1. Slide 2.4- 1 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2. OBJECTIVES Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Relations and Functions Learn the definition of a relation. Learn the definition of a function and learn how to determine which relations are functions. Learn to use functional notation and find function values. Learn to find the domain of a function. SECTION 2.4 1 2 3 4

3. OBJECTIVES Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Relations and Functions Learn to identify the graph of a function. Learn to find the average rate of change of a function. Learn to solve applied problems by using functions. SECTION 2.4 5 6 7

4. Slide 2.4- 4 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Definitions An ordered pair (a, b) is said to satisfy an equation with variables a and b if, when a is substituted for x and b is substituted for y in the equation, the resulting statement is true. An ordered pair that satisfies an equation is called a solution of the equation. Frequently, the numerical values of the variable y can be determined by assigning appropriate values to the variable x. For this reason, y is sometimes referred to as the dependent variable and x as the independent variable.

5. Slide 2.4- 5 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley DEFINTION OF A RELATION Any set of ordered pairs is called a relation.The set of all first components is called the domain of the relation, and the set of all second components is called the range of the relation.

6. Slide 2.4- 6 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 1 Finding the Domain & Range of a Relation Find the domain and range of the relation Solution The domain is the set of all first components, or {Titanic, Star Wars IV, Shrek 2, E.T., Star Wars I, Spider-Man}. { (Titanic, $600.8), (Star Wars IV, $461.0), (Shrek 2, $441.2), (E.T., $435.1), (Star Wars I, $431.1), (Spider-Man, $403.7)}. The range is the set of all second components, or {$600.8, $461.0, $441.2, $435.1, $431.1, $403.7)}.

7. Slide 2.4- 7 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley DEFINTION OF A FUNCTION A function from a set X to a set Y is a relation in which each element of X corresponds to one and only one element of Y.

8. Slide 2.4- 8 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley CORRESPONDENCE DIAGRAM A relation may be defined by a correspondence diagram, in which an arrow points from each domain element to the element or elements in the range that correspond to it.

9. Slide 2.4- 9 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Determining Whether a Relation is a Function Determine whether the relations that follow are functions. The domain of each relation is the family consisting of Malcolm (father), Maria (mother), Ellen (daughter), and Duane (son). a. For the relation defined by the following diagram, the range consists of the ages of the four family members, and each family member corresponds to that family member’s age.

10. Slide 2.4- 10 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Determining Whether a Relation is a Function

11. Slide 2.4- 11 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Determining Whether a Relation is a Function Solution The relation is a function, because each element in the domain corresponds to exactly one element in the range. For a function, it is permissible for the same range element to correspond to different domain elements. The set of ordered pairs that define this relation is {(Malcolm, 36), (Maria, 32), (Ellen, 11), (Duane, 11)}.

12. Slide 2.4- 12 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Determining Whether a Relation is a Function b.For the relation defined by the diagram on the next slide, the range consists of the family’s home phone number, the office phone numbers for both Malcolm and Maria, and the cell phone number for Maria. Each family member corresponds to all phone numbers at which that family member can be reached.

13. Slide 2.4- 13 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Determining Whether a Relation is a Function

14. Slide 2.4- 14 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Determining Whether a Relation is a Function The relation is not a function, because more than one range element corresponds to the same domain element. For example, both an office phone number and a home phone number correspond to Malcolm. The set of ordered pairs that define this relation is {(Malcolm, 220-307- 4112), (Malcolm, 220-527-6277 ), (Maria, 220- 527-6277), (Maria, 220-416-5204), (Maria, 220-433-8195), (Ellen, 220-527-6277), (Duane, 220-527-6277)}. Solution

15. Slide 2.4- 15 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley FUNCTION NOTATION We usually use single letters such as f, F, g, G, h, H, and so on as the name of a function. For each x in the domain of f, there corresponds a unique y in its range. The number y is denoted by f (x) read as “f of x” or “f at x”. We call f(x) the value of f at the number x and say that f assigns the f (x) value to x. Because the value of y depends on the given value of x, y is called the dependent variable and x is called the independent variable.

