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total household income. We want to do this for every year, adding only that year’s incomes and then collecting all the data in a new column. If w(y) is the wife’s income and h(y) is the husband’s 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 T = h + w 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 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. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 210 CHAPTER 3 FUNCTIONS For two functions f (x) and g(x) with real number outputs, we define new functions f + g, f − g, fg, and relations f _ g by the (f + g)(x) = f (x) + g(x) (f − g)(x) = f (x) − g(x) (fg)(x) = f (x)g(x) f g ) (x) = (# _ f (x)_ g(x) where g(x) ≠ 0 Example 1 Performing Algebraic Operations on Functions g _ ) (x), given f (x) = x − 1 and g(x) = x 2 − 1. Are they the same Find and simplify the functions (g − f )(x) and (# f function? Solution Begin by writing the general form, and then substitute the given functions. (g − f )(x) = g(x) − f (x) (g − f )(x) = x 2 − 1 − (x − 1) (g − f )(x) = x 2 − x (g − f )(x) = x(x − 1) x2 − 1_ x − 1
g(x)_ f (x) g _ ) (x) = (# f g _ ) (x) = (# f g _ ) (x) = (# f g _ ) (x) = x + 1 (# f (x + 1)(x − 1) ____________ x − 1 where x ≠ 1 where x ≠ 1 where x ≠ 1 No, the functions are not the same. g _ ) (x), the condition x ≠ 1 is necessary because when x = 1, the denominator is equal to 0, which makes Note: For (# f the function undefined. Try It #1 Find and simplify the functions (fg)(x) and (f − g)(x). f (x) = x − 1 and g(x) = x 2 − 1 Are they the same function? 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: ( f#∘#g)(x) = f (g(x)) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.4 COMPOSITION OF FUNCTIONS 211 We read the left-hand side as “f composed with g at x,” and the right-hand side as “ f of g of x.” The two sides of the equation have the same mathematical meaning and are equal. The open circle symbol ∘ 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)) ≠ 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 (g(x)). g(x), the output of g is the input of f (f ° g)(x) = f(g(x)) x is the input of g In general, f#∘#g and g#∘#f are different functions. In other words, in many cases f ( g(x)) ≠ g(f (x)) for all x. We will also see that sometimes two functions can be composed only in one specific order. For example, if 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)) = g(x2) = 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 = − #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#∘#g such that (f#∘#g)(x) = f (g(x)) The domain of the composite function f#∘#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 fg is not the same as the function composition f (g(x)), because, in general, f (x)g(x) ≠ f (g(x)). Example 2 Determining whether Composition of Functions is Commutative Using the functions provided, find f (g(x)) and g(f (x)). Determine whether the composition of the functions is commutative. f
(x) = 2x + 1 g(x) = 3 − x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 212 CHAPTER 3 FUNCTIONS Solution Let’s begin by substituting g(x) into f (x). Now we can substitute f (x) into g(x). f (g(x)) = 2(3 − x) + 1 = 6 − 2x + 1 = 7 − 2x g(f (x)) = 3 − (2x + 1) = 3 − 2x − 1 = − 2x + 2 We find that g(f (x)) ≠ f (g(x)), so the operation of function composition is not commutative. Example 3 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 4 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 (g(y)) or g(f (x))? 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 number of hours as the input. Trying to input a number of gallons does not make sense. The expression f (g(y)) 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) as an input value for g(y), where gallons of gas depends on miles driven, does make sense. The expression g(f (x)) makes sense, and will yield the number of gallons of gas used, g, driving a certain number of miles, f (x), in x hours. Q & A… 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. Try It #2 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. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.4 COMPOSITION OF FUNCTIONS 213 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 function’s 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 5 Using a Table to Evaluate a Composite Function Using Table 1, evaluate f (g(3)) and g(f (3)). x 1 2 3 4 f (x) 6 8 3 1 Table 1 g(x) 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 (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 g(f (3)) = g(3) = 2 Table 2 shows the composite functions f#∘#g and g#∘#f as tables. x 3 g(x) 2 f (g(x)) 8 Table 2 f (x) 3 g(f (x)) 2 Try It #3 Using Table 1, 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. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 214 CHAPTER 3 FUNCTIONS How To… 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 6 Using a Graph to Evaluate a Composite Function Using Figure 1, evaluate f (g(1)). g(x) f(x) 7 6 5 4 3 2 1 –1 –2 –3 –4 –5 321 1 –2 –3 –4 –5 321 4 5 6 7 x (a) (b) Figure 1 Solution To evaluate f (g(1)), we start with the inside evaluation. See Figure 2. g(x) f (x) 7 6 5 4
3 2 1 –1 –2 –3 –4 –5 (1, 3) 321 4 5 6 7 x g(1) = 3 Figure 2 7 6 5 4 3 2 1 –1 –2 –3 –4 –5 (3, 6) 321 4 5 6 7 x f (3) = 6 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. f (g(1)) = f (3) We can then evaluate the composite function by looking to the graph of f (x), finding the input of 3 on the x-axis and reading the output value of the graph at this input. Here, f (3) = 6, so f ( g(1)) = 6. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.4 COMPOSITION OF FUNCTIONS 215 Analysis Figure 3 shows how we can mark the graphs with arrows to trace the path from the input value to the output value. g(x) f(x) –10 –8 –6 –4 10 8 6 4 2 –2–2 –4 –6 –8 –10 10 8 6 4 2 –2–2 –4 –6 –8 –10 42 6 8 10 x 42 6 8 10 x –10 –8 –6 –4 Figure 3 Try It #4 Using Figure 1, 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 will calculate the result of a composition f ( g(x)). To do this, we will extend our idea of function evaluation. Recall that, when we evaluate a function like f (t) = t 2 − t, we substitute the value inside the parentheses into the formula wherever we see the input variable. How To… 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 7 Evaluating a Composition of Functions Expressed as Formulas with a Numerical Input Given f (t) = t2 − t and h(x) = 3x + 2, evaluate f (h(1)). 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 − 5 f (h(1)) = 20 Analysis specific numerical values. It makes no difference what the input variables t and x were called in this problem because we evaluated for Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 216 CHAPTER 3 FUNCTIONS Try It #5 Given f (t) = t 2 − t and h(x) = 3x + 2, evaluate a. h(f (2)) b. h(f (−2)) Finding the Domain of a Composite Function As we discussed previously, the domain of a composite function such as f#∘#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 ∘ g. Let us assume we know the domains of the functions f and g separately. If we write the composite function for an input x as f (g(x)), we can see right away that x must be a member of the domain of g in 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 (g(x)) cannot be completed, and the expression is still undefined. Thus the domain of f ∘ 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. domain of
