\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}