Thomas Calculus 13th [Solutions]

The line x 2 is a vertical asymptote. Dividing numerator by
the denominator gives y x 4 9 which shows that the line
x 2
y x 4 is an oblique asymptote. (See Section 2.6.) The y-
intercept is 1 . The graph has a hole at the point (1, 0).
2
Section 4.4 Concavity and Curve Sketching 275
98.
y
3
x x 2
2
x x
2
( x 1)( x x 2)
( x 1)( x)
Since 1 and 0 are roots of the denominator, the domain is
( , 0) (0,1) (1, ).
2
y x 2 4
2
, x 1; y 2
, x 1
x
x
There is a critical point at x 2 where the function has a local
minimum, and a critical point at x 2 where the function has a
local maximum. The function is increasing on 2, 0 0, 2
and decreasing on , 2 2, . There are no inflection
points. The function is concave up on ( , 0) and concave down
on (0,1) (1, ). The y-axis is a vertical asymptote. Dividing
numerator by denominator gives y x 1 2
x
which shows that
the line y x 1 is an oblique asymptote. (See Section 2.6.)
The graph has a hole at the point (1, 4).
99. y
x
x
2 1
Since 1 and 1 are roots of the denominator, the domain is
( , 1) ( 1,1) (1, ).
y
2 3
x 1 2 2
; y
2 x 6 x
2 3
( x 1) ( x 1)
There are no critical points. The function is decreasing on its
domain. There is an inflection point at x 0. The function is
concave up on ( 1, 0) (1, ) and concave down on
( , 1) (0,1). The lines x 1 and x 1 are vertical
2
asymptotes. Dividing numerator and denominator by x gives
y 1/ x
2
which show that the x-axis is a horizontal asymptote.
1 (1/ x )
The x-intercept is 0 and the y-intercept is 0.
100. y x 1
2
x ( x 2)
Since 0 and 2 are roots of the denominator, the domain is
( , 0) (0, 2) (2, ).
y
2 3 2
3 2
; y
4 3
2x 5x 4 6x 24x 40x
24
x ( x 2) x ( x 2)
There are no critical points. The function is increasing on ( , 0)
and decreasing on (0, 2) (2, ). There is an inflection point at
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