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CHAPTER
3
Applicationsof Differentiation
Extrema on an -Interval
Section 3.1
Solutions to Exercises
x2
2. f(x)
1. f(x) = x2 + 4
'
f ( ) = (x2 + 4)(2x) - (x2)(2x)= ~
x
(x2+ 4)2
(x2+ 4)2
1'(0) = 0
= cos E.2
I'(x)
= _!!2 sin~2
1'(0) = 0
1'(2) = 0
32
3. f(x) = x + x2
4.
64
I'(x) = 1 - x3
f(x) = -3x£+1
f'(x) = -3X[~(X + 1)-1/2] + v'X+T(-3)
3
= --(x
2 + 1)-I/2(x + 2(x + 1)]
1'(4) = 0
3
= --(x
2 + 1)-1/2(3x+ 2)
1'( -~) = 0
5. f(x) = (x + 2)2/3
2
I'(x) = -(x
3 + 2)-1/3
6. Using the limit definition of the derivative,
lim f(x)
- f(O) = x-+olim (4 - Ixl)X
x-+o- x - 0
1'(- 2) is undefined.
lim f(x) - f(O)
x-+o. x - 0
4
-
1
= x-+o.
Jim (4 - Ixl)-:-4 = -1
x' - 0
1'(0) does not exist, since the one-sided derivatives are not
equal.
7. f(x) = x2(x- 3) = x3 - 3x2
I'(x)
=
3x2
-
6x
= 3x(x - 2)
Critical numbers: x
= 0,x = 2
- 4) =x4 - 4x2
g '(x) = 4x3 - 8x = 4x(x2 - 2)
Critical numbers: x
= 0,x = :f:,J2
4x
9. g(t) = t.J4=t
10. f(x)
g'(t) = {~(4 - t)-1/2(_1)] + (4 - t)I/2
8 - 3t
= ¥4 - t)-I/2(-t + 2(4 - e)]= 2~4 - t
Critical numbers: t
8. g(x) = x2(x2
= x2
+ 1
'
f ( ) = (x2 + 1)(4) T"(4x)(2x) = 4(1 -,.,x2)
x
(x2 + 1)2
(x2 + 1)2
Critical numbers: x = :f:1
= 4,t = i
203
204
Chapter 3
11. h(x)
Applications
= sin2x +
of Differentiation
cosx,O ~ x < 217
12. f(O) = 2 see 0 +tan 0, 0 ~ 0 < 217
h'(x) = 2sinxcosx - sin x = sinx(2cosx - 1)
17
517
..
CotICal numbers. x = 0 x = - x = 17X= .
,
3'
,
3
f'( 0) = 2 sec 0 tan 0 + sec20
= sec 8(2 tan 0 + sec 0) .
1
sin 0
:+cos 0 . cos 0
[( )
=sec02-
]
= see28(2 sin 0 + 1)
C
..
otic
13. f(x) = x2log2(x2+ 1) = x2ln(x2
In2+ 1)
al
num
be
rs:
0
717
= 6'
0
1117
=""6
14. g(x) = 4x2(3x)
g '(x) = 8x(3x) + 4x23xIn 3 = 4x(Y)(2 + x In 3) = 0
f'(x) = 2xIn(x2+ 1) + x2
2x
In 2
In 2(X2+ 1)
-
x2
2x
- In 2 In(x2 + 1) + x2 + 1
[
Critical number: x
15. f(x)
= 2(3 -
=
]=
x = 0,
0
~
x
=0
2
x = -In 3 = -1.82
Critical numbers: x = 0, - 1.82
0
2x + 5
16. f(x) = ~'
[0,5]
x), [-1,2]
f'(x) = - 2 =>Nocriticalnumbers
Left endpoint: (-1,8)
Maximum
f'(x) =
j => No critical numbers
Right endpoint: (2, 2) Minimum
Left endpoint: (O,~) Minimum
Right endpoint: (5, 5) Maximum
17. f(x) = - x2 + 3x, [0,3]
18. f(x) = x2 + 2x - 4, [-1, 1]
f'(x) = -2x + 3
f'(x) = 2x + 2 = 2(x + 1)
Left endpoint: (0, 0) Minimum
Left endpoint: (-1, - 5) Minimum
Critical number: G,~) Maximum
Rightendpoint: (1, -1) Maximum
Right endpoint: (3, 0) Minimum
19. f(x) = x3 - 3x2, [-1,3]
20. f(x) = x3 - 12x, [0, 4]
f'(x) = 3X2- 6x = 3x(x - 2)
f'(x) = 3x2 - 12 = 3(X2- 4)
Left endpoint: (-1, -4) Minimum
Left endpoint: (0,0)
Critical number: (0,0) Maximum
Critical number: (2, -16) Minimum
Critical number: (2, -4) Minimum
Right endpoint: (4,16) Maximum
Right endpoint: (3, 0) Maximum
Note: x =
- 2 is not in the interval.
Extrema on an Interval
Section 3.1
= 3x2/3 - lx, [-1, 1]
21. f(x)
22. g(x) = Vx, [-1,1]
f'(x) = lx-I/3 - 2 = 2(1 -VxVx)
g'(x) = 3~/3
Leftendpoint:(-1,5) Maximum
Left endpoint:(- I, - 1) Minimum
Critical number: (0,0) Minimum
Criticalnumber: (0,0)
Right endpoint: (1,1)
Right endpoint: (1,1) Maximum
t2
24. g(t) = ~t + 3' [-1,1]
23. h(t) = 4 - It- 41,[I, 6]
From the graph of the function on the interval [1, 6] you
can determine the following,
6t
'
()
g t = (f + 3)2
Left endpoint: (1, 1) Minimum
Critical number: (4,4) Maximum
Left endpoint: (-1,~)
Maximum
Right endpoint: (6,2)
Critical number: (0,0) Minimum
Right endpoint: (1,~) Maximum
25. h(s) =
_s1
_
t
26. h(t) = t- 2' [3, 5]
2' [0, 1]
-2
h'(t) = (t - 2)2
-1
h'(s) = (s - 2)2
Left endpoint:(3,3) Maximum
Left endpoint:(0, -i) Maximum
Rightendpoint: (5,~) Minimum
Right endpoint: (1, -1) Minimum
27. y = e" sin x,
28. y = xln(x + 3), [0,3]
[0, 7T]
y' = e" sin x + e" cos x = e"(sinx + cosx)
Left endpoint: (0,0)
y' = x(x ~ 3) + In(x + 3)
Minimum
Leftendpoint:(0, 0) Minimum
Critical number:
37T
(
-
../2
37T
) (
)
-e3'fT/4 = - 7 46
4' 2
4' .
Rightendpoint:(3, 3 In6) = (3, 5.375) Maximum
,
Maximum
Rightendpoint:(7T,O) Minimum
29. f(x) = COS7TX,[0, iJ
f'(x) = -7Tsin TTX
30.
g(x) = csc x,
[~, ~J
g'(x) = -cscxcotx
Left endpoint: (0, 1) Maximum
'
'
R h d
19 t en point:
'
,
1 .J3
M
6' "'2
lmmum
( )
31. f(x) = tanx
lis continuous on [0, 7T/4] but not on [0, 7T].
lim
x -+'fT/2-
tan x = 00
Left endpoint:(~, 2) Maximum
'
R h
d
.
19 t en point:
2.J3 M'
( )
7T
3' 3
.
lmmum
205