\begin{frame} \frametitle{Maximum and Minimum Values} \begin{exampleblock}{} Assume that an object is moving with speed \begin{talign} v(t) = (t-1)^3 - 4t^2 + 9t + 5 \quad\quad\quad 0\le t \le 5 \end{talign}% Find the absolute minimum and maximum acceleration. \bigskip \only<-11>{% \pause The acceleration is: \begin{talign} a(t) = v'(t) = \mpause[1]{ 3(t-1)^2 - 8t + 9} \mpause[2]{= 3t^2 - 14t + 12} \end{talign} \pause\pause\pause Since $a$ is cont. on $[0,5]$ we can use Closed Interval Method. \begin{talign} &a'(t) = \mpause[1]{ 6t - 14 } &&\mpause[2]{a'(t) = 0 \;\iff\; }\mpause[3]{t = \frac{7}{3}} \end{talign} \pause\pause\pause\pause The only critical number is $\frac{7}{3}$. \pause Note that $\frac{7}{3}$ is in $[0,5]$.\\\pause No other critical numbers since $a'(t)$ is defined everywhere. }% \only<12-26>{ \pause[12] The acceleration is: \begin{talign} &a(t) = v'(t) = 3t^2 - 14t + 12\\[-1ex] &a'(t) = 6t - 14 & a'(t) = 0 \;\iff\; t = \frac{7}{3} \end{talign} \pause The values at critical numbers and end points of the interval:\vspace{-1ex} \begin{talign} &a\left(\frac{7}{3}\right) = \mpause[1]{3\left(\frac{7}{3}\right)^2 - 14\frac{7}{3} + 12} \mpause[2]{= \frac{7\cdot 7}{3} - \frac{14\cdot 7}{3} + \frac{36}{3}} \mpause[3]{= -\frac{13}{3}} \\ &\mpause[4]{a(0) =}\mpause[5]{ 12}\\[-.5ex] &\mpause[6]{a(5) =}\mpause[7]{ 3\cdot 5^2 - 14 \cdot 5 + 12} \mpause[8]{= 15\cdot 5 - 14 \cdot 5 + 12} \mpause[9]{= 17} \end{talign} \pause\pause\pause\pause\pause\pause\pause\pause\pause\pause }% \only<27->{ \begin{center}\vspace{-2ex} \scalebox{.7}{ \begin{tikzpicture}[default,baseline=1cm] {\def\diaborderx{1cm} \def\diax{t} \def\diay{} \diagram{-1}{5}{-1.5}{4}{1}} \diagramannotateyy{-1/-5,1/5,2/10,3/15} \diagramannotatex{-1,1,2,3,4} \diagramannotatez \begin{scope}[cgreen,ultra thick] \draw plot[smooth,domain=0:5,samples=30] function{ ((x-1)**3 - 4*x**2 + 9*x + 5) / 5} node[right] {$v(t)$}; \mpause[27]{ \draw[cred] plot[smooth,domain=0:5,samples=30] function{ (3*x**2 - 14*x + 12) / 5} node[above] {$a(t)$}; \node[include=cred] at (7/3,-13/3/5) {}; \node[include=cred] at (5,17/5) {}; } \end{scope} \end{tikzpicture} } \end{center} }% \pause[23]% The absolute minimum acceleration is \pause$a\left(\frac{7}{3}\right) = -\frac{13}{3}$.\\\pause The absolute maximum acceleration is \pause$a(5) = 17$.\pause\pause \end{exampleblock} \vspace{10cm} \end{frame}