
=
(3)(-D-cl)(-~
32
+1)
-6 + 2-6
10
27
27
Exercises 85 and 86 explore the radius of curvature of curves. There can be many circles that are tangent to a curve at
a particular point, but there is one that provides a "best fit" (Figure 17). This circle is called an osculating circle of the
curve. We define it formally in Section 13 .4. The radius of the osculating circle is called the radius of curvature of the
curve and can be shown to be
r=
(1 + (dy/dx)2)3/2
ld2y/dx21
FIGURE 17 The osculating circle (red) at a point is the tangent circle that best fits the curve.
186
CHAPTER 3
I
DIFFERENTIATION
85. Consider the ellipse x2 + 4y2 = 16.
(a) Compute the radius of curvature in terms of x and y.
(b) Compute the radius of curvature at (4, 0), (2, .../3), and (0, 2). Sketch the ellipse, plot these three points, and label
them with the corresponding radius of curvature.
SOLUTION
(a) Let x2 + 4y2 = 16. Then
2x + 8yy'
=0
y
so
I
X
= -4y
Additionally,
y" =
4y-x(-~)
4y(l) - x(4y')
l6y2
=-
l6y2
4y+f
l6y2
4y2+x2
= -----i6Y3
Thus,
r=
(1 +
J6y2 +x2 )3/2
x2 )3/2
i6Y2
(
(J6y2 +x2)3/2
16y2
I
I
7
64y3
(16y2 + x2)3/2
I
64
7
7
(b) At the point (4, 0), the radius of curvature is
r=
(0 + 16)
64
3 2
1
= 64 = 1
64
at the point (2, .../3), the radius of curvature is
r=
(16(3) + 4) 312
52312
13312
64
= 64 = -8-
at the point (0, 2), the radius of curvature is
r=
(16(4) + 0) 312
64312
=-=VM=8
64
64
A sketch of the ellipse with these three points labeled with the radius of curvature is shown in the figure below.
y
nm
2 r=8
--r--~' = - 8 -
1
r= I
-<--t----t---t----i,..........x
-1
-2
In Exercises 87-89, x and y are functions of a variable t. Use implicit differentiation to express dy/dt in terms of dx/dt,
x, andy.
87. x2y
=3
SOLUTION
Let x2y = 3. Assuming that x and y are function of a variable t, it follows that
( 2x ddxt ) y + x2 ddyt = 0
so
dy - - 2y dx
dt x dt
89. y 4 + 2x2 = xy
SOLUTION
Let y 4 + 2x2
= xy. Assuming that x and y are function of a variable t, it follows that
dy
dx dx
dy
4y3 - +4x- = -y+xdt
dt
dt
dt
dy
dx
3
(4y - x) dt = (y- 4x) dt
dy
dt
y-4x dx
4y -x dt
- = -3- - -
sEcT IoN
3.8
I
Implicit Differentiation
187
Further Insights and Challenges
91. Show that if Plies on the intersection of the two curves x2
the curves at P are perpendicular.
- y2 = c and xy = d (c, d constants), then the tangents to
Let Cl be the curve described by x2 - y 2 = c, and let C2 be the curve described by xy = d. Suppose that
P = (x0 , y 0 ) lies on the intersection of the two curves x2 - y2 = c and xy = d. Since x2 - y2 = c, the chain rule gives us
2x - 2yy' = 0, so that y' = ~ = ~. The slope to the tangent line to C 1 is ~. On the curve C2, since xy = d, the product
Therefore the slope to the tangent line to C2 is _ l:l!..
The two slopes are
rule yields that .xy' + y = 0, so that y' = - r.
x
~
negative reciprocals of one another, hence the tangents to the two curves are perpendicular.
SOLUTION
93. Divide the curve in Figure 19 into five branches, each of which is the graph of a function. Sketch the branches.
y
FIGURE 19 Graph of y5
SOLUTION
-
y
= :x2y + x + 1.
The branches are:
• Upper branch:
y
2
--t~--+-~-r-~1----+-x
-4
-2
2
4
-2
• Lower part of lower left curve:
y
-4
-3
-2
-1
• Upper part of lower left curve:
y
-4
-3
-2
-1
-1
-2
• Upper part of lower right curve:
y
2
4
--ir---+~_.._~-t--~+--x
-1
-2
188
CH APT E R 3
I
DIFFERENTIATION
• Lower part of lower right curve:
y
2
4
3.9 Derivatives of General Exponential and Logarithmic
Functions
Preliminary Questions
1. What is the slope of the tangent line toy = 4x at x = O?
SOLUTION
The slope of the tangent line toy = 4x at x = 0 is
2. What is the rate of change of y = ln x at x = 10?
SOLUTION
The rate of change of y = ln x at x = 10 is
I
d
-lnx
=-1
dx
x=IO
X
I
x=IO
1
=10
3. What is b > 0 if the tangent line toy = bx at x = 0 has slope 2?
SOLUTION
The tangent line toy= bx at x = 0 has slope
= bxlnbl
= lnb
d bx'
dX x=O
x=O
This slope will be equal to 2 when
ln b = 2
or
b = e2
4. What is b if (logb x)' = __!__?
3x
SOLUTION
(logb x)' =
G:: )'
= x
1~ b. This derivative will equal :}; when
lnb=3
or
b=e3
5. What are y< 100l and y< 101 l for y = cosh x?
SOLUTION Let y = cosh x. Then y' = sinh x, y" = cosh x, and this pattern repeats indefinitely. Thus,
y< 101 l = sinhx.
Exercises
In Exercises 1-20,find the derivative.
1. y = xlnx
SOLUTION
d
x
-d xlnx = lnx + - = lnx + 1
x
x
3. y = 2x3
SOLUTION
!
2.x3
= 3x2(ln 2)2.x3
y< 100l
= cosh x and
s E c T 1o N
3.9
I
Derivatives of General Exponential and Logarithmic Functions
189
5. y = ln(9.x2 - 8)
d
SOLUTION
-
dx
18x
ln(9.x2 - 8) = - 2 8
9x -
= (lnx) 2
7. y
SOLUTION
d
1 2
-(lnx)2 = (2lnx)- = - lnx
dx
x
x
SOLUTION
d
2
-e(lnx)
dx
= e(Inxl2 · 2 · -lnx
x
11. y = ln(ln x)
d
SOLUTION
-
dx
1
ln(lnx) = -1x nx
13. y = ( ln(ln x))
SOLUTION
3
3
1
~(ln(ln x))3 = 3(ln(ln x))2 (1nx
- ) (~) = 0nin x))
dx
x
x nx
2
15. y = ln ((x + 1)(2x + 9))
SOLUTION
d
1
4x + 11
dx In ((x + 1)(2x + 9)) = (x + l)( 2x + 9) · ((x + 1)2 + (2x + 9)) = (x + l)(2x + 9)
Alternately, because ln((x + 1)(2x + 9)) = ln(x + 1) + ln(2x + 9),
d
1
2
4x+ 11
dx ln((x + l)(2x + 9)) = x + 1 + 2x + 9 = (x + 1)(2x + 9)
17. y
= lP
SOLUTION
~lP=lnll·llx
dx
2x - 3-x
19. y = - x
d 2x - 3-x
x(2x ln 2 + 3-x ln 3) - (2x - 3-x)
SOLUTION dx - - x - =
x2
In Exercises 2I-24, compute the derivative.
21. f'(x),
f(x) = log2 x
lnx
SOLUTION
23.
f (x) = log2 x = ln 2 . Thus, f' (x) =
1
1
·
2
~ · ln
fr log (sin t)
3
SOLUTION
d
.
d (ln(sint))
1
1
cott
lofo(Sm t) = - - - - = . -.- . cost = dt
dt
ln3
ln3 smt
ln3
-
In Exercises 25-36,find an equation of the tangent line at the point indicated.
25. f (x)
= 6x,
x
=2
SOLUTION Let f(x) = 6x. Then f(2) = 36, f'(x) = 6xln6 and f'(2) = 36ln6. The equation of the tangent line is
therefore y = 36 ln 6(x - 2) + 36.
27. s(t)
= 39',
t
=2
Let s(t) = 391 . Then s(2) = 3 18 , s'(t) = 391 9ln3, and s'(2) = 3 18 · 9ln3 = 320 In3. The equation of the
tangent line is therefore y = 320 ln 3(t - 2) + 3 18 •
SOLUTION
29. f (x)
= 5x2-2x,
x
=1
SOLUTION Let f(x) = 5x2-2x+ 9 • Then f(l) = 58. f'(x) = ln 5 · 5x2- 2x+9 (2x - 2), so f'(l) = ln5(0) = 0. Therefore, the
equation of the tangent line is y = 5 8 •
190
CH APT E R 3
I
DIFFERENTIATION
31. s(t)
= ln(8 -
4t),
t
=1
SOLUTION Let s(t) = ln(8 - 4t). Then s(l) = ln(8 - 4) = ln4. s'(t)
equation of the tangent line is y = -1 (t - 1) + In 4.
33. R(z) = log5(2z2 + 7), z = 3
SOLUTION Let R(z) = log5(2z2 + 7). Then R(3) = log5(25) = 2,
I
R (z)
4z
= (2z2 + 7) In 5
and
= 8::_!,, so s'(l) = -4/4 = -1. Therefore the
I 3
12
R( )= 25ln5
The equation of the tangent line is therefore
12
y= --(z-3)+2
25ln5
35. f(w) = log2 w, w =
SOLUTION
Let f(w)
k
= log2 w. Then
(1)
1
-
f g = log2 8 = log2 2 3 = -3
f'(w) = wl~2' and
f'
(.!.)8 = __!__
ln2
The equation of the tangent line is therefore
In Exercises 37-44,find the derivative using logarithmic differentiation as in Example 5.
37. y
= (x + 5)(x + 9)
SOLUTION
Lety = (x + 5)(x + 9). Then lny = ln((x + 5)(x + 9))
y'
y
1
x+5
= ln(x + 5) + ln(x + 9). By logarithmic differentiation
1
x+9
-=--+-or
y'
39. y
= (x -
SOLUTION
1
= (x + 5)(x + 9) (x+5
--
1
+ - -)
x+9
= (x + 9) + (x + 5) = 2x + 14
= ln(x -
1) + ln(x - 12) + ln(x + 7). By logarithmic differentiation,
l)(x - 12)(x + 7)
Let y
= (x -
l)(x - 12)(x + 7). Then lny
y'
y
1
x- 1
1
x - 12
I
x +7
-=--+--+-or
y'
41. y
= (x -
12)(x + 7) + (x - l)(x + 7) + (x - l)(x - 12)
12x + 79
= x(x2 + 1)
Vx+I
SOLUTION
Let y
= x~). Then lny = lnx + ln(x2 + 1) y'
I
2x
! ln(x + 1). By logarithmic differentiation
1
-=-+-----y
x
x 2 +I
2(x+l)
so
= 3x2 -
s EcT Io N
43. y
I
3.9
Derivatives of General Exponential and Logarithmic Functions
x(x+ 2)
(2x + 1)(3x + 2)
=
SOLUTION
Let y
=
(2x::\~g;+2l. Then lny
= ~ [ln(x) + ln(x + 2) -
ln(2x + 1) - ln(2x + 2)]. By logarithmic differenti-
ation
y' - 1(1- +1- - - 2 - - -2 - )
y - 2 x x+2 2x+l 2x+2
so
f
y
= -1
x(x + 2)
(1
1
2
1 )
(2x+ 1)(2x+2). :;: + x+2 - 2x+ 1 - x+ 1
2
In Exercises 45-50,find the derivative using either method of Example 6.
45.
f (x) = x3x
SOLUTION
Method 1: x3x = e3x1"X, so
d
-x3x
dx
Method 2: Let y = x3x. Then, In y
= e3xlnx(3 + 3 ln x) = x3x(3 + 3 ln x)
= 3x In x. By logarithmic differentiation
y'
1
= 3x · - + 3 ln x
y
x
-
so
y'
47. f(x)
=y(3 + 3lnx) = x3x(3 + 3lnx)
= r' .
SOLUTION
Method 1: f
Method 2: Let y
= f.
= ee'lnx, so
Then In y
= ex In x. By logarithmic differentiation
-y'y = ex · -x1 +exlnx
so
49. J(x)
191
= ;rosx
SOLUTION
Method 1: ;rosx
= ecosxlnx, so
d xcosx - ecosxlnx(cosx si In )
dx
---:;:-- - nx x
= ;rosx(cosx
-x- -
. l )
srnx nx
Method 2: Let y = ;rosx. Then lny = cosx lnx. By logarithmic differentiation
y'
1
.
- = cos x · - +In x(- srn x)
y
x
so
f
Y
= Y (cosx.
-x- - srnxlnx) = x cosx(cosx
--:;:--- -
In Exercises 5I-74, calculate the derivative.
51. y = sinh(9x)
SOLUTION
:/x sinh(9x) = 9 cosh(9x)
sinxlnx)
192
CH APTER 3
I
DIFFERENTIATION
53. y = cosh2(9 - 3t)
!!:_ cosh2(9 - 3t) = 2 cosh(9 - 3t) · (-3 sinh(9 - 3t)) = -6 cosh(9 - 3t) sinh(9 - 3t)
SOLUTION
55. y
dt
= vcoshx + 1
57
!(coshx + lf 112 sinhx
2
:!:._ Vcoshx + 1 =
SOLUTION
dx
cotht
• y = 1 +tanht
d
dt
SOLUTION
59. y
= sinh(ln x)
d
-d sinh(lnx)
SOLUTION
61. y
coth t
- csch2 t(l + tanh t) - coth t(sech2 t)
1 + tanh t =
(1 + tanh t)2
x
d
tanh(ex)
x
= ex sech2(ex)
-
63. y = sech(
vx)
SOLUTION
-d sech( Vx)
x
.
dx
d
l
112
= --xsech Vxtanh Vx
2
= sech x coth x
SOLUTION
67. y
cosh(lnx)
= tanh(ex)
SOLUTION
65. y
=
d
d
dx sechxcothx = dx cschx = -cschxcothx
= cosh- 1 (3x)
SOLUTION
d
-
dx
cosh- 1(3x)
3
=--v9x2 -1
69. y = (sinh- 1(x2))3
SOLUTION
dd sinh- 1(x2) = -~
x
vx4 + 1
SOLUTION
d
h-l x
-ecos
dx
= ecosh-1 x ( - -1 - )
vx2 -1
1
73. y = tanh- (lnt)
SOLUTION
d
-d tanh- 1(lnt)
t
=
1
2
t(l -(lnt))
In Exercises 75-77, prove the formula.
d
75. d)sechx)
SOLUTION
= -sechxtanhx
1
d
l
77. - cosh- 1 t = _r;;--c;
dt
vt2 - 1
SOLUTION
1
1
dd sechx = :!:._ (-- ) = - - --sinhx
= - ( --)(sinhx) = -sechxtanhx
x
dx coshx
cosh2 x
coshx coshx
Let x
fort> 1
= cosh- 1 t. Then t = coshx and
.nh dx
xdt
1 =SI
or
Thus, fort> 1,
d
1
- cosh- t
dt
1
= -----,-1
sinh(cosh- t)
Now, fort;:::: 1, note
d [ SI"nh(cosh-I t) ] = cosh(cosh- 1 t ) -1- - = - t - = -d _vt2
,-:;;---;-d
_ 1
t
~
~
dt
s E c T 1o N 3.9 I Derivatives of General Exponential and Logarithmic Functions 193
so the functions sinh(cosh- 1 t) and ~differ by a constant; substituting t = 1 we find that the constant is 0. Therefore,
fort~ 1,
sinh(cosh- 1 t) = Vt2 - 1
and
-1
1
d
-cosh t=
= --dt
sinh(cosh- 1 t)
~
79. Over rough uneven terrain, the log wind profile from Example 3 is expressed as
=Vo
V
ln(h/0.4)
ln(ho/0.4)
With v0 = 10 mis at ho = 10 m, determine v and dv/dh at h = 60.
SOLUTION
Let Vo = 10 mis, ho = 10 m, and
v
= v ln(h/0.4) = 10 ln(h/0.4) = _!_Q_ ln(h/0.4)
0
ln(ho/0.4)
ln(l0/0.4)
ln25
Then
dv
dh
10
= ln25.
1
1
10
h/0.4. 0.4 = hln25
Thus,
10
10
v(60) = Jn(60/0.4) = ln 150 "' 15.5664 m/s
1n 25
ln25
and
dvl
dh
h=60
10
= 60ln25 "'0.0518 m/s perm
d';:
1
and ln a were
investigated. Here, it is established that they are equal. Let f(x) =ax. We have f'(x) = (lna)ax. Set up and simplify the
expression for lim f(x + h~ - f<x>, and use the resulting expression to show that lim ah h- 1 = ln a.
81. In Exercise 46 in Section 2.3 and Exercises 41 and 42 in Section 2.5, similarities between lim
h->0
h->0
SOLUTION
h->0
Letf(x) =ax. Then
f'(x) =Jim
h->0
f(x + h)
h
-
f(x)
= lim
ax+h - ax
h->O
h
= lim
But, f'(x) = (lna)aX, from which it follows that
ah - 1
Jim-- =lna
h-+0
h
83. Show that for any constants M, k, and a, the function
satisfies the logistic equation:
SOLUTION
Then
Let
~ = k( 1 - 2'...).
y
M
h->O
axah - ax
h
ah - 1
= axlim - h-+0
h
194
CHAPTER 3
I
DIFFERENTIATION
and
ky(t)(1-Yii) = ~Mk(l -tanh2 (k(t~a)))
~Mksech2 (k(t~a))
=
Finally,
85. The Palermo Technical Impact Hazard Scale P is used to quantify the risk associated with the impact of an asteroid
colliding with the earth:
where p; is the probability of impact, T is the number of years until impact, and E is the energy of impact (in megatons
of TNT). The risk is greater than a random event of similar magnitude if P > 0.
(a) Calculate dP/dT, assuming that p; = 2 x 10-5 and E = 2 megatons.
(b) Use the derivative to estimate the change in P if T increases from 8 to 9 years.
SOLUTION
(a) Observe that
so
dP
dT
Tln 10
(b) If T increases to 26 years from 25 years, then
11P"'
~~IT=25·11T =
(25y:)ln10. (1 yr)= -0.017
Further Insights and Challenges
87. For x > 0 and any real number n, x" = enlnx. Show that fxenlnx = nx"- 1, thereby proving the power rule for all
exponents (as long as x > 0).
SOLUTION
Let x > 0 and n be any real number. Then x" = enlnx, so that
!!:_x" = !!:_enlnx = enlnx!!:_(nlnx) = enlnx. ~ = x". ~ = nx"-1
dx
dx
dx
x
x
3.1 O Related Rates
Preliminary Questions
1. If~= 3 andy = x2, what is 7Ji whenx = -3,2,5?
SOLUTION
Lety = x2. Then
dy
dt
= 2xdx
dt
Given ~ = 3, it follows that
ddyl
t
= 2(-3)(3) = -18
x=-3
-dyl
= 2(2)(3) = 12
dt x=2
ddy
t
I
x=S
= 2(5)(3) = 30
SECTION 3.10
2. If ~ = 2 and y =
SOLUTION
Let y
x3, what is ~
Related Rates
195
when x = -4, 2, 6?
= x3. Then
= 3x2dx
dy
dt
Given ~
I
dt
= 2, it follows that
= 3(-4)2 (2) = 96
dyl
dt
x=-4
I
dy I
dt
dy
dt
= 3(2)2 (2) = 24
=2
= 3(6)2 (2) = 216
=6
3. Assign variables and restate the following problem in terms of known and unknown derivatives (but do not solve it):
How fast is the volume of a cube increasing if its side increases at a rate of 0.5 emfs?
SOLUTION
!!Jf if ~
Let s and V denote the length of the side and the corresponding volume of a cube, respectively. Determine
= 0.5 emfs.
4. What is the relation between dV / dt and dr / dt if V
SOLUTION
= (~ )irr3 ?
Applying the general power rule, we find !!Jf
= 4irr
2
~.
In Questions 5 and 6, water pours into a cylindrical glass of radius 4 cm. Let V and h denote the volume and water level,
respectively, at time t.
5. Restate this question in terms of dV I dt and dh/ dt: How fast is the water level rising if water pours in at a rate of
2cm3 /min?
SOLUTION
Determine
!!jf
if
%= 2 cm3/min.
6. Restate this question in terms of dV I dt and dh/ dt: At what rate is water pouring in if the water level rises at a rate of
1 cm/min?
SOLUTION
Determine !!Jf if !!jf
= 1 cm/min.
Exercises
In Exercises 1 and 2, consider a rectangular bathtub whose base is 18 ff.
1. How fast is the water level rising if water is filling the tub at a rate of 0. 7 ft3 /min?
SOLUTION
Leth be the height of the water in the tub and Vbethe volume of the water. Then V
Thus
dh
ldV
1
dt =
dt = l8 (0.7) "' 0.039 ft/min
18
3. The radius of a circular oil slick expands at a rate of 2 m/min.
(a) How fast is the area of the oil slick increasing when the radius is 25 m?
(b) If the radius is 0 at time t = 0, how fast is the area increasing after 3 min?
Let r be the radius of the oil slick and A its area.
dA
dr
(a) Then A= irr2 and -d = 2irr-. Substituting r = 25 and !Ldr = 2 we find
t
dt
t
'
SOLUTION
dA
dt ""2rr(25)(2)
= 10011""' 314.16 m2 /min
(b) Since ~ = 2 and r(O) = 0, it follows that r(t) = 2t. Thus, r(3) = 6 and
dA
dt = 2ir (6) (2) = 24ir "' 75.40 m2 /min
= 18h and dV = 18dh.
dt
dt
196
CHAPTER 3
I
DIFFERENTIATION
In Exercises 5-8, assume that the radius r of a sphere is expanding at a rate of 30 cm/min. The volume of a sphere is
3
and its surface area is 41fr2. Determine the given rate.
V=
1nr
5. Volume with respect to time when r
= 15 cm
As the radius is expanding at 30 centimeters per minute, we know that ~ = 30 cm/min. Talcing
3
yields
equation V =
SOLUTION
11fr
dV
dt
fr of the
= ~1f(3r2dr) = 47rr2dr
3
dt
dt
Substituting r = 15 and ~ = 30 yields
dV
dt
= 41f(15)2(30) = 270007r cm3 /min
7. Surface area with respect to time when r = 40 cm
SOLUTION
Talcing the derivative of both sides of A = 41fr2 with respect tot yields ~ = 87rr~. ~ = 30, so
dA = 87r(40)(30) = 96001f cm2/min
dt
9. A conical tank (as in Example 2) has height 3 m and radius 2 mat the base. Water flows in at a rate of 2 m3/min.
How fast is the water level rising when the level is 1 m and when the level is 2 m?
Let V denote the volume of the water in the tank in m3, and let h denote the height of the water in the
tank in m. Now, the volume of water in the tank can be calculated as the difference between the volume of the tank and
the volume of the conical space in the tank above the water. The volume of the conical tank is ~1f(2)2 (3) = 41f, and the
volume of the conical space is t1fr2(3 - h), where r and 3 - h are the base radius and the height, respectively, of the
conical space. Note that the triangles highlighted in the figure below are similar; therefore, 3 ~h = ~-Thus, r = 2<3;h), so
SOLUTION
v = 41f -
h))
1 ( 2(3 -1f - -
3
3
2
(3 - h)
= 41f -
4
-1f(3 - h) 3
27
Differentiating with respect to t, we obtain
4
4
2 ( dh)
2dh
-dV
= 0- -1f(3(3-h))
- - = -1f(3-h)
dt
27
dt
9
dt
Substituting 2 for !!Ji- and then solving for !lJf yields
dh
9
=----=dt
21f(3 - h)2
When the water level is 1 m, the water level is rising at a rate of
dhl
dt
9
h=I
9
= 27r(2)2 = 87r ""0.3581 m/min
when the water level is 2 m, the water level is rising at a rate of
dhl
dt
9
h=2
9
= 27r(1)2 = 27r
"" 1.4324 m/min
In Exercises 11-14, refer to a 5-m ladder sliding down a wall, as in Figures 5 and 6. The variable h is the height of the
ladder's top at time t, and xis the distance from the wall to the ladder's bottom.
11. Assume the bottom slides away from the wall at a rate of 0.8 mis. Find the velocity of the top of the ladder at t
if the bottom is l .5 m from the wall at t = 0 s.
=2 s
SECTION 3.10
I
Related Rates
197
Let x denote the distance from the base of the ladder to the wall, and h denote the height of the top of the
ladder from the floor. The ladder is 5 m long, so h2 + x2 = 52 • At any time t, x = 1.5 + 0.8t. Therefore, at time t = 2, the
base is x = 1.5 + 0.8(2) = 3.1 m from the wall. Furthermore, we have
SOLUTION
dh
dx
2h- + 2x- = 0
dt
dt
Substituting x
= 3.1, h = V5 2 -
3.l2 and~
= 0.8, we obtain
dh
dt
13. Suppose that h(O)
xdx
h dt
dh
dt
so
3.1
----;::;==-(0.8)
"' -0.63 2 m/s
V5 2 - 3.l2
= 4 and the top slides down the wall at a rate of 1.2 m/s. Calculate x and dx/dt at t = 2 s.
SOLUTION Leth and x be the height of the ladder's top and the distance from the wall of the ladder's bottom, respec2
2
tively. After 2 seconds, h = 4 + 2 (-1.2) = 1.6 m. Since h + x2 = 5 ,
x
dh
dx
Furthermore we have 2h- + 2x,
dt
dt
= V5 2 -
1.62
= 4.737 m
dh
. .
dh
fi d
= 0, so that -dx
= --hx -.
Subst:Itutmg h = 1.6, x = 4.737, and di = - 1.2, we n
dt
dt
dx
1.6
dt = - 4.737 (-1.2) "' 0.405 m/s
15. The radius r and height h of a circular cone change at a rate of 2 emfs. How fast is the volume of the cone increasing
when r = 10 and h = 20?
SOLUTION
Let r be the radius, h be the height, and V be the volume of a right circular cone. Then V = ~nr h, and
2
1 ( r2dh + 2hrdr)
-dV = -n
dt
When r = 10, h
3
dt
dt
= 20, and ~ = !!Ji = 2, we find
dV
dt
= 3n ( 102 · 2 + 2 · 20 · 10 · 2) =
lOOOn
3
- - "' 1047.20 cm /s
3
17. A man of height 1.8 m walks away from a 5-m lamppost at a speed of 1.2 mis (Figure 10). Find the rate at which
his shadow is increasing in length.
... ...
... ...
...... ...
' ' ',
'
FIGURE 10
SOLUTION Since the man is moving at a rate of 1.2 mis, his distance from the light post at any given time is x = l.2t.
Knowing the man is 1.8 meters tall and that the length of his shadow is denoted by y, we set up a proportion of similar
triangles from the diagram:
y
1.8
l.2t+ y
5
Clearing fractions and solving for y yields
y = 0.615t
Thus, dy/dt = 0.675 meters per second is the rate at which the length of the shadow is increasing.
19. At a given moment, a plane passes directly above a radar station at an altitude of 6 km.
(a) The plane's speed is 800 km/h. How fast is the distance between the plane and the station changing half a minute
later?
(b) How fast is the distance between the plane and the station changing when the plane passes directly above the station?
198
CHAPTER 3
I
DIFFERENTIATION
Let x be the distance of the plane from the station along the ground and h the distance through the air.
SOLUTION
(a) By the Pythagorean Theorem, we have
h
dh
Thus 2hdt
ifJ,
= x2 + 6 2 = x2 + 36
dx and dh
dx After h alf a mmute,
.
= 2x-,
- = -x -.
x = 21 x
dt
dt
h dt
h
and
2
20)
(3
=
2
601 x 800
20
= T20 kil ometers. w·th
1
x = J'
= 3I Y724 = 32 Vi8l "' 8.969 km
+ 36
= 800,
dh
dt
= 20
3
_
_
3 2 Vi8l
x 800 =
8000
"' 594.64 km/h
Vi8l
(b) When the plane is directly above the station, x = 0, so the distance between the plane and the station is not changing,
for at this instant we have
0
dh
-dt = -6 x 800 = 0 km/h
21. A hot air balloon rising vertically is tracked by an observer located 4 km from the lift-off point. At a certain moment,
the angle between the observer's line of sight and the horizontal is
fast is the balloon rising at this moment?
SOLUTION
~,
and it is changing at a rate of 0.2 rad/min. How
Let y be the height of the balloon (in miles) and fl the angle between the line-of-sight and the horizontal.
Via trigonometry, we have tan fl =
~. Therefore,
2
dfl
1 dy
sec f l · - = - dt
4 dt
and
dy
dfl
2
= 4-sec fl
dt
dt
-
Using 1f£ = 0.2 and fl= ~yields
dr =4 (0.2) cos ~71/5 ) "' 1.22 km/min
2
23. A rocket travels vertically at a speed of 1200 km/h. The rocket is tracked through a telescope by an observer located
16 km from the launching pad. Find the rate at which the angle between the telescope and the ground is increasing 3 min
after liftoff.
SOLUTION
we have tan fl
Let y be the height of the rocket and fl the angle between the telescope and the ground. Using trigonometry,
= f6. Therefore,
2
dfl
1 dy
sec f l · - = - dt
16 dt
and
dfl
cos 2 fl dy
dt = 16 dt
After the rocket has traveled for 3 minutes (or
15/4 and thus
fa
hour), its height is
cos fl=
4
.Yl5 2 + 4 2
-
fox 1200 = 60 km. At this instant, tan fl= 60/16 =
4
Y24T
Finally,
dfl
dt
16/241
- - (1200)
16
=-
=
1200
"'4.98 rad/hr
241
25. A police car traveling south toward Sioux Falls, Iowa, at 160 km/h pursues a truck traveling east away from Sioux
Falls at 140 km/h (Figure 12). At time t = 0, the police car is 20 km north and the truck is 30 km east of Sioux Falls.
Calculate the rate at which the distance between the vehicles is changing:
SECTION 3.10
(a) At time t
I
Related Rates
199
=0
(b) 5 min later
y
FIGURE 12
SOLUTION Let y denote the distance the police car is north of Sioux Falls and x the distance the truck is east of Sioux
Falls. Then y = 20 - l60t and x = 30 + 140t. If l denotes the distance between the police car and the truck, then
e2 = x2 + y2
= (30 + 140t)2 + (20 - 160t)2
and
edt = 140(30 + 140t) - 160(20- 160t) = 1000 + 45200t
dt
(a) At t =
o, e =
-../302
+ 20 = 10 ...ff3. so
2
dl = 1000 = 100 ...ff3 "'27.735 km/h
dt
10...[f3
13
(b) At t = 5 minutes= .&_ hour,
e=
1
(30 + 140. 12r +
(20-
1
160. 12r "'42.197 km
and
dl
dt
1000 + 45200 . .&_
42.197
"' 112.962 km/h
27. In the setting of Example 5, at a certain moment, the tractor's speed is 3 mis and the bale is rising at 2 mis. How far
is the tractor from the bale at this moment?
SOLUTION
From Example 5, we have the equation
x~
vx2 + 4_52
dh
dt
where x denotes the distance from the tractor to the bale and h denotes the height of the bale. Given
dx=3
dt
and
dh=2
dt
it follows that
3x
---;::::===2
v4.5 2 + x 2
which yields x = -../16.2 "'4.025 m.
29. Julian is jogging around a circular track of radius 50 m. In a coordinate system with its origin at the center of the track,
Julian's x-coordinate is changing at a rate of -1.25 mis when his coordinates are (40, 30). Find dy / dt at this moment.
SOLUTION We have x2 + y2 = 502, so
dx
dy
2x dt + 2y dt = 0 or
dy
dt
Given x = 40, y = 30 and dx/dt = -1.25, we find
dy
dt
40
5
= - 30 (-1. 25 ) = 3 m/s
xdx
y dt
200
CH APTER 3
I
DIFFERENTIATION
In Exercises 31and32, assume that the pressure P (in kilopascals) and volume V (in cubic centimeters) of an expanding
gas are related by PVb = C, where band Care constants (this holds in an adiabatic expansion, without heat gain or
loss).
= 1.2, P = 8 kPa, V = 100 cm2 , and dV / dt = 20 cm3/min.
Let PVb = C. Then
31. Find dP / dt if b
SOLUTION
Pbvb-1 dV + vbdP
dt
dt
=0
and
dP
dt
Substituting b
Pb dV
v dt
= 1.2, P = 8, V = 100, and !!fr = 20, we find
(8) (1.2)
dP
di = -----ioo (20) = -1.92 kPa/min
33. The base x of the right triangle in Figure 15 increases at a rate of 5 emfs, while the height remains constant at h
How fast is the angle e changing when x = 20?
0
x
FIGURE 15
SOLUTION
We have cot 8 =
x
.
, from which
20
de
1 dx
-csc2 B · - = - dt
20 dt
and thus
sin2 B dx
20 dt
dB
dt
We are given ~ = 5 and when x
= h = 20, e =
~. Hence,
dB
sin (~)
1
= - - - (5) = - - rad/s
dt
20
8
2
-
35. A particle travels along a curve y = f(x) as in Figure 16. Let L(t) be the particle's distance from the origin.
dL (x + f(x)f'(x)) dx .
. ,
.
.
.
(a) Show that -d =
t
-Vx2 + f(x)2 -dt 1f the particle s location at time t 1s P
(b) Calculate L'(t) when x = 1 and x
= 2 if f(x) =
V3x2
-
8x + 9 and dx/dt = 4.
y
--+''---l----l----+x
0
1
= (x, f(x)). .
2
FIGURE 16
= 20.
SECTION 3.10
I
Related Rates
201
SOLUTION
(a) If the particle's location at time tis P
= (x, f(x)), then
L(t)
= ,,,/x2 + f(x) 2
Thus,
(b) Given f(x) = V3x2
-
8x + 9, it follows that
3x-4
f'(x) = -v-;::3=x=
2 =-=8x=+=9
Let's start with x
= 1. Then f(l) = 2, f'(l) = -~ and
dL
dt
With x
=(
1 - 1 ) (4 )
...; 12 + 22
=O
= 2, f(2) = VS, f'(2) = 2/ VS and
dL
dt
2+2
= ~22 +
(4)
= 16
vs2
3
Exercises 37 and 38 refer to the baseball diamond (a square of side 90 ft) in Figure 17.
FIGURE 17
37. A baseball player runs from home plate toward first base at 20 ft/s. How fast is the player's distance from second
base changing when the player is halfway to first base?
Let x be the distance of the player from home plate and h the player's distance from second base. Using
the Pythagorean Theorem, we have h2 = 902 + (90 - x)2. Therefore,
SOLUTION
dh
2hdt
=2(90-x) ( -
-dx)
dt
and
dh
dt
We are given
4Ji
90-x dx
---h
dt
= 20. When the player is halfway to first base, x = 45 and h = V902 + 452 , so
45
--:::== == (20)
v902 + 4s2
= -4 Vs "" -8.94 ft/s
39. The conical watering pail in Figure 18 has a grid of holes. Water flows out through the holes at a rate of kA m3 /min,
where k is a constant and A is the surface area of the part of the cone in contact with the water. This surface area is
A = 1rr Vh2 + r 2 and the volume is V = !7rr2h. Calculate the rate dh/dt at which the water level changes at h = 0.3 m,
assuming that k = 0.25 m.
202
CH APTER 3
I
DIFFERENTIATION
0.45 m
FIGURE 18
SOLUTION
By similar triangles, we have
r
h
0.15
=
0.45 3
so
l
r = -h
3
Substituting this expression for r into the formula for V yields
From here and the problem statement, it follows that
1 2 dh
-~
-dV = -7rh
- = -kA = -0.257rr vh• + r
dt
9
dt
Solving for dh/dt gives
When h = 0.3, r
= 0.1 and
-dh = -9-0.1
dt
4 0.3 2
v0.3
2
.
+O.l 2 =-0 .79 m/mm
Further Insights and Challenges
41. A roller coaster has the shape of the graph in Figure 20. Show that when the roller coaster passes the point (x, f(x)),
the vertical velocity of the roller coaster is equal to f'(x) times its horizontal velocity.
FIGURE 20 Graph off as a roller coaster track.
Let the equation y = f(x) describe the shape of the roller coaster track. Taking ~ of both sides of this
equation yields ;J:i = f'(x)!if,. In other words, the vertical velocity of a car moving along the track, ~,is equal to f'(x)
times the horizontal velocity, ~.
SOLUTION
43. As the wheel of radius r cm in Figure 21 rotates, the rod of length L attached at point P drives a piston back and
forth in a straight line. Let x be the distance from the origin to point Q at the end of the rod, as shown in the figure.
(a) Use the Pythagorean Theorem to show that
(b) Differentiate Eq. (1) with respect tot to prove that
8
2(x- rcose)(dx + rsined ) + 2r2 sinBcosede = O
dt
dt
dt
SECTION 3.10
I
Related Rates
203
(c) Calculate the speed of the piston when() = ~, assuming that r = 10 cm, L = 30 cm, and the wheel rotates at 4
revolutions per minute.
Piston moves
back and forth
I
x
FIGURE 21
SOLUTION
From the diagram, the coordinates of Pare (r cos 8, r sin 8) and those of Qare (x, 0).
(a) The distance formula gives
L=
~(x- rcos8) 2 + (-rsin8) 2
Thus,
Note that L (the length of the fixed rod) and r (the radius of the wheel) are constants.
(b) From (a) we have
0 = 2 (x - r cos 8) (ddx + r sin ()d()) + 2r2 sin Bcos ()d()
t
dt
dt
(c) Solving for dx/dt in (b) gives
r 2 sin() cos()~
dx
dt
-----"d~t
r cos() - x
dB
rx sin()~
-rsin()- =
dt
dt
r cos() - x
With()= ~, r = 10, L = 30, and !£1 =Sn,
dx
(lO)(x){sin ~)(Sn)
.
dt =
(lO) (O) _ x
= -80n"' -251.33 cm/mm
45. A cylindrical tank of radius R and length L lying horizontally as in Figure 22 is filled with oil to height h.
(a) Show that the volume V(h) of oil in the tank is
V(h) =
L(R cos- (1-M-(R-h)V2hR-h2)
2
1
(b) Show that ~ = 2L vh(2R - h).
(c) Suppose that R = 1.5 m and L = 10 m and that the tank is filled at a constant rate of 0.6 m3 /min. How fast is the
height h increasing when h = 0.5?
FIGURE 22 Oil in the tank has level h.
204
CH APTER 3
I
DIFFERENTIATION
SOLUTION
(a) From Figure 22, we see that the volume of oil in the tank, V(h), is equal to L times A(h}, the area of that portion of
the circular cross section occupied by the oil. Now,
R2 sin()
2
R2 0
2 --
A(h) = area of sector - area of triangle =
where () is the central angle of the sector. Referring to the diagram below,
()
R-h
cos-= - -
2
R
.
and
()
sm2
=
V2hR-h2
R
~
R-h~
Thus,
() = 2cos- 1
.
.
sm 0 = 2 sm
(1- ~)
o o (R-h)V2hR-hz
R2
2 cos 2 = 2
and
(b) Recalling that .!L
cos- 1 u = -~dx'
-1 - rE!.
dx
~~ = L ( :h (R
2
=L(-R
cos-
1
(
1-
~))- ~ ((R -
-l
../1 - (1- (h/R))2
h) V2hR - hz))
+ V2hR-h2-
2
(R-h) )
V2hR- h2
=L (
R2
+ V2hR - h2 - R2 - 2Rh + h2)
V2hR - h2
V2hR - h2
2
2
2
= L(R + (2hR - h ) - (R - 2Rh + h2))
V2hR-h2
= L( 4hR - 2h2) = L(2(2hR- h2)) = 2L V2hR- h2
V2hR - h2
V2hR - h2
dV
(c)
dt
dV dh
dh
1 dV
= dh dt, so dt = dV/dh di" From part (b) with R = 4, L = 30 and h = 5,
dV
dh = 2(30) ../2(5)(4) - 52 = 60
m ft
2
Thus,
dh
dt =
1
60
-v'TS(lO) =
m
90 ""0.043 ft/min
Chapter Review Exercises
205
CHAPTER REVIEW EXERCISES
In Exercises 1-4, refer to the function f whose graph is shown in Figure 1.
0.5
1.5
1.0
2.0
FIGURE 1
1. Compute the average rate of change off(x) over [O, 2]. What is the graphical interpretation of this average rate?
The average rate of change of f(x) over [O, 2] is
SOLUTION
f (2) - f (0) = 7 2-0
l
= 3.
2-0
Graphically, this average rate of change represents the slope of the secant line through the points (2, 7) and (0, 1) on the
graph of f(x).
·
f(0. 7 + h)h - f(0. 7) 1or
" h
3. Estimate
. greater than or less than f'(O .7)?.
= 0. 3. Is th'is d'fti
i erence quotient
For h = 0.3,
SOLUTION
f(0.7 + h)- f(0.7)
h
f(l) - f(0.7)
0.3
2.8 - 2
8
""Q.3 = 3
Because the curve is concave up, the slope of the secant line is larger than the slope of the tangent line, so the value of
the difference quotient should be larger than the value of the derivative.
In Exercises 5-8, compute f'(a) using the limit definition and find an equation of the tangent line to the graph off at
x=a.
5. f(x)
= x2 -
SOLUTION
x,
a
=l
Let f (x) =
x2 - x and a = 1. Then
= lim f(a + h) -
f'(a)
.
= lim (1 + h)2- (1 + h)- (1 2 -
h
h-->O
= h-->O
hm
f(a)
1 + 2h + h2
-
1- h
h
and the equation of the tangent line to the graph off(x) at x
Y = f'(a)(x - a)+ f(a)
7. f(x)
= x- 1 ,
SOLUTION
1)
h
h-->O
= lim(l
+ h) = 1
h-->0
= a is
= l(x -
1) + 0
= x- 1
a= 4
Let f(x) = x- 1 and a = 4. Then
f'(a)
= lim f(a + h) .
/(a)
= lim
h
h-->0
-1
4lh - ~
h-->0
I
= lim 4 -
h
h-->0
(4 + h)
4h(4 + h)
I
=hm--- = - - - - =-h-->O 4(4 + h)
4(4 + 0)
16
and the equation of the tangent line to the graph off(x) at x
Y
=f
I
(a)(x-a) + f(a)
= a is
1
I
1
I
= --(x-4)
+ - = --x+ 16
4
16
2
206
C HAP TE R 3
I
DIFFERENTIATION
In Exercises 9-/2, compute dy/dx using the limit definition.
9. y =
4-.x2
Lety = 4-x2. Then
SOLUTION
dy
. 4- (x + h) 2 - (4 - x2)
. 4 - x2 - 2xh - h2 -4 + x2
_ 2
. (
h) _ 2
1
2
-=Inn
=hm
=1m-x--x- 0 --x
dx h-->O
h
h-->O
h
h-->O
I
ll. y = 2-x
SOLUTION
I
Lety =--.Then
2-x
I
I
1
dy = lim ~ - ~ = lim (2 - x) - (2 - x - h) = lim
dx h-->O
h
h-->O h(2 - x - h)(2 - x)
h->O (2 - x - h)(2 - x) = (2 - x) 2
In Exercises 13-16, express the limit as a derivative.
13. lim
Vi+h-1
h-->O
SOLUTION
h
Let f(x) =
Vx· Then
lim ...ff+h- 1 = lim f(l + h) - f(l) = f'(l)
h-->0
h
h-->0
h
sin tcos t
15. l i m - - ,__,,rr
t - 1f
SOLUTION Let f(t) = sintcos t and note that f(n) = sinncosn = 0. Then
Iim
f-'llr
sin t cost
t-1f
. f(t) - f (n) f'( )
=1m
1
=
7r
t-1f
t-:nr
17. Findf(4) andf'(4) ifthe tangent line to the graph off atx = 4has equationy = 3x-14.
The equation of the tangent line to the graph of f(x) at x = 4 is y = f'(4)(x - 4) + /(4) = /'(4)x + (/(4) 4/'(4)). Matching this toy= 3x - 14, we see that f'(4) = 3 and /(4)-4(3) = -14, so /(4) = -2.
SOLUTION
19. Is (A), (B), or (C) the graph of the derivative of the function f shown in Figure 3?
y
(A)
(B)
(C)
FIGURE 3
The graph off has four horizontal tangent lines on [-2, 2], so the graph of its derivative must have four
x-intercepts on [-2, 2]. This eliminates (B). Moreover, f is increasing at both ends of the interval, so its derivative must
be positive at both ends. This eliminates (A) and identifies (C) as the graph off'.
SOLUTION
21. Sketch the graph of a continuous function f if the graph off' appears as in Figure 5 and f (0) = 0.
y
2
3
FIGURE 5
4
Chapter Review Exercises
207
Examine Figure 5. For x < 1, f'(x) = 1, so that the graph off must be a line with slope 1. Because f(O) = 0,
it follows that f(x) = x for x < 1. For x 2: 1, f' is zero and then steadily increases, indicating that the graph off must
"emerge" from the point (1, 1) with zero slope and then curve upward. One possible graph off is shown below.
SOLUTION
y
23. A girl's height h(t) (in centimeters) is measured at time t (in years) for 0::; t::; 14:
52,
75.1, 87.5, 96.7, 104.5, 111.8, 118.7, 125.2,
131.5, 137.5, 143.3, 149.2, 155.3, 160.8, 164.7
(a) What is the average growth rate over the 14-yr period?
(b) Is the average growth rate larger over the first half or the second half of this period?
(c) Estimate h'(t) (in cm/yr) fort= 3, 8.
SOLUTION
(a) The average growth rate over the 14-year period is
164.7 - 52 = 8.05 cmlyear
14
(b) Over the first half of the 14-year period, the average growth rate is
125.2 - 52 -- 10.46 cmlyear
7
which is larger than the average growth rate over the second half of the 14-year period:
164.7 - 125.2 5 6 ml
"" . 4 c year
7
(c) Fort= 3,
h'(3) "" h(4) -_ h(3) = 104.5 - 96.7 = 7 .8 cmlyear
1
4 3
fort= 8,
h'(8)"" h(9) - h(8) = 137.5 - 131.5 = 6 0 ml
9-8
1
.cyear
In Exercises 25 and 26, use the following table of values for the number A(t) of automobiles (in millions) manufactured
in the United States in year t.
t
1970
1971
1972
1973
1974
1975
1976
A(t)
6.55
8.58
8.83
9.67
7.32
6.72
8.50
25. What is the interpretation of A'(t)? Estimate A'(1971). Does A'(1974) appear to be positive or negative?
Because A(t) measures the number of automobiles manufactured in the United States in year t, A'(t) measures the rate of change in automobile production in the United States. Fort= 1971,
SOLUTION
I
A (1971)""
A(1972) -A(1971)
8.83 - 8.58
_
=
= 0.25 million automobiles/year
1972 1971
1
Because A(t) decreases from 1973 to 1974 and from 1974 to 1975, it appears that A'(1974) would be negative.
27. Which of the following is equal to .!!.._ 2x?
dx
(a) 2x
(b) (ln2)2X
(c) x2x-I
(d) _1 2x
ln2
208
CH APTER 3
I
DIFFERENTIATION
The derivative of f(x) = 2x is
SOLUTION
Hence, the correct answer is (b).
In Exercises 29-80, compute the derivative.
29. y = 3.x5-7x2 +4
Let y = 3x5 - 7x2 + 4. Then
SOLUTION
dy
4
= 15x -14x
dx
-
31. y = r 13
Lety = r
SOLUTION
73
· •
Then
dy = -7 3rs.3
dt
.
x+l
33. y= -2-1
x +
x+l
Lety = - 2- - . Then
x +1
SOLUTION
dy
dx
35. y = (x4
(x2 + 1)(1) - (x + 1)(2x)
(x 2 + 1)2
1 - 2x - x2
(x2 + 1)2
9x) 6
-
Lety = (x4 - 9x) 6 • Then
SOLUTION
dy
dx
= 6(x4 -
5d 4
9x) dx (x
-
9x)
= 6(4x3 -
4
9)(x
-
9x) 5
37. y = (2 + 9:x2) 312
Lety = (2 + 9x2) 3i 2. Then
SOLUTION
dy
d
3
-dx = -(2
+ 9x2) 112 -(2 + 9x2) = 27x(2 + 9x2) 112
2
dx
z
-z
39. y = - ,,...---:
vl
SOLUTION
z
Lety =--.Then
~
dy
dz
41 ·Y-
x4
SOLUTION
~-z(-~) _ 1-z+ ~
1- z
- (1 - z)3/2
+ ...fX
x2
Let
y
= x4 + ...[X = x2 + x-3/2
x2
Then
-dy = 2x dx
3
-x-512
2
2-z
2(1 - z)3/2
Chapter Review Exercises
43. y ==
~x + ~x + Vx
SOLUTION
Lety ==
~x+ ~x+ Vx· Then
d
d~
==
1(
2
-~Vx)
112
x+ -vx+
d (
-~
dx x+ -vx+ Vx)
45. y == tan(t- 3 )
SOLUTION
Lety == tan(t- 3 ). Then
47. y == sin(2x) cos 2 x
SOLUTION
Let y == sin(2x) cos 2 x == 2 sin x cos3 x. Then
dy == -6 sin2 x cos2 x + 2 cos4 x
dx
49
t
• y == 1 + sect
SOLUTION
t
Lety == -1--· Then
+sect
dy
dt
l+sect-tsecttant
(1 +sec t)2
8
51 · y == _l_+_c_o_t_e
SOLUTION
Lety== 1
8
==8(1+coter 1.Then
+cote
dy
de
d
== -8(1 + cote)-2 -d (1 +cote)==
e
8csc2 e
(1 +cote)2
53. y == cos(x100 )
SOLUTION
Let y == cos(x 100 ). Then
-dy
dx
55. f(x) == 9e-4x
SOLUTION
~9e-4x ==
dx
-36e-4 x
57. g(t) == e41-t2
SOLUTION
59. f(x)
!!:...e4 t-t2 == (4 - 2t)e 41-t2
dt
= ln(4x2 + 1)
SOLUTION
d
8x
-d ln(4x2+1) == - 2 x
4x + 1
61. G(s) == (ln(s)) 2
SOLUTION
!£(ln s) 2 == 2 ln s
ds
s
. 100 d 100
= - sm(x
)-x = -100x99 sin(x100 )
dx
209
210
CH APTER 3
I
DIFFERENTIATION
63. f(8) = ln(sin8)
SOLUTION
d
.
cose
ln(sm 8) = - . - = cot 8
de
sme
-
65. h(z) = sec(z + In z)
SOLUTION
67. f(x) =
SOLUTION
!!_ sec(z + lnz) = sec(z + lnz) tan(z + lnz)
~
r2
(1 + ~)z
x
.!!:_ T 2x = ( - 2 ln 7)(T2x)
dx
69. g(x) = tan- 1(lnx)
SOLUTION
d
-I
1
dx tan (lnx) = 1 + (lnx)2
x
71. f(x) = ln(csc 1 x)
SOLUTION
d
-
dx
ln(csc- 1 x) =
lxl vx2 - 1 csc-1 x
73. R(s) = sins
SOLUTION
Rewrite
Then
dR = e(lns)2. 2lns. ~ = 2lns sins
ds
s
s
Alternately, R(s) = s1ns implies that lnR =In (s1ns) =(In s) 2. Thus,
1 dR
1
--=2lns·R ds
s
or
dR
2lns Ins
- = --s
ds
s
75. G(t) = (sin2 t)1
SOLUTION
Rewrite
G(t) = (elnsin
'Y =
2
e2tlnsint
Then
dG
2I . (
cos t
. )
. 2
-d
=et nsmt 2t· - . - + 2lnsmt = 2(sm t)'(tcott+lnsint)
t
smt
Alternately, G(t) = (sin2 t)' implies that In G = t In sin2 t = 2t In sin t. Thus,
1 dG
cost
.
- - = 2t· +2lnsmt
G dt
sint
and
~~
= 2(sin2 t)'(t cot t +In sin t)
77. g(t) = sinh(t2)
SOLUTION
fr
sinh(r) = 2tcosh(r)
79. g(x) = tanh- 1(ex)
SOLUTION
1
d
-d tanh- 1(ex) =
X
1-
(ex) 2
ex
ex
= -1 - e2x
81. For which values of a is f(x) = lxl" differentiable at x = O?
Chapter Review Exercises
Let f(x) = Jxr. If a < 0, then f(x) is not continuous at x = 0 and therefore cannot be differentiabl~ at
SOLUTION
x=
211
o. If a = O, then the function reduces to f(x) = 1, which is differentiable at x = 0. Now, suppose a > 0 and consider
the limit
lim f (x) - f (0) = lim Jxr
x-->0
x-0
x-->0 x
IfO<a<l,then
. Jxr
hm =-oo
x--+0-
. Jxl"'
hm =oo
while
x--+0+
X
X
and f' (0) does not exist. If a = l, then
lim lxl = -1
while
x-+0- X
and f'(O) again does not exist. Finally, if a> 1, then
. lxl"
hm-=0
x-->0
X
so f' (0) does exist.
In summary, f (x) = Jxr is differentiable at x = 0 when a = 0 and when a > 1.
Jn Exercises 83-88, use the following table of values to calculate the derivative of the given function at x = 2:
x
f (x)
g(x)
f'(x)
g'(x)
2
5
4
-3
9
4
3
2
-2
3
83. S(x) = 3f(x) - 2g(x)
Let S (x) = 3f(x)- 2g(x). Then S'(x) = 3f'(x) - 2g'(x) and
SOLUTION
S'(2) = 3f'(2) - 2g'(2) = 3(-3)- 2(9) = -27
f(x)
85. R(x)
= g(x)
LetR(x) = f(x)/g(x). Then
SOLUTION
R'(x) = g(x)f'(x) - f(x)g'(x)
g(x)2
and
/ 2) = g(2)f'(2) - f(2)g'(2) = 4(-3)- 5(9) = - 57
R(
g(2)2
42
16
87. F(x) = f(g(2x))
Let F(x) = f(g(2x)). Then F'(x) = 2f'(g(2x))g'(2x) and
SOLUTION
F'(2) = 2f'(g(4))g'(4) = 2f'(2)g'(4) = 2(-3)(3) = -18
89. Find the points on the graph of f(x) = x3 - 3x2 + x + 4 where the tangent line has slope 10.
Let f(x) = x3 - 3x2 + x + 4. Then f'(x) = 3x2 - 6x + 1. The tangent line to the graph off will have slope
10 when f'(x) = 10. Solving the quadratic equation 3.x2 - 6x + 1 = 10 yields x = -1 and x = 3. Thus, the points on the
graph off where the tangent line has slope 10 are (-1, -1) and (3, 7).
SOLUTION
91. Find a such that the tangent lines to y =
SOLUTION
f'(a) = 3a2
-
x3 - 2x2 + x + 1 at x
= a and x = a + 1 are parallel.
Let f(x) = x3 - 2x2 + x + 1. Then f'(x) = 3x2 - 4x + 1 and the slope of the tangent line at x = a is
4a + 1, while the slope of the tangent line at x =a+ 1 is
f' (a+ 1) = 3(a + 1)2 -
4(a + 1) + 1 = 3(a2 + 2a + 1) - 4a - 4 + 1 = 3a2 + 2a
In order for the tangent lines at x =a and x =a+ 1 to have the same slope, we must have f'(a) = f'(a + 1), or
3a2 -4a+ 1=3a2 +2a
The only solution to this equation is a
= ~.
212
CH APT E R 3
I
DIFFERENTIATION
In Exercises 93-98, calculate y".
