For what values of x does the function f(x) = | vedclass

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(A) To find the intervals where the function $f(x) = x^2 e^{-x}$ is strictly increasing,we calculate its derivative $f'(x)$.Using the product rule:$f'

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$0 < x < 2$

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(A) To find the intervals where the function $f(x) = x^2 e^{-x}$ is strictly increasing,we calculate 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})$$f'(x) = 2x e^{-x} - x^2 e^{-x}$$f'(x) = x e^{-x} (2 - x)$For the function to be strictly increasing,we must have $f'(x) > 0$.Since $e^{-x} > 0$ for all $x \in \mathbb{R}$,the inequality $x e^{-x} (2 - x) > 0$ simplifies to:$x(2 - x) > 0$Multiplying by $-1$ reverses the inequality sign:$x(x - 2) The roots of the quadratic expression are $x = 0$ and $x = 2$. The expression is negative between the roots.Therefore,the function is strictly increasing for $0 < x < 2$.

For what values of $x$ does the function $f(x) = x^2 e^{-x}$ strictly increase?

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