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

  \begin{block}{Proof of the Mean Value Theorem}
    Let $f$ be a function satisfying the all of the following:
    \begin{itemize}
      \item $f$ is continuous on $[a,b]$
      \item $f$ is differentiable on $(a,b)$
    \end{itemize}
    \pause
    
    \begin{center}\vspace{-1ex}
    \scalebox{.8}{
    \begin{tikzpicture}[default,baseline=1cm]
      \diagram{-.5}{4}{-.2}{2.5}{1}
      \diagramannotatez
      \draw[gray] (.5,2) -- node [black,at end,below] {$a$} (.5,-.2);
      \draw[gray] (3.5,1) -- node [black,at end,below] {$b$} (3.5,-.2);
      \mpause[4]{ \draw[gray] (2,2.5) -- node [black,at end,below] {$c$} (2,-.2); }
      \begin{scope}[ultra thick]
        \draw[cgreen] plot[smooth,domain=.5:3.5,samples=30] function{3-((x-2)/1.5)**2 - (x-.5)/3} node[above,yshift=4mm,xshift=-2mm] {$f$};
      \end{scope}
      \mpause[1]{
      \draw[cblue] (.5,2) -- node[below,pos=.8] {$L$} (3.5,1);
      \node[include=cblue] at (.5,2) {};
      \node[include=cblue] at (3.5,1) {};
      }
      \mpause[2]{
      \begin{scope}[ultra thick]
        \draw[cred] plot[smooth,domain=.5:3.5,samples=30] function{3-((x-2)/1.5)**2 - (x-.5)/3 - 2 + (x-.5)/3} node[above,yshift=4mm,xshift=-2mm] {$g$};
      \end{scope}
      }
%       \node (a) at (2,2.5) {};
%       \draw[cred] ($(a)+(-1.5,1.5/3)$) -- +(3,-3/3);
    \end{tikzpicture}\vspace{-1ex}
    }
    \end{center}
    \pause
    
    Let \structure{$L = mx + n$} be the line through \structure{$(a,f(a))$} and \structure{$(b,f(b))$}.
    \pause\smallskip
    
    Define \alert{$g = f - L$}. \pause Then $g(a) = 0$ and $g(b) = 0$. 
    \pause\smallskip
    
    By Rolle's Theorem there is $c$ in $(a,b)$ such that \alert{$g'(c) = 0$}.
    \pause\smallskip

    Since $f = g + L$ \pause we get \alert{$f'(c) = g'(c) + m \pause = m \pause = \frac{f(b)-f(a)}{b-a}$}.    
  \end{block}
\end{frame}