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

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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

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