1. Slide 1- 1 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2. Functions and Graphs 7.1Introduction to Functions 7.2Domain and Range 7.3Graphs of Functions 7.4The Algebra of Functions 7.5Formulas, Applications, and Variation 7

3. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Introduction to Functions Correspondences and Functions Functions and Graphs Function Notations and Equations Applications 7.1

4. Slide 7- 4 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Function A function is a special kind of correspondence between two sets. The first set is called the domain. The second set is called the range. For any member of the domain, there is exactly one member of the range to which it corresponds. This kind of correspondence is called a function. DomainRange Correspondence

5. Slide 7- 5 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Determine if the correspondence is a function. 8 0 –5 2 17 Solution The correspondence is a function because each member of the domain corresponds to exactly one member of the range. Example

6. Slide 7- 6 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Determine if the correspondence is a function. Jackson Max Cade Football Wrestling Soccer Solution The correspondence is not a function because a member of the domain (Jackson) corresponds to more than one member of the range. Example

7. Slide 7- 7 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Function A function is a correspondence between a first set called the domain, and a second set, called the range, such that each member of the domain corresponds to exactly one member of the range.

8. Slide 7- 8 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Determine whether the correspondence is a function. A set of rectangles Range Solution The correspondence is a function, because each rectangle has only one area. Each rectangle’s area A set of numbers DomainCorrespondence Example

9. Slide 7- 9 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Determine if the correspondence is a function. Famous singers Range Solution The correspondence is not a function, because some singers have recorded more than one song. A song that the singer has recorded A set of song titles DomainCorrespondence Example

10. Slide 7- 10 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Relation A relation is a correspondence between a first set called the domain, and a second set, called the range, such that each member of the domain corresponds to at least one member of the range.

11. Slide 7- 11 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Functions and Graphs When a function is given as a set of ordered pairs, the domain is the set of all first coordinates and the range is the set of all second coordinates. Function names are generally represented by lower- or upper-case letters.

12. Slide 7- 12 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Find the domain and range of the function f below. Example

13. Slide 7- 13 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Here f can be written {(–5, 1), (1, 0), (3, –5), (4, 3)}. The domain is the set of all first coordinates, {–5, 1, 3, 4}, and the range is the set of all second coordinates, {1, 0, –5, 3}.

14. Slide 7- 14 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley For the function f represented below, determine each of the following. a) What member of the range is paired with -2 c) What member of the domain is paired with 6 y x -5 -4 -3 -2 -1 1 2 3 4 5 f 4 -2 -4 -3 3 2 5 1 6 7 Example

15. Slide 7- 15 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley a) What member of the range is paired with -2 Solution Locate -2 on the horizontal axis (this is where the domain is located). Next, find the point directly above -2 on the graph of f. From that point, look to the corresponding y- coordinate, 3. The “input” -2 has the “output” 3. x y -5 -4 -3 -2 -1 1 2 3 4 5 f 4 -2 -4 -3 3 2 5 1 6 Input Output 7

16. Slide 7- 16 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley b) What member of the domain is paired with 6 Solution Locate 6 on the vertical axis (this is where the range is located). Next, find the point to the right of 6 on the graph of f. From that point, look to the corresponding x-coordinate, 2.5. The “output” 6 has the “input” 2.5. x y -5 -4 -3 -2 -1 1 2 3 4 5 f 4 -2 -4 -3 3 2 5 1 6 Input Output 7

17. Slide 2- 17 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Vertical-Line Test If it is possible for a vertical line to cross a graph more than once, then the graph is not the graph of a function.

18. Slide 2- 18 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphs that do not represent functions still do represent relations. A functionNot a function. Two y-values correspond to one x-value Not a function. Three y-values correspond to one x-value

19. Slide 7- 19 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Function Notation and Equations To understand function notation, it helps to imagine a “function machine.” Think of putting a member of the domain (an input) into the machine. The machine is programmed to produce the appropriate member of the range (the output)

20. Slide 7- 20 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Function Notation and Equations The function pictured has been named f. Here x represents an arbitrary input, and f (x) – read “f of x,” “f at x,” or “the value of f at x” represents the corresponding output.

