Answer
Absolute maximum: $(4,0)$
Absolute minimum: $(0,-2)$
Work Step by Step
$ f(x)=x^{1/2}-2 \quad$ ... defined on $[0,4]$
$ f^{\prime}(x)=\displaystyle \frac{1}{2}x^{-1/2}\quad$ ... defined on $(0,4)$,
$f^{\prime}(x)=0 \quad$ for no $\mathrm{x}\in(0,4)$
no critical numbers on $(0,4)$,
$\left[\begin{array}{lccccc}
x, \text{ interval} & 0 & (0,4) & 4\\
t=\text{ test number} & & 1 & \\
f^{\prime}(t) & & +1/2 & \\
f(x) & -2 & \nearrow & 0\\
& min & & max
\end{array}\right]$
Absolute maximum: $(4,0)$
Absolute minimum: $(0,-2)$
