\begin{frame} \frametitle{Linear Approximation and Differentials} \begin{exampleblock}{Final Exam 2003 (Fall)} Use differentials to approximate $\sqrt[3]{999}$. \pause\bigskip We have $f(x) = \sqrt[3]{x}$.\pause\bigskip We choose where to compute the linearization: $a = \pause 1000$.\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:\vspace{-.7ex} \begin{talign} L(x) = \mpause[1]{10 + \frac{1}{300}(x-1000)} \end{talign} \pause\pause 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}} \mpause[4]{= \frac{2999}{300}} \end{talign} \end{exampleblock} \end{frame}