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

  \meanvalueshort
  
  \begin{exampleblock}{}
    Suppose that $f(0) = -3$ and $f'(x) \le 5$ for all $x$.\pause\\
    How large can $f(2)$ possibly be?
    \pause\medskip
    
    By assumption, $f$ is differentiable, and hence continuous.
    \pause\medskip
    
    By the Mean Value Theorem for the interval $[0,2]$:
    \begin{talign}
      \text{There exists $c$ in $(0,2)$ such that } f'(c) = \frac{f(2) - f(0)}{2-0} \mpause[1]{= \frac{f(2) + 3}{2}.}
    \end{talign}
    \pause\pause
    We have:
    \begin{talign}
      5 \ge f'(c)\mpause[1]{ = \frac{f(2) + 3}{2}} 
      \mpause[2]{\;\implies\; 10 \ge f(2) + 3}
      \mpause[3]{\;\implies\; 7 \ge f(2)}
    \end{talign}
    \pause\pause\pause\pause
    Thus the largest possible value for $f(2)$ is $7$.
  \end{exampleblock}
  \vspace{10cm}
\end{frame}