Answer
$$\eqalign{
& \left( {\text{a}} \right)x = - \frac{{\sqrt {15} }}{6},{\text{ }}x = \frac{{\sqrt {15} }}{6} \cr
& \left( {\text{b}} \right){\text{Decreasing on: }}\left( { - \frac{{\sqrt {15} }}{6},\frac{{\sqrt {15} }}{6}} \right) \cr
& {\text{Increasing on: }}\left( { - \infty , - \frac{{\sqrt {15} }}{6}} \right),\left( {\frac{{\sqrt {15} }}{6},\infty } \right) \cr
& \left( {\text{c}} \right){\text{: Relative maximum at }}\left( { - \frac{{\sqrt {15} }}{6},\frac{{5\sqrt {15} }}{9}} \right) \cr
& \left( {\text{c}} \right){\text{: Relative minimum at }}\left( {\frac{{\sqrt {15} }}{6}, - \frac{{5\sqrt {15} }}{9}} \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = 4{x^3} - 5x \cr
& \left( {\text{a}} \right){\text{Calculating the first derivative}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {4{x^3} - 5x} \right] \cr
& f'\left( x \right) = 12{x^2} - 5 \cr
& {\text{Calculating the critical points, set }}f'\left( x \right) = 0 \cr
& f'\left( x \right) = 0 \cr
& 12{x^2} - 5 = 0 \cr
& {x^2} = \frac{5}{{12}} \cr
& x = - \frac{{\sqrt {15} }}{6},{\text{ }}x = \frac{{\sqrt {15} }}{6} \cr
& \cr
& \left( {\text{b}} \right){\text{Set the intervals }}\left( { - \infty , - \frac{{\sqrt {15} }}{6}} \right),\left( { - \frac{{\sqrt {15} }}{6},\frac{{\sqrt {15} }}{6}} \right),\left( {\frac{{\sqrt {15} }}{6},\infty } \right) \cr
& {\text{Making a table of values }}\left( {{\text{See examples on page 180 }}} \right) \cr} $$
\[\boxed{\begin{array}{*{20}{c}}
{{\text{Interval}}}&{{\text{Test Value}}}&{{\text{Sign of }}f'\left( x \right)}&{{\text{Conclusion}}} \\
{\left( { - \infty , - \frac{{\sqrt {15} }}{6}} \right)}&{x = - 1}&{7 > 0}&{{\text{Increasing}}} \\
{\left( { - \frac{{\sqrt {15} }}{6},\frac{{\sqrt {15} }}{6}} \right)}&{x = 0}&{ - 5 < 0}&{{\text{Decreasing}}} \\
{\left( {\frac{{\sqrt {15} }}{6},\infty } \right)}&{x = 1}&{7 > 0}&{{\text{Increasing}}}
\end{array}}\]
$$\eqalign{
& \left( {\text{c}} \right){\text{ By Theorem 3}}{\text{.6}} \cr
& *f'\left( x \right){\text{changes from positive to negative at }}x = - \frac{{\sqrt {15} }}{6},{\text{ so }}f\left( x \right) \cr
& {\text{has a relative maximum at }}\left( { - \frac{{\sqrt {15} }}{6},f\left( { - \frac{{\sqrt {15} }}{6}} \right)} \right) \cr
& f\left( { - \frac{{\sqrt {15} }}{6}} \right) = 4{\left( { - \frac{{\sqrt {15} }}{6}} \right)^3} - 5\left( { - \frac{{\sqrt {15} }}{6}} \right) = \frac{{5\sqrt {15} }}{9} \cr
& *f'\left( x \right){\text{changes from negative to positive at }}x = \frac{{\sqrt {15} }}{6},{\text{ so }}f\left( x \right) \cr
& {\text{has a relative minimum at }}\left( {\frac{{\sqrt {15} }}{6},f\left( {\frac{{\sqrt {15} }}{6}} \right)} \right) \cr
& f\left( {\frac{{\sqrt {15} }}{6}} \right) = 4{\left( {\frac{{\sqrt {15} }}{6}} \right)^3} - 5\left( {\frac{{\sqrt {15} }}{6}} \right) = - \frac{{5\sqrt {15} }}{9} \cr
& \cr
& \left( {\text{d}} \right){\text{Graph}} \cr} $$
