ISMT12_C04_A

CHAPTER 4 APPLICATIONS OF DERIVATIVES 4.1 EXTREME VALUES OF FUNCTIONS

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 because h is continuous on [a ß 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 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. 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 satisfied then extrema are guaranteed, but when the hypothesis is not satisfied, absolute extrema may or may not occur. 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. extrema. When the hypothesis of Theorem 1 is satisfied satisfied then extrema are guaranteed, but when the hypothesis is not satisfied, absolute extrema may or may not occur.

a b ac#ß !b a#ß !b a!ß &b ac$ß !b

a b a!ß #b

7. Local minimum at c"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 c"

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, c1  x  2, is an open interval, we do not meet the conditions conditions of Theorem 1.

Copyright © 2010 Pearson Education, Education, Inc. Publishing as Addison-Wesley. Addison-Wesley.

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, c1  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 c 1. Since f is continuous on the closed

interval on which it is defined, c1 Ÿ x Ÿ 2 1, we do meet the conditions of Theorem 1.

21. f(x) œ

2 3

x c 5 Ê f w (x) œ

f(c2) œ c

19 , 3

2 3

Ê no critical points;

f(3) œ c3 Ê the absolute maximum

is c3 at x œ 3 and the absolute minimum is c

19 3

at

x œ c2

Copyright © 2010 Pearson Education, Inc. Publishing as Addison-Wesley.

Section 4.1 Extreme Values of Functions 22. f(x) œ cx c 4 Ê f w (x) œ c1 Ê no critical points; f(c4) œ 0, f(1) œ c5 Ê the absolute maximum is 0 at x œ c4 and the absolute minimum is c5 at x œ "

23. f(x) œ x# c 1 Ê f w (x) œ 2x Ê a critical point at x œ 0; f(c1) œ 0, f(0) œ c1, f(2) œ 3 Ê the absolute maximum is 3 at x œ 2 and the absolute minimum is c1 at x œ 0

24. f(x) œ % c x# Ê f w (x) œ c2x Ê a critical point at x œ 0; f(c3) œ c5, f(0) œ 4, f(1) œ 3 Ê the absolute maximum is 4 at x œ 0 and the absolute minimum is c5 at x œ c 3

"

25. F(x) œ c

x

œ cxc# Ê Fw (x) œ 2xc$ œ

2 x

, however

x œ 0 is not a critical point since 0 is not in the domain; F(0.5) œ c4, F(2) œ c0.25 Ê the absolute maximum is

c0.25 at x œ 2 and the absolute minimum is c4 at x œ 0.5

"

26. F(x) œ c

œ cxc" Ê Fw (x) œ xc# œ

x

" x

, however

x œ 0 is not a critical point since 0 is not in the domain; F(c2) œ

" #

, F(c1) œ 1 Ê the absolute maximum is 1 at

x œ c1 and the absolute minimum is

27. h(x) œ

È œ

$

x

x"Î$ Ê hw (x) œ

" 3

" #

at x œ c2

x c#Î$ Ê a critical point

at x œ 0; h(c1) œ c1, h(0) œ 0, h(8) œ 2 Ê the absolute maximum is 2 at x œ 8 and the absolute minimum is c1 at x œ c 1


169

170


28. h(x) œ c3x#Î$ Ê hw (x) œ c#x c"Î$ Ê a critical point at x œ 0; h(c1) œ c3, h(0) œ 0, h(1) œ c 3 Ê the absolute maximum is 0 at x œ 0 and the absolute minimum is c3 at x œ 1 and at x œ c 1

29. g(x) œ

È c 4

" #

Ê gw (x) œ

a

x# œ 4 c x #

ac b

c"Î#

x#

4

b

"Î#

(c2x) œ È cx 4cx

Ê critical points at x œ c2 and x œ 0, but not at x œ 2 because 2 is not in the domain; g( c2) 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 c2

È c œ c a& c b a c b œcˆ ‰ œ Ê È œ œÈ ŠcÈ ‹ œ œ cÈ œ c È cÈ œ

30. g(x) œ c

" #

Ê gw (x) and x

x#

x#

5

x

5

x#

5

5 because

0, f(0)

5 at x

c1 #

0

)

1

œ

#

is a critical point, c1 #

is not interior to

ˆ ‰œc ˆ ‰œ ˆ ‰œ c1 #

1, f

1

#

1, f 561

Ê the absolute maximum is 1 at ) œ minimum is c1 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

