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

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

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$e^{1/e}$

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(B) Let $f(x) = (1/x)^x$.Taking the natural logarithm on both sides:$\ln f(x) = x \ln(1/x) = -x \ln x$.Differentiating with respect to $x$:$\frac{f'(x)}{f(x)} = -(\ln x + x \cdot \frac{1}{x}) = -(1 + \ln x)$.For critical points,set $f'(x) = 0$:$-(1 + \ln x) = 0 \Rightarrow \ln x = -1 \Rightarrow x = e^{-1} = 1/e$.To check for maxima,we observe the sign of $f'(x)$ around $x = 1/e$:For $x  0$.For $x > 1/e$,$\ln x > -1$,so $1 + \ln x > 0$,which means $f'(x) Since $f'(x)$ changes from positive to negative at $x = 1/e$,the function attains a maximum at $x = 1/e$.The maximum value is $f(1/e) = (1/(1/e))^{1/e} = e^{1/e}$.

The maximum value of $(1/x)^x$ is -

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