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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


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: $$\exists x\in (0,\frac{1}{e}),f'(x)=\frac{f(\frac{1}{e})-f(0)}{\frac{1}{e}-0}=\frac{(\frac{1}{e})^\frac{1}{e}}{\frac{1}{e}}=e\cdot(e^{-\frac{1}{e}})$$

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up vote 2 down vote accepted

You can't use MVT in $[0,\frac{1}{e}]$ as $f$ is not defined at $0$ (we need continuity there for MVT). You can avoid this shortcoming by extending $f$ continuously. Because $$\lim_{x\to 0^+}x^x=\lim_{x\to 0^+}e^{x\ln x}=1$$ the function: $$g(x)=\begin{cases} x^x &\mbox{if } x\in (0,\frac{1}{e}]\\ 1&\mbox{if } x=0\end{cases}$$ is continuous and differentiable in $(0,\frac{1}{e})$ with derivative $$g^{\prime}(x)=x^x(\ln x+1)$$ If you apply MVT on $[0,\frac{1}{e}]$ you won't get any good results as you have seen. Instead you will get $$\xi^\xi =e^{\frac{-1}{e}+1}-1$$ for some $\xi\in (0,\frac{1}{e})$ which is nothing like what you want to show. Instead you should apply MVT in this interval: $[x,\frac{1}{e}]$.

Here is what I propose:

Since $x\in (0,\frac{1}{e})$, $\ln x<-1$ and $$f^{\prime}(x)<0$$ If we apply the MVT now on $[x,\frac{1}{e}]$ (for fixed $0<x<\frac{1}{e}$) we have that $\exists \xi \in (x,\frac{1}{e})$ so that $$\frac{e^{\frac{-1}{e}}-x^x}{\frac{1}{e}-x}=f^{\prime}(\xi)<0\Rightarrow e^{\frac{-1}{e}}<x^x$$ Note that in this approach you don't need to define $g$ as $f$ is continuous in $[x,\frac{1}{e}]$ for fixed $0<x<\frac{1}{e}$

Moral: When you want to prove an inequality using MVT the interval that you should use may be of the form $[a,x]$ or $[x,b]$

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Take $log$, it is equivalent to prove $xln(x)>-\frac{1}{e}$. Using the points $x, \frac{1}{e}$, then mean value theorem tells us that $\frac{xlnx+\frac{1}{e}}{x-\frac{1}{e}}=1+ln(a)$, for some $a\in [x, \frac{1}{e}]$, so we need to show $(\frac{1}{e}-x)(1+ln(a))<0$, which is clear since $\frac{1}{e}>x>0, ln(a)<ln(\frac{1}{e})<-1$

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I get this: $$\frac{f(\frac{1}{e})-f(x)}{\frac{1}{e}-x}=\frac{\frac{1}{e} \ln{\frac{1}{e}}-x \ln x}{\frac{1}{e}-x}=\frac{-\frac{1}{e} - x \ln x}{\frac{1}{e}-x}=\ln (a) + 1$$ – Kasper Dec 14 '12 at 18:41
Thanks, fixed the error. – ougao Dec 14 '12 at 18:44

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