\begin{frame}
  \frametitle{Review - Midterm Exam 3}

  \begin{block}{Idea of Linearization}
    If $f(z)$ is difficult to compute, we instead compute:
    \begin{itemize}
      \pause
      \item tangent $L$ at a point $a$ near $z$ where $f(a)$ and $f'(a)$ are easy
      \pause
      \item then we approximate $f(z)$ by $L(z)$
    \end{itemize}
  \end{block}\pause
  \begin{exampleblock}{}
    Use differentials to approximate $\sqrt[3]{999}$.
    \pause\bigskip

    We have $f(x) = \sqrt[3]{x}$. \pause
    %
    We compute linearization at
    $a = \pause 1000$.\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
    $
      L(x) = \pause 10 + \frac{1}{300}(x-1000)
    $
    \pause\medskip
    
    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}}
    \end{talign}
  \end{exampleblock}
\end{frame}