a composite function The domain of a composite function f (g(x)) is the set of those inputs x in the domain of g for which g(x) is in the domain of f. How To… 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#∘#g. Example 8 Finding the Domain of a Composite Function Find the domain of (f#∘#g)(x) where f (x) = and g(x) = 5 ____ x − 1 4 _____ 3x − 2 2 __ Solution The domain of g(x) consists of all real numbers except x =, since that input value would cause us to divide by 3 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 − 2 = 1 4 = 3x − 2 6 = 3x x = 2 2 __ So the domain of f#∘#g is the set of all real numbers except and 2. This means that 3 We can write this in interval notation as 2 __ or x ≠ 2 x ≠ 3 2 2 __ __, 2 ) ∪ (2, ∞) ) ∪ (# (#− ∞, 3 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.4 COMPOSITION OF FUNCTIONS 217 Example 9 Finding the Domain of a Composite Function Involving Radicals Find the domain of (f#∘#g)(x) where f (x) = √ — x + 2 and g(x) = √ — 3 − x Solution Because we cannot take the square root of a negative number, the domain of g is (−∞, 3]. Now we check the domain of the composite function (f#∘#g)(x) = √ — 3 − x + 2 or (f#∘#g)(x) = √ — 5 − x The domain of
this function is (−∞, 5]. To find the domain of f#∘#g, we ask ourselves if there are any further restrictions offered by the domain of the composite function. The answer is no, since (−∞, 3] is a proper subset of the domain of f#∘#g. This means the domain of f#∘#g is the same as the domain of g, namely, (−∞, 3]. Analysis 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#∘#g can contain values that are not in the domain of f, though they must be in the domain of g. Try It #6 Find the domain of (f#∘#g)(x) where f (x) = 1 ____ x − 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 10 Decomposing a Function Write f (x) = √ 5 − x 2 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 − x2 is the inside of the square root. We could then decompose the function as h(x) = 5 − x2 and g(x) = √ — x We can check our answer by recomposing the functions. g(h(x)) = g(5 − x2) = √ — 5 − x2 Try It #7 Write f (x) = 4 __ — 4 + x2 3 − √ as the composition of two functions. Access these online resources for additional instruction and practice with composite functions. • Composite Functions (http://openstaxcollege.org/l/compfunction) • Composite Function Notation Application (http://openstaxcollege.org/l/compfuncnot) • Composite Functions Using Graph
s (http://openstaxcollege.org/l/compfuncgraph) • Decompose Functions (http://openstaxcollege.org/l/decompfunction) • Composite Function Values (http://openstaxcollege.org/l/compfuncvalue) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 218 CHAPTER 3 FUNCTIONS 3.4 SECTION EXERCISES VERBAL 1. 2. f∘g f g 3. 4. f∘g ALGEBRAIC 5. fx=x+xgx=−xf+g 6. fx=−x+xgx=f+g f−gfg, f g f−gfg, f g 7. fx=x +xgx= f+g x f−gfg, f g 9. fx=x gx=√ f g f−gfg, 8. fx= gx= x− f g f+gf−gfg, −x,fi — x− f+g 10. fx=√ — g x gx=x− f 11. fx=x+gx=x− a. fg b. fgx c gfx d. g"∘"gx e. f"∘"f − o fif gxgfx 12. fx=x +gx=√ — x+ 13. fx=√ — x +gx=x + 14. fx=xgx=x+ 15. fx= √ — x gx= x+ x 16. fx= x − x + gx= 17. fx= gx= " ___ x− x + o fif ghx 18. fx=x +gx=x−hx=√ — x x hx=x+ 19. fx=x +gx= x,gx=x− 20. fx= a. f"∘"gx b. f"∘"gx c. g"∘"f x d. g"∘"f x e. ( f g ) x — −x gx=− " 21. fx=√ a. g"∘"f x b. g"∘"f x x
Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.4 SECTION EXERCISES 219 22. fx= gx= − x x +x a. g"∘"f x b. g"∘"f 23. px = " mx=x − — x √ a. px mx b. pmx c. mpx 24. qx= hx=x − — x √ x gx=√ 25. fx= — x− f"∘"gx a. qx hx b. qhx c. hqx, fif xgxhx=f x 26. hx=x+ 27. hx=x− 30. hx=+ √ — x √ 31. hx= _______ x− 28. hx= x − 32. hx= x −− 29. hx= x+ √ 33. hx= _______ x− x+ 34. hx = ( +x −x ) 35. hx=√ — x+ 36. hx=x− 37. hx= √ — x− 38. hx=x + 39. hx= x− 40. hx=( ) x− 41. hx=√ x− x+ GRAPHICAL fFigure 4 g,Figure 5 f(x) – – – x (x) – – – x Figure 4 Figure 5 42. fg 46. ff 43. fg 47. ff 44. gf 48. gg 45. gf 49. gg Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 220 CHAPTER 3 FUNCTIONS f x,Figure 6 gx,Figure 7hx,Figure 8 fx f x f x gx hx Figure 6 Figure 7 Figure 8 50. gf 51. gf 52. fg 53. fg 54. fh 55. hf 56. fgh 57. fgf− NUMERIC fgTable 3 x fx gx Table 3 58. fg 62. ff 59. fg 63. ff 60. gf 64. gg 61. gf 65. gg fgTable 4 − − x f (x) g(x) −
− − Table 4 − − − 66. f"∘"g 69. g"∘"f 67. f"∘"g 70. g"∘"g 68. g"∘"f 71. f"∘"f o fif ggf 72. fx=x+gx=−x 73. fx=x+gx=−x 74. fx=√ — x+gx=−x 75. fx= gx=x+ _ x +" f x=x+gx=x +fi 76. fg 77. fgx 78. g(f− 79. g"∘"g x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.4 SECTION EXERCISES 221 EXTENSIONS f x=x+x= √ — x − 80. f"∘"gxg"∘"fx 81. f"∘"gg"∘"f 82. g"∘"fx 83. f"∘"gx __ 84. fx= x a. f"∘"fx b. f"∘"f xf F x=x+f x=xgx=x + 85. g"∘"fx=Fx 86. f"∘"gx=Fx, fif x=x+x ≥ gx=√ — x − 87. f"∘"gg"∘"f 88. g"∘"faf"∘"ga 89. f"∘"gg"∘"f REAL-WORLD APPLICATIONS 90. ThDp p. Th Cxx a. DC b. CD c. DCx= d. CDp= 92. ff xs. Thff Px x 94. rt=t+ t 96. Thr ___ "V ". π √ VrV= ftt Vt=+t a. rVt b. exact 10 in 91. ThAd d mi em. Th ftt m(t) a. Am b. mA c. Amt= d. mAd= 93. — t+ rt= t= √ 95. ft 97. Th NT=T −T+<T< T Tt=t+t a
. NTt b. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 222 CHAPTER 3 FUNCTIONS LEARNING OBJECTIVES In this section, you will: • ecreate the library of functions. • Graph and describe functions using ertical and horiontalshifts. • Graph and describe functions using reflections about the x-ais and the y-ais. • Graph and describe functions using compressions and stretches. • Combine transformations. • Gien a graph or description write the equation as a transformation of a basic function. • Perform transformations on a gien graph. 3.5 TRANSFORMATION OF FUNCTIONS Figure 1 (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 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 2 for an example. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.5 TRANSFORMATION OF FUNCTIONS 223 f(x) f(x)+1 f(x)
