Applications of the Derivative - Penn Math

CHAPTER Applications of the
Derivative
4

4.1 Linear Approximation and Applications

Preliminary Questions

1. How large is the difference g(1) − g(1.2) (approximately) if g (1) = 4?

2. Estimate f (2.1), assuming that f (2) = 1, f (2) = 3.

3. Can you estimate f (2), assuming that f (2.1) = 4, f (2) = 1?

4. The instantaneous velocity of a train at a given instant is 110 ft/sec. How far does the train
travel during the next half-second (use the Linear Approximation)?

5. Discuss how the Linear Approximation makes the following statement more precise: the
sensitivity of the output to a small change in input depends on the derivative.

Exercises

In Exercises 1–6, calculate the actual change f and the Linear Approximation f (a) x.

1. x4, a = 2, x = .3
Let f (x) = x4, a = 2, and x = .3. Then f (x) = 4x3 and f (a) = f (2) = 32.

The Linear Approximation is f ≈ f (a) x = 32(.3) = 9.6.

The actual change is
f = f (a + x) − f (a) = f (2.3) − f (2) = 27.9841 − 16 = 11.9841.

2. x = −.4
√x − 2x2, a = 5, x =1

3. 1 + x, a = 8,

Let f (x) = (1 + x)1/2, a = 8, and x = 1. Then f (x) = 1 (1 + x )−1/2 and
2
1
f (a) = f (8) = 6 .

1

2 Chapter 4 Applications of the Derivative

The Linear Approximation is f ≈ f (a) x = 1 (1) = 1 ≈ .166667.
6 6 √

The actual change is f = f (a + x) − f (a) = f (9) − f (8) = 10 − 3 ≈ .162278.

4. a = 0, x = .02
sin x, π
4
5. tan x, a = , x = .013

Let f (x) = tan x, a = π , and x = .013. Then f (x) = sec2 x and f (a) = f ( π ) = 2.
4 4

The Linear Approximation is f ≈ f (a) x = 2(.013) = .026.

The actual change is

f = f (a + x) − f (a) = f ( π + .013) − f ( π ) ≈ 1.026344 − 1 = .026344.
4 4

6. a = π , x = .03 √ √√
cos x, 4

7. Use the Linear Approximation for f (x) = x at a = 25 to estimate 26 − 25. Use a

calculator to compute the error in this estimate.

Let f (x) = √ a = 25, and x = 1. Then f (x) = 1 x −1/2 and f (a) = f (25) = 1 .
x, 2 10

The Linear Approximation is f ≈ f (a) x = 1 (1) = .1.
10

The actual change is f = f (a + x) − f (a) = f (26) − f (25) ≈ .0990195.

The error in this estimate is |.0990195 − .1| = .000980486.

8.

How would you use the Linear Approximation to estimate the quantities?

9. Use the Lin1ear App1roximation to estimate 16.51/4 − 161/4. Use a calculator to compute the
error(ain) th√is10es1ti−ma1te0.

Let (bf ()x)√=1 x1/−4, a1= 16, and x = .5. Then f (x) = 1 x −3/4 and f (a) = f (16) = 1 .
4 32

103 10

UsTehaecLailncuealartAorptporocxoimmpautitoenthies errfor≈infea(ach) esxti=ma3t12e(..5) = .015625.

The actual change is

f = f (a + x) − f (a) = f (16.5) − f (16) ≈ 2.015445 − 2 = .015445 which gives

an error of |.015625 − .015445| ≈ .00018.

10.
Apply the Linear Approximation to f (x) = x1/4 at a = 16 to estimate (15.8)1/4.

11. Use the Linear Approximation to sin x at a = 0 to estimate sin(.023).

Let f (x) = sin x, a = 0, and x = .023. Then f (x) = cos x and f (a) = f (0) = 1 and

note that sin(.023) = sin(.023) − 0 = f (a + x) − f (a). The Linear Approximation to

12. sin(.023) is f ≈ f (a) x = 1(.023) = .023. estimate 1 − tan(π/4 − .2)? Carry out
How would you use the Linear Approximation to

13. tThheeecsutibmeartoeo. t of 27 is 3. How much larger is the cube root of 27.2? Estimate using the

Linear Approximation.

Let f (x) = x 1/3, a = 27, and x = .2. Then f (x) = 1 x −2/3 and f (a) = f (27) = 1 .
3 27
1
14. The Linear Approximation is f ≈ f (a) x = 27 (.2) = .0074074.

Use the Linear Approximation to estimate sin 61◦ − sin 60◦. Note: you must express θ in
π π π
15. rAapdpiarnosx.imate sin( 4 + .01) using that sin 4 = cos 4 ≈ .7071.

Let f (x) = sin x, a = π , and x = .01. Then f (x) = cos x and
√ 4

f (a) = f ( π ) = 2 ≈ .7071.
4 2

√
2
The Linear Approximation is f ≈ f (a) x= 2 (.01) = .007071.

The actual change is

f = f (a + x) − f (a) = f ( π + .01) − f ( π ) ≈ sin( π + .01) − sin( π ) ≈ .007036.
4 4 4 4

4.1 Linear Approximation and Applications 3

16.

17. LToetd(efat(e)xr)mW=inhea√ttxhiseatahnredeaaaroe=fa a square carpet, you measure its side as 6 feet.
o1f6.the carpet, base√d on this measurement?

(a) F(bin)dEthsteimlianteearthizeastiiozne oLf(txh)eteorrfo(rxi)n=yourxaraetathce√alpcouilnattiaon=if1t6h.e measurement is off by at
(b) Use mL(oxst) .t5o icnocmhepsu.te the approximate value of 16.2. Compare with the value

obtained from a calculator.

Let f (x) = x1/2, a = 16, and x = .2. Then f (x) = 1 x −1/2 and f (a) = f (16) = 1 .
2 8

(a) The linearization to f (x) is L(x) = f (a)(x − a) + f (a) = 1 (x − 16) + 4 = 1 x + 2.
8 8

18. (b) We have L(16.2) = 4.025 compared with f (16.2) ≈ 4.024922.

Compute the linearization of f (x) = cos x sin x at a = 0 and at a = π .
4
f (x) = √1 = √1 .
19. Compute the linearization of x at a 16 and use it to approximate 15

Let f (x) = √1 and a = 16. Then f (x) = − 1 x −3/2 and f (a) = f (16) = − 1 . The
x 2 128

linearization is L(x) = f (a)(x − a) + f (a) = − 1 (x − 16) + 1 . Then
128 4
√1 1 1 33
15 = L (15) = − 128 (−1) + 4 = 128 = .257813.

20.
Use the linearization of f (x) = (1 + x)5/3 at a = 0 to approximate (1.2)5/3.

In Exercises 21–29, use the linearization to approximate the quantity and compare the result
with the value obtained from a calculator.

√
21. 17

Let f (x) = x1/2, a = 16, and x = 1. Then f (x) = 1 x −1/2 and f (a) = f (16) = 1 .
2 8

The linearization to f (x) is L(x) = f (a)(x − a) + f (a) = 1 (x − 16) + 4 = 1 x + 2.
8 8

We have L(17) = 4.125, compared with f (17) ≈ 4.123106.

22. x = .03. Then f (x) = 1 x −2/3 and f (a) = f (27) = 1 .
√1 3 27

23. (2177.03)1/3
Let f (x) = x1/3, a = 27, and

The linearization to f (x) is L(x) = f (a)(x − a) + f (a) = 1 (x − 27) + 3 = 1 x + 2.
27 27

We have L(27.03) ≈ 3.0011111, compared with f (27.03) ≈ 3.0011107.

24. x = .001. Then f (x) = 1 x −2/3 and f (a) = f (27) = 1 .
(27.02)1/3 3 27

25. (27.001)1/3
Let f (x) = x1/3, a = 27, and

The linearization to f (x) is L(x) = f (a)(x − a) + f (a) = 1 (x − 27) + 3 = 1 x + 2.
27 27

We have L(27.001) ≈ 3.0000370370370, compared with

f (27.001) ≈ 3.0000370365798.

26.
(17)1/4
1
27. 5.1

Let f (x) = x−1, a = 5, and x = .1. Then f (x ) = −x −2 and f (a) = f (5) = − 1 .
25

4 Chapter 4 Applications of the Derivative

The linearization to f (x) is L(x) = f (a)(x − a) + f (a) = − 1 (x − 5) + 1 = 2 − 1 x .
25 5 5 25

We have L(5.1) = .196, compared with f (5.1) ≈ .19608.

28.
1/(5.1)2

29. (31)3/5

Let f (x) = x3/5, a = 32, and x = −1. Then f (x) = 3 x −2/5 and f (a) = f (32) = 3 .
5 20

The linearization to f (x) is L(x) = f (a)(x − a) + f (a) = 3 (x − 32) + 8 = 3 x + 16 .
20 20 5

We have L(31) = 7.85, compared with f (31) ≈ 7.84905.

30.
A spherical balloon has radius 6 inches.

31. If a (sato)nUe sies tthoessLeidnevaerrAticpaplrlyoxiinmtahteioanitrowesitthimaanteinthiteiaclhvanelgoeciintyvovluftm/seecif, tthheenraditiurseaischinecsreaased
maximumbyh.e3iginhct hoefsv. 2/64 feet.
(a) U(bs)e UthseeLtihneeaLriAnepaprroAxpimpraotxioimn atotieosntitmoaetesttihmeaeteffethcet ocnhtahnegme ainxismuurfmacheeiagrheat oiff athceharnagdeius is
in iniinticarlevaesleodcibtyy .f3roimnc2h5est.o 26 ft/sec.
(b) Use the Linear Approximation to estimate the effect on the maximum height of a change
in initial velocity from 30 to 31 ft/sec.

(c) Does a one foot per second increase in initial velocity have a bigger effect at low veloc-
ities or at high velocities? Explain.

(d) Does a one foot per second increase in initial velocity have a bigger percentage effect at
low velocities or at high velocities?

A stone tossed vertically with initial velocity v ft/s attains a maximum height of
h = v2/64 ft.

(a) If v = 25 and v = 1, then h ≈ h (v) v = 1 (25)(1) = .78125 ft.
32

(b) If v = 30 and v = 1, then h ≈ h (v) v = 1 (30)(1) = .9375 ft.
32

(c) A one foot per second increase in initial velocity v increases the maximum height by

v/32 ft. Accordingly, there is a bigger effect at higher velocities.

(d) The percentage increase on maximum height of a one foot per second increase in initial

velocity v is v/32 = 1 Thus there is the same effect at all velocities.
v .
32.
32

A bus company finds that if it charges a fare of x dollars on the New York–Boston route,
then it will earn a monthly revenue of R(x) = 3250x − 50x2 dollars.

(a) Use the Linear Approximation to determine the effect on revenue of raising the fare
from $24 to $25.

(b) The fare is set at $27, but the company is considering changing it by an amount x.
Use the Linear Approximation to determine the effect of such a change on revenue.

4.1 Linear Approximation and Applications 5

In Exercises 33–34, use the following fact derived from Newton’s laws: an object that is released

with initial velocity v and angle θ, travels a total distance of 1 v2 sin 2θ feet.
32

θ

Figure 1 Trajectory of an object released at an angle θ .

33. A player located 18.1 feet from the basket launches a successful jump shot from a height of
10 feet (level with the rim of the basket), at an angle of 34◦ and initial velocity of 25 ft/s.

(a) Show that the distance of the shot changes by approximately 0.255 θ feet if the angle
is changed by an amount θ (degrees).

(b) Would the shot likely have been successful if the angle were off by 2◦?

Using Newton’s laws, given initial velocity v = 25 ft/s, the shot travels

x = 1 v2 sin 2t = 625 sin 2t ft, where t is in radians.
32 32

(a) If θ = 34◦ (i.e., t = 17 π ), then
90
625 17 625 17 π
x ≈ x (t) t = 16 cos( 45 π ) t = 16 cos( 45 π ) θ · 180 ≈ 0.255 θ.

(b) If θ = 2◦, this gives x ≈ 0.51 ft, in which case the shot would not have been

successful, having been off half a foot.
34.

Continuing with the above notation, estimate the change in the distance of the shot if the

angle changes from 50◦ to 51◦ when
Exercises(a3)5–v306=ar2e5bfat/sseedc on Example ??.

35. (a) E(bs)timv0at=e 30 ft/sec loss per mile of altitude for a 130-lb pilot.
the weight
(b) E(cs)timIsattheethsehoatltmituodree saetnwsihtiivche tsohethweoaunlgdlewweihgehn1t2h7e lvbe.locity is large or small? Explain.

From the discussion in the text, the weight loss W at altitude h (in miles) for a person
weighing W0 at the surface of the earth is approximately W ≈ −.0005W0h.

(a) The astronaut weighs W0 = 130 lb at the surface of the earth. Accordingly, her weight
loss at altitude h is approximately W ≈ −.065h.

(b) An estimate of the altitude at which she would weigh 127 lb is given by

h ≈ W = −3 ≈ 46.15 miles.
−.065 −.065
36.

A satellite orbits the earth in a circular orbit 2000 miles high (5960 miles from the center of
37. tThheeesatrothp)p.ing distance for an automobile (after applying the brakes) is approximately

F(s)(a=) 1H.1osw+m.u0c5h4sw2owulhderae1s60is-ltbheasstpreoendauint wmepighh. in the satellite?
(a) F(bin)dFtihnedLthineeLarinAeparprAopxpimroaxtiimonattioonFtoattshe=w3e5igahntdfusn=cti5o5n.W (x) with m = 160 at
(b) Use x(a)=to41es6t0im. ate the change in stopping distance per additional mph for s = 35 and

s(c=) U55s.e (b) to estimate the astronaut’s weight loss per additional mile of altitude beyond

Let F(s)2=0010..1s + .054s2.

(a) The Linear Approximation at s = 35 mph is
F ≈ F (35) s = (1.1 + .108 × 35) s = 4.88 s ft.

The Linear Approximation at s = 55 mph is
F ≈ F (55) s = (1.1 + .108 × 55) s = 7.04 s ft.

6 Chapter 4 Applications of the Derivative

(b) The change in stopping distance per additional mph for s = 35 mph is
38. approximately 4.88 ft.

The change in stopping distance per additional mph for s = 55 mph is
approximately 7.04 ft.

Let f (x) = 3x − 4.
39. The (lean) gWthhsaot fisatshiedelinoefaaribzoatxioinn tahteas=hap0e? oAfnaswcuebrewiisthmoeuatsmuraekdinagt 2anfyeectaalcnudlathtieonvos!lume

V =(bs3) iCs othmenpuetsetitmheatleindetaoribzeat8iofnt3a. t a = 2.
(a) E±(cs.)t2imPfLtra.(otxev)et=htehaget(rrixfo)rg. i(nx)thies vaolliunmeaer cfuanlccutiloatnioanndifLth(ex)miesatshuerelimneeanrtizoaftsioins aint axcc=uraa,tethbeyn
(b) What is the maximum allowable error in the measurement of s in inches, if the volume

calculation is to have an error of at most 1 ft3?

The volume V of a cube having side length s is V = s3.

(a) For s = 2 ft and s = ±.2 ft, we have
V ≈ V (s) s = 3s2 s = 12(±.2) ft3 = ±2.4 ft3.

(b) If the volume computation is to have an error of at most 1 ft3, then V = 12 s = 1

40. ft3, which gives s = 1 ft = 1in.
12

The volume of a certain gas (in liters) is related to pressure P (in atmospheres) by the

formula V = 24 . The volume is measured as V = 5 with a possible error of ±.5 liters.
P

Further Insights and Challenges (a) Compute P based on this measurement and estimate the possible error.

(b) What is the maximum allowable error in the measurement of V if the calculation of P

41. Show thaist tfoorhaavneyarneaelrrnourmofbeart mk,o(s1t .+5 axtm)ko≈sph1er+es?kx for small x. Estimate (1.02).7 and

(1.02)−.3.

Let f (x) = (1 + x)k. Then for small x, we have

f (x) ≈ L(x) = f (0)(x − 0) + f (0) = k(1 + 0)k−1(x − 0) + 1 = 1 + kx

Let k = .7 and x = .02. Then L(.02) = 1 + (.7)(.02) = 1.104.
Let k = −.3 and x = .02. Then L(.02) = 1 + (−.3)(.02) = .994.

