\begin{frame}
  \frametitle{Curve Sketching}

  \begin{exampleblock}{}
    Sketch the curve of $f(x) = \frac{2x^2}{x^2-1}$.
    \pause\bigskip
    
    \only<-18>{
    The domain is \pause $\{x \mid x \ne \pm 1\}$\pause , that is, $(-\infty,-1) \cup (-1,1) \cup (1,\infty)$
    \pause\bigskip

    We have \quad $f(0) = \pause 0$ \pause\quad and \quad $f(x) = 0 \;\iff\;\pause x = 0$    
    \pause\bigskip

    The vertical asymptotes are \pause\quad $x = -1$ \quad and \quad $x = 1$
    \begin{talign}
      \lim_{x \to -1^-} = \mpause[1]{\infty}
      && \lim_{x \to -1^+} = \mpause[2]{-\infty}
      && \lim_{x \to 1^-} = \mpause[3]{-\infty}
      && \lim_{x \to 1^+} = \mpause[4]{\infty}
    \end{talign}
    \pause\pause\pause\pause\pause

    The horizontal asymptotes are \mpause[3]{ \quad $y = 2$}
    \begin{talign}
      \lim_{x \to \infty} f(x) = \mpause[1]{2}
      && \lim_{x \to -\infty} f(x) = \mpause[2]{2}
    \end{talign}
    \pause\pause\pause\pause
    }
    \only<19-30>{
    \pause[19]
    The derivative is:
    \begin{talign}
      f'(x) = \mpause[1]{ \frac{4x(x^2-1) - 2x^2(2x)}{(x^2-1)^2} }
      \mpause[2]{= \frac{-4x}{(x^2-1)^2} }
    \end{talign}
    \pause\pause\pause
    Thus 
    \begin{itemize}
      \item increasing ($f'(x) > 0$) on \pause $(-\infty,-1)\cup (-1,0)$
    \pause
      \item decreasing on  ($f'(x) < 0$) on \pause $(0,1)\cup (1,\infty)$
    \end{itemize}
    \pause\bigskip
    
    The critical numbers are \pause\quad $x = 0$ \quad (since $f'(0) = 0$)
    \pause
    \begin{itemize}
      \item $f'(x)$ \pause changes from $+$ to $-$ at $0$ $\;\implies\;$ \pause local maximum $(0,0)$
    \end{itemize}
    \pause
    }
    \only<31->{
    \pause[31]\vspace{-2ex}
    \begin{talign}
      f'(x) = \frac{-4x}{(x^2-1)^2}
    \end{talign}
    The second derivative is:
    \begin{talign}
      f''(x) 
      &= \mpause[1]{ \frac{-4(x^2-1)^2 - (-4x)\cdot 2(x^2-1)\cdot 2x}{(x^2-1)^4} } \\
      &\mpause[2]{= \frac{-4(x^2-1) + 16x^2}{(x^2-1)^3} }
      \mpause[3]{= \frac{12x^2 + 4}{(x^2-1)^3} }
    \end{talign}\vspace{-2ex}
    \pause\pause\pause\pause
    \begin{talign}
      &12x^2 + 4 > 0 \quad \text{for all $x$} \\
      &\mpause[1]{f''(x) > 0 \;\iff\; } \mpause[2]{(x^2-1)^3 > 0 \;\iff\; }
      \mpause[3]{x^2-1 > 0 \;\iff\; }
      \mpause[4]{|x| > 1} %\\
%       &\mpause[5]{f''(x) < 0 \;\iff\; } 
%       \mpause[6]{|x| < 1}
    \end{talign}\vspace{-3ex}
    \pause\pause\pause\pause\pause%\pause\pause
    \begin{itemize}
      \item concave upward on \pause $(-\infty,-1)\cup (1,\infty)$
    \pause
      \item concave downward on \pause $(-1,1)$
    \pause
      \item inflection points: \pause none ($-1$ and $1$ not in the domain)
    \end{itemize}
    }
  \end{exampleblock}
  \vspace{10cm}
\end{frame}