2. Copyright © 2004 Pearson Education, Inc. Chapter 2 Graphs and Functions

3. Copyright © 2004 Pearson Education, Inc. 2.1 Graphs of Equations

4. Slide 2-4 Copyright © 2004 Pearson Education, Inc. Rectangular Coordinate System The horizontal line is called the x-axis. The vertical line is called the y-axis. The point of intersection is the origin. x-axis y-axis origin Quadrant I Quadrant II Quadrant III Quadrant IV

5. Slide 2-5 Copyright © 2004 Pearson Education, Inc. Plotting Points Each point in the xy-plane corresponds to a unique ordered pair (a, b). Plot the point (2, 4).  Move 2 units right  Move 4 units up 2 units 4 units

6. Slide 2-6 Copyright © 2004 Pearson Education, Inc. Distance Formula Suppose that P(x 1, y 1 ) and R(x 2, y 2 ) are two points in a coordinate plane. Then the distance between P and R, written d(P, R), is given by the distance formula,

7. Slide 2-7 Copyright © 2004 Pearson Education, Inc. Example Find the distance between P(  7, 3) and Q(4,  5).

8. Slide 2-8 Copyright © 2004 Pearson Education, Inc. Midpoint Formula The midpoint of a line segment with endpoints (x 1, y 1 ) and (x 2, y 2 ) is Find the midpoint M of the segment with endpoints (10,  5) and (  6, 4).

9. Slide 2-9 Copyright © 2004 Pearson Education, Inc. Finding Ordered Pairs that are Solutions of Equations For each equation find three ordered pairs that are solutions. Step: Choose any real number of x or y and substitute into the equation to get the corresponding value of the other variable. y = 5x + 2 Let y =  3  3 = 5x + 2  5 = 5x  1 = x (  1,  3) Let x = 0 y = 5(0) + 2 y = 2 (0, 2) Let x = 1 y = 5(1) + 2 y = 7 (1, 7)

10. Slide 2-10 Copyright © 2004 Pearson Education, Inc. Graphing The graph of an equation is found by plotting points that are solutions of the equation. The intercepts of the graph are good points to find first. x-intercept is an x-value where the graph intersects the x-axis. y = 0 y-intercept is a y-value where the graph intersects the y-axis. x = 0

11. Slide 2-11 Copyright © 2004 Pearson Education, Inc. Graphing an Equation by Point Plotting Step 1Find the intercepts. Step 2Find as many additional ordered pairs as needed. Step 3Plot the ordered pairs from Steps 1 and 2. Step 4Connect the points from Step 3 in a smooth line or curve.

12. Slide 2-12 Copyright © 2004 Pearson Education, Inc. Example Graph the equation y = 5x + 2 33 11 0  2/5 20 yx

13. Slide 2-13 Copyright © 2004 Pearson Education, Inc. Example Graph 62 31 20 yx

14. Slide 2-14 Copyright © 2004 Pearson Education, Inc. Example Graph x 2 + y 2 = 16  3 33 0 44 44 0 yx

15. Slide 2-15 Copyright © 2004 Pearson Education, Inc. Center-Radius Form of the Equation of a Circle The circle with center (h, k) and radius r has the equation (x  h) 2 + (y  k) 2 = r 2, the center-radius form of the equation of a circle (See Figure). A circle with center (0, 0) and radius r has equation x 2 + y 2 = r 2.

16. Slide 2-16 Copyright © 2004 Pearson Education, Inc. Graphing a Circle Graph the circle with equation (x + 2) 2 + (y  1) 2 = 9 The center is (  2, 1). The radius is 3. 3 (-2, 1)

17. Slide 2-17 Copyright © 2004 Pearson Education, Inc. Finding the Center and Radius by Completing the Square Decide whether or not the equation has a circle as its graph x 2 + 8x + y 2  6y + 16 = 0. Since 9 > 0, the equation represents a circle with center at (  4, 3) and radius 3.

