3 CHAPTER Applications of Differentiation
Section 3.1 Section 3.2 Section 3.3 Section 3.4 Section 3.5 Section 3.6 Section 3.7 Section 3.8 Section 3.9 Extrema on an Interval . . . . . . . . . . . . . . 378 . 381
Rolles Theorem and the Mean Value Theorem
Increasing and Decreasing Functions and the First Derivative Test . . . . . . . . . . . . . . 387 Concavity and the Second Derivative Test . . . . 394 Limits at Infinity . . . . . . . . . . . . . . . . . 402 . . . . . . . . . 410
A Summary of Curve Sketching
Optimization Problems . . . . . . . . . . . . . . 419 Newtons Method . . . . . . . . . . . . . . . . . 429 Differentials . . . . . . . . . . . . . . . . . . . . 434
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . 437 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . 445
CHAPTER 3 Applications of Differentiation
Section 3.1 Extrema on an Interval
Solutions to Even-Numbered Exercises
x 2 sin x 2
2. f x fx f0 f2
cos
4.
fx fx
3x x 3x 1 x 2 1 1
1 1
12 12
x 2x 2 1
1
3
2 0 0
3 x 2 3 x 2 f 2 3 0
x 3x
12
6. Using the limit definition of the derivative,
x0
8. Critical number: x 4 1 1 x 0: neither
0.
lim
fx x fx x
f0 0 f0 0
x0
lim
4 4 x
x x x 0
x0
lim
x0
lim
4
f 0 does not exist, since the one-sided derivatives are not equal. 10. Critical numbers: x x x 2: neither 5: absolute maximum 4x x2 x2 1 14 x2 4x 2x 1
2
2, 5
12. g x gx
x2 x2 4x3
4 8x
x4 4x x2 0, x
4x2 2
Critical numbers: x
2
14. f x fx
16. f 41 x2 x2 12 f
2 sec
tan , 0 < sec2 sec 1 cos 1
<2
2 sec tan sec sec sec2 2 tan 2 sin cos
Critical numbers: x
1
2 sin
On 0, 2 , critical numbers:
7 , 6
11 6
378
S ection 3.1 2x 3 5
Extrema on an Interval
379
18. f x fx
, 0, 5
20. f x fx
x2 2x
2x 2
4, 2x
1, 1 1 5 Minimum 1 Maximum
2 No critical numbers 3 0, 5 Minimum 3
Left endpoint: ( 1, Right endpoint: 1,
Left endpoint:
Right endpoint: 5, 5 Maximum 22. f x fx x3 3x2 12x, 0, 4 12 3 x2 4 24. g x gx
3
x,
3
1, 1
Left endpoint: 0, 0 Critical number: 2, 16 Minimum
1 3x2
Left endpoint:
1,
1 Minimum
Right endpoint: 4, 16 Maximum Note: x 2 is not in the interval.
Critical number: 0, 0 Right endpoint: 1, 1 Maximum
26. y
3
t
3,
1, 5 3 is a critical number.
28. h t ht
t t t 2
, 3, 5
From the graph, you see that t
4
2 2
2
1
Left endpoint: 3, 3 Maximum
5
Right endpoint:
4
5,
5 Minimum 3
Left endpoint:
1,
1 Minimum
Right endpoint: 5, 1 Critical number: 3, 3 Maximum
30. g x gx
sec x, sec x tan x
, 63
32. y y
x2 2x
2
cos x,
1, 3
sin x 1, 1.5403
Left endpoint: Right endpoint:
6 3
,
2 3
Left endpoint: 6 , 1.1547
Right endpoint: 3, 7.99 Maximum Critical number: 0, 3 Minimum
,2
Maximum
Critical number: 0, 1 Minimum 34. (a) Minimum: 4, 1 Maximum: 1, 4 (b) Maximum: 1, 4 (c) Minimum: 4, 1 (d) No extrema 36. (a) Minima: 2, 0 and 2, 0
Maximum: 0, 2 (b) Minimum: 2, 0
(c) Maximum: 0, 2 (d) Maximum: 1, 3
380
Chapter 3
Applications of Differentiation
38. f x
2 2
x2, 1 x < 3 3x, 3 x 5
40. f x
Left endpoint: 1, 1 Maximum Right endpoint: 5,
3 3
2 , 0, 2 2x Left endpoint: 0, 1 Minimum
3
13 Minimum
(1, 1) (3, 7) (5, 13)
15
12
1 1
(0, 1)
5
42. (a)
3
Maximum:
2,
8 3
44. f x fx fx
1 x2 x2 1
,
1 ,3 2
2
(2, 8 ) 3
Minimum: 0, 0 , 3, 0
3
2x 1
0 0
21 x2 24x x2
3x2 13 24x3 14
(b) f x fx
4 x3 3
x, 0, 3 x
12 12
fx 1 x 23 22 3 3 x x x x
12
4 1 x 3 3 2 4 3 3 x
1
Setting f f1
0, we have x
0, 1.
1 2
1 is the maximum value. 2
2 6 3x 33 x Critical number: x f0 f3 f2 0 Minimum 0 Minimum 8 3 2, 8 3 2
62 x 33 x
Maximum:
46.
fx
1 x2 24x x2 1
,
1, 1
fx f f f
4
24x3 See Exercise 44. 14 1
48. Let f x 1 x. f is continuous on 0, 1 but does not have a maximum. f is also continuous on 1, 0 but does not have a minimum. This can occur if one of the endpoints is an infinite discontinuity.
y 2
x x 0
24 5x4 10x2 x2 1 5 240x 3x4 x2 1
6
5
10x2
3
2 1
1 x 1 1 2
4
24 is the maximum value.
Section 3.2 50.
5 4 3 2 x 1 2 3 4 5 6 y
Rolles Theorem and the Mean Value Theorem 54. (a) No (b) Yes
381
52. (a) No (b) Yes
f
2 1 2 3
56. x
v 2 sin 2 , 32 4
3 4
58.
C C1 C 300 C 2x2 x2 x
2x
300,000 , 1 x 300 x
d is constant. dt dx dt dx d by the Chain Rule d dt v 2 cos 2 d 16 dt In the interval 4, 3 4 , 4, 3 4 indicate minimums for dx dt and 2 indicates a maximum for dx dt. This implies that the sprinkler waters longest when 4 and 3 4. Thus, the lawn farthest from the spinkler gets the most water. 60. f x x
300,002 1600 2 300,000 x2 0
300,000 150,000 100 15 387 > 300 outside of interval 300 units. 387 would minimize C.
C is minimized when x
Yes, if 1 x 400, then x
y 3 2 1 2 1 x 1 2 3
The derivative of f is undefined at every integer and is zero at any noninteger real number. All real numbers are critical numbers.
2
62. True. This is stated in the Extreme Value Theorem.
64. False. Let f x gx fx x x k
x2. x k
2
0 is a critical number of f.
k is a critical number of g.
Section 3.2
Rolles Theorem and the Mean Value Theorem
4. f x xx 3
2. Rolles Theorem does not apply to f x cot x 2 over , 3 since f is not continuous at x 2 .
x-intercepts: 0, 0 , 3, 0 fx 2x 3 0 at x 3 . 2
6. f x
3x x
1 1, 0 , 0, 0 1 1
12
x-intercepts: fx fx 1 3x x 2 3x
3x 1
1
12
3x 2 . 3
1
12
x 2
x
1
12
3 x 2
0 at x
382 8. f x f1
Chapter 3 x2 f4 5x 0
Applications of Differentiation 4, 1, 4 10. f fx 1 x f3 3x 0 1, 3 . 1 2, 1, 3
f is continuous on 1, 4 . f is differentiable on 1, 4 . Rolles Theorem applies. fx 2x 5 5 2 x 0 3 , 0, 6 14. 2x 5 5 2
f is continuous on 1, 3 . f is differentiable on Rolles Theorem applies. fx x x x c value: fx f 1 5 3 x2 x f1 1 , 0 1, 1 32x 1 2x 1 3x 6 5 1 x x 1 1
2
0x
c value:
12. f x f0
3 f6
f is continuous on 0, 6 . f is not differentiable on 0, 6 since f 3 does not exist. Rolles Theorem does not apply.
f is not continuous on 1, 1 since f 0 does not exist. Rolles Theorem does not apply. 18. f f f fx cos 2x, 3 2 1 2 f 0 , 12 6
16. f x f0
cos x, 0, 2 f2 1
f is continuous on 0, 2 . f is differentiable on 0, 2 . Rolles Theorem applies. fx c value: sin x
12 6 12
Rolles Theorem does not apply. 22. f x f0 x f1 x1 3, 0, 1 0
20. f
fx
sec x, f
, 44 2
4
4
4, 4 . f is differentiable on f is continuous on 4, 4 . Rolles Theorem applies. fx sec x tan x x c value: 0 sec x tan x 0 0
f is continuous on 0, 1 . f is differentiable on 0, 1 . (Note: f is not differentiable at x 0.) Rolles Theorem applies. fx 1
3
1 1 3 1 3 1 27
3
1 3 x2
3
x2
0
x2 x2 x
1 27 3 9 0.1925
3 9
c value:
1
0
1
1
S ection 3.2 x 2 f0 x , 6
Rolle s Theorem and the Mean Value Theorem 1 x C6 Cx 3 6x 6x x x 3 25 3 10 1 x2 0 6 4 66 3 4 In the interval 3, 6 : c 3 33 3 2 4.098. 108 1 x2 3 x 3
2
383
24. f
fx 1
sin 0
1, 0
26. C x (a) C 3 1, 0 . (b) 0 x2 2x2
10
f is continuous on 1, 0 . f is differentiable on Rolles Theorem applies. fx x cos 6 x 1 2 3 6 3 x cos 6 6
0
9 9 x
arccos
Value needed in
1, 0 .
0.5756 radian c value: 0.5756
0.02
33 2
1
0
0.01
28.
y
30. f x
x
3 , 0, 6 3.
f is not differentiable at x
f
x
a
b
32. f x x x2 x differentiable on f1 1 f 1 fx 3x 1x 1 1
2 is continuous on 1, 1 . 1 3x2 0 1 3 2x 2
1, 1 and
34. f x x 1 x is continuous on 1 2, 2 and differentiable on 1 2, 2 . f2 2 f12 12 fx x2 c 32 3 32 1 x2 1 1 1 1
1
c
384
Chapter 3
Applications of Differentiation 38. f x 2 sin x sin 2x ferentiable on 0, . f fx 2 cos x 2 is continuous on 0, f0 0 2 cos 2x cos2 x 1 1 cos x cos x x In the interval 0, :c . 0 0 0 0 1 2 1 ,, 5 3 0 and dif-
36. f x x3 is continuous on 0, 1 and differentiable on 0, 1 . f1 1 f0 0 fx x 1 1 3x2
0 1
1
0
3 3 3 . 3
2 cos x 2 2 cos x
1 cos x
In the interval 0, 1 : c
3
3
40. f x (a)
x
2 sin x on
2
, (c) f x 1 0
,
2 cos x
1
tangent
2
f
secant 2 tangent
cos x c
2
f f
2 2 2 2
2 2
2 2 1x
2
(b) Secant line: slope y y f f 2 1x x y 1 Tangent lines: y
2 2 2 2
y 2 2 y
x 1x x
42. f x m (a) 80 5
150
x4 5 0
4x3 15
8x2
5, 0, 5 , 5, 80
(c) First tangent line: y
tangent f secant tangent
fc 9.59 0
mx 15 x 15x mx
c 0.67 y c 3.79 y 74.5 0.46
y
0 0
5
Second tangent line: y 15 x 15x 4x3 15 15 4c3 12c2 16c 3.79 15 0 y 5 12x2 16x y
fc 131.35 0
(b)
Secant line: y
5 0 fx
15 x 15x
f5 5 4c3 12c2
f1 1 16c 0 c
0.67 or c
S ection 3.2 9 2 t 200 5 9 2 t
2
Rolle s Theorem and the Mean Value Theorem
385
44. S t (a) (b)
200 5 S 12 12 St 1 2 2 t
2
S0 0 200 1 28 27 27
9 14 12 450 7
200 5
92
450 7
t t
2
3.2915 months
S t is equal to the average value in April. 46. f a f b and f c fx gb k fa k 0 gc 0 where c is in the interval a, b . (b) ga gx k gx k fx gb fx fc k, b k k k 0 k k Interval: fa (c) gx g a k f kx g b k fa
(a) g x ga gx
f x g c
Interval: a, b Critical number of g: c
gx g c k
kf kx kf c ab , kk c k 0
Interval: a
Critical number of g: c
Critical number of g:
48. Let T t be the temperature of the object. Then T 0 interval 0, 5 is 390 5 1500 0 222 F hr.
1500 and T 5
390 . The average temperature over the
By the Mean Value Theorem, there exists a time to, 0 < t0 < 5, such that T t0 x , 2 x 2 x 2
222.
50. f x
3 cos2
fx
6 cos
sin
2
3 cos (a)
2 7
x x sin 2 2 (b) f and f are both continuous on the entire real line.
2
7
(c) Since f 1 f1 0, Rolles Theorem applies on 1, 1 . Since f 1 0 and f 2 3, Rolles Theorem does not apply on 1, 2 .
