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

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(C) 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 re

Vedclass · Accepted Answer

$e^{1/e}$

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(C) 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 $\frac{1}{f(x)} f'(x) = -(\ln x + x \cdot \frac{1}{x}) = -(1 + \ln x)$.Thus,$f'(x) = -f(x)(1 + \ln x)$.For critical points,set $f'(x) = 0$,which implies $1 + \ln x = 0$,so $\ln x = -1$,which gives $x = e^{-1} = \frac{1}{e}$.To check for maximum,we observe the sign of $f'(x)$. For $x  0$,and for $x > \frac{1}{e}$,$f'(x) The maximum value is $f\left(\frac{1}{e}\right) = \left(\frac{1}{1/e}\right)^{1/e} = e^{1/e}$.

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

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