\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}