Answer
The absolute maximum is $f(-1) = e$
The absolute minimum is $f(0) = 0$
$f(2) = \frac{4}{e^2}$ is a local maximum.
$f(0) = 0$ is a local minimum.
Work Step by Step
$f(x) = x^2~e^{-x}$
Note that this function is continuous on the interval $[-1,3]$ and differentiable on the interval $(-1,3)$.
We can find the points where $f'(x) = 0$:
$f'(x) = 2x~e^{-x}-x^2~e^{-x}=0$
$2x~e^{-x}= x^2~e^{-x}$
$x^2 = 2x$
$x^2 - 2x = 0$
$x(x-2) = 0$
$x = 0,2$
We can verify the values of the function when $f'(x) = 0$ and at the endpoints of the interval $[-1,3]$:
$f(-1) = (-1)^2~e^{-(-1)} = e \approx 2.72$
$f(0) = (0)^2~e^{-(0)} = 0$
$f(2) = (2)^2~e^{-(2)} = \frac{4}{e^2} \approx 0.54$
$f(3) = (3)^2~e^{-(3)} = \frac{9}{e^3} \approx 0.45$
The absolute maximum is $f(-1) = e$
The absolute minimum is $f(0) = 0$
$f(2) = \frac{4}{e^2}$ is a local maximum.
$f(0) = 0$ is a local minimum.