# How do you find the local extrema for  (x^2)(e^-x)  from [-2,4]?

Jun 3, 2018

$P M I N \left[0 , 0\right] , P M A X \left[- 2 , 4 {e}^{2}\right]$

#### Explanation:

We get
$f ' \left(x\right) = {e}^{- x} \left(2 x - {x}^{2}\right)$
$f ' ' \left(x\right) = {e}^{- x} \left({x}^{2} - 4 x + 2\right)$
solving

$f ' \left(x\right) = 0$
we get
$x = 0$
$f ' ' \left(0\right) > 0$
$x = 2$
$f ' ' \left(2\right) = {e}^{- 2} \left(- 2\right) < 0$

since

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

we get the Points as above.