Calculus Solutions 4 1

f x=c f (c)f f c f (c) x

c

b d b e ds

a c r t

e t c e s bc d r

a

f (4)=4 f (7)=0f (4)=4 f (6)=3 f (2)=1 f (5)=2

f (8)=5 f (2)=0f (1)=2 f (4)=4 f (6)=3 f (2)=0 f (5)=2 f (7)=1

1. A function has an absolute minimum at if is the smallest function value on the entiredomain of , whereas has a local minimum at if is the smallest function value when isnear .

2. (a) The Extreme Value Theorem(b) See the Closed Interval Method.

3. Absolute maximum at ; absolute minimum at ; local maxima at and ; local minima at and ;

neither a maximum nor a minimum at , , , and .

4. Absolute maximum at ; absolute minimum at ; local maxima at , , and ; local minima at , , , and ;

neither a maximum nor a minimum at .

5. Absolute maximum value is ; absolute minimum value is ; local maximum valuesare and ; local minimum values are and .

6. Absolute maximum value is ; absolute minimum value is ; local maximum valuesare , , and ; local minimum values are , , and .

7. Absolute minimum at 2, absolute maximum at 3, local minimum at 4

8. Absolute minimum at 1, absolute maximum at 5, local maximum at 2, local minimum at 4

9. Absolute maximum at 5, absolute minimum at 2, local maximum at 3, local minima at 2 and 4

1

Stewart Calculus ET 5e 0534393217;4. Applications of Differentiation; 4.1 Maximum and Minimum Values

f10. has no local maximum or minimum, but 2 and 4 are critical numbers

11. (a)

(b)

(c)

12. (a) Note that a local maximum cannot occur at an endpoint.

2


f

f

(b)

Note: By the Extreme Value Theorem, must not be continuous.

13. (a) Note: By the Extreme Value Theorem, must not be continuous; because if it were, it wouldattain an absolute minimum.

(b)

14. (a)

3


f (x)=8 3x x 1 f (1)=5

f (x)=3 2x x 5 f (5)= 7

f (x)=x2

0<x<2

(b)

15. , . Absolute maximum ; no local maximum. No absolute or localminimum.

16. , . Absolute minimum ; no local minimum. No absolute or localmaximum.

17. , . No absolute or local maximum or minimum value.

4


f (x)=x2

0<x 2 f (2)=4

f (x)=x2

0 x<2 f (0)=0

f (x)=x2

0 x 2 f (2)=4 f (0)=0

f (x)=x2

3 x 2 f ( 3)=9f (0)=0

18. , . Absolute maximum ; no local maximum. No absolute or localminimum.

19. , . Absolute minimum ; no local minimum. No absolute or localmaximum.

20. , . Absolute maximum . Absolute minimum . No local maximumor minimum.

21. , . Absolute maximum . No local maximum. Absolute and localminimum .

5


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

2 x<5 f ( 1)=1

f (t)=1/t 0<t<1

f (t)=1/t 0<t 1 f (1)=1

f ( )=sin 2 2 f32

= f2

=1

f2

= f32

= 1

22. , . No absolute or local maximum. Absolute and local minimum .

23. , . No maximum or minimum.

24. , . Absolute minimum ; no local minimum. No local or absolutemaximum.

25. , . Absolute and local maxima . Absolute and

local minima .

6


f ( )=tan4

<2

f4

= 1

f (x)=1 x f (0)=1

f (x)=ex

f (x)= 1 x2x 4

if 0 x<2if 2 x 3{f (3)=2

f (x)= x2

2 x2

if 1 x<0

if0 x 1{

26. , . Absolute minimum ; no local minimum. No absolute

or local maximum.

27. . Absolute maximum ; no local maximum. No absolute or local minimum.

28. . No absolute or local maximum or minimum value.

29.

Absolute maximum ; no local maximum. No absolute or local minimum.

30.

7


f (0)=2

f (x)=5x2+4x f

/(x)=10x+4 f

/(x)=0 x=

25

25

f (x)=x3+x

2x f

/(x)=3x

2+2x 1 f

/(x)=0 (x+1)(3x 1)=0 x= 1

13

f (x)=x3+3x

224x f

/(x)=3x

2+6x 24=3 x

2+2x 8( )

f/(x)=0 3(x+4)(x 2)=0 x= 4 2

f (x)=x3+x

2+x f

/(x)=3x

2+2x+1 f

/(x)=0 3x

2+2x+1=0 x=

2 4 126

s(t)=3t4+4t

36t

2s

/(t)=12t

3+12t

212t s

/(t)=0 12t t

2+t 1( ) t=0 t

2+t 1=0

t=1 1

24(1)( 1)

2(1)=

1 52

0.618

1.618 01 5

2

f (z)=z+1

z2+z+1

f/(z)=

z2+z+1( )1 (z+1)(2z+1)

z2+z+1( ) 2

=z2

2z

z2+z+1( ) 2

=0 z(z+2)=0 z=0 2

z2+z+1 0 <0

g(x)= 2x+3 = 2x+3(2x+3)

if 2x+3 0if 2x+3<0{ g

/(x)=

2

2

if x>32

if x<32

{ g/(x) 0

g/(x)

Absolute and local maximum .No absolute or local minimum.

