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\begin{frame}
  \frametitle{Linear Approximation and Differentials}

  \begin{exampleblock}{Final Exam 2003 (Fall)}
    Use differentials to approximate $\sqrt[3]{999}$.
    \pause\bigskip

    We have $f(x) = \sqrt[3]{x}$.\pause\bigskip
    
    We choose where to compute the linearization:
    $a = \pause 1000$.\pause
    \begin{talign}
      f(1000) &= \mpause[1]{10} \\
      \mpause[2]{f'(x) }&\mpause[2]{= }\mpause[3]{\frac{1}{3}x^{-\frac{2}{3}} = \frac{1}{3(\sqrt[3]{x})^2}} & 
      \mpause[4]{f'(1000) = \frac{1}{3\cdot 10^2}}\mpause[5]{= \frac{1}{300}}
    \end{talign}
    \pause\pause\pause\pause\pause\pause
    The linearization of $f$ at $1000$ is:\vspace{-.7ex}
    \begin{talign}
      L(x) = \mpause[1]{10 + \frac{1}{300}(x-1000)} 
    \end{talign}
    \pause\pause
    Then the approximation of $\sqrt[3]{999}$ is:\vspace{-.7ex}
    \begin{talign}
      \sqrt[3]{999} \approx
      \mpause[1]{L(999)}
      &\mpause[2]{= 10 + \frac{1}{300}(999-1000)}
      \mpause[3]{= 10 - \frac{1}{300}}
      \mpause[4]{= \frac{2999}{300}}
    \end{talign}
  \end{exampleblock}
\end{frame}