42.

GU Let f (x) = sin x and g(x) = tan x.

(a) Show that the functions f and g have the same Linear Approximation a = 0.

(b) Use a graphing calculator to graph f (x) and g(x) over the interval [0, π ]. Which
6
function is more accurately approximated by L(x)? Explain your answer.

4.2 Extreme Values (c) Calculate the error in the Linear Approximation for f and g at the value x = π . Does
Preliminary Questions 6
the answer confirm your conclusion in (b)?

1. The following questions refer to Figure 1.

(a) How many critical points does f (x) have?
(b) What is the maximum value of f (x) on [0, 8]?
(c) What are the local maximum values of f (x)?
(d) Find an interval on which both the minimum and maximum values of f (x) occur at

critical points.
(e) Find an interval on which the minimum value occurs at an endpoint.

4.2 Extreme Values 7

6
5
4
3
2
1

12345678

Figure 1

2. Which of the following statements is correct?

(a) If f (x) is not continuous on [0, 1], then it has no minimum or maximum value on [0, 1].
(b) If f (x) is not continuous on [0, 1], then it might not have a minimum or maximum value

on [0, 1].

3. Which of the following statements is correct?
(a) A continuous function f (x) does not have a minimum value on the open interval (0, 1).
(b) A continuous function f (x) might not have a minimum value on the open interval (0, 1).

4. State whether true or false and explain why.

(a) The function f (x) = 1 has no minimum or maximum value on the interval (0, 2).
x2

(b) The function f (x) = 1 has no minimum or maximum value on the interval (1, 2).
x2

(c) The function f (x) = 1 has no minimum or maximum value on the interval [1, 2].
x2

5. Which of the following statements is correct?

(a) If f (x) has no critical points in [0, 1], then it has no minimum or maximum value on
[0, 1].

(b) If f (x) has no critical points in [0, 1], then either f (0) or f (1) is the minimum value of
f on [0, 1].

6. Let f (x) = ax + b be a linear function with positive slope (a > 0). State whether true or
false and explain why.

(a) f (x) has no critical points.
(b) f has neither a minimum nor a maximum value on a closed interval [c, d].
(c) The maximum value of f (x) on [0, 1] is a + b.

Exercises

1. Let f (x) = 3x2 − 12x + 3.
(a) Find the critical point c of f (x) and compute f (c).
(b) Compute the value of f (x) at the endpoints of the interval [0, 4].
(c) Determine the extreme values (the minimum and maximum) of f (x) on [0, 4].
(d) Find the extreme values of f (x) on [0, 1].

Let f (x) = 3x2 − 12x + 3.
(a) Then f (c) = 6c − 12 = 0 implies that c = 2 is the sole critical point of f . We have

f (2) = −9.
(b) Since f (0) = f (4) = 3, the maximum value of f on [0, 4] is 3 and the minimum

value is −9.
(c) We have f (1) = −6. Hence the maximum value of f on [0, 1] is 0 and the minimum

value is −6.
2.

Find the extreme values of f (x) = 2x2 − 4x + 2 on [0, 5].

8 Chapter 4 Applications of the Derivative

In Exercises 3–10, find all critical points of the following functions.

3. x2 − 2x + 4

Let f (x) = x2 − 2x + 4. Then f (x) = 2x − 2 = 0 implies that x = 1 is the lone critical

point of f .
4.

−x2 + 4x + 8
5. x3 − 9 x2 − 54x + 2

2

Let f (x) = x3 − 9 x 2 − 54x + 2. Then f (x) = 3x2 − 9x − 54 = 3(x + 3)(x − 6) = 0
2
implies that x = −3 and x = 6 are the critical points of f .
6.
8t 3 − t 2
x
7.
x2 + 1

Let f (x) = x 1 − x2
. Then f (x) = = 0 implies that x = ±1 are the critical
x2 + 1 (x 2 + 1)2

points of f .
8. √

4t − t2 + 1
9. x 1/3

Let f (x) = 1 . Then f (x) = 1 x −2/3 = 0 implies that x = 0 is the critical point of f.
3 3
10.
x2

11. Lxe+t f1(x) = sin x + cos x.
(a) Find the critical points of f (x).

(b) Determine the maximum value of f (x) on [0, π ].
2

(a) Let f (x) = sin x + cos x 0. Taht exn=onπ4t,htehienotenrlvyalcri0ti,cπa2l , we have
f (x) = cos x − sin x = point of f
in this interval. √
√
(b) Since f ( π ) = 2 and f (0) = f ( π ) = 0, the maximum value of f on 0, π is 2.
4 2 2
12.

Let h(t) = (t2 − 1)1/3.
In Exerci(sae)s 1C3o–m4p0u, fitentdhtehcermitiacxailmpuomintasnodfmhi(nti)m. Cumhevcaklutheastoyfotuhreafnusnwcteior nisoinn tahgeregeivmeennitnwteirtvhal.

Figure 2.
13. 2x2 −(b)4xF+ind2;the [e0x,tr3e]me values of f (x) on [0, 1] and [0, 2].

Let f (x) = 2x2 − 4x + 2. Then f (x) = 4x − 4 = 0 implies that x = 1 is a critical point
of f . On the interval [0, 3F],igthuerme i2nimGumrapvhaloufeho(ft)f=is(tf2(1−) 1=)10/3,.whereas the maximum
value of f is f (3) = 8. (Note: f (0) = 2.)
14.
−x2 + 10x + 43; [3, 8]
15. x2 − 6x − 1; [−2, 2]

Let f (x) = x2 − 6x − 1. Then f (x) = 2x − 6 = 0, whence x = 3 is a critical point of f .
The minimum of f on the interval [−2, 2] is f (2) = −9, whereas its maximum is
f (−2) = 15. (Note: critical point x = 3 is outside the interval.)
16.
x2 − 6x − 1; [−2, 0]
17. −4x2 + 3x + 4; [−1, 1]

Let f (x) = −4x 2 + 3x + 4. Then f (x) = −8x +3 = 0, whence x = 3 is a critical point
8
of f . The minimum of f on the interval [−1, 1] is f (−1) = −3, whereas its maximum is

18. f ( 3 ) = 4.5625. (Note: f (1) = 3.)
8

x3 − 3x + 1; [0, 2]

4.2 Extreme Values 9

19. x3 − 6x + 1; [−1, 1]
√

Let f (x) = x3 − 6x + 1. Then f (x) = 3x2 − 6 = 0, whence x = ± 2 are the critical
points of f . The minimum of f on the interval [−1, 1√] is f (1) = −4, whereas its
maximum is f (−1) = 6. (Note: critical points x = ± 2 are not in the interval.)
20.
x3 − 12x2 + 21x; [0, 2]
21. x3 + 3x2 − 9x + 2; [−1, 1]

Let f (x) = x3 + 3x2 − 9x + 2. Then f (x) = 3x2 + 6x − 9 = 3(x + 3)(x − 1) = 0,
whence x = 1 and x = −3 are the critical points of f . The minimum of f on the interval
[−1, 1] is f (1) = −3, whereas its maximum is f (−1) = 13. (Note: the critical point
x = −3 is not in the interval.)
22.
x3 + 3x2 − 9x + 2; [0, 2]
23. x3 + 3x2 − 9x + 2; [−4, 4]

Let f (x) = x3 + 3x2 − 9x + 2. Then f (x) = 3x2 + 6x − 9 = 3(x + 3)(x − 1) = 0,
whence x = 1 and x = −3 are the critical points of f . The minimum of f on the interval
[−4, 4] is f (1) = −3, whereas its maximum is f (4) = 78. (Note: f (−4) = 22 and
f (−3) = 29.)
24.
x5 − x; [0, 2]
25. x5 − 3x2; [−1, 5]

Let f (x) = x5 − 3x2. Then f (x) = 5x4 − 6x = 0, whence x = 0 and

x = 1 (150)1/3 ≈ 1.06 are critical points of f . Over the interval [−1, 5], the minimum
5
value of f is f (−1) = −4, whereas its maximum value is f (5) = 3050. (Note: f (0) = 0

26. and f ( 1 (150)1/3) = − 9 (150)2/3 ≈ −2.03).
5 125

2s3 − 3s2 + 3; [−4, 4]

x2 + 1
27. ; [5, 6]

x −4

Let f (x) = x2 + 1 f (x) = (x − 4) · 2x − (x2 + 1) · 1 = x2 − 8x − 1 =0
. Then (x − 4)2 (x − 4)2

x −√4
implies x = 4 ± 17. Neither critical point lies in the interval [5, 6]. In this interval, the

28. minimum of f is f (5) = 25, while its maximum is f (6) = 37 = 18.5.
2

1−x [.5, 1]
x3x−+ x42x; [0, 3]
29. x +1 ;

4x 4x (x − 1)(x + 3) = 0 whence
Let f (x) = x − . Then f (x) = 1 − =
x +1 (x + 1)2 (x + 1)2
x = 1, x = −1 and x = −3 are critical points of f . The minimum of f on the interval

[0, 3] is f (1) = −1, whereas its maximum is f (0) = f (3) = 0. (Note: critical points

x = 1 and x = −3 are not in the interval.)
30.

(1 + x) 1 + (1 − x)2; [0, 1]

31. (2 + x) 2 + (2 − x)2; [0, 2]

Let f (x) = (2 + x) 2 + (2 − x)2. Then

f (x) = 2 + (2 − x)2 − (2 + x)(2 + (2 − x)2)−1/2(2 − x) = 2(x − 1)2 = 0 whence
2 + (2 − x)2

x = 1 is t√he critical point of f . In the interval [0, 2], the minimum occur√s at
f (1) = 3 3 ≈ 5.196152 and the maximum occurs at f (0) = f (2) = 4 2 ≈ 5.656854.
32. √
√1 + x2 − 2x√; [3, 6]
33. x + x2 − 2 x [0, 4]
√√
Let f (x) = x + x2 − 2 x. Then
√
1 + 2x −√2 1+ x
f (x) = 1 (x + x 2)−1/2(1 + 2x) − x −1/2 = √ +x = 0, whence x = 0,
2 2x1

10 Chapter 4 Applications of the Derivative

√
3
x = −1 and x = ± 2 are the critical points of f . On the interval [0, 4], the minimum of f

√
3
is f( 2 ) ≈ −.589980 and√the maximum is f (4) ≈ .472136. (Note: f (0) = 0 and critical

points x = −1 and x = − 3/2 are not in the interval.)

34. (t − t 2)1/3; [−1, 2]

35. sin x cos x; [0, π ]
2

Let f (x) = sin x cos x = 1 sin 2x. On the interval 0, π , f (x) = cos 2x = 0 when
x= minimum of 2 on this interval is f (0) = 2 ) = 0, whereas the maximum
π . The ( π is
4 f f
2
π 1
36. f ( 4 ) = 2 .

x − sin x; [0, 2π ]

37. x + sin x; [0, 2π ]

Let f (x) = x + sin x. On the interval [0, 2π ], f (x) = 1 + cos x = 0 at x = π , a critical
point of f . The minimum value of f on this interval is f (0) = 0, whereas the maximum
value over this interval is f (2π ) = 2π . (Note: f (π ) = π .)
38. θ − 2 sin θ ; [0, 2π ]
39. sin3 θ − cos2 θ ; [0, 2π ]

Let h(θ ) = sin3 θ − cos2 θ . On [0, 2π ], we have h (θ ) = 0 at

θ = 0, π , π, π + tan−1( √2 ), 3 π, 2π − tan−1( √2 ), and 2π . From the table of function
2 5 2 5
values below, the minimum of h on [0, 2π ] is −1 and the maximum is 1.

θ 0 π π π + tan−1( √2 ) 3 π 2π − tan−1( √2 ) 2π
2 5 2 5

h(θ ) −1 1 −1 − 23 −1 − 23 −1
40. tan x − 2x [0, 1] 27 27

41. Find the critical points of f (x) = |x − 2|.

Let f (x) = |x − 2|. For x < 2, we have f (x) = −1. For x > 2, we have f (x) = 1. Now
f (x) − f (2) (2 − x) − 0
as x → 2−, we have x −2 = x −2 → −1; whereas as x → 2+, we have

f (x) − f (2) (x − 2) − 0 f (x) − f (2)
= → 1. Therefore, f (2) = lim does not exist
x −2 x −2 x→2 x − 2
and the lone critical point of f is x = 2.
42. Find the critical points of f (x) = |3x − 9|.

43. Find the critical points of f (x) = |x2 + 4x − 12| (refer to Figure 3 if necessary) and

determine the extreme values of f (x) on [0, 3].

30
20
10

–6 2

Figure 3

Let f (x) = |x2 + 4x − 12| = |(x + 6)(x − 2)|. From the graph of f below, the critical
points of f are x = −6 and x = 2. On the interval [0, 3] the minimum value of f occurs at
f (2) = 0 and the maximum value occurs at f (0) = 12. (Note: f (3) = 9 and the critical
point x = −6 is not in the interval.)

4.2 Extreme Values 11

20

10

–8 –6 –4 –2 0 2 4

44.

45. Let a(a>) fSbohr>oawn0ytahnnadt≥isfe1tf. (fx()x )is=difxfaer−enxtibabwlhe,ertheean <f (x) and f (x )n have the same critical points
b.

(a) F(bin)dUthsee (ma)axtoimfiunmd tvhaelumeaoxfimf u(xm) voanlu[0e,o1f]gin(xt)er=ms(xo2f +a a2nxd−b.8)6 on [0, 3].
(b) Calculate the maximum value of f (x) = x5 − x10.

(a) Let f (x ) = x a − x b. Then f (x ) = ax a−1 − bx b−1. Since a < b,

f (x) = xa−1(a − bxb−a) = 0 implies critical points x = 0 and x = ( ab)1)/1(/b(−b−a)a)<, w1h.ich is
in the
interval [0, 1] as a < b implies a < 1 and consequently x = ( a
b b
Also, f (0) = f (1) = 0 and a < b implies xa > xb on the interval [0, 1], which gives

f (x) > 0 and thus the maximum value of f on [0, 1] is

f (( a )1/(b−a) ) = ( a )a/(b−a) − ( a )b/(b−a) .
b b b

(b) Let f (x) = x5 − x10. Then by part (a), the maximum value of f on [0, 1] is

f (( 1 )1/5 ) = ( 1 ) − ( 1 )2 = 1 − 1 = 1 .
2 2 2 2 4 4
46.
True or false: if f (x) is a continuous function on [0, 2] but is not differentiable for all

47. Sxk∈et[c0h,t2h]e, igtrmapahyonfoat hfuanvcetiaomn aoxnim[0u,m4]vhaalvuiengon [0, 2]. Explain.

(a) two local maxima and one local minimum

(b) absolute minimum that occurs at an endpoint, and an absolute maximum that occurs at
a critical point

(a) Here is the graph of a function on [0, 4] that has two local maxima and one local
minimum.

10

0 24

(b) Here is the graph of a function on [0, 4] that has its absolute minimum at its left
endpoint and its absolute maximum at an interior critical point.

0.4 24
0.2

0
–0.2
–0.4

48.

Sketch the graph of a function f (x) on [0, 4] such that
(a) f (x) has a discontinuity
(b) f (x) has an absolute minimum but no absolute maximum

12 Chapter 4 Applications of the Derivative

49. Sketch the graph of a function on (0, 4) with a minimum value but no maximum value.
Here is the graph of a function f on (0, 4) with a minimum value [at x = 2] but no
maximum value [since f (x) → ∞ as x → 0+].

4

2

0 24
50. Sketch the graph of a function on (0, 4) having a local minimum but no absolute minimum.

Further Insights and Challenges

The graph of a quadratic function f (x) = ax2 + bx + c is a parabola. The function f (x) has a
unique absolute minimum value if a > 0 and a unique maximum value if a < 0.