(d) lim f x
x3 x3
0 0
lim f x
386
Chapter 3
Applications of Differentiation 5, 5 . x x 0 0 54. False. f must also be continuous and differentiable on each interval. Let fx x3 x2 4x . 1
52. f is not continuous on Example: f x
y
1 x, 0,
4 2
f ( x) = 1 x
(5, 1 ) 5
x 2 4
( 5, 1 ) 5
5
56. True
58. Suppose f x is not constant on a, b . Then there exists x1 and x2 in a, b such that f x1 Theorem, there exists c in a, b such that fc f x2 x2 f x1 x1 0. 0 for all x in a, b .
f x2 . Then by the Mean Value
This contradicts the fact that f x
60. Suppose f x has two fixed points c1 and c2. Then, by the Mean Value Theorem, there exists c such that fc f c2 c2 f c1 c1 c2 c2 c1 c1 1.
This contradicts the fact that f x < 1 for all x.
cos x. f is continuous and differentiable for all real numbers. By the Mean Value Theorem, for any interval a, b , 62. Let f x there exists c in a, b such that fb b cos b b cos b cos b cos b fa a cos a a cos a cos a cos a b fc
sin c sin c b sin c b a since a a sin c 1.
Section 3.3
Increasing and Decreasing Functions and the First Derivative Test
387
Section 3.3
2. y x 1
2
Increasing and Decreasing Functions and the First Derivative Test
4. f x , 1, 1 x4 2x2 1, 0 , 1, , 1 , 0, 1
Increasing on: Decreasing on: x2 x xx x 1 2 12
Increasing on: Decreasing on:
6. y y
Critical numbers: x Test intervals: Sign of f x : Conclusion: Increasing on Decreasing on 8. h x hx hx 27x 27 0 x3 3x2 , 2,
0,
2
Discontinuity: x 2 2<x< y <0 Decreasing 1
1 1<x<0 y <0 Decreasing 0<x< y >0 Increasing
<x<
y >0 Increasing 2 , 0, 1, 1, 0
33
x3
x
Critical numbers: x Test intervals: Sign of h x : Conclusion: Increasing on Decreasing on 4 x 2x x2 2
3
<x<
3
3<x<3 h >0 Increasing
3<x< h <0 Decreasing
h <0 Decreasing 3, 3 , 3 , 3,
10. y y
x x
Critical numbers: x Test intervals: Sign of y : Conclusion: Increasing: Decreasing: ,
2
Discontinuity: 0 2 2<x<0 y <0 Decreasing 0<x<2 y <0 Decreasing 2<x< y >0 Increasing
<x<
y >0 Increasing 2 , 2,
2, 0 , 0, 2
388 12. f x fx
Chapter 3 x2 2x 8x 8
Applications of Differentiation 10 0 4 <x< f <0 Decreasing 4, , 4 4, 6 4 4<x< f >0 Increasing 14. f x fx x2 2x 8x 8 0 4
<x<
12
Critical number: x Test intervals: Sign of f x : Conclusion: Increasing on: Decreasing on: Relative minimum:
Critical number: x Test intervals: Sign of f x : Conclusion: Increasing on: Decreasing on:
4
4<x< f <0 Decreasing
f >0 Increasing , 4, 4, 4 4
Relative maximum:
16. f x fx
x3 3x2
6x2 12x
15 3x x 0, 4
<x<0
4
Critical numbers: x Test intervals: Sign of f x : Conclusion: Increasing on Decreasing on 0, 4
0<x<4 f <0 Decreasing
4<x< f >0 Increasing
f >0 Increasing , 0 , 4,
Relative maximum: 0, 15 Relative minimum: 4, 18. f x fx x 3x x 2
2
17
x
1
2 2, 0
<x<
Critical numbers: x Test intervals: Sign of f x : Conclusion: Increasing on: Decreasing on:
2
2<x<0 f <0 Decreasing
0<x< f >0 Increasing
f >0 Increasing , 2, 0 2, 0 4 2 , 0,
Relative maximum: Relative minimum: 0,
S ection 3.3 20. f x fx x4 4x3 32x 32 4 4 x3 2 <x<2 f <0 Decreasing 8
Increasing and Decreasing Functions and the First Derivative Test 22. f x fx x2 2 x 3
3
389
4
13
Critical number: x Test intervals: Sign of f x : Conclusion: Increasing on: 2, Decreasing on:
2 3x1
3
Critical number: x 2<x< f >0 Increasing Test intervals: Sign of f x : Conclusion: ,2 44 Increasing on: 0, Decreasing on:
0
<x<0
0<x< f >0 Increasing
f <0 Decreasing
Relative minimum: 2,
,0 4
Relative minimum: 0,
24. f x fx
x 3x
1 1
13
26. f x fx 1 <x<1 f >0 Increasing , 1<x< f >0 Increasing
x x x
3 3 3
1 1, 3 <x< f <0 Decreasing 3, , 3 3, 1 3 3<x< f >0 Increasing x> 1, x < 3 3
1
23
Critical number: x Test intervals: Sign of f x : Conclusion: Increasing on: No relative extrema
Critical number: x Test intervals: Sign of f x : Conclusion: Increasing on: Decreasing on: Relative minimum:
28. f x fx
x x x 1 11 x1 x1
2
1 x 1
2
Discontinuity: x Test intervals: Sign of f x : Conclusion: Increasing on: No relative extrema ,
1 <x< f >0 Increasing 1, 1, 1 1<x< f >0 Increasing
390
Chapter 3 x x2 1 x2 3 1 x 6 x3
Applications of Differentiation 3 x2 x6 x3 6 0 <x< f <0 Decreasing 6, 0 , 6 , 0, 6, 1 12 6 6<x<0 f >0 Increasing 0<x< f <0 Decreasing
30. f x fx
Critical number: x Discontinuity: x Test intervals: Sign of f x : Conclusion: Increasing on: Decreasing on: Relative minimum:
32. f x fx
x2 x x
3x 2 2 2x
4 3 x 2 <x<2 f >0 Increasing , 2 , 2, 2<x< f >0 Increasing x2 22 3x 41 x2 x 4x 2
2
10
Discontinuity: x Test intervals: Sign of f x : Conclusion: Increasing on: No relative extrema
34. f x fx
sin x cos x cos 2x 0
1 sin 2x, 0 < x < 2 2
Critical numbers: x
357 ,,, 4444 0<x< f >0 Increasing <x< f <0 Decreasing 3 4 3 5 <x< 4 4 f >0 Increasing 5 7 <x< 4 4 f <0 Decreasing 7 <x<2 4 f >0 Increasing
Test intervals: Sign of f x : Conclusion:
4
4
Increasing on: Decreasing on:
0,
4
,
35 7 , , ,2 44 4
3 57 , , , 44 44 1 51 ,, , 42 42 3 , 4 1 7 , , 2 4 1 2
Relative maxima: Relative minima:
S ection 3.3 sin x ,0<x<2 cos2x 0 3 22 , 3 ,2 2
Increasing and Decreasing Functions and the First Derivative Test
391
36. f x fx
1
Test intervals: Sign of f x : Conclusion:
0<x< f >0
2
2
<x<
3 2
3 <x<2 2 f >0 Increasing
cos x 2 sin2x 1 cos2x 2
f <0 Decreasing
Increasing
Critical numbers: x
Increasing on: Decreasing on:
0,
2
,
Relative maximum: Relative minimum:
2
,1 1
3 , 22
3 , 2
38. f x (a) f x
10 5
x2
3x
16 , 0, 5 (b)
15 12 y
5 2x 3 x2 3x 16
f
6 3 1
f
x 1 3 4
3
(c)
5 2x 3 x2 3x 16 Critical number: x
0 3 2
(d) Intervals: 0, 3 2 3 ,5 2 f x <0 Decreasing
f x >0 Increasing
f is increasing when f is positive and decreasing when f is negative. x 2 x cos , 0, 4 2 1 2 1 x sin 2 2 (b)
8 6 y
40. f x (a) f x
f
4 2
f
2 3 4 x
(c)
1 2
1 x sin 2 2 sin x 2 x 2
0 1
(d) Intervals: 0, f x >0 Increasing ,4 f x >0 Increasing
2
f is increasing when f is positive.
Critical number: x
392 42. f t ft
Chapter 3 cos2t
Applications of Differentiation sin2t 2 sin2t 2 sin 2t gt, 2<t<2 44. f x is a line of slope
6
2f x
2.
4 sin t cos t
f symmetric with respect to y-axis zeros of f:
6 6 2
4
Relative maximum: 0, 1 Relative minimum:
2
2
,
1,
2
,
1
3
3
2
46. f is a 4th degree polynomial f is a cubic polynomial.
y 6
48. f has positive slope
y 4 3 2
f
6 4 2
x 2 4 6 3 2 1 2
f
x 1 2 3
In Exercises 5054, f x > 0 on 50. gx gx g 5 3f x 3f x 3f 5 >0 5 3
,
4 , f x < 0 on 52. g x gx g0
4, 6 and f x > 0 on 6, fx fx f 0 >0 58. s t 4.9 sin t2
. 54. g x gx g8 fx fx f 10 10 2 <0
56. Critical number: x f4 f6
2.5 f is decreasing at x 3 f is increasing at x
4. 6.
(a) v t (b) If vt
9.8 sin
t
speed
9.8 sin
t
2, the speed is maximum, 9.8 t.
5, f 5 is a relative minimum. 3t 27 t Ct t3
60. C (a)
,t 0 0 0 0.5 0.055 1 0.107 1.5 0.148 2 0.171 2.5 0.176 3 0.167
The concentration seems greater near t (b)
0.25
2.5 hours. (c) C 27 3 27 27 t3 3 3t 3t 2 32 27 t 2t 3 t3 2 3
3
0 0
3
C 2.38 hours.
0 when t
2
2.38 hours.
The concentration is greatest when t
By the First Derivative Test, this is a maximum.
S ection 3.3 x2 20,000 x 10,000
Increasing and Decreasing Functions and the First Derivative Test
393
62. P P x
2.44x 2.44 24,400
5000, 0 x 35,000 0
Test intervals: Sign of P :
0 < x < 24,400 P >0
24,400 < x < 35,000 P <0
Increasing when 0 < x < 24,400 hamburgers. Decreasing when 24,400 < x < 35,000 hamburgers. 64. R (a) R T 0.001T 4 4T 100 0 (b)
125
0.004T 3 4 2 0.001T 4 4T 100 10 , R 8.3666
100 25
100
The minimum resistance is approximately R 8.37 at T 10 .
66. f x
2 sin 3x
6
4 cos 3x
6
The maximum value is approximately 4.472. You could use calculus by finding f x and then observing that the maximum value of f occurs at a point where f x 0. For instance, f 0.154 0, and f 0.154 4.472. 68. (a) Use a cubic polynomial fx a3x3 a2x2 a1x (b) f x 0, 0 : 3a3 x 2 0 0 4, 1000 : 2a 2 x a0 a1 64a3 8a2 0, a2 375 2 x. 2 16a2 a1 f0 f0 f4 f4 375 , a3 2 0 0 1000 0 125 4
a0.
1000 0
48a3 a1
(c) The solution is a0 fx (d)
1200
125 3 x 4
(4, 1000)
3
(0, 0)
400
8
394
Chapter 3
Applications of Differentiation a4 x4 a3 x3 a2x2 a1x a0.
70. (a) Use a fourth degree polynomial f x (b) f x 1, 2 : 4a4x3 2 0 1, 4 : 4 0 3, 4 : 4 0 (c) The solution is a0 fx (d)
(1 , 4)
6
3a3x2 a4 4a4 a4 4a4 81a4 108a4
23 8, 13 2x
2a2x a3 3a3 a3 a2 3a3 27a3 27a3 a1
12 4x
a1 a1 2a2 a1 2a2 9a2 6a2
3 2,
a2
a0 a1 a0 a1 3a1 a1
1 4, 23 8.
f1 f1 f f a0 f3 f3
1 2,
2 0 1 1 4 0 4 0
a2
a3
a4
1 8
14 8x
3 2x
(3, 4)
4 2
(1, 2)
6
72. False f x g x where f x Let h x hx x2 is decreasing on 76. False. The function might not be continuous. gx ,0 . x. Then
74. True If f x is an nth-degree polynomial, then the degree of f x is n 1.
78. Suppose f x changes from positive to negative at c. Then there exists a and b in I such that f x > 0 for all x in a, c and f x < 0 for all x in c, b . By Theorem 3.5, f is increasing on a, c and decreasing on c, b . Therefore, f c is a maximum of f on a, b and thus, a relative maximum of f.
Section 3.4
2. y x3 3x2
Concavity and the Second Derivative Test
2, y ,1 6x 6 4. f x x2 2x 1 ,y 1 6 2x ,
1 2,
1
1 2
3
Concave upward:
Concave upward: Concave downward: 2 xx 9
Concave downward: 1,
6. y
1 270
3x5
40x3 ,
135x , y 2 , 0, 2
2x
2
8. h x hx hx
x5 5x4 20x3
5x 5
2
Concave upward: Concave downward:
2, 0 , 2,
Concave upward: 0, Concave downward: ,0
S ection 3.4 10. y y y 2 x 1 2 csc x, 2 csc x cot x 2 csc x csc3 x csc2 x csc x 2 cot x csc x cot x , 12. f x fx fx fx
Concavity and the Second Derivative Test 2x3 6x2 12x 12x 3x2 6x 6 6 0 when x
1 2. 1 2 1 2
395
12x 12
5
cot2 x
Concave upward: 0, Concave downward: ,0 Test interval Sign of f x Conclusion Point of inflection: 14. fx fx fx 2x4 8x3 24x 2 8x 8 0 when x 0. 3 <x< f x <0
<x< f x >0
Concave downward
1 2, 13 2
Concave upward
However, 0, 3 is not a point of inflection since f x 0 for all x. Concave upward on 16. f x fx x3 x x3 x2 x fx fx 4x 2 12x x 4 3x2 x 3x 8x x 2 4 4 3 4x2 x 4x x 3 2x 0, 2. 0<x<2 f x <0 Concave downward 2<x< f x >0 Concave upward 3 12x x 2 0 ,
0 when x
Test interval Sign of f x Conclusion
<x<0 f x >0 Concave upward 16 1, x 2x 1 1
Points of inflection: 0, 0 , 2, 18. f x fx fx x x 1, Domain: 1 x x 1 12 2 6x 1 3x 4x
3x 2 2x 1 1 12 3x 4 4x 13
2
f x > 0 on the entire domain of f (except for x There are no points of inflection. Concave upward on x 1 x 1,
1, for which f x is undefined).
