The maximum value of x^4 e^{-x^2} is | vedclass

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(D) Let $f(x) = x^4 e^{-x^2}$.Taking the derivative with respect to $x$:$f'(x) = 4x^3 e^{-x^2} + x^4 e^{-x^2}(-2x)$$f'(x) = 2x^3 e^{-x^2}(2 - x^2)$Set

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$4 e^{-2}$

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(D) Let $f(x) = x^4 e^{-x^2}$.Taking the derivative with respect to $x$:$f'(x) = 4x^3 e^{-x^2} + x^4 e^{-x^2}(-2x)$$f'(x) = 2x^3 e^{-x^2}(2 - x^2)$Setting $f'(x) = 0$ gives critical points at $x = 0$ and $x = \pm \sqrt{2}$.Analyzing the sign of $f'(x)$:For $x  0$.For $-\sqrt{2} For $0  0$.For $x > \sqrt{2}$,$f'(x) Thus,$f(x)$ has local maxima at $x = \pm \sqrt{2}$.The maximum value is $f(\pm \sqrt{2}) = (\pm \sqrt{2})^4 e^{-(\pm \sqrt{2})^2} = 4 e^{-2}$.

The maximum value of $x^4 e^{-x^2}$ is

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