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

  \begin{exampleblock}{Final Exam 2004}
    Use the linearization method to approximate $(1.98)^4$.
    \pause\bigskip
    
    We have $f(x) = x^4$.\pause\bigskip
    
    We need to choose where to compute the linearization:
    $a = \pause 2$.\pause
    \begin{talign}
      f(2) &= \mpause[1]{16} \\
      \mpause[2]{f'(x) }&\mpause[2]{= }\mpause[3]{4x^3} & 
      \mpause[4]{f'(2) = }
      \mpause[5]{4\cdot 2^3}
      \mpause[6]{= 4 \cdot 8}
      \mpause[7]{= 32}
    \end{talign}
    \pause\pause\pause\pause\pause\pause\pause\pause
    The linearization of $f$ at $2$ is:\vspace{-.7ex}
    \begin{talign}
      L(x) = \mpause[1]{16 + 32(x-2)}
    \end{talign}
    \pause\pause
    Then the approximation of $(1.98)^4$ is:\vspace{-.7ex}
    \begin{talign}
      (1.98)^4 \approx
      \mpause[1]{L(1.98)}
      &\mpause[2]{= 16 + 32(1.98 - 2)}
      \mpause[3]{= 16 + 32(-0.02)}\\
      &\mpause[4]{= 16 + 32(-\frac{1}{50})}
      \mpause[5]{= 16 - \frac{16}{25}}
      \mpause[6]{= \frac{16\cdot 24}{25}}
    \end{talign}
  \end{exampleblock}  
\end{frame}