Answer
(a)
Increasing on $(-\infty,-\sqrt{3}) \cup(\sqrt{3},\infty)$
Decreasing on $(-\sqrt{3}, \sqrt{3})$
(b)
No absolute maximum.
Local maximum at $(-\sqrt{3},12\sqrt{3})$.
No absolute minimum.
Local minimum at $(\sqrt{3},-12\sqrt{3})$.
Work Step by Step
$h $ is defined everywhere.
$ h'(x)=6x^{2}-18=6(x^{2}-3),\quad$ defined everywhere.
$h'(x)=0$ for $x=\pm\sqrt{3}$
Critical points at $x=\pm\sqrt{3}$
Calculate $g'$ at test points in the intervals created by the critical points:
$(-\infty, -\sqrt{3}),\qquad h'(-2)=+2$,
$(-\sqrt{3},\sqrt{3}),\qquad h'(0)=-18$,
$(\sqrt{3}, \infty),\qquad h'(2)=-2$,
Evaluate $h$ at the critical point; observe the behavior at the far ends of the graph
$\displaystyle \lim_{x\rightarrow-\infty}h(x)=-\infty,\quad $
$h(-\sqrt{3})=12\sqrt{3},\quad $
$ h(\sqrt{3})=-12\sqrt{3}\quad$
$\displaystyle \lim_{t\rightarrow\infty}h(x)=+\infty$
Tabular view:
$ \begin{array}{l}
h':\\
\\
\\
h:\\
\end{array} \quad \begin{array}{ccccccccccc}
-\infty& &-\sqrt{3} & &\sqrt{3}& &\infty \\
{(} &-- &| &++ &| & -- &) \\ \hline
&\nearrow & 12\sqrt{3} & \searrow & & \nearrow & (+\infty) \\
(-\infty)& & & & -12\sqrt{3} & & \end{array}$
(a)
Increasing on $(-\infty,-\sqrt{3}) \cup(\sqrt{3},\infty)$
Decreasing on $(-\sqrt{3}, \sqrt{3})$
(b)
No absolute maximum.
Local maximum at $(-\sqrt{3},12\sqrt{3})$.
No absolute minimum.
Local minimum at $(\sqrt{3},-12\sqrt{3})$.
