Solucionario Zill Cap 4

Chapter 4

Applications of the Derivative

4.1 Rectilinear Motion

1. s(1/2) = −1, s(3) = 19; v(t) = 8t − 6, v(1/2) = −2, v(3) = 18, |v(1/2)| = 2, |v(3)| = 18;a(t) = 8, a(1/2) = 8, a(3) = 8

2. s(1) = 16, s(4) = 4; v(t) = 2(2t − 6)(2) = 8t − 24, v(1) = −16, v(4) = 8, |v(1)| = 16,|v(4)| = 8; a(t) = 8, a(1) = 8, a(4) = 8

3. s(−2) = 18, s(2) = 6; v(t) = −3t2 + 6t+ 1, v(−2) = −23, v(2) = 1, |v(−2)| = 23, |v(2)| = 1;a(t) = −6t+ 6, a(−2) = 18, a(2) = −6

4. s(−1) = 1, s(3) = 57; v(t) = 4t3 − 3t2 + 1, v(−1) = −6, v(3) = 82, |v(−1)| = 6, |v(3)| = 82;a(t) = 12t2 − 6t, a(−1) = 18, a(3) = 90

5. s(1/4) = −15/4, s(1) = 0; v(t) = 1 + 1/t2, v(1/4) = 17, v(1) = 2, |v(1/4)| = 17, |v(1)| = 2;a(t) = −2/t3, a(1/4) = −128, a(1) = −2

6. s(−1) = −1, s(0) = 0; v(t) = 2/(t + 2)2, v(−1) = 2, v(0) = 1/2, |v(−1)| = 2, |v(0)| = 1/2;a(t) = −4/(t+ 2)3, a(−1) = −4, a(0) = −1/2

7. s(1) = 1, s(3/2) = 1/2; v(t) = 1 + π cosπt, v(1) = 1 − π, v(3/2) = 1, |v(1)| = π − 1,|v(3/2)| = 1; a(t) = −π2 sinπt, a(1) = 0, a(3/2) = π2

8. s(1/2) = 0, s(1) = −1; v(t) = −πt sinπt+ cosπt, v(1/2) = −π/2, v(1) = −1, |v(1/2)| = π/2,|v(1)| = 1; a(t) = −π2t cosπt− 2π sinπt, a(1/2) = −2π, a(1) = π2

9. v(t) = 2t− 4

(a) Solving t2 − 4t − 5 = 0 gives t = −1, 5. The velocity when s(t) = 0 is v(−1) = −6,v(5) = 6.

(b) Solving t2 − 4t − 5 = 7 gives t = −2, 6. The velocity when s(t) = 7 is v(−2) = −8,v(6) = 8.

10. v(t) = 2t+ 6; a(t) = 2

194

4.1. RECTILINEAR MOTION 195

(a) Setting t2 + 6t+ 10 = 2t+ 6 we obtain t2 + 4t+ 4 = 0 or (t+ 2)2 = 0. Thus t = −2 ands(−2) = 2.

(b) Solving 2t+ 6 = −2 gives t = −4. The velocity when v(t) = −a(t) is v(−4) = −2.

11. v(t) = 3t2 − 4; a(t) = 6t

(a) Solving 3t2 − 4 = 2 gives t = ±√

2. When v(t) = 2, a(−√

2) = −6√

2, a(√

2) = 6√

2.(b) Solving 6t = 18 gives t = 3. Then s(3) = 15.(c) Solving t3 − 4t = t(t + 2)(t − 2) = 0 gives t = 0,±2. Then v(0) = −4, v(−2) = 8,

v(2) = 8.

12. v(t) = 3t2 − 6t; a(t) = 6t− 6

(a) Solving 3t2 − 6t = 0 gives t = 0, 2. Then s(0) = 8, s(2) = 4.(b) Solving 6t− 6 = 0 gives t = 1. Then s(1) = 6.(c) The particle is slowing down when its velocity and acceleration have opposite algebraic

signs. Since two numbers will have opposite signs when their product is negative, weconsider v(t)a(t) = (3t2 − 6t)(6t− 6) = 18t(t− 2)(t− 1) < 0. Solving this inequality, wesee that v(t) and a(t) will have opposite signs when t < 0 and 1 < t < 2.The particle is speeding up when its velocity and acceleration have the same algebraicsign. Since two numbers will have the same sign when their product is positive, weconsider v(t)a(t) = 18t(t − 2)(t − 1) > 0. Solving this inequality, we see that v(t) anda(t) will have the same signs when t > 2 or 0 < t < 1.

13. v(t) = 3t2 − 27 = 3(t− 3)(t+ 3); a(t) = 6t

v < 0 v > 0v > 0

a < 0 a > 0

t–3 0 3

The particle is slowing down on (−∞,−3) and on (0, 3); it is speeding up on (−3, 0) and on(3,∞).

14. v(t) = 4t3 − 3t2 = t2(4t− 3); a(t) = 12t2 − 6t = 6t(2t− 1)

v > 0v < 0

a < 0 a > 0a > 0

0 1/2 3/4t

The particle is slowing down on (−∞, 0) and on (1/2, 3/4); it is speeding up on (0, 1/2) andon (3/4,∞).

196 CHAPTER 4. APPLICATIONS OF THE DERIVATIVE

In order to draw the graphs in Problems 15–28 we need to determine when the particle changesdirection. For a continuous position function, this will occur when the velocity is 0. This isa necessary condition; it is not sufficient . That is, the velocity may be 0 without the particlechanging direction (see, for example, Problem 16). The arrows → and ← in the charts indicatethe direction of motion on the specified interval, as determined by the sign of the velocity on thatinterval.

15. v(t) = 2t; a(t) = 2. Solving v = 0 we obtain t = 0.

t −1 0 3s 1 ← 0 → 9v − 0 +a +

s0 1 9

t = 0t = –1

t = 3

The particle is slowing down on (−∞, 0) and speeding up on (0,∞).

16. v(t) = 3t2; a(t) = 6t. Solving v = 0 we obtain t = 0; solving a = 0 we obtain t = 0.

t −2 0 2s −8 → 0 → 8v + 0 +a − 0 +

s–8 0 8

t = 0t = –2 t = 2


17. v(t) = 2t− 4; a(t) = 2. Solving v = 0 we obtain t = 2.

t −1 2 5s 3 ← −6 → 3v − 0 +a +

s–6 3

t = 2t = –1

t = 5


18. s(t) = t2 + 2t− 3; v(t) = 2t+ 2; a(t) = 2. Solving v = 0 we obtain t = −1.

t −3 −1 1s 0 ← −4 → 0v − 0 +a +

s–4 0

t = –1t = –3

t = 1


The particle is slowing down on (−∞,−1) and speeding up on (−1,∞).

19. v(t) = 6t2− 12t = 6t(t− 2); a(t) = 12t− 12. Solving v = 0 we obtain t = 0, 2; solving a = 0we obtain t = 1.

t −2 0 1 2 3s −40 → 0 ← −4 ← −8 → 0v + 0 − 0 +a − 0 +

s–40 0

t = –2t = 0

t = 3

–8

t = 2

–4

t = 1

The particle is slowing down on (−∞, 0) and (1, 2); it is speeding up on (0, 1) and (2,∞).

20. v(t) = (t− 1)2 + (t− 2)[2(t− 1)] = (t− 1)(3t− 5); a(t) = (t− 1)(3) + (3t− 5)(1) = 6t− 8.Solving v = 0 we obtain t = 1, 5/3; solving a = 0 we obtain t = 4/3.

t −2 1 4/3 5/3 3s −36 → 0 ← −2/27 ← −4/27 → 4v + 0 − 0 +a − 0 +

s–36 4

t = –2t = 1

t = 3

–2/27

t = 5/3

0

t = 4/3

–4/27

The particle is slowing down on (−∞, 1) and (0, 5/3); it is speeding up on (1, 4/3) and(5/3,∞).

21. v(t) = 12t3 − 24t2 = 12t2(t− 2); a(t) = 36t2 − 48t. Solving v = 0 we obtain t = 0, 2.

t −1 0 2 3s 11 ← 0 ← −16 → 27v − 0 − 0 +

s–16 27

t = 2

t = 3

t = 0 t = –1

0 11

22. v(t) = 4t3 − 12t2 − 16t = 4t(t + 1)(t− 4); a(t) = 12t2 − 24t− 16. Solving v = 0 we obtaint = −1, 0, 4.


t −2 −1 0 4 5s 76 ← 57 → 60 ← −68 → −15v − 0 + 0 − 0 +

s–68 76

t = –2

t = 0

t = 5

57

t = 4

60

t = –1

–15

23. v(t) = 1− 2t−1/2; a(t) = t−3/2. Setting v = 0 we obtain 2/√t = 1, so

√t = 2 and t = 4.

t 1 4 9s −3 ← −4 → −3v − 0 +

s–4 –3

t = 4

t = 9

t = 1

24. v(t) = −π sinπt; a(t) = −π2 cosπt. Setting v = 0 we obtain sinπt = 0. Thus, for −1/2 ≤t ≤ 5/2, we have t = 0, 1, 2.

t −1/2 0 1 2 5/2s 1 → 2 ← 0 → 2 ← 1v + 0 − 0 + 0 −

s0 2

t = –1/2t = 0

t = 5/2

1

t = 2

t = 1

25. v(t) =π

2cos

π

2t; a(t) = −π

2

4sin

π

2t. Setting v = 0 we obtain cos

π

2t = 0. Thus, for

0 ≤ t ≤ 4, we have t = 1, 3.


t 0 1 3 4s 0 → 1 ← −1 → 0v + 0 − 0 +

s–1 1

t = 4

t = 0

0

t = 1

t = 3

26. v(t) = π cosπt + π sinπt; a(t) = π2 sinπt + π2 cosπt. Setting v = 0 we obtain cosπt =− sinπt or tanπt = −1. Thus, for 0 ≤ t ≤ 2, we have t = 3/4, 7/4.

t 0 3/4 7/4 2s −1 →

√2 ← −

√2 → −1

v + 0 − 0 +

s–√2 √2

t = 2

t = 0

0

t = 7/4

t = 3/4

27. v(t) = t3(−e−t)+3t2(e−t) = e−t(3t2−t3) = t2e−t(3−t); a(t) = e−t(6t−3t2)+(−e−t)(3t2−t3) = te−t(t2 − 6t+ 6). Setting v = 0 we obtain t = 0, 3. In addition, lim

t→∞v(t) = 0.

t 0 3 ∞s 0 → 27e−3 ← 0v 0 + 0 −

s0 27e–3

t = 0

t = 3

t → ∞

28. v(t) = 2t− 12t+ 1

; a(t) = 2 + 12(t+ 1)−2. Setting v = 0 we obtain t = 2 for 0 ≤ t <∞. In

addition, limt→∞

v(t) =∞.

t 0 2 ∞s 0 ← 4− 12 ln 3 → ∞v − 0 +

t → ∞

s4 – 12 ln 3 0

t = 2t = 0


29.Interval v(t) a(t)

(a, b) + −(b, c) 0 0(c, d) + +(d, e) + −(e, f) − −(f, g) − +

The particle is slowing down on (a, b), (d, e), and (f, g); it is speeding up on (c, d) and (e, f).

30.

a b c

31. (a) v(t) = −32t+48. Solving v = 0 we obtain t = 3/2. The velocity is positive on (−∞, 3/2)and negative on (3/2,∞).

(b) The maximum is attained when the velocity is 0. This height is then s(3/2) = 42 ft.

32. v(t) = −2t + 10. Solving v = 0 we obtain t = 5. For t < 5, we have v > 0 and the particleis moving to the right. On [1, 5] it moves |s(5)− s(1)| = |5− (−11)| = 16 cm. For t > 5, wehave v < 0 and the particle is moving to the left. On [5, 6] it moves |s(6)− s(5)| = |4− 5| = 1cm. Thus, the total distance traveled is 17 cm.

33. s(t) = 16t2 sin 30◦ = 8t2; v(t) = 16t; a(t) = 16. At the bottom of the hill, s = 8t2 = 256and t =

√32 = 4

√2. The velocity is v(4

√2) = 64

√2 ft/s and the acceleration is a(4

√2) =

16 ft/s2.

34. s(t) = 16t2 sin θ. Since sin θ = 3/5 in this case, s(t) =485t2. Then v(t) =

965t and a(t) =

965

.

At the bottom of the hill, s =485t2 = 500 ft and t =

504√

3=

252√

3s. The velocity is

v(25/2√

3) = 240/√

3 ft/s and the acceleration is a(25/2√

3) = 96/5 ft/s2.

35. We are given θ = 16t2. Since the circle has radius 1, y = sin θ = sin 16t2 and dy/dt =32t cos 16t2. For t =

√π/4, dy/dt = 8

√π cosπ = −8

√π ft/s. Since dy/dt is negative, the

y-coordinate is decreasing.

36. F =d

dt(mv) =

d

dt

(m0v√

1− v2/c2

)

=(1− v2/c2)1/2m0

dv

dt−m0v

[12

(1− v2/c2)−1/2(−2v/c2)dv

dt

]

1− v2/c2

=(1− v2/c2)m0a−m0v(v/c2)a

(1− v2/c2)3/2=

m0a

(1− v2/c2)3/2=

m0a√(1− v2/c2)3

4.2. RELATED RATES 201

4.2 Related Rates

1. Let V be the volume and x the length of a side. Then V = x3 anddV

dt= 3x2 dx

dt.

2. From V = xyz we find

dV

dt= x

d

dt(yz) + yz

dx

dt= x

(ydz

dt+ z

dy

dt

)+ yz

dx

dt= xy

dz

dt+ xz

dy

dt+ yz

dx

dt.

Using the given sides and rates, we obtaindV

dt= 1(2)(10) + 1(3)(10) + 2(3)(10) = 110 cm3/s.

3. Let A be the area and let x be the length of a side. Then A =√

3x2

4and

dA

dt=√

32xdx

dt.

Whendx

dt= 2 and x = 8 we have

dA

dt=√

32

(8)(2) = 8√

3 cm2/h.

4. From Problem 3 we have A =√

3x2

4and

dA

dt=√

32xdx

dt. Since A =

√75, we have x2 = 20

and x = 2√

5. Whendx

dt= 2 cm/h and x = 2

√5 cm,

dA

dt=√

32

(2√

5)(2) = 2√

15 cm2/h.

5. Let x be the length, y the width, and s the diagonal of the rectangle. Then s2 = x2 + y2 or

y2 = s2 − x2, and 2ydy

dt= 2s

ds

dt− 2x

dx

dtor

dy

dt=s

y

(ds

dt

)− x

y

(dx

dt

). When x = 8 in and

y = 6 in, s = 10 in. Thendy

dt=

106

(1)− 86

(14

)=

43

in/h.

6. Let x be the side of a cube and s the diagonal. Then s2 = 3x2 and 2sds

dt= 6x

dx

dtor

ds

dt= 3

(xz

)(dxdt

). When

dx

dt= 5 cm/h,

ds

dt= 3

(x√3x

)(5) = 5

√3 cm/h.

7. sin θ = x/s or x = s sin θ. Differentiating with respect to t givesdx

dt= s

d

dtsin θ+ (sin θ)

ds

dt=

s cos θdθ

dt+ sin θ

ds

dt.

8.dy

dt= 2x

dx

dt+ 4

dx

dt. When x = 2 cm and

dx

dt= 3 cm/min,

dy

dt= 2(2)(3) + 4(3) = 24 cm/min.

When y = 6, we have x2 +4x+1 = 6, x2 +4x−5 = 0, and (x+5)(x−1) = 0. Thus x = −5, 1.

Sincedy

dt= 2x

dx

dt+ 4

dx

dt, then for

dx

dt= 3,

dy

dt

∣∣∣∣x=1

= 2(1)(3) + 4(3) = 18 cm/min anddy

dt

∣∣∣∣x=−5

= 2(−5)(3) + 4(3) = −18 cm/min.

9. 2ydy

dt=dx

dt;

dy

dt=dx/dt

2y. From y2 = x+1 we see that for x = 8, y = ±3. Since

dx

dt= 4x+4,

we havedy

dt=

4x+ 42y

. Thusdy

dt

∣∣∣∣y=3

=4(8) + 4

2(3)= 6 and

dy

dt

∣∣∣∣y=−3

=4(8) + 42(−3)

= −6.


10. 4dy

dt= 2x

dx

dt+dx

dt. Since

dx

dt=

dy

dt, we cancel the derivatives and obtain 4 = 2x + 1 or

x =32

. From 4y = x2 + x we see that for x =32

, y =1516

. Hence the point on the graph is

(3/2, 15/16).

11. If T is the area of the triangle then T =12xy =

12x4/3 and

dT

dt=

23x1/3 dx

dt. When x = 8,

thendT

dt=

23

(8)1/3

(13

)=

49

cm2/h.

10 y

40

s

12. Using similar triangles, y/s = 10/40 or y = s/4. Thendy

dt=

14· dsdt

and sinceds

dt= 2 ft/s,

dy

dt=

14

(2) =12

ft/s.

20

l5

xs

13. (a) Since the lengths of corresponding sides in similar triangles are

proportional,520

=l

l + xor l = x/3. When

dx

dt= 3 ft/s, differ-

entiating givesdl

dt=

13· dxdt

=13

(3) = 1 ft/s.

(b) Differentiating s = l+x givesds

dt=dl

dt+dx

dt. Since

dx

dt= 3 ft/s and from (a)

dl

dt= 1 ft/s,

the tip of the shadow is moving at a rate ofds

dt= 1 + 3 = 4 ft/s.

14. (a) Since D = 2r, we havedD

dt= 2

dr

dt= 2 · 2 = 4 ft/s.

(b) Since C = 2πr, we havedC

dt= 2π

dr

dt= 2π · 2 = 4π ft/s.

(c) Since A = πr2 we havedA

dt= 2πr

dr

dt. When

dr

dt= 2 ft/s and r = 3 ft,

dA

dt= 2π · 3 · 2 =

12π ft2/s.

(d) Since A = πr2 = 8π we have r = 2√

2. Whendr

dt= 2 ft/s and r = 2

√2 ft,

dA

dt=

2π(2√

2)(2) = 8π√

2 ft2/s.

h15

x

15. From the Pythagorean Theorem, 152 = h2 + x2. Differentiating gives

0 = 2hdh

dt+ 2x

dx

dtor

dh

dt= −x

h· dxdt.

When x = 5 ft, h = 10√

2 ft, anddx

dt= 2 ft/min, we have

dh

dt= − 5

10√

2(2) = − 1√

2ft/min.

h20

x

16. From the Pythagorean Theorem, 202 = h2 + x2. Differentiating gives

0 = 2hdh

dt+ 2x

dx

dtor

dx

dt= −h

x· dhdt.


When h = 18 ft, x = 2√

19 ft, anddh

dt= −1

2ft/min, we have

dx

dt= − 18

2√

19

(−1

2

)=

92√

19ft/min.

17. Since θ1 =π

2− θ2,

dθ1

dt= −dθ2

dtand θ1 is increasing at the same rate θ2 is decreasing.

200l

x

18. From the Pythagorean Theorem, l2 = x2 + 2002. Differentiating gives

2ldl

dt= 2x

dx

dtor

dx

dt=l

x· dldt.

We are givendl

dt= 3 ft/s, and when l = 400 ft we have x2 = 4002 − 2002 or x = 200

√3 ft.

Then the kite moves at a rate ofdx

dt=

400200√

3(3) = 2

√3 ft/s.

yl

x

19. From the Pythagorean Theorem, x2 + y2 = l2. Differentiating gives

2xdx

dt+ 2y

dy

dt= 2l

dl

dt.

We are givendx

dt= 10 knots and

dy

dt= 15 knots, so

dl

dt=

10x+ 15yl

. At 2:00 PM, x = 20

nautical miles, y = 15 nautical miles, and l =√

202 + 152 = 25 nautical miles. Thus

dl

dt=

10(20) + 15(15)25

= 17 knots.

yl

x

20. From the Pythagorean Theorem, x2 + y2 = l2. Differentiating gives

2xdx

dt+ 2y

dy

dt= 2l

dl

dt.

We are givendy

dt= −9 km/h and

dx

dt= 12 km/h. Then

dl

dt=

12x− 9yl

. At 9:20 AM,

y = 20− 43

(9) = 8 km, x =43

(12) = 16 km, and l =√

162 + 82 = 8√

5 km. Therefore

dl

dt=

12(16) + 9(8)8√

5= 3√

5 km/h.

12 y

x

21. Let x be the distance from the boat to the base of the dock and y be the distancefrom the boat to the pulley. From the Pythagorean Theorem, x2 + 144 = y2.

Thus, 2xdx

dt= 2y

dy

dtand

dx

dt=(yx

) dydt

. We are givendy

dt= −1 and x = 16. At

this time, y = 20 anddx

dt=

2016

(−1) = −54

ft/s. Hence, the boat is approaching

the dock at a rate of54

ft/s.


10h

θ22. We are given

dh

dt= −1 and we want to find

dθ

dtwhen h = 30. Then

θ = csc−1 h

10,dθ

dt=

−1/10(h/10)

√(h/10)2 − 1

· dhdt

, and

dθ

dt

∣∣∣∣h=30

=−1

30√

9− 1(−1) =

√2

120rad/s.

x

1/2θ

23. We are givendx

dt= 15 and we want to find

dθ

dtwhen x = 1/2. Then

θ = tan−1 x

1/2= tan−1 2x,

dθ

dt=

21 + 4x2

· dxdt

, anddθ

dt

∣∣∣∣x=1/2

=

1(15) = 15 rad/h.

90 hT

s

f

24. From the Pythagorean Theorem, s2 = f2 + 902. Differentiating gives

2sds

dt= 2f

df

dtor

ds

dt=f

s· dfdt.

We are givendf

dt= −20 ft/s, and when the runner is 60 ft from home

base, f = 90− 60 = 30 and s =√

302 + 902 =√

9000 = 30√

10. Then

ds

dt=

3030√

10(−20) = − 20√

10≈ −6.325 ft/s.

The distance from the runner to second base is decreasing at a rate of approximately 6.325 ft/s.

Similarly, T 2 = h2 + 902 anddT

dt=

h

T· dhdt

. We are givendh

dt= 20 ft/s, and when h = 60,

T =√

602 + 902 =√

11,700 = 30√

13. Then

dT

dt=

6030√

13(20) =

40√13≈ 11.094 ft/s.

2l

x

25. From the Pythagorean Theorem, l2 = x2 + 4. Differentiating gives

2ldl

dt= 2x

dx

dtor

dl

dt=x

l· dxdt.

We are givendx

dt= −600 mi/h, and when x = 1.5 mi, l = 2.5 mi. Then

dl

dt=

1.52.5

(−600) = −360 mi/h.

Thus the distance is decreasing at a rate of 360 mi/h.


2

l

x120°

26. From the law of cosines, x2 = 22 + l2 − 2(2)l cos 120◦ = 4 + l2 + 2l. Differ-entiating gives

2xdx

dt= (2l + 2)

dl

dtor

dx

dt=l + 1x· dldt.

We are givendl

dt= 600 mi/h. After one minute, the plane has travelled 10

miles. When l = 10 mi, x2 = 4 + 102 + 2(10) and x = 2√

31 mi. Thus,

dx

dt=

10 + 12√

31(600) =

3300√31≈ 592.70 mi/h.

x

4

θ

27. Differentiating x = 4 cot θ, we obtaindx

dt= −4 csc2 θ

dθ

dt. Converting 30◦

to π/6 radians, we are givendθ

dt= −π

6. Thus, when θ = 60◦,

dx

dt= −4(csc2 60◦)

(−π

6

)=

8π9≈ 2.79 km/min.

1200

h

θ

28. Differentiating x = 1200 tan θ givesdx

dt= 1200 sec2 θ

dθ

dt. We are given

dθ

dt= 0.1.

When θ =π

6,dx

dt= 1200

(sec2 π

6

)(0.1) = 1200(4/3)(0.1) = 160 km/s.

29. Let y be the altitude of the rocket, x the distance along the ground from the point of launch,and s the distance the rocket has travelled.

(a) y = s sin 60◦ =√

32s;

dy

dt=√

32· dsdt

. Whends

dt= 1000,

dy

dt=√

32

(1000) =

500√

3 mi/h.

(b) x = s cos 60◦ =12s;

dx

dt=

12· dsdt

. Whends

dt= 1000,

dx

dt=

12

(1000) = 500 mi/h.

30. V = πr2h. Since r is a constant, differentiating with respect to t givesdV

dt= πr2h

dh

dt. When

r = 40/2 = 20 ft anddh

dt= −3/2 ft/min,

dV

dt= π(20)2(−3/2) = −600π ft3/min. Thus the

volume is decreasing at a rate of 600π ft3/min.

31. V = πr2h. Since r is a constant, differentiating with respect to t givesdV

dt= πr2h

dh

dtor

dh

dt=

1πr2· dVdt

. When r = 8 m anddV

dt= 10 m3/min, the oil level rises at a rate of

dh

dt=

10π(8)2

=5

32πm/min.


32. The volume of water is V = 5xh, sodV

dt= 5x

dh

dt+ 5h

dx

dt. We are given

dV

dt= 1,

dx

dt=

112

,

and x = 4. From V = 40 and x = 4 we see that h = 2. Then 1 = 5(4)dh

dt+ 5(2)

112

anddh

dt=

1120

ft/min =110

in/min. The water is rising at this instant.

r

3

h9

33. (a) Since the lengths of corresponding sides in similar triangles are pro-

portional,h

9=r

3or h = 3r. The volume of water is V =

13πr2h =

13π

(h

3

)2

h =127πh3. Differentiating gives

dV

dt=

19πh2 dh

dtor

dh

dt=

9πh2· dVdt.

We are givendV

dt= −1. When h = 6, the water level is changing at a rate of

dh

dt=

9π(36)

(−1) = − 14π

ft/min.

(b) From h = 3r we havedh

dt= 3

dr

dt, so when h = 6, the radius of water is changing at a

rate ofdr

dt=

13· dhdt

=13

(− 1

4π

)= − 1

12πft/min.

(c) The initial volume of water is V0 =13π(3)2(9) = 27π ft3. At time t the volume of water

is V (t) = V0 − t = 27π − t. We also have from part (a) that V =13πr2h =

13πr2(3r) =

πr3. Thus, πr3 = 27π − t or r =(

27− t

π

)1/3

. Thendr

dt= − 1

3π

(27− t

π

)−2/3

and

dr

dt

∣∣∣∣t=6

= − 13π(27− 6/π)2/3

≈ −0.0124 ft/min.

1

21/2

a1h

l

34. Since the lengths of corresponding sides in similar triangles are pro-portional, a/h = 1/2 and l = 1 + 2a = 1 + h. The volume of water is

V =h(1 + l)

2(4) = 2h(2 + h) = 4h+ 2h2. Differentiating gives

dV

dt= (4 + 4h)

dh

dtor

dh

dt=

14 + 4h

· dVdt.

When h =14

m anddV

dt=

12

m3/s, the rate at which the water level rises isdh

dt=

14 + 4(1/4)

·12

=110

m/s.


s/2

h s

35. (a) From the Pythagorean Theorem, s2 = h2 + (s/2)2 or s =2h√

3. The

volume of the water is V =12sh(20) = 10

2h√3h =

20h2

√3

. Differentiat-

ing with respect to t givesdV

dt=

40h√3· dhdt

sodh

dt=√

340h· dVdt

. When

h = 1 ft anddV

dt= 4 ft3/min, the rate at which the water level rises

isdh

dt=√

310

ft/min.

(b) From part (a) we see that the initial volume of water is V0 =20h2

0√3

. At time t, the

volume of water is V = 4t + V0 = 4t +20h2

0√3

. In terms of h, we saw in part (a) that

V =20h2

√3

. Thus,20h2

√3

= 4t+20h2

0√3

. Solving for h and differentiating, we find

h =

√√3

5t+ h2

0 anddh

dt=

12

(√3

5t+ h2

0

)−1/2 √3

5=√

310

(√3

5t+ h2

0

)−1/2

.

(c) Setting h = 5 and h0 =12

in part (b), we have 5 =

√√3

5t+

14

or t =165√

34

≈ 71.45 min.

The rate at which the water is rising when t =165√

34

is

dh

dt

∣∣∣∣t=165

√3/4

=√

310

(√3

5· 165

√3

4+

14

)−1/2

=√

350≈ 0.035 ft/min.

36. The volume between the spheres is V =43πr3

0 −43πr3i . Differentiating gives

dV

dt= 4πr2

0

dr0

dt−

4πr2i

dridt

. Fordr0

dt= 2 m/hr,

dridt

= −12

m/hr, r0 = 3 m, and ri = 1 m, we havedV

dt=

4π(9)(2)− 4π(1)(−1/2) = 74π m3/h.

37. The volume of a sphere is V =43πr3. Differentiating gives

dV

dt= 4πr2 dr

dt. The surface area

of a sphere is S = 4πr2, sodV

dt= S

dr

dt. Since we are given that

dV

dt= kS, we have

dr

dt= k.

Thus, the radius changes at a constant rate.

38. The volume is V =43πr3, so

dV

dt= 4πr2 dr

dt= k, where k is a constant. Now, the surface

area is S = 4πr2 anddS

dt= 8πr

dr

dt. From the formula for

dV

dt, we have

dr

dt=

k

4πr2. Thus,

dS

dt= 8πr

(k

4πr2

)=

2kr

, and the rate of change of the surface area is inversely proportional

to the radius.


