The function

Solution:

The values of $x$ for which $f^{\prime}(x)=0$ are called stationary values.

Let $f(x)=x^{x}$.

$f(x)=x^{x}$

$\Rightarrow \log f(x)=\log x^{x}$

$\Rightarrow \log f(x)=x \log x$

Differentiating both sides with respect to x, we get

$\frac{1}{f(x)} \times f^{\prime}(x)=x \times \frac{1}{x}+\log x \times 1$

 

$\Rightarrow f^{\prime}(x)=x^{x}(1+\log x)$

For stationary point, we have

$f^{\prime}(x)=0$

 

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

$\Rightarrow 1+\log x=0 \quad\left[x^{x}>0\right]$

$\Rightarrow \log x=-1$

 

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

Thus, the function $f(x)=x^{x}$ has a stationary point at $x=\frac{1}{e}$.

Hence, the correct answer is option (b).