# `f(x) = x^2/(x^2 - 9)` Find the critical numbers, open intervals on which the function is increasing or decreasing, apply first derivative test to identify all relative extrema.

### Textbook Question

Chapter 3, 3.3 - Problem 35 - Calculus of a Single Variable (10th Edition, Ron Larson).
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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).