20. f x fx fx
Domain: x > 0
Test intervals Sign of f x Conclusion
0<x<3 f >0 Concave upward
3<x< f <0 Concave downward
x1 2x3 2 3x 4x5 2 3, 4 3 3, 43 3
Point of inflection:
396
Chapter 3
Applications of Differentiation
22. f x fx
2 csc
3x ,0 < x < 2 2 3x 3x cot 2 2 csc 0, 3x 2 3x cot 2 2 0 for any x in the domain of f.
3 csc
fx
9 3x csc3 2 2
Concave upward:
2 4 , ,2 3 3 24 , 33
Concave downward: No points of inflection 24. f x fx fx fx sin x cos x sin x 0 when x
cos x, 0 x 2 sin x cos x 37 ,. 44 0<x< 3 4 3 7 <x< 4 4 f x >0 Concave upward 7 <x<2 4 f x <0 Concave downward
Test interval: Sign of f x : Conclusion:
f x <0 Concave downward 3 7 ,0 , ,0 4 4
Points of inflection:
26. f x fx fx fx
x 1
2 cos x, 0, 2 2 sin x 2 cos x
0 when x
3 , 22 0<x< f <0 Concave downward ,, 22 33 , 22 30. f x fx fx
3 2
Test intervals: Sign of f x : Conclusion:
2
2
<x<
3 2
3 3 <x< 2 2 f <0 Concave downward
f >0 Concave upward
Points of inflection:
28. f x fx fx
x2 2x 2
3x 3
8
x 2x 2
5
2
5
Critical number: x f
3 2
Critical number: x f 5 <0
5
>0
3 2, 41 4
Therefore,
is a relative minimum.
Therefore, 5, 0 is a relative maximum.
S ection 3.4
Concavity and the Second Derivative Test 1 x 8 x
397
32. f x fx fx
x3 3x 2 6x
9x2 18x 3
27x 27 3x 3
2
34. g x gx
2
2
x
4 1x
2
4x 2 32 x 2
2
Critical number: x
3
gx
3
3x
However, f 3 0, so we must use the First Derivative Test. f x 0 for all x and, therefore, there are no relative extrema.
Critical numbers: x g 2 9<0
2, 1, 4
2, 0 is a relative maximum. g1 1, 92>0
10.125 is a relative minimum. g4 9<0
4, 0 is a relative maximum. x x x 1 1 1
2
36. f x fx
x2 x
1
38. f x fx
x2 1 Critical number: x 0 1 fx x2 1 3 2 f0 1>0
There are no critical numbers and x 1 is not in the domain. There are no relative extrema.
Therefore, 0, 1 is a relative minimum. 40. f x fx fx f f f f 6 2 5 6 3 2 2 sin x 2 cos x 2 sin x <0 >0 <0 >0 3 53 ,, , 62 62 2 ,1 , 3 , 2 3 cos 2x, 0 x 2 2 sin 2x 4 cos 2x 2 cos x 4 sin x cos x 2 cos x 1 2 sin x 0 when x 53 ,, , . 62 6 2
Relative maxima: Relative minima:
398 42. f x
Chapter 3 x2 6
Applications of Differentiation x2, x2 x2 0, x 12
32
3 3
6,
6 (c)
6 y
(a) f x fx fx fx
3x 4 6
f
0 when x 6 x4 6 0 when x 9x2 x2
2.
x
9 2
33
.
f ''
f'
6
(b) f 0 > 0 0, 0 is a relative minimum. f 2 < 0 2, 4 2 are relative maxima. Points of inflection: 1.2758, 3.4035 44. f x (a) f x 2x sin x, 0, 2 2x cos x sin x 2x 1.84, 4.82 cos x 2x cos x 2x sin x 2x 2x
2 4
The graph of f is increasing when f > 0 and decreasing when f < 0. f is concave upward when f > 0 and concave downward when f < 0.
(c)
4
y
Critical numbers: x fx 2x sin x 2cos x 2x 4x cos x
f f'
2
2
x
f ''
4x2 1 sin x 2x 2x 4x2 2x 2x 1 sin x
(b) Relative maximum: 1.84, 1.85 Relative minimum: 4.82, 3.09 0.72 (b)
f is increasing when f > 0 and decreasing when f < 0. f is concave upward when f > 0 and concave downward when f < 0.
Points of inflection: 0.75, 0.83 , 3.42, 46. (a)
4 3 2 1 x 1 2 3 4 y
f < 0 means f decreasing f decreasing means concave downward
y 4 3 2 1 x 1 2 3 4
f > 0 means f increasing f decreasing means concave downward
48. (a) The rate of change of sales is increasing. S >0 (b) The rate of change of sales is decreasing. S > 0, S < 0 (c) The rate of change of sales is constant. C, S 0 S (d) Sales are steady. 0, S S C, S 0
50.
f f ''
3
y
f'
2
1 1
x 3
(e) Sales are declining, but at a lower rate. S < 0, S > 0 (f) Sales have bottomed out and have started to rise. S >0
S ection 3.4 52.
3 2 1 1 1 2 3 1 y
Concavity and the Second Derivative Test
y 2
399
54.
f ''
1
f
x 3
(0, 0)
1 1
(2, 0)
x 3
f'
56.
3 2
y
58. (a)
12
d
(0, 0)
1 1 1
(2, 0)
x 3 t 10
(b) Since the depth d is always increasing, there are no relative extrema. f x > 0 (c) The rate of change of d is decreasing until you reach the widest point of the jug, then the rate increases until you reach the narrowest part of the jugs neck, then the rate decreases until you reach the top of the jug. 60. (a) f x fx fx
3
x
3 2 1
y
1 23 3x 2 53 9x
6 4 2
(0, 0)
x 2 4 6
Inflection point: 0, 0 (b) f x does not exist at x 0.
2 3
62. f x
ax3
bx 2
cx
d
Relative maximum: 2, 4 Relative minimum: 4, 2 Point of inflection: 3, 3 fx f2 f4 f2 28a 12a 16a a fx
1 2,
3ax 2 8a 64a 12a 6b 4b 2b
9 2,
2bx
c, f x
6ax
2b 12b 8b 0 1 1 2c c 2 28a 0, f 3 18a 6b 2b c 0 1
4b 2c d 4 56a 16b 4c d 2 4b c c c 1 0 1 c 12, d 12x 6
92 2x
0, f 4 18a 16a 2a 6
48a 2b 2b
b
13 2x
400
Chapter 3
Applications of Differentiation 0.06x 0.04x bx 2 2bx 60 0.06 50 cx c 1000
3a
64. (a) line OA: y line CB: y fx fx ax3 3ax2
slope:
0.06
150
y
slope: 0.04 d
(1000, 60) A
100
(1000, 90) B
C (0, 50)
1000, 60 :
1000 2000b
2b
1000c c
d
1000
1000 2 3a 1000 3a 1000
2 3a
O
x 1000
1000, 90 :
90 0.04
1000 2b 2000b
1000c c 1.25 50
d
The solution to this system of 4 equations is a (b) y 1.25 10 8x3
100
10 8, b (c)
0.000025, c
0.1
0.0275, and d
50.
0.000025x2
0.0275x
1100
1100
0.1 1100 10 1100
(d) The steepest part of the road is 6% at the point A. 5.755T 3 108 8.521T 2 106 0.654T 104
66. S
0.99987, 0 < T < 25 4 and S 0.999999. (c) S 20 0.9982
(a) The maximum occurs when T (b)
1.001 1.000 0.999 0.998 0.997 0.996
T S
5
10
15
20
25
30
68. C C
2x 2
300,000 x 300,000 x2 0 when x 100 15 387
70. S (a)
100t2 ,t>0 65 t2
100
By the First Derivative Test, C is minimized when x 387 units.
0 0 35
(b) S t St
13,000t 65 t2 2 13,000 65 3t2 65 t2 3 0t 4.65
S is concave upwards on 0, 4.65 , concave downwards on 4.65, 30 . (c) S t > 0 for t > 0. As t increases, the speed increases, but at a slower rate.
S ection 3.4 72. fx fx fx P1 x P1 x P2 x P2 x P2 x 2 sin x 2 cos x 2 2 2 2 2 2 2x 2x 0
1 2
Concavity and the Second Derivative Test
401
cos x , sin x , cos x , 0 21 x
f0 f0 f0
2 2 2
6
4
f
P2
6
sin x 2x
P1
4
2x
0
2
2
2x
x2
The values of f, P1, P2, and their first derivatives are equal at x 0. The values of the second derivatives of f and P2 are equal at x 0. The approximations worsen as you move away from x 0. 74. fx fx fx P1 x P1 x P2 x P2 x P2 x x , 1 1 , 12 f2 f2 f2 x 2 2 3 22 23 82 32 x 4 32 4 23 2 16 52 2
3 P1 P2 f 1 1 5
x
x 2 xx 3x2 4x3 2 32 4 2 32 4 23 2 16
2
6x 1 , x 13 32 4
32 4
x
2 2
1 23 2 2 16
x
2
2
2
32 x 4
2
23 2 x 32
2
2
23 2 x 16
The values of f, P1, P2 and their first derivatives are equal at x 2. The values of the second derivatives of f and P2 are equal at x 2. The approximations worsen as you move away from x 2. 76. f x fx fx xx 3x 6x
2
6
2
x3 36 6x
12x2 3x 4 0
36x 2x 6 0
24x 24
Relative extrema: 2, 32 and 6, 0 Point of inflection 4, 16 is midway between the relative extrema of f.
402 78.
Chapter 3 px px px 6ax 2b x ax3 3ax2 6ax 0 b 3a
Applications of Differentiation bx2 2bx 2b cx c d
The sign of p x changes at x p b 3a a x3 3 31 2 33 27 1 2 b3 27a3 3x2 1 30 31 2 x3 b 2, a b2
b 3a. Therefore, 9a 2 1, b c b 3a 3, c d
b 3a, p 2b3 27a 2 2.
b 3a is a point of inflection. bc 3a d
When p x x0 y0
0, and d
2 3x2
0
2
0 1, 0 .
The point of inflection of p x 80. False. f x 82. True y Slope: y sin bx b cos bx
2 is x0, y0 0.
1 x has a discontinuity at x
by b
Assume b > 0 x 2 4.
84. False. For example, let f x
Section 3.5
2. f x 2x x2 2
Limits at Infinity
4. f x 2 x2 x4 1 6. f x 2x2 x2 3x 5 1
No vertical asymptotes Horizontal asymptotes: y Matches (c) 2x2 x 1 100 1 101 18.18 102 198.02 103 1998.02
2
No vertical asymptotes Horizontal asymptote: y Matches (a) 2
No vertical asymptotes Horizontal asymptote: y Matches (e) 2
8. f x x fx
x
20
104 19,998
105 199,998
106 1,999,998
0 2 10
lim f x
(Limit does not exist.)
S ection 3.5 8x x2 101 8.12 8 3 102 8.001 103 8 104 8 105 8 106 8 107 8
0 0
Limits at Infinity
403
10. f x
10
x fx
x
15
lim f x
12. f x
4 100 5
3 x2 2 101 4.03 4 102 4.0003 103 4.0 104 4.0 105 4 106 4
0
10
x fx lim f x
15 0
x
14. (a) h x
x
fx x
5x2
3x x
7
5x
3
7 x
16. (a) lim
x
3 2x 3x3 1 3 2x 3x 1 3 2x2 3x 1
0 2 3 (Limit does not exist)
lim h x fx x2 5
(Limit does not exist) 5x2 3x x2 7 5 3 x 7 x2
(b) lim
x
(b) h x
x
(c) lim
x
lim h x fx x3
(c) h x
x
5x2 0
3x x3
7
5 x
3 x2
7 x3
lim h x
18. (a) lim
x
5x3 2 4x2 1 5x3 4x 3 2 5x3 4x 3 x x
2 2
0 5 4 (Limit does not exist)
20. lim
x
3x3 9x3 2x2
2 7
3 9
1 3
(b) lim
x
1 1
(c) lim
x
22. lim 4
x
4
0
4
24. lim
x
1 x 2
4 x2
(Limit does not exist)
26. lim
x
x2
1
x
lim
1 x2 x2
1
for x < 0, x
x2
x
lim
x
1 (1 x
1
404
Chapter 3 3x x2 1 x
Applications of Differentiation 3 x2 3 1 1x , for x < 0 we have x x2 1x 1x 3
28.
x
lim
x
lim
x2
x
x
lim
30. lim
x
x
cos x x
x
lim 0
1 1
cos x x
32. lim cos
x
1 x
cos 0
1
1 Note:
x
lim
cos x x
0 by the Squeeze Theorem since
cos x 1 1 . x x x 3x x2 2 3 3
6 9
34. f x
x x
6
36. lim x tan
x
9
1 x
t 0
lim
tan t t
t 0
lim 11
sin t t 1
1 cos t
lim f x lim f x
Let x
1 t.
