Answer
(a)
Increasing on $(0,1)$.
Decreasing on $(1,\infty)$
(b)
Absolute maximum: $(1,6)$.
No other local maxima.
No absolute minimum.
Local minimum $(0,3)$
Work Step by Step
$g$ is defined on $[0,\infty)$.
$g'(x)=4\displaystyle \cdot\frac{1}{2}x^{-1/2}-2x=2x^{-1/2}(1-x^{3/2})=\frac{2(1-x^{3/2})}{\sqrt{x}}$
which is not defined at $ x=0\qquad$: critical point.
$g'(0)=0$ for $x=1 \quad$: critical point.
$g(0)=3,\qquad g(1)=6$
Use testpoints in the intervals between critical points,
$g'(\displaystyle \frac{1}{4})=\frac{79}{16} \gt 0$
$g'(4)=-5 \lt 0$
Create a table using the above:
$\begin{array}{l}
g':\\
\\
\\
g:
\end{array} \begin{array}{lllll}
0 & & 1 & & \infty\\
{[} & ++ & | & -- & )\\
& & & & \\
& \nearrow & 6 & \searrow & \\
3 & & & & \\
& & & &
\end{array}$
(a)
Increasing on $(0,1)$.
Decreasing on $(1,\infty)$
(b)
Absolute maximum: $(1,6)$.
No other local maxima.
No absolute minimum.
Local minimum $(0,3)$
