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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}
\vspace{0.3cm}

\begin{example}
\( f(x)=\sin (x)+\cos (x) \), \( x\in [0,2\pi ] \). 
\end{example}

\subsubsection*{Solution}


\subparagraph*{Intercepts}

Here the intercepts might be harder to find in general, since we are now working
with a {}``transcendental{}'' function. However, it works out without too
much trouble in this case. To find the \( y \)-intercept, \( f(0)=1 \), so
\( (0,1) \) is the \( y \)-intercept. For the \( x \)-intercepts, you have
to solve \( 0=\sin (x)+\cos (x), \) or \( \cos (x)=-\sin (x) \). If you think
about the unit circle, you can see that this happens when the angle is \( -\pi /4,\, \textrm{or}\, 3\pi /4 \).
But, here we need to keep within the domain, so this occurs when \( x=3\pi /4 \)
and \( x=7\pi /4 \). 

Since this graph has a bounded domain, it also helps to find the endpoints,
\( (0,1) \), which we already found since it was the \( y \)-intercept, and
\( (2\pi ,1) \).


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

\( f'(x)=\cos (x)-\sin (x) \), so critical points are where \( \sin (x)=\cos (x) \),
which occurs, in the domain, at \( x=\pi /4 \) and \( x=5\pi /4 \). The critical
values are \( f(\pi /4)=\sqrt{2} \) and \( f(5\pi /4)=-\sqrt{2} \). It's a
little harder to see the sign of \( f'(x) \) since the function is not factorable,
but remember that the only places that \( f'(x)=0 \) is at these two points,
so if it's positive at some point in between, it's positive at \emph{all} points
in between. I prefer to think about how the various parts of \( f'(x) \) are
changing to see the signs. Since \( f'(0)=1>0 \), then \( f'(x) \) is positive
on the whole interval \( [0,\pi /4] \), and as you pass \( \pi /4 \), \( \sin (x) \)
increases, while \( \cos (x) \) decreases, and \( f'(\pi /4)=0 \), so in the
next interval, \( (\pi /4,5\pi /4), \) \( f'(x)<0 \). As you cross \( 5\pi /4 \),
\( \sin (x) \) is decreasing (getting more negative), but \( \cos (x) \) is
increasing (less negative), so since \( f'(5\pi /4)=0 \), \( f'(x)>0 \) on
the interval \( (5\pi /4,2\pi ] \). 


\subparagraph*{Draw the graph}

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

{*} The first information you plot are the intercepts and endpoints; {[}Link
to ex3-1.gif{]}

{*} 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 ex3-2.gif{]}

{*} 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. {[}Link to ex3-3.gif{]}

{[}Each of these items should trigger the appearance of a new drawing with that
information added.{]}

\vspace{0.3cm}
{\par\centering \includegraphics{ex3-3.ps} \par}\vspace{0.3cm}

\end{document}
