CHAPTER 4 APPLICATIONS OF DERIVATIVES 4.1 EXTREME VALUES OF FUNCTIONS
1. 1. An absolu absolute te minim minimum um at x œ c# , an absolu absolute te maxim maximum um at x œ b. Theore Theorem m 1 guaran guarantee teess the existe existence nce of such such extreme values because h is continuous on [aß b]. 2. An absolu absolute te minim minimum um at x œ b, an absolu absolute te maxim maximum um at x œ c. Theore Theorem m 1 guar guarant antees ees the existe existence nce of such such extreme values because f is continuous on [a ß b]. 3. No absol absolute ute minimum minimum.. An absol absolute ute maximum maximum at x œ c. Since Since the the functi function's on's domain domain is is an open interval, interval, the function does not satisfy the hypotheses of Theorem 1 and need not have absolute extreme values. 4. No absolute extrema. extrema. The function is is neither continuous continuous nor defined defined on a closed closed interval, so it it need not fulfill fulfill the conclusions of Theorem 1. 5. 5. An abso absolu lute te mini minimu mum m at x œ a and and an abso absolu lute te maxi maximu mum m at x œ c. Note Note that that y œ g(x) g(x) is not not cont contin inuo uous us but but still has extrema. When the hypothesis of Theorem 1 is satisfied then extrema are guaranteed, but when when the the hypothesis is not satisfied, absolute extrema may or may not occur. 6. 6. Abso Absolu lute te mini minimu mum m at x œ c and and an abso absolu lute te maxi maximu mum m at x œ a. Note Note that that y œ g(x) g(x) is not not cont contin inuo uous us but but stil stilll has absolute extrema. When the hypothesis of Theorem 1 is satisfied then extrema are guaranteed, but when the hypothesis is not satisfied, absolute extrema may or may not occur.
a b a#ß !b a#ß !b a!ß &b a$ß !b
a b a!ß #b
7. Local minimum at "ß ! , local maximum at "ß ! 8. Minima at
9. Maximum at
10. Local maximum at
and
, maximum at
a b a"ß #b
. Note that there is no minimum since the endpoint #ß ! is excluded from the graph.
a b
, local minimum at #ß ! , maximum at
a b
, minimum at !ß "
11. Graph (c), since this the the only graph that has positive positive slope at c. 12. Graph (b), since this is the only graph that represents a differentiable differentiable function at a and b and has negative slope at c. 13. Graph (d), since this is the only graph representing a funtion that is differentiable differentiable at b but not at a. 14. Graph (a), since this is the only graph that represents a function that is not differentiable differentiable at a or b. 15. f has has an an abso absolut lutee min min at at x œ 0 but but does does not have have an absolu absolute te max. ax. Sinc Sincee the the inte interv rval al on whic which h f is defi define ned, d, 1 x 2, is an open interval, we do not meet the conditions of Theorem 1.
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168
Chapter 4 Applications of Derivatives
16. f has an absolute max at x œ 0 but does not have an absolute min. Since the interval on which f is defined, 1 x 1, is an open interval, we do not meet the conditions of Theorem 1.
17. f has an absolute max at x œ 2 but does not have an absolute min. Since the function is not continuous at x œ 1, we do not meet the conditions of Theorem 1.
18. f has an absolute max at x œ 4 but does not have an absolute min. Since the function is not continuous at x œ 0, we do not meet the conditions of Theorem 1.
19. f has an absolute max at x œ xœ
31 . 2
1
2
and an absolute min at
Since the interval on which f is defined,
0 x 21, is an open interval, we do not meet the conditions of Theorem 1.
20. f has an absolute max at x œ 0 and an absolute min at xœ
1
2
and x œ 1. Since f is continuous on the closed
interval on which it is defined, 1 Ÿ x Ÿ 21, we do meet the conditions of Theorem 1.
