1. Copyright © Cengage Learning. All rights reserved.3
Polynomial and Rational
Functions
Copyright © Cengage Learning. All rights reserved.

2. 3.7 Polynomial and Rational Inequalities
Copyright © Cengage Learning. All rights reserved.

3. Objectives
Polynomial Inequalities
Rational Inequalities

4. Polynomial Inequalities

5. Polynomial Inequalities
An important consequence of the Intermediate Value Theorem is that the values of a polynomial function P do not change sign between successive zeros. In other words, the values of P between successive zeros are either all positive or all negative. Graphically, this means that between successive x-intercepts, the graph of P is entirely above or entirely below the x-axis.

6. Polynomial Inequalities
Figure 1 illustrates this property of polynomials.
This property of polynomials allows us to solve polynomial inequalities like P(x)  0 by finding the zeros of the polynomial and using test points between successive zeros to determine the intervals that satisfy the inequality.
P(x) > 0 or P(x) < 0 for x between successive zeros of P
Figure 1

7. Polynomial Inequalities
We use the following guidelines.

8. Example 1 – Solving a Polynomial Inequality
Solve the inequality 2x3 + x2 + 6  13x.
Solution:
We follow the preceding guidelines.
Move all terms to one side. We move all terms to the left-hand side of the inequality to get 2x3 + x2 – 13x + 6  0
The left-hand side is a polynomial.

9. Example 1 – Solution
cont’d
Factor the polynomial. Factoring the polynomial, we get (x – 2)(2x – 1)(x + 3)  0 The zeros of the polynomial are –3, , and 2.
Find the intervals. The intervals determined by the zeros of the polynomial are

10. Example 1 – Solution
cont’d
Make a table or diagram. We make a diagram indicating the sign of each factor on each interval.
Solve. From the diagram we see that the inequality is satisfied on the intervals (–3, ) and (2, ).

11. Example 1 – Solution
cont’d
Checking the endpoints, we see that –3, , and 2 satisfy the inequality, so the solution is [–3, ]  [2, ).
The graph in Figure 2 confirms our solution.
Figure 2

12. Rational Inequalities

13. Rational Inequalities
Unlike polynomial functions, rational functions are not necessarily continuous. The vertical asymptotes of a rational function r break up the graph into separate “branches.” So the intervals on which r does not change sign are determined by the vertical asymptotes as well as the zeros of r. This is the reason for the following definition: If r(x) = P(x)/Q(x) is a rational function, the cut points of r are the values of x at which either P(x) = 0 or Q(x) = 0.

14. Rational Inequalities
In other words, the cut points of r are the zeros of the numerator and the zeros of the denominator (see Figure 4). So to solve a rational inequality like r(x)  0, we use test points between successive cut points to determine the intervals that satisfy the inequality.
r(x) > 0 or r(x) < 0 for x between successive cut points of r
Figure 4

15. Rational Inequalities
We use the following guidelines.

16. Example 3 – Solving a Rational Inequality
Solve the inequality
Solution:
We follow the preceding guidelines.
Move all terms to one side. We move all terms to the left-hand side of the inequality.
Move terms to LHS

17. Example 3 – Solution
cont’d
The left-hand side of the inequality is a rational function.
Factor numerator and denominator. Factoring the numerator and denominator, we get
Common denominator
Simplify

18. Example 3 – Solution
cont’d
The zeros of the numerator are 2 and –2, and the zeros of the denominator are –1 and 3, so the cut points are –2, –1, 2, and 3.
Find the intervals. The intervals determined by the cut points are

19. Example 3 – Solution Make a table or diagram. We make a sign diagram.
cont’d
Make a table or diagram. We make a sign diagram.
Solve. From the diagram we see that the inequality is satisfied on the intervals (–2, –1) and (2, 3).

20. Example 3 – Solution
cont’d
Checking the endpoints, we see that –2 and 2 satisfy the inequality, so the solution is [–2, –1)  [2, 3).
The graph in Figure 5 confirms our solution.
Figure 5