Answer
$ a.\quad$ critical points at $x=0$ and $x=4$
$b.\quad $f is increasing on $ (4, \infty)$, decreasing on $(0,4)$.
$ c.\quad$ local maximum at $x=4$
Work Step by Step
$(a)$
$f'(x)=3-\displaystyle \frac{6}{\sqrt{x}}=\frac{3\sqrt{x}-6}{\sqrt{x}}=\frac{3(\sqrt{x}-2)}{\sqrt{x}}$
$f'$ is not defined at $x=0\qquad $... critical point
$f'(x)=0$ for $x=4\qquad $... critical point
Critical points at $x=0$ and $x=4$
$(b)$
$\left[\begin{array}{ccccc}
& & (0,4) & (4,\infty)\\
\text{test point, }t & & 1 & 5\\
\text{evaluate }f'(t) & & -3 & 0.317
\end{array}\right]$
$f':\quad \stackrel{0}{(} \stackrel{\searrow}{- - -} \stackrel{4}{|}$ $\stackrel{\nearrow}{+++}$
f is increasing on $ (4, \infty)$, decreasing on $(0,4)$.
$(c)$
From the table, we see that f has: a local minimum at $x=4$.
