\begin{frame}
  \frametitle{Derivatives and the Shape of a Graph}

  \begin{exampleblock}{}
    What are the local extrema of $f(x) = 3x^4 - 4x^3 -12x^2 + 5$?
    \begin{talign}
      f'(x) = 12x(x-2)(x+1)
    \end{talign}
    \pause
    The critical numbers are: \pause $-1$, $0$ and $2$.
    \pause\medskip
    
    \begin{overlayarea}{\textwidth}{3cm}
    \only<-10>{
    We have already seen that:\smallskip
    
    {\small
    \begin{tabular}{|c|c|c|c|c|l|}
      \hline
      Interval & $12x$ & $x-2$ & $x+1$ & $f'(x)$ & \\
      \hline
      x < -1 & - & - & - & - & decreasing on $(-\infty,-1)$ \\
      \hline
      -1 < x < 0 & - & - & + & + & increasing on $(-1,0)$ \\
      \hline
      0 < x < 2 & + & - & + & - & decreasing on $(0,2)$ \\
      \hline
      2 < x & + & + & + & + & increasing on $(2,\infty)$ \\
      \hline
    \end{tabular}
    }
    }
    \only<11>{
    \begin{center}\vspace{-1ex}
    \scalebox{.9}{
    \begin{tikzpicture}[default,baseline=1cm]
      \diagram{-2}{4}{-1}{1.3}{1}
      \diagramannotatez
      \diagramannotateyy{-1/-30,1/30}
      \diagramannotatex{-2,-1,1,2,3}
      \begin{scope}[ultra thick]
        \draw[cgreen] plot[smooth,domain=-2:3,samples=30] function{(3*x**4 - 4*x**3 - 12*x**2 + 5)/30};
      \end{scope}
    \end{tikzpicture}\vspace{-1ex}
    }
    \end{center}
    }
    \end{overlayarea}
    \pause\medskip
    
    We have:
    \begin{itemize}
      \item $f(-1) = 0$ is \pause a local minimum \quad\textcolor{gray}{($f'$ changes from $-$ to $+$)} 
    \pause
      \item $f(0) = 5$ is \pause a local maximum \quad\textcolor{gray}{($f'$ changes from $+$ to $-$)} 
    \pause
      \item $f(2) = -27$ is \pause a local minimum \quad\textcolor{gray}{($f'$ changes from $-$ to $+$)} 
    \end{itemize}
  \end{exampleblock}
\end{frame}