21. f(x) œ
2 3
x 5 Ê f w (x) œ
f(2) œ
19 , 3
2 3
Ê no critical points;
f(3) œ 3 Ê the absolute maximum
is 3 at x œ 3 and the absolute minimum is
19 3
at
x œ 2
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Section 4.1 Extreme Values of Functions 22. f(x) œ x 4 Ê f w (x) œ 1 Ê no critical points; f(4) œ 0, f(1) œ 5 Ê the absolute maximum is 0 at x œ 4 and the absolute minimum is 5 at x œ "
23. f(x) œ x# 1 Ê f w (x) œ 2x Ê a critical point at x œ 0; f(1) œ 0, f(0) œ 1, f(2) œ 3 Ê the absolute maximum is 3 at x œ 2 and the absolute minimum is 1 at x œ 0
24. f(x) œ % x# Ê f w (x) œ 2x Ê a critical point at x œ 0; f(3) œ 5, f(0) œ 4, f(1) œ 3 Ê the absolute maximum is 4 at x œ 0 and the absolute minimum is 5 at x œ 3
"
25. F(x) œ
œ x# Ê Fw (x) œ 2x$ œ
x
2 x
, however
x œ 0 is not a critical point since 0 is not in the domain; F(0.5) œ 4, F(2) œ 0.25 Ê the absolute maximum is
0.25 at x œ 2 and the absolute minimum is 4 at x œ 0.5
"
26. F(x) œ
x
œ x" Ê Fw (x) œ x# œ
" x
, however
x œ 0 is not a critical point since 0 is not in the domain; F(2) œ
" #
, F(1) œ 1 Ê the absolute maximum is 1 at
x œ 1 and the absolute minimum is
27. h(x) œ
È œ
$
x
x"Î$ Ê hw(x) œ
" 3
" #
at x œ 2
x #Î$ Ê a critical point
at x œ 0; h(1) œ 1, h(0) œ 0, h(8) œ 2 Ê the absolute maximum is 2 at x œ 8 and the absolute minimum is 1 at x œ 1
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169
170
Chapter 4 Applications of Derivatives
28. h(x) œ 3x#Î$ Ê hw (x) œ #x "Î$ Ê a critical point at x œ 0; h(1) œ 3, h(0) œ 0, h(1) œ 3 Ê the absolute maximum is 0 at x œ 0 and the absolute minimum is 3 at x œ 1 and at x œ 1
29. g(x) œ
È 4
" #
Ê gw (x) œ
a
x# œ 4 x #
a b
"Î#
x#
4
b
"Î#
( 2x) œ È x 4x
Ê critical points at x œ 2 and x œ 0, but not at x œ 2 because 2 is not in the domain; g( 2) œ 0, g(0) œ 2, g(1) œ
È Ê
the absolute maximum is 2 at x œ 0 and the
3
absolute minimum is 0 at x œ 2
È œ a& b a b œˆ ‰ œ Ê È œ œÈ ŠÈ ‹ œ œ È œ È È œ
30. g(x) œ
" #
Ê gw (x) and x
x#
x#
5
x
5
x#
5
5 because
0, f(0)
5 at x
1 #
0
)
1
œ
#
is a critical point, 1 #
is not interior to
ˆ ‰œ ˆ ‰œ ˆ ‰œ 1 #
1, f
1
#
1, f 561
Ê the absolute maximum is 1 at ) œ minimum is 1 at ) œ
5
5 and the absolute
is not a critical point because
the domain; f
È
5
31. f()) œ sin ) Ê f w ()) œ cos ) Ê but ) œ
( 2x)
5 is not in the
Ê the absolute maximum is 0 at x minimum is
"Î#
critical points at x œ
È & x
0, but not at x
domain; f
"Î#
1
#
" #
and the absolute
1 #
32. f()) œ tan ) Ê f w ()) œ sec # ) Ê f has no critical points in
ˆ ߉ 1
1
3
4
. The extreme values therefore occur at the
endpoints: f
ˆ ‰ œ È œ È œ 1
3 and f
3
maximum is 1 at ) minimum is
3 at
1
4
)
ˆ ‰œ 1
4
1 Ê the absolute
and the absolute 1 3
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Section 4.1 Extreme Values of Functions
171
33. g(x) œ csc x Ê gw (x) œ (csc x)(cot x) Ê a critical point at x œ
1
#
;g
ˆ ‰œ
ˆ‰
2 1 È3 , g # œ 1, g
1
3
1
2 absolute maximum is È at x œ 3
absolute minimum is 1 at x œ
3
ˆ ‰œ 21 3
and x œ
2 È 3 Ê the
21 , 3
and the
1
#
34. g(x) œ sec x Ê gw (x) œ (sec x)(tan x) Ê a critical point at
ˆ ‰œ
x œ 0; g
1
3
2, g(0) œ 1, g 1
maximum is 2 at x œ
ˆ ‰œ 1
6
2 È 3 Ê the absolute
and the absolute minimum is 1
3
at x œ 0
kk
35. f(t) œ 2 t œ #
ab
Ê f w (t) œ "# t #
È œ # a b t#
t#
"Î#
"Î#
(2t) œ È t œ kttk t
Ê a critical point at t œ 0; f( 1) œ 1, f(0) œ 2, f(3) œ 1 Ê the absolute maximum is 2 at t œ 0 and the absolute minimum is 1 at t œ 3
k k È œ a b Ê a b œ œ
36. f(t) œ t 5 œ
œ
" #
(t
5)#
5)#
(t
"Î#
(2(t
(t
5))
5) #
t5
È (t 5)
"Î#
f w(t)
t5
kt 5k
Ê a critical point at t œ 5; f(4) œ 1, f(5) œ 0, f(7) œ 2 Ê the absolute maximum is 2 at t œ 7 and the absolute minimum is 0 at t œ 5 37. f(x) œ x%Î$ Ê f w (x) œ
4 3
x "Î$ Ê a critical point at x œ 0; f(1) œ 1, f(0) œ 0, f(8) œ 16 Ê the absolute
maximum is 16 at x œ 8 and the absolute minimum is 0 at x œ 0 38. f(x) œ x&Î$ Ê f w (x) œ
5 3
x #Î$ Ê a critical point at x œ 0; f(1) œ 1, f(0) œ 0, f(8) œ 32 Ê the absolute
maximum is 32 at x œ 8 and the absolute minimum is 1 at x œ 1 39. g()) œ )$Î& Ê gw () ) œ
3 #Î& ) 5
Ê a critical point at ) œ 0; g(32) œ 8, g(0) œ 0, g(1) œ 1 Ê the absolute
maximum is 1 at ) œ 1 and the absolute minimum is 8 at
)
œ 32
40. h()) œ 3)#Î$ Ê hw() ) œ 2) "Î$ Ê a critical point at ) œ 0; h(27) œ 27, h(0) œ 0, h(8) œ 12 Ê the absolute maximum is 27 at
)
œ 27 and the absolute minimum is 0 at
)
œ0
41. y œ x2 6x 7 Ê y w œ 2x 6 Ê 2x 6 œ 0 Ê x œ 3. The critical point is x œ 3.
