# The local maximum value of the function

Question:

The local maximum value of the function

$f(x)=\left(\frac{2}{x}\right)^{x^{2}}, x>0$, is

1. $(2 \sqrt{\mathrm{e}})^{\frac{1}{c}}$

2. $\left(\frac{4}{\sqrt{\mathrm{e}}}\right)^{\frac{c}{4}}$

3. $(\mathrm{e})^{\frac{2}{e}}$

4. 1

Correct Option: , 3

Solution:

$f(x)=\left(\frac{2}{x}\right)^{x^{2}} ; x>0$

$\ell \mathrm{n} \mathrm{f}(\mathrm{x})=\mathrm{x}^{2}(\ell \mathrm{n} 2-\ell \mathrm{n} \mathrm{x})$

$f^{\prime}(x)=f(x)\{-x+(\ln 2-\ln x) 2 x\}$

$f^{\prime}(x)=\underbrace{f(x)}_{+} \cdot \underbrace{x}_{+} \underbrace{(2 \ell \operatorname{n} 2-2 \ell n x-1)}_{g(x)}$

$\mathrm{g}(\mathrm{x})=2 \ell \mathrm{n}^{2}-2 \ell \mathrm{n} \mathrm{x}-1$

$=\ell \mathrm{n} \frac{4}{\mathrm{x}^{2}}-1=0 \Rightarrow \mathrm{x}=\frac{2}{\sqrt{\mathrm{e}}}$

$\mathrm{LM}=\frac{2}{\sqrt{\mathrm{e}}}$

Local maximum value $=\left(\frac{2}{2 / \sqrt{\mathrm{e}}}\right)^{\frac{4}{\mathrm{e}}} \Rightarrow \mathrm{e}^{\frac{2}{\mathrm{c}}}$

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