Answer
(a)
Increasing on $(0, \displaystyle \frac{4}{3})$
Decreasing on $(-\displaystyle \infty, 0)\cup(\frac{4}{3}, \infty)$
(b)
No absolute maximum.
Local maximum at $(\displaystyle \frac{4}{3},\frac{47}{4})$.
No absolute minimum.
Local minimum at $(0,0)$.
Work Step by Step
$h $ is defined everywhere.
$ h'(x)=-3x^{2}+4x=x(4-3x),\quad$ defined everywhere.
$h'(x)=0$ for $x=0,\displaystyle \ \frac{4}{3}$
Critical points at $x=0,\displaystyle \ \frac{4}{3}$
Calculate $g'$ at test points in the intervals created by the critical points:
$(-\infty, 0),\qquad h'(-1)=-7$,
$(0,\displaystyle \frac{4}{3}),\qquad h'(1)=+1$,
$(\displaystyle \frac{4}{3}, \infty),\qquad h'(2)=-4$,
Evaluate g at the critical point; observe the behavior at the far ends of the graph
$\displaystyle \lim_{x\rightarrow-\infty}h(x)=+\infty,\quad $
$ h(0)=0,\displaystyle \quad h(\frac{4}{3})=\frac{32}{27}\quad$
$\displaystyle \lim_{t\rightarrow\infty}h(x)=-\infty$
Tabular view:
$ \begin{array}{l}
h':\\
\\
\\
h:\\
\end{array} \quad \begin{array}{ccccccccccc}
-\displaystyle \infty& &0& &\displaystyle \frac{4}{3}& &\displaystyle \infty \\
{(} &-- &| &++ &|& -- &) \\ \hline
(+\displaystyle \infty)&\displaystyle \searrow & & \displaystyle \nearrow &\displaystyle \frac{32}{27}& \displaystyle \searrow & \displaystyle \\
& &0 & & & & (-\infty) \end{array}$
(a)
Increasing on $(0, \displaystyle \frac{4}{3})$
Decreasing on $(-\displaystyle \infty, 0)\cup(\frac{4}{3}, \infty)$
(b)
No absolute maximum.
Local maximum at $(\displaystyle \frac{4}{3},\frac{47}{4})$.
No absolute minimum.
Local minimum at $(0,0)$.
