Thomas Calculus 13th [Solutions]

274 Chapter 4 Applications of Derivatives
95.
y
2
x x 1
x 1
Since 1 is a root of the denominator, the domain is
( ,1) (1, ).
y
2
x 2x
; y 2
x 1 ( x 1)
2 3
There is a critical point at x 0, where the function has a local
maximum, and a critical point at x 2 where the function has a
local minimum. The function is increasing on ( , 0) (2, )
and decreasing on (0,1) (1, 2). 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
x 1
which shows that
the line y x is an oblique asymptote. (See Section 2.6.) The y-
intercept is 1.
96.
y
2
x x 1
x 1
Since 1 is a root of the denominator, the domain is
( ,1) (1, ).
y
2
2x x ; y 2
x 1 ( x 1)
2 3
There is a critical point at x 0, 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 (0,1) (1, 2) and
decreasing on ( , 0) (2, ). 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
x 1
which shows that the line
y x is an oblique asymptote. (See Section 2.6.) The y-
intercept is 1.
97.
y
3 2
x 3x 3x
1
2
x x 2
3
( x 1)
( x 1)( x 2)
Since 1 and 2 are roots of the denominator, the domain is
( , 2) ( 2,1) (1, ).
y ( x 1)( x 5) 18
2
, x 1; y 3
, x 1
( x 2) ( x 2)
Since 1 is not in the domain, the only critical point is at 5, x
where the function has a local maximum. The function is
increasing on ( , 5) (1, ) and decreasing on
( 5, 2) ( 2,1). There are no inflection points. The function is
concave up on ( 2, 1) (1, ) and concave down on ( , 2).
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