\chapter{Polynomial and rational functions}
\section{Polynomial Inequalities}
A polynomial inequality is an inequality involving polynomials, such as
\[
\begin{array}{ccccc}
x^2-9<0 &
\mbox{or} &
x^3-x^2\geq 12x &
\mbox{or} &
x^4+1\leq 2x^2
\end{array}
\]
Our main motivation to study polynomial inequalities is to find the domain of functions
that contain the square root of a polynomial function. As we know, the input of a square root
cannot be negative, and so to find the domain we need to solve a polynomial inequality.
Just like for the inequalities we studied in Unit 1.5, the solutions will typically be intervals,
or unions of intervals, and we need to write that in interval notation.
To solve a polynomial inequality, we proceed as in the following example.
\begin{example}
Suppose we want to solve the polynomial inequality
\(3x^2\geq 12x\). First, we move all terms to the left, in order to have only 0 on the right:
\[3x^2-12x \geq 0.\]
So we need to find all the values of \(x\) that make the quadratic function either positive
or zero. We now change the inequality to an equation, simply by replacing \(\geq \) with \(=\),
and solve it:
\[
\begin{array}{rcll}
3x^2-12x & = & 0 & \mbox{Change the inequality to an equation}\\[1ex]
3x(x-4) & = & 0 & \mbox{Solve by factoring}\\[1ex]
3x=0 \mbox{ or } x-4=0 &&&\mbox{Set each factor \(=0\)} \\[1ex]
x=0 \mbox{ or } x=4 &&&
\end{array}
\]
Next, we plot the solutions we have found on a number line.
\[
\begin{array}{c}
\img{U2_3F1.png}{}{20em}{}
\end{array}
\]
Note that the two points divide the number line into three different intervals,
\((-\infty,0)\), \((0,4)\), and \((4,\infty)\).
The plotted points \(x=0\) and \(x=4\) of course tell us where \(f(x)=0\). To find out
where \(f(x)\) is positive or negative, we select a test point in each of the three intervals.
Any point for a given interval will work. We choose \(x=-1\), \(x=1\) and \(x=5\), shown in red
in the picture below:
\[
\img{U2_3F2.png}{}{20em}{}
\]
Now we can test the inequality \(3x^2-12x\geq 0\)
and find out which of the three intervals will be part of the
solution. We want to exclude the \(x\)-values that give a negative answer.
If we substitute \(x=-1\), we find
\[3(-1)^2 -12(-1)=3(1)+12=15,\]
a positive number. So the interval \((-\infty, 0]\) is part of the solution.
If we substitute \(x=1\), we find
\[3(1)^2 -12(1)=3(1)-12=-9,\]
a negative number. So the interval \((0, 4)\) is not part of the solution.
If we substitute \(x=5\), we find
\[3(5)^2 -12(5)=75-60=15,\]
a positive number. So the interval \([4, \infty)\) is part of the solution.
We conclude that the solution of the inequality, in interval notation, is
\[(-\infty,0]\cup [4,\infty).\]
Note the use of brackets \( [ \ \ ]\) around 0 and 4, because the inequality was of type \(\geq\),
meaning that the endpoints are included.
\end{example}
We can also rely on our knowledge of the graph of polynomial functions in order to solve polynomial inequalities, as in
the next example.
\begin{example}
We solve the inequality
\[ x^3 - 2x^2 + 9x > 12x. \]
First we subtract 12 from both side, to get a zero on one side of the inequality:
\[ x^3 - 2x^2 - 3x > 0. \]
Then we factor the polynomial. As we know, this step may be hard or even impossible. The rational zero theorem is often helpful. In our examples, we will
confine ourselves to simple cases that can be easily factored. In this example, we first factor out \(x\):
\[ x(x^2-2x+9) > 0. \]
Then we factor the quadratic polynomial in the parentheses:
\[ x(x - 3)(x + 1) > 0. \]
So the zeros of the function are \(x=0, x = -1, x=3\), corresponding to the \(x\)-intercepts of \(g(x) = x^3 - 2x^2 - 3x\).
