The interval in which $y=x^2 e^{-x}$ is increasing is

1. $y'=2x e^{-x}-x^2 e^{-x}=e^{-x}x(2-x)$. 2. Since $e^{-x}>0$, sign of $y'$ follows $x(2-x)$. 3. $x(2-x)>0$ when $0<x<2$, so $y$ is increasing on $(0,2)$. 4. Outside $[0,2]$ the product is negative (decreasing), ruling out the other intervals. _Source: syllabus Mathematics Ch 6 "Application of Derivatives", p.12_