Nonlinear Inequalities

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Presentation transcript:

Nonlinear Inequalities MATH 109 - Precalculus S. Rook

Overview Section 2.7 in the textbook: Solving polynomial inequalities

Solving Polynomial Inequalities

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

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)

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

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

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)