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output, the number of police officers (N), is 300. Remember, N = f (y). The statement f (2005) = 300 tells us that in the year 2005 there were 300 police officers in the town. 3.2 Use function notation to express the weight of a pig in pounds as a function of its age in days d. Instead of a notation such as y = f(x), could we use the same symbol for the output as for the function, such as y = y(x), meaning βy is a function of x?β Yes, this is often done, especially in applied subjects that use higher math, such as physics and engineering. However, in exploring math itself we like to maintain a distinction between a function such as f, which is a rule or procedure, and the output y we get by applying f to a particular input x. This is why we usually use notation such as y = f (x), P = W(d), and so on. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 231 Representing Functions Using Tables A common method of representing functions is in the form of a table. The table rows or columns display the corresponding input and output values. In some cases, these values represent all we know about the relationship; other times, the table provides a few select examples from a more complete relationship. Table 3.3 lists the input number of each month (January = 1, February = 2, and so on) and the output value of the number of days in that month. This information represents all we know about the months and days for a given year (that is not a leap year). Note that, in this table, we define a days-in-a-month function f where D = f (m) identifies months by an integer rather than by name. Month number, m (input) Days in month, D (output) Table 3. 10 11 12 31 28 31 30 31 30 31 31 30 31 30 31 Table 3.4 defines a function Q = g(n). Remember, this notation tells us that g is the name of the function that takes the input n and gives the output Table 3.4 Table 3.5 displays the age of children in years and their corresponding heights. This table displays just some of the data available for the heights and ages of children. We can see right away that this table does not represent a function because the same
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input value, 5 years, has two different output values, 40 in. and 42 in. Age in years, a (input) 5 5 6 7 8 9 10 Height in inches, h (output) 40 42 44 47 50 52 54 Table 3.5 Given a table of input and output values, determine whether the table represents a function. 1. Identify the input and output values. 2. Check to see if each input value is paired with only one output value. If so, the table represents a function. Example 3.5 Identifying Tables that Represent Functions Which table, Table 3.6, Table 3.7, or Table 3.8, represents a function (if any)? 232 Chapter 3 Functions Input Output 2 5 8 1 3 6 Table 3.6 Input Output β3 0 4 5 1 5 Table 3.7 Input Output 1 5 5 0 2 4 Table 3.8 Solution Table 3.6 and Table 3.7 define functions. In both, each input value corresponds to exactly one output value. Table 3.8 does not define a function because the input value of 5 corresponds to two different output values. When a table represents a function, corresponding input and output values can also be specified using function notation. The function represented by Table 3.6 can be represented by writing Similarly, the statements represent the function in Table 3.7. f (2) = 1, f (5) = 3, and f (8) = 6 g(β3) = 5, g(0) = 1, and g(4) = 5 Table 3.8 cannot be expressed in a similar way because it does not represent a function. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 233 3.3 Does Table 3.9 represent a function? Input Output 1 2 3 10 100 1000 Table 3.9 Finding Input and Output Values of a Function When we know an input value and want to determine the corresponding output value for a function, we evaluate the function. Evaluating will always produce one result because each input value of a function corresponds to exactly one output value. When we know an output value and want to determine the input values that would produce that output value, we set the output equal to the functionβs formula and solve for the input. Solving can produce more than one solution because different input values can produce the same output value. Evaluation of Functions in Algebraic Forms When we have a
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function in formula form, it is usually a simple matter to evaluate the function. For example, the function f (x) = 5 β 3x2 can be evaluated by squaring the input value, multiplying by 3, and then subtracting the product from 5. Given the formula for a function, evaluate. 1. Replace the input variable in the formula with the value provided. 2. Calculate the result. Example 3.6 Evaluating Functions at Specific Values Evaluate f (x) = x2 + 3x β 4 at a. 2 b. a c. a + h d. f (a + h) β f (a) h Solution Replace the x in the function with each specified value. a. Because the input value is a number, 2, we can use simple algebra to simplify. 234 Chapter 3 Functions f (2) = 22 + 3(2. In this case, the input value is a letter so we cannot simplify the answer any further. c. With an input value of a + h, we must use the distributive property. f (a) = a2 + 3a β 4 f (a + h) = (a + h)2 + 3(a + h) β 4 = a2 + 2ah + h2 + 3a + 3h β 4 d. In this case, we apply the input values to the function more than once, and then perform algebraic operations on the result. We already found that f (a + h) = a2 + 2ah + h2 + 3a + 3h β 4 and we know that Now we combine the results and simplify. f (a) = a2 + 3a β 4 f (a + h) β f (a) h β βa2 + 2ah + h2 + 3a + 3h β 4 β β β β β βa2 + 3a β 4 β = = 2ah + h2 + 3h h h(2a + h + 3) h = h Factor out h. = 2a + h + 3 Simplify. Example 3.7 Evaluating Functions Given the function h(p) = p2 + 2p, evaluate h(4). Solution To evaluate h(4), we substitute the value 4 for the input variable p in the given function. h(p) = p2 + 2p h(4) = (4
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)2 + 2(4) = 16 + 8 = 24 Therefore, for an input of 4, we have an output of 24. 3.4 Given the function g(m) = m β 4, evaluate g(5). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 235 Example 3.8 Solving Functions Given the function h(p) = p2 + 2p, solve for h(p) = 3. Solution h(p) = 3 p2 + 2p = 3 p2 + 2p β 3 = 0 (p + 3)(p β 1) = 0 β = 0, either β βp + 3β Substitute the original function h(p) = p2 + 2p. Subtract 3 from each side. Factor. βp β 1β If β βp + 3β β β βp β 1β β = 0 or β β = 0 (or both of them equal 0). We will set each factor equal to 0 and solve for p in each case. (p + 3) = 0, (p β 1) = 0, p = β3 p = 1 This gives us two solutions. The output h(p) = 3 when the input is either p = 1 or p = β 3. We can also verify by graphing as in Figure 3.5. The graph verifies that h(1) = h(β3) = 3 and h(4) = 24. Figure 3.5 3.5 Given the function g(m) = m β 4, solve g(m) = 2. Evaluating Functions Expressed in Formulas Some functions are defined by mathematical rules or procedures expressed in equation form. If it is possible to express the function output with a formula involving the input quantity, then we can define a function in algebraic form. For example, the equation 2n + 6p = 12 expresses a functional relationship between n and p. We can rewrite it to decide if p is a function of n. 236 Chapter 3 Functions Given a function in equation form, write its algebraic formula. 1. Solve the equation to isolate the output variable on one side of the equal sign, with the other side as an expression that involves only the input variable.
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2. Use all the usual algebraic methods for solving equations, such as adding or subtracting the same quantity to or from both sides, or multiplying or dividing both sides of the equation by the same quantity. Example 3.9 Finding an Equation of a Function Express the relationship 2n + 6p = 12 as a function p = f (n), if possible. Solution To express the relationship in this form, we need to be able to write the relationship where p is a function of n, which means writing it as p = [expression involving n]. 2n + 6p = 12 Subtract 2n from both sides. Divide both sides by 6 and simplify. 6p = 12 β 2n p = 12 β 2n 6 β 2n 6 n p = 12 6 p = 2 β 1 3 Therefore, p as a function of n is written as p = f (n) = 2 β 1 3 n Example 3.10 Expressing the Equation of a Circle as a Function Does the equation x2 + y2 = 1 represent a function with x as input and y as output? If so, express the relationship as a function y = f (x). Solution First we subtract x2 from both sides. We now try to solve for y in this equation. y2 = 1 β x2 y = Β± 1 β x2 = + 1 β x2 and β 1 β x2 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 237 We get two outputs corresponding to the same input, so this relationship cannot be represented as a single function y = f (x). If we graph both functions on a graphing calculator, we will get the upper and lower semicircles. 3.6 If x β 8y3 = 0, express y as a function of x. Are there relationships expressed by an equation that do represent a function but that still cannot be represented by an algebraic formula? Yes, this can happen. For example, given the equation x = y + 2 y, if we want to express y as a function of x, there is no simple algebraic formula involving only x that equals y. However, each x does determine a unique value for y, and there are mathematical procedures by which y can be found to any desired accuracy. In this case, we say that the equation gives an implicit (implied) rule for y as a function of x, even though the formula cannot be
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written explicitly. Evaluating a Function Given in Tabular Form As we saw above, we can represent functions in tables. Conversely, we can use information in tables to write functions, and we can evaluate functions using the tables. For example, how well do our pets recall the fond memories we share with them? There is an urban legend that a goldfish has a memory of 3 seconds, but this is just a myth. Goldfish can remember up to 3 months, while the beta fish has a memory of up to 5 months. And while a puppyβs memory span is no longer than 30 seconds, the adult dog can remember for 5 minutes. This is meager compared to a cat, whose memory span lasts for 16 hours. The function that relates the type of pet to the duration of its memory span is more easily visualized with the use of a table. See Table 3.10.[2] Pet Memory span in hours Puppy Adult dog Cat Goldfish Beta fish Table 3.10 0.008 0.083 16 2160 3600 At times, evaluating a function in table form may be more useful than using equations. Here let us call the function P. The domain of the function is the type of pet and the range is a real number representing the number of hours the petβs memory span lasts. We can evaluate the function P at the input value of βgoldfish.β We would write P(goldfis ) = 2160. Notice that, to evaluate the function in table form, we identify the input value and the corresponding output value from the 2. http://www.kgbanswers.com/how-long-is-a-dogs-memory-span/4221590. Accessed 3/24/2014. 238 Chapter 3 Functions pertinent row of the table. The tabular form for function P seems ideally suited to this function, more so than writing it in paragraph or function form. Given a function represented by a table, identify specific output and input values. 1. Find the given input in the row (or column) of input values. 2. Identify the corresponding output value paired with that input value. 3. Find the given output values in the row (or column) of output values, noting every time that output value appears. 4. Identify the input value(s) corresponding to the given output value. Example 3.11 Evaluating and Solving a Tabular Function Using Table 3.11, a. Evaluate g(3
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). b. Solve g(n) = 6. n g(n Table 3.11 Solution a. Evaluating g(3) means determining the output value of the function g for the input value of n = 3. The table output value corresponding to n = 3 is 7, so g(3) = 7. b. Solving g(n) = 6 means identifying the input values, n, that produce an output value of 6. Table 3.12 shows two solutions: n = 2 and n = 4. n g(n Table 3.12 When we input 2 into the function g, our output is 6. When we input 4 into the function g, our output is also 6. 3.7 Using Table 3.11, evaluate g(1). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 239 Finding Function Values from a Graph Evaluating a function using a graph also requires finding the corresponding output value for a given input value, only in this case, we find the output value by looking at the graph. Solving a function equation using a graph requires finding all instances of the given output value on the graph and observing the corresponding input value(s). Example 3.12 Reading Function Values from a Graph Given the graph in Figure 3.6, a. Evaluate f (2). b. Solve f (x) = 4. Figure 3.6 Solution a. To evaluate f (2), locate the point on the curve where x = 2, then read the y-coordinate of that point. The point has coordinates (2, 1), so f (2) = 1. See Figure 3.7. 240 Chapter 3 Functions Figure 3.7 b. To solve f (x) = 4, we find the output value 4 on the vertical axis. Moving horizontally along the line y = 4, we locate two points of the curve with output value 4: (β1, 4) and (3, 4). These points represent the two solutions to f (x) = 4: x = β1 or x = 3. This means f (β1) = 4 and f (3) = 4, or when the input is β1 or 3, the output is 4. See Figure 3.8. Figure 3.8 3.8 Using Figure 3.6, solve f (x) = 1. This content is available for free at https://cnx.org
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/content/col11758/1.5 Chapter 3 Functions 241 Determining Whether a Function is One-to-One Some functions have a given output value that corresponds to two or more input values. For example, in the stock chart shown in 51260 (https://cnx.org/content/51260/latest/#Figure_01_00_001) at the beginning of this chapter, the stock price was $1000 on five different dates, meaning that there were five different input values that all resulted in the same output value of $1000. However, some functions have only one input value for each output value, as well as having only one output for each input. We call these functions one-to-one functions. As an example, consider a school that uses only letter grades and decimal equivalents, as listed in Table 3.13. Letter grade Grade point average A B C D Table 3.13 4.0 3.0 2.0 1.0 This grading system represents a one-to-one function because each letter input yields one particular grade-point average output and each grade-point average corresponds to one input letter. To visualize this concept, letβs look again at the two simple functions sketched in Figure 3.2(a) and Figure 3.2(b). The function in part (a) shows a relationship that is not a one-to-one function because inputs q and r both give output n. The function in part (b) shows a relationship that is a one-to-one function because each input is associated with a single output. One-to-One Function A one-to-one function is a function in which each output value corresponds to exactly one input value. There are no repeated x- or y-values. Example 3.13 Determining Whether a Relationship Is a One-to-One Function Is the area of a circle a function of its radius? If yes, is the function one-to-one? Solution A circle of radius r has a unique area measure given by A = Οr 2, so for any input, r, there is only one output, A. The area is a function of radius r. If the function is one-to-one, the output value, the area, must correspond to a unique input value, the radius. Any area measure A is given by the formula A = Οr 2. Because areas and radii are positive numbers, there is exactly one solution:
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r = A Ο. So the area of a circle is a one-to-one function of the circleβs radius. 242 Chapter 3 Functions 3.9 a. b. c. Is a balance a function of the bank account number? Is a bank account number a function of the balance? Is a balance a one-to-one function of the bank account number? 3.10 a. If each percent grade earned in a course translates to one letter grade, is the letter grade a function of the percent grade? b. If so, is the function one-to-one? Using the Vertical Line Test As we have seen in some examples above, we can represent a function using a graph. Graphs display a great many inputoutput pairs in a small space. The visual information they provide often makes relationships easier to understand. By convention, graphs are typically constructed with the input values along the horizontal axis and the output values along the vertical axis. The most common graphs name the input value x and the output value y, and we say y is a function of x, or y = f (x) when the function is named f. The graph of the function is the set of all points (x, y) in the plane that satisfies the equation y = f (x). If the function is defined for only a few input values, then the graph of the function consists of only a few points, where the x-coordinate of each point is an input value and the y-coordinate of each point is the corresponding output value. For example, the black dots on the graph in Figure 3.9 tell us that f (0) = 2 and f (6) = 1. However, the set of all points (x, y) satisfying y = f (x) is a curve. The curve shown includes (0, 2) and (6, 1) because the curve passes through those points. Figure 3.9 The vertical line test can be used to determine whether a graph represents a function. If we can draw any vertical line that intersects a graph more than once, then the graph does not define a function because a function has only one output value for each input value. See Figure 3.10. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 243 Figure 3.10 Given a graph, use the vertical line test to determine if the graph represents a function. 1. 2. Inspect
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the graph to see if any vertical line drawn would intersect the curve more than once. If there is any such line, determine that the graph does not represent a function. Example 3.14 Applying the Vertical Line Test Which of the graphs in Figure 3.11 represent(s) a function y = f (x)? Figure 3.11 Solution If any vertical line intersects a graph more than once, the relation represented by the graph is not a function. Notice that any vertical line would pass through only one point of the two graphs shown in parts (a) and (b) of Figure 3.11. From this we can conclude that these two graphs represent functions. The third graph does not represent a function because, at most x-values, a vertical line would intersect the graph at more than one point, as shown in Figure 3.12. 244 Chapter 3 Functions Figure 3.12 3.11 Does the graph in Figure 3.13 represent a function? Figure 3.13 Using the Horizontal Line Test Once we have determined that a graph defines a function, an easy way to determine if it is a one-to-one function is to use the horizontal line test. Draw horizontal lines through the graph. If any horizontal line intersects the graph more than once, then the graph does not represent a one-to-one function. Given a graph of a function, use the horizontal line test to determine if the graph represents a one-to-one function. 1. 2. Inspect the graph to see if any horizontal line drawn would intersect the curve more than once. If there is any such line, determine that the function is not one-to-one. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 245 Example 3.15 Applying the Horizontal Line Test Consider the functions shown in Figure 3.11(a) and Figure 3.11(b). Are either of the functions one-to-one? Solution The function in Figure 3.11(a) is not one-to-one. The horizontal line shown in Figure 3.14 intersects the graph of the function at two points (and we can even find horizontal lines that intersect it at three points.) Figure 3.14 The function in Figure 3.11(b) is one-to-one. Any horizontal line will intersect a diagonal line at most once. 3.12 Is the graph shown in Figure 3.12 one-to
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-one? Identifying Basic Toolkit Functions In this text, we will be exploring functionsβthe shapes of their graphs, their unique characteristics, their algebraic formulas, and how to solve problems with them. When learning to read, we start with the alphabet. When learning to do arithmetic, we start with numbers. When working with functions, it is similarly helpful to have a base set of building-block elements. We call these our βtoolkit functions,β which form a set of basic named functions for which we know the graph, formula, and special properties. Some of these functions are programmed to individual buttons on many calculators. For these definitions we will use x as the input variable and y = f (x) as the output variable. We will see these toolkit functions, combinations of toolkit functions, their graphs, and their transformations frequently throughout this book. It will be very helpful if we can recognize these toolkit functions and their features quickly by name, formula, graph, and basic table properties. The graphs and sample table values are included with each function shown in Table 3.14. 246 Chapter 3 Functions Toolkit Functions Name Function Graph Constant f (x) = c, where c is a constant Identity f (x) = x Absolute value f (x) = |x| Table 3.14 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 247 Toolkit Functions Name Function Graph Quadratic f (x) = x2 Cubic f (x) = x3 Reciprocal f (x) = 1 x Table 3.14 248 Chapter 3 Functions Toolkit Functions Name Function Graph Reciprocal squared f (x) = 1 x2 Square root f (x) = x Cube root f (x) = x3 Table 3.14 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 249 Access the following online resources for additional instruction and practice with functions. β’ Determine if a Relation is a Function (http://openstaxcollege.org/l/relationfunction) β’ Vertical Line Test (http://openstaxcollege.org/l/vertlinetest) β’ Introduction to Functions (http://openstaxcollege.org/l/introtofunction) β’ Vertical Line Test on Graph (http://openstaxcollege.org/l/vertlinegraph) β’ One-
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to-one Functions (http://openstaxcollege.org/l/onetoone) β’ Graphs as One-to-one Functions (http://openstaxcollege.org/l/graphonetoone) 250 Chapter 3 Functions 3.1 EXERCISES Verbal 1. What function? is the difference between a relation and a What is the difference between the input and the output 2. of a function? Why does the vertical line test tell us whether the graph 3. of a relation represents a function? How can you determine if a relation is a one-to-one 4. function? Why does the horizontal line test tell us whether the 5. graph of a function is one-to-one? Algebraic For the following exercises, determine whether the relation represents a function. 6. 7. {(a, b), (c, d), (a, c)} {(a, b), (b, c), (c, c)} For the following exercises, determine whether the relation represents y as a function of x. 8. 9. 5x + 2y = 10 y = x2 10. x = y2 3x2 + y = 14 2x + y2 = 6 y = β 2x2 + 40x y = 1 x x = 3y + 5 7y β 1 x = 1 β y2 y = 3x + 5 7x β 1 x2 + y2 = 9 11. 12. 13. 14. 15. 16. 17. 18. 19. This content is available for free at https://cnx.org/content/col11758/1.5 2xy = 1 20. x = y3 21. y = x3 22. 23. 24. 25. y = 1 β x2 y2 = x2 26. y3 = x2 For the following exercises, evaluate the function f at the indicated values f (β3), f (2), f (βa), β f (a), f (a + h). 27. f (x) = 2x β 5 28. 29. 30. f (x) = β 5x2 + 2x β 1 f (x) = 2 β x + 5 f (x) = 6x β 1 5x + 2 31. f (x) = |x β 1| β |x + 1| 32. Given the function g(x) = 5 β x2, simplify g(x
