Thomas Calculus 13th [Solutions]

276 Chapter 4 Applications of Derivatives
approximately x 1.223. The function is concave up on
( , 0) (0,1.223) (2, ) and concave down on (1.223, 2).
The lines x 0 (the y-axis) and x 2 are vertical asymptotes.
3
Dividing numerator and denominator by x gives
2 3
(1/ x ) (1/ x )
y which shows that the x-axis is a horizontal
1 (2/ x)
asymptote. The x-intercept is 1.
101. y 8
2
x 4
The domain is ( , ).
y
16x
16(3x
4)
2 2
; y
2 3
( x 4) ( x 4)
2
There is a critical point at x 0, where the function has a local
maximum. The function is increasing on ( , 0) and decreasing
on (0, ). There are inflection points at x 2 / 3 and at
x 2 / 3. The function is concave up on
, 2 / 3 2 / 3, and concave down on
2 / 3, 2 / 3 . Dividing numerator and denominator by
2
gives y 8/ x
2
which shows that the x-axis is a horizontal
1 (4/ x )
asymptote. The y-intercept is 2.
102. y 4x
2
x 4
The domain is ( , ).
y
2 2
4( x 4) 8 x ( x 12)
2 2
; y
2 3
( x 4) ( x 4)
There is a critical point at x 2, where the function has a local
minimum, and at x 2, where the function has a local maximum.
The function is increasing on ( 2, 2) and decreasing on
( , 2) (2, ). There are inflection points at
x 2 3, x 0, and x 2 3. The function is concave up on
2 3, 0 2 3, and concave down on
, 2 3 0, 2 3 . Dividing numerator and denominator by
2
x gives y 4/ x
2
which shows that the x-axis is a horizontal
1 (4/ x )
asymptote. The x-intercept is 0 and the y-intercept is 0.
2
x
103.
Point y y
P
Q 0
R
S 0
T
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