Answer
$$\eqalign{
& {\text{Increasing on }}\left( {0,3} \right),\left( {5,\infty } \right) \cr
& {\text{Decreasing on }}\left( { - \infty ,0} \right),\,\left( {3,5} \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{{{x^4}}}{4} - \frac{{8{x^3}}}{3} + \frac{{15{x^2}}}{2} + 8 \cr
& {\text{Derivative}} \cr
& f'\left( x \right) = {x^3} - 8{x^2} + 15x \cr
& {\text{Set the derivative to 0}} \cr
& {x^3} - 8{x^2} + 15x = 0 \cr
& x\left( {{x^2} - 8x + 15} \right) = 0 \cr
& x\left( {x - 5} \right)\left( {x - 3} \right) = 0 \cr
& {\text{Solving the equation we obtain}} \cr
& x = 0,\,\,\,x = 3{\text{ and }}x = 5 \cr
& {\text{From the critical values and the domain }}\left( { - \infty ,\infty } \right){\text{ we have}} \cr
& \left( { - \infty ,0} \right),\,\,\left( {0,3} \right),\,\,\left( {3,5} \right),\,\,\left( {5,\infty } \right) \cr
& {\text{Now}}{\text{, we will evaluate the critical value and resume in a table}} \cr} $$
\[\begin{array}{*{20}{c}}
{{\rm{Interval}}}&{{\rm{Test \ value }}\left( x \right)}&{{\rm{Sign of \ }}f'\left( x \right)}&{{\rm{Behavior \ of }}f\left( x \right)}\\
{\left( { - \infty ,0} \right)}&{ - 1}& - &{{\rm{Decreasing}}}\\
{\left( {0,3} \right)}&1& + &{{\rm{Increasing}}}\\
{\left( {3,5} \right)}&4& - &{{\rm{Decreasing}}}\\
{\left( {5,\infty } \right)}&6& + &{{\rm{Increasing}}}
\end{array}\]
$$\eqalign{
& {\text{From the table we can conlude that the function is:}} \cr
& {\text{Increasing on }}\left( {0,3} \right),\left( {5,\infty } \right) \cr
& {\text{Decreasing on }}\left( { - \infty ,0} \right),\,\left( {3,5} \right) \cr} $$