21. Slide 7- 21 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Function Notation and Equations Most functions are described by equations. For example, f (x) = 5x +2 describes the function that takes an input x, multiplies it by 5 and then adds 2. f (x) = 5x + 2 To calculate the output f (3), take the input 3, multiply it by 5, and add 2 to get 17. That is, substitute 3 into the formula for f (x). Input f (3) = 5(3) + 2 = 17

22. Slide 7- 22 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Function Notation and Equations Sometimes in place of f (x) = 5x +2, we write y = 5x +2, where it is understood that the value of y, the dependent variable, depends on our choice of x, the independent variable. a) If f (x) = 5x +2, then f(3) = 17. b) If y = 5x +2, then the value of y is 17 when x is 3.

23. Slide 7- 23 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Find each indicated function value. a) f (–2), for f (x) = 3x 2 + 2x b) g(4), for g(t) = 6t + 9 c) h(m +2), for h(x) = 8x + 1 a) f (–2) = 3(–2) 2 + 2(–2) = 12 – 4 = 8 b) g(4) = 6(4) + 9 = 24 + 9 = 33 c) h(m +2) = 8(m+ 2) + 1 = 8m + 16 + 1 = 8m + 17 Example

24. Slide 7- 24 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Note that whether we write f (x) = 5x +2 or f (m) = 5m +2, we still have f (3) = 17. Thus the independent variable can be thought of as a dummy variable. When a function is described by an equation, the domain is often unspecified. In such cases, the domain is the set of all numbers for which function values can be calculated.

25. Slide 7- 25 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley A small business started out in the year 1996 with 10 employees. By the start of 2000 there were 28 employees, and by the beginning of 2004 the business had grown to 34 employees. Estimate the number of employees in 1998 and also predict the number of employees in 2007. Example

26. Slide 7- 26 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution 1. and 2. Familiarize. and Translate. Plot the points and connect the three points. Let the horizontal axis represent the year and the vertical axis the number of employees. Label the function itself E. 10 30 20 40 50 200420001996 Number of Employees Year

27. Slide 7- 27 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 3. Carry out. To estimate the number of employees in 1998, locate the point directly above the year 1998. Then estimate its second coordinate by moving horizontally from that point to the y-axis. We see that 10 30 20 40 50 200420001996 19 1998 Year Number of Employees

28. Slide 7- 28 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 3. Carry out (continued). To predict the number of employees in 2007, extend the graph and extrapolate. We see that 10 30 20 40 50 2004200019962007 Year Number of Employees

29. Slide 7- 29 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 4. Check. A precise check would involve knowing more information. Since 19 is between 10 and 28 and 40 is greater than 34, the estimate seems plausible. 5. State. In 1997, there were about 19 employees at the small business. By 2007, the number of employees should grow to 40.

30. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Domain and Range Determining the Domain and Range Restrictions on Domain Functions Defined Piecewise 7.2

31. Slide 7- 31 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley When a function is given as a set of ordered pairs, the domain is the set of all first coordinates and the range is the set of all second coordinates. Find the domain and range for the function f given by f = {(–5, 1), (1, 0), (3, –5), (4, 3)}. Solution The domain is the set of all first coordinates: {–5, 1, 3, 4}. The range is the set of all second coordinates: {1, 0, –5, 3}. Example

32. Slide 7- 32 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Find the domain and range of the function f shown here. y x -5 -4 -3 -2 -1 1 2 3 4 5 f 4 -2 -4 -3 3 2 5 1 6 7 Example