51. Show that the function f (x) = x2 − 2x + 3 takes on only positive values. Hint: consider the
minimum value of f .

52. Observe that f (x) = x2 − 2x + 3 = (x − 1)2 + 2 > 0 for all x.

Let f (x) = x2 + r x + s.
53. Figu(rae)4Fsihnodwcsotnwdoitigornasphosnothf ecucboiecffipcoileynntosmr iaanlsd. sIttshhaotwinssuthraettahactubf i(cxp)otalykneosmoniaol nmlyaypositive

thchoaavetfe(fifbac)ihleoavGncsataisnlvlueaemeitsaaih.nnneHidremixabnuatlomm:ofcfipatanlhneldedmoccmfiouncnbaioxmdieciifmutpifimooucnlminyseonnuortonsrmadirtelioaarmlcnwaadfyhl(simxhcfah)aovx=rtehiwmen31heumxiimct3ihhn.+eiHrmf.12iFutnaamtixk:ne2davsp+acoplounlbnyexdbEiio+stxtiphoeconrp,cssowiisotsiehinvtiie5ctv.h2hee(iaan)nsdtuorneegative
f (x). values. Draw the graph of f , labeling the roots and the minimum value.

(c) Show that if f (x) takes on both positive and negative values, then it takes on a

minimum value at the midpoint between it6s0two roots.

20 30

–4 –2 2 4 –2 2 4
(b)
–20
(a)

Figure 4 (a) Graph of x3 − 6x + 1. (b) Graph of x3 − 3x2 + 9x + 10.

Let f (x) = 1 x 3 + 1 x 2 + bx + c. Using Exercise ??, we have
3 2
1
g(x) = f (x) = x2 + ax +b > 0 for all x provided b > 4 a2, in which case f has no

critical points and hence no local extrema. (Actually b ≥ 1 a2 will suffice, since in this case
4
54. [as we’ll see in a later section] f has an inflection point but no local extrema.)
Let c be a constant and let f (x) = x2 + r x + s be a quadratic polynomial. Suppose that the
55. hFoinridzothnetaml liinniemyum=acndinmterasxeicmtsumthevgalruaepshooff ff((xx))=at xtwp(o1p−oinxt)sq (oxn1,[0f,(1x]1,))wahnedre(xp2,,qf a(rxe2))
(ptohsaittiivse, nf u(xm1b)e=rs.f (x2) = c). Show that f (x) takes its minimum value at the midpoint

LMet=f (x1) += xx2p)(/12.−Hxin)qt:, s0h≤owxth≤at1,f w((hxe1re+pxa2)n/d2q) =are0.positive numbers. Then

f (x ) = x pq(1 − x )q−1(−1) + (1 − x )q px p−1 x = p
= x p−1(1 − x )q−1( pF(i1g−urxe)5− q x ) = 0 at p+q

The minimum value of f on [0, 1] is f (0) = f (1) = 0, whereas its maximum value is
p pp
f( ) = .
p +q (p + q)p+q

4.2 Extreme Values 13

56.

Figure 6 shows the graph of f (x) = 5|2x + 10| − 5|2x − 4| + |3x − 36|.
57. R &(Wa) ShCoownvtihnactethyeoucristeiclfatlhpaotitnhtesraeredoveasluneost cexfoisrt wa hciocnhtionnueouosf fthuencthtiroenewabitsholtuwtoe vloacluales is

minima bzuetron.o local maximum.
(a) S(bk)etFchintdhethgerampihniomf uam(nveacleusesaorfilyf (dxis).continuous) function with two local minima but

n(co)loDcoalesmaf x(xim) uhmav.e a maximum value?
(b) Let f (x) be a continuous function. Suppose that f (a) and f (b) are local minima

where a < b. Prove that there exists a value c between a and b such that f (c) is a local
maximuFmi.gHuirnet:6ApGplryapThheoofrefm(x?) ?=to5|t2hxe i+nt1er0v|a−l [5a|,2bx].− 4| + |3x − 36|.

(a) The function graphed here is discontinuous at x = 0.

10
9
8
7
6
5
4
3
2
−8 −6 −4 −2 0 2 4 6 8

(b) 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.
58.
Find the minimum value of

59. (Adapted from Calculus Problemsfor a New Century.) Use Rolle’s Theorem to prove the
 −2x + 1 x < 2
following statements:
(a) If f (x) is the function shfo(wxn) in F3xixg−ur5e 7(A), 2xth≤>enx5f<(x5) = 0 has at most one solution.
(b) If f (x) is the function shown in Figure 7(B), then f (x) = 0 cannot have two positive

Hints:otlhueticornitsi.cal points are points where f (x) does not exist. If necessary, draw a graph.

(A) (B)

Figure 7 Graphs of the derivatives.

(a) Let the graph of f (x) be as shown in the text; note that f (x) > 0 for all x. Assume
that f (x) = 0 has two distinct solutions, say a and b. By Rolle’s Theorem, there is a
number c between a and b such that f (c) = 0, contradicting the fact that f is
everywhere positive. Accordingly, f (x) = 0 has at most one solution.

14 Chapter 4 Applications of the Derivative

(b) Let the graph of f (x) be as shown in the text; note that f (x) > 0 for x > 0. Assume
that f (x) = 0 has two distinct positive solutions, say a and b with a < b. By Rolle’s
Theorem, there is a number c with 0 < a < c < b such that f (c) = 0, contradicting
the fact that f (x) is positive for positive values of x. Accordingly, f (x) = 0 has at
most one positive solution.

4.3 The Mean Value Theorem and Monotonicity

Preliminary Questions

1. Which value of m makes the following statement correct? If f (2) = 3 and f (4) = 9, where
f (x) is differentiable, then the graph must have a tangent line whose slope is m.

2. Which of the following conclusions does not follow from the Mean Value Theorem (assume
f is differentiable)?
(a) If f has a secant line whose slope is 0, then f (x) = 0 for some value of x.
(b) If f (x) = 0 for some value of x, then there is a secant line whose slope is 0.
(c) If f (x) > 0 for all x, then every secant line has positive slope.

3. State whether true or false and explain why (assume f is differentiable).

(a) If f (1) = f (3) = 5, then f (c) = 0 for at least one value 1 < c < 3.
(b) If f (c) = 0 for some c ∈ [a, b], then f (a) = f (b).
(c) If f (5) < f (9), then f (c) > 0 for some c ∈ [5, 9].

4. Can a function that takes on only negative values have a positive derivative? Give an example
or explain why no such functions exist.

5. Which of the following two situations corresponds to a negative first derivative? Explain.

(a) A car slowing down.
(b) A car driving in reverse.

6. (a) Use the graph of f (x) in Figure 1 to determine whether f (c) is a local minimum or
maximum.

(b) Is it correct to conclude from Figure 1 that f (x) is a decreasing function?

c

Figure 1 Graph of derivative f (x).

7. The graph of the derivative f (x) is shown in Figure 2. Does f (x) have two local minima
and one local maximum or does it have two local maxima and one local minimum?

Exercises 4.3 The Mean Value Theorem and Monotonicity 15

Figure 2 Graph of derivative f (x).
8. Sam made two statements which Deborah found dubious.

(a) “Although my average velocity for the trip was 70 mph, at no point in time did my
speedometer read 70 miles per hour.”

(b) “Although a policemen clocked me going 70 mph, my speedometer never read 65 mph.”
Which theorem did Deborah apply in each case: the Intermediate Value Theorem or the
Mean Value Theorem?

1. Determine the intervals on which f (x) is positive and negative if f (x) is the function
graphed in Figure 3.

123456

Figure 3

The derivative of f is positive on the intervals (0, 1) and (3, 5); it is negative on the (1, 3)
and (5, 6).
In Exercises 2–6, sketch graphs of functions whose derivatives have the following descriptions.
2. f > 0 for x > 3 and f < 0 for x < 3.
3. f is negative for all x.
Here is the graph of a function f for which f is negative for all x.

6
4
2
–2 2 4 6
–2
4. f > 0 for x < 1 and f < 0 for x > 1.

16 Chapter 4 Applications of the Derivative

5. f is negative on (1, 3) and positive everywhere else.
Here is the graph of a function f for which f is negative on (1, 3) and positive elsewhere.

8
6
4
2

–2 1 2 3 4 5

6.
f (x) makes the sign transitions +, − + −.

In Exercises 7–9, use the First Derivative Test to determine whether the given critical point is a
local minimum or local maximum (or neither).

7. 7 + 4x − x2; c = 2

Let f (x) = 7 + 4x − x2. Then f (c) = 4 − 2c = 0 implies c = 2 is a critical point of f .

Since f makes the sign transition +, − as x increases through c = 2, we conclude that

8. f (2) = 11 is a local maximum of f .
π
9. sxi3n−x c2o7sxx+; 2c; c==4−3

Let f (x) = x3 − 27x + 2. Then f (c) = 3c2 − 27 = 0 at c = −3. Since f makes the sign
transition +, − as x increases through c = −3, we conclude that f (−3) = 56 is a local
10. maximum of f .
The critical points of f (x) = x4 − 20x2 − 12x + 4 are x = −3, −.303, and 3.303. Use
11. TFihgeugrera4phtoodfettheermdeinreiviaftitvheeyofarfe(lxo)ca=l m.2ixn5im−a.o1r2l5oxc4al−m.1axxi3m−a .(2o5rxn2ei+thxeri)s. shown in
Figure 5. Find the critical points of f (x) and determine whether they are local minima,
maxima, or neither.

Figure 4 Graph of the derivative of f (x) = x4 − 20x2 − 12x + 4.

6

-2 -1 -2 .5 23

Figure 5 Graph of the derivative of f (x) = .2x5 − .125x4 − .1x3 − .25x2 + x.

Let f (x) = 1 x5 − 1 x4 − 1 x3 − 1 x 2 + x . Then f (x) = x4 − 1 x3 − 3 x2 − 1 x + 1. Using
5 8 10 4 2 10 2
Maple 8, the roots of f are all complex. (Their approximate values are

.89 ± .45i, −.64 ± .77i.) This tells us that f is nonzero for all real values of x. Moreover,

since f , a polynomial, is everywhere continuous and because f (0) = 1 > 0, we conclude

that f is positive for all real values of x. Accordingly, there are no real critical points of f .

In Exercises 12–18, find a point c satisfying the conclusion of the Mean Value Theorem for the
given function and values a, b.
12.

f (x)√= x−1, a = 1, b = 4
13. f (x) = x, a = 4, b = 9

Let f (x) = x1/2, a = 4, b = 9. By the MVT, there exists a c ∈ (4, 9) such that

1 c−1/2 = f (c) = f (b) − f (a) = 3 − 2 = 1 .
2 b−a 9−4 5

Thus √1 = 2 whence c = 25 = 6.25 ∈ (4, 9).
c 5 4

4.3 The Mean Value Theorem and Monotonicity 17

14. f (x) = (x − 1)(x − 3), a = 1, b = 3
15. f (x) = cos x − sin x, a = 0, b = 2π

Let f (x) = cos x − sin x, a = 0, b = 2π . By the MVT, there exists a c ∈ (0, 2π ) such that

− sin c − cos c = f (c) = f (b) − f (a) = 1−1 = 0.
b − a 2π − 0

16. Thus − sin c = cos c. Choose c = 7 π ∈ (0, 2π ).
x 4

17. ff ((xx)) == xx3+, ar1b,itara=ry6a, ban=d b3

Let f (x) = x3 and let a < b be arbitrary.

Case 1. If a = 0, then by the MVT there exists a c ∈ (0, b) such that

3c2 = f (c) = f (b) − f (0) = b3 − 0 = b2, whence 3c2 = b2 implies c = ± b .
b − 0 b−0 √
3
Choose c = √b ∈ (0, b).
3

Case 2. If a = 0, consider g(x) = f (x + a) on [0, B], where A = 0 and B = b − a > 0.
g(B) − g(0)
By Case 1, C = √B is such that g (C ) = . By the Chain Rule, we have
3 B−0
g (x) = f (x + a). Choose

c = a + C = a + B = a + b− a ∈ (a, b).
√ √
33

Then

g(B) − g(0)
f (c) = f (a + C) = g (C) = B − 0

= f (B + a) − f (0 + a) = f (b) − f (a)
B−0 b−a

In other words, f (c) = f (b) − f (a)
, which is the conclusion of the MVT, as required.
b−a

18. g(x) = sin x, a = 0, b = π (use a calculator)
4
19. Let f (x) = 1 − |x|.

(a) Show that f (x) does not satisfy the conclusion of the Mean Value Theorem for the
points a = −2, b = 1.

(b) Why does the Mean Value Theorem not apply in this case?

(c) Show that the Mean Value Theorem does hold if a and b are both positive or both

negative.

Let f (x) = 1 − |x|.

(a) For a = −2 and b = 1, we have f (b) − f (a) = 0 − (−1) = 1 Yet there is no point
.
b−a 1 − (−2) 3

c ∈ (−2, 1) such that f (c) = 1 . Indeed, f (x) = 1 for x < 0, f (x) = −1 for x > 0,
3
and f (0) is undefined.

(b) The MVT does not apply in this case, since f is not differentiable on the open interval
(−2, 1).

(c) If a and b (where a < b) are both positive (or both negative), then f is continuous on
[a, b] and differentiable on (a, b). Accordingly, the hypotheses of the MVT are met
and the theorem does apply. Indeed, in these cases, any point c ∈ (a, b) satisfies the
conclusion of the MVT (since f is constant on [a, b] in these instances).

18 Chapter 4 Applications of the Derivative

In Exercises 20–35, (1) determine the intervals on which the function is monotonic increasing
or decreasing, and (2) use the First Derivative Test to determine if the local extrema are local
minima or maxima (or neither).

Here is a table legend for Exercises 20–35.
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.

20.
−x2 + 7x − 17

21. 5x2 + 6x − 4

Let f (x) = 5x 2 + 6x − 4. Then f (x) = 10x +6 = 0 yields the critical point c = − 3 .
5

x −∞, − 3 −3/5 − 3 , ∞
5 5
f− 0+

f m
22.

x3 − 6x2
23. x(x + 1)3

Let f (x) = x (x + 1)3. Then f (x) = x · 3 (x + 1)2 + (x + 1)3 · 1 = (4x + 1) (x + 1)2 = 0

yields critical points c = −1, − 1 .
4

x (−∞, −1) −1 (−1, −1/4) −1/4 (−1/4, ∞)

f−0 − 0 +

f¬ m
24.

3x4 + 8x3 − 6x2 − 24x
25. x2 + (10 − x)2

Let f (x) = x2 + (10 − x)2. Then f (x) = 2x + 2 (10 − x) (−1) = 4x − 20 = 0 yields the

critical point c = 5.

x (−∞, 5) 5 (5, ∞)

f − 0+

26. fm

1 x 3+ 3 x 2 + 2x + 4
3 1 2

27.
x2 + 1

Let f (x) = x2 + 1 −1. Then f (x) = −2x x2 + 1 −2 = 0 yields critical point c = 0.

x (−∞, 0) 0 (0, ∞)

f + 0−

f M
28.

x5 + x3 + 1
29. x 4 + x 3

Let f (x) = x4 + x3. Then f (x) = 4x3 + 3x2 = x2(4x + 3) yields critical points

c = 0, − 3 .
4

x −∞, − 3 − 3 − 3 , 0 0 (0, ∞)
4 4 4
f − 0 + 0+

f mM
30. 2x + 1

x2 + 1

4.3 The Mean Value Theorem and Monotonicity 19

31. x 2 − x 4

Let f (x) = x2 − x4. Then f (x) = 2x − 4x3 = 2x 1 − 2x2 = 0 yields critical points

c = 0, ± √1 .
2

x −∞, − √1 − √1 − √1 , 0 0 0, √1 √1 √1 , ∞
222 222

f+ 0 − 0+ 0 −

32. f M mM
33.
cxo+s θx1+ s(ixn >θ , 0[0), 2π ]

Let f (θ ) = cos θ + sin θ . Then f (θ ) = cos θ − sin θ , which yields c = π , 5π on the
4 4
interval [0, 2π ].

x 0, π π π , 5π 5π 5π , 2π
4 4 4 4 4 4
f +0 − 0 +

fM m
34. θ + cos θ , [0, 2π ]
35. θ − 2 cos θ , [0, 2π ]

Let f (θ ) = θ − 2 cos θ . Then f (θ ) = 1 + 2 sin θ , which yields c = 7π , 11π on the interval
6 6
[0, 2π ].

x 0, 7π 7π 7π , 11π 11π 11π , 2π
6 6 6 6 6 6
f +0 − +
0

36. fM m

Show that the quadratic polynomial f (x) = x2 + bx +c is decreasing on (−∞, − b ) and
iLnectrefa(sxin)g=onx3(−−b22,x∞2 +). 2x. 2
37. (a) Find the minimum value of

f (x ).

