The curve $$y = x{e^x}$$   has minimum value equal to :

Solution :
Let $$y = x{e^x}$$
Differentiate both side w.r.t. $$‘x’.$$
$$\eqalign{ & \Rightarrow \frac{{dy}}{{dx}} = {e^x} + x{e^x} = {e^x}\left( {1 + x} \right) \cr & {\text{Put }}\frac{{dy}}{{dx}} = 0 \cr & \Rightarrow {e^x}\left( {1 + x} \right) = 0 \Rightarrow x = - 1 \cr & {\text{Now, }}\frac{{{d^2}y}}{{d{x^2}}} = {e^x} + {e^x}\left( {1 + x} \right) = {e^x}\left( {x + 2} \right) \cr & {\left( {\frac{{{d^2}y}}{{d{x^2}}}} \right)_{\left( {x = - 1} \right)}} = \frac{1}{e} + 0 > 0 \cr} $$
Hence, $$y = x{e^x}$$   is minimum function and $${y_{\min }} = - \frac{1}{e}.$$