Answer
(a)
Increasing on $(-\displaystyle \infty, \frac{3}{2})$
Decreasing on $(\displaystyle \frac{3}{2}, \infty).$
(b)
Absolute maximum at $(\displaystyle \frac{3}{2},\frac{47}{4})$.
No other local maxima.
No absolute minimum.
No local minima.
Work Step by Step
$g$ is defined everywhere.
$ g'(t)=-6t+9,\quad$ defined everywhere.
$g'(t)=0$ for $t=\displaystyle \frac{3}{2}$
Critical point at $t=\displaystyle \frac{3}{2} .$
Calculate $g'$ at test points in the intervals created by the critical points:
$(-\displaystyle \infty, \frac{3}{2}),\qquad g'(0)=9$,
$(\displaystyle \frac{3}{2}, \infty),\qquad g'(2)=-3$,
Evaluate g at the critical point; observe behavior at far ends of the graph
$\displaystyle \lim_{t\rightarrow-\infty}g(t)=-\infty,\quad$
$ g(\displaystyle \frac{3}{2})=\frac{47}{4},\quad$
$\displaystyle \lim_{t\rightarrow\infty}g(t)=-\infty$
Tabular view:
$ \begin{array}{l}
g':\\
\\
\\
g:\\
\end{array} \quad \begin{array}{ccccccccccc}
-\displaystyle \infty& &\displaystyle \frac{3}{2}& &\displaystyle \infty \\
{(} &++ &| &--&) \\ \hline
&\displaystyle \nearrow &\displaystyle \frac{47}{4}&\displaystyle \searrow & \displaystyle \\
(-\infty)& & & & (-\infty) \end{array}$
