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

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Answer 1 · 2016-02-03T14:01:14

This function has no local extreme values.

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

$f'$ 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.

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