\begin{frame} \frametitle{Linear Approximation and Differentials} \begin{exampleblock}{Final Exam 2004} Use the linearization method to approximate $(1.98)^4$. \pause\bigskip We have $f(x) = x^4$.\pause\bigskip We need to choose where to compute the linearization: $a = \pause 2$.\pause \begin{talign} f(2) &= \mpause[1]{16} \\ \mpause[2]{f'(x) }&\mpause[2]{= }\mpause[3]{4x^3} & \mpause[4]{f'(2) = } \mpause[5]{4\cdot 2^3} \mpause[6]{= 4 \cdot 8} \mpause[7]{= 32} \end{talign} \pause\pause\pause\pause\pause\pause\pause\pause The linearization of $f$ at $2$ is:\vspace{-.7ex} \begin{talign} L(x) = \mpause[1]{16 + 32(x-2)} \end{talign} \pause\pause Then the approximation of $(1.98)^4$ is:\vspace{-.7ex} \begin{talign} (1.98)^4 \approx \mpause[1]{L(1.98)} &\mpause[2]{= 16 + 32(1.98 - 2)} \mpause[3]{= 16 + 32(-0.02)}\\ &\mpause[4]{= 16 + 32(-\frac{1}{50})} \mpause[5]{= 16 - \frac{16}{25}} \mpause[6]{= \frac{16\cdot 24}{25}} \end{talign} \end{exampleblock} \end{frame}