Problem 30 from Section 4.4
April 26, 2005
30.
f(x) = 2x2−5x+ 7
x2−9
The vertical asymptotes will occur where the denominator equals 0. This
occurs when x2−9 = 0 or when x= 1 and x=−1. The horizontal asymptotes
will occur where the function has a limiting value in the long run. That is, what
is
lim
x→∞
f(x)
and
lim
x→−∞
f(x)
In both cases, the limit is 2, so there will be a horizontal asymptote at y= 2.
Now we investigate the derivative
f0(x) = 5(x−9)(x−1)
(x2−9)2
The critical numbers occur where the derivative equals 0 or does not exist. In
this case, the critical numbers are
x= 1,9,3 and −3
These 4 numbers split the real number line into 5 intervals
(−∞,−3) ,(−3,1) ,(1,3) ,(3,9) ,(9,∞)
By testing the first derivative on each interval we see that the function is
Increasing on (−∞,−3) ∪(−3,1) ∪(9,∞)
Decreasing on (1,3) ∪(3,9)
Now we investigate the second derivative
f00(x) = −10(x3−15x2+ 27x−45
(x2−9)3
which has second order critical points at
x= 3,−3 and x≈13.2
By testing the second derivative on the appropriate intervals we find the function
is
Concave Up on (−∞,−3) ∪(3,13,2)
Concave Down on (3,3) ∪(13.2,∞)
