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CHAPTER 3 Functions, Equations, and Inequalities Section 3.5 - Inequalities for values of x in which everything is defined Example: If -3x < 15 then … x > -5 If a function f is always either increasing or decreasing (not both) on its domain, then it’s one-to-one. Proof by Contraposition: This is new!! If a function f is always either increasing or decreasing (not both) on its domain, then it’s one-to-one. Proof by Contraposition: This is new!! What is the contrapositive of the expression p q ? ~q ~p What is the contraposition of the conditional expression that we’re trying to prove? De Morgan’s Law ~(the function is one-to-one) ~(either increasing or decreasing on its domain) If f is not one-to-one, it’s not increasing and not decreasing on its domain. If a function f is always either increasing or decreasing (not both) on its domain, then it’s one-to-one. Proof by Contraposition: (show that if f is not one-to-one, it’s not increasing and not decreasing). Assume f is not 1-1. If a function f is always either increasing or decreasing (not both) on its domain, then it’s one-to-one. Proof by Contraposition: (show that if f is not one-to-one, it’s not increasing and not decreasing). If a function f is always either increasing or decreasing (not both) on its domain, then it’s one-to-one. Proof by Contraposition: (show that if f is not one-to-one, it’s not increasing or decreasing). So if f is not one-to-one, it is not increasing and not decreasing. Therefore, by contraposition: If f is either increasing or decreasing then it is one-to-one. Q.E.D. = “Quod Erat Demonstratum” means “So it is demonstrated” So, for every function h(x) which is either increasing or decreasing on its domain… •Is one-to-one •Passes the horizontal line test •Has inverse function h-1(x) •f(x) = g(x) if and only if h f(x) = h g(x) (This is biconditional) Section 3.5 - Inequalities A function f(x) is one-to-one iff it has an inverse f -1(x) If a function f is always either increasing or decreasing (not both) on its domain, it has an inverse f -1(x) Section 3.5 - Inequalities If a function f is always increasing on its domain, then f –1 is increasing on its domain. If a function f is always decreasing on its domain, then f –1 is decreasing on its domain. Proof by Contradiction!! If a function f is always increasing on its domain, then f –1 is increasing on its domain. Assume f is always increasing but f -1 is not always increasing. Then f -1must be not increasing somewhere. Proof by Contradiction: Assume the premise is true and conclusion is false and reach a contradiction If a function f is always increasing on its domain, then f –1 is increasing on its domain. This contradicts our assumption that f is increasing on its domain. Thus, f -1 must be increasing on its domain. Section 3.5 - Inequalities If a h(x) increasing on its domain, then If a h(x) decreasing on its domain, then Section 3.5 - Examples Solve: Section 3.5 - Homework Read Section 3.5 and do 1 – 19 Section 3.6 - Examples Solve: 1. “factor” by chunking 2. Using the Zero-Product Property Section 3.6 - Warm Up The sum of a positive number and its reciprocal is less than two. What is the number? Section 3.4 - Quick Review: What does it mean to say a function is continuous on an interval? Give an example of a function which is continuous. Give an example of a function which is not continuous. What does the Intermediate Value Theorem say?

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posted: | 9/20/2011 |

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