Question

Find the critical points of the following function and test them for their maxima and minima.
$y=x\cdot e^{-x2}$

Answer 1

$y=x.e^{-x^2}$

$\frac{dy}{dx}=e^{-x^2}+x{e}^{-x^2}(-2x)=e^{-x^2}(1-2x^2)=0$

$e^{-x^2}\ne 0andx=\pm \frac{1}{\sqrt{2}}$

$\frac{d^{2}y}{d{x}^2}=e^{-x^2}(-6x+4x^3)$

For $x=-\frac{1}{\sqrt{2}},\frac{d^{2}y}{d{x}^2}>0$,Hence y is minimum here

For $x=+\frac{1}{\sqrt{2}},\frac{d^{2}y}{d{x}^2}<0$,Hence y is maximum here