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) = 4{x^3}{e^{ - {x^2}}} + {x^4}{e^{ - {x^2}}}( - 2

Vedclass · Accepted Answer

$4{e^{ - 2}}$

Vedclass · Answer

(D) Let $f(x) = {x^4}{e^{ - {x^2}}}$.To find the maximum value,we find the derivative $f'(x)$:$f'(x) = 4{x^3}{e^{ - {x^2}}} + {x^4}{e^{ - {x^2}}}( - 2x) = 2{x^3}{e^{ - {x^2}}}(2 - {x^2})$.For critical points,set $f'(x) = 0$:$2{x^3}{e^{ - {x^2}}}(2 - {x^2}) = 0$.Since ${e^{ - {x^2}}} 
eq 0$,we have $x = 0$ or ${x^2} = 2$,which means $x = \pm \sqrt{2}$.Evaluating $f(x)$ at these 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 $x 	o \pm \infty$,$f(x) 	o 0$.Therefore,the maximum value is $4{e^{ - 2}}$.

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

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