If the function f(x) = (frac{1}{x})^{2x} for | vedclass

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(C) Let $f(x) = (\frac{1}{x})^{2x}$.Taking the natural logarithm on both sides,we get $\ln(f(x)) = 2x \ln(\frac{1}{x}) = -2x \ln(x)$.To find the criti

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$e^\pi > \pi^e$

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(C) Let $f(x) = (\frac{1}{x})^{2x}$.Taking the natural logarithm on both sides,we get $\ln(f(x)) = 2x \ln(\frac{1}{x}) = -2x \ln(x)$.To find the critical points,we differentiate with respect to $x$:$\frac{f'(x)}{f(x)} = -2(1 \cdot \ln(x) + x \cdot \frac{1}{x}) = -2(\ln(x) + 1)$.Setting $f'(x) = 0$,we get $\ln(x) = -1$,which implies $x = \frac{1}{e}$.Since $f'(x) > 0$ for $x  \frac{1}{e}$,the function attains its maximum value at $x = \frac{1}{e}$.Thus,$f(x) \leq f(\frac{1}{e})$ for all $x > 0$.$f(\frac{1}{e}) = (\frac{1}{1/e})^{2(1/e)} = e^{2/e}$.For any $x$,$(\frac{1}{x})^{2x} \leq e^{2/e}$.Taking $x = \frac{1}{\pi}$,we have $(\frac{1}{1/\pi})^{2(1/\pi)} \leq e^{2/e}$.$\pi^{2/\pi} \leq e^{2/e}$.Raising both sides to the power of $\frac{\pi e}{2}$,we get $(\pi^{2/\pi})^{\pi e/2} \leq (e^{2/e})^{\pi e/2}$.$\pi^e \leq e^\pi$.Since $\pi 
eq e$,we have $e^\pi > \pi^e$.

If the function $f(x) = (\frac{1}{x})^{2x}$ for $x > 0$ attains the maximum value at $x = \frac{1}{e}$,then:

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