\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}