(b) Use (a) to show that f is an increasing function.

Let f (x) = x3 − 2x2 + 2x.

(a) For all x, we have f (x) = 3x2 − 4x + 2 = 3 x − 2 2 + 2 ≥ 2 > 0. The minimum
3 3 3
2 2
value of f (x) is f 3 = 3 .

(b) Since f (x) > 0 for all x, the function f is everywhere increasing.
38.
Find conditions on a and b that ensure that x3 + ax + b is increasing on (−∞, ∞).

Further Insights and Challenges

39. Which values of c satisfy the conclusion of the Mean Value Theorem on the interval [a, b]
if f (x) is a linear function?

Let f (x) = px + q, where p and q are constants. Then the slope of every secant line and

tangent line of f is p. Accordingly, considering the interval [a, b], every point c ∈ (a, b)
f (b) − f (a)
satisfies f (c) = p= b − a , the conclusion of the MVT.
Show that if f is
40. a quadratic polynomial, then the midpoint c = (a + b)/2 satisfies the

41. cFoinndclausviaolnueofotfhcesMateisafnyiVnaglutheeTchoenocrleumsioonn o[af ,thbe].Mean Value Theorem for f (x) = xn on

the interval [0, b].

Let f (x) = xn and consider the interval [0, b]. By the MVT we have
f (b) − f (0) bn
ncn−1 = f (c) = = = bn−1. Thus ncn−1 = bn−1 and hence
b−0 b
bb
c= √ = .
nn−1 n1/(n−1)
42.
Let f be a differentiable function on (−∞, ∞) whose derivative is never equal to 0. Use

the Mean Value Theorem (or Rolle’s Theorem) to show that the graph of f crosses the

x-axis at most once.

20 Chapter 4 Applications of the Derivative

43. Use the previous exercise to show that the polynomial f (x) = x5 + 2x + 9 has at most one
real root.

Let f (x) = x5 + 2x + 9. Then f (x) = 5x4 + 2 > 0 for all x. By Exercise ?? the graph of
f crosses the x-axis at most once. Accordingly, f has at most one real root. (Indeed, its
44. lone root is approximately equal to −1.437.)
Show that the cubic function f (x) = x3 + ax2 + bx + c is increasing on (−∞, ∞) if
45. bPr>ovae2t/h3e.following assertion: if f (0) = g(0) and f (x) ≤ g (x) for x, then f (x) ≤ g(x)
for all x.

Hint: show that f (x) − g(x) is nondecreasing.

The assertion is false. Let f (x) = 2x and g(x) = 5x. Then f (0) = g(0) = 0. Moreover,
f (x) = 2 ≤ 5 = g (x) for all x. But f (−1) = −2 > −5 = g(−1). Therefore, it is not
true that f (x) ≤ g(x) for all x. (Note that f and g are odd functions.)

Suppose instead that f (0) = g(0) and f (x) ≤ g (x) for x ≥ 0. Let
h(x) = g(x) − f (x). Then h(0) = f (0) − g(0) = 0 and h (x) = g (x) − f (x) ≥ 0 for
x ≥ 0. Thus h is nondecreasing. Accordingly, h(x) = g(x) − f (x) ≥ 0 for x ≥ 0.
Therefore, f (x) ≤ g(x) for x ≥ 0.

46. Use the previous exercise to prove that sin x ≤ x for all x.

47. Use Exercise 45 to establish the following assertions for all x. Each assertion follows from

the previous one.

(a) cos x ≥1− 1 x 2 (use Exercise 46).
2

(b) sin x ≥x − 1 x 3
6

(c) cos x ≤1− 1 x2 + 1 x 4
2 24

(d) Can you guess the next inequality in the series?

(a) This assertion is true because the functions involved are even.

Let f (x) = cos x and g(x) = 1− 1 x 2. Then f (0) = g(0) = 1 and
2
f (x) = − sin x ≥ −x = g (x) for x ≥ 0 by the alternative formulation of Exercise

??. Now apply the alternative formulation of Exercise 45 to conclude that

cos x ≥ 1 − 1 x 2 for x ≥ 0.
2

For x < 0, we have −x > 0 and thus cos x = cos (−x) ≥ 1 − 1 (−x )2 = 1 − 1 x 2 ;
2 2
1
i.e., cos x ≥ 1 − 2 x 2 for x < 0.

Hence for all x we have cos x ≥ 1 − 1 x 2.
2

(b) This assertion is false because the functions involved are odd. We prove the usual

alternative instead. Let f (x) = sin x and g(x) = x − 1 x 3. Then f (0) = g(0) = 0 and
6
1
f (x) = cos x ≥ 1− 2 x 2 = g (x) for x ≥ 0 by part (a). Now apply the alternative

formulation of Exercise 45 to conclude that sin x ≥ x − 1 x 3 for x ≥ 0.
6

(c) This assertion is true because the functions involved are even.

Let f (x) = 1 − 1 x2 + 1 x4 and g(x) = cos x. Then f (0) = g(0) = 1 and
2 24
1 2
f (x) = −x + 6 x ≥ − sin x = g (x) for x ≥ 0 by part (b). Now apply the

alternative formulation of Exercise 45 to conclude that 1 − 1 x2 + 1 x4 ≥ cos x or
2 24
1 1
cos x ≤ 1 − 2 x 2 + 24 x 4 for x ≥ 0.

For x < 0, we have −x > 0 and thus

cos x = cos (−x) ≤ 1 − 1 (−x )2 + 1 (−x )4 = 1 − 1 x 2 + 1 x 2 ; i.e.,
2 24 2 24
1 2 1 2
cos x ≤ 1 − 2 x + 24 x for x < 0.

4.3 The Mean Value Theorem and Monotonicity 21

Hence for all x we have cos x ≤ 1 − 1 x 2 + 1 x 4 .
2 24

(d) The next inequality in the series is sin x ≤ x − 1 x3 + 1 x 5, valid for x ≥ 0. We
6 120
gradually construct so-called Maclaurin series expansions for sin x and cos x; i.e.,

sin x = ∞ (−1)k x 2k+1 and cos x = ∞ (−1)k x 2k
k=0 (2k + 1)! k=0 . Can you also see from these

(2k)!

48. expansions why sin x is an odd function and cos x is an even one?

Let f (x) be a differentiable function such that f (x) = 0 for all x and set a = f (0) and
49. bG=U f (L0)e.t Tfh(ex)fo=lloxw3isnign(s1xte)p. s show that f is a linear function.

(a) S(ah)owShthoawt txha=t 0f i(sxa) c=ritaicfaolrpaollinxt.of f (x).

(b) E(bx)amShinoewththeagtrafp(hxs) o−f afx(xis) caonndstfan(txa).nCd acnonthceluFdierstthDaterfi(vxa)tiv=e aTxes+t bbe faoprpalileldx?.

(c) Show that f (0) is neither a local minimum nor maximum.

(a) Let f (x) = x 3 sin( 1 ). Then
x
1 1 −2) 1 1
f (x ) = 3x 2 sin( x ) + x 3 cos( x )(−x = x (3x sin( x ) − cos( x )), which yields the

critical point x = 0.

(b) The first figure is f (x) and the second is f (x). Note how the two functions oscillate
near x = 0, which implies that the First Derivative Test cannot be applied.

x 10−3
4
2
0
−2
−4
−6
−8
−10
−12
−0.25 −0.2 −0.15 −0.1 −0.05 0 0.05 0.1 0.15 0.2 0.25

0.2
0.15
0.1
0.05

0
−0.05
−0.1
−0.15
−0.2

−0.25 −0.2 −0.15 −0.1 −0.05 0 0.05 0.1 0.15 0.2 0.25

(c) As x approaches 0 from either direction, f (x) alternates between positive and
negative, which implies that f (0) alternates between being a local maximum and a
local minimum and is therefore considered neither.

22 Chapter 4 Applications of the Derivative

4.4 The Shape of a Graph

Preliminary Questions

1. Choose the correct answer: If f is concave up, then f is:

(a) increasing
(b) decreasing

2. If x0 is a critical point and f is concave down, then f (x0) is a:
(a) local minimum
(b) local maximum
(c) cannot be determined

Exercises In Questions 3–8, state whether true or false and explain.

3. If f (c) = 0 and f (c) < 0, the f (c) must be a local minimum.
4. A function that is concave down on (−∞, ∞) can have no minimum value.
5. If f (c) = 0, then f must have a point of inflection at x = c.
6. If f has a point of inflection at x = c, then f (c) = 0.
7. If f is concave up and f changes sign at x = c, then f must go from negative to positive

at x = c.
8. If f (c) is a local maximum, then f (c) must be negative.
9. Suppose that f (c) = 0 and that f (x) goes from positive to negative at x = c. Which of

the following statements are correct?
(a) f (x) has a local maximum at x = a.
(b) f (x) has a local minimum at x = a.
(c) f (x) has a local maximum at x = a.
(d) f (x) has a point of inflection at x = a.

1. Match the graphs in Figure 1 with the description:
(a) f (x) < 0 for all x
(b) f (x) goes from + to −
(c) f (x) > 0 for all x
(d) f (x) goes from − to +

(A) (B) (C) (D)

Figure 1

(a) In C, we have f (x) < 0 for all x.
(b) In A, f (x) goes from + to −.

4.4 The Shape of a Graph 23

(c) In B, we have f (x) > 0 for all x.
(d) In D, f (x) goes from − to +.

In Exercises 2–7, determine the intervals on which the given function is concave up and concave
do2w. n and find the points of inflection.

x2 + 7x + 10
3. t3 − 3t2 + 1

Let f (t) = t3 − 3t2 + 1. Then f (t) = 3t2 − 6t and f (t) = 6t − 6 = 0 at t = 1.

Thus f is concave up on (1, ∞), since f (t) > 0 there.
Moreover, f is concave down on (−∞, 1), since f (t) < 0 there.

4. x − 2 cos x
5. (x − 2)(1 − x3)

Let f (x) = (x − 2) 1 − x3 = x − x4 − 2 + 2x3. Then f (x) = 1 − 4x3 + 6x2 and
f (x) = 12x − 12x2 = 12x(1 − x) = 0 at x = 0 and x = 1.

Thus f is concave up on (0, 1) since f (x) > 0 there.
Moreover, f is concave down on (−∞, 0) ∪ (1, ∞) since f (x) < 0 there.

6.
x 7/4

7. x 7/4 − x 2

Let f (x ) = x 7/4 − x 2 and note that f exists for x ≥ 0 only. Then f (x) = 7 x 3/4 − 2x and
4
4 ≈ .185472. (Note:
f (x) = 21 x −1/4 −2 = 0 at x = 21 f (x) is undefined at x = 0.)
16 32

Thus f is concave up on 0, 21 4 since f (x) > 0 there.
32

Moreover, f is concave down on 21 4,∞ since f (x) < 0 there.
32

In Exercises 8–17, find the critical points and apply the Second Derivative Test (if possible) to
de8t.ermi7ne+ifxea−ch3xc2ri−tic8axl 3point is a local minimum or maximum.

9. x3 − 3x2 − 9x

Let f (x) = x3 − 3x2 − 9x. Then f (x) = 3x2 − 6x − 9 = 3(x − 3)(x + 1) = 0 at x = 3,
−1 and f (x) = 6x − 6. Thus, f (3) > 0, which implies f (3) is a minimum and
10. f (−1) < 0, which implies f (−1) is a maximum.
3x4 − 8x3 + 6x2
11. x 5 − x 2

Let f (x) = x5 − x2. Then f (x) = 5x4 − 2x = x(5x3 − 2) = 0 at x = 0, 2 1
5
3 and

f (x) = 20x3 − 2. Thus, f (0) < 0, which implies f (0) is a maximum and f (( 2 ) 1 ) > 0,
5 3

which implies f (( 2 ) 1 ) is a minimum.
5 3

12. x5 − x3

13. sin2 x + cos x

Let f (x) = sin2 x + cos x. Then f (x) = 2 sin x cos x − sin x = sin x(2 cos x − 1) = 0 at

x = 0, π, 2π, . . . , and x = π , 5π , 7π/3, . . . , and f (x) = 2 cos2 x − 2 sin2 x − cos x.
3 3
f (x) > 0 for x = 0, π, 2π, . . . , which implies that f (0), f (π ), f (2π ) and so on are all

minima. f (x) < 0 for x = π , 5π , 7π , . . . , which implies that f ( π ), f ( 5π ), f ( 7π ) and so
3 3 3 3 3 3
on are all maxima.

24 Chapter 4 Applications of the Derivative

14.
1

15. cos x1+ 2
x2 − x + 2

Let f (x) = x2 1 . Then f (x) = −2x + 1 = (x −2x + 1 = 0 at x = 1
−x +2 (x 2 − x + 2)2 − 2)2(x − 1)2 2

and f (x) = −2(x − 2)(x − 1) + 2(2x − 1)2 . Thus f 1 < 0, which implies that f 1
2 2
(x − 2)3(x − 1)3
is a maximum. (Note: f (x), f (x), and f (x) are undefined at x = 1, 2.)
16.

3x 3/2 − x 1/2
17. x 7/4 − x

Let f (x ) = x 7/4 − x . Then f (x) = 7 x 3/4 −1 = 0 at x = ( 4 )4/3 and f (x) = 21 x −1/4 .
4 7 16
4
Thus, f (( 4 )4/3) > 0 which implies f (( 4 ) 3 ) is a minimum.
7 7
18.
Sketch the graph of f (x) = x4 and state whether f has any points of inflection. Verify

19. Tyohuergcroanpchluosfitohnebdyersihvoawtivinegfth(axt) fon(x[0),d1o.2es] insosthcohwangien sFigignu.re 2.

(a) Determine the intervals on which f (x) is increasing and decreasing.

.4 .64 1 1.2
.17

Figure 2

(b) Determine the intervals on which f (x) is concave up and concave down.
(c) Determine the points of inflection of f .
(d) Determine the location of the local minima and maxima of f (x).

Recall that the graph is that of f , not f .
(a) f is increasing on the intervals (0, .17) and (.64, 1) and decreasing on the intervals

(.17, .64) and (1, 1.2).
(b) Now f is concave up where f is increasing. This occurs on (0, .17) ∪ (.64, 1).

Moreover, f is concave down where f is decreasing. This occurs on
(.17, .64) ∪ (1, 1.2).
(c) The inflection points of f occur where f changes from increasing to decreasing or
vice versa. These occur at x = .17, x = .64, and x = 1.
(d) The local extrema of f occur where f changes sign. These occur at x = 0 and x = .4.

In Exercises 20–31, determine the intervals on which the function is concave up and concave
down, find the points of inflection, and determine if the critical points are local minima or max-
ima.

Here is a table legend for Exercises 20–31.

4.4 The Shape of a Graph 25

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.
M The function ( f , g, etc.) is decreasing on the given interval.
m The function ( f , g, etc.) is concave up on the given interval.
I The function ( f , g, etc.) is concave down on the given interval.
¬ The function ( f , g, etc.) has a local maximum at the specified point.
The function ( f , g, etc.) has a local minimum at the specified point.
The function ( f , g, etc.) has an inflection point here.
There is no local extremum or inflection point here.

20.
x3 − 2x2 + x

21. x2(x − 4)

Let f (x) = x2 (x − 4) = x3 − 4x2.

Then f (x) = 3x 2 − 8x = x (3x − 8) = 0 yields x = 0, 8 as candidates for extrema.
3

Moreover, f (x) = 6x − 8 = 0 gives x = 4 as an inflection point candidate.
3

x (−∞, 0) 0 (0, 8 ) 8 ( 8 , ∞) x (−∞, 4 ) 4 ( 4 , ∞)
3 3 3 3 3 3
f + 0−0 + f − 0+

f M m fI
22.

t2 − t3
23. 2x4 − 3x2 + 2

Let f (x) = 2x4 − 3x2 + 2.

