589. $y=x^{2} e^{-x}$.
Solution. $1^{0} . y^{\prime}=2 x e^{-x}-x e^{-x}=x^{2} e^{-x}(2-x)$.

$2^{0} . y^{\prime}=0$ at $x_{1}=0$ and $x_{2}=2$.

$3^{0} . y^{\mu}=(2-2 x) e^{-x}-\left(2 x-x^{2}\right) e^{-x}-e^{-x}\left(2-4 x+x^{2}\right)$.

$4^{0} . y^{\prime \prime}(0)=2>0, y^{\prime \prime}(2)=e^{-2}(-2)<0 ; y_{\min }=y(0)=0, y_{\max }=y(2)=$ $=4 e^{-2}$.