3 4 5 21 x –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 Figure 2 Vertical shift by k = 1 of the cube root function f ( x) = 3 √ — x. To help you visualize the concept of a vertical shi ft, 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 1 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 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. V 300 250 200 150 100 50 – –4 50– 4 8 12 16 20 24 28 t Figure 3 Solution 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 up, as shown in Figure 4. V 300 250 240 200 150 100 50 20 – –4 50– Up 20 4 8 12 16 20 24 28 t Figure 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 224 CHAPTER 3 FUNCTIONS Notice that in Figure 4, 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 1. t V(t) S(t) 0 0 20 8 0 20 17 220 240 19 0 20 24 0 20 10 220 240 Table 1 How To… 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 2 Shifting a Tabular Function Vertically A function f (x) is given in Table 2. Create a table for the function g(x) = f (x) − 3. x f (x) 2 1 4 3 Table 2 6 7 8 11 Solution The formula g (x) = f (x) − 3 tells us that we can find the output values of g by subtracting 3 from the output values of f. For example: Given Given transformation f (2) = 1 g(x) = f (x) − 3 g(2) = f (2) − 3 = 1 − 3 = −2 Subtracting 3 from each f (x) value, we can complete a table of values for g(x) as shown in Table 3. x f (x) g(x) 2 1 −2 4 3 0 Table 3 6 7 4 8 11 8 Analysis As with the earlier vertical shift, notice the input values stay the same and only the output values change. Try It #1 The function h(t) = –4.9t2 + 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-m building. Relate this new height function b(t) to h(t), and then find a formula for b(t). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.5 TRANSFORMATION OF FUNCTIONS 225 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 5. f(x) –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 f(x+1) f(x) 21 3 4 5 x Figure 5 Horizontal shift of the function f (x) = 3 √ — x. 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 − 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 − 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 Adding a Constant to an Input Returning to our building airflow example from Figure 3, 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 6. Notice also that the vents first opened to 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 = −2. That means that the same output values are reached when F (t) = V(t − (−2)) = V
(t + 2). V 300 250 240 200 150 100 50 –2 0 50– – –4 Left 2 4 8 12 16 20 24 28 t Figure 6 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 226 CHAPTER 3 FUNCTIONS Analysis Note that V(t + 2) has the effect of shifting the graph to the left. Horizontal changes or “inside changes” 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). How To… 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 4 Shifting a Tabular Function Horizontally A function f (x) is given in Table 4. Create a table for the function g (x) = f (x − 3). x f (x) 2 1 4 3 Table 4 6 7 8 11 Solution The formula g(x) = f (x − 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 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. We continue with the other values to create Table 5. g (5) = f (5 − 3) = f (2) = 1 x x − 3 f (x) g(x) 5 2 1 1 7 4 3 3 Table 5 9 6 7 7 11 8 11 11 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 7 represents both of the functions. We can see the horizontal shift in each point. y 12 10 8 6 4 2 –2–2 –4 –6 –8 –10 –12 –12 –10 –8 –6 –4 42 6 8 10 12 x f (x) g (x) = f(x – 3) Figure 7 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.5 TRANSFORMATION OF FUNCTIONS 227 Example 5 Identifying a Horizontal Shift of a Toolkit Function Figure 8 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). f (x) 6 5 4 3 2 1 – –1 1– 1 2 3 4 5 6 7 x Figure 8 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 − 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 − 2). f(x) = x2 g(x) = f (x − 2) g (x) = f (x − 2) = (x − 2)2 Analysis To determine whether the shift is +2 or −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 − 2) = f (0) = 0. For this to work, we will need to subtract 2 units from
our input values. Example 6 Interpreting Horizontal versus Vertical Shifts 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. Try It #2 Given the function f (x) = √ Is this a horizontal or a vertical shift? Which way is the graph shifted and by how many units? — x, graph the original function f (x) and the transformation g (x) = f (x + 2) on the same axes. 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-) axis values and shift the function up or down. Horizontal shifts are inside changes that affect the input (x-) axis 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 right or left. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 228 CHAPTER 3 FUNCTIONS How To… 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 7 Graphing Combined Vertical and Horizontal Shifts Given f (x) =. x., sketch a graph of h (x) = f (x + 1) − 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) − 3 is a change to the outside of the function, giving a vertical shift down by 3. The transformation of the graph is illustrated in Figure 9. Let us follow one point of the graph of f (x) =. x.. • The point (0, 0) is transformed first by shifting left 1 unit: (0, 0) → (−1, 0) • The point (−1, 0) is transformed next by shifting down 3 units: (−1, 0) → (−1, −3) y = \|x + 1\| y = \|x\| y = \|x + 1\| – 3 21 1 –1 –2 –3 –4 –5 –5 –4 –3 –2 Figure 10 shows the graph of h. Figure 9 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 y = \|x + 1\| – 3 21 3 4 5 x Figure 10 Try It #3 Given f (x) =. x., sketch a graph of h(x) = f (x − 2) + 4. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.5 TRANSFORMATION OF FUNCTIONS 229 Example 8 Identifying Combined Vertical and Horizontal Shifts Write a formula for the graph shown in Figure 11, which is a transformation of the toolkit square root function. h(x) 6 5 4 3 2 1 – –1 1– 1 2 3 4 5 6 7 x Figure 11 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(x) = f (x − 1) + 2 Analysis Note that this transformation has changed the domain and range of the function. This new graph has domain [1, ∞) and range [2, ∞). Try It #4 1 __ Write a formula for a transformation of the toolkit reciprocal function f (x) = that shifts the function’s graph one x 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 12. y Horizontal reflection f(x) Original function f(–x) x Vertical reflection –f(x) Figure 12 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. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 230 CHAPTER 3 FUNCTIONS reflections Given a function f (x), a new function g(x) = −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 (−x) is a horizontal reflection of the function f (x), sometimes called a reflection about the y-axis. How To… Given a function, reflect the graph both vertically and horizontally. 1. Multiply all outputs by –1 for a vertical reflection. The new graph is a reflection of the original graph about the x-axis. 2. Multiply all inputs by –1 for a horizontal reflection. The new graph is a reflection of the original graph about the y-axis. Example 9 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 13. s(t) 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 t –5 –4 –3 –2 v(t) 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 t Figure 13 Vertical reﬂection of the square root function Because each output value is the opposite of the original output value, we can write V(t) = −s(t) or
