The maximum value of f(x) = (1/x)^x is

The correct answer is Option (3) → $e^{1/e}$

Given: Function $f(x) = \left( \frac{1}{x} \right)^x = x^{-x}$

Let: $y = x^{-x}$

Taking natural log:

$\ln y = -x \ln x$

Differentiating both sides:

$\frac{1}{y} \cdot \frac{dy}{dx} = -\ln x - 1$

$\frac{dy}{dx} = y(-\ln x - 1) = x^{-x}(-\ln x - 1)$

Set $\frac{dy}{dx} = 0$:

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

Maximum value:

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

∴ Maximum value of $f(x) = x^{-x}$ is: $e^{1/e}$