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+ h) β g(x) h, h β 0. 33. Given g(x) β g(a) x β a the function g(x) = x2 + 2x, simplify, x β a. 34. Given the function k(t) = 2t β 1: a. Evaluate k(2). b. Solve k(t) = 7. 35. Given the function f (x) = 8 β 3x: a. Evaluate f ( β 2). b. Solve f (x) = β1. 36. Given the function p(c) = c2 + c: Chapter 3 Functions 251 a. Evaluate p(β3). b. Solve p(c) = 2. 37. Given the function f (x) = x2 β 3x: a. Evaluate f (5). b. Solve f (x) = 4. 38. Given the function f (x) = x + 2: a. Evaluate f (7). b. Solve f (x) = 4. 39. Consider the relationship 3r + 2t = 18. a. Write the relationship as a function r = f (t). b. Evaluate f (β3). c. Solve f (t) = 2. Graphical For the following exercises, use the vertical line test to determine which graphs show relations that are functions. 40. 41. 42. 43. 44. 45. 252 46. 47. 48. 49. Chapter 3 Functions 50. 51. 52. Given the following graph, β’ Evaluate f (β1). β’ Solve for f (x) = 3. 53. Given the following graph, β’ Evaluate f (0). β’ Solve for f (x) = β3. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 253 54. Given the following graph, β’ Evaluate f (4). β’ Solve for f (x) = 1. 58. 59. For the following exercises, determine if the given graph is a one-to-one function. 55. 56. 57. Numeric For the following exercises, determine whether the relation represents a function. 60. 61. 62. {(β1, β1), (β2, β2), (β3, β3)} {(3, 4), (4, 5), (5, 6
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)} β§ β¨(2, 5), (7, 11), (15, 8), (7, 9)β« β¬ β β© For the following exercises, determine if the relation represented in table form represents y as a function of x. x y 5 3 10 8 15 14 63. 64. 254 65. x y 5 3 10 15 8 8 x y 5 3 10 8 10 14 For the following exercises, use the function f represented in Table 3.15(x) 74 28 1 53 56 3 36 45 14 47 Table 3.15 66. Evaluate f (3). 67. Solve f (x) = 1. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions For the following exercises, evaluate the function f at the values f (β2), f (β1), f (0), f (1), and f (2). 68. f (x) = 4 β 2x 69. f (x) = 8 β 3x 70. 71. 72. f (x) = 8x2 β 7x + 3 f (x) = 3 + x + 3 f (x) = x β 2 x + 3 73. f (x) = 3 x For the following exercises, evaluate the expressions, given functions f, g, and h: β’ β’ β’ f (x) = 3x β 2 g(x) = 5 β x2 h(x) = β2x2 + 3x β 1 74. 3 f (1) β 4g(β2) 75(β2) Technology For the following exercises, graph y = x2 on the given viewing window. Determine the corresponding range for each viewing window. Show each graph. 76. 77. 78. [ β 0.1, 0.1] [ β 10, 10] [ β 100, 100] For the following exercises, graph y = x3 on the given viewing window. Determine the corresponding range for each viewing window. Show each graph. 79. 80. 81. [ β 0.1, 0.1] [ β 10, 10] [ β 100, 100] For the following exercises, graph y = x on the given viewing window. Determine the corresponding range for each viewing window. Show each graph. 255 Chapter 3 Functions 82. [0, 0.01] 83. [0, 100] 84
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. [0, 10,000] For the following exercises, graph y = x3 viewing window. Determine the corresponding range for each viewing window. Show each graph. on the given 85. 86. 87. [β0.001, 0.001] [β1000, 1000] [β1,000,000, 1,000,000] Real-World Applications The amount of garbage, G, produced by a city with 88. population p is given by G = f (p). G is measured in tons per week, and p is measured in thousands of people. a. The town of Tola has a population of 40,000 and produces 13 tons of garbage each week. Express this information in terms of the function f. b. Explain the meaning of the statement f (5) = 2. The number of cubic yards of dirt, D, needed to cover 89. a garden with area a square feet is given by D = g(a). a. A garden with area 5000 ft2 requires 50 yd3 of dirt. Express this information in terms of the function g. b. Explain the meaning of the statement g(100) = 1. 90. Let f (t) be the number of ducks in a lake t years after 1990. Explain the meaning of each statement: a. b. f (5) = 30 f (10) = 40 91. Let h(t) be the height above ground, in feet, of a rocket t seconds after launching. Explain the meaning of each statement: a. b. h(1) = 200 h(2) = 350 92. Show that the function f (x) = 3(x β 5)2 + 7 is not one-to-one. 256 Chapter 3 Functions 3.2 | Domain and Range Learning Objectives In this section, you will: 3.2.1 Find the domain of a function defined by an equation. 3.2.2 Graph piecewise-defined functions. If youβre in the mood for a scary movie, you may want to check out one of the five most popular horror movies of all timeβI am Legend, Hannibal, The Ring, The Grudge, and The Conjuring. Figure 3.15 shows the amount, in dollars, each of those movies grossed when they were released as well as the ticket sales for horror movies in general by year. Notice that we can use the data to create a function of the amount each movie earned or the
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total ticket sales for all horror movies by year. In creating various functions using the data, we can identify different independent and dependent variables, and we can analyze the data and the functions to determine the domain and range. In this section, we will investigate methods for determining the domain and range of functions such as these. Figure 3.15 Based on data compiled by www.the-numbers.com.[3] Finding the Domain of a Function Defined by an Equation In Functions and Function Notation, we were introduced to the concepts of domain and range. In this section, we will practice determining domains and ranges for specific functions. Keep in mind that, in determining domains and ranges, we need to consider what is physically possible or meaningful in real-world examples, such as tickets sales and year in the horror movie example above. We also need to consider what is mathematically permitted. For example, we cannot include any input value that leads us to take an even root of a negative number if the domain and range consist of real numbers. Or in a function expressed as a formula, we cannot include any input value in the domain that would lead us to divide by 0. We can visualize the domain as a βholding areaβ that contains βraw materialsβ for a βfunction machineβ and the range as another βholding areaβ for the machineβs products. See Figure 3.16. 3. The Numbers: Where Data and the Movie Business Meet. βBox Office History for Horror Movies.β http://www.thenumbers.com/market/genre/Horror. Accessed 3/24/2014 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 257 Figure 3.16 We can write the domain and range in interval notation, which uses values within brackets to describe a set of numbers. In interval notation, we use a square bracket [ when the set includes the endpoint and a parenthesis ( to indicate that the endpoint is either not included or the interval is unbounded. For example, if a person has $100 to spend, he or she would need to express the interval that is more than 0 and less than or equal to 100 and write (0, 100]. We will discuss interval notation in greater detail later. Letβs turn our attention to finding the domain of a function whose equation is provided. Oftentimes, finding the domain of such functions involves remembering three different
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forms. First, if the function has no denominator or an even root, consider whether the domain could be all real numbers. Second, if there is a denominator in the functionβs equation, exclude values in the domain that force the denominator to be zero. Third, if there is an even root, consider excluding values that would make the radicand negative. Before we begin, let us review the conventions of interval notation: β’ The smallest number from the interval is written first. β’ The largest number in the interval is written second, following a comma. β’ Parentheses, ( or ), are used to signify that an endpoint value is not included, called exclusive. β’ Brackets, [ or ], are used to indicate that an endpoint value is included, called inclusive. See Figure 3.17 for a summary of interval notation. 258 Chapter 3 Functions Figure 3.17 Example 3.16 Finding the Domain of a Function as a Set of Ordered Pairs Find the domain of the following function: {(2, 10), (3, 10), (4, 20), (5, 30), (6, 40)}. Solution First identify the input values. The input value is the first coordinate in an ordered pair. There are no restrictions, as the ordered pairs are simply listed. The domain is the set of the first coordinates of the ordered pairs. {2, 3, 4, 5, 6} This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 259 3.13 Find the domain of the function: β§ β¨(β5, 4), (0, 0), (5, β4), (10, β8), (15, β12)β« β¬ β β© Given a function written in equation form, find the domain. 1. 2. Identify the input values. Identify any restrictions on the input and exclude those values from the domain. 3. Write the domain in interval form, if possible. Example 3.17 Finding the Domain of a Function Find the domain of the function f (x) = x2 β 1. Solution The input value, shown by the variable x in the equation, is squared and then the result is lowered by one. Any real number may be squared and then be lowered by one, so there are no restrictions on the domain of this function. The domain is the set of real numbers. In interval form,
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the domain of f is (ββ, β). 3.14 Find the domain of the function: f (x) = 5 β x + x3. Given a function written in an equation form that includes a fraction, find the domain. 1. 2. Identify the input values. Identify any restrictions on the input. If there is a denominator in the functionβs formula, set the denominator equal to zero and solve for x. If the functionβs formula contains an even root, set the radicand greater than or equal to 0, and then solve. 3. Write the domain in interval form, making sure to exclude any restricted values from the domain. Example 3.18 Finding the Domain of a Function Involving a Denominator Find the domain of the function f (x) = x + 1 2 β x. Solution When there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to 0 and solve for x. 260 Chapter 3 Functions 2 β x = 0 βx = β2 x = 2 Now, we will exclude 2 from the domain. The answers are all real numbers where x < 2 or x > 2 as shown in Figure 3.18. We can use a symbol known as the union, βͺ, to combine the two sets. In interval notation, we write the solution: (ββ, 2) βͺ (2, β). Figure 3.18 3.15 Find the domain of the function: f (x) = 1 + 4x 2x β 1. Given a function written in equation form including an even root, find the domain. 1. Identify the input values. 2. Since there is an even root, exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve for x. 3. The solution(s) are the domain of the function. If possible, write the answer in interval form. Example 3.19 Finding the Domain of a Function with an Even Root Find the domain of the function f (x) = 7 β x. Solution When there is an even root in the formula, we exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve for x. 7 β x β₯ 0 βx β₯
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β7 x β€ 7 Now, we will exclude any number greater than 7 from the domain. The answers are all real numbers less than or equal to 7, or ( β β, 7]. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 261 3.16 Find the domain of the function f (x) = 5 + 2x. Can there be functions in which the domain and range do not intersect at all? Yes. For example, the function f (x) = β 1 x has the set of all positive real numbers as its domain but the set of all negative real numbers as its range. As a more extreme example, a functionβs inputs and outputs can be completely different categories (for example, names of weekdays as inputs and numbers as outputs, as on an attendance chart), in such cases the domain and range have no elements in common. Using Notations to Specify Domain and Range In the previous examples, we used inequalities and lists to describe the domain of functions. We can also use inequalities, or other statements that might define sets of values or data, to describe the behavior of the variable in set-builder notation. For example, {x|10 β€ x < 30} describes the behavior of x in set-builder notation. The braces {} are read as βthe set of,β and the vertical bar | is read as βsuch that,β so we would read {x|10 β€ x < 30} as βthe set of x-values such that 10 is less than or equal to x, and x is less than 30.β Figure 3.19 compares inequality notation, set-builder notation, and interval notation. Figure 3.19 To combine two intervals using inequality notation or set-builder notation, we use the word βor.β As we saw in earlier examples, we use the union symbol, βͺ, to combine two unconnected intervals. For example, the union of the sets {2, 3, 5} and {4, 6} is the set {2, 3, 4, 5, 6}. It is the set of all elements that belong to one or the other (or both) of the original two sets. For sets with a finite number of elements like these, the elements do not have to be listed in ascending 262 Chapter 3 Functions order of numerical value. If the original two sets have some elements in common,
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those elements should be listed only once in the union set. For sets of real numbers on intervals, another example of a union is {x| |x| β₯ 3} = (ββ, β 3] βͺ [3, β) Set-Builder Notation and Interval Notation Set-builder notation is a method of specifying a set of elements that satisfy a certain condition. It takes the form {x| statement about x} which is read as, βthe set of all x such that the statement about x is true.β For example, {x|4 < x β€ 12} Interval notation is a way of describing sets that include all real numbers between a lower limit that may or may not be included and an upper limit that may or may not be included. The endpoint values are listed between brackets or parentheses. A square bracket indicates inclusion in the set, and a parenthesis indicates exclusion from the set. For example, (4, 12] Given a line graph, describe the set of values using interval notation. 1. Identify the intervals to be included in the set by determining where the heavy line overlays the real line. 2. At the left end of each interval, use [ with each end value to be included in the set (solid dot) or ( for each excluded end value (open dot). 3. At the right end of each interval, use ] with each end value to be included in the set (filled dot) or ) for each excluded end value (open dot). 4. Use the union symbol βͺ to combine all intervals into one set. Example 3.20 Describing Sets on the Real-Number Line Describe the intervals of values shown in Figure 3.20 using inequality notation, set-builder notation, and interval notation. Figure 3.20 Solution To describe the values, x, included in the intervals shown, we would say, β x is a real number greater than or equal to 1 and less than or equal to 3, or a real number greater than 5.β This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 263 Inequality 1 β€ x β€ 3 or x > 5 Set-builder notation β§ β¨x|1 β€ x β€ 3 or x > 5β« β¬ β β© Interval notation [1, 3] βͺ (5, β) Remember that, when writing
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or reading interval notation, using a square bracket means the boundary is included in the set. Using a parenthesis means the boundary is not included in the set. 3.17 Given Figure 3.21, specify the graphed set in a. words b. c. set-builder notation interval notation Figure 3.21 Finding Domain and Range from Graphs Another way to identify the domain and range of functions is by using graphs. Because the domain refers to the set of possible input values, the domain of a graph consists of all the input values shown on the x-axis. The range is the set of possible output values, which are shown on the y-axis. Keep in mind that if the graph continues beyond the portion of the graph we can see, the domain and range may be greater than the visible values. See Figure 3.22. 264 Chapter 3 Functions Figure 3.22 We can observe that the graph extends horizontally from β5 to the right without bound, so the domain is β‘ β£β5, β). The β¦. Note that the domain and range are vertical extent of the graph is all range values 5 and below, so the range is (ββ, 5β€ always written from smaller to larger values, or from left to right for domain, and from the bottom of the graph to the top of the graph for range. Example 3.21 Finding Domain and Range from a Graph Find the domain and range of the function f whose graph is shown in Figure 3.23. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 265 Figure 3.23 Solution We can observe that the horizontal extent of the graph is β3 to 1, so the domain of f is (β3, 1]. The vertical extent of the graph is 0 to β4, so the range is [β4, 0). See Figure 3.24. Figure 3.24 Example 3.22 Finding Domain and Range from a Graph of Oil Production Find the domain and range of the function f whose graph is shown in Figure 3.25. 266 Chapter 3 Functions Figure 3.25 (credit: modification of work by the U.S. Energy Information Administration)[4] Solution The input quantity along the horizontal axis is βyears,β which we represent with the variable t for time. The output quantity is βthousands of barrels of oil per day,β which we represent with the variable b for
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barrels. The graph may continue to the left and right beyond what is viewed, but based on the portion of the graph that is visible, we can determine the domain as 1973 β€ t β€ 2008 and the range as approximately 180 β€ b β€ 2010. In interval notation, the domain is [1973, 2008], and the range is about [180, 2010]. For the domain and the range, we approximate the smallest and largest values since they do not fall exactly on the grid lines. 3.18 Given Figure 3.26, identify the domain and range using interval notation. Figure 3.26 Can a functionβs domain and range be the same? Yes. For example, the domain and range of the cube root function are both the set of all real numbers. Finding Domains and Ranges of the Toolkit Functions We will now return to our set of toolkit functions to determine the domain and range of each. 4. http://www.eia.gov/dnav/pet/hist/LeafHandler.ashx?n=PET&s=MCRFPAK2&f=A. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 267 Figure 3.27 For the constant function f (x) = c, the domain consists of all real numbers; there are no restrictions on the input. The only output value is the constant c, so the range is the set {c} that contains this single element. In interval notation, this is written as [c, c], the interval that both begins and ends with c. Figure 3.28 For the identity function f (x) = x, there is no restriction on x. Both the domain and range are the set of all real numbers. 268 Chapter 3 Functions Figure 3.29 For the absolute value function f (x) = |x|, there is no restriction on x. However, because absolute value is defined as a distance from 0, the output can only be greater than or equal to 0. Figure 3.30 For the quadratic function f (x) = x2, the domain is all real numbers since the horizontal extent of the graph is the whole real number line. Because the graph does not include any negative values for the range, the range is only nonnegative real numbers. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 269 Figure 3
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.31 For the cubic function f (x) = x3, the domain is all real numbers because the horizontal extent of the graph is the whole real number line. The same applies to the vertical extent of the graph, so the domain and range include all real numbers. Figure 3.32 For the reciprocal function f (x) = 1 x, we cannot divide by 0, so we must exclude 0 from the domain. Further, 1 divided by any value can never be 0, so the range also will not include 0. In set-builder notation, we could also write {x| x β 0}, the set of all real numbers that are not zero. 270 Chapter 3 Functions Figure 3.33 For the reciprocal squared function f (x) = 1 x2, we cannot divide by 0, so we must exclude 0 from the domain. There is also no x that can give an output of 0, so 0 is excluded from the range as well. Note that the output of this function is always positive due to the square in the denominator, so the range includes only positive numbers. Figure 3.34 For the square root function f (x) = x, we cannot take the square root of a negative real number, so the domain must be 0 or greater. The range also excludes negative numbers because the square root of a positive number x is defined to be positive, even though the square of the negative number β x also gives us x. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 271 Figure 3.35 For the cube root function f (x) = x3, the domain and range include all real numbers. Note that there is no problem taking a cube root, or any odd-integer root, of a negative number, and the resulting output is negative (it is an odd function). Given the formula for a function, determine the domain and range. 1. Exclude from the domain any input values that result in division by zero. 2. Exclude from the domain any input values that have nonreal (or undefined) number outputs. 3. Use the valid input values to determine the range of the output values. 4. Look at the function graph and table values to confirm the actual function behavior. Example 3.23 Finding the Domain and Range Using Toolkit Functions Find the domain and range of f (x) = 2x3 β x. Solution There are no restrictions on the domain, as any
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real number may be cubed and then subtracted from the result. The domain is (ββ, β) and the range is also (ββ, β). Example 3.24 Finding the Domain and Range Find the domain and range of f (x) = 2 x + 1. 272 Chapter 3 Functions Solution We cannot evaluate the function at β1 because division by zero is undefined. The domain is (ββ, β1) βͺ (β1, β). Because the function is never zero, we exclude 0 from the range. The range is (ββ, 0) βͺ (0, β). Example 3.25 Finding the Domain and Range Find the domain and range of f (x) = 2 x + 4. Solution We cannot take the square root of a negative number, so the value inside the radical must be nonnegative. The domain of f (x) is [ β 4, β). x + 4 β₯ 0 when x β₯ β 4 We then find the range. We know that f (β4) = 0, and the function value increases as x increases without any upper limit. We conclude that the range of f is β‘ β£0, β). Analysis Figure 3.36 represents the function f. Figure 3.36 3.19 Find the domain and range of f (x) = β 2 β x. Graphing Piecewise-Defined Functions Sometimes, we come across a function that requires more than one formula in order to obtain the given output. For example, in the toolkit functions, we introduced the absolute value function f (x) = |x|. With a domain of all real numbers and a range of values greater than or equal to 0, absolute value can be defined as the magnitude, or modulus, of a real number This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 273 value regardless of sign. It is the distance from 0 on the number line. All of these definitions require the output to be greater than or equal to 0. If we input 0, or a positive value, the output is the same as the input. f (x) = x if x β₯ 0 If we input a negative value, the output is the opposite of the input. f (x) = β x if x < 0 Because this requires two different processes or pieces, the absolute value function is an example of a piecewise function.
