Answer
$(a)$
Local maximum: $(0,5)$
Local minimum: $(-5,0),\ (5,0)$
$(b)$
Absolute maximum: $(0,5)$
Absolute minimum: $(-5,0),\ (5,0)$
$(c)$
See graph.
Work Step by Step
$f(x)=(25-x^{2})^{1/2},\quad x\in[-5,5]$
$f'(x)=\displaystyle \frac{1}{2}(25-x^{2})^{-1/2}\cdot(-2x)=-\frac{x}{\sqrt{(5-x)(5+x)}}$
$f'$ is undefined for $ x=\pm 5\qquad$ ... critical point$s$.
$f'(x)=0$ for $ x=0\qquad$ ... critical point.
$\left[\begin{array}{cccccccc}
interval & [ & (-5,0) & & (0,5) & ]\\
t & -5 & -3 & 0 & 3 & 5\\
f'(t) & & 0.75 & & -0.75 & \\
f(t) & 0 & \nearrow & 5 & \searrow & 0\\
& & & & &
\end{array}\right]$
