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

  \begin{exampleblock}{Final Exam 2005}
    Use differential approximation, or the linearization method, to approximate $\sqrt[4]{15.5}$.
    \pause\bigskip
    
    We have $f(x) = \sqrt[4]{x}$.\pause\bigskip
    
    We need to choose where to compute the linearization:
    $a = \pause 16$.\pause
    \begin{talign}
      f(16) &= \mpause[1]{2} \\
      \mpause[2]{f'(x) }&\mpause[2]{= }\mpause[3]{\frac{1}{4}x^{-\frac{3}{4}}} & 
      \mpause[4]{f'(16) = }
      \mpause[5]{\frac{1}{4}16^{-\frac{3}{4}}}
      \mpause[6]{= \frac{1}{4}\sqrt[4]{16}^{-3} }
      \mpause[7]{= \frac{1}{4}\cdot \frac{1}{8} }
      \mpause[8]{= \frac{1}{32} }
    \end{talign}
    \pause\pause\pause\pause\pause\pause\pause\pause\pause
    The linearization of $f$ at $16$ is:\vspace{-.7ex}
    \begin{talign}
      L(x) = \mpause[1]{2 + \frac{1}{32}(x-16)}
    \end{talign}
    \pause\pause
    Then the approximation of $\sqrt[4]{15.5}$ is:\vspace{-.7ex}
    \begin{talign}
      \sqrt[4]{15.5} \approx
      \mpause[1]{L(15.5)}
      \mpause[2]{= 2 + \frac{1}{32}(15.5-16)}
      \mpause[3]{= 2 - \frac{1}{64}}
      \mpause[4]{= \frac{127}{64}}
    \end{talign}
  \end{exampleblock}  
\end{frame}