18. Copyright © 2004 Pearson Education, Inc. 2.2 Functions

19. Slide 2-19 Copyright © 2004 Pearson Education, Inc. Definitions Relation—A relation is a set of ordered pairs. Function—A function is a relation in which, for each value of the first component of the ordered pairs, there is exactly one value of the second component. If the value of the variable y depends on the value of the x variable, then y is the dependent variable and x is the independent variable. (x, y)

20. Slide 2-20 Copyright © 2004 Pearson Education, Inc. Example Decide whether each relation defines a function. a) H = {(2, 3), (4,  1), (  3, 5)}  H is a function, because for each different x-value there is exactly one y-value. b) J = {(3, 1), (3, 4), (3, 6), (4, 5)}  J is not a function because one first component corresponds to more than one second component.

21. Slide 2-21 Copyright © 2004 Pearson Education, Inc. Domain and Range In a relation, the set of all values of the independent variable (x) is the domain; the set of all values of the dependent variable (y) is the range.

22. Slide 2-22 Copyright © 2004 Pearson Education, Inc. Example Give the domain and range of the relation. The domain is {4, 5, 6, 8}; the range is {C, D, E}. The mapping defines a function— each x-value corresponds to exactly one y-value. Give the domain and range of the relation. The domain is {3, 4, 2} and the range is {6}. The table defines a function because each different x-value corresponds to exactly one y-value. 45684568 CDECDE 62 64 63 yx

23. Slide 2-23 Copyright © 2004 Pearson Education, Inc. Finding Domains and Ranges from Graphs Find the domain and range of the relation. The x-values of the points on the graph include all numbers between  3 and 3, inclusive. The y-values include all numbers between  2 and 2, inclusive. Domain = [  3, 3] Range = [  2, 2] range domain

24. Slide 2-24 Copyright © 2004 Pearson Education, Inc. Example Find the domain and range of the relation. The arrowheads indicate that the line extends indefinitely left and right, as well as up and down. Therefore, both the domain and range include all real numbers, written ( ,  ).

25. Slide 2-25 Copyright © 2004 Pearson Education, Inc. Definitions Agreement on Domain Unless specified otherwise, the domain of a relation is assumed to be all real numbers that produce real numbers when substituted for the independent variable. Vertical Line Test If each vertical line intersects a graph in at most one point, then the graph is that of a function.

26. Slide 2-26 Copyright © 2004 Pearson Education, Inc. Example Use the vertical line test to determine whether each relation graphed is a function. The graph of (a) fails the vertical line test, it is not the graph of a function. Graph (b) represents a function. a.b.

27. Slide 2-27 Copyright © 2004 Pearson Education, Inc. Identifying Function, Domains, and Ranges from Equations Decide whether each relation defines a function and give the domain and range. a) y = x + 5 b) c)

28. Slide 2-28 Copyright © 2004 Pearson Education, Inc. Solutions a) y = x + 5  Function: Each value of x corresponds to just one value of y so the relation defines a function.  Domain: Any real number of ( ,  )  Range: Any real number ( ,  )

29. Slide 2-29 Copyright © 2004 Pearson Education, Inc. Solutions continued b)  Function: For any choice of x there is exactly one corresponding y-value. So the relation defines a function.  Domain: [1/3,  )  Range: y  0, or [0,  ) Range Domain

30. Slide 2-30 Copyright © 2004 Pearson Education, Inc. Solutions continued c)  Function: There is exactly one value of y for each value in the domain, so this equation defines a function.  Domain: All real numbers except those that make the denominator 0. ( ,  1)  (  1,  ).  Range: Values of y can be positive or negative, but never 0. The range is the interval ( , 0)  (0,  ). Domain Range

31. Slide 2-31 Copyright © 2004 Pearson Education, Inc. Variations of the Definition of Function A function is a relation in which, for each value of the first component of the ordered pairs, there is exactly one value of the second component. A function is a set of ordered pairs in which no first component is repeated. A function is a rule or correspondence that assigns exactly one range value to each domain value.