39. V = x3,dV

dt= 3x2 dx

dt. The surface area of the cube is S = 6x2, so when S = 54, x = 3. Now

whendV

dt= −1

4we have −1

4= 3(3)2 dx

dtand

dx

dt= − 1

108. From

dS

dt= 12x

dx

dtwe use x = 3

anddx

dt= − 1

108to compute

dS

dt= 12(3)

(− 1

108

)= −1

3in2/min.

(x, y)

y

x

–64

60

θ

ground

40. Place the origin at the center of the ferris wheel with the x-axisparallel to the ground. To find the vertical and horizontal rates,we use y = 60 sin θ and x = 60 cos θ. Differentiating, we havedy

dt= 60 cos θ

dθ

dtand

dx

dt= −60 sin θ

dθ

dt. Assuming the wheel re-

volves counter-clockwise once every two minutes,dθ

dt= π radians per

minute. Thus,dy

dt= 60π cos θ and

dx

dt= −60π sin θ.

When the passenger is 64 feet above the ground, θ = 0, sin θ = 0, and cos θ = 1. Thus, the

passenger is risingdy

dt= 60π ft/min, and is moving horizontally

dx

dt= 0 ft/min.

(x, y)

y

x

–64

60

θ

ground

41. Place the origin at the center of the ferris wheel with the x-axis parallelto the ground. The coordinates of the point P are x = 60 cos θ andy = 60 sin θ. If the coordinates of the point Q are (q,−64) then

the slope of the line through PQ is60 sin θ + 6460 cos θ − q . Since this line is

perpendicular to the line through the center of the wheel and P , its

slope is also − 1tan θ

.

Solving60 sin θ + 6460 cos θ − q = − 1

tan θfor q we obtain q = 60 cos θ + 64 tan θ + 60 sin θ tan θ and

dq

dt=(−60 sin θ + 64 sec2 θ + 60 sin θ sec2 θ + 60 sin θ

) dθdt

. When θ = π/4 anddθ

dt= π we

havedq

dt

∣∣∣∣θ=π/4

= (60√

2 + 128)π ≈ 668.7 radians/min.

sθ/2

15

15

42. (a) From the figure, we can see that tan θ/2 =15

s− 15and so

θ/2 = tan−1 15s− 15

and θ = 2 tan−1 15s− 15

.

(b) Differentiating the expression in (a) yields

dθ

dt=

2

1 +(

15s− 15

)2 ·−15

(s− 15)2· dsdt

= − 30(s− 15)2 + 225

· dsdt.

From s = −16t2 − t+ 200 we getds

dt= −32t− 1, so when t = 3, s = 53 and

ds

dt= −97.

Thus,dθ

dt

∣∣∣∣t=3

= − 30(53− 15)2 + 225

(−97) ≈ 1.74 rad/s.


(c) As s→ 15, θ → 2 lims→15

tan−1 15s− 15

= 2 limx→∞

tan−1 x = 2(π/2) = π.

(d) To find when the diver hits the water, we solve −16t2−t+200 = 15, obtaining t ≈ 3.37 s.

Thendθ

dt

∣∣∣∣t=3.37; s=15

= − 30225

[−32(3.37)− 1] ≈ 14.51 rad/s.

43. R−1 = R−11 +R−1

2 . Differentiating with respect to t gives

−R−2 dR

dt= −R−2

1

dR1

dt−R−2

2

dR2

dtand

1R2· dRdt

=1R2

1

· dR1

dt+

1R2

2

· dR2

dt

sodR

dt=R2

R21

· dR1

dt+R2

R22

· dR2

dt.

44. From PV 1.4 = k we obtain Pd

dtV 1.4 + V 1.4 dP

dt= 0 and P (1.4V 0.4)

dV

dt+ V 1.4 dP

dt= 0 so

dP

dt= −1.4PV 0.4

V 1.4· dVdt

. When P = 100 lb/in2, V = 32 in3 anddV

dt= −2 in3/s, so

dP

dt= −1.4(100)(32)0.4

(32)1.4(−2) = 8.75 lb/in2/s.

45. (a) From R =C

T=

0.493T − 0.913T

= 0.493 − 0.913T

we finddR

dt=

0.913T 2

· dTdt

> 0. Thus,the ratio increases.

(b) To find the value of T when C =T

3, we solve

T

3= 0.493T − 0.913, obtaining T ≈ 5.718.

ThendR

dt

∣∣∣∣T=5.718

≈ 0.913(5.718)2

(1) ≈ 0.028 = 2.8%/day.

46. The rate of change of length isdL

dt=

18− 1020

= 0.4 cm per million years. From E =

0.007P 2/3 = 0.007(0.12L2.53)2/3 ≈ 0.0017L1.68667 we obtaindE

dt≈ 0.0029L0.68667 dL

dt=

0.0029L0.68667(0.4) = 0.00115L0.68667. To determine the value of L when the fish was halfits final body weight, we note that the final body weight is P = 0.12(18)2.53 and solve12

(0.12)(18)2.53 = 0.12L2.53. This gives L ≈ 13.69 mm. Thus, the rate at which the species’sbrain was growing is

dE

dt

∣∣∣∣L=13.69

≈ 0.00115(13.69)0.68667 ≈ 0.0069 g/million years.

47. (a) FromdP

dt= 800

dm

dtand

dm

dt= 30 kg/h, we see that the momentum is changing at a

rate of 800(400) = 24, 000 kg km/h.

(b) In this case, both m and v are variables sodP

dt= m

dv

dt+ v

dm

dt. At t = 1 hour the mass

of the airplane is 105 + 30 = 100,030 kg and the velocity is 750 km/h. Thus,

dP

dt

∣∣∣∣t=1

= 100,030(20) + 750(30) = 2,023,100 kg km/h2.


4.3 Extrema of Functions

-5 5

-5

51. (a) Absolute maximum: f(2) = −2; absolute minimum: f(−1) = −5

(b) Absolute maximum: f(7) = 3; absolute minimum: f(3) = −1

(c) No extrema

(d) Absolute maximum: f(4) = 0; absolute minimum: f(1) = −3

-5 5

-5

52. (a) Absolute maximum: f(−1) = 5; absolute minimum: f(2) = 2

(b) Absolute maximum: f(7) = 3; absolute minimum: f(4) = 0

(c) Absolute minimum: f(4) = 0

(d) Absolute maximum: f(1) = 3; absolute minimum: f(4) = 0

-5 5

-5

53. (a) Absolute maximum: f(4) = 0; absolute minimum: f(2) = −4

(b) Absolute maximum: f(1) = f(3) = −3;absolute minimum: f(2) = −4

(c) Absolute minimum: f(2) = −4

(d) Absolute maximum: f(5) = 5

-5 5

-5

54. (a) Absolute maximum: f(0) = 3; absolute minimum: f(−3) = f(3) = 0

(b) Absolute maximum: f(0) = 3

(c) Absolute minimum: f(0) = 3

(d) Absolute maximum: f(0) = 3;absolute minimum: f(−1) = f(1) = 2

√2

-3

3

–π

π2

π2

π–π

5. (a) No extrema

(b) Absolute maximum: f(π/4) = 1; absolute minimum: f(−π/4) = −1

(c) Absolute maximum: f(π/3) =√

3; absolute minimum: f(0) = 0

(d) No extrema

-3

3

–π

π2

π2

π–π

6. (a) Absolute maximum: f(0) = 2; absolute minimum: f(−π) = f(π) = −2

(b) Absolute maximum: f(0) = 2;absolute minimum: f(−π/2) = f(π/2) = 0

(c) Absolute maximum: f(π/3) = 1; absolute minimum: f(2π/3) = −1

(d) Absolute maximum: f(0) = 2; absolute minimum: f(π) = −2

7. Solving f ′(x) = 4x− 6 = 0 we obtain critical number 3/2.

8. Since f ′(x) = 3x2 + 1 > 0 for all x, the function has no critical numbers.

4.3. EXTREMA OF FUNCTIONS 211

9. Solving f ′(x) = 6x2 − 30x − 36 = 6(x − 6)(x + 1) = 0 we obtain the critical numbers 6 and−1.

10. Solving f ′(x) = 4x3 − 12x2 = 4x2(x− 3) = 0 we obtain the critical numbers 0 and 3.

11. Solving f ′(x) = (x−2)2(1)+(x−1)[2(x−2)] = (x−2)[(x−2)+2(x−1)] = (x−2)(3x−4) = 0we obtain the critical numbers 2 and 4/3.

12. Solving f ′(x) = x2[3(x+1)2]+(x+1)3(2x) = x(x+1)2[3x+2(x+1)] = x(x+1)2(5x+2) = 0we obtain the critical numbers 0, −1, and −2/5.

13. Solving f ′(x) =x1/2 − (1 + x)

(12x−1/2

)

x=

x− 12x3/2

we obtain the critical number 1. f ′(x)

does not exist when x = 0, but 0 is not in the domain of f(x), so the only critical number is1.

14. Solving f ′(x) =2− x2

(x2 + 2)2= 0 we obtain the critical numbers

√2 and −

√2.

15. We note that f ′(x) =4

3(4x− 3)2/36= 0 for all x and f ′(3/4) does not exist. Since 3/4 is in

the domain of f(x), it is a critical number.

16. Solving f ′(x) =23x−1/3 + 1 = 0 we obtain the critical number −8/27. We note that f ′(x)

does not exist when x = 0. Since 0 is in the domain of f(x), it is a critical number.

17. Solving f ′(x) =13

(x−1)2(x+2)−2/3 +2(x+2)1/3(x−1) = 0 we observe (x−1)2(x+1)−2/3 +

6(x + 2)1/3(x − 1) = 0 or (x + 2)1/3(x − 1)[(x − 1)(x + 2)−1 + 6] = 0. Thus, −2 and 1 are

critical numbers. Since we also havex− 1x+ 2

+ 6 = 0 or x − 1 = −6(x + 2), then x = −11/7

and −11/7 is also a critical number.

18. Solving

f ′(x) =(x+ 1)1/3 − (x+ 4)

(13

)(x+ 1)−2/3

(x+ 1)2/3=

3(x+ 1)− (x+ 4)3(x+ 1)4/3

=2x− 1

3(x+ 1)4/3= 0

we obtain the critical number x = 1/2. The value x = −1 is not in the domain of f(x), sothe only critical number is 1/2.

19. Solving f ′(x) = −1 + cosx = 0 we obtain the critical numbers 2nπ where n = 0,±1,±2, . . . .

20. Solving f ′(x) = −4 sin 4x = 0 we obtain the critical numbers nπ/4, where n = 0,±1,±2, . . . .

21. Solving f ′(x) = 2x − 8/x = 0 we obtain the critical number 2. f ′(x) = 0 when x = −2, but−2 is not in the domain of f(x), so it is not a critical number. f ′(x) does not exist whenx = 0, but 0 is not in the domain of f(x), so the only critical number is 2.


22. Solving f ′(x) = e−x + 2 = 0 we obtain the critical number ln 1/2 ≈ −0.693.

23. Solving f ′(x) = −2x + 6 = 0 we obtain the critical number 3. The absolute maximum isf(3) = 9 and the absolute minimum is f(1) = 5.

x 1 3 4f(x) 5 9 8

24. Solving f ′(x) = 2(x − 1) = 0 we obtain the critical number 1 which is not in [2, 5]. Theabsolute maximum is f(5) = 16 and the absolute minimum is f(2) = 1.

x 2 5f(x) 1 16

25. We note that f ′(x) =23x−1/3 does not exist at x = 0. Since 0 is in the domain, it is a critical

number. The absolute maximum is f(8) = 4 and the absolute minimum is f(0) = 0.

x −1 0 8f(x) 1 0 4

26. We note that f ′(x) =83x5/3 − 2

3x−1/3 = (8x2 − 2)/3x1/3 does not exist when x = 0, but 0 is

in the domain of f(x) so it is a critical number. Solving f ′(x) = 0, we obtain x = −1/2, 1/2.The absolute maximum is 0 and occurs at −1, 0, and 1. The absolute minimum is −3/28/3

and occurs at −1/2 and 1/2.

x −1 −1/2 0 1/2 1f(x) 0 −3/28/3 0 −3/28/3 0

27. Solving f ′(x) = 3x2 − 12x = 0 we obtain the critical numbers 0 and 4. However, only 0 is in[−3, 2]. The absolute maximum is f(0) = 2 and the absolute minimum is f(−3) = −79.

x −3 0 2f(x) −79 2 −14

28. Solving f ′(x) = −3x2 − 2x+ 5 = 0 we obtain the critical numbers −5/3 and 1. The absolutemaximum is f(1) = 3 and the absolute minimum is f(−5/3) = −175/27.

x −2 −5/3 1 2f(x) −6 −175/27 3 −2

29. Solving f ′(x) = 3x2 − 6x+ 3 = 0 we obtain the critical number 1. The absolute maximum isf(3) = 8 and the absolute minimum is f(−4) = −125.

x −4 1 3f(x) −125 0 8

30. Solving f ′(x) = 4x3 + 12x2 = 0 we obtain the critical numbers 0 and −3. However, only 0 isin [0, 4]. The absolute maximum is f(4) = 502 and the absolute minimum is f(0) = −10.

x 0 4f(x) −10 502


31. Write f(x) = x6− 2x5 + x4. Then solving f ′(x) = 6x5− 10x4 + 4x3 = 2x3(3x− 2)(x− 1) = 0we obtain 0, 2/3, and 1. The absolute maximum is f(2) = 16 and the absolute minimum is0 and occurs at x = 0 and x = 1.x −1 0 2/3 1 2

f(x) 4 0 16/729 0 16

32. Solving f ′(x) =1− 3x2

2√x(x2 + 1)2

= 0 we obtain the critical numbers ±√

1/3. Neither is in

[1/4, 1/2]. The absolute maximum is f(1/2) = 2√

2/5 and the absolute minimum is f(1/4) =8/17.

x 1/4 1/2f(x) 8/17 2

√2/5

33. Solving f ′(x) = −4 sin 2x + 4 sin 4x = −4 sin 2x + 8 sin 2x cos 2x = 4 sin 2x(2 cos 2x − 1) = 0on [0, 2π] we obtain the critical numbers 0, π/2, π, 3π/2, 2π, π/6, 5π/6, 7π/6, and 11π/6.The absolute maximum is 3/2 and occurs at x = π/6, 5π/6, 7π/6, and 11π/6. The absoluteminimum is −3 and occurs at x = π/2 and 3π/2.

x 0 π/6 π/2 5π/6 π 7π/6 3π/2 11π/6 2πf(x) 1 3/2 −3 3/2 1 3/2 −3 3/2 1

34. Solving f ′(x) = 15 cos 3x = 0 on [0, π/2] we obtain the critical numbers π/6 and π/2. Theabsolute maximum is f(π/6) = 6 and the absolute minimum is f(π/2) = −4.

x 0 π/6 π/2f(x) 1 6 −4

35. Solving f ′(x) = 96 sin 24x cos 24x = 48 sin 48x = 0 on [0, π] we obtain the critical numberskπ/48, where k is an integer from 0 to 48. The absolute maximum is 5 and occurs when k isodd. The absolute minimum is 3 and occurs when k is even.x 0 π/48 π/24 . . . 23π/24 47π/48 π

f(x) 3 5 3 . . . 3 5 3

36. Solving f ′(x) = 2 − sec2 x = 0 we obtain the critical numbers −π/4 and π/4. The absolutemaximum is f(π/4) = π/2− 1 ≈ 0.57 and the absolute minimum is f(1.5) ≈ −11.10.

x −1 −π/4 π/4 1.5f(x) −0.44 −0.57 0.57 −11.10

37. Solving f ′(x) =

{2x+ 2, x < 02x− 2, x > 0

= 0 we obtain the critical numbers 1 and −1. We note

that f ′(x) does not exist when x = 0. Since 0 is in the domain of f(x), it is also a criticalnumber. The absolute minimum is f(−1) = f(1) = −1, the endpoint absolute maximum isf(3) = 3, and the relative maximum is f(0) = 0.

x −2 −1 0 1 3f(x) 0 −1 0 −1 3


38. Solving f ′(x) =

{4, −5 ≤ x < −2

2x, −2 < x ≤ 1= 0 we obtain the critical number 0. We note that f ′(x)

does not exist when x = −2. Since −2 is in the domain of f(x), it is also a critical number.The absolute minimum is f(−2) = 4, the endpoint absolute maximum is f(−5) = −8, andthe relative maximum is f(0) = 0.

x −5 −2 0 1f(x) −8 4 0 1

39. (a) c1, c3, c4, c10

(b) c2, c5, c6, c7, c8, c9(c) Endpoint absolute maximum: f(b); absolute minimum: f(c7)

(d) Relative maxima: f(c3), f(c5), f(c9); relative minima: f(c2), f(c4), f(c7), f(c10)

-5 5

-5

5

–π

π2

π2

40. Solving f ′(x) = 1 − 1x2

= 0 we obtain the critical numbers 1 and −1.

From the graph of f(x) we see that f(−1) = −2 is a relative minimum andf(1) = 2 is a relative maximum. Thus, the relative minimum is greaterthan the relative maximum.

41. (a) −16t2 + 320t is negative outside the interval [0, 20].

(b) Solving s′(t) = −32t+ 320 = 0 we obtain the critical num-ber t = 10. From the data in the accompanying table, wesee that the maximum height attained by the projectile on[0, 20] is 1600 ft.

t 0 10 20s(t) 0 1600 0

42. (a) From the diagram, we see that v is defined for r in [0, R].

(b) Holding R constant we have v′(r) = −Pr/2vl. Solvingv′(r) = 0 we obtain the critical number r = 0. Fromthe data in the accompanying table, we see that the max-imum velocity is PR2/4vl cm/s and the minimum velocityis 0 cm/s.

r 0 Rv(r) PR2/4vl 0

43.

44. f(x) = c, where c is a constant.

45. For every x that is not an integer, f ′(x) = 0, and for every integer value of x, f ′(x) does notexist. Therefore, since f(x) is defined for all real x, every value of x is a critical number.

46. Setting f ′(x) =(cx+ d) · a− (ax+ b) · c

(cx+ d)2=

ad− bc(cx+ d)2

= 0, we obtain ad − bc = 0. Hence,

if ad − bc 6= 0, there are no critical points. When ad − bc = 0, b/a = d/c and f(x) =a(x+ b/a)x(x+ d/c)

=a

c. Thus f(x) is a constant function of (−∞,−d/c) ∪ (−d/c,∞).


47. Solving f ′(x) = nxn−1 = 0, we see that 0 is a critical number and f(0) = 0 is the only possiblerelative extremum. When n is even, f(x) is positive for all non-zero x, and f(0) = 0 is arelative minimum. When n is odd, f(x) < 0 for x < 0 and f(x) > 0 for x > 0, so f(0) = 0 isnot a relative extremum in this case.

48. The derivative of an n-th degree polynomial is a polynomial of degree n−1, and hence has atmost n− 1 zeros. The n-th degree polynomial therefore has at most n− 1 critical numbers.

49. Since f(a) is a relative minimum, there is an interval (c1, c2) around a in which f(x) ≥ f(a).Consider the interval (−c2,−c1) around −a. Since f(x) is even, f(−x) = f(x), and for −x in(−c2,−c1), f(−x) = f(x) ≥ f(a) = f(−a). Therefore f(−a) is a relative minimum.

50. Since f(a) is a relative maximum, there is an interval (c1, c2) around a in which f(x) ≤ f(a).Consider the interval (−c2,−c1) around −a. Since f(x) is odd, f(−x) = −f(x), and for −xin (−c2,−c1), f(−x) = −f(x) ≥ −f(a) = f(−a). Therefore f(−a) is a relative minimum.

51. Since f(x) is even and everywhere differentiable, we have f(−x) = f(x) and −f ′(−x) = f ′(x)through implicit differentiation. When x = 0, we have −f ′(0) = f ′(0) and so f ′(0) = 0. Thus,0 is a critical number of f .

52. (a) Dx(k + f(x)) = f ′(x); c

(b) Dx(kf(x)) = kf ′(x); c

(c) Dx(f(x+ k)) = f ′(x+ k); c− k(d) Dx(f(kx)) = kf ′(kx); c/k

53. (a)

-3

3

–π π 2π

(b) Solving f ′(x) = 2 sinx − 2 sin 2x = 2 sinx − 4 sinx cosx = 2 sinx(1 − 2 cosx) = 0 on[0, 2π], we obtain the critical numbers 0, π/3, π, 5π/3, and 2π.

(c) Computing f(0) = f(2π) = −1, f(π/3) = f(−5π/3) = −3/2, and f(π) = 3, we see thatthe absolute maximum is f(π) = 3 and the absolute minimum is f(π/3) = f(−5π/3) =−3/2.

54. (a)

5 10 15 20

1

–π

π2

π2


(b) Setting I ′(t) =bω

2cosωt +

4bω3π

sin 2ωt =bω

2cosωt +

8bω3π

sinωt cosωt = 0 we obtain

cosωt = 0 and sinωt = −3π16

. In the former case, we have ωt =π

2and

3π2

. Setting

ω =π

12we find t = 6 and t = 18. From sinω = −3π

16we find ωt ≈ −0.6299. We

want ωt > 0 so we use the fact that sin(π − ωt) = sin(2π + ωt) = sinωt. Then ωt ≈π+ 0.6299 ≈ 3.7715 and t ≈ 14.406, and ωt ≈ 2π− 0.6299 ≈ 5.6533 and t ≈ 21.594. Thecritical numbers are 6, 14.406, 18, and 21.594.

4.4 Mean Value Theorem

1. f(x) is continuous and differentiable on [−2, 2] and f(−2) = f(2) = 0, so Rolle’s Theoremapplies. Solving f ′(c) = 2c = 0 we obtain c = 0.

2. f(x) is continuous and differentiable on [1, 5] and f(1) = f(5) = 0, so Rolle’s Theorem applies.Solving f ′(c) = 2c− 6 = 0 we obtain c = 3.

3. Since f(−2) = 19 6= 0, Rolle’s Theorem does not apply.

4. f(x) is continuous and differentiable on [0, 4] and f(0) = f(4) = 0, so Rolle’s Theorem applies.Solving f ′(c) = 3c2 − 10c + 4 = 0 we obtain c = (5 ±

√13)/3. Both of these values are in

(0, 4).

5. f(x) is continuous and differentiable on [−1, 0] and f(−1) = f(0) = 0, so Rolle’s Theoremapplies. Solving f ′(c) = 3c2 + 2c = 0 we obtain c = 0,−2/3. Only c = −2/3 is in the interval(−1, 0).

6. f(x) is continuous and differentiable on [0, 1] and f(0) = f(1) = 0, so Rolle’s Theorem applies.Writing f(x) = x3 − 2x2 + x we obtain f ′(c) = 3c2 − 4c+ 1 = (3c− 1)(c− 1). Thus f ′(c) = 0on (0, 1) for c = 1/3.

7. f(x) is continuous and differentiable on [−π, 2π] and f(−π) = f(2π) = 0, so Rolle’s Theoremapplies. Solving f ′(c) = cos c = 0 on (−π, 2π), we obtain c = −π/2, π/2, 3π/2.

8. f(x) is not continuous at π/2 so Rolle’s Theorem does not apply.

9. Since f ′(x) = x−1/3, f(x) is not differentiable on (−1, 1) and Rolle’s Theorem does not apply.

10. f(x) is continuous and differentiable on [1, 8] and f(1) = f(8) = 0, so Rolle’s Theorem applies.

Solving f ′(c) =23c−1/3 − c−2/3 = 0 we obtain

23c1/3 − 1 = 0 or c =

278

.

11. f(a) 6= 0

12. f is not differentiable at every point in (a, b).

13. f(x) is continuous and differentiable on [−1, 7], so the Mean Value Theorem applies. Setting

f ′(c) = 2c =f(7)− f(−1)

7 + 1= 6, we obtain c = 3.

4.4. MEAN VALUE THEOREM 217

14. f(x) is continuous and differentiable on [2, 3], so the Mean Value Theorem applies. Setting

f ′(c) = −2c+ 8 =f(3)− f(2)

3− 2= 3, we obtain c = 5/2.


f ′(c) = 3c2 + 1 =f(5)− f(2)

5− 2= 40, we obtain 3c2 = 39. Then on (2, 5), c =

√13.

16. f(x) is continuous and differentiable on [−3, 3], so the Mean Value Theorem applies. Setting

f ′(c) = 4c3 − 4c =f(3)− f(−3)

3 + 3= 0, we obtain 4c(c2 − 1) = 0. Then c = −1, 0, 1.

17. f(x) is not continuous at 0, so the Mean Value Theorem does not apply.


f ′(c) = 1 − 1/c2 =f(5)− f(1)

5− 1= 4/5, we obtain 1/5 = 1/c2 or c2 = 5. Then on (1, 5),

c =√

5.


f ′(c) = 1/2√c =

f(9)− f(0)9− 0

= 1/3, we obtain c = 9/4.


f ′(c) = 2/√

4c+ 1 =f(6)− f(2)

6− 2= 1/2, we obtain 4 =

√4c+ 1 or 16 = 4c + 1. Then

c = 15/4.

21. f(x) is continuous and differentiable on [−2,−1], so the Mean Value Theorem applies. Setting

f ′(c) = −2/(c − 1)2 =f(−1)− f(−2)−1 + 2

= −1/3, we obtain (c − 1)2 = 6. Then on [−2,−1],

c = 1−√

6.

22. Since f ′(x) =13x−2/3 − 1, f(x) is not differentiable at 0 and the Mean Value Theorem does

not apply.

23. f is not continuous at b.

24. f is not differentiable at every point in (a, b).

25. f ′(x) = 2x. Solving f ′(x) = 0, we obtain the critical number 0. Thefunction is decreasing on (−∞, 0] and increasing on [0,∞). x 0

f ↘ ↗f ′ − 0 +

26. f ′(x) = 3x2. Solving f ′(x) = 0, we obtain the critical number 0. Thefunction is decreasing on (−∞, 0] and increasing on [0,∞). x 0

f ↗ ↗f ′ + 0 +


27. f ′(x) = 2x+ 6. Solving f ′(x) = 0, we obtain the critical number −3.The function is decreasing on (−∞,−3] and increasing on [−3,∞). x −3

f ↘ ↗f ′ − 0 +

28. f ′(x) = −2x + 10. Solving f ′(x) = 0, we obtain the critical number5. The function is decreasing on [5,∞) and increasing on (−∞, 5]. x 5

f ↗ ↘f ′ + 0 −

29. f ′(x) = 3x2− 6x = 3x(x− 2). Solving f ′(x) = 0, we obtainthe critical numbers 0 and 2. The function is decreasing on[0, 2] and increasing on (−∞, 0] and [2,∞).

x 0 2f ↗ ↘ ↗f ′ + 0 − 0 +

30. f ′(x) = x2 − 2x − 8 = (x − 4)(x + 2). Solving f ′(x) = 0,we obtain the critical numbers −2 and 4. The function isdecreasing on [−2, 4] and increasing on (−∞,−2] and [4,∞).

x −2 4f ↗ ↘ ↗f ′ + 0 − 0 +

31. f ′(x) = 4x3− 12x2 = 4x2(x− 3). Solving f ′(x) = 0, we ob-tain the critical numbers 0 and 3. The function is decreasingon (−∞, 0] and [0, 3], and increasing on [3,∞).

x 0 3f ↘ ↘ ↗f ′ − 0 − 0 +

32. f ′(x) = 20x4 − 40x3 = 20x3(x − 2). Solving f ′(x) = 0,we obtain the critical numbers 0 and 2. The function isdecreasing on [0, 2] and increasing on (−∞, 0] and [2,∞).

x 0 2f ↗ ↘ ↗f ′ + 0 − 0 +

33. f ′(x) = −13x−2/3. The only critical number is at 0 where

f ′(x) does not exist. The function is decreasing on (−∞, 0]and [0,∞).

x 0f ↘ ↘f ′ − undefined −

34. f ′(x) =23x−1/3− 2

3x−2/3 =

23

(x1/3− 1)/x2/3. Solv-

ing f ′(x) = 0, we obtain the critical number 1. Sincef ′(x) does not exist at 0, the critical numbers are 0and 1. The function is decreasing on (−∞, 0] and[0, 1), and increasing on [1,∞).

x 0 1f ↘ ↘ ↗f ′ − undefined − 0 +

35. f ′(x) = 1 − 1/x2 = (x2 − 1)/x2. Solv-ing f ′(x) = 0, we obtain the criticalnumbers −1 and 1. At x = 0, f(x) isundefined. The function is decreasingon [−1, 0) and (0, 1], and increasing on(−∞, 1] and [1,∞).

x −1 0 1f ↗ ↘ undefined ↘ ↗f ′ + 0 − undefined − 0 +


36. f ′(x) = −1/x2 − 2/x3 = −(x + 2)/x3. Solvingf ′(x) = 0, we obtain the critical number −2. Atx = 0, f(x) is undefined. The function is decreas-ing on (−∞,−2] and (0,∞), and increasing on[−2, 0).

x −2 0f ↘ ↗ undefined ↘f ′ − 0 + undefined −

37. f ′(x) = x

( −2x2√

8− x2

)+√

8− x2 =−x2 + 8− x2

√8− x2

=8− 2x2

√8− x2

. Solving f ′(x) = 0, we obtain

the critical numbers −2 and 2. Also, f(x) is only defined for −√

8 ≤ x ≤√

8 and f ′(x) isonly defined for −

√8 < x <

√8. The function is decreasing on [−

√8,−2] and [2,

√8], and

increasing on [−2, 2].

x −√

8 −2 2√

8f ↘ ↗ ↘f ′ undefined − 0 + 0 − undefined

38. f ′(x) =√x2 + 1− (x+ 1)(2x/2

√x2 + 1)

x2 + 1=

x2 + 1− (x+ 1)x(x2 + 1)

√x2 + 1

=

1− x(x2 + 1)

√x2 + 1

. Solving f ′(x) = 0, we obtain the critical number 1.