V(t) = − √ — t 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. b. Reflecting horizontally means that each input value will be reflected over the vertical axis as shown in Figure 14. s(t) 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 t –5 –4 –3 –2 H(t) 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 t Figure 14 Horizontal reﬂection of the square root function Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.5 TRANSFORMATION OF FUNCTIONS 231 Because each input value is the opposite of the original input value, we can write H(t) = s(−t) or H(t) = √ — −t 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, ∞) and range [0, ∞), the vertical reflection gives the V(t) function the range (−∞, 0] and the horizontal reflection gives the H(t) function the domain (−∞, 0]. Try It #5 Reflect the graph of f (x) = \|x − 1\| a. vertically and b. horizontally. Example 10 Reﬂecting a Tabular Function Horizontally and Vertically A function f (x) is given as Table 6. Create a table for the functions below. a. g(x) = −f (x) b. h(x) = f (−x) x f (x) 2 1 4 3 Table 6 6 7 8 11 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 7. x g (x) 2 –1 6 –7 8 –11 4 –3 Table 7 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 8. x h(x) –2 1 –4 3 Table 8 –6 7 –8 11 Try It #6 A function f (x) is given as Table 9. Create a table for the functions below. x f(x) −2 5 0 10 Table 9 2 15 4 20 a. g(x) = −f (x) b. h(x) = f (−x) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 232 CHAPTER 3 FUNCTIONS Example 11 Applying a Learning Model Equation A common model for learning has an equation similar to k(t) = −2−t + 1, where k is the percentage of mastery that can be achieved after t practice sessions. This is a transformation of the function f (t) = 2t shown in Figure 15. Sketch a graph of k(t). f (t) –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 t Solution This equation combines three transformations into one equation. Figure 15 • A horizontal reflection: f (−t) = 2−t • A vertical reflection: −f (−t) = −2−t • A vertical shift: −f (−t) + 1 = −2−t + 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 the points (0, 1) and (1, 2). 1. First, we apply a horizontal reflection: (0, 1) (−1, 2). 2. Then, we apply a vertical reflection: (0, −1) (1, −2). 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 16, 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. f (t) f (t) –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2
–3 –4 –5 (a) 1 2 3 4 5 t –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 t –5 –4 –3 –2 (b) Figure 16 k(t) 5 4 3 2 1 –1 –1 –2 –3 –4 –5 (c) 21 3 4 5 t Analysis As a model for learning, this function would be limited to a domain of t ≥ 0, with corresponding range [0, 1). Try It #7 Given the toolkit function f (x) = x 2, graph g(x) = −f (x) and h(x) = f (−x). Take note of any surprising behavior for these functions. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.5 TRANSFORMATION OF FUNCTIONS 233 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. 1 _ If the graphs of f (x) = x3 or f (x) = x were reflected over both axes, the result would be the original graph, as shown in Figure 17. y y f (x) = x3 f(x) = x3 y y f (−x) f(−xf (−x) −f (−x5 –5 –4 –4 –3 –3 –2 –2 –1 –1 –1 –1 –2 –2 –3 –3 –4 –4 –5 –5 –5 –5 –4 –4 –3 –3 –2 –2 –1 –1 –1 –1 –2 –2 –3 –3 –4 –4 –5 –5 21 21 3 3 4 4 5 5 x x 21 21 3 3 4 4 5 5 x x –5 –5 –4 –4 –3 –3 –2 –2 –1 –1 –1 –1 –2 –2 –3 –3 –4 –4 –5 –5 (a) (a) (b) (b
) (c) (c) Figure 17 (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) = 2x is neither even nor odd. Also, the only function that is both even and odd is the constant function f (x) = 0. even and odd functions A function is called an even function if for every input x: f (x) = f (−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: f (x) = −f (−x) The graph of an odd function is symmetric about the origin. How To… Given the formula for a function, determine if the function is even, odd, or neither. 1. Determine whether the function satisfies f (x) = f (−x). If it does, it is even. 2. Determine whether the function satisfies f (x) = −f (−x). If it does, it is odd. 3. If the function does not satisfy either rule, it is neither even nor odd. Example 12 Determining whether a Function Is Even, Odd, or Neither Is the function f (x) = x 3 + 2x even, odd, or neither? 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. Let’s begin with the rule for even functions. f (−x) = (−x)3 + 2(−x) = −x 3 − 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 −f (−x) = f (x), this is an odd function. −f (−x) = −(−x 3 − 2x) = x 3 + 2x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 234 CHAPTER 3 FUNCTIONS Analysis Consider the graph of f
in Figure 18. Notice that the graph is symmetric about the origin. For every point (x, y) on the graph, the corresponding point (−x, −y) is also on the graph. For example, (1, 3) is on the graph of f, and the corresponding point (−1, −3) is also on the graph. f (x) f 5 4 3 2 1 –1 –2 –3 –4 –5 (−1, −3) –1 –2 –3 –4 –5 (1, 3) 21 3 4 5 x Figure 18 Try It #8 Is the function f (s) = s 4 + 3s 2 + 7 even, odd, or neither? Graphing Functions Using Stretches and Compressions Adding a constant to the inputs or outputs of a function changed the 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. 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 19 shows a function multiplied by constant factors 2 and 0.5 and the resulting vertical stretch and compression. y Vertical stretch Vertical compression 2 f(x) f(x) 0.5 f(x) x Figure 19 Vertical stretch and compression vertical stretches and compressions Given a function f (x), a new function g(x) = af (x), where a is a constant, is a vertical stretch or vertical compression of the function f (x). • 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. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.5 TRANSFORMATION OF FUNCTIONS 235 How To… 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 13 Graphing a Vertical Stretch A function P(t) models the population of fruit flies. The graph is shown in Figure 20. P(t) 6 5 4 3 2 1 – –1 1– 1 2 3 4 5 6 7 t Figure 20 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 population’s output values are always twice the original function’s output values. Graphically, this is shown in Figure 21. If we choose four reference points, (0, 1), (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) → (0, 2) (3, 3) → (3, 6) (6, 2) → (6, 4) (7, 0) → (7, 0) Q(t) 6 5 4 3 2 1 – –1 1– 1 2 3 4 5 6 7 t Figure 21 Symbolically, the relationship is written as Q(t) = 2P(t) 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. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 236 CHAPTER 3 FUNCTIONS How To… 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 14 Finding a Vertical Compression of a Tabular Function A function f is given as Table 10. Create a table
