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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}
\( f(x)=x^{4}+4x^{3}+4x^{2}-2 \). 
\end{example}

\subsubsection*{Solution}


\subparagraph*{Intercepts}

Not every calculation can be equally easy for these functions. Typically, the
problems you see in the texts have perhaps one, or two, of: intercepts, or critical
points, or inflection points, easy to find. The others are usually difficult
to evaluate. That is especially true with this problem. I do not want to find
some 4-degree formula, like the quadratice formula, to solve it. Such a formula
does exist, but you don't want to see it.

The only intercept we can easily find is the \( y \)-intercept at \( (0,-2) \).
I used Maple tro draw a graph, and saw that there are \( x \)-intercepts at
about \( 1/2 \) and \( -5/2 \) (they are not exactly at those points, but
close). 


\subparagraph*{Increasing/decreasing, critical points}

\( f'(x)=4x^{3}+12x^{2}+8x \), so critical points occur when\begin{eqnarray*}
0 & = & f'(x)\\
 & = & 4x^{3}+12x^{2}+8x\\
 & = & 4x(x^{2}+3x+2)\\
 & = & 4x(x+1)(x+2),
\end{eqnarray*}
so the only places where \( f'(x)=0 \) are \( x=0,-1,-2 \). The values are:
\( f(0)=-2 \), \begin{eqnarray*}
f(-1) & = & (-1)^{4}+4(-1)^{3}+4(-1)^{2}-2\\
 & = & 1-4+4-2\\
 & = & -1,
\end{eqnarray*}
and\begin{eqnarray*}
f(-2) & = & (-2)^{4}+4(-2)^{3}+4(-2)^{2}-2\\
 & = & 16-32+16-2\\
 & = & -2,
\end{eqnarray*}
or the critical points and values are \( (0,-2), \) \( (-1,-1) \), and \( (-2,-2) \).

In between, the sign of \( f'(x) \) is given as follows:

\vspace{0.3cm}
{\par\centering \includegraphics{ex5-mum-line-1.ps} \par}
\vspace{0.3cm}


\subparagraph*{Concavity and points of inflection}

\( f''(x)=12x^{2}+24x+8 \), so inflection points occur at solutions to \begin{eqnarray*}
0 & = & f''(x)\\
 & = & 12x^{2}+24x+8\\
 & = & 4(3x^{2}+6x+2),
\end{eqnarray*}
which has roots at \[
x=-1\pm \frac{1}{\sqrt{3}},\]
using again the quadratic formula, or, really, Maple. Finding the \( y \)-values
of these points is a hassle, let's not. 

Here is the number line indicating the sign of the second derivative:

\vspace{0.3cm}
{\par\centering \includegraphics{ex5-num-line-2.ps} \par}
\vspace{0.3cm}


\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 intercepts and endpoints; {[}Link to
ex5-1.gif{]}
\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 ex5-2.gif{]}
\item Plot the inflection points. I mark them with an {}``I{}'' to indicate that
they are inflection points. {[}Link to ex5-3.gif{]}
\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 ex5-4.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{ex5-4.ps} \par}\vspace{0.3cm}

\end{document}