The function is decreasing on [1,∞) and increasing on (−∞, 1].

x 1f ↗ ↘f ′ + 0 −

39. f ′(x) = −10x/(x2 + 1)2. Solving f ′(x) = 0, we obtain the criticalnumber 0. The function is decreasing on [0,∞) and increasing on(−∞, 0].

x 0f ↗ ↘f ′ + 0 −

40. f ′(x) =(x+ 1)(2x)− x2

(x+ 1)2=

x2 + 2x(x+ 1)2

=x(x+ 2)(x+ 1)2

. Solving f ′(x) = 0, we obtain the critical

numbers 0 and −2. At x = −1, f(x) is undefined. The function is decreasing on [−2,−1)and (−1, 0], and increasing on (−∞,−2] and [0,∞).

x −2 −1 0f ↗ ↘ undefined ↘ ↗f ′ + 0 − undefined − 0 +

41. f ′(x) = 3x2− 12x+ 9 = 3(x− 1)(x− 3). Solving f ′(x) = 0,we obtain the critical numbers 1 and 3. The function isdecreasing on [1, 3] and increasing on (−∞, 1] and [3,∞).

x 1 3f ↗ ↘ ↗f ′ + 0 − 0 +

42. f ′(x) = 3(x2 − 1)2(2x). Solving f ′(x) = 0, we obtain thecritical numbers 0 and 1. The function is decreasing on(−∞, 0] and increasing on [0, 1] and [1,∞).

x 0 1f ↘ ↗ ↗f ′ − 0 + 0 +


43. f ′(x) = cosx. Solving f ′(x) = 0, we obtain the critical numbers π/2+kπ for k = 0,±1,±2, . . . .The sign of f ′(x) = cosx is positive on (−π/2 + 2kπ, π/2 + 2kπ) for k = 0,±1,±2, . . . , andnegative on the other intervals. Thus, f(x) = sinx is increasing on [−π/2 + 2kπ, π/2 + 2kπ]and decreasing on [π/2 + 2kπ, 3π/2 + 2kπ] for k = 0,±1,±2, . . . .

44. f ′(x) = −1 + sec2 x. When x = π/2 + kπ, k = 0,±1,±2, . . . , f(x) is undefined. Solvingf ′(x) = 0, we obtain the critical numbers kπ for k = 0,±1,±2, . . . . Since sec2 x ≥ 1 for all x inthe domain of f(x), f ′(x) is always nonnegative. Thus, f(x) is increasing on (−π/2 + kπ, kπ]and [kπ, π/2 + kπ) for k = 0,±1,±2, . . . .

45. f ′(x) = 1− e−x. Solving f ′(x) = 0, we obtain the critical number 0.The function is decreasing on (−∞, 0] and increasing on [0,∞). x 0

f ↘ ↗f ′ − 0 +

46. f ′(x) = e−x(−x2 + 2x) = e−xx(2 − x). Solving f ′(x) = 0,we obtain the critical numbers 0 and 2. The function isdecreasing on (−∞, 0] and [2,∞), and increasing on [0, 2].

x 0 2f ↘ ↗ ↘f ′ − 0 + 0 −

47. Since f ′(x) = 12x2 + 1 > 0 for all x, the function is increasing and has relative extrema.

48. Since f(x) = −1 − 1/2√

2− x < 0 for all x, the function is decreasing and has no relativeextrema.

49. Let s(t) denote the distance travelled since 1:15 P.M. At 2:15 P.M., we have t = 1. By the

Mean Value Theorem applied to the interval [0, 1], we have v(c) = s′(c) =s(1)− s(0)

1− 0= 70,

for some c between 0 and 1. That is, at some time between 1:15 and 2:15 P.M. the motoristwas speeding at 70 mi/h.

50. V ′(r) = −kr4 + 4kr3(r0 − r) = −kr3(5r − 4r0). Solv-ing V ′(r) = 0 we obtain the critical number r = 4r0/5on [r0/2, r0]. V is increasing on [r0/2, 4r0/5] and de-creasing on [4r0/5, r0]. The volume of air flow will bemaximum for r = 4r0/5.

r r0/2 4r0/5 r0

V ↗ ↘V ′ + 0 −

51. f ′(x) = 4x3 + 3x2 − 1. Since f and f ′ are continuous and differentiable on [−1, 1] andf(−1) = f(1) = 0, Rolle’s Theorem applies. Thus, there exists c in (−1, 1) such that f ′(c) =4c3 + 3c2 − 1 = 0.

52. Suppose x1 < x2 on [a, b]. Since f ′(x) > 0 and g′(x) > 0 for all x in (a, b), then both fand g are increasing on [a, b], and so f(x1) < f(x2) and g(x1) < g(x2). Then (f + g)(x1) =f(x1) + g(x1) < f(x2) + g(x2) = (f + g)(x2) and f + g is increasing on [a, b].

53. We want (fg)′(x) = f(x)g′(x)+f ′(x)g(x) > 0 for all x in (a, b). Since f ′(x) > 0 and g′(x) > 0,the condition will hold if f(x) > 0 and g(x) > 0 for all x in (a, b).

54. We have f ′(x) = 3ax2 + b > 0 since a > 0 and b > 0. If there exist r1 and r2 such thatf(r1) = f(r2) = 0, then the hypotheses of Rolle’s Theorem are satisfied. Thus, there exists cbetween r1 and r2 such that f ′(c) = 0. Since f ′(x) > 0 for all x, the function f(x) = ax3+bx+ccannot have two distinct real roots.


55. We have f ′(x) = 2ax + b. If there exist numbers r1 < r2 < r3 such that f(r1) = f(r2) =f(r3) = 0, then by Rolle’s Theorem there exist c1 in (r1, r2) and c2 in (r2, r3) such thatf ′(c1) = f ′(c2) = 0. But this is impossible since f ′(x) has only the single real root −b/2a.Therefore f(x) can have at most two real roots.

56. We have f ′(x) = 2ax + b. Since f(x) is continuous and differentiable on [x1, x2], the MeanValue Theorem applies. Now,

f ′(x3) = f ′(x1 + x2

2

)= a(x1 + x2) + b

and

f(x2)− f(x1)x2 − x1

=ax2

2 + bx2 + c− (ax21 + bx1 + c)

x2 − x1=a(x2

2 − x21) + b(x2 − x1)x2 − x1

= a(x2 + x1) + b, so f ′(x3) =f(x2)− f(x1)

x2 − x1.

57. The polynomial function f is continuous and differentiable everywhere; if it has four distinctx-intercepts, then there exist values x1 < x2 < x3 < x4 such that f(x1) = f(x2) = f(x3) =f(x4) = 0. Since the values are distinct, there are three distinct intervals (x1, x2), (x2, x3),and (x3, x4) for which f satisfies the hypotheses of Rolle’s Theorem, and so there exist c1 in(x1, x2), c2 in (x2, x3), and c3 in (x3, x4) such that f ′(c1) = f ′(c2) = f ′(c3) = 0. Becausex1 < x2 < x3 < x4, then c1 < c2 < c3, and as such there are at least three points at which atangent line to the graph of f is horizontal.

58. Any quadratic function with an interval centered around its vertex satisfies the given condi-tions. As a specific example, consider f(x) = x2 + 6x + 10 on [−6, 0]. f(x) is continuous on[−6, 0] and differentiable on (−6, 0), and f(−6) = f(0) = 10. Solving f ′(x) = 2x+ 6 = 0, weget c = −3.

59. f ′(x) = x cosx + sinx. The hypotheses of Rolle’s Theorem apply on [0, π], so for some x in(0, π), x cosx+ sinx = 0 or cotx = −1/x.

60. (a)

-8 8

-8

8

–π

π2

π2

(b) f is continuous on [−8, 8], and f(−8) = f(8) = 0. However, f is not differentiable on(−8, 8), since there is a vertical tangent at the origin.

(c) f ′(x) = 1− 43x−2/3. Solving f ′(c) = 1− 4

3c−2/3 = 0 we obtain

13√c2

=34

, c2 =6427

, and

c = ±8/3√

3 ≈ ±1.5396.

61. We have f ′(x) = −2 sin 2x. Setting f ′(c) = −2 sin 2c =f(π/4)− f(0)

π/4− 0= −4/π, we obtain

sin 2c = 2/π. Then 2c ≈ 0.6901 and c ≈ 0.3451.


62. We have f ′(x) = cosx. Setting f ′(c) = cos c =f(π/2)− f(π/4)

π/2− π/4 ≈ 2− 1.7071π/4

≈ 0.3729, we

obtain c ≈ 1.1886.

4.5 Limits Revisited — L’Hopital’s Rule

In this exercise set, the symbol “ h=” is used to denote the fact that L’Hopital’s Rule was applied toobtain the equality.

1. limx→0

cosx− 1x

h=− sinx

1= 0

2. limt→3

t3 − 27t− 3

h= limh→3

3t2

1= 27

3. limx→1

2x− 2lnx

h= limx→1

21/x

= 2

4. limx→0+

ln 2xln 3x

h= limx→0+

1/x1/x

= 1

5. limx→0

e2x − 13x+ x2

h= limx→0

2e2x

3 + 2x=

23

6. limx→0

tanx2x

h= limx→0

sec2 x

2=

12

7. limt→π

5 sin2 t

1 + cos th= limt→π

10 sin t cos t− sin t

= limt→π

10 cos t−1

= 10

8. limθ→1

θ2 − 1eθ2 − e

h= limθ→1

2θ2θeθ2

=1e

9. limx→0

6 + 6x+ 6x2 − 6ex

x− sinxh= limx→0

6 + 6x− 6ex

1− cosxh= limx→0

6− 6ex

sinxh= limx→0

−6ex

cosx= −6

10. It is not necessary to use L’Hopital’s Rule here. Instead, multiply the numerator and denom-inator by 1/x3.

limx→∞

3x2 − 4x3

5x+ 7x3= limx→∞

3/x− 45/x2 + 7

= −47

11. It is not necessary to use L’Hopital’s Rule here.

limx→0+

cot 2xcotx

= limx→0+

cos 2x sinxsin 2x cosx

= limx→0+

(cos2 x− sin2 x) sinx2 sinx cosx cosx

= limx→0+

cos2 x− sin2 x

2 cos2 x=

12

12. limx→0

arcsin(x/6)arctan(x/2)

h= limx→0

1/√

36− x2

2/(4 + x2)=

1/62/4

=13

4.5. LIMITS REVISITED — L’HOPITAL’S RULE 223

13. limt→2

t2 + 3t− 10t3 − 2t2 + t− 2

h= limt→2

2t+ 33t2 − 4t+ 1

=75

14. limr→−1

r3 − r2 − 5r − 3(r + 1)2

h= limr→−1

3r2 − 2r − 52(r + 1)

h= limr→−1

6r − 22

= −4

15. limx→0

x− sinxx3

h= limx→0

1− cosx3x2

h= limx→0

sinx6x

h= limx→0

cosx6

=16

16. L’Hopital’s Rule does not apply. limx→1

x2 + 4x2 + 1

=1 + 41 + 1

=52

17. limx→0

cos 2xx2

has the form 1/0 and does not exist.

18. limx→∞

2e4x + x

e4x + 3xh= limx→∞

8e4x + 14e4x + 3

h= limx→∞

32e4x

16e4x= 2

19. limx→1+

ln√x

x− 1= limx→1+

12

lnx

x− 1h= limx→1+

1/2x1

=12

20. limx→∞

ln(3x2 + 5)ln(5x2 + 1)

h= limx→∞

6x/(3x2 + 5)10x/(5x2 + 1)

= limx→∞

3(5x2 + 1)5(3x2 + 5)

h= limx→∞

30x30x

= limx→∞

1 = 1

21. limx→2

ex2 − e2x

x− 2h= limx→2

2xex2 − 2e2x

1= 2e4

22. limx→0

4x − 3x

x

h= limx→0

4x ln 4− 3x ln 31

= ln43

23. limx→∞

x lnxx2 + 1

h=1 + lnx

2xh= limx→∞

1/x2

= 0

24. limt→0

1− cosh tt2

h= limt→0

− sinh t2t

h= limt→0

− cosh t2

= −12

25. limx→0

1− tan−1 x

x3

h= limx→0

1− 11 + x2

3x2= limx→0

x2

3(x2 + x4)= limx→0

13(1 + x2)

=13

26. limx→0

(sin 2x)2

x2

h= limx→0

4 sin 2x cos 2x2x

= limx→0

sin 4xx

h= limx→0

4 cos 4x1

= 4

27. limx→∞

ex

x4

h= limx→∞

ex

4x3

h= limx→∞

ex

12x2

h= limx→∞

ex

24xh= limx→∞

ex

24The limit has the form ∞/24 and does not exist.

28. limx→∞

e1/x

sin(1/x)has the form 1/0 and does not exist.


29. limx→0

x− tan−1 x

x− sin−1 x

h= limx→0

1− 11 + x2

1− 1√1− x2

= limx→0

1− (1 + x2)−1

1− (1− x2)−1/2

h= limx→0

2x(1 + x2)−2

−x(1− x2)−3/2= limx→0

2(1− x2)3/2

−(1 + x2)2= −2

30. limt→1

t1/3 − t1/2t− 1

h= limt→1

13t−2/3 − 1

2t−1/2

1= limt→1

2t1/2 − 3t2/3

6t7/6= −1

6

31. limu→π/2

ln(sinu)(2u− π)2

h= limu→π/2

cotu4(2u− π)

h= limu→π/2

− csc2 u

8= −1

8

32. limθ→π/2

tan θln(cos θ)

h= limθ→π/2

sec2 θ

− tan θ= limθ→π/2

−1cos θ sin θ

The limit has the form −1/0 and does not exist.

33. limx→−∞

1 + e−2x

1− e−2x

h= limx→−∞

−2e−2x

2e−2x= −1

34. limx→0

ex − x− 12x2

h= limx→0

ex − 14x

h= limx→0

ex

4=

14

35. limr→0

r − cos rr − sin r

has the form −1/0 and does not exist.

36. limt→π

csc 7tcsc 2t

= limt→π

sin 2tsin 7t

h= limt→π

2 cos 2t7 cos 7t

= −27

37. limx→0+

x2

ln2(1 + 3x)h= limx→0+

2x6 ln(1 + 3x)

1 + 3x

= limx→0+

x(1 + 3x)3 ln(1 + 3x)

h= limx→0+

1 + 6x9

1 + 3x

= limx→0+

(1 + 6x)(1 + 3x)9

=19

38. limx→3

(lnx− ln 3x− 3

)2

= limx→3

(lnx)2 − 2 ln 3 lnx− (ln 3)2

x2 − 6x+ 9h= limx→3

(2 lnx)/x− (2 ln 3)/x2x− 6

= limx→3

lnx− ln 3x(x− 3)

h= limx→3

1/x2x− 3

=19

Alternatively, note that

limx→3

(lnx− ln 3x− 3

)2

=[

limx→3

lnx− ln 3x− 3

]2

=(d

dxlnx∣∣∣∣x=3

)2

=(

1x

∣∣∣∣x=3

)2

=19.

39. limx→0

3x2 + ex − e−x − 2 sinxx sinx

h= limx→0

6x+ ex + e−x − 2 cosxx cosx+ sinx

h= limx→0

6 + ex − e−x + 2 sinx−x sinx+ 2 cosx

= 3

40. limx→8

√x+ 1− 3x2 − 64

h= limx→8

1/2√x+ 1

2x=

1/616

=196


41. Indeterminate form: ∞−∞

limx→0

(1

ex − 1− 1x

)= limx→0

x− ex + 1xex − x

h= limx→0

1− exxex + ex − 1

h= limx→0

−exxex + 2ex

= −12


limx→0+

(cotx− cscx) = limx→0+

(cosxsinx

− 1sinx

)= limx→0+

cosx− 1sinx

h= limx→0+

− sinxcosx

= 0


limx→∞

x(e1/x − 1) = limx→∞

e1/x − 11/x

h= limx→∞

−e1/x/x2

−1/x2= limx→∞

e1/x = 1

44. Indeterminate form: 0 · ∞

limx→0+

x lnx = limx→0+

lnx1/x

h= limx→0+

1/x−1/x2

= limx→0+

(−x) = 0

45. Indeterminate form: 00. Set y = xx. Then ln y = x lnx and limx→0+

x lnx = limx→0+

lnx1/x

h=

limx→0+

1/x−1/x2

= limx→0+

(−x) = 0. Thus, limx→0+

xx = e0 = 1.

46. Indeterminate form: 1∞. Set y = x1/(1−x). Then ln y =lnx

1− x and limx→1−

lnx1− x

h= limx→1−

1/x−1

=

−1. Thus, limx→1−

x1/(1−x) = e−1.


limx→0

(1x− 1

sinx

)= limx→0

sinx− xx sinx

h= limx→0

cosx− 1x cosx+ sinx

h= limx→0

− sinx−x sinx+ 2 cosx

= 0


limx→0

(1x2− cos 3x

x2

)= limx→0

1− cos 3xx2

h= limx→0

3 sin 3x2x

h= limx→0

9 cos 3x2

=92


limt→3

(√t+ 1t2 − 9

− 2t2 − 9

)= limt→3

√t+ 1− 2t2 − 9

h= limt→3

1/2√t+ 1

2t= limt→3

14t√t+ 1

=124


limx→0+

[1x− 1

ln(x+ 1)

]= limx→0+

ln(x+ 1)− xx ln(x+ 1)

h= limx→0+

1/(x+ 1)− 1x/(x+ 1) + ln(x+ 1)

= limx→0+

1− (x+ 1)x+ (x+ 1) ln(x+ 1)

h= limx→0+

−11 + 1 + ln(x+ 1)

= −12



limθ→0

θ csc 4θ = limθ→0

θ

sin 4θh= limθ→0

14 cos 4θ

=14

52. Indeterminate form: 1∞. Set y = (sin2 x)tan x = (sinx)2 tan x. Then ln y = 2 tanx ln sinx and

limx→π/2−

2 tanx ln sinx = limx→π/2−

2 ln sinxcotx

h= limx→π/2−

2 cosx sinx− csc2 x

= limx→π/2−

(−2 cosx sinx) = 0.

Thus, limx→π/2−

(sin2 x)tan x = e0 = 1.

53. Indeterminate form: ∞0. Set y = (2 + ex)e−x

. Then ln y = e−x ln(2 + ex) and

limx→∞

e−x ln(2 + ex) = limx→∞

ln(2 + ex)ex

h= limx→∞

ex/(2 + ex)ex

= limx→∞

12 + ex

= 0.

Thus, limx→∞

(2 + ex)e−x

= e0 = 1.

54. Indeterminate form: 00. Set y = (1− ex)x2. Then ln y = x2 ln(1− ex) and

limx→0−

x2 ln(1− ex) = limx→0−

ln(1− ex)x−2

h= limx→0−

−ex/(1− ex)−2/x3

= limx→0−

x3ex

2(1− ex)

h= limx→0−

x3ex + 3x2ex

−2ex= 0.

Thus, limx→0−

(1− ex)x2

= e0 = 1.

55. Indeterminate form: 1∞. Set y = (1 + 3/t)t. Then ln y = t ln(1 + 3/t) and

limt→∞

t ln(1 + 3/t) = limt→∞

ln(1 + 3/t)1/t

h= limt→∞

−3/t2

1 + 3/t−1/t2

= limt→∞

31 + 3/t

= 3.

Thus, limt→∞

(1 + 3/t)t = e3.

56. Indeterminate form: 1∞. Set y = (1+2h)4/h. Then ln y =4h

ln(1+2h) and limh→0

4 ln(1 + 2h)h

h=

limh→0

8/(1 + 2h)1

= 8. Thus, limh→0

(1 + 2h)4/h = e8.

57. Indeterminate form: 00. Set y = x(1−cos x). Then ln y = (1− cosx) lnx and

limx→0

(1− cosx) lnx = limx→0

lnx1/(1− cosx)

h= limx→0

1/x− sinx/(1− cosx)2

= limx→0

−(1− cosx)2

x sinxh= limx→0

−2(1− cosx) sinxx cosx+ sinx

= limx→0

−2 sinx+ sin 2xx cosx+ sinx

h= limx→0

−2 cosx+ 2 cos 2x−x sinx+ 2 cosx

= 0.

Thus, limx→0

x(1−cos x) = e0 = 1.


58. Indeterminate form: 1∞. Set y = (cos 2θ)1/θ2 . Then ln y =1θ2

ln(cos 2θ) and

limθ→0

ln(cos 2θ)θ2

h= limθ→0

−2 sin 2θ/ cos 2θ2θ

= limθ→0

− tan 2θθ

h= limθ→0

−2 sec2 2θ1

= −2.

Thus, limθ→0

(cos 2θ)1/θ2 = e−2.


limx→∞

1x2 sin2(2/x)

= limx→∞

1/x2

sin2(2/x)h= limx→∞

−2/x3

[2 sin(2/x) cos(2/x)](−2/x2)

= limx→∞

12x sin(2/x) cos(2/x)

= limx→∞

1/xsin(4/x)

h= limx→∞

−1/x2

cos(4/x)(−4/x2)

= limx→∞

14 cos(4/x)

=14

60. limx→1

(x2 − 1)x2

does not exist since for x < 1, x2 − 1 is negative and even roots of negativenumbers do not exist.


limx→1

(1

x− 1− 5x2 + 3x− 4

)= limx→1

[1

x− 1− 5

(x− 1)(x+ 4)

]= limx→1

x+ 4− 5(x− 1)(x+ 4)

= limx→1

x− 1(x− 1)(x+ 4)

= limx→1

1x+ 4

=15

62. Indeterminate form: ∞−∞. limx→0

(1x2− 1x

)= lim

x→0

1− xx2

. The limit has the form 1/0 and

does not exist.


limx→∞

x5e−x = limx→∞

x5

exh= limx→∞

5x4

exh= · · · h= lim

x→∞5!ex

= 0

64. Indeterminate form: ∞0. Set y = (x+ ex)2/x. Then ln y =2x

ln(x+ ex) and

limx→∞

2 ln(x+ ex)x

h= limx→∞

2(1 + ex)/(x+ ex)1

= limx→∞

2(1 + ex)x+ ex

h= limx→∞

2ex

1 + exh= limx→∞

2ex

ex= 2.

Thus, limx→∞

(x+ ex)2/x = e2.


limx→∞

x(π

2− arctanx

)= limx→∞

π/2− tan−1 x

1/xh= limx→∞

−1/(1 + x2)−1/x2

= limx→∞

x2

1 + x2

h= limx→∞

2x2x

= 1



limt→π/4

(t− π/4) tan 2t = limt→π/4

t− π/4cot 2t

h= limt→π/4

1−2 csc2 2t

= −12


limx→∞

x tan(

5x

)= limx→∞

tan(5/x)1/x

h= limx→∞

sec2(5/x)(−5/x2)−1/x2

= 5 limx→∞

sec2(5/x) = 5


limx→0+

x ln(sinx) = limx→0+

ln(sinx)1/x

h= limx→0+

cosx/ sinx−1/x2

= limx→0+

( −x2

tanx

)h= limx→0+

( −2xsec2 x

)= 0

69. Indeterminate form: ∞−∞. limx→−∞

(1ex− x2

)= limx→−∞

1− x2ex

ex. Now

limx→−∞

x2ex = limx→−∞

x2

e−xh= limx→−∞

2x−e−x

h= limx→−∞

2e−x

= 0,

so limx→−∞

1− x2ex

exhas the form 1/0 and lim

x→−∞

(1ex− x2

)does not exist.

70. Indeterminate form: 1∞. Set y = (1 + 5 sinx)cot x. Then ln y = cotx ln(1 + 5 sinx) and

limx→0

cotx ln(1 + 5 sinx) = limx→0

ln(1 + 5 sinx)tanx

h= limx→0

5 cosx/(1 + 5 sinx)sec2 x

= limx→0

5 cos3 x

1 + 5 sinx= 5.

Thus, limx→0

(1 + 5 sinx)cot x = e5.

71. Indeterminate form: 1∞. Set y =(

3x3x+ 1

)x. Then ln y = x ln

(3x

3x+ 1

)and

limx→∞

x ln(

3x3x+ 1

)= limx→∞

ln(

3x3x+ 1

)

1/xh= limx→∞

(3x+ 1

3x

)[3(3x+ 1)− 3(3x)

(3x+ 1)2

]

−1/x2

= limx→∞

( −x3x+ 1

)h= limx→∞

−13

= −13.

Thus, limx→∞

(3x

3x+ 1

)x= e−1/3.


limθ→π/2−

(sec3 θ − tan3 θ) = limθ→π/2−

(1

cos3 θ− sin3 θ

cos3 θ

)= limθ→π/2−

1− sin3 θ

cos3 θ

h= limθ→π/2−

−3 sin2 θ cos θ−3 cos2 θ sin θ

= limθ→π/2−

sin θcos θ

This limit has the form 1/0 and does not exist.


73. Indeterminate form: 00. Set y = (sinhx)tan x. Then ln y = tanx ln(sinhx) and

limx→0

tanx ln(sinhx) = limx→0

ln(sinhx)cotx

h= limx→0

coshx/ sinhx− csc2 x

= limx→0

− sin2 x

tanhxh= limx→0

−2 sinx cosxsech2 x

= 0.

Thus, limx→0

(sinhx)tan x = e0 = 1.

74. Set y = xln2 x. Then ln y = ln2 x lnx = ln3 x and limx→0+

ln3 x = −∞. Thus, limx→0+

xln2 x =

limt→−∞

et = 0.

75. Since limx→0+

ex − 1x

h= limx→0+

ex

1= 1, lim

x→0+lnex − 1x

= 0, and1x

ln(ex − 1x

)has the form 0/0

as x→ 0. Now

limx→0+

ln[(ex − 1)/x]x

= limx→0+

ln(ex − 1)− lnxx

h= limx→0+

ex

ex − 1− 1x

1= limx→0+

xex − ex + 1xex − x

h= limx→0+

xex

xex + ex − 1h= limx→0+

xex + ex

xex + 2ex=

12.

76. Since limx→∞

ex − 1x

h= limx→∞

ex

1= ∞, lim

x→0+lnex − 1x

= ∞, and1x

ln(ex − 1x

)has the form

∞/∞ as x→∞. Now

limx→∞

ln(ex − 1x

)

x

h= limx→∞

(x− 1)ex + 1x(ex − 1)

h= limx→∞

xex

xex + ex − 1h= limx→∞

xex + ex

xex + 2ex= limx→∞

x+ 1x+ 2

= 1.

77.

5 10 15 20 25

500

1000

In all three cases, it appears that limx→∞

f(x) =∞.

78.

5 10 15 20

10

20

In all three cases, it appears that limx→∞

f(x) = 0.


79. Since n is a positive integer, then by repeated applications of L’Hopital’s Rule,

limx→∞

xn

exh= limx→∞

nxn−1

exh= limx→∞

n(n− 1)xn−2

exh= · · · h= lim

x→∞n!ex

= 0.

80. Since n is a positive integer, then by repeated applications of L’Hopital’s Rule,

limx→∞

ex

xnh= limx→∞

ex

nxn−1

h= limx→∞

ex

n(n− 1)xn−2

h= · · · h= limx→∞

ex

n!=∞.

81. (a) Letting r be the radius of the circle, we see that the area of the circular sector is AC =12r2(2θ) = r2θ and the area of the isosceles triangle is AT = 2

(12r sin θ · r cos θ

)=

12r2 sin 2θ. Then the area of the shaded region is A = AC − AT = r2

(θ − 1

2sin 2θ

).

Now the length of the arc is 5, so rθ = 5 and r = 5/θ. Thus, A = 25(θ − 1

2sin 2θ

)/θ2.

(b) limθ→0

Ah= limθ→0

25(1− cos 2θ)2θ

h= limθ→0

50 sin 2θ2

= 0

(c)dA

dθ= 25

θ2(1− cos 2θ)− 2θ

(θ − 1

2sin 2θ

)

θ4

= 25

(−θ − θ cos 2θ + sin 2θθ3

)

limθ→0

dA

dh

h= 25 limθ→0

[−1− (−2θ sin 2θ + cos 2θ) + 2 cos 2θ3θ2

]

= 25 limθ→0

(−1 + 2θ sin 2θ + cos 2θ3θ2

)h= 25 lim

θ→0

(4θ cos 2θ + 2 sin 2θ − 2 sin 2θ

6θ

)

= 25 limθ→0

(2 cos 2θ

3

)=

503

82. (a) x(t) = limγ→ω

F0

ω(ω2 − γ2)(−γ sinωt+ ω sin γt) =

F0

ωlimγ→ω

−γ sinωt+ ω sin γtω2 − γ2

h=F0

ωlimγ→ω

− sinωt+ ωt cos γt−2γ

=F0(− sinωt+ ωt cosωt)

−2ω2

(b) Plotting x(t) =2(− sin t+ t cos t)

−2= sin t − t cos t, we get the

graph on the right. As t → ∞, the spring/mass oscillates togreater and greater displacements.