31. . , so is the only critical number.

32. . , . These are the only

critical numbers.

33. .

, . These are the only critical numbers.

34. . . Neither of these is

a real number. Thus, there are no critical numbers.

35. . or . Using the

quadratic formula to solve the latter equation gives us ,

. The three critical numbers are , .

36. , are the

critical numbers. (Note that since the discriminant .)

37. is never , but

does not exist for 8


x=32

32

g(x)=x1/3

x2/3

g/(x)=

13

x2/3

+23

x5/3

=13

x5/3

(x+2)=x+2

3x5/3

g/( 2)=0 g

/(0) 0 g 2

g(t)=5t2/3

+t5/3

g/(t)=

103

t1/3

+53

t2/3

g/(0) t=0

g/(t)=

53

t1/3

(2+t)=0 t= 2 t= 2

g(t)= t (1 t)=t1/2

t3/2

g/(t)=

1

2 t

32

t g/(0) t=0

0=g/(t)=

1 3t

2 tt=

13

t=13

F(x)=x4/5

(x 4)2

F/(x) =x

4/52(x 4)+(x 4)

2 45

x1/5

=15

x1/5

(x 4) 5 x 2+(x 4) 4

=(x 4)(14x 16)

5x1/5

=2(x 4)(7x 8)

5x1/5

=0 x=487

F/(0)

087

4

G(x)=3

x2

x G/(x)=

13

x2

x( ) 2/3(2x 1) G

/(x) x

2x=0

x=0 1 G/(x)=0 2x 1=0 x=

12

x=012

1

f ( )=2cos +sin2

f/( )= 2sin +2sin cos f

/( )=0 2sin (cos 1)=0 sin =0

cos =1 =n n =2n =n =2n=n

g( )=4 tan g/( )=4 sec

2g

/( )=0 sec

2=4 sec = 2 cos =

12

=3

+2n

, so is the only critical number.

38. .

and does not exist, but is not in the domain of , so the only critical number is .

39. . does not exist, so is a critical number.

, so is also a critical number.

40. . does not exist, so is a critical number.

, so is also a critical number.

41.

when , ; and does not exist.

Critical numbers are , , .

42. . does not exist when , that is, when

or . . So the critical numbers are , , .

43. . or ( an integer) or . The solutions include the solutions , so the

critical numbers are .

44. . ,

9


53

+2n23

+2n43

+2n

g/( ) g

f (x)=xln x f/(x)=x(1/x)+(ln x) 1=ln x+1 f

/(x)=0 ln x= 1 x=e

1=1/e

x=1/e

f (x)=xe2x

f/(x)=x(2e

2x)+e

2x=e

2x(2x+1) e

2x0 f

/(x)=0

2x+1=0 x=12

12

f (x)=3x2

12x+5 0,3 f/(x)=6x 12=0 x=2

f (0)=5 f (2)= 7 f (3)= 4 f (0)=5 f (2)= 7

f (x)=x3

3x+1 0,3 f/(x)=3x

23=0 x= 1 1 0,3 f (0)=1 f (1)= 1

f (3)=19 f (3)=19 f (1)= 1

f (x)=2x3

3x2

12x+1 [ 2,3] f/(x)=6x

26x 12=6 x

2x 2( )=6 x 2( ) x+1( )=0 x=2, 1

f ( 2)= 3 f ( 1)=8 f (2)= 19 f (3)= 8 f ( 1)=8f (2)= 19

f (x)=x3

6x2+9x+2 [ 1,4] f

/(x)=3x

212x+9=3 x

24x+3( )=3(x 1)(x 3)=0 x=1,3 f ( 1)= 14

f (1)=6 f (3)=2 f (4)=6 f (1)= f (4)=6 f ( 1)= 14

f (x)=x4

2x2+3 [ 2,3]. f

/(x)=4x

34x=4x x

21( )=4x(x+1)(x 1)=0 x= 1 0 1 f ( 2)=11

f ( 1)=2 f (0)=3 f (1)=2 f (3)=66 f (3)=66 f ( 1)=2

f (x)= x2

1( ) 3[ 1,2] f

/(x)=3 x

21( ) 2

2x( )=6x x+1( )2

x 1( )2=0 x= 1,0,1 f ( 1)=0 f (0)= 1

f (2)=27 f (2)=27 f (0)= 1

f (x)=x

x2+1

0,2 f/(x)=

x2+1( ) x(2x)

x2+1( ) 2

=1 x

2

x2+1( ) 2

=0 x= 1 1 0,2 f (0)=0

f (1)=12

f (2)=25

f (1)=12

f (0)=0

, , and are critical numbers.