(c 2x)

5 is not in the

Ê the absolute maximum is 0 at x minimum is

c"Î#

critical points at x œ c

È & c x

0, but not at x

domain; f

"Î#

1

#

" #

and the absolute

c1 #

32. f()) œ tan ) Ê f w ()) œ sec # ) Ê f has no critical points in

ˆ ß‰ c1

1

3

4

. The extreme values therefore occur at the

endpoints: f

ˆ ‰ œ cÈ œ cÈ œ c1

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 c1 3


Section 4.1 Extreme Values of Functions

171

33. g(x) œ csc x Ê gw (x) œ c(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 c

1

3

2, g(0) œ 1, g 1

maximum is 2 at x œ c

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 c t œ # c

ab

Ê f w (t) œ c "# t #

È œ # c a b t#

t#

c"Î#

"Î#

(2t) œ c È t œ c kttk t

Ê a critical point at t œ 0; f( c1) œ 1, f(0) œ 2, f(3) œ c1 Ê the absolute maximum is 2 at t œ 0 and the absolute minimum is c1 at t œ 3

k k È c œ a c b Ê ac b c œ œ

36. f(t) œ t c 5 œ

œ

" #

(t

5)#

5)#

(t

c"Î#

(2(t

(t

5))

5) #

tc5

È (t c 5)

"Î#

f w(t)

tc5

kt c 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(c1) 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(c1) œ c1, f(0) œ 0, f(8) œ 32 Ê the absolute

maximum is 32 at x œ 8 and the absolute minimum is c1 at x œ c1 39. g()) œ )$Î& Ê gw () ) œ maximum is 1 at

)

3 c#Î& ) 5

Ê a critical point at ) œ 0; g(c32) œ c8, g(0) œ 0, g(1) œ 1 Ê the absolute

œ 1 and the absolute minimum is c8 at

)

œ c32

40. h()) œ 3)#Î$ Ê hw() ) œ 2) c"Î$ Ê a critical point at ) œ 0; h(c27) œ 27, h(0) œ 0, h(8) œ 12 Ê the absolute maximum is 27 at

)

œ c 27 and the absolute minimum is 0 at

)

œ0

41. y œ x2 c 6x b 7 Ê y w œ 2x c 6 Ê 2x c 6 œ 0 Ê x œ 3. The critical point is x œ 3.

ab

ab

a b

42. f x œ 6x2 c x3 Ê f w x œ 12x c 3x2 Ê 12x c 3x2 œ 0 Ê 3x 4 c x œ 0 Ê x œ 0 or x œ 4. The critical pointss are x œ 0 and x œ 4.


172


a b a b Ê a b œ < a c b ac b‘ b a c b œ a c b
43. f x œ x 4 c 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 b

2 x

Ê y w œ 2x c x22 œ

a

2x3 c 2 x2

2x3 c 2 x2

Ê

œ ! Ê 2x3 c 2 œ ! Ê x œ 1;

b a b

2x3 c 2 x2

œ undefined Ê x2 œ 0 Ê x œ 0.

The domain of the function is c_, 0 r 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 c b œ Ê œ

46. f x œ

x2 xc2

x

2

2 2 2 ax c 2†b 2x c x2 a1 b œ axx cc24x Ê axx cc24x œ ! Ê x2 c 4x œ ! Ê x œ 0 or x œ 4; axx cc24x œ undefined ax c 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 c 32

È œ a bœÈ

Ê

x

48. g x

x

b a b

2. The domain of the function is c_, 2 r 2, _ , thus x œ 2 is not in the domain, so the only

16 œ y w œ 2x c È x

2x3

2

c 16

Èx

Ê

2 x3

2

c 16

Èx

œ ! Ê 2x3Î2 c 16 œ 0 Ê x œ 4;

2x3

2

ab

2x c x2 Ê g w x œ È1 c x 2 Ê È1 c x 2 œ 0 Ê 1 c x œ 0 Ê x œ 1; È 1 c x 2 œ undefined Ê 2x c x 2x c x 2x c x

49. Minimum value is 1 at x œ 2.

50. To find the exact values, note that yw œ 3x2 c 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 c 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.

