\begin{frame} \frametitle{Newton's Method} \begin{center} \scalebox{.8}{ \begin{tikzpicture}[default,baseline=1cm] \diagram{-.5}{4}{-.5}{4}{1} \diagramannotatez \begin{scope}[ultra thick] \draw[cgreen,ultra thick] plot[smooth,domain=-0:3.3,samples=200] function{-.5 + (.5*x)**3}; \node[include=cgreen] (r) at (1.58,0) {}; \node[anchor=south east] at (r) {$r$}; \end{scope} \draw[gray] (3.2,-.2) -- node[at start,below,black] {$x_1$} node[include=cred,at end] {} (3.2,{-.5 + (.5*3.2)^3}); \tangent{4cm}{.5cm}{-.5 + (.5*\x)^3}{3.2} \draw[gray] (2.28,-.2) -- node[at start,below,xshift=1mm,black] {$x_2$} (2.28,.2); \end{tikzpicture} } \end{center} How can we compute $x_2$? \pause The tangent at $(x_1,f(x_1))$ is \begin{talign} y = \mpause[1]{f(x_1) + f'(x_1)(x-x_1) } \end{talign} \pause\pause For the $x$-intercept $x_2$ of the tangent, we have: \begin{talign} 0 = f(x_1) + f'(x_1)(x_2-x_1) \mpause[1]{ \implies \alert{x_2 = x_1 - \frac{f(x_1)}{f'(x_1)}}} \end{talign} \pause\pause We can repeat this process to get $x_3$, $x_4$, $x_5$\ldots \vspace{10cm} \end{frame}