The difference between the maximum and minimu | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

$A$ solid hemisphere is mounted on a solid cylinder,both having equal radii. If the whole solid is to have a fixed surface area and the maximum possible volume,then the ratio of the height of the cylinder to the common radius is

The displacement of a particle at time $t$ is $x$, where $x = t^4 - k t^3$. If the velocity of the particle at time $t = 2$ is minimum, then

At which point in the interval $[0, 1]$ is the function $f(x) = x^{25}(1 - x)^{75}$ maximum?

Find two positive numbers $x$ and $y$ such that $x+y=60$ and $x y^{3}$ is maximum.

If $f(x)=a \log |x|+b x^2+x$ has extreme values at $x=-1$ and $x=2$,then the ordered pair $(a, b)=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module