93. y
= 12x3 -
SOLUTION
5x2 + 3x
Let y
= 12x3 -
5x2 + 3x. Then
y'
= 36x2 -
lOx + 3 and y"
= nx -
10
95. y = V2x + 3
SOLUTION
y'
97. y
Let y
= -../2x + 3 = (2x + 3) 1' 2. Then
1
d
= -(2x
+ 3)- 112 -(2x + 3) = (2x + 3)- 112
2
~
and
y"
1
d
= --(2x
+ 3)-3 ' 2 -(2x + 3) = -(2x + 3)-3' 2
2
~
= tan(x2)
SOLUTION
Let y
= tan(x2). Then
y' = 2x sec2 (x2 )
y"
.
and
= 2x (2 sec(x2) ~ sec(x2)) + 2 sec2 (x2) = 8x2 sec2 (x2) tan(x2) + 2 sec2 (x2)
d
In Exercises 99-104, computed~.
99. x 3 -y3 = 4
SOLUTION
Consider the equation x3
- y3
= 4. Differentiating with respect to x yields
3x2 - 3y2 dy = 0
dx
Therefore,
x2
dy
dx
y2
101. y = xy2 + 2x2
SOLUTION
Consider the equation y
= xy2 + 2x2. Differentiating with respect to x yields
dy
dx
dy
dx
- = 2.xy- + y 2 + 4x
Therefore,
dy
dx
103. y
y2 +4x
1-2.xy
= sin(x + y)
SOLUTION
Consider the equation y
= sin(x + y). Differentiating with respect to x yields
dy = cos(x + y) ( 1 + -dy)
-dx
dx
Therefore,
dy
dx
cos(x + y)
1-cos(x+y)
Chapter Review Exercises
213
d2 y
dy
In Exercises I 05 and 106 compute dx and dx2 •
105.
:x2 - 4y2
SOLUTION
= 8
Let :x2
- 4y2 = 8. Then
dy
2x-8ydx=0
so
dy
x
-=dx 4y
Moreover,
f
d2y
4y(l) - x(4~)
4y 4y2 - :x2
-8
dx2 =
l6y2
= 16y2 = ~ = 16y3 = - 2y3
107. In Figure 7, for the three graphs on the left, identify f, f', and f". Do the same for the three graphs on the right.
y
FIGURE 7
SOLUTION First consider the plot on the left. Observe that the green curve is nonnegative whereas the red curve is
increasing, suggesting that the green curve is the derivative of the red curve. Moreover, the green curve is linear with
negative slope for x < 0 and linear with positive slope for x > 0 while the blue curve is a negative constant for x < 0 and
a positive constant for x > 0, suggesting the blue curve is the derivative of the green curve. Thus, the red, green, and blue
curves, respectively, are the graphs off, f' and f".
Now consider the plot on the right. Because the red curve is decreasing when the blue curve is negative and increasing
when the blue curve is positive and the green curve is decreasing when the red curve is negative and increasing when the
red curve is positive, it follows that the green, red, and blue curves, respectively, are the graphs off, f', and f".
In Exercises 109-114, use logarithmic differentiation to find the derivative.
109
•y
= (x+1)3
(4x-2)2
SOLUTION
(x + 1)3
Let y = (4x - 2)2. Then
3
(x + 1) )
lny =In (
= ln(x+ 1)3 -ln(4x-2)2 = 3ln(x+ l)-2ln(4x-2)
(4x - 2) 2
By logarithmic differentiation,
y'
3
2
3
4
-=-----·4=----y
x + 1 4x - 2
x + 1 2x - 1
so
3
(x+l) ( 3
4)
y = (4x-2) 2 x+ 1 - 2x- l
I
111. y
= e<x-1)2 e<x-3)2
SOLUTION
Let y = e<x-l)2e<x- 3)2 • Then
2
2
lny = ln(e<x-1) e<x- 3l
)
2
2
= ln(e<x-1) +<x- 3l
)
= (x - 1)2 + (x - 3)2
By logarithmic differentiation,
y'
-y = 2(x -
1) + 2(x - 3)
= 4x -
so
y' = 4e<x-1)2 e<x-3)2 (x - 2)
8
214
CH AP TE R 3
I
DIFFERENTIATION
e3x(x -
113. y=
2)2
(x+1)2
e3x(x -
SOLUTION
Let y =
2)2
(x + 1)2 . Then
lny=ln
2)2)
=lne3x+ln(x-2)2 -ln(x+1)2
(x+ 1)2
e3x(x
(
= 3x + 2 ln(x - 2) - 2 ln(x + 1)
By logarithmic differentiation,
y'
2
2
-=3+----y
x-2 x+l
so
3
y= e x(x - 2)
2
(x
+ 1)
2
(
3 +2- - - -2 - )
x - 2
x +1
115. How fast does the water level rise in the tank in Figure 9 when the water level is h = 4 m and water pours in at
20m3/min?
FIGURE 9
When the water level is at height h, the length of the upper surface of the water is 24 + ~ h and the volume
of water in the trough is
SOLUTION
v=
15 2
1 ( 24 + 24 + -zh
3 ) (10) = 240h + 2h
-zh
Therefore,
dV
dh
.
- = (240 + 15h)-d = 20 m3/mm.
dt
t
When h = 4, we have
dh
dt
20
1 ml .
240 + 15(4) = 15
mm
117. Chloe and Bao are in motorboats at the center of a lake. At time t = 0, Chloe begins traveling south at a speed of
50 km/h. One minute later, Bao takes off, heading east at a speed of 40 km/h. At what rate is the distance between them
increasing at t = 12 min?
SOLUTION Take the center of the lake to be origin of our coordinate system. Because Chloe travels at 50 km/h= ~
km/min due south, her position at time t > 0 is (0, ~t); because Bao travels at 40 km/h= ~ km/min due east, her position
at time t > 1 is (~(t - 1), 0). Thus, the distance between the two motorboats at time t > 1 is
s=
4
9
25
36
1
6
-Ct - 1)2 + - t2 = - v41t2 - 32t + 16
and
ds
41t-16
dt
6vf4lt2 -32t+16
At t = 12, it follows that
ds
476
-d = _~ ~ 1.066 km/min
t
6 v5536
Chapter Review Exercises
215
119. In Figure 10, xis increasing at 2 cm/s, y is increasing at 3 cm/s, and fJ is decreasing such that the area of the triangle
has the constant value 4 cm2 .
(a) How fast is fJ decreasing when x = 4, y = 4?
(b) How fast is the distance between P and Q changing when x = 4, y = 4?
p
L
Q
y
FIGURE 10
SOLUTION
(a) The area of the triangle is
A
1
.
= -xysmB
=4
2
Differentiating with respect to t, we obtain
1
dfJ 1 . dx 1 . dy
-dA
= -xycosfJ+ -ysmB- + -xsmfJ- = 0
dt
2
dt
2
dt
2
dt
When x
= y = 4, we have i(4)(4) sinfJ = 4, so sinfJ = i· Thus, fJ = ~and
1
.../3 -dfJ + -(4)
1
( -1 ) (2) + -(4)
1 ( -1 ) (3)
-(4)(4)2
2 dt 2
2
2
2
=0
Solving for dfJ/dt, we find
dB
5
- = - - - ""-0.72 rad/s
dt
4.../3
(b) By the Law of Cosines, the distance D between P and Q satisfies
D 2 = x2 + y2 - 2xy cos fJ
so
dD
2D-d
t
With x
dx
dy
.
dfJ
= 2x-dt +2y+ 2xysmfJ- dt
dt
dy
dx
2xcosfJ- -2ycosfJdt
dt
= y = 4 and fJ = ~,
Therefore,
dD
dt
16 + 24 - ~ - 12 .../3 - 8-../3
c:--;;
8-yL-
vj
"" -1.50 cm/s
4 APPLICATIONS OF THE
DERIVATIVE
4.1 Linear Approximation and Applications
Preliminary Questions
1. True or False? The Linear Approximation says that the V<?rtical change in the graph is approximately equal to the
vertical change in the tangent line.
This statement is true. The linear approximation does say that the vertical change in the graph is approximately equal to the vertical change'in the tangent line.
SOLUTION
2. Estimate g(l.2) - g(l) if g'(l)
SOLUTION
= 4.
Using the Linear Approximation,
g(l.2)- g(l) ""g'(l)(l.2- 1) =4(0.2)
3. Estimate /(2.1) if /(2)
SOLUTION
=0.8
= 1 and f'(2) =3.
Using the Linearization,
/(2.1) ""/(2) + /'(2)(2.1 - 2)
= 1 + 3(0.1) = 1.3
4. Complete the following sentence: The Linear Approximation shows that up to a small error, the change in output /::,,f
is directly proportional to ------------- .
SOLUTION The Linear Approximation tells us that up to a small error, the change in output /::,,f is directly proportional
to the change in input /::,,x when /::,,x is small.
Exercises
In Exercises 1-6, use Eq. (1) to estimate /::,,f =/(3.02) - /(3).
1. f(x)
=i2
SOLUTION
3. f(x)
=x- 1
SOLUTION
5. f(x)
Let f(x) = i2. Then f'(x) = 2x and/::,,/"" f'(3)/::,,x = 6(0.02) = 0.12.
Let f(x) = x- 1• Thenf'(x) = -x- 2 and/::,,/"" f'(3)dx
1
= -9(0.02) =
-0.00222.
= vx+6
SOLUTION
Let f(x) = e2x. Then f'(x) = 2e2x and
/::,,f"" f'(3)fa = 2e6 (0.02) = 0.04e6 = 16.137
7. The cube root of 27 is 3. How much larger is the cube root of 27.2? Estimate using the Linear Approximation.
SOLUTION Let f(x)
=x
113
,
a
=27, and tu =0.2. Then f'(x) = ~x-
213
and f'(a)
=f'(27) = 'ii· The Linear
Approximation is
/::,,/"" f'(a)dx
Thus, -</27.2 is larger than
= 271 (0.2) =0.0074074
m by approximately 0.0074074.
217
218
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
In Exercises 9-12, use Eq. (1) to estimate !if. Use a calculator to compute both the error and the percentage error.
9. f (x) =
SOLUTION
-Yl+X,
a = 3,
Ax = 0.2
Let f(x) = (1 + x) 112 , a = 3, and Ax= 0.2. Then f' (x) =
!CI + xt 1t2 , f'(a) = f'(3) = ~ and
!if~ J'(a)!ix = ~(0.2) = 0.05
The actual change is
!if= f(a +Ax) - f(a) = f(3.2) - f(3) =
V4.2 - 2 ~ 0.049390
The error in the Linear Approximation is therefore 10.049390 - 0.051 = 0.000610; in percentage terms, the error is
0.000610
0.049390 x IOO% ~ 1. 24 %
1
11. f(x) = - - 2 ,
1 +x
SOLUTION
a = 3,
Let f(x)
Ax= 0.5
= 1.;.2 , a= 3, and Ax= 0.5. Then f'(x)
!if~
=-
(I;;i)l, f'(a) = f'(3) = -0.06 and
f'(a)!ix = -0.06(0.5) = -0.03
The actual change is
!if= f(a +Ax) - f(a) = f(3.5) - f(3)
~
-0.0245283
The error in the Linear Approximation is therefore I - 0.0245283 - (-0.03)1 = 0.0054717; in percentage terms, the
error is
0.00547171
I-0.0245283
x lOO% ~ 2231 %
In Exercises 13-20, using Linear Approximation, estimate !if for a change in xfrom x =a to x
approximate f(b), and.find the error using a calculator.
= yx, a = 25, b = 26
SOLUTION Let f(x) = yx, a= 25, and Ax= I. Then f'(x) = 4x- 112 and f'(a)
• The Linear Approximation is !if~ f'(a)!ix = fo(l) = 0.1.
= b. Use the estimate to
13. f(x)
• Using the Linear Approximation, f(b)
~
f(a) +!if= 5 + 0.1
= f'(25) = fo·
= 5.1.
• Using a calculator, the error in this estimate is I '126- 5.11~15.0990195 - 5.11=0.0009805.
15. f(x) =
SOLUTION
}x, a= 100, b = 101
Let f(x) =
}x, a= 100, and Ax= I. Then f'(x) =
fx(x- 112 ) = -tx-312 and f'(a) =
-!( 1o\io) = -0.0005.
• The Linear Approximation is !if~ f'(a)!ix = -0.0005(1) = -0.0005.
• Using the Linear Approximation,
f(b)
~
1
f(a) +!if= IO+ (-0.0005) = 0.0995
• Using a calculator, the error in this estimate is
I~ - 0.09951 ~ 10.0995037 _ 0.09951 = 3.7 x 10-
6
17. f(x) = x 113 , a= 8, b = 9
SOLUTION
Let f(x) = x 113 , a= 8, and !ix= 1. Then f'(x) = ~x-213 and f'(a) = f'(8) = fz.
• The Linear Approximation is !if,,, f'(a)!ix = fz(l) = 0.083333.
• Using the Linear Approximation, f(b) ~ f(a) +!if= 2 + 0.083333 = 2.083333.
• Using a calculator, the error in this estimate is 1-W - 2.0833331 ,,, 12.080084 - 2.0833331 = 0.003249.
S E CT I 0 N 4.1
I
Linear Approximation and Applications
219
19. f(x) = eX, a= 0, b = -0.1
SOLUTION
Let f(x) = eX, a= 0, and /:u = -0.1. Then f'(x) =ex and f'(a) = f'(O) = 1.
• The Linear Approximation is h.f,,, f'(a)l:1x = 1(-0.1) = -0.1.
• Using the Linear Approximation, f(b),,, f(a) + l:1f = 1+(-0.1)=0.9.
• Using a calculator, the error in this estimate is le-0· 1 - 0.91 ,,, I0.904837 - 0.91
=0.004837.
In Exercises 21-28,find the linearization at x =a and then use it to approximate f(b).
21. f(x) = x4,
SOLUTION
a = 1,
Let f(x) =
b = 0.96
x4. Then f'(x) = 4x3, f(a) = 1, and f'(a) = 4. The linearization ata = 1 is
L(x) = f'(a)(x - a)+ f(a) = 4(x - 1) + 1 = 4x - 3
and
f(b) = f(0.96) ,,, L(0.96)
a= ~.
2
23. f(x) = sin x,
SOLUTION
=4(0.96) - 3 =0.84
'1"
b=
Letf(x) = sin x. Thenf'(x) = 2sinxcosx = sin2x, f(a) = ~. andf'(a) = 1. Thelinearization ata = ~
2
is
L(x) = f ' (a)(x - a)+ f(a) = 1 ( x - -1C) + -1 = x - -1C + -1
4
2
4 2
and
l.11C 1C 1 0.11C 1
f(b)=f ( -l.11C) ,,,L ( -l.11C) =---+-=-+-,,,0578540
4
4
4
42
4
2.
25. f(x) = (1 + x)- 112 ,
SOLUTION
a= 0,
b = 0.08
Let f(x) = (1 + x)-
112
•
Then f'(x) = -~(1 + xt 312 , f(a) = 1, and f'(a) = -~.The linearization at a= 0 is
1
L(x) = f'(a)(x -a)+ f(a) = -2x + 1
and
f(b) = f(0.08),,, L(0.08) =
27. f(x) = eVx,
SOLUTION
a= 1,
-~(0.08) + 1 = 0.96
b = 0.85
Letf(x) = eVx. Then
f'(x) =
1
2
Vxe'1X, f(a) = e, and f'(a) =
1
2
e
so the linearization off(x) at a = 1 is
L(x) = f'(a)(x - a)+ f(a) = !e(x - 1) + e
2
= !e(x + 1)
2
and
1
f(b) = f(0.85),., L(0.85) = 2e(0.85 + 1) =0.925e,,, 2.514411
In Exercises 29~32, estimate l:1y using differentials [Eq. (3)].
a=
29. y = cosx,
SOLUTION
E• dx =0.014
Let f(x)
=cosx. Then f'(x) =-sinx and
l:1y,,, dy = f'(a)dx = -sinrn)<o.014) = -0.001
10-.x2
31. y= - ,
2+x2
a= 1,
dx=0.01
220
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
SOLUTION
Let f(x)
• Then
= -10-x2
2 +x2
f'(x) = (2 + x2)(-2x) - (10- x2)(2x) = _
(2 + x2)2
24x
(2 + x2)2
and
.1y "'dy
= f'(a)dx = -
24
= -0.026667
(0.01)
9
33. Estimate /(4.03) for f(x) as in Figure 9.
y
y
=
tcx) /
I
\"
(10, 4)
. / T'""'"'""
/(4,2)
~r--~--'-~~~~~~---x
FIGURE 9
SOLUTION
Using the Linear Approximation, /(4.03) "'f(4) + f'(4)(0.03). From the figure, we find that /(4)
= 2 and
4-2
1
f'( 4 ) = 10-4 = 3
Thus,
1
/(4.03) "'2 + 3(0.03)
= 2.01
fil - ...fi or Y9J - .../9? Explain using the Linear Approximation.
Letf(x) = yx, and .1x = 0.1. Then f'(x) = !x- 112 and the Linear Approximation at x=a gives
35. Which is larger:
SOLUTION
-~
.1f = va + u.1
-
Va"' J'(a)(0.1) = -1 a- 112 (0.l) = -0.05
...;a
2
We see that df decreases as a increases. In particular
is larger than
37. Box office revenue at a multiplex cinema in Paris is R(p) = 3600p - 10p3 euros per showing when the ticket price is
p euros. Calculate R(p) for p = 9 and use the Linear Approximation to estimate M if p is raised or lowered by 0.5 euro.
LetR(p) = 3600p- 10p3. Then R(9)
so by the Linear Approximation,
SOLUTION
= 3600(9) -
M "' R' (9).1p
10(9)3 = 25110 euros. Moreover, R'(p)
= 3600 -
30p2 ,
= l 170.1p
If p is raised by 0.5 euros, then M "' 585 euros; on the other hand, if p is lowered by 0.5 euros, then M "' -585 euros.
39. A thin silver wire has length L = 18 cm when the temperature is T = 30°C. Estimate .1L when T decreases to 25°C
if the coefficient of thermal expansion is k = 1.9 x 10-5 oc- 1 (see Example 3).
SOLUTION
We have
:~ = kL = (1.9 x
The change in temperature is .1T
10-5 )(18)
= 3.42 x
10-4 cm/°C
= -5°C, so by the Linear Approximation, the change in length is approximately
.1L"' 3.42 x 10-4 .1T
= (3.42 x 10-4 )(-5) = -0.00171 cm
At T = 25°C, the length of the wire is approximately 17 .99829 cm.
41. The atmospheric pressure at altitude h (kilometers) for 11 :::; h:::; 25 is approximately
P(h) = 128e-0· 157h kilopascals
SE
cT I0 N
4.1
I
Linear Approximation and Applications
221
(a) Estimate 11P at h = 20 when 11h = 0.5.
(b) Compute the actual change, and compute the percentage error in the Linear Approximation.
SOLUTION
(a) Let P(h)
= 128e-o.157h. Then P'(h) = -20.096e-0 ·151h. Using the Linear Approximation,
11P::; P'(h)/1h
= P'(20)(0.5) = -0.434906 kilopascals
(b) The actual change in pressure is
P(20.5) - P(20)
= -0.418274 kilopascals
The percentage error in the Linear Approximation is
-0.434906 - (-0.418274) I
,,,
100910 3 98910
•
-0.418274
x
1
43. Newton's Law of Gravitation shows that if a person weighs
weight at distance x from the center of the earth is
wR2
x2
W(x) = -
w pounds on the surface of the earth, then his or her
(for x
~
R)
where R = 3960 miles is the radius of the earth (Figure 10).
(a) Show that the weight lost at altitude h miles above the earth's surface is approximately /1 W ::; -(0.0005w)h. Hint:
Use the Linear Approximation with dx = h.
(b) Estimate the weight lost by a 200-lb football player flying in a jet at an altitude of 7 miles.
FIGURE 10 The distance to the center of the earth is 3960 + h miles.
SOLUTION
(a) Using the Linear Approximation
,
11W::; W (R)l1x
(b) Substitute w
2wR2
2wh
= ---wh = -R "'-0.0005wh
= 200 and h = 7 into the result from part (a) to obtain
11W"' -0.0005(200)(7) = -0.7 pounds
45. A stone tossed vertically into the air with initial velocity v emfs reaches a maximum height of h = v2 /1960 cm.
(a) Estimate 11h if v = 700 emfs and /1v = 1 emfs.
(b) Estimate 11h if v = 1000 emfs and /1v = 1 emfs.
(c) In general, does a 1-cmfs increase in v lead to a greater change in hat low or high initial velocities? Explain.
SOLUTION A stone tossed vertically with initial velocity v cm/s attains a maximum height of h(v) = v2 /1960 cm.
Thus, h'(v) = v/980.
(a) If v = 700 and /1v = 1, then l1h::; h'(v)/1v = 9 ~0 (700)(1) "'0.71 cm.
= 1, then l1h "' h' (v)/1v = 9 ~0 (1000)(1) =1.02 cm.
(c) A 1 centimeter per second increase in initial velocity v increases the maximum height by approximately v/980 cm.
Accordingly, there is a bigger effect at higher velocities.
(b) If v = 1000 and /1v
In Exercises 47 and 48, use the following fact derived from Newton's Laws: An object released at an angle() with initial
velocity v ft/s travels a horizontal distance
s
= 312 v2 sin W ft (Figure 11)
222
C HAP TE R 4
I
APPLICATIONS OF THE DERIVATIVE
s
FIGURE 11 Trajectory of an object released at an angle 8.
47. A player located 18. l ft from the basket launches a successful jump shot from a height of l 0 ft (level with the rim of
the basket), at an angle 8 = 34° and initial velocity v = 25 ft/s.
(a) Show that 11s "" 0.255118 ft for a small change of 118.
(b) Is it likely that the shot would have been successful if the angle had been off by 2°?
(c) Estimate l:!.s if 8 = 34°, v = 25 ft/s, and 11v = 2.
SOLUTION
Using Newton's laws and the given initial velocity of v = 25 ft/s, the shot travels s
fi sin 2t ft, where tis in radians.
~v2 sin2t
=
6
(a) If 8
= 34° (i.e., t = ~7r), then
11s"" s' (t)l:!.t
625 cos ( 17 7r) l:!.t = l6
625 cos ( 17 7r) 118 · 1T ""0.255118
= l6
45
45
180
(b) If 118 = 2°, this gives 11s "" 0.51 ft, in which case the shot would not have been successful, having been off half a
foot.
(c) Using Newton's laws and the fixed angle of 8 = 34° = ~7r, the shot travels
1
s
With v = 25 ft/sand 11v
= 32 v
2
.
17
sm 45 7r
= 2 ft/s, we find
11s"" s'(v)l1v =
l
. I 77r
(25) sm 4S · 2
16
= 2.897 ft
49. The radius of a spherical ball is measured at r == 25 cm. Estimate the maximum error in the volume and surface area
if r is accurate to within 0.5 cm.
SOLUTION The volume and surface area of the sphere are given by V == ~7rr 3 and S
11r = ±0.5, then
= 47rr2, respectively. If r = 25 and
11 V "" V' (25)1:!.r = 47r(25) 2 (0.5) "" 3927 cm3
and
11S ""S'(25)M = 87r(25)(0.5)"" 314.2 cm2
51. The volume (in liters) and pressure P (in atmospheres) of a certain gas satisfy PV = 24. A measurement yields V == 4
with a possible error of ±0.3 L. Compute P and estimate the maximum error in this computation.
SOLUTION
Thus, P'
Given PV == 24 and V == 4, it follows that P = 6 atmospheres. Solving PV
= -24v-2 and
11P"" P'(4)11 V == -24(4)- 2 (±0.3)
53. Approximate f(2) if the linearization of f(x) at a
SOLUTION
/(2) "" L(2)
Let f(x)
= ±0.45 atmospheres
= 2 is L(x) = 2x + 4?
= 2(2) + 4 == 8.
55. Estimate .../16.2 using the linearization L(x) of f(x}
determine whether the estimate is too large or too small.
SOLUTION
= 24 for P yields P == 24v- 1 •
= x 112 , a=
16, and 11x
= -./X at a = 16. Plot f
= 0.2. Then f'(x) = !x- 112 and f'(a)
and Lon the same set of axes and
== f'(l6) == ~·The linearization to
f(x) is
1
I
L(x) == f'(a)(x- a)+ f(a) == g(x- 16) + 4 = gx + 2
Thus, we have V16.2"" L(l6.2) == 4.025. Graphs of f(x) and L(x) are shown below. Because the graph of L(x) lies above
the graph off(x), we expect that the estimate from the Linear Approximation is too large.
s E c TI o N
4.1
I
Linear Approximation and Applications
223
In Exercises 57--65, approximate using linearization and use a calculator to compute the percentage error.
1
57. - ,,.-;;
v17
SOLUTION
to f(x) is
Let f(x) = x- 112 , a= 16, and /':i.x = 1. Then f'(x) = -~x- 3 1 2 , f'(a) = f'(16) = - 1 ~ 8 and the linearization
I
1
1
1
3
L(x) = f (a)(x - a)+ f(a) = --(x - 16) + - = - - x + 128
4
128
8
Thus, we have
Ju z L(l 7) z 0.24219. The percentage error in this estimate is
...L
0.242191
Vfi - ...L
x 100%,,, 0.14%
Vfi
I
1
59. (10.03)2
SOLUTION
Let f(x) = x-2 , a= 10 and -6.x = 0.03. Then f'(x) = -2x-3, f'(a) = f'(lO)
= -0.002 and the linearization
to f(x) is
L(x) = J'(a)(x - a)+ f(a)
= -0.002(x -
10) + O.Ql = -0.002x + 0.03
Thus, we have
1
(10.03)2 ""L(l0.03)
= -0.002(10.03) + 0.03 = 0.00994
The percentage error in this estimate is
-
1
- 2 -0009941
.X 100%;;:, 0.0027%
1_·
(IO.Q3) _
1
(10.03)2
61. (64.1) 113
SOLUTION
Let f(x) = x 113 , a = 64, and ~x = 0.1. Then f'(x) = ~x-2 13, f'(a) = f'(64) =
f(x) is
L(x)=f'(a)(x-a)+f(a)=
1
(x-64)+4=
48
;8 x+~
Thus, we have (64.1) 113 ""L(64.1) ""4.002083. The percentage error in this estimate is
(64.1) 113 - 4.0020831
(64.1)1/3
x 100% ""0.000019%
I
63. cos- 1(0.52)
SOLUTION
Let f(x) = cos- 1 x and a = 0.5. Then
1
2
f'(x) = - -- , f'(a) = f'(O) = - V3
../1 - x 2
3
and the linearization to f(x) is
L(x)
7r
= f'(a)(x- a)+ f(a) == -2.../3
--(x- 0.5) + 3
3
Thus, we have cos- 1(0.52) "" L(0.02) == 1.024104. The percentage error in this estimate is
cos- 1(0.52)- 1.0241041
cos-l(0.52)
x 100% ""0.015%
l
.jg
and the linearization to
224
CH APT E R 4
I
APPLICATIONS OF THE DERIVATIVE
65• e-0.012
SOLUTION
Let f(x) =ex and a= 0. Then f'(x)
= f'(a)(x -
L(x)
Thus, we have e- 0·012 ""L(-0.012)
= e"', f'(a) = f'(O) =
=1-
0.012
a)+ f(a)
= l(x -
I and the linearization to f(x) is
0) + 1 = x + 1
= 0.988. The percentage error in this estimate is
e-0.012 - 0.9881
e_0_012
x 100% "" 0.0073%
I
67. Show that the Linear Approximation to f(x) = Vi at x = 9 yields the estimate .../9 + h - 3"" kh. Set K = 0.01 and
show that lf"(x)I::; K for x;:::: 9. Then verify numerically that the error E satisfies Eq. (5) for h = 10-n, for I ::; n::; 4.
SOLUTION
Let f(x)
= '{i. Then f(9) = 3, f'(x) = tx- 112 and f'(9) = i· Therefore, by the Linear Approximation,
J(9+h)-f(9)
=
v9 +h-3"" 61h
Moreover, f"(x) = -ix- 312 , so lf"(x)I = ix- 312 • Because this is a decreasing function, it follows that for x ~ 9,
K =max lf"(x)I ::; 1/"(9)1 =
1
< 0.01
108
From the following table, we see that for h = 10-n, 1 ::; n ::; 4, E ::; ~ Kh 2 .
h
E
= I .../9 + h -
10-1
10-2
10-3
10-4
~Kh 2
3 - ihl
4.604 x 10-5
4.627 x 10-7
4.629 x 10-9
4.627 x 10- 11
5.00 x 10-5
5.00 x 10-7
5.00 x 10-9
5.00x 10- 11
Further Insights and Challenges
69. Compute dy/dx at the point P = (2 I) on the curve y 3 + 3.xy = 7 and show that the Iinearization at Pis L(x)
1
-tx +~-Use L(x) to estimate they-coordinate of the point on the curve where x = 2.1.
SOLUTION
=
Differentiating both sides of the equation y3 + 3.xy = 7 with respect to x yields
3y2 dy + 3x dy + 3y = 0
dx
dx
so
dy
dx
y
- y 2 +x
Thus,
dy'
- __
1_ dx (2,1) - 12 + 2 -
_!
3
and the linearization at P = (2, 1) is
L(x)
I
1
5
= 1- -(x-2)
= --x+
3
3
3
Finally, when x = 2.1, we estimate that the y-coordinate of the point on the curve is
y ""L(2.I) = -
71. Apply the method of Exercise 69 to P
SOLUTION
1
5
3(2.1) + j = 0.967
= (-1,2) ony4 + 7.xy = 2 to estimate the solution ofy4 -
Differentiating both sides of the equation y4 + 7.xy = 2 with respect to x yields
4ydy +7xdy +7y=O
dx
dx
so
dy
dx
7y
4y3 +·7x
7.7y = 2 near y
= 2.
s Ec T IO N
4.2
I
Extreme Values
225
Thus,
dy
dx
and the linearization at P
I
-
= f(5 + h) -
SOLUTION
Let f(x)
14
7.7y
14
25
14
25
36
25
= 2- -(x+ 1) = --x+ -
= 2 corresponds to x = -1.1, so we estimate the solution of this equation near y = 2 is
y ""L(-1.1)
73. Let t::..f
7(2)
= - 4(2)3 + 7(-1) =- 25
=(-1, 2) is
L(x)
Finally, the equation y4
(-l,2)
f(5), where f(x)
14
36
= - 25(-1.1) + 25 = 2.056
= x2. Verify directly that E = \Af -
f'(5)h\ satisfies (5) with K
= 2.
= x2. Then
Af = f(5 + h) - f(5) = (5 + h) 2
-
52
= h2 + lOh
and
E = \t::..f- J'(5)h\
= \h2 + lOh -
lOh\ = h2 ~ 2h2
4.2 Extreme Values
Preliminary Questions
1. What is the definition of a critical point?
A critical point is a value of the independent variable x in the domain of a function fat which either
f'(x) = 0 or f'(x) does not exist.
SOLUTION
In Questions 2 and 3, which is the correct conclusion, (a) or (b)?
2. If f is not continuous on [0, 1], then
f has no extreme values on [0, l].
f might not have any extreme values on [O, 1].
SOLUTION The correct response is (b): f might not have any extreme values on [0, l]. Although [O, 1] is closed,
(a)
(b)
because f is not continuous, the function is not guaranteed to have any extreme values on [O, 1].
3. If f is continuous but has no critical points in [0, l], then
(a)
f
has no min or max on [O, 1].
(b) Either f(O) or f(l) is the minimum value on [0, l].
SOLUTION The correct response is (b): either f(O) or f(l) is the minimum value on [O, 1]. Remember that extreme
values occur either at critical points or endpoints. If a continuous function on a closed interval has no critical points, the
extreme values must occur at the endpoints.
4. For each statement, indicate whether it is true or false. If false, correct the statement or explain why it is false.
(a) If f'(c) = 0, then f(c) is either a local minimum or a local maximum.
(b) If f(c) is the absolute maximum off on an interval I, then f'(c) = 0.
(c) If f is differentiable and f(c) is a local minimum off, then f'(c) = 0.
(d) If there is one local minimum off on an interval /,then it is the absolute minimum on/.
SOLUTION
(a) This statement is false. The condition f'(c) = 0 is not sufficient to guarantee that f(c) is either a local minimum or a
local maximum. For example, consider the function f(x) = x3. Then f'(O) = 0, but f(O) is neither a local minimum nor
a local maximum.
(b) This statement is false. If f(c) is the absolute maximum off on an interval/, it could be that f'(c) does not exist or
that f'(c) is any real number if c happens to be an endpoint of the interval/. For example, consider the function f(x) = x2
on the interval [0, 2]. The absolute maximum value off on [O, 2] is f(2) = 4, but f' (2) = 4 0.
(c) This statement is true.
(d) This statement is false. If there is one local minimum off on an interval I, then the absolute minimum might occur
at an endpoint of the interval. For example, consider the function f(x) = x3 - x2 on the interval [-1, l]. The function f
has only one local minimum on this interval, at x = 2/3, but the absolute minimum occurs at the endpoint x = -1.
*
226
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
Exercises
1.
(a)
(b)
(c)
The following refer to Figure 18.
What are the critical points off on [0, 8]?
What are the maximum and minimum values off on [O, 8]?
What are the local maximum and minimum values off, and where do they occur?
(d) Find a closed interval on which both the minimum and maximum values off occur at critical points.
(e) Find an open interval on which f has neither a minimum nor a maximum value.
(f) Find an open interval on which f has a maximum value but no minimum value.
y
6
5
4
; y =f(x)
3
2
1 2
3 4 5 6 7
FIGURE 18
8
SOLUTION
(a) The function f has three critical points on the interval [0, 8]: at x = 3, x = 5, and x = 7. 1\vo of these, x = 3 and
x = 5, are where the derivative is zero and one; x = 7 is where the derivative does not exist.
(b) The maximum value off on [0, 8] is 6; the function takes this value at x = 0. The minimum value is 1; the function
takes this value at x = 7.
(c) The function f achieves a local maximum of 5 at x = 5. On the other hand, the function f achieves a local minimum
value of 3 at x = 3 and a local minimum value of 1 at x = 7.
(d) Answers may vary. One example is the interval [4, 8]. Another is [2, 6].
(e) Answers may vary. One example is (0, 2).
(f) Answers may vary. One example is (4, 6).
In Exercises 3-20,find all critical points of thefu,nction.
3. f(x) = x2
2x + 4
Let f(x)
SOLUTION
5. f(x) =
-
.x3 -
~x
2
-
= x2 -
2x + 4. Then f' (x)
54x + 2
Let f(x) = .x3 - ~x2 = 6 are the critical points off.
SOLUTION
x
= 2x - 2 = 0 implies that x = 1 is the lone critical point off.
54x + 2. Then f'(x)
= 3x2 -
9x- 54
= 3(x + 3)(x -
6)
= 0 implies that x = -3 and
7. f(x) = x- 1 - x- 2
SOLUTION
Let J(x) = x- 1 - x- 2. Then
2-x
= -x-2 + 2x-3 = =0
x3
'
f (x)
implies that x = 2 is the only critical point off. Though f' does not exist at x = 0, this is not a critical point off because
x = 0 is not in the domain off.
x
9. f(x) = x2 +
1
x
1-x2
SOLUTION Let f(x) = - 2- - . Then f'(x) = 2 l) 2 = 0 implies that x = ±1 are the critical points off.
x +1
(x +
11. f(t) = t - 4 Yt+I
SOLUTION
Let f(t) = t - 4
w+r. Then
f
implies that t
13. f(x)
I
(t)= I - -2- =0
w+I
= 3 is a critical point off. Because f' does not exist at t = -1, this is another critical point off.
= xe2x
SOLUTION
Let J(x)
= xe2x. Then f'(x) = (2x + l)e2x = 0 implies that x = -~ is the only critical point off.
S E C T I O N 4.2
15. g(B)
I
Extreme Values
227
= sin2 (}
SOLUTION
Let g(B)
= sin2 B. Then g'(B) = 2 sinBcos (} = sin2B = 0 implies that
nn
2
B= -
is a critical value of g for all integer values of n.
17. f(x) = xlnx
SOLUTION
Let f(x) = x In x. Then f'(x) = I +In x = 0 implies that x = e- 1 = ~ is the only critical point off.
19. f(x) = sin- 1 x - 2x
SOLUTION
Let f(x) = sin- 1 x - 2x. Then
f'(x)
implies that x = ±
21. Let f(x) =
f
2x2 -
=-
1
-
-Vt - x 2
-2 = 0
are the critical points off.
Sx + 7.
(a) Find the critical point c off and compute f(c).
(b) Find the extreme values off on (0, 5].
(c) Find the extreme values off on (-4, l].
SOLUTION
(a) Let f(x) = 2x2 - Sx + 7. Then f'(x) = 4x - 8 = 0 implies that c = 2 is the critical point off. At the critical point,
/(2) = -1.
(b) First note that the critical point c = 2 lies in the interval [O, 5]. Comparing the value of f at the endpoints of the ,
interval (0, 5], f(O) = 7 and /(5) = 17, with the value off at the critical point, f (2) = -1, we see that the maximum
value off on (0, 5] is 17 and the minimum value is -1.
(c) First note that the critical point c = 2 does not lie in the interval [-4, l]. Comparing the value off at the endpoints
of the interval [-4, 1], f ( -4) = 71 and /(1) = 1, we see that the maximum value off on [-4, 1] is 71 and the minimum
value is 1.
23. Find the critical points of f(x) = sin x +cos x and determine the extreme values on [O,
fl.
SOLUTION
• Let f(x) = sinx + cosx. Then on the interval
point off in this interval.
• Since/(~)=
[o, ~],we have f'(x) = cosx -
sinx = 0 at x = ~.the only critical
Y2 and /(0) = f(~) = 0, the maximum value off on [o, ~]is V2, while the minimum value is 0.
!Gq
Plot f(x) = 4 -{X - 2x + 3 on (0, 3] and indicate where it appears that the minimum and maximum occur.
25.
Then determine the minimum and maximum using calculus.
From the graph off(x) = 4 -{X - 2x + 3, which is shown below, the minimum value off on the interval
[O, 3] appears to occur at x = 0 while the maximum value appears to occur at x "' 1.
SOLUTION
y
Maximum
5
4
2
x
2
To determine the minimum and maximum values using calculus, first note that the domain off is the set {xix
!
,(x)
= -
2
Vx
c 0). Now,
-2
We see that f'(x) = 0 when x = 1, and, on the domain off, f'(x) does not exist when x = 0. Thus, x = 0 and x = 1 are
the critical points off. With f(O) = 3, f(l) = 5, and f(3) = 4 ../3 - 3 "'3.9282, it follows that the minimum value off,
on the interval [O, 3] is /(0) = 3, and the maximum value is f(l) = 5.
In Exercises 27-60,find the minimum and maximum values of the function on the given interval by comparing values at
the critical points and endpoints.
27. y
= 2x2 + 4x + 5,
(-2,2]
228
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
SOLUTION Let f(x) = 2.x2 + 4x + 5. Then f'(x) = 4x + 4 = 0 implies that x = -1 is the only critical point off. The
minimum off on the interval [-2, 2] is f(-1) = 3, whereas its maximum is f(2) = 21. (Note: f(-2) = 5.)
29. y
= 6t -
t2 ,
[O, 5]
Let f(t) = 6t - t2. Then f'(t) = 6 - 2t = 0 implies that t = 3 is the only critical point off. The minimum
off on the interval [0, 5] is f(O) = 0, whereas the maximum is f(3) = 9. (Note: f(5) = 5.)
SOLUTION
31. y
= x3 -
6.x2
+ 8,
[l, 6]
Let f(x) = x3 - 6x2 + 8. Then f'(x) = 3.x2 - l2x = 3x(x - 4) = 0 implies that x = 0 and x = 4 are the
critical points off. The minimum off on the interval [l, 6] is f(4) = -24, whereas the maximum is /(6) = 8. (Note:
f(l) = 3 and the critical point x = 0 is not on the interval [l, 6].)
SOLUTION
33. y
= x3 -
6x2 + 8,
[l, 3]
SOLUTION Let f(x) = x3 - 6.x2 + 8. Then f'(x) = 3.x2 - l2x = 3x(x - 4) = 0 implies that x = 0 and x = 4 are the
critical points off. The minimum off on the interval [l, 3] is /(3) = -19, whereas the maximum is f(l) = 3. (Note:
Neither of the critical points are on the interval [1, 3].)
35. y
= 2t3 + 3t2,
[1,2]
Let f(t) = 2t3 + 3t2 . Then f'(t) = 6t2 + 6t = 6t(t + 1) = 0 implies that t = O and t = -1 are the critical
points off. The minimum off on the interval [1, 2] is f(l) = 5, whereas the maximum is f(2) = 28. (Note: Neither of
the critical points are on the interval [l, 2].)
SOLUTION
37. y
= z5 -
80z,
[-3,3]
Let f(z) = z5 - 80z. Then f'(z) = 5z4 - 80 = 5(z4 - 16) = 5(z2 + 4)(z + 2)(z - 2) = 0 implies that z = ±2
are the critical points off. The minimum value off on the interval [-3,3] is f(2) = -128, whereas the maximum is
f(-2) = 128. (Note: f(-3) = 3 and f(3) = -3.)
SOLUTION
x2 + 1
39. y = --4'
xSOLUTION
[5,6]
Let f(x)
=
.x2 + 1
x - 4 . Then
,
f
(x) =
(x - 4) · 2x - (x2
(x - 4)2
+ 1) · l
.x2 - 8x - l
=
=0
(x - 4)2
implies x = 4 ± Yf7 are critical points off. x = 4 is not a critical point because x = 4 is not in the domain off. On the
= 18.5, whereas the maximum off is f(5) = 26. (Note: The critical points
interval [5, 6], the minimum off is f(6) =
x = 4 ± Yf7 are not on the interval [5, 6].)
¥
4x
41. y = x - --1,
x+
SOLUTION
[0, 3]
Let f (x)
=x -
4x
--1 . Then
x+
f'(x) = 1 _ _4_ = (x- l)(x+ 3) =
0
(x + 1)2
(x + 1)2
implies that x = 1 and x = -3 are critical points off. x = -1 is not a critical point because x = -1 is not in the domain
off. The minimum off on the interval [0, 3] is f(l) = -1, whereas the maximum is f(O) = f(3) = 0. (Note: The critical
point x = -3 is not on the interval [0, 3].)
= (2 + x) -../2 + (2 - x)2, [0, 2]
SOLUTION Let f(x) = (2 + x) ,,j2 + (2 43. y
f' (x) =
-../2 + (2 - x)2
x) 2. Then
-
(2 + x)(2 + (2 - x)2r 112(2 - x) =
2
<x - l)
=
-../2 + (2 - x) 2
2
implies that x = 1 is the critical point off. On the interval [O, 2], the minimum is f (0) = 2 V6
is /(2) = 4 -fi. "' 5.66. (Note: f(l) = 3 V3"' 5.2.)
45. y = .Yx + xL 2-{X,
SOLUTION
Let f(x)
o
"' 4.9 and the maximum
[0,4]
= .Yx + x 2 - 2 Vx· Then
f'(x) = !(x + .x2)-112(l + 2x) - x-112 = 1+2x - 2
2
implies that x = 0 and x =
f
~=0
2-{XV!+X
are the critical points of f. Neither x = -1 nor x = -
neither is in the domain off. On the interval [0,4], the minimum off is f(
/(4) "'0.472136. (Note: f(O) = 0.)
f) "'
f
is a critical point because
-0.589980 and the maximum is
I
SECT I o N 4.2
229
[O, ~]
47. y = sinxcosx,
[o.
Let f(x) = sinxcosx = ~ sin2x. On the interval
~]. f'(x) = cos2x = 0 when x =%·The minimum of
on this interval is f(O) = /(~) = 0, whereas the maximum is/(%)= ~-
SOLUTION
f
Extreme Values
49. y=
Vie-secB,
[O,!f]
Let f(B) =
SOLUTION
...fi.e- sec e. On the interval [O, :O.J'(B) = V2 - sec Btan () = 0 at()= %·The minimum value
off on this interval is f(O) = -1, whereas the maximum value over this interval is/(%} = Vi.(% - 1) ,.,, -0.303493.
(Note: f(}) = .../2} - 2,.,, -0.519039.)
[-2,2]
51. y=x3+x2-x,
Let f(x) = x3 + x2 - x. Then f'(x) = 3x2 + 2x - 1 = (3x - l)(x + 1) = 0 implies that x = 1/3 and x = -1
are critical points off. The minimum off on the interval [-2, 2] is f (-2) = -2, whereas the maximum is f (2) = l 0.
(Note: f(-1) = 1 and f(l/3) = -5/27.)
SOLUTION
53. y = () - 2 sin(),
[O, 2n]
SOLUTION Letg(B) = B-2sinB. On the interval [0,2n], g'(B) = l -2cosB = 0 at()=} and()= ~n. The minimum
of g on this interval is g(}) = } - .../3 ,.,, -0.685 and the maximum is g( ~n) = ~n + .../3 ,., 6.968. (Note: g(O) = 0 and
g(2n) = 2n ,.,, 6.283.)
55. y = tanx-2x,
SOLUTION
/(%) = l 57. y
~
lnx
x
= -,
SOLUTION
[O, l]
Let f(x) = tanx- 2x. Then on the interval [0, l], f'(x) = sec2 x- 2 = 0 at x = %·The minimum off is
,.,, -0.570796 and the maximum is f(O) = 0. (Note: f(l) =tan 1 - 2,.,, -0.442592.)
[1,3]
:x. Then, on the interval [l, 3],
Let f(x) = 1
f'(x)= 1-lnx =O
x2
at x = e. The minimum off on this interval is f(l) = 0 and the maximum is f(e) = e- 1
ln 3 ,.,, 0.366204.)
t
59. y
= 3ex -
e2x,
,.,,
0.367879. (Note: f(3) =
[-~, l]
Let f(x) = 3~ - e2x. Then, on the interval[-~. l], f'(x) = 3ex - 2e 2x = ex(3 - 2ex) = 0 at x = ln(3/2).
The minimum off on this interval is f(l) = 3e - e2 ,.,, 0.765789 and the maximum is f(ln(3/2)) = 2.25. (Note:
f(-1/2) = 3e- 112 - e- 1 ,.,, 1.451713.)
SOLUTION
:x2
61. !Guij Plot f(x) = 2
on (0, 5) and use the graph to explain why there is a minimum value, but no maximum
value, off on (0, 5). Use calculus to find the minimum value.
The figure below shows the graph of f(x) = 2+/ on (0, 5). As x ~ o+, f(x) ~ oo. Therefore, f has no
absolute maximum value on (0, 5) but the graph suggests there is an absolute minimum at x ,.,, 1.4. To determine the
absolute minimum value off on (0, 5), first rewrite
SOLUTION
2 + x2
2
f(x) = - - = - + x = 2x- 1 + x
x
x
Then
f'(x)
2
= -2x-2 + l = -+l
x2
On the interval (0, 5), the derivative is defined and is equal to zero when x =
off on (0,5) is
2
.../2. Therefore, the absolute minimum value
f ( .../2) = 2 + ( .../2) = _i__ = 2 Vi
V2
V2
y
x
I
2
3
4
5
230
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
63. Letf(8)=2sin28+sin48.
(a) Show that 8 is a critical point if cos 48 = - cos 28.
(b) Show, using a unit circle, that cos 81 = - cos 82 if and only if 81 = n ± 82 + 2nk for an integer k.
(c) Show thatcos48 = -cos28ifand only if8 = ~ +nkor8 = ~ + CVk.
(d) Find the six critical points off on [0, 2n] and find the extreme values off on this interval.
(e)
!GUii
Check your results against a graph off.
f(8) = 2 sin 28 + sin 48 is differentiable at all 8, so the way to find the critical points is to find all points
such that f'(8) = 0.
(a) f'(8) = 4cos 28 + 4cos48. If f'(8) = 0, then 4cos 48 = -4cos 28, so cos 48 = -cos 28.
(b) Given the point (cos 8, sin 8) at angle 8 on the unit circle, there are two points with x coordinate - cos 8. The graphic
shows these two points, which are:
SOLUTION
• The point (cos(8 + n), sin(8 + n)) on the opposite end of the unit circle.
• The point (cos(n - 8), sin(8- n)) obtained by reflecting through they axis.
If we include all angles representing these points on the circle, we find that cos 81 = - cos 82 if and only if 8 1 = (n + 82) +
2nk or 81 = (n - 82) + 2nk for integers k.
(c) Using (b), we recognize that cos48 = -cos28 if 48 = 28 + n + 2nk or 48 = n - 28 + 2nk. Solving for 8, we obtain
8 = ~ +kn or 8 = ~ + ~k.
(d) To find all 8, 0 ~ 8 < 2n indicated by (c), we use the following table:
k
0
l
~+kn
2"
T
~
6"
2"
+ ~k
2
3
4
5
3,,
5,,
3,,
71f
6
6
T
111'
6
The critical points in the range [0, 2n] are~. ~. ~' 7tf ,~,and '!".On this interval, the maximum value is f<V = f(7tf) =
3
f,3 and the minimum value is f (~) = f ( 1!") = - 3 f,3.
(e) The graph of f(8) = 2 sin W +sin 48 is shown here:
y
We can see that there are six fiat points on the graph between 0 and 2n, as predicted. There are 4 local extrema, and two
points at ( ~, 0) and ( ~, 0) where the graph has neither a local maximum nor a local minimum.
In Exercises 65--68,find the critical points and the extreme values on [0, 4]. In Exercises 67 and 68, refer to Figure 2I.
y
y
y
-6 '
2
y = lx 2 + 4x - 121
=lcosxl
-rr-i---;~~7r~~rr~~3~rr~x
-2
FIGURE 21
2
2
s Ec T Io N
65. y
= Ix -
4.2
I
Extreme Values
231
21
Let f(x) = Ix - 21. For x < 2, we have f'(x) = -1. For x > 2, we have f'(x) = 1. Now as x --4 2-,
f(x) - f (2)
(2 - x) - 0
f (x) - f (2)
(x - 2) - 0
we have
= · x-2 --4 -1; whereas as x --4 2+, we have x-2 = x-2 --4 1. Therefore,
x-2
2
f'(2) = lim f(x) - f( ) does not exist and the lone critical point off is x = 2. Alternately, we examine the graph of
x--+2
x-2
f(x) = Ix - 21 shown below.
To find the extremum, we check the values off at the critical point and the endpoints. f(O) = 2, f(3) = 1, and
f(2) = 0. f takes its minimum value of 0 at x = 2, and its maximum of 2 at x = 0.
SOLUTION
y
2
1.5
0.5
2
0
4
67. y = lx2 + 4x - 121
SOLUTION Let f(x) = lx2 + 4x - 121 = l(x + 6)(x - 2)1. From the graph off in Figure 21, we see that f'(x) does not
exist at x = -6 and at x = 2, so these are critical points of f. There is also a critical point between x = -6 and x = 2 at
which f'(x) = 0. For -6 < x < 2, f(x) = -x2 - 4x + 12, so f'(x) = -2x - 4 = 0 when x = 2. On the interval [0, 3] the
minimum value off is f(2) = 0 and the maximum value is f(O) = 12. (Note: f(3) = 9 and the critical points x = -6 and
x = -2 are not in the interval.)
In Exercises 69-72, verify Rolle's Theorem for the given interval by checking f(a)
(a, b) such that f'(c) = 0.
[~,2]
69. f(x) = x + x- 1 ,
SOLUTION
Because f is continuous on[~, 2], differentiable on(~, 2) and
f
(!)2 = !2 + .!. = ~2 = + !2 =
x2
= Sx _ 15 ,
2
~
we may conclude from Rolle's Theorem that there exists a c
we may take c = 1.
71. f(x)
= f(b) and then finding a value c in
E
f(2)
(~, 2) at which f'(c)
= 0. Here, f'(x) = 1 -
x-2
= x:~ 1 , so
[3, 5]
SOLUTION Because f is continuous on [3, 5], differentiable on (3, 5) and f(3)
Rolle's Theorem that there exists a c E (3, 5) at which f'(c) = 0. Here,
f'(
= f(5) = 1, we may conclude from
_ (8x-15)(2x)- 8x2 _ 2x(4x-15)
(8x - 15)2
(8x - 15)2
x) -
so we may take c =
lf.
73. Prove that f(x)
= x5 + 2x3 + 4x -
12 has precisely one real root.
SOLUTION Let's first establish that f(x) = x5 + 2x3 + 4x - 12 has at least one root. Because f is a polynomial, it is
continuous for all x. Moreover, f(O) = -12 < 0 and f (2) = 44 > 0. Therefore, by the Intermediate Value Theorem, there
exists a c E (0, 2) such that f(c) = 0.
Next, we prove that this is the only root. We will use proof by contradiction. Suppose f(x) = x5 + 2x3 + 4x - 12 has
two real roots, x = a and x = b. Then f(a) = f(b) = 0 and Rolle's Theorem guarantees that there exists a c E (a, b) at
which f'(c) = 0. However, f'(x) = 5x4 + 6x2 + 4 ~ 4 for all x, so there is no c E (a,b) at which f'(c) = 0. Based on
this contradiction, we conclude that f(x) = x5 + 2x3 + 4x - 12 cannot have more than one real root. Finally, f must have
precisely one real root.
75. Prove that f(x)
Theorem.
= x4 + 5x3 + 4x has no root c satisfying c > 0. Hint: Note that x = 0 is a root and apply Rolle's
SOLUTION We will proceed by contradiction. Note that f(O) = 0 and suppose that there exists a c > 0 such that
f(c) = 0. Then f(O) = f(c) = 0 and Rolle's Theorem guarantees that there exists ad E (0, c) such that f'(d) = 0.
However, f'(x) = 4.x3 + 15x2 + 4 > 4 for all x > 0, so there is no d E (0, c) such that f'(d) = 0. Based on this
contradiction, we conclude that f (x) = x4 + 5.x3 + 4x has no root c satisfying c > 0.
232
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
77. The position of a mass oscillating at the end of a spring is s(t) = A sin wt, where A is the amplitude and w is the
angular frequency. Show that the speed lv(t)I is at a maximum when the acceleration a(t) is zero and that la(t)I is at a
maximum when v(t) is zero.
SOLUTION
Let s(t) =A sin wt. Then
ds
v(t) = - =Aw cos wt
dt
and
a(t)
dv
.
= dt = -Aw2 sm wt
Thus, the speed
lv(t)I
= IAw cos wtl
is a maximum when Icos wtl = 1, which is precisely when sin wt
acceleration a(t) is zero. Similarly,
= O; that is, the speed lv(t)I is at a maximum when the
la(t)I = 1Aw2 sinwtl
is a maximum when Isin wtl = 1, which is precisely when cos wt = 0; that is, la(t)I is at a maximum when v(t) is zero.
79. [ R 5 Antibiotic Levels A study shows that the concentration C(t) (in micrograms per milliliter) of antibiotic in
a patient's blood serum after t hours is C(t) = 120(e-0·21 - e-h1), where b ~ 1 is a constant that depends on the particular
combination of antibiotic agents used. Solve numerically for the value of b (to two decimal places) for which maximum
concentration occurs at t = 1 h. You may assume that the maximum occurs at a critical point, as suggested by Figure 22.
C(mcg/mL)
100
80
60
40
20
->-T-1-+-i>-T-t--+-i>-+-+-+-i-
2
FIGURE 22 Graph of C(t)
SOLUTION
= 120(e-0·2' -
.4
6
8
10
t (h)
lf
e-bt) with b chosen so that the maximum occurs at t
= 1 h.
Answer is b = 2.86. The max of C(t) occurs at t = ln(5b)/(b - 0.2) so we solve ln(5b)/(b - 0.2) = 1
numerically.
