Answer
Absolute maximum: $(9/4,9/4)$
Absolute minimum: $(0,0)$ and $(9,0)$
Work Step by Step
$ h(x)=3x^{1/2}-x \quad$ ... defined on $[0,9]$
$ h^{\prime}(x)=\displaystyle \frac{3}{2}x^{-1/2}-1\quad$ ... defined on $(0,9)$,
$h^{\prime}(x)=0 \quad$ for
$\displaystyle \frac{3}{2}x^{-1/2}-1=0$
$\displaystyle \frac{3}{2}x^{-1/2}=1$
$x^{-1/2}=\displaystyle \frac{2}{3}$
$x=(\displaystyle \frac{2}{3})^{-2}$
$ x=\displaystyle \frac{9}{4}\in(0,9)\quad$ ...critical number
$\left[\begin{array}{llllll}
x, \text{ interval} & 0 & (0,9/4) & 9/4 & (9/4,9) & 9\\
t=\text{ test number} & & 1 & & 4 & \\
h^{\prime}(t) & & +2 & & -1/4 & \\
h(x) & 0 & \nearrow & 9/4 & \searrow & 0\\
& min & & max & & min
\end{array}\right]$
Absolute maximum: $(9/4,9/4)$
Absolute minimum: $(0,0)$ and $(9,0)$