ab
ab
a b
42. f x œ 6x2 x3 Ê f w x œ 12x 3x2 Ê 12x 3x2 œ 0 Ê 3x 4 x œ 0 Ê x œ 0 or x œ 4. The critical pointss are x œ 0 and x œ 4.
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172
Chapter 4 Applications of Derivatives
a b a b Ê a b œ a b a b‘ a b œ a b a b‘ œ a b a b œ a b a bÊ a b a bœ Ê œ œ œ œ a b œ a b a b Ê a b œ a b † a ba b a ba b † a b œ a ba ba b a b‘ œ a ba ba b Ê a ba ba b œ Ê œ 3
43. f x œ x 4 x 4 4
x
44. g x
2
x
2 x
fw x
1
1
x
2
3 x
x 3 4
4 4
x
3
1
x
2
2
x
1
4
0
x
x
1
x
gw x
1
2
x
1
3
x
x
2
4
1 or x
2 x
4 x
3
3 1
3 x
x
2
3x
4
x
4
4. The critical points are x
2 x
1 1
2
4 x
1 x
x
3
3 x
x
2
4
4x
1 and x
4.
2
1 x
2
0
x
3 or x œ 1 or
x œ 2. The critical points are x œ 1, x œ 2, and x œ 3. 45. y œ x2
2 x
Ê y w œ 2x x22 œ
a
2x3 2 x2
2x3 2 x2
Ê
œ ! Ê 2x3 2 œ ! Ê x œ 1;
b a b
2x3 2 x2
œ undefined Ê x2 œ 0 Ê x œ 0.
The domain of the function is _, 0 0, _ , thus x œ 0 is not in the domain, so the only critical point is x œ 1.
a b Ê a bœ Êa b œ Ê œ
46. f x œ
x2 x2
x
2
2 2 2 ax 2†b 2x x2 a1 b œ axx 24x Ê axx 24x œ ! Ê x2 4x œ ! Ê x œ 0 or x œ 4; axx 24x œ undefined ax 2b2 b2 b2 b2
fw x
2
0
x
a
critical points are x œ 0 and x œ 4
È Ê
47. y œ x2 32
È œ a bœÈ
Ê
x
48. g x
x
b a b
2. The domain of the function is _, 2 2, _ , thus x œ 2 is not in the domain, so the only
16 œ y w œ 2x È x
2x3
2
16
Èx
Ê
2 x3
2
16
Èx
œ ! Ê 2x3Î2 16 œ 0 Ê x œ 4;
2x3
2
ab
2x x2 Ê g w x œ È1 x 2 Ê È1 x 2 œ 0 Ê 1 x œ 0 Ê x œ 1; È 1 x 2 œ undefined Ê 2x x 2x x 2x x
49. Minimum value is 1 at x œ 2.
50. To find the exact values, note that yw œ 3x2 2, which is zero when x œ „ 2 3
œ undefined
0 Ê x œ 0. The critical points are x œ 4 and x œ 0.
2x x2 œ 0 Ê x œ 0 or x œ 2. The critical points are x œ 0, x œ 1, and x œ 2.
ŠÉ ß
16
Èx
2 . 3
Local maximum at
‹ ¸ a Þ ß Þ b ŠÉ ß % ‹ ¸ a ß
4
minimum at
É
4È 6 9
# $
0 816 5 089 ; local
%È ' *
b
0.816 2.911
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È 2x
x2 œ 0
Section 4.1 Extreme Values of Functions 51. To find the exact values, note that that y w œ 3x 2 2x 8
a
ba b a ß b
œ 3x 4 x 2 , which is zero when x œ 2 or x œ $% . Local maximum at
2 17 ; local minimum at
a ba b œ a ba b
ˆ ß ‰ % $
52. Note that yw œ 5x# x 5 x 3 , which is zero at x œ 0, x œ 3, and x
a
%" #(
b
5. Local maximum at 3, 108 ;
local minimum at 5, 0 ; 0, 0 is neither a maximum nor a minimum.
