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

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(D) Let $f(x) = x^4 e^{-x^2}$.To find the maximum value,we find the derivative $f'(x)$:$f'(x) = 4x^3 e^{-x^2} + x^4 e^{-x^2}(-2x)$$f'(x) = 2x^3 e^{-x^

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$4e^{-2}$

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(D) Let $f(x) = x^4 e^{-x^2}$.To find the maximum value,we find the derivative $f'(x)$:$f'(x) = 4x^3 e^{-x^2} + x^4 e^{-x^2}(-2x)$$f'(x) = 2x^3 e^{-x^2}(2 - x^2)$Setting $f'(x) = 0$,we get $x = 0$ or $x^2 = 2$,which means $x = 0, \pm \sqrt{2}$.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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