25 50

-50

-25

25

50

4.6. GRAPHING AND THE FIRST DERIVATIVE 231

83. (a) From p1vγ1 = k = p2v

γ2 we have p2 = p1(v1/v2)γ . Then

W =p2v2 − p1v1

1− γ =p1(v1/v2)γv2 − p1v1

1− γ = p1v1

[(v1/v2)γ−1 − 1

1− γ

]

= p1v1

[(v2/v1)1−γ − 1

1− γ

].

(b) limγ→1

{p1v1

[(v2/v1)1−γ − 1

1− γ

]}h= p1v1 lim

γ→1

(v2/v1)1−γ ln(v2/v1)(−1)−1

= p1v1 ln(v2

v1

).

84. (a) Letting p = 2 we obtain σ =1− 10−0.05(4)

0.115(4)× 100 ≈ 80.23%

(b) limp→0

1− 10−0.05p2

0.115p2× 100 h= lim

p→0

0.1p10−0.05p2 ln 100.23p

× 100

= limp→0

0.1 · 10−0.05p2 ln 100.23

× 100 =0.1 ln 10

0.23× 100 ≈ 100.11%

A possible explanation for why this percentage is more than 100% is that the formulamay be only an approximation with diminished accuracy near p = 0.

85. limh→0

f(x+ h)− 2f(x) + f(x− h)h2

h= limh→0

f ′(x+ h)− f ′(x− h)2h

h= limh→0

f ′′(x+ h) + f ′′(x− h)2

= f ′′(x)

86. (a)

-5 5-1

1

(b) limx→∞

f(x) = 0

(c) limx→∞

x sinx does not exist.

4.6 Graphing and the First Derivative

1. The x-intercepts are 1 ±√

2. The y-intercept is 1.Solving f ′(x) = −2x + 2 = 0 we obtain the criticalnumber 1. The relative maximum is f(1) = 2.

x 1f ↗ 2 ↘f ′ + 0 − -3 -2 -1 1 2 3

-2

2

–π

π2

π2

2. The x-intercepts are −3 and 1. The y-intercept is−3. Solving f ′(x) = 2x+2 = 0 we obtain the criticalnumber −1. The relative maximum is f(−1) = −4.

x −1f ↘ −4 ↗f ′ − 0 + -4 -2 2 4

-3

3

–π

π2

π2


3. The x-intercepts are 0 and ±√

3. The y-intercept is 0. Solving f ′(x) = 3x2−3 = 0we obtain the critical numbers −1 and 1.The relative maximum is f(−1) = 2 andthe relative minimum is f(1) = −2.

x −1 1f ↗ 2 ↘ −2 ↗f ′ + 0 − 0 +

-2 -1 1 2

-2

2

–π

π2

π2

4. The y-intercept is 1. (The x-intercepts arenot easily determined.) Solving f ′(x) =x2−x = x(x−1) = 0 we obtain the criticalnumbers 0 and 1. The relative maximumis f(0) = 1 and the relative minimum isf(1) = 5/6.

x 0 1f ↗ 1 ↘ 5/6 ↗f ′ + 0 − 0 + -2 -1 1 2

-2

2

–π

π2

π2

5. The x-intercepts are 0 and 2. The y-intercept is 0. Writing f(x) = x3−4x2+4xand solving f ′(x) = 3x2 − 8x+ 4 = (3x−2)(x − 2) = 0 we obtain the critical num-bers 2/3 and 2. The relative maximum isf(2/3) = 32/27 and the relative minimumis f(2) = 0.

x 2/3 2f ↗ 32/27 ↘ 0 ↗f ′ + 0 − 0 + -1 1 2 3 4

-2

2

–π

π2

π2

6. The y-intercept is −1. Solving f ′(x) =−3x2 + 6x + 9 = −3(x − 3)(x + 1) = 0we obtain the critical numbers −1 and 3.The relative maximum is f(3) = 26 andthe relative minimum is f(−1) = −6.

x −1 3f ↘ −6 ↗ 26 ↘f ′ − 0 + 0 −

-3 3

-10

10

20

30

–π

π2

π2

7. There is no easily determined x-intercept. The y-intercept is −3. Sincef ′(x) = 3x2 + 1 > 0 for all x, there are no critical numbers and no relativeextrema.

-2 2

-4

-2–π

π2

π2

8. Solving f(x) = (x + 1)3 − 4 = 0 we see that the x-intercept is 3

√4 − 1. The y-intercept is −3. Solving

f ′(x) = 3x2 + 6x+ 3 = 3(x+ 1)2 = 0 we obtain thecritical number −1. There are no relative extrema.

x −1f ↗ −4 ↗f ′ + 0 +

-2 2

-6

-4

-2

2

–π

π2

π2

9. The x-intercepts are 0 and −41/3. The y-intercept is0. Solving f ′(x) = 4x3 +4 = 0 we obtain the criticalnumber −1. The relative minimum is f(−1) = −3.

x −1f ↘ −3 ↗f ′ − 0 + -2 2

-2

2

4

–π

π2

π2


10. The x-intercepts are ±1. The y-intercept is 1. Solving f ′(x) = 4x(x2 − 1) = 0 we obtain thecritical numbers −1, 0, and 1. The relative maximum is f(0) = 1 and the relative minima aref(−1) = f(1) = 0.

x −1 0 1f ↘ 0 ↗ 1 ↘ 0 ↗f ′ − 0 + 0 − 0 +

-2 -1 1 2

2

–π

π2

π2

11. The x- and y-intercepts are 0. Solvingf ′(x) = x3 + 4x2 + 4x = x(x + 2)2 = 0 weobtain the critical numbers −2 and 0. Therelative minimum is f(0) = 0.

x −2 0f ↘ 4/3 ↘ 0 ↗f ′ − 0 − 0 +

-4 -2 2

2

4

–π

π2

π2

12. The x-intercepts are ±√

4±√

58/2. The y-intercept is 3. Solving f ′(x) = 8x3 − 32x =8x(x2 − 4) = 0 we obtain the critical numbers −2, 0, and 2. The relative maximum isf(0) = 3 and the relative minima are f(−2) = f(2) = −29.

x −2 0 2f ↘ −29 ↗ 3 ↘ −29 ↗f ′ − 0 + 0 − 0 +

-4 -2 2 4

-30

-15

15

–π

π2

π2

13. The x-intercepts are 0 and 3. The y-intercept is 0. Solving f ′(x) = −2x2(x − 3) + (x −3)(−2x) = −2x(x− 3)(2x− 3) = 0 we obtain the critical numbers 0, 3/2, and 3. The relativemaxima are f(0) = f(3) = 0 and the relative minimum is f(3/2) = −81/16.

x 0 3/2 3f ↗ 0 ↘ −81/16 ↗ 0 ↘f ′ + 0 − 0 + 0 −

-2 2 4

-4

-2–π

π2

π2

14. There are no easily determined x-intercepts. The y-intercept is −2. Solv-ing f ′(x) = −12x3 + 24x2 − 12x =−12x(x − 1)2 = 0 we obtain the criti-cal numbers 0 and 1. The relative max-imum is f(0) = −2.

x 0 1f ↗ −2 ↘ −3 ↘f ′ + 0 − 0 −

-2 2

-4

-2–π

π2

π2


15. The x-intercepts are 0 and 5/4. The y-intercept is 0. Solving f ′(x) = 20x4 −20x3 = 20x3(x− 1) = 0 we obtain the crit-ical numbers 0 and 1. The relative maxi-mum is f(0) = 0 and the relative minimumis f(1) = −1.

x 0 1f ↗ 0 ↘ −1 ↗f ′ + 0 − 0 + -2 2

-2

2

–π

π2

π2

16. The x-intercepts are −3 and 2. The y-intercept is 108. Solving f ′(x) = 3(x − 2)2(x + 3)2 +2(x − 2)(x + 3)3 = 5x(x − 2)(x + 3)2 = 0 we obtain the critical numbers −3, 0, and 2. Therelative maximum is f(0) = 108 and the relative minimum is f(2) = 0.

x −3 0 2f ↗ 0 ↗ 108 ↘ 0 ↗f ′ + 0 + 0 − 0 +

-3 3

-50

50

100

–π

π2

π2

17. The y-intercept is 3. The function is undefined for x = −1. Solving f ′(x) = (x+ 2x−3)/(x+1)2 = (x + 3)(x − 1)/(x + 1)2 = 0 we obtain the critical numbers −3 and 1. The relativemaximum is f(−3) = −6 and the relative minimum is f(1) = 2.

x −3 −1 1f ↗ −6 ↘ undefined ↘ 2 ↗f ′ + 0 − undefined − 0 + -10 10

-10

10

π2

π2

18. There are no intercepts. The function is undefined for x = 0. Solving f ′(x) = 1−25/x2 = (x−25)/x2 = 0 we obtain the critical numbers −5 and 5. The relative maximum is f(−5) = −10and the relative minimum is f(5) = 10.

x −5 0 5f ↗ −10 ↘ undefined ↘ 10 ↗f ′ + 0 − undefined − 0 + -40 -20 20 40

-40

-20

20

40

–π

π2

π2

19. We write f(x) = (x2 − 1)/x3. The x-intercepts are ±1. The function is undefined for x = 0.Solving f ′(x) = −1/x2 +3/x4 = (3−x2)/x4 = 0 we obtain the critical numbers −

√3 and

√3.

The relative maximum is f(√

3) = 2/3√

3 and the relative minimum is f(−√

3) = −2/3√

3.

x −√

3 0√

3f ↘ −2/3

√3 ↗ undefined ↗ 2/3

√3 ↘

f ′ − 0 + undefined + 0 − -5 5

-5

5

–π

π2

π2


20. The x- and y-intercepts are 0. The function is undefined for x = ±2. Solving f ′(x) =−8/(x2 − 4)2 = 0 we obtain the critical number 0. The relative maximum is f(0) = 0.

x −2 0 2f ↗ undefined ↗ 0 ↘ undefined ↘f ′ + undefined + 0 − undefined − -5 5

-5

5

–π

π2

π2

21. The y-intercept is 10. Solving f ′(x) = −20x/(x2 +1) = 0 we obtain the critical number 0. The relativemaximum is f(0) = 10.

x 0f ↗ 10 ↘f ′ + 0 −

-5 5

10

–π

π2

π2

22. The x- and y-intercepts are 0. Solving f ′(x) = 2x(1 − x4)/(x4 + 1)2 = 0 we obtain thecritical numbers −1, 0, and 1. The relative maxima are f(−1) = f(1) = 1/2 and the relativeminimum is f(0) = 0.

x −1 0 1f ↗ 1/2 ↘ 0 ↗ 1/2 ↘f ′ + 0 − 0 + 0 −

-5 5-2

2

π2

π2

23. The x-intercepts are 2 and−2. The y-intercept is 2 3√

2. Solving f ′(x) = (4/3)x(x2−4)−1/3 = 0we obtain the critical number 0. Also, f ′(x) is undefined for x = ±2, which are criticalnumbers. The relative maximum is f(0) ≈ 2.5 and the relative minima are f(±2) = 0.

x −2 0 2f ↘ 0 ↗ 2.5 ↘ 0 ↗f ′ − undefined + 0 − undefined + -4 -2 2 4

2

24. The x-intercepts are ±1. The y-intercept is −1. Solving f ′(x) = 2x/3(x2 − 1)2/3 = 0 weobtain the critical number 0. Also, f ′(x) is undefined for x = ±1, which are critical numbers.The relative minimum is f(0) = −1.

x −1 0 1f ↘ 0 ↘ −1 ↗ 0 ↗f ′ − undefined − 0 + undefined +

-4 -2 2 4-2

2

25. The x-intercepts are −1, 0, and 1. The y-intercept is 0. The function is undefined forx < −1 and x > 1. Solving f ′(x) = (1 − 2x2)/

√1− x2 = 0 we obtain the critical numbers

−1/√

2 and 1/√

2. The relative maximum is f(1/√

2) = 1/2 and the relative minimum isf(−1/

√2) = −1/2.


x −1 −1/√

2 1/√

2 1f 0 ↘ −1/2 ↗ 1/2 ↘ 0f ′ undefined − 0 + 0 − undefined -1 1

-1

1

26. The x-intercepts are −√

5, 0, and√

5. The y-intercept is 0. Solving f ′(x) = (5x2− 15)/(x2−5)2/3 = 0 we obtain the critical numbers −

√3 and

√3. Also, f ′(x) is undefined for x = ±

√5,

which are critical numbers. The relative maximum is f(−√

3) = 1081/6 and the relativeminimum is f(

√3) = −1081/6.

x −√

5 −√

3√

3√

5f ↗ 0 ↗ 1081/6 ↘ −1081/6 ↗ 0 ↗f ′ + undefined + 0 − 0 + undefined + -3 3

-3

3

27. The x-intercepts are −24√

3, 0, and 24√

3. The y-intercept is 0. Solving f ′(x) = 1−4x−2/3 =(x2/3 − 4)/x2/3 = 0 we obtain the critical numbers −8 and 8. Also, f ′(x) is undefined forx = 0, which is a critical number. The relative maximum is f(−8) = 16 and the relativeminimum is f(8) = −16.

x −8 0 8f ↗ 16 ↘ 0 ↘ −16 ↗f ′ + 0 − undefined − 0 +

-40 -20 20 40-20

20

28. We have f(x) = x1/3(x+ 32). The x-intercepts are 0 and −32. The y-intercept is 0. Solving

f ′(x) =43x1/3 +

323x−2/3 =

4x+ 323x2/3

= 0

we obtain the critical number −8. Also, f ′(x) is undefined for x = 0, which is a criticalnumber. The relative minimum is f(−8) = −48.

x −8 0f ↘ −48 ↗ 0 ↗f ′ − 0 + undefined +

-60 -40 -20 20

-50

50

100

29. We have f(x) =

{x3 − 24 lnx, x > 0

x3 − 24 ln(−x), x < 0. There are no easily-determined x-intercepts. The

function is undefined for x = 0. Solving f ′(x) = 3x2−24/x = 0 we obtain the critical number2. Also, f ′(x) is undefined for x = 0, which is a critical number. The relative minimum isf(2) = 8− 24 ln 2 ≈ −8.6355.


x 0 2f ↗ undefined ↘ −8.6355 ↗f ′ + undefined − 0 +

-3 3

-100

100

200

30. The x-intercept is 1. The function isundefined for x ≤ 0. Solving f ′(x) =(1 − lnx)/x2 = 0 we obtain the criti-cal number e. The relative maximum isf(e) = 1/e.

x 0 ef undefined ↗ 1/e ↘f ′ undefined + 0 −

5 10

-10

-5

31. The x-intercept is 3. The y-intercept is9. Solving

f ′(x) = −(x+ 3)2e−x + 2e−x(x+ 3)

= e−x(−x2 − 4x− 3)

= −e−x(x+ 1)(x+ 3) = 0

x −3 −1f ↘ 0 ↗ 4e ↘f ′ − 0 + 0 −

-5 5

5

10

15

we obtain the critical numbers −3 and −1. The relative maximum is f(−1) = 4e and therelative minimum is f(−3) = 0.

32. The x-intercept is 0. The y-intercept is 0. Solving f ′(x) = 8x2[(−2x)e−x2] + 16x(e−x

2) =

16xe−x2(1 − x2) = 16xe−x

2(1 + x)(1 − x) = 0 we obtain the critical numbers −1, 0, and 1.

The relative maxima are f(−1) = f(1) = 8/e and the relative minimum is f(0) = 0.

x −1 0 1f ↗ 8/e ↘ 0 ↗ 8/e ↘f ′ + 0 − 0 + 0 − -3 3

3

33.

a

34.

ab

35.

a b

36. 37.a b c

38. 39.

5

540.

-1 1

41.

-4 -2 2 4

3

6 42.

-2 2 4 6

-3

3

43. The slopes of the tangent lines are given by f ′(x) = 3x2 + 12x− 1.Solving f ′′(x) = 6x+ 12 = 0 we obtain the point −2. The relativeminimum is f ′(−2) = −13.

x −2f ′ ↘ −13 ↗f ′′ − 0 +


44. The slopes of the tangent lines are given by f ′(x) = 4x3−12x. Solving f ′′(x) = 12x2− 12 = 0 we obtain the points−1 and 1. The relative maximum is f ′(−1) = 8 and therelative minimum is f ′(1) = −8.

x −1 1f ′ ↗ 8 ↘ −8 ↗f ′′ + 0 − 0 +

45. (a) g(x) > 0 for x in (kπ, kπ+π/2), where k is an integer. g(x) < 0for x in (kπ − π/2, kπ), where k is an integer.

π-1

1

(b) Solving f ′(x) = 2 sinx cosx = sin 2x = g(x) = 0 we obtain the critical numbers kπ/2,where k is an integer. The relative maxima of f(x) occur at kπ + π/2, where k is aninteger. The relative minima of f(x) occur at kπ, where k is an integer.

(c)

π-1

1

46. (a) Solving f ′(x) = 1− cosx = 0 we obtain the critical numbers 2πk, where k is an integer.(b) Since f ′(x) ≥ 0 for all x, the sign of f ′(x) cannot change around a critical number. Thus

f(x) has no relative extrema.(c)

-2π 2π

-2π

2π

47. (a) Setting f ′(x) = 2(x−a1)+2(x−a2)+· · ·+2(x−an) = 0 we obtain nx = a1 +a2 +· · ·+anor x = (a1 + a2 + · · ·+ an)/n = x. Thus, x is a critical number of f(x).

(b) To see that f(x) is a relative minimum write f(x) = nx2− 2(a1 +a2 + · · ·+an)x+ (a21 +

a22 + · · · + a2

n). Then f ′(x) = 2nx − 2(a1 + a2 + · · · + an) = 2n(x − x). When x < x,f ′(x) < 0 and when x > x, f ′(x) > 0. Thus, f(x) is a relative minimum.

48. (a) T (1/r) =4/r

(1/r + 1)2=

4r[(1 + r)/r]2

=4

(1 + r)2/r=

4r(r + 1)2

= T (r). Since r is a

ratio, T (r) = T (1/r) means that the fraction of energy transmitted is the same in bothdirections.

(b) Solving T ′(r) =4(r + 1)2 − 8r(r + 1)

(r + 1)4= −4(r − 1)

(r + 1)3= 0 we obtain the critical number 1.

Since T ′(0) > 0 and T ′(2) < 0 we see that T (1) = 1 is a relative maximum.(c)

5 10

1

4.7. GRAPHING AND THE SECOND DERIVATIVE 239

49. Solving f ′(x) = 2ax + b = 0 we obtain the critical number −b/2a. We want −b/2a = 2 orb = −4a. We are given f(2) = 4a+2b+c = 6 and f(0) = c = 4. Solving these three equations,

we obtain a = −1/2, b = 2, and c = 4. Thus f(x) = −12x2 + 2x+ 4.

50. f ′(x) = 3ax2 + 2bx + c. Since 0 is a critical number, f ′(0) = c = 0 and f ′(x) = 3ax2 + 2bx.Since 1 is a critical number, f ′(1) = 3a + 2b = 0. We also have f(0) = d = −3 andf(1) = a+ b+ c+ d = a+ b− 3 = 4. Solving the system

3a+ 2b = 0a+ b = 7,

we obtain a = −14 and b = 21. Thus f(x) = −14x3 + 21x2 − 3.

51. If f ′(0) > 0, then f ′(x) > 0 on some interval (−a, a). But then f(x) is increasing on (−a, a)and cannot be symmetric about the y-axis. Similarly, if f ′(0) < 0, then f(x) is decreasingon some open interval containing 0 and cannot be symmetric about the y-axis. Since f ′(0)exists, we must have f ′(0) = 0. Since f(x) is neither increasing nor decreasing in an openinterval around 0, it must have a relative extremum at 0.

52. Solving f ′(x) = xm[n(x − 1)n−1] + (x − 1)n(mxm−1) =xm−1(x − 1)n−1[(m + n)x − m] = 0 we see that f(x) hascritical numbers at 0,

m

m+ n, and 1. If n is odd, then

(x − 1)n−1 > 0 for all x 6= 1 and the relative minimum is

f

(m

m+ n

).

x mm+n 1

f ↘ ↗f ′ − 0 + 0

If n is even, then (x− 1)n−1 < 0 for x < 1 and the relativeminimum is f(1). x m

m+n 1f ↘ ↗f ′ 0 − 0 +

53. (a) Since f and g are differentiable and have a critical number at c, f ′(c) = g′(c) = 0.Then (f + g)′(c) = f ′(c) + g′(c) = 0, (f − g)′(c) = f ′(c) − g′(c) = 0, and (fg)′(c) =f(c)g′(c) + f ′(c)g(c) = 0. Thus, f + g, f − g, and fg have critical numbers c.

(b) Suppose f ′(x) > 0 and g′(x) > 0 for a < x < c, and f ′(x) < 0 and g′(x) < 0 forc < x < b. Then (f + g)′(x) = f ′(x) + g′(x) > 0 for a < x < c and (f + g)′(x) < 0 forc < x < b. Thus, f + g has a relative maximum at c.To see that neither f − g nor fg necessarily have relative maxima at c let f(x) = −x4

and g(x) = −x2. Both have relative maxima at c = 0. However (f − g)(x) = −x4 + x2

and (fg)(x) = x6 both have relative minima at c = 0.

4.7 Graphing and the Second Derivative

1. f ′(x) = −2x + 7; f ′′(x) = −2. Since f ′′(x) < 0 for all x, the graph is concave downward on(−∞,∞).


2. f ′(x) = −2(x+ 2); f ′′(x) = −2. Since f ′′(x) < 0 for all x, the graph is concave downward on(−∞,∞).

3. f ′(x) = −3x2 +12x+1; f ′′(x) = −6x+12. Solving f ′′(x) = 0 we obtainx = 2. The graph is concave upward on (−∞, 2) and concave downwardon (2,∞).

x 2f ′′ + 0 −

4. f ′(x) = 3(x+ 5)2; f ′′(x) = 6(x+ 5). Solving f ′′(x) = 0 we obtain x =−5. The graph is concave upward on (−5,∞) and concave downwardon (−∞,−5).

x −5f ′′ − 0 +

5. f ′(x) = (x − 4)2(4x − 4); f ′′(x) = 12(x − 4)(x − 2). Solvingf ′′(x) = 0 we obtain x = 2 and 4. The graph is concave upwardon (−∞, 2), (4,∞), and concave downward on (2, 4).

x 2 4f ′′ + 0 − 0 +

6. f ′(x) = 24x3 + 6x2 − 24x; f ′′(x) = 72x2 + 12x − 24 =12(3x+2)(2x−1). Solving f ′′(x) = 0 we obtain x = −2/3and 1/2. The graph is concave upward on (−∞,−2/3),(1/2,∞), and concave downward on (−2/3, 1/2).

x −2/3 1/2f ′′ + 0 − 0 +

7. f ′(x) =13x−2/3 + 2; f ′′(x) = −2

9x−5/3. f ′′(x) is undefined for

x = 0. The graph is concave upward on (−∞, 0) and concavedownward on (0,∞).

x 0f ′′ + undefined −

8. f ′(x) =83x5/3−40

3x−1/3; f ′′(x) =

409x2/3+

409x−4/3 =

409

(x2+

1)/x4/3. f ′′(x) is undefined for x = 0. The graph is concaveupward on (−∞, 0) and (0,∞).

x 0f ′′ + undefined +

9. f ′(x) = 1− 9/x2; f ′′(x) = 18/x3. f ′′(x) is undefined for x = 0.The graph is concave upward on (0,∞) and concave downwardon (−∞, 0).

x 0f ′′ − undefined +

10. f ′(x) = x/√x2 + 10; f ′′(x) = 10/(x2 + 10)3/2. Since f ′′(x) > 0 for all x, the graph is concave

upward on (−∞,∞).

11. f ′(x) =−2x

(x2 + 3)2; f ′′(x) =

(x2 + 3)2(−2) + 2x[4x(x2 + 3)](x2 + 3)4

=6(x2 − 1)(x2 + 3)3

.

Solving f ′′(x) = 0 we obtain x = ±1. The graph is concaveupward on (−∞,−1) and (1,∞), and concave downward on(−1, 1).

x −1 1f ′′ + 0 − 0 +

12. f ′(x) = 3/(x + 2)2; f ′′(x) = −6/(x + 2)3. f ′′(x) is undefinedfor x = −2. The graph is concave upward on (−∞,−2) andconcave downward on (−2,∞).

x −2f ′′ + undefined −

13. f ′ is increasing on (−2, 2).f ′ is decreasing on (−∞,−2) and (2,∞). x −2 2

f ′′ − 0 + 0 −


14. f ′ is increasing on (−∞, 0).f ′ is decreasing on (0,∞). x 0

f ′′ + 0 −

15. f ′ is increasing on (−∞,−1) and (4,∞).f ′ is decreasing on (−1, 4). x −1 4

f ′′ + 0 − 0 +

16. f ′ is increasing on (−2, 0) and (2,∞).f ′ is decreasing on (−∞,−2) and (0, 2). x −2 0 2

f ′′ − 0 + 0 − 0 +

17. f ′(x) = secx tanx;

f ′′(x) = (secx)(sec2 x) + (tanx)(secx tanx) = (secx)(sec2 x + tan2 x) = (1 + sin2 x)/ cos3 x.f ′′(x) is positive when cosx > 0 and negative when cosx < 0. Thus, the graph of f(x) = secxis concave upward when cosx > 0 and concave downward when cosx < 0.

18. f ′(x) = − cscx cotx;

f ′′(x) = −(cscx)(− csc2 x)+(cotx)(cscx cotx) = (cscx)(csc2 x+cot2 x) = (1+cos2 x)/ sin3 x.f ′′(x) is positive when sinx > 0 and negative when sinx < 0. Thus, the graph of f(x) = cscxis concave upward when sinx > 0 and concave downward when sinx < 0.

19. f ′(x) = 4x3−24x+1; f ′′(x) = 12x2−24. Solving f ′′(x) =0 we obtain x = −

√2 and

√2. The inflection points are

(−√

2,−21−√

2) and (√

2,−21 +√

2).x −

√2

√2

f ′′ + 0 − 0 +

20. f ′(x) =53x2/3 + 4; f ′′(x) =

109x−1/3. f ′′(x) is undefined for

x = 0. The inflection point is (0, 0).x 0f ′′ − undefined +

21. f ′(x) = cosx; f ′′(x) = − sinx. Solving f ′′(x) = 0 we obtainx = kπ, where k is an integer. Since f(x) = sinx is 2π-periodic, the graph has inflection points at (kπ, 0), where k isan integer.

x 0 π 2πf ′′ 0 − 0 + 0

22. f ′(x) = − sinx; f ′′(x) = − cosx. Solv-ing f ′′(x) = 0 we obtain x = π/2 + kπ,where k is an integer.

x −π/2 π/2 3π/2 5π/2f ′′ 0 − 0 + 0 − 0

Since f(x) = cosx is 2π-periodic, the graph has inflection points at (π/2 + kπ, (−1)k), wherek is an integer.

23. f ′(x) = 1− cosx; f ′′(x) = sinx. Solving f ′′(x) = 0 we obtain x = kπ, where k is an integer.Since the sign of f ′′(x) = sinx changes around each kπ, the graph of f(x) = x − sinx hasinflection points at (kπ, kπ), where k is an integer.

24. f ′(x) = sec2 x; f ′′(x) = 2 sec2 x tanx = 2 sinx/ cos3 x. Solving f ′′(x) =0 on (−π/2, π/2) we obtain x = 0. The graph of f(x) has an inflectionpoint at (0, 0), and since f(x) = tanx is π-periodic, it has inflectionpoints at (kπ, 0), where k is an integer.

x 0f ′′ − 0 +


25. f ′(x) = 1− xe−x + e−x = 1 + (1− x)e−x; f ′′(x) = −(1− x)e−x− e−x =(x− 2)e−x. Solving f ′′(x) = 0 we obtain x = 2. The inflection point is(2, 2 + 2/e2).

x 2f ′′ − 0 +

26. f ′(x) = −2x2e−x2

+ e−x2

= (1− 2x2)e−x2;

f ′′(x) = −2x(1− 2x2)e−x2 − 4xe−x

2

= (4x3 − 6x)e−x2

= (4x2 − 6)xe−x2.

x −√

3/2 0√

3/2f ′′ − 0 + 0 − 0 +

Solving f ′′(x) = 0 we obtain x = 0, −√

3/2, and√

3/2. The inflection points are (0, 0),(−√

3/2,−√

3/2e−3/2), and (√

3/2,√

3/2e−3/2).

