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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}