Note: The values of that make undefined are not in the domain of .

45. . . Therefore, the onlycritical number is .

46. . Since is never , we have only when

. So is the only critical number.

47. , . . Applying the Closed Interval Method, we findthat , , and . So is the absolute maximum value and is theabsolute minimum value.

48. , . , but is not in . , , and . So is the absolute maximum value and is the absolute minimum value.

49. , . . , , , and . So is the absolute maximum value and is the absolute minimum value.

50. , . . , , , and . So is the absolute maximum value and is the

absolute minimum value.

51. , , , . , , , , . So is the absolute maximum value and is the

absolute minimum value.

52. , . . , , and . So is the absolute maximum value and is the absolute minimumvalue.

53. , . , but is not in .

, , . So is the absolute maximum value and is the absolute

10


f (x)=x

24

x2+4

[ 4,4] f/(x)=

x2+4( )(2x) x

24( )(2x)

x2+4( ) 2

=16x

x2+4( ) 2

=0 x=0 f ( 4)=1220

=35

f (0)= 1 f ( 4)=35

f (0)= 1

f (t)=t 4 t2

1,2

f/(t)=t

12

4 t2( ) 1/2

2t( )+ 4 t2( ) 1/2

1=t2

4 t2

+ 4 t2

=t2+ 4 t

2( )4 t

2=

4 2t2

4 t2

f/(t)=0

4 2t2=0 t

2=2 t= 2 t= 2 1,2 f

/(t)

4 t2=0 t= 2 2 f ( 1)= 3 f 2( )=2 f (2)=0

f 2( )=2 f ( 1)= 3

f (t)=3

t 8 t( ) [0,8] f (t)=8t1/3

t4/3

f/(t)=

83

t2/3 4

3t1/3

=43

t2/3

(2 t)=4(2 t)

33

t2

f/(t)=0 t=2

f/(t) t=0 f (0)=0 f (2)=6

32 7.56 f (8)=0

f (2)=63

2 f (0)= f (8)=0

f (x)=sin x+cos x 0,3

f/(x)=cos x sin x=0 sin x=cos x

sin xcos x

=1 tan x=1

x=4

f (0)=1 f4

= 2 1.41 f3

=3 +12

1.37 f4

= 2

f (0)=1

f (x)=x 2cos x , f/(x)=1+2sin x=0 sin x=

12

x=56 6

f ( )=2 1.14

f56

= 356

0.886 f6

=6

3 2.26 f ( )= +2 5.14 f ( )= +2

f6

=6

3

f (x)=xex

0,2 f/(x)=x( e

x)+e

x=e

x(1 x)=0 x=1

f (0)=0 f (1)=e1=1/e 0.37 f (2)=2/e

20.27 f (1)=1/e

f (0)=0

minimum value.

54. , . . and

. So is the absolute maximum value and is the absolute minimum value.

55. , .

.

, but is not in the given interval, . does not exist if

, but is not in the given interval. , , and . So is the absolute maximum value and is the absolute minimum value.

56. , . .

. does not exist if . , , and .

So is the absolute maximum value and is the absolute minimum value.

57. , .

. , , . So is the absolute

maximum value and is the absolute minimum value.

58. , . , . ,

, , . So is the

absolute maximum value and is the absolute minimum value.

59. , . .

, , . So is the absolute maximum value and is the absolute minimum value.

