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\begin{frame}
  \frametitle{Continuity: Intermediate Value Theorem}

  \begin{alertblock}{}
    Whenever applying the Intermediate Value Theorem, it is \emph{important} to check
    that the function is \emph{continuous} on the interval.
  \end{alertblock}
  \pause\medskip
  
  \begin{center}
    \scalebox{.8}{
    \begin{tikzpicture}[default]
      \diagram{-.5}{7}{-.5}{3.2}{1}
      \diagramannotatez
      \diagramannotatex{1,2,3,4,5,6}
      \diagramannotatey{1,2}
      \draw[cred] (-.5,0) to[out=51,in=180] (3,1);
      \draw[cred] (3,2) to[out=0,in=180] (7,3);
      
      \node[include={cred}] at (3,1) {};
      \node[exclude={cred}] at (3,2) {};
    \end{tikzpicture}
    }
  \end{center}
  \pause
  
  Here we have:
  \begin{itemize}
    \item $f(2) < 1$
    \item $f(4) > 2$
  \end{itemize}
  \pause
  But there exists no $2 < c < 4$ such that $f(c) = 1.5$! 
  \vspace{10cm}
\end{frame}