ProofOfDarbouxsTheorem 2002-06-06 03:56:10 2008-06-05 06:18:41 Proof Darboux's theorem (analysis) 0 % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newcommand{\Prob}[2]{\mathbb{P}_{#1}\left\{#2\right\}} \newcommand{\norm}[1]{\left\|#1\right\|} % Some sets \newcommand{\Nats}{\mathbb{N}} \newcommand{\Ints}{\mathbb{Z}} \newcommand{\Reals}{\mathbb{R}} \newcommand{\Complex}{\mathbb{C}} Without loss of generality we migth and shall assume $f'_{+}(a)>t>f'_{-}(b)$. Let $g(x):=f(x)-tx$. Then $g'(x)=f'(x)-t$, $g'_{+}(a)>0>g'_{-}(b)$, and we wish to find a zero of $g'$. Since $g$ is a continuous function on $[a,b]$, it attains a maximum on $[a,b]$. Since $g'_+(a)>0$ and $g'_+(b)<0$ \PMlinkname{Fermat's theorem}{FermatsTheoremStationaryPoints} states that neither $a$ nor $b$ can be points where $f$ has a local maximum. So a maximum is attained at some $c \in (a,b)$. But then $g'(c)=0$ again by \PMlinkname{Fermat's theorem}{FermatsTheoremStationaryPoints}.