If $p$ and $q$ are respectively the global maximum and global minimum of the function $f\left(x\right)=x^2{\mathrm{e}}^{2x}$ on the interval $[-2,2]$, then $p{e}^{-4}+q{e}^4=$

It is given that,

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

Differentiating the above equation we get,

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

$\Rightarrow 0=2e^{2x}\left(x^2+x\right)$

$\Rightarrow x^2+x=0$

$\Rightarrow x=0, -1$

The maxima and minima is obtained at $-2, -1, 0, 2$ as the
function bound.

Thus,

$f(-2)=(-2{)}^2{e}^{-4}=4e^{-4}$

$f(-1)=(1{)}^2{e}^{-2}$

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

$f\left(2\right)=(2{)}^2{e}^4=4e^4$

Thus, $p=4{\mathrm{e}}^4$ and $q=0$.

$p{e}^{-4}+q{e}^4=4{\mathrm{e}}^4\left(e^{-4}\right)+0=4{\mathrm{e}}^0=4$