Now draw a quick graph of the same function. We are only interested in the \(x\)-intercepts
so it is not necessary to find the \(y\)-intercept or any other points. All we need to know is that the end behavior is like \(x^3\)
(because the Leading Coefficient is positive), and the \(x\)-intercepts are \( (0,0), (-1,0)\) and \((3,0)\) in order to draw the following graph:
\[\img{U2_3F7a.png}{}{12em}{}.\]
Since the inequality is asking us to find the \(x\)-values for which the function is greater than zero,
we need to find the \(x\) values for which the graph is above the \(x\)-axis, where \(y\) is positive.
This problem was discussed in Unit 1.8 (see Example 1.8.1). The solid part of the graph in the
following picture corresponds to the solution.
Note that we do not include the \(x\)-intercepts because the function is equal to zero at those points:
\[\img{U2_3F8a.png}{}{12em}{}.\]
The solution set \(S\) consists of those parts of the \(x\)-axis corresponding to where the function is positive:
\[\img{U2_3FAa.png}{}{20em}{}\]
In interval notation, the solution is
\[ (-1,0)\cup(3, \infty). \]
We are using parentheses \( (\ \ )\) because the inequality is of type \(> \), so the end-points are excluded.
\end{example}
\begin{example}
Solve the inequality, graph the solution on a number line, and write the answer in interval notation:
\[ x^{2} - 2x + 9 \geq 12. \]
This time we do include the zeros
of the function in the solution because the inequality is now asking us to find the \(x\)-values
for which the function is greater than or equal to zero:
\[\img{U2_3FB.png}{}{20em}{}\]
In interval notation, the solution is
\[ (-\infty, -1] \cup [3, \infty). \]
\end{example}
\subsection{Domain of functions involving square roots}
We can now find the domain of functions that have a quadratic function inside a square root.
\begin{example} Consider the function \(f(x) = \sqrt{9 - 4x^{2}}\).
The domain will consist of all the \(x\) values
that make the quadratic function
inside the square root positive or zero. In other words, we need
\[ 9 - 4x^{2} \geq 0. \]
We already have a zero on one side of the inequality, so we factor the quadratic function
and set it equal to zero:
\[ (3 - 2x)(3 + 2x)= 0 \]
The zeros are: \(x = -\dfrac{3}{2}=-1.5\), \(x=\dfrac{3}{2}=1.5\). Draw a quick graph of the
function \(g(x) = 9 - 4x^{2}\), that is a parabola opening down:
\[\img{U2_3F11.png}{}{15em}{}.\]
Since the square root is only defined when its input is non-negative, we need to find the \(x\)-values
for which the \(9-4x^2\) is greater than or equal to zero, and that means
where the graph is above or on
the \(x\)-axis. Looking at the graph,
\[\img{U2_3F12.png}{}{15em}{}.\]
we see that the domain is
\[ \mbox{Domain } = \left[-1.5, 1.5\right]. \]
\end{example}
Problems
\problem
Solve the inequalities, and write the answer in interval notation.
\begin{enumerate}
\item
\( x^2 -4x -12 < 0\)
\item
\(2x^2 -3x-9 > 0\)
\item
\(4x^2+4x+1 \leq 0\)
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
\( x^2 -4x -12 < 0\) means that we want the quadratic function \(x^2 -4x -12\) to be negative.
\[
\begin{array}{rcll}
x^2-4x-12 & < & 0 & \mbox{The given inequality}\\[1ex]
x^2 -4x -12 & = & 0 & \mbox{Change the inequality to an equation}\\[1ex]
(x-6)(x+2) & = & 0 & \mbox{Solve by factoring}\\[1ex]
x-6=0 \mbox{ or } x+2=0 &&& \mbox{Set each factor \(=0\)} \\[1ex]
x=6 \mbox{ or } x=-2 &&& \mbox{Find \(x\)}
\end{array}
\]
Plot the solutions on a number line (shown in black), and choose test points (shown in red):
\[
\begin{array}{c}
\img{U2_3F3.png}{}{20em}{}
\end{array}
\]
Test the inequality with \(x=-3\):
\[(-3)^2 -4(-3) -12 = 9+12-12=9\]
So the interval \( (-\infty,-2)\) is not part of the solution
(because we are looking for negative numbers).
