Answer
$${\text{Increasing on: }}\left( { - \infty ,\infty } \right)$$
Work Step by Step
$$\eqalign{
& h\left( x \right) = {\left( {x + 2} \right)^{1/3}} + 8 \cr
& {\text{Calculate the first derivative}} \cr
& h'\left( x \right) = \frac{d}{{dx}}\left[ {{{\left( {x + 2} \right)}^{1/3}} + 8} \right] \cr
& h'\left( x \right) = \frac{1}{3}{\left( {x + 2} \right)^{ - 2/3}} \cr
& {\text{Find the critical points, set the first derivative to }}0 \cr
& h'\left( x \right) = \frac{1}{3}{\left( {x + 2} \right)^{ - 2/3}} \cr
& \frac{1}{{3{{\left( {x + 2} \right)}^{2/3}}}} = 0 \cr
& h'\left( x \right){\text{ is never 0, and }}h'\left( x \right){\text{is not defined at }}x = - 2 \cr
& {\text{We have the critical point }}x = - 2 \cr
& {\text{Set the intervals }}\left( { - \infty , - 2} \right){\text{ and }}\left( { - 2,\infty } \right) \cr
& {\text{Making a table of values }}\left( {{\text{See examples on page 178 }}} \right) \cr} $$
\[\boxed{\begin{array}{*{20}{c}}
{{\text{Interval}}}&{\left( { - \infty , - 2} \right)}&{\left( { - 2,\infty } \right)} \\
{{\text{Test Value}}}&{x = - 3}&{x = 6} \\
{{\text{Sign of }}h'\left( x \right)}&{h'\left( { - 3} \right) = \frac{1}{3} > 0}&{h'\left( 3 \right) = \frac{1}{{12}} > 0} \\
{{\text{Conclusion}}}&{{\text{Increasing}}}&{{\text{Increasing}}}
\end{array}}\]
$$\eqalign{
& {\text{By Theorem 3}}{\text{.5 }}h{\text{ is:}} \cr
& {\text{Increasing on: }}\left( { - \infty ,\infty } \right) \cr} $$