Therefore, y 3 and y horizontal asymptotes.
3 are both
38. lim 2x
x
4x2
1
x
lim
2x
4x2
1
2x 2x 3x 3x
4x2 4x2 9x2 9x2
1 1 x x
x
lim
2x
1 4x2
1
0
40.
x
lim
3x
9x2
x
x
lim
3x
9x2 x 9x2 1 9x2 x2 1 9 1x
x
x
lim
3x
x for x < 0 we have x x2
x
lim 3
x
x
lim
3
1 6
42.
x fx lim x2
100 1.0 x x2 1
101 5.1 x x2 x2
102 50.1
103 500.1 x x
104 5000.1 lim
105 50,000.1 x3 x x2
106 500,000.1
0
30
x
x x2 x x2
50 0
x
x2
x
Limit does not exist.
S ection 3.5 44.
3
Limits at Infinity
405
x fx lim x
100 2.000 1 0
101 0.348
102 0.101
103 0.032
104 0.010
105 0.003
106 0.001
0
25
x
xx
1
46. x
2 is a critical number.
48. (a) The function is even: (b) The function is odd:
x x
lim f x
5 5
f x < 0 for x < 2. f x > 0 for x > 2.
x
lim f x
lim f x
x
lim f x
6 6 2 6.
For example, let f x
y
0.1 x
2
1
8
4
x
2 2 4 6
50. y
x x
3 2 3 2
52. y
2x 9 x2
Intercepts: 3, 0 , 0, Symmetry: none
Intercept: 0, 0 Symmetry: origin 1 since x lim x x 3 . 2 Horizontal asymptote: y Vertical asymptote: x
y 6 5 4 3 2 1 5 4 1 2 3 4 5 6 x 12 6
Horizontal asymptote: y x lim x x 3 2 1
0
3
Discontinuity: x
y 5 4 3 2 4 3 2 1 2 3 4 5
2 (Vertical asymptote)
x 1 3456
54. y
x2 x2 9
5 4 3 2 5 4 1 2 3 4 5
y
Intercept: 0, 0 Symmetry: y-axis Horizontal asymptote: y
x
x 1 45
1 since
x
lim
x2 x2 9
1
lim
x2 x2 9
.
Discontinuities: x
3 (Vertical asymptotes)
Relative maximum: 0, 0
406
Chapter 3 2x2 x2 4
Applications of Differentiation
56. y
58. x2y
4
Intercept: 0, 0 Symmetry: y-axis Horizontal asymptote: y Relative minimum: 0, 0
y 5 4 3 1 5 4 3 2 1 2 3 4 5 x 12345
Intercepts: none Symmetry: y-axis 2 Horizontal asymptote: y
x
0 since 4 . x2
lim
4 x2
0
x
lim
Discontinuity: x
y
0 (Vertical asymptote)
4 3 2 1 5 4 3 2 1 x 12345
60. y
2x 1 x2
62. y
1
1 x 1, 0
Intercept: 0, 0 Symmetry: origin Horizontal asymptote: y
x
Intercept:
Symmetry: none 0 since
x
Horizontal asymptote: y x2 .
x
1 since
x
lim
2x 1 x2
0
lim
2x 1
lim
1
1 x
1
lim
1
1 . x
Discontinuities: x
y 5 4 3 2 1 5 4 3 2
1 (Vertical asymptotes)
Discontinuity: x
y 5 4 3 2
0 (Vertical asymptote)
x 2
5 4 3 2 1 3 4 5
x 12345
64. y
41
1 x2
66. y
x x2
4 , 2 , 2,
Intercepts: 1, 0 Symmetry: y-axis Horizontal asymptote: y Vertical asymptote: x
y 5 3 2 1 5 4 3 2 x 2345
Domain:
Intercepts: none 4 0 Symmetry: origin Horizontal asymptotes: y
x
1 since
lim
x x2
4
1, lim
x
x x2
4
1.
Vertical asymptotes: x
y 5 4 3 2 5 4 3 1 2 3 4 5 x 1 345
2 (discontinuities)
S ection 3.5 x2 x2 x2 x2 1 1 2x x2 2x 2 2 x 1 1
2
Limits at Infinity
407
68. f x
x = 1
4
x=1
fx fx
x2
2x 1
2
0 when x 2 3x2 1 x2 1 3
0.
3
y=1
4
(0, 0)
3
2 x2
2x 2 x2 14
1 2x
Since f 0 < 0, then 0, 0 is a relative maximum. Since f x of inflection. Vertical asymptotes: x Horizontal asymptote: y 1 x 2x x x x x
1
0, nor is it undefined in the domain of f, there are no points
1 1 1x 0 when x 2
70. f x fx fx
x2 x
2
2 1 2 2
2 2
x
2 1 . 2 x2 x 2 2x 1
3
x = 1
2
x=2
( 1 , 4( 2 9
3
x2 6 x2 x2
2x 1 2 x2 x 2 4
y=0
2
1 23
1 2, 4 9
Since f 1 < 0, then 2 points of inflection. Vertical asymptotes: x
is a relative maximum. Since f x 2
0, nor is it undefined in the domain of f, there are no
1, x 0
Horizontal asymptote: y x x2 x2 x xx x 1 1 2 1
2
72. f x fx fx
( 0.6527, 0.4491)
2
(0.5321, 0.8440)
0 when x 0 when x
0,
2. 0.5321, 0.6527, 2.8794.
3
3
2 x3 x2
3x2 1 x 13
(
2, 1
3
(
2
(0, 1)
( 2.8794, 0.2931)
f 0 <0 Therefore, 0, 1 is a relative maximum. f 2 >0
Therefore, 2, 1 3
is a relative minimum. Points of inflection: 0.5321, 0.8440 , Horizontal asymptote: y 0 0.6527, 0.4491 and 2.8794, 0.2931
408
Chapter 3 2x 3x2 2 3x2 1
Applications of Differentiation
74. g x gx gx
4
1
6
y= 2
3
6
32
18x 3x2 1 5
y= 2
2
3 4
No relative extrema. Point of inflection: 0, 0 . Horizontal asymptotes: y No vertical asymptotes 2 sin 2x Hole at 0, 4 x 4x cos 2x x2 2 sin 2x
2 3
76. f x fx
6
2
2
5 2
There are an infinite number of relative extrema. In the interval 2 , 2 , you obtain the following. Relative minima: 2.25, 0.869 , 5.45, 0.365
Relative maxima: 3.87, 0.513 Horizontal asymptote: y No vertical asymptotes x3
f=g
0
78. f x (a)
6
2x2 2x2
4
2
,g x
1 x 2
1
1 x2 (c)
70
6
80
80
4
70
(b) f x
x3 x3 2x2 1 x 2
2x2 2x2 2x2 2x2 1
2 2 2x2 gx
The graph appears as the slant asymptote y
1 2x
1.
1 x2
80.
v1 v2
lim
100 1
1 v1 v2
c
100 1
0
100%
82. y (a)
3.351t2
5
42.461t t2
543.730
20 0
100
(b) Yes. lim y
t
3.351
Section 3.5 100t2 ,t>0 65 t2
120
Limits at Infinity
409
84. S (a)
5 0
30
(b) Yes. lim S
t
100 1 an xn bm xm
100
86. lim
x
px qx
x
lim
... ...
a1x b1x
a0 b0
Divide p x and q x by xm. an xm n bm an
x
... ... ... ...
px Case 1: If n < m: lim x qx
x
lim
a1 xm 1 b1 xm a1 xm b1 xm ...
1 1 1
Case 2: If m
n: lim
x
px qx
lim bm an x n
a0 xm 0 ... 0 0 0 b0 bm . . . 0 0 bm xm a0 an . . . 0 0 an xm . b0 bm . . . 0 0 bm xm a0 xm b0 xm
0
m
px Case 3: If n > m: lim x qx
x
lim bm ...
a1 xm 1 b1 xm
1
bm
... ...
0 0
.
88. False. Let y1 y1 Let p ax2 Therefore, fx 4x bx
x 1 1
1, then y1 0
32
1. Thus, y1 1 . 4 1. Thus, p 1, x<0 x0
12x
1 and y1 0
1 2. Finally,
and y1 0
1, then p 0 1 2x
2ax
b and p 0
1 2
b
1 2.
Finally, p
2a and p 0
1 4
a
1 8.
1 8 x2 x 12 12x 1,
and f 0 1 2
1,
fx
1 4 x, 1, 1 4,
x<0 x>0
and f 0
, and 1 4
fx
x<0 x>0
1 4x
1
32
,
and f 0
.
f x < 0 for all real x, but f x increases without bound.
410
Chapter 3
Applications of Differentiation
Section 3.6
A Summary of Curve Sketching
4. The slope is positive up to approximately x Matches (B) (b) x2, x3 (d) x1 1.5.
2. The slope of f approaches as x 0 , and approaches as x 0 . Matches (C) 6. (a) x0, x2, x4 (c) x1 (e) x2, x3 x x2 1 x2 1 x2 12 x2 13
8. y y y
y 1
1
xx x2 1
1
2
0 when x 0, 3.
1.
(1, 1 ) 2 (1, 1 ) 2
(
3, 3 4
)
x
2x 3 x2
1
2
0 when x 0 y y
(0, 0)
(
y Conclusion Decreasing, concave down 3 4 1 1 2 0 0 Point of inflection Decreasing, concave up Relative minimum Increasing, concave up 0 0 Point of inflection Increasing, concave down 1 2 0 Relative maximum Decreasing, concave down 3 4 0 Point of inflection Decreasing, concave up
Horizontal asymptote: y
3, 3 4
)
<x< x 3 3<x< x 1 1<x<0 x 0
3
0<x<1 x 1 3
1<x< x 3 3<x<
S ection 3.6 x2 x2 x2 1 9 20x 9
2
A Summary of Curve Sketching 2 2 x 0.
411
10. y y y
12. f x 0 when x 0 0 fx fx
x x
1
2 < 0 when x x2 4 x3 0 2, 0 0
60 x2 3 < 0 when x x2 9 3
Therefore, 0, Intercept: 0,
1 is a relative maximum. 9 1 9
3
Intercept:
Vertical asymptote: x Horizontal asymptote: y
y 5 4 3
1
Vertical asymptotes: x Horizontal asymptote: y Symmetric about y-axis
y 5 4 3 2
1
y=1
2 x 1 2 3 4
(2, 0)
x=0
y=1
x
5 4
x = 3
1 2 3 4 5
(
0, 9 x=3
45 1
)
14. f x fx fx
x 1
32 x2 64 x3 x 4 x2 x3 0. 4x 16 0 when x 4.
16. f x fx fx
x3 x2 x2 x2 x2 8x x2 x2 4
x 12 42 12 43
4x x2 0 4 when x 0 0, 2 3
192 > 0 if x x4
0 when x
Therefore, 4, 6 is a relative minimum. Intercept; 2
3
Intercept: 0, 0 Relative maximum: 2 3, 33
4, 0 0 x
Vertical asymptote: x Slant asymptote: y
y
Relative minimum: 2 3, 3 3 Inflection point: 0, 0 Vertical asymptotes: x
2
(2 4, 0)
3
8 6 4 2
(4, 6)
Slant asymptote: y
y
x
y=x
x 2 4 6 8
8 6 4 6
x=0
8 6 4
8 6 4 x 4 6 8 10
412
Chapter 3 2x2 x 2 6 x 2
3
Applications of Differentiation 5 3 x 5
2
18. y
5x 2 3 x 2
2x 2x2
1 8x x2
2 0 when x 4 6 . 2
12 8
y
(4 +2 6, 7.899 (
y = 2x 1
x 4 8 12
y y
2
(
4+ 6 , 1.899 2
8 4
(
0 4 2 4 2 6 6 , 1.8990
x=2
8
Relative maximum: Relative minimum: Intercept: 0, 52
, 7.8990
Vertical x asymptote: Slant asymptote: y 20. g x gx gx x9 2x
2 1 22. y y 6 y x 16 28 16 x2 x2 x2 Domain: 0 when x 0 when x 4x4
2
x Domain: x 9 0 when x 6
36 x 29 x 3x 49
2
12 < 0 when x x32
2x x2 24 16 x2 3 2
0
Relative maximum: 6, 6 3 Intercepts: 0, 0 , 9, 0 Concave downward on
y 10 8 6 4 0) 2
Relative maximum: 2 2, 8 Relative minimum: 2 2, 8
,9
Intercepts: 0, 0 , 4, 0 Symmetric with respect to the origin Point of inflection: 0, 0
y
(6, 6
3)
(0,
(9, 0)
2 4 6 8 10
8 6 4 2
x
( 4, 0)
5
10 8 6 4 2
(2
2 , 8)
(0, 0)
(4, 0)
3 2 1 6 8 10
x
12345
(2
2 , 8)
24. y y
3x x 2 1
1
23
x 2x
1 1 1
2
y
2
13
2x 14 x 113
3
0 when x
0, 2
5 4 3
(2, 0)
5 4 3
(2, 2)
x
y undefined for x y 3x 2 1
43
2 < 0 for all x
1
1 2 3 4 5
12345
(1, 0)
Concave downward on
, 1 and 1,
Relative maximum: 0, 2 , 2, 2 Relative minimum: 1, 0 Intercepts: 0, 2 , 1, 0 , 1.280, 0 , 3.280, 0
S ection 3.6
1 3
A Summary of Curve Sketching
413
26. y y y
x3
2
3x 1
2 0 when x 0 y y y Conclusion Decreasing, concave up
4 3
y
x
1
(2, 0)
1
2 1
2x
0 when x
(1, 0)
x
( (1, 4 ( 3
2
0,
1 2 3
2
(
<x< x 1 1<x<0 x 0
1 0
Relative minimum Increasing, concave up
2 3
0
Point of inflection Increasing, concave down
0<x<1 x 1 0 0
Relative maximum Decreasing, concave down
1<x<
28. f x fx fx
1 3
x
1 1
2
3
2 0 when x 0 when x fx fx 0 1. 1. fx 0 Conclusion Increasing, concave down 2 Point of inflection Increasing, concave up
4 3
y
x 2x
1
(1, 2)
2
( 5)
0, 3 ( 0.817, 0)
x 1 2 3
<x<1 x 1 1< x <
30. f x fx
x x 3 x2
1x 1x 4x 2
2x 2 1 x
5
12
y
1x
5 2
x 3.