for the function g(x) = #1 __ f (x). 2 x f (x) 2 1 6 7 8 11 4 3 Table 10 Solution The formula g(x) =# #1 __ f (x) tells us that the output values of g are half of the output values of f with the same 2 inputs. For example, we know that f (4) = 3. Then We do the same for the other values to produce Table 11. (3) =# #3 f (4) =# #1 g(4) =# #1 __ __ __ 2 2 2 x g(x) 2 1 __ 2 4 3 __ 2 Table 11 6 7 __ 2 8 11 __ 2 1 __ Analysis The result is that the function g(x) has been compressed vertically by. Each output value is divided in half, 2 so the graph is half the original height. Try It #9 A function f is given as Table 12. Create a table for the function g(x) =# #3 __ f (x). 4 x f(x) 2 12 4 16 Table 12 6 20 8 0 Example 15 Recognizing a Vertical Stretch The graph in Figure 22 is a transformation of the toolkit function f (x) = x3. Relate this new function g(x) to f (x), and then find a formula for g(x). g(x) –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Figure 22 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.5 TRANSFORMATION OF FUNCTIONS 237 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. the outputs of the function f because g(2) =# #1 1 __ __ f (2). From this we Based on that, it appears that the outputs of g are 4 4 can fairly safely conclude that g(x) =# #1 __ f (x). 4 We can write a formula for g by using the definition of the function f. f (x) =# #1 g(x) =# #1 __ __ x3 4
4 Try It #10 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 function’s 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. Horizontal compression y = x2 y = (0.5 x)2 Horizontal stretch y = (2 x)2 y 10 5 –4 –3 –2 –1 1 2 3 4 5 x –1 Figure 23 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 23. The graph of y = (0.5x)2 is a horizontal stretch of the graph of the function y = x 2 by a factor of 2. The graph of y = (2x)2 is a horizontal compression of the graph of the function y = x 2 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). 1 __ • If b > 1, then the graph will be compressed by. b 1 __ • If 0 < b < 1, then the graph will be stretched by. b • If b < 0, then there will be combination of a horizontal stretch or compression with a horizontal reflection. How To… 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. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 238 CHAPTER 3 FUNCTIONS Example 16 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 24 for a graphical comparison of the original population and the compressed population. f (x) 6 5 4 3 2 1 – –1 1– Original population, P(t) Transformed population, R(t) f (x1 1a) (b) Figure 24 (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 17 Finding a Horizontal Stretch for a Tabular Function 1 __ x ). A function f (x) is given as Table 13. Create a table for the function g(x) = f (# 2 x f (x) 2 1 6 7 8 11 4 3 Table 13 1 __ x ) tells us that the output values for g are the same as the output values for Solution The formula g(x) = f (# 2 the function f at an input half the size. Notice that we do not have enough information to determine g(2) because 1 __ ċ 2 ) = f (1), and we do not have a value for f (1) in our table. Our input values to g will need to be twice as g(2) = f (# 2 large to get inputs for f that we can evaluate. For example, we can determine g(4). We do the same for the other values to produce Table 14. 1 __ ċ 4 ) = f (2) = 1 g(4) = f (# 2 x g(x) 4 1 8 3 12 7 16 11 Table 14 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.5 TRANSFORMATION OF FUNCTIONS 239 Figure 25 shows the graphs of both of these sets of points. –20 –16 –12 –8 y 12 10 8 6 4 2 –4 –2 –4 –6 –8 –10 –12 (a) 84 12 16 20 x –20 –16 –12 –8 Figure 25 y 12 10 8 6 4 2 –
4 –2 –4 –6 –8 –10 –12 (b) 84 12 16 20 x 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. Example 18 Recognizing a Horizontal Compression on a Graph Relate the function g(x) to f (x) in Figure 26. f (x) 6 5 4 3 2 1 – –1 1 Figure 26 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 1 1 __ __ ) = 2. We might also notice that g(2) = f (6), because 6 (# (2, 4), we can see that the x-values have been compressed by 3 3 1 __ and g(1) = f (3). Either way, we can describe this relationship as g(x) = f (3x). This is a horizontal compression by. 3 Analysis Notice that the coefficient needed for a horizontal stretch or compression is the reciprocal of the stretch or 1 1 __ __ x ). in our function: f (# compression. So to stretch the graph horizontally by a scale factor of 4, we need a coefficient of 4 4 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. Try It #11 Write a formula for the toolkit square root function horizontally stretched by a factor of 3. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 240 CHAPTER 3 FUNCTIONS 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 2f (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 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 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. Let’s work through an example. We can factor out a 2. p_ f (bx + px) = (2x + 4)2 f (x) = (2(x + 2))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 af (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 1 __ then horizontally stretch by. b 1_ When combining horizontal transformations written in the form f (b(x + h)), first horizontally stretch by b and then horizontally shift by h. Horizontal and vertical transformations are independent. It does not matter whether horizontal or vertical transformations are performed first. Example 19 Finding a Triple Transformation of a Tabular Function Given Table 15 for the function f (x), create a table of values for the function g(x) = 2f (3x) + 1. x f (x) 6 10 12 14 Table 15 18 15 24 17 Solution There are three steps to this transformation, and we will work from the inside out. Starting with the horizontal 1 1 __ __. See Table 16., which means we multiply each x-value by transformations, f (3x) is a horizontal compression by 3 3 x f (3x) 2 10 4 14 Table 16 6 15 8 17 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION
3.5 TRANSFORMATION OF FUNCTIONS 241 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 17. x 2f (3x) 2 20 4 28 Table 17 6 30 8 34 Finally, we can apply the vertical shift, which will add 1 to all the output values. See Table 18. x g(x) = 2f (3x) + 1 4 29 6 31 8 35 2 21 Table 18 Example 20 Finding a Triple Transformation of a Graph 1 __ Use the graph of f (x) in Figure 27 to sketch a graph of k(x) = f (# x + 1 ) − 3. 2 f (x) –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Figure 27 Solution To simplify, let’s start by factoring out the inside of the function. 1 1 __ __ (x + 2 (# f (# 2 2 1 __ on the inside of the function. By factoring the inside, we can first horizontally stretch by 2, as indicated by the 2 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 28. f (x) –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x Figure 28 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 242 CHAPTER 3 FUNCTIONS Next, we horizontally shift left by 2 units, as indicated by x + 2. See Figure 29. f (x) –6 –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 6 x Figure 29 Last, we vertically shift down by 3 to complete our sketch, as indicated by the −3 on the outside of the function. See Figure 30. f (x) –6 –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 6 x Figure 30 Access this online resource for additional instruction and practice with transformation of functions. • Function Transformations (http://open