32. Slide 2-32 Copyright © 2004 Pearson Education, Inc. Function Notation y = f(x) is called function notation We read f(x) as “f of x” (The parentheses do not indicate multiplication.) The letter f stand for function. f(x) is just another name for the dependent variable y. Example: We can write y = 10x + 3 as f(x) = 10x + 3

33. Slide 2-33 Copyright © 2004 Pearson Education, Inc. Example Let f(x) = x 2  3x + 4. Find f(2) and f(a). Solutions: Thus, f(2) = 2; (2, 2)

34. Slide 2-34 Copyright © 2004 Pearson Education, Inc. Example using function notation Let g(x) = 3x + 1. Find and simplify g(a + 2).

35. Slide 2-35 Copyright © 2004 Pearson Education, Inc. For each function, find f(4) a) f(x) = 5x  2 f(4) = 5(4)  2 f(4) = 18 b) f = {(  3, 5), (2, 3), (4, 1)} Look at the ordered pairs for the y value when paired with x = 4. (4, 1), so f(4) = 1. Find 4 on the x-axis and move up until the graph is reached. The value on the y-axis is 2, thus f(4) = 2. y = f(x) c)

36. Slide 2-36 Copyright © 2004 Pearson Education, Inc. Finding an Expression for f(x) Step 1Solve the equation of y. Step 2Replace y with f(x). Example: Rewrite the equation x + 5y = 3. Solve for y.

37. Slide 2-37 Copyright © 2004 Pearson Education, Inc. Increasing, Decreasing, and Constant Functions Suppose that a function f is defined over an interval I. If x 1 and x 2 are in I, (a) f increases on I if, whenever x 1 < x 2, f(x 1 ) < f(x 2 ); (b) f decreases on I if, whenever x 1 f(x 2 ); (c) f is constant on I if, for every x 1 and x 2, f(x 1 ) = f(x 2 ).

38. Slide 2-38 Copyright © 2004 Pearson Education, Inc. Example Determine the intervals over which the function is increasing, decreasing, or constant. On the interval ( ,  1) the y-values are decreasing. On the interval [0.5, 1], the y-values are increasing. On the interval [1,  ) the y-values are constant.

39. Copyright © 2004 Pearson Education, Inc. 2.3 Linear Functions

40. Slide 2-40 Copyright © 2004 Pearson Education, Inc. Definition of a Linear Function A function f is a linear function if, for real numbers a and b, f(x) = ax + b.

41. Slide 2-41 Copyright © 2004 Pearson Education, Inc. X and Y-Intercepts The x-intercept is found by letting f(x) = 0 and solving for x.  Example: f(x) =  2x + 6 0 =  2x + 6  6 =  2x 3 = x The y-intercept is f(0).  Example: f(x) =  2x + 6 f(x) =  2(0) + 6 f(x) = 6

42. Slide 2-42 Copyright © 2004 Pearson Education, Inc. Graphing a Horizontal Line Graph f(x) =  3 y is always equal to  3, the value of y can never be 0. The graph is parallel to the x-axis.

43. Slide 2-43 Copyright © 2004 Pearson Education, Inc. Graphing a Vertical Line Graph x =  3 x always equals  3, the value of x can never be 0. The graph is parallel to the y-axis.

44. Slide 2-44 Copyright © 2004 Pearson Education, Inc. Standard Form Ax + By = C  Example: 4x  5y = 0

45. Slide 2-45 Copyright © 2004 Pearson Education, Inc. Slope A numerical measure of the steepness of a line. The slope m of the line through the points (x 1,y 1 ) and (x 2,y 2 ) is

46. Slide 2-46 Copyright © 2004 Pearson Education, Inc. Finding Slope Example 1: Find the slope of the line through the points (5,  3) and (  2,  3)

47. Slide 2-47 Copyright © 2004 Pearson Education, Inc. Finding Slope Example 2: Find the slope of the line through the points (6,  4) and (2,  5)

48. Slide 2-48 Copyright © 2004 Pearson Education, Inc. Types of Slope The slope of a vertical line is undefined. The slope of a horizontal line is zero. zero negative undefined positive

49. Slide 2-49 Copyright © 2004 Pearson Education, Inc. Graphing a Line Using a Point and Slope Example: Graph the line passing through the point (  1, 5) and having slope Plot the point (  1, 5). A change of  5 units vertically and 3 units horizontally results in the point (2, 0). Plot the point and draw the line. 5 down 3 right

50. Slide 2-50 Copyright © 2004 Pearson Education, Inc. Average Rate of Change Slope is a ratio of the vertical change in y to the horizontal change in x. Average rate of change in y per unit of change in x. The value of y depends on the value of x.