2x
5
(2
3, 6 3 ) (2, 0)
0 when x 2. fx 0
fx
6x
0 when x fx
12 9 6 3
x 3 6 9 12
fx
Conclusion Increasing, concave down Relative maximum Decreasing, concave down
(2 +
3, 6 3)
<x<2 x 2 x x 2 2 3 3 3<x< 2 2 3 3<x<2
3 63 0 63 0
0
Point of inflection Decreasing, concave up Relative minimum Increasing, concave up
2<x<2
Intercepts: 0, 10 ,
1, 0 , 2, 0 , 5, 0
414
Chapter 3
Applications of Differentiation 5 3 12x x2 12 3x2 y 1 1 y 0 when x 0 when x y 0, x
1.
32. y y y
3x4 12x3 36x2
6x2 12x 12
y 7 6 5 4 3 2
(0, 5 ) 3
( 33 , 0(
x 2345 4 1, 3
3 . 3
( 33 , 0(
5 4 3 2 4 1, 3
Conclusion Decreasing, concave up
(
)3 (
)
<x< x 1 1<x< x 3 3
1 43 3 3 0 0 0
Relative minimum Increasing, concave up Point of inflection Increasing, concave down
3 <x<0 3 x 0 3 3 0 0 53 0
Relative maximum Decreasing, concave down Point of inflection Decreasing, concave up
0<x< x 3 3
3 <x<1 3 x 1 43 0
Relative minimum Increasing, concave up
1<x<
34. f x fx fx
x4 4x3 12x 2
8x3 24x2 48x
18x2 36x 36 fx
16x 16 12 x fx 0
5 4x 3x fx 0 0 4x 1 1
2
y 4 4
0 when x 3, x
1, x 1.
4.
(0, 5) (1, 0)
1 3 4
(5, 0)
6 7
x
0 when x
12
Conclusion Decreasing, concave up Point of inflection Decreasing, concave down Point of inflection Decreasing, concave up Relative minimum Increasing, concave up
20 28
(3, 16) (4, 27)
<x<1 x x x 1 3 4 0 16 27 0 1<x<3 3<x<4 4<x<
S ection 3.6 36. y y y x 5x 20 x 1
5 4
A Summary of Curve Sketching
415
y
1
0 when x
3
1. 1. y 0 Conclusion Increasing, concave down Point of inflection Increasing, concave up
1
2 1
1
0 when x y y 0
(1, 0)
x 1 1 2 3
<x<1 x 1 0 1<x<
38. y y
x2 2x
6x
2
5 5 2x x 3x 5x 5x 1 1, x 1 5. 1, x 5.
6 5 4 3 2
y
3 x2 6x x 6x 5
(3, 4)
0 when x y 2 x2 x2 6x 6x
3 and undefined when x 5 5 y 2x x 5x 5x y Undefined 0 Undefined
1 undefined when x 1 y Undefined
1
(1, 0)
1 2 3 4 5
(5, 0)
6
x
Conclusion Decreasing, concave up Relative minimum, point of inflection Increasing, concave down Relative maximum Decreasing, concave down
<x<1 x x x 1 3 5 0 4 0 1<x<3 3<x<5 Undefined 5<x<
Relative minimum, point of inflection Increasing, concave up
40. y y y
cos x sin x cos x 4 cos x
2
1 cos 2x, 0 x 2 2
2
y
sin 2x 2 cos 2x cos x 2
sin x 1 cos x
2 cos x 2 2 cos2 x
0 when x 1
5 . 0, , , 33
1 x
1
2
0 when cos x
1 8
33
0.8431,
0.5931.
2
Therefore, x
0.5678 or 5.7154, x 53 3 ,, , 34 34 , 3 2
2.2057 or 4.0775.
Relative maxima: Relative minimum:
Inflection points: 0.5678, 0.6323 , 2.2057,
0.4449 , 5.7154, 0.6323 , 4.0775,
0.4449
416 42. y y y
Chapter 3 2x 2 2 csc2 x
Applications of Differentiation cot x, 0 < x < 0 when x 0 when x 33 , 42 , 42 2 , 0, x 8 8 1 3 4 5 3 44 , 2
5 4 3 2 1 1 2 3 4 5 y
2 csc2 x cot x
x
Relative maximum: Relative minimum: Point of inflection:
Vertical asymptotes: x x 8
44. y y
sec2 2 sec2
2 tan
1,
3<x<3 x 8 0x 2
5 4 3 2
y
x x tan 8 8
2 sec2
8
x
Relative minimum: 2,
5 4 3 2 1 2 3 4 5
(2, 1)
345
46. g x gx
x cot x, sin x cos x sin2x
2
<x<2
y 4 3 2
x x x 0 tan x lim
g 0 does not exist. But lim x cot x
x0
1.
2
1 2 3 4
2
x
Vertical asymptotes: x Intercepts: 3 ,0 , 2
2 ,
2
,0 ,
2
,0 ,
3 ,0 2
Symmetric with respect to y-axis. Decreasing on 0, 1 x
6
and 1
,2 4x x2
6
48. f x
5
4
x
2
50. f x
15
52. f is constant. f is linear. f is quadratic.
9
9
8
8
y
f
6
6
f ''
x y
2, 4 vertical asymptote 0 horizontal asymptote
y
4 horizontal asymptotes
x
0, 0 point of inflection
f'
S ection 3.6 54.
120 100 80 60 10 y y
A Summary of Curve Sketching
417
f
8 6 4 2 x
f
x 4 2 2 2 4 6 8 10
6 3
3
6
9
12 15
(any vertical translate of f will do) 56.
2 1 x 2 1 2 1 2 2 1 1 2 1 2 1 x y y
(any vertical translate of the 3 segments of f will do)
58. If f x 2 in 5, 5 , then f x 2x 3 and f 2 7 is the least possible value of f 2 . If f x fx 4x 3 and f 2 11 is the greatest possible value of f 2 . 3x4 x4 5x 3 1 x2 x x 3 x 1 2x x1 1 2
4 in
5, 5 , then
60. g x
62. g x
Vertical asymptote: none Horizontal asymptote: y
7
x 2, if x Undefined, if x
1 1
The rational function is not reduced to lowest terms.
4
6 1
8 6
4
4
The graph crosses the horizontal asymptote y 3. If a function has a vertical asymptote at x c, the graph would not cross it since f c is undefined. 2x2 x
18
hole at 1, 3
64. g x
8x 5
15
2x
2
5 x 5
10 2
20
The graph appears to approach the slant asymptote y 2x 2.
418
Chapter 3
Applications of Differentiation
66. f x (a)
2
tan sin x
3
(b) f
2
x
tan sin
x
tan fx
sin x
tan sin x
Symmetry with repect to the origin
3
1, 1 , there is a relative maximum at (d) On 1 and a relative minimum at tan 1 . 2,
1 2,
tan 1
(c) Periodic with period 2 (e) On 0, 1 , the graph of f is concave downward.
68. Vertical asymptote: x
3
70. Vertical asymptote: x Slant asymptote: y y x 1 x 1
0 x x2 x
Horizontal asymptote: none y x2 x 3 1 ax ax 2 1
72. f x fx fx
1 ax 2 a2x
2
ax a a ax
2 ,a
0 1 . a
0 when x
a2 > 0 for all x. 2 ,0 a 1 , a 1 2 (b)
a=2
5 y
(a) Intercepts: 0, 0 , Relative minimum:
a = 2
4
Points of inflection: none
3
a=1
x
a = 1
1
2
3
74. Tangent line at P: y (a) Let y 0: y0
y0
f x0 x x0 y0
x0 (b) Let x 0: y y0 y x0 f x0 f x0 y y-intercept: 0, f x0 (d) Let x x0 0: y y0 y y-intercept: (f) PC PC (h) AP f x0 f x0
2
f x0 x x0 f x0 x0 y0 f x0
f x0 y0 f x0
x0 x0 f x0 x0 f x0
f x0 x x x-intercept: x0
x0 f x0 1 f x0 y0 x0 f x0 f x0 2 f x0 2 f x0 f x0 2
2 2
f x0 ,0 f x0 y0 1 x f x0 1 x f x0 x x0 x0 y0 f x0 x0 f x0 f x0 x0
x0 x0 f x0
(c) Normal line: y Let y 0: y0
0, y0 f x0 f x0
y0 f x0 x x-intercept: x0 (e) BC (g) AB x0 x0
y02 f x0
f x0 f x0 , 0 x0 f x0 f x0 f x0 f x0
2
f x0 f x0 x0
1 f x0 f x0
2
2
f x0 2 f x0 f x0 1
y02 f x0
2
AP
Section 3.7
Optimization Problems
419
Section 3.7
Optimization Problems
y S. 4. Let x and y be two positive numbers such that xy S dS dy d 2S dy 2 x 3 3y 192 y2 192 y 3y 8. 192.
2. Let x and y be two positive numbers such that x P dP dx d dx 2
2P
xy S
xS 2x
x
Sx
x2 S . 2
0 when x S . 2
0 when y 8.
2 < 0 when x y
384 > 0 when y y3
P is a maximum when x
S 2.
S is minimum when y
8 and x
24.
6. Let x and y be two positive numbers such that x 2y 100. P dP dy d 2P dy 2 xy 100 y 100 4y 2y 100y 25. 2y2
0 when y 25.
4 < 0 when y
P is a maximum when x
50 and y
25.
8. Let x be the length and y the width of the rectangle. 2x 2y y A dA dx d 2A dx2 P P 2 xy P 2 x 2x P 2 2x P 2 x x P x 2 x2
x y
0 when x P . 4
P . 4
2 < 0 when x y
A is maximum when x
P 4 units. (A square!)
10. Let x be the length and y the width of the rectangle. xy y P dP dx d 2P dx2 A A x 2x 2 2y 2A x2 2x 2 A x 2x A. A. A centimeters. 2A x
x y
0 when x
4A > 0 when x x3 y
P is minimum when x (A square!)
420
Chapter 3 x
Applications of Differentiation 8, 2, 0 14. f x d x x x2 x2 x4 1 2, 5, 3 5
2
12. f x
From the graph, it is clear that 8, 0 is the closest point on the graph of f to 2, 0 .
y
x 25 25 x2
1
2
3
2
10x 10x 4x3
x2 x4 18x
2x 4x3 29
2
2
8x
4
4
2
Since d is smallest when the expression inside the radical is smallest, you need to find the critical numbers of
x 2 4 6 8 10 12
gx gx
x4 4x3 2x
4x3 12x2 1 2x
x2 2x
2
18x 18 9
29
8x
By the First Derivative Test, x 1 yields a minimum. Hence, 1, 4 is closest to 5, 3 . v 0.02v 2 0.02v 2 0.02v 2 2 1100 33.166.
16.
F dF dv
22 22 22
18. 4x
3y A dA dx d 2A dx2
200 is the perimeter. (see figure) 2xy 8 50 3 2x 200 3 0 when x 25.
100 3
4x
8 50x 3 25.
x2
0 when v
2x
By the First Derivative Test, the flow rate on the road is maximized when v 33 mph.
16 < 0 when x 3
A is a maximum when x
25 feet and y
feet.
y x x
20. (a)
Height, x 1 2 3 4 5 6
Length & Width 24 24 24 24 24 24 21 22 23 24 25 26 2 x 24 1 24 2 24 3 24 4 24 5 24 6 24 2x
2
Volume 21 22 23 24 25 26 24
2 2 2 2 2 2
(b) V 484 800 972 1024 980 864
0 0
x 24
2x 2, 0 < x < 12
(d)
1200
12
The maximum volume seems to be 1024.
(c)
dV dx
2x 24 12 12
2x x4 16 4.
2x 24
6x
0 when x
12, 4 12 is not in the domain .
d dx2
2V
12 2x
d 2V < 0 when x dx2 When x 4, V
1024 is maximum.
S ection 3.7 22. (a) P 2x 2x 2x y (b) Length, x 10 20 30 40 50 60 2 2 2 2 2 2 Width, y 100 100 100 100 100 100 10 20 30 40 50 60 10 20 30 40 50 60 2 2 2 2 2 2 Area, xy 100 100 100 100 100 100 10 20 30 40 50 60 573 1019 1337 1528 1592 1528 200 2r 2 y y 2
x y 2
Optimization Problems
421
y
200 2x 2 100 x
The maximum area of the rectangle is approximately 1592 m2. (c) A (e)
2000
xy
x
2
100
x
2
100x
x2
0 0
100
Maximum area is approximately 1591.55 m2 x 50 m . 6 2 x
24. You can see from the figure that A A dA dx d 2A dx2 x 6 2 2x x 1 6x 2 x.
2
xy and y
.
