# Question #87d49

Mar 4, 2017

$2$ and $6$

#### Explanation:

$f ' \left(x\right) = \frac{2 x \left(x - 4\right) - \left({x}^{2} - 12\right) \left(1\right)}{x - 4} ^ 2$

$= \frac{{x}^{2} - 8 x + 12}{x - 4} ^ 2$

$f '$ exists for all $x$ in the domain of $f$ and

$f ' \left(x\right) = 0$ at $2$ and at $6$