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

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

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

The correct answer is Option (2) → $x=\frac{1}{e}$ $f(x) =x^x$ $f'(x)=x^x(1+\log x)=0$ $⇒\log x=-1$ $x=\frac{1}{e}$ → critical point