Thomas Calculus 13th [Solutions]

Section 4.4 Concavity and Curve Sketching 273
92.
y
2
2
x
x
4
2
Since 2 and 2 are roots of the denominator, the domain is
, 2 2, 2 2, .
2
4
4(3x
2)
y x
2 2
; y
2 3
( x 2) ( x 2)
There is a critical point at x 0, where the function has a local
minimum. The function is increasing on 0, 2 2, and
decreasing on , 2 2, 0 . There are no inflection
points. The function is concave up on 2, 2 and concave
down on , 2 2, . The lines x 2 and x 2
are vertical asymptotes. Dividing numerator and denominator by
2
2
1 (4/ x )
x gives y
2
which shows that the line y 1 is a
1 (2/ x )
horizontal asymptote. The x-intercepts are 2 and the y-
intercept is 2 .
93.
y
x
x
2
1
Since 1 is a root of the denominator, the domain is
( , 1) ( 1, ).
2
y x 2 x 2
2
; y
3
( x 1) ( x 1)
There is a critical point at x 0, where the function has a local
minimum, and a critical point at x 2 where the functions has a
local maximum. The function is increasing on ( , 2) (0, )
and decreasing on ( 2, 1) ( 1, 0). There are no inflection
points. The function is concave up on ( 1, ) and concave down
on ( , 1) . The line x 1 is a vertical asymptote. Dividing
numerator by denominator gives y x 1 1
1
, which shows
x
that the line y x 1 is an oblique asymptote. (See Section 2.6.)
The x-intercept is 0 and the y-intercept is 0.
94.
y
2 4
1
x
x
Since 1 is a root of the denominator, the domain is
( , 1) ( 1, ).
2
y x 2 x 4 6
2
; y
3
( x 1) ( x 1)
There are no critical points. The function is decreasing on its
domain. There are no inflection points. The function is concave up
on ( 1, ) and concave down on ( , 1) . The line x 1 is a
vertical asymptote. Dividing numerator by denominator gives
y 1 x 3 , which shows that the line y 1 x is an oblique
x 1
asymptote. (See Section 2.6.) The x-intercepts are 2 and the y-
intercept is 4.
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