The interval in which y=x^{2} e^{-x} is incre | vedclass

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(D) We have,$y = x^{2} e^{-x}$.To find the intervals where the function is increasing,we calculate the derivative:$\frac{dy}{dx} = 2x e^{-x} - x^{2} e

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$(0, 2)$

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(D) We have,$y = x^{2} e^{-x}$.To find the intervals where the function is increasing,we calculate the derivative:$\frac{dy}{dx} = 2x e^{-x} - x^{2} e^{-x} = x e^{-x}(2 - x)$.Setting $\frac{dy}{dx} = 0$ gives critical points:$x e^{-x}(2 - x) = 0 \Rightarrow x = 0$ or $x = 2$.These points divide the real line into three intervals: $(-\infty, 0)$,$(0, 2)$,and $(2, \infty)$.We test the sign of $f'(x) = x e^{-x}(2 - x)$ in each interval:$1$. For $x \in (-\infty, 0)$,let $x = -1$: $f'(-1) = (-1)e^{1}(2 - (-1)) = -3e $2$. For $x \in (0, 2)$,let $x = 1$: $f'(1) = (1)e^{-1}(2 - 1) = e^{-1} > 0$. Thus,$f$ is increasing.$3$. For $x \in (2, \infty)$,let $x = 3$: $f'(3) = (3)e^{-3}(2 - 3) = -3e^{-3} Therefore,the function is increasing in the interval $(0, 2)$.The correct answer is $D$.

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

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