1. Nonlinear Inequalities
MATH Precalculus
S. Rook

2. Overview
Section 2.7 in the textbook:
Solving polynomial inequalities

3. Solving Polynomial Inequalities

4. Solving Inequalities
An inequality is similar to an equality except instead of =, we have >, <, ≤ or ≥.
Essentially solved in the same way as an equality
Solution set – all values that satisfy an inequality
Whereas an equality has at most 1 solution, the solution to an inequality is a set – possibly with infinitely many elements

5. Solving Polynomial Inequalities
We discussed how to solve polynomial equalities
Only difference is that the solution set now consists of intervals instead of a single real number
To solve an inequality such as (x + 1)(x – 1) > 0:
By the Zero Product Principle, x = -1 or x = 1
This subdivides the interval (-oo, +oo) into three subintervals:
(-oo, -1), (-1, 1), (1, +oo)

6. Solving Polynomial Inequalities (Continued)
Pick one value in each subinterval and test it in the inequality to find the sign of that subinterval
Recall the sign property of polynomials
The sign in each subinterval is the SAME
Keep only those intervals that satisfy the inequality
If more than one interval satisfies the inequality, union them

7. Solving Polynomial Inequalities (Example)
Ex 1: Write the solution set in interval notation: a) x2 + 4x + 4 ≥ 9 b) 3x2 + 7x < 6 c) 2x2 – 7x ≤ -2 d) x3 – 4x > 0

8. Summary After studying these slides, you should be able to:
Solve polynomial inequalities, graph the solution set on a number, and write the solution set in interval notation
Additional Practice
See the list of suggested problems for 2.7
Next lesson
Exponential Functions and Their Graphs (Section 3.1)