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A piecewise function is a function in which more than one formula is used to define the output over different pieces of the domain. We use piecewise functions to describe situations in which a rule or relationship changes as the input value crosses certain βboundaries.β For example, we often encounter situations in business for which the cost per piece of a certain item is discounted once the number ordered exceeds a certain value. Tax brackets are another real-world example of piecewise functions. For example, consider a simple tax system in which incomes up to $10,000 are taxed at 10%, and any additional income is taxed at 20%. The tax on a total income S would be 0.1S if S β€ $10,000 and $1000 + 0.2(S β $10,000) if S > $10,000. Piecewise Function A piecewise function is a function in which more than one formula is used to define the output. Each formula has its own domain, and the domain of the function is the union of all these smaller domains. We notate this idea like this: f (x) = β§ β¨ β© formula 1 if x is in domain 1 formula 2 if x is in domain 2 formula 3 if x is in domain 3 In piecewise notation, the absolute value function is |x| = x if x β₯ 0 β§ β¨ βx if x < 0 β© Given a piecewise function, write the formula and identify the domain for each interval. 1. Identify the intervals for which different rules apply. 2. Determine formulas that describe how to calculate an output from an input in each interval. 3. Use braces and if-statements to write the function. Example 3.26 Writing a Piecewise Function A museum charges $5 per person for a guided tour with a group of 1 to 9 people or a fixed $50 fee for a group of 10 or more people. Write a function relating the number of people, n, to the cost, C. Solution Two different formulas will be needed. For n-values under 10, C = 5n. For values of n that are 10 or greater, C = 50. C(n) = Analysis β§ β¨5n if 0 < n < 10 50 if β© n β₯ 10 274 Chapter 3 Functions The function is represented in Figure 3.37. The graph is a diagonal line from n = 0 to n = 10 and a constant
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after that. In this example, the two formulas agree at the meeting point where n = 10, but not all piecewise functions have this property. Figure 3.37 Example 3.27 Working with a Piecewise Function A cell phone company uses the function below to determine the cost, C, in dollars for g gigabytes of data transfer. C(g) = β§ 25 β¨ 25 + 10(g β 2) if β© if 0 < g < 2 g β₯ 2 Find the cost of using 1.5 gigabytes of data and the cost of using 4 gigabytes of data. Solution To find the cost of using 1.5 gigabytes of data, C(1.5), we first look to see which part of the domain our input falls in. Because 1.5 is less than 2, we use the first formula. C(1.5) = $25 To find the cost of using 4 gigabytes of data, C(4), we see that our input of 4 is greater than 2, so we use the second formula. C(4) = 25 + 10(4 β 2) = $45 Analysis The function is represented in Figure 3.38. We can see where the function changes from a constant to a shifted and stretched identity at g = 2. We plot the graphs for the different formulas on a common set of axes, making sure each formula is applied on its proper domain. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 275 Figure 3.38 Given a piecewise function, sketch a graph. 1. Indicate on the x-axis the boundaries defined by the intervals on each piece of the domain. 2. For each piece of the domain, graph on that interval using the corresponding equation pertaining to that piece. Do not graph two functions over one interval because it would violate the criteria of a function. Example 3.28 Graphing a Piecewise Function Sketch a graph of the function. f (x) = β§ β¨ β© x β€ 1 x2 if 3 if 1 < x β€ 2 x if x > 2 Solution Each of the component functions is from our library of toolkit functions, so we know their shapes. We can imagine graphing each function and then limiting the graph to the indicated domain. At the endpoints of the domain, we draw open circles to indicate where the endpoint is not included because of a less-than
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or greater-than inequality; we draw a closed circle where the endpoint is included because of a less-than-or-equal-to or greater-than-or-equalto inequality. Figure 3.39 shows the three components of the piecewise function graphed on separate coordinate systems. 276 Chapter 3 Functions Figure 3.39 (a) f (x) = x2 if x β€ 1; (b) f (x) = 3 if 1< x β€ 2; (c) f (x) = x if x > 2 Now that we have sketched each piece individually, we combine them in the same coordinate plane. See Figure 3.40. Figure 3.40 Analysis Note that the graph does pass the vertical line test even at x = 1 and x = 2 because the points (1, 3) and (2, 2) are not part of the graph of the function, though (1, 1) and (2, 3) are. 3.20 Graph the following piecewise function. f (x) = β§ β¨ β© x3 if β2 if β if x > 4 Can more than one formula from a piecewise function be applied to a value in the domain? No. Each value corresponds to one equation in a piecewise formula. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 277 Access these online resources for additional instruction and practice with domain and range. β’ Domain and Range of Square Root Functions (http://openstaxcollege.org/l/domainsqroot) β’ Determining Domain and Range (http://openstaxcollege.org/l/determinedomain) β’ Find Domain and Range Given the Graph (http://openstaxcollege.org/l/drgraph) β’ Find Domain and Range Given a Table (http://openstaxcollege.org/l/drtable) β’ Find Domain and Range Given Points on a Coordinate Plane (http://openstaxcollege.org/ l/drcoordinate) 278 Chapter 3 Functions 3.2 EXERCISES Verbal 93. Why does the domain differ for different functions? How do we determine the domain of a function defined 94. by an equation? 95. Explain why the domain of f (x) = x3 is different from the domain of f (x) = x. 96. When describing sets of numbers using interval notation, when do you
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use a parenthesis and when do you use a bracket? 97. How do you graph a piecewise function? Algebraic For the following exercises, find the domain of each function using interval notation. 98. f (x) = β 2x(x β 1)(x β 2) 99. f (x) = 5 β 2x2 100. 101. 102. 103. 104. 105. 106. 107. 108. f (x) = 3 x β 2 f (x) = 3 β 6 β 2x f (x) = 4 β 3x f (x) = x2 + 4 3 f (x) = 1 β 2x 3 f (x) = x β 1 f (x) = 9 x β 6 f (x) = 3x + 1 4x + 2 f (x) = x + 4 x β 4 109. f (x) = x β 3 x2 + 9x β 22 110. f (x) = 1 x2 β x β 6 111. This content is available for free at https://cnx.org/content/col11758/1.5 f (x) = 2x3 β 250 x2 β 2x β 15 112. 113. 114. 5 x β 3 2x + 1 5 β x f (x) = 115. f (x) = 116. f (xx) = x2 β 9x x2 β 81 Find the domain of the function f (x) = 2x3 β 50x 117. 118. by: a. using algebra. b. graphing the radicand and function in the determining intervals on the x-axis for which the radicand is nonnegative. Graphical For the following exercises, write the domain and range of each function using interval notation. 119. 120. 121. Chapter 3 Functions 279 122. 123. 124. 125. 126. 127. 128. 129. 280 Chapter 3 Functions For the following exercises, given each function f, evaluate f (β1), f (0), f (2), and f (4). 141. 142. 143. f (x) = β¨7x + 3 if x < 0 β§ 7x + 6 if x β₯ 0 β© f (x) = β§ β¨ x2 β 2 if x < 2 4 + |x β 5| if x β₯ 2 β© f (x) = β§ οΏ½
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οΏ½οΏ½ β© x < 0 5x if 3 if 0 β€ x β€ 3 x2 if x > 3 For the following exercises, write the domain for the piecewise function in interval notation. 144. 145. f (x) = x + 1 if x < β 2 β§ β¨ β2x β 3 if x β₯ β 2 β© f (x) = β§ x2 β 2 if x < 1 β¨ βx2 + 2 if x > 1 β© 146. f (x) = β§ β¨2x β 3 β3x2 β© if x < 0 if x β₯ 2 Technology 147. Graph y = 1 x2 on the viewing window [β0.5, β0.1] and [0.1, 0.5]. Determine corresponding range for the viewing window. Show the graphs. the 148. Graph y = 1 x on the viewing window [β0.5, β0.1] and [0.1, 0.5]. Determine the corresponding range for the viewing window. Show the graphs. Extension Suppose the range of a function f is [β5, 8]. What 149. is the range of | f (x)|? Create a function in which the range is all 150. nonnegative real numbers. 151. Create a function in which the domain is x > 2. Real-World Applications The height h of a projectile is a function of the time 152. t it is in the air. The height in feet for t seconds is given by the function h(t) = β16t 2 + 96t. What is the domain of the function? What does the domain mean in the context of the problem? For the following exercises, sketch a graph of the piecewise function. Write the domain in interval notation. 130. 131. 132. 133. 134. 135. 136. f (x) = if 2x β 3 if x β₯ β 2 β© f (x) = β¨2x β 1 if x < 1 β§ if x β₯ 1 1 + x β© f (x) = x + 1 if x < 0 β§ β¨ x β 1 if x > 0 β© f (x) = β§ β¨ 3 if x < 0 x if x β₯ 0 β© f (x) = β§ β¨x2 if x
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< 0 1 β x if x > 0 β© f (x) = β§ x2 β¨ x + 2 β© if x < 0 if x β₯ 0 f (x) = x + 1 if x < 1 β§ β¨ x3 if x β₯ 1 β© β§ β¨|x| 1 β© if x < 2 if x β₯ 2 137. f (x) = Numeric For the following exercises, given each function f, evaluate f (β3), f (β2), f (β1), and f (0). 138. 139. 140. f (x) = if 2x β 3 if x β₯ β 2 β© f (x) = β§ β¨1 if x β€ β 3 0 if x > β 3 β© f (x) = β§ β¨β2x2 + 3 if x β€ β 1 if x > β 1 β© 5x β 7 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 281 153. The cost in dollars of making x items is given by the function C(x) = 10x + 500. a. The fixed cost is determined when zero items are produced. Find the fixed cost for this item. b. What is the cost of making 25 items? c. Suppose the maximum cost allowed is $1500. What are the domain and range of the cost function, C(x)? 282 Chapter 3 Functions 3.3 | Rates of Change and Behavior of Graphs Learning Objectives In this section, you will: 3.3.1 Find the average rate of change of a function. 3.3.2 Use a graph to determine where a function is increasing, decreasing, or constant. 3.3.3 Use a graph to locate local maxima and local minima. 3.3.4 Use a graph to locate the absolute maximum and absolute minimum. Gasoline costs have experienced some wild fluctuations over the last several decades. Table 3.17[5] lists the average cost, in dollars, of a gallon of gasoline for the years 2005β2012. The cost of gasoline can be considered as a function of year. y 2005 2006 2007 2008 2009 2010 2011 2012 2.31 2.62 2.84 3.30 2.41 2.84 3.58 3.68 C(y) Table 3.17
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If we were interested only in how the gasoline prices changed between 2005 and 2012, we could compute that the cost per gallon had increased from $2.31 to $3.68, an increase of $1.37. While this is interesting, it might be more useful to look at how much the price changed per year. In this section, we will investigate changes such as these. Finding the Average Rate of Change of a Function The price change per year is a rate of change because it describes how an output quantity changes relative to the change in the input quantity. We can see that the price of gasoline in Table 3.17 did not change by the same amount each year, so the rate of change was not constant. If we use only the beginning and ending data, we would be finding the average rate of change over the specified period of time. To find the average rate of change, we divide the change in the output value by the change in the input value. Average rate of change = = = = Change in output Change in input Ξy Ξx y2 β y1 x2 β x1 β β f β f β βx2 x2 β x1 βx1 β β β The Greek letter Ξ (delta) signifies the change in a quantity; we read the ratio as βdelta-y over delta-xβ or βthe change in y divided by the change in x. β Occasionally we write Ξ f instead of Ξy, which still represents the change in the functionβs output value resulting from a change to its input value. It does not mean we are changing the function into some other function. In our example, the gasoline price increased by $1.37 from 2005 to 2012. Over 7 years, the average rate of change was Ξy Ξx = $1.37 7 years β 0.196 dollars per year On average, the price of gas increased by about 19.6Β’ each year. Other examples of rates of change include: 5. http://www.eia.gov/totalenergy/data/annual/showtext.cfm?t=ptb0524. Accessed 3/5/2014. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 283 β’ A population of rats increasing by 40 rats per week β’ A car traveling 68 miles per hour
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(distance traveled changes by 68 miles each hour as time passes) β’ A car driving 27 miles per gallon (distance traveled changes by 27 miles for each gallon) β’ The current through an electrical circuit increasing by 0.125 amperes for every volt of increased voltage β’ The amount of money in a college account decreasing by $4,000 per quarter Rate of Change A rate of change describes how an output quantity changes relative to the change in the input quantity. The units on a rate of change are βoutput units per input units.β The average rate of change between two input values is the total change of the function values (output values) divided by the change in the input values. Ξy Ξx = f (x2) β f (x1) x2 β x1 (3.1) Given the value of a function at different points, calculate the average rate of change of a function for the interval between two values x1 and x2. 1. Calculate the difference y2 β y1 = Ξy. 2. Calculate the difference x2 β x1 = Ξx. 3. Find the ratio Ξy Ξx. Example 3.29 Computing an Average Rate of Change Using the data in Table 3.17, find the average rate of change of the price of gasoline between 2007 and 2009. Solution In 2007, the price of gasoline was $2.84. In 2009, the cost was $2.41. The average rate of change is Ξy Ξx = y2 β y1 x2 β x1 $2.41 β $2.84 2009 β 2007 = = β$0.43 2 years = β$0.22 per year Analysis Note that a decrease is expressed by a negative change or βnegative increase.β A rate of change is negative when the output decreases as the input increases or when the output increases as the input decreases. 3.21 Using the data in Table 3.17, find the average rate of change between 2005 and 2010. 284 Chapter 3 Functions Example 3.30 Computing Average Rate of Change from a Graph Given the function g(t) shown in Figure 3.41, find the average rate of change on the interval [β1, 2]. Figure 3.41 Solution At t = β 1, Figure 3.42 shows g(β1) = 4. At t = 2, the graph shows g(2) = 1. Figure 3.42 The horizontal change Ξt = 3 is shown by
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the red arrow, and the vertical change Ξg(t) = β 3 is shown by the turquoise arrow. The average rate of change is shown by the slope of the orange line segment. The output changes by β3 while the input changes by 3, giving an average rate of change of 1 β 4 2 β (β1) = β3 3 = β1 Analysis Note that the order we choose is very important. If, for example, we use y2 β y1 x1 β x2, we will not get the correct answer. Decide which point will be 1 and which point will be 2, and keep the coordinates fixed as (x1, y1) and (x2, y2). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 285 Example 3.31 Computing Average Rate of Change from a Table After picking up a friend who lives 10 miles away and leaving on a trip, Anna records her distance from home over time. The values are shown in Table 3.18. Find her average speed over the first 6 hours. t (hours) 0 D(t) (miles) 10 1 55 2 3 4 5 6 7 90 153 214 240 292 300 Table 3.18 Solution Here, the average speed is the average rate of change. She traveled 282 miles in 6 hours. The average speed is 47 miles per hour. 292 β 10 6 β 0 = 282 6 = 47 Analysis Because the speed is not constant, the average speed depends on the interval chosen. For the interval [2,3], the average speed is 63 miles per hour. Example 3.32 Computing Average Rate of Change for a Function Expressed as a Formula Compute the average rate of change of f (x) = x2 β 1 x on the interval [2, 4]. Solution We can start by computing the function values at each endpoint of the interval. f (2) = 22 β 4) = 42 β 1 4 = 16 β 1 4 = 63 4 Now we compute the average rate of change. 286 Chapter 3 Functions Average rate of change = f (4) β f (2) 4 β 2 63 = 4 β 7 2 4 β 2 49 4 2 = 49 8 = 3.22 Find the average rate of change of f (x) = x β 2 x on the interval [1, 9]. Example 3.33 Finding the Average Rate of Change of a Force The electrostatic force F
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, measured in newtons, between two charged particles can be related to the distance d 2. Find the average rate of change of force if in centimeters, by the formula F(d) = 2 between the particles d, the distance between the particles is increased from 2 cm to 6 cm. Solution We are computing the average rate of change of F(d) = 2 d 2 on the interval [2, 6]. Average rate of change = 2 = F(6) β F(2) 6 β 2 62 β 2 22 6 β 2 36 β 2 2 4 β 16 36 4 = β 1 9 = = 4 Simplify. Combine numerator terms. Simplify The average rate of change is β 1 9 newton per centimeter. Example 3.34 Finding an Average Rate of Change as an Expression This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 287 Find the average rate of change of g(t) = t 2 + 3t + 1 on the interval [0, a]. The answer will be an expression involving a in simplest form. Solution We use the average rate of change formula. Average rate of change = = g(a) β g(0) a β 0 β βa2 + 3a + 1 β β02 + 3 β β β a β 0 β β0 β β β + 1 β Evaluate. Simplify. = a2 + 3a + 1 β 1 a a(a + 3) = a = a + 3 Simplify and factor. Divide by the common factor a. This result tells us the average rate of change in terms of a between t = 0 and any other point t = a. For example, on the interval [0, 5], the average rate of change would be 5 + 3 = 8. Find the average rate of change of f (x) = x2 + 2x β 8 on the interval [5, a] in simplest forms in terms 3.23 of a. Using a Graph to Determine Where a Function is Increasing, Decreasing, or Constant As part of exploring how functions change, we can identify intervals over which the function is changing in specific ways. We say that a function is increasing on an interval if the function values increase as the input values increase within that interval. Similarly, a function is decreasing on
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an interval if the function values decrease as the input values increase over that interval. The average rate of change of an increasing function is positive, and the average rate of change of a decreasing function is negative. Figure 3.43 shows examples of increasing and decreasing intervals on a function. 288 Chapter 3 Functions Figure 3.43 The function f (x) = x3 β 12x is increasing on (ββ, β 2) βͺ (2, β) and is decreasing on ( β 2, 2). While some functions are increasing (or decreasing) over their entire domain, many others are not. A value of the input where a function changes from increasing to decreasing (as we go from left to right, that is, as the input variable increases) is called a local maximum. If a function has more than one, we say it has local maxima. Similarly, a value of the input where a function changes from decreasing to increasing as the input variable increases is called a local minimum. The plural form is βlocal minima.β Together, local maxima and minima are called local extrema, or local extreme values, of the function. (The singular form is βextremum.β) Often, the term local is replaced by the term relative. In this text, we will use the term local. Clearly, a function is neither increasing nor decreasing on an interval where it is constant. A function is also neither increasing nor decreasing at extrema. Note that we have to speak of local extrema, because any given local extremum as defined here is not necessarily the highest maximum or lowest minimum in the functionβs entire domain. For the function whose graph is shown in Figure 3.44, the local maximum is 16, and it occurs at x = β2. The local minimum is β16 and it occurs at x = 2. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 289 Figure 3.44 To locate the local maxima and minima from a graph, we need to observe the graph to determine where the graph attains its highest and lowest points, respectively, within an open interval. Like the summit of a roller coaster, the graph of a function is higher at a local maximum than at nearby points on both sides. The graph will also be lower at a local minimum than at neighboring points. Figure 3.45 illustrates these ideas for a local maximum. Figure
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3.45 Definition of a local maximum These observations lead us to a formal definition of local extrema. Local Minima and Local Maxima A function f is an increasing function on an open interval if f (b) > f (a) for any two input values a and b in the given interval where b > a. A function f is a decreasing function on an open interval if f (b) < f (a) for any two input values a and b in the given interval where b > a. A function f has a local maximum at x = b if there exists an interval (a, c) with a < b < c such that, for any x f has a local minimum at x = b if there exists an interval (a, c) with in the interval (a, c), a < b < c such that, for any x in the interval (a, c), f (x) β€ f (b). Likewise, f (x) β₯ f (b). 290 Chapter 3 Functions Example 3.35 Finding Increasing and Decreasing Intervals on a Graph Given the function p(t) in Figure 3.46, identify the intervals on which the function appears to be increasing. Figure 3.46 Solution We see that the function is not constant on any interval. The function is increasing where it slants upward as we move to the right and decreasing where it slants downward as we move to the right. The function appears to be increasing from t = 1 to t = 3 and from t = 4 on. In interval notation, we would say the function appears to be increasing on the interval (1,3) and the interval (4, β). Analysis Notice in this example that we used open intervals (intervals that do not include the endpoints), because the function is neither increasing nor decreasing at t = 1, t = 3, and t = 4. These points are the local extrema (two minima and a maximum). Example 3.36 Finding Local Extrema from a Graph Graph the function f (x) = 2 x + determine the intervals on which the function is increasing. x 3. Then use the graph to estimate the local extrema of the function and to Solution Using technology, we find that the graph of the function looks like that in Figure 3.47. It appears there is a low point, or local minimum, between x = 2 and x = 3, and a mirror-image high point, or local maximum, somewhere
