Answer
$(a)$
Local maximum: none.
local minimum: $(0,0)$.
$(b)$
Absolute maximum: none.
Absolute minimum: $(0,0)$.
$(c)$
See graph.
Work Step by Step
$h(x)=\displaystyle \frac{x^{3}}{3}-2x^{2}+4x,\quad x\in[0,\infty)$
$h'(x)=x^{2}-4x+4=(x-2)^{2}$
$f$ and $f'$ are defined on $[0,\infty)$
$h'(x)=0$ for $ x=2\qquad$ ... critical point.
$\left[\begin{array}{cccccc}
t & 0 & 1 & 2 & & 3 & \infty\\
interval & [ & (0,2) & & (2,,\infty) & & )\\
h'(x) & & 1 & 0 & 12 & & \\
h(x) & 0 & \nearrow & 8/3 & \nearrow & & \\
& & & & & &
\end{array}\right]$
