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

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Question

(A) To find the interval where the function $y = x^2 e^{-x}$ is decreasing, we calculate its derivative $\frac{dy}{dx}$.Using the product rule: $\frac

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$(-\infty, 0) \cup (2, \infty)$

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(A) To find the interval where the function $y = x^2 e^{-x}$ is decreasing, we calculate its derivative $\frac{dy}{dx}$.Using the product rule: $\frac{dy}{dx} = \frac{d}{dx}(x^2) \cdot e^{-x} + x^2 \cdot \frac{d}{dx}(e^{-x}) = 2x e^{-x} - x^2 e^{-x} = x e^{-x} (2 - x)$.$A$ function is decreasing when $\frac{dy}{dx} Since $e^{-x} > 0$ for all real $x$, the inequality $\frac{dy}{dx} Multiplying by $-1$ reverses the inequality: $x(x - 2) > 0$.This inequality holds when $x  2$.Thus, the function is decreasing on the interval $(-\infty, 0) \cup (2, \infty)$.

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

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