ŠcÉ ß

c 16

Èx

2 . 3

Local maximum at

‹ ¸ ac Þ ß Þ b ŠÉ ß % c ‹ ¸ a ß

4b

minimum at

É

4È 6 9

# $

0 816 5 089 ; local

%È ' *

b

0.816 2.911


È c 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 b 2x c 8

a

ba b ac ß b

œ 3x c 4 x b 2 , which is zero when x œ c 2 or x œ %$ . Local maximum at

2 17 ; local minimum at

a ba b œ a ba b

ˆ ßc ‰ % $

52. Note that yw œ 5x# x c 5 x c 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 c" or x œ ".

54. Note that yw œ

È x c 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, c4

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

.


173

174


56. Maximum value is 2 at x œ " ; minimum value is 0 at x œ c" and x œ $.

" at x œ "à # " c # as x œ c".

57. Maximum value is minimum value is

" at x œ 0à # " c # as x œ c 2.

58. Maximum value is minimum value is

ab

a b

59. yw œ x#Î$ " b #$ xc"Î$ x b # œ crit. pt.

c %&

xœ x œ!

&x b % $È x

derivative

extremum

value

!

local max local min

"# "Î$ #& "!

undefined

œ "Þ!

%$0



a b

a

b

60. yw œ x#Î$ #x b #$ xc"Î$ x# c % œ crit. pt. x œ c" x œ! xœ"

derivative

)x c ) $È x

extremum minimum local max minimum

! undefined

!

a b a bÈ % c

61. yw œ x È " c #x b " # %cx % c #x œ cxÈb a% c x b œ È %cx %cx

value

c$ 0

$

x#

crit. pt. x œ c#

derivative undefined

extremum local max

xœc

! !

minimum

È # œ È #

x x œ#

undefined

maximum local min

value

! c# # !

a b È

c 1 b #x $ c x 62. yw œ x# È " # $cx a%xba$ c xb b "#x œ cx b œ _5x #È$ c x #È $ c x crit. pt. xœ0 x œ "# & x œ$

63. yw œ

! ! undefined

œ c#"

, ,

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

#

c", x  ! # c #x, x  ! derivative undefined

!

extremum local min local max

value

$ %


175

176


65. yw œ

œ cc

2x c 2, 2x b 6,

crit. pt. x œ c1 xœ1 xœ3

x1 x1

derivative

extremum maximum local min maximum

! undefined

!

value 5 1 5

ab

a b œ c c 'c b )b a b œ $ c "# b ) a"b œ c"

66. We begin by determining whether f w x is defined at x œ " , where f x œ

ab a" b b œ c" a b œ œ $ c c"# bc)

c "# x c "#

Clearly, f w x œ lim f w w

f x

x

#

Note that c "# x c But # c

#È$ $

crit. pt. x œ c" x ¸ $Þ"&&

if x  ", and lim f w " b h œ c" . 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 c #b

67. (a) No, since f w x œ

# $

x

c"Î$

#

" x #

x

"& , %

xŸ" x"

x,

if x  " , and

. Thus,

"# „ È"# c %a$ba)b #a$b

¸ !Þ)%&  " , so the critical points occur at x œ c" and x œ # b extremum local max local min

x

x#

œ ! when x œ c" , and $ x# c "# x b ) œ ! when x œ

derivative

$

" # x %

#È $ $

œ

"# „ È%) '

œ#„

#È $ $ .

¸ $Þ"&& .

value 4

¸ c$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 œc

68. Note that f x œ

x$ b *x, x$ c * x,

a b œ c$$

x Ÿ c$ or ! Ÿ x  $ . Therefore, f w x œ c$  x  ! or x  $

x$ b * , x$ c * ,

x  c$ or !  x  $ . c$  x  ! or x  $

(a) No, since the left- and right-hand derivatives at x œ !, are c* and *, respectively. (b) No, since the left- and right-hand derivatives at x œ $, are c") and "), respectively. (c) No, since the left- and right-hand derivatives at x œ c$, are ") and c"), 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 c$, at x œ !, and at x œ $; local maxima occur at c


and

.


k k È œ a b x#

x#

69. Yes, since f(x) œ x œ

"Î#

Ê f w (x) œ

" #

ab x#

c"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(cx) œ f(x) Ÿ f(c) œ f(cc) for all cx in the open interval ( cb ß ca) containing cc. That is, f assumesa local maximum at the point cc. 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(cx) œ cg(x) Ÿ cg(c) œ g(cc) for all cx in the open interval (cbß ca) containing cc. That is, g assumes a local maximum at the point cc. 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 c_  x  _. (Any other linear function f(x) œ mx b b with m Á 0 will do as well.)

ab a b œ "'! c "!% b "#

73. (a) V x œ "'!x c &# x# b % x$ Vw x

x

a ba a!ß &b

x# œ % x c # $ x c #!

