The set of all points,for which f(x) = x^2 e^ | vedclass

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(A) Given $f(x) = x^2 e^{-x}$.To find the interval where $f(x)$ is strictly increasing,we calculate the derivative $f'(x)$:$f'(x) = \frac{d}{dx}(x^2)

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$(0, 2)$

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(A) Given $f(x) = x^2 e^{-x}$.To find the interval where $f(x)$ is strictly increasing,we calculate the derivative $f'(x)$:$f'(x) = \frac{d}{dx}(x^2) e^{-x} + x^2 \frac{d}{dx}(e^{-x}) = 2x e^{-x} - x^2 e^{-x} = x e^{-x}(2 - x)$.For $f(x)$ to be strictly increasing,we must have $f'(x) > 0$.Since $e^{-x} > 0$ for all real $x$,the inequality $x e^{-x}(2 - x) > 0$ simplifies to $x(2 - x) > 0$.Multiplying by $-1$ reverses the inequality: $x(x - 2) This inequality holds when $x$ lies between the roots $0$ and $2$.Thus,$x \in (0, 2)$.

The set of all points,for which $f(x) = x^2 e^{-x}$ strictly increases,is

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