The function f(x) = x e^{-x}, forall x in R a | vedclass

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

Question

(A) Given the function $f(x) = x e^{-x}$.To find the critical points,we calculate the first derivative $f'(x)$ using the product rule:$f'(x) = (1)e^{-

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given the function $f(x) = x e^{-x}$.To find the critical points,we calculate the first derivative $f'(x)$ using the product rule:$f'(x) = (1)e^{-x} + x(-e^{-x}) = e^{-x}(1 - x)$.Setting $f'(x) = 0$ for maxima or minima:$e^{-x}(1 - x) = 0$.Since $e^{-x} 
eq 0$ for any $x \in R$,we have $1 - x = 0$,which gives $x = 1$.Now,we find the second derivative $f''(x)$ to check for maxima:$f''(x) = \frac{d}{dx}[e^{-x} - x e^{-x}] = -e^{-x} - [e^{-x} - x e^{-x}] = e^{-x}(x - 2)$.Evaluating at $x = 1$:$f''(1) = e^{-1}(1 - 2) = -e^{-1} Since the second derivative is negative at $x = 1$,the function attains a maximum value at $x = 1$.

The function $f(x) = x e^{-x}, \forall x \in R$ attains a maximum value at $x$ equal to:

Similar Questions

The function $f(x) = x^5 - 5x^4 + 5x^3 - 10$ has a local maximum when $x$ is equal to:

What is the maximum slope of the curve $y = -x^3 + 3x^2 + 9x - 27$?

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

Let $p(x)$ be a polynomial of degree $4$ having extrema at $x=1$ and $x=2$,and $\lim_{x \rightarrow 0} \left(1+\frac{p(x)}{x^2}\right) = 2$. Then the value of $p(2)$ is

The maximum value of the function $f(x) = -|x+1| + 3, x \in R$ is . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module