53. Minimum value is 0 when x œ " or x œ ".
54. Note that yw œ
È x 2 È x , which is zero at x œ 4 and is
a b
undefined when x œ 0. Local maximum at 0, 0 ;
a b
absolute minimum at 4, 4
55. The actual graph of the function has asymptotes at x œ „ ", so there are no extrema near these values. (This is an example of grapher failure.) There is a local minimum at
a!ß "b
.
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173
174
Chapter 4 Applications of Derivatives
56. Maximum value is 2 at x œ " ; minimum value is 0 at x œ " and x œ $.
" at x œ "à # " # as x œ ".
57. Maximum value is minimum value is
" at x œ 0à # " # as x œ 2.
58. Maximum value is minimum value is
ab
a b
59. yw œ x#Î$ " #$ x"Î$ x # œ crit. pt.
%&
xœ x œ!
&x % $È x
derivative
extremum
value
!
local max local min
"# "Î$ #& "!
undefined
œ "Þ!
%$0
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Section 4.1 Extreme Values of Functions
a b
a
b
60. yw œ x#Î$ #x #$ x"Î$ x# % œ crit. pt. x œ " x œ! xœ"
derivative
)x ) $È x
extremum minimum local max minimum
! undefined
!
a b a bÈ %
61. yw œ x È " #x " # %x % #x œ xÈ a% x b œ È %x %x
value
$ 0
$
x#
crit. pt. x œ #
derivative undefined
extremum local max
xœ
! !
minimum
È # œ È #
x x œ#
undefined
maximum local min
value
! # # !
a b È
1 #x $ x 62. yw œ x# È " # $x a%xba$ xb "#x œ x œ _5x #È$ x #È $ x crit. pt. xœ0 x œ "# & x œ$
63. yw œ
! ! undefined
œ #"
, ,
crit. pt. x œ"
64. yw œ
derivative
œ
crit. pt. x œ! xœ"
extremum minimum local max minimum
value
extremum minimum
value
! "%% "Î# "#& "&
¸ %Þ%'#
!
x" x "
derivative undefined
#
", x ! # #x, x ! derivative undefined
!
extremum local min local max
value
$ %
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175
176
Chapter 4 Applications of Derivatives
65. yw œ
œ
2x 2, 2x 6,
crit. pt. x œ 1 xœ1 xœ3
x1 x1
derivative
extremum maximum local min maximum
! undefined
!
value 5 1 5
ab
a b œ ' ) a b œ $ "# ) a"b œ "
66. We begin by determining whether f w x is defined at x œ " , where f x œ
ab a" b œ " a b œ œ $ "# )
"# x "#
Clearly, f w x œ lim f w w
f x
x
#
Note that "# x But #
#È$ $
crit. pt. x œ " x ¸ $Þ"&&
if x ", and lim f w " h œ " . Also, f w x hÄ!
. Since f is continuous at x œ ", we have that f w
h
hÄ!
a b
" #
x
" x #
" , #
xŸ"
x
,
x"
! !
a b a #b
67. (a) No, since f w x œ
# $
x
"Î$
#
" x #
x
"& , %
xŸ" x"
x,
if x " , and
. Thus,
"# „ È"# %a$ba)b #a$b
¸ !Þ)%& " , so the critical points occur at x œ " and x œ # extremum local max local min
x
x#
œ ! when x œ " , and $ x# "# x ) œ ! when x œ
derivative
$
" # x %
#È $ $
œ
"# „ È %) '
œ#„
#È $ $ .
¸ $Þ"&& .
value 4
¸ $Þ!(*
, which is undefined at x œ #.
ab
ab
(b) The derivative is defined and nonzero for all x Á #. Also, f # œ ! and f x ! for all x Á #.
ab
(c) No, f x need not have a global maximum because its domain is all real numbers. Any restriction of f to a closed interval of the form Òa, bÓ would have both a maximum value and minimum value on the interval. (d) The answers are the same as (a) and (b) with 2 replaced by a.
a b œ
68. Note that f x œ
x$ *x, x$ * x,
a b œ $$
x Ÿ $ or ! Ÿ x $ . Therefore, f w x œ $ x ! or x $
x$ * , x$ * ,
x $ or ! x $ . $ x ! or x $
(a) No, since the left- and right-hand derivatives at x œ !, are * and *, respectively. (b) No, since the left- and right-hand derivatives at x œ $, are ") and "), respectively. (c) No, since the left- and right-hand derivatives at x œ $, are ") and "), respectively.
ab
(d) The critical points occur when f w x œ ! (at x œ „
È $
ab
) and when f w x is undefined (at x œ ! and x œ „ $). The
Š È$ß 'È$ ‹ ŠÈ$ß ' È$ ‹
minimum value is ! at x œ $, at x œ !, and at x œ $ ; local maxima occur at
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
and
.