81. In 1919, physicist Alfred Betz argued that the maximum efficiency of a wind turbine is around 59%. If wind enters
a turbine with speed v1 and exits with speed v2 , then the power extracted is the difference in kinetic energy per unit time:
P
1 2
= 2mv
1-
1 2
2mv2
watts
where mis the mass of wind flowing through the rotor per unit time (Figure 23). Betz assumed that m = pA(v 1 + v2 )/2,
where p is the density of air and A is the area swept out by the rotor. Wind flowing undisturbed through the same area A
would have mass per unit time pAv1 and power Po = tpAvi. The fraction of power extracted by the turbine is F =Pf P 0 •
(a) Show that F depends only on the ratio r = v 2 /v 1 and is equal to F(r) =
r 2 )(1 + r), where O ~ r ~ 1.
(b) Show that the maximum value of F, called the Betz Limit, is 16/27 ""0.59.
to -
(c) ~ Explain why Betz's formula for Fis not meaningful for r close to zero. Hint: How much wind would pass
through the turbine if v2 were zero? Is this realistic?
F
0.6
0.5
0.4
0.3
0.2
0.1
0.5
(A) Wind flowing through a turbine.
(B) Fis the fraction of energy
extracted by the turbine as a
function of r = v21v1.
FIGURE 23
S E CT I o N 4.2
I
Extreme Values
233
SOLUTION
(a) We note that
F
p
4pA(vrv2\vi - vi)
Po
4pAvt
= - = -=---=--'--1
vf - v~
v 1 + v2
=1~·-V-1-
v2)
1 ( 1 -v~)(
=- 1+-
vi
2
VJ
(b) Based on part (a),
1
= -(1
2
I
F (r)
2
r ) - r(l
+ r)
The roots of this quadratic are r = -1 and r =~-Now, F(O) =
32
= --r
2
1
2
+-
r
1, F(l) = 0 and
Thus, the Betz Limit is 16/27 ""0.59.
(c) If v2 were 0, then no air would be passing through the turbine, which is not realistic.
83. The response of a circuit or other oscillatory system to an input of frequency w ("omega") is described by the function
<fi(w) =
1
----;:::=====
~(w6 - w2)2 + 4D2w2
Bothw0 (the natural frequency of the system) and D (the damping factor) are positive constants. The graph of<{! is called a
resonance curve, and the positive frequency Wr > 0, where<{! takes its maximum value, if it exists, is called the resonant
frequency. Show that Wr = ~w6 - 2D2 if 0 < D < w / V2 and that no resonant frequency exists otherwise (Figure 25).
0
2
Wr
(A)D
=
0.01
(l)r
(B)D=0.2
2
2
(C) D
=
3
0.75 (no resonance)
FIGURE 25 Resonance curves with w 0 = 1.
and the nonnegative critical points are w = 0 and w =
w6 - 2D
2
~w6 -
2D2 • The latter critical point is positive if and only if
> 0, and since we are given D > 0, this is equivalent to 0 < D <
Define Wr =
~w6 -
2D2 • Now, <fi(O) =
l/w6 and <fi(w)
w Vi.
0/
--+ 0 as w--+ oo. Finally,
1
<fi(wr) = - - - - -
2D~w6-D2
which, for 0 < D < w 0 /
Vi, is larger than 1/ w6. Hence, the point w = ~w~ -
2D2 , if defined, is a local maximum.
234
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
85. Find the maximum of y = X' - X1 on [O, 1], where 0 < a < b. In particular, find the maximum of y =
[0, l].
x5 -
x 10 on
SOLUTION
• Let f(x) = X' - X1. Then f'(x) = ax:i- 1 - bX1- 1 • Since a < b, f'(x) = x:i- 1(a - bX1-a) = 0 implies critical points
x = 0 and x = <V 1f<h-a>, which is in the interval [O, l] as a < b implies ~ < 1 and consequently x = (~) 1 /(b-aJ < 1.
Also, f(O) = f(l) = 0 and a < b implies X' > X1 on the interval [0, l], which gives f(x) > 0 and thus the maximum
value off on [0, l] is
• Let f(x) = x5 - x 10 • Then by part (a), the maximum value off on [0, l] is
2
4
4
In Exercises 86--88, plot the function using a graphing utility and find its critical points and extreme values on [-5, 5].
1
lmTII
87• ~
SOLUTION
1
y = 1 + lx-11 + -1-+-lx---4-1
Let
1
f(x)
= 1 +Ix -
1
11 + -l-+-,-lx-----,,41
The plot follows:
-5 -4 -3 -2 -I
I
2
3 4
5
We can see on the plot that the critical points of f (x) lie at the cusps at x = 1 and x = 4 and at the location of the
it follows that the maximum
horizontal tangent line at x = ~-With f(-5) = ~,f(l) = f(4) = ~,f(~) = ~ and f(5) =
value off on [-5,5] isf(l) = f(4) =~and the minimum value isf(-5) = ~-
fo,
89. (a) Use implicit differentiation to find the critical points on the curve 27x2 = (x2 + y2) 3 .
(b) ~GUii
Plot the curve and the horizontal tangent lines on the same set of axes.
SOLUTION
(a) Differentiating both sides of the equation 27 x2 = (x2 + y2) 3 with respect to x yields
54x
= 3(x2 + y2) 2 ( 2x + 2y :~)
Solving for dy/dx we obtain
dy
dx
27x - 3x(x2 + y 2 ) 2
3y(x2 + y2)2
x(9 - (x2 + y2 ) 2 )
y(x2 + y2)2
Thus, the derivative is zero when x2 + y2 = 3. Substituting into the equation for the curve, this yields x2 = 1, or x = ±1.
There are therefore four points at which the derivative is zero:
(-1, -v'z), (-1, v'z), (1, -v'z), (1, v'z)
There are also critical points where the derivative does not exist. This occurs when y = 0 and gives the following points
with vertical tangents:
co, o), (± m, o)
(b) The curve 27x2 = (x2 + y2) 3 and its horizontal tangents are plotted below.
s Ec T Io N
4.2
I
Extreme Values
235
y
I
91. Sketch the graph of a continuous function on (0, 4) having a local minimum but no absolute minimum.
SOLUTION Here is the graph of a function f on (0, 4) with a local minimum value [between x = 2 and x = 4) but no
absolute minimum [since f(x)---+ -oo as x---+ O+].
y
10
2
-10
93. Sketch the graph of a function f on [O, 4) with a discontinuity such that f has an absolute minimum but no absolute
maximum.
Here is the graph of a function f on [0,4) that (a) has a discontinuity [at x = 4) and (b) has an absolute
minimum [at x = OJ but no absolute maximum [since f(x) ---+ oo as x ---+ 4-).
SOLUTION
y
4
2
0
4
2
Further Insights and Challenges
95. Show that the extreme values of f(x) = a sin x + b cos x are ± Ya2 + b2.
SOLUTION If f(x) = asinx + bcosx, then f'(x) = acosx - bsinx, so that f'(x) = 0 implies acosx - bsinx = 0.
This implies tan x = ~. Then,
±a
sinx=--Ya2 +b2
and
±b
cosx= - - Ya2 + b2
Therefore
±a
f(x) =a sinx + bcos x =a ....r;;r+b2 + b
a2 + b2
±b
...r;;r+b2 =
a2
+ b2
a 2 + b2
a2 + b2
± ~ = ± Ya2 + b2
97. Show thatifthe quadratic polynomial f(x) = x2 + rx +stakes on both positive and negative values, then its minimum
value occurs at the midpoint between the two roots.
SOLUTION Let f(x) = x2 + rx + s and suppose that f takes on both positive and negative values. This will guarantee
that f has two real roots. By the quadratic formula, the roots off are
X=
-r±~
2
Observe that the midpoint between these roots is
~(-r+ ~ +
2
Next, f'(x) = 2x + r = 0 when x =
-~
2
-r- ~)=-:_
2
2
and, because the graph of f(x) is an upward opening parabola, it follows that
f (- ~) is a minimum. Thus, f takes on its minimum value at the midpoint between the two roots.
236
C HAP TE R 4
I
APPLICATIONS OF THE DERIVATIVE
99. A cubic polynomial may have a local min and max, or it may have neither (Figure 28). Find conditions on the
coefficients a and b of
f(x) =
1
1
3x3 + 2ax2 +bx+ c
that ensure f has neither a local min nor a local max. Hint: Apply Exercise 96 to f'(x).
y
y
60
(A)
(B)
FIGURE 28 Cubic polynomials.
SOLUTION Let f(x) = ~x3 + ~ax2 +bx+ c. Using Exercise 96, we have g(x) = f'(x) = x2 +ax+ b > 0 for all x
provided b > 2 , in which case f has no critical points and hence no local extrema. (Actually b ~ 2 will suffice, since
in this case [as we'll see in a later section] f has an inflection point but no local extrema.)
ia
ia
lSJ
101.
Prove that if f is continuous and f(a) and f(b) are local minima where a < b, then there exists a value c
between a and b such that f(c) is a local maximum. (Hint: Apply Theorem I to the interval [a, b].) Show that continuity
is a necessary hypothesis by sketching the graph of a function (necessarily discontinuous) with two local minima but no
local maximum.
SOLUTION
• Let f(x) be a continuous function with f(a) and f(b) local minima on the interval [a, b]. By Theorem 1, f(x) must
take on both a minimum and a maximum on [a, b]. Since local minima occur at f(a) and f(b), the maximum must
occur at some other point in the interval, call it c, where f(c) is a local maximum.
• The function graphed here is discontinuous at x = 0.
y
-+--+--+--+--+--+--+--+--+-x
2 4 6 8
-8 -6 -4 -2
4.3 The Mean Value Theorem and Monotonicity
Preliminary Questions
1. For which value of mis the following statement correct? If /(2)
a tangent line of slope m.
SOLUTION
is differentiable, then f has
The Mean Value Theorem guarantees that the function has a tangent line with slope equal to
_f(_4)_-_f_(2_)
4-2
Hence, m
= 3 and f(4) = 9, and f
= 9_-_3 = 3
4-2
= 3 makes the statement correct.
2. Assume f is differentiable. Which of the following statements does not follow from the MVT?
(a) If f has a secant line of slope 0, then f has a tangent line of
slope 0.
(b) If f(5) < f(9), then f'(c) > 0 for some c
E
(5, 9).
(c) If f has a tangent line of slope 0, then f has a secant line of slope 0.
(d) If f'(x) > 0 for all x, then every secant line has positive slope.
SOLUTION
f(x)
Conclusion (c) does not follow from the Mean Value Theorem. As a counterexample, consider the function
= x3. Note that f'(O) = 0, but no secant line has zero slope.
s E CT I O N
I
4.3
The Mean Value Theorem and Monotonicity
237
3. Can a function with the real numbers as its domain that takes on only negative values have a positive derivative? If
so, sketch an example.
SOLUTION
Yes. The figure below displays a function that takes on only negative values but has a positive derivative.
y
4. For f with derivative as in Figure 14:
(a) Is f(c) a local minimum or maximum?
(b) Is fa decreasing function?
y
FIGURE 14 Graphofderivativef'.
SOLUTION
(a) To the left of x =c, the derivative is positive, sojis increasing; to the right of x = c, the derivative is negative, sof is
decreasing. Consequently, f(c) must be a local maximum.
(b) No. The derivative is a decreasing function, but as noted in part (a), f(x) is increasing for x < c and decreasing for
x> c.
5. Which of the six standard trigonometric functions have infinitely many local minima and infinitely many local
maxima but no absolute maximum and no absolute minimum over their whole domain?
SOLUTION The trigonometric functions _sec x and csc x have infinitely many local minima and infinitely many local
maxima but no absolute maximum and no absolute minimum over their whole domain.
6. Compose the absolute value with a familiar function to define a function f that
• has infinitely many local maxima, all of which occur where f' = 0, and
• has infinitely many local minima, all of which occur where f' is undefined.
The figure below shows the graph of f(x) =I sin xi. We see that f has infinitely many local maxima, all of
which occur where f' = 0, and has infinitely many local minima, all of which occur where f' is undefined. The function
f(x) = I cos xi also satisfies these conditions.
SOLUTION
y
-2rr
2rr
Exercises
In Exercises /-8,find a point c satisfying the conclusion of the MVT for the given.function and interval.
1. y
= x- 1,
SOLUTION
[2,8]
Let f(x)
= x- 1, a= 2, b = 8. Then f'(x) = -x-2, and by the MVT, there exists a c E (2, 8) such that
-~ = f'(c) = f(b)- f(a)
c2
Thus c2
= 16 and c = ±4. Choose c = 4 E (2, 8).
3. y = cos x - sin x,
[O, 2n]
b-a
_
f - ~ = __
8 -2
16
238
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
SOLUTION Let f(x) = cosx - sinx, a = 0, b = 2:1r. Then f'(x) = -sinx - cosx, and by the MVT, there exists a
c E (0, 2:1r) such that
.
,
f(b) - f(a)
1- 1
-smc-cosc=f(c)=
= - - =0
b-a
2n-O
Thus - sin c = cos c. Choose either c = ~ or c =
5. y = x
3
,
SOLUTION
f
E
(0, 2:1r).
[-4, 5]
Let f(x) =
x3, a= -4, b = 5. Then f'(x) = 3x2, and by the MVT, there exists a c E (-4, 5) such that
3c 2 = f'(c) = f(b) - f(a) = 189 = 21
b-a
9
Solving for c yields c2 = 7, so c = ±
0. Both of these values are in the interval [-4, 5], so either value can be chosen.
7. y = e- 2x, [0,3]
SOLUTION
Let f(x) = e- 2x, a= 0, b = 3. Then f'(x) = -2e-2X, and by the MVT, there exists a c E (0, 3) such that
6
6
_ 2e_zc = J'(c) = f(b) - f(a) = e- - 1 = e- - 1
b-a
3-1
2
Solving for c yields
C
=
+ "'
-21 ln ( 1
-
6
)
0.6944
E
(0, 3)
In Exercises 9-12,find a point c satisfying the conclusion of the MVT for the given junction and interval. Then draw the
graph of the function, the secant line between the endpoints of the graph and the tangent line at (c,f(c)), to see that the
secant and tangent lines are, in fact, parallel.
9. y =
x2,
[0, 1]
Letf(x) = x2, a= 0, andb = 1. Thenf'(x) = 2x. Because! is continuous on the closed interval [O, 1] and
differentiable on the open interval (0, 1), by the MVT, there exists a c E (0, 1) such that
SOLUTION
2c = J'(c) = f(b)- f(a) = 1 - 0 = 1
b-a
1-0
Solving for c yields c = ~. The figure below displays the graph of y = x2 as the solid curve, the secant line between the
points (0, 0) and (1, 1) as the dotted line, and the tangent line at x = ~ as the dashed line. It is clear that the secant line
and the tangent line are parallel.
y
2.0
1.5
-·
1.0
0.5
11. y =ex,
1.0
1.5
[0, 1]
Let f(x) = eX, a= 0, and b = 1. Then f'(x) =ex. Because f is continuous on the closed interval [0, 1] and
differentiable on the open interval (0, 1), by the MVT, there exists a c E (0, 1) such that
SOLUTION
ec = J'(c) = f(b)- f(a) = e - 1 = e _ 1
b-a
l-0
Solving for c yields c = ln(e - 1). The figure below displays the graph of y = ex as the solid curve, the secant line
between the points (0, 1) and (1, e) as the dotted line, and the tangent line at x = ln(e - 1) as the dashed line. It is clear
that the secant line and the tangent line are parallel.
y
0.2
0.2
0.4
0.6
0.8
1.0
I
S EC TIO N 4.3
The Mean Value Theorem and Monotonicity
239
IGUll
13.
Letf(x) = x5 +.x2. Thesecantlinebetween(O,O)and(l,2)hasslope2(checkthis),sobytheMVT,f'(c) = 2
for some c E (0, I). Plot f and the secant line on the same axes. Then plot y = 2x + b for different values of b until the
line becomes tangent to the graph off. Zoom in on the point of tangency to estimate the x-coordinate c of the point of
tangency.
SOLUTION
Let f(x)
= x5 + .x2. The slope of the secant line between x = 0 and x = l is
f(l) - f(O)
=2-
l -0
0
l
=2
A plot off, the secant line between x = 0 and x = I, and the line y = 2x - 0.764 is shown below at the left. The line
y = 2x - 0.764 appears to be tangent to the graph of y = f(x). Zooming in on the point of tangency (see below at the
right), it appears that the x-coordinate of the point of tangency is approximately 0.62.
y
0.6
y=x5+x2
y
0.5
0.4
0.56
0.6
0.64
15. Determine the intervals on which f'(x) is positive and negative, assuming that Figure 15 is the graph off.
¥"
~
12
3456
FIGURE 15
SOLUTION The derivative off is positive on the intervals (0, I) and (3, 5) where f is increasing; it is negative on the
intervals (1, 3) and (5, 6) where f is decreasing.
17. State whether f(2) and /(4) are local minima or local maxima, assuming that Figure 15 is the graph off'.
SOLUTION
• f' makes a transition from positive to negative at x = 2, so f (2) is a local maximum.
• f' makes a transition from negative to positive at x = 4, so f(4) is a local minimum.
In Exercises 19-22, sketch the graph of a function
f whose derivative f' has the given description.
19. f'(x) > 0 for x > 3 and f'(x) < 0 for x < 3
SOLUTION
Here is the graph of a function f for which f'(x) > 0 for x > 3 and f'(x) < 0 for x < 3.
y
IO
6
4
2
0
2
4
21. f'(x) is negative on (1, 3) and positive everywhere else.
SOLUTION
Here is the graph of a function f for which f'(x) is negative on (I, 3) and positive elsewhere.
240
CH APT E R 4
I
APPLICATIONS OF THE DERIVATIVE
y
In Exercises 23-26, find all critical points off and use the First Derivative Test to determine whether they are local
minima or maxima.
23. f (x) = 4 + 6x - x2
SOLUTION Let f(x) = 4 + 6x - x2. Then f'(x) = 6 - 2x = 0 implies that x = 3 is the only critical point off. As x
increases through 3, f'(x) makes the sign transition+,-. Therefore, f(3) = 13 is a local maximum.
25. f(x)
x2
x+
= - -1
.SOLUTION
x2
Letf(x) =--.Then
x+l
f' (x)
= x(x + 2) = 0
(x + 1)2
implies that x = 0 and x = -2 are critical points. Note that x = -1 is not a critical point because it is not in the domain
off. As x increases through -2, f'(x) makes the sign transition+, - so f(-2) = -4 is a local maximum. As x increases
through 0, f'(x) makes the sign transition-,+ so f(O) = 0 is a local minimum.
In Exercises 27-58,find the critical points and the intervals on which the function is increasing or decreasing. Use the
First Derivative Test to determine whether the critical point yields a local min or max (or neither).
SOLUTION Here is a table legend for Exercises 27-48.
SYMBOL
MEANING
-
The entity is negative on the given interval.
0
The entity is zero at the specified point.
+
The entity is positive on the given interval.
u
The entity is undefined at the specified point.
/'
f is increasing on the given interval.
\.
f is decreasing on the given interval.
M
f has a local maximum at the specified point.
m
f has a local minimum at the specified point.
-,
There is no local extremum here.
27. y=-x2+7x-l7
SOLUTION
29. y
= x3 -
SOLUTION
Let f(x) = -x2 + 7x - 17. Then f'(x) = 7 - 2x = 0 yields the critical point c = ~·
x
(-oo,
!'
f
n
7/2
n.oo)
+
0
-
/'
M
\.
l2.x2
Let f(x) =
x3 -
12x2. Then f'(x) = 3x2 - 24x = 3x(x- 8) = 0 yields critical points c = 0, 8.
x
(-oo, 0)
0
(0,8)
8
(8, oo)
f'
+
0
-
0
+
f
/'
M
\.
ID
/'
s E c TIO N
31. y
= 3x4 + 8.x3 -
= 12.x3 + 24.x2 - 12x - 24
= 12x2(x + 2) - 12(x + 2) = 12(x + 2)(.x2 = 12 (x -
yields critical points c
1) (x + l)(x + 2)
1)
=0
= -2, -1, 1.
x
(-oo, -2)
-2
(-2,-1)
-1
(-1, 1)
1
(1, CXl)
f'
-
0
+
0
-
0
+
m
/'
M
m
/'
f
3
241
6x2 - 24x
f'(x)
= ~x
The Mean Value Theorem and Monotonicity
Let f(x) = 3x4 + 8.x3 - 6.x2 - 24x. Then
SOLUTION
33. y
I
4.3
"""
"""
+ ~x2 + 2x + 4
SOLUTION
Let f(x)
=
~x3 + ~x2 + 2x + 4. Then f'(x)
= x2
+ 3x + 2
= (x + 1) (x + 2) = 0 yields critical points
c = -2,-1.
x
(-oo, -2)
-2
(-2,-1)
-l
(-1,oo)
f'
+
0
-
0
+
f
/'
M
m
/'
"""
35. y
= x5 + x3 + 1
SOLUTION
Let f(x)
= x5 + x3 + 1. Then f'(x) = 5x4 + 3.x2 = x2(5x2 + 3) yields a single critical point: c = 0.
x
(-oo, 0)
f'
f
0
(0, CXl)
+
0
+
/'
..,
/'
.,
37. y = x4
- 4x312 (x > 0)
SOLUTION Let f(x) = x4 - 4x312 for x > 0. Then f'(x) = 4x3 - 6x 112 = 2x 112 (2x51 2
critical point c = (~) 2 15 . (Note: c = 0 is not in the interval under consideration.)
x
(0, (~)215)
f'
-
3 2/5
((~)2/5,
2
f
0
+
m
/'
-
3)
oo)
"""
39. y = x+x- 1
SOLUTION
Let f(x) = x + x- 1• Then
f
yields the critical points c
I
(x)
=1-
x
-2
x2 - I
= I - -x21 = - =0
x2
= ±1. c = 0 is not a critical point because it is not in the domain off.
x
(-oo, -1)
-1
(-1,0)
0
(0, I)
1
(1,oo)
f'
+
0
-
CXl
-
0
+
f
/'
M
m
/'
..,
"""
"""
1
x +
41. y = -2-1
SOLUTION
Letf(x)
= (x2+1f1. Thenf'(x) = -2x(x2+1( = Oyieldscriticalpointc = 0.
= 0,
which gives us the
242
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
x
(-oo, 0)
0
(0, oo)
!'
+
0
-
f
/'
M
'\,
x3
43. y= -2-1
x +
SOLUTION
Letf(x)
x3
= - 2- . Then
x +1
f'(x)
= (x2 + 1)(3x2) -x3(2x) = x2(x2 + 3) = 0
(x2
+ 1)2
+ 1)2
(x2
yields the single critical point c = 0.
45. y = e +sine+ cose,
SOLUTION
Let f(e)
47. y = sin2 e +sine,
SOLUTION
c = ~'
x
(-oo, 0)
0
(0, oo)
f'
+
0
+
f
/'
~
/'
[O, 2rr]
= e + sine+ cos e. Then f' (e) = 1 +cos e -
sine
e (o, ~)
= 0 yields the critical points c = ~
"
2
(~,rr)
7r
(rr, 2rr)
f'
+
0
-
0
+
f
/'
M
'\,
m
/'
!".
= 0 yields the critical points
1
(}
(o, ~)
2"
(1'2, 11!)
6
6
(11!6, ~)
2
T
(~
2 , lli!)
6
!'
+
0
-
0
+
0
f
/'
M
'\,
m
/'
M
71'
3ir
6
!Irr
(1~", 2rr)
-
0
+
'\,
m
/'
49. y = x+e-x
SOLUTION
Let f(x) = x + e-x. Then f'(x) = 1 - e-x, which yields c = 0 as the only critical point.
51. y = e-x COSX,
SOLUTION
[ -
x
(-oo, 0)
0
(0, oo)
f'
-
0
+
f
'\,
m
/'
~, ~]
Let f(x) = e-x COS X. Then
f' (x) = -e-x sin x - e-x cos x = -e-x(sin x +cos x),
which yields c = -
~
= 7r.
[0, 2rr]
Let f(B) = sin2 e +sine. Then f'(e) = 2 sinBcos e +cos e =cos 8(2 sine+ 1)
If, ~'and
and c
as the only critical point on the interval [-
~, ~].
x
[-~,-~)
-4"
(-~,~]
f'
+
0
-
f
/'
M
\,
s E c TI ON
53. y
I
4.3
The Mean Value Theorem and Monotonicity
243
= tan- 1 x- !x
SOLUTION
Let f(x)
= tan- 1 x - ! x. Then
f
1
I
1
(x) = 1 + x2 -
2'
which yields c = ±1 as critical points.
x
(-oo,-1)
-1
(-1, 1)
1
(1,oo)
f'
-
0
+
0
-
f
"\,
m
/'
M
'-,.
55. y = x - ln x2
SOLUTION
Let f(x) = x - ln(x2). Then
I
f (x)
yields the critical point c
1
x 2 (2x)
=1-
= 1-
2
~
x-2
= -x- = 0
=2. c =0 is not a critical point because it is not in the domain off.
x
(-oo, 0)
0
(0,2)
2
(2, oo)
f'
+
00
-
0
+
...,
'-,.
m
/'
f
/'
57. y = x'i 3
SOLUTION
Let f(x)
= x 113 • Then j'(x) = ix-2 13 , which yields c = 0 as the only critical point.
59. Find the maximum value of f(x)
SOLUTION
x
(-oo, 0)
0
(0, oo)
!'
+
00
+
f
/'
-,
/'
= x-x for x > 0.
Let f(x) = x-x. To find the derivative off, first rewrite the function as f(x) = e-xlnx. Then
J'(x)
= e-xlnx (- Inx + (-x)~) = -x-x(I +In x) = 0
yields the critical point c = e- 1 • For 0 < x < e- 1 , f'(x) > 0 and f is increasing, while for x > e- 1, f'(x) < 0 and f is
decreasing. Thus, f achieves its maximum value at x = e- 1, and that maximum value is
f(e- 1) = (e- 1)-·-' = e 11•
= x3 - 2x2 + 2x is an increasing function. Hint: Find the minimum value off'.
Let f(x) = x3 - 2x2 + 2x. For all x, we have
61. Show that f(x)
SOLUTION
2
f'(x)
= 3x2 -
4x + 2
= 3 ( x - 32)
2 <': 3
2 >0
+3
Since f'(x) > 0 for all x, the function f is everywhere increasing.
63. Ron's toll pass recorded him entering the tollway at mile 0 at 12:17 PM He exited at mile 115 at 1:52 PM, and soon
thereafter he was pulled over by the state police. "The speed limit on the tollway is 65 miles per hour," the trooper told
Ron. "You exceeded that by more than five miles per hour this afternoon." "No way!" responded Ron. "I glance at the
speedometer frequently, and not once did it read over 65!" How did the trooper use the Mean Value Theorem to support
her claim that Ron must have gone more than 70 miles per hour at some point?
SOLUTION
Ron traveled 115 miles in 95 minutes, so his average velocity was
115 miles 60 minutes
.
.
·
h
""' 72 .63 rm1es/hour
95 rmnutes
1 our
The Mean Value theorem guarantees that if the average velocity was 72.63 miles/hour, then at some time during the trip
the instantaneous velocity was 72.63 miles/hour. At that time, Ron's speed exceeded 70 miles/hour.
244
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
65. Determine where f(x) = (1,000 - x) 2 + x2 is decreasing. Use this to decide which is larger: 8002 + 2002 or 6002 +
4002 .
SOLUTION If f(x) = (1000 - x) 2 + x2, then f'(x) = -2(1000 - x) + 2x = 4x - 2000. f'(x) < 0 as long as x < 500.
Therefore, 8002 + 2002 = /(200) > /(400) = 6002 + 4002 •
67. Which values of c satisfy the conclusion of the MVT on the interval [a, b] if f is a linear function?
SOLUTION
Let f(x) = px + q, where p and q are constants. Then the slope of every secant line and tangent line off is
p. Accordingly, considering the interval [a,b], every point c
E
(a, b) satisfies f'(c) = p =
f(b~ -
f(a), the conclusion of
-a
theMVT.
69. Suppose that f(O) = 2 and f'(x) s 3 for x > 0. Apply the MVT to the interval [0,4] to prove that /(4)
more generally that f (x) s 2 + 3x for all x > 0.
SOLUTION
s
14. Prove
The MVT, applied to the interval [0, 4], guarantees that there exists a c E (0, 4) such that
f'(c) = f(4) - f(O)
4-0
or
/(4) - f(O) = 4f'(c)
Because c > 0, f'(c) s 3, so f(4) - f(O) s 12. Finally, /(4) s f(O) + 12 = 14.
More generally, let x > 0. The MVT, applied to the interval [O, x], guarantees there exists a c
f'(c) = f(x) - f(O)
or
x-0
Because c > 0, f'(c)
s
3, so f(x) - f(O)
s
3x. Finally, f(x)
f(x)- f(O)
E
(0, x) such that
= f'(c)x
s f(O) + 3x = 3x + 2.
71. Show that if /(2) = 5 and f'(x) ;::: 10 for x > 2, then f(x) ;::: lOx - 15 for all x > 2.
SOLUTION
Let x > 2. The MVT, applied to the interval [2, x], guarantees there exists a c E (2, x) such that
f'(c) = f(x)- /(2)
or
x-2
f(x) - /(2) = (x - 2)f'(c)
Because f'(x) ;::: 10, it follows that f(x) - f(2) ;::: IO(x - 2), or f(x) ;::: f(2) + lO(x - 2) = IOx - 15.
Further Insights and Challenges
73. Prove that if f(O) = g(O) and f'(x) s g'(x) for x ;::: 0, then f(x)
given by y = f(x) - g(x) is nonincreasing.
s
g(x) for all x ;::: 0. Hint: Show that the function
SOLUTION Let h(x) = f(x) - g(x). By the sum rule, h'(x) = f'(x) - g'(x). Since f'(x) s g'(x) for all x;::: 0, h'(x) s 0
for all x ;::: 0. This implies that h is nonincreasing. Since h(O) = f(O) - g(O) = 0, h(x) s 0 for all x ;::: 0 (as h is
nonincreasing, it cannot climb above zero). Hence f(x) - g(x) s 0 for all x;::: 0, and so f(x) s g(x) for x;::: 0.
75. Use Exercises 73 and 74 to prove the following assertions for all x;::: 0 (each assertion follows from the previous one):
(a)
COSX;:::
1- ~x2
(b) sinx;::: x (c)
COSX
s
ix3
1 - ~x2+1.ix4
Can you guess the next inequality in the series?
SOLUTION
!x2.
(a) Let g(x) = cos x and f(x) = 1 Then f(O) = g(O) = 1 and g'(x) = - sinx;::: -x = f'(x) for x;::: 0 by Exercise
74. Now apply Exercise 73 to conclude that cosx;::: 1 for x;::: 0.
(b) Let g(x) = sinx and f(x) = x Then /(0) = g(O) = 0 and g'(x) = cosx;::: 1 = f'(x) for x;::: 0 by part (a).
Now apply Exercise 73 to conclude that sin x;::: x for x;::: 0.
(c) Let g(x) = 1 - ~x2 + 1.ix4 and f(x) =cos x. Then /(0) = g(O) = 1 and g'(x) = -x +
sinx = f'(x) for x;::: 0
by part (b ). Now apply Exercise 73 to conclude that cos x s 1 - ! x2 + x4 for x ;::: 0.
The next inequality in the series is sin x s x + 1 0 .x5, valid for x;::: 0.
ix3.
!x2
!x2
ix3
ix3 i
f4
ix3 ;: : -
77. Assume that f" exists and f"(x) = 0 for all x. Prove that f(x) = mx + b, where m = f'(O) and b = f(O).
SOLUTION
• Let f"(x) = 0 for all x. Then f'(x) =constant for all x. Since f'(O) = m, we conclude that f'(x) = m for all x.
• Let g(x) = f(x) - mx. Then g'(x) == f'(x) - m = m - m = 0 which implies that g(x) = constant for all x and
consequently f(x) - mx = constant for all x. Rearranging the statement, f(x) = mx +constant. Since /(0) = b, we
conclude that f(x) = mx + b for all x.
79. Suppose that f(x) satisfies the following equation (an example of a differential equation):
f" (x) = - f (x)
S E CT I O N 4.4
(a) Show that f(x) 2 + f'(x) 2
I
The Second Derivative and Concavity
245
= f(0) 2 + f'(0) 2 for all x. Hint: Show that the function on the left has zero derivative.
(b) Verify that sinx and cos x satisfy Eq. (1), and deduce that sin2 x + cos 2 x
= 1.
SOLUTION
(a) Let g(x) = f(x)2 + f' (x) 2 . Then
g' (x) = 2f(x)f' (x) + 2f' (x)f" (x)
= 2f(x)f' (x) + 2f' (x)(-f(x)) = 0
where we have used the fact that f"(x) = -f(x). Because g'(O) = 0 for all x, g(x) = f(x)2 + f'(x) 2 must be a constant
function. In other words, f(x) 2 + f'(x) 2 = C for some constant C. To determine the value of C, we can substitute any
number for x. In particular, for this problem, we want to substitute x = 0 and find C = f(0)2 + f' (0) 2 • Hence,
f (x)2 + f' (x)2
(b) Let f(x) = sinx. Then f'(x) = cosx and f"(x)
- sinx, f"(x) = -cos x, and we again have f"(x)
guarantees that
sin2 x + cos2 x
= f (0)2 + f' (0)2 •
= -sinx, so f"(x) = -f(x). Next, let f(x) = cosx. Then f'(x) =
= -f(x). Finally, if we take f(x) = sinx, the result from part (a)
= sin2 0 + cos2 0 = 0 + 1 = 1
81. Use Exercise 80 to prove f(x) = sinx is the unique solution of Eq. (1) such that f(O) = 0 and f'(O) = 1; and
g(x) = cos x is the unique solution such that g(O) = 1 and g' (0) = 0. This result can be used to develop all the properties
of the trigonometric functions "analytically"-that is, without reference to triangles.
SOLUTION
In part (b) of Exercise 79, it was shown that f(x) = sin x satisfies Eq. (1), and we can directly calculate that
= sin 0 = 0 and f' (0) = cos 0 = 1. Suppose there is another function, call it F(x), that satisfies Eq. (1) with the same
initial conditions: F(O) = 0 and F'(O) = 1. By Exercise 80, it follows that F(x) = sinx for all x. Hence, f(x) = sinx is
f (0)
the unique solution of Eq. (1) satisfying f(O) = 0 and f'(O) = 1. The proof that g(x)
Eq. (1) satisfying g(O) = 1 and g'(O) = 0 is carried out in a similar manner.
= cos xis the unique solution of
4.4 The Second Derivative and Concavity
Preliminary Questions
1. If f is concave up, then f' is (choose one)
(b) decreasing
(a) increasing
SOLUTION The correct response is (a): increasing. If the function is concave up, then f" is positive. Since f" is the
derivative off', it follows that the derivative off' is positive and f' must therefore be increasing.
2. What conclusion can you draw if f'(c) = 0 and f"(c) < O?
SOLUTION
If f'(c) = 0 and f"(c) < 0, then f(c) is a local maximum.
3. True or false? If f(c) is a local min, then f"(c) must be positive.
SOLUTION
False. f"(c) could be zero.
4. True or false? If f"(c)
= 0, then f
has an inflection point at x
= c.
SOLUTION This statement is false. The condition f"(c) = 0 is not sufficient to conclude that f has an inflection point
at x = c. For example, consider the function f (x) = x 4 • Then f" (x) = 12x2, so that f" (0) = 0. However, f" (x) is positive
for any nonzero value of x. This means that f is concave up on (-oo, 0) and on (0, oo ). Because the concavity off did not
change at x = 0, f does not have an inflection point at x = 0.
5. The function f (x)
Explain.
= xi-; 1 is concave down for x < 0 and concave up for x > 0. Is there an inflection point at x = 0?
xi-;
1 does not have an inflection point at x
No, the function f(x) =
off. Thus, there is no point on the graph at x = 0.
SOLUTION
= 0 because x = 0 is not in the domain
6. Can a function have an inflection point at a critical point? Explain.
Yes, a function can have an inflection point at a critical point. Consider the function f(x) = x3. Then
f'(x) = 3x2 and f"(x) = 6x. It follows that f'(O) = 0, so x = 0 is a critical point off. Moreover, f"(x) < 0 for x < 0 and
f"(x) > O for x > 0. This means that f is concave down for x < 0 and concave up for x > 0, so that f has an inflection
point at x = 0.
SOLUTION
246
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
Exercises
1. Match the graphs in Figure 13 with the description:
(b) f"(x) goes from+ to-.
(a) f"(x) < 0 for all x.
(c) f"(x) > 0 for all x.
(d) f"(x) goes from - to+.
(C)
(B)
(A)
(D)
FIGURE 13
SOLUTION
(a) In C, we have f"(x) < 0 for all x.
(b) In A, f"(x) goes from+ to-.
(c) In B, we have f"(x) > 0 for all x.
(d) In D, f"(x) goes from - to+.
3. jGUll Plot f(x) = (2x - x2)ex and indicate on the graph where it appears that inflection points occur. Then find the
inflection points using calculus.
SOLUTION The figure below shows the graph of f(x) = (2x - .x2)ex with the approximate location of inflection points
indicated by a square.
y
-3
To determine the inflection points, first calculate
f'(x) = (2 - 2x)ex + (2x - x2)ex = (2-x2)ex
and
f" (x) = -2xex + (2 -
x2)ex
= (2 -
2x - x2)ex
The second derivative exists for all x and is equal to zero when
x= 2± "4-4(-1)(2) = 2±2-./3 =-l±
-2
V3
-2
Now, f is concave down for x < -1 - V3 and for x > -1 + V3 because f" (x) < 0 there. Moreover, f is concave up for
-1 - V3 < x < -1 + V3 because f"(x) > 0 there. Concavity changes at x = -1 - V3 and at x = -1 + -./3, so f has
inflection points at both x = -1 - V3 and x = -1 + V3. The inflection points are
(-1- -./3,f(-1- -./3))= (-1- -./3,(-6-4-.f3V 1 -V3) and (-1 + -./3,J(-1 + -./3))= (-1 + -./3,(-6+4\f3V 1 +V3)
In Exercises 5-24, determine the intervals on which thefanction is concave up or down and find the points of inflection.
5. y = :x:2 - 4x + 3
SOLUTION Let f(x) = x2 - 4x + 3. Then f'(x) = 2x - 4 and f"(x) = 2 > 0 for all x. Therefore, f is concave up
everywhere, and there are no points of inflection.
7. y = 10x3 - x5
x5. Then f'(x) = 30.x2 - 5x4 and f"(x) = 60x - 20x3 = 20x(3 - .x2). Now, f is concave
up for x < -V3 and for 0 < x < V3 since f"(x) ::> Othere. Moreover, f is concave down for --1/3 < x < Oand for
SOLUTION
Let f(x) = 10x3 -
x > V3 since f"(x) < 0 there. Finally, because f"(x) changes sign at x = 0 and at x = ± -./3, f(x) has a point of inflection
at x = 0 and at x = ± V3. The points of inflection are
(0,f(O)) = (0,0),
9. y=8-2sin8,
[0,2ir]
( -./3,J( \13)) = ( -./3,21 \13),
and
(-\13,J(-\13)) = (-\13,-21
V3)
S E CT I O N 4.4
I
The Second Derivative and Concavity
247
Letf(O) = 8-2sin8. Thenf'(O) = l -2cos8andf"(O) = 2sin8.Now,f isconcaveupforO < 8<1f
since f"(O) > 0 there. Moreover, f is concave down for 1f < e < 21f since f"(O) < 0 there. Finally, because f"(O) changes
sign ate= lf, j(O) has a point of inflection ate = lf. The point of inflection is (lf, f(7r)) = (lf, lf).
SOLUTION
11. y
= x(x- 8yx)
(x ~ 0)
Let f(x) = x(x - 8 yx) = x2 - S.x3'2. Then f'(x) = 2x - 12x1' 2 and f"(x) = 2 - 6x- 112 . Now, f is concave
down for 0 < x < 9 since f"(x) < 0 there. Moreover, f is concave up for x > 9 since f"(x) > 0 there. Finally, because
j"(x) changes sign at x = 9, f(x) has a point of inflection at x = 9. The point of inflection is (9,f(9)) = (9, -135).
SOLUTION
13. y = (x - 2)(1 -
.x3)
Let f(x) = (x - 2) ( 1 - .x3) = x - x4 - 2 + 2.x3. Then f'(x) = 1 - 4.x3 + 6x2 and f"(x) = 12x - 12x2 =
12x(l - x) = 0 at x = 0 and x = 1. Now, f is concave up on (0, 1) since f"(x) > 0 there. Moreover, f is concave down
on (-oo,O) u (l,oo) sincef"(x) < 0 there. Finally, becausej"(x) changes sign at both x = 0 andx = 1, f(x) has a point
of inflection at both x = 0 and x = 1. The points of inflections are (0,f(O)) = (0, -2) and (l,f(l)) = (I, 0).
SOLUTION
1
15. y= ~3
x +
Let f(x) = x2
SOLUTION
~ 3. Then J' (x) =
!
(x2
~3)2 and
" x = - 2(x2 + 3)2 - 8x2(x2 + 3) = 6x2 - 6
( )
(x2 + 3)4
(x2 + 3)3
Now, f is concave up for lxl > 1 since f"(x) > 0 there. Moreover, f is concave down for lxl < 1 since f"(x) < 0 there.
Finally, because f" (x) changes sign at both x = -1 and x = 1, f(x) has a point of inflection at both x = -1 and x = 1.
The points of inflection are
(-l,f(-1))=(-l,~)
17. f(x)
=
and
(l,f(l))=(1,~)
l:x
SOLUTION
.x3
Letf(x) =--.Then
l+x
' '(x) = (1 + x)(3x2) - .x3(1) = 3x2 + 2.x3
f
(1 + x) 2
(1 + x) 2
and
J"(x) = (1+x) 2 (6x+6x2)- (3x2 + 2.x3)(2)(1 + x) = 2x(x2 + 3x + 3)
(1 + x)4
(I + x) 3
Now, f is concave up for x < -1 and for x > 0 since f"(x) > 0 there. Moreover, f is concave down for -1 < x < 0
since f"(x) < 0 there. Finally, because f"(x) changes sign at x = 0, f(x) has a point of inflection at x = 0. The point of
inflection is (O,f(O)) = (0, 0). Though the sign f"(x) does change sign at x = -1, there is no point of inflection at x = -I
because x = -1 is not in the domain off.
19. y = xe-3-"
SOLUTION Let j(x) = xe- 3-". Then f'(x) = -3xe- 3-" + e- 3x = (1 - 3x)e- 3-" and f"(x) = -3(1 - 3x)e-3-" - 3e- 3-" =
(9x - 6)e-3-". Now, f is concave down for x < ~ since f" (x) < 0 there. Moreover, f is concave up for x > since
j"(x) > 0 there. Finally, because f"(x) changes sign at x =
f(x) has a point of inflection at x =
The point of
inflection is
1·
21. y = 2x2 + In x
1·
1
(x > 0)
Let f(x) = 2x2 + Inx. Then f'(x) = 4x + x- 1 and f"(x) = 4 - x- 2. Now, f is concave down for x < k
since f"(x) < 0 there. Moreover, f is concave up for x > since f"(x) > 0 there. Finally, because j"(x) changes sign at
x = f has a point of inflection at x = The point of inflection is
SOLUTION
k,
k.
1
= (_!_ _!_+In_!_)=(_!_ _!_ - ln2)
(_!2'_ 1(_!_))
2
2' 2
2
2' 2
23. f(t)
= te-r2
248
c HAP TE R
4
I
APPLICATIONS OF THE DERIVATIVE
SOLUTION
Let f(t) = te-
12
•
Then
and
j"(t)
= -4te-t2 + (l -2t2y12 (-2t) =2t(2t2- 3)e-t2
Now, f is concave up for -.../6/2 < t < 0 and fort > .../6/2 since f"(t) > 0 there. Moreover,
t < - .../6/2 and for 0 < t < .../6/2 since f"(t) < 0 there. Finally, because f"(t) changes sign at t
f(t) has points of inflection at t = - .../6/2, t = 0, and t = .../6/2. The points of inflection are
f
is concave down for
= ± .../6/2 and at t = 0,
25. The position of an ambulance in kilometers on a straight road over a period of 4 hours is given by the graph in
Figure 15.
(a) Describe the motion of the ambulance.
(b) Explain what the fact that this graph is concave up tells us about the speed of the ambulance.
y
100
4
FIGURE 15
SOLUTION
(a) Suppose, for point of reference, the straight road along which the ambulance is driving travels north and south,
with north being the positive direction. The ambulance travels south for 100 kilometers, turns around, and travels 100
kilometers north, returning to its original position.
(b) The fact that the graph is concave up indicates that the velocity of the ambulance is always increasing. On the first
part of its trip, the ambulance is traveling south, so the velocity of the ambulance is negative and increasing, which means
the speed of the ambulance is decreasing. On the second part of the trip, the ambulance is traveling north, so the velocity
of the ambulance is positive and increasing, which means the speed of the ambulance is increasing.
27. ~ The growth of a sunflower during the first 100 days after sprouting is modeled well by the logistic curve
y = h(t) shown in Figure 17. Estimate the growth rate at the point of inflection and explain its significance. Then make a
rough sketch of the first and second derivatives of h.
Height (cm)
250 +·••"""'""''""'"""""''" .......... ,. . . . ",_.
200+· .................,....................
150 "t .............,
.............. .
100+··............,.........../
50+·.................,,,,...............;....................
c._--1----1---+---+----1-....
20
40
60
80
FIGURE 17
t (days)
100
The point of inflection in Figure 17 appears to occur at t = 40 days. The graph below shows the logistic
curve with an approximate tangent line drawn at t = 40. The approximate tangent line passes roughly through the points
(20, 20) and (60, 240). The growth rate at the point of inflection is thus
SOLUTION
240-20
_
60 20
220
= 4o = 5.5 cm/day
Because the logistic curve changes from concave up to concave down at t = 40, the growth rate at this point is the
maximum growth rate for the sunflower plant.
S E CT I O N 4.4
I
The Second Derivative and Concavity
249
Height(cm)
300
250
,, ,,,,..
200
/
150
)
100
50
/
[.......--"'(
t (days)
20
40
60
80
100
Sketches of the first and second derivative of h(t) are shown below at the left and at the right, respectively.
h'
h"
6
5
I
4
2
\
II
\
J
\
/
'\.
20
40
60
,-......._
80
t
100
29. Repeat Exercise 28 but assume that Figure 18 is the graph of the derivative f'.
Points of inflection occur when f"(x) changes sign. Consequently, points of inflection occur when f'(x)
changes from increasing to decreasing or from decreasing to increasing. In Figure 18, this occurs at x = b and at x = e;
therefore, j(x) has an inflection point at x = b and another at x = e. The function f(x) will be concave down when
f"(x) < 0 or when j'(x) is decreasing. Thus, f(x) is concave down for b < x < e.
SOLUTION
31. Figure 19 shows the derivative f' on [0, 1.2]. Locate the points of inflection off and the points where the local
minima and maxima occur. Determine the intervals on which f has the following properties:
(b) Decreasing
(a) Increasing
(d) Concave down
(c) Concave up
FIGURE 19
SOLUTION Recall that the graph is that off', not f. The inflection points off occur where/' changes from increasing
to decreasing or vice versa because it is at these points that the sign off" changes. From the graph we conclude that f
has points of inflection at x = 0.17, x = 0.64, and x = I. The local extrema off occur where/' changes sign. This occurs
at x = 0.4. Because the sign off' changes from + to-, f(0.4) is a local maximum. There are no local minima.
(a) f is increasing when f' is positive. Hence, f is increasing on (0, 0.4).
(b) f is decreasing when f' is negative. Hence, f is decreasing on (0.4, 1) U (I, 1.2).
(c) Now f is concave up where f' is increasing. This occurs on (0, 0.17) U (0.64, I).
(d) Moreover, f is concave down where f' is decreasing. This occurs on (0.17, 0.64) u (I, 1.2).
In Exercises 33-46,find the critical points and apply the Second Derivative Test (or state that it fails).
33. f(x) = x3 - 12x2 + 45x
Let f(x) = x3 - 12x2 + 45x. Then f'(x) = 3x2 - 24x + 45
x = 3 and x = 5. Moreover, f"(x) = 6x - 24, so /"(3) = -6 < 0 and /"(5)
Test, /(3) = 54 is a local maximum, and /(5) = 50 is a local minimum.
SOLUTION
35. f(x) = 3x4
-
= 3(x - 3)(x - 5), and the critical points are
= 6 > 0. Therefore, by the Second Derivative
8x3 + 6x2
SOLUTION Let f(x) = 3x4 - 8x3 + 6x2. Then f'(x) = 12x3 - 24x2 + 12x = 12x(x - 1)2 = 0 at x = 0, I and
f"(x) = 36x2 - 48x + 12. Thus, f"(O) > 0, which implies /(0) is a local minimum; however, f"(I) = 0, which is
inconclusive.
37. /(x)
x2 - 8x
= -x +1
250
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
SOLUTION
x2 - 8x
Letf(x) =~·Then
!
'( ) _ x2 + 2x -
8
and
(x + 1)2
x -
f" (x) = 2(x + 1)2 - 2(x2 + 2x - 8)
(x + 1)3
Thus, the critical points are x = -4 and x = 2. Moreover, f" (-4) < 0 and f" (2) > 0. Therefore, by the second derivative
test, f(-4) = -16 is a local maximum and f(2) = -4 is a local minimum.
4x 112
39. y = 6x3i 2
-
SOLUTION
Let f(x)
= 6x31 2 -
4x 1' 2 • Then f'(x)
= 9x112 -
2x- 1l 2 = x- 112(9x - 2), so there are two critical points: x = 0
and x =~-Now,
f"(x)
n
Thus, f" ( > 0, which implies f
cannot be applied there.
41. f(x)=sin2 x+cosx,
(
n
9
1
= -x-112
+ x-3/2 = -x-3f2(9x
+ 2)
2
is a local minimum.
2
f" (x) is undefined at x = 0, so the Second Derivative Test
[O,rr]
Let f(x) = sin2 x + cosx. Then f'(x) = 2sinxcosx - sinx = sinx(2cosx - 1). On the interval [O,rr],
f'(x) = 0 at x = 0, x = !f and x = 11'. Now,
SOLUTION
f" (x) = 2 cos 2 x - 2 sin2 x - cos x
Thus, f"(O) > 0, so f(O) is a local minimum. On the other hand, f"(D < 0, so f(!p is a local maximum. Finally,
f"(rr) > 0, so f(rr) is a local minimum.
43. f(x) = xe-:?
SOLUTION
Let f(x) = xe-:?. Then f'(x) = -2x2e-x2- + e-:? = (1 - 2x2)e-x2-, so there are two critical points: x = ±
f.
Now,
J"(x) = (4x3 - 2x)e-x2- - 4xe-x2-
Thus, f"
= (4x3 -
6x)e-x2-
( f) < 0, so f ( f) is a local maximum. On the other hand, f" (- f) > 0, so f (- f) is a local minimum.
45. f(x) = x3 ln x
SOLUTION
(x > 0)
Let f(x) = x3 ln x. Then f'(x) = x2 + 3x2 ln x = x2(1 + 3 ln x), so there is only one critical point: x = e- 113.
Now,
f" (x) = 3x + 2x(l + 3 ln x) = x(5 + 6 ln x)
Thus, f"
(e- 113 ) > 0, so f (e- 1! 3 ) is a local minimum.
In Exercises 47-62,find the intervals on which f is concave up or down, the points of inflection, the critical points, and
the local minima and maxima.
SOLUTION
Here is a table legend for Exercises 47-59.
SYMBOL
MEANING
-
The entity is negative on the given interval.
0
The entity is zero at the specified point.
+
The entity is positive on the given interval.
u
The entity is undefined at the specified point.
/'
The function (f, g, etc.) is increasing on the given interval.
The function (f, g, etc.) is decreasing on the given interval.
""'~
The function (f, g, etc.) is concave up on the given interval.
The function (f, g, etc.) is concave down on the given interval.
M
The function (f, g, etc.) has a local maximum at the specified point.
m
The function (f, g, etc.) has a local minimum at the specified point.
I
The function (f, g, etc.) has an inflection point here.
--,
There is no local extremum or inflection point here.
s E CT I 0 N
47. f(x)
= x3 -
SOLUTION
2x2
4.4
I
The Second Derivative and Concavity
251
+x
Let f(x) =
x3 -
+ x.
2x2
1
• Then f' (x) = 3.x2 - 4x + 1 = (x - 1)(3x - 1) = 0 yields x = 1 and x = as candidates for extrema.
• Moreover, f" (x) = 6x - 4 = 0 gives a candidate for a point of inflection at x = ~.
1)
3
I
ct, 1)
1
(1, 00)
x
(-oo, ~)
3
2
(~,oo)
+
0
-
0
+
!"
-
0
+
/'
M
'\,
m
/'
f
I
-
x
(-oo,
!'
f
~
49. f(t) = t2 - t 3
SOLUTION
t3.
Let f(t) = t2 -
• Then f'(t) = 2t - 3t2 = t(2 - 3t) = 0 yields t = 0 and t = ~ as candidates for extrema.
• Moreover, f" (t) = 2 - 6t = 0 gives a candidate for a point of inflection at t =
1.
51. f(x)
= x2 -
SOLUTION
t
(-oo,O)
0
(0, ~)
3
2
(~,oo)
t
!'
-
0
+
0
-
!"
f
'\,
m
/'
M
'\,
f
8x 1' 2
(-oo,
1)
I
3
ct,oo)
+
0
-
-
I
-
(x ;::.: 0)
Let f(x) = x2 - 8x 112 • Note that the domain off is x;::.:. 0 .
4x- 112 = x- 112 (2.x312 - 4)
= 2x -
• Then f'(x)
• Moreover, f"(x) = 2 +
2x-312
= 0 yields x = 0 and x = (2) 213 as candidates for extrema.
> 0 for all x ;::-: 0, which means there are no inflection points.
x
0
(o, (2)213)
(2)2/3
( (2)2/3 , 00)
!'
u
-
0
+
f
M
'\,
m
/'
x
53. f(x)
= x2 + 27
x
SOLUTION
Let f(x) = x2 + 27.
• Then f' (x)
=
27-x2
• Moreover, f"(x)
inflection at x
55. f(x) = x513
SOLUTION
-
=
-2x(x2
+ 21)
2
-
(27 - .x2)(2) (x2 + 27) (2x)
(x2
= 0 and at x = ±9.
+ 27)
4
=
2x(x2 - 81)
(x2
+ 27)
3
.
x
(-oo, -3 .../3)
-3.../3
(-3 .../3, 3 .../3)
3 V'3
(3 .../3, oo)
!'
-
0
+
0
-
f
'\,
m
/'
M
'\,
x
(-oo, -9)
-9
(-9,0)
0
(0,9)
9
(9, oo)
!"
-
0
+
0
-
0
+
f
-
I
-
I
-
I
-
= x513 -
= ~_x2!
• Moreover, f"(x)
3
-
x.
1 = 0 yields x
=± ( ~)
3/2
as a candidate for an extremum.
= 1J1x- 1 , which yields x = 0 as a candidate for a point of inflection.
13
.
.
= 0 gives candidates for a pomt of
x
Let f(x)
• Then f'(x)
= 0 yields x = ±3 ...f3 as candidates for extrema.
(x2 + 27) 2
252
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
x
-Gt2) -ot2
f'
+
0
f
/'
M
57. f(e) = e + sine,
SOLUTION
(-oo,
(-(~t2
,G)3/2) m3/2 mt2 ,oo)
-
0
+
m
/'
""'
x
(-oo, 0)
0
(0, oo)
f"
-
u
+
f
-
I
-
[O, 2n]
Let f(O) = 8 +sin 8 on [0, 2n].
• Then f' (8) = 1 + cos e = 0 yields e = n as a candidate for an extremum.
• Moreover, f"(O) = - sine= 0 gives candidates for a point of inflection ate= 0, ate= n, and ate= 2n.
59. f(x)=tanx,
SOLUTION
e
(0, n)
1l"
(1!", 21!")
e
0
(0,1!")
