\begin{frame} \frametitle{Mean Value Theorem} \rolle \only<1>{ \begin{center} \scalebox{.8}{ \begin{tikzpicture}[default,baseline=1cm] \diagram{-.5}{4}{-.5}{4}{1} \diagramannotatez \draw[gray] (.5,2) -- node [black,at end,below] {$a$} (.5,-.2); \draw[gray] (3.5,2) -- node [black,at end,below] {$b$} (3.5,-.2); \draw[gray] (2.5,3) -- node [black,at end,below] {$c$} (2.5,-.2); \begin{scope}[ultra thick] \draw[cgreen,ultra thick] (.5,2) to[out=45,in=180] (2.5,3) to[out=0,in=110] (3.5,2); \end{scope} \node (a) at (2.5,3) {}; \draw[cred] ($(a)+(-1,0)$) -- +(2,0); \end{tikzpicture} }\hspace{1cm} \scalebox{.8}{ \begin{tikzpicture}[default,baseline=1cm] \diagram{-.5}{4}{-.5}{4}{1} \diagramannotatez \draw[gray] (.5,2) -- node [black,at end,below] {$a$} (.5,-.2); \draw[gray] (3.5,2) -- node [black,at end,below] {$b$} (3.5,-.2); \draw[gray] (1.3,2.5) -- node [black,at end,below] {$c_1$} (1.3,-.2); \draw[gray] (2.8,1.5) -- node [black,at end,below] {$c_2$} (2.8,-.2); \begin{scope}[ultra thick] \draw[cgreen,ultra thick] (.5,2) to[out=45,in=180] (1.3,2.5) to[out=0,in=180] (2.8,1.5) to[out=0,in=-110] (3.5,2); \end{scope} \node (a) at (1.3,2.5) {}; \draw[cred] ($(a)+(-1,0)$) -- +(2,0); \node (a) at (2.8,1.5) {}; \draw[cred] ($(a)+(-1,0)$) -- +(2,0); \end{tikzpicture} } \end{center} } \pause \begin{proof} \begin{itemize} \pause \item If $f$ is constant\pause, then $f'(c) = 0$ for all $c$ in $(a,b)$. \pause \item If $f$ is not constant\pause, then there is $x$ in $(a,b)$ such that \vspace{-1ex} \begin{talign} f(x) > f(a) &&\text{ or }&& f(x) < f(a) \end{talign} \pause Assume $f(x) > f(a)$. \pause By the Extreme Value Theorem there is a $c$ in $[a,b]$ such that $f(c)$ is the absolute maximum. \pause\medskip Then $c$ must be in $(a,b)$ and hence is a local maximum. \pause Hence $f'(c) = 0$ by Fermat's Theorem.\vspace{-3.8ex} \end{itemize} \end{proof} \vspace{10cm} \end{frame}