Answer
$ a.\quad$ critical points at $x=-4,-1,2,3$
$ b.\quad$ f is increasing on $(-\infty, -4)\cup(-1,2)\cup(3, \infty)$, decreasing on $(-4,-1)$.
$ c.\quad$ local maxima at $x=-4$ and $x=2$
Work Step by Step
$(a)$
$f'$ is defined everywhere except at $ x=-1,3\qquad$ ... critical points
$f'(x)=0$ for $ x=-4,2\qquad$ ... critical points
Critical points at $x=-4,-1,2,3$
$(b)$
$\left.\begin{array}{cccccccccc}
& (-\infty,-4) & (-4,-1) & (-1,2) & (2,3) & (3,\infty)\\
\text{test point} & -5 & -2 & 0 & 2.5 & 4\\
\text{evaluate }f' & 0.2188 & -1.6 & 2.667 & -1.857 & 3.2\\
\text{sign of }f' & + & - & + & - & +\\
\text{behavior of }f(x) & \nearrow & \searrow_{....} & {}_{...}\nearrow & \searrow_{...} & {}_{...}\nearrow\\
& & & & &
\end{array}\right.$
(the "$\searrow_{....},\ {}_{...}\nearrow$" indicate that $f'$ is undefined at the border.)
f is increasing on $(-\infty, -4)\cup(-1,2)\cup(3, \infty)$, decreasing on $(-4,-1)$.
$(c)$
From the table, we see that f has:
local maxima at $x=-4$ and $x=2$