√
Then f (x) = 8x3 − 6x = 2x 4x2 − 3 3
= 0 yields x = 0, ± 2 as candidates for

extrema.

Moreover, f (x) = 24x 2 − 6 = 6(4x 2 − 1) = 0 gives x = ± 1 as inflection point
2
candidates.

√√√ √√√
3 3 3 3 3 3
x (−∞, − 2 ) − 2 (− 2 , 0) 0 (0, 2 ) 2 ( 2 , ∞)

f− 0 + 0− 0 +

f m Mm

x (−∞, − 1 ) − 1 (− 1 , 1 ) 1 ( 1 , ∞)
2 2 2 2 2 2
f + 0 − 0+

f I I
24.

x 2 − x 1/2
x

25.
x2 + 2

Let f (x) = x .

x2 + 2

Then f (x) = 2 − x2 = 0 yields x = √ as candidates for extrema.
(x 2 + 2)2 ±2

−2x x2 + 2 − (2 − x2)(2) x2 + 2 (2x) 2x x2 − 6
Moreover, f (x) = = (x 2 + 2)3 = 0
√ (x 2 + 2)4

gives x = 0, ± 6 as inflection point candidates.

26 Chapter 4 Applications of the Derivative

√ √ √√ √ √
x −∞, − 2 − 2 − 2, 2 2 2, ∞

f− 0 + 0−
f mM

√√√ √√√
x −∞, − 6 − 6 − 6, 0 0 0, 6 6 6, ∞

f − 0 + 0− 0 +

f I II
26.

1
27. t2 +1 1

x4 + 1

Let f (x) = 1 .

x4 + 1

Then f (x ) = − (x 4x 3 = 0 yields x = 0 as a candidate for an extremum.
4 + 1)2

Moreover,

f (x) = x 4 + 1 2 −12x 2 − (−4x 3) · 2 x 4 + 1 4x 3

(t 2 + 1)4

4x2 5x4 − 3
= (x 4 + 1)3 = 0

gives x = 0 and x = 3 1/4 ≈ .880111 as inflection point candidates.
5

x (−∞, 0) 0 (0, ∞)
g + 0−
gM

x (−∞, 0) 0 (0, .880111) .880111 (.880111, ∞)
0 +
g +0 − I

gI
28. θ + sin θ for 0 ≤ θ ≤ 2π

29. sin2 t for 0 ≤ t ≤ π

Let f (t) = sin2 t on [0, π ].

Then f (t ) = 2 sin t cos t = sin 2t = 0 yields t = π as a candidate for an extremum.
2

Moreover, f (t) = 2 cos 2t = 0 gives t = π , 3π as inflection point candidates.
4 4

t 0, π π π , π
2 2 2
f +0 −

fM

x 0, π π π , 3π 3π 3π , π
4 4 4 4 4 4
f +0 − 0 +

fI I
30. x − sin x for 0 ≤ x ≤ 2π
π π
31. f (x) = tan x for − 2 <x < 2

Let f (x) = tan x on − π , π .
2 2

Then f (x) = sec2 x ≥ 1 > 0 on − π , π .
2 2

4.4 The Shape of a Graph 27

Moreover, f (x) = 2 sec x · sec x tan x = 2 sec2 x tan x = 0 gives x = 0 as an inflection
point candidate.

x − π , π x − π , 0 0 0, π
2 2 2 2
f+ f − 0+

f fI
32. Explain the statement: the derivative f (x) takes on a local minimum or maximum value at
33. aRp&oiWnt of Ainnfleinctfieocnti.ous 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 the end of the epidemic.

(b) Write a one-sentence news bulletin describing the status of the epidemic on the day
that R(t) has a point of inflection.

(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.”
34. It was shown in Example ?? that x = 0 is a critical point but is not an extreme value for

35. Sketch the graph of a function f (x) such that f (x) > 0 and f (x) < 0 for all x.
s51atxis5f−yin45gx 4
Here is the graph of a functiofn(xf )(x=) 4 +(x )3 >x 3 0+fo1r0all x and f (x) < 0 for all x.

f

Prove that x = 0 is a point of inflection by showing that f (x) changes sign at 0. Hint: find
the zeroes of f (x) and use test values to determine the sign of f (x).

0.5

0 12
–0.5

36.

Sketch the graph of a function f (x) satisfying the two conditions:
37. Sketc(hi)thfe (gxra)p>h o0ffaorfuanllcxtion f (x) satisfying the conditions:

(i) (ifi)(xf) <(x)0<for0xfo<r x1<an0d axn>d f4 (x) > 0 for x > 0
(ii) f (x) > 0 for 1 < x < 4
(iii) f (x) < 0 for x < 2 and f (x) > 0 for x > 2

There is no function f (x) that satisfies the following conditions.

f (x) < 0 for x < 1 and x > 4
f (x) > 0 for 1 < x < 4
f (x) < 0 for x < 2 and f (x) > 0 for x > 2

Suppose such a function exists, and let g(x) = f (x). Then g satisfies the following
conditions.

g(x) < 0 for x < 1 and x > 4
g(x) > 0 for 1 < x < 4
g (x) < 0 for x < 2 and g (x) > 0 for x > 2

28 Chapter 4 Applications of the Derivative

Since g (x) > 0 for x > 2, we have that g(x) is continuous for x > 2. Since g(x) > 0 for
1 < x < 4 and g(x) < 0 for x > 4, we must have g(4) = 0 by continuity. Hence as x
increases through 4, g(x) is decreasing, contradicting the fact that g (x) > 0 for x > 2.
Therefore, no such function f exists.
However, the following graph of f almost fits the bill (except for the fact that f does not
exist at x = 1 or x = 4).

1 246
0.5
–2 0

Further Insights and Challenges

38.
It is clear graphically that a parabola has no points of inflection. Prove this by showing

39. tRha&t tWhe seSchoonwd dthearitveavtievrey ocfubaicqupaodlyrantoicmpiaollyfn(oxm) i=al afx(3x+) =bxa2 x+2 c+x b+xd+(wc i(thwiath=a0=) ha0s)
pnreevceirseclhyanognespsoiignnt.of inflection. Under what conditions (on the coefficients) is the transition

at the inflection point from concave up to concave down (instead of from concave down to

concave up)?

Let f (x) = ax3 + bx2 + cx + d, with a = 0. Then f (x) = 6ax + 2b = 0 implies

f (− b ) = 0. If a < 0, then f changes sign from + to −, changing from concave up to
3a
b
concave down, as x increases through − 3a . If a > 0, then f changes sign from − to +,

changing from concave down to concave up, as x increases through − b . In either case,
3a
b
40. − 3a is an inflection point for f.

Concavity and Tangent Lines This exercise shows that if f is concave up, then its graph

41. SliecsoanbdovDeearlilviatstitvaengTeenst linGesiv(eaasifmorimlaar laprgrouomfeonft tshheoSwesctohnadt tDhergivratpihveliTeessbt:elIofwc istsa
ctarnitgiceanltplionientifsufchistchoant cfav(ec)do>w0n,).thAesnsufm(ce)thisaat lfoc(axl)m>in0imfourmx. iHninant: oupsenthinetfearcvtatlh(aat,tbh)e
afunndclteiot nc F∈((xa),ibn).the previous exercise is positive. Assume that f (x) is continuous, so that
f (x) > 0 for all x in a small open interval containing c.

Suppose that f (c) = 0 and f (c) = LF>ig0u.rTeh3en

(a) Let f (c) = lim f (x) − f (c) = lim f (x) = L > 0.

x→c x − c x→c x − c

Hence there is a δ > F(x) = f (x) − f (c)(x x>f−−(0cx))c)s−h>ofw0(.cth)at
Compute F 0 such that 0 < |x − c| < δ implies
(x) and (under the hypothesis f (x) F (x) < 0 for

If c <ax <<xc <+ cδ a(annddFhe(nxc)e>x 0−focr>c 0<),xth<is bm.eans f (x) > 0.
If (cb−) Aδ p<plxy t<hecM(aenadnhVeanlcuee xTh−eocr<em0t)o, tthhies dmifefaenresnfce(xqu) o<tie0n.t

F(x) − F(c)
Accordingly, f changes sign from − to 0 to + asxx−inccreases through c. By the First
Derivative Test, f (c) is a local minimum value of f . (The case for a local maximum is
similar.) together with part (a) to conclude that F(x) ≥ 0 for all x ∈ (a, b).
42. Tang(ecn) tPCrorvoesstihnagt thUe gseratphhe lmieesthaobdovoef tEhxeetracnisgeen4t0litnoeshatowx =thact.if P = (c, f (c)) is a point
43. oLfetinCfl(exc)tiobne,ththeecnotshteotfapnrgoednutclineg Lx uant iPts mofuastccerrotassinthgeoogdra.pAhs.sHuminet:tShheogwratphhatotfhCe is
cfuonncctaivoen uFp.must change sign at c.

(a) Show that the average cost A(x) = C(x)/x is minimized at that production level x0 for
which average cost equals marginal cost.

Figure 4 Tangent line crosses graph at point of inflection.

4.4 The Shape of a Graph 29

(b) Explain why the line through (0, 0) and (x0, C(x0)) is tangent to the graph of C(x).

Let C(x) be the cost of producing x units of a commodity. Assume the graph of C is

concave up.

(a) Let A(x) = C(x)/x be the average cost and let x0 be the production level at which
(x0) − C (x0)
average cost is minimized. Then A (x0) = x0C x02 = 0 implies

x0C (x0) − C (x0) = 0, whence C (x0) = C (x0) /x0 = A (x0). In other words,

A (x0) = C (x0) or average cost equals marginal cost at production level x0.

(b) The line between (0, 0) and (x0, C(x0) is

C (x0) − 0 (x − x0) + C (x0) = C (x0) (x − x0) + C (x0)
x0 − 0 x0

= A (x0) (x − x0) + C (x0)

= C (x0) (x − x0) + C (x0)

44. which is the tangent line to C at x0.

Let f (x) be a polynomial of degree n.
45. Crit(icaa)l SPhooinwtsthaantdifInnflisecotdidonanPdonin>ts 1,Itfhcenis fa(cxr)ithicaasl aptolienatsbt uotnef (pco)inist noof tinaflleocctailon.

“exretraes(mbon)eaSvbahlelou”wee,xbmaymugspitvlxeins=g(aacnndbeeixtaaimsptopruilneettohofaftainlilffltnehceitsieoexnva?emnApillttehnsoeiuendgthhneothttiehsxaitvs)e,trtahuiepsooeifxnmet roocfissitenflsehcotwiosn.

that it is not true in general. Let

f (x) = x2 sin 1 for x = 0
x for x = 0
0

(a) Show that f (0) exists and is equal to 0. Hint: Show that f (0) = lim h sin 1 and use
h
h→0

the Squeeze Theorem to show that f (0) = 0.

(b) Show that f does not have a local minimum or maximum value at x = 0.

(c) Show that f does not have a point of inflection at x = 0.

Therefore x = 0 is a critical point but f (0) is neither a local extreme value nor a point of

inflection.

Let f (x) = x2 sin (1/x) for x = 0 .
0 for x = 0

(a) Now f (0) = lim f (x) − f (0) = lim x2 sin (1/x) = lim x sin 1 = 0 by the
x→0 x − 0 x→0 x x→0 x

Squeeze Theorem: as x → 0 we have

x sin 1 − 0 = |x| sin 1 → 0,
xx

since the | sin u| ≤ 1.

(b) Since sin( 1 ) oscillates through every value between −1 and 1 with increasing
x
frequency as x → 0, in any open interval (−δ, δ) there are points a and b such that

f (a) = a2 sin( 1 ) < 0 and f (b) = b2 sin( 1 ) > 0. Accordingly, f (0) = 0 can neither
a b
be a local minimum value nor a local maximum value of f .

(c) In part (a) it was shown that f (0) = 0. For x = 0, we have

f (x) = x2 cos 1 1 1 11
− + 2x sin = 2x sin − cos
x x2 x x x

As x → 0, the difference quotient f (x) − f (0) = 2 sin 1 1 1
x − 0 x − cos x

x

oscillates increasingly rapidly between values of increasingly greater magnitude.

30 Chapter 4 Applications of the Derivative

Accordingly, f (0) = lim f (x) − f (0) f does not have a point
does not exist. Hence
x→0 x − 0
of inflection at x = 0.

4.5 Graph Sketching and Asymptotes

Preliminary Questions

1. Match the statement with the appropriate graph and explain.
(a) The outlook is great: the growth rate keeps increasing.
(b) We’re losing money, but not as quickly as before.
(c) We’re losing money, and it’s getting worse as time goes on.
(d) We’re doing well but out growth rate is leveling off.
(e) Business had been cooling off but now it’s picking up.
(f) Business had been picking up but now it’s cooling off.

Exercises Figure 1
1. Determine the sign combinations of f and f for each interval A–G in Figure 2.

A BC D E F G

Figure 2

In A, f < 0 and f > 0.
In B, f > 0 and f > 0.
In C, f > 0 and f < 0.

4.5 Graph Sketching and Asymptotes 31

In D, f < 0 and f < 0.
In E, f < 0 and f > 0.
In F, f > 0 and f > 0.
In G, f > 0 and f < 0.

2. State the sign transition that takes place at each of the points A-G in Figure 3. Example:
f (x) goes from + to − at A.

In Exercises 3–6, draw the graph of a function for which f and f take on the given sign
combinations.

3. ++, +−, −− Figure 3

This function changes from concave up to concave down at x = −1 and from increasing to
decreasing at x = 0.

0

−0.5

−1

−1.5
−2 −1.5 −1 −0.5 0 0.5 1 1.5

4. +−, −−, −+
5. −+, −−, −+

The function is decreasing everywhere and changes from concave up to concave down at

x = −1 and from concave down to concave up at x = − 1 .
2

0.07 0
0.06
0.05
0.04
0.03
0.02
0.01

0
−1.4 −1.2 −1 −0.8 −0.6 −0.4 −0.2

6. −+, ++, +−
7. R & W Are all sign transitions possible for a differentiable function? Explain with a

sketch why the sign transitions ++ → −+ and −− → +− do not occur if the function is
differentiable. (See Ex. 62 for a proof).

In both cases, there is a point where f is not differentiable at the transition from increasing
to decreasing or decreasing to increasing.

32 Chapter 4 Applications of the Derivative

0
−0.2
−0.4
−0.6
−0.8

−1
−1.2
−1.4
−1.6

−4 −3 −2 −1 0 1 2 3 4

1.6
1.4
1.2

1
0.8
0.6
0.4
0.2

0
−4 −3 −2 −1 0 1 2 3 4

In8.Exercises 8–10, sketch the graph of the quadratic polynomial.
x2 − 2x + 3

9. x2 − 4x − 12
Let f (x) = x2 − 4x − 12. Then f (x) = 2x − 4 and f (x) = 2. Hence f has a local
minimum at x = 2 and is concave up everywhere.

50

40

30

20

10

0

−10

−20
−6 −4 −2 0 2 4 6 8 10

10.
3 + 5x − 2x2

11. Sketch the graph of the cubic f (√x) = x3 − 3x2 + 2. For extra accuracy, plot the zeroes of
f (x), which are x = 1 and x = 3 ± 1 or x ≈ −.73, 2.73.
Let f (x) = x3 − 3x2 + 2. Then f (x) = 3x2 − 6x = 3x(x − 2) = 0 yields x = 0, 2and
f (x) = 6x − 6. Thus f has an inflection point at x = 1, a local maximum at x = 0, and a
local minimum at x = 2.

4.5 Graph Sketching and Asymptotes 33

2 12 3
1 x
–1
–1
–2

12.
Show that the cubic x3 − 3x2 + 6x has a point of inflection but no local extreme values.

13. ESkxetetcnhd tthhee gsrkaeptchh. of the graph of f (x) = cos x + 1 x over [0, π ] developed in Example ??
2
to the interval [0, 5π ].

