The maximum value of (1/x)^x for x 0 is | vedclass

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(C) Let $f(x) = (1/x)^x = x^{-x}$.Taking the natural logarithm on both sides: $\ln(f(x)) = -x \ln(x)$.To find the critical points, differentiate with

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$e^{1/e}$

Vedclass · Answer

(C) Let $f(x) = (1/x)^x = x^{-x}$.Taking the natural logarithm on both sides: $\ln(f(x)) = -x \ln(x)$.To find the critical points, differentiate with respect to $x$: $\frac{d}{dx}(\ln(f(x))) = -[\ln(x) + x \cdot \frac{1}{x}] = -(\ln(x) + 1)$.Set the derivative to $0$: $-(\ln(x) + 1) = 0 \implies \ln(x) = -1 \implies x = e^{-1} = 1/e$.Check the second derivative: $\frac{d^2}{dx^2}(\ln(f(x))) = -\frac{1}{x}$. At $x = 1/e$, this is $-e The maximum value is $f(1/e) = (1/(1/e))^{1/e} = e^{1/e}$.

The maximum value of $(1/x)^x$ for $x > 0$ is

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