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between x = β3 and x = β2. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 291 Figure 3.47 Analysis Most graphing calculators and graphing utilities can estimate the location of maxima and minima. Figure 3.48 provides screen images from two different technologies, showing the estimate for the local maximum and minimum. Figure 3.48 Based on these estimates, the function is increasing on the interval ( β β, β 2.449) and (2.449,β). Notice that, while we expect the extrema to be symmetric, the two different technologies agree only up to four decimals due to the differing approximation algorithms used by each. (The exact location of the extrema is at Β± 6, but determining this requires calculus.) 3.24 Graph the function f (x) = x3 β 6x2 β 15x + 20 to estimate the local extrema of the function. Use these to determine the intervals on which the function is increasing and decreasing. Example 3.37 Finding Local Maxima and Minima from a Graph 292 Chapter 3 Functions For the function f whose graph is shown in Figure 3.49, find all local maxima and minima. Figure 3.49 Solution Observe the graph of f. The graph attains a local maximum at x = 1 because it is the highest point in an open interval around x = 1. The local maximum is the y -coordinate at x = 1, which is 2. The graph attains a local minimum at x = β1 because it is the lowest point in an open interval around x = β1. The local minimum is the y-coordinate at x = β1, which is β2. Analyzing the Toolkit Functions for Increasing or Decreasing Intervals We will now return to our toolkit functions and discuss their graphical behavior in Figure 3.50, Figure 3.51, and Figure 3.52. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 293 Figure 3.50 Figure 3.51 294 Chapter 3 Functions Figure 3.52 Use A Graph to Locate the Absolute Maximum and Absolute Minimum There is a difference between locating the highest and lowest points on a graph in a region around an open interval (locally) and locating the highest and lowest points on the graph for the entire domain. The y
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- coordinates (output) at the highest and lowest points are called the absolute maximum and absolute minimum, respectively. To locate absolute maxima and minima from a graph, we need to observe the graph to determine where the graph attains it highest and lowest points on the domain of the function. See Figure 3.53. Figure 3.53 Not every function has an absolute maximum or minimum value. The toolkit function f (x) = x3 is one such function. Absolute Maxima and Minima The absolute maximum of f at x = c is f (c) where f (c) β₯ f (x) for all x in the domain of f. The absolute minimum of f at x = d is f (d) where f (d) β€ f (x) for all x in the domain of f. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 295 Example 3.38 Finding Absolute Maxima and Minima from a Graph For the function f shown in Figure 3.54, find all absolute maxima and minima. Figure 3.54 Solution Observe the graph of f. The graph attains an absolute maximum in two locations, x = β2 and x = 2, because at these locations, the graph attains its highest point on the domain of the function. The absolute maximum is the y-coordinate at x = β2 and x = 2, which is 16. The graph attains an absolute minimum at x = 3, because it is the lowest point on the domain of the functionβs graph. The absolute minimum is the y-coordinate at x = 3, which is β10. Access this online resource for additional instruction and practice with rates of change. β’ Average Rate of Change (http://openstaxcollege.org/l/aroc) 296 Chapter 3 Functions 3.3 EXERCISES Verbal Can the average rate of change of a function be 154. constant? a If function f is on (a, b) and 155. decreasing on (b, c), then what can be said about the local extremum of f on (a, c)? increasing How are the absolute maximum and minimum similar 156. to and different from the local extrema? How does the graph of the absolute value function 157. compare to the graph of the quadratic function, y = x2, in terms of increasing and decreasing intervals? Algebraic
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For the following exercises, find the average rate of change of each function on the interval specified for real numbers b or h in simplest form. 158. 159. f (x) = 4x2 β 7 on [1, b] g(x) = 2x2 β 9 on β‘ β£4, bβ€ β¦ 160. p(x) = 3x + 4 on [2, 2 + h] 161. k(x) = 4x β 2 on [3, 3 + h] 162. 163. 164. 165. 166. 167. 168. f (x) = 2x2 + 1 on [x, x + h] g(x) = 3x2 β 2 on [x, x + h] a(t) = 1 t + 4 on [9, 9 + h] b(x) = 1 x + 3 on [1, 1 + h] j(x) = 3x3 on [1, 1 + h] r(t) = 4t 3 on [2, 2 + h] f (x + h) β f (x) h given f (x) = 2x2 β 3x on [x, x + h] Graphical For the following exercises, consider the graph of f shown in Figure 3.55. This content is available for free at https://cnx.org/content/col11758/1.5 Figure 3.55 Estimate the average rate of change from x = 1 to 169. x = 4. Estimate the average rate of change from x = 2 to 170. x = 5. For the following exercises, use the graph of each function to estimate the intervals on which the function is increasing or decreasing. 171. 172. 173. Chapter 3 Functions 297 Figure 3.57 177. If the complete graph of the function is shown, estimate the intervals where the function is increasing or decreasing. If the complete graph of the function is shown, 178. estimate the absolute maximum and absolute minimum. Numeric 179. Table 3.19 gives the annual sales (in millions of dollars) of a product from 1998 to 2006. What was the average rate of change of annual sales (a) between 2001 and 2002, and (b) between 2001 and 2004? 174. For the following exercises, consider the graph shown in Figure 3.56. Figure 3.56 Estimate the intervals where the function is increasing 175. or
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decreasing. 176. Estimate the point(s) at which the graph of f has a local maximum or a local minimum. For the following exercises, consider the graph in Figure 3.57. 298 Chapter 3 Functions Year Sales (millions of dollars) Year Population (thousands) 1998 1999 2000 2001 2002 2003 2004 2005 2006 201 219 233 243 249 251 249 243 233 2000 2001 2002 2003 2004 2005 2006 2007 2008 87 84 83 80 77 76 78 81 85 Table 3.19 Table 3.20 180. Table 3.20 gives the population of a town (in thousands) from 2000 to 2008. What was the average rate of change of population (a) between 2002 and 2004, and (b) between 2002 and 2006? For the following exercises, find the average rate of change of each function on the interval specified. 181. 182. 183. 184. 185. 186. 187. f (x) = x2 on [1, 5] h(x) = 5 β 2x2 on [β2, 4] q(x) = x3 on [β4, 2] g(x) = 3x3 β 1 on [β3, 3] y = 1 x on [1, 3] p(t) = β β βt 2 β 4 β (t + 1) t 2 + 3 on [β3, 1] k(t) = 6t 2 + 4 t 3 on [β1, 3] Technology For the following exercises, use a graphing utility to estimate the local extrema of each function and to estimate the intervals on which the function is increasing and decreasing. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 299 f (x) = x4 β 4x3 + 5 Real-World Applications At the start of a trip, the odometer on a car read 197. 21,395. At the end of the trip, 13.5 hours later, the odometer read 22,125. Assume the scale on the odometer is in miles. What is the average speed the car traveled during this trip? A driver of a car stopped at a gas station to fill up his 198. gas tank. He looked at his watch, and the time read exactly 3:40 p.m. At this time, he started pumping gas into the tank. At exactly 3:44, the tank was
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full and he noticed that he had pumped 10.7 gallons. What is the average rate of flow of the gasoline into the gas tank? falls is a function of Near the surface of the moon, the distance that an 199. is given by object d(t) = 2.6667t 2, where t is in seconds and d(t) is in feet. If an object is dropped from a certain height, find the average velocity of the object from t = 1 to t = 2. time. It The graph in Figure 3.59 illustrates the decay of a 200. radioactive substance over t days. 188. 189. h(x) = x5 + 5x4 + 10x3 + 10x2 β 1 190. g(t) = t t + 3 191. 192. 193. k(t) = 3t 2 3 β t m(x) = x4 + 2x3 β 12x2 β 10x + 4 n(x) = x4 β 8x3 + 18x2 β 6x + 2 Extension 194. The graph of the function f is shown in Figure 3.58. Figure 3.58 on shot, Based (1.333, 5.185) is which of the following? calculator screen the the point Figure 3.59 Use the graph to estimate the average decay rate from t = 5 to t = 15. A. a relative (local) maximum of the function B. C. the vertex of the function the absolute maximum of the function D. a zero of the function 195. Let f (x) = 1 x. Find a number c such that the average rate of change of the function f on the interval (1, c) is β 1 4. 196. Let f (x) = 1 x. Find the number b such that the average rate of change of f on the interval (2, b) is β 1 10. 300 Chapter 3 Functions 3.4 | Composition of Functions Learning Objectives In this section, you will: 3.4.1 Combine functions using algebraic operations. 3.4.2 Create a new function by composition of functions. 3.4.3 Evaluate composite functions. 3.4.4 Find the domain of a composite function. 3.4.5 Decompose a composite function into its component functions. Suppose we want to calculate how much it costs to heat a house on a particular day of the year. The cost to heat a house will depend on
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the average daily temperature, and in turn, the average daily temperature depends on the particular day of the year. Notice how we have just defined two relationships: The cost depends on the temperature, and the temperature depends on the day. Using descriptive variables, we can notate these two functions. The function C(T) gives the cost C of heating a house for a given average daily temperature in T degrees Celsius. The function T(d) gives the average daily temperature on day d of the year. For any given day, Cost = Cβ β means that the cost depends on the temperature, which in turns depends on the day of the year. Thus, we can evaluate the cost function at the temperature T(d). For example, we could evaluate T(5) to determine the average daily temperature on the 5th day of the year. Then, we could evaluate the cost function at that temperature. We would write Cβ βT(d)β βT(5)β β . By combining these two relationships into one function, we have performed function composition, which is the focus of this section. Combining Functions Using Algebraic Operations Function composition is only one way to combine existing functions. Another way is to carry out the usual algebraic operations on functions, such as addition, subtraction, multiplication and division. We do this by performing the operations with the function outputs, defining the result as the output of our new function. Suppose we need to add two columns of numbers that represent a husband and wifeβs separate annual incomes over a period of years, with the result being their total household income. We want to do this for every year, adding only that yearβs incomes and then collecting all the data in a new column. If w(y) is the wifeβs income and h(y) is the husbandβs income in year y, and we want T to represent the total income, then we can define a new function. T(y) = h(y) + w(y) If this holds true for every year, then we can focus on the relation between the functions without reference to a year and write Just as for this sum of two functions, we can define difference, product, and ratio functions for any pair of functions that have the same kinds of inputs (not necessarily numbers) and also the same kinds of outputs (which do have to be numbers so that the usual
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operations of algebra can apply to them, and which also must have the same units or no units when we add and subtract). In this way, we can think of adding, subtracting, multiplying, and dividing functions. T = h + w This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 301 For two functions f (x) and g(x) with real number outputs, we define new functions f + g, f β g, f g, and f relations g by the ( f + g)(x) = f (x) + g(x) ( f β g)(x) = f (x) β g(x) ( f g)(x) = f (x)g(x) f g β β β β (x) = f (x) g(x) where g(x) β 0 Example 3.39 Performing Algebraic Operations on Functions Find and simplify the functions β βg β f β β (x) and β β g f β β (x), given f (x) = x β 1 and g(x) = x2 β 1. Are they the same function? Solution Begin by writing the general form, and then substitute the given functions. (g β f )(x) = g(x) β f (x) (g β f )(x) = x2 β 1 β (x β 1) = x2 β x = x(x β 1) β β β β g f g f β β (x) = β β (x) = = g(x) f (x) x2 β 1 x β 1 (x + 1)(x β 1) x β 1 = x + 1 where x β 1 No, the functions are not the same. Note: For β β g f β β (x), the condition x β 1 is necessary because when x = 1, the denominator is equal to 0, which makes the function undefined. 3.25 Find and simplify the functions β β f gβ β (x) and β β f β gβ β
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(x). Are they the same function? f (x) = x β 1 and g(x) = x2 β 1 Create a Function by Composition of Functions Performing algebraic operations on functions combines them into a new function, but we can also create functions by composing functions. When we wanted to compute a heating cost from a day of the year, we created a new function that takes a day as input and yields a cost as output. The process of combining functions so that the output of one function becomes the input of another is known as a composition of functions. The resulting function is known as a composite function. We represent this combination by the following notation: 302 Chapter 3 Functions β β f β gβ β (x) = f β βg(x)β β We read the left-hand side as β f composed with g at x,β and the right-hand side as β f of g of x.β The two sides of the equation have the same mathematical meaning and are equal. The open circle symbol β is called the composition operator. We use this operator mainly when we wish to emphasize the relationship between the functions themselves without referring to any particular input value. Composition is a binary operation that takes two functions and forms a new function, much as addition or multiplication takes two numbers and gives a new number. However, it is important not to confuse function composition with multiplication because, as we learned above, in most cases f (g(x)) β f (x)g(x). It is also important to understand the order of operations in evaluating a composite function. We follow the usual convention with parentheses by starting with the innermost parentheses first, and then working to the outside. In the equation above, the function g takes the input x first and yields an output g(x). Then the function f takes g(x) as an input and yields an output f β βg(x)β β . In general, f β g and g β f are different functions. In other words, in many cases f β see that sometimes two functions can be composed only in one specific order. βg(x)β β β gβ β f (x)β β for all x. We will also For example, if
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f (x) = x2 and g(x) = x + 2, then but f (g(x)) = f (x + 2) = (x + 2)2 = x2 + 4x + 4 g( f (x)) = gβ βx2β β = x2 + 2 These expressions are not equal for all values of x, so the two functions are not equal. It is irrelevant that the expressions happen to be equal for the single input value x = β 1 2. Note that the range of the inside function (the first function to be evaluated) needs to be within the domain of the outside function. Less formally, the composition has to make sense in terms of inputs and outputs. Composition of Functions When the output of one function is used as the input of another, we call the entire operation a composition of functions. For any input x and functions f and g, this action defines a composite function, which we write as f β g such that β β f β gβ β (x) = f β βg(x)β β (3.2) The domain of the composite function f β g is all x such that x is in the domain of g and g(x) is in the domain of f. It is important to realize that the product of functions f g is not the same as the function composition f β in general, f (x)g(x) β f β βg(x)β β . βg(x)β β , because, This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 303 Example 3.40 Determining whether Composition of Functions is Commutative Using the functions provided, find f β commutative. βg(x)β β and gβ β f (x)β β . Determine whether the composition of the functions is f (x) = 2x + 1 g(x) = 3 β x Solution Letβs begin by substituting g(x) into f (x). Now we can substitute f (x) into g(x). f (g(x))
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= 2(3 β x) + 1 = 6 β 2x + 1 = 7 β 2x g( f (x)) = 3 β (2x + 1) = 3 β 2x β 1 = β2x + 2 We find that g( f (x)) β f (g(x)), so the operation of function composition is not commutative. Example 3.41 Interpreting Composite Functions The function c(s) gives the number of calories burned completing s sit-ups, and s(t) gives the number of sit-ups a person can complete in t minutes. Interpret c(s(3)). Solution The inside expression in the composition is s(3). Because the input to the s-function is time, t = 3 represents 3 minutes, and s(3) is the number of sit-ups completed in 3 minutes. Using s(3) as the input to the function c(s) gives us the number of calories burned during the number of sit-ups that can be completed in 3 minutes, or simply the number of calories burned in 3 minutes (by doing sit-ups). Example 3.42 Investigating the Order of Function Composition Suppose f (x) gives miles that can be driven in x hours and g(y) gives the gallons of gas used in driving y miles. Which of these expressions is meaningful: f β β or gβ βg(y)β β f (x)β β ? 304 Chapter 3 Functions Solution The function y = f (x) is a function whose output is the number of miles driven corresponding to the number of hours driven. number of miles = f (number of hours) The function g(y) is a function whose output is the number of gallons used corresponding to the number of miles driven. This means: number of gallons = g (number of miles) The expression g(y) takes miles as the input and a number of gallons as the output. The function f (x) requires a β number of hours as the input. Trying to input a number of gallons does not make sense. The expression f β βg(y)β is meaningless. The expression f (x) takes hours as input and a number of miles driven as the output. The function g(y) requires a number of miles as the input. Using f (x) (miles driven
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) as an input value for g(y), where gallons of gas β makes sense, and will yield the number of depends on miles driven, does make sense. The expression gβ gallons of gas used, g, driving a certain number of miles, f (x), in x hours. β f (x)β Are there any situations where f(g(y)) and g( f(x)) would both be meaningful or useful expressions? Yes. For many pure mathematical functions, both compositions make sense, even though they usually produce different new functions. In real-world problems, functions whose inputs and outputs have the same units also may give compositions that are meaningful in either order. 3.26 The gravitational force on a planet a distance r from the sun is given by the function G(r). The acceleration of a planet subjected to any force F is given by the function a(F). Form a meaningful composition of these two functions, and explain what it means. Evaluating Composite Functions Once we compose a new function from two existing functions, we need to be able to evaluate it for any input in its domain. We will do this with specific numerical inputs for functions expressed as tables, graphs, and formulas and with variables as inputs to functions expressed as formulas. In each case, we evaluate the inner function using the starting input and then use the inner functionβs output as the input for the outer function. Evaluating Composite Functions Using Tables When working with functions given as tables, we read input and output values from the table entries and always work from the inside to the outside. We evaluate the inside function first and then use the output of the inside function as the input to the outside function. Example 3.43 Using a Table to Evaluate a Composite Function Using Table 3.21, evaluate f (g(3)) and g( f (3)). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 305 x f(x) g(x) 1 2 3 4 6 8 3 1 Table 3.21 3 5 2 7 Solution To evaluate f (g(3)), we start from the inside with the input value 3. We then evaluate the inside expression g(3) using the table that defines the function g : g(3) = 2. We can then use that result as the input to the function f, so g(3) is replaced by 2 and we get f (
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2). Then, using the table that defines the function f, we find that f (2) = 8. g(3) = 2 f (g(3)) = f (2) = 8 To evaluate g( f (3)), we first evaluate the inside expression f (3) using the first table: f (3) = 3. Then, using the table for g, we can evaluate Table 3.22 shows the composite functions f β g and g β f as tables. g( f (3)) = g(3) = 2 g(x) f β βg(x)β β f (x) β f (x)β gβ β 2 8 3 2 x 3 Table 3.22 3.27 Using Table 3.21, evaluate f (g(1)) and g( f (4)). Evaluating Composite Functions Using Graphs When we are given individual functions as graphs, the procedure for evaluating composite functions is similar to the process we use for evaluating tables. We read the input and output values, but this time, from the x- and y- axes of the graphs. 306 Chapter 3 Functions Given a composite function and graphs of its individual functions, evaluate it using the information provided by the graphs. 1. Locate the given input to the inner function on the x- axis of its graph. 2. Read off the output of the inner function from the y- axis of its graph. 3. Locate the inner function output on the x- axis of the graph of the outer function. 4. Read the output of the outer function from the y- axis of its graph. This is the output of the composite function. Example 3.44 Using a Graph to Evaluate a Composite Function Using Figure 3.60, evaluate f (g(1)). Figure 3.60 Solution To evaluate f (g(1)), we start with the inside evaluation. See Figure 3.61. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 307 Figure 3.61 We evaluate g(1) using the graph of g(x), finding the input of 1 on the x- axis and finding the output value of the graph at that input. Here, g(1) = 3. We use this value as the input to the function f. We can then evaluate the composite function by looking to the
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graph of f (x), finding the input of 3 on the xaxis and reading the output value of the graph at this input. Here, f (3) = 6, so f (g(1)) = 6. f (g(1)) = f (3) Analysis Figure 3.62 shows how we can mark the graphs with arrows to trace the path from the input value to the output value. 308 Chapter 3 Functions Figure 3.62 3.28 Using Figure 3.60, evaluate g( f (2)). Evaluating Composite Functions Using Formulas When evaluating a composite function where we have either created or been given formulas, the rule of working from the inside out remains the same. The input value to the outer function will be the output of the inner function, which may be a numerical value, a variable name, or a more complicated expression. While we can compose the functions for each individual input value, it is sometimes helpful to find a single formula that β . To do this, we will extend our idea of function evaluation. Recall that, will calculate the result of a composition f β when we evaluate a function like f (t) = t 2 β t, we substitute the value inside the parentheses into the formula wherever we see the input variable. βg(x)β Given a formula for a composite function, evaluate the function. 1. Evaluate the inside function using the input value or variable provided. 2. Use the resulting output as the input to the outside function. Example 3.45 Evaluating a Composition of Functions Expressed as Formulas with a Numerical Input Given f (t) = t 2 β t and h(x) = 3x + 2, evaluate f (h(1)). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 309 Solution Because the inside expression is h(1), we start by evaluating h(x) at 1. Then f (h(1)) = f (5), so we evaluate f (t) at an input of 5. h(1) = 3(1) + 2 h(1) = 5 f (h(1)) = f (5) f (h(1)) = 52 β 5 f (h(1)) = 20 Analysis It makes no difference what the input variables t and x were called in this problem because we evaluated for specific numerical values. 3.29 Given
