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\begin{document}


\title{On-line Math 21}


\email{david.johnson@lehigh.edu}


\urladdr{http://www.lehigh.edu/\textasciitilde{}dlj0}

\maketitle

\subsection{Curve Sketching}

\begin{example}
\[
f(x)=\frac{2x+3}{x+1}\]

\end{example}

\subsubsection*{Solution}


\subparagraph*{Asymptotes}

Here it is easy to see, since the limit at \( -1 \) is infinte, that \( x=-1 \)
is a vertical asymptote. Also, \( y=2 \) is a horizontal asymptote, and these
are clearly the only possibile asymptotes. 

Now, to get the tails of the graph, as indicated, you have to look more closely.
For \( x \) large, the numerator is 1 more than twice the denominator, so \( f(x) \)
will be a bit more than 2 for \( x>>0 \). But on the other hand, for \( x<<0 \)
(that means {}``\( x \) much less than 0{}''), the denominator is also 1
more than twice the denominator. However, this time, both the numerator and
denominator are \emph{negative}, so 1 more than twice the denominator has absolute
value less than twice the absolute value of the denominator. So, the fraction
is a bit less than 2 for \( x<<0 \).

For the vertical asymptote, since for \( x>-1 \), but \( x \) near \( -1 \),
\( f(x) \) is positive, since both numerator (near +1) and denominator (near
0, but positive) are positive. But it is becoming infinite as \( x \) approaches
\( 0 \), so it must be that \[
\lim _{x\rightarrow -1^{+}}\frac{2x+3}{x+1}=+\infty ,\]
so the curve approaches the asymptote at the top on the right side. Similarly,
\[
\lim _{x\rightarrow -1^{-}}\frac{2x+3}{x+1}=-\infty ,\]
so the curve approaches the asymptote at the bottom on the left side.


\subparagraph*{Intercepts}

\( f(0)=3 \), so \( (0,3) \) is the \( y \)-intercept. There is only one
\( x \)-intercept, where the numerator is \( 0 \), at \( x=-3/2 \), so (-3/2,0)
is the \( x- \)intercept.


\subparagraph*{Increasing/decreasing, and critical points}

\begin{eqnarray*}
f'(x) & = & \frac{2(x+1)-1(2x+3)}{(x+1)^{2}}\\
 & = & \frac{-1}{(x+1)^{2}},
\end{eqnarray*}
so the curve is always decreasing. No critical points.


\subparagraph*{Concavity, and inflection points}

\begin{eqnarray*}
f''(x) & = & \left( \frac{-1}{(x+1)^{2}}\right) '\\
 & = & (-1)(-2)(x+1)^{-3}\\
 & = & \frac{2}{(x+1)^{3}},
\end{eqnarray*}
so the curve is concave up for \( x>-1 \) and concave down for \( x<-1 \). 


\subparagraph*{Draw the graph}

Use this information to draw a fair representation of the graph. 

\begin{itemize}
\item The first information you plot are the asymptotes and {}``tails{}'' {[}Link
to ex6-1.gif{]}
\item Then find intercepts and endpoints; {[}Link to ex6-2.gif{]}
\item Then plot the critical points (and values). Of course, there are none here.
\item Plot the inflection points. None of these in this example, either.
\item Then fill in the graph, connecting the dots and making sure that the graph you
draw is increasing/decreasing where the number line indicates it should be,
and concave-up or concave-down where the sign of the second derivative indicates.
{[}Link to ex6-3.gif{]}
\end{itemize}
{[}Each of these items should trigger the appearance of a new drawing with that
information added.{]}

\vspace{0.3cm}
{\par\centering \includegraphics{ex6-3.ps} \par}\vspace{0.3cm}

\end{document}
