Find the values of x for which f ( x )= x^x, x >0 is increasing.

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

Put, $\displaystyle f^{\prime}(x)=0 \Rightarrow x^{x}(1+\log x)=0$

As $\displaystyle x^{x}>0$ for $\displaystyle x>0,(1+\log x)=0 \Rightarrow x=\dfrac{1}{e}$

$\displaystyle \therefore f(x)$ is increasing when $\displaystyle x>\dfrac{1}{e}$ or $\displaystyle x \in\left(\dfrac{1}{e}, \infty\right)$ or $\displaystyle x \in\left[\dfrac{1}{e}, \infty\right)$