staxcollege.org/l/functrans) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.5 SECTION EXERCISES 243 3.5 SECTION EXERCISES VERBAL 1. 2. x y 3. When examining the formula of a function that 4. When examining the formula of a function that is is the result of multiple transformations, how can you tell a horizontal compression from a vertical compression? 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? 5. How can you determine whether a function is odd or even from the formula of the function? ALGEBRAIC ft 6. fx) =√ — x hift 8. fx=!! __ hift x 7. fx=xhift 9. fx= __ x hift f 10. y = f x − 11. y = f x + 12. y = f x + 13. y = f x − 14. y = f x + 15. y = f x + 16. y = f x − 17. y = f x − 18. y = f x − + 19. y = f x + − s — — 20. f x = x + − 21. gx=x+− 22. ax=√ −x+ 23. kx = −√ x − GRAPHICAL f x=xFigure 31 f x 24. gx = x + 25. hx = x − 26. wx = x − 27. f t = t + − 28. hx = \|x − \| + – – – – 29. kx = x − − 30. mt = + √ — Figure 31 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 244 CHAPTER 3 FUNCTIONS NUMERIC 31. 34. 37. 33. 36. 32. 35. y f – – – – –– – – – – x y x f – – – – –– – – – – y f x – – – – –– – – – – – – – – – – – – 38 –– – – – –– – – – – – – – – – –– – – – – fi 39. y x f – – – – ––
– – – – 40. y f – – – – –– – – – – x Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.5 SECTION EXERCISES 245 41. 44. y f – – – – –– – – – – 42. x y f x – – – – –– – – – – 43. y f – – – – –– – – – – x y – – – – –– – – – – f x 45. fx=x 48. fx=x− 46. gx=√x — 49. gx=x 47. hx=__ x +x 50. hx=x−x f 51. gx=−fx 52. gx=f−x 53. gx=fx 54. gx=fx 55. gx=fx 56. gx=fx 59. gx=f−x 60. gx=−fx 57. __ x ) gx=f (! 58. __ x ) gx=f (! g 61. Thfx=∣ x ∣y __ 63. __ Thfx= x __ hift 3 uni 62. 64. Thfx=√x x — Thfx=__ x hift 65. Thfx=x __ hift 66. Thfx=x hift Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 246 CHAPTER 3 FUNCTIONS Th 67. gx=x+− 68. gx=x+− 70. kx=−√x − — 71. mx=!!__ x 73. px=! (!!__ x ) − 74. qx= (!!__ x ) + 69. hx=−∣!x−∣+ 72. nx=!__ x− 75. — ax=√−x+ 78. ()=(+)+ y f x 81. ()= () y f x x x ()=( )+ y 77. ()=(+) 76. 79. f ()=( ) y f 82. ()=( ) y f y f 80. ()= () y f x x x Download the OpenStax text for free at http://cnx.org/content/col11
759/latest. SECTION 3.6 ABSOLUTE VALUE FUNCTIONS 247 LEARNING OBJECTIVES In this section you will: • Graph an absolute value function. • Solve an absolute value equation. 3.6 ABSOLUTE VALUE FUNCTIONS Figure 1 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 x { f (x) = ∣ x ∣ #= −x if x ≥ 0 if x < 0 Example 1 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 vary somewhat from piece to piece, even when they are supposed to be the same. The best that manufacturers can do is to try to guarantee that the variations will stay within a specified range, often ±1%, ±5%, or ±10%. Suppose we have a resistor rated at 680 ohms, ±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 − 680 \| ≤ 34 Try It #1 Students who score within 20 points of 80 will pass a test. Write this as a distance from 80 using absolute value notation. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 248 CHAPTER 3 FUNCTIONS 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 2. –5 –4 –3 –2 y = \|x\| 21 1–1 –2 –3 –4 –5 Figure 2 Figure 3 shows the graph of y = 2\| x − 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. y = 2\|x − 3\| Vertical stretch y = \|x\| y 10 \|x − 3\| + 4 Up 4 y = \|x − 3\| Right 3 –6 –5 –4 –3 –2 –1–1 1 2 3 4 5 6 x Figure 3 Example 2 Writing an Equation for an Absolute Value Function Given a Graph Write an equation for the function graphed in Figure 4. y 5 4 3 2 1 –1–1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x Figure 4 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 5. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.6 ABSOLUTE VALUE FUNCTIONS 249 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x (3, –2) Figure 5 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 6. Ratio 2/1 y
5 4 3 2 1 –5 –4 –3 –2 –1–1 –2 –3 –4 –5 Ratio 1/1 x 21 3 4 5 (3, −2) From this information we can write the equation Figure 6 f (x) = 2\| x − 3 \|#− 2, treating the stretch as a vertical stretch, or f (x) = \| 2(x − 3) \|#− 2, treating the stretch as a horizontal compression. Analysis Note that these equations are algebraically equivalent—the stretch for an absolute value function can be written interchangeably as a vertical or horizontal stretch or compression. Q & A… If we couldn’t 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 − 3 \|#− 2 2 = a\| 1 − 3 \|#− 2 4 = 2a a = 2 Try It #2 Write the equation for the absolute value function that is horizontally shifted left 2 units, is vertically flipped, and vertically shifted up 3 units. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 250 CHAPTER 3 FUNCTIONS Q & A… 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 7). y 5 4 3 2 1 –1 –1 –2 –3 –4 (a) –5 –4 –3 –2 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1–1 –2 –3 –4 (b) 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1–1 –2 –3 –4 (c) 21 3 4 5 x Figure 7 (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 Types 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 − 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 − 6 = 8 or 2x − 6 = −8 2x = −2 x = −1 2x =#14 x = 7 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 at a specified distance from a given reference point. An absolute value equation is an equation in which the unknown variable appears in absolute value bars. For example, \| x \|#= 4, \| 2x − 1 \|#= 3 \| 5x + 2 \|#− 4 = 9 solutions to absolute value equations For real numbers A and B, an equation of the form \| A \|#= B, with B ≥ 0, will have solutions when A = B or A = −B. If B <#0, the equation \| A \|#= B has no solution. How To… 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 −A = B, assuming B > 0. 3. Solve for x. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.6 ABSOLUTE VALUE FUNCTIONS 251 Example 3 Finding the Zeros of an Absolute Value Function For the function f (x) = \| 4x + 1 \|#− 7, find the values of x such that f (x) = 0. Solution 0 = \| 4x +#1 \|#−#7 7 = \| 4x +#1 \| 7 =#4x +#1 or −7 =#