16. Slide 2.4- 16 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley FUNCTION NOTATION We can describe functions in many ways: A table

17. Slide 2.4- 17 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley FUNCTION NOTATION A graph

18. Slide 2.4- 18 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley FUNCTION NOTATION An equation

19. Slide 2.4- 19 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 3 Evaluating a Function Solution Let g be the function defined by the equation y = x 2 – 6x + 8. Evaluate each function value.

20. Slide 2.4- 20 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 3 Evaluating a Function Solution continued

21. Slide 2.4- 21 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 4 Determining Whether an Equation Defines a Function Determine whether each equation determines y as a function of x. a. 6x 2 – x – 2 b. y 2 – x 2 = 4 One value of y corresponds to one value of x so it defines y as a function of x. Solution a.

22. Slide 2.4- 22 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 4 Determining Whether an Equation Defines a Function Solution continued b. y 2 – x 2 = 4 Two values of y correspond to the same value of x so it does not define y as a function of x.

23. Slide 2.4- 23 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley AGREEMENT ON DOMAIN If the domain of a function that is defined by an equation is not explicitly specified, then we take the domain of the function to be the largest set of real numbers that result in real numbers as outputs.

24. Slide 2.4- 24 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Finding the Domain of a Function Find the domain of each function. Solution a. f is not defined when the denominator is 0. Domain: {x|x ≠ –1 and x ≠ 1}

25. Slide 2.4- 25 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Finding the Domain of a Function Solution continued The square root of a negative number is not a real number and is excluded from the domain. Domain: {x|x ≥ 0}, [0, ∞) The square root of a negative number is not a real number and is excluded from the domain, so x – 1 ≥ 0. However, the denominator ≠ 0.

26. Slide 2.4- 26 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Finding the Domain of a Function Solution continued So x – 1 > 0 so x > 1. Domain: {x|x > 1}, or (1, ∞) Any real number substituted for t yields a unique real number. Domain: {t|t is a real number}, or (–∞, ∞)

27. Slide 2.4- 27 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EQUALITY OF FUNCTIONS Two functions f and g are equal if and only if 1.f and g have the same domain and 2.f (x) = g (x) for all x in the domain.

28. Slide 2.4- 28 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley VERTICAL LINE TEST If no vertical line intersects the graph of a relation at more than one point, then the graph is the graph of a function.

29. Slide 2.4- 29 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 6 Use the vertical-line test to determine which graphs are graphs of functions. Identifying the Graph of a Function Solution Not a function Does not pass the vertical line test

30. Slide 2.4- 30 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 6 Use the vertical-line test to determine which graphs are graphs of functions. Identifying the Graph of a Function Solution Not a function Does not pass the vertical line test

31. Slide 2.4- 31 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 6 Use the vertical-line test to determine which graphs are graphs of functions. Identifying the Graph of a Function Solution Is a function Does pass the vertical line test

32. Slide 2.4- 32 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 6 Use the vertical-line test to determine which graphs are graphs of functions. Identifying the Graph of a Function Solution Is a function Does pass the vertical line test

33. Slide 2.4- 33 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 7 Let a.Is the point (1, –3) on the graph of f ? b.Find all values of x such that (x, 5) is on the graph of f. c.Find all y–intercepts of the graph of f. d.Find all x–intercepts of the graph of f. Examining the Graph of a Function Solution a. Check whether (1, –3) satisfies the equation. (1, –3) is not on the graph of f. ?

34. Slide 2.4- 34 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 7 b.Find all values of x such that (x, 5) is on the graph of f. Substitute 5 for y and solve for x. Examining the Graph of a Function Solution continued (–2, 5) and (4, 5) are on the graph of f.

35. Slide 2.4- 35 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 7 c.Find all y–intercepts of the graph of f. Substitute 0 for x and solve for y. Examining the Graph of a Function Solution continued The only y-intercept is –3.