Let f (x) = cos x + 1 x . Then f (x) = − sin x + 1 = 0 yields critical points at x = π , 5π ,
2 2 6 6
13π 17π 25π 29π π 3π 5π 7π 9π
6 , 6 , 6 , and 6 and f (x) = − cos x = 0 yields x = 2 , 2 , 2 , 2 , and 2 as points

of inflection.

7
6
5
4
3
2
1

0 2 4 6 8 10 12 14
x

14.
Sketch the graphs of x 2/3 and x 4/3.

In Exercises 15–24, sketch the graph of the function. Indicate the transition points (local
extrema and points of inflection).

15. x3 + 3 x 2
2

Let f (x) = x3 + 3 x 2 . Then f (x) = 3x2 + 3x = 3x (x + 1) and f (x) = 6x + 3. Sign
2
analyses reveal a local maximum at x = −1, a local minimum at x = 0, and a critical point

at x = − 1 . Here is a graph of f with these transition points highlighted as in the graphs in
2
the textbook.

2

–2 –1 1

–2

16. f (x) = 1 x 3 + x2 + 3x −2
3

34 Chapter 4 Applications of the Derivative

17. 4 − 2x2 + 1 x 4
6

Let f (x) = 1 x4 − 2x 2 + 4. Then f (x) = 32√x 3 − 4x = 23√x x2 − 6 and f (x) = 2x2 − 4.
6

Sign analyses reveal local min√ima at x = − 6 and x = 6, a local maximum at x = 0,
and inflection points at x = ± 2. Here is a graph of f with transition points highlighted.

5

–2 0 2

18.
7x4 − 6x2 + 1

19. 6x7 − 7x6

Let f (x) = 6x7 − 7x6. Then f (x) = 42x6 − 42x5 = 42x5 (x − 1) and

f (x) = 252x5 − 210x4 = 42x4 (6x − 5). Sign analyses reveal a local maximum at x = 0,

a local minimum at x = 1, and an inflection point at x = 5 . Here is a graph of f with
6
transition points highlighted.

1
–0.5 0.5 1

–1

–2

20.
√x 6 − 3√x 4

21. x + 9 − x

Let f (x) = √ √ = x 1/2 + (9 − x )1/2. Then f (x) = 1 x −1/2 − 1 (9 − x )−1/2 and
x + 9−x 2 2

f (x) = − 1 x −3/2 − 1 (9 − x )−3/2 < 0 on (0, 9). Sign analyses reveal a local maximum at
4 4
9
x = 2 . Here is a graph of f with the transition point highlighted.

4
3

02 4 6 8

22.
2 sin x − cos2 x

23. f (x) = x(1 − x)1/3

Let f (x) = x (1 − x)1/3. Then 3 − 4x
3 (x − 1)2/3
f (x) = x · 1 (1 − x )−2/3 (−1) + (1 − x )1/3 ·1 = and similarly
3

6 − 4x
f (x ) = 9 (x − 1)5/3 .

4.5 Graph Sketching and Asymptotes 35

Sign analyses reveal a local maximum at x = 3 and inflection points at x = 1 and x = 3 .
4 2
Here are two graphs of f with the transition points highlighted.

–5
–1 1 2

–10

–1 –15

–20
–2 4 8

24. cos x + x
25. Let f (x) = 6 + 2(x − 3)(x − 1)2/3.

(a) The first two derivatives of f (x) are

f (x) = 10 (x − 9 ) , f (x) = 20 (x − 3 )
3 − 5 9 5

(x 1)1/3 (x − 1)4/3

Use this to determine the transition points and the sign combinations of f and f on
the intervals between the transition points.
(b) Sketch the graph of f (x).

(a) Let f (x) = 6 + 2(x − 3)(x − 1)2/3. Then f (x) = d 10 (x − 9 ) yields critical points at
3 5
(x −1)1/3

x= 9 , 1 and f (x) = 20 (x − 3 ) yields potential inflection points at x = 3 , 1.
5 9 5 5

(x − 1)4/3

Interval Signs of f and f

−∞, 3 +−
5 ++
3 −+
5 , 1 ++

1, 9
5
9
5 , ∞

(b) Here is the graph of f .

6 x 3
12
4

2
–1

0
–2

–4

–6

26.
Which of the graphs in Figure 4 cannot be the graph of a polynomial? Explain.

In Exercises 27–36, calculate the following limits by dividing the numerator and denominator
by the highest power of x appearing in the dFeingoumriena4tor.

x
27. lim

x→∞ x + 9
By the Theorem regarding horizontal asymptotes of a rational function (THARF), we have
lim x = 1 lim 1 = 1.
x→∞ x + 9 1 x→∞

36 Chapter 4 Applications of the Derivative

28. 3x2 + 20x

lim 43xx2 2++920x
29. xl→im∞
x→∞ 2x 4 + 3x 3 − 29

Via THARF, lim 3x2 + 20x = 3 lim x−2 = 0.
30. x→∞ 2x 4 + 3x 3 − 29 2 x→∞
4
lim
31. xl→im∞ 7xx+−59

x→∞ 4x + 3

Via THARF, lim 7x − 9 = 7 lim 1 = 7
.
x→∞ 4x + 3 4 x→∞ 4
32.
7x2 − 9
lim
33. x→lim∞ 47xx+2 −3 9

x→−∞ 4x + 3

7x2 − 9 7
Via THARF, lim = lim x = −∞.
x→−∞ 4x + 3 4 x→−∞
34. 5x − 9

35. lim 4xx 23 −+ 11
x→lim∞
x→−∞ x + 4

x2 − 1 1
Via THARF, lim = lim x = −∞.
36. x→−∞ x + 4 1 x→−∞
2x5 + 3x4 − 31x
37. Dxl→iem∞ter8mxi4n−e w31hixc2h+cu1r2ve in Figure 5 is the graph of f (x) = (2x4 − 1)/(1 + x4) on the basis
of horizontal asymptotes. Explain your reason.

22

–4 –2 2 4 –4 –2 2 4
–1.5 –1.5
(i) Figure 5 (ii)

Since (via THARF)

2x4 − 1 2
lim = lim 1 = 2 and f (0) = −1,
x→±∞ 1 + x 4 1 x→±∞

the left curve is the picture of the graph of f (x) = 2x4 − 1 .

38. 1 + x4
Match the graphs in Figure 6 with the two functions 3x/(x2 − 1) and 3x2/(x2 − 1).

39. MExaptclahinthyeofuurnrcetaiosonnsiwngit.h their graphs in Figure 7.

1 Figure 6
(a)

x2 − 1
x2

(b)
x2 + 1

1
(c)

x2 + 1
x

(d)
x2 − 1

4.5 Graph Sketching and Asymptotes 37

(i) (ii)

(iii) (iv)

Figure 7

(a) The graph of 1/x2 − 1 is (iv).
(b) The graph of x2/x2 + 1 is (i).
(c) The graph of 1/x2 + 1 is (ii).
(d) The graph of x/x2 − 1 is (iii).
40.

Let f (x) = 2x − 1 .
Let f (x) = +x .
41. x 1
(a) Showx2th+at1the line y = 2 is a horizontal asymptote for f .
(a) S(bh)owWthhaatt itshethleinveeryti=cal0aissyamhpotoritzeofnotralfa?symptote for f .
(b) F(cin)dFtihnedctrhiteiccarlitpicoailnptsooinftsf oafndf tahnedinthteerivnatlesrvoanlswohnicwhhficihs ifncisreianscirnegasoinr gdeocrredaescirnega.sing.
(c) S(dh)owDethteartmine the points of inflection and the intervals on which f is concave up or
√ √
(e) concave down. s=ke2tcxh(xth+e(xgr2a+3p)h(1xo)3f−f 3) the behavior as |x | tends
Use the above informatifon(xto)
, indicating

to infinity.

Conclude that the sign of f is equal to the sign of the numerator. Make a sign chart

and find the intervals on which f is positive or negative.

(d) Use the above information to sketch of the graph of f by indicating the behavior as |x|

tends to infinity.

Let f (x) = x .

x2 + 1

(a) Via THARF, lim f (x) = 1 lim x −1 = 0. Hence y = 0 is a horizontal asymptote for f.
x →∞ 1
x →∞

(b) Now f (x) = x2 + 1 · 1 − x · 2x 1 − x2
= (x2 + 1)2 is negative for x < −1 and x > 1,
(x 2 + 1)2
positive for −1 < x < 1, and 0 at x = ±1. Accordingly, f has a local minimum value

at x = −1 and a local maximum value at x = 1.

(c) Moreover,

f (x) = x 2 + 1 2 (−2x ) − 1 − x 2 ·2 x2 + 1 · 2x = 2x3 − 6x
(x + 1)4 (x + 1)3
√√
2x x − 3 x + 3

= (x + 1)3 .

38 Chapter 4 Applications of the Derivative

Here is a sign chart for the second derivative, similar to those constructed in various
exercises in Secti√on 4.4. (T√he legend√is on page 25.) √ √ √

x −∞, − 3 − 3 − 3, 0 0 0, 3 3 3, ∞

f− 0 + 0− 0 +
I I
fI

(d) Here is a graph of f (x) = x .

x2 + 1

0.5 10

–10 0
–0.5

In Exercises 42–57, sketch the graph of the function. Indicate the asymptotes, local extrema,

and points of inflection.
42.

1
43. 2x x− 1

2x − 1

x −1 4
Let f (x) = . Then f (x) = and f (x) = . Sign analyses
2x − 1 (2x − 1)2 (2x − 1)3

reveal no extrema and no points of inflection. Limit analyses give vertical asymptote x = 1
2
1
and horizontal asymptote y = 2 .

10 24 6
8 x
6

y
4
2

–6 –4 –2 0
–2
–4
–6
–8

–10

44. x + 1
45. 2xx+−31

x −2

Let f (x) = x +3 . Then f (x) = −5 and f (x) = 10 . Sign analyses reveal

x −2 (x − 2)2 (x − 2)3
no extrema and no points of inflection. Limit analyses give vertical asymptote x = 2 and

horizontal asymptote y = 1.

20 5 10
y 10 x
–10 –5 0

–10
–20

4.5 Graph Sketching and Asymptotes 39

46. 1
x
47. x1 + 1
+
x x−1

f (x) = 11 2x2 − 2x + 1
Let + . Then f (x) = − and
x x −1 x 2 (x − 1)2

f (x) = 2 2x3 − 3x2 + 3x − 1 . Sign analyses reveal an inflection point at x = 1 . Limit
x 3 (x − 1)3 2

analyses give vertical asymptotes x = 0 and x = 1 and horizontal asymptote y = 0.

5

–1 0 12
–5

48.
49. 2x1−−xx+−1 x1+1 1

Let f (x) = 2−x + 1 f (x) = − x 2 + 2x + 2 = − (x + 1)2 + 1 and
+ 1 . Then (x + 1)2 (x + 1)2
x

2
f (x) = (x + 1)3 . Sign analyses reveal no transition points. Limit analyses give a vertical
asymptote at x = −1.

5

–4 –2 0 24
–5

50. 1
51. x12 −+6x +18

x 2 (x − 2)2

Let f (x) = 1 + (x 1 f (x ) = −2x −3 − 2 (x − 2)−3 and
x2 − 2)2 . Then

f (x) = 6x−4 + 6 (x − 2)−4. Sign analyses reveal a local minimum at x = 1. Limit

analyses give vertical asymptotes x = 0 and x = 2 and horizontal asymptote y = 0.

6

4

2

–2 0 24
–2

40 Chapter 4 Applications of the Derivative

52.
11

53. x 2 4− (x − 2)2
x2 − 9

4 8x 24 x2 + 3
Let f (x) = . Then f (x ) = − (x 2 − 9)2 and f (x) = (x2 − 9)3 . Sign analyses
x2 − 9

reveal a local maximum at x = 0. Limit analyses give x = −3 and x = 3 as vertical

asymptotes and y = 0 as a horizontal asymptote.

2 5

–5 0
–2

54.
1

(x 2 + 1)2

x2
55.

(x2 − 1)(x2 + 1)

x 2 2(x + x 5)
Let f (x) = . Then f (x) = − and
(x2 − 1)(x2 + 1) (−1 + x 4)2

f (x) = 2 + 24x4 + 6x8 . Sign analyses reveal a local minimum at x = 0 and no points of
(−1 + x 4)3

inflection (as the second derivative is always negative).

1
0.8
0.6
0.4
0.2

–10 –8 –6 –4 –2 2 4 6 8 10
x

56. x − 1
57. √x2 −x 1

x2 + 1

Let f (x) = √ x . Then f (x ) = (x 2 + 1)−3/2 and f (x) = −3x . Sign analyses

x2 + 1 (x 2 + 1) 5
2

reveal no extrema and a point of inflection at x = 0. Limit analyses give horizontal

asymptotes y = 1 and y = −1.

4.5 Graph Sketching and Asymptotes 41

1
0.5

–10 –8 –6 –4 –2 2 4 6 8 10
–0.5 x

–1

Further Insights and Challenges

Slant Asymptotes The next exercises explore a function whose graph approaches a
non-horizontal line as x → ∞. A line y = ax + b is a called a slant asymptote if
limx→∞( f (x ) − (ax + b)) = 0 or limx→−∞( f (x ) − (ax + b)) = 0.
58.

Let

59. Sketch the graph of x2 1 . Profc(exe)d=as inx t2he previous exercise (show that
y = x −1 is a slant f (x) = x −1

x+
asymptote).

Let((baf))(xCC)oo=nnfifixrrmmx+2tthh1aa.ttTffhehisnacsfoan(lcxoa)cva=el umx(pix(nox+inm+(1u12)m,2)∞aatn)xdan=fd 2(cxoa)nnc=daav(elxodc+o2awl1nm)3oa.xnSi(mig−un∞man,aa1ty)xlsa=essin0re.Fviegaulre 8.
a loc(acl)mSinhiomwutmhaatt lxim= 0f ,(xa)lo=ca∞l maanxdimluimm aft(x )==−−2∞and that f is concave down on
(−∞, −1) and conxc→av1+e up on (−1, ∞).xL→i1m−it analyses give a vertical asymptote at
(d) Show that y = x + 1 is a slant asymptote of 1f (x) as x → ±∞.

x = (−e)1.SBhyowpotlhyantotmheiaslladnivt iassioynm,pfto(txe)l=iesxb−elo1w+thxe +gra1phanodf f (x) for x > 1 and above the
−gr1ap+h xfo+1r x1<−1(xas−in1d)ic=at0e,dwinhiFcihguimrep8li.es
lim x that the slant asymptote is x − 1. Notice

x →±∞

that f approaches the slant asymptote as in exercise 59.

Figure 8

10 2468
8 x
6

y
4
2

–8 –6 –4 –2 0
–2
–4
–6
–8

–10

60.
Show that y = 3x is a slant asymptote for f (x) = 3x + x−2. Determine whether f (x)

61. aSpkpertcohacthheesgtrhaephslaonft fa(sxy)m=pto1te−frxo2m. above or below and make a sketch of the graph.
2−x

Let f (x) = 1 − x2 . Using polynomial division, f (x) = x + 2 + 3 . Then

2−x x −2

lim ( f (x) − (x + 2)) = lim (x + 2) + 3 − (x + 2)
x →±∞ x →±∞ x −2

= lim 3 = 3 lim x−1 = 0
x→±∞ x − 2 1 x→±∞

42 Chapter 4 Applications of the Derivative

which implies that x + 2 is the slant asymptote of f (x). Since f (x) − (x + 2) = x 3 >0
−2
as x → ∞, f (x) approaches the slant asymptote from above for x > 2 and similarly,

x 3 < 0 as x → −∞ so f (x) approaches the slant asymptote from below for x < 2.
−2
x2 − 4x + 1 −6
Then f (x) = and f (x) = . Sign analyses reveal a local minimum
√ (2 − x)2 (2 −√x)3
at x = 2 + 3, a local maximum at x = 2 − 3 and that f is concave down on (−∞, 2)

and concave up on (2, ∞). Limit analyses give a vertical asymptote at x = 2.