Test the inequality with \(x=0\):
\[(0)^2 -4(0) -12 = -12\]
So the interval \((-2,6)\) is part of the solution.
Test the inequality with \(x=7\):
\[(7)^2 -4(7) -12 = 49-28-12=9\]
So the interval \((6,\infty)\) is not part of the solution.
The solution of the inequality is
\[(-2,6).\]
Note the parentheses \( ( \ \ )\), because the inequality is of type \( < \), meaning
that the endpoints are excluded.
\item
\(2x^2 -3x-9 > 0\) means that we want the quadratic function to be positive.
\[
\begin{array}{rcll}
2x^2 -3x-9 & > & 0 & \mbox{The given inequality}\\[1ex]
2x^2 -3x-9 & = & 0 & \mbox{Change the inequality to an equation}\\[1ex]
2x^2-6x+3x-9 & = & 0 & \mbox{ \(-18=(-6)(3), -3=-6+3\)}\\[1ex]
2x(x-3)+3(x-3)& = & 0 & \mbox{Group factor}\\[1ex]
(x-3)(2x+3)&=&0&\\[1ex]
2x+3=0 \mbox{ or } x-3=0 &&&\mbox{Set each factor\(=0\)} \\[1ex]
x=-3/2 \mbox{ or } x=3 &&&
\end{array}
\]
Plot the solutions on a number line, and choose test points:
\[
\begin{array}{c}
\img{U2_3F4.png}{}{15em}{}
\end{array}
\]
Test the inequality with \(x=-2\):
\[2(-2)^2 -3(-2) -9 = 8+6-9=5\]
So the interval \( (-\infty,-3/2)\) is part of the solution.
Test the inequality with \(x=0\):
\[2(0)^2 -3(0) -9 = -9\]
So the interval \( (-3/2,3)\) is not part of the solution.
Test the inequality with \(x=4\):
\[2(4)^2 -3(4) -9 = 32-12-9=11\]
So the interval \( (3,\infty)\) is part of the solution.
The solution of the inequality is
\[\left(-\infty,-\frac{3}{2}\right)\cup (3,\infty).\]
\item
\(4x^2+4x+1 \leq 0\) means that we want the quadratic function to be negative or zero.
\[
\begin{array}{rcll}
4x^2 +4x+1 & \leq & 0 & \mbox{The given inequality}\\[1ex]
4x^2+4x+1 & = & 0 & \mbox{Change the inequality to an equation}\\[1ex]
(2x+1)(2x+1) & = & 0 & \mbox{Solve by factoring}\\[1ex]
2x+1=0 & & \\[1ex]
x=-1/2 & & & \mbox{There is only one solution}
\end{array}
\]
Plot the solution on a number line, and choose test points:
\[
\begin{array}{c}
\img{U2_3F5.png}{}{12em}{}
\end{array}
\]
Test the inequality with \(x=-2\):
\[4(-2)^2 +4(-2) +1 = 16-8+1=9\]
So the interval \( (-\infty,-1/2)\) is not part of the solution.
Test the inequality with \(x=0\):
\[4(0)^2 +4(0) +1 = 1\]
So the interval \( (-1/2,\infty)\) is not part of the solution.
