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\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}