15

10
y

5

–10 –8 –6 –4 –2 2 4 6 8 10
–5 x

–10

62.
Let c be a critical point of a function f (x). Suppose that f (x) and f (x) exist for all x and

that f (x) > 0, so that f (x) is concave up. Show that f cannot make a transition from +

to − at c. Hint: since c is a critical point, the tangent line at x = c is the horizontal line

y = f (c). Now use Exercise 40 in Section ?? which asserts that if a curve is concave up,

4.6 Applied Optimization then it lies above its tangent lines.

Exercises

1. A 50-inch piece of wire is to be bent into a rectangular shape in such a way as to maximize
the area of the rectangle.
(a) Suppose the sides of the rectangle are called x and y. What is the constraint equation
relating x and y?
(b) Find a formula for the area in terms of x alone.
(c) Does this problem require optimization over an open interval or a closed interval?
(d) Solve the optimization problem.

(a) The perimeter of the rectangle is 50 inches, which gives 50 = 2x + 2y, equivalent to

y = 25 − x.

(b) Using part (a), A = x y = x(25 − x) = 25x − x2.

(c) This problem requires an open interval. If x = 0 or 50 there would not be a rectangle at

all.

(d) A (x) = 25 − 2x = 0, which yields x = 25 and consequently, y = 25 and the
2 2
maximum area is A = 156.25 square inches.
2.

A 100-inch piece of wire is divided into two pieces and each piece is bent into a square.
3. OThuer goal is to mfininditmheizpeotshietivsuemnuomf btheer axresuaschofthtahtetthweosusqmuaorfexs.and its reciprocal is as

smal(laa)s Lpoetssxibalned. y be the lengths of the two pieces. Express the sum of the areas of the
(a) Doesqthuiasrepsroinblteemrmrseqoufixreaonpdtiym.ization over an open interval or a closed interval?

(b) S(bo)lvWe thaetoispttihmeiczoatnisotnrapinrot belqeuma.tion relating x and y?

Let x(c>) D0 oaensdthfi(sxp)r=oblxem+ rxe−q1u.ire optimization over an open interval or a closed interval?
(d) Solve the optimization problem.

4.6 Applied Optimization 43

(a) Here we required optimization over an open interval, since x ∈ (0, ∞).

(b) Solve f (x) = 1 − x−2 = 0 for x > 0 to obtain x = 1. Since f (x) = 2x−3 > 0 for all
x > 0, f has an absolute minimum of f (1) = 2 at x = 1.

4. Consider a right triangle whose sides other than the hypotenuse have lengths a, b such that
5. Fai+ndbp=osi1ti0v.eHnouwmbsheorsuxld, ay asuncdhbthbaetcxhyos=en1t6oamndaxxim+izye itsheasasremaaollfaths epotrsisainbglele.?

Let x, y > 0. Now x y = 16 implies y = 16 . Let f (x) = x + y = x + 16x−1. Solve
f (x ) = 1 − 16x −2 = x 4. Since f (x) = 32x−2 > 0 for x >
0 for x > 0 to obtain x = 0, f

6. has an absolute minimum of f (4) = 8 at x = 4.
A 20-inch piece of wire is bent into the shape of an L (it is bent into a right angle). Where

7. Cshoonusliddethrethbeesnedt boef malladreecitfanthgeledsiswtaitnhcearbeeatw10e0e.n the two ends is as small as possible?

(a) What are the dimensions of the rectangle with least perimeter?

(b) Is there a rectangle with greatest perimeter? Explain.

Consider the set of all rectangles with area 10.

(a) Let x, y > 0 be the lengths of the sides. Now x y = 100, whence y = 100/x. Let
p(x) = 2x + 2y = 2x + 200x−1 be the perimeter. Solve p (x) = 2 − 200x−2 = 0 for
x > 0 to obtain x = 10. Since p (x) = 40x−3 > 0 for x > 0, the least perimeter is
p (10) = 40 when x = 10 and y = 10.

(b) There is no rectangle in this set with greatest perimeter. For as x → ∞, we have
p(x) = 2x + 20x−1 → ∞.

8.

A box has a square base of side x and height y.
9. Supp(oas)eFthinadt 6th0e0dftimofenwsiiroenasrxe,uysefdortowbhuicilhdtahecovrorlaulminetihse1s2haanpde othfeasruercftaacnegalerewa iitshaas small as

semicirclpeowsshibolsee. diameter is a side of the rectangle as in Figure 1. Find the dimensions of
the c(obr)raFl iwndiththme adximimeunmsioanrseaf.or which the surface area is 20 and the volume is as large as

possible.

Figure 1

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 + π x = 2y + (1 + π )x = 600 ft or equivalently, y = 1 600 − (1 + π )x .
2 2 2 2

The area of the corral is the sum of the area of the rectangle and semicircle,

A = xy + π x2. Making the substitution for y from the constraint equation,
8
1 π π 1 π π
A(x ) = 2 x 600 − (1 + 2 )x + 8 x2 = 300x − 2 1 + 2 x2 + 8 x2.

A (x) = 300 − 1 + π x + π x = 0 implies x = 300 ≈ 168.029746.
2 4
(1+ π )
4

The optimization is over an open interval and sign analyses reveal that x = 300 is a

(1+ π )
4
local maximum. Thus, the corral of maximum area has dimensions x and

y = 84.01487303.

10.
Find the rectangle of maximum area that can be inscribed in a right triangle with sides of
length 3 and 4 and hypotenuse 5 (the sides of the rectangle should be parallel to the sides of
the triangle).

Figure 2

44 Chapter 4 Applications of the Derivative

11. A rectangular garden of total area 100 ft2 is to be enclosed with a brick wall costing $30
per foot on one side and a metal fence costing $10 per foot on the remaining sides. Find the
dimensions of the garden that minimize the cost.

Let x be the length of the brick wall and y the length of an adjacent side with x, y > 0.

With xy = 100 or y = 100 , the total cost is C(x) = 30x + 10√(x + 2y) = 40x + 2000x −1.
x

Solve C (x) = 40 − 2000x−2 = 0 for x > 0 to obtain x = 5 √2. Since √
C (x) = 400√0x−3 > 0 for all x > 0, t√he minimum cost is C(5 2) = 400 2 ≈ $565.69
when x = 5 2 ≈ 7.07 ft and y = 10 2 ≈ 14.14 ft.
12.
Find the point on the line y = x closest to the point (1, 0).

13. Find the point P on the parabola y = x2 closest to the point (1, 0).

P

1

Figure 3

With y = x2, let’s equivalently minimize the square of the distance,
f (x) = (x − 1)2 + y2 = x4 + x2 − 2x + 1. Solve f (x) = 4x3 + 2x − 2 = 0 to obtain

√
(54 + 6 87)2/3 − 6
x = x0 = √ ≈ 0.59
6(54 + 6 87)1/3

(plus two complex sol√utions, which we discard). Since f (x) = 12x2 + 2 > 0 for all x, the
minimum distance is f (x0) ≈ 0.54 (the exact answer is rather lengthy) when x = x0,
14. y = x02 ≈ 0.35, and P ≈ (0.59, 0.35).
A box is constructed out of two different types of metal. The metal for the top and bottom

15. cFoinstdst$h1e pdeimr seqnusaiorensfoooftthanedretchteanmgeletaol fomr athxeimsiudmesacroeasttsh$a2t cpaenr bsqeuianrsecrfioboetd. Iifntahecibroclxeiosfto

hraadvieusa rv.olume of 20 cubic feet, how should the box be designed to minimize cost?

Figure 4

Place the center of the circle at the origin with the sides of the rectangle (of lengths 2x > 0

and 2y > 0) p√arallel to the coordinate axes. By the Pythagorean Theorem, x2 +√y2 = r2,
whence y = r2 − x2. Thus the area of the rectangle is A(x) = 2x · 2y = 4x r2 − x2.

4x 2 = 0 for x > 0 to obtain x = √r . Since
Solve A (x) = 4 r2 − x2 − √ 2
r2 − x2
4x 3
A (x) = − √ 12x − (r 2 − x 2)3/2 < 0 for x > 0, the maximum area is A( √r ) = 2r 2
r2 − x2 2

when x = y = √r .
2

4.6 Applied Optimization 45

16.
Rice production requires both labor and capital investment in equipment and land. Suppose

17. Ktheapt ilferx’sdWollianres BpearrarcerleParroebilnevmesteTdhine flaoblloorwaindg yprdooblleamrs wpears asctareteadreanindvseosltveeddinin a work

epPnqutub=iitlplies1mhd0ee0Ndn√otivnaxan1+ds6tl1ea5rne0dbo√,ymtytheh.eternIifaatashdterfooaylrniimooerlmeudrmeiPrnvJvoioenfhsartarsinci$one4repu0semKrpeae(rScpraoleeclrirides(1,Gg5hiev7ooew1mn–s1ebh6tyro3yut0hlo)de.f faoWrmiunleaBarrel),
Wtheha$t4a0rebethdeivided

bdeimtweenesniolnasbofr tahnedccyalpinitdaelrinovfelsatrmgeesnttvionluomrderthtoatmcanxibmeizinestchreibaemdoinuntht eofsprihcerperoofdruacdeidu?s
R? Hint: Show that the volume of the cylinder is 2π(R2 − x2)x where x is one-half the

height of a cylinder.

Place the center of the sphere at the origin in three-dimensional space. Let the cylinder be

of radius x and half-height y. The Pythagorean Theorem states, x2 + y2 = R2, whence

x2 = R2 − y2. The volume of the cylinder is

V (y) = π x2 (2y) = 2π R2 − y2 y = 2π R2 y − 2π y3. Allowing for degenerate cylinders,

we have 0 ≤ y ≤ R. Solve V (y) = 2π R2 − 6π y2 = 0 for y ≥ 0 to obtain y = √R . Since
√3
V (0) = V (R) = 0, the largest volume is V ( √R ) = 4 π 3R3when y = √R and x = 2 R .
18. 3 9 3 3

Find the dimensions x, y of a rectangle inscribed in a circle in such a way as to maximize
19. tFhiendquthaentaitnyglxeyθ2.that maximizes the area of the quadrilateral with base of length 4 and sides

of length 2, as in Figure 5.

22
θ

4

Figure 5

Allowing for degenerate quadrilaterals, we have 0 ≤ θ ≤ π . Via trigonometry and surgery

(slice off a right triangle and rearrange the quadrilateral into a rectangle), we have that the

area of the quadrilateral is equivalent to the area of a rectangle of base 4 − 2 cos θ and

height 2 sin θ ; i.e,

A(θ ) = (4 − 2 cos θ ) · 2 sin θ = 8 sin θ − 4 sin θ cos θ = 8 sin θ − 2 sin 2θ , where

0 ≤ θ ≤ π . Solve A (θ ) = 8 cos θ − 4 cos 2θ = 4 + 8 cos θ − 8 cos2 θ = 0 for 0 ≤ θ ≤ π

to obtain

π √ 3 − 1 ≈ 1.94553.
2 2
θ = θ0 = + sin−1

√√
Since A(0) = A(π ) = 0, the maximum area is A(θ0) = 31/4(3 + 3) 2 or approximately
20. 8.80734 when θ = θ0.

Consider a rectangular industrial warehouse consisting of three separate spaces of equal
21. Ssiuzpepaossien, iFnigthuerep6re. vAiossuusmexeetrhcaist eth, ethwatatlhl emwataerreihalosucsoesits$t2o0c0onpseirsltionfeanr sfet paanrdattehespcaocmespaonfy

eaqlluoaclatseizse$.2F,4in0d0,a0f0o0rmfourltahienptreorjmecsto. f n for the maximum possible area of the warehouse.

For (na)coWmhpiacrhtmdeimntesn, wsioitnhsxmaanxdimyiazse btheefotroet,alcoasret a=of2,t4h0e0w,0a0re0h=ous2e0?0((n + 1)x + 2ny)
and (yb=) W12h0a0t 0is−th(ena+rea1)oxf .eTachhencompartment in this case?

2n

A = nxy = 12000 − (n + 1)x = 6000x − n + 1x2
x Fi2gure 6 2

6000 3000
and A = 6000 − (n + 1)x = 0 yields x = n + 1 and consequently y = . Thus the
maximum area is given by n

A=n 6000 3000 = 18,000,000 .
n+1 n n+1

46 Chapter 4 Applications of the Derivative

22.
The amount of light that reaches a point located at a distance r from a light source of

23. iAnctecnosridtiyngI tios proosptaolrtrieognuallatoionI /s,ra2.cSaurtpopnocsaentwboe shoiuprpceds ofvelirgshetasAoannlyd iBf tohfeisnutemnsoiftyitsIA
ahnedighIBt ,arnedspgeirctthivdeolye,sanroetsexpcaeraetded74byinachdeist(atnhceegoirfth10ofeaept.aFciknadgethies pthoeindtiostnanthce saergomunednthe
jmoidndinleg).AFaind tBhewdhimerentshieontostoalf athmeoruenctaonfgluiglahrt cisarmtoiniwmiitzhetdh.e maximum volume,

assuming that the base of the carton is a square.

Let the height of the carton be y and the length of a side of its square base be x. The

carton’s girth is 2x + 2y. Clearly the volume will be maximized when the sum of the height

and girth equals 74; i.e., 2x + 3y = 74, whence y = 74 − 2 x . Allowing for degenerate
3 3
74 2
cartons, the carton’s volume is V (x) = x2y = 3 x 2 − 3 x 3, where 0 ≤ x ≤ 37. Solve

V (x) = 148 x − 2x2 = 0 for x to obtain x = 0 or x = 74 . Since V (0) = V (37) = 0, the
3 3
74 74
maximum volume is V 3 = 405224/81 ≈ 5002.77 in3 when x = 3 ≈ 24.67 in and

24. y= 74 ≈ 8.22 in.
9

(Problems 24–25 and 45 from Calculus Problems for a New Century.) Find the area of the

25. Alarrgeecsttatnrgialenghlaestdhiamt ceannsiboensfoLrmaned iWn .thWe hfiartstisqtuhaedraarneat boyf the lxa-ragxeisst, rye-catxanisg,laentdhaat tcaanngebnet

ctoirtchuemgsrcarpibheodfayro=un(dx2it?+H1i)n−t1:.Express the areas of the circumscribed rectangle in terms of

the angle θ . Observe that each side of the inner rectangle is the hypotenuse of a right

triangle in Figure 7.

θ
R

L

Figure 7

Position the L × W rectangle in the first quadrant of the x y-plane with its “southwest”

corner at the origin. Let θ be the angle the base of the circumscribed rectangle makes with

the positive x -axis, where 0 ≤ θ ≤ π . Then the area of the circumscribed rectangle is
2
1 1 1 2 W 2) sin 2θ ,
A = LW + 2 · 2 (W sin θ )(W cos θ) + 2 · 2 ( L sin θ )(L cos θ ) = LW + 2 ( L +

26. which has a maximum value of L W + 1 (L2 + W 2) when θ = π .
2 4

Optimal Price Let r be the monthly rent per unit in an apartment building with 100

27. Gunivitesn. Snunpupmosbeetrhsaxt1a,l.l .u.n,itxsnc,alnetbxe brenttheednautmr b=er$f3o0r0wahnidchththatewsuitmh each additional $10 in

rent, two units become vacant. Each occupied unit requires $30 in service per month.
(a) Show that the num(xb−er xo1f)u2 n+its(xre−ntexd2)i2s+16·0· ·−+5r(fxo−r 3x0n0)2≤ r ≤ 800.

is mi(nbi)mFailn. Sdhaofwortmhautlax fiosretqhueaml toonthelyacvaesrahgientvaakleue(revenue minus service).

(c) Which rent r maximizes cash intake?
1
n (x1 + x2 + · · · + xn )

Let f (x) = nk=nk=11x(kx. − xk )2. Solve f (x) = 2 n (x − xk) = 0 to obtain
Since f n 1 k=1 for all x, we conclude
= x¯ = 1 (x) = 2 k=1 = 2n > 0
x n that f is

28. minimal when x = x¯.

Let P = (a, b) be a point in the first quadrant.
(a) Find the slope of the line through P such that the triangle bounded by this line and the
axes in the first quadrant has smallest possible area.

(b) Show that P is the midpoint of the hypotenuse of this triangle with smallest area.

