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

  \meanvalueshort

  \begin{exampleblock}{}
    Consider the function
    \begin{talign}
      f(x) = x^3 - x
    \end{talign}
    on the interval $[a,b]$ with $a = 0$ and $b = 2$.
    \pause\medskip
    
    This is a polynomial, thus continuous and differentiable on $[0,2]$.\hspace{-2ex}
    \pause\medskip
    
    By the Mean Value Theorem, there is a $c$ in $(0,2)$ such that
    \begin{talign}
      f'(c) = \frac{f(2) - f(0)}{2-0} \mpause[1]{= \frac{6}{2} = 3}
    \end{talign}
    \pause\pause
    Indeed, we can find such a $c$, namely: $f'\left(\frac{2}{\sqrt{3}}\right) = 3$.
  \end{exampleblock}
  \vspace{10cm}
\end{frame}