Question

How do you find the local maximum and minimum values of `f(x) = (3x) / (x² - 1)` using both the First and Second Derivative Tests?

Answer 1

This function has no local extreme values.

Explanation:

`f'(x) = ((3)(x^2-1)-(3x)(2x))/(x^2-1)^2 = (-3(x^2+1))/(x^2-1)^2`

`f'` is never `0` and is defined on the whole domain of `f`.

Therefore, `f` has no critical numbers, so `f` has no local extreme values.

Because it has no critical numbers, there is nothing to apply the tests to.