27. The x-intercept is 5/2. The y-intercept is −25.f ′(x) = −4(2x − 5); f ′′(x) = −8. Solvingf ′(x) = 0 we obtain the critical number 5/2.The relative maximum is (5/2, 0).

x 5/2f 0f ′ 0f ′′ − − −

-2 2 4 6

-6

-4

-2

2

28. The x-intercepts are 3− 3√

5, 0, and 3 + 3√

5. The y-intercept is 0. f ′(x) = x2 − 4x− 12 =(x − 6)(x + 2); f ′′(x) = 2x − 4. Solving f ′(x) = 0 we obtain the critical numbers −2 and6. Solving f ′′(x) = 0 we obtain x = 2. The relative maximum is (−2, 40/3), the relativeminimum is (6,−72), and the inflection point is (2,−88/3).

x −2 2 6f 40/3 −88/3 −72f ′ 0 0f ′′ − − − 0 + + +

-10 -5 5 10

-75

-50

-25

25

50

29. f(x) = (x + 1)3. The x-intercept is −1. They-intercept is 1. f ′(x) = 3(x + 1)2; f ′′(x) =6(x+1). Solving f ′(x) = 0 we obtain the criticalnumber −1. Solving f ′′(x) = 0 we obtain x =−1. Since f ′′(−1) = 0 the second derivative testdoes not apply. There are no relative extrema.The inflection point is (−1, 0).

x −1f ↗ 0 ↗f ′ + 0 +f ′′ − 0 + -3 3

-3

3

30. f(x) =14x2(x2 − 8). The x-intercepts are −

√8, 0, and

√8. The y-intercept is 0. f ′(x) =

x3 − 4x = x(x2 − 4); f ′′(x) = 3x2 − 4. Solving f ′(x) = 0 we obtain the critical numbers −2,0, and 2. Solving f ′′(x) = 0 we obtain −2/

√3 and 2/

√3. The relative minima are (−2,−4)

and (2,−4). The relative maximum is (0, 0). The inflection points are (−2/√

3,−20/9) and(2/√

3,−20/9).


x −2 −2/√

3 0 2/√

3 2f −4 −20/9 0 −20/9 −4f ′ 0 0 0f ′′ + + + 0 − − − 0 + + + -4 4

-4

4

31. f(x) = 2x3(3x2 − 5). The x-intercepts are −√

5/3, 0, and√

5/3. The y-intercept is 0.f ′(x) = 30x4 − 30x2 = 30x2(x2 − 1); f ′′(x) = 120x3 − 60x = 60x(2x2 − 1). Solving f ′(x) = 0we obtain the critical numbers −1, 0, and 1. Solving f ′′(x) = 0 we obtain x = −1/

√2, 0,

and 1/√

2. The second derivative test does not apply at 0. The relative maximum is (−1, 4)and the relative minimum is (1,−4). The inflection points are (−1/

√2, 7/√

8), (0, 0), and(1/√

2,−7/√

8).

x −1 −1/√

2 0 1/√

2 1f 4 7/

√8 ↘ 0 ↘ −7/

√8 −4

f ′ 0 − 0 − 0f ′′ − − − 0 + 0 − 0 + + + -2 -1 1 2

-4

4

32. The x-intercepts are −1 and 0. The y-intercept is 0. f ′(x) = x2(x + 1)(5x + 3); f ′′(x) =2x(10x2+12x+3). Solving f ′(x) = 0 we obtain the critical numbers −1, −3/5, and 0. Solvingf ′′(x) = 0 we obtain (−6 −

√6)/10, (−6 +

√6)/10, and 0. The second derivative test does

not apply at 0. The relative maximum is (−1, 0) and the relative minimum is (−3/5,−0.03).The inflection points are (0, 0), ((−6−

√6)/10,−0.01), and ((−6 +

√6)/10,−0.02).

x −1−6−

√6

10−3/5

−6 +√

610

0

f 0 ↗ 0 ↗f ′ 0 0 + 0 +f ′′ − − − 0 + + + 0 − 0 +

-1

-0.2

-0.1

0.1

0.2

33. The x- and y-intercepts are 0. f ′(x) =2− 2x2

(x2 + 2)2; f ′′(x) =

2x(x2 − 6)(x2 + 2)3

. Solving f ′(x) = 0 we

obtain the critical numbers −√

2 and√

2. Solving f ′′(x) = 0 we obtain −√

6, 0, and√

6. Therelative maximum is (

√2,√

2/4) and the relative minimum is (−√

2,−√

2/4). The inflectionpoints are (−

√6,−√

6/8), (0, 0), and (√

6,√

6/8).

x −√

6 −√

2 0√

2√

6f −

√6/8 −

√2/4 0

√2/4

√6/8

f ′ 0 0f ′′ − 0 + + + 0 − − − 0 +

-5 5-0.5

0.5


34. There are no x- or y-intercepts. f ′(x) = 2x− 2/x3 = 2(x4− 1)/x3; f ′′(x) = 2 + 6/x4. Solvingf ′(x) = 0 we obtain the critical numbers −1 and 1. The function is undefined at x = 0.f ′′(x) = 0 has no real solution. The relative minima are (−1, 2) and (1, 2).

x −1 0 1f 2 undefined 2f ′ 0 undefined 0f ′′ + + + undefined + + +

-3 3

5

10

35. The domain of the function is [−3, 3]. The x-intercepts are −3 and 3. The y-intercept is3. f ′(x) = −x/

√9− x2; f ′′(x) = −9/(9 − x2)3/2. Solving f ′(x) = 0 we obtain the critical

number 0. f ′′(x) = 0 has no real solution. The relative maximum is (0, 3).

x −3 0 3f 0 3 0f ′ undefined 0 undefinedf ′′ undefined − − − undefined

-3 3

-2

2

4

36. The domain of the function is [6,∞). The x-intercept is 0. f ′(x) =3x− 122√x− 6

; f ′′(x) =

3x− 244(x− 6)3/2

. f ′(x) = 0 has no solution in [6,∞). Solving f ′′(x) = 0 we obtain x = 8. The

inflection point is (8, 8√

2).

x 6 8f 0 8

√2

f ′ undefined + + +f ′′ undefined − 0 +

2 4 6 8

5

10

15

37. The x-intercepts are −1 and 0. The y-intercept is 0. f ′(x) =4x+ 13x2/3

; f ′′(x) =4x− 29x5/3

.

Solving f ′(x) = 0 we obtain the critical point −1/4. Also, f ′(x) is undefined for x = 0.Solving f ′′(x) = 0 we obtain x = 1/2. The relative minimum is (−1/4,−3/44/3). Theinflection points are (1/2, 3/24/3) and (0, 0).


x −1/4 0 1/2f −3/44/3 0 3/24/3

f ′ 0 undefinedf ′′ + + + undefined − 0 + -1 1

-1

1

38. The domain of the function is [0,∞). The x-intercepts are 0 and 16. The y-intercept is 0.

f ′(x) = 1/2√x− 1/4 =

2−√x4√x

; f ′′(x) = −1/4x3/2. Solving f ′(x) = 0 we obtain the critical

number 4. f ′′(x) = 0 has no solution. The relative maximum is (4, 1).

x 0 4f 1f ′ undefined 0f ′′ undefined − − − 5 10 15

-1

1

39. The x-intercepts are (kπ/6, 0) for k = 1, 3, 5, 7, 9, and 11. The y-intercept is 1. f ′(x) = −3 sin 3x; f ′′(x) = −9 cos 3x. Solving f ′(x) =0 we obtain the critical numbers kπ/3 for k = 0, 1, 2, 3, 4, 5, and6. Computing f ′′(x) at these values we see that f(x) has relativemaxima at (2π/3, 1) and (4π/3, 1), and relative minima at (π/3,−1),(π,−1), and (5π/3,−1).

π 2π-1

1

Solving f ′′(x) = 0 we obtain (kπ/6, 0) for k = 1, 3, 5, 7, 9, and 11. These are all points ofinflection since the sign of f ′′(x) changes around each one.

40. There are no x-intercepts. The y-intercept is 2. f ′(x) = 2 cos 2x;f ′′(x) = −4 sin 2x. Solving f ′(x) = 0 we obtain the critical numbersπ/4, 3π/4, 5π/4, and 7π/4. Computing f ′′(x) at these values we seethat f(x) has relative maxima at (π/4, 3) and (5π/4, 3), and relativeminima at (3π/4, 1) and (7π/4, 1). π 2π

2

Solving f ′′(x) = 0 we obtain (kπ/2, 2) for k = 0, 1, 2, 3, and 4. These are all points ofinflection since the sign of f ′′(x) changes around each one.

41. Solving sinx = − cosx or tanx = −1 we obtain the x-intercepts 3π/4 and 7π/4. The y-intercept is 1. f ′(x) = − sinx + cosx; f ′′(x) = − cosx − sinx. Solving f ′(x) = 0 we obtainthe critical numbers π/4 and 5π/4. Solving f ′′(x) = 0 we obtain x = 3π/4 and 7π/4. Therelative maximum is (π/4,

√2) and the relative minimum is (5π/4,−

√2). The inflection

points are (3π/4, 0) and (7π/4, 0).


x 0 π/4 3π/4 5π/4 7π/4 2πf 1

√2 0 −

√2 0 1

f ′ 0 0f ′′ − − − 0 + + + 0 −

π 2π-1

1

42. Solving 2 sinx+ sin 2x = 2 sinx+ 2 sinx cosx = (2 sinx)(1 + cosx) = 0we obtain the x-intercepts 0, π, and 2π. The y-intercept is 0. f ′(x) =2 cosx + 2 cos 2x = 2 cosx + 2(2 cos2 x − 1) = 2(2 cos2 x + cosx − 1) =2(2 cosx − 1)(cosx + 1). f ′′(x) = −2 sinx − 4 sin 2x = −2(sinx +4 sinx cosx) = −2(sinx)(1 + 4 cosx). Solving f ′(x) = 0 we obtain thecritical numbers π/3, π, and 5π/3. Solving f ′′(x) = 0 we obtain x = 0,π, 2π, and c, where cos c = −1/4. Using a calculator, we find c ≈ 1.82and 4.46. The relative maximum is (π/3, 3

√3/2) and the relative min-

imum is (5π/3,−3√

3/2). The inflection points are (0, 0), (1.82, 1.45),and (4.46,−1.45).

π 2π

-3

3

x 0 π/3 1.82 π 4.46 5π/3 2πf 0 3

√3/2 1.45 0 −1.45 −3

√3/2 0

f ′ 0 − − − 0 − − − 0f ′′ 0 − − − 0 + 0 − 0 + + + 0

43. The domain of the function is (0,∞). Solving 2x = x lnx or lnx = 2 we obtain the x-intercepte2. There is no y-intercept. f ′(x) = 1− lnx; f ′′(x) = −1/x. Solving f ′(x) = 0 we obtain thecritical number e. f ′′(x) = 0 has no solution. The relative maximum is (e, e).

x 0 e e2

f undefined e 0f ′ undefined 0f ′′ undefined − − − − 5

-3

3

44. There are no x-intercepts. The y-intercept is ln 2. f ′(x) =2x

x2 + 2; f ′′(x) =

4− 2x2

(x2 + 2)2=

2(2− x2)(x2 + 2)2

. Solving f ′(x) = 0 we obtain the critical number 0. Solving f ′′(x) = 0 we obtain

x = −√

2 and√

2. The relative minimum is (0, ln 2). The inflection points are (−√

2, ln 4)and (

√2, ln 4).

x −√

2 0√

2f ln 4 ln 2 ln 4f ′ 0f ′′ − 0 + + + 0 − -5 5

2

4


45. f(x) =12

sin 2x; f ′(x) = cos 2x; f ′′(x) = −2 sin 2x. Since f ′′(π/4) = −2 < 0, the function has

a relative maximum at (π/4, 1/2).

46. f ′(x) = x cosx + sinx; f ′′(x) = −x sinx + 2 cosx. Since f ′′(0) = 2 > 0, the function has arelative minimum at (0, 0).

47. f ′(x) = 2 tanx sec2 x; f ′′(x) = 2 sec4 x + 4 tan2 x sec2 x. Since f ′′(π) = 2 > 0, the functionhas a relative minimum at (π, 0).

48. f ′(x) = 12(cos 4x)(1 + sin 4x)2;

f ′′(x) = 12(cos 4x)[2(1 + sin 4x)(4 cos 4x)] + (1 + sin 4x)2(−48 sin 4x).

Since f ′′(π/8) = −192 < 0, the function has a relative maximum at (π/8, 8).

49.

5

5 50.

5

551.

-2π 2π

-5

552.

-4 -2 2 4

-2

2

53. f ′(x) = 3ax2 + 2bx + c; f ′′(x) = 6ax + 2b. Since the graph has an inflection point at (1, 1),f ′′(1) = 6a + 2b = 0. Using the fact that (−1, 0) and (1, 1) lie on the graph, we have−a+ b− c = 0 and a+ b+ c = 1. Solving these three equations, we obtain a = −1/6, b = 1/2,and c = 2/3.

54. f ′(x) = 3ax2 + 2bx+ c; f ′′(x) = 6ax+ 2b. Since the graph has a horizontal tangent at (1, 1),f ′(1) = 3a+ 2b+ c = 0. Since the graph has an inflection point at (1, 1), f ′′(1) = 6a+ 2b = 0.Using the fact that (1, 1) is on the graph, we have f(1) = a+ b+ c = 1. Solving these threeequations, we obtain a = 1, b = −3, and c = 3.

55. Since f(x) is an odd function, the graph is symmetric with re-spect to the origin. f ′(x) = − cos(1/x)/x2; f ′′(x) = [2x cos(1/x) −sin(1/x)]/x4. Solving f ′(x) = 0 we obtain the positive critical num-bers 2/π, 2/3π, 2/5π, 2/7π, . . . . Since the sign of f ′′(x) alternatesat these points, they are alternately relative maxima and relativeminima. For x > 3π, f ′′(x) > 0 and the graph is concave upwards.

-3 3-1

1

56. f ′(x) = nanxn−1 + (n− 1)an−1x

n−2 + . . .+ a2x+ a1;

f ′′(x) = n(n− 1)anxn−2 + (n− 1)(n− 2)an−1xn−3 + . . .+ a2.

Since a polynomial of degree n−2 can have at most n−2 zeroes, f(x) can have at most n−2points of inflection.

57. f ′(x) = n(x− x0)n−1; f ′′(x) = n(n− 1)(x− x0)n−2.

(a) If n is odd, then the sign of f ′′(x) changes around x0 and (x0, 0) is a point of inflection.


(b) If n is even, then f ′′(x) > 0 for x 6= x0 and (x0, 0) is not a point of inflection. Using thefirst derivative test, we see that f ′(x) < 0 for x < x0 and f ′(x) > 0 for x > x0. Thus(x0, 0) is a relative minimum.

58. f ′(x) = 2ax + b; f ′′(x) = 2a. If a > 0 then f ′′(x) > 0 and the graph of f(x) is concaveupward on (−∞,∞). If a < 0 then f ′′(x) < 0 and the graph of f(x) is concave downward on(−∞,∞).

59. Since f ′′′(c) 6= 0, the graph of f ′′(x) is either increasing (when f ′′′(c) > 0) or decreasing(when f ′′′(c) < 0) through (c, 0). Thus, the sign of f ′′(x) changes around x = c, and (c, f(c))is a point of inflection.

60. f(x) = x

61. The statement is false: Consider f(x) = x1/3. Since f ′(x) = 1/3x2/3 is not defined at x = 0,f ′ cannot have a critical number at 0. (Recall that a critical number must be in the domainof the function.) From f ′′(x) = −2/9x5/3 we see that f ′′(x) > 0 for x < 0, and f ′′(x) < 0 forx > 0. Thus (0, 0) is a point of inflection.

62. f ′(x) = 20x− 1 + ex; f ′′(x) = 20 + ex. f ′′(x) = 0 has no solution, and therefore f(x) cannothave a point of inflection.

63. The graph of f(x) =

{4x2 − x, x ≤ 0

−x3, x > 0is shown on the right. The

function has no tangent line at (0, 0) since f ′+(0) = 0 and f ′−(0) = −1.Although concavity changes at (0, 0), according to Definition 4.7.2 thereis no point of inflection because there is no tangent line at the point. -2 2

-2

2

64. (a) Since f is a polynomial function of degree 3, then f ′ is a polynomial function of degree2, and its graph is a parabola. We are given that c1 and c2 are distinct critical numbers;thus, c1 and c2 are not on the f ′ parabola’s vertex. Since f ′′ = 0 only at the f ′ parabola’svertex, then f ′′(c1) and f ′′(c2) cannot be zero, and therefore f(c1) and f(c2) must berelative extrema.

(b) Let c be the x coordinate of the f ′ parabola’s vertex; thus, f ′′(c) = 0, and (c, f(c)) is

the point of inflection for the graph of f . c must therefore bec1 + c2

2.

65. As stated in Notes from the Classroom, textbooks disagree on the precise definition of a pointof inflection. With sufficient research, the student should find definitions for which somefunction f will have different points of inflection.

4.8 Optimization

1. Let x and 60−x be the two numbers. We want to maximize P (x) = x(60−x) = 60x−x2 on[0, 60]. Solving P ′(x) = 60− 2x = 0 we obtain the critical number 30. Since P ′′(x) = −2 < 0,the product is maximized by the numbers 30, 30.

4.8. OPTIMIZATION 249

2. Let x and 50/x be the two numbers. We want to minimize S(x) = x+50/x on (0,∞). SolvingS′(x) = 1 − 50/x2 = 0 we obtain the critical number

√50. Since S′′(

√50) > 0, the sum is

minimized by the numbers√

50,√

50.

3. Let x be the number. We want to maximize f(x) = x − x2. Solving f ′(x) = 1 − 2x = 0we obtain the critical number 1/2. Since f ′′(x) = −2 < 0, the product is maximized by thenumber 1/2.

4. Let x and S−x be the numbers. We want to maximize P (x) = xm(S−x)n on [0, S]. Solving

P ′(x) = xm[n(S − x)n−1(−1)] + (S − x)n(mxm−1) = xm−1(S − x)n−1[mS − (m+ n)x] = 0

we obtain the critical numbers 0, S, andmS

m+ n. Since P (0) = P (S) = 0, the product is

maximized by the numbers x =mS

m+ nand S − x =

nS

m+ n.

5. Let x and 1−x be the two numbers. We want to maximize S(x) = x2+2(1−x)2 = 2−4x+3x2.Solving S′(x) = −4 + 6x = 0 we obtain the critical number 2/3. Since S′′(x) = 6 > 0, thesum is minimized by x = 2/3 and 1− x = 1/3.

6. Let x and 1/x be the numbers. We want to maximize S(x) = x + 1/x on (0,∞). SolvingS′(x) = 1 − 1/x2 = 0 we obtain the critical number 1. Since S′′(1) = 2 > 0, the minimumsum is S(1) = 2.

7. Let (x,√

6x) be on the graph. We will minimize the square of the distance.

For (5, 0), D(x) = (x − 5)2 + (√

6x − 0)2 = x2 − 4x + 25. Solving D′(x) = 2x − 4 = 0 weobtain the critical number 2. Since D′′(x) = 2 > 0, the distance is minimized by the points(2,±2

√3) on the graph.

For (3, 0), D(x) = (x − 3)2 + (√

6x − 0)2 = x2 + 9. Solving D′(x) = 2x = 0 we obtain thecritical number 0. Since D′′(x) = 2 > 0, the distance is minimized by the point (0, 0) on thegraph.

8. Let (x, 1− x) be a point on the graph. We will minimize the square of the distance to (2, 3):D(x) = (x − 2)2 + (1 − x − 3)2 = 2x2 + 8. Solving D′(x) = 4x = 0 we obtain the criticalnumber 0. Since D′′(x) = 4 > 0, the distance is minimized by the point (0, 1) on the graph.

9. The slope of the tangent line at x is s(x) = 3x2 − 8x. To minimize s(x), we solve s′(x) =6x − 8 = 0. This gives x = 4/3. Since s′′(x) = 6 > 0, the slope is minimized at the point(4/3,−128/27).

10. The slope of the tangent line at x is s(x) = 16x− 1/x2. Since limx→∞

s(x) = +∞, the tangentline to the graph does not have a maximum slope.

11. Let (x, y) be the corner of the rectangle lying on the line. Then y = 2− 23x, and we want to

maximize A(x) = xy = 2x− 23x2 on [0, 3]. Solving A′(x) = 2− 4

3x = 0 we obtain the critical

number 3/2. Since A′′(x) = −4/3 < 0, the area is maximized when the base of the rectangleis 3/2 and the height is 1.


12. Let the width of the rectangle be 2x and the height be y. Then y = 24− x2, and we want tomaximize A(x) = 2xy = 2x(24− x2) on [0,

√24]. Solving A′(x) = 48− 6x2 = 0 we obtain the

critical number 2√

2. Since A′′(2√

2) < 0 the area is maximized when the rectangle is 4√

2units wide and 16 units high.

13. Triangle OXY is similar to triangle AXP , so

y

x=

4x− 2

and y =4xx− 2

.

We want to minimize the area of the triangle A(x) =12xy =

12x

(4xx− 2

)=

2x2

x− 2on (2,∞).

33. We want to maximize area. The perimeter of the rectangleis p = 2x + 2y and the area is A = xy. Solving p = 2x + 2y

for y, we obtain 2y = p ! 2x or y = 12 p ! x. Then, the

objective function is

A(x) = xy = x

!12

p! x

"=

12

px! x2.

In this problem, since A(x) must be positive, 0 " x " 12p.

x

y

34. We see from the figure that triangle OXY is similar to tri-angle AXP , so

y

x=

4x! 2

and y =4x

x! 2.

We want to maximize the area of the triangle, so the objec-tive function is

A =12xy =

12x

!4x

x! 2

"=

2x2

x! 2.

The domain of A(x) is (2,#).

Y!0,y"

P!2,4"

X!x,0"A!2,0"O

35. See the figure in the text. The volume of the box is V = x2y = 32,000, so y = 32,000/x2.

(a) We want to minimize the surface area of the open box, so the objective function is

A = x2 + 4xy = x2 + 4x

!32,000

x2

"= x2 +

128,000x

.


(b) We want to minimize the surface area of the closed box, so the objective function is

A = 2x2 + 4xy = 2x2 + 4x

!32,000

x2

"= 2x2 +

128,000x

.


36. The figure for Problem 35 in the text applies, except that the top of the box is closed. In thiscase we want to maximize volume. The total cost to construct each box is

2x2 + xy + xy + xy + xy = 2x2 + 4xy = 36.

The volume is V = x2y. Solving the cost equation for y, we obtain

2x2 + 4xy = 36

4xy = 36! 2x2

y =36! 2x2

4x=

18! x2

2x.

2.8 Translating Words into Functions 155

Solving A′(x) =4x(x− 2)− 2x2

(x− 2)2=

2x(x− 4)(x− 2)2

= 0 we obtain the critical number 4. Since

A′′(4) > 0, the area is minimized when the vertices are (4, 0) and (0, 8).

14. We want to maximize s(x) = 1−x−(x2−1) = 2−x−x2 on [−2, 1]. Solving s′(x) = −1−2x = 0we obtain the critical number −1/2. Since s′′(x) = −2 < 0, the distance is maximized whenx = −1/2. The maximum distance is s(−1/2) = 9/4.

15. Let x and 1500 − x be the two sides of the corral. We want to maximizeA(x) = x(1500− x) on [0, 1500]. Solving A′(x) = 1500− 2x = 0 we obtainthe critical number 750. Since A′(x) = −2 < 0, the area is maximizedwhen the corral is 750 ft× 750 ft. x

1500− x

16. Let x and y be the lengths shown in the figure. Since xy = 4000,y = 4000/x, and we want to minimize P (x) = 4x + 2y = 4x + 8000/x.Solving P ′(x) = 4 − 8000/x2 we obtain the critical number 20

√5. Since

P ′′(20√

5) > 0, the amount of fence is minimized when x = 20√

5 m andy = 40

√5 m.

y

x

-�

17. Let x and y be as shown in the figure. Since 4x+2y = 8000, y = 4000−2x,and we want to maximize A(x) = xy = 4000x − 2x2. Solving A′(x) =4000− 4x = 0 we obtain the critical number 1000. Since A′′(x) = −4 < 0,the area is maximized when x = 1000 m and y = 2000 m. y

x

-�

18. Let x and y be as shown in the figure. Then 2x + 40 + 2y = 160 andy = 60− x. We want to maximize A(x) = (x+ 40)y = (x+ 40)(60− x) =2400 + 20x−x2 for x in [0, 60]. Solving A′(x) = 20− 2x = 0 we obtain thecritical number 10. Since A′′(x) = −2 < 0, the maximum area is obtainedwhen x = 10 ft and y = 50 ft. Thus, the enclosed yard will be a square 50feet on each side.

y40x

x + 40

y

19. Let x and y be as shown in the figure. Then 2x+40+2y = 80 and y = 20−x.We want to maximize A(x) = (x+40)y = (x+40)(20−x) = 800−20x−x2

for x in [0, 20]. Solving A′(x) = −20−2x = 0 we obtain the critical number−10. Comparing A(0) = 800 and A(20) = 0 we see that the maximumarea is obtained when x = 0 and y = 20. Thus, the yard will be a rectangle40 feet long by 20 feet wide.

y40x

x + 40

y


20. Let x be the lengths of the sides perpendicular to the cliff and y the lengthof the sides parallel to the cliff. Then xy = 128, 000 and y = 128, 000/x.We want to minimize C(x) = 2(2.5x) + 2.5y + 1.5y = 5x + 512, 000/x.Solving C ′(x) = 5 − 512, 000/x2 = 0 we obtain the critical number 320.Since C ′′(320) > 0, the cost is minimized when x = 320 ft and y = 400 ft.

x

ycliff

21. Let x be a side of the base and y the height of the box. Then x2y =32, 000 and y = 32, 000/x2. We want to minimize A(x) = x2 + 4xy =x2 + 128, 000/x. Solving A′(x) = 2x − 128, 000/x2 = 0 we obtain thecritical number 40. Since A′′(40) > 0, the amount of material is minimizedwhen x = 40 cm and y = 20 cm. x

x

y

22. In this case, we want to minimize A(x) = 2x2 + 4xy = 2x2 + 128, 000/x. Solving A′(x) =4x − 128, 000/x2 we obtain the critical number 20 3

√4. Since A′′(20 3

√4) > 0, the amount of

material is minimized when x = y = 20 3√

4 cm.

23. Let x be the side of the square cut-out. We want to maximize V (x) =x(40−2x)2 on [0, 20]. Solving V ′(x) = 12x2−320x+1600 = 4(3x−20)(x−20) = 0 we obtain the critical numbers 20/3 and 20. Since V (20) = 0and V ′′(20) < 0, the volume is maximized when the height is 20/3 cmand the base is 80/3 cm × 80/3 cm. The maximum volume is V (20/3) =128, 000/27 cm3.

x

40

24. V (x) = x(30−4x)(20−2x) = 600x−140x2+8x3. Solving V ′(x) = 600−280x+24x2 = 8(3x2−35x+ 75) = 0 we obtain the critical number (35− 5

√13)/6 ≈ 2.83 and (35 + 5

√13)/6 ≈ 8.84.

Since 0 ≤ x ≤ 7.5 and V ′′(2.83) = −280 + 48(2.83) < 0, we see that the volume is maximizedwhen x = (35− 5

√13)/6 ≈ 2.83. The dimensions of this box are 20/3 + 10

√13/3 ≈ 18.69 in

long, 25/3+5√

13/3 ≈ 14.34 in wide by 35/6−5√

13/6 ≈ 2.83 in high. The maximum volumeis the product of these dimensions or approximately 758.08 in3.

25. Let x be the height of the gutter. We want to maximize A(x) = x(30−2x)on [0, 15]. Solving A′(x) = 30−4x = 0 we obtain the critical number 15/2.Since A′′(x) = −4 < 0, the cross-sectional area and hence the volume, ismaximized when the gutter is 7.5 cm high and 15 cm wide.

x

30 – 2x

26. The total area of the two triangles is 100 sin θ cos θ. We want to maximizeA(θ) = area of triangles+area of a rectangle = 100 sin θ cos θ+100 sin θon [0, π/2]. Solving

A′(θ) = 100(cos2 θ − sin2 θ) + 100 cos θ

= 100(2 cos2 θ − 1) + 100 cos θ = 100(2 cos θ − 1)(cos θ + 1) = 010

10θ

10 cos θ

10 sin θ

on [0, π/2] we obtain the critical number π/3. Comparing A(0) = 0, A(π/3) = 75√

3, andA(π/2) = 100, we see that the cross-sectional area and hence the volume, is maximized whenθ = π/3.


27. Let x be the distance from the 10-foot flagpole. We want to minimizeL(x) =

√400 + (30− x)2 +

√100 + x2 on [0, 30]. Setting

L′(x) =−(30− x)√

400 + (30− x)2+

x√100 + x2

= 0 30 – x x

1020

we obtain x2+20x−300 = (x+30)(x−10) = 0. The positive critical number is 10. ComparingL(0) = 10 + 10

√13, L(10) = 30

√2, and L(30) = 20 + 10

√10, we see that the length of wire

is minimized when it is attached 10 feet from the 10-foot flagpole.

28. Let x be the radius of the semicircle and y the length of the rectangle. Then 2y + 2πx = 2and y = 1 − πx. We want to maximize A(x) = 2xy = 2x(1 − πx) = 2x − 2πx2. SolvingA′(x) = 2 − 4πx = 0 we obtain the critical number 1/2π. Since A′′(x) = −4π < 0, the areais maximized when the semicircle has radius 1/2π ≈ 0.1592 km = 159.2 m and the length ofthe rectangle is 1− π(1/2π) = 0.5 km = 500 m.

29. Let the radius of the semicircle be r and the height of the rectangle h. Then the

perimeter is 2r+ 2h+ πr = 10 and h =12

[10− (2 + π)r]. We want to maximize

A(r) = 2rh+πr2

2= 10r −

(2 +

π

2

)r2.