51. Copyright © 2004 Pearson Education, Inc. 2.4 Equations of Lines; Curve Fitting

52. Slide 2-52 Copyright © 2004 Pearson Education, Inc. Point-Slope Form The line with slope m and passing through the point (x 1, y 1 ) has an equation y  y 1 = m(x  x 1 ). Example: Write an equation of the line through (  2, 3) having slope 4. Solution: Here x 1 =  2, y 1 = 3, and m = 4.

53. Slide 2-53 Copyright © 2004 Pearson Education, Inc. Two-Points Find the equation of the line through (3, 4)and (  2, 6). Solution: Find the slope first. Use either point. We choose (3, 4).

54. Slide 2-54 Copyright © 2004 Pearson Education, Inc. Slope-Intercept Form The line with slope m and y-intercept b has an equation y = mx + b. Example: Find an equation of the line with slope and y-intercept 4. m = and b = 4. substitute these values into the slope-intercept form.

55. Slide 2-55 Copyright © 2004 Pearson Education, Inc. Using Slope-Intercept to Graph Find the slope and y- intercept of 4x  y = 3. Then graph the line. First, write the equation in slope-intercept form. y = 4x  3 slope y-intercept y changes 4 units x changes 1 unit

56. Slide 2-56 Copyright © 2004 Pearson Education, Inc. Equations of Horizontal and Vertical Lines An equation of a vertical line through point (a, b) is x = a.  x = 2 An equation of a horizontal line through point (a, b) is y = b.  y = 3

57. Slide 2-57 Copyright © 2004 Pearson Education, Inc. Parallel and Perpendicular Lines Parallel Lines Two distinct nonvertical lines are parallel if and only if they have the same slope. Perpendicular Lines Two lines, neither of which is vertical, are perpendicular if and only if their slopes have a product of  1. Thus, the slopes of perpendicular lines, neither of which is vertical, are negative reciprocals.

58. Slide 2-58 Copyright © 2004 Pearson Education, Inc. Example: Find the equation in slope-intercept form of the line that passes through the point (2, 4) and satisfy the given conditions. Parallel to the line 3x + 4y = 12 Solve for y, to find the slope. Use the point-slope form. Perpendicular to the line 3x + 4y = 12. The slope is  3/4, so the slope of a line perpendicular to it is 4/3.

59. Slide 2-59 Copyright © 2004 Pearson Education, Inc. Curve Fitting Step 1: Make a scatter diagram of the data. Step 2:Find an equation that models the data. For a line, this involves selecting two data points and find the equation of the line through them.

60. Slide 2-60 Copyright © 2004 Pearson Education, Inc. Finding an Equation of a Line that Models Data Average annual tuition and fees for in-state students at public 4-year colleges are shown in the table for selected years and graphed as ordered pairs of points in the figure shown, where x = 0 represents 1990, x = 4 represents 1994, and so on, and y represents the cost in dollars. This graph of ordered pairs of data is called a scatter diagram. a) Find an equation that models the data. b) Use the equation from part (a) to approximate the cost of tuition and fees at public 4-year colleges in 2002.

61. Slide 2-61 Copyright © 2004 Pearson Education, Inc. Table and Scatter Diagram Table Scatter Diagram Source: U.S. National Center for Education Statistics 37742000 34861998 31511996 28201994 20351990 Cost (in dollars) Year

62. Slide 2-62 Copyright © 2004 Pearson Education, Inc. Solution a) Choose two data points (0, 2035) and (10, 3774), find the slope. The slope 173.9 indicates that the cost of tuition and fees for in-state students increased by about $174 per year from 1990 to 2000. Use the slope and y-intercept 2035, to write the equation of the line.

