# Chapter 3 Polynomial and Rational Functions 3.6 Polynomial and ```Chapter 3
Polynomial and
Rational Functions
3.6 Polynomial and
Rational Inequalities
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1
Objectives:
•
•
Solve polynomial inequalities.
Solve rational inequalities.
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Definition of a Polynomial Inequality
A polynomial inequality is any inequality that can be
put into one of the forms
f ( x)  0, f ( x)  0, f ( x)  0, or f ( x)  0,
where f is a polynomial function.
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Procedure for Solving Polynomial Inequalities
1. Express the inequality in the form
f(x) &lt; 0 or f(x) &gt; 0,
where f is a polynomial function.
2. Solve the equation f(x) = 0. The real solutions are
boundary points.
3. Locate these boundary points on a number line,
thereby dividing the number line into intervals.
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Procedure for Solving Polynomial Inequalities
(continued)
4. Choose one representative number, called a test
value, within each interval and evaluate f at that
number.
a. If the value of f is positive, then f(x) &gt; 0 for all
numbers, x, in the interval.
b. If the value of f is negative, then f(x) &lt; 0 for all
numbers, x, in the interval.
5. Write the solution set, selecting the interval or
intervals that satisfy the given inequality.
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Procedure for Solving Polynomial Inequalities
(continued)
This procedure is valid if &lt; is replaced by  or &gt; is
replaced by  . However, if the inequality involves
 or , include the boundary points [the solutions of
f(x) = 0] in the solution set.
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Example: Solving a Polynomial Inequality
Solve and graph the solution set: x 2  x  20
Step 1 Express the inequality in the form f(x) &gt; 0 or
f(x) &lt; 0
x 2  x  20
x 2  x  20  0
Step 2 Solve the equation f(x) = 0.
x 2  x  20  0
( x  5)( x  4)  0
x5
x  4
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Example: Solving a Polynomial Inequality
(continued)
Solve and graph the solution set: x 2  x  20
Step 3 Locate the boundary points on a number line
and separate the line into intervals.
The boundary points divide the line into three intervals
(, 4)
(4,5)
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(5, )
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Example: Solving a Polynomial Inequality
(continued)
Solve and graph the solution set: x 2  x  20
Step 4 Choose one test value within each interval and
evaluate f at that number.
Interval
Test Substitute into
2
f
(
x
)

x
 x  20
Value
(, 4) –5
f (5)  (5) 2  (5)  20
 10
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Conclusion
f(x) &gt; 0
for all x in
(, 4)
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Example: Solving a Polynomial Inequality
Solve and graph the solution set: x 2  x  20
Step 4 Choose one test value within each interval
and evaluate f at that number.
Interval
(4,5)
Test
Value
0
Substitute into
f ( x)  x 2  x  20
f (0)  (0) 2  (0)  20
 20
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Conclusion
f(x) &lt; 0
for all x in
(4,5)
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Example: Solving a Polynomial Inequality
Solve and graph the solution set: x 2  x  20
Step 4 Choose one test value within each interval
and evaluate f at that number.
Interval
(5, )
Test
Value
6
Substitute into
f ( x)  x 2  x  20
f (6)  (6) 2  (6)  20
 10
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Conclusion
f(x) &gt; 0
for all x in
(5, )
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Example: Solving a Polynomial Inequality
Solve and graph the solution set: x 2  x  20
Step 5 Write the solution set, selecting the interval
or intervals that satisfy the given inequality.
Based on Step 4, we see that f(x) &gt; 0 for all x in (, 4)
and (5, ).
Thus, the solution set of the given inequality x 2  x  20
is (, 4) (5, ).
The graph of the solution set on a number line is shown
as follows:
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Definition of a Rational Inequality
A rational inequality is any inequality that can be put
into one of the forms
f ( x)  0, f ( x)  0, f ( x)  0, or f ( x)  0,
where f is a rational function.
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Example: Solving a Rational Inequality
2
x
Solve and graph the solution set:
1
x 1
Step 1 Express the inequality so that one side is zero
and the other side is a single quotient.
2x
2x
2 x 1( x  1)
1
1  0

0
x 1
x 1
x 1 x 1
2 x  ( x  1)
0
x 1
2x  x 1
0
x 1
x 1
0
x 1
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Example: Solving a Rational Inequality (continued)
Solve and graph the solution set: 2 x  1
x 1
Step 1 (cont) Express the inequality so that one side
is zero and the other side is a single quotient.
2x
x 1
is
equivalent
to
1
0
x 1
x 1
x 1
It is in the form f ( x)  0 where f ( x) 
.
x 1
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Example: Solving a Rational Inequality (continued)
2x
1
Solve and graph the solution set:
x 1
2x
x 1
We have found that
 1 is equivalent to f ( x) 
.
x 1
x 1
Step 2 Set the numerator and the denominator of f
equal to zero.
x 1  0  x  1
x  1  0  x  1
We will use these solutions as boundary points on a
number line.
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Example: Solving a Rational Inequality (continued)
2x
1
Solve and graph the solution set:
x 1
Step 3 Locate the boundary points on a number line
and separate the line into intervals.
The boundary points divide the line into three intervals
(, 1)
(1,1]
[1,  )
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Example: Solving a Rational Inequality (continued)
Solve and graph the solution set: 2 x  1
x 1
2x
x 1
 1 is equivalent to f ( x) 
.
We have found that
x 1
x 1
Step 4 Choose one test value within each interval
and evaluate f at that number.
Interval
Test
Value
(, 1)
–2
Substitute into
Conclusion
x 1
f ( x) 
.
x 1
2  1 3
f(x) &gt; 0 for all
f (2) 

3
x in (, 1)
2  1 1
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Example: Solving a Rational Inequality (continued)
Solve and graph the solution set:
2x
1
x 1
2x
 1 is equivalent to f ( x)  x  1.
We have found that
x 1
x 1
Step 4 (cont) Choose one test value within each
interval and evaluate f at that number.
Interval
Test
Value
(1,1]
0
Substitute into
x 1
f ( x) 
x 1
0  1 1
f (0) 
 1
0 1 1
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Conclusion
f(x) &lt; 0 for all
x in (1,1]
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Example: Solving a Rational Inequality (continued)
Solve and graph the solution set:
2x
1
x 1
2x
 1 is equivalent to f ( x)  x  1.
We have found that
x 1
x 1
Step 4 (cont) Choose one test value within each
interval and evaluate f at that number.
Interval
Test
Value
[1,  )
2
Substitute into
x 1
f ( x) 
x 1
2 1 1
f (2) 

2 1 3
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Conclusion
f(x) &gt; 0 for all
x in [1,  )
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Example: Solving a Rational Inequality (continued)
Solve and graph the solution set:
2x
1
x 1
Step 5 Write the solution set, selecting the interval
or intervals that satisfy the given inequality.
The solution set of the given inequality is
(,1) [1, ) or  x x  1 or x  1
The graph of the solution set on a number line is shown
as follows:
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