Section 4.1 Extreme Values of Functions
k k È œ a b x#
x#
69. Yes, since f(x) œ x œ
"Î#
Ê f w (x) œ
" #
ab x#
"Î#
(2x) œ
x
ax b
177
œ kxxk is not defined at x œ 0. Thus it is
not required that f w be zero at a local extreme point since f w may be undefined there. 70. If f(c) is a local maximum value of f, then f(x) Ÿ f(c) for all x in some open interval (a ß b) containing c. Since f is even, f(x) œ f(x) Ÿ f(c) œ f(c) for all x in the open interval ( b ß a) containing c. That is, f assumesa local maximum at the point c. This is also clear from the graph of f because the graph of an even function is symmetric about the y-axis. 71. If g(c) is a local minimum value of g, then g(x) g(c) for all x in some open interval (a ß b) containing c. Since g is odd, g(x) œ g(x) Ÿ g(c) œ g(c) for all x in the open interval (bß a) containing c. That is, g assumes a local maximum at the point c. This is also clear from the graph of g because the graph of an odd function is symmetric about the origin. 72. If there are no boundary points or critical points the function will have no extreme values in its domain. Such functions do indeed exist, for example f(x) œ x for _ x _. (Any other linear function f(x) œ mx b with m Á 0 will do as
well.)
ab a b œ "'! "!% "#
73. (a) V x œ "'!x &# x# % x$ Vw x
x
a ba a!ß &b
x# œ % x # $ x #!
The only critical point in the interval
b
ab
is at x œ #. The maximum value of V x is 144 at x œ #.
(b) The largest possible volume of the box is 144 cubic units, and it occurs when x œ # units.
ab
74. (a) f w x œ $ax# #bx c is a quadratic, so it can have 0, 1, or 2 zeros, which would be the critical points of f.The
ab
ab
function f x œ x$ $x has two critical points at x œ " and x œ ". The function f x œ x $ " has one critical point
ab
$
at x œ !Þ The function f x œ x x has no critical points.
(b) The function can have either two local extreme values or no extreme values. (If there is only one critical point, the cubic function has no extreme values.) 75. s œ "# gt# v! t s ! Ê
ds dt
œ gt v ! œ ! Ê t œ
Š ‹ œ Š ‹ Š ‹
Thus s
76.
dI dt
v g
" g vg #
2
v g
v0
œ #sin t #cos t, solving
dI dt
s0 œ
v2 2g
v g
È #
ˆ
gt 2
‰
v0 œ 0 Í t œ 0 or t œ
s0 s0 is the maximum height over the interval 0 Ÿ t Ÿ
œ ! Ê tan t œ " Ê t œ
never negative) Ê the peak current is #
ab
. Now s t œ s0 Í t
1
%
2v0 . g 2v0 . g
n1 where n is a nonnegative integer (in this exercise t is
amps.
77. Maximum value is 11 at x œ 5; minimum value is 5 on the interval Ò3 ß 2Ó;
a b
local maximum at 5ß 9
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178
Chapter 4 Applications of Derivatives
78. Maximum value is 4 on the interval Ò5ß 7Ó; minimum value is 4 on the interval Ò2ß 1Ó.
79. Maximum value is 5 on the interval Ò3ß _Ñ; minimum value is 5 on the interval Ð_ß 2Ó.
80. Minimum value is 4 on the interval Ò"ß $Ó
81-86.
Example CAS commands:
Maple:
with(student): f := x -> x^4 - 8*x^2 + 4*x + 2; domain := x=-20/25..64/25; plot( f(x), domain, color=black, title="Section 4.1 #81(a)" ); Df := D(f); plot( Df(x), domain, color=black, title="Section 4.1 # 81(b)" ) StatPt := fsolve( Df(x)=0, domain ) SingPt := NULL; EndPt := op(rhs(domain)); Pts :=evalf([EndPt,StatPt,SingPt]); Values := [seq( f(x), x=Pts )]; Maximum value is 2.7608 and occurs at x=2.56 (right endpoint). %
Minimum value $ is -6.2680 and occurs at x=1.86081 (singular point). Mathematica : (functions may vary) (see section 2.5 re. RealsOnly ): <
Clear[f,x] a = 1; b = 10/3; Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
Section 4.2 The Mean Value Theorem
179
f[x_] =2 2x 3 x2/3
f'[x] Plot[{f[x], f'[x]}, {x, a, b}] NSolve[f'[x]==0, x]
{f[a], f[0], f[x]/.%, f[b]//N In more complicated expressions, NSolve may not yield results. In this case, an approximate solution (say 1.1 here) is observed from the graph and the following command is used: FindRoot[f'[x]==0,{x, 1.1}]
4.2 THE MEAN VALUE THEOREM
1. When f(x) œ x# 2x 1 for 0 Ÿ x Ÿ 1, then 2. When f(x) œ x#Î$ for 0 Ÿ x Ÿ 1, then 3. When f(x) œ x
4. When f(x) œ
" x
for
È x
" #
fa1b fa0b 10
1 for 1 Ÿ x Ÿ 3, then
È 7
¸ 1.22 and 1 3
œ
ab
œ f w c Ê 3 œ 2c 2 Ê c œ #" .
ab
œ fw c Ê 1œ
ˆ‰ 2 3
ab œ a b Ê œ a b Ê
c"Î$ Ê c œ
fa1Î2b Ÿ x Ÿ 2, then fa22b œ fw c Ê 0œ " 1Î2
fa3b fa1b 31
5. When f(x) œ x3 x2 for 1 Ÿ x Ÿ 2, then 1 È7 3
fa1b fa0b 10
È 2
fw c
fa2b fa1b 2 a1b
#
fw c
" c
8
#7
.
