The difference between the maximum and minimu | vedclass

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(C) Given the function $f(x) = x^4e^{-x^2}$.To find the critical points,we calculate the derivative $f'(x)$:$f'(x) = 4x^3e^{-x^2} + x^4e^{-x^2}(-2x) =

Vedclass · Accepted Answer

$\frac{4}{e^2}$

Vedclass · Answer

(C) Given the function $f(x) = x^4e^{-x^2}$.To find the critical points,we calculate the derivative $f'(x)$:$f'(x) = 4x^3e^{-x^2} + x^4e^{-x^2}(-2x) = 2x^3e^{-x^2}(2 - x^2)$.Setting $f'(x) = 0$,we get $x = 0$ or $x^2 = 2$,which implies $x = 0, \sqrt{2}, -\sqrt{2}$.Evaluating $f(x)$ at these points:$f(0) = 0^4e^0 = 0$.$f(\sqrt{2}) = (\sqrt{2})^4e^{-(\sqrt{2})^2} = 4e^{-2} = \frac{4}{e^2}$.$f(-\sqrt{2}) = (-\sqrt{2})^4e^{-(-\sqrt{2})^2} = 4e^{-2} = \frac{4}{e^2}$.As $x 	o \pm \infty$,$f(x) 	o 0$.Thus,the maximum value is $\frac{4}{e^2}$ and the minimum value is $0$.The difference between the maximum and minimum values is $\frac{4}{e^2} - 0 = \frac{4}{e^2}$.

The difference between the maximum and minimum values of $f(x) = x^4e^{-x^2}$ for all $x \in R$ is:

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