The maximum value of {left( {frac{1}{x}} righ | vedclass

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(B) 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

Vedclass · Accepted Answer

${e^{1/e}}$

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(B) 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 $\frac{f'(x)}{f(x)} = -(\ln(x) + x \cdot \frac{1}{x}) = -(\ln(x) + 1)$.Thus,$f'(x) = -f(x)(\ln(x) + 1) = -{\left( {\frac{1}{x}} \right)^x}(\ln(x) + 1)$.For critical points,set $f'(x) = 0$,which implies $\ln(x) + 1 = 0$,so $\ln(x) = -1$,which gives $x = {e^{-1}} = \frac{1}{e}$.Evaluating the function at $x = \frac{1}{e}$,we get $f\left( \frac{1}{e} \right) = {\left( {\frac{1}{1/e}} \right)^{1/e}} = {e^{1/e}}$.Therefore,the maximum value of the function is ${e^{1/e}}$.

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

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