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 $x=-2,0,2$
$(b)$
$\left[\begin{array}{lllll}
& (-\infty,-2) & (-2,0) & (0,2) & (2,\infty)\\
\text{test point, }t & -4 & -1 & 1 & 4\\
\text{evaluate }f'(t) & 0.75 & -3 & -3 & 0.75\\
\text{sign of }f' & + & - & - & +\\
\text{behavior of }f(x) & \nearrow & \searrow_{....} & {}^{....}\searrow & \nearrow
\end{array}\right]$
(the "$\searrow_{....},\ {}^{....}\searrow$" indicate that $f'$ is undefined at the border )
$f':\quad +++\stackrel{-2}{|}- - - \stackrel{0}{)(}- - - \stackrel{2}{|}+++$
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 $x=2$