33. Slide 7- 33 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution x y -5 -4 -3 -2 -1 1 2 3 4 5 f 4 -2 -4 -3 3 2 5 1 6 The domain of f 7 The domain of f is the set of all x-values that are used in the points on the curve. These extend continuously from -5 to 3 and can be viewed as the curve’s shadow, or projection, on the x-axis. Thus the domain is

34. Slide 7- 34 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The range of f Solution The range of f is the set of all y-values that are used in the points on the curve. These extend continuously from -1 to 7 and can be viewed as the curve’s shadow, or projection, on the y-axis. Thus the range is

35. Slide 7- 35 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Determine the domain of Solution We ask, “Is there any number x for which we cannot compute 3x 2 – 4?” Since the answer is no, the domain of f is the set of all real numbers. Example

36. Slide 7- 36 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Determine the domain of Solution We ask, “Is there any number x for which cannot be computed?” Since cannot be computed when x – 8 = 0 the answer is yes. x – 8 = 0, x = 8 Thus 8 is not in the domain of f, whereas all other real numbers are. The domain of f is To determine what x-value would cause x – 8 to be 0, we solve an equation: Example

37. Slide 7- 37 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Functions Defined Piecewise Some functions are defined by different equations for various parts of their domains. Such functions are said to be piecewise defined. To evaluate a piecewise-defined function for an input a, we first determine what part of the domain a belongs to. The we use the appropriate formula for that part of the domain.

38. Slide 7- 38 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Find each function value for the function f given by a) f(  3)b) f(2)c) f(7) Solution a) f(x) = x + 3: f(  3) =  3 + 3 = 0 b)f(x) = x 2 ; f(2) = 2 2 = 4 c)f(7) = 4x = 4(7) = 28 Example

39. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graphs of Functions Linear Functions Nonlinear Functions 7.3

40. Slide 7- 40 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Equations of Lines Standard form: Ax + By = C Slope-intercept form: y = mx + b Point-slope form: y  y 1 = m(x  x 1 )

41. Slide 7- 41 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph 5x + 2y = 10. Solution Let x = 0 to find the y-intercept: 5 0 + 2y = 10 2y = 10 y = 5 The y-intercept is (0, 5). Let y = 0 to find the x-intercept: 5x + 2 0 = 10 5x = 10 x = 2 The x-intercept is (2, 0). 5x + 2y = 10 x-intercept (2, 0) y-intercept (0, 5) Example

42. Slide 7- 42 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph: 3x + 4y = 12 Solution Rewrite the equation in slope-intercept form. Example

43. Slide 7- 43 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution The slope is  3/4 and the y-intercept is (0, 3). We plot (0, 3), then move down 3 units and to the right 4 units. An alternate approach would be to move up 3 units and to the left 4 units. up 3 down 3 left 4 right 4 (0, 3) (4, 0) (  4, 6)

44. Slide 7- 44 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Linear Function A function described by an equation of the form f(x) = mx + b is a linear function. Its graph is a straight line with slope m and y-intercept at (0, b). When m = 0, the function is described by f(x) = b is called a constant function. Its graph is a horizontal line through (0, b).

45. Slide 7- 45 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph: Solution The slope is 4/3 and the y-intercept is (0,  2). We plot (0,  2), then move up 4 units and to the right 3 units. We could also move down 4 units and to the left 3 units. Then draw the line. up 4 units right 3 down 4 left 3 (  3,  6) (3, 2) (0,  2) Example

46. Slide 7- 46 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph y = 2 Solution This is a constant function. For every input x, the output is 2. The graph is a horizontal line. y = 2 (  4, 2) (0, 2) (4, 2) Example

47. Slide 7- 47 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Domain and Range of a Linear Function The domain of any linear function f(x) = mx + b is {x|x is a real number}. The range of any linear function f(x) = mx + b, m  0 is {y|y is a real number}. The range of a constant function f(x) = b is b.