1l"
(1!", 21!")
21!"
f'
+
0
+
f"
0
-
0
+
0
f
/'
..,
/'
f
..,
-
I
-
..,
(-~,D
Letf(x)=tanxon(-~,~).
• Then f'(x) = sec2 x ~ 1 > 0 on(-~,~).
• Moreover, f" (x) = 2 sec x · sec x tan x = 2 sec2 x tan x = 0 gives a candidate for a point of inflection at x = 0.
61. y
= (x2 -
SOLUTION
x
(-~,I)
x
(-~,o)
0
(o,~)
f'
+
f"
-
0
+
f
/'
f
-
I
-
(x > 0)
2)e-x
Let f(x) = (r - 2)e-x.
• Then f'(x) = -(x2 - 2x - 2)e-x = 0 gives x = 1 + '13 as a candidate for an extrema.
• Moreover, f" (x) = (x2 - 4x)e-x = 0 gives x = 4 as a candidate for a point of inflection.
x
(o, 1 + '13)
f'
+
f
/'
V3
(1 + ,,.;3,oo)
x
(0,4)
4
(4, oo)
0
-
f"
-
0
+
M
~
f
-
I
-
1+
63. Sketch the graph of an increasing function such that f"(x) changes from + to - at x = 2 and from - to + at x = 4.
Do the same for a decreasing function.
SOLUTION The graph shown below at the left is an increasing function which changes from concave up to concave
down at x = 2 and from concave down to concave up at x = 4. The graph shown below at the right is a decreasing
function which changes from concave up to concave down at x = 2 and from concave down to concave up at x = 4.
y
y
2
4
2
4
s E CT I o N
4.4
I
The Second Derivative and Concavity
253
In Exercises 64-66, sketch the graph of a fanction f satisfying all of the given conditions.
65. (i) f'(x) > 0 for all x, and
(ii) f"(x) < 0 for x < 0 and f"(x) > 0 for x > 0
SOLUTION Here is the graph of a function f(x) satisfying (i) f'(x) > 0 for all x and (ii) f"(x) < 0 for x < 0 and
f"(x) > 0 for x > 0.
y
-5
-10
67. ~ An infectious flu spreads slowly at the beginning of an epidemic. The infection process accelerates until a
majority of the susceptible individuals are infected, at which point the process slows down.
(a) If R(t) is the number of individuals infected at time t, describe the concavity of the graph of R near the beginning and
end of the epidemic.
(b) Describe the status of the epidemic on the day that R has a point of inflection.
SOLUTION
(a) Near the beginning of the epidemic, the graph of R is concave up. Near the epidemic's end, R is concave down.
(b) "Epidemic subsiding: number of new cases declining."
69. ~ Water is pumped into a sphere of radius R at a variable rate in such a way that the water level rises at a
constant rate (Figure 20). Let V(t) be the volume of water in the tank at time t. Sketch the graph V (approximately, but
with the correct concavity). Where does the point of inflection occur?
r
IR
ih i
FIGURE 20
SOLUTION Because water is entering the sphere in such a way that the water level rises at a constant rate, we expect the
volume to increase more slowly near the bottom and top of the sphere where the sphere is not as "wide" and to increase
more rapidly near the middle of the sphere. The graph of V(t) should therefore start concave up and change to concave
down when the sphere is half full; that is, the point of inflection should occur when the water level is equal to the radius
of the sphere. A possible graph of V(t) is shown below.
v
71. Image Processing The intensity of a pixel in a digital image is measured by a number u between 0 and
1. Often,
images can be enhanced by rescaling intensities, as in the images of Amelia Earhart in Figure 21. When rescaling, pixels
of intensity u are displayed with intensity g(u) for a suitable function g. One common choice is the sigmoidal correction,
defined for constants a, b by
f(u)- f(O)
g(u) = /(1) - f(O)'
where
/(u)
= {l + eb(a-uJr 1
254
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
Original
Sigmoidal correction
both: Library of Congress Prints
and Photographs Division
FIGURE 21
Figure 22 shows that g(u) reduces the intensity of low-intensity pixels [where g(u) < u] and increases the intensity of
high-intensity pixels.
(a) Verify that f'(u) > 0 and use this to show that g(u) increases from 0 to 1for0
~
u
~
(b) Where does g(u) have a point of inflection?
y
1.0
/
/
/
0.8
/
/
// y = u
/
0.6
/
/
/
0.4
0.2
0.2 0.4 0.6 0.8 1.0
FIGURE 22 Sigmoidal correction with
a= 0.47, b = 12.
SOLUTION
(a) With f(u) = (1 +
eb(a-u))- 1,
it follows that
f'(u)
=-(1 + eb(a-u))-2. -beb(a-u) =
(1
b b(a-u)
e
+ eb(a-u))2
>0
for all u. Next, observe that
(O)
g
= f(O) -
f(O)
f(l) - f(O)
=O
l _ f(l) - f(O) _
g( ) - f(l) - f(O) - l,
'
and
1
I
g (u)
~
for all u. Thus, g(u) increases from 0 to 1for0
= f(l) -
I
f(O/ (u) > 0
u ~ 1.
(b) Working from part (a), we find
b2eb(a-u)(2eb(a-u) _
f
11
(u)
=
(1 + eh(a-•))3
Because
g"(u)
= f(l) ~ f(o/"Cu)
it follows that g(u) has a point of inflection when
2eb(a-u) -
1 =0
1)
1.
S EC T I 0 N 4.4
I
The Second Derivative and Concavity
255
Further Insights and Challenges
In Exercises 73-75, assume that f is differentiable.
73. Proof of the Second Derivative Test Let c be a critical point such that f"(c) > 0 [the case f"(c) < 0 is similar].
f'(c + h)
(a) Showthatf"(c)=lim
h-.O
h
·
(b) Use (a) to show that there exists an open interval (a, b) containing c such that f'(x) < 0 if a< x < c and f'(x) > 0 if
c < x < b. Conclude that f(c) is a local minimum.
SOLUTION
(a) Because c is a critical point, either f'(c) = 0 or f'(c) does not exist; however, f"(c) exists, so f'(c) must also exist.
Therefore, f'(c) = 0. Now, from the definition of the derivative, we have
f"(c) = lim f'(c + h) - f'(c) = lim f'(c + h)
h-.O
h
h-.O
h
(b) We are given that f"(c) > 0. By part (a), it follows that
. f'(c+h)
1Im
>0
h-.O
h
in other words, for sufficiently small h,
f'(c+h) >
0
h
Now, if h is sufficiently small but negative, then f'(c + h) must also be negative (so that the ratio f'(c + h)/h will be
positive) and c + h < c. On the other hand, if his sufficiently small but positive, then f'(c + h) must also be positive and
c + h > c. Thus, there exists an open interval (a,b) containing c such that f'(x) < 0 for a < x < c and f'(c) > 0 for
c < x < b. Finally, because f'(x) changes from negative to positive at x = c, f(c) must be a local minimum.
75.
~ Assume that f" exists and let c be a point of inflection off.
(a) Use the method of Exercise 74 to prove that the tangent line at x = c crosses the graph (Figure 23). Hint: Show that
G(x) changes sign at x = c.
(b) !GUij
Verify this conclusion for f(x) = - 2x
by graphing f and the tangent line at each inflection point on the
3x + 1
same set of axes.
I
FIGURE 23 Tangent line crosses graph at point of inflection.
SOLUTION
(a) Let G(x) = f(x) - f'(c)(x - c) - f(c). Then, as in Exercise 74, G(c) = G'(c) = 0 and G"(x) = f"(x). If f"(x)
changes from positive to negative at x = c, then so does G"(x) and G'(x) is increasing for x < c and decreasing for x > c.
This means that G'(x) < 0 for x < c and G'(x) < 0 for x > c. This in tum implies that G(x) is decreasing, so G(x) > O for
x < c but G(x) < 0 for x > c. On the other hand, if f"(x) changes from negative to positive at x = c, then so does G"(x)
and G'(x) is decreasing for x < c and increasing for x > c. Thus, G'(x) > 0 for x < c and G'(x) > 0 for x > c. This in
tum implies that G(x) is increasing, so G(x) < 0 for x < c and G(x) > 0 for x > c. In either case, G(x) changes sign at
x = c, and the tangent line at x = c crosses the graph of the function.
x
(b) Let f(x) = x2 + . Then
3
1
l -3x2
f'(x)
= (3x2 + 1)2
and
J"(x) = -18x(l - x2)_
(3x2 + 1)3
Therefore f(x) has a point of inflection at x = 0 and at x = ±1. The figure below shows the graph of y = f(x) and its
tangent lines at each of the points of inflection. It is clear that each tangent line crosses the graph of f(x) at the inflection
point.
256
CH APT E R 4
I
APPLICATIONS OF THE DERIVATIVE
y
77. Let f be a polynomial of degree n 2: 2. Show that f has at least one point of inflection if n is odd. Then give an
example to show that f need not have a point of inflection if n is even.
Let f(x) == anx" + an-1x"- 1 + · · · + a1x + ao be a polynomial of degree n. Then f'(x) == nanx"- 1 + (n 2
l)an-ix"- + · · · + 2a2x + a1 and f"(x) == n(n - l)anx"-2 + (n - l)(n - 2)an-1x"- 3 + · · · + 6a3x + 2a2. If n 2: 3 and is
odd, then n - 2 is also be odd and f"(x) is a polynomial of odd degree. Therefore f"(x) must take on both positive and
negative values. it follows that f"(x) has at least one root c such that f"(x) changes sign at c. The function f(x) will then
have a point of inflection at x == Ci, On the other hand, the functions f (x) == x2, x4 and x 8 are polynomials of even degree
that do not have any points of inflection.
SOLUTION
4.5 L:Hopital's Rule
Preliminary Questions
2
1. What is wrong with applying L'Hopital's Rule to lim x2 - x?
X--'>0
SOLUTION
As x
~
3X - 2
0,
x2-2x
3x-2
is not of the form§ or~, so L'Hopital's Rule cannot be used.
2. Does L'Hopital's Rule apply to limf(x)g(x) if f(x) and g(x) both approach oo as x
~a?
X--'>a
SOLUTION
No. L'Hopital's Rule only applies to limits of the form§ or ~·
3. What is wrong with saying, "To apply L'Hopital's Rule to
ln(l - x) , use th e Quotient
.
Ru1e to d"ffi
.
ln(l - x) and then
1 erentiate
lim
X
X--'>0
X
take the limit"?
SOLUTION To apply L'Hopital's Rule, you do not apply the Quotient Rule. Rather, you differentiate the numerator and
the denominator separately and then take the limit of the rational function obtained in this way.
4. What is wrong with applying L'Hopital 's Rule to lim x ~?
X-4Q+
The function f(x) == x fx is not indeterminate as x ~ o+. The function is of the form 0 as x ~ O+, which
corresponds to a limit with the value 0.
1
SOLUTION
5. What property of the function f(x)
00
==ex
allows us to say limeg(x) == el~g(x)?
X--'>G
SOLUTION
The fact that the function f(x) ==
ex
is continuous allows us to say lim eg(x) == el~g(x).
X--'>Q
Exercises
In Exercises 1-10, use L'Hopital's Rule to evaluate the limit, or state that L'Hopital's Rule does not apply.
1
•
r 2x2 - 5x- 3
x~
x-4
SOLUTION
Because the quotient is not indeterminate at x == 3,
2x2 - 5x x-4
L'Hopital's Rule does not apply.
x3 -64
3. lim~
X-t4 X
+ 16
31
== 18 - 15 x=3
3-4
3== ~
-1
s E c TI o N
SOLUTION
4.5
I
t:Hopital's Rule
257
Because the quotient is not indeterminate at x = 4,
x3-641
64-64
x 2 + 16 = 4
0
= 16 + 16 = 32
L'Hopital's Rule does not apply.
5. lim
X--+9
x112 +x-6
3/2
27
X
-
SOLUTION
Because the quotient is not indeterminate at x = 9,
x 112 + x - 61
3+9- 6 6
x312 - 27 x=9 = 27 - 27 = 0
L'Hopital's Rule does not apply.
7
'
r
sin4x
x~ x 2 + 3x+ 1
SOLUTION
Because the quotient is not indeterminate at x = 0,
sin4x
x 2 + 3x + 1
I
0
x=O
0
= 0 +0 + 1 =l
L'Hopital's Rule does not apply.
cos2x- l
9. lim--.-x ..... o sm5x
SOLUTION
The functions cos 2x - 1 and sin 5x are differentiable, but the quotient is indeterminate at x = 0,
cos2x-11
=1-l=Q
sin5x x=O
0
0
so L'Hopital's Rule applies. We find
Jim cos2x- 1 =Jim -2sin2x = Q = 0
x ..... o sin 5x
x ..... o 5 cos 5x
5
In Exercises 11-16, use L'Hopital's Rule to evaluate the limit.
9x+4
11. I i m 3 2X
X--+00
-
SOLUTION
As x
--+
4
9
oo, the quotient x + is of the form '.:"'., so L'Hopital's Rule applies. We find
3 - 2x
oo
. 9x+4
hm - 3 --' 2x
x__,oo
13. Jim
X--+00
.
9
9
= x__,oo
hm - = --2
2
lnx
~
12
X
SOLUTION
As x--+ oo, the quotient
1
~1~ is of the form.'.:'., so L'Hopital's Rule applies. We find
x
00
. lnx
.
~
.
1
hm- =hm-- = hm-- =0
x..... oo ~ x-112
x__,oo 2x112
x__,oo xI/2
. ln(x4 + 1)
15. hm
X--+-00
X
.
ln(x4 + I) .
oo
As x--+ oo, the quotient
is of the form-, so L'Hop1tal's Rule applies. Here, we use L Hop1tal's
x
00
Rule twice to find
A
SOLUTION
Jim ln(x4 + 1) = Jim
x--+oo
X
x--+oo
~X:.1
1
•
= Jim 12x2 = Jim
x--+oo
4x3
x--HXl
•
~ =0
X
,
A
•
258
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
In Exercises 17-54, evaluate the limit.
vs+ x- 3x113
17. lim-.,...---x--+1 x2 - 3x + 2
VS+ x - 3xl/3
l(S + x)-112 - x-2/3
= lim =2- - - - - - SOLUTION lim
x--+1 X2 - 3X + 2
X--+1
2X - 3
~-1
-1
5
6
3x-2
19. lim X-t---00 1 - 5 X
3x-2
lim - x--+-oo 1 - 5x
SOLUTION
21
lim 2_ =
x--+-oo -5
-~
5
7x2 + 4x
· E1!1oo 9 - 3x2
lim 7x2 +4x = lim 14x+4 = lim ~ =
x--+-00 9 - 3x2 x--+-00 -6x
x--+-00 -6
SOLUTION
_2
3
3x) 1' 2 -
(1 +
2
23. lim ---~­
x--+I (1 + 7x) 113 - 2
SOLUTION
Apply L'Hopital's Rule once:
. (1 + 3x) 112 - 2
. 1(1 + 3x)- 1' 2
hm
= hm -,,2- - - x--+1 (1 + 7x)l/3 - 2 X--+1 ~(1 + 7x)-2/3
r
25
•
(~)!
9
(~)(i)
7
sin2x
x~ sin7x
SOLUTION
. sin2x
. 2cos2x 2
hm - - = hm - - - = x--+0 sin 7x
x--+07cos1x
7
27. lim tanx
x--+0
X
2
SOLUTION
lim tanx = lim sec x = 1
x--+0 X
x--+0
1
. sin x - x cos x
29• hm
.
x--+0
x- smx
SOLUTION
. sinx+xcosx
xsinx
. sinx-xcosx
.
. cosx+cosx-xsinx
1!ID
1
=!ID
=Im
1
=Im
=2
1
X--+0
X - Sin X
X--+0 1 - COS X
X--+0
Sin X
X--+0
COS X
cos(x + ~)
31. lim --.-~
x--+0
SlnX
. cos(x+
SOLUTION hm
.
x--+0
sm x
33. lim
~)
. -sin(x+
= hm
x--+0
cos x
D = -1
~osx
x--+1'/2 sm(2x)
lim
SOLUTION
cos x
x--+1'/2 sin(2x)
lim
- sinx
x--+1'/2 2 cos(2x)
2
35. lim (sec x - tan x)
x-nr/2
SOLUTION
.
. ( -1- - sinx)
. (1-sinx)
hm(secx-tanx)=hm
- =hm
- - =lim (-cosx)
- - - =0
x--+~
cos x
cos x
x--+~
cos x
x--+l
- sinx
x--+~
SOLUTION
ex - 1
39. lim-.x--+O
SIIlX
2 2
~ 2 (Z~) =limsin (~x)=-~
x--+1
2
limtan(nx)lnx=lim ln:x =lim "
2
x--+1
x--+I COt( 2)
x--+l -2 CSC
1fX
1f
s E c TI o N
SOLUTION
. ex - 1
hm -.sm x
x->0
4.5
I
L'Hopital's Rule
1·
~
=x--.0
Im - - = l
cos x
e2x-1-x
41. l i m - - - x-.o
x2
SOLUTION
e2x - 1 - x
:i2
Jim
x-+0
2eh - 1
= lim - - - which does not exist.
x-+0
2X
43. lim (sin t )(In t)
t-+O+
SOLUTION
.
.
.
Int
45. lim <i' x-+0
1
i
.
hm(smt)(lnt) = hm - - = hm
1--.0+
HO+ csc t
HO+
-
.
-sin2 t
. -2sintcost __
= hm - - - = hm - - - - 0
csc t cot t HO+ t cos t
HO+ cos t - t sin t
(a > 0)
X
ax-l
SOLUTION
lna·ax
lim - - = lim - - x--.0
X
1
x--.0
=Ina
47. lim(l + Inx) 1l<x-I)
x->I
SOLUTION
Use the change-of-base formula to write
(I+ lnx)l/x =
ei--f ln(l+lnx)
Now,
1
I
ln(l + In x)
- - •1
1
lim - - ln(l + lnx) = lim
=Jim l+lnx x = lim
x--.1 x-1
x-.J
x-1
x--.1
1
x--.1 x(l +lnx)
=1
Hence,
lim(l + lnx)lf(x-1)
lim .!±!!!..:!
= Iimex-l ln(l+lnx) =ex-I
x-l
1
x-+1
=el
=e
x-+I
49. Iim(cosx)31x2
x->0
SOLUTION
Use the change-of-base formula to write
(cos x)3/x2 = e:;t ln(cosx)
Now,
3 sec 2 x
3
.
3 tan x Ii
. 3 (
)
. 3 ln cos x
== 1im---== m----==--.
1
IIm-2 1ncosx =Im
x->O x
x--.0
x2
x_,O
2x
x-.O
2
2
Hence,
x2
3
Jim(cos x )31 = lime :;z
x-+O
x-+0
. -I
51. Jim sm x
x-+0
X
. -I
SOLUTION
53. Jim
~--.1
_1_
Jim~ = lim ~
x->0
X
x-.O
1
=1
tan-IX-!!:4
"
tan 4 x - 1
tan-IX-!!:
SOLUTION
I
rx1!f tan(nx/4) -4 1 = x->I
lim
J+x2
~ sec 2 (nx/4)
I
l
"2
I
.
n(cos x)
Jim 3 ln(cosx)
== eHO --;:;:- = e- 312
259
260
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
. cosmx h
55. Evaluate lim - - - , w ere m, n
x-+1'/2 cos nx
SOLUTION
.
* 0 are mtegers.
Suppose m and n are even. Then there exist integers k and l such that m
lim cosmx = coshr
cos nx
cos l7r
= 2k and n = 2l and
= (-li_1
x-+1f/2
Now, suppose m is even and n is odd. Then
cosmx
.
11m - - X-+1f/2
cos nx
does not exist (from one side the limit tends toward -oo, while from the other side the limit tends toward +oo ). Third,
suppose m is odd and n is even. Then
lim cosmx
cos nx
=0
x->1'/2
Finally, suppose m and n are odd. This is the only case when the limit is indeterminate. Then there exist integers k and l
such that m = 2k + 1, n = 21+1 and, by L'Hopital's Rule,
. cosmx _ . -msinmx _ ( l)k-lm
- 1l m - - - - 1im
x->1f/2
cos nx
x->1f/2
-n sin nx
n
To summarize,
( - l)(m-n)/2,
lim cos mx
x-+"12 cos nx
=
m,neven
meven,n odd
modd,neven
m,nodd
does not exist,
0
{
( -1 )(m-n)/2 ~,
57. Evaluate each of the following limits.
(1 + .!.);i.
lim (1 + ~)x
(a) lim
(b)
X--+OO
X
X--+OO
X
SOLUTION
(a) Use the change-of-base formula to write
(
I
l + ~1 )i'- = ei'-1n(1+,)
Now,
Hence,
. ( 1 + -l)i'- = hm
. ei'-ln(l+x)
l =ex~=
limi'-1n1+l
hm
( x)
X
X--+OO
= 00
X--+OO
(b) Use the change-of-base formula to write
Now,
lim
x---+oo
Hence,
1)
xln (1 + 2
X
I
=
lim
x---+oo
ln(l+*)
1
X
=
lim
x--+oo
2
1+:;h-·-~
-
1
~
=
lim X--+00
X
2
- = O
+ !x
s E c T 1o N
4.5
I
t:Hopital's Rule
261
In Exercises 59-60, a projectile is launched straight up in the air and is acted on by air resistance and gravity as
in Example 5. The fanction M gives the maximum height that the projectile attains as a fanction of the air resistance
parameter k. In each case, determine the maximum height as we let the air resistance term go to zero; that is, determine
limM(k).
k->0
59. A ball with a mass of I kilogram is launched upward with an initial velocity of 60 mis, and
M(k)
=
300k +I )
60k- 9.8ln ( 49
k2
(Compare with Exercises 37 in Section 2.5 and 29 in Section 3.4.)
SOLUTION
The quotient
M(k)
has the indeterminate form 0/0 ask
.
l~
=
60k- 9.8ln 300k + 1)
( 49
k2
0. To evaluate the limit, we need to use L'Hopital's Rule twice. We find
-t
)
300k
60k-9.8ln ( - - +I
49
k2
.
= l~
60
60- ~
49 + I
18000/49
= lim
2k
(300k + l)2
49
2
k->0
= 9000
,.., 183.67 m
49
Thus, the maximum height as air resistance goes to zero is approximately 183.67 m. This result is in agreement with the
results from Exercises 37 in Section 2.5 and 29 in Section 3.4.
·61. In each case, show that the form is indeterminate by showing that if lim f (x )8 (x) has the form, then the limit in the
X->C
.
lim g(x)lnf(x)
.
.
f,
exponent m ex~c
has a known mdetermmate orm.
(a) 100
(b)
000
SOLUTION
(a) Suppose Iimf(x) 8 (x) has the form 100 • Then limf(x) =I and limg(x) = oo. It follows that limlnf(x) = 0, so
x--1-c
X--JC
X--+c
X--+C
limg(x) lnf(x) has the indeterminate form 0 · oo.
X->C
(b) Suppose Iimf(x) 8<x> has the form
00°.
x-+c
Then limf(x)
x-+c
=oo
and limg(x)
x--+c
=0.
It follows that limlnf(x)
x--+c
=oo,
so
Iimg(x) lnf(x) has the indeterminate form 0 · oo.
X->C
63. Let f(x)
= x 11x for x > 0.
(a) Calculate lim f(x) and lim f(x).
x-+O+
x-+oo
(b) Find the maximum value off and determine the intervals on which f is increasing or decreasing.
SOLUTION
(a) Let f(x)
= x 1lx. Note that x-+0+
lim x 1fx is not indeterminate. As x -t O+, the base of the function tends toward 0 and the
exponent tends toward +oo. Both of these factors force x 1fx toward 0. Thus, lim f(x) = 0. On the other hand, lim f(x) is
x-+0+
indeterminate. To calculate this limit, use the change-of-base formula to write
xlfx
= e~ lnx
Now,
. 1
. lnx
hm - lnx = hm X--JOO
x
X--+OO
x
. ~
= lllll
- =0
1
x-+oo
Hence,
. J( X ) = l"lfil exl 1nx = exJim lxJnx
1Im
00
x-+oo
x--+oo
=
eo = l
X-+OO
262
CH APT E R 4
I
APPLICATIONS OF THE DERIVATIVE
(b) Again, let f (x) = x 1f x, so that Inf (x) = ~ In x. To find the derivative f', we apply the derivative to both sides:
(!
.!!:.._ lnf(x) = .!!:.._
lnx)
dx
dx x
I ,
In x
I
-f(x)=--+f(x)
x2
x2
lnx
I ) x 1fx
f'(x) = f(x) ( - - + = -(1-lnx)
x2
x2
x2
Thus,
f is increasing for 0 <
f(e) = e 1fe
::e
x < e, is decreasing for x > e, and has a maximum at x = e. The maximum value is
1.444668.
65. Determine whether f « g or g « f (or neither) for the functions f(x) = log 10 x and g(x) = Inx.
Because
SOLUTION
lim f(x) = lim log10 x = lim i~(o = _I_
g(x) x-+co In X
x-+co In X
In 10
x-+co
neither g «
f
nor f « g is satisfied.
67. Just as exponential functions are distinguished by their rapid rate of increase, the logarithm functions grow particularly slowly. Show that In x « x" for all a > 0.
SOLUTION
Using L'Hopital's Rule:
x- 1
1 a
. In x
.
•
= 1im - - = lim -x- = 0
Iim x--+oo ;x'l
x--+oo axa- J
x--+oo a
hence, In x « x".
69. Determine whether
SOLUTION
Yx «
yx. Hint: Use the substitution u = In x instead of L'Hopital's Rule.
e V1IlX ore V1IlX «
Let u = In X, then x = e", and as x
~ oo,
u
~ oo.
So
. eV!IlX
. e-Vu
.
u
hm - - = lim = hm e-Vu-1
X--+OO
Yx
U--J.OO
eu/2
We need to examine lim( -./U - ~).Since
U-+CO
lim u/
U-JoCO
-./U «
2
Vu
= lim
Yu
U--+CO
2
= oo
u/2 and lim ( ...fU-1!:.) = -oo. Thus
2
U-HX>
lim e-Vu-7_ = e- = 0
00
so
U-+co
and e V1IlX «
e V1IlX
lim
X-+OO
.r.. = 0
vX
yx.
71. Assumptions Matter Suppose f (x) = x(2 + sin x) and let
g(x) = x2 + 1.
(a) Show directly that lim f(x)/g(x) = 0.
x-+co
(b) Show that lim f(x) = lim g(x) = oo, but lim f' (x)/ g' (x) does not exist.
X-+CO
X--+00
x--+co
Do (a) and (b) contradict L'Hopital's Rule? Explain.
SOLUTION
(a) 1
s
2 + sin x
s
3, so
x
x(2 + sinx)
3x
--<
<-x2+1x 2 +1
-x2 +1
Since,
.
x
.
3x
Ii m2- - = 1im--=0
x + 1 X-+CO x2 + 1
X-+CO
it follows by the Squeeze Theorem that
. x(2 + sinx)
Iim
=0
x2 + 1
x-+co
s Ec T I o N
(b) lim f(x)
X---tDO
= x->oo
lim x(2 + sinx) ~
4.5
I
L'Hopital's Rule
263
lim x = oo and lim g(x) = lim(x2 + 1) = oo, but
x-i.oo
X--HX)
. !' (x)
11m - X-->00 g' ( X)
X---tOO
. x(cos x) + (2 +sin x)
= X-->OO
11m - - - - - - - 2x
does not exist since cos x oscillates. This does not violate L'Hopital's Rule since the theorem clearly states
lim f(x) = lim f'(x)
X-->OO g(x)
X-->00 g'(x)
"provided the limit on the right exists."
73. Let G(b) = lim(l + bx)lfx.
X-->OO
(a) Use the result of Exercise 72 to evaluate G(b) for all b > 0.
(b) !Guij Verify your result graphically by plotting y = (I + bx)Ifx together with the horizontal line y
= G(b) for the
values b = 0.25, 0.5, 2, 3.
SOLUTION
(a) Using Exercise 72, we see that G(b)
(b)
= eH(bJ. Thus, G(b) = 1ifO:5b:51 and G(b) = b if b >I.
y
y
4
4
3
2
2
b=0.25
5
b=0.5
15
10
5
15
10
y
42\~
__
_
h=2.0
b=3.0
2
5
In Exercises
5
15
10
10
15
75-77, let
f(x)
={
e-lfx'-
for x
*0
0
forx
=0
These exercises show that f has an unusual property: All of its derivatives at x
= 0 exist and are equal to zero.
75. Show that lim f~) = 0 for all k. Hint: Lett= x- 1 and apply the result of Exercise 74.
x-i.O
SOLUTION
...t."
. f (x) = l.Im - -I2 . Let t = I /x. As x
lIm---:;-x-->O x·
x-->O
xkelfx
~
0, t
~DO.
Thus,
. m1 - = l"Im-=
f
0
II
2
x-->0 x.kelfx
by Exercise 74.
77. Show that fork
;?:
I and x
for some polynomial P(x) and some exponent r
to zero for all k ;?: I.
For x
eil
* 0,
fkl(x) =
SOLUTION
t-->oo
~
P( ) -11x2
x e
x'
I. Use the result of Exercise 75 to show that fkl(O) exists and is equal
* 0, f'(x) = e-i;x2 ( x23 ) . Here P(x) = 2 and r = 3. Assume fk\x) = P(x)e-lfx2
x'
. Then
fk+ll(x) = e-l/x2
(x3 P'(x) + (2 - rx2)P(x))
x+3
which is of the form desired.
264
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
Moreover, from Exercise 76, f'(O)
fk+l\O)
= 0. Suppose J<kl(O) = 0. Then
= lim J<kl(x) x--.O
X -
J<kl(O)
0
= lim P(x)e-1fx2 = P(O) lim f(x) = 0
x--.O
xr+I
x--.O xr+I
Further Insights and Challenges
79. The Second Derivative Test for critical points fails if f"(c)
based on the sign of the first nonzero derivative. Suppose that
f'(c)
= 0. This exercise develops a Higher Derivative Test
= f"(c) = · · · = fn-ll(c) = 0,
but fnl(c)
-:f:.
0
(a) Show, by applying L'Hopital's Rule n times, that
lim f(x) - f(c) = _..!:._ fnl(c)
x--.c
(x - c)n
n!
where n! = n(n - l)(n - 2) · · · (2)(1).
(b) Use (a) to show that if n is even, then f(c) is a local minimum if J<nl(c) > 0 and is a local maximum if J<nl(c) < 0.
Hint: If n is even, then (x - c)n > 0 for x -:f:. a, so f(x) - f(c) must be positive for x near c if J<nl(c) > 0.
(c) Use (a) to show that if n is odd, then f(c) is neither a local minimum nor a local maximum.
SOLUTION
(a) Repeated application ofL'Hopital's rule yields
. f(x) - f(c)
1tm
x--.c
(x - c)n
f'(x)
= lix--.c
m---n(x - c)n-1
.
f"(x)
=Iim-----x--.c n(n - l)(x - c)n-2
.
f"'(x)
= ltm
---.,.-,---::-:-------::x--.c n(n - l)(n - 2)(x - c)n-3
(b) Suppose n is even. Then (x - c)n > 0 for all x -:f:. c. If J<nl(c) > 0, it follows that f(x) - f(c) must be positive for x
near c. In other word~, f(x) > f(c) for x near c and f(c) is a local minimum. On the other hand, if J<nl(c) < 0, it follows
that f(x) - f(c) must be negative for x near c. In other words, f(x) < f(c) for x near c and f(c) is a local maximum.
(c) If n is odd, then (x - c)n > 0 for x > c but (x - c)n < 0 for x < c. If J<nl(c) > 0, it follows that f(x) - f(c) must be
positive for x near c and x > c but is negative for x near c and x < c. In other words, f(x) > f(c) for x near c and x > c
but f(x) < f(c) for x near c and x < c. Thus, f(c) is neither a local minimum nor a local maximum. We obtain a similar
result if J<nl(c) < 0.
~
We expended a lot of effort to evaluate lim sin x in Chapter 2. Show that we could have evaluated it easily
x--.o x
using L'Hopital's Rule. Then explain why this method would involve circular reasoning.
81.
SOLUTION
. sinx
11m
-x--.O
X
. cosx
.,.,
L'HA . 1,
li sinx
th th d .
= 1x--.O
1m - - = 1. 10 use
op1ta s Rule to eva1uate m - - , we must know at e envative
x--.O X
1
of sin x is cos x, but to determine the derivative of sin x, we must be able to evaluate lim sin x.
x-+0
83. Patience Required
sinx)l/x2
( x
. ( -1- - 1)
(b) 11m
2
2
(a) lim - x-+o+
X
Use L'Hopital's Rule to evaluate and check your answers numerically:
x--.o
sin x
x
SOLUTION
(a) Use the change-of-base formula to write
Now, repeatedly using L'Hopital's Rule, we find
.
1 (sinx)
ltm - 1n - x--.o+ x2
X
1
. xcosx - sinx
. ln(sinx)- lnx
.
-xsinx
. cotx - x11m
11m
= x--.o+
= x--.O+
= x--.0+
= x--.O+
lim --=-----,-----,--11m
x2
2x
2x2 sin X
2x2 COS X + 4x sin X
.
-xcosx-sinx
-2cosx+xsinx
.
Irm
1tm -:--::-----:-::-----:----:-= x--.o+
= x--.O+
8XCOS X + 4 sin X - 2x2 sin X
12 COS X - 2x2 COS X - 12x sin X
2
12
1
6
s E cT IO N
4.6
I
Analyzing and Sketching Graphs of Functions
265
Therefore,
. (sinx)11X1
. exZ
1...1n(''"x)
hm
-= hm
x
=
x-+O+
X
lim
-;t1n(''~x)
ex~o+ x
=
e-1/6
x-+O+
Numerically we find:
x
ci:xtX1
1
0.1
O.DI
0.841471
0.846435
0.846481
Note that e- 116 ""0.846481724.
(b) Repeatedly using L'Hopital's Rule and simplifying, we find
. Y?-sin2 x
1
.
1)
2x-2sinxcosx
.
2x-2sin2x
=
I
m
1
=Im
=hm
1
lim
2
2
2
2
(
2
2
2
x-+O sin x
x
x-+O x (2 sin x cos x) + 2x sin x
x-+O x sin 2x + 2x sin x
x-+O x2 sin x
2- 2cos2x
.
= lI m - - - - - - - - - - - - - - -22
x-+0 2x cos 2x + 2x sin 2x + 4x sin x cos x + 2 sin x
= lim
x-+O
2 - 2cos2x .
2x2 cos 2x + 4x sin 2x + 2 sin2 x
-r
-
4sin2x
x~ -4x2 sin2x + 4xcos2x + 8xcos 2x + 4sin2x + 4sinxcosx
.
4sin2x
= lI m - -------x-+O (6 - 4x2 ) sin 2x + 12XCOS 2x
.
8 cos2x
= lI m - ---------------x-+O (12 - 8x2 ) cos2x - 8x sin 2x + 12cos 2x - 24xsin 2x
3·
Numerically we find:
x
l
l
sin2 x - x2
1
0.1
O.Dl
0.412283
0.334001
0.333340
4.6 Analyzing and Sketching Graphs of Functions
Preliminary Questions
1. Sketch an arc where f' and f" have the sign combination++. Do the same for-+.
An arc with the sign combination++ (increasing, concave up) is shown below at the left. An arc with the
sign combination-+ (decreasing, concave up) is shown below at the right.
SOLUTION
2. If the sign combination off' and f" changes from ++ to +- at x = c, then (choose the correct answer)
(a) f(c) is a local min.
(b) f(c) is a local max.
(c) (c, f(c)) is a point of inflection.
SOLUTION
Because the sign of the second derivative changes at x = c, the correct response is (c): (c, f(c)) is a point
of inflection.
3. The second derivative of the function f(x) = (x - 4)- 1 is
f"(x) = 2(x - 4)- 3 . Although f"(x) changes sign at x = 4, f does not have a point of inflection at x = 4. Why not?
SOLUTION
The function! does not have a point of inflection at x
= 4 because x = 4 is not in the domain off
266
CH APT E R 4
I
APPLICATIONS OF THE DERIVATIVE
Exercises
1. Determine the sign combinations off' and f" for each interval A--G in Figure 15.
y
A
BCD EFG
FIGURE 15
SOLUTION
•
•
•
•
•
•
•
In A, f is decreasing and concave up, so f' < 0 and f" > 0.
In B, f is increasing and concave up, so f' > 0 and f" > 0.
In C, f is increasing and concave down, so f' > 0 and f" < 0.
In D, f is decreasing and concave down, so f' < 0 and f" < 0.
In E, f is decreasing and concave up, so f' < 0 and f" > 0.
In F, f is increasing and concave up, so f' > 0 and f" > 0.
In G, f is increasing and concave down, so f' > 0 and f" < 0.
In Exercises 3-6, draw the graph of a function for which f' and f" take on the given sign combinations in order.
3. ++,
+-, --
SOLUTION
X=
This function changes from concave up to concave down at x
= -1
and from increasing to decreasing at
0.
y
5. -+,
--, -+
SOLUTION The function is decreasing everywhere and changes from concave up to concave down at x
concave down to concave up at x = - ~.
0.05
-1
0
7. Sketch the graph of a function that could have the graphs off' and f" appearing in Figure 17.
1
3
4
I
5\
I
y
5
y
FIGURE 17
= f"(x)
= -1 and from
S E CT I O N 4.6
I
Analyzing and Sketching Graphs of Functions
267
Based on the graphs off' and f" appearing in Figure 17, the graph off needs to be increasing for x < 1
and 3 < x < 5, decreasing for 1 < x < 3 and x > 5, concave up for 2 < x < 4, and concave down for x < 2 and x > 4.
One possible graph for f is:
SOLUTION
y
6
5
4
3
2
2
9. Investigate the behavior and sketch the graph of y
4
= x2 -
5x + 4.
SOLUTION Letf(x) = x2 -5x+4. Thenf'(x) = 2x-5 andf"(x) = 2. Hencef is decreasing for x < 5/2, is increasing
for x > 5/2, has a local minimum at x = 5/2 and is concave up everywhere.
y
11. Investigate the behavior and sketch the graph off(x) =
1 ± Y3 (approximately -0.73, 2.73).
x3 - 3x2 + 2. Include the zeros off, which are x
= 1 and
Let f(x) = x3 - 3x2 + 2. Then f'(x) = 3x2 - 6x = 3x(x - 2) = 0 yields x = 0, 2 and f"(x) = 6x- 6. Thus
f is concave down for x < 1, is concave up for x > I, has an inflection point at x = 1, is increasing for x < 0 and for
x > 2, is decreasing for 0 < x < 2, has a local maximum at x = 0, and has a local minimum at x = 2.
SOLUTION
13. Extend the sketch of the graph of f(x) = cosx + !x in Example 4 to the interval (0, Sn].
SOLUTION
= cos x + !x. Then f' (x) = - sin x + ! = 0 yields critical points at x
= - cos x so there are points of inflection at x = ~, ~, ~, 1f, and llf.
Let f(x)
2
~". Moreover, f" (x)
1
1
= ~, ~, ~", ~", 2 ~", and
6
4
2
0
2
4
6
8 10 12 14
In Exercises 15-36,find the transition points, intervals of increase/decrease, concavity, and asymptotic behavior. Then
sketch the graph, with this information indicated.
15. y = x3 + 24x2
SOLUTION Let f(x) = x3 + 24x2. Then f'(x) = 3x2 + 48x = 3x(x + 16) and f"(x) = 6x + 48. This shows that f has
critical points at x = 0 and x = -16 and a candidate for an inflection point at x = -8.
Interval
Signs off' and f"
(-oo,-16)
(-16,-8)
(-8, 0)
(0, oo)
+-
--
-+
++
268
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
Thus, there is a local maximum at x = -16, a local minimum at x = 0, and an inflection point at x = -8. Moreover,
lim f(x)
X--+-00
= -oo
and lim f(x)
X--+OO
= oo
Here is a graph off with these transition points highlighted as in the graphs in the textbook.
y
3000
17. y = x2 - 4x3
SOLUTION
and x =
Let f(x) = x2 - 4x3. Then f' (x) = 2x - 12.x2 = 2x(I - 6x) and f"(x) = 2 - 24x. Critical points are at x = 0
k, and the sole candidate point of inflection is at x = ii.
(-oo, 0)
Interval
Signs off' and f"
Thus, f (0) is a local minimum, f(
-+
(0,
12) (tz,k) Ci. oo)
++
+-
--
i) is a local maximum, and there is a point of inflection at x = rz. Here is the graph of
f with transition points highlighted as in the textbook:
y
19. y = 4 - 2x2 + kx4
SOLUTION
Let f(x) = ix4
-
2.x2 + 4. Then f'(x) = ~x3 - 4x = ~x(x2 -
has critical points at x = 0 and x = ±
Interval
Signs off' and f"
6) and f"(x)
= 2x2 - 4. This shows that f
'16 and has candidates for points of inflection at x = ± 0.
(-oo, -../6)
(- ../6, - 1/2)
(- 1/2, 0)
-+
++
+-
(0,
1/2)
--
(0,
../6)
-+
( ../6, oo)
++
Thus, f has local minima at x = ± ../6, a local maximum at x = 0, and inflection points at x = ± 1/2. Here is a graph off
with transition points highlighted.
y
21. y
= x5 +5x
Let f(x) = x5 + 5x. Then f'(x) = 5x4 + 5 = 5(x4 + 1) and f"(x) = 20x3. f'(x) > 0 for all x, so the graph
has no critical points and is always increasing. f"(x) = 0 at x = 0. Sign analyses reveal thatf"(x) changes from negative
to positive at x = 0, so that the graph of f(x) has an inflection point at (0, 0). Here is a graph off with transition points
highlighted.
SOLUTION
SECT I 0 N 4.6
I
Analyzing and Sketching Graphs of Functions
269
23. y = x4 - 3.x3 + 4x
SOLUTION
Let f(x)
= x 4 - 3.x3 + 4x. Then f'(x) = 4.x3 - 9.x2 + 4 = (4.x2 1
x - 2)(x - 2) and j"(x)
= 12.x2 -
18x
=
Y33 and candidate points of inflection at x = O
8
and x = ~- Sign analyses reveal that f'(x) changes from negative to positive at x = 1
from positive to negative at
= 2 and x =
6x(2x - 3). This shows that f has critical points at x
±
-'1!3,
= 2. Therefore, j(1-'f3) and f(2) are local minima of f(x), and
f (I+ '/,33) is a local maximum. Further sign analyses reveal that f" (x) changes from positive to negative at x = O and from
x = 1+'f3, and again from negative to positive at x
negative to positive at x = ~'so that there are points of inflection both at x
transition points highlighted.
= 0 and x
= ~·Here is a graph of f(x) with
y
25. y
= x7 -
l 4.x6
SOLUTION Letf(x) = x 7 - 14.x6. Then f'(x) = 7x6 - 84.x5 = 7.x5 (x- 12) and f"(x) = 42.x5 -420x4 = 42x4 (x- 10).
Critical points are at x = 0 and x = 12, and candidate inflection points are at x = 0 and x = 10. Sign analyses reveal
that f'(x) changes from positive to negative at x = 0 and from negative to positive at x = 12. Therefore f(O) is a local
maximum and f(l2) is a local minimum. Also, f"(x) changes from negative to positive at x = 10. Therefore, there is a
point of inflection at x = 10. Moreover,
= -oo
lim f(x)
X--l'-00
and
lim f(x)
X--l'OO
= oo
Here is a graph off with transition points highlighted.
y
Ix 107
5x 106
-5
-5x106
21.y=x-4yx
SOLUTION Let f(x) = x - 4 yx = x - 4x1' 2 . Then f'(x) = I - zx- 1' 2 • This shows that f has critical points at x = 0
(where the derivative does not exist) and at x = 4 (where the derivative is zero). Because f'(x) < 0 for 0 < x < 4 and
f'(x) > 0 for x > 4, f (4) is a local minimum. Now f"(x) = x- 312 > 0 for all x > 0, so the graph is always concave up.
Moreover,
limf(x) = oo
x-->oo
Here is a graph off with transition points highlighted.
y
29. y
= x(8 -
SOLUTION
x) 113
Let f(x) = x (8 - x) 113 . Then
f'(x)
= x.
l (8 -x)-2/3 (-1) + (8 - x)I/3. I=
3
and similarly
,,
f
4x-48
(x) = 9 (8 - x)5/3
24-4x
3 (8 _ x)2f3
270
CH APT E R 4
I
APPLICATIONS OF THE DERIVATIVE
Critical points are at x = 8 and x = 6, and candidate inflection points are x = 8 and x = 12. Sign analyses reveal that
f'(x) changes from positive to negative at x = 6 and f'(x) remains negative on either side of x = 8. Moreover, f"(x)
changes from negative to positive at x = 8 and from positive to negative at x = 12. Therefore, f has a local maximum at
x = 6 and inflection points at x = 8 and x = 12. Moreover,
Iim f(x)
X-+±00
= -'-OO
Here is a graph off with the transition points highlighted.
y
-30
31. y = xe-x2
SOLUTION
Let f(x)
= xe-x2. Then
and
f"(x)
= (4x3 -
2x)e-x2 - 4xe-x2
= 2x(2x2 -
3)e-x2
= ± f, and x = 0 and x = ± Y,f are candidates for inflection points. Sign analysis shows that
f'(x) changes from negative to positive at x = - f and from positive to negative at x = f. Moreover, f"(x) changes
from negative to positive at both x = ± Yf and from positive to negative at x = 0. Therefore, f has a local minimum
at x = - f, a local maximum at x = V{ and inflection points at x = 0 and at x = ± Yf. Here is a graph off with the
There are critical points at x
transition point.s highlighted.
y
33. y = x - 2 ln x
SOLUTION
Let f(x) = x - 2 ln x. Note that the domain off is x > 0. Now,
f'(x) = 1 -
~
and
x
f"(x)
2
= 2x
The only critical point is x = 2. Sign analysis shows that f'(x) changes from negative to positive at x
local minimum. Further, f"(x) > 0 for x > 0, so the graph is always concave up. Moreover,
limf(x) = oo
x->oo
Here is a graph off with the transition points highlighted.
y
8
6
4
2
2
4
6
= 2, so f
(2) is a
S EC TIO N 4.6
35. y
= x2 -
SOLUTION
I
Analyzing and Sketching Graphs of Functions
271
Z ln x
Let f(x) =
x2 -
Z ln x. Note that the domain off is x > 0. Now,
f'(x)
= 2x -
z
and
-
x
f"(x)
z
= Z + 2x
The only critical point is x = 1. Sign analysis shows that f'(x) changes from negative to positive at x
local minimum. Further, f"(x) > 0 for x > 0, so the graph is always concave up. Moreover,
Jim f(x)
x--+oo
= 1, so f
(l) is a
= oo
Here is a graph off with the transition points highlighted.
y
IO
6
2
4
37. Investigate the behavior and sketch the graph of f(x) = 18(x - 3)(x - 1)213 using the formulas
,
f
20(x - 1)
3o(x-V
(x) = (x - 1)1/3'
f" (x)
= (x - 1)4~3
SOLUTION
yields critical points at x
30(x- 2.)
f '(x)
5
- (x - 1)1/3
f "(x)
ZO(x - 1)
s
- (x - 1)4/3
= ~, x = 1.
yields potential inflection points at x
= ~, x = 1.
Interval
signs off' and f"
(-oo, ~)
+-
(~, 1)
++
(1, ~)
-+
(~, oo)
++
The graph has an inflection point at x = ~, a local maximum at x = 1 (at which the graph has a cusp), and a local
minimum at x = ~. The sketch looks something like this.
272
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
In Exercises 39-42, sketch the graph of the function, indicating all transition points. If necessary, use a graphing
utility or computer algebra system to locate the transition points numerically.
[RS
39. y =
x2 - 10ln(x2 +I)
SOLUTION
Let f(x)
_?
,
_
= x- -
IO ln(x-7 + I). Then f (x)
J"(x)
=2 _
= 2x -
20x
x 2 + , and
1
(:x2 + 1)(20) - (20x)(2x)
(x2 + 1)2
= x4 + 12x2 (x2
9
+ 1)2
There are critical points at x = 0 and x = ±3, and x = ± ~-6 + 3 "15 are candidates for inflection points. Sign analysis
shows that f'(x) changes from negative to positive at x = ±3 and from positive to negative at x = 0. Moreover, f"(x)
~-6 + 3 "15 and from negative to positive at x = ~-6 + 3 "15. Therefore, f
has a local maximum at x = 0, local minima at x = ±3 and points of inflection at x = ± ~-6 + 3 "15. Here is a graph of
changes from positive to negative at x = -
f with the transition points highlighted.
y
41. y = x 4 -4x2 +x+ I
SOLUTION
Let f(x) = x 4
-
4x2 + x + 1. Then f'(x) = 4.x3 - 8x + I and f"(x) = 12x2 - 8. The critical points are
-Jf.
= --Jf
x = -1.473, x = 0.126, and x = 1.347, while the candidates for points of inflection are x = ±
Sign analysis reveals
that f'(x) changes from negative to positive at x = -1.473, from positive to negative at x = 0.126 and from negative to
positive at x
= 1.347. For the second derivative, f"(x) changes from positive to negative at x
to positive at x =
-Jf.
points of inflection at x
and from negative
Therefore, f has local minima at x = -1.473 and x = 1.347, a local maximum at x = 0.126 and
=±
-Jf.
Moreover,
lim f(x) = oo
x-+±oo
Here is a graph off with the transition points highlighted.
y
In Exercises 43-48, sketch the graph over the given interval, with all transition points indicated.
43. y = x +sin x,
[0, 27T]
Let f(x) = x + sinx. Setting f'(x) = I+ cosx = 0 yields cosx = -1, so that x = 7T is the Ione critical
point on the interval [O, 27T]. Setting f"(x) = - sin x = 0 yields potential points of inflection at x = 0, 7T, 27T on the interval
[0, 27T].
SOLUTION
The graph has an inflection point at x
highlighted.
Interval
signs off' and f"
(0, 7r)
+-
(7T, 27T)
++
= 7T, and no local maxima or minima. Here is a graph off without transition points
SECT I 0 N 4.6
0
45. y
= 2 sin x -
cos2 x,
I
Analyzing and Sketching Graphs of Functions
273
23456
[O, 2:ir]
Letf(x) = 2sinx-cos2 x. Thenf'(x) = 2cosx-2cosx(-sinx) = sin2x+2cosxandf"(x) = 2cos2x2 sin x. Setting f' (x) = 0 yields sin 2x = -2 cos x, so that 2 sin x cos x = -2 cos x. This implies cos x = 0 or sin x = -1,
so that x = ~ or ~. Setting f" (x) = 0 yields 2 cos 2x = 2 sin x, so that 2 sin(~ - 2x) = 2 sin x, or ~ - 2x = x ± 2n:ir. This
yields 3x = ~ + 2n:ir, or x = ~' ~' 9f = ~·
SOLUTION
Interval
signs off' and f"
(O,V
++
(~,D
+-
?f)
--
(~,~)
-+
(f,2:ir)
++
(~,
The graph has a local maximum at x = ~,a local minimum at x = ~,and inflection points at x = ~and x = ~·Here is
a graph off without transition points highlighted.
47. y
=sinx+ -.f3cosx,
SOLUTION
[0,:ir]
-./3 cos x. Setting f'(x) = cos x - -./3 sin x = 0 yields tan x = ~. In the interval [0, :ir],
~. Setting f"(x) = - sin x - -./3 cos x = 0 yields tan x = - -./3. In the interval [0, :ir], the lone solution
Let f(x) = sin x +
the solution is x =
isx = ~-
The graph has a local maximum at x
highlighted.
Interval
signs off' and f"
(0, :ir/6)
+-
(:ir/6, 2:ir/3)
--
(2:ir/3, :ir)
-+
= ~ and a point of inflection at x = ~. Here is a graph off without transition points
274
CH APTER 4
f
APPLICATIONS OF THE DERIVATIVE
49. ~ Are all sign transitions possible? Explain with a sketch why the transitions ++
occur if the function is differentiable. (See Exercise 80 for a proof.)
~ -+ and -- ~ +- do not
SOLUTION In both cases, there is a point where f is not differentiable at the transition from increasing to decreasing
or decreasing to increasing.
y
y
In Exercises 5I-52, draw the graph of a function f having the given limits at ±oo and for which f' and f" take on the
given sign combinations in order.
51. lim f(x) = -oo, lim f(x) = O;
x-+oo
x-+-oo
+-, --,
-+, ++, +-
The graph of one possible function f satisfying lim f(x) = -oo and lim f(x) = 0 and for which f' and
SOLUTION
X--+-oo
f" take on the sign combinations
+-,
-+,
++,
X-HX>
+-
is shown below.
y
2
53. Match the graphs in Figure 19 with the two functions y =
3
x and y = ;x2 l . Explain.
x2 - 1
x -
y
y
iL_jiiL
I
I
I
I
I
I
~~~~--t--'k--t-~~~---...x
11
-11
I
I
I
I
I
I
I
11
I
I
I
I
I
I
I
(A)
x
I
I
I
I
I
I
I
(B)
FIGURE 19
3x2
3
3.x2
Since Iim - 2- - = - · Iim I = 3, the graph of y = - 2- - has a horizontal asymptote of y = 3; hence,
X--+±00 x
- 1 I X-+±00
x - 1
SOLUTION
3x2
the right curve is the graph of f(x) = - 2- - . Since
x - 1
.
3X
3 .
-I
hm--=-·hmx =0
X-+±00
x2 -
1
1
X-+±00
3x
.
.
3x
the graph of y = - 2- - has a honzontal asymptote of y = O; hence, the left curve is the graph of f(x) = -::;- .
x - 1
...- - 1
In Exercises 55-72, sketch the graph of the function. Indicate the transition points and asymptotes.
1
55· y = 3x- 1
s EC TI ON
4.6
I
Analyzing and Sketching Graphs of Functions
275
* t· Moreover, f"(x) =
1
3
Let f(x) = - - . Then f'(x) =
, so that f is decreasing for all x
3x - 1
(3x- 1)2
1
18
, so that f is concave up for x >
and concave down for x <
Because lim - (3x - 1)3
x->±oo 3x - 1
asymptote at y = 0. Finally, f has a vertical asymptote at x = with
SOLUTION
t
t.
= 0, f
has a horizontal
t
.
1
11m - - =-oo
x->j-
3x - 1
and
lim - - =oo
3x - 1
x->j+
y
x+3
57. y= --2
xSOLUTION
x+ 3
Letf(x) = - - . Thenf'(x)
x- 2
-5
.
"
10
=- - , sothatf 1s. decreasmg
for all x * 2. Moreover, f (x) = - (3 ,
(x - 2) 2
x - 2)
so that f is concave up for x > 2 and concave down for x < 2. Because lim x +
x-Hoo X-
y = 1. Finally, f has a vertical asymptote at x = 2 with
lim x+
x->2- X-
1
3
= -oo
2
and
3
2
= 1, f
has a horizontal asymptote at
x+3
lim - - = oo
x- 2
x->2+
1
59. y= - + --1
x x1
,
2x2 -2x+l
, so that f is decreasing for all x
+--.Then f (x) =
0, 1. Moreover,
x- 1
xz (x - 1)2
2 ( 2x3 - 3x2 + 3x - 1)
f"(x) =
, so that f is concave up for 0 < x < ~ and x > 1 and concave down for x < 0 and
3
SOLUTION
Let f(x)
x 3 (x-1)
~
< x < 1. Because lim
X-4±00
x = 0 and x
*
l
= -x
(!
