Thomas Calculus 13th [Solutions]

Section 4.4 Concavity and Curve Sketching 271
86.
87.
88.
y
2
x 49
2
x 5x
14
Since 7 and 2 are roots of the denominator, the domain is
( , 7) ( 7, 2) (2, ).
y
5 10
; y
2 3
( x 2) ( x 1)
( x 7)
There are no critical points. The function is increasing on its
domain. There are no inflection points. The function is concave
up on ( , 7) ( 7, 2) and concave down on (2, ). The
numerator and denominator share a factor of x 7. Dividing out
this common factor gives y x 7
x 2 ( x 7), which shows that
x 1 is a vertical asymptote. Now dividing numerator and
1 (7/ x)
denominator by x gives y
1 (2/ x) , which shows that y 1 is a
horizontal asymptote. The graph will have a hole at x 7,
y ( 1) 7
( 7) 2
14
9 . 2 .
4
2
y x 1
x
Since 0 is a root of the denominator, the domain is
( , 0) (0, ).
4
2x
2 6
y ; y 2
3 4
x
x
There are critical points at x 1. The function is increasing on
( 1, 0) (1, ) and decreasing on ( , 1) (0,1). There are no
inflection points. The function is concave up on its domain. The
y-axis is a vertical asymptote. Dividing numerator and
2
2 2
denominator by x gives y x 1/ x , which shows that there
1
are no horizontal asymptotes. For large x , the graph is close to
the graph of
y x
2 .
y x 2 4
2 x
Since 0 is a root of the denominator, the domain is
( , 0) (0, ).
2
4 2
; y 4
3
y x
2x
x
There are no critical points at x 2. The function is increasing
on ( , 2) (2, ) and decreasing on ( 2, 0) (0, 2). There
are no inflection points. The function is concave down on ( , 0)
and concave up on (0, ). The y-axis is a vertical asymptote.
Dividing numerator and denominator by x gives y x 4/ x , which
2
shows that the line y x is an asymptote.
2
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