Slide 1.7- 1 Copyright 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley.

Slide 1.7- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

OBJECTIVES

Polynomial & Rational Inequalities

Learn to solve quadratic inequalities.

Learn to solve polynomial inequalities.

Learn to solve rational inequalities.

SECTION 1.7

1

2

3

Slide 1.7- 3 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 1 Using the Test-point Method to Solve a Quadratic Inequality

Solve x2 2x 7. Write the solution ininterval notation and graph the solution set.Solutionx2 2x 7 2x 7 2x 7

x2 2x 7 0First, solve the associated equation. x2 2x 7 0

x 2 2 2 4 1 7

2 1

Slide 1.7- 4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 1 Using the Test-point Method to Solve a Quadratic Inequality

This divides the number line into 3 intervals.

Solution continued

x 2 32

212 2

x 1.8x 3.8so

We select the “test points” –3, 0 and 4.

1 20 4 53–1–2–3– 41 2 2 1 2 2

0 0

Slide 1.7- 5 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 1 Using the Test-point Method to Solve a Quadratic Inequality

Solution continuedInterval Point Value Result

–3 8 +

0 –7 –

4 1 +

,1 2 2 1 2 2,1 2 2

1 2 2,

1 20 4 53–1–2–3– 41 2 2 1 2 2

0 0– – – – – – – – – + + + + + + + + + + + + +

x2 2x 7

Slide 1.7- 6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 1 Using the Test-point Method to Solve a Quadratic Inequality

Solution continued

1 20 4 53–1–2–3– 41 2 2 1 2 2

0 0– – – – – – – – – + + + + + + + + + + + + +

,1 2 2 It is positive in the intervals

1 2 2, .()

1 20 4 53–1–2–3– 41 2 2 1 2 2

Slide 1.7- 7 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 2 Calculating Speeds from Telltale Skid Marks

In the introduction to this section, a car involved in an accident left skid marks over 75 feet long. Under the road conditions at the accident, the distance d (in feet) it takes a car traveling v miles per hour to stop is given by the equation

d 0.05v2 v.The accident occurred in a 25-mile-per-hour speed zone. Was the driver going over the speed limit?

Slide 1.7- 8 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 2 Calculating Speeds from Telltale Skid Marks

Solve the inequality

Solution

(stopping distance) > 75 feet, or 0.05v2 v 75.

0.05v2 v 75

0.05v2 v 75 0

0.05v2 v 75 0

5v2 100v 7500 0

v2 20v 1500 0

v 50 v 30 0v 50 0 or v 30 0v 50 or v 30

These divide the number line into 3 intervals.

Slide 1.7- 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 2 Calculating Speeds from Telltale Skid Marks

Solution continued

0.05v2 v

0 4030–50– 60

0 0– – – – – – – – – – – – – – – + + + + + + + + + + + +

Interval Point Value Result

(–∞, –50) – 60 45 +

(–50, 30) 0 –75 –

(30, ∞) 40 45 +

Slide 1.7- 10 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 2 Calculating Speeds from Telltale Skid Marks

Solution continued

For this situation, we look at only the positive values of v. Note that the numbers corresponding to speeds between 0 and 30 miles per hour (that is, 0 ≤ v ≤ 30 ) are not solutions ofThus, the car was traveling more than 30 miles per hour. The driver was going over the speed limit.

0.05v2 v 75.

0 4030–50– 60

0 0– – – – – – – – – – – – – – – + + + + + + + + + + + +

Slide 1.7- 11 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

ONE SIGN THEOREM

If a polynomial equation has no real solution, then the polynomial is either always positive or always negative.

Slide 1.7- 12 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 3Using the One-sign Theorem to solve a Quadratic Inequality

Solve: x2 – 2x 2 0.

Since there are no obvious factors, evaluate the discriminant to see if there are any real roots.

Solve the equation Solution

x2 – 2x 2 0.

b2 – 4ac 2 2 4 1 2 4Since the discriminant is negative, there are no real roots. Use 0 as a test point, which yields 2. The inequality is always positive, the solution set is (–∞, ∞).

Slide 1.7- 13 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 4 Solving a Polynomial Inequality

Solve x4 1. Write the answer in intervalnotation and graph the solution set.Solution

x4 1

x4 1 0

x2 1 x2 1 0

x 1 x 1 x2 1 0

x 1 x 1 x2 1 0

x 1 0, x 1 0, x2 1 0x 1, x 1

These divide the number line into 3 intervals.

Slide 1.7- 14 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 4 Solving a Polynomial Inequality

Solution continued

Interval Point Value of Result

(–∞, –1) – 2 15 +

(–1, 1) 0 –1 –

(1, ∞) 2 15 +

x 1 x 1 x2 1

1 20 43–1–2–3

0 0– – – + + + + + + + + + + + + + + +

Slide 1.7- 15 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 4 Solving a Polynomial Inequality

Solution continued

The solution set consists of all x between –1 and 1, including both –1 and 1.

1 20–1–2][

–1 ≤ x ≤ 1, or [–1, 1]

1 20 43–1–2–3

0 0– – – + + + + + + + + + + + + + + +

Slide 1.7- 16 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 5 Solving a Rational Inequality

Solve3

x 11. Write the solution in interval

notation and graph the solution set.

Solution3

x 11

3x 1

1 0

3 x 1 x 1

0

4 xx 1

0

4 x 0x 4

x 1 0x 1

Num = 0 and Den = 0Solve

Slide 1.7- 17 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

EXAMPLE 5 Solving a Rational Inequality

Solution continuedWe have 3 intervals (–∞, 1), (1, 4), and (4, ∞).

3 40 6521–1

0 0 – – – – – + + + + + + + – – – – – –

We know it’s positive in the interval (1, 4) and it is undefined for x = 1 and is 0 for x = 4. The solution set is {x | 1 < x ≤ 4}, or (1, 4].

1 < x ≤ 4, or (1, 4]

](3 40 6521–1