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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}