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

The greatest value of the function $f(x) = -1 + \frac{2}{2^{x^2} + 1}$ is:

The number of integers $n$ for which $3x^3-25x+n=0$ has three real roots is

The vertices of a triangle are $(0, 0)$,$(x, \cos x)$,and $(\sin^3 x, 0)$ where $0 < x < \frac{\pi}{2}$. The maximum area for such a triangle in sq. units is:

Observe the statements given below :
Assertion $(A)$ : $f(x)=x e^{-x}$ has the maximum at $x=1$
Reason $(R)$ : $f^{\prime}(1)=0$ and $f^{\prime \prime}(1) < 0$
Which of the following is correct?

The minimum value of the function $y = 2x^3 - 21x^2 + 36x - 20$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module