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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}
\[
\displaystyle y=xe^{1/x}.\]

\end{example}

\subsubsection*{Solution}


\subparagraph*{Asymptotes}

As \( x\rightarrow 0 \), this function gets a little complicated. For \( x>0 \),
yet \( x \) near 0, this one-sided limit \[
\lim _{x\rightarrow 0^{+}}xe^{1/x}\]
 is an <a href="hospital.html">indeterminate form</a>, of type \( 0\cdot \infty  \).
In order to convert this to something that l'H\^{o}pital's rule can handle,
it needs to be switched to either \( \infty /\infty  \) or \( 0/0 \). Since
I don't want to invert the exponential term, we'll convert to \( \infty /\infty  \):\begin{eqnarray*}
\lim _{x\rightarrow 0^{+}}xe^{1/x} & = & \lim _{x\rightarrow 0^{+}}\frac{e^{1/x}}{1/x}\\
 & = & \lim _{x\rightarrow 0^{+}}\frac{e^{1/x}(-1/x^{2})}{-1/x^{2}}\\
 & = & \lim _{x\rightarrow 0^{+}}e^{1/x}\\
 & = & \infty ,
\end{eqnarray*}
so from the right side, the line \( x=0 \) is a vertical asymptote. However,
if you approach from the other side, \[
\lim _{x\rightarrow 0^{-}}xe^{1/x}\]
is not an indeterminate form at all, since \[
\lim _{x\rightarrow 0^{-}}e^{1/x}=0,\]
because \( 1/x \) is headed to \emph{negative} infinity, and so \[
\lim _{x\rightarrow 0^{-}}xe^{1/x}=0\]
 as well. So, the line \( x=0 \) is a vertical asymptote on one side only,
and the graph approaches the origin from the other side (although \( (0,0) \)
is of course not on the graph).

As \( x\rightarrow \pm \infty  \), then \( e^{1/x}\rightarrow e^{0}=1 \),
and so \( xe^{1/x} \) approaches \( \pm \infty  \) as \( x \) does. But,
for \( x \) large, since that exponential term is nearly 1, the graph ought
to look like a line, and in fact\[
\lim _{x\rightarrow \infty }xe^{1/x}-(x+1)=\lim _{x\rightarrow \infty }x\left( e^{1/x}-(1+1/x)\right) ,\]
which is an indeterminate form of type \( \infty \cdot 0 \), so converting
to \( 0/0 \),\begin{eqnarray*}
\lim _{x\rightarrow \infty }xe^{1/x}-x & = & \lim _{x\rightarrow \infty }\frac{\left( e^{1/x}-(1+1/x)\right) }{1/x}\\
 & = & \lim _{x\rightarrow \infty }\frac{e^{1/x}(-1/x^{2})-(-1/x^{2})}{-1/x^{2}},\, \textrm{using l}'\textrm{Hopital}'\textrm{s}\, \textrm{rule}\\
 & = & \lim _{x\rightarrow \infty }e^{1/x}-1\\
 & = & 0,
\end{eqnarray*}
so that the graph has a slant-asymptote \( y=x+1 \).


\subparagraph*{Intercepts}

Since \( 0 \) is not in the domain, there can be no \( y \)-intercept, even
though there is a one-sided limit towards a point on the \( y \)-axis. For
\( x\neq 0 \), though, there is no solution to \( f(x)=0 \), so that there
are no \( x \)-intercepts, either.


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

\begin{eqnarray*}
f'(x) & = & 1\cdot e^{1/x}+x\cdot e^{1/x}(-1/x^{2})\\
 & = & e^{1/x}\left( 1-\frac{1}{x}\right) ,
\end{eqnarray*}
so there is a critical point at \( x=1 \) only, with critical value \( f(1)=e \),
and the curve is increasing (\( f'(x)>0 \)) for \( x>1 \), and for \( x<0 \).
The curve is decreasing, \( f'(x)<0 \), only when \( 0<x<1 \). 


\subparagraph*{Concavity, and inflection points}

\begin{eqnarray*}
f''(x) & = & \left( e^{1/x}\left( 1-\frac{1}{x}\right) \right) '\\
 & = & e^{1/x}(-1/x^{2})\left( 1-\frac{1}{x}\right) +e^{1/x}(+1/x^{2})\\
 & = & \frac{e^{1/x}}{x^{2}}\left( \left( \frac{1}{x}-1\right) +1\right) \\
 & = & \frac{e^{1/x}}{x^{3}},
\end{eqnarray*}
so there are no inflection points, the graph is concave up for \( x>0 \) and
concave down for \( x<0 \).


\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 ex8-1.gif{]}
\item Then find intercepts and endpoints. But there are none for this graph.
\item Then plot the critical points (and values). I always mark it with a horizontal
line to remind myself that it is a critical point. {[}Link to ex8-2.gif{]}
\item Plot the inflection points. None of these, at least
\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 ex8-3.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{ex8-3.ps} \par}\vspace{0.3cm}

\end{document}
