Calculus of a Single Variable, Chapter 3, 3.3, Section 3.3, Problem 35

Given: f(x)=x^2/(x^2-9)
Find the critical value(s) of the function by setting the first derivative equal to zero and solving for the x value(s).
f'(x)=[(x^2-9)(2x)-(x^2)(2x)]/(x^2-9)^2=0
2x^3-18x-2x^3=0
-18x=0
x=0
A critical value is at 0. Critical values also exist where f(x) is not defined. Therefore there are also critical values at 3 and -3.
If f'(x)>0 the function is increasing on the interval.
If f'(x)<0 the function is decreasing on the interval.
Choose a value for x that is less than -3.
f'(-4)=1.469
Choose a value for x that is between -3 and 0.
f'(-2)=1.440
Choose a value for x that is between 0 and 3.
f'(2)=-1.440
Choose a value for x that is greater than 3.
f'(4)=-1.469
The function increases in the interval (-oo,-3).
The function increases in the interval (-3, 0).
The function decreases in the interval (0, 3).
The function decreases in the interval (3, oo).
Because the function changes direction from increasing to decreasing, there is a relative maximum at x=0. The relative maximum occurs at the point (0, 0).