Answer
$$\eqalign{
& {\text{Inflection points: }}\left( { - 2, - 8} \right){\text{ and }}\left( {0,0} \right) \cr
& {\text{Concave downward}}:{\text{ }}\left( { - 2,0} \right) \cr
& {\text{Concave upward}}:{\text{ }}\left( { - \infty ,2} \right){\text{ and }}\left( {0,\infty } \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \frac{1}{2}{x^4} + 2{x^3} \cr
& {\text{Calculate the second derivative}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\frac{1}{2}{x^4} + 2{x^3}} \right] \cr
& f'\left( x \right) = 2{x^3} + 6{x^2} \cr
& f''\left( x \right) = \frac{d}{{dx}}\left[ {2{x^3} + 6{x^2}} \right] \cr
& f''\left( x \right) = 6{x^2} + 12x \cr
& {\text{Set }}f''\left( x \right) = 0 \cr
& 6{x^2} + 12x = 0 \cr
& 6x\left( {x + 2} \right) = 0 \cr
& 6x = 0,{\text{ }}x + 2 = 0 \cr
& x = 0,{\text{ }}x = - 2 \cr
& {\text{Set the intervals }}\left( { - \infty , - 2} \right),\left( { - 2,0} \right),\left( {0,\infty } \right) \cr
& {\text{Making a table of values }}\left( {{\text{See examples on page 188 }}} \right) \cr} $$
\[\boxed{\begin{array}{*{20}{c}}
{{\text{Interval}}}&{\left( { - \infty , - 2} \right)}&{\left( { - 2,0} \right)}&{\left( {0,\infty } \right)} \\
{{\text{Test Value}}}&{x = - 4}&{x = - 1}&{x = 2} \\
{{\text{Sign of }}f''\left( x \right)}&{48 > 0}&{ - 6 < 0}&{48 > 0} \\
{{\text{Conclusion}}}&{{\text{C}}{\text{. upward}}}&{{\text{C}}{\text{. downward}}}&{{\text{C}}{\text{. upward}}}
\end{array}}\]
$$\eqalign{
& {\text{The inflection points occurs at }}x = - 2{\text{ and }}x = 0 \cr
& f\left( { - 2} \right) = \frac{1}{2}{\left( { - 2} \right)^4} + 2{\left( { - 2} \right)^3} = - 8,{\text{ }}\left( { - 2, - 8} \right) \cr
& f\left( 0 \right) = \frac{1}{2}{\left( 0 \right)^4} + 2{\left( 0 \right)^3} = 0,{\text{ }}\left( {0,0} \right) \cr
& {\text{Inflection points: }}\left( { - 2, - 8} \right){\text{ and }}\left( {0,0} \right) \cr
& {\text{Concave downward}}:{\text{ }}\left( { - 2,0} \right) \cr
& {\text{Concave upward}}:{\text{ }}\left( { - \infty ,2} \right){\text{ and }}\left( {0,\infty } \right) \cr} $$