X
+
_!__I)
= 0, f
X -
0. Finally, f
has a horizontal asymptote at y =
has vertical asymptotes at
= 1 with
. ( -1 + -1- ) = -oo
hm
x x-1
x--.>0-
and
lim
x->O+
(! + X
1
x-1
)
= oo
and
Jim
x->1-
(! +
X
_I)
X - 1
= -oo
and
. (1 + -I - ) =oo
hm
x->1+
X
X -
1
I
61.y=--
x(x-2)
. )
1
Th
,
2(1 - x)
. .
Let f(x = x(x _ ).
en f (x) = xZ(x _ ) 2 , so that f 1s mcreasing for x < 0 and 0 < x < 1 and
2
2
,,
2(3x2 - 6x + 4)
, so that f is concave up for x < 0 and x > 2 and
decreasing for 1 < x < 2 and x > 2. Moreover, f (x) =
3
3
SOLUTION
x (x-2)
276
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
1
concave down for 0 < x < 2. Because Jim ( - -- ) = 0, f has a horizontal asymptote at y = 0. Finally, f has vertical
x-+±oo
x(x - 2)
asymptotes at x = 0 and x = 2 with
r1m ( 1 - ) =+oo
x(x - 2)
r1m ( l - ) =-oo
X(X - 2)
and
x-+0-
x--.O+
and
r!ill (1 - ) =-00
x(x - 2)
rJill ( l - ) =oo
x(x - 2)
and
x--.2-
x-+2+
y
5
x
-2
4
-5
l
63.y=26
x - x+ 8
SOLUTION
1
Let f(x) = x2 - 6x + 8
1
h
I
- - - - - . T en f (x) =
~-~~-~
6-2X
~-fu+~
2
,
. .
.
<'
so that f 1s rncreasrng 1or x < 2
and for 2 < x < 3, is decreasing for 3 < x < 4 and for x > 4, and has a local maximum at x = 3. Moreover,
2 (3x2 - l 8x + 28)
, so that f is concave up for x < 2 and for x > 4 and is concave down for 2 < x < 4. Because
3
(x2 - 6x + 8)
1
lim
= 0, f has a horizontal asymptote at y = 0. Finally, f has vertical asymptotes at x = 2 and x = 4, with
X-+±00 X2 - 6X + 8
f"(x) =
. (
hm
x-+2-
= oo
x2
-
I
)
6x + 8
x2
-
I
) = -oo
6x + 8
. (
hm
and
x-+2+
l
x2 - 6x + 8
)
= -oo
and
. (
hm
x-+4-
. (
l !ill
and
x-+4+
x2
-
I
)
6x + 8
= 00
y
65. y
3
4
= 1- -x + -x3
SOLUTION
3
4
Letf(x) = 1 - - + 3"· Then
x x
f'(x) = ~ _ 12 = 3(x-2)(x+2)
x2
x4
x4
so that f is increasing for lxl > 2 and decreasing for -2 < x < 0 and for 0 < x < 2. Moreover,
f"(x) =
-~ + 48
x3
x5
= 6(8 - x2)
x5
so that f is concave down for -2 V2 < x < 0 and for x > 2 Vi, while f is concave up for x < -2 V2 and for 0 < x < 2 Vi.
Because
. ( 3 4)
hm 1 - - + - =l
X4±00
X
X3
f has a horizontal asymptote at y = l. Finally, f has a vertical asymptote at x = 0 with
. ( 3 4)
hm 1 - - + - = -oo
x x3
x-+0-
and
lim
x-+0+
(1 - ~x + ~)
= oo
x3
I
SECTION 4.6
Analyzing and Sketching Graphs of Functions
277
y
6
4
2
-6 -4 -2
-2
2
4
6
-4
-6
1
67. y
= x2
1
- (x - 2)2
Let f(x)
SOLUTION
= 21
x
1
- - - -2 . Then f'(x)
(x- 2)
= -2x-3 + 2 (x -
2)- 3 , so that f is increasing for x < 0 and for x > 2
and is decreasing for 0 < x < 2. Moreover,
f" (x) = 6x-4 -
6 (x - 2)-4
= -48(x -
1)(x2 - 2x + 2)
so that f is concave up for x < 0 and for 0 < x < 1, is concave down for 1 < x < 2 and for x > 2, and has a point
1
of inflection at x = 1. Because lim (~ - - -- 2 ) = 0, f has a horizontal asymptote at y = 0. Finally, f has vertical
X
X->±oo
2)
(X -
asymptotes at x = 0 and x = 2 with
1
-)
lim (_!__ - -
= oo
and
) = -oo
and
(x - 2)2
x2
x->0-
1
lim (_!__2 - ) = oo
x
(x - 2)2
x->O+
and
lim (_!__2 - -
x->2-
x
1
(x - 2)2
y
1
69. y
= (x2 + 1)2
Let f(x)
SOLUTION
-4x
1
= (x2 + l) 2 . Then f' (x) = (x2 + l)3 , so that f
is increasing for x < 0, is decreasing for x > 0 and
has a local maximum at x = 0. Moreover,
f"(x)
= -4(x2 + 1)3 + 4x · 3(x2 + 1)2 · 2x = 20x2 (x2
so that
f
+ 1)6
(x2
4
+ 1)4
is concave up for lxl > 1I -../5, is concave down for lxl < 1I -../5, and has points of inflection at x = ± 1/
1
l) 2 = 0, f has a horizontal asymptote at y = 0. Finally, f has no vertical asymptotes.
2
Because lim (
X-t±OO
X
+
y
-4
1
71. y= -~
2
vx + 1
-2
2
4
-../5.
278
C HAP TE R 4
I
APPLICATIONS OF THE DERIVATIVE
SOLUTION
Letf(x)
=
~-Then
V'x'·+J
f'(x)
x
=
-;:::=== = -x(x2 + 1)-3/2
""(x2 + J)3
so that f is increasing for x < 0 and decreasing for x > 0. Moreover,
J"(x)
so that f is concave down for
= -~x(x2 + 1r5/2(-2x)- (.x2 + 1)-3/2 = (2.x2 2
Jxl <
f
and concave up for
Ix! >
f. Because
1
lim - -..,/x2
x_,±""
f has a horizontal asymptote at y
= 0. Finally, f
l)(.x2 + o-5/2
+I
=0
has no vertical asymptotes.
y
-5
-IO
73. Consider the general logistic function, P(x) =
(a) P'(x)
= (IMAke-kx
+Ae-kx)2
(b) lim P(x)
x--+-oo
and P"(x)
1
5
+!-kx,
IO
with A, M, and k all positive. Show that
= MAk(I2e-kx(Ae-kx
-1)
+Ae-kx)3
= 0 and Jim P(x) = M, and therefore P = 0 and P =Mare horizontal asymptotes of P.
x--+oo
(c) Pis increasing for all x.
(d) The only inflection point of Pis at (1"t, ~).To the left of it Pis concave up, and to the right of it Pis concave down.
SOLUTION
Let P(x)
= 1 +.7.-kx, with A, M, and k all positive.
(a) Then
P'(x) =
and
(b) Because k is positive,
lim
e-kx
= oo
and
lim e-kx
=0
X-HX>
It then follows that
Jim P(x)
x-;-oo
= I +A
M
Jim
e-kx
=O
and
}~~ P(x)
M
M
= I +Alim e-kx = I + A(O) = M
x-;oo
Therefore, P = 0 and P = M are horizontal asymptotes of P.
(c) Because M, A, and k are all positive,
MAke-kx
P'( )-
x - (I + Ae-kx)2
for all x. Therefore, Pis increasing for all x.
>0
s E C T I o N 4.6 I Analyzing and Sketching Graphs of Functions
279
(d) The second derivative of Pis defined for all x and is equal to zero when
Ae-kx - 1 = 0
x
or
= -lnA
k
Note
p(~A)-
k
-
M
_
M
2
M
1 +Ae-lnA - 1 +A·A- 1
Now, for x < lnkA, P"(x) > 0, while for x > l~A, P"(x) < 0. Thus, the only inflection point of Pis at (1n/, -¥).To the left
of this point P is concave up, and to the right P is concave down.
Further Insights and Challenges
In Exercises 75-79, we explore functions whose graphs approach a nonhorizontal line as x
called a slant asymptote if
lim(f(x) - (ax+ b))
X->OO
-4
oo. A line y = ax+ b is
=0
or
lim (f(x) - (ax+ b))
x-+-oo
75. Let f(x)
=0
= _!___
(Figure 21). Verify the following:
x-l
(a) f(O) is a local max and f(2) a local min.
(b) f is concave down on (-oo, 1) and concave up on (1, oo ).
(c) lim f(x) = -oo and lim f(x) = oo.
X-+1-
X-+1+
(d) y = x + 1 is a slant asymptote off as x -4 ±oo.
(e) The slant asymptote lies above the graph off for x < 1 and below the graph for x > 1.
y
10
x2
1D(x)=
x-1
I
~
I
I1
......... ,..
I
/
..........,
-t/
--
y=x+ I
-10
x
10
-10
FIGURE 21
x2
x(x- 2)
2
. Then f'(x) = (x _ l) 2 and f"(x) = (x _ l) 3 ·
1
(a) Critical points of j'(x) occur at x = 0 and x = 2. x = 1 is not a critical point because it is not in the domain off.
Sign analysis reveals that f' changes from positive to negative at x =0 and from negative to positive at x = 2. Thus, f(O)
is a local maximum, and f (2) is a local minimum.
(b) Sign analysis of f"(x) reveals that f"(x) < 0 and f is concave down on (-oo, 1) whereas f"(x) > 0 and f is concave
up on (1, oo).
(c)
SOLUTION
Let f(x) = x _
lim f(x)
x->I-
(d) Note
that
using
1
= -1 x->Iiim - - = -oo
1- X
polynomial
division,
and
x-+±oo
(e) For x > 1, f(x) - (x + 1)
1
X-
= x _1 1
1
= 1 x-+1+
lim - - = oo
X - 1
x2
f(x)
lim (f(x) - (x + 1)) = lim x + 1 + _!_l - (x + 1) = lim -
x-+±oo
lim j(x)
'x-+l+
x_
1
X-+±oo X -
1
x
+
1
+
x-1
Then
- = 0.
I
> 0, so f(x) approaches x + 1 from above. Similarly, for x < 1, f(x) - (x + 1)
- - < 0, so f (x) approaches x + 1 from below.
x-1
77. Sketch the graph of
x2
f(x)= x+l
Proceed as in the previous exercise to find the slant asymptote.
=
280
CH APT E R 4
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APPLICATIONS OF THE DERIVATIVE
x2
x(x+ 2)
2
Let f(x) = x + . Then f'(x) = (x + l) 2 and f"(x) = (x + 1)3 • Thus, f is increasing for x < -2 and for
1
x > 0, is decreasing for -2 < x < -1 and for -1 < x < 0, has a local minimum at x = 0, has a local maximum at x = -2,
is concave down on (-oo, -1) and concave up on (-1, oo). Limit analyses give a vertical asymptote at x = -1, with
SOLUTION
.
1!ID
x2
=-oo
--
x~-1-x+l
.
x2
11m--=oo
x+ 1
and
x~-1+
1
By polynomial division, f(x) = x - 1 + - - and
x+l
lim
x~±oo
(x - 1 + - +1-1 - (x - 1)) = 0
X
which implies that the slant asymptote is y = x - 1. Notice that f approaches the slant asymptote as in Exercise 78.
y
4
-t---+-+-''+""..,__t---+-x
-4
-2 : ,. /
....... ~ ...... -2
_Ai
4
-4
li
7
2
-6
l -x2
79. Sketch the graph of f(x) = - - .
2-x
SOLUTION
Let f(x) =
3
- x2. Using polynomial division, f(x) = x + 2 + - -. Then .
2 -x
x-2
1
lim (f(x) - (x + 2)) = lim (ex+ 2) + X-+±oo
which implies that y
3
X-
x-+±oo
3
- - (x + 2)) = lim - - = ~ . lim x- 1 = O
2
x-+±oo X - 2
1 x-+±oo
= x + 2 is the slant asymptote of f(x). Since f(x) -
(x + 2) = -
3
- > 0 for x > 2, f(x) approaches
x-2
3
the slant asymptote from above for x > 2; similarly, - - < 0 for x < 2 so f(x) approaches the slant asymptote
x-2
x2 -4x+ 1
-6
from below for x < 2. Moreover, f'(x) =
and f"(x) = - (3 . Sign analyses reveal a local minimum at
( 2 -x) 2
2 -x)
x = 2 + '13, a local maximum at x = 2 - V3 and that f is concave down on (-oo, 2) and concave up on (2, oo ). Limit
analyses give a vertical asymptote at x = 2.
y
10
5
-10
81.
~
f'(b)
-5
il/
J,,,//
/:
5
10
x
Assume that f" exists and f"(x) > 0 for all x. Show that f(x) cannot be negative for all x. Hint: Show that
* 0 for some band use the result of Exercise 74 in Section 4.4.
SOLUTION Let f(x) be a function such that f"(x) exists and f"(x) > 0 for all x. Since j"(x) > 0, there is at least one
point x = b such that f'(b)
0. If not, f'(x) = 0 for all x, so f"(x) = 0. By the result of Exercise 74 in Section 4.4,
f(x) ~ f(b) + f'(b)(x - b). Now, if f'(b) > 0, we find that f(b) + f'(b)(x - b) > 0 whenever
*
bf'(b) - f(b)
x>----f'(b)
a condition that must be met for some x sufficiently large. For such x, f(x) > f (b) + f' (b )(x - b) > 0. On the other hand,
if f'(b) < 0, we find that f(b) + f'(b)(x - b) > 0 whenever
x<
For such an x, f(x) > f(b) + f'(b)(x - b) > 0.
bf'(b) - f(b)
f'(b)
S E CT I O N 4. 7
I
Applied Optimization
281
4.7 Applied Optimization
Preliminary Questions
1. The problem is to find the right triangle of perimeter I 0 whose area is as large as possible. What is the constraint
equation relating the base b and height h of the triangle?
SOLUTION
The perimeter of a right triangle is the sum of the lengths of the base, the height, and the hypotenuse. If the
base has length b and the height is h, then the length of the hypotenuse is Yb 2 + h2 and the perimeter of the triangle is
P = b + h + Yb 2 + h 2 • The requirement that the perimeter be 10 translates to the constraint equation
b + h + Yb 2 + h 2 = IO
2. Describe a way of showing that a continuous function on an open interval (a, b) has a minimum value.
SOLUTION If the function tends to infinity at the endpoints of the interval, then the function must take on a minimum
value at a critical point.
3. Is there a rectangle of area I 00 of largest perimeter? Explain.
No. Even by fixing the area at 100, we can take one of the dimensions as large as we like, thereby allowing
the perimeter to become as large as we like.
SOLUTION
Exercises
1. Find the dimensions x and y of the rectangle of maximum area that can be formed using 3 m of wire.
(a) What is the constraint equation relating x and y?
(b) Find a formula for the area in terms of x alone.
(c) What is the interval of optimization? Is it open or closed?
(d) Solve the optimization problem.
SOLUTION
(a) The perimeter of the rectangle is 3 meters, so 3 = 2x + 2y, which is equivalent toy = ~ - x.
= ~x - x2.
(c) This problem requires optimization over the closed interval [0, ~],since both x and y must be nonnegative.
(d) A'(x) = ~ - 2x = 0, which yields x = and consequently, y =
Because A(O) = A(3/2) = 0 and A(i) = 0.5625,
the maximum area 0.5625 m2 is achieved with x = y = ~ m.
(b) Using part (a), A= xy = x(~ - x)
i
i·
3. A rectangular bird sanctuary is being created with one side along a straight riverbank. The remaining three sides
are to be enclosed with a protective fence. If there are 12 km of fence available, find the dimension of the rectangle to
maximize the area of the sanctuary.
SOLUTION Let x denote the length of the rectangular sanctuary running perpendicular to the riverbank, and y denote
the length of the sanctuary parallel to the riverbank. The area enclosed by the sanctuary is A = xy, and the amount
of protective fencing used to construct the sanctuary is 2x + y km. There are 12 km of protective fencing available, so
2x + y = 12, or y = 12 - 2x. Substituting for yin the formula for the area yields
A = x(l2 - 2x) = l2x - 2x2
Because both x and y must be nonnegative, the domain for the area function is the closed interval [O, 6]. Now,
A'=12-4x=0
when x = 3. Evaluating A at the endpoints of the domain and at the critical point x
A(O) = 0,
A(3)
= 18,
and
A(6)
= 3 yields
=0
Thus, the area of the sanctuary is maximized when 3 km of fencing is used perpendicular to the riverbank and 6 km of
fencing is used parallel to the riverbank.
5. Find two positive real numbers such that the sum of the first number squared and the second number is 48 and their
product is a maximum.
SOLUTION Let x denote the first number, and y denote the second number. The objective is to maximize P = xy subject
to the condition that x2 + y = 48, or y = 48 - x2. Substituting for y in the formula for P yields
P
= x(48-x2) = 48x-x3
Because x and y must both be positive, the domain for the product function is the open interval (0, 4 -./3). Now,
P'
= 48 -
3x2
=0
when x = 4. Because P(4) = 128, while P ~ 0 as x ~ o+ and as x ~ 4-,/3-, we conclude that P achieves its absolute
maximum when x = 4. Thus, the two positive real numbers such that the sum of the first number squared and the second
number is 48 and their product is a maximum are 4 and 32.
282
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
7. A wire of length 12 mis divided into two pieces and the pieces are bent into a square and a circle. How should this
be done in order to minimize the sum of their areas?
Suppose the wire is divided into one piece of length x m that is bent into a circle and a piece of length
12 - x m that is bent into a square. Because the circle has circumference x, it follows that the radius of the circle is x/2Tr;
therefore, the area of the circle is
SOLUTION
1f
x2
(2Trx)2 = 4Tr
As for the square, because the perimeter is 12 - x, the length of each side is 3 - x/4 and the area is (3 - x/4) 2 • Then
)2
1
A(x) = -x2 + ( 3 - -x
4Tr
4
Now
when
12Tr
x
Because A(O)
.
=- m "" 5.28 m
4+Tr
= 9 m 2 , A(12) = 36/n "" 11.46 m2 , and
A(
12
Tr ) "" 5.04 m 2
4 +Tr
we see that the sum of the areas is minimized when approximately 5.28 m of the wire is allotted to the circle.
9. Find two positive real numbers such that they add to 40 and their product is as large as possible.
SOLUTION
x +y
Let x and y denote the two numbers. The objective is to maximize P
x. Substituting for y in the formula for P yields
= 40, or y = 40 -
= xy subject to the condition that
P = x(40-x) = 40x-x2
Because x and y must both be positive, the domain for the product function is the open interval (0, 40). Now,
P' =40-2x=O
when x = 20. Because P(20) = 400, while P --+ 0 as x --+ o+ and as x --+ 40-, we conclude that P achieves its absolute
maximum when x = 20. Thus, the two positive real numbers such that they add to 40 and their product is as large as
possible are 20 and 20.
11. Find two positive real numbers x and y such that their product is 800 and x + 2y is as small as possible.
SOLUTION
The objective is to minimize S
= x + 2y subject to the condition that xy = 800, or y = si:i. Substituting for
y in the formula for S yields
S =x+ 1600
x
The domain for this function is x > 0. Now,
S'
=l
_ 1600
=O
x2
when x = 40. As x --+ o+ and as x --+ oo, S --+ oo, so S achieves its absolute minimum when x = 40. Thus, the two
positive real numbers x and y such that their product is 800 and x + 2y is as small as possible are x = 40 and y = 20.
13. Find the dimensions of the box with square base with
(a) Volume 12 and the minimal surface area.
(b) Surface area 20 and maximal volume.
SOLUTION
A box has a square base of side x and height y where x, y > 0. Its volume is V = x2y and its surface area is
s = 2x2 +4xy.
(a) If V = x2y = 12, then y = 12/ x2 and S (x) = 2x2 + 4x ( 12/x2) = 2x2 + 48x- 1 • Solve S '(x) = 4x - 48x-2 = 0 to obtain
x = 12113 • Since S (x) --+ oo as x --+ O+ and as x --+ oo, the minimum surface area is S (12 1' 3 ) = 6 (12) 213 "" 31.45, when
x = 12113 andy = 12 113 •
I
SECTION 4.7
Applied Optimization
283
(b) If S = 2x2 + 4.xy = 20, then y = sx- 1 - !x and V(x) = x 2y = 5x - ~x3. Note that x must lie on the closed interval
[0,
VW]. Solve V'(x)
1,5. Since V(O) = V( VW)
1,5 and y = 1,5.
= 5 - ~x2 for x > 0 to obtain x =
maximum volume is V( 1,5) =
lf V30 ~ 6.086, when x =
= O and V (
1,5)
=
10
f35, the
15. A rancher will use 600 m of fencing to build a corral in the shape of a semicircle on top of a rectangle (Figure 11).
Find the dimensions that maximize the area of the corral.
FIGURE 11
SOLUTION Let x be the width of the corral and therefore the diameter of the semicircle, and let y be the height of the
rectangular section. Then the perimeter of the corral can be expressed by the equation 2y + x + ~ x = 2y + (1 + ~ )x = 600
m or equivalently, y = ~ (600 - (1 + Dx). Since x and y must both be nonnegative, it follows that x must be restricted to
the interval [0, 1 ~<;;>12 ]. The area of the corral is the sum of the area of the rectangle and semicircle, A = .xy + ~ x2. Making
the substitution for y from the constraint equation,
Now,A'(x)
= 300-(l + ~)x+ ~x = 0 impliesx = (;~) ~
! ~14 ) ~ 25204.5 m2
0
A(
1
l68.029746m. WithA(O)
= O m2 ,
2
600 )
A ( l + n/ 2 ~ 21390.8 m
and
it follows that the corral of maximum area has dimensions
300
1 +n/4
x=---m
and
17. Find the dimensions of the rectangle of maximum area that can be inscribed in a circle of radius r
= 4 (Figure 13).
FIGURE 13
SOLUTION
Place the center of the circle at the origin with the sides of the rectangle (of lengths 2x > 0 and 2y > 0)
parallel to the coordinate axes. By the Pythagorean Theorem, x2 + y2 = r 2 = 16, so thaty = V16 -x2 • Thus the area of
the rectangle is A(x) = 2x · 2y = 4x V16 - x 2. To guarantee both x and y are real and nonnegative, we must restrict x to
the interval [0, 4]. Solve
A'(x)
}z
= 2y = 4 -../2.
for x > 0 to obtain x =
dimensions 2x
= 4 V16 -
x2 -
4
x2
v16-x2
=O
= 2 Vi. Since A(O) = A(4) = 0 and A(2 -../2) = 32, the rectangle of maximum area has
19. In the setting of Examples 2 and 3, let r denote the speed along the road, and h denote the speed along the highway.
(a) Show that the travel-time function T(x) has a critical point at x = ~ and explain why this indicates that if
(h/r)2 -1
r
~
h there is no critical point.
(b) Explain why there cannot be a critical point at x
close to 0.
= 0, but depending on the speeds, the critical point can be arbitrarily
284
CH APT E R 4
I
APPLICATIONS OF THE DERIVATIVE
SOLUTION
(a) In the setting of Examples 2 and 3, if r is the speed along the road and h is the speed along the highway, then the
travel-time function is
T(x)=
.Y900 + x2
r
50 - x
+-h-
Therefore,
T'(x) =
1
x
r\1900 + x
2
h
Setting T'(x) = 0 and solving for x yields
x
-
r\1900 + x2
h
xh = .Y900 + x 2
r
x2h2
- 2 =900+x2
r
x=
30
-;:==:::==
2
...j(h/r) - 1
If r:::: h, then (h/r) 2 - 1 ::; 0, implying there is no solution to T'(x) = 0 and therefore no critical point.
(b) Because the numerator in the expression for the critical point is nonzero, the critical point must be nonzero. Note
that for fixed r,
30
lim
=0
h->oo ...j(h/r)2 - 1
so by making h large enough, the critical point can be made arbitrarily close to 0. Also, for fixed h,
lim
r->O
30
...j(h/r)2 - 1
=0
so by making r small enough, the critical point can be made arbitrarily close to 0.
21. In the article "Do Dogs Know Calculus?" the author Timothy Pennings explained how he noticed that when he threw
a ball diagonally into Lake Michigan along a straight shoreline, his dog Elvis seemed to pick the optimal point in which
to enter the water so as to minimize his time to reach the ball, as in Figure 14. He timed the dog and found Elvis could
run at 6.4 mis on the sand and swim at 0.91 mis. If Tim stood at point A and threw the ball to a point Bin the water,
which was a perpendicular distance 10 m from point Con the shore, where C is a distance 15 m from where he stood, at
what distance x from point C did Elvis enter the water if the dog effectively minimized his time to reach the ball?
B
FIGURE 14
Based on the diagram in Figure 14, it follows that Elvis runs a distance of 15 - x m along the shoreline and
swims a distance of .Yx 2 + 100 m. The time it takes for Elvis to reach point B is then
SOLUTION
15-x
.Yx2 +100
T=--+---6.4
0.91
The domain of this function is the closed interval [O, 15]. Now,
T' = - __!__ +
x
6.4 0.91.Yx2 +100
=0
SECTION 4.7
I
Applied Optimization
285
when
0.91
x
or
x
9.1
= --,:.=.~===;;:
"" 1.44
2
.../6.4 + 0.9!2
Evaluating T at the endpoints of the domain and at the critical point x "" 1.44 yields
T(O) "" 13.33,
and
T(l.44) "" 13.22,
T(15) "" 19.81
Therefore, Elvis entered the water a distance of 1.44 m from point C.
23. Find the point on the line y = x closest to the point (1, 0). Hint: It is equivalent and easier to minimize the square of
the distance.
Withy = x, let's equivalently minimize the square of the distance, f(x) = (x - 1)2 + y2 = 2x2 - 2x + I,
which is computationally easier (when working by hand). Solve f'(x) = 4x - 2 = 0 to obtain x =
Since f(x) ~ oo as
x ~ ±oo, ( ~, is the point on y = x closest to (I, 0).
SOLUTION
4.
4)
25. [A 5 Find a good numerical approximation to the coordinates of the point on the graph of y = In x - x closest to
the origin (Figure 16).
y
\
\
K
FIGURE 16
The distance from the origin to the point (x, In x - x) on the graph of y = In x - xis d = 1/x2 + (In x - x) 2 •
As usual, we will minimize d2. Let d2 = f(x) = x2 +(In x - x)2. Then
SOLUTION
f' (x)
= 2x + 2(ln x - x)
G-
1)
To determine x, we need to solve
2lnx
4x + - - - 2 ln x - 2
x
=0
This yields x "" 0.632784. Thus, the point on the graph of y = In x - x that is closest to the origin is approximately
(0.632784, -1.090410).
27. Find the angle(} that maximizes the area of the isosceles triangle whose legs have length t (Figure 17), using the fact
the area is given by A = 2 sin 6.
4e
f)
FIGURE 17
SOLUTION
The area of the triangle is
A(6) =
~t2 sin 6
where 0 :o; (} :o; 7r. Setting
A'(6)
yields(}= ~. Since A(O) = A(7r) = 0 and A(~) =
4e
2
,
1
2
= -t
cose = 0
2
the angle that maximizes the area of the isosceles triangle is
e= ~.
286
CH APT E R 4
I
APPLICATIONS OF THE DERIVATIVE
29. Find the area of the largest isosceles triangle that can be inscribed in a circle of radius 1 (Figure 19).
FIGURE 19
SOLUTION
A= x Yl - xz for O ~ x
0, and x
= 1. Hence, we wish to maximize
= '1fi/ = 0. Thus, (1 - 2x)(l + x) =
The area of the isoceles triangle is given by A= (x + l)y where x2 + y2
~ 1. Then A'(x)
= 1/2. A(l /2) = 3 f,
= Vl - x + (1 + x)~(l - :x?-r 1 (-2x)
= 1 and A(l) = 0, so the maximum area is 3 f.
12
2
whereas A(O)
31. A poster of area 6000 cm2 has blank margins of width 10 cm on the top and bottom and 6 cm on the sides. Find the
dimensions that maximize the printed area.
Let x be the width of the printed region, and let y be the height. The total printed area is A = xy. Because the
total area of the poster is 6000 cm2 , we have the constraint (x + 12)(y + 20) = 6000, so that xy + 12y + 20x + 240 = 6000,
-zox Therefore ' A(x) = 20 288x+l2
x-x2 ' where 0 <
x<
288 •
or y -- 5760
x+12 •
A(O) = A(288) = 0, so we are looking for a critical point on the interval [O, 288]. Setting A'(x) = 0 yields
SOLUTION
20
(x + 12)(288 - 2x) - (288x - x2) =
(x + 12)2
0
-x2 - 24x + 3456
-----=0
(x + 12)2
x2 + 24x - 3456 = 0
(x - 48)(x + 72) = 0
Therefore x = 48 or x = - 72. x = 48 is the only critical point of A(x) in the interval [O, 288], so A(48) = 3840 is the
maximum value of A(x) in the interval [O, 288]. Now, y = 20~888;08 = 80 cm, so the poster with maximum printed area is
48 + 12 = 60 cm. wide by 80 + 20 = 100 cm. tall.
33. Kepler's Wine Barrel Problem In his work Nova stereometria doliorum vinariorum (New Solid Geometry of a
Wine Barrel), published in 1615, astronomer Johannes Kepler stated and solved the following problem: Find the dimensions of the cylinder of largest volume that can be inscribed in a sphere of radius R. Hint: Show that an inscribed cylinder
has volume 2nx(R2 - x2), where xis one-half the height of the cylinder.
Place the center of the sphere at the origin in three-dimensional space. Let the cylinder be of radius y
and half-height x. The Pythagorean Theorem states, x2 + y2 = R2 , so that y2 = R2 - x2. The volume of the cylinder
is V(x) = ny2 (2x) = 2n(R 2 - :x?-)x = 2nR2 x - 2nx3. Allowing for degenerate cylinders, we have 0 ~ x ~ R. Solve
SOLUTION
V'(x)
= 2nR2 -
when x
6n:x?-
= ~ andy =
= 0 for x;:::
0 to obtain x
=
~-Since V(O)
= V(R) = 0, the largest volume is V( ~) = ~7!' -../3R3
jfR.
35. A landscape architect wishes to enclose a rectangular garden of area 1000 m2 on one side by a brick wall costing
$90/m and on the other three sides by a metal fence costing $30/m. Which dimensions minimize the total cost?
SOLUTION Let x be the length of the brick wall and y the length of an adjacent side with x,y > 0. With xy
y = 1 ~, the total cost is
= 1000 or
= 90x + 30 (x + 2y) = 120x + 60000x- 1
0 for x > 0 to obtain x = 10 VS. Since C(x) ~
C(x)
Solve C'(x) = 120 - 6000ox-2 =
oo as x ~ O+ and as x ~ oo, the
minimum cost is C( 10 .../5) = 2400 .../5 "" $5366.56 when x = 10 -VS "" 22.36 m and y = 20 .../5 "" 44. 72 m.
37. Find the maximum area of a rectangle inscribed in the region bounded by the graph of y
21).
y
4
FIGURE 21
=4-
x and the axes (Figure
2+x
S EC TI 0 N 4.7
SOLUTION
I
Applied Optimization
287
Lets be the width of the rectangle. The height of the rectangle is h == i~:. so that the area is
4- s 4s - s 2
A(s) == s - - == - 2+s
2+s
We are maximizing on the closed interval [0,4]. It is obvious from the pictures that A(O) = A(4) = 0, so we look for
critical points of A.
A'(s) = (2 + s)(4 - 2s) - (4s - s
(2 + s) 2
2
)
= _ s2 + 4s - 8
(s + 2) 2
The only point where A'(s) doesn't exist is s = -2 which isn't under consideration.
Setting A'(s) = 0 gives, by the quadratic formula,
s=
-4 ± V48
./;;
== -2 +- 2 v3
2
Of these, only -2 + 2 VJ is positive, so this is our lone critical point. A(-2 + 2 VJ) "" 1.0718 > 0. Since we are finding
the maximum over a closed interval and -2 + 2 VJ is the only critical point, the maximum area is A(-2 + 2 VJ) "" 1.0718.
39. Find the maximum area of a rectangle circumscribed around a rectangle of sides L and H. Hint: Express the area in
terms of the angle e (Figure 22).
FIGURE 22
SOLUTION Position the L x H rectangle in the first quadrant of the .xy-plane with its "northwest" comer at the origin.
Let e be the angle the base of the circumscribed rectangle makes with the positive x-axis, where 0 $ e $ ~. Then the area
of the circumscribed rectangle is A= LH + 2 · ~(Hsin8)(Hcos8) + 2 · ~(LsinO)(Lcose) = LH + ~(L2 + H 2 )sin28,
which has a maximum value of LH + ~(L2 + H 2 ) when e =~because sin28 achieves its maximum when e = ~-
41. Find the equation of the line through P == (4, 12) such that the triangle bounded by this line and the axes in the first
quadrant has minimal area.
SOLUTION Let P == (4, 12) be a point in the first quadrant and y - 12 == m(x - 4), -oo < m < 0, be a line through P
that cuts the positive x- and y-axes. Then y = L(x) = m(x - 4) + 12. The line L(x) intersects the y-axis at H (0, 12 - 4m)
and the x-axis at W (4 - -!: , 0). Hence the area of the triangle is
I (12 - 4m) ( 4 - 12)
A(m) = 2
m = 48 - Sm - nm- 1
Solve A'(m) = 72m-2 - 8 = 0 form< 0 to obtain m = -3. Since A--> oo as m--> -oo or m--> 0-, we conclude that the
minimal triangular area is obtained when m = -3. The equation of the line through P = (4, 12) is y = -3(x - 4) + 12 =
-3x+ 24.
43. Archimedes's Problem A spherical cap (Figure 24) of radius rand height h has volume V == nh2 (r - th) and
surface area S == 2nrh. Prove that the hemisphere encloses the largest volume among all spherical caps of fixed surface
areas.
FIGURE 24
288
CH APT E R 4
I
APPLICATIONS OF THE DERIVATIVE
SOLUTION
Consider all spherical caps of fixed surface area S. Because S = 27rrh, it follows that
s
r
= 27rh
and
V(h) = Trh 2 ( - S
27rh
I ) = S-h - 7r
- -h
-h 3
3
2
3
Now
V'(h)
= ~2 -Trh2 = 0
when
s
or h = - = r
27rh
Hence, the hemisphere encloses the largest volume among all spherical caps of fixed surface area S .
45. A box of volume 72 m 3 with a square bottom and no top is constructed out of two different materials. The cost of
the bottom is $40/m2 and the cost of the sides is $30/m2 . Find the dimensions of the box that minimize total cost.
SOLUTION
Lets denote the length of the side of the square bottom of the box and h denote the height of the box. Then
V
= s2h = 72
72
or h = s2
The cost of the box is
C = 40s2 + 120sh
8640
= 40s2 + s
so
C'(s)
when s = 3 f4 m and h
the minimum cost.
8640
= 80s- T
=0
= 2 f4 m. Because C ~ oo as s ~ 0- and as s ~ oo, we conclude that the critical point gives
47. Your task is to design a rectangular industrial warehouse consisting of three separate spaces of equal size as in
Figure 26. The wall materials cost $500 per linear meter and your company allocates $2,400,000 for that part of the
project involving the walls.
(a) Which dimensions maximize the area of the warehouse?
(b) What is the area of each compartment in this case?
I I I I
FIGURE 26
SOLUTION Let one compartment have length x and width y. Then total length of the wall of the warehouse is P =
4x + 6y and the constraint equation is cost= 2,400,000 = 500(4x + 6y), which gives y = 800 - ~x.
(a) Area is given by A= 3.xy = 3x(800 - ~x) = 2400x - 2x2, where 0 :<;; x :<;; 1200. Then A'(x) = 2400 - 4x = 0 yields
= 600 and consequently y = 400. Since A(O) = A(l200) = 0 and A(600) = 720, 000, the area of the warehouse is
maximized when each compartment has length of 600 m and width of 400 m.
x
(b) The area of one compartment is 600 · 400 = 240, 000 square meters.
49.
According to a model developed by economists E. Heady and J. Pesek, if fertilizer made from N pounds of nitrogen
and P lb of phosphate is used on an acre of farmland, then the yield of corn (in bushels per acre) is
Y = 7.5 + 0.6N + 0.7P- O.OOIN2
-
0.002p2 + 0.001NP
A farmer intends to spend $30/acre on fertilizer. If nitrogen costs 25 cents/lb and phosphate costs 20 centsflb, which
combination of N and P produces the highest yield of corn?
SECTION 4.7
SOLUTION
I
Applied Optimization
289
The farmer's budget for fertilizer is $30 per acre, so we have the constraint equation
0.25N + 0.2P
= 30
or
P
= 150 -
l.25N
Substituting for P in the equation for Y, we find
Y(N) = 7.5 + 0.6N + 0.7(150 - 1.25N) - 0.001N2
-
0.002(150 - 1.25N)2 + 0.001N(150 - 1.25N)
= 67.5 + 0.625N - 0.005315N2
Both N and P must be nonnegative. Since P
:~ = 0.625 Now, Y(O) = 67.5, Y(120)
and P"' 77.33 pounds.
= 150 -
1.25N:::: 0, we require that 0 5 N 5 120. Next,
=>
0.01075N = 0
N =
0.625
_
"'58.14 pounds
0 01075
= 65.1 and Y(58.14) "' 85.67, so the maximum yield of corn occurs for N
"' 58.14 pounds
51. All units in a 100-unit apartment building are rented out when the monthly rent is set at r = $900/month. Suppose
that one unit becomes vacant with each $10 increase in rent and that each occupied unit costs $80/mon in maintena~ce.
Which rent r maximizes monthly profit?
SOLUTION
Let n denote the number of $10 increases in rent. Then the monthly profit is given by
P(n)
= (100 -
n)(900 + lOn - 80)
= 82000 + 180n -
10n2
and
P'(n) = 180 - 20n = 0
when n = 9. We know this results in maximum profit because this gives the location of vertex of a downward opening
parabola. Thus, monthly profit is maximized with a rent of $990.
53. The monthly output of a Spanish light bulb factory is P = 2LK2 (in millions), where Lis the cost of labor and K is
the cost of equipment (in millions of euros). The company needs to produce 1.7 million units per month. Which values
of L and K would minimize the total cost L + K?
5
SOLUTION Since P = 1.7 and P = 2LK2 , we have L =
. Accordingly, the cost of production is
o-;
C(K)
Solve C'(K)
=L + K =K +
= 1 - ~ for K ~ 0 to obtain K = {!G. Since C(K)
of production is achieved for K
and 600, 000 euros in labor.
= {fG "'
1.2 and L
0.85
K2
----7
oo as K
----7
O+ and as K
----7
oo, the minimum cost
= 0.6. The company should invest 1.2 million euros in equipment
55. Brandon is on one side of a river that is 50 m wide and wants to reach a point 200 m downstream on the opposite
side as quickly as possible by swimming diagonally across the river and then running the rest of the way. Find the best
route if Brandon can swim at 1.5 mis and run at 4 mis.
Let lengths be in meters, times in seconds, and speeds in m/s. Suppose that Brandon swims diagonally to
a point located x meters downstream on the opposite side. Then Brandon then swims a distance x2 + 502 and runs a
distance 200 - x. The total time of the trip is
SOLUTION
v
f(x)
=
Vx2 1.+ 2500 + -200-x
-,
5
4
0 5 x 5 200
Solve
f'(x)
=
3
vx 2x+2500 - !4 = 0
2
to obtain x = 30-fr "' 20.2 and /(20.2) "' 80.9. Since /(0) "' 83.3 and /(200) "' 137.4, we conclude that the minimal
time is 80.9 s. This occurs when Brandon swims diagonally to a point located 20.2 m downstream and then runs the rest
of the way.
57. Vascular Branching A small blood vessel of radius r branches off at an angle O from a larger vessel of radius R to
supply blood along a path from A to B. According to Poiseuille's Law, the total resistance to blood flow is proportional to
T
=(a - R4bcotO + bcscO)
r4
where a and b are as in Figure 29. Show that the total resistance is minimized when cos B = (r I R)4 .
290
CH APT E R 4
I
APPLICATIONS OF THE DERIVATIVE
a
FIGURE 29
SOLUTION
With a, b, r,R > 0 and R > r, let T(8) = (
a - bcot8 bcsc8)
R4
+ ~ . Set
2
, -(bcsc 8 _ bcsc8cot8)- 0
T (8) R4
r4
Then
b(r4 - R4 cos 8)
--'----.,--=0
2
4
R r4 sin 8
4
so that cos 8 = ( !._ ) • Since Jim T (8)
R
o~o+
= oo and o~~Jim T (8) = oo, the minimum value of T (8) occurs when cos 8 = ( !._ )
R
4
•
In Exercises 58-59, a box (with no top) is to be constructed from a piece of cardboard with sides of length A and B by
cutting out squares of length hfrom the corners and folding up the sides (Figure 30).
'
'
_J
L-
B
r-
h T -1
'
~------~·~
A
FIGURE 30
59. Which values of A and B maximize the volume of the box if h
SOLUTION
With h = 10 and AB= 900 (which means that B
V(A) =
= 10 cm and AB = 900 cm2 ?
= 900/A), the volume of the box is
IO(A-20)( 9~0 -20) = 13000-200A- 18 ~00
where 20 ::; A ::; 45. Now, solving
V'(A)
yields A= 30. Because V(20)
= -200 +
180000
A2
=0
= V(45) = 0 and V(30) = 1000 cm3 , maximum volume is achieved with A= B = 30 cm.
61. Given n numbers x 1 , .•. , Xn, find the value of x minimizing the sum of the squares:
(x - xi)2 + (x - x2) 2 + · · · + (x First, solve for n
SOLUTION
Xn)
2
= 2, 3 and then try it for arbitrary n.
Note that the sum of squares approaches oo as x
~
±oo, so the minimum must occur at a critical point.
• For n = 2: Let f(x) = (x- xi)2 + (x - x2 )2. Then setting f'(x) = 2(x - x 1) + 2(x - x 2 ) = 0 yields x = ~(x 1 + x 2).
• For n = 3: Let f(x) = (x - x1)2 + (x - Xz) 2 + (x - X3) 2 , so that setting f'(x) = 2(x- x 1 ) + 2(x - Xz) + 2(x - x 3 ) = 0
yields x = 1Cx1 + Xz + x3).
• Let f(x) = LZ=i (x - xk) 2 • Solve f'(x) = 2 I:~=I (x - xk) = 0 to obtain x = x = ~ I:~=I Xk·
Note that the optimum value for xis the average of xi. ... , Xn.
63. Solve Exercise 62 again using geometry rather than calculus. There is a unique circle passing through points B and
C that is tangent to the street. Let R be the point of tangency. Note that the two angles labeled lfJ in Figure 31 are equal
because they subtend equal arcs on the circle.
(a) Show that the maximum value of 8 is 8 = l/J. Hint: Show that
if;= 8 + LPBA, where A is the intersection of the circle with PC.
(b) Prove that this agrees with the answer to Exercise 62.
(c) Show that LQRB = LRCQ for the maximal angle If!.
S E CT I O N 4. 7
I
Applied Optimization
291
SOLUTION
(a) We note that LPAB is supplementary to both if! and(}+ LPBA; hence, if!=(}+ LPBA. From here, it is clear that(} is
at a maximum when LPBA = O; that is, when A coincides with P. This occurs when P =R.
(b) To show that the two answers agree, let 0 be the center of the circle. One observes that if d is the distance from R to
the wall, then 0 has coordinates (-d, § + h). This is because the height of the center is equidistant from points Band C
and because the center must lie directly above R if the circle is tangent to the floor.
Now we can solve ford. The radius of the circle is clearly § + h, by the distance formula:
This gives
or d = ../bh + h2 as claimed.
(c) Observe that the arc RB on the dashed circle is subtended by LQRB and also by LRCQ. Thus, both are equal to
one-half the angular measure of the arc.
65. Victor Klee's Endpoint Maximum Problem Given 40 m of straight fence, your goal is to build a rectangular
enclosure using 80 additional meters of fence that encompasses the greatest area. Let A(x) be the area of the enclosure,
with x as in Figure 32.
(a) Find the maximum value of A(x).
(b) Which interval of x values is relevant to our problem? Find the maximum value of A(x) on this interval.
40
x
:
I
j
20-xl
I
I!,,
120-x
____ _
40+x
FIGURE 32
SOLUTION
(a) From the diagram, A(x)
= (40 + x)(20 -
x)
= 800 -
20x -
x2 = 900 -
(x + 10)2 • Thus, the maximum value of A(x)
is 900 square meters, occurring when x = -10.
(b) For our problem, x E [O, 20]. On this interval, A(x) has no critical points and A(O) = 800, while A(20) = 0. Thus, on
the relevant interval, the maximum enclosed area is 800 square meters.
67. The force F (in Newtons) required to move a box of mass m kg in motion by pulling on an attached rope (Figure 33) is
F((}) =
fmg
cos9+fsin9
where(} is the angle between the rope and the horizontal, f is the coefficient of static friction, and g = 9.8 m/s 2 . Find the
angle (}that minimizes the required force F, assuming f = 0.4. Hint: Find the maximum value of cos(} + f sin(}.
F
FIGURE 33
SOLUTION
Let F(9)
= sm(}
. +\
5 cos(}
, where k > 0 is a proportionality constant and O ::; (} ::; ~.Solve
F'(9)=
for O ::; (} ::; ~ to obtain tan(}
kn sine- cose)
(sin(}+~ cos(})
2
=0
= ~. From the diagram below, we note that when tan (} = ~,
. (}
Sill
=
5
-=
v29
and
cos(}=
2
V29
292
CH APT E R 4
I
APPLICATIONS OF THE DERIVATIVE
Therefore, at the critical point the force is
k
~+~~
Since F(O) = ~ k = 2.5k and F ( ~)
= 5 V29 k
0 93k
';:j
29
.
= lk, we conclude that the minimum force occurs when tan 0 = ~.
5 .
2
69. Bird Migration Ornithologists have found that the power (in joules per second) consumed by a certain pigeon
flying at velocity v mis is described well by the function P(v) = l 7v- 1 + 10-3v3 joules/s. Assume that the pigeon can
store 5 x 104 joules of usable energy as body fat.
(a) Show that at velocity v, a pigeon can fly a total distance of D(v) = (5 x 104 )v/P(v) ifit uses all of its stored energy.
(b) Find the velocity vP that minimizes P.
(c) Migrating birds are smart enough to fly at the velocity that maximizes distance traveled rather than minimizes power
consumption. Show that the velocity vd which maximizes D(v) satisfies P'(vd) = P(vd)/vd. Show that vd is obtained
graphically as the velocity coordinate of the point where a line through the origin is tangent to the graph of P (Figure
34).
(d) Find vd and the maximum distance D(vd).
Power (joules/s)
Minimum power /
consumptiy
4
I
5
r'-..._ Maximum
J
distance
I
traveled
15
10
Velocity (mis)
FIGURE 34
SOLUTION
4
(a) Flying at a velocity v, the birds will exhaust their energy store after T
.
.
distance traveled 1s then D(v)
5 · 10 joules
= --~--­
P(v) joules/sec
5 · 104 sec
P( v)
. The total
5 · 104 v
= vT = ---.
P(v)
(b) Let P(v) = 11v- 1 + 10-3 v3 • Then P'(v) = -l1v-2 + 0.003v2 = 0 implies vP = (o.~S ';:j 8.676247. This critical
point is a minimum, because it is the only critical point and P(v) ~ oo both as v ~ O+ and as v ~ oo.
P(v) · 5 · 104 - 5 · 104 v · P'(v)
P(v) - vP'(v)
P(v)
(c) D'(v) =
(P(v)) 2
= 5 · 104 (P(v)) 2 = 0 implies P(v) - vP'(v) = 0, or P'(v) = v
14
Since D(v) ~ 0 as v ~ 0 and as v
Graphically, the expression
~
oo, the critical point determined by P'(v)
P(v)
v
= P(v)/v corresponds to a maximum.
P(v)-0
v-0
is the slope of the line passing through the origin and (v, P(v)). The condition P'(v) = P(v)/v which defines vd therefore
indicates that vd is the velocity component of the point where a line through the origin is tangent to the graph of P(v).
.
( )
P(v) .
(d) Usmg P' v = -- gives
v
-17v-2 + 0.0Q3v2
which simplifies to 0.002v4
191.741 kilometers.
= 34 and thus vd
';:j
=
l 7v-l + O.OQl V
v
3
= 17v-2 + 0.QQl v2
11.418583. The maximum total distance is given by D(vd)
=5·
104
· vd
P(vd)
I
S EC TIO N 4.7
Applied Optimization
293
71. Redo Exercise 70 for arbitrary b and h.
SOLUTION
Consider the right triangle formed by the right half of the rectangle and its "roof." This triangle has hy-
potenuses.
x
b/2
As shown, let y be the height of the roof, and let x be the distance from the right base of the rectangle to the base of the
roof. By similar triangles applied to the smaller right triangles at the top and right of the larger triangle, we get:
y-h
b/2
h
or
x
bh
y=-+h
2x
s,y, and x are related by the Pythagorean Theorem:
2
2
2
b
) +l= (b
s2= ( :z+x
:z+x) + (bh
Zx +h)
Since s > 0, s 2 is least whenever sis least, so we can maximize s 2 instead of s. Setting the derivative equal to zero yields
2(~+x)+2(~: +h)(-:~)=o
2
(~ + x) + 2~ (~ + x) (-::
=0
2)
2(~+x)(1-~~:)=o
The zeros are x = -~ (irrelevant) and
bl/3h2/3
x=~
Since this is the only critical point of s with x > 0, and since s
where s attains its minimum. For this value of x,
-t
oo as
x
-t
0 and s
-t
oo as
x
-t
oo, this is the point
2 - (~ bl/3h2/3 )2 (b2/3hl/3
)2
s - 2 + 21/3
+
22/3 + h
2
2
- b2f3 ( b2/3
2/3 )
2/3 ( b2f3
2/3 ) - ( b2f3
2/3 )3
- 22/3 2213 + h
+h
22/3 + h
- 22/3 + h
so the smallest roof length is
b2/3
)3/2
h2/3
s - ( 22/3 +
73. Redo Exercise 72 for arbitrary widths a and b.
If the corridors have widths a and b, and if Bis the angle between the beam and the line perpendicular to
the corridor of width a, then we have to minimize
SOLUTION
a
b
f(B)= - + cosB sinB
Setting the derivative equal to zero,
a sec Btan B- b·cotBcsc B = 0
we obtain the critical value B0 defined by
b)l/3
tan Bo= ( ;
and from this we conclude (witness the diagram below) that
cos Bo =
1
--;::=====
,./1 + (b/a)2f3
and
(b/a)If3
sin Bo =
--;::=====
,./1 + (b/a) 1
2 3
294
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
This gives the minimum value as
f(00) =a
.../I + (b/a) 213 + b(b/a)- 113 .../1 + (b/a)2f3
= a2f3 Va2/3 + b2/3 + b2/3 Va2/3 + b2/3
= (a2/3 + b2/3)3/2
75. ~ A basketball player stands d feet from the basket. Let h and a be as in Figure 38. Using physics, one can
show that if the player releases the ball at an angle e, then the initial velocity required to make the ball go through the
basket satisfies
v
2
16d
= cos2 (}(tan(}
-
tan a)
(a) Explain why this formula is meaningful only for a < (} < ~-Why does v approach infinity at the endpoints of this
interval?
(b)
Take a = ~ and plot v2 as a function of (} for ~ < (} < ~. Verify that the minimum occurs at (} = ~.
(c) Set F(O) = cos 2 O(tan (} - tan a). Explain why vis minimized for(} such that F(O) is maximized.
(d) Verify that F' ((}) = cos(a - 28) sec a (you will need to use the addition formula for cosine) and show that the
maximum value of F on [a, ~J occurs at 00 = ~ + ~.
(e) For a given a, the optimal angle for shooting the b~sket is (}0 because it minimizes v2 and therefore minimizes the
energy required to make the shot (energy is proportional to v2 ). Show that the velocity Vopt at the optimal angle 00 satisfies
IGq
32dcosa
32d2
2
vopt= 1-sina = -h+ vd2+h2
IGU[I
(f)
Show with a graph that for fixed d (say, d = 15 ft, the distance of a free throw), v~pt is an increasing function
of h. Use this to explain why taller players have an advantage and why it can help to jump while shooting.
:n,'I
1,r:
i!
I
I
!j
h
u
FIGURE 38
SOLUTION
(a) a= 0 corresponds to shooting the ball directly at the basket while a= ;r/2 corresponds to shooting the ball directly
upward. In neither case is it possible for the ball to go into the basket.
If the angle a is extremely close to 0, the ball is shot almost directly at the basket, so that it must be launched with
great speed, as it can only fall an extremely short distance on the way to the basket.
On the other hand, if the angle a is extremely close to ;r/2, the ball is launched almost vertically. This requires the
ball to travel a great distance upward in order to travel the horizontal distance. In either one of these cases, the ball has to
travel at an enormous speed.
(b)
7!
6
The minimum clearly occurs where e = ;r/3.
u
7r
7!51(7!
4
3
12
2
SECTION 4.7
I
Applied Optimization
295
(c) If F(B) = cos2 (}(tan(} - tan a:),
2
16d
v =------cos2 (}(tan(} - tan a:)
16d
F(B)
Since a: :5 (}, F(B) > 0, hence v2 is smallest whenever F(B) is greatest.
(d) F'(B) = -2 sin Bcos (}(tan(} - tan a:) + cos2 (} (sec 2 e) = -2 sin Bcos Btan (} + 2 sin (}cos (}tan a:+ 1. We will apply all
the double angle formulas:
cos(28) = cos 2 (} - sin2 (} = 1 - 2 sin2 8; sin 28 = 2 sin(} cos(}
getting:
F'(B) = 2sin8cos8tana:-2sin8cos8tan(} + 1
.
sin a:
.
sin (}
= 2sm8cos8-- -2sm8cos8-- + 1
cos a:
cos(}
= sec a: (-2 sin2 (}cos a:+ 2 sin Bcos (}sin a:+ cos a:)
= sec a: (cos a: ( 1 - 2 sin2 e) +sin a: (2 sin Bcos 8))
= sec a: (cos a:(cos 28) + sin a:( sin 28))
= sec a: cos( a: - 28)
A critical point of F(B) occurs where cos(a: - 28) = 0, so that a: - 28 = -~ (negative because 28 >(}>a:), and this gives
us(}= a:/2 + 7r/4. The minimum value F(Bo) takes place at 80 = a:/2 + 7r/4.
(e) Plug in 80 = a:/2 + 7r/4. To find v~pt we must simplify
cos
2n
u0
(
D
tan uo
-
)
cos Bo(sin Bo cos a: - cos Bo sin a:)
tan a: = - - - - - - - - - - - cos a:
By the addition law for sine:
sin Bo cos a: - cos Bo sin a: = sin(80
-
a:) = sin( -a: /2 + 7r / 4)
and so
cos8o(sin80 cos a: - cos80 sin a:)= cos(a:/2 + 7r/4) sin(-a:/2 + 7r/4)
Now use the identity (that follows from the addition law):
.
1 .
)
.
smxcosy= z-Csm(x+y +sm(x-y))
to get
cos(a:/2+7r/4)sin(-a:/2+7r/4) = (1/2)(1-sina:)
So we finally get
(1 /2)(1 - sin a:)
cos 2 Bo(tanBo - tan a : ) = - - - - - cos a:
and therefore
32d cos a:
1 - sin a:
V~pt
as claimed. From Figure 38 we see that
d
cosa:= - - - -
vd2
+ h2
h
and
sina:= - - +h2
vd2
Substituting these values into the expression for v~P' yields
32J2
v2 - - - - - opt - -h +
+ h2
vd2
(t) A sketch of the graph of ~pt versus h ford = 15 feet is given below: v~P' increases with respect to basket height
relative to the shooter. This shows that the minimum velocity required to launch the ball to the basket drops as shooter
height increases. This shows one of the ways height is an advantage in free throws; a taller shooter need not shoot the
ball as hard to reach the basket.