48. Slide 7- 48 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Determine the domain and range of each of the following functions. a) f, where f(x) = 3x + 2b) h, where h(x) = 5 Solution a) f(x) = 3x + 2 describes a linear function, but not a constant function, Domain =  All R and Range = All R  b) h(x) = 5 is a constant function. Domain = All R and Range = {5} Example

49. Slide 7- 49 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Nonlinear Functions A function for which the graph is not a straight line is a nonlinear function. Type of functionExample Absolute-valuef(x) = |x + 2| Polynomialp(x) = x 4 + 3x 2 – 2 Quadratich(x) = x 2 – 4x + 2 Rational

50. Slide 7- 50 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Graph the function given by g(x) = |x + 2|. Solution Calculate function values for several choices of x and list the results in a table. xg(x) = |x + 2|(x, f(x)) 13(1, 3) 24(2, 4) –11(–1, 1) –20(–2, 0) –31(–3, 1) 02(0, 2) Domain: Range: {y|y > 0} Example

51. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Algebra of Functions The Sum, Difference, Product, or Quotient of Two Functions Domains and Graphs 7.4

52. Slide 7- 52 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Sum, Difference, Product, or Quotient of Two Functions Suppose that a is in the domain of two functions, f and g. The input a is paired with f (a) by f and with g(a) by g. The outputs can then be added to get f (a) + g(a).

53. Slide 7- 53 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Algebra of Functions If f and g are functions and x is in the domain of both functions, then:

54. Slide 7- 54 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution For a) ( f + g)(4)b) ( f – g)(x) c) ( f /g)(x)d) find the following. a) Since f (4) = –8 and g(4) = 13, we have ( f + g)(4) = f (4) + g(4) = –8 + 13 = 5. Example

55. Slide 7- 55 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution b) We have, d) Since f (–1) = –3 and g(–1) = –2, we have c) We have, We assume

56. Slide 7- 56 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Domains and Graphs we must first be able to find f (a) and g(a). This means a must be in the domain of both f and g.

57. Slide 7- 57 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley find the domains of Thus the domain of f + g, f – g, and Solution The domain of f is The domain of g is Example

58. Slide 7- 58 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued To find the domain of f /g, note that can not be evaluated if x + 1 = 0 or x – 2 = 0. Thus the domain of f /g is

59. Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Formulas, Applications and Variation Formulas Direct Variation Inverse Variation Joint and Combined Variation 7.5

60. Slide 7- 60 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Formulas Formulas occur frequently as mathematical models. Many formulas contain rational expressions, and to solve such formulas for a specified letter, we proceed as when solving rational equations.

61. Slide 7- 61 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley In a hydraulic system, a fluid is confined to two connecting chambers. The pressure in each chamber is the same and is given by finding the force exerted (F) divided by the surface area (A). Therefore, we know Solve for A 2. Example

62. Slide 7- 62 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Multiplying both sides by the LCD Simplifying by removing factors Dividing both sides by F 1 This formula can be used to calculate A 2 whenever A 1, F 2, and F 1 are known.

63. Slide 7- 63 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley To Solve a Rational Equation for a Specified Variable 1.If necessary, multiply both sides by the LCD to clear fractions. 2.Multiply, as needed, to remove parentheses. 3.Get all terms with the specified variable alone on one side. 4.Factor out the specified variable if it is in no more than one term. 5.Multiply or divide on both sides to isolate the specified variable.

64. Slide 7- 64 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Variation To extend our study of formulas and functions, we now examine three real-world situations: direct variation, inverse variation, and combined variation.

65. Slide 7- 65 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Direct Variation A mass transit driver earns $17 per hour. In 1 hr, $17 is earned. In 2 hr, $34 is earned. In 3 hr, $51 is earned, and so on. This gives rise to a set of ordered pairs:(1, 22), (2, 34), (3, 51), (4, 68), and so on. Note that the ratio of earnings E to time t is 17/1 in every case. If a situation gives rise to pairs of numbers in which the ratio is constant, we say that there is direct variation. Here earnings vary directly as the time: We have E/t = 17, so E = 17t, or using function notation, E(t) = 17t.

