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\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}