#### Question

f(x) = xe^{x}.

#### Solution

\[\text { Given: } \hspace{0.167em} f\left( x \right) = x e^x \]

\[ \Rightarrow f'\left( x \right) = e^x + x e^x \]

\[\text { For the local maxima or minima, we must have }\]

\[ f'\left( x \right) = 0\]

\[ \Rightarrow e^x + x e^x = 0\]

\[ \Rightarrow e^x \left( 1 + x \right) = 0\]

\[ \Rightarrow e^x \neq 0 , x = - 1\]

\[ \Rightarrow x = - 1\]

\[\text { Thus, x = - 1 is the possible point of local maxima or local minima } . \]

\[\text { Now,} \]

\[f''\left( x \right) = e^x + e^x + x e^x \]

\[\text { At } x = - 1: \]

\[ f''\left( - 1 \right) = e^{- 1} + e^{- 1} - e^{- 1} = e^{- 1} > 0\]

\[\text { So, x = - 1 is the point of local minimum } . \]

\[\text { The local minimum value is given by }\]

\[f\left( - 1 \right) = - e^{- 1} = - \frac{1}{e}\]