6 5 4 3
y
y=
6x 2 ( x, y )
x
1 6 2
0 when x 3. 3 and y
3.
2 1 1 2 3
4
5
6
1 < 0 when x
A is a maximum when x
3 2.
422
Chapter 3 1 base 2 1 2 16 2 16 dA dh 1 16 2 16
Applications of Differentiation
26. (a)
A
height
4
h2 4 h
12
h
h 4 16 h 2
h2 4 h2 h2
2h 4 h4 h
h 16 16
16 h2
h2
12
12
2 h2 2h 8 16 h2
2h
4h 2 h2
dA 0 when h 2, which is a maximum by the First Derivative Test. dh Hence, the sides are 2 16 h2 4 3, an equilateral triangle. Area (b) cos tan Area 2 4 4h 84 h 16 h2 4h 1 2 16 h 2 tan tan sec2 4 1 1 4 sin (c) Equilateral triangle cos3 4 cos3 sin sin sin tan tan tan 0 h2 4 h 4 8 h
12 3 sq. units.
4 4 16 h 2
8
4+h
h
64 cos4 A cos4 64 cos4 sec2
4 cos sin2 1 2
30 and A
12 3.
28. A dA dx
2xy
2x r 2
x2
see figure 2r . 2
y
2 r2 2x2 r2 x2
0 when x
(x,
r 2 x2
( ( x,
r 2 x2
(
x
By the First Derivative Test, A is maximum when the rectangle has dimensions 2r by 2r 2.
(r, 0) (r, 0)
S ection 3.7 36 y x 36 dA dx 108 x2 3y 108 x 3 9 V0 r2 V0 r
3
Optimization Problems
423
30. xy A
36 x 3 3x x 9 108 x 6, y 6 36 3 x 3
x+3 x
y
y+3
0 3x2
Dimensions: 9
32. V S dS dr h
r 2h 2 r2 22 r
V0 cubic units or h 2 rh V0 r2
2
2
r2
0 when r V0 2 2 3 V02 3
V0 units. 2 2r
3
V0 V0 2
2V01 3 2 13
By the First Derivative Test, this will yield the minimum surface area. 34. x V 2r V dV dr r 2x 108 x r2 108 108 2r 6 r2 36 2r see figure 2 r3 r
r x
108 r 2 6 r 36 36.
216r 0 when r
and x
d 2V dr 2
216
12 r < 0 when r
36
. 36 x2 x2
r h
Volume is maximum when x 36. V dV dx 2 x 2h x 2 2 r2 x2 x2
12
36 inches and r 2 x2 r2 2x 2x r 2
11.459 inches. see figure
x
(x,
r2 x2
(
12 x2 r 2 2x 2r 2 r2 x2
3x2 2r 2 x 3 6r . 3
0 when x
0 and x 2
(x,
r2 x2
(
By the First Derivative Test, the volume is a maximum when x 6r and h 3 2r . 3
Thus, the maximum volume is V 22 r 3 2r 3 4 r3 . 33
424
Chapter 3
Applications of Differentiation
38. No. The volume will change because the shape of the container changes when squeezed. 43 r 3 3000 r2
40. V
3000
r 2h 4 r 3 cost per square foot of the hemispherical ends. k 16 2 r 3 6000 r
h Let k C dC dr
cost per square foot of the surface area of the sides, then 2k 2k 4 r 2 32 r 3 k 2 rh 6000 r2 k 8 r2 2r 3000 r2 4 r 3
k
0 when r
3
1125 2
5.636 feet and h
22.545 feet.
By the Second Derivative Test, we have d 2C dr 2 k 32 3 12,000 r3 > 0 when r
3
1125 . 2
Therefore, these dimensions will produce a minimum cost. 42. (a) Let x be the side of the triangle and y the side of the square. 3 cot x 2 4 3 32 x 4 3 x 2 4 cot y 2 where 3x 4 4 32 20 x ,0x . 4 3 3 x 4 3 4 (b) Let x be the side of the square and y the side of the pentagon. A 4y 20 4 cot x 2 4 4 x2 5 A x 0 When x 0, A 27.528, when x 2.62, A 13.102, and when x 5, A 25. Area is maximum when all 20 feet are used on the pentagon. 2x 2.62 5 cot y 2 where 4x 4 5 5y 20
A
1.7204774 4 2.75276384 4
42 x , 0 x 5. 5 4 x 5 0
A
25
x
60 43 9
When x 0, A 25, when x 60 4 3 9 , A 10.847, and when x 20 3, A 19.245. Area is maximum when all 20 feet are used on the square. (c) Let x be the side of the pentagon and y the side of the hexagon. A 5 cot x 2 4 5 5 cot x2 4 5 A 5 cot x 2 5 2.0475 6 cot y 2 where 5x 4 6 3 2 20 6 5 6 20 6 5x
2
(d) Let x be the side of the hexagon and r the radius of the circle. A 6 cot x2 4 6 3 32 x 2 A 33 1.748 6 10 10 3x r 2 where 6x 3x
2
6y
20
2r 10 . 3
20
3
, 0 x 4. 0 x
,0 x
33
5x
0
x
When x 0, A 28.868, when x 2.0475, A 14.091, and when x 4, A 27.528. Area is maximum when all 20 feet are used on the hexagon
When x 0, A 31.831, when x 1.748, A 15.138, and when x 10 3, A 28.868. Area is maximum when all 20 feet are used on the circle. In general, using all of the wire for the figure with more sides will enclose the most area.
S ection 3.7 44. Let A be the amount of the power line. A dA dy d 2A dy 2 h 1 2x2 y2 y 2 x2 2y x2
32
y
Optimization Problems
425
y2 0 when y x 3 x 3 . x 3. .
(0, h) hy y
x
y2
x2
> 0 for y
(x, 0)
(x, 0)
The amount of power line is minimum when y 46. f x (a)
12 2x
9
gx
14 16 x
12 2x
on 0, 4 (c) f x x, Tangent line at 2 2, 4 is y 4 y gx
13 4x
f g
1 3 4
2 2x 2 2x
22 4.
x, Tangent line at 2 2, 0 is
1 4
(b) d x dx
fx 2x
gx
13 4x
12 2x
14 16 x
12 2x
x2
14 16 x
y
0 y
22
3
22x
22
0 8x x
x3 0, 2 2 in 0, 4 4 when x 2 2.
2 2x
8. gx, gx
The tangent lines are parallel and 4 vertical units apart. fx (d) The tangent lines will be parallel. If d x 0 fx g x implies that f x then d x at the point x where the distance is maximum.
The maximum distance is d
48. Let F be the illumination at point P which is x units from source 1. F dF dx
3 3 3 3
kI1 x2 2kI1 x3 x d x x
3 3 3
kI2 d d x
2
2kI2 x
3
0 when
2kI1 x3
d
2kI2 . x3
I1 I2 I1 I1 x
x I2 I1
3 3
d
x d
I2
d I1
I1
3
I2
4
dF dx2
2
6kI1 x4
6kI2 dx
> 0 when x
3
d I1
3
I1
3
I2
.
This is the minimum point. x2 2
2 x 3x Q
50. (a) T
4
3 4
x
(b) x x2
dT dx 4 2x2 x2 x T2 2
x 2 x2 1 2 x2 4 2 4
4
1 4
0
x2 + 4
CONTINUED
1 hours 4
426
Chapter 3
Applications of Differentiation
50. CONTINUED (c) T dT dx x x2 sin 4 x2 4 v1 v1 v1 v2 v1 v2 v1 only. v2 d22 v2 x d1
2
3 v2 4 1 v2
x
(d) Cost 0
x2
4 C1
3 3x 1 C2 1 C1 1 C2
x C2
x x2
x2 4 1 C1 From above, sin
C2 C1
depends on
52.
T dT dx Since
x2 v1 x v1 x2
d12
a
x
2
54. C x Cx x
2
2k x2 2xk x2 4 x2 x2 4 2 3 4 4
4 k
k4 0
x
v2 d2
2
a a
0 2x 4x2
x x2 we have sin v1 Since
1
d12
sin
1
and
x d22
a a
x
2
sin
2
3x2 x
sin v2
2
0
sin v1
1
sin 2 . v2
Or, use Exercise 50(d): sin Thus, x 2 . 3
x2 + 4 x 4x
C2 C1
1 2
30 .
d dx2
2T
d1 v1 x2
2
d12
32
v2 d22
d2 a
2
x
232
>0
2
this condition yields a minimum time.
56.
V dV dr
12 rh 3 1 3 1 3 r2
12 r 144 3 1 144 2 r2
12
r2 2r 2r 144 r2 0, 4 6. 4 3.
288r 3r3 144 r 2
r 96 r 2 144 r 2
0 when r
By the First Derivative Test, V is maximum when r Area of circle: A 12
2
4 6 and h
144 46 12 r 2 46 72 6 1.153 radians or 66
2
Lateral surface area of cone: S Area of sector: 144 144 48 6 48 6 72
43
2
48 6
2 3 3
S ection 3.7
Optimization Problems
427
58. Let d be the amount deposited in the bank, i be the interest rate paid by the bank, and P be the profit. P d P dP di d 2P di 2 0.12 d id
k i 2 since d is proportional to i 2 0.12 ki 2 k 0.24i k 0.24 3i 2 i ki 2 k 0.12i 2 0.24 3 i3 0.08.
0 when i
6i < 0 when i
0.08 Note: k > 0 . 8%.
The profit is a maximum when i 13 s 10 dP ds d 2P ds 2
60.
P (a)
6s 2
400 12s 3 ss 10 40 0 when x 0, s 40.
32 s 10 3 s 5
12 0 yields a minimum. 40 yields a maximum. 40, which corresponds to $40,000 P $3,600,000 .
d 2P 0 > 0s ds 2 d 2P 40 < 0 s ds2
The maximum profit occurs when s (b) d 2P ds 2 3 s 5 12 0 when s 20.
The point of diminishing returns occurs when s 62. S2 4m 1 5m 6 10m 3
20, which corresonds to $20,000 being spent on advertising.
Using a graphing utility, you can see that the minimum occurs when m Line y S2
S2
0.3.
0.3x 4 0.3 1 5 0.3 6 10 0.3 3 4.7 mi.
30
20
10
(0.3, 4.5)
m 1 2 3
428
Chapter 3
Applications of Differentiation h x and cos . The area A of the 2 r 2 r cross equals the sum of two large rectangles minus the common square in the middle.
2
64. (a) Label the figure so that r2 x2 h2. Then, the area A is 8 times the area of the region given by OPQR:
h
(b) Note that sin
A
R h
2 2x 2h 8r2 sin 4r2 sin 2 cos
4h2 2 sin2
8xh 4r2 sin2
4h2 2
O
x Pr Q
2 sin 2 cos 2 0
A
8 8
12 h 2 12 r 2
x x2 x2 Ax
hh x 4x2 8 r2 8x x r2 x r2 x2 r2 r2 4r2 x2 8 r2 x2 x2 x2 x2 r2 x2 8x2 r2 x2 8x 0 x2 r2 x2
A cos tan
4r2 cos sin 2 arctan 2 cos
2
2
2 sin
8x r2
1.10715
or 63.4
8x2 r2 x2 x2 2x2 4x4 5x4 x2 4x2r2 5x2r2 5r2 r2 r4 r4
0 Quadratic in x2. 20r4 r2 5 10 5.
25r4 10 Take positive value. x r 5 10 r2 5 10 x2 5
12
5
0.85065r
Critical number r2 5 10
(c) Note that x2 Ax 8x r2 8 8 r2 5 10
5 and r2 4x2 r2 5 10 2r2 2 5 4r2
x2
5.
12
5 5r2
4 4r2
r2 5 10
5
4r2
r4 20 10
82 r5 5 2r2 4 5 5
2r2 1
2 52 r 5 5 5 2r2 5 1 2, sin 2 and sin2 2 5 1 1 2 cos 1 1 2 1 . 5
Using the angle approach, note that tan
Thus, A
4r2 sin 4r2 4r2 2 5 5 2
sin2 1 1 2 1
2 1 5 2r2 5 1
Section 3.8
Newtons Method
429
Section 3.8
2. f x fx x1 1 2x2 4x 3
Newtons Method
f xn f xn 1 4 0.025 f xn f xn 5 4 1.225
n 1 2 5 4
xn 1 1.25
f xn 1 0.125
f xn 4 5.0
xn
4. f x fx x1
tan x sec2 x 0.1 n 1 2 xn 0.1000 0.0007 f xn 0.1003 0.0007 f xn 1.0101 1.0000
f xn f xn 0.0993 0.0007
xn
f xn f xn 0.0007 0.0000
6. f x fx
x5 5x
4
x 1
1 n 1 2 3 4 xn 0.5000 0.8571 0.7707 0.7553 f xn 0.4688 0.3196 0.0426 0.0011 f xn 1.3125 3.6983 2.7641 2.6272
f xn f xn 0.3571 0.0864 0.0154 0.0004
xn
f xn f xn 0.8571 0.7707 0.7553 0.7549
Approximation of the zero of f is 0.755.
8. f x fx
x 1
2x 1 x
1 n 1 1 2 xn 5 4.8293 f xn 0.1010 0.0005 f xn 0.5918 0.5858
f xn f xn 0.1707 .00085
xn
f xn f xn 4.8293 4.8284
Approximation of the zero of f is 4.8284.