296
CH APT E R 4
I
APPLICATIONS OF THE DERIVATIVE
600
500
400
300
200
100
0
2
4
5
Further Insights and Challenges
77. Tom and Ali drive along a highway represented by the graph off in Figure 40. During the trip, Ali views a billboard
represented by the segment BC along the y-axis. Let Q be they-intercept of the tangent line toy = f(x). Show that e is
maximized at the value of x for which the angles LQPB and LQCP are equal. This generalizes Exercise 63 (c) [which
corresponds to the case f(x) = 0]. Hints:
(a) Show that dB/dx is equal to
cx2 + (xf'(x)) 2) - (b- (f(x) - xf'(x)))(c- (f(x) - xf'(x)))
(b - c) . ---'--'--,,------'----,---,,....-----'-..,.----'----(x2 + (b - f(x))2)(x2 + (c - f(x))2)
(b) Show that they-coordinate of Q is f(x) - xf'(x).
(c) Show that the condition dB/dx = 0 is equivalent to
PQ2 = BQ·CQ
(d) Conclude that 6QPB and 6QCP are similar triangles.
y
C=(O,c)
Billboard
=f(x)
Q
FIGURE 40
SOLUTION
(a) From the figure, we see that
B(x) = tan-1 c - f(x) - tan-1 b - f(x)
x
x
Then
B'(x)
=bxz
(f(x)- xf'(x)) _ c - (f(x)- xf'(x))
+ (b - f(x))Z
x2 + (c - f(x))Z
= (b _ c) x2 -
be+ (b + c)(f(x) - xf'(x)) - (f(x))2 + 2xf(x)f'(x)
(x2 + (b - f(x))2)(x 2 + (c - f(x))2)
= (b _ c) (x2 + (xf'(x)) 2 - (be - (b
+ c)(f(x) - xf'(x)) + (f(x) - xf'(x)) 2 )
2
(x + (b - f(x)) 2)(x2 + (c - f(x)) 2)
= (b _ c) (x2 + (xf'(x))
2
(x2
(b - (f(x) - xf'(x)))(c - (f(x) - xf'(x)))
+ (b - f(x)) 2)(x2 + (c - f(x))2)
-
(b) The point Q is they-intercept of the line tangent to the graph of f(x) at point P. The equation of this tangent line is
Y - f(x) = f'(x)(X - x)
They-coordinate of Q is then f(x) - xf'(x).
SECTION 4.7
I
Applied Optimization
297
(c) From the figure, we see that
= b- (f(x)- xf'(x)),
CQ = c - (j(x) - xf'(x))
BQ
and
PQ
= ../x2 + (f(x) -
= ../x2 + (xf'(x))2
(f(x) - xf'(x)))2
Comparing these expressions with the numerator of d()I dx, it follows that
~~
= 0 is equivalent to
PQ2 =BQ·CQ
(d) The equation PQ2 = BQ · CQ is equivalent to
PQ
BQ
CQ
PQ
In other words, the sides CQ and PQ from the triangle ll.QCP are proportional in length to the sides PQ and BQ from
the triangle ll.QPB. As LPQB = LCQP, it follows that triangles ll.QCP and ll.QPB are similar.
Seismic Prospecting Exercises 78-80 are concerned with determining the thickness d of a layer of soil that lies on top
of a rock formation. Geologists send two sound pulses from point A to point D separated by a distance s. The first pulse
travels directly from A to D along the suiface of the earth. The second pulse travels down to the rock formation, then
along its suiface, and then back up to D (path ABCD ), as in Figure 41. The pulse travels with velocity v 1 in the soil and
v2 in the rock.
A
s
D
FIGURE 41
79. In this exercise, assume that v2 /v 1
;:.::
../1 + 4(d/ s)2.
(a) Show that inequality (1) holds if sin()= vifv2 •
(b) Show that the minimal time for the second pulse is
2d
tz = -(1 - k2)1/2
VJ
s
+-
V2
where k = V1 /v2.
t1
(c) Conclude that - =
t1
2d(l - k2) 1' 2
+ k.
s
SOLUTION
(a) If sin() = ~, then
tan()=
Because
?i- ;:.: ~ 1 + 4( ~ )2, it follows that
Hence, tan ()
~
-fJ as required.
vi
~v~ -vi
~U~)
2
-1
298
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
(b) For the time-minimizing choice of(}, we have sin(} =
~ from which sec(} =
v2
vi
~v~-vi
and tan(} =
~. Thus
I
s-2d-v_,_
2d
s - 2dtan(}
t 2 =-sec(}+---V1
V2
2d
V1
V1
yv~-..:f
---+----~
V2
s
(c) Recall that t 1 = -. We therefore have
VJ
2d(!-k2)''2 +
.L
v2
vi
.L
v1
81. ~ In this exercise, we use Figure 42 to prove Heron's principle of Example 7 without calculus. By definition,
C is the reflection of B across the line MN (so that BC is perpendicular to MN and BN = CN). Let P be the intersection
of AC and MN. Use geometry to justify the following:
(a) t:,PNB and t:,PNC are congruent and er =
e1.
(b) The paths APB and APC have equal length.
(c) Similarly, AQB and AQC have equal length.
(d) The path APC is shorter than AQC for all Q
*
P.
Conclude that the shortest path AQB occurs for Q = P.
A
M
FIGURE 42
SOLUTION
(a) By definition, BC is orthogonal to QM, so triangles t:,PNB and t:,PNC are congruent by side-angle-side. Therefore
(}I
= (}i
(b) Because t:,PNB and t:,PNC are congruent, it follows that PB and PC are of equal length. Thus, paths APB and APC
have equal length.
(c) The same reasoning used in parts (a) and (b) lead us to conclude that t:,QNB and t:,QNC are congruent and that PB
and PC are of equal length. Thus, paths AQB and AQC are of equal length.
(d) Consider triangle t:,AQC. By the triangle inequality, the length of side AC is less than or equal to the sum of the
lengths of the sides AQ and QC. Thus, the path APC is shorter than AQC for all Q P.
*
Finally, the shortest path AQB occurs for Q = P.
SECT Io N 4.8
I
Newton's Method
299
4.8 Newton's Method
Preliminary Questions
1. How many iterations of Newton's Method are required to compute a root if f is a linear function?
SOLUTION Remember that Newton's Method uses the linear approximation of a function to estimate the location of a
root. If the original function is linear, then only one iteration of Newton's Method will be required to compute the root.
2. What happens in Newton's Method if your initial guess happens to be a zero off?
SOLUTION
If x0 happens to be a zero off, then
X1
= Xo -
f(xo)
f'(xo)
--
= Xo - 0 = Xo
in other words, every term in the Newton's Method sequence will remain Xo.
3. What happens in Newton's Method if your initial guess happens to be a local min or max off?
SOLUTION Assuming that the function is differentiable, then the derivative is zero at a local maximum or a local
minimum. If Newton's Method is started with an initial guess such that f' (x0 ) = 0, then Newton's Method will fail in the
sense that x 1 will not be defined. That is, the tangent line will be parallel to the x-axis and will never intersect it.
4. Is the following a reasonable description of Newton's Method: "A root of the equation of the tangent line to the graph
off is used as an approximation to a root off itself''? Explain.
SOLUTION Yes, that is a reasonable description. The iteration formula for Newton's Method was derived by solving
the equation of the tangent line toy = f (x) at x0 for its x-intercept.
Exercises
In this exercise set, all approximations should be carried out using Newton's Method.
In Exercises 1-6, apply Newton's Method to f and initial guess x0 to calculate XJ. x2 , x 3 .
1. f (x)
= x2 -
SOLUTION
6,
xo
=2
Let f(x) =
x2 - 6 and define
With x 0 = 2, we compute
3.
f (x) = x3 - 10,
SOLUTION
=2
= x3 -
n
1
2
3
Xn
2.5
2.45
2.44948980
Xo
Let f(x)
10 and define
With x 0 = 2 we compute
5.
f (x)
= cos x - 4x,
SOLUTION
n
1
2
3
Xn
2.16666667
2.15450362
2.15443469
xo = 1
Let f(x) = cos x - 4x and define
Xn+I
With x 0
=Xn -
f (Xn)
f'( Xn )
=Xn -
COS Xn - 4xn
.
-SlilXn -
4
= 1 we compute
n
1
2
3
Xn
0.28540361
0.24288009
0.24267469
300
CH APT E R 4
I
APPLICATIONS OF THE DERIVATIVE
7. Use Figure 6 to choose an initial guess x 0 to the unique real root of x3 + 2x + 5 = 0 and compute the first three
Newton iterates.
y
FIGURE 6 Graph ofy
SOLUTION
Let f(x) =
= x3 + 2x + 5.
x3 + 2x + 5 and define
f (xn)
~ + 2xn + 5
Xn+I = Xn - f'(Xn) = Xn 3 X~ + 2
We take xo = -1.4, based on the figure, and then calculate
n
1
2
3
Xn
-1.330964467
-1.328272820
-1.328268856
9. Approximate both solutions of ex
= 5x to three decimal places (Figure 7).
y
20
10
2
3
FIGURE 7 Graphs ofy =ex andy = 5x.
SOLUTION
We need to solve ex - 5x
= 0, so let f(x) =ex -
5x. Then f'(x) =ex - 5. With an initial guess of x0
we calculate
Newton's Method (First root)
x = 0.2 - f (0.2)
xo = .2 (guess)
"='
.25859
f (.25859)
f'(.25859)
Xz "='
.25917
x3 = .25917 - f(.25917)
X3 "='
.25917
I
X1
f'(0.2)
x 2 = .25859 -
f'(.25917)
For the second root, we use an initial guess of x 0 = 2.5.
Newton's Method (Second root)
x = 2.5 - f (2.5)
I
f'(2.5)
xo
= 2.5 (guess)
X1 "='
2.54421
JC2-54421 )
!'(2.54421)
X2 "='
2.54264
x 3 = 2.54264 - f(2.54264)
!'(2.54264)
X3 "='
2.54264
x 2 = 2.54421 -
Thus the two solutions of ex = 5x are approximately r 1
""
.25917 and r2
""
2.54264.
= 0.2,
cT I0 N
SE
4.8
I
Newton's Method
301
In Exercises I I-14, approximate to three decimal places using Newton's Method and compare with the value from a
calculator.
11.
vTI
SOLUTION
Let f(x)
= x2 -
11, and let x 0 = 3. Newton's Method yields:
n
1
2
3
Xn
3.33333333
3.31666667
3.31662479
A calculator yields 3.31662479.
13. 2113
SOLUTION
Note that 2713 = 4 · 2 113 • Let f(x) =
x3 - 2, and let Xo = 1. Newton's Method yields:
n
1
2
3
Xn
1.33333333
1.26388889
1.25993349
Thus, 2713 ""4 · 1.25993349 = 5.03973397. A calculator yields 5.0396842.
15. Approximate the largest positive root of f(x) = x4
SOLUTION
4
f(x) = x
-
-
6x2 + x + 5 to within an error of at most 10-4 • Refer to Figure 5.
Figure 5 from the text suggests the largest positive root of f(x) = x4
6x2 + x + 5 and take x0 = 2.
-
6x2 + x + 5 is near 2. So let
n
1
2
3
4
Xn
2.111111111
2.093568458
2.093064768
2.093064358
The largest positive root of x4 - 6x2 + x + 5 is approximately 2.093064358.
!GUij
In Exercises 16-21, approximate the value specified to three decimal places using Newton's Method. Use a plot
to choose an initial guess.
17. Negative root of f(x) =
x5 - 20x + 10
Let f(x) = x5 - 20x + 10. The graph of f(x) shown below suggests taking x0
x0 = -2.2, the first three iterates of Newton's Method are:
SOLUTION
n
1
2
3
Xn
-2.22536529
-2.22468998
-2.22468949
Thus, to three decimal places, the negative root of f(x) =
x5 -
-2.2. Starting from
20x + 10 is -2.225.
y
19. Positive solution of 2 tan- 1 x
=x
From the graph below, we see that the positive solution to the equation 2 tan- 1 x = xis approximately
x = 2.3. Now, let f(x) = 2 tan- 1 x - x and define
SOLUTION
Xn+I
=
Xn -
f(Xn)
f'(Xn)
=
Xn -
2 tan-I Xn _z__ l
Xn
I+~
With x 0
= 2.3, we find
n
1
2
3
Xn
2.33128574
2.33112237
2.33112237
Thus, to three decimal places, the positive solution to the equation 2 tan- 1 x =xis 2.331.
302
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
y=x
4
21. Solution ofln(x + 4) = x
SOLUTION From the graph below, we see that the positive solution to the equation ln(x + 4)
x = 2. Now, let f(x) = ln(x + 4) - x and define
=x
is approximately
f(xn)
ln(Xn + 4) - Xn
Xn+I = Xn - f'(Xn) = Xn _I_ - 1
Xn+4
With x0 = 2 we find
n
1
2
3
Xn
1.750111363
1.749031407
1.749031386
Thus, to three decimal places, the positive solution to the equation ln(x + 4) =xis 1.749.
2.5
2.0
1.5
1.0
0.5
0.5
23. !GU~
SOLUTION
1.0
1.5
Find the smallest positive value of x at which y
2.0
2.5
= x and y = tan x intersect. Hint: Draw a plot.
Here is a plot of tan x and x on the same axes:
y
The first intersection with x > 0 lies on the second "branch" of y = tan x, between x =
The graph suggests an initial guess x0 =
from which we get the following table:
¥-,
n
1
2
3
4
Xn
6.85398
21.921
4480.8
7456.27
¥- and x = ~. Let f (x) = tan x - x.
This is clearly leading nowhere, so we need to try a better initial guess. Note: This happens with Newton's Method-it is
and ~, x0 = 1~" :
sometimes difficult to choose an initial guess. We try the point directly between
¥-
n
1
2
3
4
5
6
7
Xn
4.64662
4.60091
4.54662
4.50658
4.49422
4.49341
4.49341
The first point where y
= x and y = tan x cross is at approximately x = 4.49341, which is approximately 1.4303n.
25. Find (to two decimal places) the coordinates of the point Pin Figure 9 where the tangent line toy = cosx passes
through the origin.
y
y=
FIGURE 9
COSX
SECTION 4.8
I
Newton's Method
303
Let (xn cos(xr )) be the coordinates of the point P. The slope of the tangent line is - sin(xr ), so we are
looking for a tangent line:
SOLUTION
such that y
= 0 when x = 0. This gives us the equation:
- sin(Xr )(-Xr) + cos(Xr)
=0
Let f(x) = cos x + x sin x. We are looking for the first point x = r where f(r) = 0. The sketch given indicates that
x 0 = 3n /4 would be a good initial guess. The following table gives successive Newton's Method approximations:
n
1
2
3
4
Xn
2.931781309
2.803636974
2.798395826
2.798386046
The point P has approximate coordinates (2.7984, -0.941684).
Newton's Method is often used to determine interest rates in financial calculations. In Exercises 26-28, r denotes a yearly
interest rate expressed as a decimal (rather than as a percent).
27. If you borrow L dollars for N years at a yearly interest rate r, your monthly payment of P dollars is calculated using
the equation
L=P (
(a) Find P if L = $5000, N = 3, and r
1-b-12N)
,
b-1
where b
= 1 + 12r
= 0.08 (8%).
(b) You are offered a loan of L = $5000 to be paid back over 3 years with monthly payments of P = $200. Use Newton's
Method to compute b and find the implied interest rate r of this loan. Hint: Show that
(L/ P)b 12N+I - (1 + L/ P)b 12N + 1 = 0
SOLUTION
(a) b = (1 + 0.08/12) = 1.00667
p
b- 1 )
=L ( 1-
b-12N
= 5000
( 1.00667 - 1 )
1 - 1.00667-36 "" $156.69
(b) Starting from
L=P (
1 _ b-12N)
b-1
divide by P, multiply by b - 1, multiply by b12N and collect like terms to arrive at
(L/ P)b 12N+I - (1 + L/ P)b 12N + 1 = 0
Since L/ P = 5000/200 = 25, we must solve
25b37 - 26b36 + 1 = 0
Newton's Method gives b"" 1.02121 and
r
= 12(b -
1)
= 12(0.02121) "" 0.25452
So the interest rate is around 25.45%.
29. There is no simple formula for the position at time t of a planet Pin its orbit (an ellipse) around the sun. Introduce
the auxiliary circle and angle (} in Figure 10 (note that P determines (} because it is the central angle of point B on the
circle). Let a= OA and e =OS /OA (the eccentricity of the orbit).
(a) Show that sector BSA has area (a2 /2)((} - e sine).
(b) By Kepler's Second Law, the area of sector BSA is proportional to the time t elapsed since the planet passed point
A, and because the circle has area na2 , BSA has area (na 2 )(t/T), where Tis the period of the orbit. Deduce Kepler's
Equation:
2m
.
=6-esm 6
T
304
CH APT E R 4
I
APPLICATIONS OF THE DERIVATIVE
(c) The eccentricity of Mercury's orbit is approximately e = 0.2. Use Newton's Method to find (} after a quarter of
Mercury's year has elapsed (t = T /4). Convert(} to degrees. Has Mercury covered more than a quarter of its orbit at
t = T/4?
Auxiliary circle
___ B
'::----,,. ,,.
/
/
/
I
I
I
I
Sun
0
A
I
I
\
I
\
\
Elliptical orbit
' '...... ...... _
------
/
_....
//
I
/
FIGURE 10
SOLUTION
(a) The sector SAE is the slice OAB with the triangle OPS removed. OAB is a central sector with arc(} and radius
OA = a, and therefore has area 0~ 8 • 0 PS is a triangle with height a sin(} and base length OS = ea. Hence, the area of the
sector is
a2
1
a2
-{}- -ea2 sin(}=-((}- esinO)
2
2
2
(b) Since Kepler's second law indicates that the area of the sector is proportional to the time t since the planet passed
point A, we get
;ra2 (t/T) = a2 /2 (0 - e sin(})
t
2;rT
.
= (} - e sm (}
(c) If t = T /4, the last equation in (b) gives:
z
7r
.
= (} - e sm
(}
=
(}
.
- .2 sm (}
Let f(O) = (} - .2 sin(} - 1. We will use Newton's Method to find the point where f(O) = 0. Since a quarter of the year on
Mercury has passed, a good first estimate 00 would be 1·
n
1
2
3
4
Xn
1.7708
1.76696
1.76696
1.76696
From the point of view of the Sun, Mercury has traversed an angle of approximately 1.76696 radians = 101.24°. Mercury
has therefore traveled more than one-fourth of the way around (from the point of view of central angle) during this time.
31. Let f(x) = xe-x.
(a) Show that a sequence obtained by applying Newton's Method to f satisfies
Xn+l
=
x!"- •
1
(b) CR 5 Compute x 1, x 2 , ••• ,x10 separately with x 0 = 0.8 and x 0 = 5. Discuss what appears to be happening in each
case. (Note: the only root off is at x = 0.)
SOLUTION
(a) Let f(x) = xe-x. Then f'(x) = e-x - xe-x, and the sequence obtained by applying Newton's Method to f is
(b) Taking x 0 = 0.8, we have
n
1
2
3
4
5
6
Xn
-3.200000
-2.438095
-1.728954
-1.095395
-0.572632
-0.208509
n
7
8
9
IO
Xn
-0.035975
-0.001249
-1.5587 x 10-6
-2.4294 x 10- 12
SECTION 4.8
I
Newton's Method
305
This sequence appears to be converging toward the root at zero.
On the other hand, taking x0 = 5, we have
n
1
2
3
4
5
6
Xn
6.250000
7.440476
8.595744
9.727397
10.841979
11.943584
n
7
8
9
10
Xn
13.034962
14.118053
15.194284
16.264735
This sequence appears to be diverging.
33. What happens when you apply Newton's Method to find a zero of f(x)
SOLUTION
Letf(x)
= x 113? Note that x = 0 is the only zero.
= x 113 • Define
Take x 0 = 0.5. Then the sequence of iterates is -1, 2, -4, 8, -16, 32, -64, ... That is, for any nonzero starting value, the
sequence of iterates diverges spectacularly, since Xn = (-2)n xo. Thus lim Jxnl = lim 2n Jxol = oo.
n--i-oo
n~oo
Further Insights and Challenges
35. Newton's Method can be used to compute reciprocals without performing division. Let c > 0 and set f(x)
(a) Show that x - (f(x)/ f'(x))
= 2x -
c.
ci2.
(b) Calculate the first three iterates of Newton's Method with c
(c) Explain graphically why x 0
= x- 1 -
= 10.3 and the two initial guesses x0 = 0.1 and x 0 = 0.5.
= 0.5 does not yield a sequence converging to 1/10.3.
SOLUTION
(a) Let f (x)
= ~ - c. Then
f(x)
x - -j'(x)
(b) For c
= 10.324, we have f(x) = ~ -
• Take xo
10.324 and thus
~ - c
-2
-x-
Xn+I
= 2x -
= 2xn -
ci2
10.324~.
= 0.1.
n
1
0.09676
Xn
• Take x 0
=x -
2
0.096861574
3
0.096861681
= 0.5.
2
n
I
(c) The graph is disconnected. If x0
any point under Newton's Method.
Xn
I
-1.581
3
-8720.949819
-28.96746776
= .5, (x 1,f(x1)) is on the other portion of the graph, which will never converge to
In Exercises 36 and 37, consider a metal rod of length L fastened at both ends. If you cut the rod and weld on an
additional segment of length m, leaving the ends fixed, the rod will bow up into a circular arc of radius R (unknown), as
indicated in Figure I 2.
,_ _ _L_ _ _<
I
I
:/,/~I
/'
I
-F=---~
',
R' '
I
I
I
I
,/
\
!-,.()'-/
I
',1/
,
\
I
I
I
\
I
I
\
' '......
,__ -
....~
/
/
FIGURE 12 The bold circular arc has length L + m.
306
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
37. Let L = 3 and m = 1. Apply Newton's Method to Eq. (1) to estimate 8, and use this to estimate h.
SOLUTION
We let L = 3 and m = 1. We want the solution of:
sine
8
L
L+m
sine _ _ L_ =
0
8
L+m
sine_~ =
e
Letf(e)
= si~B -
4
0
~y
The figure above suggests that 80 = 1.5 would be a good initial guess. The Newton's Method approximations for the
solution follow:
The angle where si~e =
n
1
2
3
4
en
1.2854388
1.2757223
1.2756981
1.2756981
L:m is approximately 1.2757. Hence
h=Ll-cosB~lll181
2 sin8
·
In Exercises 39-41, a flexible chain of length Lis suspended between two poles of equal height separated by a distance
2M (Figure 13). By Newton's laws, the chain describes a catenary
y=
acosh(~)
where a is the number such that
L = 2asinh( ~)
The sag s is the vertical distance from the highest to the lowest point on the chain.
y
= a cosh(xla)
s
x
FIGURE 13 Chain hanging between two poles.
39. Suppose that L = 120 and M = 50.
(a) Use Newton's Method to find a value of a (to two decimal places) satisfying L = 2a sinh(M/a).
(b) Compute the sags.
SOLUTION
(a) Let
5
f(a) = 2a sinh ( aO )- 120
The graph off shown below suggests a~ 47 is a root off. Starting with a0 = 47, we find the following approximations
using Newton's method:
a 1 = 46.95408
Thus, to two decimal places, a= 46.95.
and
a2 = 46.95415
SECT Io N 4.8
I
Newton's Method
307
y
40
42
44
48
46
(b) The sag is given by
s =.y(M)-y(O) = (acosh
~ + c)-(acosh ~ + c) = acosh ~-a
Using M = 50 and a= 46.95, we finds= 29.24.
41.
(a)
(b)
/1L
(c)
Suppose that L = 160 and M = 50.
Use Newton's Method to find a value of a (to two decimal places) satisfying L = 2a sinh(M/a).
Use Eq. (2) and the Linear Approximation to estimate the increase in sag /1s for changes in length /1L
= 5.
CR S Compute s(l61) - s(l60) and s(l65) - s(l60) directly and compare with your estimates in (b).
= 1 and
SOLUTION
(a) Let f(x) = 2xsinh(50/x) - 160. Using the graph below, we select an initial guess of x 0
yields:
n
1
2
3
Xn
28.30622107
28.45653356
28.45797517
= 30. Newton's Method then
Thus, to two decimal places, a "" 28.46.
250
200
150
100
50
10
(b) With M
20
30
40
= 50 and a"" 28.46, we find using equation (2) that
ds
dL = 0.61
By the Linear Approximation,
ds
11s"" - · 11L
dL
If L increases from 160 to 161, then 11L
= 1 foot and /1s
"" 0.61; if L increases from 160 to 165, then /1L = 5 and
11s"" 3.05.
(c) When L = 160, a"" 28.46 and
s(160) = 28.46 cosh (
2:~6 )- 28.46 "" 56.45
whereas, when L = 161, a"" 28.25 and
s(l61) = 28.25 cosh(
2:_~5 )- 28.25 ""57.07
Therefore, s(161) - s(160) = 0.62, very close to the approximation obtained from the Linear Approximation. Moreover,
when L = 165, a"" 27.49 and
s(l65) = 27.49cosh(
thus, s(l65) - s(160)
2~-~9 )- 27.49 ""59.47
= 3.02, again very close to the approximation obtained from the Linear Approximation.
308
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
CHAPTER REVIEW EXERCISES
In Exercises J-6, estimate using the Linear Approximation or linearization, and use a calculator to estimate the error.
1. 8.1 113 - 2
SOLUTION
Let f(x)
= x 113 , a= 8 and tu= 0.1. Then f'(x) = ~x!J,.f
Using a calculator, 8.1 113
-
= 8.1 1' 3 -
2 ""f'(a)/J,.x
2 3
1 , f'(a)
=
nand, by the Linear Approximation,
1
= 12 (0.1) = 0.00833333
2 = 0.00829885. The error in the Linear Approximation is therefore
10.00829885 - 0.008333331 = 3.445 x 10-5
3. 625 1' 4
-
SOLUTION
624 114
Letf(x)
= x 1' 4 , a= 625 and /J,.x = -1. Thenf'(x) = ~x= 624 114 -
!J,.f
625 114 ""f'(a)/J,.x
314
, f'(a)
= 5fxi and, by the Linear Approximation,
1
= -500
-(-1) = -0.002
Thus 625 114 - 624 1' 4 ""0.002. Using a calculator,
625 1' 4
-
624 114
= 0.00200120
The error in the Linear Approximation is therefore
10.00200120 - (0.002)1 = 1.201
s.
x 10-6
1
1.02
SOLUTION
Let f(x)
= x- 1 and a=
1. Then f(a)
= 1, f'(x) = -x-2 and f'(a) = -1. The linearization of f(x) at a = 1
is therefore
L(x)
and l.~2 ""L(l.02)
= f(a) + f'(a)(x -
a)= 1 - (x - 1)
=2 -
x
= 0.98. Using a calculator, l.~2 = 0.980392, so the error in the Linear Approximation is
10.980392 - 0.981
= 3.922 x
10-4
In Exercises 7-12,find the linearization at the point indicated.
7. y
= ..fX,
SOLUTION
= 25
Let y =
a
vx and a = 25. Then y(a) = 5, y' = ~x- 1
1 2
and y'(a)
= fo· The linearization of y at a = 25 is
therefore
L(x) = y(a) + y'(a)(x - 25) = 5 +
9. A(r)
= ~nr3,
SOLUTION
=3
Let A(r) = ~nr
a
3
and a
a = 3 is therefore
L(r)
11. P(x) = e-Xl-12 ,
1~ (x - 25)
= 3. Then A(a) = 36n, A'(r) = 4nr2 and A'(a) = 36n. The linearization of A(r) at
= A(a) +A' (a)(r -
a)
= 36n + 36n(r -
3)
= 36n(r -
2)
a= 1
Let P(x) = e-X1-t2 and a= 1. Then P(a)
P(x) at a= 1 is therefore
SOLUTION
L(x)
= P(a) + P'(a)(x -
= e- 112, P'(x) = -xe-x212, and P'(a) = -e- 112 • The linearization of
a)
= e- 112 -
e- 112 (x - 1)
1
= -(2
-
-..je
x)
Chapter Review Exercises
309
Jn Exercises 13-16, use the Linear Approximation.
13. The position of an object in linear motion at time tis s(t) = 0.4t2 + (t + 1)- 1• Estimate the distance traveled over the
time interval [4, 4.2].
SOLUTION Let s(t) = 0.4t2 + (t + 1)- 1, a= 4and11t = 0.2. Then s'(t) = 0.8t - (t + 1)-2 and s'(a) = 3.16. Using the
Linear Approximation, the distance traveled over the time interval [4, 4.2] is approximately
11s = s(4.2) - s(4) "" s'(a)l1t = 3.16(0.2) = 0.632
15. When a bus pass from Albuquerque to Los Alamos is priced at p dollars, a bus company takes in a monthly revenue
of R(p) = 1.5p - 0.0lp2 (in thousands of dollars).
(a) Estimate /1R if the price rises from $50 to $53.
(b) If p = 80, how will revenue be affected by a small increase in price? Explain using the Linear Approximation.
SOLUTION
(a) If the price is raised from $50 to $53, then 11p = 3 and
l1R ""R'(50)11p = (1.5 - 0.02(50))(3) = 1.5
We therefore estimate an increase of $1500 in revenue.
(b) Because R'(80) = 1.5 - 0.02(80) = -0.1, the Linear Approximation gives /1R"" -0.l/1p. A small increase in price
would thus result in a decrease in revenue.
17. Use the Intermediate Value Theorem to prove that sinx - cosx = 3x has a solution, and use Rolle's Theorem to
show that this solution is unique.
SOLUTION Let f(x) = sin x - cos x - 3x, and observe that each root of this function corresponds to a solution of the
equation sin x - cos x = 3x. Now,
and
/(0) = -1<0
Because f is continuous on (- ~, 0) and f (- ~) and /(0) are of opposite sign, the Intermediate Value Theorem guarantees
there exists a c E
0) such that f(c) = 0. Thus, the equation sinx - cosx = 3x has at least one solution.
Next, suppose that the equation sin x - cos x = 3x has two solutions, and therefore f has two roots, say a and b.
Because f is continuous on [a, b], differentiable on (a, b) and /(a) = f(b) = 0, Rolle's Theorem guarantees there exists
c E (a, b) such that f'(c) = 0. However,
<-¥,
J'(x) = cosx+ sinx-3 :$ -1
for all x. We have reached a contradiction. Consequently, f has a unique root, and the equation sin x - cos x = 3x has a
unique solution.
19. Verify the MVT for f(x) =In x on [1, 4].
SOLUTION Let f(x) = lnx. On the interval [1,4], this function is continuous and differentiable, so the MVT applies.
Now, f'(x) = ~.so
!
c
=f'(c)= f(b)-f(a) = ln4-lnl = ! 1n 4
b-a
4-1
3
or
3
c= ln ""2.164E(l,4)
4
21. Use the MVT to prove that if f'(x)
:$
2 for x > 0 and /(0) = 4, then f(x) :$ 2x + 4 for all x
~
0.
SOLUTION Let x > 0. Because f is continuous on [O, x] and differentiable on (0, x), the Mean Value Theorem guarantees there exists a c E (0, x) such that
f'(c)
= f(x) -
f(O)
x-0
or
f(x)
= f(O) + xf'(c)
Now, we are given that f (0) = 4 and that f' (x) :$ 2 for x > 0. Therefore, for all x
f(x) :$ 4 + x(2) = 2x + 4
~
0,
31 O
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
In Exercises 23-28,find the critical points and determine whether they are minima, maxima, or neither.
23. f(x) =
x3 - 4x2 + 4x
SOLUTION Let f(x) = x3 - 4x2 + 4x. Then f' (x) = 3x2 - 8x + 4 = (3x - 2)(x - 2), so that x = %and x = 2 are critical
points. Next, f"(x) = 6x - 8, so f"C%) = -4 < 0 and f"(2) = 4 > 0. Therefore, by the Second Derivative Test, JC%) is a
local maximum while f (2) is a local minimum.
25. f(x) = x2(x + 2) 3
SOLUTION
Let f(x) = x2(x + 2) 3 • Then
j'(x) = 3.x2(x + 2)2 + 2x(x + 2) 3 = x(x + 2) 2 (3x + 2x + 4) = x(x + 2) 2 (5x + 4)
so that x = 0, x = -2 and x = -~ are critical points. The sign of the first derivative on the intervals surrounding the
critical points is indicated in the table below. Based on this information, f (-2) is neither a local maximum nor a local
minimum, f (- ~) is a local maximum and f (0) is a local minimum.
Interval
Sign off'
(-oo, -2)
(-2, -~)
(-~, 0)
(0, oo)
+
+
-
+
27. g(B) = sin2 e + ()
SOLUTION
Let g(O) = sin2 () + e. Then
g'(e) = 2sinecose + 1 = 2 sinW + 1
so the critical points are
3
for all integers n. Because g'(e) ;:::: 0 for all e, it follows that g ( ; +
mr) is neither a local maximum nor a local minimum
for all integers n.
In Exercises 29-36,find the extreme values on the interval.
29. f(x)
= x(lO -
[-1, 3]
x),
SOLUTION Let f(x) = x(lO - x) = lOx - x2. Then f'(x) = 10 - 2x, so that x = 5 is the only critical point. As this
critical point is not in the interval [-1, 3], we only need to check the value off at the endpoints to determine the extreme
values. Because f(-1) = -11 and f(3) = 21, the maximum value of f(x) = x(lO - x) on the interval [-1, 3] is 21 while
the minimum value is - 11.
31. g(e) = sin 2 e - cos e,
SOLUTION
Let g(B)
[0, 2n]
= sin2 () -
cos e. Then
g'(()) = 2 sin ecos e +sine= sin0(2 cos e + l) = 0
when e = 0, ~, JT, ~, 2n. The table below lists the value of g at each of the critical points and the endpoints of the interval
[0, 2n]. Based on this information, the minimum value of g on the interval [0, 2n] is -1 and the maximum value is ~.
e
g(B)
33. f(x) = x21 3 - 2x 113,
0
2n/3
:IT
4n/3
2JT
-1
5/4
1
5/4
-1
[-1, 3]
Let f(x) = x2 13 - 2x 113 . Then f'(x) = %x- 1l 3 - %x-213 = %x- 213(x 113 - 1), so that the critical points are
x = 0 and x = 1. With f(-1) = 3, f(O) = 0, f(l) = -1 and f(3) = -W - 2 'if3 "" -0.804, it follows that the minimum
value off on the interval [-1, 3] is -1 and the maximum value is 3.
SOLUTION
35. f(x) = x - 12 ln x,
[5, 40]
SOLUTION Let f(x) = x - 12lnx. Then f'(x) = 1 - lf, whence x = 12 is the only critical point. The minimum
value off is then 12 - 12ln 12 "" -17.818880, and the maximum value is 40 - 12ln40 "" -4.266553. Note that
f(5) = 5 -12ln5 ""-14.313255.
Chapter Review Exercises
311
37. Find the critical points and extreme values of f(x) = Ix - 11+12x - 61 in [0, 8].
SOLUTION
Let
7-3x,
f (x)
= Ix -
11 + 12x - 61
=
5 - x,
{
3x-7,
x<1
1 ::; x < 3
x <:: 3
The derivative off is never zero but does not exist at the transition points x = 1 and x = 3. Thus, the critical points off
are x = 1 and x = 3. With f(O) = 7, f(l) = 4, f(3) = 2, and f(8) = 17, it follows that the minimum value off on the
interval [O, 8] is 2 and the maximum value is 17.
In Exercises 39-44, find the points of inflection.
39. y
= x3 - 4x2 + 4x
Let y = x3 - 4x2 + 4x. Then y' = 3x2 - 8x + 4 and y" = 6x - 8. Thus, y" > 0 and y is concave up for
x > ~, while y" < 0 and y is concave down for x < ~. Hence, there is a point of inflection at x = ~.
SOLUTION
x2
41. y= ~4
x +
SOLUTION
Let y
x2
= x2 + 4 = 1 y
4
I
x 2 + 4 . Then y
,,
=
8x
= (x 2 + 4)2
and
(x2 + 4)2 (8) - 8x(2)(2x)(x2 + 4)
(x2 + 4)4
8(4 - 3x2)
(x2
+ 4)3
Thus, y" > 0 and y is concave up for
2
2
--<x<-
V3
.../3
while y" < 0 and y is concave down for
Hence, there are points of inflection at
2
x=±-
-\13
43. f(x) = (x2 - x)e-x
SOLUTION
Let j(x)
= (x2 -
x)e-x. Then
f'(x)
= -(x2 -
x)e-x + (2x - l)e-x
= -(x2 -
3x + l)e-x
and
J"(x)
= (x2 -
3x + l)e-x - (2x- 3)e-x
= e-x(x2 -
5x + 4)
= e-x(x -
l)(x - 4)
Thus, f"(x) > 0 and f is concave up for x < 1 and for x > 4, while f"(x) < 0 and f is concave down for 1 < x < 4.
Hence, there are points of inflection at x = 1 and x = 4.
In Exercises 45-54, sketch the graph, noting the transition points and asymptotic behavior.
45. y = 12x - 3x2
SOLUTION Let y = 12x - 3x2. Then y' = 12 - 6x and y" = -6. It follows that the graph of y = l2x - 3x2 is increasing
for x < 2, decreasing for x > 2, has a local maximum at x = 2 and is concave down for all x. Because
lim (12x - 3x2) = -oo
X-+±CO
the graph has no horizontal asymptotes. There are also no vertical asymptotes. The graph is shown below.
y
312
CHAPTER 4
I
APPLICATIONSOFTHEDERIVATIVE
47. y = x3
- 2x2 + 3
Let y = .x3 - 2x2 + 3. Then y' = 3x2 - 4x and y" = 6x - 4. It follows that the graph of y = .x3 - 2x2 + 3
is increasing for x < 0 and x > ~.is decreasing for 0 < x < ~.has a local maximum at x = 0, has a local minimum at
x = ~, is concave up for x > ~, is concave down for x < ~ and has a point of inflection at x = ~. Because
SOLUTION
lim
x-+-oo
c.x3 -
2Y!- + 3) = -oo
lim(x3
and
-
X->OO
2Y + 3) = oo
the graph has no horizontal asymptotes. There are also no vertical asymptotes. The graph is shown below.
y
49. y=
x
-3-1
x +
SOLUTION
x
Let y = -3--1. Then
x +
,
y =
.x3 + 1 - x(3x2)
(x 3
1 - 2.x3
+ 1)2
+ 1)2
(x 3
and
(.x3 + 1)2 (-6x2) - (1 - 2.x3)(2)(.x3 + 1)(3x2)
II
y =
(x3
+ 1)4
6x2(2 - .x3)
(x3
+ 1)3
=~
is increasing for x < -1 and -1 < x < 3 ~, is decreasing for x > 3 ~, has a local
x+l
'/2
'f'i
maximum at x = ;jf. is concave up for x < -1 and x > f/2, is concave down for -1 < x < 0 and O < x < {/2 and has
a point of inflection at x = i/2. Note that x = -1 is not an inflection point because x = -1 is not in the domain of the
It follows that the graph of y
function. Now,
lim _x_ =0
+1
X->±oo X3
so y = 0 is a horizontal asymptote. Moreover,
lim _x_ = oo
x->-1-
x3
+1
and
lim _x_ =-oo
+1
X->-1+ X3
so x = -1 is a vertical asymptote. The graph is shown below.
1
Sl. y = -lx_+_2_1_+_1
SOLUTION
Let y = Ix +
1
21
+ . Because
1
.
l Im
x->±oo
1
Ix + 21 + 1
=0
the graph of this function has a horizontal asymptote of y = 0. The graph has no vertical asymptotes as Ix + 21 + 1 ~ 1
for all x. The graph is shown below. From this graph we see there is a local maximum at x = -2.
-8 --{; -4 -2
2
4
Chapter Review Exercises
313
53. y= "3sinx-cosxon[0,2n]
SOLUTION Let y = Y3 sin x - cos x. Then y' = V3 cos x + sin x and y" = - V3 sin x + cos x. It follows that the graph
ofy = ...;3sinx-cosxis increasing forO < x < 5n/6 and lln/6 < x < 2n, is decreasing for 5n/6 < x < lln/6, has a
local maximum at x = 5n/6, has a local minimum at x = lln/6, is concave up for 0 < x < n/3 and 4n/3 < x < 2n, is
concave down for n/3 < x < 4n/3 and has points of inflection at x = n/3 and x = 4n/3. The graph is shown below.
y
55. Draw a curve y
= f (x) for which f'
and f" have signs as indicated in Figure 2.
I
-2
SOLUTION
I
+++
-+
•x
0 1
3
5
FIGURE 2
-+
The figure below depicts a curve for which f' and/" have the required signs.
y
57. A rectangular open-topped box of height h with a square base of side b has volume V
2
= 4 m3 . Two of the side faces
2
are made of material costing $40/m • The remaining sides cost $20/m • Which values of b and h minimize the cost of the
box?
SOLUTION
Because the volume of the box is
V
= b2 h = 4
it follows that h
=~
b2
Now, the cost of the box is
C
= 40(2bh) + 20(2bh) + 20b2 = I20bh + 20b2 = 4!0 + 20b2
Thus,
C'(b)
= - ~20 + 40b = 0
m meters. Because C(b) ~ oo as b ~ O+ and as b ~ oo, it follows that cost is minimized when b = {'12
= ~ m meters.
when b =
meters and h
59. Let N(t) be the size of a tumor (in units of 106 cells) at time t (in days). According to the Gompertz Model,
dN / dt = N(a - b In N), where a, b are positive constants. Show that the maximum value of N is eE and that the tumor
increases most rapidly when N = ei- 1•
SOLUTION Given dN I dt = N(a - b In N), the critical points of N occur when N = O and when N = eafb. The sign of
N'(t) changes from positive to negative at N = eafb so the maximum value of N is eafb. To determine when N changes
most rapidly, we calculate
N"(t)
= N(-~) +a -blnN =(a -
Thus, N'(t) is increasing for N < ~/h- 1 , is decreasing for N >
Therefore, the tumor increases most rapidly when N = ei- 1•
eafb-I
b)-blnN
and is therefore maximum when N =
eafb- 1.
314
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
61. Find the maximum volume of a right-circular cone placed upside-down in a right-circular cone of radius R = 3 and
height H = 4 as in Figure 3. A cone ofradius rand height h has volume ~nr2 h.
FIGURE 3
Let r denote the radius and h the height of the upside down cone. By similar triangles, we obtain the relation
SOLUTION
4-h
r
4
3
so
dV =
~n(2r -
and the volume of the upside down cone is
forO
~
r
~
3. Thus,
dr
3
r 2)
and the critical points are r = 0 and r = 2. Because V(O) = V(3) = 0 and
V(2)=
~n(4-~)=
3
3
16
n
9
the maximum volume of a right-circular cone placed upside down in a right-circular cone of radius 3 and height 4 is
63. Show that the maximum area of a parallelogram ADEF that is inscribed in a triangle ABC, as in Figure 4, is equal
to one-half the area of D.ABC.
B
A
SOLUTION
c
F
FIGURE 4
Let 8 denote the measure of angle BAC. Then the area of the parallelogram is given by AD ·AF sin 8. Now,
suppose that
BE/BC =x
Then, by similar triangles, AD = (1 - x)AB, AF = DE = xAC, and the area of the parallelogram becomes AB· ACx(l x) sin 8. The function x(l - x) achieves its maximum value of when x = ~.Thus, the maximum area of a parallelogram
inscribed in a triangle !:!ABC is
i
1- -
1 (1- -
) = 1 (areaofMBC)
4AB·ACsin8= "2 2AB·ACsin8
2
65. Let f be a function whose graph does not pass through the x-axis and let Q = (a, 0). Let P = (x0 , f(x0 )) be the point
on the graph closest to Q (Figure 5). Prove that PQ is perpendicular to the tangent line to the graph of x 0 • Hint: Find the
minimum value of the square of the distance from (x, f(x)) to (a, 0).
Chapter Review Exercises
y
y
315
= f(x)
Q =(a, 0)
FIGURE 5
Let P = (a, 0) and let Q
segment joining P and Q is then
= (x0 , f(x 0 )) be the point on the graph of y = f(x) closest to P. The slope of the
SOLUTION
f(Xo)
x0 -a
Now, let
q(x)
= ~(x -
a)2
the distance from the arbitrary point (x, f(x)) on the graph of y
P, we must have
'(
q Xo)
= 2(xo -
+ (f(x)) 2
=f(x) to the point P. As (x0 , f(xo)) is the point closest to
a)+ 2f(xo)f'(xo)
~(xo - a)2 + (f(xo)) 2
=0
Thus,
f'(xo)
= _Xo -
a
= -( f(xo)
Xo -a
f(Xo)
)-I
In other words, the slope of the segment joining P and Q is the negative reciprocal of the slope of the line tangent to the
graph of y = f(x) at x = x 0 ; hence; the two lines are perpendicular.
67. Use Newton's Method to estimate
SOLUTION
Let f(x) ~
Ws to four decimal places.
x3 - 25 and define
~ -25
f(xn)
Xn+I
=
Xn -
f'(Xn)
=
Xn -
3 X~
With Xo = 3, we find
Thus, to four decimal places
n
l
2
3
Xn
2.925925926
2.924018982
2.924017738
Ws = 2.9240.
+1
e2x
69. Find the local extrema of f(x)
SOLUTION
= -ex+-1- .
To simplify the differentiation, we first rewrite f(x)
f(x)
= _e2x
ex+I
+_
l
ex+I
= -;:;1 using the Laws of Exponents:
= e2x-(x+I) + e-(x+I) =ex-I + e-x-1
Now,
f'(x)
=ex-I - e-x-1
Setting the derivative equal to zero yields
Thus,
x- 1 ""-x -1 or x = 0
Next, we use the Second Derivative Test. With f"(x)
f"(O)
Hence, x
=ex-I
+ e-x- 1, it follows that
= e- 1 +e- 1 =
2
- >0
e
= 0 is a local minimum. Since f(O) = e0 - 1 + e-0- 1 = ~,we conclude that the point (0, ~)is a local minimum.
316
CHAPTER 4
I
APPLICATIONSOFTHEDERIVATIVE
In Exercises 71-74,find the local extrema and points of inflection, and sketch the graph. Use L'Hopital's Rule to determine the limits as x - O+ or x - ±oo if necessary.
71. y = x ln x
SOLUTION
(x > 0)
Lety
= xlnx. Then
y' = lnx + xG) = 1 + lnx
and y" = ~-Solving y' = 0 yields the critical point x = e- 1 • Since y"(e- 1) = e > 0, the function has a local minimum
at x = e- 1 . y" is positive for x > 0, hence the function is concave up for x > 0 and there are no points of inflection. As
x - O+ and as x - oo, we find
lim xlnx
X--+0+
lnx
x- 1
.
•
= x--+0+
lim = x--+0+
hm - -2 = lim(-x) = 0;
x-1
- xx--+0+
lim xlnx = oo
X->OO
The graph is shown below:
y
6
4
73. y
=x(ln x) 2
SOLUTION
(x > 0)
Lety
= x(lnx) 2. Then
y'
= x2lnx
- - + (lnx) 2 = 2lnx + (lnx) 2 = lnx(2 + lnx)
x
and
y
,,
2
= -x
2lnx
+ -x
2
= -(1
+ lnx)
x
Solving y' = 0 yields the critical points x = e- 2 and x = 1. Since y"(e- 2 ) = -2e2 < 0 and y"(l) = 2 > 0, the function
has a local maximum at x = e- 2 and a local minimum at x = 1. y" < 0 and the function is concave down for x < e- 1,
whereas y" > 0 and the function is concave up for x > e- 1 ; hence, there is a point of inflection at x = e- 1 • As x - O+
and as x - oo, we find
=x--+0+
hm - -1- = hm
xx--+0+
lim x(ln x) 2
= oo
x--+O+
X->OO
.
(lnx) 2
.
hm x(ln x) 2
.
2lnx·x- 1
-x-2
1
•
.
2lnx
. 2x=x--+0+
hm - -1 = hm - - = hm 2x = O;
-xx--+0+ x-2
x--+O+
The graph is shown below:
0.8
0.6
0.4
0.2
0.5
75.
1.0
1.5
($1 Explain why L'Hopital's Rule gives no information about lim
x->oo
method.
SOLUTION
As x -
2
x - sin x . Evaluate the limit by another
3x + cos 2x
oo, both 2x - sin x and 3x + cos 2x tend toward infinity, so L'Hopital's Rule applies to
. l"nmt,
. l"1m 2 - cos
. d ue to the osc1·11 at10n
. of sin x and cos x.
. x , does not exist
1im 2x - sin x ; however, the resu1tmg
x->oo 3x + cos 2x
x->oo 3 - 2 sm 2 x
To evaluate the limit, we note
lim
x .... oo
2x-sinx
2-sinx
= lim
x
3x + cos 2x x .... oo 3 + cos 2x
x
2
3
Chapter Review Exercises
317
In Exercises 77-88, verify that L'Hopital's Rule applies and evaluate the limit.
4x-12
77. l i m
---x--+3 x 2 - 5x+ 6
The given expression is an indeterminate form of type
SOLUTION
§, therefore L'Hopital's Rule applies. We find
4x - 12
Ii
4
.
= m--= 4
11m
x--+3 x2 - 5x + 6
x--+3 2x - 5
79. lim x 112 ln x
X--iO+
First rewrite
SOLUTION
x 112 lnx
The rewritten expression is an indeterminate form of type
as
lnx
;rl/2
~,therefore
~x
L'Hopital's Rule applies. We find
1~
~~
lim x 112 lnx = lim - 1-2 = lim - - - = lim - - = 0
x--+0+ - 1/2 -3/2
x--+0+ x- 1
x--+0+
2
x--+0+
x
2sinB- sinW
81. lim -.- - - O--+O smB-BcosB
The given expression is an indeterminate form of type §;hence, we may apply L'Hopital's Rule. We find
SOLUTION
. 2sin0-sin20
hm
.
n
0--+0 sm{;I- {;lcosu
-r
-
2cos8-2cos28
-r 2cos8-2cos2e
1m cos{;I- (cosu,, - uSlilu
,, . ll) - 0--+0
1m
. n
Osmu
0--+0
.
= 1!ID
0--+0
.
-2sinB + 4sin20
-2cose + 8cos28
= 11m - - - - - - - sinB + Bcos e
o--+o cos e +cos B - BsinB
-2+8
---=3
1+1-0
. ln(t + 2)
83• lim--1--+oo
log2 t
The limit is an indeterminate form of type
SOLUTION
~;hence,
we may apply L'Hopital's Rule. We find
I
. <+Z
. ln 2
. ln(t + 2)
. t ln 2
11m--- =lim- = 11m-- = 11m- =ln2
t--+oo
log2 t
t--+oo t
1~ 2
t--+oo
t+2
t--+oo
1
. -1
85. lim sm y - y
y--+0
SOLUTION
y3
The limit is an indeterminate form of type §;hence, we may apply L'Hopital's Rule. We find
. -1
-
1
--1
lim sm y - y = lim _p_,_--;
__ = lim y(l -y2)y--+0
y3
y--+O
3y2
y--+0
6y
312
2
= lim (1 - Y t
y--+O
6
32
'
1
6
.
sinh(x2)
87• hm---x--+O coshx- 1
SOLUTION
The limit is an indeterminate form of type §;hence, we may apply L'Hopital's Rule. We find
lim sinh(x2) = lim 2xcosh(x2) = lim 2cosh(x2) + 4x2 sinh(x2) = 2 + 0 =
2
x--+O cosh x - 1
x--+O
sinh x
x--+O
cosh x
1
89. Let f(x) = e-AX11 2 , where A> 0 is a constant. Given any n numbers a 1, a 2 , .. ., an, set
«l>(x)
= f(x -
al)f(x - az) · · · f(x - an)
(a) Assume n = 2 and prove that Ill attains its maximum value at the average x = ~(a 1 + a 2 ). Hint: Calculate <I>'(x) using
logarithmic differentiation.
(b) Show that for any n, «I> attains its maximum value at x = ~(a 1 + a2 +···+an). This fact is related to the role of f(x)
(whose graph is a bell-shaped curve) in statistics.
318
CH APTER 4
I
APPLICATIONS OF THE DERIVATIVE
SOLUTION
(a) For n
= 2 we have,
<l>(x)
=f
(x - a1) f (x - a2)
= e-4<x-a1)2 . e-4<x-a2i2 = e-H<x-a1)2+(x-a2l2)
Since e-4y is a decreasing function of y, it attains its maximum value where y is minimum. Therefore, we must find the
minimum value of
Now, y' = 4x - 2(a 1 + a 2 ) = 0 when
a1 +a2
x=---
2
We conclude that <l>(x) attains a maximum value at this point.
(b) Wehave
Since the function e-4y is a decreasing function of y, it attains a maximum value where y is minimum. Therefore we
must minimize the function
We find the critical points by solving:
y'
= 2 (x -
a1) + 2 (x - a2) + · · · + 2 (x - an)
=0
2nx = 2 (a1 + a2 + · · · + an)
a1 +···+an
x=-----
n
We verify that this point corresponds the minimum value of y by examining the sign of y" at this point: y" = 2n > O. We
conclude that y attains a minimµm value at the point x = ai +~+a. , hence <l>(x) attains a maximum value at this point.
SECTION 5.4
3. f(x)=x- 2 ,
I
The Fundamental Theorem of Calculus, Part I
[l,2]
SOLUTION
'
We have the area
r.
A=
x-•1'
x- 2 dx = : } = - 1
2
1
+ = Z1
1
In Exercises 5-40, evaluate the integral using FTC I.
S. {' xdx
J,
I
SOLUTION
7.
I
1
(4x-9x2)dx
l
SOLUTION
9.
J:
f
(4x-9x2)dx=(:zx2-3i3)[ =:(2-3)-(0-0)=--l
[<12x" + 3x2 - 4x)dx = (2x6 +
x' - 2x2)1: = (128 + 8 - 8) - (0 + 0- 0) = 128
(2r' - 6t2) dt
fc2t3-61
SOLUTION
1
(~t4 -2t~)[ =(0-0)-(~1 -s4)= ~
)dt=
.../Ydy
13. [
f
~y1f2r = ~(4)3/2 - ~(0)lf2 =
Jor4 y1f2 dy = 3
lo3
3
r.-yYdy =
SOT,lJTION
15.
1
(12x" + 3x2 - 4x)dx
SOLUTION
11.
I i .=I-(6) 2 - -(3)
I 2 = -27
xdx = -x2
2 3 2
2
2
16
3
1
1 114 dt
Jl/16
SOLUTION
.
1
l/l~
t
•
•
•
1
31
= - - - =1l/16 5 40 40
114 di= -t514
5
~
17. [
SOLUTION
r.
J
=1
3
dt
2
t
3
t- 2 dt=-t- 1/ =
I
I
_! +] = ~
3
3
f ~ dx
1
19.
J112
OOWTID•
J:
1
!/2
21, [
8
3 dx
x
=
f
1
J112
Kx-
3
1
dx = -4x- 2/
= -4 + 16 = 12
1/2
(x2 - x- 2 )dx
soLunoN
_(cr-x- 2 )Jx=-(~x>+x- 1 )r =(j+~}-{~+1)= 1;
355
356
CHAPTER 5
I
INTEGRATION
27t+l
23.
J,
--:-r. dt
I
V'
SOLUTION
sinBd6
25. [
' -1'/4
j
SOLUTION
sinBd0=-cos61"
J-,,/4
27.
["
I'"
!'
