The maximum value of left(frac{1}{x}right)^x | vedclass

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

Question

(A) Let $f(x) = (\frac{1}{x})^x = x^{-x}$.Taking the natural logarithm on both sides,we get $\ln(f(x)) = -x \ln(x)$.To find the critical points,we dif

Vedclass · Accepted Answer

$e^{1/e}$

Vedclass · Answer

(A) Let $f(x) = (\frac{1}{x})^x = x^{-x}$.Taking the natural logarithm on both sides,we get $\ln(f(x)) = -x \ln(x)$.To find the critical points,we differentiate with respect to $x$:$\frac{d}{dx}(\ln(f(x))) = \frac{d}{dx}(-x \ln(x))$$\frac{f'(x)}{f(x)} = -[\ln(x) + x \cdot \frac{1}{x}] = -(\ln(x) + 1)$.Setting $f'(x) = 0$,we get $\ln(x) + 1 = 0$,which implies $\ln(x) = -1$,so $x = e^{-1} = \frac{1}{e}$.Now,we check the second derivative or the sign of $f'(x)$ around $x = \frac{1}{e}$.For $x  0$ and for $x > \frac{1}{e}$,$f'(x) The maximum value is $f(\frac{1}{e}) = (\frac{1}{1/e})^{1/e} = e^{1/e}$.

The maximum value of $\left(\frac{1}{x}\right)^x$ is

Similar Questions

The number of real roots of the equation $e^{4x} + 2e^{3x} - e^{x} - 6 = 0$ is:

Number of solutions of the equation $3 \tan x + x^3 = 2$ in the interval $(0, \frac{\pi}{4})$ is:

Prove that the function $h(x) = x^{3} + x^{2} + x + 1$ does not have any local maxima or local minima.

$A$ manufacturer sells $x$ items at a price of rupees $\left(6-\frac{x}{40}\right)$ each. The cost price of $x$ items is Rs. $\left(\frac{x}{5}+193\right)$. The maximum profit in Rs. is

In the interval $(0, 1)$,the maximum value of the function $f(x) = |x \ln x|$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module