11


f (x)=ln x

x1,3 f

/(x)=

x(1/x) ln x

x2

=1 ln x

x2

=0 1 ln x=0 ln x=1 x=e f (1)=0/1=0

f (e)=1/e 0.368 f (3)=(ln 3)/3 0.366 f (e)=1/e f (1)=0

f (x)=x 3ln x 1,4 f/(x)=1

3x

=x 3x

=0 x=3 f/

x=0 0

f f (1)=1 f (3)=3 3ln 3 0.296 f (4)=4 3ln 4 0.159 f (1)=1f (3)=3 3ln 3 0.296

f (x)=ex

e2x

0,1 f/(x)=e

x( 1) e

2x( 2)=

2

e2x

1

ex

=2 e

x

e2x

=0 ex=2 x=ln 2 0.69 f (0)=0

f (ln 2)=eln 2

e2ln 2

= eln 2( ) 1

eln 2( ) 2

=21

22=

12

14

=14

f (1)=e1

e2

0.233 f (ln 2)=14

f (0)=0

f (x)=xa(1 x)

b0 x 1 a>0 b>0

f/(x) =x

ab(1 x)

b 1( 1)+(1 x)

bax

a 1=x

a 1(1 x)

b 1x b( 1)+(1 x) a

=xa 1

(1 x)b 1

(a ax bx)

f (0)= f (1)=0 f (0,1) f/(x)=0

x=a

a+b.

fa

a+b=

aa+b

a1

aa+b

b=

aa

(a+b)a

a+b aa+b

b=

aa

(a+b)a

bb

(a+b)b

=a

ab

b

a+b( )a + b

fa

a+b=

aab

b

a+b( )a + b

f/(x)=0 x=0.0 2.0 f

/(x) x= 0.7 1.0

2.7 f 0.7 0.0 1.0 2.0 2.7

60. , . . ,

, . So is the absolute maximum value and isthe absolute minimum value.

61. , . . does not exist for , but is not in the

domain of . , , . So is the absolutemaximum value and is the absolute minimum value.

62. , . .

, , . So

is the absolute maximum value and is the absolute minimum value.

63. , , , .

At the endpoints, we have [ the minimum value of ]. In the interval ,

.

So is the absolute maximum value.

64.

We see that at about and , and that does not exist at about , , and , so the critical numbers of are about , , , , and .

12


f ( 1.63)=9.71f (1.63)= 7.71

0,1( ) y=x3

8x

f (x)=x3

8x+1 f/(x)=3x

28 f

/(x)=0 x=

83

f83 =

83

3

883

+1=83

83

883

+1

=163

83

+1=132 6

9163

83

+1=1+32 6

9

f ( 0.58)=1.47f ( 1)= f (0)=1.00

f (x)=ex

3x

f/(x)=e

x3

x3x

21( ) f

/(x)=0 1,0 x= 1/3 f ( 1)= f (0)=1

f 1/3( )=e3 /9+ 3 /3

=e2 3 /9

f (0.75)=0.32f (0)= f (1)=0

65. (a) From the graph, it appears that the absolute maximum value is about , and the absoluteminimum value is about . These values make sense because the graph is symmetric

about the point . ( is symmetric about the origin.)

(b) . So .

or

(From the graph, we see that the extreme values do not occur at the endpoints.)

66. (a) From the graph, it appears that the absolute maximum value is about , and the absoluteminimum value is about ; that is, at both endpoints.

(b) . So on . (minima)

and (maximum).

67. (a) From the graph, it appears that the absolute maximum value is about , and the absoluteminimum value is ; that is, at both endpoints.

13


f (x)=x x x2

f/(x)=x

1 2x

2 x x2

+ x x2

=x 2x

2( )+ 2x 2x2( )

2 x x2

=3x 4x

2

2 x x2

f/(x)=0

3x 4x2=0 x(3 4x)=0 x=0

34

f (0)= f (1)=0

f34

=34

34

34

2=

3 316

f (5.76)=0.58f (3.67)= 0.58

f (x)=cos x

2+sin xf

/(x)=

(2+sin x)( sin x) (cos x)(cos x)

2+sin x( )2

=1 2sin x

2+sin x( )2

f/(x)=0 sin x=

12

x=76

116

f76

=3 /2

3/2=

1

3

f11

6=

3 /23/2

=1

3

=mass

volume=

1000V (T )

/3

VddT

= 1000V2 dV

dTV 0 V

V (T )=999.87 0.06426T +0.0085043T2

0.0000679T3

V/(T )= 0.06426+0.0170086T 0.0002037T

20

T T =0.0170086 0.0170086

24 0.0002037 0.06426

2( 0.0002037)3.9665 79.5318

0 T 30

3.9665 (0)1000

999.871.00013 (30)

10001003.7628

0.99625

(b) . So

or . ,

and .

68. (a) From the graph, it appears that the absolute maximum value is about , and the absoluteminimum value is about .

(b) .

So or . Now ,

and .

69. The density is defined as (in g cm ). But a critical point of will also be a

critical point of since [ and is never ], and is easier to differentiate than

.

. Setting this equal to and using the quadratic formulato

find , we get C or C.

