Answer
\[{\text{Relative minimum at }}\left( {0, - 4} \right)\]
Work Step by Step
$$\eqalign{
& {\text{We have }}f\left( x \right) = {x^{2/3}} - 4 \cr
& {\text{Calculating the first derivative}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {{x^{2/3}} - 4} \right] \cr
& f'\left( x \right) = \frac{3}{2}{x^{ - 1/3}} - 0 \cr
& f'\left( x \right) = \frac{3}{2}{x^{ - 1/3}} \cr
& {\text{Calculating the critical points, set }}f'\left( x \right) = 0 \cr
& f'\left( x \right) = 0 \cr
& \frac{3}{2}{x^{ - 1/3}} = 0 \cr
& \frac{3}{{2{x^{1/3}}}} = 0 \cr
& {\text{The derivative is not defined at }}x = 0,{\text{ we obtain the critical }} \cr
& {\text{point }}x = 0 \cr
& {\text{Set the intervals }}\left( { - \infty ,0} \right),\left( {0,\infty } \right) \cr
& {\text{Making a table of values }}\left( {{\text{See examples on page 180 }}} \right) \cr} $$
\[\begin{array}{*{20}{c}}
{{\text{Interval}}}&{\left( { - \infty ,0} \right)}&{\left( {0,\infty } \right)} \\
{{\text{Test Value}}}&{x = - 1}&{x = 1} \\
{{\text{Sign of }}f'\left( x \right)}&{{\text{ }}f'\left( { - 1} \right) < 0}&{{\text{ }}f'\left( 1 \right) < 0} \\
{{\text{Conclusion}}}&{{\text{Decreasing}}}&{{\text{Increasing}}}
\end{array}\]
$$\eqalign{
& {\text{By Theorem 3}}{\text{.6}} \cr
& f'\left( x \right){\text{ changes from negative to positive at }}x = 0,{\text{ so }}f\left( x \right) \cr
& {\text{has a relative minimum at }}\left( {0,f\left( 0 \right)} \right) \cr
& f\left( 0 \right) = {\left( 0 \right)^{2/3}} - 4 \cr
& f\left( 0 \right) = - 4 \cr
& {\text{Relative minimum at }}\left( {0, - 4} \right) \cr} $$
