Mathematics - Application of Derivative Question with Solution | TestHub

We have, 

$f\left(x\right)=x^2{e}^{-x^2}$

$f^{\text{'}}\left(x\right)=2x{e}^{-x^2}-2x^3{e}^{-x^2}$

$f^{\text{'}}\left(x\right)=2x{e}^{-x^2}\left(1-x^2\right)$

$f^{\text{'}}\left(x\right)=0,x=0,-1,1$

$f^{\text{'}}\left(-h\right)<0,f^{\text{'}}\left(h\right)>0$, So minimum at $x=0$

$f^{\text{'}}\left(1-h\right)>0,f^{\text{'}}\left(1+h\right)<0$, So maximum at $x=1$

$f^{\text{'}}\left(-1-h\right)>0,f^{\text{'}}\left(-1+h\right)<0$, So maximum at $x=-1$

Minimum $f\left(x\right)=f\left(0\right)=0$

Maximum $f\left(x\right)=e^{-1}=\frac{1}{e}$

Since, $f\left(x\right)\ge 0\forall x$

So, difference between maximum and minimum values

$=\frac{1}{e}-0=\frac{1}{e}$