If f(x) = x^2e^{-2x}, x 0,then the maximum | vedclass

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(C) Given $f(x) = x^2 e^{-2x}$.First,find the derivative $f'(x)$ using the product rule:$f'(x) = x^2 \frac{d}{dx}(e^{-2x}) + e^{-2x} \frac{d}{dx}(x^2)

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$1/e^2$

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(C) Given $f(x) = x^2 e^{-2x}$.First,find the derivative $f'(x)$ using the product rule:$f'(x) = x^2 \frac{d}{dx}(e^{-2x}) + e^{-2x} \frac{d}{dx}(x^2)$$f'(x) = x^2(-2e^{-2x}) + e^{-2x}(2x)$$f'(x) = 2xe^{-2x}(1 - x) = \frac{2x(1 - x)}{e^{2x}}$.To find the critical points,set $f'(x) = 0$:Since $x > 0$ and $e^{2x} > 0$,we have $1 - x = 0$,which gives $x = 1$.Now,check the second derivative $f''(x)$ to confirm the maximum:$f''(x) = \frac{d}{dx}[2(x - x^2)e^{-2x}] = 2[(1 - 2x)e^{-2x} + (x - x^2)(-2e^{-2x})]$$f''(1) = 2[(1 - 2)e^{-2} + (1 - 1)(-2e^{-2})] = 2[-e^{-2}] = -2/e^2 Since $f''(1) The maximum value is $f(1) = (1)^2 e^{-2(1)} = 1/e^2$.

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

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