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f (t) = t 2 β t and h(x) = 3x + 2, evaluate a. h( f (2)) b. h( f ( β 2)) Finding the Domain of a Composite Function As we discussed previously, the domain of a composite function such as f β g is dependent on the domain of g and the domain of f. It is important to know when we can apply a composite function and when we cannot, that is, to know the domain of a function such as f β g. Let us assume we know the domains of the functions f and g separately. If we write β , we can see right away that x must be a member of the domain of g in the composite function for an input x as f β order for the expression to be meaningful, because otherwise we cannot complete the inner function evaluation. However, we also see that g(x) must be a member of the domain of f, otherwise the second function evaluation in f β β cannot be completed, and the expression is still undefined. Thus the domain of f β g consists of only those inputs in the domain of g that produce outputs from g belonging to the domain of f. Note that the domain of f composed with g is the set of all x such that x is in the domain of g and g(x) is in the domain of f. βg(x)β βg(x)β Domain of a Composite Function The domain of a composite function f β domain of f. βg(x)β β is the set of those inputs x in the domain of g for which g(x) is in the Given a function composition f(g(x)), determine its domain. 1. Find the domain of g. 2. Find the domain of f. 3. Find those inputs x in the domain of g for which g(x) is in the domain of f. That is, exclude those inputs x from the domain of g for which g(x) is not in the domain of f. The resulting set is the domain of f β g. 310 Chapter 3 Functions Example 3.46 Finding the Domain of a Composite Function Find the domain of Solution β β f β gβ β (x) where f (x) = 5 x β 1 and g(x) = 4 3x
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β 2 The domain of g(x) consists of all real numbers except x = 2 3, since that input value would cause us to divide by 0. Likewise, the domain of f consists of all real numbers except 1. So we need to exclude from the domain of g(x) that value of x for which g(x) = 1. 4 3x β 2 = 1 4 = 3x β 2 6 = 3x x = 2 So the domain of f β g is the set of all real numbers except 2 3 and 2. This means that We can write this in interval notation as x β 2 3 or x β 2 β βββ, 2, β), 2 Example 3.47 Finding the Domain of a Composite Function Involving Radicals Find the domain of β β f β gβ β (x) where f (x) = x + 2 and g(x) = 3 β x Solution Because we cannot take the square root of a negative number, the domain of g is (ββ, 3]. Now we check the domain of the composite function β β f β gβ β (x) = 3 β x + 2 or β β f β gβ β (x) = 5 β x The domain of this function is (ββ, 5β€ β¦. To find the domain of f β g, we ask ourselves if there are any further restrictions offered by the domain of the composite function. The answer is no, since (ββ, 3] is a proper subset of the domain of f β g. This means the domain of f β g is the same as the domain of g, namely, (ββ, 3]. Analysis This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 311 This example shows that knowledge of the range of functions (specifically the inner function) can also be helpful in finding the domain of a composite function. It also shows that the domain of f β g can contain values that are not in the domain of f, though they must be in the domain of g. 3.30 Find the domain of β β f β gβ β (x) where f (x) = 1 x
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β 2 and g(x) = x + 4 Decomposing a Composite Function into its Component Functions In some cases, it is necessary to decompose a complicated function. In other words, we can write it as a composition of two simpler functions. There may be more than one way to decompose a composite function, so we may choose the decomposition that appears to be most expedient. Example 3.48 Decomposing a Function Write f (x) = 5 β x2 as the composition of two functions. Solution We are looking for two functions, g and h, so f (x) = g(h(x)). To do this, we look for a function inside a function in the formula for f (x). As one possibility, we might notice that the expression 5 β x2 is the inside of the square root. We could then decompose the function as We can check our answer by recomposing the functions. h(x) = 5 β x2 and g(x) = x g(h(x)) = gβ β5 β x2β β = 5 β x2 3.31 Write f (x) = 4 3 β 4 + x2 as the composition of two functions. Access these online resources for additional instruction and practice with composite functions. β’ Composite Functions (http://openstaxcollege.org/l/compfunction) β’ Composite Function Notation Application (http://openstaxcollege.org/l/compfuncnot) β’ Composite Functions Using Graphs (http://openstaxcollege.org/l/compfuncgraph) β’ Decompose Functions (http://openstaxcollege.org/l/decompfunction) β’ Composite Function Values (http://openstaxcollege.org/l/compfuncvalue) 312 Chapter 3 Functions 3.4 EXERCISES Verbal e. β β f β f β β (β2) How does one find the domain of the quotient of two 201. functions, f g? 202. What is the composition of two functions, f β g? If the order is reversed when composing two 203. functions, can the result ever be the same as the answer in the original order of the composition? If yes, give an example. If no, explain why not. How do you find the domain for the composition of 204. two functions, f β g? Algebra
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ic For the following exercises, determine the domain for each function in interval notation. 205. Given f (x) = x2 + 2x and g(x) = 6 β x2, find f + g, f β g, f g, and f g. 206. Given f (x) = β 3x2 + x and g(x) = 5, find f + g, f β g, f g, and f g. 207. Given f (x) = 2x2 + 4x and g(x) = 1 2x, find f + g, f β g, f g, and f g. 208. Given f (x) = 1 x β 4 and g(x) = 1 6 β x, find f + g, f β g, f g, and f g. 209. Given f (x) = 3x2 and g(x) = x β 5, find f + g, f β g, f g, and f g. 210. Given f (x) = x and g(x) = |x β 3|, find g f. For the following exercise, find the indicated function 211. given f (x) = 2x2 + 1 and g(x) = 3x β 5. For the following exercises, use each pair of functions to find f β β . Simplify your answers. β and gβ β f (x)β βg(x)β 212. 213. f (x) = x2 + 1, g(x) = x + 2 f (x) = x + 2, g(x) = x2 + 3 214. f (x) = |x|, g(x) = 5x + 1 215. f (x) = x3, g(x) = x + 1 x3 216. 217. f (x) = 1 x β 6, g(x) = 7 x + 6 f (x) = 1 x β 4, g(x) = 2 x + 4 For the following exercises, use each set of functions to find f β β . Simplify your answers. βh(x)β βgβ β β 218. 219. 220. f (x) = x4 + 6, g(x)
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= x β 6, and h(x) = x f (x) = x2 + 1, g(x) = 1 x, and h(x) = x + 3 Given f (x) = 1 x and g(x) = x β 3, find the following: a. b. c. d. e. ( f β g)(x) the domain of ( f β g)(x) in interval notation (g β f )(x) the domain of (g β f )(x) f g β β β β x 221. Given f (x) = 2 β 4x and g(x) = β 3 x, find the following: a. b. (g β f )(x) the domain of (g β f )(x) in interval notation a. b. c. d. f (g(2)) f (g(x)) g( f (x)) (g β g)(x) Given 222. f (x) = 1 β x x and g(x) = a. b. (g β f )(x) (g β f )(2) This content is available for free at https://cnx.org/content/col11758/1.5 the 1 1 + x2, find the following: functions Chapter 3 Functions 313 223. Given functions p(x) = 1 x and m(x) = x2 β 4, state the domain of each of the following functions using interval notation: 236. h(x) = (5x β 1)3 237. 3 h(x) = x β 1 a. b. p(x) m(x) p(m(x)) c. m(p(x)) 224. Given functions q(x) = 1 x and h(x) = x2 β 9, state the domain of each of the following functions using interval notation. a. b. c. q(x) h(x) qβ βh(x)β β βq(x)β hβ β 225. For f (x) = 1 x and g(x) = x β 1, write the domain of ( f β g)(x) in interval notation. For the following exercises, find functions f (x
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) and g(x) so the given function can be expressed as h(x) = f β βg(x)β β . 238. 239. 240. 241. h(x) = |x2 + 7| h(x) = 1 (x β 2)3 h(x) = 2 β β 1 2x β 3 β β h(x) = 2x β 1 3x + 4 Graphical For the following exercises, use the graphs of f, shown in Figure 3.63, and g, shown in Figure 3.64, to evaluate the expressions. 226. h(x) = (x + 2)2 227. h(x) = (x β 5)3 228. h(x) = 3 x β 5 229. h(x) = 4 (x + 2)2 230. h(x) = 4 + x3 231. h(x) = 3 1 2x β 3 232. h(x) = 1 (3x2 β 4)β3 233. 234. 4 h(x) = 3x β 2 x + 5 h(x) = β 8 + x3 β 8 β x3 4 β β 235. h(x) = 2x + 6 Figure 3.63 Figure 3.64 242. βg(3)β f β β 243. βg(1)β f β β 244. β f (1)β gβ β 245. β f (0)β gβ β 314 246. β f (5)β f β β 247. β f (4)β f β β 248. βg(2)β gβ β 249. βg(0)β gβ β For the following exercises, use graphs of f (x), shown in Figure 3.65, g(x), shown in Figure 3.66, and h(x), shown in Figure 3.67, to evaluate the expressions. Figure 3.65 Figure 3.66 Figure 3.67 250. β f
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(1)β gβ β 251. β f (2)β gβ β 252. βg(4)β f β β This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 253. βg(1)β f β β 254. βh(2)β f β β 255. β f (2)β hβ β 256. 257. f β βgβ βh(4)β β β β βgβ f β β f (β2)β β β β Numeric For the following exercises, use the function values for f and g shown in Table 3.23 to evaluate each expression(x) g(x Table 3.23 258. βg(8)β f β β 259. βg(5)β f β β 260. β f (5)β gβ β 261. β f (3)β gβ β 262. Chapter 3 Functions 315 β f (4)β f β β 263. β f (1)β f β β 264. βg(2)β gβ β 265. βg(6)β gβ β For the following exercises, use the function values for f and g shown in Table 3.24 to evaluate the expressions. x f(x) g(x) β3 β2 β1 0 1 2 3 11 9 7 5 3 1 β1 Table 3.24 β8 β3 0 1 0 β3 β8 266. ( f β g)(1) 267. ( f β g)(2) 268. (g β f )(2) 269. (g β f )(3) 270. (g β g)(1) 271. ( f
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β f )(3) For the following exercises, use each pair of functions to find f β β and gβ β f (0)β β . βg(0)β 274. 275. f (x) = x + 4, g(x) = 12 β x3 f (x) = 1 x + 2, g(x) = 4x + 3 the For following f (x) = 2x2 + 1 and g(x) = 3x + 5 the composite function as indicated. exercises, use the functions to evaluate or find 276. βg(2)β f β β 277. βg(x)β f β β 278. β f ( β 3)β gβ β 279. (g β g)(x) Extensions For the following exercises, use f (x) = x3 + 1 and 3 g(x) = x β 1. 280. Find ( f β g)(x) and (g β f )(x). Compare the two answers. 281. Find ( f β g)(2) and (g β f )(2). 282. What is the domain of (g β f )(x)? 283. What is the domain of ( f β g)(x)? 284. Let f (x) = 1 x. a. Find ( f β f )(x). b. Is ( f β f )(x) for any function f the same result as the answer to part (a) for any function? Explain. For the following exercises, let F(x) = (x + 1)5, f (x) = x5, and g(x) = x + 1. 285. True or False: (g β f )(x) = F(x). 286. True or False: ( f β g)(x) = F(x). For the following exercises, find the composition when f (x) = x2 + 2 for all x β₯ 0 and g(x) = x β 2. 287. ( f β g)(6); (g β f )(6) 288. (g β f )(a); ( f β g)(a)
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f (x) = 4x + 8, g(x) = 7 β x2 272. 273. f (x) = 5x + 7, g(x) = 4 β 2x2 289. ( f β g)(11); (g β f )(11) 316 Chapter 3 Functions Real-World Applications a. Find the composite function r(V(t)). b. Find the exact time when the radius reaches 10 inches. The number of bacteria in a refrigerated food product 297. is given by N(T) = 23T 2 β 56T + 1, 3 < T < 33, where T is the temperature of the food. When the food is removed from the refrigerator, the temperature is given by T(t) = 5t + 1.5, where t is the time in hours. a. Find the composite function N(T(t)). b. Find the time (round to two decimal places) when the bacteria count reaches 6752. The function D(p) gives the number of items that 290. will be demanded when the price is p. The production cost C(x) is the cost of producing x items. To determine the cost of production when the price is $6, you would do which of the following? a. Evaluate Dβ βC(6)β β . b. Evaluate Cβ βD(6)β β . c. Solve D(C(x)) = 6. d. Solve Cβ βD(p)β β = 6. 291. The function A(d) gives the pain level on a scale of 0 to 10 experienced by a patient with d milligrams of a painreducing drug in her system. The milligrams of the drug in the patientβs system after t minutes is modeled by m(t). Which of the following would you do in order to determine when the patient will be at a pain level of 4? a. Evaluate A(m(4)). b. Evaluate m(A(4)). c. Solve A(m(t)) = 4. d. Solve mβ βA(d)β β = 4. 292. A store offers customers a 30% discount on the price x of selected items. Then, the store takes off
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an additional 15% at the cash register. Write a price function P(x) that computes the final price of the item in terms of the original price x. (Hint: Use function composition to find your answer.) 293. A rain drop hitting a lake makes a circular ripple. If the radius, in inches, grows as a function of time in minutes according to r(t) = 25 t + 2, find the area of the ripple as a function of time. Find the area of the ripple at t = 2. 294. A forest fire leaves behind an area of grass burned in an expanding circular pattern. If the radius of the circle of burning grass is increasing with time according to the formula r(t) = 2t + 1, express the area burned as a function of time, t (minutes). Use the function you found in the previous exercise to 295. find the total area burned after 5 minutes. 296. The radius r, in inches, of a spherical balloon is related to the volume, V, by r(V) = 3V 4Ο into the balloon, so the volume after t seconds is given by V(t) = 10 + 20t.. Air is pumped 3 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 317 3.5 | Transformation of Functions Learning Objectives In this section, you will: 3.5.1 Graph functions using vertical and horizontal shifts. 3.5.2 Graph functions using reflections about the x-axis axis and the y-axis. 3.5.3 Determine whether a function is even, odd, or neither from its graph. 3.5.4 Graph functions using compressions and stretches. 3.5.5 Combine transformations. Figure 3.68 (credit: "Misko"/Flickr) We all know that a flat mirror enables us to see an accurate image of ourselves and whatever is behind us. When we tilt the mirror, the images we see may shift horizontally or vertically. But what happens when we bend a flexible mirror? Like a carnival funhouse mirror, it presents us with a distorted image of ourselves, stretched or compressed horizontally or vertically. In a similar way, we can distort or transform mathematical functions to better adapt them to describing objects or processes in the real world. In this section, we will take a look at several kinds of transformations. Graphing Functions Using Vertical and Horizontal Shifts Often when
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given a problem, we try to model the scenario using mathematics in the form of words, tables, graphs, and equations. One method we can employ is to adapt the basic graphs of the toolkit functions to build new models for a given scenario. There are systematic ways to alter functions to construct appropriate models for the problems we are trying to solve. Identifying Vertical Shifts One simple kind of transformation involves shifting the entire graph of a function up, down, right, or left. The simplest shift is a vertical shift, moving the graph up or down, because this transformation involves adding a positive or negative constant to the function. In other words, we add the same constant to the output value of the function regardless of the input. For a function g(x) = f (x) + k, the function f (x) is shifted vertically k units. See Figure 3.69 for an example. 318 Chapter 3 Functions Figure 3.69 Vertical shift by k = 1 of the cube root function f (x) = x3. To help you visualize the concept of a vertical shift, consider that y = f (x). Therefore, f (x) + k is equivalent to y + k. Every unit of y is replaced by y + k, so the y-value increases or decreases depending on the value of k. The result is a shift upward or downward. Vertical Shift Given a function f (x), a new function g(x) = f (x) + k, where k is a constant, is a vertical shift of the function f (x). All the output values change by k units. If k is positive, the graph will shift up. If k is negative, the graph will shift down. Example 3.49 Adding a Constant to a Function To regulate temperature in a green building, airflow vents near the roof open and close throughout the day. Figure 3.70 shows the area of open vents V (in square feet) throughout the day in hours after midnight, t. During the summer, the facilities manager decides to try to better regulate temperature by increasing the amount of open vents by 20 square feet throughout the day and night. Sketch a graph of this new function. Figure 3.70 Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 319 We can sketch a graph of this new function by adding 20 to each of the output values of the original function. This will have the effect of shifting the graph vertically
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up, as shown in Figure 3.71. Figure 3.71 Notice that in Figure 3.71, for each input value, the output value has increased by 20, so if we call the new function S(t), we could write S(t) = V(t) + 20 This notation tells us that, for any value of t, S(t) can be found by evaluating the function V at the same input and then adding 20 to the result. This defines S as a transformation of the function V, in this case a vertical shift up 20 units. Notice that, with a vertical shift, the input values stay the same and only the output values change. See Table 3.25. 0 0 8 0 10 17 220 220 20 20 240 240 19 0 20 24 0 20 t V(t) S(t) Table 3.25 Given a tabular function, create a new row to represent a vertical shift. 1. Identify the output row or column. 2. Determine the magnitude of the shift. 3. Add the shift to the value in each output cell. Add a positive value for up or a negative value for down. Example 3.50 Shifting a Tabular Function Vertically 320 Chapter 3 Functions A function f (x) is given in Table 3.26. Create a table for the function g(x) = f (x) β 3. x f(x) 2 1 4 3 6 7 8 11 Table 3.26 Solution The formula g(x) = f (x) β 3 tells us that we can find the output values of g by subtracting 3 from the output values of f. For example: Given f (2) = 1 g(x) = f (x) β 3 Given transformation g(2) = f (2) β 3 = 1 β 3 = β2 Subtracting 3 from each f (x) value, we can complete a table of values for g(x) as shown in Table 3.27. x f(x) 2 1 g(x) β2 4 3 0 6 7 4 8 11 8 Table 3.27 Analysis As with the earlier vertical shift, notice the input values stay the same and only the output values change. 3.32 The function h(t) = β 4.9t 2 + 30t gives the height h of a ball (in meters) thrown upward from the ground after t seconds. Suppose the ball was instead thrown from the top of a 10
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-m building. Relate this new height function b(t) to h(t), and then find a formula for b(t). Identifying Horizontal Shifts We just saw that the vertical shift is a change to the output, or outside, of the function. We will now look at how changes to input, on the inside of the function, change its graph and meaning. A shift to the input results in a movement of the graph of the function left or right in what is known as a horizontal shift, shown in Figure 3.72. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 321 Figure 3.72 Horizontal shift of the function f (x) = x3. Note that h = + 1 shifts the graph to the left, that is, towards negative values of x. For example, if f (x) = x2, then g(x) = (x β 2)2 is a new function. Each input is reduced by 2 prior to squaring the function. The result is that the graph is shifted 2 units to the right, because we would need to increase the prior input by 2 units to yield the same output value as given in f. Horizontal Shift Given a function f, a new function g(x) = f (x β h), where h is a constant, is a horizontal shift of the function f. If h is positive, the graph will shift right. If h is negative, the graph will shift left. Example 3.51 Adding a Constant to an Input Returning to our building airflow example from Figure 3.70, suppose that in autumn the facilities manager decides that the original venting plan starts too late, and wants to begin the entire venting program 2 hours earlier. Sketch a graph of the new function. Solution We can set V(t) to be the original program and F(t) to be the revised program. V(t) = the original venting plan F(t) = starting 2 hrs sooner In the new graph, at each time, the airflow is the same as the original function V was 2 hours later. For example, in the original function V, the airflow starts to change at 8 a.m., whereas for the function F, the airflow starts to change at 6 a.m. The comparable function values are V(8) = F(6). See Figure 3.73. Notice also that the vents first opened to
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220 ft2 at 10 a.m. under the original plan, while under the new plan the vents reach 220 ft2 at 8 a.m., so V(10) = F(8). In both cases, we see that, because F(t) starts 2 hours sooner, h = β 2. That means that the same output values are reached when F(t) = V(t β (β2)) = V(t + 2). 322 Chapter 3 Functions Figure 3.73 Analysis Note that V(t + 2) has the effect of shifting the graph to the left. Horizontal changes or βinside changesβ affect the domain of a function (the input) instead of the range and often seem counterintuitive. The new function F(t) uses the same outputs as V(t), but matches those outputs to inputs 2 hours earlier than those of V(t). Said another way, we must add 2 hours to the input of V to find the corresponding output for F : F(t) = V(t + 2). Given a tabular function, create a new row to represent a horizontal shift. 1. Identify the input row or column. 2. Determine the magnitude of the shift. 3. Add the shift to the value in each input cell. Example 3.52 Shifting a Tabular Function Horizontally A function f (x) is given in Table 3.28. Create a table for the function g(x) = f (x β 3). x f(x) 2 1 4 3 6 7 8 11 Table 3.28 Solution The formula g(x) = f (x β 3) tells us that the output values of g are the same as the output value of f when the input value is 3 less than the original value. For example, we know that f (2) = 1. To get the same output from This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 323 the function g, we will need an input value that is 3 larger. We input a value that is 3 larger for g(x) because the function takes 3 away before evaluating the function f. g(5) = f (5 β 3) = f (2) = 1 We continue with the other values to create Table 3.29. 7 4 3 3 9 6 7 7 11 8 11 11 x x β 3 f(x) g(x) 5
