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

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

Question

(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

Similar Questions

In $\triangle ABC$,$\frac{r_1-r}{a}+\frac{r_2-r}{b}+\frac{r_3-r}{c} =$

Which of the following characteristics is false for $Bryophytes$?

Which of the following is a plasmid?

The vapour pressure in $mm$ of $Hg$,of an aqueous solution obtained by adding $18 \ g$ of glucose $(C_6H_{12}O_6)$ to $180 \ g$ of water at $100^{\circ} C$ is

If $\frac{1-x+6x^2}{x-x^3} = \frac{A}{x} + \frac{B}{1-x} + \frac{C}{1+x}$,then $A$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module