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%TCIDATA{<META NAME="Title" CONTENT="Nonlinear Inequalities">}
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\section{\protect\vspace{1pt}Nonlinear Inequalities\label%
{Nonlinear Inequalities}}

To solve inequalities involving squares and higher powers of the variable
you can use factoring, together with your knowledge of the sign of a product
or quotient. Using the addition and subtraction rules for inequalities, you
can always get all of the nonzero terms on one side, leaving a statement
like $2x^{3}-2x\geq 0$; this is just a statement about the sign of the
nonzero side (in this case, saying that $2x^{3}-2x$ is either a positive
number or $0$). If you can write the nonzero side in factored form, like $%
2x\left( x+1\right) \left( x-1\right) \geq 0$, then the problem is reduced
to deciding when a product of several numbers is positive or $0$.

\subsection{The Sign of a Product or Quotient\label%
{The Sign of a Product or Quotient}}

If a product or quotient has an \emph{even} number of negative factors, then
its value is positive.

If a product or quotient has an \emph{odd} number of negative factors, then
its value is negative.

\subsection{Solving Nonlinear Inequalities\label%
{Solving Nonlinear Inequalities}}

\begin{enumerate}
\item  If necessary, rewrite the inequality so that all nonzero terms appear
on one side of the inequality sign.

\item  If the nonzero side of the inequality involves quotients, bring them
to a common denominator.

\item  Factor the nonzero side of the inequality.

\item  List the intervals determined by factorization.

\item  Make a table or diagram of the signs of each factor on each interval.
In the last row of the table, determine the sign of the product (or
quotient) of these factors.

\item  Determine the solution set from the last row of the table. Be sure to
check whether the inequality is satisfied by some or all of the endpoints of
the intervals (this may happen if the inequality involves $\leq $ or $\geq $%
).
\end{enumerate}

\begin{itemize}
\item[Example]  Solve the inequality 
\[
x^{2}-5x+6\leq 0
\]

\item[Solution]  All the nonzero terms are already on one side of the
inequality sign, and there are no quotients, so the first two steps are
taken care of. Now factor 
\[
\left( x-3\right) \left( x-2\right) \leq 0
\]
According to your knowledge of the sign of a product or quotient, you can
solve this if you know exactly when each of the factors is positive or
negative. Since $x-3$ is negative for $x<3$ and positive for $x>3$, and $x-2$
is negative for $x<2$ and positive for $x>2$, the problem is reduced to
examining what happens in the four cases $x<3$, $x>3$, $x<2$ and $x>2$, and
checking the values $x=3$ and $x=2$ as well. So the intervals you need to
examine are 
\begin{eqnarray*}
x &<&2\text{ or the interval }\left( -\infty ,2\right)  \\
2 &<&x<3\text{ or the interval }\left( 2,3\right)  \\
x &>&3\text{ or the interval }\left( 3,\infty \right) 
\end{eqnarray*}
Now make the table for step 5: 
\[
\begin{tabular}{|l||l|l|l|}
\hline
Interval & $\left( -\infty ,2\right) $ & $\left( 2,3\right) $ & $\left(
3,\infty \right) $ \\ \hline\hline
Sign of $x-2$ & $-$ & $+$ & $+$ \\ \hline
Sign of $x-3$ & $-$ & $-$ & $+$ \\ \hline\hline
Sign of $\left( x-2\right) \left( x-3\right) $ & $+$ & $-$ & $+$ \\ \hline
\end{tabular}
\]
From the table, you can easily see that $\left( 2,3\right) $ is in the
solution set, and that $\left( -\infty ,2\right) $ and $\left( 3,\infty
\right) $ are not. As for the endpoints, you see that both $x=2$ and $x=3$
satisfy $\left( x-3\right) \left( x-2\right) \leq 0$, so they both belong in
the set. So the final solution is 
\[
2\leq x\leq 3\text{ or the interval }\left[ 2,3\right] 
\]
\end{itemize}

\noindent \hrulefill

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