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2 1 1 Table 3.29 The result is that the function g(x) has been shifted to the right by 3. Notice the output values for g(x) remain the same as the output values for f (x), but the corresponding input values, x, have shifted to the right by 3. Specifically, 2 shifted to 5, 4 shifted to 7, 6 shifted to 9, and 8 shifted to 11. Analysis Figure 3.74 represents both of the functions. We can see the horizontal shift in each point. 324 Chapter 3 Functions Example 3.53 Identifying a Horizontal Shift of a Toolkit Function Figure 3.75 represents a transformation of the toolkit function f (x) = x2. Relate this new function g(x) to f (x), and then find a formula for g(x). Figure 3.75 Solution Notice that the graph is identical in shape to the f (x) = x2 function, but the x-values are shifted to the right 2 units. The vertex used to be at (0,0), but now the vertex is at (2,0). The graph is the basic quadratic function shifted 2 units to the right, so g(x) = f (x β 2) Notice how we must input the value x = 2 to get the output value y = 0; the x-values must be 2 units larger because of the shift to the right by 2 units. We can then use the definition of the f (x) function to write a formula for g(x) by evaluating f (x β 2). f (x) = x2 g(x) = f (x β 2) g(x) = f (x β 2) = (x β 2)2 Analysis To determine whether the shift is + 2 or β 2, consider a single reference point on the graph. For a quadratic, looking at the vertex point is convenient. In the original function, f (0) = 0. In our shifted function, g(2) = 0. To obtain the output value of 0 from the function f, we need to decide whether a plus or a minus sign will work to satisfy g(2) = f (x β 2) = f (0) = 0. For this to work, we will need to subtract 2 units from our input values. Example 3.54 Interpreting Horizontal versus Vertical Shifts This content is available for free at https://cnx.org
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/content/col11758/1.5 Chapter 3 Functions 325 The function G(m) gives the number of gallons of gas required to drive m miles. Interpret G(m) + 10 and G(m + 10). Solution G(m) + 10 can be interpreted as adding 10 to the output, gallons. This is the gas required to drive m miles, plus another 10 gallons of gas. The graph would indicate a vertical shift. G(m + 10) can be interpreted as adding 10 to the input, miles. So this is the number of gallons of gas required to drive 10 miles more than m miles. The graph would indicate a horizontal shift. 3.33 Given the function f (x) = x, graph the original function f (x) and the transformation g(x) = f (x + 2) on the same axes. Is this a horizontal or a vertical shift? Which way is the graph shifted and by how many units? Combining Vertical and Horizontal Shifts Now that we have two transformations, we can combine them. Vertical shifts are outside changes that affect the output (y-) values and shift the function up or down. Horizontal shifts are inside changes that affect the input (x-) values and shift the function left or right. Combining the two types of shifts will cause the graph of a function to shift up or down and left or right. Given a function and both a vertical and a horizontal shift, sketch the graph. 1. Identify the vertical and horizontal shifts from the formula. 2. The vertical shift results from a constant added to the output. Move the graph up for a positive constant and down for a negative constant. 3. The horizontal shift results from a constant added to the input. Move the graph left for a positive constant and right for a negative constant. 4. Apply the shifts to the graph in either order. Example 3.55 Graphing Combined Vertical and Horizontal Shifts Given f (x) = |x|, sketch a graph of h(x) = f (x + 1) β 3. Solution The function f is our toolkit absolute value function. We know that this graph has a V shape, with the point at the origin. The graph of h has transformed f in two ways: f (x + 1) is a change on the inside of the function, giving a horizontal shift left by 1, and the subtraction by 3 in f (x + 1) β 3 is a change to the outside of
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the function, giving a vertical shift down by 3. The transformation of the graph is illustrated in Figure 3.76. Let us follow one point of the graph of f (x) = |x|. β’ The point (0, 0) is transformed first by shifting left 1 unit: (0, 0) β (β1, 0) 326 Chapter 3 Functions β’ The point (β1, 0) is transformed next by shifting down 3 units: (β1, 0) β (β1, β3) Figure 3.76 Figure 3.77 shows the graph of h. Figure 3.77 3.34 Given f (x) = |x|, sketch a graph of h(x) = f (x β 2) + 4. Example 3.56 Identifying Combined Vertical and Horizontal Shifts Write a formula for the graph shown in Figure 3.78, which is a transformation of the toolkit square root function. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 327 Figure 3.78 Solution The graph of the toolkit function starts at the origin, so this graph has been shifted 1 to the right and up 2. In function notation, we could write that as Using the formula for the square root function, we can write h(x) = f (x β 1) + 2 h(x) = x β 1 + 2 Analysis Note that this transformation has changed the domain and range of the function. This new graph has domain [1, β) and range [2, β). 3.35 Write a formula for a transformation of the toolkit reciprocal function f (x) = 1 x that shifts the functionβs graph one unit to the right and one unit up. Graphing Functions Using Reflections about the Axes Another transformation that can be applied to a function is a reflection over the x- or y-axis. A vertical reflection reflects a graph vertically across the x-axis, while a horizontal reflection reflects a graph horizontally across the y-axis. The reflections are shown in Figure 3.79. 328 Chapter 3 Functions Figure 3.79 Vertical and horizontal reflections of a function. Notice that the vertical reflection produces a new graph that is a mirror image of the base or original graph about the x-axis. The horizontal reflection produces a new graph that is a mirror image of the base or original graph about the y-axis. Reflections Given a function f (x), a
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new function g(x) = β f (x) is a vertical reflection of the function f (x), sometimes called a reflection about (or over, or through) the x-axis. Given a function f (x), a new function g(x) = f ( β x) is a horizontal reflection of the function f (x), sometimes called a reflection about the y-axis. Given a function, reflect the graph both vertically and horizontally. 1. Multiply all outputs by β1 for a vertical reflection. The new graph is a reflection of the original graph about the x-axis. 2. Multiply all inputs by β1 for a horizontal reflection. The new graph is a reflection of the original graph about the y-axis. Example 3.57 Reflecting a Graph Horizontally and Vertically Reflect the graph of s(t) = t (a) vertically and (b) horizontally. Solution a. Reflecting the graph vertically means that each output value will be reflected over the horizontal t-axis as shown in Figure 3.80. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 329 Figure 3.80 Vertical reflection of the square root function Because each output value is the opposite of the original output value, we can write (3.3) Notice that this is an outside change, or vertical shift, that affects the output s(t) values, so the negative sign belongs outside of the function. V(t) = β s(t) or V(t) = β t b. Reflecting horizontally means that each input value will be reflected over the vertical axis as shown in Figure 3.81. Figure 3.81 Horizontal reflection of the square root function Because each input value is the opposite of the original input value, we can write H(t) = s( β t) or H(t) = βt 330 Chapter 3 Functions Notice that this is an inside change or horizontal change that affects the input values, so the negative sign is on the inside of the function. Note that these transformations can affect the domain and range of the functions. While the original square root function has domain [0, β) and range [0, β), the vertical reflection gives the V(t) function the range (ββ, 0] and the horizontal reflection gives the H(t) function the domain (ββ, 0]. 3.36 Reflect the graph of
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f (x) = |x β 1| (a) vertically and (b) horizontally. Example 3.58 Reflecting a Tabular Function Horizontally and Vertically A function f (x) is given as Table 3.30. Create a table for the functions below. a. b. g(x) = β f (x) h(x) = f (βx) x f(x) 2 1 4 3 6 7 8 11 Table 3.30 Solution a. For g(x), the negative sign outside the function indicates a vertical reflection, so the x-values stay the same and each output value will be the opposite of the original output value. See Table 3.31. x 2 4 6 8 g(x) β1 β3 β7 β11 Table 3.31 b. For h(x), the negative sign inside the function indicates a horizontal reflection, so each input value will be the opposite of the original input value and the h(x) values stay the same as the f (x) values. See Table 3.32. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 331 x β2 β4 β6 β8 h(x) 1 3 7 11 Table 3.32 3.37 A function f (x) is given as Table 3.33. Create a table for the functions below. a. g(x) = β f (x) b. h(x) = f (βx) x β2 0 2 4 f(x) 5 10 15 20 Table 3.33 Example 3.59 Applying a Learning Model Equation A common model for learning has an equation similar to k(t) = β 2βt mastery that can be achieved after t practice sessions. This is a transformation of the function f (t) = 2 in Figure 3.82. Sketch a graph of k(t). + 1, where k is the percentage of t shown 332 Chapter 3 Functions Figure 3.82 Solution This equation combines three transformations into one equation. β’ A horizontal reflection: f (βt) = 2βt β’ A vertical reflection: β f (βt) = β 2βt β’ A vertical shift: β f (βt) + 1 = β 2βt + 1 We can sketch a graph by applying these transformations one at a time to the original function. Let us follow two points through each of the three transformations. We will choose
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the points (0, 1) and (1, 2). 1. First, we apply a horizontal reflection: (0, 1) (β1, 2). 2. Then, we apply a vertical reflection: (0, β1) (1, β2). 3. Finally, we apply a vertical shift: (0, 0) (1, 1). This means that the original points, (0,1) and (1,2) become (0,0) and (1,1) after we apply the transformations. In Figure 3.83, the first graph results from a horizontal reflection. The second results from a vertical reflection. The third results from a vertical shift up 1 unit. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 333 Figure 3.83 Analysis As a model for learning, this function would be limited to a domain of t β₯ 0, with corresponding range [0, 1). 3.38 Given the toolkit function f (x) = x2, graph g(x) = β f (x) and h(x) = f ( β x). Take note of any surprising behavior for these functions. Determining Even and Odd Functions Some functions exhibit symmetry so that reflections result in the original graph. For example, horizontally reflecting the toolkit functions f (x) = x2 or f (x) = |x| will result in the original graph. We say that these types of graphs are symmetric about the y-axis. A function whose graph is symmetric about the y-axis is called an even function. If the graphs of f (x) = x3 or f (x) = 1 Figure 3.84. x were reflected over both axes, the result would be the original graph, as shown in 334 Chapter 3 Functions Figure 3.84 (a) The cubic toolkit function (b) Horizontal reflection of the cubic toolkit function (c) Horizontal and vertical reflections reproduce the original cubic function. We say that these graphs are symmetric about the origin. A function with a graph that is symmetric about the origin is called an odd function. Note: A function can be neither even nor odd if it does not exhibit either symmetry. For example, f (x) = 2 nor odd. Also, the only function that is both even and odd is the constant function f (x) = 0. x is neither even Even and Odd
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Functions A function is called an even function if for every input x f (x) = f ( β x) The graph of an even function is symmetric about the y- axis. A function is called an odd function if for every input x The graph of an odd function is symmetric about the origin. f (x) = β f ( β x) Given the formula for a function, determine if the function is even, odd, or neither. 1. Determine whether the function satisfies f (x) = f ( β x). If it does, it is even. 2. Determine whether the function satisfies f (x) = β f ( β x). If it does, it is odd. 3. If the function does not satisfy either rule, it is neither even nor odd. Example 3.60 Determining whether a Function Is Even, Odd, or Neither Is the function f (x) = x3 + 2x even, odd, or neither? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 335 Solution Without looking at a graph, we can determine whether the function is even or odd by finding formulas for the reflections and determining if they return us to the original function. Letβs begin with the rule for even functions. f ( β x) = ( β x)3 + 2( β x) = β x3 β 2x This does not return us to the original function, so this function is not even. We can now test the rule for odd functions. Because β f ( β x) = f (x), this is an odd function. β f ( β x) = β ββx3 β 2xβ β β = x3 + 2x Analysis Consider the graph of f in Figure 3.85. Notice that the graph is symmetric about the origin. For every point (x, y) on the graph, the corresponding point (βx, β y) is also on the graph. For example, (1, 3) is on the graph of f, and the corresponding point (β1, β3) is also on the graph. Figure 3.85 3.39 Is the function f (s) = s4 + 3s2 + 7 even, odd, or neither? Graphing Functions Using Stretches and Compressions Adding a constant to the inputs or outputs of a function changed the
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position of a graph with respect to the axes, but it did not affect the shape of a graph. We now explore the effects of multiplying the inputs or outputs by some quantity. We can transform the inside (input values) of a function or we can transform the outside (output values) of a function. Each change has a specific effect that can be seen graphically. 336 Chapter 3 Functions Vertical Stretches and Compressions When we multiply a function by a positive constant, we get a function whose graph is stretched or compressed vertically in relation to the graph of the original function. If the constant is greater than 1, we get a vertical stretch; if the constant is between 0 and 1, we get a vertical compression. Figure 3.86 shows a function multiplied by constant factors 2 and 0.5 and the resulting vertical stretch and compression. Figure 3.86 Vertical stretch and compression Vertical Stretches and Compressions Given a function f (x), a new function g(x) = a f (x), where a is a constant, is a vertical stretch or vertical compression of the function f (x). β’ β’ β’ If a > 1, then the graph will be stretched. If 0 < a < 1, then the graph will be compressed. If a < 0, then there will be combination of a vertical stretch or compression with a vertical reflection. Given a function, graph its vertical stretch. 1. Identify the value of a. 2. Multiply all range values by a. 3. If a > 1, the graph is stretched by a factor of a. If 0 < a < 1, the graph is compressed by a factor of a. If a < 0, the graph is either stretched or compressed and also reflected about the x-axis. Example 3.61 Graphing a Vertical Stretch A function P(t) models the population of fruit flies. The graph is shown in Figure 3.87. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 337 Figure 3.87 A scientist is comparing this population to another population, Q, whose growth follows the same pattern, but is twice as large. Sketch a graph of this population. Solution Because the population is always twice as large, the new populationβs output values are always twice the original functionβs output values. Graphically, this is shown in Figure 3.88. If we choose four reference points, (0, 1),
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(3, 3), (6, 2) and (7, 0) we will multiply all of the outputs by 2. The following shows where the new points for the new graph will be located. (0, 1) β (0, 2) (3, 3) β (3, 6) (6, 2) β (6, 4) (7, 0) β (7, 0) Figure 3.88 Symbolically, the relationship is written as Q(t) = 2P(t) 338 Chapter 3 Functions This means that for any input t, the value of the function Q is twice the value of the function P. Notice that the effect on the graph is a vertical stretching of the graph, where every point doubles its distance from the horizontal axis. The input values, t, stay the same while the output values are twice as large as before. Given a tabular function and assuming that the transformation is a vertical stretch or compression, create a table for a vertical compression. 1. Determine the value of a. 2. Multiply all of the output values by a. Example 3.62 Finding a Vertical Compression of a Tabular Function A function f is given as Table 3.34. Create a table for the function g(x) = 1 2 f (x). x f(x) 2 1 4 3 6 7 8 11 Table 3.34 Solution The formula g(x) = 1 2 f (x) tells us that the output values of g are half of the output values of f with the same inputs. For example, we know that f (4) = 3. Then We do the same for the other values to produce Table 3.35. g(4) = 1 2 f (4) = 1 2 (3) = 3 2 x g(x 11 2 Table 3.35 Analysis The result is that the function g(x) has been compressed vertically by 1 2 the graph is half the original height.. Each output value is divided in half, so This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 339 3.40 A function f is given as Table 3.36. Create a table for the function g(x) = 3 4 f (x). x 2 4 6 f (x) 12 16 20 8 0 Table 3.36 Example 3.63 Recognizing a Vertical Stretch The graph in Figure 3.89 is a transformation of
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the toolkit function f (x) = x3. Relate this new function g(x) to f (x), and then find a formula for g(x). Figure 3.89 Solution When trying to determine a vertical stretch or shift, it is helpful to look for a point on the graph that is relatively clear. In this graph, it appears that g(2) = 2. With the basic cubic function at the same input, f (2) = 23 = 8. Based on that, it appears that the outputs of g are 1 4 f (x). the outputs of the function f because g(2) = 1 4 this we can fairly safely conclude that g(x) = 1 4 f (2). From We can write a formula for g by using the definition of the function f. g(x) = 1 4 f (x) = 1 4 x3 340 Chapter 3 Functions 3.41 Write the formula for the function that we get when we stretch the identity toolkit function by a factor of 3, and then shift it down by 2 units. Horizontal Stretches and Compressions Now we consider changes to the inside of a function. When we multiply a functionβs input by a positive constant, we get a function whose graph is stretched or compressed horizontally in relation to the graph of the original function. If the constant is between 0 and 1, we get a horizontal stretch; if the constant is greater than 1, we get a horizontal compression of the function. Figure 3.90 Given a function y = f (x), the form y = f (bx) results in a horizontal stretch or compression. Consider the function y = x2. Observe Figure 3.90. The graph of y = (0.5x)2 is a horizontal stretch of the graph of the function y = x2 by a factor of 2. The graph of y = (2x)2 is a horizontal compression of the graph of the function y = x2 by a factor of 2. Horizontal Stretches and Compressions Given a function f (x), a new function g(x) = f (bx), where b is a constant, is a horizontal stretch or horizontal compression of the function f (x). β’ β’ β’ If b > 1, then the graph will be compressed by 1 b. If 0 < b < 1, then the graph will be stretched by 1 b. If b < 0, then there will be combination
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of a horizontal stretch or compression with a horizontal reflection. Given a description of a function, sketch a horizontal compression or stretch. 1. Write a formula to represent the function. 2. Set g(x) = f (bx) where b > 1 for a compression or 0 < b < 1 for a stretch. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 341 Example 3.64 Graphing a Horizontal Compression Suppose a scientist is comparing a population of fruit flies to a population that progresses through its lifespan twice as fast as the original population. In other words, this new population, R, will progress in 1 hour the same amount as the original population does in 2 hours, and in 2 hours, it will progress as much as the original population does in 4 hours. Sketch a graph of this population. Solution Symbolically, we could write R(1) = P(2), R(2) = P(4), and in general, R(t) = P(2t). See Figure 3.91 for a graphical comparison of the original population and the compressed population. Figure 3.91 (a) Original population graph (b) Compressed population graph Analysis Note that the effect on the graph is a horizontal compression where all input values are half of their original distance from the vertical axis. Example 3.65 Finding a Horizontal Stretch for a Tabular Function A function f (x) is given as Table 3.37. Create a table for the function g(x) = f β β xβ β . 1 2 342 Chapter 3 Functions x f(x) 2 1 4 3 6 7 8 11 Table 3.37 1 2 Solution xβ The formula g(x) = f β β tells us that the output values for g are the same as the output values for the β function f at an input half the size. Notice that we do not have enough information to determine g(2) because g(2) = f β β be twice as large to get inputs for f that we can evaluate. For example, we can determine g(4). β β = f (1), and we do not have a value for f (1) in our table. Our input values to g will need to β
2 1 2 We do the same for the other values
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to produce Table 3.38. g(42) = 1 x g(x) 4 1 8 3 12 16 7 11 Table 3.38 Figure 3.92 shows the graphs of both of these sets of points. Figure 3.92 Analysis Because each input value has been doubled, the result is that the function g(x) has been stretched horizontally by a factor of 2. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 343 Example 3.66 Recognizing a Horizontal Compression on a Graph Relate the function g(x) to f (x) in Figure 3.93. Figure 3.93 Solution The graph of g(x) looks like the graph of f (x) horizontally compressed. Because f (x) ends at (6, 4) and g(x) ends at (2, 4), we can see that the x- values have been compressed by 1 3 β β = 2. We might also notice that g(2) = f (6) and g(1) = f (3). Either way, we can describe this relationship as g(x) = f (3x). This is a horizontal compression by 1 3, because 6 1 3 β β. Analysis Notice that the coefficient needed for a horizontal stretch or compression is the reciprocal of the stretch or compression. So to stretch the graph horizontally by a scale factor of 4, we need a coefficient of 1 4 in our function: f β β 1 4 xβ β . This means that the input values must be four times larger to produce the same result, requiring the input to be larger, causing the horizontal stretching. 3.42 Write a formula for the toolkit square root function horizontally stretched by a factor of 3. Performing a Sequence of Transformations When combining transformations, it is very important to consider the order of the transformations. For example, vertically shifting by 3 and then vertically stretching by 2 does not create the same graph as vertically stretching by 2 and then vertically shifting by 3, because when we shift first, both the original function and the shift get stretched, while only the original function gets stretched when we stretch first. When we see an expression such as 2 f (x) + 3, which transformation should we start with? The answer here follows nicely from the order of operations. Given the output value of f (x), we first multiply
