\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}