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# b #bx b 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$ c $x has two critical points at x œ c" and x œ ". The function f x œ x $ c " has one critical point

ab

$

at x œ !Þ The function f x œ x b 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 c "# gt# b v! t b s ! Ê Thus s

76.

dI dt

ds dt

œ c gt b v ! œ ! Ê t œ

Š ‹ œ c Š ‹ b Š ‹b v g

" g vg #

2

v g

v0

œ c#sin t b #cos t, solving

dI dt

s0 œ

v2 2g

v g

È #

ˆ

gt 2

‰

b v0 œ 0 Í t œ 0 or t œ

b 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 c

1

%

2v0 . g 2v0 . g

b 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 Òc3 ß 2 Ó;

a b

local maximum at c5 ß 9


178


78. Maximum value is 4 on the interval Ò5ß 7Ó; minimum value is c4 on the interval Òc2 ß 1Ó.

79. Maximum value is 5 on the interval Ò3ß _Ñ; minimum value is c5 on the interval Ðc_ß c2Ó.

80. Minimum value is 4 on the interval Òc"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 ): <
Section 4.2 The Mean Value Theorem

179

f[x_] =2 b 2x c 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# b 2x c 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 b

4. When f(x) œ

" x

for

È c x

" #

fa1b c fa0b 1c0

1 for 1 Ÿ x Ÿ 3, then

È 7

¸ 1.22 and 1 c3

œ

ab

œ f w c Ê 3 œ 2c b 2 Ê c œ #" .

ab

œ fw c Ê 1œ

ˆ‰ 2 3

ab œ a b Ê œ a b Ê

cc"Î$ Ê c œ

fa1Î2b Ÿ x Ÿ 2, then fa22b c œ fw c Ê 0œ"c c 1Î2

fa3b c fa1b 3c1

5. When f(x) œ x3 c x2 for c1 Ÿ x Ÿ 2, then 1 b È7 3

fa1b c fa0b 1c0

È 2

fw c

fa2b c fac1b 2 c ac1b

#

fw c

" c

8

#7

.

Ê c œ 1.

œ #È "cc1 Ê c œ 3# .

2 œ 3c2 c 2c Ê c œ

1 „ È 7 . 3

¸ c0.549 are both in the interval c 1 Ÿ x Ÿ 2.

ab

x3 x2

ab

ab

c2 Ÿ x Ÿ 0 ga2b c gac2b , then 2 c ac2b œ g w c Ê 3 œ g w c . If c 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 c 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 ( c"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). xÄ!

11. Does not; f is not differentiable at x œ c1 in (c2ß 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 xÄ!

lim f(x) œ lim f(x) Ê c1 b 3 b a œ m b b Ê 5 œ m b b. Since f(x) must also be differentiable at

xÄ1

xÄ1

x œ 1 we have lim f w(x) œ lim f w(x) Ê c2x b 3 xÄ1

xÄ1

k

x=1

œm

k

x=1

Ê 1 œ m. Therefore, a œ 3, m œ 1 and b œ 4.


180


15. (a) i ii iii iv (b) Let r" and r# be zeros of the polynomial P(x) œ x n b a n-1x n-1 b á b a "x b 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 b (n c 1) an-1 xn-2 b á b 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 $) œ 0. To generalize, if f has nb1 zeros in [aß b] and f ÐnÑ is continuous on [aß b], then f ÐnÑ has at least one zero between a and b. 17. Since f ww exists throughout [aß b] the derivative function f w is continuous there. If f w has more than one zero in [a ß b], sa y f w (r") œ f w(r #) œ 0 for r " Á r #, then by Rolle's Theorem there is a c between r " and r # such that f ww(c) œ 0, contrary to f ww  0 throughout [aß b]. Therefore f w has at most one zero in [aß b]. The same argument holds if f ww  0 throughout [aß b]. 18. If f(x) is a cubic polynomial with four or more zeros, then by Rolle's Theorem f w(x) has three or more zeros, f ww(x) has 2 o r more zeros and f www(x) has at least one zero. This is a contradiction since f www(x) is a non-zero constant when f(x) is a cubi c polynomial. 19. With f(c2) œ 11  0 and f(c1) œ c1  0 we conclude from the Intermediate Value Theorem that f(x) œ x % b 3x b 1 has at least one zero between c2 and c1. Then c2  x  c 1 Ê c)  x$  c 1 Ê c 32  4x$  c 4

