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\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{{[}Mean Value Theorem (MVT){]}}} If \( f \) is continuous on
a closed interval \( [a,b] \), and differentiable on the open interval \( (a,b) \),
then there is some point \( c\in (a,b) \) so that \[
f'(c)=\frac{f(b)-f(a)}{b-a}.\]

\end{thm}
\begin{proof}
The (venerable) trick is to construct from \( f \) another function, one which
satisfies Rolle's theorem, and see what it tells us. Set \[
g(x):=f(x)-f(a)-\frac{f(b)-f(a)}{b-a}(x-a).\]
Then, because \( f \) is, \( g \) is continuous on the closed interval \( [a,b] \),
and differentiable on the open interval \( (a,b) \). Also, \( g(a)=g(b)=0 \),
so we satisfy Rolle's theorem's hypothesis. That means that there is some \( c\in (a,b) \)
so that \( g'(c)=0 \). But, \[
g'(x)=f'(x)-\frac{f(b)-f(a)}{b-a},\]
 so \[
0=g'(c)=f'(c)-\frac{f(b)-f(a)}{b-a}\Rightarrow f'(c)=\frac{f(b)-f(a)}{b-a},\]
 as desired. \end{proof}

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