%% LyX 1.1 created this file.  For more info, see http://www.lyx.org/.
%% Do not edit unless you really know what you are doing.
\documentclass[11pt,oneside]{amsart}
\usepackage[T1]{fontenc}
\usepackage{geometry}
\geometry{verbose,letterpaper,tmargin=1in,bmargin=1in,lmargin=1in,rmargin=1in}
\setlength\parskip{\medskipamount}
\setlength\parindent{0pt}

\makeatletter


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% LyX specific LaTeX commands.
\providecommand{\LyX}{L\kern-.1667em\lower.25em\hbox{Y}\kern-.125emX\@}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Textclass specific LaTeX commands.
 \theoremstyle{plain}    
 \newtheorem{thm}{Theorem}[section]
 \numberwithin{equation}{section} %% Comment out for sequentially-numbered
 \numberwithin{figure}{section} %% Comment out for sequentially-numbered

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% User specified LaTeX commands.
\setcounter{section}{4}
\setcounter{subsection}{0}

\makeatother

\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{The Mean Value Theorem}

\begin{thm}
\textbf{\textup{{[}Rolle's Theorem{]}.}} If \( f \) is continuous on a closed
interval \( [a,b] \), and differentiable on the open interval \( (a,b) \),
and if \( f(a)=f(b) \), then there is some point \( c\in (a,b) \) so that
\( f'(c)=0 \). 
\end{thm}
\begin{proof}
If, for some \( x\in (a,b) \), \( f(x)>f(a) \), then there is a maximum point
by the theorem we proved a while back about the existence of a maximum of a
continuous function. That maximum value has to be greater than \( f(a) \),
so it isn't going to happen at \( x=b \), since \( f(a)=f(b) \). So, it's
at some \( c\in (a,b) \). Our assumption was that the function was differentiable
on \( (a,b) \), so \( f'(c) \) exists. By the previous result, then, \( f'(c)=0 \).
\\
The same argument works if, for some \( x \), \( f(x)<f(a) \) (all the inequalities
are reversed). The only other possibility is that \( f(x) \) is never bigger
or smaller than \( f(a) \), which says that \( f(x) \) is constant, so has
0 derivative at all \( x\in (a,b) \). \end{proof}

\end{document}