4x +#1 6 = 4x −8 =#4x −8 ___ 4 6 __ x = = 1.5 4 The function outputs 0 when x = 1.5 or x = −2. See Figure 8. = −2 x = Substitute 0 for f (x). Isolate the absolute value on one side of the equation. Break into two separate equations and solve. y 10 8 6 4 2 –0.5 –2 –4 –6 –8 –10 (−2, 0) –1 –1.5 –2 –2.5 x where f(x) = 0 f x (1.5, 0) 0.5 1 1.5 2 2.5 x where f(x) = 0 Try It #3 For the function f (x) = \| 2x − 1 \|#− 3, find the values of x such that f (x) = 0. Figure 8 Q & A… 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 − 5 \|#= 1. Access these online resources for additional instruction and practice with absolute value. • Graphing Absolute Value Functions (http://openstaxcollege.org/l/graphabsvalue) • Graphing Absolute Value Functions 2 (http://openstaxcollege.org/l/graphabsvalue2) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 252 CHAPTER 3 FUNCTIONS 3.6 SECTION EXERCISES VERBAL 1. How do you solve an absolute value equation? 3. 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? ALGEBRAIC 2. How can you tell whether an absolute value function has two x-intercepts without graphing the function? 4. How can you use the graph of an absolute value function to determine the x-values for which the function values are negative? 5. Describe all numbers x that are at a distance of 4 from the number 8. Express this using absolute value notation. 6. Describe all numbers x that are at a distance of 1__ 2 from the number −4. Express this using absolute value notation. 7. Describe the situation in
which the distance that point x is from 10 is at least 15 units. Express this using absolute value notation. 8. Find all function values f (x) such that the distance from f (x) to the value 8 is less than 0.03 units. Express this using absolute value notation. For the following exercises, find the x- and y-intercepts of the graphs of each function. 9. f (x)#= 4 ∣ x − 3 ∣ + 4 10. f (x)#= −3 ∣ x − 2 ∣ − 1 11. f (x)#= −2 ∣ x + 1 ∣ + 6 12. f (x)#= −5 ∣ x + 2 ∣ + 15 13. f (x)#= 2 ∣ x − 1 ∣ − 6 14. f (x)#= ∣ −2x + 1 ∣ − 13 15. f (x)#= − ∣ x − 9 ∣ + 16 GRAPHICAL For the following exercises, graph the absolute value function. Plot at least five points by hand for each graph. 16. y = \| x − 1 \| 17. y = \| x + 1#\|# 18. y = \| x \|#+ 1 For the following exercises, graph the given functions by hand. 19. y = \| x \|#− 2 20. y = −\| x#\|# 21. y = −\| x \| −#2 22. y = −\| x − 3 \|#−#2 23. f (x) = −\| x − 1 \|#−#2 24. f (x) = −\| x + 3 \|#+ 4 25. f (x) = 2\| x + 3 \|#+ 1 26. f (x) = 3\| x − 2 \|#+ 3 27. f (x) = \| 2x − 4 \|#− 3 28. f (x) = \| 3x + 9 \|#+ 2 29. f (x) = −\| x − 1 \|#− 3 30. f (x) = −\| x + 4 \| −3 1 __ \| x + 4 \|#− 3 31. f (x) = 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.6 SECTION EXERCISES 253 TECHNOLOGY 32. Use a
graphing utility to graph f (x) = 10\| x − 2\| on the viewing window [0, 4]. Identify the corresponding range. Show the graph. 33. Use a graphing utility to graph f (x) = −100\| x\|#+ 100 on the viewing window [−5, 5]. Identify the corresponding range. Show the graph. For the following exercises, graph each function using a graphing utility. Specify the viewing window. 35. f (x) = 4 × 109 ∣ x − (5 × 109) ∣ #+ 2 × 109 34. f (x) = −0.1\| 0.1(0.2 − x)\|#+ 0.3 EXTENSIONS For the following exercises, solve the inequality. 36. If possible, find all values of a such that there are no x-intercepts for f (x) = 2\| x + 1\|#+ a. 37. If possible, find all values of a such that there are no y-intercepts for f (x) = 2\| x + 1\|#+ a. REAL-WORLD APPLICATIONS 38. Cities A and B are on the same east-west line. 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. 39. 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. 40. Students who score within 18 points of the number 82 will pass a particular test. Write this statement using absolute value notation and use the variable x for the score. 41. 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. 42. 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. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 254 CHAPTER 3 FUNCTIONS LEARNING OBJECTIVES In this section, you will: • etermine whether a
function is one-to-one. • erify the inerse using composition. • Find the inerse of a function. • etermine the domain and range of an inerse function and restrict the domain of a function to mae it one-to-one. • Use the graph of a one-to-one function to graph its inerse function on the same aes. • Use a graph or table to ealuate a function and its inerse. 3.7 INVERSE FUNCTIONS 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 “machines” we have been studying can also run backwards. Figure 1 provides a visual representation of this question. In this section, we will consider the reverse nature of functions. x y f? y x Figure 1 Can a function “machine” 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 5 __ C = (F − 32) 9 5 __ (75 − 32) ≈ 24°C. 9 Knowing that a comfortable 75 degrees Fahrenheit is about 24 degrees Celsius, he sends his assistant the week’s weather forecast from Figure 2 for Milan, and asks her to convert all of the temperatures to degrees Fahrenheit. Mon Tue Web Thu 26°C \| 19°C 29°C \| 19°C 30°C \| 20°C 26°C \| 18°C Figure 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.7 INVERSE FUNCTIONS 255 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 26 =# #5 __ (F − 32) 9 9 __ = F − 32 26 ċ 5 9 __ + 32 ≈ 79 F = 26 ċ 5 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 −1(x), read as “ f inverse of x.” The raised −1 is part of the notation. It is not an exponent; it does not imply a power of −1. In other words, f −1(x) does not mean is the reciprocal of f and not the inverse. The “exponent-like” notation comes from an analogy between function composition and multiplication: just as a−1 a = 1 (1 is the identity element for multiplication) for any nonzero number a, so f −1 ∘ f equals the identity function, that is, 1 ___ f (x) 1 ___ f (x) because (f −1 ∘ f )(x) = f −1( f (x)) = f −1 (y) = x This holds for all x in the domain of f. Informally, this means that inverse functions “undo” 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 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 __ x are inverse functions. 4 and (f −1 ∘ f )(x) = f −
1(4x) =# #1 __ (4x) = x 4 1 1 __ __ (f ∘ f −1)(x) = f (# x ) = x x ) = 4 (# 4 4 A few coordinate pairs from the graph of the function y = 4x are (−2, −8), (0, 0), and (2, 8). A few coordinate pairs 1 __ x are (−8, −2), (0, 0), and (8, 2). If we interchange the input and output of each from the graph of the function y = 4 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 −1 (x) is an inverse function of f if f −1(y) = x. This can also be written as f −1( f (x)) = x for all x in the domain of f. It also follows that f ( f −1(x)) = x for all x in the domain of f −1 if f −1 is the inverse of f. The notation f −1 is read “ f inverse.” Like any other function, we can use any variable name as the input for f −1, so we will often write f −1(x), which we read as “ f inverse of x.” Keep in mind that and not all functions have inverses. f −1(x) ≠ 1 ___ f (x) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 256 CHAPTER 3 FUNCTIONS Example 1 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 −1 (4) = 2; f (5) = 12, then f −1 (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 1.
