\begin{frame} \frametitle{Mean Value Theorem} \begin{block}{Proof of the Mean Value Theorem} Let $f$ be a function satisfying the all of the following: \begin{itemize} \item $f$ is continuous on $[a,b]$ \item $f$ is differentiable on $(a,b)$ \end{itemize} \pause \begin{center}\vspace{-1ex} \scalebox{.8}{ \begin{tikzpicture}[default,baseline=1cm] \diagram{-.5}{4}{-.2}{2.5}{1} \diagramannotatez \draw[gray] (.5,2) -- node [black,at end,below] {$a$} (.5,-.2); \draw[gray] (3.5,1) -- node [black,at end,below] {$b$} (3.5,-.2); \mpause[4]{ \draw[gray] (2,2.5) -- node [black,at end,below] {$c$} (2,-.2); } \begin{scope}[ultra thick] \draw[cgreen] plot[smooth,domain=.5:3.5,samples=30] function{3-((x-2)/1.5)**2 - (x-.5)/3} node[above,yshift=4mm,xshift=-2mm] {$f$}; \end{scope} \mpause[1]{ \draw[cblue] (.5,2) -- node[below,pos=.8] {$L$} (3.5,1); \node[include=cblue] at (.5,2) {}; \node[include=cblue] at (3.5,1) {}; } \mpause[2]{ \begin{scope}[ultra thick] \draw[cred] plot[smooth,domain=.5:3.5,samples=30] function{3-((x-2)/1.5)**2 - (x-.5)/3 - 2 + (x-.5)/3} node[above,yshift=4mm,xshift=-2mm] {$g$}; \end{scope} } % \node (a) at (2,2.5) {}; % \draw[cred] ($(a)+(-1.5,1.5/3)$) -- +(3,-3/3); \end{tikzpicture}\vspace{-1ex} } \end{center} \pause Let \structure{$L = mx + n$} be the line through \structure{$(a,f(a))$} and \structure{$(b,f(b))$}. \pause\smallskip Define \alert{$g = f - L$}. \pause Then $g(a) = 0$ and $g(b) = 0$. \pause\smallskip By Rolle's Theorem there is $c$ in $(a,b)$ such that \alert{$g'(c) = 0$}. \pause\smallskip Since $f = g + L$ \pause we get \alert{$f'(c) = g'(c) + m \pause = m \pause = \frac{f(b)-f(a)}{b-a}$}. \end{block} \end{frame}