Find the maximum value of (1/x)^x. | vedclass

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Question

(B) Let $f(x) = (1/x)^x = x^{-x}$.Taking the natural logarithm on both sides,$\ln(f(x)) = -x \ln(x)$.Differentiating with respect to $x$,$\frac{f'(x)}

Vedclass · Accepted Answer

$(e)^{1/e}$

Vedclass · Answer

(B) Let $f(x) = (1/x)^x = x^{-x}$.Taking the natural logarithm on both sides,$\ln(f(x)) = -x \ln(x)$.Differentiating with respect to $x$,$\frac{f'(x)}{f(x)} = -(\ln(x) + x \cdot \frac{1}{x}) = -(\ln(x) + 1)$.So,$f'(x) = -(1/x)^x (\ln(x) + 1)$.For critical points,set $f'(x) = 0$,which implies $\ln(x) + 1 = 0$,so $\ln(x) = -1$,which gives $x = 1/e$.To check for maxima,we evaluate the second derivative or observe the sign change of $f'(x)$.For $x  0$.For $x > 1/e$,$\ln(x) > -1$,so $f'(x) Since $f'(x)$ changes from positive to negative at $x = 1/e$,$f(x)$ has a local maximum at $x = 1/e$.The maximum value is $f(1/e) = (1/(1/e))^{1/e} = e^{1/e}$.

Find the maximum value of $(1/x)^x$.

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