The local maximum value of the function f(x) | vedclass

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(C) Given $f(x) = \left(\frac{2}{x}\right)^{x^{2}}$.Taking natural logarithm on both sides:$\ln f(x) = x^{2} (\ln 2 - \ln x)$.Differentiating with res

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$(e)^{\frac{2}{e}}$

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(C) Given $f(x) = \left(\frac{2}{x}\right)^{x^{2}}$.Taking natural logarithm on both sides:$\ln f(x) = x^{2} (\ln 2 - \ln x)$.Differentiating with respect to $x$:$\frac{f'(x)}{f(x)} = 2x(\ln 2 - \ln x) + x^{2} \left(-\frac{1}{x}\right) = 2x \ln 2 - 2x \ln x - x = x(2 \ln 2 - 2 \ln x - 1)$.$f'(x) = f(x) \cdot x \cdot (2 \ln 2 - 2 \ln x - 1)$.For critical points,set $f'(x) = 0$. Since $f(x) > 0$ and $x > 0$,we have:$2 \ln 2 - 2 \ln x - 1 = 0$$2 \ln \left(\frac{2}{x}\right) = 1$$\ln \left(\frac{2}{x}\right) = \frac{1}{2}$$\frac{2}{x} = e^{1/2} = \sqrt{e}$$x = \frac{2}{\sqrt{e}}$.At $x = \frac{2}{\sqrt{e}}$,the function attains a local maximum.The local maximum value is $f\left(\frac{2}{\sqrt{e}}\right) = \left(\frac{2}{2/\sqrt{e}}\right)^{(2/\sqrt{e})^{2}} = (\sqrt{e})^{4/e} = (e^{1/2})^{4/e} = e^{2/e}$.

The local maximum value of the function $f(x) = \left(\frac{2}{x}\right)^{x^{2}}$,$x > 0$,is

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