Ê c29  4x$ b 3  c 1 Ê f w (x)  0 for c 2  x  c 1 Ê f(x) is decreasing on [c#ß c 1] Ê f(x) œ 0 has exactly one solution in the interval ( c#ßc1). 20. f(x) œ x$ b

4 x

b 7 Ê f w(x) œ 3x # c

8 x

 0 on (c_ß 0) Ê f(x) is increasing on (c_ß 0). Also, f(x)  0 if x  c 2 and

f(x)  0 if c2  x  0 Ê f(x) has exactly one zero in (c_ß !).

È b È b c œ È  Ê œ b È b c

21. g(t) œ

t

g(15)

t

15

0

" 4 Ê gw (t) œ È b 2È "tb1  0 Ê g(t) is increasing for t in (!ß _); g(3) œ # t

È c 3

2  0 and

g(t) has exactly one zero in (!ß _)Þ

b 2È "1bt  0 Ê g(t) is increasing for t in (c1ß 1); g(c0.99) œ c2.5 and g(0.99) œ 98.3 Ê g(t) has exactly one zero in (c1ß 1).

22. g(t)

" "ct

1

23. r() ) œ ) b sin#

1

t

ˆ ‰c )

3

3.1 Ê gw (t) œ

8 Ê rw () ) œ 1 b

(c_ß _); r(0) œ c8 and r(8) œ sin # 24. r()) œ 2) c cos# ) b

"

("ct)

È Ê 2

2 3

sin

ˆ ‰ 8 3

ˆ ‰ ˆ ‰œ )

)

cos

3

3

1b

" 3

sin

ˆ ‰ 2) 3

0 on (c_ß _ ) Ê r() ) is increasing on

0 Ê r( )) has exactly one zero in (c_ß _).

rw ()) œ 2 b 2 sin ) cos ) œ 2 b sin 2)  0 on (c_ß _ ) Ê r() ) is increasing on (c_ß _ );

r(c#1) œ c41 c cos (c#1) b

È

È

2 œ c41 c 1 b

0 and r(21) œ 41 c 1 b

2

È  2

0 Ê r()) has exactly one zero in

(c_ß_). 25. r()) œ sec ) c

" )

b 5 Ê rw( )) œ (sec ))(tan )) b

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

.


1

#

; r(0.1) ¸ c994 and

Section 4.2 The Mean Value Theorem

ˆ ‰Ê

26. r()) œ tan ) c cot ) c ) Ê rw ()) œ sec# ) b csc # ) c 1 œ sec # ) b cot # )  0 on !ß r

ˆ ‰œc 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(c1) œ 3 we have C œ 3 Ê f(x) œ 3 for all x. 28. g(x) œ 2x b 5 Ê gw (x) œ 2 œ f w(x) for all x. By Corollary 2, f(x) œ g(x) b C for some constant C. Then f(0) œ g(0) b C Ê 5 œ 5 b C Ê C œ 0 Ê f(x) œ g(x) œ 2x b 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) b C. (a) f(0) œ 0 Ê 0 œ g(0) b C œ 0 b 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) b C œ 1 b C Ê C œ c 1 Ê f(x) œ x# c 1 Ê f(2) œ 3 (c) f(c2) œ 3 Ê 3 œ g(c2) b C Ê 3 œ 4 b C Ê C œ c 1 Ê f(x) œ x# c 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 b œ mx b b where b is a constant. Therefore all functions whose derivatives are constant can be graphed as straight lines y œ mx b b. 31. (a) y œ

x

#

bC

(b) y œ

32. (a) y œ x# b C

" #

(c) y œ 2x c 2 35. (a) y œ c (c) y œ c

" x

bC

" # " #

È b x

"

