slide 1.7- 1 copyright 2007 pearson education, inc. publishing as pearson addison-wesley
Post on 18-Jan-2018
227 Views
Preview:
DESCRIPTION
TRANSCRIPT
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
top related