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

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Answer 1 · 2018-06-03T13:08:57

$PMIN[0,0],PMAX[-2,4e^2]$

We get
$f'(x)=e^(-x)(2x-x^2)$
$f''(x)=e^(-x)(x^2-4x+2)$
solving

$f'(x)=0$
we get
$x=0$
$f''(0)>0$
$x=2$
$f''(2)=e^(-2)(-2)<0$

since

$f(-2)=4e^2>4e^(-2)$

we get the Points as above.

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