66. Slide 7- 66 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Direct Variation When a situation gives rise to a linear function of the form f (x) = kx, or y = kx, where k is a nonzero constant, we say that there is direct variation, that y varies directly as x, or that y is proportional to x. The number k is called the variation constant, or constant of proportionality.

67. Slide 7- 67 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Find the variation constant and an equation of variation if y varies directly as x, and y = 15 when x = 3. The variation constant is 5. The equation of variation is y = 5x. The notation y(x) = 5x or f (x) = 5x is also used. We know that (3, 15) is a solution of y = kx. Therefore, Substituting Solving for k Example

68. Slide 7- 68 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Inverse Variation To see what we mean by inverse variation, suppose it takes one person 8 hours to paint the baseball fields for the local park district. If two people do the job, it will take only 4 hours. If three people paint the fields, it will take only 2 and 2/3 hours, and so on. This gives rise to pairs of numbers, all have the same product: (1, 8), (2, 4), (3, 8/3), (4, 2), and so on.

69. Slide 7- 69 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Inverse Variation Note that the product of each pair of numbers is 8. Whenever a situation gives rise to pairs of numbers for which the product is constant, we say that there is inverse variation. Since pt = 8, the time t, in hours, required for the fields to be painted by p people is given by t = 8/p or, using function notation, t(p) = 8/p.

70. Slide 7- 70 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Inverse Variation When a situation gives rise to a rational function of the form f (x) = k/x, or y = k/x, where k is a nonzero constant, we say that there is inverse variation, that is y varies inversely as x, or that y is inversely proportional to x. The number k is called the variation constant, or constant of proportionality.

71. Slide 7- 71 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution The time, t, required to empty a tank varies inversely as the rate, r, of pumping. If a pump can empty a tank in 90 minutes at the rate of 1080 kL/min, how long will it take the pump to empty the same tank at the rate of 1500 kL/min? 1. Familiarize. Because of the phrase “... varies inversely as the rate, r, of pumping,” we express the amount of time needed to empty the tank as a function of the rate: t(r) = k/r Example

72. Slide 7- 72 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution (continued) 2. Translate. We use the given information to solve for k. Then we use that result to write the equation of variation. The equation of variation is t(r) = 97,200/r. This is the translation. Using function notation Replacing r with 1080 Replacing t(90) with 1080 Solving for k, the variation constant

73. Slide 7- 73 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley 3. Carry out. To find out how long it would take to pump out the tank at a rate of 1500 mL/min, we calculate t(1500). 4. Check. We could now recheck each step. Note that, as expected, as the rate goes up, the time it takes goes down. t = 64.8 when r = 1500 Solution (continued) 5. State. If the pump is emptying the tank at a rate of 1500 mL/min, then it will take 64.8 minutes to empty the entire tank.

74. Slide 7- 74 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Joint and Combined Variation When a variable varies directly with more than one other variable, we say that there is joint variation. For example, in the formula for the volume of a right circular cylinder, V = πr 2 h, we say that V varies jointly as h and the square of r.

75. Slide 7- 75 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Joint Variation y varies jointly as x and z if, for some nonzero constant k, y = kxz.

76. Slide 7- 76 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Find an equation of variation if a varies jointly as b and c, and a = 48 when b = 4 and c = 2. We have a = kbc, so The equation of variation is a = 6bc. The variation constant is 6. Example

77. Slide 7- 77 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Joint variation is one form of combined variation. In general, when a variable varies directly and/or inversely, at the same time, with more than one other variable, there is combined variation.

78. Slide 7- 78 Copyright © 2010 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution Find an equation of variation if y varies jointly as x and z and inversely as the product of w and p, and y = 60 when x = 24, z = 5, w = 2, and p = 3. The equation if variation is of the form so, substituting, we have: Thus, Example