Ê c œ 1.
œ #È "c1 Ê c œ 3# .
2 œ 3c2 2c Ê c œ
1 „ È 7 . 3
¸ 0.549 are both in the interval 1 Ÿ x Ÿ 2.
ab
x3 x2
ab
ab
2 Ÿ x Ÿ 0 ga2b ga2b , then 2 a2b œ g w c Ê 3 œ g w c . If 2 Ÿ x 0, then gw (x) œ 3x2 Ê 3 œ g w c ! x Ÿ 2 Ê 3c2 œ 3 Ê c œ „ 1. Only c œ 1 is in the interval. If ! x Ÿ 2, then gw (x) œ 2x Ê 3 œ g w c Ê 2c œ 3 Ê c œ 32 .
6. When g(x) œ
ab
7. Does not; f(x) is not differentiable at x œ 0 in ( "ß 8). 8. Does; f(x) is continuous for every point of [0 ß 1] and differentiable for every point in (0 ß 1). 9. Does; f(x) is continuous for every point of [0 ß 1] and differentiable for every point in (0 ß 1). 10. Does not; f(x) is not continuous at x œ 0 because lim f(x) œ 1 Á 0 œ f(0). x Ä
!
11. Does not; f is not differentiable at x œ 1 in (2ß 0). 12. Does; f(x) is continuous for every point of [0ß 3] and differentiable for every point in (0 ß 3). 13. Since f(x) is not continuous on 0 Ÿ x Ÿ 1, Rolle's Theorem does not apply:
lim f(x) œ lim x œ 1 Á 0 œ f(1).
xÄ1
x
Ä
1
14. Since f(x) must be continuous at x œ 0 and x œ 1 we have lim f(x) œ a œ f(0) Ê a œ 3 and x Ä
!
lim f(x) œ lim f(x) Ê 1 3 a œ m b Ê 5 œ m b. Since f(x) must also be differentiable at
xÄ1
x
Ä
1
x œ 1 we have lim f w(x) œ lim f w(x) Ê 2x 3 xÄ1
x
Ä
1
k
x=1
œm
k
x=1
Ê 1 œ m. Therefore, a œ 3, m œ 1 and b œ 4.
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180
Chapter 4 Applications of Derivatives
15. (a) i ii iii iv (b) Let r" and r# be zeros of the polynomial P(x) œ x n a n-1x n-1 á a "x a !, then P(r ") œ P(r #) œ 0. Since polynomials are everywhere continuous and differentiable, by Rolle's Theorem P w(r) œ 0 for some r between r" and r#, where Pw (x) œ nxn-1 (n 1) an-1 xn-2 á a". 16. With f both differentiable and continuous on [a ß b] and f(r ") œ f(r #) œ f(r $) œ 0 where r ", r # and r $ are in [a ß b], then by Rolle's Theorem there exists a c " between r " and r # such that f w(c") œ 0 and a c # between r # and r $ such that f w (c# ) œ 0. Since f w is both differentiable and continuous on [a ß b], Rolle's Theorem again applies and we have a c$ between c" and c# such that f ww (c$ ) œ 0. To generalize, if f has n1 zeros in [aß b] and f ÐnÑ is continuous on [aß b], then f ÐnÑ has at least one zero between a and b. 17. Since f ww exists throughout [aß b] the derivative function f w is continuous there. If f w has more than one zero in [a ß b], sa y f w (r" ) œ f w(r #) œ 0 for r " Á r #, then by Rolle's Theorem there is a c between r " and r # such that f ww(c) œ 0, contrary to f ww 0 throughout [aß b]. Therefore f w has at most one zero in [aß b]. The same argument holds if f ww 0 throughout [aß b]. 18. If f(x) is a cubic polynomial with four or more zeros, then by Rolle's Theorem f w(x) has three or more zeros, f ww(x) has 2 o r more zeros and f www(x) has at least one zero. This is a contradiction since f www(x) is a non-zero constant when f(x) is a cubi c
polynomial.
19. With f(2) œ 11 0 and f(1) œ 1 0 we conclude from the Intermediate Value Theorem that f(x) œ x % 3x 1 has at least one zero between 2 and 1. Then 2 x 1 Ê ) x$ 1 Ê 32 4x$ 4
Ê 29 4x$ 3 1 Ê f w (x) 0 for 2 x 1 Ê f(x) is decreasing on [#ß 1] Ê f(x) œ 0 has exactly one solution in the interval ( #ß1). 20. f(x) œ x$
4 x
7 Ê f w(x) œ 3x #
8 x
0 on (_ß 0) Ê f(x) is increasing on (_ß 0). Also, f(x) 0 if x 2 and
f(x) 0 if 2 x 0 Ê f(x) has exactly one zero in (_ß !).