63. Slide 2-63 Copyright © 2004 Pearson Education, Inc. Solution continued b) The value x = 12 corresponds to the year 2002, so we substitute 12 for x. According to the model, average tuition and fees for in-state students at public 4-year colleges in 2002 were about $4122.

64. Copyright © 2004 Pearson Education, Inc. 2.5 Graphs of Basic Functions

65. Slide 2-65 Copyright © 2004 Pearson Education, Inc. Continuity A function is continuous over an interval of its domain if its hand-drawn graph over that interval can be sketched without lifting the pencil from the paper. If a function is not continuous at a point, then it has a discontinuity there.

66. Slide 2-66 Copyright © 2004 Pearson Education, Inc. Examples The function is continuous over its entire domain, ( ,  ). The function has a point of discontinuity at x = 4. Thus it is continuous over the intervals ( , 4) and (4,  ).

67. Slide 2-67 Copyright © 2004 Pearson Education, Inc. The Identity Function

68. Slide 2-68 Copyright © 2004 Pearson Education, Inc. The Squaring Function

69. Slide 2-69 Copyright © 2004 Pearson Education, Inc. Cubing Function

70. Slide 2-70 Copyright © 2004 Pearson Education, Inc. Square Root Function

71. Slide 2-71 Copyright © 2004 Pearson Education, Inc. Cube Root Function

72. Slide 2-72 Copyright © 2004 Pearson Education, Inc. Absolute Value Function

73. Slide 2-73 Copyright © 2004 Pearson Education, Inc. Piecewise-Defined Functions Graph the function. Graph each interval of the domain separately. If x  2, the graph of f(x) =  2x + 5 has an endpoint at x = 2. We find the corresponding y-value by substituting 2 for x to get y = 1. We select other points on this part of the graph and draw the graph as a partial line with endpoint (2, 1)

74. Slide 2-74 Copyright © 2004 Pearson Education, Inc. Piecewise-Defined Functions continued Graph the function for x > 2 similarly, using f(x) = x + 1. This partial line has an open endpoint at (2, 3). Use y = x + 1 to find another point with x-value greater than 2 to complete the graph.

75. Slide 2-75 Copyright © 2004 Pearson Education, Inc. Greatest Integer Function

76. Copyright © 2004 Pearson Education, Inc. 2.6 Graphing Techniques

77. Slide 2-77 Copyright © 2004 Pearson Education, Inc. Stretching and Shrinking The graph of g(x) = a f(x) has the same general shape as the graph of f(x). If |a| > 1, then the graph is stretched vertically (narrower) compared to the graph of f(x). If 0 < |a| < 1, then the graph is shrunken vertically (wider) compared to the graph of f(x).

78. Slide 2-78 Copyright © 2004 Pearson Education, Inc. Example Graph the function. Compare the tables and graphs. The graph g(x) = 3|x| is narrower (blue) than that of f(x) (red). 622 311 000 31 11 62 22 3|x||x||x|x

79. Slide 2-79 Copyright © 2004 Pearson Education, Inc. Example Graph the function The coefficient 1/3 causes the graph to be wider. 133 1/311 000 1 11 13 33 |x||x|x

80. Slide 2-80 Copyright © 2004 Pearson Education, Inc. Reflecting Across an Axis The graph of y =  f(x) is the same as the graph of y = f(x) reflected across the x-axis. The graph of y = f(  x) is the same as the graph of y = f(x) reflected across the y-axis.

81. Slide 2-81 Copyright © 2004 Pearson Education, Inc. Example Graph 22 22 1 000 x

82. Slide 2-82 Copyright © 2004 Pearson Education, Inc. Example Graph undefined 11 2  2 undefined22 1 000 x

83. Slide 2-83 Copyright © 2004 Pearson Education, Inc. Symmetry with Respect to an Axis The graph of an equation is symmetric with respect to the y-axis if the replacement of x with  x results in an equivalent equation. The graph of an equation is symmetric with respect to the x-axis if the replacement of y with  y is an equivalent equation.

