The maximum value of $(\frac{1}{x})^x$ for $x > 0$ is

$e^{1/e}$

The correct answer is Option (2) → $e^{1/e}$

Let

$y = \left( \frac{1}{x} \right)^x = x^{-x}$

Take log:

$\ln y = -x \ln x$

Differentiate and set to zero:

$-\ln x - 1 = 0 \Rightarrow \ln x = -1 \Rightarrow x = \frac{1}{e}$

Substitute back:

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