È È œ È Ê œ È
21. g(t) œ
t
g(15)
t
15
0
" 4 Ê gw (t) œ È 2È "t1 0 Ê g(t) is increasing for t in (!ß _); g(3) œ # t
È 3
2 0 and
g(t) has exactly one zero in (!ß _)Þ
2È "1t 0 Ê g(t) is increasing for t in (1ß 1); g(0.99) œ 2.5 and g(0.99) œ 98.3 Ê g(t) has exactly one zero in (1ß 1).
22. g(t)
" "t
1
23. r() ) œ ) sin#
1
t
ˆ ‰ )
3
3.1 Ê gw (t) œ
8 Ê rw () ) œ 1
(_ß _); r(0) œ 8 and r(8) œ sin # 24. r()) œ 2) cos# )
"
("t)
È Ê 2
2 3
sin
ˆ ‰ 8 3
ˆ ‰ ˆ ‰œ )
)
cos
3
3
1
" 3
sin
ˆ ‰ 2) 3
0 on (_ß _ ) Ê r() ) is increasing on
0 Ê r( )) has exactly one zero in (_ß _).
rw ()) œ 2 2 sin ) cos ) œ 2 sin 2) 0 on (_ß _ ) Ê r() ) is increasing on (_ß _ );
r(#1) œ 41 cos (#1)
È
È
2 œ 41 1
0 and r(21) œ 41 1
2
È 2
0 Ê r()) has exactly one zero in
(_ß_). 25. r()) œ sec )
" )
5 Ê rw( )) œ (sec ))(tan ))
3
r(1.57) ¸ 1260.5 Ê r()) has exactly one zero in !ß
ˆ ‰Ê
0 on !ß
ˆ ‰ )
1
#
1
#
ˆ ‰
r()) is increasing on !ß
.
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
1
#
; r(0.1) ¸ 994 and
Section 4.2 The Mean Value Theorem
ˆ ‰Ê
26. r()) œ tan ) cot ) ) Ê rw ()) œ sec# ) csc # ) 1 œ sec # ) cot # ) 0 on !ß r
ˆ ‰œ 1
1
4
4
ˆ ‰
0 and r(1.57) ¸ 1254.2 Ê r()) has exactly one zero in !ß
1
#
1
#
181
ˆ ‰
r()) is increasing on 0ß
1
#
.
27. By Corollary 1, f w (x) œ 0 for all x Ê f(x) œ C, where C is a constant. Since f(1) œ 3 we have C œ 3 Ê f(x) œ 3 for all x. 28. g(x) œ 2x 5 Ê gw (x) œ 2 œ f w(x) for all x. By Corollary 2, f(x) œ g(x) C for some constant C. Then f(0) œ g(0) C Ê 5 œ 5 C Ê C œ 0 Ê f(x) œ g(x) œ 2x 5 for all x. 29. g(x) œ x# Ê gw (x) œ 2x œ f w(x) for all x. By Corollary 2, f(x) œ g(x) C.
(a) f(0) œ 0 Ê 0 œ g(0) C œ 0 C Ê C œ 0 Ê f(x) œ x# Ê f(2) œ 4
(b) f(1) œ 0 Ê 0 œ g(1) C œ 1 C Ê C œ 1 Ê f(x) œ x# 1 Ê f(2) œ 3
(c) f(2) œ 3 Ê 3 œ g(2) C Ê 3 œ 4 C Ê C œ 1 Ê f(x) œ x# 1 Ê f(2) œ 3
30. g(x) œ mx Ê gw (x) œ m, a constant. If f w(x) œ m, then by Corollary 2, f(x) œ g(x) b œ mx b where b is a constant. Therefore all functions whose derivatives are constant can be graphed as straight lines y œ mx b. 31. (a) y œ
x
#
C
(b) y œ
32. (a) y œ x# C
" #
(c) y œ 2x 2 35. (a) y œ (c) y œ
" x
C
" # " #
È x
"
(b) y œ x
x"Î# Ê y œ x"Î# C Ê y œ #
C
(c) y œ
(b) y œ x# x C
33. (a) yw œ x# Ê y œ 34. (a) yw œ
x 3
È x
x
C
(c) y œ x$ x# x C
C
(c) y œ 5x
È
(b) y œ 2
C
x 4
x
C
cos 2t C cos 2t 2 sin
(b) y œ 2 sin t
#
t
C
#
C (b) yw œ
) "Î#
Ê yœ
2 $Î# ) 3
C
(c) y œ
2 $Î# ) 3
37. f(x) œ x# x C; 0 œ f(0) œ 0# 0 C Ê C œ 0 Ê f(x) œ x # x " x
x# C; 1 œ g( 1) œ
39. r()) œ 8) cot ) C; 0 œ r
" 1
( 1)# C Ê C œ 1 Ê g(x) œ x" x# 1
ˆ ‰ œ ˆ ‰ ˆ ‰ 1
8
4
1
4
cot
1
4
C Ê 0 œ 21 1 C Ê C œ 21 1
Ê r()) œ 8) cot ) 21 1 40. r(t) œ sec t t C; 0 œ r(0) œ sec (0) 0 C Ê C œ 1 Ê r(t) œ sec t t 1 41. v œ
ds dt
œ *Þ) t & Ê s œ %Þ* t# & t C; at s œ "! and t œ ! we have C œ "! Ê s œ %Þ* t# & t "!