[
~"
10
ln/3
0
1::
0
-./3
1f
-{j
=sin3-sin0=2-0=2·
4
3
3
3
(2-csc2 x)dx= (Lt+ cotx)i::: = 2 · : +cot : -(2 ·~+cot~)= ; -
I-(~+ I)= ir- 2
dx
-
x
l
SOLUTION
'"dx
-=lnlxl
x
I'"
=lnl0-ln2=ln5
'
t+ I dt
37. [
I
'
SOLUTION
39.
costdt=sint
14
J,
v;
(2-csc 2 x)dx
SOLUTION
35.
=-cosir-(-cos(-~))=-(-1)-(-v;)=l+
costdt
SOLUTION
29.
-1'/4
f
r:l dt= [(1+7)dt=(t+lnltl)i: =e+lnlel-(l+lnlll)=e+l-(l+O)=e
£
(3x - 9e '') dx
SOLUTION
£(3x-9e 3x)dx==
{~x2-3e)x)[
2
=(0-3)-(6-3e-6)=3e-6 -9
41. In Example 5, what is tht: probability that the Cool-Lite bulb lasts for more than 100 h of use? More than 1000 h of
use?
SOLUTION
The integral
l
/00
lino
0.00le-O.UOli dt gives [he probability Ulat a Cool-Lite bulb fails bt:fvre 100 h of use. Now,
1'00 = -e-0001 1'00 = -e--0.1 _ (-1) = 1 _ e-0.1
j
QJ){)\e--0.001r dt = 0.(JO\ --e-ooou
-0.001
0
0
It follows that the probability that a Cool-I .ite bulb la.~ts for more than 100 h of use is 1 - (I - e-o.t) = e- 0 1
1
""
0.9048.
1000
The integral
O.OOle-o.ooii dt gives the probability that a Cool-Litt: bulb fails before 1000 h of use. Now,
{1000 O.OOle--0.0011 dt = 0.001-'-e-0.001111000 = -e--0.00111000 = -e-1 - (-1) = 1 - e-1
Jo
-0.001
o
o
It follows that the probability that a Cool-Lite bulb lasts for more than 1000 h of use is 1 - (l - e- 1) = e- 1 "" 0.3679.
I
SECTION 5.4
The Fundamental Theorem of Calculus, Part I
In Exercises 43-48, write the imegral as a ~·um of integrals without absolute values and evaluate.
43.
J_: )x) dx
soLunoN
J,
2
45.
lx)dx=
£I
-2
(-x)dx+
0
'l"'I' (')
1
xdx==--:x!- +-x2 ==0- --(4) +-::0'
2-220
2
22
L'
1x31dx
SOLUTION
0
r
x3dx;;:-!.x4J +!x41'
Jo
4-240
3
r3lx1 ldx== r(-x3)dx+
J_2
J_2
1414
4
4
97
4
==0+-(-2) +-3 -0==47.
I
icosxjdx
SOLUTION
12
[ 0 icosxldx"" [ 0
12
cosxdx + (" (-cosx)dx = sinxJ" - sinx/" ;;: 1 - 0- (-1 - 0);;: 2
J,,12
0
tr/2
In Exercises 49-54, evaluate tire integral in ternu of the conl·tants.
49. [
i!dx
SOLUTION
f
x3 dx ==
1
1
!b4
4
-
!(1)4 ;;: _!_ (b4
4
4
-
1) for any number b.
x~dx
51. [
SOWTION
53.
!x4 1~ =
4
ra
J"
SOLUTION
[
j 6
x5 dx = 6x
I'
1
I 6 - I (1)6 = '6(b
I 6 - 1) for any number b.
== fib
6
dx
x
J:Q d; == In lxlr = In 15al - In 1a1 ;;: Ins
55. Calculate£ f(x) dx, where
12 -x2
f(x) = ( x'
forx:S2
forx>2
SOLUTION
r f(x)dx==. J_2r f(x)dx+f,f(x)dx=
r2(12-i2)dx+l X'dx
J_2
2
J_z
2
3
• (12x- ~x')I', + ~x•r,
• (12(2)- !23 )-(12(-2) - _!_(-2) 3) + !34
3
3
4
-
!24
4
128 65
707
•-+-=-
3
J_,
4
12
1
57. Use FfC I to show that
SOLUTION
x" dx == 0 ifn is an odd whole number. Explain graphically.
We have
r1 x"cJx= x""'I 11 ""(l)""'I
J_1
n+l-1
n+1
- (-J)"+l
n+I
357
358
CHAPTER 5
I
INTEGRATION
Because n is odd, n + l is even, which means that (- l)n+l = (l)n+l = 1. Heni,;e
(l)n+l
(-1)"+1
J
I
-----=---=0
n+l
n+l
n+l
n+1
Graphically speaking, for an odd function such as
x3 shown here, the positively signed area from x = 0 to x
= I cancels
the negatively signed area fro1n x = -1 to x "' 0.
59. Calculate F(4) given that F(l) = 3 and F'(x) =
SOLUTION
x?. Hint: Express F(4) - F(l) as a deftnitc integral.
By FTC I,
F(4)- F(l) = [
43 -13
..?dx = - - = 21
3
Therefore f(4) = F(l) + 21 "'3 + 21 = 24.
61.
($1
SOLUTION
1
1
With n > 0, does
Let n >
;;(''
dx get larger or smaller as n increases? Explain graphically.
o. Then
r'X'dx=(-' ,... )l'=(-1
(l)"")-(-1
1or')=-I
b
n+I
n+1
Jo
n+l
n+l
J,'
which decreases as n in1.:rea~es. Reca!l that
X' dx represents lhe area between the positive curve f(x) = X' and the
x-axis over the interval [O, I]. Accordingly, this area gets sinaller as n gets larger. This is readily evident in the following
graph, which shows curves for several valui;:s of n.
63. Theorem I is stated with the assumption that a < b. Provi: Uiat the FTC I relationship
J: f(x) dx = F(b) - F(a) also
holds for a = b and for b < a assuming that Fis an anti derivative off on fb, a].
SOLUTION
When a = b, we have
[.t(x)dx=
I'
f(x)dx=O
and
F(b)-F(a)=F(b)-F(b)=O
so the FfC I relationship .lb f(x)dx = F(b) - t'(a) hokb. Now, suppose b <a. Then
f'
so the FfC I rclaliunship
f(x)dx = - [
f(x)dx = -(F(a)- F(h)) = F(h) - F(a)
£' j(x)dx = F(b) - F(a) again holds.
Further Insights and Challenges
65. Prove a famous resnlt of Archimedes (gcncrali;i.:ing Ex1;:r1,;i~1;: 64): For r < s, the ari:a uf the shaded region in Figure 7
is equal to four-thirds the area of triangle MCE, where C is the point on the parahola at which the tangent line is parallel
to secant line AE.
SECT I 0 N 5.4
( The Fundamental Theorem of Galculus,
Part I 359
(a) Show that Chas x-coordinate (r + s)/2.
(b) Show that ARDE has area (s - r) 3 /4 by viewing it as a parallelogram of height.~ - rand base of length CF.
(c) Show that 6ACE has area (s - r) 3/8 by observing Lb.at it has the same base and height as !he parallelogram.
(d) Con1pute the shaded area as the area under the graph minus the area of a trapezoid, and prove Archimedes's result.
= (x - a)(b- x).
FIGURE 7 Graph of f(x)
SOLUTION
(a) The slope of the secant line AE is
.f(.~)
- f(r)
~~~~=
(s - a)(b- s) -(r - a)(b- r)
s-r
s-r
=a+b-(r+s)
and the slope of the tangent line along the parabola is
f'(x)=a+b-2x
If C is the point on the parabola at which the tangent line is parallel to lhe secant line AE, then its x-coordinate must
satisfy
,.,
x=-,-
a+ b- 2x =a+ b- (r + s)
(h) Parallelogram ABDE has heighl s - rand base of length CF. Since the equation of the secant lineAE is
y =[a+ b -(r + s)] (x- r) + (r - a)(b - r)
the length of the segment er· is
'"
(<-,)'
- "
--a)(b 2) -[a+b-(r+s)] ( " " -r) -(r-a)(b-r)=("
2
2
Thus, the area of ABDE is
ts-;;
4
1' .
(c) Triangle ACE is comprised of !!ACF and ACEF. Each of these smaller triangles has height
.r;c and ba~e of length
«-;;P. Thus, the area of !!ACE is
2
I s-r (s-r)
I s-r (s-r)
(~·-r)3
--. --+--. --= --22
4
22
4
8
2
(d) 1·he area under the graph of the parabola between x = r and x = s is
f'cx -a)(b - x)dx = (-abx +~(a+ b):2 - ~x3 )[
1
I
I
1
=-•·-~·~--«+•-~+~+-~
2
3
2
3
= ab(r - s) +
3
I
2(a + b)(s -
r)(s + r) +
I
3
(r - s)(r2 +rs+.~)
while the area of the trapezoid under the shade<l region is
I
z<s
- r) [(s -
a)(b- s) + (r -a)(b - r)l
I
.v2]
= 2(s- r)[-2ab +(a+ b)(r+ s)-r2 _
= ab(r - s) ~·
I
2ca + b)(s -
r)(r + s) +
I
(r - s)(r 2 + .?)
2
Thus, the area of the shaded region is
(r -
s)(irz +~rs+~.~
-
~r2 ~s2 ) = (s- r)(~r2-irs+ ~·~) = ~(s-r)~
which is four-thirds the area of the triangle ACF:.
-
360
CHAPTER 5
I
INTEGRATION
67. Use the method ot· Exercise 66 to prove that
r
r
i'
i'
i'
,.<
l--<cosx<J--+2 2
24
x - - <sinx;5;x--+fi 6
120
(forx<:;O)
Verify these inequalities for x = 0.1. Why have we specified x ;:.: 0 for sin x but not for cos x?
SOLUTION
By Exercise 66, t- ~1 3
::;
sint ::; r fort> 0. Integrating this inequality over the interval [O,x],
and
then
solving for cos x, yields:
!i'--_!_x4
2
24
::;
1-cosx< _!_xz
- 2
1,_
1?
14
I - -r < cosx::; 1 - - r + - x
2 2
24
These inequalities apply for x ;:.: O. Since cos x, 1 - ~, and 1 Having estahlishcd that
? + ~ are all even functions, they also apply for x S 0.
!2
t2
t4
2
24
x3
x;
l--<cost<l--+-
2-
-
for all t ;:.: O, we integrate over the interval [0, x], to obtain:
x3
x--<sinx<x--+6 6
120
tx3
k
The functions sin x, x - x 3 and x + 1 ~0 x' are all odd functions, so the inequalities are reversed for x < 0.
Evaluating these inequalities at x = . I yields
0.995000000 :S 0.995004165
s 0.995004167
0.0998333333 :S 0.0998334166 :S 0.0998334167
both of which are true.
69. Use FI'C l to prove that if lf'(x)I ::; K for x E La, b], then lf(x) - f(a)I :S Klx - al for x E [a, b].
SOLUTION
Let a> b he real numbers, and let f be such that lf'(x)I SK for x
L'
E
[u,b]. By FTC I,
f'(t)dt = f(x)- f(a)
Since f'(x):,:.: -K for all x E [a,b], we get:
f(x)-f(a)= Ixf(t)dt;o:-K(x-a)
Since f'(x} :SK for all x c [u,b], we get:
f(x)- f(a) =Ix f'(t)dt::; K(x - a)
Combining these two inequalities yields
-K(x - a) :S f(x) - f(u)::; K(x - a)
so that, hy definition,
lf(x) - f(a)] :i K]x - al
5.5 The Fundamental Theorem of Calculus, Part II
Preliminary Questions
1. Let G(x) = [
....fi3+l. dt.
(a) Is the FTC II needed to calculate G{4)?
(b) Ts the F<TC II nced1;:d to calculate G'(4)?
S EC TI O N 5.5
I
The Fundamental Theorem of Calculus, Part II
361
SOLUTION
(a) No. G(4) =
J. Yif+i dt = 0.
(b) Yes. By the FTC II, G'(x) =
yx3 +I, so G'(4) =
r.
-.(65.
2. Which of the following is an anti derivative F of f(x) =
(a) [ , 2tdt
SOLUTION
(b)
The correct answer is (e): [ ,
x2
satisfying F(2) = O?
t'dt
(c) [ , t'dt
t2 dt.
3. Does every continuous function have an antiderivative? Explain.
SOLUTION
Yes. All continuous functions have an antiderivative, namely
4. Let G(x)
=I,,
ix
f(t) dt.
sintdt. Which of the following statements are correct?
(a) G is the composite function sin(r').
(b) G is the composite function A(x'), where
A(x) = [
sintdt
(c) G(x) is too complicated to differentiate.
(d) The Product Rule is used to differentiate G.
(e) The Chain Rule is used to differentiate G.
(f) G'(x) = 3x' sin(x').
SOLUTION
Statements (b), (e), and (f) are correct.
Exercises
In Exercises 1-8 compute an area function A(x) of f(x) with lower limit a. Then, to verify the FTC II inverse relationship,
compute A'(x) and show that it equals f(x).
1. f(x) = 4 - 2x,
soLunoN
a=0
Let f(x) = 4 - 2x and a :::: 0. Then an area function off with lower limit a is
A(x)=
J.'
f(t)dt= [(4-2t)dt=(4t-t2{ =4x-x'-(0-0)=4x-x'
To verify the FTC II inverse relationship, observe that
A'(x)
3. f(x)=4x+6x1,
SOLUTION
= 4- 2x = f(x)
a=-1
Let f(x) = 4x + 6x2 and a= -1. Then an area function off with lower limit a is
A(x) = [
f(t) dt =
l~ (4t + 612) dt = (2t2 + 2r')L = 2x' + 2x' -
(2 - 2) = 2x2 + 2x'
To verify the FTC Il inverse relationship, observe that
A'(x) = 4x + 6x' = f(x)
5. f(x) = x' - sinx,
SOLUTION
a= 0
Letf(x) = :x? - sinx and a:::: 0. Then an area function off with lower limit a is
A(x)= [f(t)dt=
[ct2-sint)dt=(~+cost}[=;
To verify the FTC Il inverse relationship, observe that
A'(x) = x' - sinx = f(x)
+cosx-(0+1)=; +cosx-1
362
C H APT E R 5
I
INTEGRATION
7. f(x) = e"',
a= 0
Let f(x) = e2x and a = 0. Then an area function off with lower limit a is
SOLUTION
A(x) =
l,
f(t) dt =
"
l'
o
I'
I 21 = -ezx
I
I 0 = -e
I 2x - -1
e2' dt = -e
- -e
2o2
2
2
2
To verify the FfC II inverse relationship, observe that
A'(x)
= e"' = f(x)
Jn Exercises 9-12, compute or approximate the corresponding function values and derivative values for the given area
function. In some cases, approximations will need to be done via a Riemann sum.
9. F(x) =
J.' Vt
2
+ tdt. Find F(O), F(3), F'(O), and F'(3).
Let F(x) =
SOLUTION
l"'
Yii+t dt. Then
F(O)
= [ Vt2 + tdt =0
and
F(3)
=
r
Yii+tdt"' 5.784
where the latter value was obtained from the right-endpoint approximation R 100 • By FfC Il, F'(x) =
F'(O)
11. F(x) = ("
= VO' +O =0
and
F'(3)
"l/:x2 + x, so
= v32 +3 =2'13
J_2 u2du+ 1 . Find F(-2). F(2). F'(O), and F'(2).
LetF(x) =
SOLUTION
,
L
2
du
- 2- - . Then
u +1
2
F(-2)=[
-2
~
=0
u2 +1
2
and
F(2) =
1
2
du
- 2"'2.214
u +1
where the latter value was obtained from the right-endpoint approximation R 100 . By FrC II, F'(x) = x2~I' so
F'(O)
= 02 ~ I = 1
and
F'(2)
= 22 ~ I = ~
In Exercises 13-22, find formulas for the functions represented by the integrals.
13. [
u4 du
F(x) =
SOLUTION
15. [
F(x) =
f
!,,
4
11'1
du= -u5 = -x5
5 2 5
- -32
5
ix
sin u du = (-cos u)I: = 1 - cos x
3
F(x) =
3
e "
[
I 3"
du= -e
3
I'
I 3x - -e12
I
= -e
3
4
3
tdt
SOLUTION
21.
u
e "du
SOLUTION
19.
x
sinudu
SOLunON
17. [
f,
F(x) =
,+z
SOLUTION
f
1
tdt = -i'
2
Ir =
I
1 4 - -1
-x
2
2
e-u du
F(x) =
l
9x+2
3x
e-u du= -e-u
19x+2
3x
= -e-9x-2 + e-3x
S EC TI O N 5.5
23. Verify [
ltldt =
Let f(t)
SOLUTION
I
The Fundamental Theorem of Calculus, Part II
363
~xix!. Hint: Consider x;,: 0 and x 5 0 separately.
= ltl = {
fort~O
t
-t fort<O
. Then
lx
{[-tdt
for x ~ 0
tdt
=[
F(x)
f(t)dt
=
0
for x < 0
!x2
For x ~ 0, we have F(x) =
= !xlxl since lxl = x, while for x < 0, we have F(x) =
Therefore, for all real x we have F(x) = !x lxl.
-!x2 = !xlxl since Jxl = -x.
In Exercises 25-28, calculate the derivative.
25. -d
dx
lx
SOLUTION
27.
(t' - 9i') dt
0
By FfC II,
!!_ (
dt
J 100
SOLUTION
:X [(r -9i')dt = x'-9x3.
sec(5x - 9) dx
By FTC II, _dd (
t
J100
29. Let A(x) = [
sec(5x - 9) dx = sec(5t - 9).
f(t) dt for f(x) in Figure 8.
(a) Calculate A(2). A(3), A'(2), and A'(3).
(b) Find formulas for A(x) on [O, 2] and [2, 4], and sketch the graph of A.
y
4
"
3
2
"'"
./
2
./
3
4
FIGURE 8
SOLUTION
=
=
=
(a) A(2)
2 · 2 4, the area under the graph of y
f(x) from x
under the graph of y f(x) from x 0 to x 3. By the FTC, A'(x)
=
=
=
= 0 to x = 2, while A(3) = 2 · 3 + ~ = 6.5, the area
= f(x) so A'(2) =f(2) =2 and A'(3) =f(3) =3.
(b) For each x E [O, 2], the region under the graph of y = f(x) is a rectangle oflength x and height 2; for each x e [2,4],
the region is comprised of a square of side length 2 and a trapezoid of height x - 2 and bases 2 and x. Hence,
A(x) = {
2x,
!x2+2,
0.:5x<2
2Sx.:::;4
A graph of the area function A(x) is shown below.
y
10
Area Function
A(x)
2
3
4
In Exercises 31-32, do the following:
• For x = 0, 1,2, ... , 10, approximate A(x) = J;,x f(t)dt.
• For x = 1, 2, 3, ... , 9, approximate A'(x) using 8.x = I and the symmetric difference quotient approximation, A'(x) :::::
A(x + L\.x) -A(x- L\.x)
2L\.x
• Plot the values of A'(x) on a graph off to demonstrate A':::::/.
364
C H APT E R 5
I
INTEGRATION
y
y
4
8
2
4
x
0
0
-2
-4
-4
-8
(B)
(A)
FIGURE 10
31. Use /(x) from Figure I O(A).
SOLUTION
Jo
• First, note that A(O) =
f(t) dt = 0. For other x, we can estimate A(x) by counting 1 x 1 squares (and partial
squares) between the graph and the x-axis from 0 to x. For example, there are roughly 3.3 squares between the
graph and the x-axis from 0 to 1, so A(l) = 3.3. Continuing in this fashion, we obtain the approximate values of
A(x) listed in the table below.
x
0
I
2
3
4
5
6
7
8
9
10
A(x)
0
3.3
6.4
6.9
3.2
-0.6
-3.l
-3.7
-2.1
-0.2
1.6
• For x = 1,
A'(l)= A(2)-A(O) = 6.4-0 = 32
2
2
.
and for x = 2,
A'( 2)
=A(3) - A(l) = 6.9 2
2
3.3 = I 8
.
Continuing in this fashion, we obtain the approximate values of A'(x) listed in the table below.
x
I
2
3
4
5
6
7
8
9
A'(x)
3.2
1.8
-1.6
-3.75
-3.15
-1.55
0.5
1.75
1.85
• The values of A' are plotted together with the graph off below.
y
4
2
o+-+~-+1-~~x
-2
-4
In Exercises 33-38, calculate the derivative.
33• .!:.._ I
dx)0
SOLUTION
,?
tdt
t+I
.
d [
By the Chain Rule and the FfC, dx 0
£
tdt
x2- . 2x = 2x3- .
-- = r+l
x2 +I
x2+I
0
d
35. -d
s
SOLUTION
"
4
u du
-6
By the Chain Rule and the FfC, -d
ds
Hint for Exercise 36: F(x) = A(x') - A(x').
I:o"
_6
u4 du
= cos4 s(- sins) = - cos4 s sins.
I
S EC TI O N 5.5
37•
.!__
dx
r
J.,rx
SOLUTION
The Fundamental Theorem of Calculus, Part II
365
tantdt
Let
G(x);
f" tantdt;l?
tantdt-l'"tantdt
0
0
)yx
Applying the Chain Rule combined with FTC twice, we have
G'(x); tan(x') · 2x- tan( vx).
~x-11' ;
2
2xtan(x')- tan( Vx)
2,.'X
In Exercises 39-42, with y ; f(x) as in Figure 11, let
A(x); [
f(t)dt
and
B(x) ;
l'
f(t) dt
y
Zt·r······················~;;•···········;
i
O+-~,_,'-t~-+~-<-_,,.._+-~1-+X
-I t--'-f ......,
-2+···················'-···
·+ ........ ' ... t-_,6
•...............•.............•
.. ··'
FIGURE 11
39. Find the min and max of A on [O, 6].
soLVTION The minimum values of A(x) on [0, 6] occur where A'(x) :::. f(x) goes from negative to positive. This occurs
at one place, where x = 1.5. The minimum value of A(x) is therefore A(l .5) = -1.25. The maximum values of A(x) on
[O, 6] occur where A'(x) = f(x) goes from positive to negative. This occurs at one place, where x = 4.5. The maximum
value of A(x) is therefore A(4.5); 1.25.
41. Find formulas for A(x) and B(x) valid on [2,4].
SOLUTION
On the interval [2,4], A'(x); B'(x); f(x); I. A(2);
f.
f(t)dt; -I and 8(2); [
f(t)dt; 0. Hence
A(x) ; (x - 2) - I and B(x) ; (x - 2).
43. Let A(x) ; [
f(t)dt, with y; f(x) as in Figure 12.
(a) Does A have a local maximum at P?
(b) Where does A have a local minimum?
(c) Where does A have a local maximum?
(d) True or false? A(x) < 0 for all x in the interval shown.
y
Q
FIGURE 12 Graph ofy; f(x).
SOLVTION
(a) In order for A to have a local maximum, A'(x) = f(x) must transition from positive to negative. As this does not
happen at P, A does not have a local maximum at P.
(b) A will have a local minimum when A'(x) = f(x) transitions from negative to positive. This happens at R, so A has a
local minimum at R.
(c) A will have a local maximum when A'(x) = f(x) transitions from positive to negative. This happens at S, so A has a
local maximum at S.
(d) It is true that A(x) < 0 on the interval shown since the signed area from 0 to x is clearly always negative from the
figure.
366
C H A PT E R 5
I
INTEGRATION
In Exercises 45-46, let A(x)
45. r=§;1
=I'
f(t)dt.
Area Functions and Concavity
Explain why the following statements are true. Assume f is differentiable.
(a) If c is an inftection point of A, then f'(c) = 0.
(b) A is concave up if f is increasing.
(c) A is concave down if f is decreasing.
SOLUTION
(a) If x = c is an inftection point of A, then A"(c) = f'(c) = 0.
(b) If A is concave up, then A"(x) > 0. Since A(x) is the area function associated with f(x), A'(x) = f(x) by FTC II, so
A"(x) = f'(x). Therefore f'(x) > 0, so f is increasing.
(c) If A is concave down, then A"(x) < 0. Since A(x) is the area function associated with f(x), A'(x) = f(x) by FTC II,
so A"(x) = f'(x). Therefore, f'(x) < 0 and so f is decreasing.
47. Let A(x) =
l'
f(t) dt, with y = f(x) as in Figure 13. Determine:
(a) The intervals on which A is increasing and decreasing
(b) The values x where A has a local min or max
(c) The inflection points of A
(d) The intervals where A is concave up or concave down
y
Y = f(x)
FIGURE 13
SOLUTION
(a) A is increasing when A'(x) = f(x) > 0, which corresponds to the intervals (0,4) and (8, 12). A is decreasing when
A'(x) = f(x) < 0, which corresponds to the intervals (4, 8) and (12, oo).
(b) A has a local minimum when A'(x) = f(x) changes from - to+, corresponding to x = 8. A has a local maximum
when A'(x) = f(x) changes from+ to-, corresponding to x == 4 and x::; 12.
(c) Inflection points of A occur where A"(x)::; f'(x) changes sign, or where f changes from increasing to decreasing or
vice versa. Consequently, A has inflection points at x::; 2, x::; 6, and x::; 10.
(d) A is concave up when A"(x) ::; f'(x) is positive or f is increasing, which corresponds to the intervals (0, 2) and
(6, 10). Similarly, A is concave down when f is decreasing, which corresponds to the intervals (2, 6) and (10, oo).
49. Sketch the graph of an increasing function f such that both f'(x) andA(x) ==Lr f(t)dt are decreasing.
SOLUTION
If f'(x) is decreasing, then f"(x) must be negative. Furthermore, if A(x) == [
f(t)dt is decreasing, then
A'(x) = f(x) must also be negative. Thus, we need a function which is negative but increasing and concave down. The
graph of one such function is shown below.
y
51.
!Gull
Find the smallest positive critical point of
F(x) = [
cos(t'12 ) dt
and determine whether it is a local min or max. Then find the smallest positive inflection point of F(x) and use a graph
of y ::; cos(x3 12 ) to determine whether the concavity changes from up to down or from down to up.
I
s E c T 1o N 5.5
The Fundamental Theorem of Calculus, Part II
367
A critical point of F occurs where F'(x) = cos(x312 ) = 0. The smallest positive critical points occurs where
x312 = n/2, so that x = (n/2) 213. F'(x) goes from positive to negative at this point, so x = (n/2) 213 corresponds to a local
maximum.
Candidate inflection points of F occur where F"(x) = 0. By FTC, F'(x) = cos(x312), so F"(x) = -(3/2)x 112 sin(x3'2J.
Finding the smallest positive solution of F"(x) = 0, we get:
SOLUTION
-(3/2)x 112 sin(x3 12 ) = 0
(since x > 0)
sin(x'''J = 0
x''' ="
x = 7'1 13 "'2.14503
From the plot below, we see that F'(x) = cos(x312) changes from decreasing to increasing at il-13, so F changes from
concave down to concave up at that point.
y
0.5
2
3
--05
-1
53. The Gauss error function is defined by erf(x) = .}n:-
foe-fl dt.
(a) Explain why erf(x) is an increasing function.
(b) Explain why erf(x) is an odd function.
(c) CRS
Use Riemann sums to approximate erf(x) for x = 1/2, I, 3/2, 2, and 5/2.
(d) From (b) and (c), where does it appear that erf(x) has horizontal asymptotes?
(e) From the information in (a)-(d), sketch a graph of erf(x).
SOLUTION
(a) Because
d
dx
2
-erf(x) = -e-
i'
>0
Vii
for all x, it follows that erf(x) is an increasing function for all a.
(b) Note
erf(-x)
= -2
Vii
[ ' e-? dt
=- 2
i
Vii _,
0
e-? dt
Because f(t) = e-r2 is an even function, it follows by symmetry that
_r=;
2
y1l
i
e-? dt = - 2
Vii
-x
l'
e-? dt
0
Thus,
erf(-x) =
-~
Vii
f
e-? dt
~ -erf(x)
0
and erf(x) is an odd function.
(c) For x = 1/2, 1, 3/2, 2, and 5/2, we calculate the right-endpoint Riemann sum using l:!t = 0.01 to approximate the
value of each definite integral. For example, to approximate A(l/2), we compute R50 for the definite integral over [O, 1/2],
while to approximate A(2), we compute R2oo for the definite integral over [0, 2J. The values obtained in this fashion are
given in the table below.
x
1/2
I
3/2
2
5/2
erf(x)
0.5192
0.8391
0.9611
0.9898
0.9940
(d) Based on the results obtained in part (c), it would appear that erf(x) has a horizontal asymptote ofy = 1 as x ~ oo,
Using the result from part (b) that erf(x) is an odd function would then imply a horizontal asymptote of y = -1 as
x
~-co.
368
C H A PT E R 5
I
INTEGRATION
(e) Using the infonnation in parts (a)-(d), we obtain the following sketch of the graph of erf(x).
y
Further Insights and Challenges
55. Proof of FTC I
f(t)dt = F(b)- F(a) if F'(x) = f(x). Use FTC II to give a new proof of
FTC I asserts that [
FTC I as follows. Set A(x) =
I'
f(t) dt.
(a) Show that F(x) = A(x) + C for some constant.
(b) Show that F(b)- F(a) = A(b)- A(a) = [
f(t) dt.
Let F'(x) = f(x) and A(x) = .[," f(t)dt.
(a) Then by the FTC, Part II, A'(x) = f(x) and thus A and Fare both antiderivatives off. Hence F(x) = A(x)
some constant C.
SOLUTION
+ C for
(b)
F(b) - F(a) = (A(b) + C) - (A(a) + C) = A(b) -A(a)
= [
f(t)dt = [
f(t)dt- [
f(t)dt-0 = [
f(t)dt
which proves the FfC, Part I.
57. Prove the formula
-d
dx
SOLUTION
Write
l
~x)
=
f(x) dx
l'"'
[,
M
f(t) dt
= f(v(x))v'(x) - f(u(x))u'(x)
u(.r)
f(x) dx +
l~x)
f(x) dx
o
M
=
l~x)
f(x) dx -
o
J:(x) f(x) dx
O
Then, by the Chain Rule and the FTC,
-d
l'"'
dxu(x)
f(x)dx= -d
l~., f(x)dx--dd
I"'
Xo
dxo
f(x)dx
= f(v(x))v'(x) - f(u(x))u'(x)
5.6 Net Change as the Integral of a Rate of Change
Preliminary Questions
1. A hot metal object is submerged in cold water. The rate at which the object cools (in degrees per minute) is a function
f(t) of time. Which quantity is represented by the integral
Lr
f(t) dt?
fo7
The definite integral
f(t) dt represents the total drop in temperature of the metal object in the first T minutes
after being submerged in the cold water.
SOLUTION
2. A plane travels 560 km from Los Angeles to San Francisco in 1 hour (h). If the plane's velocity at time tis v(t)
what is the value of
SOLUTION
it
km/h,
v(t) dt?
The definite integral
.£ v(t) dt represents the total distance traveled by the airplane during the 1-hour flight from
1
Los Angeles to San Francisco. Therefore, the value of
fo1 v(t) dt is 560 km.
s EcT IO N
5.6
I
Net Change as the Integral of a Rate of Change
369
3. Which of the following quantities would be naturally represented as derivatives and which as integrals?
(a)
(b)
(c)
(d)
Velocity of a train
Rainfall during a 6-month period
Mileage per gallon of an automobile
Increase in the U.S. population from 1990 to 2010
Quantities (a) and (c) involve rates of change, so these would naturally be represented as derivatives. Quantities
(b) and (d) involve an accumulation, so these would naturally be represented as integrals.
SOLUTION
Exercises
1. Water flows into an empty reservoir at a rate of 3000 + 20t L per hour (tis in hours). What is the quantity of water
in the reservoir after 5 h?
SOLUTION
The quantity of water in the reservoir after five hours is
L
5
(3000 + 20t) dt
= (3000t + 10t2) 1: = 15,250 gallons
3. A survey shows that a mayoral candidate is gaining votes at a rate of 2000t + 1000 votes per day, where t is the
number of days since she announced her candidacy. How many supporters will the candidate have after 60 days, assuming
that she had no supporters at t = O?
SOLUTION
The number of supporters the candidate has after 60 days is
f
(2000t + IOOO)dt
= (1000t2 + lOOOt)I~ = 3,660,000
5. Find the displacement of a particle moving in a straight line with velocity v(t) = 4t - 3 rn/s over the time interval
[2, 5].
SOLUTION
The displacement is given by
ls
t
(4t- 3)dt = (2t2- 3t)1: = (50- 15)-(8 - 6) = 33 m
7. A cat falls from a tree (with zero initial velocity) at time t
= 1 s? Use Galileo's formula v(t) = -9.8t rn/s.
SOLUTION
Given v(t)
= 0. How far does the cat fall between t = 0.5 second and
= -9.8t m/s, the total distance the cat falls during the interval[!, l] is
(1
lv(t)I dt = (' 9.8tdt = 4.9rl'
J112
J112
= 4.9 - 1.225 = 3.675 m
1/2
In Exercises 9-12, a particle moves in a straight line with the given velocity (in meters per second). Find the displacement
and distance traveled over the time interval, and draw a motion diagram like Figure 3 (with distance and time labels).
9. v(t) = 12 - 4t,
[O, 5]
1
5
SOLUTION
Displacement is given by
( 12
- 4t) dt
L
3
[
112-4tl dt =
(12-4t)dt+
=(12t -
l
2t2 {
= 10 ft, while total distance is given by
5
(4t-12)dt = (12t-2r)I: + (2r -12t)I: = 26ft
The displacement diagram is given here.
f;5
/;3
t;O
Distance
0
11. v(t) = r
SOLUTION
2 -
1,
IO
18
[0.5, 2]
Displacement is given by
i
2
.5
(t-2
-
1) dt
= (-r' -
2
t)l
.5
= 0 m, while total distance is given by
370
CH APTER 5
I
INTEGRATION
The displacement diagram is given here.
1=2
::::::::::::::~
I=
1
1=0
1--------+--Distance
0.5
0
13. Find the net change in velocity over [1, 4] of an object with a(t) = St - t2 m/s 2 •
SOLUTION
The net change in velocity is
[
(4r-~t3)[ =39m/s
a(t)dt= [<st-r)dt=
15. The traffic flow rate past a certain point on a highway is q(t) = 3000 + 2000t - 300t2 (tin hours), where t = 0 is
8 AM. How many cars pass by in the time interval from 8 to 10 AM?
SOLUTION
The number of cars is given by
fo
2
q(t) dt =
fo
2
(3000 + 2ooot - 300t2) dt = (3ooot + 10oot2 - 10ot3)
1:
= 3000(2) + 1000(4) - 100(8) = 9200 cars
17. A small boutique produces wool sweaters at a marginal cost of 40 - 5Lx/5J for 0::; x::; 20, where LxJ is the greatest
integer function. Find the cost of producing 20 sweaters. Then compute the average cost of the first 10 sweaters and the
last 10 sweaters.
SOLUTION
The total cost of producing 20 sweaters is
1
20
(40-5[[x/5]])dx=
0
15
40dx+
0
llO
35dx+
115 30dx+ I.20 25dx
5
IO
15
= 40(5) + 35(5) + 30(5) + 25(5) = 650 dollars
From this calculation, we see that the cost of the first 10 sweaters is $375 and the cost of the last 10 sweaters is $275;
thus, the average cost of the first 10 sweaters is $37.50 and the average cost of the last 10 sweaters is $27.50.
19. The velocity of a car is recorded at half-second intervals (in feet per second). Use the average of the left- and
right-endpoint approximations to estimate the total distance traveled during the first 4 s.
SOLUTION
t
0
0.5
1
1.5
2
2.5
3
3.5
4
v(t)
0
12
20
29
38
44
32
35
30
Let M = 0.5. Then
RN = 0.5 · (12 + 20 + 29 + 38 + 44 + 32 + 35 + 30) = 120 ft
LN = 0.5 · (0 + 12 + 20 + 29 + 38 + 44 + 32 + 35) = 105 ft
The average of RN and LN is 112.5 ft.
21. The snowfall rate R (in inches per hour) was tracked during a major 24-hour lake effect snowstorm in Buffalo, New
4
York. The graph in Figure 5 shows Ras a function oft (hours) during the storm. Wh.at quantity does Jo2
R(t) dt represent?
0
Approximate the integral.
R
4
3
2
0
5
10
15
20
FIGURE 5 The snowfall rate during a major 24-h storm.
I
SE c TI 0 N 5.6
SOLUTION
The integral
fo
24
Net Change as the Integral of a Rate of Change
R(t) dt represents the total amount of snow that fell during the snowstorm. Using M
left- and right-endpoint approximations to the value of
fo
24
371
= 1, the
R(t) dt are
L"" 1(2.1+1.6 + 1.2 + 0.3 + 0.3 + 0 + 0 + 0 + 1.3 + 1.9 + 2.6 + 3.1+3.2 + 4.1+3.3 + 2.3
+ 1.1 + 0.4 + 0 + 0 + 2.0 + 3.2 + 3.6 + 1.4) = 39.0 in.
and
R"" 1(1.6 + 1.2 + 0.3 + 0.3 + 0 + 0 + 0 + 1.3 + 1.9 + 2.6 + 3.1+3.2 + 4.1+3.3 + 2.3 + 1.1
+0.4 + 0 + 0 + 2.0 + 3.2 + 3.6 + 1.4 + 0.5) = 37.4 in.
The average of these values is 38.3 in.
23. Let N(d) be the number of asteroids of diameter
~ d
kilometers. Data suggest that the diameters are distributed
according to a piecewise power law:
N'(d) =
1.9 x 109 d-2·3
ford< 70
{ 2.6 x 1012 a-4
ford ~ 70
(a) Compute the number of asteroids with a diameter between 0.1 and 100 km.
(b) Using the approximation N(d + 1) - N(d) ""N'(d), estimate the number of asteroids of diameter 50 km.
SOLUTION
(a) The number of asteroids with diameter between 0.1 and 100 km is
f
100
Jo.1
N'(d)dd =
r7° 1.9 x 10 a
9
23
Jo.1
i.9 x 109 a-1.3 170
1.3
0.1
dd +
_
f
100
J10
2.6 x l0 12 d-4 dd
2.6 x 1012 d_ 3 1100
3
70
= 2.916 x 1010 + 1.66 x 106 "" 2.916 x 1010
(b) Taking d
= 49.5,
N(50.5) - N(49.5) ""N'(49.5)
= 1.9 x 109 49.5-2·3 = 240,525.79
Thus, there are approximately 240,526 asteroids of diameter 50 km.
25. Figure 7 shows the rate R(t) of natural gas consumption (billions of cubic feet per day) in the mid-Atlantic states
(New York, New Jersey, Pennsylvania). Express the total quantity of natural gas consumed in 2009 as an integral (with
respect to time tin days). Then estimate this quantity, given the following monthly values of R(t):
3.18,
1.01,
2.86,
0.89,
2.39,
0.89,
1.49,
1.20,
1.08, 0.80,
1.64, 2.52
Keep in mind that the number of days in a month varies with the month.
Natural gas consumption (109 ft 3/day)
3
2
J F M A M J J A S 0 N D
FIGURE 7 Natural gas consumption in 2009 in the mid-Atlantic states.
SOLUTION
The total quantity of natural gas consumed is given by
r365
Jo
R(t)dt
372
CH APT E R 5
I
INTEGRATION
With the given data, we find
Jor365 R(t) dt"' 31(3.18) + 28(2.86) + 31(2.39) + 30(1.49) + 31(1.08) + 30(0.80)
+31(1.01) + 31(0.89) + 30(0.89) + 31(1.20) + 30(1.64) + 31(2.52)
= 605.05 billion cubic feet
Exercises 27 and 28: A study suggests that the extinction rate r(t) of marine animal families during the Phanerozoic Eon
can be modeled by the function r(t) = 3130/(t + 262) for 0 s t s 544, where tis time elapsed (millions of years) since
the beginning of the eon 544 million years ago. Thus, t = 544 refers to the present time, t = 540 is 4 million years ago,
and soon.
27. Compute the average of RN and LN with N
periods 100 st s 150 and 350 st s 400.
= 5 to estimate the total number of families that became extinct in the
SOLUTION
• (100
sts
150) For N = 5,
= 10
At= 150 - 100
5
The table of values (r(t;) };=o.. .s is given below:
t;
100
110
120
130
140
150
r(t;)
8.64641
8.41398
8.19372
7.98469
7.78607
7.59709
The endpoint approximations are:
RN= 10(8.41398 + 8.19372 + 7.98469 + 7.78607 + 7.59709)"' 399.756 families
LN = 10(8.64641+8.41398 + 8.19372 + 7.98469 + 7.78607) "'410.249 families
The right endpoint approximation estimates 399.756 families became extinct in the period 100 s t s 150, the
left endpoint approximation estimates 410 .249 families became extinct during this time. The average of the two is
405.362 families.
• (350 s t s 400) For N = 5,
= 400 -
!:it
350
= 10
5
The table of values (r(t;));=o...s is given below:
t;
350
360
370
380
390
400
r(t;)
5.11438
5.03215
4.95253
4.87539
4.80061
4.72810
The endpoint approximations are:
RN= 10(5.03215 + 4.95253 + 4.87539 + 4.80061 + 4.72810) "'243.888 families
LN
= 10(5.11438 + 5.03215 + 4.95253 + 4.87539 + 4.80061) "'247.751 families
The right endpoint approximation estimates 243.888 families became extinct in the period 350 s t s 400, the
left endpoint approximation estimates 247. 751 families became extinct during this time. The average of the two is
245.820 families.
Further Insights and Challenges
= 1 and moving along the x-axis with velocity v(t) = r
29. Show that a particle, located at the origin at t
pass the point x = 2.
The particle's velocity is v(t) = s'(t) = r
position at time t is
SOLUTION
s(t) =
It
I
2
,
an antiderivative for which is F(t) =
2
,
will never
-r 1• Hence, the particle's
It
1
s'(u)du = F(u) = F(t)- F(l) = 1 - - < 1
I
t
for all t:::: 1. Thus, the particle will never pass x = 1, which implies it will never pass x = 2 either.
s EcT ION
5.6
I
Net Change as the Integral of a Rate of Change
373
31. In a free market economy, the demand curve is the graph of the function D that represents the demand for a specific
product by the consumers in the economy at price q. It is not surprising that the curve is decreasing, as the demand
drops as the price goes up. The supply curve is the graph of the function S that represents the supply of the product
that the producers are willing to produce as a function of the price q. As the price goes up, the producers are willing
to produce more of the item, and therefore, this curve is increasing. The point (p*, q*) at which the two curves cross is
called the equilibrium point, where the supply and demand balance. Tradition in economics is to make the horizontal
axis the quantity q of the item and the vertical axis the price p. We define p = S (q) to correspond to the supply curve and
p = D(q) to correspond to the demand curve. In other words, we have inverted the formula for these two functions from
giving quantity in terms of price to giving price in terms of quantity. The areas depicted in Figure 9 represent the excess
supply and excess demand.
(a) The consumer surplus represents the savings on the part of consumers if they pay price p* rather than the price
greater than p* that many were willing to pay. Write a formula for this consumer surplus. The formula will include a
definite integral and it will depend on p* and q*.
(b) The producer surplus represents the savings on the part of producers if they sell at price p* rather than the price less
than p* that some producers were willing to accept. Write a formula for this producer surplus.
(c) A variety of coffee shops in a town sell mocha latte supreme coffees. If the supply curve is given by p
and the demand curve is given by p
= l~O + 1
= q/l~~ + 1 , determine the equilibrium point (p*, q*) and the consumer surplus and
producer surplus when the mocha latte supreme coffees are sold at price p*.
Price p
~---~----+-Quantity
q*
q
FIGURE 9 The supply and demand curves.
SOLUTION
(a) The consumer surplus is the area under the graph of the demand function from q = O to q = q* minus the area of the
rectangle formed by the vertical and horizontal line segments through the equilibrium point (p*, q*). Thus, a formula for
consumer surplus is
[ ' D(q)dq- p*q*
(b) The producer surplus is the area of the rectangle formed by the vertical and horizontal line segments through the
equilibrium point (p*, q*) minus the area under the graph of the supply function from q = O to q = q*. Thus, a formula
for producer surplus is
(c) Given S (q)
= ibo + 1 and D(q) = qfl~+i, equilibrium occurs when
Thus, q*
1)
2
_!j_+l=
10
100
q/100 + 1
or
+
( _!j_
100
= 10
= 100( ....ffO- 1) and p* = ....ffO. By part (a), the consumer surplus is
r·
Jo
10
rlOO(VfQ-1)
D(q) dq - p'q*
=Jo
1000 In
= 1000 ln
q/lOO +
VlO -
dq - 100 VW( VW- 1)
lOO(Yl0-1)
I Il
l~ + 1
1
-
1000 + 100 v'iQ
0
1000 + 100 VlO = 500 ln 10 - 1000 + 100 v'iO
374
CHAPTER 5
I
INTEGRATION
Finally, by part (b), the producer surplus is
•
p*q*- [
r100(Yl0-1)
S(q)dq=lOO-v'iO(-v'iO-l)0
Jo
2
= 1000-100-v'iO-(:Oo
+q)
{i6o+1)dq
llOO(Yl0-1)
0
= 1000 -
100-v'iO- 50(-v'iO - 1)2 + 100(-v'iO - 1)
= 1000 -
100 -v'iO - 450
= 550 -
100 -v'iO
5. 7 The Substitution Method
Preliminary Questions
1. Which of the following integrals is a candidate for the Substitution Method?
4
(a) J5x sin(.x5)dx
(b)
f
sin5 xcosxdx
(c) J .x5 sinxdx
The function in (c): x5 sin xis not of the form g(u(x))u'(x). The function in (a) meets the prescribed pattern with
x5. Similarly, the function in (b) meets the prescribed pattern with g(u) = u 5 and u(x) = sin x.
SOLUTION
g(u) = sin u and u(x) =
2. Find an appropriate choice of u for evaluating the following integrals by substitution:
(a) f x(x2 + 9)4 dx
(c)
I
(b) f
x2 sin(x3) dx
sinx cos2 xdx
SOLUTION
(a) x(x2
+ 9)4 = !<2x)(x2 + 9) 4 ; hence, c = !, f(u) = u4 , and u(x) = x2 + 9.
(b) x2 sin(.x3) = t<3x2) sin(.x3); hence, c = ~' f(u) =sin u, and u(x) = x3.
(c) sinxcos2 x = -(-sinx)cos2 x; hence, c = -1, f(u) = u2 , and u(x) = cosx.
1
2
3. Which of the following is equal to
(a)
! f
3
Jo
SOLUTION
x2(.x3 + l)dx for a suitable substitution?
2
9
(b)
udu
( udu
Jo
(c)
2
Exercises
In Exercises I-6, calculate du.
= x3 -
SOLUTION
x2
Let u =
x3 -
x2. Then du = (3x2 - 2x) dx.
3. u = cos(x2)
SOLUTION
5.
U
Let u = cos(x2). Then du= -sin(x2) · 2xdx = -2xsin(x2)dx.
= e4x+I
SOLUTION
Let u = e4x+I. Then du= 4e4x+I dx.
In Exercises 7-24, write the integral in terms of u and du. Then evaluate.
7. f<x+8) 4 dx,
SOLUTION
3
I
1
With the substitution u =x3 + 1, the definite integral ~ x2(.x3 + 1) dx becomes t
is {c).
1. u
!
u=x+8
Let u = x + 8. Then du = dx. Hence,
4
f (x + 8) dx
=
I
4
u du=
~u5 + C = ~(x + 8)5 + C
9
udu
Ji9 u du. The correct answer
s E c TI o N
9.
f
Let u
u.
f
= 3t -
4. Then du
= 3 dt and
t...fi2+1 dt, u = t2 + 1
Let u =
SOLUTION
t2 + 1. Then du = 2t dt. Hence,
I ...fi2+i
t
13•
I
t3
dt
(4 - 2t4) 11 ,
Let u
SOLUTION
=4 -
u
=4 -
dt
= -21
x(x + 1)9 dx,
Let u
SOLUTION
u
-~
8
= x + 1. Then x = u -
1 and du
x(x + 1)9 dx =
f
x2.../4-xdx,
Let u
SOLUTION
=4 -
u- 11 du= _!_u- 10 + C = _!_(4 - 2t4t 10 + C
80
80
= dx. Hence
f
1
- u 10 +C
(u
1 11
= -u
-
u and du
= -dx. Hence,
10
10
-
u9 )du
1
= -(x+
1) 11 -
11
1
-(x+ 1) 10 +C
10
u=4-x
x. Then x
=4 -
x2 V4- xdx = -
J
1 + 1)3' 2 + C
= -(t2
3
J
(u - l)u 9 du=
11
f
3
=x + 1
f
17.
1 3' 2 + C
u 112 du= -u
= -8t3 dt. Hence,
2t4 • Then du
t3_ _ dt =
(4 - 2t4) 11
f
I
2t4
f __
15.
The Substitution Method
u = 3t - 4
(3t _ 4)5 dt,
SOLUTION
I
5.7
=-
J (4- u) 2u 112 du= -
f
2 12
(16 - 8u + u )u ' du
J(16u112 -8u3/2 + us12)du
32
3
16
5
= -(16. ~u3/2 -
8.
~us12 + ~u112) +C
2
7
= --(4 - x) 312 + -(4 - x) 512 - -(4 - x)7' 2 + C
19.
f
sinBcos 3 BdB,
Let u
SOLUTION
u=cose
= cos B. Then du = - sin Bde. Hence,
f
21.
J
xe-x2 dx,
Let u
SOLUTION
23.
J
SOLUTION
f
3
u du=
-~u4 + C = -~ cos4 (J + C
-x2
= -x2. Then du= -2xdx or-~ du= xdx. Hence,
(lnx;2dx,
Let u
u=
3
sinBcos Ode= -
u = lnx
= ln x. Then du = ~ dx, and
f
2
(lnx)
--dx =
x
I
u2 du
1 3
1
3
= -u
+ C = -(lnx)
+C
3
3
375
376
CH APTER 5
I
INTEGRATION
Jn Exercises 25-28, evaluate the integral in the form a sin(u(x)) + C for an appropriate choice of u(x) and constant a.
25.
J
4
x3 cos(x ) dx
Let u = x4 • Then du= 4x3 dx or~ du= x3dx. Hence
SOLUTION
4
3
J x cos(x ) dx
27.
=~ J
cos u du
= ~ sin u + C = ~ sin(x4 ) + C
J
x 112 cos(x3i 2) dx
SOLUTION
= x312. Then du = ~ x 1i2 dx or ~ du = x1i2 dx. Hence
Let u
J
x 112 cos(x3 12 )dx
= ~3
f
cos udu
= ~3 sinu + C = ~3 sin(x3i2) + C
Jn Exercises 29-76, evaluate the indefinite integral.
9
29. Jc4x + 5) dx
SOLUTION
Let u = 4x + 5. Then du = 4 dx and
J
(4x+ 5)9 dx
31.
= -41
J
u9 du= -1u 10 + C
40
1
10 +C
= -(4x+5)
40
f-d_t_
vt+ 12
SOLUTION
Let u = t + 12. Then du= dt and
J
33.
I
_d_t_
vt+ 12
x+l
(x2 + 2x)3 dx
SOLUTION
Let u
= x2 + 2x. Then du = (2x + 2) dx or tdu = (x + 1) dx. Hence
J
x+1
1
(x2 + 2x)3 dx = 2
35.
= Ju- 112 du= 2u 112 + C = 2-Vt+li + C
J
1
1 ( 1 -2)
1
u3 du= 2 -2u
+ C = -4(x2 + 2x)-2 + C
= 4(x2 -1
+ 2x)2
J -~dx
vx2 + 9
SOLUTION
Let u
= x2 + 9. Then du = 2x dx or tdu = x dx. Hence
J
x
- - - dx
vx2 + 9
= -l f - l du = -lyu
- + C = Vx2 + 9 + C
2
Vu
2 t
37. Jc2x3 - 7) 2 dx
SOLUTION
Multiplying out the integrand,
2
Jc2x3 - 7) dx
39.
= Jc4x 6 -
28x3 + 49) dx
= ~x1 -
7x4 + 49x +
c
J
SOLUTION
x2(2x3 - 7) 3 dx
Let u = 2x3 - 7. Then du=
J
6x2 dx. Hence,
3
x2(2x3 -7) dx
= !6 Ju 3 du=
2-u4 + C
24
= 2-c2x3
-7)4 + c
24
+C
s E c TI o N
41.
Let u = 3x + 8. Then du = 3 dx and
f
I
Let u = x3 + 1. Then du = 3x2 dx and
I
1 12 + C = -1 (3x + 8) 12 + C
u11 du= -u
36
36
x2.Yx3 + ldx = -1
3
I
3 2
2 3i 2 +C = -(x3+1)
2
u112 du = ,-u
1 +C
9
9
I (x~~)3
Let u = x + 5. Then du= dx and
SOLUTION
J
i(z3 +1) 12 dz
Let u = z3 + 1. Then du = 3z2 dz and
SOLUTION
I
49.
(3x + 8) 11 dx = -1
3
x2.../x3 + ldx
SOLUTION
47.
The Substitution Method
(3x + 8) 11 dx
I
45.
I
J
SOLUTION
43.
5.7
i(z3 + 1)12 dz= -1
3
I
1 13 + C = -(z
1 3 + 1) 13 + C
u12 du= -u
39
39
J
(x + 2)(x + 1) 114 dx
Let u = x + 1. Then x = u - 1, du = dx and
SOLUTION
f
(x + 2)(x + 1) 114 dx =
=
f
(u + l)u 1i 4 du=
4
4
+ -u5f4
+c
-u9f4
9
4
f
(u 514 + u 114 )du
5
4
= -(x + 1)9/4 + -(x + 1)5/4 + C
9
5
51.
J
sin(8 - 3B) dB
SOLUTION
Let u = 8 - 3B. Then du= -3 dB and
I
53.
sin(8 - 3B) dB = _
f
sin u du =
!3 cos u + C = !3 cos(8 -
30) + C
I co~
Vt dt
Let u = Vt=
SOLUTION
t 1i 2 • Then
I
55.
!3
du= ~r 1 i 2 dt and
cos
w
~ dt = 2
I
cos u du = 2 sin u + C = 2 sin Vt + C
J
SOLUTION
tan(49 + 9) dB
Let u =49 + 9. Then du
I
=4 de and
1
tan(49+9)d9= 4
I
tanudu= 1 Inlsecul+C= 1 Inisec(4B+9)l+C
4
4
377
378
CH APTER 5
I
INTEGRATION
57.
J
cotxdx
Let u = sin x. Then du = cos x dx, and
SOLUTION
I
59.
f
cosx
. - dx =
smx
J
sec
2
(;)
sec2(4x + 9)dx = 1
4
I
sec2 udu = 1 tanu + C = 1 tan(4x + 9) + C
4
4
Vx· Then du = 2 ~ dx or 2 du = }x dx. Hence,
I
2
sec ( -{i) dx
Vx
=2
I
sec2 u dx = 2 tan u + C = 2 tan( -y'X) + C
J
sin4x.Ycos4x+ldx
Let u = cos 4x + 1. Then du = -4 sin 4x or -idu = sin 4x. Hence
SOLUTION
J
sin4xYcos4x+ ldx = -~
J
u
112
du = -~
(~u 3 ' 2 )+c = -~(cos4x+ 1)3' 2 +C
J
seclJtanlJ(seclJ- l)dlJ
SOLUTION
Let u = sec e - 1. Then du = sec e tan e de and
J
seclJtanlJ(seclJ- l)dlJ =
67.
. xi + C
-du = In lul + C = In I sm
u
dx
Let u =
SOLUTION
65.
I
Let u = 4x + 9. Then du= 4dx or i du= dx. Hence
I
63.
