\begin{frame}
  \frametitle{Mean Value Theorem}
  
  \rolle
  \only<1>{
  \begin{center}
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  \scalebox{.8}{
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    \diagram{-.5}{4}{-.5}{4}{1}
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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}