Answer
$ a.\quad$ critical points at $x==-2,0,2$
$ b.\quad$ f is increasing on $(-\infty, -2)\cup(2, \infty)$, decreasing on $(-2,0)\cup(0,2)$.
$ c.\quad$ local maximum at $x=-2,$
local minimum at $x=2$
Work Step by Step
$(a)$
$f'$ is defined everywhere except at $ x=0\qquad$ ... critical point
$f'(x)=0$ for $ x=\pm 2\qquad$ ... critical points
Critical points at $x=-2,0,2$
$(b)$
$\left[\begin{array}{ccccc}
& (-\infty,-2) & (-2,0) & (0,2) & (2,\infty)\\
\text{test point, }t & -4 & -1 & 1 & 4\\
\text{evaluate }f'(t) & 0.75 & -3 & -3 & 0.75
\end{array}\right]$
$f':\quad ++\stackrel{-2}{|}--\stackrel{0}{)(}--\stackrel{2}{|}++$
f is increasing on $(-\infty, -2)\cup(2, \infty)$, decreasing on $(-2,0)\cup(0,2)$.
$(c)$
From the table, we see that f has:
local maximum at $x=-2,$
local minimum at $x=2$