I
sec2 (4x+9)dx
SOLUTION
61 •
cot x dx =
I
SOLUTION
J
udu =
~u2 + C = ~(sece- 1)2 + C
el4x-1 dx
Let u = l4x - 7. Then du = 14 dx or
I
f.i du = dx. Hence,
e14x-? dx = _.!._I e" du= _.!._e" + C = _.!._e14x-1 + C
14
14
14
exdx
69. I
(eX + 1)4
SOLUTION
Let u
=
ex + 1. Then du = ex dx, and
I
71 I
•
SOLUTION
ex
(ex+ 1)4 dx =
f-4u du= -
1
3u3
1
+ C = 3(ex + 1)3 + C
e'dt
e2'+2e'+l
Let u = e1• Then du = e' dt, and
I
e' dt
e2' + 2e' + 1 =
I
du
u2 + 2u + 1
=
I
du
1
1
(u + 1)2 = - u + 1 + C = - e' + 1 + C
s E c TI o N
73.
I
Let u = ln x. Then du = ~ dx, and
I
I
The Substitution Method
379
(lnx)4 dx
x
SOLUTION
75.
I
5.7
I
4
(lnx)
-x-dx=
u4 du=
1
5
5u
1
5
+C=:s(lnx) +C
tan(lnx) d
--- x
x
SOLUTION
Let u = cos(ln x). Then du = - ~ sin(ln x) dx or -du = ~ sin(ln x) dx. Hence,
I
77. Evaluate f
SOLUTION
(1
tan(ln
x) dx =
-x
dx
+ yx)3
Let u = 1 +
I
sin(ln x) dx = xcos(lnx)
using u = 1 +
I
-du = - ln lul + C = - ln Icos(ln x)I + C
u
yx. Hint: Show that dx = 2(u -
l)du.
Vx· Then
du=
2
1
_r;;.dx
or
dx=2../Xdu=2(u-l)du
yX
Hence,
I
dx
(1 + yx)3
= 2 f uu~ 1du=2f(u-2 -u-3)du
2
1
= -2u- 1 + u-2 + C = - - - - +
+C
1 + Vx (1 + yx)2
J
79. Evaluate sin x cos x dx using substitution in two different ways: first using u = sin x and then using u = cos x.
Reconcile the two different answers.
SOLUTION
First, let u = sin x. Then du = cos x dx and
Next, let u = cos x. Then du = - sin x dx or -du = sin x dx. Hence,
f sinxcosxdx = - f udu =
-~u2 + C2 = -~ cos2 x + C2
To reconcile these two seemingly different answers, recall that any two antiderivatives of a specified function differ by a
constant. To show that this is true here, note that(~ sin2 x + C 1) - (-~ cos2 x + C2 ) = ~ + C 1 - C2 , a constant. Here we
used the trigonometric identity sin2 x + cos2 x = 1.
81. What are the new limits of integration if we apply the substitution u = 3x + n to the integral
SOLUTION
fo sin(3x + n) dx?
The new limits of integration are u(O) = 3 · 0 + 7r: = n and u(n) = 3n + n = 4n.
In Exercises 83-84, compute the definite integral two ways:
• Multiply out the integrand and then integrate directly without a substitution.
• Use the Change of Variables Formula with the substitution provided.
83. [
(2x - 3)3 dx,
u
= 2x -
3
SOLUTION
• Multiplying out the integrand
[
(2x - 3) 3 dx =
[(sx3 -
= 2(4)4
-
36x2 + 54x - 27)dx = {2x4
12(4)3 + 27(4) 2
-
-
12x3 + 27x2 - 27x) [
27(4) - (2 - 12 + 27 - 27) = 78
• Let u = 2x - 3. Then du= 2dx. Furthermore, when x = l, u = -1, and when x = 4, u = 5. Hence,
[
1
11
5
(2x- 3)3 dx = 2
-1
1
5
1 4
u3 du= -u4 1 = -(5)
8 -1 8
-
4 = 78
1
-(-1)
8
380
CH APTER 5
I
INTEGRATION
In Exercises 85-98, use the Change of Variables Formula to evaluate the definite integral.
85. (2
Jo
dx
.Y2x + 5
Let u = 2x + 5. Then du = 2 dx or ~du = dx. Hence
SOLUTION
87.
L1
(x2: 1)3 dx
Let u = x2 + 1. Then du = 2x dx or
SOLUTION
r1
Jo (x
2
x
dx = !
+ 1)3
2
i du = x dx. Hence
r2 .!._du=
! (-!u- )j
u
2 2
3
JI
1
25
.Yx2 + 9dx = -1
29
[0
LI
16
4
~
= 0.1875
16
Let u = x2 + 2x. Then du= (2x + 2)dx = 2(x + l)dx, and
I
L
0
93. [
25
2 312 )1 = -(125
1
-Yudu = -1 ( -u
- 27) = -98
23
9
3
3
(x + l)(x2 + 2x)5 dx
SOLUTION
(x + l)(x2 + 2x) 5 dx = -1
2
L
3
0
1
3
1 6 = -729 = -243
u5 du= -u
12 0
12
4
lnx dx
X
I
Let u = ln x. Then du = ~ dx, and
SOLUTION
dx =
[ I Inx
X
LI
o
("/2
SOLUTION
II
!u2 = !(1)2- !(0)2 = !
2 0 2
2
2
Let u = cose2. Then du= -2Bsine2 dB or -~du= Bsine2 dB. Hence,
LI
Jo
(1 udu =
Jo
Btan(e2)dB
SOLUTION
97.
_2_ + ! =
Let u = x2 + 9. Then du = 2x dx or ~ du = x dx. Hence
SOLUTION
95.
=
I
xVx2 + 9dx
89. [
91.
2
2
Btan(e2)dB=
LI --m-dB=-esin(e2)
1 [osl du
1
lcosI =--[ln(cosl)+lnl]=-ln(secl)
1
1
-=--lnlul
o cos(u-)
2
u
1
2
I
cos 3 xsinxdx
Let u = cos x. Then du = - sin x dx. Hence
L"
12
o
L ~5
cos 3 xsinxdx=-
1 LI
u3 du=
I
O
1 1 1
1
u3 du= -u4 =--0=4o
4
1 4
2
99. Evaluate
SOLUTION
2
r
- -./4- r2 dr.
Let u = 5 - .Y4 ~ r 2 • Then
rdr
rdr
du = - - - = - V4- r2 5 - u
2
s Ec T Io N
5. 7
I
The Substitution Method
381
so that
rdr
= (5 -
u)du
Hence, the integral becomes:
fo r~52
V4-r 2 dr =ls vu(5-u)du =ls (5u 112
-
u312 ) du= (13° u3' 2 -
~us 12)[
101. Wind engineers have found that wind speed v (in meters per second) at a given location follows a Rayleigh distribution of the type
W(v)
164
= 2-ve-v2
32
This means that at a given moment in time, the probability that v lies between a and b is equal to the shaded area in
Figure 3.
2
(a) Show that the probability that v e [O, b] is 1 - e-h t64 •
(b) Calculate the probability that v e [2, 5].
y
0.1
0.05
a
b
20
FIGURE 3 The shaded area is the probability that v lies between a and b.
SOLUTION
(a) The probability that v
E
[O, b] is
Let u = -v2 /64. Then du= -v/32dv and
I
0
1
-ve-v2t64 dv
32
=-
[h2
/64
e" du= -e"
0
i-h2 /64 = -e-b2f64 + 1
0
(b) The probability that v E [2, 5] is the probability that v
probability that v e [2, 5] is ·
E
[0, 5] minus the probability that v e [O, 2]. By part (a), the
In Exercises 103-106, use substitution to evaluate the integral in terms of f(x).
103.
J
f(x) 3 f'(x)dx
SOLUTION
Let u
= f(x). Then du= f'(x)dx. Hence
J
3
f(x) J'(x)dx
105 Jf'(x) d
'
SOLUTION
f(x)
=
J
3
u du=
~u4 + C = ~f(x)4 + C
x
Let u = f(x). Then du= f'(x)dx, and
I
f'(x)
f(x) dx
=
I
1
~ du
= ln Jul + C = In lf(x)J + c
382
CH APTER 5
I
INTEGRATION
107. Show that
l
7r/6
f(sinO)dO=
ll/2
o
SOLUTION
o
1
f(u) _,,--;;du.
vl - u2
Let u = sin(). Then u(lf /6) = 1/2 and u(O) = 0, as required. Furthermore, du = cos() d(), so that
du
cos()
d()=-
If sine= u, then u2 + cos2 e = 1, so that cos()= Vl - u2 • Therefore de= du/ Vl - u2 • This gives
/6
[
f(sinO)dO
o
=
ll/2
o
1
f(u) .,,.----:; du
vl - u2
Further Insights and Challenges
~
M
109. Evaluate I =
. Hint: Use substitution to show that I is equal to J
lf"/2
l+tan6000 ()
0
Jo
check that I + J =
SOLUTION
=
[P
0
M
n
1 +cot6000 u
and then
d().
To evaluate
I - [12 _ _
d_x__
0
1 + tan6000 x
we substitute t = lf/2 - x. Then dt = -dx, x = 7r/2 - t, t(O) = lf/2, and t(7r/2) = 0. Hence,
1
Let J =
("
12
dx
{'>
dt
[
1 + tan6000 x = - )"12 1 + tan6000 (1f /2 - t) =
Jo
=
12
o
dt
1 + cot6000 t
J;12 1 +cotd~ooo (t) . We know I= J, so I+ J = 2/. On the other hand, by the definition of I and J and the linearity
of the integral,
I+J=
l "/
L"/
2
0
-
-
dx
dx
+
=
1 + tan 6000 x
1 + cot6000 x
2
(
o
-I/2(
-
0
l"
12
-
=
l"/
(
0
1
1 + tan6000 x
+
1
) dx
1 + cot6000 x
1
1
) dx
+
1 + tan6000 x
1 + (1 / tan6000 x)
1
+
1
)dx
1 + tan6000 x
(tan6000 x + 1)/ tan6000 x
0
6000
1
+ tan 6000x ) dx
6000
1 + tan
x 1 + tan
x
0
1 + tan6000
dx=
1 + tan6000 x
(
2
l "/2 (
x)
l"/2
ldx=lf/2
0
Hence, I+ J = 2/ = lf/2, so I= lf/4.
L
b
111. Prove that
a
1
- dx =
X
[fa
I
1
- dx for a, b > 0. Then show that the regions under the hyperbola over the intervals
X
[1, 2], [2, 4], [4, 8], ... all have the same area (Figure 4).
y
I
I
2
4'-.......
!
8
'--+--+--'--+-----.:...:-.....;...-1--x
2
4
8
FIGURE 4 The area undery =~over [2n,2n+ 1] is the same for all n = 0, 1,2, ....
s E CT I O N
5.8
I
Further Integral Formulas
383
SOLUTION
(a) Let u
= ~·Then au= x and du=
~ dx or adu
1
[ a -X dx
Note that
[/a ~ = [la ~
du
l
U
= dx. Hence
=
lb/a -
a
du
au
I
=
[/a -
1
U
I
du
= u.
dx after the substitution x
X
I
(b) The area under the hyperbola over the interval [l, 2] is given by the definite integral Ji2 ~ dx. Denote this definite
integral by A. Using the result from part (a), we find the area under the hyperbola over the interval [2, 4] is
(4 ~ dx = [
J2
x
12
~ dx = f ~ dx = A
l x
J1 x
2
Similarly, the area under the hyperbola over the interval [4, 8] is
14
2
[ 4 ! dx = [ I ! dx = J1f ! dx = A
X
X
X
zn+ 1]
In general, the area under the hyperbola over the interval [2n,
2n+l 1
L
2•
113. Area of an Ellipse
is
l2n+l /2" 1
-dx=
X
-dx=
I
12
1
-dx=A
l
X
X
Prove the formula A = 7rab for the area of the ellipse with equation (Figure 6)
x2
y2
a2+b2=l
Hint: Use a change of variables to show that A is equal to ab times the area of the unit circle.
y
b
a
-a
-b
FIGURE 6 Graph of
SOLUTION
Consider the ellipse with equation ~ + ~
x2 y2
a2 + b2
= 1.
= 1; here a, b > 0. The area between the part of the ellipse in the
J:
f(x) dx. By symmetry, the part of the elliptical region in
upper half-plane, y = f(x) = ~b2 ( 1 - ~),and the x-axis is
the lower half-plane has the same area. Accordingly, the area enclosed by the ellipse is
Now, let u
= x/a. Then x = au and a du = dx. Accordingly,
2b [
Here we recognized that
2(*) ~·
=
f
1
~1-(~}2 dx = 2ab
I'.
Vl -u2 du
Vl - u2 du represents the area of the upper unit semicircular disk, which by Exercise 112 is
5.8 Further Integral Formulas
Preliminary Questions
1. Find b such that [
l
(a) In 3
= 2abG) =7rab
dx is equal to
x
(b) 3
384
CH APTER 5
I
INTEGRATION
For b > 0,
SOLUTION
dx
lb
[ I -X =lnlxl I =lnb-lnl =lnb
(a) For the value of the definite integral to equal In 3, we must have b = 3.
(b) For the value of the definite integral to equal 3, we must have b = e3 •
2. Find b such that
J:
1
~xx2 = ~ ·
In general,
SOLUTION
[
-dx-2 = tan- 1 x lb = tan- 1 b - tan- 1 0 = tan- 1 b
1 +x
o
0
For the value of the definite integral to equal
~,
we must have
tan- 1 b = ~
3
or b =tan~ =
3
3. Which integral should be evaluated using substitution?
9dx
(a)
1 + x2
I
(b)
I
V3
dx
1 +9x2
Use the substitution u = 3x on the integral in (b).
SOLUTION
4. Which relation between x and u yields V16 + x2 = 4 Vl + u 2 ?
To transform V16 + x2 into 4 Vl + u2 , make the substitution x = 4u.
SOLUTION
Exercises
In Exercises 1-12, evaluate the definite integral.
l.
(9
dx
x
J1
9
SOLUTION
3
3.
1
I
SOLUTION
•
l2
l
1!
3
t
3
dt
=
In itl1•
=
Let u
2
"3
1
I
SOLUTION
3
=
1;2
1
= -31 [
-du
JO
U
40
= -31 In lul 1
IO
f,
"3
dx
-1--2
+X
=
tan- 1 x
I
V'S
=
I
SOLUTION
7r
dx
--x2
l
l/2
dx
.
- - - =Sill
o
-../1 - xz
-I
11/2
x
• -
1
=Sill I -
o
2xdx
3
1
o 2x dx
=
2x 13
23
20
In 2 o = ln 2 - In 2
=
.
- Sill-I
2
3
SOLUTION
= -1 (ln 40 - In 10) = -1 ln 4
3
3
7r
tan- 1( VJ)- tan- 1(1) = - - 3 4
v1
o
11.1
- dt
3t+ 4
dx
-x2 + 1
I
•
In I
= 3t + 4. Then du = 3 dt and
1
9
-
dt
--
12
•
In e3
I
3t + 4
2
SOLUTION
7
In 1 = In 9
1
-dt
t
I
5
9
! dx = In lxll = In 9 J1 X
I
(
7
In 2
0=
7r
-
6
= -
7r
12
S E CT I O N 5.8
I
Further Integral Formulas
13. Use the substitution u = x/3 to prove
!tan- ~ +C
9+x
3
3
I ~=
1
2
Let u = x/3. Then, x = 3u, dx = 3 du, 9 + x2 = 9(1 + u2), and
SOLUTION
I
dx
9 + x2 =
I
3du
1
9(1 + u2) = 3
I
du
1
_1
1
_1 x
1 + u2 = 3 tan u + C = 3 tan 3 + C
In Exercises 15-34, calculate the integral.
15.
I
4x24+ 1 dx
Let u = 2x. Then x = u/2, dx = ~du, 4x2 + 1 = u2 + 1, and
SOLUTION
I
3
17.
4
x + dx = 2
4 2 1
dx
1
x +
0
1
dx
=
L~
x +3
_r;;
v3
o
dt
-~ =
v1-1m~
I
du
1 . 1
1
-~ = - sm- u+C= -sin- 1(4t)+C
4
4vl-u2
4
dt
3r2
Let t = Y5!3u. Then dt =
I
I
dt
v5 - 3t2 =
I vs~
1
Y5f3du
Let u = 2x. Then du = 2 dx and
I
I
Y5i3 du and
=
Y3
I Vl -
1
du
u2 =
. _1
Y3 sm
u+c =
1
. _
Y3 sm
{3
I '/
dx
xv12x2 -3
SOLUTION
x
V dx2
12x
-
3
1
= _r;;
I
v3
du
1
_1
1
-~ = _r;; sec u +C = - sec- 1(2x) + C
u vu 2 - 1
v3
Y3
dx
xvx4 - 1
SOLUTION
Let u = x2. Then du = 2x dx, and
I
27
I
du
1
'13
1
:rr
= _r;;tan- 1 u
= _r;;(tan- 1 .../3-tan- 10)= _r;;
u +1
v3
o
v3
3 v3
- 2-
Let u = 4t. Then du = 4 dt, and
I -vs -
SOLUTION
25.
o
dt
I
23.
L-fJ
.Y1 - 16t2
SOLUTION
21.
Y3 du and
Let x = Y3u. Then dx =
3
I
1
1
u2 ~ du = 2 tan- u + C = 2 tan- 2x + c
1
~3
SOLUTION
19.
I
dx
-~ =
xvr-1
I
du
1
1
1
~ = - sec- u+C = -sec- 1x2 +C
2
2u u2-1
2
r
exdx
• J-102 1 + e2
X
SOLUTION
Let u
=~-Then
du= exdx, and
[
ex dx
- -- =
In2 1 + e2x
II
du
-I 11
:rr
- - =tan u
= - - tan- 1(1/2)
1/2 1 + U 2
1/2 4
5t + c
385
386
'CH APTER 5
I
INTEGRATION
29.
tan- 1 xdx
1
+x2
I
SOLUTION
31.
l
Let u
dx
= tan- 1 x. Then du = -1--2
, and
+x
I
tan-' xdx
1 + x2
\log., 3
1
=
f
udu
= !u2+ C = (tan-' x)2
2
+C
2
log.,(3)
4x dx
0
4x
{1og.,(3)
SOLUTION
33.
I
Jo
4x dx
= ln4 o = In4 (3 -
2
1)
1
= In4 = In2
9x sin(9x) dx
SOLUTION
Let u = 9x. Then du = 9x In 9 dx and
I
9x sin(9x) dx
1
= -In9
- I
sin u du
1
cos u + C = - - - cos(9x) + C
= __!.__
In 9
In9
In Exercises 35-74, evaluate the integral using the methods covered in the text so far.
35.
f
ye2 dy
SOLUTION
Use the substitution u = y 2, du
= 2y dy. Then
I ye2 dy
37. I
Let u = 4x2 + 9. Then du= 8xdx and
--==x=-_ dx = ! Ju- 112 du= !u 112 + C = ! v4x2 + 9 + C
4
4
7"
7-x
Let u = -x. Then du= -dx and
I
I
s
v4x2 + 9
rxdx
SOLUTION
41.
= ~e" + C = ~j1 + C
xdx
I
f
e" du
v4x2 + 9
SOLUTION
39.
=~ I
7
sec 2 Otan
SOLUTION
Let u
rx dx
=-
I
7" du
= - In 7 + C = - In 7 + C
()d()
= tan(}. Then du = sec2 (} d(} and
2
I sec Otan7 ()d() =I u1 du=
~us+ C = ~tans()+ C
43. J 7 d w
SOLUTION
Let u = 1 - In w. Then du= _lw dw and
I
Yl-lnw
- - - - dw
w
=-
I
2
u112 du= --u312 + C
3
2
= --(1In w) 3' 2 + C
3
s Ec T I o N
SOLUTION
=7 -
Let u
= -2t dt and
t2. Then du
I -v1--- = -- I
1
2
t dt
t2
47.
u- 112 du= -u 112 + C
=-
5.8
I
Further Integral Formulas
_
r;:;--:;;
v7 - t2 + C
(3x+2)dx
x2 + 4
I
SOLUTION
Write
I
In the first integral, let u
=
(3x + 2) dx
x 2 +4
I
3x dx +
x 2 +4
I
2 dx
x 2 +4
= x2 + 4. Then du = 2x dx and
I - - = - J=
=
J J
3
2
3x dx
x 2 +4
For the second integral, let x
2u. Then dx
du
3
2
-
u
-lnJul+C1
3
= -ln(x2+4)+C
1
2
2 du and
2dx
xz + =
4
du
uz +
1
= tan
_1
-1
u + C2 = tan (x/2) + C2
Combining these two results yields
I
49.
I
(3x+2)dx
xz +
4
dx
v9-4x2
SOLUTION
Let u = 4x. Then du = 4 dx and
I Yl -
-;:::=d=x=::::;:
51.
f
3
= 2 ln(x2 + 4) + tan- 1(x/2) + C
16x2
1
= -4
I -~ = - sin~
v 1 - u2
1
4
1 u + C = ! sin- 1 (4x) + C
4
<e-x - 4x)dx
SOLUTION
First, observe that
In the remaining integral, use the substitution u
I
= -x, du = -dx. Then
e-x dx = -
I
eu du = -eu + C = -e-x + C
Finally,
f
53.
I
e2x -e4x
eX
<e-x - 4x)dx = -e-x - 2x2 + C
dx
SOLUTION
J(
e2x
:X e4x) dx=
J
3x
(ex-e 3x)dx=ex_e +C
3
387
388
CH APT E R 5
I
INTEGRATION
55•
I
(x + 5)dx
v4-x2
SOLUTION
Write
I
f---2!!!_2
(x + 5)dx =I~+
v4-x2
v4-x2
v4-x
In the first integral, let u = 4 - x2. Then du = - 2x dx and
=-!Ju2
I~
v4-x2
112
du=-u 112 +C 1 =-V4-x2 +C1
In the second integral, let x = 2u. Then dx = 2 du and
=5f -~ =5sin- u+C2 =5sin- (x/2)+C2
I -~
v4-x2
vl -u
1
1
2
.
Combining these two results yields
f
_(x_,+=-S)_d_x = -V4-x2 + 5 sin- 1(x/2) + C
v4-x2
57. Jex cos( ex) dx
SOLUTION
Use the substitution u = ex' du = ex dx. Then
f
59.
I
ex cos(ex) dx =
f
cos u du = sin u + C = sin(ex) + C
dx
v9 - 16x2
SOLUTION First rewrite
I
Now, let u
f
=3
J___!!!!:____
=
dx
_
(~x)2
= ~x. Then du=~ dx and
I
61.
lf ~l
dx
V9 - 16x2
!
dx
=
--/9 - 16x2
4
--/1 - u2
! sin4
1
u+C =
! sin- 1( 4x) + C
4
ex(e2x + 1)3 dx
SOLUTION
Use the substitution u = eX, du = ex dx. Then
e7x 3e5x
= - ++ e3x +ex+ C
7
63.
I
5
X2dx
x3 + 2
SOLUTION
Let u =
x3 + 2. Then du = 3x2 dx, and
J
x2dx
x3 + 2 =
lfdu
3
1
--; = 3 In ix3 + 21 + C
65. J cotxdx
SOLUTION
We rewrite Jcotxdx as J c?sx
smx dx. Let u = sinx. Then du= cosxdx, and
I
cosx
-.-dx=
smx
I
du
- =lnlsinxl+C
u
3
sEcT IoN
5.8
I
Further Integral Formulas
389
I
4lnx+5d
x
x
SOLUTION Let u = 4lnx + 5. Then du= (4/x)dx, and
67.
I
. 69.
f
4lnx+5
- - d x = -1
x
4
Let u =
x2. Then du = 2x dx, and
I
f
1 2 +C = -(4lnx+5)
1
2
udu= -u
+C
8
8
x3x2 dx
SOLUTION
71.
I
1
x3x2 dx= 2
I
3"du
1 3·
3x2
=- +C = - +C
2 ln3
2ln3
cotxln(sinx)dx
SOLUTION
Let u = ln(sinx). Then
1
du=-.-· cosxdx
smx
=cotxdx
and
I
73.
cotxln(sinx)dx=
I
u2
(ln(sinx)) 2
udu=2+C=
+C
2
f i2~dt
SOLUTION
Let u = t - 3. Then t = u + 3, du = dt and
J
i2~dt= J<u+3) 2 vudu
=
f
(u
2
2
+ 6u + 9) vu du= J (u 512 + 6u 312 + 9u 11 )du
= ~u112 + 12 us12 + 6u3/2 + C
7
5
2
12
= -(t - 3)712 + -(t - 3)512 + 6(t - 3)312 + c
7
5
75. Use Figure 4 to prove
x
L
0
1
1
~dt = -xvl -x2 + -sin- 1 x
2
2
y
I
x
FIGURE 4
SOLUTION
The definite integral
Lx ~
dt represents the area of the region under the upper half of the unit circle
from 0 to x. The region consists of a sector of the circle and a right triangle. The sector has a central angle of ~ - (}, where
cos (} = x. Hence, the sector has an area of
2
!(1)
(?:2 - cos- 1 x) = !2 sin- 1 x
2
The right triangle has a base of length x, a height of v 1 - x2 , and hence an area of ~xv 1 - x2 • Thus,
L
x
0
1
1
~dt = -xvl -x2 + -sin- 1 x
2
2
390
CH APT E R 5
I
INTEGRATION
77. Prove
J
sin- 1 ldl
SOLUTION
= ~ + lsin- 1 l + C
Let G(l) = ~ + lsin- 1 l. Then
G'(t)
= _ddt ~ + !!_
(1sin- 1 1) = -~ + (1 · _dd sin- 1 l+sin- 1 1)
dt
v 1 _ 12
t
-t
t
. -I )
. -I
= _r.---::;
+ ( _,,.----;;
+ sm
t = sm
t
2
vl-t2
This proves the formula
J
sin- 1 tdt =
vl-t
~ + tsin- 1 t + C.
Further Insights and Challenges
79. Recall that if f(l)
~
g(t)
fort~
~
0, then for all x
lx
0,
J(t) dt ;?:
lx
g(t) dt
The inequality e' ~ 1 holds fort~ 0 because e > I. Use (2) to prove that ex~ I + x for x
integration, the following inequalities (for x ~ 0):
ex> 1 + X +
-
SOLUTION
Integrating both sides of the inequality e'
~
1
~
0. Then prove, by successive
1
-x2
+ -.x3
2
6
1 yields
Integrating both sides of this new inequality then gives
~ x + x2 /2
or
ex
~
~ x + x2 /2 + .x3 /6
or
ex
~ I + x + x2 /2 + .x3 /6
Lx e' dt = ex - I
I + x + x2 /2
Finally, integrating both sides again gives
Lx e' dt = ex - I
as requested.
81. Use Exercise 79 to show that ex /x2 ~ x/6 and conclude that lim ex /x2
x->oo
generally that lim ex I .x" = oo for all n.
= oo. Then use Exercise 80 to prove more
X->OO
SOLUTION
By Exercise 79, ex~ I+ x +
4+ t· Thus
ex
1
llxx
->-+-+-+->x2- x2 x 2 6-6
Since lim x/6
X---tOO
= oo, lim ~I x2 = oo. More generally, by Exercise 80,
X-+00
x2
.x"+I
ex> I + - + · · · + - - 2
(n + l)!
Thus
~
I
x
x
->-+···+--->--xn - x"
(n + l)! - (n + I)!
.
Iim -x(
.r -- oo.
Smce
IJ' -- oo, l"1m x;;X--tOO n+ .
X-+OO
Exercises 82-84 develop an elegant approach to the exponential and logarithm functions. Define a function G(x) for
x > 0:
G(x) =
r~dt
I
t
s EcT ION
83. Defining tr
5.8
I
Further Integral Formulas
391
Use Exercise 82 to prove the following statements:
(a) G has an inverse with domain Rand range {x: x > 0}. Denote the inverse by F.
(b) F(x + y)
(c) F(r)
= F(x)F(y) for all x,y. Hint: It suffices to show that G(F(x)F(y)) = G(F(x + y)).
=E' for all numbers. In particular, F(O) = 1.
(d) F'(x) = F(x). Hint: Use the formula for the derivative of an inverse function that appears in Exercise 28 of the
Chapter 3 Review Exercises.
This shows that E = e and F(x) =ex as defined in the text.
SOLUTION
(a) The domain of G(x) is x > 0 and, by part (i) of the previous exercise, the range of G(x) is R. Now,
1
G'(x) = - > 0
x
for all x > 0. Thus, G(x) is increasing on its domain, which implies that G(x) has an inverse. The domain of the inverse
is Rand the range is {x: x > 0). Let F(x) denote the inverse of G(x).
(b) Let x and y be real numbers and suppose that x
using part (b) of the previous exercise
F(x + y)
= G(w) and y = G(z) for some positive real numbers wand z. Then,
=F(G(w) + G(z)) =F(G(wz)) =wz =F(x) + F(y)
(c) Let r be any real number. By part (k) of the previous exercise, G(E')
follows that F(r) = E'.
(d) By the formula for the derivative of an inverse function
85. The formula
= r. By definition of an inverse function, it then
x"+I
J
x" dx
= n + 1 + C is valid for n * -1. Show that the exceptional case n = -1 is a limit of the general
case by applying L'Hopital's Rule to the limit on the left:
lim
rx t' dt = J1rx t- dt
1
n--.-1J1
(for fixed x > 0)
x"+I - 1
Note that the integral on the left is equal t o - - - .
n+l
SOLUTION
. Ix
Jim
n--.-1
1
.
rn+l
t'dt= lim - n--.-1 n + 1
Ix = bm
. (x"+I
-n-->-1
n+1
1n+I )
-n+1
x"+l - 1
=Jim - - - = lim(x"+ 1)lnx=lnx=
Ix
n-->-1
I
n-->-1
1
n+1
t- 1 dt
Note that when using L'Hopital's Rule in the second line, we need to differentiate with respect to n.
87.
~
~ (a)
r1"" eY dy.
Explain why the shaded region in Figure 5 has area Jo
(b) Prove the formula
(c) Conclude that
I
I
r1"" eY dy.
lnxdx =a Ina- Jo
1
lnxdx =a Ina - a+ I.
(d) Use the result of (a) to find an antiderivative of y = In x.
y
y=lnx
FIGURE 5
392
CHAPTER 5
I
INTEGRATION
SOLUTION
(a) Interpreting the graph with y as the independent variable, we see that the function is x
gives the area of the shaded region as fo'"a eY dy.
= eY. Integrating in y then
(b) We can obtain the area under the graph of y = In x from x = I to x = a by computing the area of the rectangle
extending from x = 0 to x = a horizontally and from y = 0 to y = In a vertically and then subtracting the area of the
shaded region. This yields
[
Inxdx =a Ina -
1
r••a eY dy
Jo
(c) By direct calculation
Jo{Ina eY dy = eY 'Ina = a 0
I
Thus,
Inxdx =a Ina - (a -1) =a Ina - a+ 1
[
(d) Based on these results it appears that
f
Inxdx = xlnx-x + C
CHAPTER REVIEW EXERCISES
In Exercises I-4, refer to the function f whose graph is shown in Figure 1.
2
3
FIGURE 1
1. Estimate L 4 and
SOLUTION
M4
4
on [0, 4].
With n = 4 and an interval of [0, 4], ~x = 4 ~ 0 = 1. Then,
1
L 4 = ~x(f(O) + f(l) + f(2) + f(3)) = 1 ( 4 + 1 +
z5 + 2) = 423
and
3. Find an interval [a, b] on which R4 is larger than [
SOLUTION
In general, RN is larger than
f(x) dx. Do the same for L4 .
t f(x) dx on any interval [a, b] over which f(x) is increasing. Given the graph
of f(x), we may take [a, b] = [O, 2]. In order for L4 to be larger than
[a,b]. We may therefore take [a,b] = [2,3].
In Exercises 5--8, let f(x)
= x2 + 3x.
J: f(x) dx, f(x) must be decreasing over the interval
Chapter Review Exercises
S. Calculate R,. M 6 , aud
4
for f on the interval (2, 5]. Sketch the graph off and the corresponding rectangles for
each approximation.
SOLUTION
Let
f(x) = x' + 3x. A uniform partition of (2, 5] with N = 6 subintervals has
5 -2 1
Llx= -6- = 2'
Xj
= a+ j/lx = 2 + ~
aud
Now,
R,
=Llx tf(xj) =Ht(~)+ f(3) +tG)+ f(4) +t(~) +f(5))
=~2 (554 +18 +914 +28 +1354 +40) =6258
The rectangles corresponding to this approximation are shown below.
x
2.0
25
3.0 3.5 4.0
45
Next,
M,
=dx tf(xj) =~ (t(~) +t( 141) +t( 1}) +t( ~) +t( 1:) +t( ~))
= ~ ( 189 + 253 + 325
2
393
16
16
405 + 493 + 589) = 2254 = 1127
16 + 16
16
16
32
16
The rectangles corresponding to this ~pproximation are shown below.
y
35
30
25
20
IS
lO
x
2.0 2.5
3.0 3.5 4.0 4.5
Finally,
4= i1x ti<xj)
=
~ (1<2> + 1(~) + t<3) + 1G) + t<4> + 1(~))
55
91
135) =
505
=-I ( 10+-+18+-+28+2
4
4
4
8
The rectangles corresponding to this approximation are shown below.
y
35
30
25
20
15
lO
x
2.0 2.5
3.0 3.5 4.0 4.5
394
C H A PT E R 5
I
INTEGRATION
7. Find a formula for RN for f on [2, 5] and compute
J,'
f(x) dx by taking the limit.
SOLUTION
Let f(x) =
x2 + 3x on the interval [2, 5]. Then .6.x =
63
= 30+ N'
27
N
5-2
~ =
3
N and a = 2. Hence,
N
'[,j+ N3 L,f
fd
jo=]
(N
(N2 !!.)
!!.)
3
= 30 + 63
+
+ 27
+ N' +
N'22N33
26
141 45
9
=-+-+2
N
2N'
and
limRN= lim(141 + 45
N-+oo
2
N
N-+oc
+2:~)=
1v-
1241
9. Calculate R,, M,, and Ls for f(x) = (x' + 1)-1 on the interval [0, I].
SOLUTION
Let f(x) = (x' + 1)-1 • A uniform partition of [O, I] with N = 5 subintervals has
1-0 I
Llx=--=5
5'
.,
j
x·=a+11..;lx=
J
5-
and
Now,
= -1 (25
- + -25 + -25 + -25 + -I) ""0.733732
5 26 29 34 41 2
Next,
100
100
4 +100
++- +100)
- ""0.786231
= -I ( 5 IOI
109 5 149 181
Finally,
=
I(
S
25 25 25 25)
I+ 26 + 29 + 34 + 41 ""0.833732
11. Which approximation to the area is represented by the shaded rectangles in Figure 3? Compute R5 and L5 .
y
30-r---..,_
2
3
FIGURE 3
4
5
Chapter Review Exercises
395
There are five rectangles and the height of each is given by the function value at the right endpoint of the
subinterval. Thus, the area represented by the shaded rectangles is R5.
From the figure, we see that Ax = I. Then
SOLUTION
R5
= 1(30 + 18 + 6 + 6 + 30) = 90
and
Ls
= 1(30 + 30 + 18 + 6 + 6) = 90
In Exercises 13-16, express the limit as an integral (or multiple of an integral) and evaluate.
l3.
. 6N
" LJ
~ sm
. ("3 + 6N
1Tj)
,)i_;>!
j:=1
SOLUTION
= sin x and N be a positive integer. A uniform partition of the interval [.1l'/3, .1l'/2] with N subintervals
Let f(x)
has
Ax= _.::._
and
6N
for0:5j.:5N. Then
" f . ("3 + 1Tj)
~ f(x;) =RN
N = !ix L_,
N L_, sm
6 j::J
6
j=cd
consequently,
"
N
(" +7rj
1
1
lim-L::sin
- ) = [ " sinxdx=-cosx\"'' =0+-=6N i==I
3 6N
11/3
11/3
2 2
N--->""'
5 N
15. Jim - '\' ,/4 + 5j/N
N->00 N L..J
j::J
Let f(x) =
SOLUTION
Vx and N be a positive integer. A unifonn partition of the interval [4, 9] with N subintervals has
5
!ix= -
and
N
5j
X·=4+,
N
forO .:5 j .:5 N. Then
S
N
N
NL ,/4+5j/N= !ix Lf(x) =RN
j=l
consequently,
. -5
l!m
N
N---><X>
IN
,/4+5j/N=
_;...1
J
(4:i' -2x')dx
SOLUTION
19.
J
f
(4:i' - zx')dX = x4 -
~;i> + C
sin(O - 8) dO
SOLUTION
J
sin(O - 8) dO = - cos(O - 8) +
c
21. Jc4i-' - 12r- )dr
4
SOLUTION
23.
f
SOLUTION
f
(4C3 - l21 4 )dl = -21- 2 + 41-3 + C
sec2 x dx
f
2
sec x dx = tan x + C
r.
4
In Exercises 17-30, calculate the indefinite integral.
17.
j==l
I'
2 12 = -54 - -16 = -38
Viidx= -:i'
3
4
3
3
3
396
CH A PT E A 5
I
INTEGRATION
25.
J
4
(y+ 2) dy
J
(y + 2)4 dy =
SOLUTION
27.
f <e" -
x)dx
f<i'-x)dx=ex-~r+c
SOLUTION
29.
~(y + 2) 5 + C
J
4x- 1 dx
J
= 4 ln fx) + c
4x- 1 dx
SOLUTION
In Exercises 31-36, solve the differential equation with the given initial condition.
31. -dy
dx
=4x-,'
SOLUTION
y(I)
=4
Let ~ = 4x3. Then
y(x) =
J
4
4x' dx = x + C
Using the initial conditiony(l) = 4, we findy(l) = 14 + C = 4, so C = 3. Thus, y(x) = x4 + 3.
33.
~~ = x- 11',
SOLUTION
y(I)
=I
Let ~ = x- 112 • Then
y(x)
=
J
x- 112 dx::;: 2x 112 + c
Using the initial condition y(l) = I, we find y(l) = 2 VI+ C = I, so C = -1. Thus, y(x) = 2x112
-
I.
dy - _,
35. - -e ' y(O) = 3
dx
SOLUTION
Let ~ = e-x. Then
y(x) =
Using the initial condition y(O)
J
e-" dx = -e-• + C
= 3, we find y(O) = -e0 + C = 3, so C = 4. Thus, y(x) = 4 -
e-x.
37. Find/(I) if f"(t) =I -2t,f(O) = 2, and/'(0) = -1.
SOLUTION
Suppose f"(t) = 1 - 2t. Then
f'(t)= J f"(t)dt= J(l-2t)dt=t-t2+C
Usingtheinitialcondition/'(O) = -1, wefindf'(O) = 0-02 +C = -1, soC = -1. Thus,/'(t) = t-t2 - I.Now,
f(t) = J f'(t)dt = f<r-t2- l)dt=
Using the initial condition /(0) = 2, we find /(0) = 402 -
t<r) =
~t2- ~t'- t+C
!O' - 0 + C = 2, so C = 2. Thus,
!,,.
_!r
- 1+2
2
3
In Exercises 39-42, use the given substitution to evaluate the integral.
2
39.
1
SOLUTION
dt
- - , u=4t+l2
4t+ 12
Let u = 4t + 12. Then du= 4dt, and the new limits of integration are u = 12 and u = 20. Thus,
f
I'°
du
I
I
I
20
I 5
- dt
- = -I f , -=-lnu
=-(ln20-ln12)=-ln-=-lno 4t + 12
4 12 u
4
12
4
4
12 4 3
Chapter Review Exercises
16
41. [
sin xcos4 x dx,
u = cosx
Let u = cos x. Then du = - sin x dx and the new limits of integration are u = 1 and u =
SOLUTION
/6
I
sinxcos4 xdx=-
1/3/2. Thus,
J,Y3/2 u du
4
1
0
-~usl../312
=
5
=
I
~(15
9'321/3)
In Exercises 43-92, evaluate the integral.
43. J(20x4 -9x'-2x)dx
f<20x4 -9x'-2x)dx= 4x'-
SOLUTION
45.
f
<2x2 - 3x)2 dx
f
SOLUTION
47.
<2x2 -3x)' dx = f<4x4 -12x' + 9x2)dx = ~x' -3x4 + 3x' + c
J
3
x' :, x' dx
J
x'+3x' dx =
-x'--
SOLUTION
49.
~x4 -x2 +C
£
J
1 + x' + c
(x' + 3x')dx = 4x'
lx2-4ldx
SOLUTION
3
r'1x2-41dx= ('(x2-4)dx+ r\4-x')dx+J, (x'-4)dx
J-3
J_2
)_3
2
=(136 -3)+(136+136)+(-3+136)
46
=3
51. [
l2tJdt
SOLUTION
[
L2tJdt=
I
r~ L2tJdt+ i
l
=
r
/2
2dt+
I
=h
3/2
1I
!
2
2
Im
L2tJdt+
2
3dt+ J,5/2 4dt+
3/2
2
5/2
+Y +~
13/2
12
= (3-2) +(627
L2tJdt+
3/2
+~
I'
5/2
+&
!
[I'
L'
L2tJdt+
5/2
5dt+
~
Lm
L2tJdt+
3
l7/2 6dt+ !m 7dt
3
•
17/2
+U
~)+(10-8) +(15- 2i)+ (21-18)+ (28- ~)
£
7/2
L2tJdt
397
398
CH A PT EA 5
I
INTEGRATION
14
53. J(lor- 7) dt
Letu= IOt-7.Thendu= 10dtand
SOLUTION
I
55
(lOt- 7) 14 dt = -1
10
I
u 14 du= -1u 15 + C
150
1
=-(IOt
- 7)
150
15
+C
(2x' + 3x)dx
(3x4 + 9x 2 ) 5
· I
Let u = 3x' + 9x2. Then du= (!2x' + l&x)dx = 6(2x' + 3x)dx and
SOLUTION
(2x' + 3x)dx =~I _5 d = _..!.._ -4 C
(3x' + 9x 2 ) 5
6
u
u
24 u +
I
57. [
= _2._(3
24
4
x +9
x')-4
+
C
l5xVx+4dx
Let u = x + 4. Then x
SOLUTION
= u - 4, du = dx and the new limits of integration are u = 4 and u = 9. Thus,
L'
l5xvx+4dx= . [ l5(u-4)Vudu
151
9
=
= 15
(u
312
-4u
112
)du
(~u'P - ~u3f2 )[
4 6
= 15 (( :
-n)-(~ _~))
=506
J: cos(i(t+2))dt= ~sin(i(t+2))J:
1
SOLUTION
61.
Jt2
sec2 (9t' + I) dt
Let u
SOLUTION
= 9t3 + l. Then du = 27t2 dt and
I t2
63.
sec2 (9t' + l)dt = I
27
sec2 udu
= 27I tanu + C =
1 tan(9t3 + 1) + C
27
csc2 (9 - 28) dB
Let u = 9 - 28. Then du = -2 dB and
I
13
I
SOLUTION
2
sine
cos
Let u = cos 8. Then du = - sin BdB and when
f
f
'f
csc udu = -1 cotu + C = -1 cot(9 -28) + C
2
2
2
csc2 (9-W)dB = --
---ziJB dB
0
SOLUTION
67.
I
J
SOLUTION
65.
3
= - ;}
/3
sinB
J,112
cos2/J ()
1
---dB= -
e9 -2x dx
Let u = 9 - 2x. Then du
= -2 dx, and
(J
= 0, u = l and when B = j, u =
u- 213 du= -3u 113
11;2
1
= -3(2- 113
-
i. FinaJly,
3 {14
I)= 3 - - -
2
Chapter Review Exercises
69.
f x2e"
dx
Let u = x3. Then du = 3.x2 dx, and
SOLUTION
x2e" dx = !3 Je"du = !e"
+ C = !e" + C
3
3
J
71.
f "'10'
I
soLVTION
73
J
dx
,
e ICY dx =
I
x
(IOe)'
(IOe)'
IO'e'
(lOe) dx = ln(lOe) + C = In 10 + lne + C = In 10 + 1 + C
e-'dx
+ 2)3
•
(e-:c
Let u = e-x + 2. Then du = -e-x dx and
SOLUTION
J
_e-_'_d_x_
(e-x + 2)'
=-I
u-3du =_I_ +C =
1
+C
2u2
2(e-x + 2)2
75. [." tan 28 dO
r.
soLVTION
77
•
I
tan W dO = I In Isec 28/
/6
2
I'"
0
= I In 2
2
dt
t(l +(lnt)2 )
Let u = Int. Then, du = ~ dt and
SOLUTION
_ _d_t~~ = J-d_u_ =tan-I"+ C = tan-I(Jnt) + C
I + u'
J t(I +(In r)')
lnxdx
79• [
X
I
Let u = In x. Then du = ~ and the new limits of integration are u = In J = 0 and u == In e = I. Thus,
SOLUTION
81.
dx
4x' + 9
J
SOLUTION
!
Let u = ~. Then x = u, dx = ~ du, and
I
83.
J.12
4
SOLUTION
85.
dx
4x' + 9 =
I
~du
I
4 · ~u' + 9 = 6
I
du
I _1
I _1
u2 +I = 6tan u + C = 6tan
(2x}
3 +C
dx
xvx'-1
12
dx
J.
~ =sec-Ix
x x2-1
12
= sec- 1 12 - sec- 14
14
f' dx
J, x' +9
SOLUTION
Let u =
f. Then du = ¥, and the new limits of integration are u = 0 and u == 1. Thus,
399
400
C HAP TE R 5
I
INTEGRATION
J__!!!!.__
VI
87.
-x4
SOLUTION
.x:2. Then du = 2xdx, and -../ 1 - ..:t4 ==
Let u =
V1 -
u 2 • Thus,
dx
2x' + I
89. [
SOLUTION
1
91.
1
SOLUTION
Let u = Y2.x. Then du =
V2 dx. and the new limits of integration are u = 0 and u = 4 Vi. Thus,
(tan- 1 x) 3 dx
1 +x2
Let u = tan- 1 x. Then
1
du=--dx
I+ x'
and
l'
(ran-I x)J dx
-'-~~- =
1+.x2
0
[I'
u3 du =
0
93. Combine to write as a single integral:
{" f(x)dx+ [f(x)dx
Jo(' f(x)dx+ J_2
a
SOLUTION
First, rewrite
[
f(x)dx=
f
f(x)dx+ [
f(x)dx
and observe that
I
f(x)dx = - [ f(x)dx
Thus,
[
f(x) dx + [
f(x) dx = [
f(x) dx
Finally,
r' f(x)dx +
Jo
{" f(x)dx + (" f(x)dx = (" f(x)dx + {" f(x)dx = [,f(x)dx
J_2
Js
Jo
J_z
95• .Find the local minima, the local maxima. and the inflection points of A(x)
SOLUTION
Let
A(x)=
I'
3
tdt
t2 + I
Then
, ) =xA(x
x' + 1
-2
= f
)3
:dt .
t +1
Chapter Review Exercises
401
and
"
A (x)
=
(x'
+ 1)(1) -
I - x2
x(2x)
(x2 + 1)2
= (x' + I )2
Now, x = O is the only critical point of A~ because A"(O) > 0, it follows that A has a local minimum at x = 0. There are
no local maxima. Moreover, A is concave down for !xi > 1 and concave up for lxl < I. A therefore has inflection points at
x= ±l.
97. For the function
f
illustrated in Figure 6, do the following:
.£,"
(a) For x = o. I, 2, ... , IO, approximate A(x) = f(t)dt.
(b) For x = I, 2,3, ... , 9, approximate A'(x) using L\x = 1 and the symmetric difference quotient approximation,A'(x) ~
A(x+lil:)-A(x-.\x)
'""
(c) Plot the values of A'(x) on a graph off to demonstrate A' : : : : /.
y
...
8
I
4
.
·=
"
"
'
0
I
,JO
x
-4
-8
/
FIGURE 6
SOLUTION,
(a) First, note that A(O) = .£:' f(t) dt = 0. For other x, we can estimate A(x) by counting I x 2 squares (and partial
squares) between the graph and the x-axis from 0 to x. For example, there are roughly 6.9 squares between the graph and
the x-axis from 0 to I below the axis, so A(l) ~ -6.9. Continuing in this fashion, we obtain the approximate values of
A(x) listed in the table below.
x
0
I
2
3
4
5
6
7
8
9
10
A(x)
0
-6.9
-10.3
-10.3
-4.7
1.6
2.2
-2.6
-6.5
-4.7
-0.7
(b) For x = I,
and for x = 2,
A'( )"' A(3)-A(I) = -10.3-(-6.9)
2
2
2
-1.
7
Continuing in this fashion, we obtain the approximate values of A' (x) listed in the table below.
x
I
2
3
4
5
6
7
8
9
A'(x)
-5.15
-1.7
2.8
5.95
3.45
-2.l
-4.35
-1.05
2.9
(c) The values of A' are plotted together with the graph off below.
y
10
5
0 ,_.,,__..,._-+~~x
-5
-IO
99. On a typical day, a city consumes water at the rate of r(t) = 100 + 72t - 3t2 (in thousands of gallons per hour),
where t is the number of hours past midnight. What is the daily water consumption? How much water is consumed
between 6 PM and midnight?
SOLUTION
With a consumption rate of r(t) = 100 + 72t - 3t2 thousand gallons per hour, the daily consumption of water
is
f
(100+ 72t-3t2)dt
= (100t+36t2-t3)c = 100(24) +36(24f-(24)3 = 9312
402
CHAPTER
s I
INTEGRATION
or 9.312 million gallons. From 6 PM to midnight, the water consumption is
1
24 (100
8
+ 72t-3t')dt = (1001+ 36t'-t')
I"
IB
= 100(24) + 36(24)
2
-
(24) 3
-
(100(18) + 36(18)2
-
(18) 3)
= 9312 -7632 = 1680
or 1.68 million gallons.
101. Cost engineers at NASA have the task of projecting the cost P of major space projects. It has been found that the
cost C of developing a projection increases with Pat the rate dC/dP ~ 21p-0 ·65 , where C is in thousands of dollars and
P in millions of dollars. What is the cost of developing a projection for a project whose cost turns out to be P = $35
million?
Assuming it costs nothing to develop a projection for a project with a cost of $0, the cost of developing a
projection for a project whose cost turns out to be $35 million is
SOLUTION
l ,,
21P--0.05 dP
I"
= 601"'·35 = 60(35)0·35 "'208.245
0
0
or $208,245.
103. Evaluate
SOLUTION
I
s
x1sdx
8
3 +cos2 x
,
using the properties of odd functions.
"
Let f(x) = 3 ~os2 x and note that
x15
(-x)1'
f(-x)
= 3 + cos2 (-x) =
3 + cos2 x = -f(x)
Because f is an odd function and the interval -8 :$ x :$ 8 is symmetric about x = 0, it follows that
x 15 dx
--~--o
3+cos2 x -
£
8
105,
!GU/I
Plot the graph of f(x) = sinmx sinnx on [0,7r] for the pairs (m,n) = (2,4), (3,5) and in each case guess
the·va.Iue of I= [
f(x)dx. Experiment with a few more values (including two cases with m = n) and fonnulate a
conjecture for when I is zero.
The graphs of f(x) = sinm.xsinnx with (m,n) = (2,4) and (m, n) = (3,5) are shown below. It appears as if
the positive areas balance the negative areas, so we expect that
SOLUTION
I
=[
f(x) dx
=0
in these cases.
y
y
(2,4)
We arrive at the same conclusion for the cases (m, n) = (4, 1) and (m, n) = (5, 2).
y
y
(5, 2)
Chapter Review Exercises
However, when (m, n) = (3, 3) and when (m, n) = (5, 5), the value of
I= [
f(x)dx
is clearly not zero as there is no negative area.
y
y
0.5
I
l.5
2
2.5
3
0.511.522.53
--0.5
--0.5
(5,5)
(3, 3)
We therefore conjecture that I is zero whenever m -::/:. n.
107. Prove
and
SOLUTION
The function f(x) = 2.r is increasing, so 1 :5 x :5 2 implies that 2 = 2 1 :5
2
=
f
2dx :5
f
2x dx :5
f
zx :5 22 = 4. Consequently,
4dx = 4
On the other hand, the function f(x) = 3-x is decreasing, so 1 S x S 2 implies that
!9 = 3-2 <- 3-x <- 3-I = !3
It then follows that
2.
9
109. Find upper and lower bounds for
=
fo
f
f
f
2.dx <
3-xdx <
2.dx = 2.
19-1
-13
3
1
f(x) dx, for y = f(x) in Figure 7.
y=x2+1
y
y= f(x)
2
FIGURE 7
SOLUTION
From the figure, we see that the inequalities x2 + 1 S f(x)
s Vx + 1 hold for O :5 x :5 1. Because
and
5
3
it foHows that
34 5
l1
0
f(x)dx 5 5
3
403
404
C HA PT E R 5
I
INTEGRATION
In Exercises 110-115, find the derivative.
cost
111. A'(1'), where A(x) = [
SOLUTION
Let A(x) =
- - dt
1 +t
i
x cost
cosx
and
- - dt. Then A'(x) = - 1 +x
2 l+t
A'(1')
113. G'(x), where G(x)
1
COS1'
= -l + 7r = --l+rr
= J_,
t" i' dt
.,,
SOLUTION
Let G(x) =
[,
i' dt. Then
2
G'(x) = sin3 x!!._ sinx = Sin3 xcosx
dx
115. H'(I), where H(x) = ("
J4x2
9
SOLUTION
Let H(x) =
1
~ dt
t
-I dt = -
4_r!f
(I
- dt. Then
9
t
H'(x) =
and H'(I)
117.
Bx
4x
2
x
= -2.
l'°§J
SOLUTION
I d
4x2 dx
---4x2 = - -2 = --
Explain with a graph: If f is linear on [a, b], then the [
f(x)dx =
~(RN+ LN) for all N.
Consider the figure below. which displays a portion of the graph of a linear function.
y
---1-----~-~-x
The shaded rectangles represent the differences between the right-endpoint approximation RN and the left-endpoint approximation LN. In particular, the portion of each rectangle that lies below the graph of y = f(x) is the amount by which
LN underestimates the area under the graph, whereas the portion of each rectangle that lies above the graph of y = f(x)
is the amount by which RN overestimates the area. Because the graph of y = f(x) is a line, the lower portion of each
shaded rectangle is exactly the same size as the upper portion. Therefore, if we average LN and RN, the error in the two
approximations will exactly cancel, leaving
119. Let
F(x) = x
Vx2 -
I- 2
J" ~
dt
Prove that F(x) and y = cosh- 1 x differ by a constant by showing that they have the same derivative. Then prove they are
equal by evaluating both at x = 1.
SOLUTION
Let
F(x)=xVx'-1-2 [
~dt
Then
dF =
dx
Vx'-1 + _!:.__-2Vx'- I=
Vx'- I
_x'_ Vx' - I
Vx'-1
= -
1
-
Vx' -
I
Chapter Review Exercises
405
Also.1.;(cosh- 1x) = ~;therefore, F(x) and cosh- 1x have the same derivative. We conclude that F(x) and cosh-'x
differ by a constant:
F(x)=cosh- 1x+C
Now, let x = I. Because F(l) = 0 and cosh- 1 1 = 0, it follows that C = 0. Therefore,
F(x) = cosh- 1x
121. ""''
How can we interpret the quantity I in Eq. (2) if~ < b
~ O? Explain with a graph.
We will consider each term on the right-hand side of (2) separately. For convenience, let I, II.
denote the area of the similarly labeled region in the diagram below.
SOLUTION
y
-
f(b)
I
I
I
I Il
I
I
I
---1ii-----11v f<a!
b
a
Because b < 0, the expression bf(b) is the opposite of the area of the rectangle along the right; that is,
bf(b) =-II - IV
Similarly,
-af(a) = ill+ IV
and
- [
f(x)dx=-I-m
Therefore,
bf(b)- af(a)- [
f(x)dx =-I-II
that is, the opposite of the area of the shaded region shown below.
y
m, and IV