College Algebra 9th txtbk

SECTION 4–4 Rational Functions and Inequalities 311
Interval Test number x f(x) Sign of f
(, 4) 5 25/21
(4, 2) 3 9/5
(2, 0) 1 1
(0, 2) 1 5/3
(2, ) 3 63/5
We conclude that the solution set of the inequality is
(, 4) (2, 0) (0, 2)
MATCHED PROBLEM 8
Solve
x 2 1
x 2 9 0.
EXAMPLE 9 Solving Rational Inequalities with a Graphing Calculator
Solve
1 9x 9
x 2 x 3
to three decimal places.
SOLUTION First we convert the inequality to an equivalent inequality in which one side is 0:
10
10
Z Figure 12
10
10
(a) f (x) x2 8x 6
x 2 x 3
10
10
(b) g(x) f (x)
f (x)
10
10
x 2 x 3 0 Find a common denominator.
Subtract
9x 9
from both sides.
x 2 x 3
Simplify.
The zeros of x 2 8x 6, to three decimal places, are 0.838 and 7.162. The zeros
of x 2 x 3 are 2.303 and 1.303. These four zeros partition the x axis into five intervals:
We graph
1 9x 9
x 2 x 3
x 2 x 3 9x 9
x 2 x 3 0
x 2 8x 6
x 2 x 3 0
1 9x 9
x 2 x 3
(, 2.303), (2.303, 0.838), (0.838, 1.303), (1.303, 7.162), and (7.162, )
f (x) x2 8x 6
x 2 x 3
(Fig. 12) and observe that the graph of f is above the x axis on the intervals (, 2.303),
(0.838, 1.303), and (7.162, ). So the solution set of the inequality is
Note that the endpoints that are zeros of f are included in the solution set of the inequality,
but not the endpoints at which f is undefined.
and
g(x) f (x)
f (x)
(, 2.303) [0.838, 1.303) [7.162, )
x 3 4x 2 7
MATCHED PROBLEM 9 Solve to three decimal places.
x 2 5x 1 0