College Algebra 9th txtbk

284 CHAPTER 4 POLYNOMIAL AND RATIONAL FUNCTIONS
Interval (, 2) (2, 1) (1, 3) (3, )
Test number x 3 0 2 4
P(x) 24 6 4 18
Sign of P
We conclude that the solution set of the inequality is the intervals where P(x) is positive:
(2, 1) ´ (3, )
MATCHED PROBLEM 5 Solve the inequality x 3 x 2 x 1 0.
EXAMPLE 6 Solving Polynomial Inequalities with a Graphing Calculator
Solve 3x 2 12x 4 2x 3 5x 2 7 to three decimal places.
SOLUTION
Subtracting the right-hand side gives the equivalent inequality
P(x) 2x 3 8x 2 12x 11 0
The zeros of P(x), to three decimal places, are 1.651, 0.669, and 4.983 [Fig. 7(a)].
By inspecting the graph of P we see that P is above the x axis on the intervals
(, 1.651) and (0.669, 4.983). So the solution set of the inequality is
(, 1.651] ´ [0.669, 4.983]
The square brackets indicate that the endpoints of the intervals—the zeros of the polynomial—
also satisfy the inequality.
An alternative to inspecting the graph of P is to inspect the graph of
f (x) P(x)
P(x)
The function f(x) has the value 1 if P(x) is positive, because then the absolute value of P(x)
is equal to P(x). Similarly, f(x) has the value 1 if P(x) is negative. This technique makes
it easy to identify the solution set of the original inequality [Fig. 7(b)] and often eliminates
difficulties in choosing appropriate window variables.
100
10
10
10
10
10
100
10
(a) P(x) 2x 3 8x 2 12x 11
Z Figure 7
(b) f (x) P(x)
P(x)
MATCHED PROBLEM 6 Solve to three decimal places 5x 3 13x 4x 2 10x 5.