\begin{frame} \frametitle{Linear Approximation and Differentials} \begin{exampleblock}{} Find the linearization of $f(x) = \sqrt{x+ 3}$ at $1$ and use it to approximate $\sqrt{3.98}$. \pause\medskip We have: \begin{talign} f(1) &= \mpause[1]{\sqrt{3+1} = 2} \\ \mpause[2]{f'(x)} &\mpause[2]{= }\mpause[3]{\frac{1}{2\sqrt{x+3}}} & \mpause[4]{f'(1)} &\mpause[4]{= }\mpause[5]{\frac{1}{2\sqrt{1+3}}} \mpause[6]{= \frac{1}{4}} \end{talign} \pause\pause\pause\pause\pause\pause\pause Thus the linearization of $f$ at $1$ is: \begin{talign} L(x) = \mpause[1]{ 2 + \frac{1}{4}(x-1) } \end{talign} \pause\pause Thus for $x$ close to $1$ we approximate $f(x)$ by:\vspace{-0.5ex} \begin{talign} f(x) = \sqrt{x+3} \quad\approx\quad 2 + \frac{1}{4}(x-1) \end{talign} \pause In particular:\vspace{-1.2ex} \begin{talign} \sqrt{3.98} = \mpause[1]{\sqrt{0.98 + 3} \approx} \mpause[2]{2 + \frac{1}{4}(0.98-1) =} \mpause[3]{2 - 0.005 =} \mpause[4]{1.995} \end{talign} \end{exampleblock} \end{frame}