The interval in which the function f(x)=x^x, x0, is strictly increasing is

Question

The interval in which the function $f(x)=x^x, x>0$, is strictly increasing is

Accepted Answer

First note that

$$f(x)=x^x=e^{\,x\ln x}\,,\quad x>0.$$

Differentiating gives

$$f'(x)=x^x\bigl(\ln x+1\bigr)\,. $$

Since $x^x>0$, the sign of $f'(x)$ is the sign of $\ln x+1$.  Hence

$ f'(x)>0\quad\Longleftrightarrow\quad \ln x>-1\quad\Longleftrightarrow\quad x>\frac1e. $

So $f$ is strictly increasing for all $x>1/e$.  Among the given choices that corresponds to

Option D: $\displaystyle\bigl[\tfrac1e,\infty\bigr)$.

Solutions · 1