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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}
\[
\displaystyle y=\sqrt{x}-\sqrt{x-1}.\]

\end{example}

\subsubsection*{Solution}


\subparagraph*{Asymptotes}

Here there are no vertical asymptotes, even though there is an {}``end{}''
to the domain. This function is only defined for \( x\geq 1 \), so that the
stuff inside the square roots is always positive. 

Now, for horizontal asymptotes,\begin{eqnarray*}
\lim _{x\rightarrow \infty }\sqrt{x}-\sqrt{x-1} & = & \lim _{x\rightarrow \infty }\frac{\left( \sqrt{x}-\sqrt{x-1}\right) \left( \sqrt{x}+\sqrt{x-1}\right) }{\left( \sqrt{x}+\sqrt{x-1}\right) }\\
 & = & \lim _{x\rightarrow \infty }\frac{\left( x-(x-1)\right) }{\left( \sqrt{x}+\sqrt{x-1}\right) }\\
 & = & \lim _{x\rightarrow \infty }\frac{1}{\left( \sqrt{x}+\sqrt{x-1}\right) }\\
 & = & 0,
\end{eqnarray*}
so the \( x \)-axis is a horizontal asymptote. Since \( f(x)>0 \), the tail
is above the horizontal asymptote on the right (there is only the asymptote
on the right).


\subparagraph*{Intercepts}

Here we should also worry about endpoints. Since \( x\geq 1 \) is the domain,
we need to find \( f(1) \), which is of course \( 1 \). The only place to
look for intercepts is on the \( x \)-axis, which requires a solution of \begin{eqnarray*}
0 & = & f(x)\\
 & = & \sqrt{x}-\sqrt{x-1},\: \textrm{or}\\
\sqrt{x} & = & \sqrt{x-1},
\end{eqnarray*}
which has no solutions. So there are no intercepts. 


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

\begin{eqnarray*}
f'(x) & = & \frac{1}{2\sqrt{x}}-\frac{1}{2\sqrt{x-1}}\\
 & = & \frac{\sqrt{x-1}-\sqrt{x}}{2\sqrt{x}\sqrt{x-1}},
\end{eqnarray*}
so, just as there were no \( x \)-intercepts, there can't be any critical points.
\( f'(x)<0 \) for all \( x. \) 


\subparagraph*{Concavity, and inflection points}

\begin{eqnarray*}
f''(x) & = & \left( \frac{1}{2\sqrt{x}}-\frac{1}{2\sqrt{x-1}}\right) '\\
 & = & \left( \frac{1}{2}x^{-1/2}-\frac{1}{2}(x-1)^{-1/2}\right) '\\
 & = & -\frac{1}{4}x^{-3/2}+\frac{1}{4}(x-1)^{-3/2}\\
 & = & \frac{x^{3/2}-(x-1)^{3/2}}{4x^{3/2}(x-1)^{3/2}},
\end{eqnarray*}
which is always positive. So, there are no inflection points, and the graph
is always concave up. 


\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 ex7-1.gif{]}
\item Then find intercepts and endpoints; {[}Link to ex7-2.gif{]}
\item Then plot the critical points (and values). Again, none in this case.
\item Plot the inflection points. Also none of these.
\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 ex7-3.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{ex7-3.ps} \par}\vspace{0.3cm}

\end{document}
