In which of the following interval the function $f(x) = x^x,x>0$ is strictly increasing?

$(\frac{1}{e},∞)$

The correct answer is Option (2) → $(\frac{1}{e},∞)$

Given function

$f(x)=x^x,\ x>0$

Differentiating

$f'(x)=x^x(\log x+1)$

For strictly increasing function

$f'(x)>0$

$x^x(\log x+1)>0$

Since $x^x>0$ for $x>0$, condition becomes

$\log x+1>0$

$\log x>-1$

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

Hence $f(x)$ is strictly increasing in $\left(\frac{1}{e},\infty\right)$