The difference of maximum and minimum values of f ( x ) = x^ 2 e^ -...

We have,&

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

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

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

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

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

Maximum $f\left(x\right) =$ e^{- 1} = $\dfrac{1}{e}</p><p>Since,$f\left(x\right)$\ge 0 \forall x$

So, difference between maximum and minimum values

=\dfrac{1}{e} -0 = \dfrac{1}{e}