Let $f:(0,\infty)\to\mathbb{R}:x\mapsto x^x$
Prove, using the mean value theorem, that: $\forall x\in (0,\frac{1}{e}),(f(x) >e^\frac{-1}{e})$
Mean Value Theorem Let $f$ be a continuous function on $[a,b]$ that is differentiable on $(a,b)$. Then there exist [at least one] $x$ in $(a,b)$ such that
$f'(x)=\frac{f(b)-f(a)}{b-a}$
I think I can prove that $f$ is continuous on $[0,\frac{1}{e}]$ and differentiable on $(0,\frac{1}{e})$. By the mean value theorem: $\exists x\in (0,\frac{1}{e}),f'(x)=\frac{f(\frac{1}{e})-f(0)}{\frac{1}{e}-0}=\frac{(\frac{1}{e})^\frac{1}{e}}{\frac{1}{e}}=e\cdot(e^{-\frac{1}{e}})$