To solve a rational inequality, we can extend the graphing and algebraic techniques we used to
solve polynomial inequalities.
Interval
Test
Point
Outcome for the
Test Point
Resulting Sign
Satisfies
Inequality:
< 0
?
I:
(0, 1.75)
0
16
0
2
+ 132
(0
−
182
−
yes
II:
(1.75, 6.5)
3
−
16
(3
2
+ 132
(3
−
182
+
no
III:
(6.5, 8.25)
7
−
16
(7
2
+ 132
(7
−
182
−
yes

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80
Let us start with solving by graphing.
Example 7
The graph of
d9
=
@ ( 0
@ ( 0
is given.
-6
-4
-2
2
4
6
8
10
12
14
-6
-4
-2
2
4
6
8
10
x
y
x
= 4
Use the graph to solve the following.
a.
JH ( à
H ( à
≥
0
b.
JH ( à
H ( à
<
0
Solution
a. The given function is not defined at
9 =
4, a it has a zero at
9 =
2
. From the graph, we can see
that the function will be on or above the
x
-axis when
9 ≤ 2
or
9 > 4
. The solution in interval
notation is given by
(−
∞,
2< ∪ (4
, ∞
.
b. The graph of the function lies below the
9
-axis for
2 < 9 < 4
, or the interval
(2
, 4
.
Example 8
Solve the rational inequality by graphing:
H ( ä
H
J
( ã
>
0.
Solution
We graph
d9
=
@ (
@
+
(
. The function has vertical asymptotes at
9 = ±
3
, and a zero at
9 =
1.
-5
-4
-3
-2
-1
1
2
3
4
5
-6
-4
-2
2
4
6
x
y
x
= -3
x
= 3
The graph shows that the given function will be greater than zero for
−3 < 9 < 1
or
9 > 3
. The
solution in interval notation is
(−3
,
1
∪ (3
, ∞
.

81
Let us now solve rational inequalities algebraically.
Solving a Rational Inequality Algebraically
(1)
Write the given inequality in the formd9< 0, d9> 0, d9≤0, or d9≥0, as appropriate. If there is more than one rational expression on the left-hand side of the inequality, combine them into a single fraction.
(2)
Determine any value(s) where the numerator and the denominator are zero. These
will be the critical values that will determine the intervals on the number line.
(3)
Plot the solutions from (2) on a real number line.
For the numerator, use a solid dot if the inequality contains
≤
or
≥
; if the inequality
is < or
>, use an open circle.
Since we need to exclude any values that would make the denominator equal to
zero on any rational expression in the inequality, we always use an open circle for
critical values related to the denominator.
(4)
Select one test point within each interval and see whether it satisfies the inequality.
(5)
The solution will consist of the interval(s) that will satisfy the original inequality.
Example 9
Solve the rational inequality algebraically:
H ( ä
H
J
( ã
>
0
.
Solution
We now follow the steps for solving a rational equation algebraically.
(1) The given function already satisfies this step.
(2) We will find the critical values by setting the numerator and denominator equal to zero,
and solving for
9
.
Numerator:
9 − 1 = 0
9 = 1
Denominator:
(9 + 3(9 − 3 = 0
9 = −3
or
9 = 3
(3) These critical values divide the real number line into four intervals.
I
II
III
IV
0
3
-3
1
(−
∞,
3
)
(−3
,
1
(1
,
3
(3
,
∞
(4) We will construct the sign chart using test points
4, 0, 2, and 4.

82
(5) As we can see from the sign chart, the solution for
@ (
@
+
(
>
0 is given by
3
,
1
∪ (3
, ∞
,
which is the same solution we found solving the problem graphically in Example 8.
Example 10
Solve each the rational inequality algebraically. Confirm your answers graphically.
a. @@ ( <@ ( b. @ ( @ & ≤@ & @( 0