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

e $e^e$ $e^{\frac{1}{e}}$ $\left(\frac{1}{e}\right)^{\frac{1}{e}}$

$e^{\frac{1}{e}}$

The correct answer is Option (3) → $e^{\frac{1}{e}}$ $\frac{1}{x^x}$ $\log y=-x\log x$ differentiating wrt x $\frac{1}{y}\frac{dy}{dx}=-\log x-1⇒\frac{dy}{dx}=-\frac{(\log x+1)}{x^x}$ so at optimize point $\frac{dy}{dx}=0⇒\log x=-1$ $⇒x=\frac{1}{e}$ max. value → $y(\frac{1}{e})=e^{\frac{1}{e}}$