Neither interval is part of the solution. But the inequality is of type \(\leq\), meaning that
the \(x\)-value \(x=-1/2\) that makes it zero will be a solution. So the solution of the inequality
is a single number, or in interval notation:
\[ \{-1/2\}.\]
\end{enumerate}
\end{sol} \mproblem
Solve the inequalities, and write the answer in interval notation:
\begin{enumerate}
\item
\( 6x^2-x -1 \leq 0\)
\item
\( x^2+6x+9 \geq 0\)
\item
\(x^2+x+1 < 0\)
\end{enumerate}
\problem
Find the domain of the functions:
\begin{enumerate}
\item
\(f(x)=\sqrt{9-x^2}\)
\item
\(g(t)= t+\sqrt{t^2-1}\)
\item
\(\displaystyle h(x)=\frac{1}{\sqrt{x^2-6x+8}}\)
\item
\(\displaystyle p(x)=\frac{x}{\sqrt{x^2+9}}\)
\end{enumerate}
\begin{sol}
\begin{enumerate}
\item
\(f(x)=\sqrt{9-x^2}\). The expression under the square root cannot be negative. This means that
we need \(9-x^2\) to be either zero or positive. In other words, we need to solve the
quadratic inequality
\[9-x^2\geq 0,\]
and the solution of the inequality will be the domain of the given function.
\[
\begin{array}{rcll}
9-x^2 & \geq & 0 & \mbox{The given inequality}\\[1ex]
9-x^2 & = & 0 & \mbox{Change the inequality to an equation}\\[1ex]
(3+x)(3-x) & = & 0 & \mbox{Solve by factoring}\\[1ex]
x+3=0 \mbox{ or } x-3=0&&&\\[1ex]
x=-3 \mbox{ or } x=3&&&
\end{array}
\]
Plot the solutions on a number line, and choose test points:
\[
\begin{array}{c}
\img{U2_3F6.png}{}{20em}{}
\end{array}
\]
Test the inequality with \(x=-4\):
\[9-(-4)^2 = 9-16 = -7\]
So the interval \( (-\infty,-3]\) is not part of the domain.
Test the inequality with \(x=0\):
\[9-(0)^2 = 9\]
So the interval \( [-3,3]\) is part of the domain.
Test the inequality with \(x=4\):
\[9-(4)^2 = 9-16 = -7\]
So the interval \( [3,\infty)\) is not part of the domain.
The domain of the function is (remember to include endpoints, because the inequality
is of type \( \geq \) )
\[\mbox{Domain }=[-3,3].\]
\item
\(g(t)= t+\sqrt{t^2-1}\).
We need \(t^2-1\geq 0\).
Solving the inequality (check it!), we find that the solution, and so the domain of the function, is
\[D=(-\infty,-1]\cup[1,\infty).\]
\item
\(\displaystyle h(x)=\frac{1}{\sqrt{x^2-6x+8}}\).
This time we need \(x^2-6x +8 > 0\), because the quadratic function is in the denominator,
and zero cannot be used. So we need the quadratic function to be positive, and the endpoints
of the interval will not be part of the domain.
Solving the inequality (check!), we find
\[D=(-\infty,2)\cup (4,\infty).\]
\item
\(\displaystyle p(x)=\frac{x}{\sqrt{x^2+9}}\).
As in the previous part, we need \(x^2 + 9 > 0\). If we try to solve the equation
\(x^2+9=0\) we find \(x^2=-9\), and since we cannot take the square root of a negative number,
the equation has no solutions. This means that \(x^2+9\) can never be zero. Choosing \(x=0\) as
test point we find \(0^2 +9=9\), and so the function \(x^2+9\) is always positive. So the
domain is
\[D=(-\infty, \infty).\]
\end{enumerate}
\end{sol} \mproblem
Find the domain of the functions:
\begin{enumerate}
\item
\(f(x)=x-\sqrt{4x^2-1}\)
\item
\(\displaystyle g(x)=\frac{x+2}{\sqrt{x^2+2x+1}}\)
\item
\(h(t)= \sqrt{16-t^2}\)
\item
\(\displaystyle p(x)=\frac{x}{\sqrt{x^2+x-6}}\)
\end{enumerate}