Thomas Calculus 13th [Solutions]

272 Chapter 4 Applications of Derivatives
89. y 1
2
x 1
Since 1 and 1 are roots of the denominator, the domain is
( , 1) ( 1,1) (1, ).
y
2 x
2 2
; y 6 x 2
2 3
( x 1) ( x 1)
2
90.
There is a critical point at x 0, where the function has a local
maximum. The function is increasing on ( , 1) ( 1, 0) and
decreasing on (0,1) (1, ). The function is concave up on
( , 1) (1, ) and concave down on ( 1,1). The lines x 1
and x 1 are vertical asymptotes. The x-axis is a horizontal
asymptote.
y
2
x
x
2 1
Since 1 and 1 are roots of the denominator, the domain is
( , 1) ( 1,1) (1, ).
y
2 x
2 2
; y 6 x 2
2 3
( x 1) ( x 1)
2
There is a critical point at x 0, where the function has a local
maximum. The function is increasing on ( , 1) ( 1, 0) and
decreasing on (0,1) (1, ). There are no inflection points. The
function is concave up on ( , 1) (1, ) and concave down on
( 1,1). The lines x 1 and x 1 are vertical asymptotes.
2
Dividing numerator and denominator by x gives y 1
2
1 (1/ x )
which shows that the line y 1 is a horizontal asymptote. The x-
intercept is 0 and the y-intercept is 0.
91.
y
2
x
x
2
2
1
Since 1 and 1 are roots of the denominator, the domain is
( , 1) ( 1,1) (1, ).
y
2 2
x
2 2
; y 6 x 2
2 3
( x 1) ( x 1)
There is a critical point at x 0, where the function has a local
maximum. The function is increasing on ( , 1) ( 1, 0) and
decreasing on (0,1) (1, ). There are no inflection points. The
function is concave up on ( , 1) (1, ) and concave down on
( 1,1). The lines x 1 and x 1 are vertical asymptotes.
Dividing numerator and denominator by
2
x gives
2
1 (2/ x )
2
1 (1/ x )
which shows that the line y 1 is a horizontal asymptote. The
x-intercepts are 2 and the y-intercept is 2 .
y
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