84. Slide 2-84 Copyright © 2004 Pearson Education, Inc. Example Test for symmetry with respect to the x-axis and the y-axis. a) y = x 2  4b) x = y 2 + 2 a) Replace x with  x. The result is the same as the original equation, so the graph is symmetric with respect to the y-axis. It is NOT symmetric with respect to the x-axis.

85. Slide 2-85 Copyright © 2004 Pearson Education, Inc. Example continued b) x = y 2 + 2 Replace y with  y. The result is the same as the original equation. The graph is symmetric with respect to the x-axis. It is NOT symmetric with respect to the y-axis.

86. Slide 2-86 Copyright © 2004 Pearson Education, Inc. Symmetry with Respect to the Origin The graph of an equation is symmetric with respect to the origin if the replacement of both x with  x and y with  y results in an equivalent equation. Example: x 2 + y 2 = 9 Replace x with  x and y with  y. The graph is symmetric with respect to the origin.

87. Slide 2-87 Copyright © 2004 Pearson Education, Inc. Even and Odd Functions A function f is called an even function if f(  x) = f(x) for all x in the domain of f. (Its graph is symmetric with respect to the y-axis.) A function f is called an odd function if f(  x) =  f(x) for all x in the domain of f. (Its graph is symmetric with respect to the origin.)

88. Slide 2-88 Copyright © 2004 Pearson Education, Inc. Example Decide whether the function is even, odd, or neither. a) f(x) = 4x 4  x 2 Replacing x in f(x) = 4x 4  x 2 with  x gives Since f(  x) = f(x) for each x in the domain of the function, f is even.

89. Slide 2-89 Copyright © 2004 Pearson Education, Inc. Vertical Translations If a function g is defined by g(x) = f(x) + c, where c is a real number, then for every point (x, y) on the graph of f, there will be a corresponding point (x, y + c) on the graph of g. The graph of g will be the same as the graph of f, but translated c units up if c is positive or |c| units down if c is negative. The graph of g is called a vertical translation of the graph of f.

90. Slide 2-90 Copyright © 2004 Pearson Education, Inc. Example Graph y = |x| + 2. 422 311 200 31 11 42 22 |x|+2|x||x|x

91. Slide 2-91 Copyright © 2004 Pearson Education, Inc. Horizontal Translations If a function g is defined by g(x) = f(x  c), where c is a real number, then for every point (x, y) on the graph of f, there will be a corresponding point (x + c, y) on the graph of g. The graph of g will be the same as the graph of f, but translated c units to the right if c is positive or |c| units to the left if c is negative. The graph of g is called a horizontal translation of the graph of f.

92. Slide 2-92 Copyright © 2004 Pearson Education, Inc. Example Graph g(x) = |x + 1| 322 211 100 01 11 12 22 |x + 1||x||x|x

93. Slide 2-93 Copyright © 2004 Pearson Education, Inc. Summary of Translations Right y = f(x  c) Lefty = f(x + c) Down y = f(x)  c Upy = f(x) + c Shift the Graph of y = f(x) by c Units To Graph:

94. Slide 2-94 Copyright © 2004 Pearson Education, Inc. More than One Translation Graph The lowest point of the graph of y = |x| is translated 2 units right and 1 unit up. The graph opens down because of the negative sign in front of the absolute value expression, making the lowest point now the highest point on the graph.

95. Slide 2-95 Copyright © 2004 Pearson Education, Inc. Example Graph The graph will have the same shape as y = x 2, but it is wider (that is, shrunken vertically) and translated 2 units down.

96. Copyright © 2004 Pearson Education, Inc. 2.7 Function Operations and Composition

97. Slide 2-97 Copyright © 2004 Pearson Education, Inc. Operations on Functions Given two functions f and g, then for all values of x for which both f(x) and g(x) are defined, the functions f + g, f  g, fg, and are defined as follows:  (f + g)(x) = f(x) + g(x) Sum  (f  g)(x) = f(x)  g(x)Difference  (fg)(x) = f(x)  g(x) Product  Quotient

98. Slide 2-98 Copyright © 2004 Pearson Education, Inc. Example Let f(x) = x + 1 and g(x) = 2x 2  3. Find (f + g)(3) Find (fg)(  2)