42. v œ
ds dt
œ $#t # Ê s œ "' t# # t C; at s œ % and t œ
43. v œ
ds dt
œ sin 1t Ê s œ 1" cos 1t C; at s œ ! and t œ ! we have C œ
44. v œ
ds dt
œ 12 cos
ab ˆ ‰Ê #t 1
C
C
36. (a) y œ tan ) C
38. g(x) œ
" x
s œ sin
ab ˆ ‰ #t 1
C; at s œ " and t œ
" #
1#
we have C œ " Ê s œ "' t# # t " " 1
Ê sœ
" cosa1tb
we have C œ " Ê s œ sin
1
ˆ ‰ " #t 1
Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.
tan ) C
;
182
Chapter 4 Applications of Derivatives
45. a œ $# Ê v œ $# t C" ; at v œ #! and t œ ! we have C" œ #! Ê v œ $# t #! Ê s œ "' t# #! t C# ; at s œ & and t œ ! we have C# œ & Ê s œ "' t# #! t & 46. a œ 9.8 Ê v œ 9.8t C" ; at v œ $ and t œ ! we have C" œ $ Ê v œ *Þ) t $ Ê s œ %Þ* t# $ t C# ; at s œ ! and t œ ! we have C# œ ! Ê s œ %Þ* t# $ t
ab
ab
ab
ab
47. a œ %sin #t Ê v œ #cos #t C" ; at v œ # and t œ ! we have C" œ ! Ê v œ # cos # t Ê s œ sin # t C #; at s œ $
ab
and t œ ! we have C# œ $ Ê s œ sin # t $ 48. a œ
* 1
ˆ ‰Ê
cos
$t 1
vœ
$ 1
sin
ˆ ‰ $t 1
C" ; at v œ ! and t œ ! we have C" œ ! Ê v œ
ˆ‰
s œ " and t œ ! we have C# œ ! Ê s œ cos
$t
$ 1
sin
ˆ ‰Ê $t 1
s œ cos
ˆ ‰ $t 1
C# ; at
1
49. If T(t) is the temperature of the thermometer at time t, then T(0) œ 19° C and T(14) œ 100° C. From the Mean Value Theorem there exists a 0 t ! 14 such that
T(14) T(0) 14 0
œ 8.5° C/sec œ T w(t !), the rate at which the temperature was
changing at t œ t! as measured by the rising mercury on the thermometer. 50. Because the trucker's average speed was 79.5 mph, by the Mean Value Theorem, the trucker must have been going that speed at least once during the trip. 51. Because its average speed was approximately 7.667 knots, and by the Mean Value Theorem, it must have been going that speed at least once during the trip. 52. The runner's average speed for the marathon was approximately 11.909 mph. Therefore, by the Mean Value Theorem, the runner must have been going that speed at least once during the marathon. Since the initial speed and final speed are both 0 mph and the runner's speed is continuous, by the Intermediate Value Theorem, the runner's speed must have been 11 mph at least twice. 53. Let d(t) represent the distance the automobile traveled in time t. The average speed over 0 Ÿ t Ÿ 2 is d(2) d(0) #0 .
Value Theorem says that for some 0 t! 2, dw(t !) œ
d(2) d(0) #0 .
The Mean
T he value d w(t !) is the speed of the automobile at time t !
(which is read on the speedometer).
ab ab
ab
a b
ab
a b œ È
54. a t œ vw t œ "Þ' Ê v t œ "Þ' t C; at !ß ! we have C œ ! Ê v t œ "Þ' t. When t œ $! , then v $! œ %) m/sec.
55. The conclusion of the Mean Value Theorem yields
b
ba
œ c" Ê c#
56. The conclusion of the Mean Value Theorem yields
b a ba
œ 2c Ê c œ
a
ˆ ‰œ ab ab
ab
#
ab Ê c
ab.
.
57. f w (x) œ [cos x sin (x 2) sin x cos (x 2)] 2 sin (x 1) cos (x 1) œ sin (x x 2) sin 2(x 1)
œ sin (2x 2) sin (2x 2) œ 0. Therefore, the function has the constant value f(0) œ sin # 1 ¸ 0.7081 which explains why the graph is a horizontal line.
a b a ba b a ba b
58. (a) f x œ x # x " x x " x # œ x& & x$ % x is one possibility.
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