Let f(x) = x^2 e^{-2x}, x 0. The maximum va | vedclass

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(B) Given the function $f(x) = x^2 e^{-2x}$ for $x > 0$.To find the maximum value,we first find the derivative $f'(x)$:$f'(x) = \frac{d}{dx}(x^2) \cdo

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$\frac{1}{e^2}$

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(B) Given the function $f(x) = x^2 e^{-2x}$ for $x > 0$.To find the maximum value,we first find the derivative $f'(x)$:$f'(x) = \frac{d}{dx}(x^2) \cdot e^{-2x} + x^2 \cdot \frac{d}{dx}(e^{-2x})$$f'(x) = 2x e^{-2x} + x^2(-2)e^{-2x}$$f'(x) = 2x e^{-2x}(1 - x)$Setting $f'(x) = 0$ for critical points:$2x e^{-2x}(1 - x) = 0$Since $x > 0$ and $e^{-2x} 
eq 0$,we have $1 - x = 0$,which gives $x = 1$.Using the first derivative test:For $0  0$ (function is increasing).For $x > 1$,$f'(x) Thus,$f(x)$ has a local maximum at $x = 1$.The maximum value is $f(1) = (1)^2 e^{-2(1)} = e^{-2} = \frac{1}{e^2}$.

Let $f(x) = x^2 e^{-2x}, x > 0$. The maximum value of $f(x)$ is

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