The interval, on which the function f(x) = x^2e^-x is increasing, is…

The correct answer is Option (4) → $(0,2)$

$f(x)=x^{2}e^{-x}$

Differentiate:

$f'(x)=\frac{d}{dx}(x^{2})e^{-x}+x^{2}\frac{d}{dx}(e^{-x})$

$f'(x)=2xe^{-x}-x^{2}e^{-x}$

$f'(x)=e^{-x}(2x-x^{2})$

$2x-x^{2}=x(2-x)$

Since $e^{-x}>0$ for all $x$, the sign of $f'(x)$ depends on $x(2-x)$.

$f'(x)>0$ when $x(2-x)>0$

This holds for:

Interval of increase: $(0,2)$