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