4.6 Applied Optimization 47

29. Let be the triangle with vertices (0, 0), (5, 0), (0, 10). Determine the rectangle of largest
area whose sides are parallel to the x- and y-axes, which can be placed inside . Use the
second derivative test to insure that a minimum has been found.

The line through (5, 0) and (0, 10) is y = 10 − 2x. The area of the rectangle is

A(x ) = xy = 10x − 2x 2. Solve A (x) = 10 − 4x = 0 to obtain x = 5 . Since
2
5 25 5
30. A (x) = −4 < 0 for all x, the minimum area is A 2 = 2 when x = 2 and y = 5.

The problem is to connect three towns (A, B, and C) by an underground fiber cable in the

31. cInonthfiegupraetviioonuisllpursotbraletemd,isnhFoiwguthreat8f.oTrhtehehovarilzuoenotaflxdmistiannimceizDingantdhethlenvgetrhtiocaflthdeisctaanbcle, L

athre tghivre√een.aOngulresgoeamlains atotinfigndfrothme tvhaelumeiodfdxlebpeotiwnet ePn a0reanadllLeqtuhalt tmoi1n2im0◦i.zeAsstshuemteottahlat
aDm≤ou2nt o3f Lcasbolethuastexd.= c. Hint: let θ = A P B. By symmetry, it suffices to show that the

uppe(raa)nLgleet θf (=x)1b2e0◦t.hTeotodtoaltlheins,gothbsoefrcvaebtlheaut sDed/.2Sxh=owtatnhaθt/2f .(x√) has a unique critical point c.
Compute c and show that c ∈ [0, L√] if and only if D ≤ 2 3 L.
32. (b) Show that if the inequality D ≤ 2 3 L is satisfied, then the minimum is achieved for

A flowerxst=orec.r(eOquthireerswaisterutchke tmoimniamkuemdeolicvceurrisesa.tAosnseuomfeththeattwtoheenpdripcoeinotfsaannedwintrfuacckt, iist will
33. I$n30th,0e0s0eta-hnuedpntoohfcacthutehr epatrceoxvsi=tooufLs m)p.raoibnlteamin,inhgowa toruftcekntshyoeuarlds ias truck be purchased so as to

m28i0n0im0(i1ze+av√erta)g−e1 yearly cost assuming that after t years the old truck can be sold for
dollars? C(t) = 1280 + 360(t2 + t)

dAonlleawrs.truck costs $30,000, the cost of mFaiginutareini8ng it t years is C(t) = 1280 + 360 t2 + t

dolla(ras), aSnhdoawfttehratt yfe(atr)s=the1t (o3ld00tr0u0c+k cCan(tb))e rseoplrdefsoernts2t8h0e√0t0otadloallvaerrsa.ge yearly cost of keeping
a truck t years. 1+ t

(a) T(bh)e Atostasul mcoisntgfothrakteaepuisnegd atrutrcukckhatsyzeearros tirsa3d0e0-i0n0v+aluCe(,th)o−w o2f8te0n√00sh.oHuledncaetrtuhcek be
purchased, if the goal is to minimize average yearly cost?1 + t

average cost of keeping it t years is

A(t) = 1 30000 + C(t) − 280√00 = 360t + 360 + 31280t−1 − 2800√0
t 1+ t t 1+ t

(b) Numerically solve

A (t) = 360 − 31280t−2 + 2800√0 + 14000 =0
t2 1 + t t 3/2 √
1+ t 2

for t > 0 to obtain t ≈ 7.659 years. Now A(t) ↑ ∞ as t ↓ 0 or t ↑ ∞. Accordingly,
the minimum average yearly cost is approximately A (7.659) ≈ $6,230.96 when
t ≈ 7.659 years. A new truck should be purchased every 7.659 years to minimize costs.

The next two exercises refer to Example ?? in the text.
34.

Find the x-coordinate of the point P where the light beam strikes the mirror, assuming that
35. Sh1up=po1s0e, thh2at=th5e,paonidntLs A=a2n0d. B have the same distance from the mirror (i.e., h1 = h2).

(a) Show that θ1 = θ2.

(b) Can you find a reason why we must have θ1 = θ2 in this case, without doing any
calculation? Hint: use symmetry.

Suppose that A and B have the same distance from the mirror; i.e., h1 = h2 = h.
(a) Substitute h1 = h2 = h into

x L−x x L−x
= to obtain √ = .
x 2 + h21 x2 + h2 (L − x )2 + h2
(L − x )2 + h 2
2

Solving for x gives x = L . Similar triangles then gives θ1 = θ2.
2

(b) Since h1 = h2 = h, the diagram must look the same from the front and back of the

page. This can only occur if θ1 = θ2.

48 Chapter 4 Applications of the Derivative

36.
What is the height of the rectangular box (with no top) of greatest volume that can be
constructed from 100 square inches of cardboard if the base is twice as long as it is wide?

In the next three problems (courtesy of Kay Dundas), a pizza box (not including its top) of
height h is to be constructed from a piece of cardboard of dimensions A × B (see Figure 9).

B

h
A

Figure 9

37. (a) Show that the volume of the box constructed from cardboard as in Figure 9 is
V (h) = h(A − 2h)(B − 2h)

(b) Which value of h maximizes the volume if A = 15 and B = 24? What are the
dimensions of the resulting box?

(a) Use the diagram in Exercise ?? (except substitute h for t) with x + 2h = A = 15 and
y + 2h = B = 24, whence x = A − 2h = 15 − 2h and y = B − 2h = 24 − 2h. The

volume of the box is

V (h) = hx y = t (A − 2h) (B − 2h) = t (15 − 2h) (24 − 2h) = 4h3 − 78h2 + 360h,

where 0 ≤ h ≤ 15 (allowing for degenerate boxes). Note that V (0) = V 15 = 0, as
2 2
expected.

(b) Solve V (h) = 12h2 − 156h + 360 = 0 for h to obtain h = 3 or h = 10. Toss out
h = 10 (since it’s outside the range of h). This leaves height h = 3 with maximum
volume V (3) = 486 from a box of dimensions h = 3, x = 9, and y = 18.

38. Which value of h maximizes the volume of the box if A = B?
39. Suppose that 144 square inches of cardboard is to be used to construct the pizza box

(A B = 144).

(a) Find the values of A and B that give a box of maximum volume, assuming the height
of the box is h = 3.

(b) Suppose more generally that h = c where c is a constant. Show that the values of A

and B giving a box of maximum volume does not depend on c.

(a) Use the diagram in Exercise ?? with t = 3, x + 2t = A, y + 2t = B, and A B = 144.

From the four preceding equations we have (x + 6) (y + 6) = 144, whence
108 − 6x 6 (18 − x)
y = 144 −6= x +6 = x+6 .
x +6

The volume of the box is V (x) = txy = 18x (18 − x) where 0 ≤ x ≤ 18 (allowing
x +6 ,

for degenerate boxes). Note that V (0) = V (18) = 0 as expected.

Solve V (x) = −18 x2 + 12x − 18 = 0 to obtain x = −18 (outside the range of

(x + 6)2
x) and x = 6 at which the maximum volume V (6) = 108 occurs. Thus y = 6,

A = 12, and B = 12, with all lengths in inches.

4.6 Applied Optimization 49

(b) Use the diagram in Exercise ?? with t = c > 0, x + 2t = A, y + 2t = B, and

A B = 144. From the four preceding equations we have (x + 2c) (y + 2c) = 144,

whence y = 2 72 − 2c2 − cx .

x + 2c

The volume of the box is V (x) = t x y = 2cx 72 − 2c2 − cx , where

x + 2c
0 ≤ x ≤ 72 − 2c (allowing for degenerate boxes). Note that
c
72
V (0) = V − 2c = 0 as expected.
c

Solve V (x) = −2c2 x2 + 4cx + 4c2 − 144 = 0 to obtain x = −2c − 12 (outside

(x + 2c)2
the range of x) and x = 12 − 2c at which the maximum volume

V (12 − 2c) = 4c (6 − c)2 occurs. Thus y = 12 − 2c, A = 12, and B = 12, with all

lengths in inches.

40. The monthly output P of a light bulb factory is given by the formula P = 350L K where L
41. (isa)thSehaomwouthnattoafmmoonngeayllinrvigehstetdriainnglalebsorwaitnhdhKypiosttehneusaemoofulnetnogfthmoonnee,ythineviessotsecdeliens

equitprmianengtl.eIhfatshemcaoxmimpuanmyanreeead. s to produce 10,000 units per month, how should the

(ibnv)eCstamneynot ubesedeivmidoerde admireocntglylawbhoyr athnids emqusipt mbeentrtuteobmyirneiamsiozneinthgefcromst oFfigpurroed1u6ct?ion? The
cost of production is L + K .

(a) Position the right angle of the triangle at the origin of the x y-plane with its legs along

the positive x- and√y-axes. Since the hypotenuse has length 1, the area of the triangle is
A(x ) = 21√x y = 1 x 1 − x2, 0 ≤ x ≤ 1, allowing for degenerate tri√angles. Solve
2

A (x) = 1 − x2 − x2 = 0 for 0 ≤ x ≤ 1 to obtain x = 2
√ . Since
2 2 1 − x2 2√
√
2 1 2
A(0) = A(1) = 0, the maximum area of A( 2 ) = 4 occurs when x = 2 . The sine of

√
2
angle θ between the positive y-axis and the hypotenuse is sin θ = x/1 = 2 , whence

θ = π . Therefore, the triangle isosceles.
4

(b) In the figure, let θ be the acute angle above the label a. Let f (θ ) be the area of the right

triangle whose hypotenuse is the diameter of the semicircle. Allowing for degenerate

triangles, we have 0 ≤ θ ≤ π . Since f is symmetric with respect to the value θ = π
2 4
π π π
(i.e., for 0 ≤θ ≤ 4 , we have f 2 − θ = f (θ )) and because f (0) = f 2 = 0, the
maximum area occur when = In is
must θ π . this case the triangle in question

isosceles. 4

42. The minimum force required to drive a wedge of angle α
Coefficient of Static Friction

43. Iintothae bselottcinkg(FoifgEurxeer1c0is)eis4p2r,osphoorwtiothnatl thoe minimal force required is proportional to

1/ 1 + f 2. Hint: find the optimal angle α and then compute F(α).
1
k F(α) =
Let F (α) = sin α +f , where k > 0sisnaαp+ropfocrotisoαnality constant and 0 ≤ α ≤ π .
cos α 2

wSohlevree f is the so-called “coefficient of static friction.” Find the angle α for which the least

force is required, assuming f = .4. k ( f sin α − cos α)
F (α) =
sin α + 2 cos α 2 = 0
5

for 0 ≤ α ≤ π to obtain α = tan−1( 1 ).FSiignucereF1(00) = 1 k and F ( π ) = k, we conclude that
2 f f 2

the minimum force of F(tan−1( 1 )) = k f2 is required when α = tan−1( 1 ).
f 1+ f

50 Chapter 4 Applications of the Derivative

Further Insights and Challenges

44.

R & W Bird Migration By definition, power P (in joules per second) is the rate at

45. wRh&ichWeneCrgoynEtin(uininjgouwlietsh)tihsecnoontsautmioendopfetrheunpirtetvimioeus(ienxseerciosned, sa)e.rOodrnyintahmoliocgaisntaslyhsaivse

sfohuonwdstthhaatttihnegpeonwerearl,ctohnesupmowederb(yinaJc/esrt)acinonpsiugmeoendflbyyinagbairtdvefllyoicnigtyavt v(ienlomci/tsy) vis(dinesmcr/isb)eids
dweeslclrbibyetdhebyfuanfcutinocntioPn(vP)(=v)1=7va−v1−+1 +10b−v3v3 3wjhoeurlesap,ebr asreecopnodsi.tTivheecpoingsetoantcsatnhasttodreepeand
otontacleortfa5in· c1h0a4rajocutelerisstoifcsusoafbtlheeebnierdrg.yTahse btotdayl dfaist.tance it can travel at velocity v is
D(v)(a=) FEivn/dPth(ev)vewlohceirteyEvp(mainctohnasttmanint)imisizthees puosawbelre
that (b) Show that if the bird uses all its energy, then ceonnesrguymipttcioann.store as body fat. Show v is
the total distance it can fly at velocity

D(v) = 5 · 104(v/P(v)). Hint:vfidmnadx t>hevtpimmine T at which all the energy is used up and
then compute distance as vT .
wher(ec)vdSmhaxo,wvptmhinatatrheeavseilnotchiteypvretvhiaotums aexxiemrcizisees.total distance traveled satisfies

vLdmetax((P,deu))(sveFFP)iiP=nn(ddv(a)vttvhh)=ee−=1mvP+ea(Plovxbv(ci)vvm/it3)vyu...mTvThdhmteoiansxtagtPlhivda(eitvssmt)a−=anaxciev−m−(a2iiznv+e−ks2i3tl+oobtmva32leb=tdvei2rssat=)avnt−hc01aeivt.+mthpbelvibe3sir=vdpcmaaivnn−=fl2 y+.3abb v12/4w. Thoichfi,nd
simp(liffi) eEd,xypilealidns, u2sbivng4 =(c)2aanadnFdigfiunraell1y1v, dwmahxy=vdmabax 1i/s4o. bTthaeined as the velocity coordinate of the
vpmin = p3aobin1t/4w<hereab a1/l4in=e tvhdrmoaux.gNh othtieceortihgeindiifsfetarnengceentbteotwtheeengrtahpehtwofoPv(avlu)e. s is the 3 in
46. the denominator of vpmin.
Jane can swim 3 mph and run 8 mph. She is standing at one bank of a river that is 300 ft

47. wGaidseolained iws adnetlsivteoreredatcohaagpaosinstaltoiocnatevdFe2irg0y0utrfdetady1os1w. Snustprpeaomse othnathteheotchoesrtspideredaesliqvueircykliys das
pdolslsairbslea.nSdhtehewailvlesrwagime ddaiialygocnoasltlyofasctroorsisngthtehreivgearsoalnidnethisenstjodgolallaornsgwtherreivderanbdanska.rFeind
cthoensbteasnttsro. uStheofworthJant ef t(ot)ta=ked. t−1 + st represents the total average daily cost and find the

value of t that minimizes average daily cost.

The total cost to operate the gas station over t days is d + t · st. Thus the total average daily

cost is f (t) = d +t ·s t = d + st. Solve f (t) = s − dt−2 = 0 for t > 0 to obtain t = d .
t t s
√
Since f (t) = 2d/t3 > 0 for all t > 0, the minimum average daily cost is f ( d ) = 2 ds
s

when t = d .
48. s

Problem of Tartaglia (1500–1557) Among all numbers a, b whose sum is 8, find those

49. (foar) wFhinicdhththeerapdroiudsuacnt dofhtehieghtwt ofnaucmyblienrdsraicnadl tchaenirodfitfofetarel nsucerfiascelaargreast.AHwinhto:sTehveoqluumanetitsy

to beasmlaxrgime aizsepdoisssaibblex. where x = a − b. Write this quantity in terms of x alone and find

(ibts)mCaaxnimyuomu .design a cylinder with total surface area A and minimal total volume?

Let a closed cylindrical can be of radius r and height h.

(a) Its total surface area is S = 2πr2 + 2πr h = A, whence h = A − r. Its volume is
2π r
V (r) = πr2h = 1 − πr3. (r ) = 1 − 3πr 2 >
thus 2 Ar Solve V 2 A for r 0 to obtain

r= A (r) = −6πr < 0 for r > 0, the maximum volume is
6π . Since V

√
A 6 A3/2 A 1 6A
V = √ when r = and h = .
6π 18 π 6π 3 π

(b) For a can of total surface area A, there are cans of arbitrarily small volume since
lim V (r) = 0.

50. r→0+
Find the area of the largest isosceles triangle that can be inscribed in a circle of radius r.

51. Snell’s Law The x-axis represents the surface of a swimming pool. A light beam travels
from point A located above the pool to point B located underneath the water. Let v1 be the
velocity of light in air and let v2 be the velocity of light in water (it is a fact that v2 < v1).
Show that the path from A to B that takes the least time satisfies the relation

sin(θ1) = sin(θ2) .
v1 v2