Answer
$$\eqalign{
& {\text{Increasing on: }}\left( {1,\infty } \right) \cr
& {\text{Decreasing on: }}\left( {0,1} \right) \cr} $$
Work Step by Step
$$\eqalign{
& h\left( x \right) = \sqrt x \left( {x - 3} \right),{\text{ }}x > 0 \cr
& h\left( x \right) = {x^{3/2}} - 3{x^{1/2}} \cr
& {\text{Calculate the first derivative}} \cr
& h'\left( x \right) = \frac{d}{{dx}}\left[ {{x^{3/2}} - 3{x^{1/2}}} \right] \cr
& h'\left( x \right) = \frac{3}{2}{x^{1/2}} - 3\left( {\frac{1}{2}{x^{ - 1/2}}} \right) \cr
& h'\left( x \right) = \frac{3}{2}{x^{1/2}} - \frac{3}{2}{x^{ - 1/2}} \cr
& {\text{Find the critical points, set the first derivative to }}0 \cr
& h'\left( x \right) = \frac{3}{2}{x^{1/2}} - \frac{3}{2}{x^{ - 1/2}} \cr
& \frac{3}{2}{x^{1/2}} - \frac{3}{2}{x^{ - 1/2}} = 0 \cr
& \frac{3}{2}{x^{ - 1/2}}\left( {x - 1} \right) = 0 \cr
& {\text{We have the critical point }}x = 1 \cr
& {\text{Set the intervals }}\left( {0,1} \right){\text{ and }}\left( {1,\infty } \right) \cr
& {\text{Making a table of values }}\left( {{\text{See examples on page 178 }}} \right) \cr} $$
\[\boxed{\begin{array}{*{20}{c}}
{{\text{Interval}}}&{\left( {0,1} \right)}&{\left( {1,\infty } \right)} \\
{{\text{Test Value}}}&{x = 0.5}&{x = 2} \\
{{\text{Sign of }}h'\left( x \right)}&{h'\left( {0.5} \right) < 0}&{h'\left( 2 \right) > 0} \\
{{\text{Conclusion}}}&{{\text{Decreasing}}}&{{\text{Increasing}}}
\end{array}}\]
$$\eqalign{
& {\text{By Theorem 3}}{\text{.5 }}h{\text{ is:}} \cr
& {\text{Increasing on: }}\left( {1,\infty } \right) \cr
& {\text{Decreasing on: }}\left( {0,1} \right) \cr} $$
