\begin{frame} \frametitle{Review - Midterm Exam 3} \begin{block}{Idea of Linearization} If $f(z)$ is difficult to compute, we instead compute: \begin{itemize} \pause \item tangent $L$ at a point $a$ near $z$ where $f(a)$ and $f'(a)$ are easy \pause \item then we approximate $f(z)$ by $L(z)$ \end{itemize} \end{block}\pause \begin{exampleblock}{} Use differentials to approximate $\sqrt[3]{999}$. \pause\bigskip We have $f(x) = \sqrt[3]{x}$. \pause % We compute linearization at $a = \pause 1000$.\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 $ L(x) = \pause 10 + \frac{1}{300}(x-1000) $ \pause\medskip 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}} \end{talign} \end{exampleblock} \end{frame}