2r

h

r

Solving A′(r) = 10 − (4 + π)r = 0 we obtain the critical number10

4 + π. Since A′′(r) =

−4−π < 0, the area is maximized when the base of the window is20

4 + πm, the height of the

rectangular portion is10

4 + πm, and the radius of the circular portion is

104 + π

m.

30. Let the base of the window be 2x and the height of the rectangle h. Thenthe perimeter is 6x+2h = 10 and h = 5−3x. We want to maximize A(x) =2xh+

√3x2 = 10x−6x2 +

√3x2. Solving A′(x) = 10− (12−2

√3)x = 0 we

obtain the critical number5

6−√

3. Since A′′(x) = −12+2

√3 < 0, the area

is maximized when the base of the window is10

6−√

3≈ 2.34 m and the

height of the rectangular portion is 5−3(

56−√

3

)=

15− 5√

36−√

3≈ 1.49 m.

2x

h

2x

x

√3x

31. By the Pythagorean Theorem, L2 = (x+5)2+y2. Using similar triangles, we havey

10=x+ 5x

and y =10(x+ 5)

x. We want to minimize

L2 = (x+ 5)2 +100(x+ 5)2

x2= (x+ 5)2

(1 +

100x2

)


for x > 0. SettingdL2

dx= 0 we obtain

dL2

dx= (x+ 5)2

(−200x3

)+(

1 +100x2

)[2(x+ 5)] = 2(x+ 5)

[1 +

100x2− (x+ 5)

100x3

]

= 2(x+ 5)(

1− 500x3

)= 2(x+ 5)

(x3 − 500

x3

)= 0

so x = 3√

500. Using the first derivative test,dL2

dx< 0 for 0 < x < 3

√500 and

dL2

dx> 0

for x > 3√

500, we see that L2 and hence L is minimized when x = 3√

500. In this caseL = (5 + 3

√500)

√1 + 100/5002/3 ≈ 20.81 ft.

32. As seen in Figure 4.8.23, let the height and width of the box be x and the length y. Theny + 4x = 108, and we want to maximize V (x) = x2y = x2(108 − 4x). Solving V ′(x) =216x − 12x2 = 0 we obtain the critical numbers 0 and 18. Since V ′′(18) = −216 < 0, thevolume is maximized when the width and height are 18 in, and the length is 36 in.

33. As seen in Figure 4.8.24, let r be the radius and h the height of the cylinder. Then, using

similar triangles, we haveh

12=

8− r8

or h = 12 − 12r8

= 12 − 3r2

. We want to maximize

V (r) = πr2h = 12πr2 − 3πr3

2on [0, 8]. Solving V ′(r) = 24πr − 9πr2

2= 0 we obtain the

critical numbers 0 and 16/3. Comparing V (0), V (16/3), and V (8) we see that the volume ismaximized when r = 16/3 and h = 4.

34. As seen in Figure 4.8.25, let L be the length of the line segment touchingboth outer walls and the inner corner. Let x be the length of the segmentfrom one outer wall to the inner corner. Using similar triangles, we have

L− x8

=x√

x2 − 64.

xL – x 8

8 √x2 – 64

Solving for L we obtain L(x) = x +8x√

8x2 − 64, which is to be minimized. Solving L′(x) =

1 − 512(x2 − 64)3/2

= 0 we obtain the positive critical number 8√

2. Since L′′(8√

2) > 0, the

length is minimized when x = 8√

2 ft. Thus, the maximum length of a board that will fitaround the corner is L(8

√2) = 16

√2 ft.

35. As seen in Figure 4.8.26, let r be the radius and h the height of the can. Then πr2h = 32and h = 32/πr2. We want to minimize A(r) = 2πr2 + 2πrh = 2πr2 + 64/r on (0,∞). SolvingA′(r) = 4πr − 64/r2 = 0 we obtain the critical number 3

√16/π. Since A′′(r) > 0 for r > 0,

the surface area of the can is minimized when r = 3√

16/π in and h = 2 3√

16/π in.

36. From Problem 35, the volume of the can remains the same and so does the expression for h interms of r. However, the area that we want to minimize is A(r) = 2(4r2)+2πrh = 8r2 +64/ron (0,∞). Solving A′(r) = 16r−64/r2 = 0 we obtain the critical number 3

√4. Since A′′(r) > 0

for r > 0, the metal used for the can (including waste) is minimized when r = 3√

4 in andh = 8 3

√4/π in.


37. Let x be the distance on land across from the island to the point where the

bird first intersects the shore. We want to minimize T (x) =16√x2 + 9 +

110

(20 − x) on [0, 20]. Solving T ′(x) =16x/√x2 + 9 − 1

10= 0 we obtain

the critical number 9/4. Since T ′(0) = −1/10 < 0 and T ′(4) = 1/30 > 0,the time is minimized when x = 9/4 km. Therefore, the bird should flyover water to a point on land 20− 9/4 = 17.75 km from the nest.

3

x

20 – x

√x2 + 9

38. Let x be the distance along the opposite bank to which the pipeline isrun. Let h be the length of pipe across the swamp. Then h2 = x2 + 16.We want to minimize C(x) = 25h + 20(4 − x) = 25

√x2 + 16 + 20(4 − x)

on [0, 4], where the cost is measured in units of $1000. Solving C ′(x) =25x/

√x2 + 16 − 20 = 0 we obtain the critical number 16/3. Comparing

C(0) = 180, C(16/3) = 160, and C(4) = 100, we see that the cost isminimized when x = 4 mi.

h

x4 – x

4

39. Let x be the distance along the opposite bank to which the pipeline is run.Let h be the length of pipe across the swamp. Then h2 = x2 +16. We wantto minimize C(x) = 2h + 4 − x = 2

√x2 + 16 + 4 − x on [0, 4], where the

cost is measured in units of $1000. Solving C ′(x) = 2x/√x2 + 16− 1 = 0

we obtain the critical number 4/√

3. Comparing C(0) = 12, C(4/√

3) =4√

3 + 4, and C(4) = 8√

2, we see that the cost is minimized when x =4√

3 + 4 mi.

h

x4 – x

4

40. Let h be the distance between the ships at time t. Then h2 = x2 +y2 where50 − x = 20t and y = 10t. We will maximize h2 = (50 − 20t)2 + (10t)2 =500t2 − 2000t + 2500. Solving dh2/dt = 1000t − 2000 = 0 we obtain thecritical number 2. Since d2h2/dt2 = 1000 > 0, the distance is minimizedat 2 AM. y

xh

B

A50

41. As seen in Figure 4.8.30, let r be the radius of the cylinder and h the length of the cylinder

without the hemispherical ends. The volume of the container is πr2h+43πr3 = 30π. Solving

for h we obtain h =30r2− 4r

3. We want to minimize

C(r) = 2πrh+32

(4πr2) = 2πr(

30r2− 4r

3

)+ 6πr2 =

60πr− 10πr2

3.

Setting C ′(r) = 0 we have

C ′(r) = −60πr2

+20π3r = 0;

−3r2

+r

3= 0;

r3 − 93r2

= 0;

so r = 91/3. Since C ′′(r) = 120π/r3 + 20π/3 and C ′′(91/3) = 20π(

69

+13

)> 0, the cost is

minimized when r = 91/3 and h = 30/92/3 − 4(91/3)/3 = 2(91/3).

42. Let x be the height and y the width of the printed portion. Then xy = 32 and y = 32/x.We want to minimize A(x) = (x + 2)(y + 4) = 40 + 4x + 64/x on (0,∞). Solving A′(x) =


4− 64/x2 = 0 we obtain the critical number 4. Since A′′(4) > 0, the area is minimized whenx = 4 and y = 8. The page should then be 6 in high and 12 in wide.

43. Label the figure as shown. Note that 4ADE and 4DBC are similarand that 4ABD is a right triangle. Using similarity, we have

y

8.5=

x√x2 − (8.5− x)2

or y =√

8.5x√2x− 8.5

=√

4.25x√x− 4.25

.

We want to minimize L2 = x2 + y2 = x2 +4.25x2

x− 4.25on (4, 8].

B

AE

D

C 8.5

x

x8.5 – x

y

L

θϕ

θϕ

SolvingdL2

dx= 2x +

4.25x2 − 36.125x(x− 4.25)2

=2x2(x− 6.375)

(x− 4.25)2= 0, we obtain the critical number

6.375. SincedL2

dx

∣∣∣∣x=6

= −8.8163 < 0 anddL2

dx

∣∣∣∣x=7

= 8.0992 > 0, we see that the length is

minimized when the width of the fold is 6.375 inches.

44. We see from the figure that x2 + b2 = 4 and x2 + c2 = 9. The area of the kiteis A = xb+ xc. We want to maximize A(x) = x(

√4− x2 +

√9− x2). Setting

A′(x) = x

(− x√

4− x2− x√

9− x2

)+√

4− x2 +√

9− x2

=−x2(

√9− x2 +

√4− x2) + (4− x2)

√9− x2 + (9− x2)

√4− x2

√4− x2

√9− x2

=(4− 2x2)

√9− x2 + (9− 2x2)

√4− x2

√4− x2

√9− x2

= 0

2 2

33

x xb

c

we obtain (4− 2x2)√

9− x2 = −(9− 2x2)√

4− x2

(16− 16x2 + 4x4)(9− x2) = (81− 36x2 + 4x4)(4− x2)

144− 160x2 + 52x4 − 4x6 = 324− 225x2 + 52x4 − 4x6

65x2 − 180 = 0

5(13x2 − 36) = 0.

Thus, x = 6/√

13, b =√

4− 36/13 = 4/√

13, and c =√

9− 36/13 = 9/√

13. The crossbarshave lengths 2x = 12/

√13 ≈ 3.33 ft and b+ c =

√13 ≈ 3.61 ft.

45. We want to maximize A(θ) = (a sin θ + b cos θ)(a cos θ + b sin θ)

= (a2 + b2) sin θ cos θ + ab =12

(a2 + b2) sin 2θ + ab

on (0, π/2). Solving A′(θ) = (a2 + b2) cos 2θ = 0 we obtain the critical number π/4. SinceA′′(π/4) < 0, the area is maximized when θ = π/4. The circumscribed rectangle is a squarewhose side is a sinπ/4 + b cosπ/4 = (a+ b)/

√2.


46. As seen in Figure 4.8.35, let x be the distance from the person to the pedestal.

Then tanφ = 1/x and tan(φ−θ) = 1/2x. Since tan(φ−θ) =tanφ− tan θ

1 + tanφ tan θ,

we have

12x

=1/x− tan θ

1 +1x

tan θ=

1− x tan θx+ tan θ

.

x

θ ϕ

1/2

1/2

Solving for tan θ we obtain tan θ =x

2x2 + 1. To maximize θ, it will suffice to maximize

tan θ since tan θ is an increasing function on [0, π/2). Solvingd

dxtan θ =

1− 2x2

(2x2 + 1)2= 0,

we obtain the critical number√

2/2. Sinced2

dx2tan θ

∣∣∣∣x=√

2/2

= −√

2/2 < 0, the angle is

maximized when x =√

2/2 m.

47. As seen in Figure 4.8.36, let x be the width of the beam, y the height, and d the diameter ofthe log and diagonal of the wooden beam (shown by the dotted line). From the PythagoreanTheorem, we have d2 = x2 + y2. We want to maximize S(x) = xy2 = x(d2 − x2). SolvingS′(x) = d2 − 3x2 = 0 we obtain the critical number d/

√3. Since S′′(d/

√3) = −6d/

√3 < 0,

the strength is maximized when the length is d/√

3 and the width is d√

2/3.

48. From Problem 38 in Part C of Chapter 1 in Review we have S(θ) = 25π csc θ− 503π cot θ+40.

Setting

S′(θ) = −25π csc θ cot θ +503π csc2 θ = 25π csc θ

(23

csc θ − cot θ)

= 0

we obtain23

csc θ = cot θ or cos θ =23

and θ ≈ 0.84 radian ≈ 48.19◦. This critical numbercan be seen to be a relative minimum by the first derivative test. The minimum surface areais S(0.84) ≈ 98.54 ft2.

49. Let x be the distance from P to the bulb with intensity I1. We want to

minimize E(x) =125x2

+216

(10− x)2on (0, 10). Setting E′(x) = −250

x3+

432(10− x)3

= 0, we have(10− x)3

x3=

432250

=216125

or10− xx

=65

.

Exercises 4.8

Exercises 4.8, Page 11

48. The container shown in figure 4.8.37 is to be constructed by attaching an inverted cone (open at

its top) to the bottom of a right circular cylinder (open at its top and bottom) of radius 5 ft. The

container is to have a volume of 100 ft3. Find the value of the indicated angle so that the total

surface area of the container is a minimum. What is the minimum surface area? [Hint: See

Problem 38 in part C of Chapter 1 in Review.]

Mathematical Models

49. The illuminance E due to a light source or intensity I at a distance r from the source is given by

E = I r2 . The total illuminance from two light bulbs of intensities I1 = 125 and I2 = 216 is the

sum of the illuminances. Find the point P between the two light bulbs 10 m apart at which the

total illuminance is a minimum. See figure 4.8.38.

50. The illuminance E at any point P on the edge of a circular table caused by a light placed directly

above its center is given by E = (I cos!) r2. See figure 4.8.39. Given that the radius of the table is

1 m and I = 100, find the height at which the light should be placed so that E is a maximum.

figure 4.8.38 Light sources in Problem 49

figure 4.8.37 Container in Problem 48

x 10 – x

Thus, 50/11 is a critical number, and by the first derivative test, we see that the totalilluminance will be a minimum at 50/11 m from the bulb with intensity I1 = 125.

50. Letting r be the distance from the light to P , we have r = csc θ. We want tomaximize E(θ) = 100 cos θ/ csc2 θ = 100 sin2 θ cos θ = 100 cos θ − 100 cos3 θ forθ > 0. Solving E′(θ) = −100 sin θ+300 cos2 θ sin θ = (100 sin θ)(3 cos2 θ−1) = 0we obtain a positive critical number θc when cos θc = 1/

√3. To apply the first

derivative test, we note that 100 sin θ > 0 for any θ near θc. Now for θ < θc,cos θ > cos θc and 3 cos2 θ − 1 > 3 cos2 θc − 1 = 0, so E′(θ) > 0.

hθr

1P


Similarly, for θ > θc it can be shown that E′(θ) < 0. Therefore E has a relative maximum atθc. The corresponding height is

h = r cos θc = csc θc cos θc =cos θcsin θc

=1/√

3√1− (1/

√3)2

= 1/√

2 m.

51. Let x be the point on the x-axis where the light crosses from onemedium to the other. We want to minimize

T (x) =h1

c1+h2

c2=√x2 + a2

c1+

√(d− x)2 + b2

c2.

To find the critical numbers, we set

T ′(x) =x

c1√x2 + a2

− d− xc2√

(d− x)2 + b2= 0.

b

a

d – x

x

θ1 θ1

θ2

h1

h2

Thenx/√x2 + a2

c1=

(d− x)√

(d− x)2 + b2

c2, and since sin θ1 =

x

h1and sin θ2 =

d− xh2

, we

have sinθ1

c1= sin

θ2

c2at the critical number. To see that the time is actually minimized at

this point, we compute the second derivative T ′′(x) =a2

c1(x2 + a2)3/2+

b2

c2[(d− x)2 + b2]3/2.

Since T ′′(x) > 0 for all x, we do have a minimum at the critical number.

52. From the figure we see that y = a csc θ and l− x = y cos θ = a csc θ cos θ =a cot θ. Thus x = l − a cot θ. We want to minimize

R(θ) =kx

r41

+ky

r42

=k

r41

(l − a cot θ) +k

r42

(a csc θ).a

θ

y

l – x

Setting R′(θ) =ak

r41

(csc2 θ) − ak

r42

(csc θ cot θ) = 0 we obtaincsc θr41

=cot θr42

orr42

r41

= cos θ. To

see that this value of θ minimizes r, we compute

R′′(θ) = ak

(−2 csc2 θ cot θ

r41

+csc3 θ + csc θ cot2 θ

r42

)

= ak csc θ(

csc2 θ + cot2 θ

r42

− 2 csc θ cot θr41

)

=ak csc θr42

(csc2 θ + cot2 θ − 2r4

2

r41

csc θ cot θ)

=ak csc θr42

(csc2 θ + cot2 θ − 2 cos θ csc θ cot θ)

=ak csc θr42

(csc2 θ + cot2 θ − 2 cot2 θ) =ak csc θr42

(csc2 θ − cot2 θ) =ak csc θr42

.

Thus, R′′(θ) > 0 for 0 < θ < π and the resistance is minimum when cos θ = r42/r

41.


53. U ′(x) = −24/x13 + 6/x7 = 6(x6 − 4)/x13. Solving U ′(x) = 0 we obtain the critical numbers± 3√

2. From the first derivative test, we see that relative minima occur at both of these points.Thus, the minimum potential energy is U( 3

√2) = −1/8.

x − 3√

2 0 3√

2f ↘ ↗ ↘ ↗f ′ − 0 + undefined − 0 +

54. y′ = tan θ0−(

g

v20 cos2 θ0

)x. Solving y′ = 0 we obtain the critical number

v20 tan θ0 cos2 θ0

g=

v20 sin θ0 cos θ0

g. Since y′′ = − g

v20 cos2 θ0

< 0, the height is maximum when x =v2

0 sin θ0 cos θ0

g.

The maximum height is

h = tan θ0v2

0 sin θ0 cos θ0

g− g

2v20 cos2 θ0

(v2

0 sin θ0 cos θ0

g

)2

=v2

0 sin2 θ0

g− v2

0 sin2 θ0

2g=v2

0 sin2 θ0

2g.

55. (a) We want to maximize

y(x) =w0

24EI(L2x2 − 2Lx3 + x4) =

w0

24EIx2(x− L)2.

Now

y′(x) =w0

24EI[2x2(x− L) + 2x(x− L)2

]=

w0

12EIx(x− L)(2x− L).

Solving y′(x) = 0 we obtain the critical numbers 0, L/2, and L. Using the first derivativetest we see that y(L/2) = w0L

4/384EI.

(b)Lx

y

56. (a) The tree attains its maximum height H when h′(d) = a − 2bd = 0 or d = a/2b. ThusD = a/2b and the maximum height is H = h(D) = 137 + aD − bD2. Substitutinga = 2bD into the above equation, we obtain H = 137 + 2bD2 − bD2 = 137 + bD2. Then

b =H − 137D2

and a =2D(H − 137)

D2=

2(H − 137)D

. Thus

h(d) = 137 + 2(H − 137

D

)d− H − 137

D2d2.


(b) Letting H = 1500, D = 800, and h = 1000 we obtain

1000 = 137 + 2(

1500− 13780

)d− 1500− 137

6400d2 = 137 +

136340

d− 13636400

d2

or

13636400

d2 − 136340

d+ 863 ≈ 0.213d2 − 34.075d+ 863 = 0.

Using the quadratic formula, we find d ≈ 31.55 cm and d ≈ 128.45 cm. Assumingd < 0.8 m when h = 10 m we conclude that the diameter of the tree was approximately31.55 cm.

57. We want to minimize m(x) =πρM2/3

2K2/3

[2− x2

(1− x4)2/3

]. Solving

m′(x) = −πρM2/3x

3K2/3

[x4 − 8x2 + 3(1− x4)5/3

]= 0

we obtain the critical numbers 0,√

4−√

13, and√

4 +√

13. Since x = r/R where 0 <r < R, we must have x in the interval (0, 1). Thus, the only appropriate critical number isx =

√4−√

13 ≈ 0.63. Use the first derivative test with x = 0.6 and 0.7 to show that themass m is minimized for x = r/R ≈ 0.63 or r ≈ 0.63R.

58. We want to maximize P (I) =100I

I2 + I + 4on [0,∞). Solving P ′(I) =

100(4− I2)(I2 + I + 4)2

= 0 we

obtain the critical number 2. Since P ′(1) > 0 and P ′(3) < 0, we see by the first derivativetest that P is largest for I = 2.

59. Let x be the length of wire formed into a circle and 1− x the length formedinto a square. Since x is the circumference of the circle, its radius is x/2π.

We want to maximize A(x) = π( x

2π

)2

+(

1− x4

)2

=x2

4π+

(1− x)2

16on [0, 1].

x 1 – x

Solving A′(x) =x

2π− 1− x

8= 0 we obtain the critical number

π

4 + π. Comparing A(0) = 1/6,

A

(π

4 + π

)=

14(4 + π)

, and A(1) = 1/4π, we see that the maximum area is obtained when

the entire wire is formed into a circle of radius 1/2π m.

60. Let x be the length of wire formed into a circle and 1− x the length formedinto a triangle. Since x is the circumference of the circle, its radius is x/2π.

The perimeter of the triangle is 1 − x, so each side is1− x

3and the area is

(1− x

6

)2√3. We want to minimize and maximize

x 1 – x

A(x) = π( x

2π

)2

+(

1− x6

)2√3 =

x2

4π+(

1− x6

)2√3


on [0, 1]. Solving A′(x) =x

2π−(

1− x18

)√3 = 0 we obtain the critical number

π√

39 + π

√3

.

Comparing A(0) ≈ 0.0481, A

(π√

39 + π

√3

)≈ 0.0299, and A(1) ≈ 0.0796, we see that the sum

of the areas is minimized when x =π√

39 + π

√3

and maximized when x = 1, or when the entire

wire is formed into a circle.

61. (a) By the Pythagorean Theorem, R2 = r2 + h2 or r2 = R2 − h2. We want to maximize

V (h) =13πr2h =

13π(R2 − h2)h on [0, R]. Solving V ′(h) =

13π

(R2 − 1

3h2

)= 0 we

obtain the critical numbers h = ±R/√

3. Since V (0) = V (R) = 0 and V ′′(R/√

3) =

−29R/√

3 < 0, the volume is maximized when h = r/√

3 or

r =√R2 − h2 =

√R2 −R2/3 =

√2/3R.

(b) The maximum volume is V =13π

(23R2

)R/√

3 =29πR3/

√3.

(c) The circumference of the circular piece of paper is 2πR. The circumfer-ence of the base of the cone is 2πr = 2π

√2/3R. Thus, s = 2πR−2πr =

2πR(1−√

2/3) and θ = s/R = 2π(1−√

2/3) ≈ 1.15 radians. θR

s

62. From Figure 4.8.45, by the Pythagorean theorem L2 = x2 + y2, so y2 = L2−x2, where L is afixed number. The circumference of the base of the cylinder is y = 2πr, so r = y/2π. Thus,we want to maximize the volume of the cylinder

V (x) = πr2x = π( y

2π

)2

x =xy2

4π=

14πx(L2 − x2)

on [0, L]. Solving V ′(x) =L2 − 3x2

4π= 0 we obtain the critical number L/

√3. Since V ′′(x) =

−6x/4π < 0 for x > 0, the volume is maximized when the cylinder has a height of L/√

3 and

a radius of

√2/3L2π

.

63. Problem 27 showed that the optimal amount of wire (the leastamount) is used when it is attached 10 feet from the 10-foot flag-pole. From the figure, this means that the right triangles formed bythe flagpoles, the wire, and the ground are isosceles right triangles,and therefore similar. Thus, the non-right angles θ1 and θ2 of eachtriangle are the same.

20 10

1020

θ1 θ2

64. Using the unknown coordinates of P , the slope of the tangent line is m = −2x0. Theequation of the tangent line at P is then y − (1 − x2

0) = −2x0(x − x0). The x-intercept is


then (x2

0 + 12x0

, 0); the y-intercept is (0, 1 + x20). The area of the triangle as a function of x0 is

A(x0) =14

[(x2

0 + 1)2

x0

]. Solving A′(x0) =

14

[(x2

0 + 1)(3x20 − 1)

x20

]= 0 we obtain the critical

number 1/√

3. The first or second derivative test shows that there is a relative minimum ofthe area function at this number. Now, using y = 1 − x2 we find the coordinates of P are(1/√

3, 2/3). The slope at the point is m = −2/√

3. An equation of the tangent line is

y − 23

= − 2√3

(x− 1√

3

)or y = − 2√

3x+

43.

65. The total time it takes the swimmer to reach C from A is T =√x2 + 1

3+√x2 − 8x+ 17

2on

[0, 4]. Differentiating then solving for 0 gives

dT

dx=

x

3√x2 + 1

+x− 4

2√x2 − 8x+ 17

= 0

2x√x2 − 8x+ 17 = −3(x− 4)

√x2 + 1

With CAS help, the foregoing equation has only one real root in the interval [0, 4], namelyx ≈ 3.176. Now, T (0) ≈ 2.395, T (3.176) ≈ 1.758, and T (4) ≈ 1.874. Therefore, to minimizeher time in the race, she should swim from point A to point B about 3.18 miles down thebeach from the point of the beach closest to A, then proceed directly to C.

66. (a) L(θ) = 10 csc θ + 2 sec θ

(b) L′(θ) = −10 csc θ cot θ + 2 sec θ tan θ

(c) We see from the graph that L′(θ) < 0 for 0 < θ < θc and L′(θ) > 0 forθc < θ < L/2. By the first derivative test, L(θ) has a relative minimumat θ = θc.

(d) With CAS help to numerically solve L′(θ) = 0 we find θc ≈ 1.04. (Actu-ally, it is easily seen that θc = tan−1 3

√5.) Then L(θc) ≈ 15.55.

π / 2

67. (a) Let x be the length of the cable AB. Then L(x) = x+ 2√

(4− x)2 + 4.

(b)

2 4

2

4

6

8

10

(c) Solving L′(x) = 1+−2(4−x)/√

(4− x)2 + 4 = 0 we obtain the critical numbers 4± 23√

3.

Since 4 +23√

3 > 4, we use x = 4 − 23√

3. We know from the graph that this gives aminimum.

(d) Let x be the length of the cable AB. Then L(x) = x+√

1 + (4− x)2 +√

4 + (4− x)2.


(e)

2 4

2

4

6

8

10

(f) From the graph in part (e) we estimate that L(x) is minimized when x ≈ 3.2. (Using anumerical procedure to solve L′(x) = 0 gives x ≈ 3.1955.)

68. (a) From the figure, the distance B between the transmitter at (xt, yt) and the point (x, y)is√

(x− xt)2 + (y − yt)2, and the distance A between the second transmitter at (xi, 0)and (x, y) is

√(x− xi)2 + y2. Writing y in terms of x, we have y =

√r2 − x2, and so

the distances are:

B =

√(x− xt)2 +

(√r2 − x2 − yt

)2

=√x2 − 2xtx+ x2

t + (r2 − x2)− 2yt√r2 − x2 + y2

t

=√r2 − 2xtx+ x2

t − 2yt√r2 − x2 + y2

t

A =√

(x− xi)2 + (r2 − x2) =√x2 − 2xix+ x2

i + r2 − x2 =√r2 − 2xix+ x2

i

Let Sp(x) and Ss(x) be the primary and secondary signal strengths at (x, y), respectively.These functions are therefore

Sp(x) =1

r2 − 2xtx+ x2t − 2yt

√r2 − x2 + y2

t

Ss(x) =1

r2 − 2xix+ x2i

and the signal to noise ratio R(x) is

R(x) =Sp(x)Ss(x)

=r2 − 2xix+ x2

i

r2 − 2xtx+ x2t − 2yt

√r2 − x2 + y2

t

.

(b) Converting to meters, we have xt = 760, yt = −560, r = 1100, and xi = 12000.Substituting, we have

R(x) =11002 − 2(12000)x+ 120002

11002 − 2(760)x+ 7602 − 2(−560)√

11002 − x2 + (−560)2

and from the resulting CAS-generated graph below, the domain of R(x) appears to be[−1100, 1100], with a range of approximately (40, 275).

-1000 -500 500 1000

100

200

300

4.9. LINEARIZATION AND DIFFERENTIALS 263

(c) Still using the graph, the minimum ratio R appears to occur at x ≈ −700. This yieldsR ≈ 39.3653, which is well above the FAA’s minimum threshold.

(d) Using a CAS to differentiate R(x) and finding the root of R′(x) = 0, we get x =

−220(30409√

1740266645− 153341760)353629009

≈ −693.799, R(x) ≈ 39.3649, which comparefavorably with our estimates from the graph.

(e) Using x = −693.799, we get y =√

11002 − 693.7992 ≈ 853.60585 and the point on C isapproximately (−693.799, 853.60585).

(f) The assumption that (x, y) is in the top half plane when (xt, yt) was in the lower halfplane is correct because the points on C in the top half plane are farther from (xt, yt)than in the lower half plane, with the range of distances from (xi, 0) remaining the same,thus resulting in lower signal to noise ratios.

(g) According to Mathematica, there are two possible expressions for x that result in theminimum interference. The full expressions are too long to reproduce in a practicalmanner, but suffice it to say that they certainly justify the existence of CAS for functionssuch as these, especially in terms of multiple symbolic constants.

(h) When yt = 0, we get R(x) =r2 − 2xix+ x2

i

r2 − 2xtx+ x2t

. Since x must be on the circle C, then we

seek an extremum on the closed interval [−r, r]. By Theorem 4.3.3, candidate extrema

occur at x = −r, x = r, or at the critical number x =r2 + x2

t

2xt, if this value is in [−r, r].

4.9 Linearization and Differentials

1. Using f(9) = 3, f ′(x) = 1/2√x, and f ′(9) = 1/6, the tangent line to the graph is y − 3 =

(1/6)(x− 9). The linearization of f(x) at a = 9 is L(x) =16

(x− 9) + 3.