(b) y œ x b

xc"Î# Ê y œ x"Î# b C Ê y œ #

bC

(c) y œ

(b) y œ x# c x b C

33. (a) yw œ cxc# Ê y œ 34. (a) yw œ

x 3

Èb x

x

bC

(c) y œ x$ b x# c x b C

bC

(c) y œ 5x c

È b

(b) y œ 2

C

x 4

x

C

cos 2t b C cos 2t b 2 sin

(b) y œ 2 sin t

#

t

bC

#

bC (b) yw œ

) "Î#

Ê yœ

2 $Î# ) 3

bC

(c) y œ

2 $Î# ) c 3

37. f(x) œ x# c x b C; 0 œ f(0) œ 0# c 0 b C Ê C œ 0 Ê f(x) œ x # c x " x

b x# b C; 1 œ g(c 1) œ c

39. r()) œ 8) b cot ) b C; 0 œ r

" c1

b (c 1)# b C Ê C œ c 1 Ê g(x) œ c x" b x# c 1

ˆ ‰ œ ˆ ‰b ˆ ‰b 1

8

4

1

4

cot

1

4

C Ê 0 œ 21 b 1 b C Ê C œ c 21 c 1

Ê r()) œ 8 ) b cot ) c 21 c 1 40. r(t) œ sec t c t b C; 0 œ r(0) œ sec (0) c 0 b C Ê C œ c1 Ê r(t) œ sec t c t c 1 41. v œ

ds dt

œ *Þ) t b & Ê s œ %Þ* t# b & t b C; at s œ "! and t œ ! we have C œ "! Ê s œ %Þ* t# b & t b "!

42. v œ

ds dt

œ $#t c # Ê s œ "' t# c # t b C; at s œ % and t œ

43. v œ

ds dt

œ sin 1t Ê s œ c 1" cos 1t b C; at s œ ! and t œ ! we have C œ

44. v œ

ds dt

œ 12 cos

ab ˆ ‰Ê #t 1

bC

C

36. (a) y œ tan ) b C

38. g(x) œ c

" x

s œ sin

ab ˆ ‰b #t 1

C; at s œ " and t œ

" #

1#

we have C œ " Ê s œ "' t# c # t b " " 1

Ê sœ

" c cosa1tb

we have C œ " Ê s œ sin

1

ˆ ‰b " #t 1


tan ) b C

;

182


45. a œ $# Ê v œ $# t b C" ; at v œ #! and t œ ! we have C" œ #! Ê v œ $# t b #! Ê s œ "' t# b #! t b C# ; at s œ & and t œ ! we have C# œ & Ê s œ "' t# b #! t b & 46. a œ 9.8 Ê v œ 9.8t b C" ; at v œ c$ and t œ ! we have C" œ c$ Ê v œ *Þ) t c $ Ê s œ %Þ* t# c $ t b C# ; at s œ ! and t œ ! we have C# œ ! Ê s œ %Þ* t# c $ t

ab

ab

ab

ab

47. a œ c%sin #t Ê v œ #cos #t b C" ; at v œ # and t œ ! we have C" œ ! Ê v œ # cos # t Ê s œ sin # t b C #; at s œ c$

ab

and t œ ! we have C# œ c$ Ê s œ sin # t c $ 48. a œ

* 1

ˆ ‰Ê

cos

$t 1

vœ

$ 1

sin

ˆ ‰b $t 1

C" ; at v œ ! and t œ ! we have C" œ ! Ê v œ

s œ c" and t œ ! we have C# œ ! Ê s œ c cos

ˆ‰ $t

$ 1

sin

ˆ ‰Ê $t 1

s œ c cos

ˆ ‰b $t 1

C# ; at

1

49. If T(t) is the temperature of the thermometer at time t, then T(0) œ c19° C and T(14) œ 100° C. From the Mean Value Theorem there exists a 0  t !  14 such that

T(14) c T(0) 14 c 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) c d(0) #c0 .

Value Theorem says that for some 0  t !  2, dw(t !) œ

d(2) c d(0) #c0 .

The Mean

The 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 b C; at !ß ! we have C œ ! Ê v t œ "Þ' t. When t œ $! , then v $! œ %) m/sec. c

55. The conclusion of the Mean Value Theorem yields

b

bca

œ c c" Ê c#

56. The conclusion of the Mean Value Theorem yields

b ca bca

œ 2c Ê c œ

a

ˆ ‰œ acb ab

abb

#

acb Ê c

ab.

.

57. f w (x) œ [cos x sin (x b 2) b sin x cos (x b 2)] c 2 sin (x b 1) cos (x b 1) œ sin (x b x b 2) c sin 2(x b 1)

œ sin (2x b 2) c sin (2x b 2) œ 0. Therefore, the function has the constant value f(0) œ csin # 1 ¸ c0.7081 which explains why the graph is a horizontal line.

a b a ba b a ba b

58. (a) f x œ x b # x b " x x c " x c # œ x& c & x$ b % x is one possibility.


ISMT12_C04_A

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