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

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

Question

(B) Let $y = x e^{-x}$.On differentiating with respect to $x$,we get:$\frac{dy}{dx} = x \cdot (-e^{-x}) + e^{-x} \cdot (1) = e^{-x}(1 - x)$.For maximu

Vedclass · Accepted Answer

$\frac{1}{e}$

Vedclass · Answer

(B) Let $y = x e^{-x}$.On differentiating with respect to $x$,we get:$\frac{dy}{dx} = x \cdot (-e^{-x}) + e^{-x} \cdot (1) = e^{-x}(1 - x)$.For maximum or minimum values,we set $\frac{dy}{dx} = 0$:$e^{-x}(1 - x) = 0$.Since $e^{-x} 
eq 0$ for any real $x$,we have $1 - x = 0$,which implies $x = 1$.Now,we find the second derivative to check for maxima:$\frac{d^2y}{dx^2} = e^{-x}(-1) + (1 - x)(-e^{-x}) = -e^{-x} - e^{-x} + x e^{-x} = e^{-x}(x - 2)$.At $x = 1$:$\frac{d^2y}{dx^2} = e^{-1}(1 - 2) = -\frac{1}{e} Since the second derivative is negative at $x = 1$,the function has a local maximum at $x = 1$.The maximum value is $y(1) = 1 \cdot e^{-1} = \frac{1}{e}$.

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

Similar Questions

The local maximum of $y=x^3-3 x^2+5$ is attained at

If $y = a \log |x| + b x^2 + x$ has its extremum values at $x = -1$ and $x = 2$,then

If $f(x)=x^5-5 x^4+5 x^3-10$ has its local maxima and minima at $x=a$ and $x=b$ respectively,then $2 a+b$ is equal to

The maximum area of a triangle whose one vertex is at $(0,0)$ and the other two vertices lie on the curve $y = -2x^2 + 54$ at points $(x, y)$ and $(-x, y)$ where $y > 0$ is:

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module