2. Using f(1) = 1, f ′(x) = −2x−3, and f ′(1) = −2, the tangent line to the graph is y − 1 =−2(x− 1). The linearization of f(x) at a = 1 is L(x) = −2x+ 3.

3. Using f(π/4) = 1, f ′(x) = sec2 x, and f ′(π/4) = 2, the tangent line to the graph is y − 1 =2(x− π/4). The linearization of f(x) at a = π/4 is L(x) = 2(x− π/4) + 1.

4. Using f(π/2) = 0, f ′(x) = − sinx, and f ′(π/2) = −1, the tangent line to the graph isy − 0 = −1(x− π/2). The linearization of f(x) at a = π/2 is L(x) = π/2− x.

5. Using f(1) = 0, f ′(x) = 1/x, and f ′(1) = 1, the tangent line to the graph is y− 0 = 1(x− 1).The linearization of f(x) at a = 1 is L(x) = x− 1.

6. Using f(2) = 11, f ′(x) = 5 + ex−2, and f ′(2) = 6, the tangent line to the graph is y − 11 =6(x− 2). The linearization of f(x) at a = 2 is L(x) = 6x− 1.

7. Using f(3) = 2, f ′(x) = 1/2√

1 + x, and f ′(3) = 1/4, the tangent line to the graph is

y − 2 = (1/4)(x− 3). The linearization of f(x) at a = 3 is L(x) =14

(x− 3) + 2.


8. Using f(6) = 1/3, f ′(x) = − 12(3 + x)3/2

, and f ′(6) = −1/54, the tangent line to the graph is

y − 1/3 = (−1/54)(x− 6). The linearization of f(x) at a = 6 is L(x) = − 154

(x− 6) +13

.

9. Using f(0) = 1, f ′(x) = ex, and f ′(0) = 1, the tangent line to the graph is y − 1 = 1(x− 0),yielding L(x) = x+ 1 or ex ≈ 1 + x whenever x is close to 0.

10. Using f(0) = 0, f ′(x) = sec2 x, and f ′(0) = 1, the tangent line to the graph is y−0 = 1(x−0),yielding L(x) = x or tanx ≈ x whenever x is close to 0.

11. Using f(0) = 1, f ′(x) = 10(1 + x)9, and f ′(0) = 10, the tangent line to the graph is y − 1 =10(x− 0), yielding L(x) = 10x+ 1 or (1 + x)10 ≈ 1 + 10x whenever x is close to 0.

12. Using f(0) = 1, f ′(x) = −6(1 + 2x)−4, and f ′(0) = −6, the tangent line to the graph isy − 1 = −6(x− 0), yielding L(x) = 1− 6x or (1 + 2x)−3 ≈ 1− 6x whenever x is close to 0.

13. Using f(0) = 1, f ′(x) = −1/2√

1− x, and f ′(0) = −1/2, the tangent line to the graph is

y − 1 = −12

(x− 0), yielding L(x) = 1− 12x or

√1− x ≈ 1− 1

2x whenever x is close to 0.

14. Using f(0) = 2, f ′(x) =2x+ 1

2√x2 + x+ 4

, and f ′(0) = 1/4, the tangent line to the graph is

y − 2 =14

(x− 0), yielding L(x) =14x+ 2 or

√x2 + x+ 4 ≈ 2 +


15. Using f(0) = 1/3, f ′(x) = − 1(3 + x)2


y − 1/3 = −19

(x− 0), yielding L(x) = −19x+

13

or1

3 + x≈ 1

3− 1


16. Using f(0) = 1, f ′(x) = − 43(1− 4x)2/3


y − 1 = −43

(x− 0), yielding L(x) = 1− 43x or 3

√1− 4x ≈ 1− 4


17. From Problem 2 we have1x2≈ −2x+ 3 whenever x is close to 1. Thus,

(1.01)−2 = f(1.01) ≈ −2(1.01) + 3 = 0.98.

18. From Problem 1 we have√x ≈ 1

6(x− 9) + 3 whenever x is close to 9. Thus,

√9.05 = f(9.05) ≈ 1

6(9.05− 9) + 3 =

361120

.

19. From Problem 6 we have 5x+ ex−2 ≈ 6x− 1 whenever x is close to 2. Thus,

10.5 + e0.1 = f(2.1) ≈ 6(2.1)− 1 = 11.6.


20. From Problem 5 we have lnx ≈ x− 1 whenever x is close to 1. Thus,

ln 0.98 = f(0.98) ≈ 0.98− 1 = −0.02.

21. From Problem 12 we have (1 + 2x)−3 ≈ 1− 6x whenever x is close to 0. Thus,

(1.1)−3 = f(0.05) ≈ 1− 6(0.05) = 0.7.

22. From Problem 11 we have (1 + x)10 ≈ 1 + 10x whenever x is close to 0. Thus,

(1.02)10 = f(0.02) ≈ 1 + 10(0.02) = 1.2.

23. From Problem 16 we have 3√

1− 4x ≈ 1− 43x whenever x is close to 0. Thus,

(0.88)1/3 = f(0.03) ≈ 1− 43

(0.03) = 0.96.

24. From Problem 14 we have√x2 + x+ 4 ≈ 2 +

14x whenever x is close to 0. Thus,

√4.11 = f(0.1) ≈ 2 +

14

(0.1) = 2.025.

25. To find an approximation for (1.8)5 we choose f(x) = x5; a = 2. Using f(2) = 32, f ′(x) = 5x4,and f ′(2) = 80, the tangent line to the graph is y− 32 = 80(x− 2). The linearization of f(x)at a = 2 is L(x) = 80x− 128. Thus,

(1.8)5 = f(1.8) ≈ 80(1.8)− 128 = 16.

26. To find an approximation for (7.9)2/3 we choose f(x) = x2/3; a = 8. Using f(8) = 4,f ′(x) = 2/3 3

√x, and f ′(8) = 1/3, the tangent line to the graph is y − 4 = (1/3)(x− 8). The

linearization of f(x) at a = 8 is L(x) =13

(x− 8) + 4. Thus,

(7.9)2/3 = f(7.9) ≈ 13

(7.9− 8) + 4 =12130

.

27. To find an approximation for(0.9)4

(0.9) + 1we choose f(x) =

x4

x+ 1; a = 1. Using f(1) = 1/2,

f ′(x) =4x3(x+ 1)− x4

(x+ 1)2=

x3(3x+ 4)(x+ 1)2

, and f ′(1) = 7/4, the tangent line to the graph is

y − 12

=74

(x− 1). The linearization of f(x) at a = 1 is L(x) =74

(x− 1) +12

. Thus,

(0.9)4

(0.9) + 1= f(0.9) ≈ 7

4(0.9− 1) +

12

=1340

= 0.325.


28. To find an approximation for (1.1)3+6(1.1)2 we choose f(x) = x3+6x2; a = 1. Using f(1) = 7,f ′(x) = 3x2 + 12x, and f ′(1) = 15, the tangent line to the graph is y − 7 = 15(x − 1). Thelinearization of f(x) at a = 1 is L(x) = 15x− 8. Thus,

(1.1)3 + 6(1.1)2 = f(1.1) ≈ 15(1.1)− 8 = 8.5.

29. To find an approximation for cos(π/2−0.4) we choose f(x) = cosx; a = π/2. Using f(π/2) =0, f ′(x) = − sinx, and f ′(π/2) = −1, the tangent line to the graph is y − 0 = −1(x − π/2).The linearization of f(x) at a = π/2 is L(x) = −x+ π/2. Thus,

cos(π/2− 0.4) = f(π/2− 0.4) ≈ −(π/2− 0.4) + π/2 = 0.4.

30. To find an approximation for sin 1◦ we choose f(x) = sinx; a = 0. Using f(0) = 0, f ′(x) =cosx, and f ′(0) = 1, the tangent line to the graph is y − 0 = 1(x − 0). The linearization off(x) at a = 0 is L(x) = x. Thus,

sin 1◦ = sin(π/180) = f(π/180) ≈ π/180.

31. To find an approximation for sin 33◦ we choose f(x) = sin(x+π/6); a = 0. Using f(0) = 1/2,

f ′(x) = cos(x + π/6), and f ′(0) =√

32

, the tangent line to the graph is y − 12

=√

32

(x − 0).

The linearization of f(x) at a = 0 is L(x) =√

32x+

12

. Thus,

sin 33◦ = sin(π/6 + 3π/180) = f(π/60) ≈√

32

( π60

)+

12

=√

3π + 60120

.

32. To find an approximation for tan(π

4+ 0.1

)we choose f(x) = tanx; a =

π

4. Using f(π/4) = 1,

f ′(x) = sec2 x, and f ′(π/4) = 2, the tangent line to the graph is y − 1 = 2(x − π/4). Thelinearization of f(x) at a =

π

4is L(x) = 2(x− π/4) + 1. Thus,

tan(π

4+ 0.1

)= f(π/4 + 0.1) ≈ 2[(π/4 + 0.1)− π/4] + 1 = 1.2.

33. According to the graph, f(1) = 4 and f ′(1) = 2, so the tangent line to the graph is y − 4 =2(x−1). The linearization of f(x) at a = 1 is L(x) = 2x+2. Thus, f(1.04) ≈ 2(1.04)+2 = 4.08.

34. According to the graph, f(−2) = 5 and f ′(−2) = −1/2, so the tangent line to the graph

is y − 5 = −12

[x − (−2)]. The linearization of f(x) at a = −2 is L(x) = 4 − x/2. Thus,

f(−1.98) ≈ 4− (−1.98)/2 = 4.99.

35. ∆y = (x+ ∆x)2 + 1− (x2 + 1) = 2x∆x+ (∆x)2; dy = 2x dx

36. ∆y = 3(x+∆x)2−5(x+∆x)+6− (3x2−5x+6) = 6x∆x−5∆x+3(∆x)2; dy = (6x−5)dx

37. ∆y = (x+ ∆x+ 1)2 − (x+ 1)2 = (x+ ∆x)2 + 2(x+ ∆x) + 1− (x+ 1)2

= x2 + 2x∆x+ (∆x)2 + 2x+ 2∆x+ 1− x2 − 2x− 1 = 2x∆x+ 2∆x+ (∆x)2

dy = 2(x+ 1)dx


38. ∆y = (x+ ∆x)3 − x3 = 3x2∆x+ 3x(∆x)2 + (∆x)3; dy = 3x2 dx

39. ∆y =3(x+ ∆x) + 1

x+ ∆x− 3x+ 1

x=

3x(x+ ∆x) + x− [3x(x+ ∆x) + (x+ ∆x)]x(x+ ∆x)

= − ∆xx(x+ ∆x)

dy =x(3)− (3x+ 1)

x2dx = −dx

x2

40. ∆y =1

(x+ ∆x)2− 1x2

=x2 − [x2 + 2∆x+ (∆x)2]

x2(x+ ∆x)2= −2∆x+ (∆x)2

x2(x+ ∆x)2; dy = − 2

x3dx

41. ∆y = sin(x+ ∆x)− sinx = sinx cos ∆x+ cosx sin ∆x− sinx; dy = cosx dx

42. ∆y = −4 cos 2(x+ ∆x) + 4 cos 2x = 4 cos 2x− 4 cos(2x+ 2∆x); dy = 8 sin 2x dx

43. ∆y = 5(x+ ∆x)2 − 5x2 = 10x∆x+ 5(∆x)2

∆y|x=2 = 20∆x+ 5(∆x)2;dy = 10x dx; dy|x=2 = 20 dx

(Recall that dx = ∆x.)

x ∆x ∆y dy ∆y − dy2 1.00 25.0000 20.0 5.00002 0.50 11.2500 10.0 1.25002 0.10 2.0500 2.0 0.05002 0.01 0.2005 0.2 0.0005

44. ∆y =1

x+ ∆x− 1x− ∆xx(x+ ∆x)

∆y|x=2 = − ∆x2(2 + ∆x)

;

dy = − 1x2

dx; dy|x=2 = −dx4

(Recall that dx = ∆x.)

x ∆x ∆y dy ∆y − dy2 1.00 −0.1667 −0.2500 0.08332 0.50 −0.1000 −0.1250 0.02502 0.10 −0.0238 −0.0250 0.00122 0.01 −0.0025 −0.0025 0.0000

45. (a) df = (8x+ 5) dx. When x = 4 and dx = 0.03, df = [8(4) + 5](0.03) = 1.11.

(b) When x = 3 and dx = −0.1, df = [8(3) + 5](−0.1) = −2.9.

46. (a) f ′(x) = 3x2 + 6x; f ′(1) = 9. Since f(1) = 4 the equation of the tangent line isy − 4 = 9(x− 1) or y = 9x− 5.

(b) Letting x = 1.02 in the equation of the tangent line, we obtain y = 4.18.

(c) Identifying x = 1 and dx = 0.02 we have f(1.02) = f(1 + 0.02) ≈ f(1) + f ′(1)(0.02) =4 + 9(0.02) = 4.18. This is exactly the same as the value in part (b). This is to beexpected since formula (3) in effect uses the tangent line to approximate the functionvalue.

47. (a) A(4) = 16π cm; A(5) = 25π cm; A(5)−A(4) = 9π cm.

(b) dA = 2πr dr. When r = 4 cm and dr = 1 cm, dA = 8π cm.

48. R(r) = klr−4; dR = −4klr−5 dr. When r = 0.2 mm and dR = 0.1 mm, the approximatechange in R is dR = −4kl(0.2)−5(0.1) = −1250 kl/mm4.


49. The exact volume of the cover is

∆V =43π(r + t)3 − 4

3πr3 =

43π(3r2t+ 3rt2 + t3).

dV = 4πr2dr. When dr = t, the approximate volume of the cover is 4πr2t. If r = 0.8 in andt = 0.04 in, then the approximate volume of the cover is 4π(0.8)2(0.04) = 0.1024π in3.

50. V = πr2h = 150πr2; dV = 300πr dr. If r = 2 cm and dr = 0.25 cm, then the approximatevolume is dV = 300π(2)(0.25) = 150π ≈ 471.24 cm3.

51. A = s2; dA = 2s ds. Letting s = 10 cm and ds = 0.3 cm, we obtain dA = 2(10)(0.3) =6 cm2. The maximum error in the area is ±6 cm2. The approximate relative error is ±6/102 =±0.06 cm2 and the approximate percentage error is ±6%.

52. The volume of a cylinder is V = πr2h or V = 5πr2 when the height is 5 m. The approximateerror in the volume is dV = 10πr dr or 10π(8)(±0.25) = ±20π ≈ ±62.8318 m2 when r = 8 mand dr = ±0.25 m. The approximate volume of the tank is V = 5π(82) = 320π, so theapproximate relative error is ±20π/320π = ±0.0625 m3, and the approximate percentageerror is ±6.25%.

53. P = cV −γ ; dP = −γcV −γ−1dV . The approximate relative error in P is

dP

P=−γcV −γ−1dV

cV −γ= −γ dV

V,

so the approximate relative error in P is proportional to the approximate relative error in V .

54. R = v20(sin 2θ)g−1; dR = −v2

0(sin 2θ)g−2dg. The approximate relative error in R is

dR

R=−v2

0(sin 2θ)g−2dg

v20(sin 2θ)g−1

= −dgg,

so the approximate relative error in R is proportional to the approximate relative error in g.

55. For v0 = 256 ft/s, θ = 45◦, and g = 32 ft/s2, the range is R =2562

32sin 90◦ = 2048 ft.

The approximate change in R with respect to v0 is dV =2v0

gsin 2θdv0. For v0 = 256 ft/s,

g = 32 ft/s2, θ = 45◦, and dv0 = 10 ft/s, the approximate change in the range is dv =2(256)

32(1)(10) = 160 ft.

56. (a) Solving km1m2/r2 = m1g for g we obtain g = km2/r

2.

(b) From (a), g = km2r−2 and

dg = −2km2r−3dr =

−2km2dr

r3=−2km2

r2

dr

r= −2g

dr

r.

Thus, dg/g = −2 dr/r.


(c) Letting g = 9.8 m/s2, r = 6400 km× 100 m/km, and dr = 16 km× 100 m/km we havefrom (b)

dg

9.8=−2(16× 1000)6400× 1000

= − 1200

,

or dg = −9.8/200 m/s2 = −0.049 m/s2. Thus the approximate value of g at r = 16 kmis g + dg = 9.8− 0.049 ≈ 9.75 m/s2.

57. (a) Setting g′ = 978.0318(53.024× 10−4 × 2 sin θ cos θ − 5.9× 10−6 × 4 sin 2θ cos 2θ) = 0 weobtain sin 2θ(53.024× 10−4 − 5.9× 10−6 × 4 cos 2θ = 0. From sin 2θ = 0 we find θ = 0◦

or 90◦, and since 4 cos 2θ = 53.024 × 10−4/5.9 × 10−6 × 4 > 1, we see that these arethe only critical numbers. By inspection of g we find that g is minimum on the equator(θ = 0◦) and maximum at the poles (θ = 90◦).

(b) g(60◦) ≈ 981.9169 cm/s2

(c) dg = 978.0318 sin 2θ(53.024× 10−4 − 5.9× 10−6 × 4 cos 2θ)dθ.

Using θ = π/3 and dθ = π/180 we find dg ≈ 0.07856 cm/s2.

58. T (4) = 2π√

4/9.8 ≈ 4.0142 s; T (5) = 2π√

5/9.8 ≈ 4.4880 s.

The change in period is T (5) − T (4) ≈ 0.4738 s. From T =2π√gL1/2 we obtain dT =

π√gL−1/2dL. For g = 9.8 ft/s2, L = 4 m, and dL = 1 m, we have dT =

π√9.8

4−1/2(1) ≈0.5018 s.

59. Writing T = 2π√Lg−1/2, we have dT = −π

√Lg−3/2dg. For L = 4 m, g = 9.8 m/s2, and

dg = −0.05 m/s2, we obtain the approximate change dT = −π√

4(9.8)−3/2(−0.05) ≈ 0.0102 s.

60. (a) From Figure 4.9.9 we see that tanθ

2=

1/2D

, so that D =12

cotθ

2.

(b) Using12

◦=

12

( π

180

)=

π

360radian, D

(12

◦)= D

( π

360

)=

12

cotπ

720≈ 1

2(229.18) =

114.59 ft.

(c) Let x and y represent the distances cars A and B, respectively, have traveled from some

initial point. Then y = x + D. Since 2 seconds =130

◦we are given

dx

dt= 30 mi/h =

30(

52803600

)ft/s = 44 ft/s and

dθ

dt= − 1

30

◦/s = − 1

30

( π

180

)= − π

5400radian/s. Then

car B is moving at a rate of

dy

dt=dx

dt+dD

dt= 44− 1

4

(csc2 θ

2

)dθ

dt= 44− 1

4

(csc2 π

720

)(− π

5400

)

≈ 51.64 ft/s ≈ 35.21 mi/h.


(d) From dD = −14

csc2 θ

2dθ we obtain

dD

d=− csc2 θ

2dθ/4

cotθ

2/2

= − dθ

2 sin2 θ

2cot

θ

2

= − dθ

2 sinθ

2cos

θ

2

= − dθ

sin θ.

Letting θ =12

◦and dθ = ± 1

60

( π

180

)we find

dD

D= ± π/10800

sin(π/360)≈ 0.033 = 3.3%.

61. By the definition in (2), L(x) = f(a) + f ′(a)(x− a). Since p(a) = f(a) and p′(a) = f ′(a), wecan rewrite L(x) = p(a) + p′(a)(x − a) = (c1a + c0) + (c1)(x − a). Simplifying, c1a cancelsout, and we get L(x) = c1x+ c0, which is p(x).

62. The linearization of cosx at a = 0 is L(x) = cos 0− (sin 0)(x−0) = 1. Thus, cosx ≈ L(x) = 1for small values of x.

63. If f ′′(x) > 0 for all x in some open interval containing a, then that interval is concave up.Thus, the tangent at a will lie below f(x) for all x within that interval, and so L(x) willunderestimate f(x) for x near a.

64. If (c, f(c)) is a point of inflection for the graph of y = f(x), then the graph of f(x) changesconcavity at c. If f(x) changes from concave up to concave down at x = c, then L(x) < f(x)or f(x)− L(x) > 0 for x < c and L(x) > f(x) or f(x)− L(x) < 0 for x > c. If f(x) changesfrom concave down to concave up at x = c, then L(x) > f(x) or f(x) − L(x) < 0 for x < cand L(x) < f(x) or f(x)−L(x) > 0 for x > c. In either case, the graph of f(x)−L(x) crossesthe x-axis (i.e., changes sign) at x = c.

65. ∆A = (x+ ∆x)2 − x2 = 2x∆x+ (∆x)2

dA = 2x dx = 2x∆x (Recall that dx = ∆x.)

∆A− dA = [2x∆x+ (∆x)2]− 2x∆x = (∆x)2

Thus, ∆A is the combined area of the beige and green regions, dA is the combined area ofthe beige regions only, and ∆A− dA is the area of the green region.

4.10. NEWTON’S METHOD 271

4.10 Newton’s Method

1.

-4 -2 2 4

-4

-2

2

4

The equation has onereal root.

2.

-4 -2 2 4-2

2

4

6

The equation has threereal roots.

3.

-4 -2 2 4

-4

-2

2

4

The equation has no realroots.

4.

-4 -2 2 4

-4

-2

2

4

The equation has in-finitely many real roots.

5.

-4 -2 2 4

-4

-2

2

4

The equation has onereal root.

6.

5 10

-5

5

The equation has in-finitely many real roots.

7. Let f(x) = x2 − 10. Then f ′(x) = 2x and

xn+1 = xn −x2n − 102xn

=x2n + 102xn

.

Choosing x0 = 3 we obtain x1 ≈ 3.1667, x2 ≈ 3.1623, x3 ≈ 3.1623. Thus,√

10 ≈ 3.1623.

8. Let x = 1+√

5. Then x−1 =√

5 and x2−2x+1 = 5. We use the function f(x) = x2−2x−4.Then f ′(x) = 2x− 2 and

xn+1 = xn −x2n − 2xn − 4

2xn − 2=

x2n + 4

2xn − 2.

Choosing x0 = 3 we obtain x1 ≈ 3.2500, x2 ≈ 3.2361, x3 ≈ 3.2361. Thus, 1 +√

5 ≈ 3.2361.

9. Let f(x) = x3 − 4. Then f ′(x) = 3x2 and

xn+1 = xn −x3n − 43x2

n

=2x3

n + 43x2

n

.

Choosing x0 = 1 we obtain x1 ≈ 2.0000, x2 ≈ 1.6667, x3 ≈ 1.5911, x4 ≈ 1.5874, x5 ≈ 1.5874.Thus, 3

√4 ≈ 1.5874.


10. Let f(x) = x5 − 2. Then f ′(x) = 5x4 and

xn+1 = xn −x5n − 25x4

n

=4x5

n + 25x4

n

.

Choosing x0 = 1 we obtain x1 ≈ 1.2000, x2 ≈ 1.1529, x3 ≈ 1.1487, x4 ≈ 1.1487. Thus,5√

2 ≈ 1.1487.

11. Let f(x) = x3 + x− 1. Then f ′(x) = 3x2 + 1 and

xn+1 = xn −x3n + xn − 13x2

n + 1=

2x3n + 1

3x2n + 1

.

From the graph we see that f(x) has a single root near x0 = 1. Thenx1 ≈ 0.7500, x2 ≈ 0.6860, x3 ≈ 0.6823, x4 ≈ 0.6823, and the only realroot is approximately 0.6823.

-3 3

-3

3

12. Let f(x) = x3 − x2 + 1. Then f ′(x) = 3x2 − 2x and

xn+1 = xn −x3n − x2

n + 13x2

n − 2xn=

2x3n − x2

n − 13x2

n − 2xn.

From the graph we see that f(x) has a single root near x0 = −1. Thenx1 ≈ −0.8000, x2 ≈ −0.7568, x3 ≈ −0.7549, x4 ≈ −0.7549, and theonly real root is approximately −0.7549.

-3 3

-3

3

13. From the quadratic formula, x2 =−1±

√1 + 12

2=−1±

√13

2. Since x2 must be positive for

x real, we have x = ±√−1±

√13

2≈ ±1.1414. Newton’s Method is not necessary.

14. Let f(x) = x4 − 2x− 1. Then f ′(x) = 4x3 − 2 and

xn+1 = xn −x4n − 2xn − 1

4x3n − 2

=3x4

n + 14x3

n − 2.

From the graph we see that f(x) has two real roots. Choosing x0 = −1and then x0 = 1, we obtain x1 ≈ −0.6667, x2 ≈ −0.5000, x3 ≈ −0.4750,x4 ≈ −0.4746, x5 ≈ −0.4746, and x1 ≈ 2.0000, x2 ≈ 1.6333, x3 ≈1.4486, x4 ≈ 1.3988, x5 ≈ 1.3954, x6 ≈ 1.3953, x7 ≈ 1.3953. Thus, thetwo real roots are approximately −0.4746 and 1.3953.

-3 3

3

6

15. Let f(x) = x2 − sinx. Then f ′(x) = 2x− cosx and

xn+1 = xn −x2n + sinxn

2xn − cosxn=x2n − xn cosxn + sinxn

2xn − cosxn.

From the graph we see that f(x) has one root at x = 0 and another onenear x0 = 1. Then x1 ≈ 0.8914, x2 ≈ 0.8770, x3 ≈ 0.8767, x4 ≈ 0.8767.Thus, the two real roots are 0 and approximately 0.8767.

-2 2

-2

2


16. Let f(x) = x+ cosx. Then f ′(x) = 1− sinx and

xn+1 = xn −xn + cosxn1− sinxn

=−xn sinxn − cosxn

1− sinxn.

From the graph we see that f(x) has a single root near x0 = −1. Thenx1 ≈ −0.7504, x2 ≈ −0.7391, x3 ≈ −0.7391, and the only real root isapproximately −0.7391.

-2 2

-2

2

17. f ′(x) = −3 sinx+ 4 cosx and

xn+1 = xn −3 cosxn + 4 sinxn−3 sinxn + 4 cosxn

=(4xn − 3) cosxn − (3xn + 4) sinxn

4 cosxn − 3 sinxn.

From the graphs of 3 cosx and −4 sinx we see that the first positive rootof f(x) is near x0 = 2. Then x1 ≈ 2.5438, x2 ≈ 2.4981, x3 ≈ 2.4981.Thus, the smallest positive x-intercept is approximately 2.4981.

-4 4

-4

4

18. Let F (x) = x5 + x2 − 4. Then F ′(x) = 5x4 + 2x and

xn+1 = xn −x5n + x2

n − 45x4

n + 2xn=

4x5n + x2

n + 45x4

n + 2xn.

Since F (x) < 0 for 0 ≤ x ≤ 1, we try x0 = 1. Then x1 ≈ 1.2857, x2 ≈ 1.2139, x3 ≈1.2057, x4 ≈ 1.2056, x5 ≈ 1.2056, and the smallest positive number for which x5 + x2 = 4 isapproximately 1.2056.

19. We want to solve60x2 − x3

16000= 0.01 or x3 − 60x2 + 160 = 0. Let f(x) = x3 − 60x2 + 160.

Then f ′(x) = 3x2 − 120x and

xn+1 = xn −x3n − 60x2

n + 1603x2

n − 120xn=

2x3n − 60x2

n − 1603x2

n − 120xn.

Choosing x0 = 1 we obtain x1 ≈ 1.8632, x2 ≈ 1.6670, x3 ≈ 1.6560, x4 ≈ 1.6560. Thus,x ≈ 1.6560 ft.

20. We want to solve 10 = 35r2/3 or 103 = (35)3r2. Let f(r) = 42.875r2−1. Then f ′(r) = 85.75rand

rn+1 = rn −42.875r2

n − 185.75rn

=42.875r2

n + 185.75rn

.

Choosing r0 = 1 we obtain r1 ≈ 0.5117, r2 ≈ 0.2786, r3 ≈ 0.1812, r4 ≈ 0.1550, r5 ≈ 0.1527,r6 ≈ 0.1527. Thus, r ≈ 0.1527 m and d ≈ 0.3054 m.

21. We want to solve sin θ =1

1.5=

23

or 3 sin θ = 2. Let f(θ) = 3 sin θ − 2. Then f ′(θ) = 3 cos θand

θn+1 = θn −3 sin θn − 2

3 cos θn=

3θn cos θn − 3 sin θn + 23 cos θn

.

Choosing θ0 = 0 we obtain θ1 ≈ 0.6667, θ2 ≈ 0.7281, θ3 ≈ 0.7297, θ4 ≈ 0.7297. Thus,θ ≈ 0.7297 radian.


22. We want to solve 404 = 400 +8d2

1200− 32d4

320, 000, 000or 1 =

d2

600− d4

40, 000, 000. Let f(d) =

2.5× 10−8d4 − 1.667× 10−3d2 + 1. Then f ′(d) = 10−7d3 − 3.333× 10−3d and

dn+1 = dn −2.5× 10−8d4

n − 1.667× 10−3d2n + 1

10−7d3n − 3.333× 10−3dn

=7.5× 10−8d4

n − 1.667× 10−3d2n − 1

10−7d3n − 3.333× 10−3dn

.

Choosing d0 = 20 we obtain d1 ≈ 25.1219, d2 ≈ 24.6125, d3 ≈ 24.6074, d4 ≈ 24.6074. Thusd ≈ 24.6 ft.