99. Slide 2-99 Copyright © 2004 Pearson Education, Inc. Example continued Solution: Since f(3) = 4 and g(3) = 15, (f + g)(3) = f(3) + g(3) (f + g)(x) = f(x) + g(x) = 4 + 15 = 19 Since f(  2) =  1 and g(  2) = 5, (fg)(  2) = f(  2)  g(  2) (fg)(x) = f(x)  g(x) =  1  5 =  5

100. Slide 2-100 Copyright © 2004 Pearson Education, Inc. Domains For functions f and g, the domain of f + g, f  g, and fg include all real numbers in the intersection of the domains of f and g, while the domain of includes those real numbers in the intersection of the domains of f and g for which g(x)  0. Example: Let f(x) = 3x  5 and g(x) = Find (f  g)(x) Find Give the domains of each function.

101. Slide 2-101 Copyright © 2004 Pearson Education, Inc. Domains continued Solutions: (f  g)x = f(x) – g(x) = In part (a), the domain of f is the set of all real numbers ( , ), and the domain of g, since includes just the real numbers to make 3x  2 nonnegative. That is, so.

102. Slide 2-102 Copyright © 2004 Pearson Education, Inc. Domains continued The domain of g is. The domain of f  g is the intersection of the domains of f and g, which is. The domain of includes those real numbers in the intersection above for which ; that is, the domain of is.

103. Slide 2-103 Copyright © 2004 Pearson Education, Inc. The Difference Quotient Suppose the point P lies on the graph of y = f(x), and h is a positive number. If we let (x, f(x)) denote the coordinates of P and (x + h, f(x + h)) denote the coordinates of Q, then the line joining P and Q has slope This expression is called the difference quotient.

104. Slide 2-104 Copyright © 2004 Pearson Education, Inc. Example Let f(x) = x 2 + 4x. Find the difference quotient and simplify the expression. Solution: Step 1 Find f(x + h) Replace x in f(x) with x + h. f(x + h) = (x + h) 2 + 4(x + h) Step 2 Find f(x + h)  f(x) f(x + h)  f(x) = [ (x + h) 2 + 4(x + h)]  (x 2 + 4x) Substitute. = x 2 + 2xh + h 2 + 4(x + h)  (x 2 + 4x) Square x + h. = x 2 + 2xh + h 2 + 4x + 4h  x 2  4x = 2xh + h 2 + 4h

105. Slide 2-105 Copyright © 2004 Pearson Education, Inc. Example continued Step 3 Find the difference quotient.

106. Slide 2-106 Copyright © 2004 Pearson Education, Inc. Composition of Functions If f and g are functions, then the composite functions, or composition, of g and f is defined by The domain of is the set of all numbers x in the domain of f such that f(x) is in the domain of g.

107. Slide 2-107 Copyright © 2004 Pearson Education, Inc. Example Let f(x) = 3x  1 and g(x) = x + 5. Find each composition. Solution: First find g(2). Since g(x) = x + 5, g(2) = 2 + 5 = 7. Now find = f[g(2)] = f(7): f[g(2)] = f(7) = 3(7)  1 = 20.

108. Slide 2-108 Copyright © 2004 Pearson Education, Inc. Example continued Solution: First find f(2). Since f(x) = 3x  1 f(2) = (3(2)  1) = 5 Now find = g[f(2)] = g(5) g[f(2)] = g(5) = 5 + 5 = 10.

109. Slide 2-109 Copyright © 2004 Pearson Education, Inc. Finding Composite Functions Example: Let f(x) = x 2 and g(x) = x  2. Find and. Solution:

110. Slide 2-110 Copyright © 2004 Pearson Education, Inc. Composite Functions The previous example shows it is not always true that.  In fact, the composite functions are equal only for a special class of functions.

111. Slide 2-111 Copyright © 2004 Pearson Education, Inc. Finding Functions That Form a Given Composite Example: Find functions f and g such that Solution: Note the repeated quantity x 2 + 3. If we choose g(x) = x 2 + 3 and f(x) = x 3  2x + 7 then