36. Slide 2.4- 36 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 7 d.Find all x–intercepts of the graph of f. Substitute 0 for y and solve for x. Examining the Graph of a Function Solution continued The x–intercepts of the graph of f are –1 and 3.

37. Slide 2.4- 37 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley THE AVERAGE RATE OF CHANGE OF A FUNCTION Let (a, f (a)) and (b, f (b)) be points on the graph of a function f. Then the average rate of change of f (x) as x changes from a to b is defined by

38. Slide 2.4- 38 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 8 Find the average rate of change of f (x) = 2x 2 – 3 as x changes a.from x = 2 to x = 4; b.from x = 3 to x = 6; c.from x = c to x = c + h, h ≠ 0. Finding the Average Rate of Change Solution Average rate of change

39. Slide 2.4- 39 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 8 b. from x = 3 to x = 6; Finding the Average Rate of Change Solution continued Average rate of change

40. Slide 2.4- 40 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 8 c. from x = c to x = c + h, h ≠ 0. Finding the Average Rate of Change Solution cont’ Average rate of change

41. Slide 2.4- 41 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley DIFFERENCE QUOTIENT For a function f, the quantity is called the difference quotient.

42. Slide 2.4- 42 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 9 Let f (x) = 2x 2 – 3 + 5. Find and simplify Evaluating and Simplifying a Difference Quotient Solution Find Now substitute into the difference quotient.

43. Slide 2.4- 43 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 9 Evaluating and Simplifying a Difference Quotient Solution continued

44. Slide 2.4- 44 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 10 Many drugs used to lower high blood cholesterol levels are called statins and are very popular and widely prescribed. These drugs, along with proper diet and exercise, help prevent heart attacks and strokes. Recall from the introduction to this section that bioavailability is the amount of a drug you have ingested that makes it into your bloodstream. Cholesterol–Reducing Drugs

45. Slide 2.4- 45 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 10 A statin with a bioavailability of 30% has been prescribed for Boris to treat his cholesterol levels. Boris is to take 20 milligrams of this statin every day. During the same day, one-half of the statin is filtered out of the body. Find the maximum concentration of the statin in the bloodstream on each of the first ten days of using the drug, and graph the result. Cholesterol–Reducing Drugs

46. Slide 2.4- 46 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 10 Since the statin has 30% bioavailability and Boris takes 20 milligrams per day, the maximum concentration in the bloodstream is 30% of 20 milligrams, or 20(0.3) = 6 milligrams from each day’s prescription. Because one-half of the statin is filtered out of the body each day, the daily maximum concentration is Cholesterol–Reducing Drugs Solution

47. Slide 2.4- 47 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 10 Cholesterol–Reducing Drugs Solution continued DayMax Concentration 16.000 29.000 310.500 411.250 511.625 611.813 711.906 811.953 911.977 1011.988 Maximum concentration approaches 12 mg.

48. Slide 2.4- 48 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley FUNCTIONS IN ECONOMICS b is the fixed cost a (the cost of producing each item in dollars per item) is called the marginal cost. Linear Cost Function Average Cost

49. Slide 2.4- 49 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley FUNCTIONS IN ECONOMICS Suppose x items can be sold (demanded) at a price of p dollars per item. Linear Price–Demand Function

50. Slide 2.4- 50 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley FUNCTIONS IN ECONOMICS Revenue Function Profit Function Revenue = (Price per item) (Number of items sold) Profit = Revenue – Cost

51. Slide 2.4- 51 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 11 Metro Entertainment Co. spent $100,000 on production costs for its off-Broadway play Bride and Prejudice. Once running, each performance costs $1000 per show and the revenue from each is $2400. Using x to represent the number of shows, Breaking Even a.Write cost function C(x). b.Write revenue function R(x). c.Write profit function P(x). d.How many showings of Bride and Prejudice must be held for Metro to break even?

52. Slide 2.4- 52 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 11 Breaking Even Solution

53. Slide 2.4- 53 Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 11 Breaking Even Solution continued d. Metro will break even when P(x) = 0. So 72 is the number of shows required for Metro to break even.