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

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

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$(0, 2)$

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(A) Given the function $f(x) = x^2 e^{-x}$.To find the interval where the function is strictly increasing,we find its derivative $f'(x)$:$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 $e^{-x}$ is always positive for all real $x$,the sign of $f'(x)$ depends on $x(2 - x)$.$x(2 - x) > 0$Multiplying by $-1$ reverses the inequality:$x(x - 2) The roots are $x = 0$ and $x = 2$. Testing the intervals $(-\infty, 0)$,$(0, 2)$,and $(2, \infty)$:For $x \in (0, 2)$,$x$ is positive and $(x - 2)$ is negative,so $x(x - 2)  0$.Thus,the function $f(x)$ is strictly increasing in the interval $(0, 2)$.

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

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