\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}