10. f x fx
1
2x3 6x2 n 1 2 3 4 xn 1 0.8333 0.7955 0.7937 f xn 1 0.1573 0.0068 0.0000 f xn 6 4.1663 3.7969 3.7798 f xn f xn 0.1667 0.0378 0.0018 0.0000 xn f xn f xn 0.8333 0.7955 0.7937 0.7937
Approximation of the zero of f is 0.7937.
430
Chapter 3 14 x 2 2x3
Applications of Differentiation
12. f x fx
3x 3
3
n 1 0.8937. 2 3
xn 1 0.9 0.8937
f xn 0.5 0.0281 0.0001
f xn 5 4.458 4.4276
f xn f xn 0.1 0.0063 0.0000 f xn f xn 0.0769 0.0049 0.0000
xn
f xn f xn 0.9 0.8937 0.8937 f xn f xn
Approximation of the zero of f is
Approximation of the zero of f is 2.0720.
n 1 2 3
xn 2 2.0769 2.0720
f xn 1 0.0725 0.0003
f xn 13 14.9175 14.7910
xn
2.0769 2.0720 2.0720
14. f x fx
x3 3x2
cos x sin x n 1 2 3 xn 0.9000 0.8666 0.8655 f xn 0.1074 0.0034 0.0001 f xn 3.2133 3.0151 3.0087
f xn f xn 0.0334 0.0011 0.0000
xn
f xn f xn 0.8666 0.8655 0.8655
Approximation of the zero of f is 0.866.
16. h x hx
fx 1
gx 2x x2 1
3
x
1 x2 1 n 1 2 xn 2.9000 2.8933 h xn 0.0063 0.0000 h xn 0.9345 0.9341
h xn h xn 0.0067 0.0000
xn
h xn h xn 2.8933 2.8933
2
Point of intersection of the graphs of f and g occurs when x 2.893. x2
18. h x hx
fx 2x
gx sin x
cos x
n 1 2
xn 0.8000 0.8245
h xn 0.0567 0.0009
h xn 2.3174 2.3832
h xn h xn 0.0245 0.0004
xn
h xn h xn 0.8245 0.8241
One point of intersection of the graphs of f and g occurs x2 and g x cos x are when x 0.824. Since f x both symmetric with respect to the y-axis, the other point of intersection occurs when x 0.824. 20. f x fx xi
1
xn nx xi nxin
n
a
1
0
xin a nxin 1 xin nxin
1
a
n
1 xin nxin 1
a
S ection 3.8 xi2 5 2xi 1 2.0000 2.236 tan x sec2 x n 1 2 3 28. y y x1 f x2 n 1 2 4x3 12x2 3 2 0; therefore, the method fails. xn 3 2 1 f xn 3 2 1 f xn 3 0 f xn f xn 1 2 xn f xn f xn 1 12x2 24x 12x 12 3 fx fx xn 3.0000 3.1397 3.1416 30. f x fx x1 f xn 0.1425 0.0019 0.0000 2 sin x 2 cos x 3 2 0. f xn 1.0203 1.0000 1.0000 cos 2x 2 sin 2x 2 2.2500 3 2.2361 4 2.2361
3
Newton s Method
431
22. xi i xi
1
24. xi i xi
1
2xi3 15 3xi2 1 2.5000 2.466 f xn f xn 0.1397 0.0019 0.0000 f xn f xn 3.1397 3.1416 3.1416 2 2.4667 3 2.4662 4 2.4662
5 26. f x fx
15
xn
Approximation of the zero: 3.142
Fails because f x1 n 1 xn 3 2 f xn 3
f xn 0
32. Newtons Method could fail if f c 34. Let g x gx fx csc2 x x cot x 1. x
0, or if the initial value x1 is far from c.
n 1 2 3
xn 1.0000 0.8516 0.8603
g xn 0.3579 0.0240 0.0001
g xn 2.4123 2.7668 2.7403
g xn g xn 0.1484 0.0087 0.0000
xn
g xn g xn 0.8516 0.8603 0.8603
The fixed point is approximately 0.86.
36. f x (a)
sin x, f x
2
cos x (d)
2 1 y
(1.8, 0.974) (3, 0.141) (6.086, 0)
2
x
2
1 2
(3.143, 0)
(b) x1 x2 (c) x1 x2
1.8 x1 3 x1 f x1 f x1 3.143 f x1 f x1 6.086
The x-intercepts correspond to the values resulting from the first iteration of Newtons Method. (e) If the initial guess x1 is not close to the desired zero of the function, the x-intercept of the tangent line may approximate another zero of the function.
432
Chapter 3 xn 2 1 0.3000 0.333 x sin x, 0, x cos x
Applications of Differentiation 3xn 2 0.3300 3 0.3333 4 0.3333
1 11
38. (a) xn i xi
1 3
1
(b) xn i xi
1
xn 2 1 0.1000 0.091
11xn 2 0.0900 3 0.0909 4 0.0909
40. f x fx
y
sin x
0
4 3 2
Letting F x Fx n 1 2 xn 2.0000 2.0290
f x , we can use Newtons Method as follows. 2 cos x x sin x F xn 2.6509 2.7044 F xn F xn 0.0290 0.0002 xn F xn F xn 2.0290 2.0288
(2.029, 1.820)
1 x
F xn 0.0770 0.0007
4
2
3 4
Approximation to the critical number: 2.029 42. y d fx x x2, 4, 4
2
3 y 3
2
y
x 7x2 8
4 8x
2
x2
3
2
x4
7x2
8x
25
3 2 1 2 1 1 2 3 1
d is minimum when D gx gx n 1 2 3 x xn 0.5000 0.5294 0.5291 0.529 D 12x2 4x3 14 g xn
x4 14x
25 is minimum.
(0.529, 0.280)
x 2 3 4
(4, 3)
g xn 17.0000 17.3632 17.3594
g xn g xn 0.0294 0.0003 0.0000
xn
g xn g xn 0.5294 0.5291 0.5291
0.5000 0.0051 0.0001
Point closest to 4, 3t 2 50
3 is approximately 0.529, 0.280 . t t3 2t3 50 300t 300. 575, the solution is in the interval 4, 5 . 300t t3 2 50 50 0
44. Maximize: C C Let f x fx Since f 4 3t4 12t3
n 1 2
xn 4.5000 4.4864
f xn 12.4375 0.0658
f xn 915.0000 904.3822
f xn f xn 0.0136 0.0001
xn
f xn f xn 4.4864 4.4863
3t4 2t3 6t2
354 and f 5
Approximation: t
4.486 hours
S ection 3.8 46. 170 Let f x fx 0.808x3 17.974x2 71.248x 110.843, 1 x 5 59.157
Newton s Method
433
0.808x3 2.424x2
17.974x2 35.948x
71.248x 71.248.
From the graph, choose x1 n 1 2 3 n 1 2 3 xn 1.0000 1.1345 1.1429 xn 3.5000 3.6878 3.6762 f xn 5.0750 0.2805 0.0006 f xn 4.6725 0.3286 0.0009
1 and x1 f xn 37.7240 33.5849 33.3293 f xn 24.8760 28.3550 28.1450 1.1429 and x
3.5. Apply Newtons Method. f xn f xn 0.1345 0.0084 0.0000 f xn f xn 0.1878 0.0116 0.0000 xn f xn f xn 1.1345 1.1429 1.1429 xn f xn f xn 3.6878 3.6762 3.6762
The zeros occur when x 3676 rev min. 48. True 52. f x Domain: x 4 x2 sin x 2, 2 2
3.6762. These approximately correspond to engine speeds of 1143 rev min and
50. True
2 and x fx 4 1. xn 1.0000 1.1361 1.1416
2, x
y 1
2 are both zeros. x2 cos x 2 x 4 x2 sin x 2
Let x1 n 1 2 3 Zeros: x
f xn 0.2444 0.0090 0.0000 1.142
f xn 1.7962 1.6498 1.6422
f xn f xn 0.1361 0.0055 0.0000
xn
f xn f xn 1.1361 1.1416 1.1416
x 2 1 2
2 3
434
Chapter 3
Applications of Differentiation
Section 3.9
2. f x fx 6 x2 6x 12x
3 2
Differentials
x 12 x3 3 : 2 2 9 2 3 x 2 2 fx Tx 6 x2 3 x 2 9 2 1.9 1.6620 1.65 1.99 1.5151 1.515 2 1.5 1.5 2.01 1.4851 1.485 2.1 1.3605 1.35
Tangent line at 2, y 3 2 y
12 x 8 3 x 2 x
4. f x fx
x fx 2: 2 2 1 2 Tx x x 2 1 2
1.9 1.3784 1.3789
1.99 1.4107 1.4107
2 1.4142 1.4142
2.01 1.4177 1.4177
2.1 1.4491 1.4496
1 2x
Tangent line at 2, y y f2 2 y
f2x 1 x 22 x 22
6. f x fx
csc x csc x cot x f2 csc 2 y f2x 2 2 2 csc 2 2 1.0998 1.0998 2.01 1.1049 1.1048 2.1 1.1585 1.1501
Tangent line at 2, csc 2 : y y
csc 2 cot 2 x csc 2 cot 2 x 1.9 1.99
x fx Tx csc x csc 2 cot 2 x 2 csc 2
1.0567 1.0494
1.0948 1.0947
8. y
fx y
1 fx f 1
2x2, f x x 0.1 2 fx f0 0.1 1, f x x fx f2 1
2
4x, x
0, x
dx
0.1 dy f x dx f0 0.1 0.1 0
1 2, x
20
2
0.02 dx 0.01 dy f x dx
0
10. y
fx y
2x fx f 2.01 2 2.01
2, x
f 2 0.01 22 1 0.02 2 0.01 0.02
S ection 3.9 12. y dy 3x2 2x
3
Differentials
435
14. y dx 2 dx x1 3 dy
9 1 9 2
x2 x2
12
13
2x dx
x 9 x2
dx
16.
y
x 1 2x sec2 x x2 1 x2
1 x 1 dx 2x x x1 dx 2x x
18.
y dy
x sin x x cos x sin x dx
dy
20. y dy
22. (a) f 1.9 1 2 sec2 x tan x x2 1 2 sec2 x 2x x dx
f2
0.1
f2 1
f2 1
0.1 0.1 1.1
(b) f 2.04
f2
0.04
f2 1
f 2 0.04 1 0.04 0.96
2 sec2x x2tan x tan x x2 1 2
dx
24. (a) f 1.9
f2
0.1
f2 1 0 f2 1
f2 0.1
0.1 1
26. (a) g 2.93
g3
0.07
g3 8 3
g3 0.07
0.07 7.79
(b) f 2.04
f2
0.04
f 2 0.04 0 0.04 g3 5 0.07 1 0.07 7.65 30.
(b) g 3.1
g3
0.1
g3 8
g 3 0.1 3 0.1 8.3
28. (a) g 2.93
g3
0.07
g3 8
A db dA A
1 2 bh,
b
36, h
50
dh
1 2b
0.25
(b) g 3.1
g3
0.1
g3 8
g 3 0.1 5 0.1 8.5
dh
1 2
1 2h
db
1 2
dA
36 0.25
50 0.25
10.75 square centimeters
32.
x x
12 inches dx
0.03 inch
34. (a)
C C C
56 centimeters dC
1.2 centimeters
(a) V dV
x3 3x2 dx 3 12
2
2 rr r2 1 C dC 2 33.6 14 56 1 2 C dC 1 4 C2 C 2
0.03
C 2
2
12.96 cubic inches
A dA dA A (b) dA A
12 C 4 33.6
(b) S dS
6x2 12x dx 12 12 0.03
1 56 1.2 2
2
4.32 square inches
0.042857 2dC 0.03 C 1.5%
4.2857%
dC 0.03 C 2
0.015
436
Chapter 3
Applications of Differentiation 12 x 2 x
36.
P dP
500x 500
x2 2x
77x
3000 , x changes from 115 to 120 577 dP 100 P 3x dx 577 3 115 120 2.7% 115 1160
77 dx
Approximate percentage change:
1160 100 43517.50
38. V V
4 3
r3, r dV
100 cm, dr 4 r dr
2
0.2 cm 100
2
40. 8000 cm
3
E R dR dR R dR R
IR E I E dI I2 E I 2 dI EI dI I dI I dI I
4
0.2
42. See Exercise 41. A dA dA A 1 base height 2 45.125 csc2 d cot csc2 d d sin cos 1 9.5cot 2 9.5 45.125 cot
44.
h
50 tan 71.5 1.2479 radians d
0.06 0.06 0.018
h
dh dh x
50
sec2 sec2
50 1.2479 d 50 tan 1.2479 9.9316 d 2.9886 d
50 ft
0.25 sin 26.75 cos 26.75 0.0044 sin 0.4669 cos 0.4669 0.0109 46. Let f x fx
3 3
1.09% in radians 27, dx f x dx 1 272 1
3
x, x fx
3
x 26
x
1 3 3
3
x2
dx 2.9630
27
3
3 26
3
1
1 27
Using a calculator, 48. Let f x fx fx x3, x x x
2.9625 0.01. f x dx x3 33
2
3, dx fx 2.99
3 3
3x2 dx 0.01 27 0.27 26.73
33
Using a calculator: 2.99
26.7309
Review Exercises for Chapter 3 50. Let f x Then f 0.05 tan 0.05 y x dy dx f0 tan 0 f 0 dx sec2 0 0.05 0 1 0.05 . 56. False Let f x y and dy f x dx 1 3 21 3 . 2 fx x, x x 1, and x fx dx f4 3. Then f1 1 tan x, x 0, dx 0.05, f x sec2 x. 52. Propagated error relative error fx x f x, dy y
437
dy , and the percent error y
100.
