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

  
  \begin{exampleblock}{}
    Where are inflection points of $f(x) = x^4 - 4x^3$?
    \pause
    \begin{talign}
      f'(x) &= 4x^3 - 12x^2 \\
      \mpause[1]{f''(x) &= 12x^2 - 24x} \mpause[2]{= 12x(x - 2)}
    \end{talign}
    \pause\pause\pause
    Thus $f''(x) = 0$ for \alert{$x = \mpause[1]{0}$} and \alert{$x = \mpause[1]{2}$}.
    \begin{center}
    \pause\pause
    \begin{tabular}{|c|c|l|}
      \hline
      Interval & $f''(x)$ & \\
      \hline
      \mpause[1]{ $x < 0$ } & \mpause[4]{+} & \mpause{ concave upward on $(-\infty,0)$ } \\
      \hline
      \mpause[2]{ $0 < x < 2$ } & \mpause[6]{-} & \mpause{ concave downward on $(0,2)$ } \\
      \hline
      \mpause[3]{ $2 < x$ } & \mpause[8]{+} & \mpause{ concave upward on $(2,\infty)$ } \\
      \hline
    \end{tabular}
    \end{center}
    \pause\pause\pause\pause\pause\pause\pause\pause\pause\pause
    Thus the \alert{inflection points} are:
    \begin{itemize}
    \pause
      \item \alert{$(0,0)$} since the curve changes from concave up to down
    \pause
      \item \alert{$(2,-16)$} since the curve changes from concave down to up
    \end{itemize}
  \end{exampleblock}  
\end{frame}