\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}