54. True,
a
Thus, dy >
y in this example.
Review Exercises for Chapter 3
2. (a) f 4 (c)
6 4
f
y
4
3
(b) f
3
f3
4
4
(d) Yes. Since f 2 f2 1 1 and f1 f1 2, the Mean Value says that there exists at least one value c in 2, 1 such that
x 4 6
6 4 2 4 6
fc
x 0
f1 1
f 2
2
21 12
1.
(e) No, lim f x exists because f is continuous at 0, 0 . (f) Yes, f is differentiable at x 2. 6, 6 .
At least six critical numbers on x x2 x
4. f x fx
1
, 0, 2 1
32
6. No. f is not differentiable at x 2x x2 1
12
2.
12 x 2 1 1
32
x2
No critical numbers Left endpoint: 0, 0 Minimum Right endpoint: 2, 2 5 Maximum
8. No; the function is discontinuous at x 1 ,1x4 x 1 x2 14 1 41 1 c2 2 1 4 34 3
0 which is in the interval
2, 1 . 2x, 0 x 4 2 3 2 2 3 2
10.
fx fx fb b fa a fc c
12.
fx fx
x 1 2x
1 4
fb b
fa a fc c
60 40 1 2c 1
438 14.
Chapter 3 fx fx fb b fa a fc c 2x2 4x 21 4 4c 2 1
3
Applications of Differentiation 3x 3 1 0 3 5 5 1
Midpoint of 0, 4
16. g x gx
x 3x
Interval
2
<x< g x >0 Increasing
1
1<x< g x >0 Increasing
1
Sign of g x 1 Conclusion
Critical number: x
18. f x fx
sin x cos x
cos x, 0 x 2 sin x 4 ,x 5 4
Interval Sign of f x Conclusion
0<x<
4
4
<x<
5 4
5 <x<2 4 f x >0 Increasing
Critical numbers: x
f x >0 Increasing
f x <0 Decreasing
20. g x gx
3 x sin 2 2
1 , 0, 4 1 2 ,3 2
Test Interval Sign of g x Conclusion
0<x<1 g x >0 Increasing
2
1
2
<x<3
2
3
2
<x<4
3 x cos 22 2 0 when x 1 1 3
g x <0 Decreasing
g x >0 Increasing
Relative maximum: Relative minimum:
23 , 2 2 , 3 2 2 km 1 2 km 1 2 km
22. (a) y y
A sin
k mt
B cos k mt
k mt B k m sin k mt
(b) Period:
A k m cos 0 when
Frequency: sin k m t cos k m t A tan B k mt A . B
Therefore, sin cos When v y A k mt k mt y 0, A A2 B2 B B A2 B2 A2 B2. A A2 B2 B . A2 B2
R eview Exercises for Chapter 3 24. f x fx fx x 3x2 6x 2
2
439
x
4
x3
12x
16
Test Interval Sign of f x
<x<0 f x <0 Concave downward
0<x< f x >0 Concave upward
12 0 when x 0. 16 1, 3 28.
Conclusion
Point of inflection: 0, 26. ht ht ht h3 t 1 1 1 4t 2 t
32
1 Domain: 1 0t
y 7 6 5 4 3 2 1 1 x 1 2 3 4 5 6 7
t
1 >0 8 Q s x Qs x2 r 2 2Qs r 2Qs r 2x 3x2 5 5x2 x2 2 5x2 x2 2 r 2
3,
5 is a relative minimum.
30.
C dC dx Qs x2 x2 x
x r 2 0
32. (a) S (b)
25
0.1222t3
1.3655t2
0.9052t
4.8429
1 0
14
0 when t 3.7. This is a maximum by the (c) S t First Derivative Test. (d) No, because the t3 coefficient term is negative. 2x 5 x2 3x x2 4 3x x2 2
x
34. lim
x
x
lim
3
0
36. lim
x
x
lim
3 1 4 x2
3
38. g x
40. f x lim 5 2 x2 5 5 lim
x
lim
x
1
3x x2 2
x
lim lim lim
x2 1
3x x 2 3 2 x2
x2 3
Horizontal asymptote: y
x
x
lim
3x x2 2
x
x2 1
3
3x x 2 3 2 x2
x2 3
x
lim
Horizontal asymptotes: y
440
Chapter 3
Applications of Differentiation
2
42. f x
x3
3x2
2x
xx
1x
2
44. g x
3
4 cos x
cos 2x
Relative minima: (0, 0 , 1, 0 , 2, 0 Relative maxima: 1.577, 0.38 , 0.423, 0.38
3
Relative minima: 2 k, 0.29 where k is any integer. Relative maxima:
10
2k
1 , 8.29 where k is any integer.
2 1
4 5 2
0
5 2
46. f x
4x3
x4 ,
2
x3 4
x
30
y
Domain: fx fx 12x 24x f 3 <0
; Range: 4x
3
, 27 x x 0 when x 0 when x 0, 3. 0, 2.
2
25 20
4x 3 12x 2
2
15 10 5 x 1 2 3 5
12x 2
Therefore, 3, 27 is a relative maximum. Points of inflection: 0, 0 , 2, 16 Intercepts: 0, 0 , 4, 0 48. f x x2 4
2
y
Domain: fx fx 4x x2 4 3x2 f 0 <0
, 4 4
; Range: 0, 0 when x 0 when x 0, 2. 23 . 3
(2, 0)
24 20
(0, 16) (2, 0)
8 4 x 3 2 1 1 2 3
Therefore, 0, 16 is a relative maximum. f 2 > 0 Therefore, 2, 0 are relative minima. Points of inflection: 2 3 3, 64 9 Intercepts: 2, 0 , 0, 16 , 2, 0 Symmetry with respect to y-axis 50. f x x 3x , x 4x fx 4x 6 2x f
7 4
2
3
y
Domain: fx
; Range: 2 2
2 2
16,875 256 ,
(2, 0)
4 2 2
(3, 0)
4
x
33x 7x 72x 1x
x
2
3
(0,24)
20 40 60
0 when x x 2
2
2, 4 2,
7 4.
2 2
0 when x
1 2.
( 7 , 16.875) 4 256
>0
7 4, 16,875 256
Therefore,
is a relative minimum. 2, 0 ,
1 2, 625 16
Points of inflection: Intercepts: 2, 0 , 0,
24 , 3, 0
R eview Exercises for Chapter 3 52. f x x 2
13
441
x
1
23
y
Graph of Exercise 39 translated 2 units to the right x replaces by x 1, 0 is a relative maximum. 1,
3
2.
3 2
(1, 0)
3 2 2 3
1
(2, 0)
1 3
4 is a relative minimum.
x
2, 0 is a point of inflection. Intercepts: 2x 1 x2
3 y
(1, 41/3)
1, 0 , 2, 0
54. f x
Domain: fx fx 21
, x1 1 x2
; Range: x
2
1, 1
1.
1
2
(1, 1)
1 x 1 2 3
0 when x 0 when x
2x 3 x2 1 x2 3 f 1 <0
(1, 1)
0,
3.
2 3
Therefore, 1, 1 is a relative maximum. f 1 >0 1, 1 is a relative minimum. 3, 3 2 , 0, 0 , 3, 32
Therefore,
Points of inflection: Intercept: 0, 0
Symmetric with respect to the origin Horizontal asymptote: y x2 1 x4 , 1 1 ; Range: 0, 1 2 2x 1 x1 x1 1 x4 2 x2 0 when x 21 0, 1. 0 when x
4
0
56. f x
Domain: fx fx
x4 2x x2 4x3 42 1x x4
2
2
10x4 1
2x 2x5 2 1 x4 4
x4 4x3
12x4 3x8 1 x4 3
6 3
33
.
f 1 < 0 Therefore, 1, f 0 >0 Therefore, 0, 0 is a relative minimum. Points of inflection: Intercept: 0, 0 Symmetric to the y-axis Horizontal asymptote: y 0
4
1 are relative maxima. 2
1
y
(1, )
1 2
3 4 1 2
1 (1, 2)
(0, 0)
6 3
33
, 0.29 ,
4
6 3
33
x
, 0.40
3 2 1
1 2
1
2
3
442
Chapter 3
Applications of Differentiation 2x 4, 2, 2x 4, x1 1<x3 x>3
58. f x
x2
1 x
x3 x , 0 , 0, 1 x2
1 ; Range: 0 when x 0 when x , 1
3
60. f x
x
1 ,
x
3
Domain: fx fx 2x 2 1 3 2
Domain: 2 . Range: 2, Intercept: 0, 4
y 4
2x3 1 x2 2 x3 1 x3
2 x3 >0
1.
(0, 4)
f
3 2 1 x 1 2 3 4
13 Therefore, 3 , 3 is a relative minimum. 24 Point of inflection: Intercept: 1, 0 0 Vertical asymptote: x
y 3 2
1, 0
(1, 0)
3 2
1
( 12 , 34 )
3 3
x 1 2 3
62. f x
1
y
2 sin x
sin 2 x
1
Domain: fx
1, 1 ; Range: cos 2 x
,0
3 33 3 , 2 2 2 2 cos x 1 cos x 1 0
1 2
1 2
x 1 2 1
1 2
2 cos x
Critical Numbers: x fx 2 sin x
2 3
2 sin 2 x 2 , 3
2
sin x
1
4 cos
x
0 when x
0, 1, 0.420.
By the First Derivative Test:
33 is a relative minimum. 2
23 3 is a relative maximum. , 32 Points of inflection: 0.420, 0.462 , 0.420, 0.462 , 1, 0 , 0, 0 Intercepts: ( 1, 0 , 0, 0 , 1, 0 Symmetric with respect to the origin 64. f x (a) f x xn, n is a positive integer. nxn
1
The function has a relative minimum at 0, 0 when n is even. (b) f x nn 1 xn
2
The function has a point of inflection at 0, 0 when n is odd and n 3.
R eview Exercises for Chapter 3 x2 144 2x 4 3 y2 16 2 3 1 3 x2
443
66. Ellipse: A dA dx
1, y 144
144
x2
12
y
4 x 144 3 144 x2
x2
12
8
( x,
1 3
144 x 2
(
x2 144 x2
x 12 8 12
4 144 2x2 3 144 x2
0 when x
72
6 2. 2 3 144 72 4 2.
The dimensions of the rectangle are 2x
12 2 by y
68. We have points 0, y , x, 0 , and 4, 5 . Thus, m Let f x y 0 L2 5 4 5 4 x2 0 or y x 5x x fx x xx x2 4
3 2
5x x 4
(0, y)
.
L (4, 5) 5 (x, 0)
4 2x 0 0 when x x
2
50
x x 4
x x
4 4
x
2
0
4
100x x4
3
100 25x2 x4
0 or x
2
4
3
100 .
3
L
x
4
x
4
25
100 4 3 100
1002
3
25
12.7 feet
70. Label triangle with vertices 0, 0 , a, 0 , and b, c . The equations of the sides of the triangle are y c b x and y c b a x a . Let x, 0 be a vertex of the inscribed rectangle. The coordinates of the upper left vertex are x, c b x . The y-coordinate of the upper right vertex of the rectangle is c b x. Solving for the x-coordinate x of the rectangles upper right vertex, you get c x b b ax x c b bx b b a a x a x a a a b b x.
(0, 0) (x, 0)
a
y= cx b
(b, c) y= c ( x a) ba c (a a b b x, b x(
(
x, c x b
(
Finally, the lower right vertex is a a b b x, 0 . a b b x x
(a a b b x, 0(
(a, 0)
Width of rectangle: a Height of rectangle: A dA dx A b 2 c x b
see figure a c x b cb b2 a b a 2 c 2 a b b x x c x b a 0 when x 11 ac 22 ac xx bb b . 2 1 Area of triangle 2
Width Height a a ac x bb ab b2
ac b
2ac x b2 1 ac 4
444
Chapter 3
Applications of Differentiation
72. You can form a right triangle with vertices 0, y , 0, 0 , and x, 0 . Choosing a point a, b on the hypotenuse (assuming the triangle is in the first quadrant), the slope is m Let f x y 0 L2 b a x2 b a 0 y x y2 fx 2x a a x
3
a
bx . x
x2 2x 2
bx 2 . ax a bx x 0, a b
2
a
ab x
3 3
2
ab2
3
x
0 when x
ab2. a 2b.
Choosing the nonzero value, we have y L a2 a2
3
a
3
ab2
2 3
b
3
a2b
3
3a4 3b2 b2
3a2 3b4 meters
b2
12
332
74. Using Exercise 73 as a guide we have L1 tan
3
a csc
3
and L2 a2
3 3
b sec . Then dL d b2
3
a csc cot
b sec tan
0 when
a b, sec
a2
3
b2
b1 3 a csc b sec
, csc a2
a1 3 b2 a1
3 312
and a2
3
L
L1
L2
a
b
b2 b1
3
312
a2
3
b2
3 3 2.
This matches the result of Exercise 72. 76. Total cost T dT dv Cost per hour Number of hours v2 500 11 50 7.50 825 v2 110 v 11v 2 11v 50 825 v
41,250 50v 2 25 6 61.2 mph.
0 when v d 2T dv 2 78. f x fx x3
3750
1650 > 0 when v v3 2x 3x2 1 2 1, 0 . f xn 0.1250 0.0029 1, 0 : x
25 6 so this value yields a minimum.
From the graph, you can see that f x has one real zero. f changes sign in n 1 2 xn 0.5000 0.4545
f xn 2.7500 2.6197 0.453.
f xn f xn 0.0455 0.0011
xn
f xn f xn 0.4545 0.4534
On the interval