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Sullivan Algebra and Trigonometry: Section 4.5 Solving Polynomial and Rational Inequalities Objectives Solve Polynomial Inequalities Solve Rational Inequalities.

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Presentation on theme: "Sullivan Algebra and Trigonometry: Section 4.5 Solving Polynomial and Rational Inequalities Objectives Solve Polynomial Inequalities Solve Rational Inequalities."— Presentation transcript:

1 Sullivan Algebra and Trigonometry: Section 4.5 Solving Polynomial and Rational Inequalities Objectives Solve Polynomial Inequalities Solve Rational Inequalities

2 Steps for Solving Polynomial and Rational Inequalities 1. Write the inequality so that the polynomial or rational expression is on the left and zero is on the right. 2. Determine when the expression on the left is equal to zero or, with a rational expression, is undefined. 3. Use the numbers found in step 2 to separate the real number line into intervals. 4. Select a number in each interval and determine if the function on the left is positive (> 0) or negative (< 0). If the inequality isn’t strict, include the solutions to f(x) = 0 in the solution set.

3 Example: Solve the inequality x(x+1) > 20. So, divide the number line at x = – 5 and at x = 4 For the region Choose x = – 6, then – 6(– 6 +1) > 20 is TRUE. So, this region is included in the solution set.

4 For the region Choose x = 0, then 0(0+1) > 20 is FALSE. So, this region is not included in the solution set. For the region Choose x = 5, then 5(5+1) > 20 is TRUE. So, this region is included in the solution set. Since this is a strict inequality, we do not include the endpoints in the solution set. The solution set is

5 Example: Solve the inequality So, divide the number line at x = – 7 and at x = 4. Note that 4 is NOT in the solution set.

6 For the region Choose x = – 8, then 2( –8) + 3 > 1 (–8) – 4 So, this region is included in the solution set. For the region So, this region is not included in the solution set. Choose x = 0, then 2(0) + 3 > 1 (0) – 4

7 For the region So, this region is included in the solution set. Since this is not a strict inequality but involves an equal sign, we do include the endpoint x = – 7 in the solution set, since the left hand side is zero at x = – 7. The other endpoint, x = 4, cannot be included since it makes the denominator equal to zero. The solution set is Choose x = 5, then 2(5) + 3 > 1 (5) – 4


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