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by 2, causing the vertical stretch, and then add 3, causing the vertical shift. In other words, multiplication before addition. Horizontal transformations are a little trickier to think about. When we write g(x) = f (2x + 3), for example, we have to think about how the inputs to the function g relate to the inputs to the function f. Suppose we know f (7) = 12. What input to g would produce that output? In other words, what value of x will allow g(x) = f (2x + 3) = 12? We would need 344 Chapter 3 Functions 2x + 3 = 7. To solve for x, we would first subtract 3, resulting in a horizontal shift, and then divide by 2, causing a horizontal compression. This format ends up being very difficult to work with, because it is usually much easier to horizontally stretch a graph before shifting. We can work around this by factoring inside the function. f (bx + p) = f β βbβ βx + β β β β p b Letβs work through an example. We can factor out a 2. f (x) = (2x + 4)2 f (x) = β β2(x + 2)β β 2 Now we can more clearly observe a horizontal shift to the left 2 units and a horizontal compression. Factoring in this way allows us to horizontally stretch first and then shift horizontally. Combining Transformations When combining vertical transformations written in the form a f (x) + k, first vertically stretch by a and then vertically shift by k. When combining horizontal transformations written in the form f (bx + h), first horizontally shift by h and then horizontally stretch by 1 b. When combining horizontal transformations written in the form f (b(x + h)), first horizontally stretch by 1 b and then horizontally shift by h. Horizontal and vertical transformations are performed first. transformations are independent. It does not matter whether horizontal or vertical Example 3.67 Finding a Triple Transformation of a Tabular Function Given Table 3.39 for the function f (x), create a table of values for the function g(x) = 2 f (3x) + 1. x 6 12 18 24 f(x) 10 14 15 17 Table 3.39 Solution There are three steps to this transformation, and we will
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work from the inside out. Starting with the horizontal transformations, f (3x) is a horizontal compression by 1. See 3, which means we multiply each x- value by 1 3 Table 3.40. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 345 x 2 4 6 8 f(3x) 10 14 15 17 Table 3.40 Looking now to the vertical transformations, we start with the vertical stretch, which will multiply the output values by 2. We apply this to the previous transformation. See Table 3.41. x 2 4 6 8 2 f(3x) 20 28 30 34 Table 3.41 Finally, we can apply the vertical shift, which will add 1 to all the output values. See Table 3.42. x 2 4 6 8 g(x) = 2 f(3x) + 1 21 29 31 35 Table 3.42 Example 3.68 Finding a Triple Transformation of a Graph Use the graph of f (x) in Figure 3.94 to sketch a graph of k(x. 1 2 346 Chapter 3 Functions Figure 3.94 Solution To simplify, letβs start by factoring out the inside of the functionx + 2) β β 3 1 2 By factoring the inside, we can first horizontally stretch by 2, as indicated by the 1 2 on the inside of the function. Remember that twice the size of 0 is still 0, so the point (0,2) remains at (0,2) while the point (2,0) will stretch to (4,0). See Figure 3.95. Figure 3.95 Next, we horizontally shift left by 2 units, as indicated by x + 2. See Figure 3.96. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 347 Figure 3.96 Last, we vertically shift down by 3 to complete our sketch, as indicated by the β 3 on the outside of the function. See Figure 3.97. Figure 3.97 Access this online resource for additional instruction and practice with transformation of functions. β’ Function Transformations (http://openstaxcollege.org/l/functrans) 348 Chapter 3 Functions 3.5 EXERCISES Verbal 298. When examining the formula of a function that is the result of multiple transformations, how can you tell a horizontal shift from a vertical shift
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? 299. When examining the formula of a function that is the result of multiple transformations, how can you tell a horizontal stretch from a vertical stretch? 300. When examining the formula of a function that is the result of multiple transformations, how can you tell a horizontal compression from a vertical compression? 301. When examining the formula of a function that is the result of multiple transformations, how can you tell a reflection with respect to the x-axis from a reflection with respect to the y-axis? How can you determine whether a function is odd or 302. even from the formula of the function? Algebraic For the following exercises, write a formula for the function obtained when the graph is shifted as described. 303. f (x) = x is shifted up 1 unit and to the left 2 units. 304. unit. 305. units. 306. f (x) = |x| is shifted down 3 units and to the right 1 f (x) = 1 x is shifted down 4 units and to the right 3 f (x) = 1 x2 is shifted up 2 units and to the left 4 units. For the following exercises, describe how the graph of the function is a transformation of the graph of the original function f. 307. y = f (x β 49) 308. y = f (x + 43) 309. y = f (x + 3) 310. y = f (x β 4) 311. y = f (x) + 5 312. y = f (x) + 8 313. y = f (x) β 2 This content is available for free at https://cnx.org/content/col11758/1.5 314. y = f (x) β 7 315. y = f (x β 2) + 3 316. y = f (x + 4) β 1 For the following exercises, determine the interval(s) on which the function is increasing and decreasing. 317. 318. f (x) = 4(x + 1)2 β 5 g(x) = 5(x + 3)2 β 2 319. a(x) = βx + 4 320. k(x) = β 3 x β 1 Graphical For the following exercises, use the graph of f (x) = 2 shown in Figure 3.98 to sketch a graph of each transformation of f (x). x Figure 3.98 321. g(x) = 2 322. h(x) = 2
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x x + 1 β 3 323. w(x) = 2 x β 1 For the following exercises, sketch a graph of the function as a transformation of the graph of one of the toolkit functions. f (t) = (t + 1)2 β 3 324. 325. Chapter 3 Functions 349 h(x) = |x β 1| + 4 326. k(x) = (x β 2)3 β 1 327. m(t) = 3 + t + 2 Numeric 328. Tabular representations for the functions f, g, and Write g(x) and h(x) as below. given h are transformations of f (x). x β2 β1 0 f(x) β2 β1 β3 x β1 0 1 g(x) β2 β1 β3 x β2 β1 0 h(x) β1 0 β 329. Tabular representations for the functions f, g, and Write g(x) and h(x) as below. given h are transformations of f (x). x β2 β1 f(x) β1 β3 0 4 x β3 β2 β1 g(x) β1 β2 β1 h(x) β2 β4 0 3 1 1 2 0 For the following exercises, write an equation for each graphed function by using transformations of the graphs of one of the toolkit functions. 330. 331. 332. 333. 350 Chapter 3 Functions 337. 334. 335. 336. This content is available for free at https://cnx.org/content/col11758/1.5 the following For of exercises, transformations of the square root function to find a formula for each of the functions. graphs use the 338. 339. Chapter 3 Functions 351 For the the following exercises, use the graphs of transformed toolkit functions to write a formula for each of the resulting functions. 340. 341. 342. 343. 347. f (x) = (x β 2)2 348. g(x) = 2x4 349. h(x) = 2x β x3 For the following exercises, describe how the graph of each function is a transformation of the graph of the original function f. 350. g(x) = β f (x) 351. g(x) = f ( β x) 352. g(x) = 4 f (x) 353. g(x) = 6 f (x) 354. g
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(x) = f (5x) 355. g(x) = f (2x) 356. 357. g(x) = f β β xβ β 1 3 g(x) = f β β xβ β 1 5 358. g(x) = 3 f (βx) 359. g(x) = β f (3x) For the following exercises, write a formula for the function g that results when the graph of a given toolkit function is transformed as described. 360. The graph of f (x) = |x| is reflected over the y -axis and horizontally compressed by a factor of 1 4. 361. The graph of f (x) = x is reflected over the x -axis and horizontally stretched by a factor of 2. 362. The graph of f (x) = 1 x2 is vertically compressed by For the following exercises, determine whether the function is odd, even, or neither. a factor of 1 3 units., then shifted to the left 2 units and down 3 344. f (x) = 3x4 345. g(x) = x 346. h(x) = 1 x + 3x 363. The graph of f (x) = 1 x is vertically stretched by a factor of 8, then shifted to the right 4 units and up 2 units. 364. 352 Chapter 3 Functions Figure 3.99 375. g(x) = f (x) β 2 376. g(x) = β f (x) 377. g(x) = f (x + 1) 378. g(x) = f (x β 2) The graph of f (x) = x2 is vertically compressed by a factor of 1 2, then shifted to the right 5 units and up 1 unit. 365. The graph of f (x) = x2 is horizontally stretched by a factor of 3, then shifted to the left 4 units and down 3 units. For the following exercises, describe how the formula is a transformation of a toolkit function. Then sketch a graph of the transformation. 366. 367. g(x) = 4(x + 1)2 β 5 g(x) = 5(x + 3)2 β 2 368. h(x) = β 2|x β 4| + 3 369. k(x) = β 3 x β 1 370. 371. 372.
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373. m(x) = 1 2 x3 n(x) = 1 3|x β 2| x(x) = q(x) = 3 xβ β β β 1 4 + 1 374. a(x) = βx + 4 For the following exercises, use the graph in Figure 3.99 to sketch the given transformations. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 353 3.6 | Absolute Value Functions Learning Objectives In this section you will: 3.6.1 Graph an absolute value function. 3.6.2 Solve an absolute value equation. Figure 3.100 Distances in deep space can be measured in all directions. As such, it is useful to consider distance in terms of absolute values. (credit: "s58y"/Flickr) Until the 1920s, the so-called spiral nebulae were believed to be clouds of dust and gas in our own galaxy, some tens of thousands of light years away. Then, astronomer Edwin Hubble proved that these objects are galaxies in their own right, at distances of millions of light years. Today, astronomers can detect galaxies that are billions of light years away. Distances in the universe can be measured in all directions. As such, it is useful to consider distance as an absolute value function. In this section, we will continue our investigation of absolute value functions. Understanding Absolute Value Recall that in its basic form f (x) = |x|, the absolute value function is one of our toolkit functions. The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. Algebraically, for whatever the input value is, the output is the value without regard to sign. Knowing this, we can use absolute value functions to solve some kinds of real-world problems. Absolute Value Function The absolute value function can be defined as a piecewise function f (x) = |x| = x if x β₯ 0 β§ β¨ βx if x < 0 β© Example 3.69 Using Absolute Value to Determine Resistance Electrical parts, such as resistors and capacitors, come with specified values of their operating parameters: resistance, capacitance, etc. However, due to imprecision in manufacturing, the actual values of these parameters 354 Chapter 3 Functions vary somewhat from piece to piece, even when they are supposed to
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be the same. The best that manufacturers can do is to try to guarantee that the variations will stay within a specified range, often Β±1%, Β± 5%, or Β± 10%. Suppose we have a resistor rated at 680 ohms, Β± 5%. Use the absolute value function to express the range of possible values of the actual resistance. Solution We can find that 5% of 680 ohms is 34 ohms. The absolute value of the difference between the actual and nominal resistance should not exceed the stated variability, so, with the resistance R in ohms, |R β 680| β€ 34 Students who score within 20 points of 80 will pass a test. Write this as a distance from 80 using absolute 3.43 value notation. Graphing an Absolute Value Function The most significant feature of the absolute value graph is the corner point at which the graph changes direction. This point is shown at the origin in Figure 3.101. Figure 3.101 Figure 3.102 shows the graph of y = 2|x β 3| + 4. The graph of y = |x| has been shifted right 3 units, vertically stretched by a factor of 2, and shifted up 4 units. This means that the corner point is located at (3, 4) for this transformed function. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 355 Figure 3.102 Example 3.70 Writing an Equation for an Absolute Value Function Given a Graph Write an equation for the function graphed in Figure 3.103. Figure 3.103 Solution The basic absolute value function changes direction at the origin, so this graph has been shifted to the right 3 units and down 2 units from the basic toolkit function. See Figure 3.104. 356 Chapter 3 Functions Figure 3.104 We also notice that the graph appears vertically stretched, because the width of the final graph on a horizontal line is not equal to 2 times the vertical distance from the corner to this line, as it would be for an unstretched absolute value function. Instead, the width is equal to 1 times the vertical distance as shown in Figure 3.105. Figure 3.105 From this information we can write the equation f (x) = 2|x β 3| β 2, f (x) = |2(x β 3)| β 2, treating the stretch as a vertical stretch,or treating the stretch as a horizontal compression. Analysis Note that these equations are algebraically equivalentβthe stretch for
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an absolute value function can be written interchangeably as a vertical or horizontal stretch or compression. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 357 If we couldnβt observe the stretch of the function from the graphs, could we algebraically determine it? Yes. If we are unable to determine the stretch based on the width of the graph, we can solve for the stretch factor by putting in a known pair of values for x and f (x). Now substituting in the point (1, 2) f (x) = a|x β 3| β 2 2 = a|1 β 3| β 2 4 = 2a a = 2 Write the equation for the absolute value function that is horizontally shifted left 2 units, is vertically 3.44 flipped, and vertically shifted up 3 units. Do the graphs of absolute value functions always intersect the vertical axis? The horizontal axis? Yes, they always intersect the vertical axis. The graph of an absolute value function will intersect the vertical axis when the input is zero. No, they do not always intersect the horizontal axis. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. It is possible for the absolute value function to intersect the horizontal axis at zero, one, or two points (see Figure 3.106). Figure 3.106 (a) The absolute value function does not intersect the horizontal axis. (b) The absolute value function intersects the horizontal axis at one point. (c) The absolute value function intersects the horizontal axis at two points. Solving an Absolute Value Equation In Other Type of Equations, we touched on the concepts of absolute value equations. Now that we understand a little more about their graphs, we can take another look at these types of equations. Now that we can graph an absolute value function, we will learn how to solve an absolute value equation. To solve an equation such as 8 = |2x β 6|, we notice that the absolute value will be equal to 8 if the quantity inside the absolute value is 8 or -8. This leads to two different equations we can solve independently. 2x β 6 = 8 2x = 14 x = 7 or 2x β 6 = β8 2x = β2 x = β1 Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are
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at a specified distance from a given reference point. 358 Chapter 3 Functions An absolute value equation is an equation in which the unknown variable appears in absolute value bars. For example, |x| = 4, |2x β 1| = 3, or |5x + 2| β 4 = 9 Solutions to Absolute Value Equations For real numbers A and B, an equation of the form |A| = B, with B β₯ 0, will have solutions when A = B or A = β B. If B < 0, the equation |A| = B has no solution. Given the formula for an absolute value function, find the horizontal intercepts of its graph. 1. Isolate the absolute value term. 2. Use |A| = B to write A = B or βA = B, assuming B > 0. 3. Solve for x. Example 3.71 Finding the Zeros of an Absolute Value Function For the function f (x) = |4x + 1| β 7, find the values of x such that f (x) = 0. Solution 0 = |4x + 1| β 7 7 = |4x + 1| 7 = 4x + 1 6 = 4x x = 6 4 = 1.5 or β7 = 4x + 1 β8 = 4x x = β8 4 = β 2 Substitute 0 for f (x). Isolate the absolute value on one side of the equation. Break into two separate equations and solve. The function outputs 0 when x = 3 2 or x = β 2. See Figure 3.107. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 359 Figure 3.107 3.45 For the function f (x) = |2x β 1| β 3, find the values of x such that f (x) = 0. Should we always expect two answers when solving |A| = B? No. We may find one, two, or even no answers. For example, there is no solution to 2 + |3x β 5| = 1. Access these online resources for additional instruction and practice with absolute value. β’ Graphing Absolute Value Functions (http://openstaxcollege.org/l/graphabsvalue) β’ Graphing Absolute Value Functions 2 (http://openstaxcollege.org/l/graphabsvalue2) 360 Chapter 3 Functions 3.6 EXERCIS
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ES Verbal 379. How do you solve an absolute value equation? How can you tell whether an absolute value function 380. has two x-intercepts without graphing the function? 381. When solving an absolute value function, the isolated absolute value term is equal to a negative number. What does that tell you about the graph of the absolute value function? 382. How can you use the graph of an absolute value function to determine the x-values for which the function values are negative? Algebraic Describe all numbers x that are at a distance of 4 383. from the number 8. Express this set of numbers using absolute value notation. 384. Describe all numbers x that are at a distance of 1 2 from the number β4. Express this set of numbers using absolute value notation. Describe the situation in which the distance that point 385. x is from 10 is at least 15 units. Express this set of numbers using absolute value notation. Find all function values f (x) such that the distance 386. from f (x) to the value 8 is less than 0.03 units. Express this set of numbers using absolute value notation. For the following exercises, find the x- and y-intercepts of the graphs of each function. 387. f (x) = 4|x β 3| + 4 388. f (x) = β 3|x β 2| β 1 389. f (x) = β 2|x + 1| + 6 390. 391. 392. 393. f (x) = β 5|x + 2| + 15 f (x) = 2|x β 1| β 6 f (x) = | β 2x + 1| β 13 f (x) = β |x β 9| + 16 Graphical For the following exercises, graph the absolute value function. Plot at least five points by hand for each graph. This content is available for free at https://cnx.org/content/col11758/1.5 394. y = |x β 1| 395. y = |x + 1| 396. y = |x| + 1 For the following exercises, graph the given functions by hand. 397. y = |x| β 2 398. y = β |x| 399. y = β |x| β 2 400. y = β |x β 3| β 2 401. 402. 403. f (x) = β |
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x β 1| β 2 f (x) = β |x + 3| + 4 f (x) = 2|x + 3| + 1 404. f (x) = 3|x β 2| + 3 405. f (x) = |2x β 4| β 3 406. f (x) = |3x + 9| + 2 407. f (x) = β |x β 1| β 3 408. f (x) = β |x + 4| β 3 409. f (x) = 1 2|x + 4| β 3 Technology 410. Use a graphing utility to graph f (x) = 10|x β 2| on Identify the corresponding the viewing window [0, 4]. range. Show the graph. a Use graphing 411. f (x) = β 100|x| + 100 on the viewing window β‘ Identify the corresponding range. Show the graph. utility to graph β¦. β£β5, 5β€ For the following exercises, graph each function using a graphing utility. Specify the viewing window. 412. 413. f (x) = β 0.1|0.1(0.2 β x)| + 0.3 f (x) = 4Γ109|x β β β5Γ109β β | + 2Γ109 361 Chapter 3 Functions Extensions For the following exercises, solve the inequality. If possible, find all values of a such that there are no 414. x- intercepts for f (x) = 2|x + 1| + a. If possible, find all values of a such that there are no 415. y -intercepts for f (x) = 2|x + 1| + a. Real-World Applications 416. line. Cities A and B are on the same east-west Assume that city A is located at the origin. If the distance from city A to city B is at least 100 miles and x represents the distance from city B to city A, express this using absolute value notation. 417. The true proportion p of people who give a favorable rating to Congress is 8% with a margin of error of 1.5%. Describe this statement using an absolute value equation. Students who score within 18 points of the number 82 418. will pass a particular test. Write this statement using absolute value notation and use the variable x for the score
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. 419. A machinist must produce a bearing that is within 0.01 inches of the correct diameter of 5.0 inches. Using x as the diameter of the bearing, write this statement using absolute value notation. 420. The tolerance for a ball bearing is 0.01. If the true diameter of the bearing is to be 2.0 inches and the measured value of the diameter is x inches, express the tolerance using absolute value notation. 362 Chapter 3 Functions 3.7 | Inverse Functions Learning Objectives In this section, you will: 3.7.1 Verify inverse functions. 3.7.2 Determine the domain and range of an inverse function, and restrict the domain of a function to make it one-to-one. 3.7.3 Find or evaluate the inverse of a function. 3.7.4 Use the graph of a one-to-one function to graph its inverse function on the same axes. A reversible heat pump is a climate-control system that is an air conditioner and a heater in a single device. Operated in one direction, it pumps heat out of a house to provide cooling. Operating in reverse, it pumps heat into the building from the outside, even in cool weather, to provide heating. As a heater, a heat pump is several times more efficient than conventional electrical resistance heating. If some physical machines can run in two directions, we might ask whether some of the function βmachinesβ we have been studying can also run backwards. Figure 3.108 provides a visual representation of this question. In this section, we will consider the reverse nature of functions. Figure 3.108 Can a function βmachineβ operate in reverse? Verifying That Two Functions Are Inverse Functions Suppose a fashion designer traveling to Milan for a fashion show wants to know what the temperature will be. He is not familiar with the Celsius scale. To get an idea of how temperature measurements are related, he asks his assistant, Betty, to convert 75 degrees Fahrenheit to degrees Celsius. She finds the formula and substitutes 75 for F to calculate C = 5 9 (F β 32) (75 β 32) β 24Β°C 5 9 Knowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, he sends his assistant the weekβs weather forecast from Figure 3.109 for Milan, and asks her to convert all of the temperatures to degrees Fahrenheit. This content is available for free at https://cnx.org/content/col11
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758/1.5 Chapter 3 Functions 363 Figure 3.109 At first, Betty considers using the formula she has already found to complete the conversions. After all, she knows her algebra, and can easily solve the equation for F after substituting a value for C. For example, to convert 26 degrees Celsius, she could write (F β 32) 26 = 5 9 = F β 32 26 β
9 5 F = 26 β
