The maximum value of f(x) = x^4 e^{-x^2} is | vedclass

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(C) Let $f(x) = x^4 e^{-x^2}$.To find the maximum value,we find the derivative $f'(x)$ and set it to $0$.$f'(x) = 4x^3 e^{-x^2} + x^4 e^{-x^2} (-2x) =

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$4e^{-2}$

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(C) Let $f(x) = x^4 e^{-x^2}$.To find the maximum value,we find the derivative $f'(x)$ and set it to $0$.$f'(x) = 4x^3 e^{-x^2} + x^4 e^{-x^2} (-2x) = 2x^3 e^{-x^2} (2 - x^2)$.Setting $f'(x) = 0$,we get $x = 0$ or $x^2 = 2$,which means $x = \pm \sqrt{2}$.Evaluating $f(x)$ at these critical points:$f(0) = 0$.$f(\sqrt{2}) = (\sqrt{2})^4 e^{-(\sqrt{2})^2} = 4 e^{-2}$.$f(-\sqrt{2}) = (-\sqrt{2})^4 e^{-(-\sqrt{2})^2} = 4 e^{-2}$.As $x 	o \pm \infty$,$f(x) 	o 0$.Thus,the maximum value is $4e^{-2}$.

The maximum value of $f(x) = x^4 e^{-x^2}$ is

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