23. (a) The volume of water displaced is Vw = 3(7)(2) = 42 ft3. The volume of steel in the tubis Vs = 42− (3−2t)(7−2t)(2− t) = 4t3−28t2 + 61t. Since the weight of water displacedis equal to the weight of the tub,

62.4(42) = 490(4t3 − 28t2 + 61t) or f(t) = t3 − 7t2 +614t− 1638

1225= 0.

(b) Then f ′(t) = 3t2 − 14t+ 61/4 and

tn+1 = tn −f(t)f ′(t)

.

From the graph we see that f(t) has its only root near t0 = 0. Thent1 ≈ 0.0877, t2 ≈ 0.0915, t3 ≈ 0.0915, and t ≈ 0.915 ft.

2 4

-10

10

20

24. From the figure we see that sin θ =4r

and10

2πr=

2θ2π

or sin θ = 4 and

θ =5r

. Then r sin5r

= 4 and we identify f(r) = r sin5r− 4. Using f ′(θ) =

sin5r− 5r

cos5r

we have

rn+1 = rn −rn sin

5rn− 4

sin5rn− 5rn

cos5rn

.

r

10

4θθ

Clearly r > 4 so we choose r0 = 4. Then r1 ≈ 4.3678, r2 ≈ 4.4196, r3 ≈ 4.4205, r4 ≈ 4.4205,and r ≈ 4.4205 ft.

25. (a) Since θ subtends an arc of length L/2 on a circle of radius R, we have L/2 = Rθ andR = L/2θ. From Figure 4.10.11 we see that

sin θ =(L− l)/2

R=

(L− l)/2L/2θ

=(

1− l

L

)θ and cos θ =

R− hR

.

Then

h = R(1− cos θ) =L

2θ

(1−

√1− sin2 θ

)=

L

2θ

[1−

√1− (1− l/L)2θ2

]

=L

2θ1−

[1− (1− l/L)2θ2

]

1 +√

1− (1− l/L)2θ2=

L(1− l/L)2θ

2[1 +

√1− (1− l/L)2θ2

] .


(b) Setting L = 5280 and l = 1 we have sin θ =(

1− 15280

)θ or f(θ) = sin θ − 5279

5280θ = 0.

The formula for Newton’s Method is

θn+1 = θn −f(θn)f ′(θn)

= θn −sin θn −

52795280

θn

cos θn −52795280

=θn cos θn − sin θn

cos θn − 5279/5280.

Taking θ0 = 0.1 we obtain θ1 ≈ 0.069282, θ2 ≈ 0.050143, θ3 ≈ 0.039358, θ4 ≈ 0.034732,θ5 ≈ 0.033754, θ6 ≈ 0.033711, θ7 ≈ 0.033711, so θ ≈ 0.033711 and h ≈ 44.494 ft.

(c) From sin θ = (1− l/L)θ ≈ θ − θ3/6 we obtain l/L ≈ θ2/6 and θ ≈√

6l/L. Then

h ≈ Lθ

4≈ L

√6l/L4

≈√

3lL8.

Setting l = 1 and L = 5280 we find h ≈ 44.4972, which is very close to the result obtainedin (b).

26. The volume of the sphere is43π(2)3 =

323π and the volume of the rod is 15πr2 +

23πr3. Thus,

15πr2 +23πr3 =

323π or 2r3 +45r2−32 = 0. From f(r) = 2r3 +45r2−32 and f ′(r) = 6r2 +90r

we have

rn+1 = rn −2r3n + 45r2

n − 326r2n + 90rn

.

Taking r0 = 1 we obtain r1 ≈ 0.8437, r2 ≈ 0.8283, r3 ≈ 0.8282, r4 ≈ 0.8282 and r ≈ 0.8282 ft.

27. Setting M = 4m we have 4mgr

2sin θ−mgrθ = 0 or 2 sin θ− θ = 0. Setting f(θ) = 2 sin θ− θ

we obtain f ′(θ) = 2 cos θ − 1 and

θn+1 = θn −2 sin θn − θn2 cos θn − 1

=2θn cos θn − 2 sin θn

2 cos θn − 1.

Taking θ0 = 2 we find θ1 ≈ 1.9010, θ2 ≈ 1.8955, θ3 ≈ 1.8955, and θ ≈ 1.8955 radians.

28. (a) Using similar triangles we obtaina

x=h

xand

b

z=h

y. Adding, we find

a+ b

z=h

x+h

y,

z

z=hy + hx

xy, xy = hy + hx, (x− h)y = hx,

so y =hx

x− h . Using the Pythagorean Theorem we obtain z2 +x2 = L22

and z2 + y2 = L21. Subtracting, we have

x2 − y2 = L22 − L2

1

x2 −(

hx

x− h

)2

+ L21 − L2

2 = 0

x4 − 2hx3 + (L21 − L2

2)x2 − 2h(L21 − L2

2)x+ h2(L21 − L2

2) = 0.

z

yx

ha b

L1L2


(b) Letting h = 10, L1 = 40, and L2 = 30 we have f(x) = x4−20x3+700x2−1, 400x+70, 000and f ′(x) = 4x3 − 60x2 + 1, 400x− 14, 000. Then

xn+1 = xn −x4n − 20x3

n + 700x2n − 1, 400xn + 70, 000

4x3n − 60x2

n + 1, 400xn − 14, 000.

Since x > h = 10, we choose x0 = 11. Then x1 = −10.043, x2 ≈ −1.8771, x3 ≈ 3.9863,x4 ≈ 6.6511, x5 ≈ 7.2874, x6 ≈ 7.3299, x7 ≈ 7.3301, x8 ≈ 7.330, and x = 7.3301. Whilethis is a root of f(x), it is too small to be a solution to the problem. Trying x0 = 12we obtain x1 ≈ 22.2836, x2 ≈ 18.1499, x3 ≈ 15.8728, x4 ≈ 15.0336, x5 ≈ 14.9119,x6 ≈ 14.9094, x7 ≈ 14.9094, and x ≈ 14.9094 ft.

(c) From z2 + x2 = L22 we find z =

√L2

2 − x2 ≈ 26.0329 ft.

29. f(x) = 2x5+3x4−7x3+2x2+8x−8; f ′(x) = 10x4+12x3−21x2+4x+8

xn+1 = xn −f(xn)f ′(xn)

x0 = −1, x1 ≈ −1.3158, x2 ≈ −1.2517, x3 ≈ −1.2494, x4 ≈ −1.2494x0 = −3, x1 ≈ −2.7679, x2 ≈ −2.6776, x3 ≈ −2.6641, x4 ≈ −2.6638,x5 ≈ −2.6638x0 = 1, x1 = 1

-4 -2 2 4

-50

50

30. f(x) = 4x12 + x11 − 4x8 + 3x3 − 2x2 + x− 10f ′(x) = 48x11 + 11x10 − 32x7 + 9x2 − 4x+ 1

xn+1 = xn −f(xn)f ′(xn)

x0 = −1.2, x1 ≈ −1.2531, x2 ≈ −1.2416, x3 ≈ −1.2408, x4 ≈ −1.2408x0 = 1, x1 ≈ 1.2121, x2 ≈ 1.1468, x3 ≈ 1.1128, x4 ≈ 1.1047, x5 ≈1.1044, x6 ≈ 1.1044

-2 2

-20

20

31. (a)

-2 2

-2

2

(b)

-2 2

-0.4

-0.2

0.2

0.4

(c) The number of roots appears to be twoin part (a), but the “zoomed in” graphin part (b) shows there is only one root.

(d) y = 0.5x3 − x− cosxy′ = 1.5x2 − 1 + sinx

xn+1 = xn −0.5x3

n − xn − cosxn1.5x2

n − 1 + sinxnx0 = 1.5, x1 ≈ 1.4654, x2 ≈ 1.4645,x3 ≈ 1.4645

32. Since f(x0) = −f(x1) and f ′(x0) = f ′(x1), x1 = x0 −f(x0)f ′(x0)

= x0 +f(x1)f ′(x1)

andf(x1)f ′(x1)

=

x1 − x0. Then x2 = x1 −f(x1)f ′(x1)

= x1 − (x1 − x0) = x0.

CHAPTER 4 IN REVIEW 277

33. f ′(x) =

12√

4− x, x < 4

12√x− 4

, x > 4

Since f ′(4) does not exist, Newton’s Method will fail for x0 = 4. For any choice of x0 < 4,

x1 = x0 −f(x0)f ′(x0)

= x0 − (2x0 − 8) = 8− x0.

Since x1 = 8 − x0 > 4, f(x1) =√x1 − 4 =

√8− x0 − 4 =

√4− x0 = −f(x0), and f ′(x1) =

12√x1 − 4

=1

2√

8− x0 − 4=

12√

4− x0= f ′(x0). By Problem 32, then, x2 = x0. The same

result is obtained for any choice of x0 > 4. Newton’s Method will therefore yield the sequenceof iterates x0, x1, x0, x1, . . ., and thus fail to converge.

Chapter 4 in Review

A. True/False

1. False; the function may not be differentiable, or f ′(x) may be 0 for some x on the interval.

2. False; for f(x) = x3, f ′(0) = 0, but f(x) has no extremum at 0.

3. False; this is only true when the velocity is positive.

4. True

5. True

6. False; the concavity must change around c. Consider f(x) = x4 at x = 0.

7. False; f(x) need not be differentiable at c.

8. False; f ′′(x) need not exist at c.

9. True

10. False; if the extremum occurs at an endpoint of the interval, it cannot be a relative extremum.

11. True

12. False; x = 1 is not in the domain of√x2 − 2x.

13. True

14. False; consider f(x) = −x2, f ′(x) = −2x, f ′′(x) = −2 on (−∞, 0).

15. False; this is an indeterminate form.

16. False; see Problem 55, Section 4.5.

17. True


18. False; an expression with this form will have limit 0.

19. False; let a be ∞, f(x) = x2 and g(x) = ex.

20. False; L’Hopital’s Rule uses the quotient of the derivatives, rather than the derivative of thequotient.

B. Fill in the Blanks

1. Velocity

2. One, since the second derivative is linear and can have at most one root.

3. f(x) = x1/3

4. Let x and 8 − x be the two numbers. We want to maximize S(x) = x2 + (8 − x)2 on [0, 8).Solving S′(x) = 2x − 2(8 − x) = 4x − 16 = 0 we obtain the critical number 4. ComparingS(0) = S(8) = 64 and S(4) = 32 we see that the sum is maximized when the two numbersare 0 and 8.

5. 0

6. 0

7. 2

8. y − 13 = 5(x− 1) or L(x) = 5x+ 8; f(1.1) ≈ 13.5

9. ∆y = (x+ ∆x)2 − (x+ ∆x)− (x2 − x) = (2x− 1)∆x+ (∆x)2

10. dy = (−x3e−x + 3x2e−x)dx = (3x2 − x3)e−xdx

C. Exercises

1. Solving f ′(x) = 3x2 − 75 = 0, we obtain the critical numbers −5 and 5, neither of which isin the interval [−3, 4]. Comparing f(−3) = 348 and f(4) = −86, we see that the absolutemaximum is 348 and the absolute minimum is −86.

2. Solving f ′(x) = 8x+1/x2 = 0, we obtain the critical number −1/2 which is not in the interval[1/4, 1]. Comparing f(1/4) = −15/4 and f(1) = 3, we see that the absolute maximum is 3and the absolute minimum is −15/4.

3. Solving f ′(x) = x(x+ 8)/(x+ 4)2 = 0, we obtain the critical numbers −8 and 0. Comparingf(−1) = 1/3, f(0) = 0, and f(3) = 9/7, we see that the absolute maximum is 9/7 and theabsolute minimum is 0.

4. Solving f ′(x) =12

(2x− 3)/√x2 − 3x+ 5 = 0, we obtain the critical number 3/2. Comparing

f(1) =√

3, f(3/2) =√

11/2, and f(3) =√

5, we see that the absolute maximum is√

5 andthe absolute minimum is

√11/2.


5.

-4 -2 2 4

-3

3

6. f(x) = x+ sinx; f ′(x) = 1 + cosx; f ′′(x) = − sinx. Solving f ′(x) = 0, we obtain the criticalnumbers (2k + 1)π, for k an integer. Solving f ′′(x) = 0 we obtain the values kπ, for k aninteger.

x −π 0 π 2πf −π ↗ 0 ↗ π ↗ 2πf ′ 0 + + + 0 + +f ′′ 0 + 0 − 0 + 0 -2π -π π 2π

-2π

-π

π

2π

g(x) = x + sin 2x; g′(x) = 1 + 2 cos 2x; g′′(x) = −4 sin 2x. Solving

g′(x) = 0 we obtain the critical numbersπ

3+ πk and

2π3

+ πk, for k an

integer. Solving g′′(x) = 0 we obtain the values kπ, for k an integer.-2π -π π 2π

-2π

-π

π

2π

x − 2π3 −π2 −π3 0 π

3π2

2π3 π 4π

3

g −1.2 −π2 0.2 0 1.9 π2 1.2 π 5.1

g′ 0 − − − 0 + + + 0 − − − 0 + + + 0g′′ − − 0 + + + 0 − − − 0 + + + 0 − −

7. v(t) = −3t2 + 12t. Solving v(t) = 0 we obtain t = 0, 4. To find the maximum velocity,we solve v′(t) = −6t + 12 = 0 and obtain t = 2. Comparing v(−1) = −15, v(2) = 12, andv(5) = −15, we see that the maximum velocity is 12. Since speed is the absolute value ofvelocity, the maximum speed on the interval is 15 when t = −1 and t = 5.

t −1 0 4 5s 7 ← 0 → 32 ← 25v − 0 + 0 −

s0 32

t = –1

t = 4

t = 5

25

t = 0

7


8. Solving s′(t) = −9.8t + 14.7 = 0 we obtain t = 3/2. Since s′′(t) = −9.8 < 0, the maximumheight is s(3/2) = 60.025. Solving s(t) = −4.9(t2 − 3t − 10) = −4.9(t − 5)(t + 2) = 0, weobtain t = −2 and 5. When the projectile strikes the ground at 5 seconds, the speed is|s′(5)| = 34.3 m/s.

9. (a) f ′(x) = (x− a)2[(x− b)2g′(x) + 2(x− b)g(x)] + 2(x− a)(x− b)2g(x)= (x− a)(x− b)[(x− a)(x− b)g′(x) + 2(x− a)g(x) + 2(x− b)g(x)]

We see immediately that f(a) = f(b) = 0, so by Rolle’s Theorem there exists c in (a, b)such that f ′(c) = 0.

(b) If g(x) = C then g′(x) = 0 and f ′(x) = (x−a)(x− b)[2C(x−a+x− b)] = 2C(x−a)(x−b)(2x− a− b). Solving f ′(x) = 0 we obtain the critical numbers a, b, and

a+ b

2.

10. f ′(x) = 1/3x2/3 is not defined at x = 0, so f(x) is not differentiable on [−1, 8]. Solving

1/3x2/3 = [81/3 − (−1)1/3]/[8− (−1)] =13

, we obtain x = ±1. There is no problem with theconclusion of a theorem applying, even though the hypotheses are not satisfied, unless thetheorem is of the “if and only if” type.

11. Solving f ′(x) = 6x2 + 6x − 36 =6(x + 3)(x − 2) = 0 we obtain thecritical numbers −3 and 2. The rel-ative maximum is (−3, 81) and therelative minimum is (2,−44).

x −3 2f ↗ 81 ↘ −44 ↗f ′ + 0 − 0 +

-5 5

-50

50

12. Solving f ′(x) = 5x4−5x2 =5x2(x − 1)(x + 1) = 0 weobtain the critical numbers−1, 0, and 1. The rela-tive maximum is (−1, 8/3)and the relative minimum is(1, 4/3).

x −1 0 1f ↗ 8/3 ↘ 2 ↘ 4/3 ↗f ′ + 0 − 0 − 0 +

-2 -1 1 2

3

13. Solving f ′(x) = 4 − 4x−1/3 =4(x1/3 − 1)/x1/3 = 0 we obtainthe critical number 1. Since f ′(0)does not exist, 0 is also a criticalnumber. The relative maximum is(0, 2) and the relative minimum is(1, 0).

x 0 1f ↗ 2 ↘ 0 ↗f ′ + undefined − 0 +

-2 -1 1 2 3 4

-2

2

14. Solving f ′(x) = x(x − 2)/(x − 1)2 = 0 we obtain the critical numbers 0 and 2. The relativemaximum is (0,−2) and the relative minimum is (2, 2).


x 0 1 2f ↗ −2 ↘ undefined ↘ 2 ↗f ′ + 0 − undefined − 0 +

-5 5

-4

4

15. Solving f ′(x) = 4x3 + 24x2 + 36x = 4x(x +3)2 = 0 we obtain the critical numbers −3 and0. Solving f ′′(x) = 12x2 + 48x + 36 = 12(x +1)(x + 3) = 0 we obtain x = −1 and −3. Therelative minimum is (0, 0). The inflection pointsare (−3, 27) and (−1, 11).

x −3 −1 0f ↘ 27 ↘ 11 ↘ 0 ↗f ′ − 0 − − − 0 +f ′′ + 0 − 0 + + +

16. Solving f ′(x) = 6x5 − 12x3 = 6x3(x2 − 2) = 0 we obtain the critical numbers −√

2, 0, and√2. Solving f ′′(x) = 30x4 − 36x2 = 6x2(5x2 − 6) = 0 we obtain the values −

√6/5, 0, and√

6/5. The relative maximum is (0, 5). The relative minima are (−√

2, 1) and (√

2, 1). Theinflection points are (−

√6/5, 301/125) and (

√6/5, 301/125).

x −√

2 −√

6/5 0√

6/5√

2f ↘ 1 ↗ 301/125 ↗ 5 ↘ 301/125 ↘ 1 ↗f ′ − 0 + + + 0 − − − 0 +f ′′ + + + 0 − 0 − 0 + + +

17. Since f ′(x) = −13

(x − 3)−2/3, we see that the only critical

number is 3. f ′′(x) =29

(x − 3)−5/3 is undefined at x = 3.

There are no relative extrema. The inflection point is (3, 10).

x 3f 10f ′ − undefined −f ′′ − undefined +

18. The domain of f(x) is [1,∞). Solving f ′(x) = (x− 1)3/2

(72x− 1

)= 0 we obtain the critical

number 1. Solving f ′′(x) = (x − 1)1/2

(354x− 5

)= 0 we obtain the value 1. Since neither

f ′(x) nor f ′′(x) is 0 in (1,∞), the function has no relative extrema or inflection points.

19. (c), (d)

20. (d), (e), (f)

21. (c), (d), (e)

22. (d), (f)

23. (a), (c)

24. (a), (b), (d)


25. f ′(x) = (x−a)[(x−b)+(x−c)]+(x−b)(x−c) = (x−a)(x−b)+(x−a)(x−c)+(x−b)(x−c)f ′′(x) = x− a+ x− b+ x− a+ x− c+ x− b+ x− c = 6x− 2(a+ b+ c)

Solving f ′(x) = 0 we obtain x =13

(a + b + c). Since f ′′(16

(a + b + c)) = −(a + b + c) and

f ′′(a+ b+ c) = 4(a+ b+ c), we see that f ′′(x) has opposite signs on either side of13

(a+ b+ c).

Thus, the graph of f(x) has a point of inflection at13

(a+ b+ c).

26. Let b be the base of the triangle, h the altitude, and A the area. Then A =12bh and

dA/dt =12b dh/dt +

12h db/dt. Given dA/dt = 15, db/dt = −1

2, h = 8, and b = 6, we have

15 =12

(6)dh/dt+12

(8)(−1

2

)= 3 dh/dt− 2, so dh/dt =

173

in//min.

27. Assume that the center of the circle is at the origin so that x2 +y2 = r2. In the first quadranta vertex of the square is (x, y) = (x, x). Therefore 2x2 = r2, A = (2x)(2x), A(r) = 2r2, anddA

dt= 4r

dr

dt. At the instant when r = 2,

dA

dt= 4(2)(4) = 32 in2/min.

28. The rate of change of the volume of water isdV

dt=

110− 1

5= − 1

10m3/min. Since this is

negative, the volume of the water, and hence the height, is decreasing. We want to find dh/dt

when h = 5 m. Differentiating the volume with respect to time givesdV

dt= (20πh− πh2)

dh

dt,

and so

dh

dt=

120πh− πh2

dV

dtand

dh

dt

∣∣∣∣h=5

=1

75π

(− 1

10

)= − 1

750πm/min.

29. B′(x) =12µ0r

20I

{−3

2

[r20 +

(x+

r0

2

)2]−5/2 [

2(x+

r0

2

)]

−32

[r20 +

(x− r0

2

)2]−5/2 [

2(x− r0

2

)]}

= −32µ0r

20I

x+ r0/2[r20 +

(x+

r0

2

)2]5/2

+x− r0/2[

r20 +

(x− r0

2

)2]5/2

We see that B′(0) = 0. To apply the first derivative test, we compute

B′(−r0

4

)= −384µ0I

r20

(1

175/2− 3

255/2

)and B′

(r0

4

)= −384µ0I

r20

(3

255/2− 1

175/2

).

Since B′(−r0

4

)> 0 and B′

(r0

4

)< 0, B has a maximum at x = 0.


30. We want to maximize P (R) = RE2/(r +R)2. Solving

P ′(R) =(r +R)2E2 −RE2[2(r +R)]

(r +R)4=

(r −R)E2

(r +R)3= 0,

we obtain the critical number R = r. Since P ′(R) > 0 for R < r and P ′(R) < 0 for R > r,we see from the first derivative test that the power dissipated is maximum when R = r.

31.dx

dy=

h− 2y√y(h− y)

. Solvingdx

dy= 0 we obtain the critical number h/2. Since 0 ≤ y ≤ h, we

compare z(0) = 0, x(h/2) = h, x(h) = 0, and observe that the maximum distance of h ft isobtained for y = h/2.

32. We are given 2r + s = 60 and we want to maximize A(r) =12rs =

12r(60 − 2r) = 30r − r2.

Solving A′(r) = 30 − 2r = 0 we obtain the critical number 15. Since A′′(15) = −2 < 0, themaximum area is A(15) = 225 cm2.

x

xy – x

x45°

33. Since 585 feet of fence is available, y = 585− x. We want to maximize A(x) =

x(y − x) +12x2 = x(585− 2x) +

12x2 = 585x− 3

2x2. Solving A′(x) = 585− 3x

we obtain the critical number 195. Since A′′(x) = −3 < 0, the maximumarea is obtained when x = 195 ft and y = 390 ft. The maximum area isA(195) = 57037.5 ft2.

34. Let x and√

1002 − x2 be the lengths along the walls. We want to maximize

A(x) =12x√

1002 − x2 on [0, 100]. Solving

A′(x) = − x2

2√

1002 − x2+√

1002 − x2

2=

1002 − 2x2

2√

1002 − x2= 0

x100

1002 – x2

we obtain the critical number 50√

2. Comparing A(0) = 0, A(50√

2) = 2500, and A(100) = 0,we see that the area is maximized when the length along each wall is 50

√2 m.

35. We want to minimize T (x) =

√x2 + h2

1

c+

√(d− x)2 + h2

2

con [0, d].

Setting

T ′(x) =1c

[x√

x2 + h21

− d− x√(d− x)2 + h2

2

]= 0, x d – x

h1 h2θ1 θ1

θ2θ2

we obtain x√

(d− x)2 + h22 = (d − x)

√x2 + h2

1 or x2[(d − x)2 + h22] = (d − x)2(x2 + h2

1).

Simplifying, we haved− xh2

=x

h1or tan θ2 = tan θ1. Solving for x, we find x =

h1d

h1 + h2.

Since

T ′′(x) =1c

{h2

1

(x2 + h21)3/2

+h2

2

[(d− x)2 + h22]3/2

}> 0,

the time is minimized when tan θ2 = tan θ1.


A

C D

BE

r

36. Note that 4AEB is similar to 4ACD. Let h = AC and R = CD. Then

R

r=

h

AE=

h√(h− r)2 − r2

=h√

r2 − 2rhand R =

rh√h2 − 2rh

.

We want to minimize V (h) =13πR2h =

13π

r2h3

h2 − 2rhon (2r,∞). Solving

V ′(h) =13πr2h

h− 4r(h− 2r)2

= 0,

we obtain h = 4r. Since V ′(3r) < 0 and V ′(5r) > 0, we have by the first derivative test thatthe volume is minimized when h = 4r and R =

√2r.

hr

37. Let r be the radius and h the height. Then πr2h = 100 and h = 100/πr2. Wewant to minimize C(r) = 3πr2 + πr2 + 2πrh = 4πr2 + 200/r. Solving C ′(r) =8πr−200/r2 = 0 we obtain the critical number 3

√25/π. Since C ′′( 3

√25/π) > 0,

the cost is minimized when r = 3√

25/π and h = 4 3√

25/π. (In this case h = 4r.)

38. Let x be the side of the square cut-out. We want to maximize V (x) = (15−2x)(

30− 2x2

)x =

225x − 45x2 + 2x3 on [0, 7.5]. Solving V ′(x) = 225 − 90x + 6x2 = 0 we obtain the critical

numbers15± 5

√3

2. Only

15− 5√

32

≈ 3.17 is in [0, 7.5], so comparing V (0) = V (7.5) = 0

and V (3.17) ≈ 324.76 we see that the maximum volume of 324.76 in3 is obtained when thesquare cut-out has sides approximately 3.17 in. The dimensions of the box are approximately8.66 in by 11.83 in by 3.17 in.

39. limx→√

3

√3− tan(π/x2)x−√

3h= limx→√

3

[sec2(π/x2)](2π/x3)1

=8π

3√

3

40. limθ→0

10θ − 5 sin 2θ10θ − 2 sin 5θ

h= limθ→0

10− 10 cos 2θ10− 10 cos 5θ

h= limθ→0

20 sin 2θ50 sin 5θ

h= limθ→0

40 cos 2θ250 cos 5θ

=425

41. limx→∞

x

(cos

1x− e2/x

)= limx→∞

cos(1/x)− e2/x

1/xh= limx→∞

[sin(1/x)](1/x2)− e2/x(−2/x2)−1/x2

= limx→∞

sin(1/x) + 2e2/x

−1= −2

42. limy→0

[1y− 1

ln(y + 1)

]= limy→0

ln(y + 1)− yy ln(y + 1)

h= limy→0

1/(y + 1)− 1y/(y + 1) + ln(y + 1)

= limy→0

1− (y + 1)y + (y + 1) ln(y + 1)

h= limy→0

−11 + 1 + ln(y + 1)

= −12

43. limt→0

(sin t)2

sin t2h= limt→0

2 sin t cos t2t cos t2

= limt→0

sin 2t2t cos t2

h= limt→0

2 cos 2t−4t2 sin t2 + 2 cos t2

= 1

44. limx→0

tan 5xe3x/2 − e−x/2

h= limx→0

5 sec2 5x(3/2)e3x/2 + (1/2)e−x/2

=5

3/2 + 1/2=

52


45. Set y = (3x)−1/ ln x. Then ln y = − ln 3xlnx

and limx→0+

(− ln 3x

lnx

)h= limx→0+

−1/x1/x

= −1. Thus,

limx→0+

(3x)−1/ ln x = e−1.

46. Set y = (2x+ e3x)4/x. Then ln y =4 ln(2x+ e3x)

xand

limx→0

4 ln(2x+ e3x)x

h= limx→0

4(2 + 3e3x)/(2x+ e3x)1

= limx→0

4(2 + 3e3x)2x+ e3x

= 20.

Thus, limx→0

(2x+ e3x)4/x = e20.

47. limx→∞

ln(x+ e2x

1 + e4x

)= ln

(limx→∞

x+ e2x

1 + e4x

)h= ln

(limx→∞

1 + 2e2x

4e4x

)h= ln

(limx→∞

4e2x

16e4x

)

= ln(

limx→∞

14e2x

)= ln 0 = −∞

The limit does not exist.

48. limx→0+

x(lnx)2 = limx→0+

(lnx)2

1/xh= limx→0+

(2 lnx)/x−1/x2

= limx→0+

2 lnx−1/x

h= limx→0+

2/x1/x2

= limx→0+

2x = 0

49. Let f(x) = x3 − 4x+ 2. Then f ′(x) = 3x2 − 4 and

xn+1 = xn −x3n − 4xn + 2

3x2n − 4

=2x3

n − 23x2

n − 4.

From the graph we see that f(x) has its largest positive root near x0 = 2.Then x1 ≈ 1.7500, x2 ≈ 1.6807, x3 ≈ 1.6752, x4 ≈ 1.6751, x5 ≈ 1.6751,and the largest positive root is approximately 1.6751.

5

5

50. Write the equation as 2 sin2 x = x2 and let f(x) = 2 sin2 x − x2. Thenf ′(x) = 4 sinx cosx− 2x and

xn+1 = xn −2 sin2 xn − x2

n

4 sinxn cosxn − 2xn.

From the graph we see that f(x) has its smallest positive root near x0 = 1.

-2 2

2

Then x1 ≈ 3.2940, x2 ≈ 1.4896, x3 ≈ 1.4022, x4 ≈ 1.3917, x5 ≈ 1.3916, x6 ≈ 1.3916, and thesmallest positive root is approximately 1.3916.

Solucionario Zill Cap 4

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