Consider the following.

$h(x)=e^{x^2-4}$

Find $h^{\text{'}}(x).$

$h^{\text{'}}(x)=$

Solve $h^{\text{'}}(x)=0$ for $x.$

$x$:$=$

Find $h(0),h(-2),$ and $h(2).$

$h(0)=$

$h(-2)=$

$h(2)=$

Find the absolute extrema of the function $h(x)=e^{x^2-4}$ on $-2,2$

Absolute maximum value:

at $x=+-$

Absolute minimum value:

at $x=$