\begin{frame} \frametitle{Mean Value Theorem} \rolle \begin{exampleblock}{} Show that the function $f$ is one-to-one (never takes the same value twice):\vspace{-1.7ex} \begin{talign} f(x) = x^3 + x - 1 \end{talign} \pause Assume there would be $x_1 < x_2$ such that $f(x_1) = f(x_2)$. \pause\medskip The function $f$ is continuous and differentiable on $[x_1,x_2]$. \pause\medskip By Rolle's Theorem there exists $c$ in $(x_1,x_2)$ with $f'(c) = 0$. \pause\medskip This is a contradiction since $f'(x) = \pause 3x^2 + 1 \pause \ge 1$ for all $x$. \\\pause There no $x_1 < x_2$ such that $f(x_1) = f(x_2)$. Thus $f$ is one-to-one. \end{exampleblock} \vspace{10cm} \end{frame}