Since we are only interested in the region C C, we check the density at the endpoints

and at C: ; ;

14


(3.9665)1000

999.74471.000255 3.9665

F=W

sin +cosdFd

=( sin +cos )(0) W ( cos sin )

sin +cos( )2

=W ( cos sin )

sin +cos( )2

dFd

=0 cos sin =0 =sincos

=tan tan F

F=(tan )W

(tan )sin +cos=

W tan

sin2

cos+cos

=W tan cos

sin2

+cos2

=W sin

1=W sin

tan = sin =2+1

F=2+1

W

F F(0)= W F2

=W2+1

1

2+1

2+1

W

F(0) F2 2

+1

W

F( ) tan =

I(t)=0.00009045t5+0.001438t

40.06561t

3+0.4598t

20.6270t+99.33 0,10

I/(t)=0.00045225t

4+0.005752t

30.19683t

2+0.9196t 0.6270 I

/t

I I/(t)=0

I/(t)=0 t 29.7186 0.8231 5.1309 11.0459

0,10 II(0.8231) 99.09 I(5.1309) 100.67 II(0)=99.33 I(10) 96.86

t 5.1309 t=10

. So water has its maximum density at about C.

70. .

So . Substituting for in gives us

.

If , then (see the figure), so . We compare this with the

value of at the endpoints: and . Now because and

, we have that

is less than or equal to each of and . Hence, is the absolute minimum

value of , and it occurs when .

71. We apply the Closed Interval Method to the continuous function

on . Its derivative is

. Since exists for all , the only

critical numbers of occur when . We use a root finder on a computer algebra system (or a

graphing device) to find that when , , , or , but only thesecond and third roots lie in the interval . The values of at these critical numbers are

and . The values of at the endpoints of the interval are and . Comparing these four numbers, we see that food was most expensive

at (corresponding roughly to August, 1989) and cheapest at (midyear 1994).

15


v(t)=0.00146t3

0.11553t2+24.98169t 21.26872

a(t)=v/(t)=0.00438t

20.23106t+24.98169 a

/(t)=0.00876t 0.23106 a

/(t)=0

t1=

0.231060.00876

26.4 a(0) 24.98 a(t1) 21.93 a(125) 64.54

64.5 /2

21.93 /2

v(r)=k(r0

r)r2=kr

0r

2kr

3v

/(r)=2kr

0r 3kr

2v

/(r)=0 kr(2r

03r)=0 r=0

23

r0

0

v12

r0

23

r0

r0

v12

r0

=18

kr3

0

v23

r0

=4

27kr

3

0v(r

0)=0

427

>18

v r=23

r0

v4

27kr

3

0

g(x)=2+(x 5)3

g/(x)=3(x 5)

2g

/(5)=0 5 g(5)=2 g

>2 <2 5 g5

f (x)=x101

+x51

+x+1 f/(x)=101x

100+51x

50+1 1 x f

/(x)=0

f (x) f (x)

f c f (x) f (c) x cg(x)= f (x) f (c)=g(c) x c g(x) c

f c g(x)= f (x) c g/(c)=0

72. (a) The equation of the graph in the figure is

.

(b) .

. , , and . The maximum acceleration is

about ft s and the minimum acceleration is about ft s .

73. (a) . or (but

is not in the interval). Evaluating at , , and , we get ,

, and . Since , attains its maximum value at . This

supports the statement in the text.

(b) From part (a), the maximum value of is .

(c)

74. , so is a critical number. But and takes onvalues and values in any open interval containing , so does not have a local maximum orminimum at .

75. for all , so has no solution. Thus, has no critical number, so can have no local maximum or minimum.

76. Suppose that has a minimum value at , so for all near . Then for all near , so has a maximum value at .

77. If has a local minimum at , then has a local maximum at , so by the case 16


f/(c)= g

/(c)=0

f (x)=ax3+bx

2+cx+d a 0 f

/(x)=3ax

2+2bx+c 2 1

0 f (x) 2 1 0

f (x)=x3

3x

f/(x)=3x

23 x= 1,1 f

/(x)=3x,so

f (x)=x3+3x

f/(x)=3x

2+3

of Fermat’s Theorem proved in the text. Thus, .

78.(a) , . So is a quadratic and hence has either , ,

or real roots, so has either , or critical numbers.

Case (i) (2 critical numbers): Case (ii) (1 criticalnumber): Case (iii) (no critical number):

, so are critical numbers.

f(x)=x

x=0is the only critical number.

,so there are no real roots.

(b) Since there are at most two critical numbers, it can have at most two local extreme values andby (i) this can occur. By (iii) it can have no local extreme value. However, if there is only onecritical number, then there is no local extreme value.

17


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Calculus Solutions 4 1