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\begin{frame}
\frametitle{Mean Value Theorem}

\rolle
\only<1>{
\begin{center}
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\pause

\begin{proof}
\begin{itemize}
\pause
\item If $f$ is constant\pause, then $f'(c) = 0$ for all $c$ in $(a,b)$.
\pause
\item If $f$ is not constant\pause, then there is $x$ in $(a,b)$
such that \vspace{-1ex}
\begin{talign}
f(x) > f(a) &&\text{ or }&& f(x) < f(a)
\end{talign}
\pause
Assume $f(x) > f(a)$. \pause
By the Extreme Value Theorem there is a $c$ in $[a,b]$ such that $f(c)$ is
the absolute maximum.
\pause\medskip

Then $c$ must be in $(a,b)$ and hence is a local maximum. \pause
Hence $f'(c) = 0$ by Fermat's Theorem.\vspace{-3.8ex}
\end{itemize}
\end{proof}

\vspace{10cm}
\end{frame}