The function $f(x) = x^x$ has a stationary point at

$x = \frac{1}{e}$

The correct answer is Option (2) → $x = \frac{1}{e}$ ##

A stationary point of a function $f(x)$ is a point where the derivative of $f(x)$ is equal to zero

i.e., $f'(x) = 0 \quad \text{or} \quad \frac{dy}{dx} = 0$

We have, $f(x) = x^x$

Let $y = x^x$

Taking log on both sides

and $\log y = x \log x$

$∴\frac{1}{y} \frac{dy}{dx} = x \cdot \frac{1}{x} + \log x \cdot 1$

$\frac{dy}{dx} = (1 + \log x)y$

$\Rightarrow \frac{dy}{dx} = (1 + \log x) \cdot x^x \quad [∵y = x^x]$

$∴\frac{dy}{dx} = 0$

$\Rightarrow (1 + \log x) \cdot x^x = 0$

$\Rightarrow \log x = -1$

$\Rightarrow \log x = \log e^{-1}$

$\Rightarrow x = e^{-1} \Rightarrow x = \frac{1}{e}$

Hence, $f(x)$ has a stationary point at $x = \frac{1}{e}$.