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Question
f(x) = xex.
Solution
\[\text { Given: } \hspace{0.167em} f\left( x \right) = x e^x \]
\[ \Rightarrow f'\left( x \right) = e^x + x e^x \]
\[\text { For the local maxima or minima, we must have }\]
\[ f'\left( x \right) = 0\]
\[ \Rightarrow e^x + x e^x = 0\]
\[ \Rightarrow e^x \left( 1 + x \right) = 0\]
\[ \Rightarrow e^x \neq 0 , x = - 1\]
\[ \Rightarrow x = - 1\]
\[\text { Thus, x = - 1 is the possible point of local maxima or local minima } . \]
\[\text { Now,} \]
\[f''\left( x \right) = e^x + e^x + x e^x \]
\[\text { At } x = - 1: \]
\[ f''\left( - 1 \right) = e^{- 1} + e^{- 1} - e^{- 1} = e^{- 1} > 0\]
\[\text { So, x = - 1 is the point of local minimum } . \]
\[\text { The local minimum value is given by }\]
\[f\left( - 1 \right) = - e^{- 1} = - \frac{1}{e}\]
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