y = x^2 e^{-x} is increasing on . . . . . . | vedclass

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(D) To find the intervals where the function $y = f(x) = x^2 e^{-x}$ is increasing,we find its derivative $f'(x)$.Using the product rule,$f'(x) = \fra

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$(0, 2)$

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(D) To find the intervals where the function $y = f(x) = x^2 e^{-x}$ is increasing,we find its derivative $f'(x)$.Using the product rule,$f'(x) = \frac{d}{dx}(x^2) \cdot e^{-x} + x^2 \cdot \frac{d}{dx}(e^{-x})$.$f'(x) = 2x e^{-x} - x^2 e^{-x} = x(2 - x) e^{-x}$.For the function to be increasing,we require $f'(x) > 0$.Since $e^{-x} > 0$ for all real $x$,the sign of $f'(x)$ depends on $x(2 - x)$.$x(2 - x) > 0$ implies $x(x - 2) This inequality holds when $x$ is between the roots $0$ and $2$.Thus,the function is increasing on the interval $(0, 2)$.

$y = x^2 e^{-x}$ is increasing on . . . . . . .

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