Answer
(a)
Increasing on $(0, \displaystyle \frac{1}{2})$
Decreasing on $(-\displaystyle \infty, 0)\cup(\frac{1}{2}, \infty)$
(b)
No absolute maximum.
Local maximum at $(\displaystyle \frac{1}{2},\frac{1}{4})$.
No absolute minimum.
Local minimum at $(0,0)$.
Work Step by Step
$f$ is defined everywhere.
$ f'(\theta)=6\theta-12\theta^{2}=6\theta(1-2\theta),\quad$ defined everywhere.
$f'(\theta)=0$ for $\theta=0$ and $\displaystyle \frac{1}{2}$
Critical points: $\theta=0$ and $\displaystyle \frac{1}{2}$.
$f(0)=0,\displaystyle \qquad f(\frac{1}{2})=\frac{1}{4}.$
Using testpoints in the intervals between critical points,
$f'(-1) \lt 0$
$f'(1/4) \gt 0$
$f'(1) \lt 0$
$ \begin{array}{l}
f':\\
\\
\\
f:\\
\end{array} \quad \begin{array}{ccccccccccc}
-\infty& &0 & &\displaystyle \frac{1}{2} & &\displaystyle \infty \\
{(} &-- &| &++ &| & -- &) \\ \hline
(+\displaystyle \infty)&\displaystyle \searrow & & \displaystyle \nearrow & \displaystyle \frac{1}{4} & \displaystyle \searrow & \displaystyle \\
& & 0 & & & & (-\infty) \end{array}$
(a)
Increasing on $(0, \displaystyle \frac{1}{2})$
Decreasing on $(-\displaystyle \infty, 0)\cup(\frac{1}{2}, \infty)$
(b)
No absolute maximum.
Local maximum at $(\displaystyle \frac{1}{2},\frac{1}{4})$.
No absolute minimum.
Local minimum at $(0,0)$.
