\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}