9 5 + 32 β 79 After considering this option for a moment, however, she realizes that solving the equation for each of the temperatures will be awfully tedious. She realizes that since evaluation is easier than solving, it would be much more convenient to have a different formula, one that takes the Celsius temperature and outputs the Fahrenheit temperature. The formula for which Betty is searching corresponds to the idea of an inverse function, which is a function for which the input of the original function becomes the output of the inverse function and the output of the original function becomes the input of the inverse function. Given a function f (x), we represent its inverse as f β1(x), read as β f inverse of x.β The raised β1 is part of the notation. It is not an exponent; it does not imply a power of β1. In other words, f β1(x) does not mean 1 f (x) because 1 f (x) is the reciprocal of f and not the inverse. The βexponent-likeβ notation comes from an analogy between function composition and multiplication: just as aβ1 a = 1 (1 is the identity element for multiplication) for any nonzero number a, so f β1 β f equals the identity function, that is, β β f β1 β f β β (x) = f β1 β β f (x)β β = f β1 (y) = x This holds for all x in the domain of f. Informally, this means that inverse functions βundoβ each other. However, just as zero does not have a reciprocal, some functions do not have inverses. Given a function f (x), we can verify whether some other function g(x) is the inverse of f (x) by checking whether either g( f (x)) = x or f (g(x)) = x is true. We can test whichever equation is
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more convenient to work with because they are logically equivalent (that is, if one is true, then so is the other.) For example, y = 4x and y = 1 4 x are inverse functions. and β f β1 β f β β β (x) = f β1 (4x) = 1 4 (4x) = x β f β f β1β β β (x) = f β β 1 4 xβ β = 4 β β xβ β = x 1 4 364 Chapter 3 Functions A few coordinate pairs from the graph of the function y = 4x are (β2, β8), (0, 0), and (2, 8). A few coordinate pairs from the graph of the function y = 1 4 x are (β8, β2), (0, 0), and (8, 2). If we interchange the input and output of each coordinate pair of a function, the interchanged coordinate pairs would appear on the graph of the inverse function. Inverse Function For any one-to-one function f (x) = y, a function f β1 (x) is an inverse function of f if f β1(y) = x. This can also be written as f β1( f (x)) = x for all x in the domain of f. It also follows that f ( f β1(x)) = x for all x in the domain of f β1 if f β1 is the inverse of f. The notation f β1 is read β f so we will often write f β1(x), which we read as β f inverse of x.β Keep in mind that inverse.β Like any other function, we can use any variable name as the input for f β1, f β1(x) β 1 f (x) and not all functions have inverses. Example 3.72 Identifying an Inverse Function for a Given Input-Output Pair If for a particular one-to-one function f (2) = 4 and f (5) = 12, what are the corresponding input and output values for the inverse function? Solution The inverse function reverses the input and output quantities, so if f (2) = 4, then f β1(4) = 2; f (5) =
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12, then f β1(12) = 5. Alternatively, if we want to name the inverse function g, then g(4) = 2 and g(12) = 5. Analysis Notice that if we show the coordinate pairs in a table form, the input and output are clearly reversed. See Table 3.42. β βx, f(x)β β β βx, g(x)β β (2, 4) (4, 2) (5, 12) (12, 5) Table 3.42 3.46 Given that hβ1(6) = 2, what are the corresponding input and output values of the original function h? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 365 Given two functions f(x) and g(x), test whether the functions are inverses of each other. 1. Determine whether f (g(x)) = x or g( f (x)) = x. 2. If either statement is true, then both are true, and g = f β1 and f = gβ1. If either statement is false, then both are false, and g β f β1 and f β gβ1. Example 3.73 Testing Inverse Relationships Algebraically If f (x) = 1 x + 2 and g(x) = 1 x β 2, is g = f β1? Solution so g( f (x)) = β This is enough to answer yes to the question, but we can also verify the other formula. g = f β1 and f = gβ1 f (g(x)) = Analysis Notice the inverse operations are in reverse order of the operations from the original function. 3.47 If f (x) = x3 β 4 and g(x) = x + 4, is g = f β1? 3 Example 3.74 Determining Inverse Relationships for Power Functions If f (x) = x3 (the cube function) and g(x) = 1 3 x, is g = f β1? Solution No, the functions are not inverses. βg(x)β f β β = x3 27 β x 366 Analysis Chapter 3 Functions The correct inverse to the cube is, of course,
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the cube root x3 = x multiplier. 1 3, that is, the one-third is an exponent, not a 3.48 If f (x) = (x β 1)3 and g(x) = x3 + 1, is g = f β1? Finding Domain and Range of Inverse Functions The outputs of the function f are the inputs to f β1, so the range of f is also the domain of f β1. Likewise, because the inputs to f are the outputs of f β1, the domain of f is the range of f β1. We can visualize the situation as in Figure 3.110. Figure 3.110 Domain and range of a function and its inverse When a function has no inverse function, it is possible to create a new function where that new function on a limited domain does have an inverse function. For example, the inverse of f (x) = x is f β1(x) = x2, because a square βundoesβ a square root; but the square is only the inverse of the square root on the domain [0, β), since that is the range of f (x) = x. We can look at this problem from the other side, starting with the square (toolkit quadratic) function f (x) = x2. If we want to construct an inverse to this function, we run into a problem, because for every given output of the quadratic function, there are two corresponding inputs (except when the input is 0). For example, the output 9 from the quadratic function corresponds to the inputs 3 and β3. But an output from a function is an input to its inverse; if this inverse input corresponds to more than one inverse output (input of the original function), then the βinverseβ is not a function at all! To put it differently, the quadratic function is not a one-to-one function; it fails the horizontal line test, so it does not have an inverse function. In order for a function to have an inverse, it must be a one-to-one function. In many cases, if a function is not one-to-one, we can still restrict the function to a part of its domain on which it is oneto-one. For example, we can make a restricted version of the square function f (x) = x2 with its range limited to [0, β), which is
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a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function). If f (x) = (x β 1)2 on [1, β), then the inverse function is f β1(x) = x + 1. β’ The domain of f = range of f β1 = [1, β). β’ The domain of f β1 = range of f = [0, β). Is it possible for a function to have more than one inverse? No. If two supposedly different functions, say, g and h, both meet the definition of being inverses of another function f, then you can prove that g = h. We have just seen that some functions only have inverses if we restrict the domain of the original function. In these cases, there may be more than one way to restrict the domain, leading to different inverses. However, on any one domain, the original function still has only one unique inverse. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 367 Domain and Range of Inverse Functions The range of a function f (x) is the domain of the inverse function f β1(x). The domain of f (x) is the range of f β1(x). Given a function, find the domain and range of its inverse. 1. 2. If the function is one-to-one, write the range of the original function as the domain of the inverse, and write the domain of the original function as the range of the inverse. If the domain of the original function needs to be restricted to make it one-to-one, then this restricted domain becomes the range of the inverse function. Example 3.75 Finding the Inverses of Toolkit Functions Identify which of the toolkit functions besides the quadratic function are not one-to-one, and find a restricted domain on which each function is one-to-one, if any. The toolkit functions are reviewed in Table 3.43. We restrict the domain in such a fashion that the function assumes all y-values exactly once. Constant Identity Quadratic Cubic Reciprocal f (x) = c f (x) = x f (x) = x2 f (x) = x3 f (x) = 1 x Reciprocal squared Cube root Square root Absolute value f (
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x) = x3 f (x) = x f (x) = |x| f (x) = 1 x2 Table 3.43 Solution The constant function is not one-to-one, and there is no domain (except a single point) on which it could be oneto-one, so the constant function has no inverse. The absolute value function can be restricted to the domain [0, β), where it is equal to the identity function. The reciprocal-squared function can be restricted to the domain (0, β). Analysis We can see that these functions (if unrestricted) are not one-to-one by looking at their graphs, shown in Figure 3.111. They both would fail the horizontal line test. However, if a function is restricted to a certain domain so that it passes the horizontal line test, then in that restricted domain, it can have an inverse. 368 Chapter 3 Functions Figure 3.111 (a) Absolute value (b) Reciprocal square 3.49 The domain of function f is (1, β) and the range of function f is (ββ, β2). Find the domain and range of the inverse function. Finding and Evaluating Inverse Functions Once we have a one-to-one function, we can evaluate its inverse at specific inverse function inputs or construct a complete representation of the inverse function in many cases. Inverting Tabular Functions Suppose we want to find the inverse of a function represented in table form. Remember that the domain of a function is the range of the inverse and the range of the function is the domain of the inverse. So we need to interchange the domain and range. Each row (or column) of inputs becomes the row (or column) of outputs for the inverse function. Similarly, each row (or column) of outputs becomes the row (or column) of inputs for the inverse function. Example 3.76 Interpreting the Inverse of a Tabular Function A function f (t) is given in Table 3.44, showing distance in miles that a car has traveled in t minutes. Find and interpret f β1(70). t (minutes) 30 50 70 90 f(t) (miles) 20 40 60 70 Table 3.44 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 369 Solution The inverse function takes an output of f and returns an input for f. So in the
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expression f β1(70), 70 is an output value of the original function, representing 70 miles. The inverse will return the corresponding input of the original function f, 90 minutes, so f β1(70) = 90. The interpretation of this is that, to drive 70 miles, it took 90 minutes. Alternatively, recall that the definition of the inverse was that if f (a) = b, then f β1(b) = a. By this definition, if we are given f β1(70) = a, then we are looking for a value a so that f (a) = 70. In this case, we are looking for a t so that f (t) = 70, which is when t = 90. 3.50 Using Table 3.45, find and interpret (a) f (60), and (b) f β1(60). t (minutes) 30 50 60 70 90 f(t) (miles) 20 40 50 60 70 Table 3.45 Evaluating the Inverse of a Function, Given a Graph of the Original Function We saw in Functions and Function Notation that the domain of a function can be read by observing the horizontal extent of its graph. We find the domain of the inverse function by observing the vertical extent of the graph of the original function, because this corresponds to the horizontal extent of the inverse function. Similarly, we find the range of the inverse function by observing the horizontal extent of the graph of the original function, as this is the vertical extent of the inverse function. If we want to evaluate an inverse function, we find its input within its domain, which is all or part of the vertical axis of the original functionβs graph. Given the graph of a function, evaluate its inverse at specific points. 1. Find the desired input on the y-axis of the given graph. 2. Read the inverse functionβs output from the x-axis of the given graph. Example 3.77 Evaluating a Function and Its Inverse from a Graph at Specific Points A function g(x) is given in Figure 3.112. Find g(3) and gβ1(3). 370 Chapter 3 Functions Figure 3.112 Solution To evaluate g(3), we find 3 on the x-axis and find the corresponding output value on the y-axis. The point (3, 1) tells us that g(3) = 1. To evaluate gβ1(3), recall that
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by definition gβ1(3) means the value of x for which g(x) = 3. By looking for the output value 3 on the vertical axis, we find the point (5, 3) on the graph, which means g(5) = 3, so by definition, gβ1(3) = 5. See Figure 3.113. Figure 3.113 3.51 Using the graph in Figure 3.113, (a) find gβ1(1), and (b) estimate gβ1(4). Finding Inverses of Functions Represented by Formulas Sometimes we will need to know an inverse function for all elements of its domain, not just a few. If the original function is given as a formulaβfor example, y as a function of xβ we can often find the inverse function by solving to obtain x as a function of y. Given a function represented by a formula, find the inverse. 1. Make sure f is a one-to-one function. 2. Solve for x. 3. Interchange x and y. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 371 Example 3.78 Inverting the Fahrenheit-to-Celsius Function Find a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature. Solution C = 5 9 (F β 32) (F β 32) C = 5 9 = F β 32 C β
9 5 F = 9 5 C + 32 By solving in general, we have uncovered the inverse function. If then C = h(F) = 5 9 (F β 32), F = hβ1(C) = 9 5 C + 32 In this case, we introduced a function h to represent the conversion because the input and output variables are descriptive, and writing C β1 could get confusing. 3.52 Solve for x in terms of y given y = 1 3 (x β 5). Example 3.79 Solving to Find an Inverse Function Find the inverse of the function f (x) = 2 x β 3 + 4. Solution or f β1 (x) = 2 x = + 3 + 3. x β 4 Set up an equation. Subtract 4 from both sides. Multiply both sides by x β 3 and divide by y β 4. Add 3 to both sides. So f β1 (y) = 2 y β 4 Analysis 372
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Chapter 3 Functions The domain and range of f exclude the values 3 and 4, respectively. f and f β1 are equal at two points but are not the same function, as we can see by creating Table 3.45. x f(x) 1 3 2 2 5 5 f β1(y) y Table 3.45 Example 3.80 Solving to Find an Inverse with Radicals Find the inverse of the function f (x) = 2 + x β 4. Solution y = 2 + x β 4 (y β 2)2 = x β 4 x = (y β 2)2 + 4 So f β1 (x) = (x β 2)2 + 4. The domain of f is [4, β). Notice that the range of f is [2, β), so this means that the domain of the inverse function f β1 is also [2, β). Analysis The formula we found for f β1 (x) looks like it would be valid for all real x. However, f β1 itself must have an inverse (namely, f ) so we have to restrict the domain of f β1 to [2, β) in order to make f β1 a one-to-one function. This domain of f β1 is exactly the range of f. 3.53 What is the inverse of the function f (x) = 2 β x? State the domains of both the function and the inverse function. Finding Inverse Functions and Their Graphs Now that we can find the inverse of a function, we will explore the graphs of functions and their inverses. Let us return to the quadratic function f (x) = x2 restricted to the domain [0, β), on which this function is one-to-one, and graph it as in Figure 3.114. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 373 Figure 3.114 Quadratic function with domain restricted to [0, β). Restricting the domain to [0, β) makes the function one-to-one (it will obviously pass the horizontal line test), so it has an inverse on this restricted domain. We already know that the inverse of the toolkit quadratic function is the square root function, that is, f β1(x) = x. What happens if we graph both f and f β1
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on the same set of axes, using the x- axis for the input to both f and f β1? We notice a distinct relationship: The graph of f β1(x) is the graph of f (x) reflected about the diagonal line y = x, which we will call the identity line, shown in Figure 3.115. Figure 3.115 Square and square-root functions on the nonnegative domain This relationship will be observed for all one-to-one functions, because it is a result of the function and its inverse swapping inputs and outputs. This is equivalent to interchanging the roles of the vertical and horizontal axes. Example 3.81 Finding the Inverse of a Function Using Reflection about the Identity Line Given the graph of f (x) in Figure 3.116, sketch a graph of f β1(x). 374 Chapter 3 Functions Figure 3.116 Solution This is a one-to-one function, so we will be able to sketch an inverse. Note that the graph shown has an apparent domain of (0, β) and range of (ββ, β), so the inverse will have a domain of (ββ, β) and range of (0, β). If we reflect this graph over the line y = x, the point (1, 0) reflects to (0, 1) and the point (4, 2) reflects to (2, 4). Sketching the inverse on the same axes as the original graph gives Figure 3.117. Figure 3.117 The function and its inverse, showing reflection about the identity line 3.54 Draw graphs of the functions f and f β1 from Example 3.79. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 375 Is there any function that is equal to its own inverse? Yes. If f = f β1, then f β β f (x)β function does, and so does the reciprocal function, because β = x, and we can think of several functions that have this property. The identity Any function f (x) = c β x, where c is a constant, is also equal to its own inverse. = x 1 1 x Access these online resources for additional instruction and practice with inverse functions. β’ Inverse Functions (http://openstaxcollege.org/l/inversefunction) β’ One-to-one Functions (http
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://openstaxcollege.org/l/onetoone) β’ Inverse Function Values Using Graph (http://openstaxcollege.org/l/inversfuncgraph) β’ Restricting the Domain and Finding the Inverse (http://openstaxcollege.org/l/ restrictdomain) this website (http://openstaxcollege.org/l/PreCalcLPC01) Visit Learningpod. for additional practice questions from 376 Chapter 3 Functions 3.7 EXERCISES Verbal Describe why the horizontal line test is an effective 421. way to determine whether a function is one-to-one? Why do we restrict 422. f (x) = x2 to find the functionβs inverse? the domain of the function 423. Can a function be its own inverse? Explain. Are one-to-one functions either always increasing or 424. always decreasing? Why or why not? How do you find the inverse of a function 425. algebraically? Algebraic Show that 426. inverse for all real numbers a. the function f (x) = a β x is its own b. What does the answer tell us about the relationship between f (x) and g(x)? For the following exercises, use function composition to verify that f (x) and g(x) are inverse functions. 437. 438. 3 f (x) = x β 1 and g(x) = x3 + 1 f (x) = β 3x + 5 and g(x) = x β 5 β3 Graphical For the following exercises, use a graphing utility to determine whether each function is one-to-one. 439. f (x) = x 440. 3 f (x) = 3x + 1 For the following exercises, find f β1(x) for each function. 441. f (x) = β5x + 1 442. f (x) = x3 β 27 For the following exercises, determine whether the graph represents a one-to-one function. 443. 427. f (x) = x + 3 428. f (x) = x + 5 429. f (x) = 2 β x 430. f (x) = 3 β x 431. f (x) = x x + 2 432. f (x) = 2x + 3 5x + 4 For the following exercises, find a domain on which each function f is
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one-to-one and non-decreasing. Write the domain in interval notation. Then find the inverse of f restricted to that domain. 444. 433. f (x) = (x + 7)2 434. f (x) = (x β 6)2 435. f (x) = x2 β 5 436. Given f (x) = x3 β 5 and g(x) = 2x 1 β x : a. Find f (g(x)) and g( f (x)). This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 377 For the following exercises, use the graph of f shown in Figure 3.118. Numeric For the following exercises, evaluate or solve, assuming that the function f is one-to-one. 453. 454. 455. 456. If f (6) = 7, find f β1(7). If f (3) = 2, find f β1(2). If f β1 (β4) = β 8, find f ( β 8). If f β1 (β2) = β 1, find f ( β 1). For the following exercises, use the values listed in Table 3.46 to evaluate or solve. x f(x) Figure 3.118 445. Find f (0). 446. Solve f (x) = 0. 447. Find f β1 (0). 448. Solve f β1 (x) = 0. For the following exercises, use the graph of the one-to-one function shown in Figure 3.119. Figure 3.119 449. Sketch the graph of f β1. 450. Find f (6) and f β1(2). Table 3.46 If the complete graph of f is shown, find the domain 451. of f. 457. Find f (1). 458. Solve f (x) = 3. 452. f. If the complete graph of f is shown, find the range of 459. Find f β1 (0). 460. Solve f β1 (x) = 7. 378 Chapter 3 Functions 461. Use the tabular representation of f in Table 3.47 to create a table for f β1 (x). x f(x) 3 1 6 4 9 7 13 14 12 16 Table 3.47 Technology For the following exercises, find
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the inverse function. Then, graph the function and its inverse. 462. f (x) = 3 x β 2 463. f (x) = x3 β 1 464. Find the inverse function of f (x) = 1 x β 1. Use a graphing utility to find its domain and range. Write the domain and range in interval notation. Real-World Applications 465. To convert from x degrees Celsius to y degrees Fahrenheit, we use the formula f (x) = 9 5 x + 32. Find the inverse function, if it exists, and explain its meaning. The circumference C of a circle is a function of its 466. radius given by C(r) = 2Οr. Express the radius of a circle as a function of its circumference. Call this function r(C). Find r(36Ο) and interpret its meaning. A car travels at a constant speed of 50 miles per hour. 467. The distance the car travels in miles is a function of time, t, in hours given by d(t) = 50t. Find the inverse function by expressing the time of travel in terms of the distance traveled. Call this function t(d). Find t(180) and interpret its meaning. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 3 Functions 379 CHAPTER 3 REVIEW KEY TERMS absolute maximum the greatest value of a function over an interval absolute minimum the lowest value of a function over an interval average rate of change the difference in the output values of a function found for two values of the input divided by the difference between the inputs composite function input of another the new function formed by function composition, when the output of one function is used as the decreasing function a function is decreasing in some open interval if f (b) < f (a) for any two input values a and b in the given interval where b > a dependent variable an output variable domain the set of all possible input values for a relation even function the y- axis a function whose graph is unchanged by horizontal reflection, f (x) = f ( β x), and is symmetric about function a relation in which each input value yields a unique output value horizontal compression constant b > 1 a transformation that compresses a functionβs graph horizontally, by multiplying the input by a horizontal line test a method of testing whether a function is one-to-one by determining whether any horizontal line intersects the graph more
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