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

  \begin{exampleblock}{}
    Find the linearization of $f(x) = \sqrt{x+ 3}$ at $1$
    and use it to approximate $\sqrt{3.98}$.
    \pause\medskip
    
    We have:
    \begin{talign}
      f(1) &= \mpause[1]{\sqrt{3+1} = 2} \\
      \mpause[2]{f'(x)} &\mpause[2]{= }\mpause[3]{\frac{1}{2\sqrt{x+3}}} &
      \mpause[4]{f'(1)} &\mpause[4]{= }\mpause[5]{\frac{1}{2\sqrt{1+3}}} \mpause[6]{= \frac{1}{4}}
    \end{talign}
    \pause\pause\pause\pause\pause\pause\pause
    Thus the linearization of $f$ at $1$ is:
    \begin{talign}
      L(x) = \mpause[1]{ 2 + \frac{1}{4}(x-1) }
    \end{talign}
    \pause\pause
    Thus for $x$ close to $1$ we approximate $f(x)$ by:\vspace{-0.5ex}
    \begin{talign}
      f(x) = \sqrt{x+3} \quad\approx\quad  2 + \frac{1}{4}(x-1)
    \end{talign}
    \pause
    In particular:\vspace{-1.2ex}
    \begin{talign}
      \sqrt{3.98} = \mpause[1]{\sqrt{0.98 + 3} \approx} \mpause[2]{2 + \frac{1}{4}(0.98-1) =}
      \mpause[3]{2 - 0.005 =} \mpause[4]{1.995} 
    \end{talign}
  \end{exampleblock}
\end{frame}