Given $\mathrm{f}(\mathrm{x})=\mathrm{x} \mathrm{e}^{x}$
$
\begin{array}{l}
f(x)=e^{x}+x e^{i}=e^{x}(x+1) \\
f^{\prime}(x)=e^{2}(x+1)+e^{t} \\
=e^{2}(x+2)
\end{array}
$
For maxima and minima,
$
\begin{array}{l}
\mathrm{f}(\mathrm{x})=0 \\
\operatorname{el}^{x}(x+1)=0 \\
\mathrm{x}=-1
\end{array}
$
Now $f^{\prime}(-1)=e^{-1}=1 / e>0$
$x=-1$ is point of local minima
Hence, local $\min =\mathrm{f}(-1)=-1 / \mathrm{e}$