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\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}