The maximum value of {left( {frac{1}{x}} righ | vedclass

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

Question

(B) Let $f(x) = {\left( {\frac{1}{x}} \right)^x} = {x^{-x}}$.Taking the natural logarithm on both sides,we get $\ln(f(x)) = -x \ln(x)$.Differentiating

Vedclass · Accepted Answer

${e^{1/e}}$

Vedclass · Answer

(B) Let $f(x) = {\left( {\frac{1}{x}} \right)^x} = {x^{-x}}$.Taking the natural logarithm on both sides,we get $\ln(f(x)) = -x \ln(x)$.Differentiating with respect to $x$,we have $\frac{f'(x)}{f(x)} = -(\ln(x) + x \cdot \frac{1}{x}) = -(\ln(x) + 1)$.Thus,$f'(x) = -f(x)(\ln(x) + 1) = -{\left( {\frac{1}{x}} \right)^x}(\ln(x) + 1)$.For critical points,set $f'(x) = 0$,which implies $\ln(x) + 1 = 0$,so $\ln(x) = -1$,which gives $x = {e^{-1}} = \frac{1}{e}$.Evaluating the function at $x = \frac{1}{e}$,we get $f\left( \frac{1}{e} \right) = {\left( {\frac{1}{1/e}} \right)^{1/e}} = {e^{1/e}}$.Therefore,the maximum value of the function is ${e^{1/e}}$.

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

Similar Questions

$A$ rectangle with one side lying along the x-axis is to be inscribed in the closed region of the $xy$ plane bounded by the lines $y = 0$,$y = 3x$,and $y = 30 - 2x$. The largest area of such a rectangle is

$A$ manufacturer produces $x$ items per week at a total cost of Rs $(x^2+78x+2500)$. The price per unit is given by $8x = 600 - p$,where $p$ is the price of each unit. Then the maximum profit obtained is

Find the maximum area of an isosceles triangle inscribed in the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with its vertex at one end of the major axis.

The maximum value of $xy$ subject to $x+y=7$ is

Slope of the tangent to the curve $y=2 e^x \sin \left(\frac{\pi}{4}-\frac{x}{2}\right) \cos \left(\frac{\pi}{4}-\frac{x}{2}\right)$ where $0 \leq x \leq 2 \pi$ is minimum at $x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module