(x, f (x)) (2, 4) (5, 12) (x, g (x)) (4, 2) (12, 5) Table 1 Try It #1 Given that h−1(6) = 2, what are the corresponding input and output values of the original function h? How To… 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 −1 and f = g −1. If either statement is false, then both are false, and g ≠ f −1 and f ≠ g−1. Example 2 Testing Inverse Relationships Algebraically If f (x) = 1 ____ x + 2 and g (x) =# #1 __ − 2, is g = f −1? x Solution so 1 _ g (f (x)) = − 2 1 _ ) (# 1 and f = g −1 This is enough to answer yes to the question, but we can also verify the other formula. f (g (x)) = 1 _ 1 __ − 2 + 2 x 1 _ 1 __ x = x = Analysis Notice the inverse operations are in reverse order of the operations from the original function. Try It #2 If f (x) = x3 − 4 and g (x) = # 3 √ — x − 4, is g = f −1? Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.7 INVERSE FUNCTIONS 257 Example 3 Determining Inverse Relationships for Power Functions If f (x) = x3 (the cube function) and g (x) =# #1 __ x, is g = f −1? 3 x3 __ Solution 27 f (g (x)) = ≠ x No, the functions are not inverses. Analysis The correct inverse to the cube is, of course, the cube root not a multiplier. 3 √ — x = x1/3 that is, the one-third is an exponent, Try It #3 If f (x) = (x − 1)3 and g (x) = 3 √ — x + 1, is g = f −1
? Finding Domain and Range of Inverse Functions The outputs of the function f are the inputs to f −1, so the range of f is also the domain of f −1. Likewise, because the inputs to f are the outputs of f −1, the domain of f is the range of f −1. We can visualize the situation as in Figure 3. Domain of f Range of f f (x) a b Range of f –1 f –1(x) Domain of f –1 Figure 3 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 x is f −1(x) = x2, because a square “undoes” domain does have an inverse function. For example, the inverse of f (x) = √ a square root; but the square is only the inverse of the square root on the domain [0, ∞), 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 −3. 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 “inverse” 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 one-to-one. For example, we can make a restricted version of the square function f (x) = x2 with its range limited to [0, ∞), which is a one-to-one function (it passes the horizontal line test) and which has an inverse (the square-root function
). If f (x) = (x − 1)2 on [1, ∞), then the inverse function is f −1(x) = √ x + 1. — • The domain of f = range of f −1 = [1, ∞). • The domain of f −1 = range of f = [0, ∞). Q & A… 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. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 258 CHAPTER 3 FUNCTIONS domain and range of inverse functions The range of a function f (x) is the domain of the inverse function f −1(x). The domain of f (x) is the range of f −1(x). How To… Given a function, find the domain and range of its inverse. 1. 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. 2. 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 4 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 2. We restrict the domain in such a fashion that the function assumes all y-values exactly once. Constant f (x) = c Identity f (x) = x Quadratic f (x) = x 2 Cubic f (x) = x 3 Reciprocal f (x) = 1_ x Reciprocal squared Cube root Square root Absolute value f (x) = 1 _ x2 f (x) = 3 √ —
x f (x) = √ — x f (x) =.x. Table 2 Solution The constant function is not one-to-one, and there is no domain (except a single point) on which it could be one-to-one, so the constant function has no meaningful inverse. The absolute value function can be restricted to the domain [0, ∞), where it is equal to the identity function. The reciprocal-squared function can be restricted to the domain (0, ∞). Analysis We can see that these functions (if unrestricted) are not one-to-one by looking at their graphs, shown in Figure 4. 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. f (x) f (x) –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 (a) 21 3 4 5 x –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 (b) 321 4 5 x Figure 4 (a ) Absolute value (b ) Reciprocal squared Try It #4 The domain of function f is (1, ∞) and the range of function f is (−∞, −2). Find the domain and range of the inverse function. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.7 INVERSE FUNCTIONS 259 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 5 A function f (t) is given in Table 3, showing distance in miles that a car has traveled in t minutes. Find and interpret f −1(
70). Interpreting the Inverse of a Tabular Function t (minutes) f (t) (miles) 30 20 50 40 70 60 90 70 Table 3 Solution The inverse function takes an output of f and returns an input for f. So in the expression f −1(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 −1(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 −1(b) = a. By this definition, if we are given f −1(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. Try It #5 Using Table 4, find and interpret a. f (60), and b. f −1(60). t (minutes) f (t) (miles) 30 20 60 50 70 60 90 70 50 40 Table 4 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 function’s graph. How To… 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 function’s output from the x-axis of the given graph. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 260 CHAPTER 3 FUNCTIONS Example 6 Evaluating a Function and Its Inverse from a Graph at Speciﬁc Points A function
g(x) is given in Figure 5. Find g(3) and g−1(3). g(x) 6 5 4 3 2 1 – –1 1– 1 2 3 4 5 6 7 x Figure 5 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 g −1(3), recall that by definition g −1(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, g −1(3) = 5. See Figure 6. g(x) 6 5 4 3 2 1 – –1 1– (5, 3) (3, 1) 1 2 3 4 5 6 7 x Figure 6 Try It #6 Using the graph in Figure 6, a. find g−1(1), and b. estimate g−1(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 formula—for example, y as a function of x—we can often find the inverse function by solving to obtain x as a function of y. How To… 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. Example 7 Inverting the Fahrenheit-to-Celsius Function Find a formula for the inverse function that gives Fahrenheit temperature as a function of Celsius temperature. 5 __ C = (F − 32) 9 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.7 INVERSE FUNCTIONS 261 Solution 5 __ (F − 32) C = 9 9 __ = F − 32 C ċ 5 9 __ C + 32 F = 5 By solving in general, we have uncovered the inverse function. If then 5 __ C = h(F) = (F − 32), 9 9 __ F = h−1(C) = C + 32. 5 In this case, we introduced a function h
to represent the conversion because the input and output variables are descriptive, and writing C−1 could get confusing. Try It #7 1 __ (x − 5) Solve for x in terms of y given y = 3 Example 8 Solving to Find an Inverse Function Find the inverse of the function f (x) = 2 ____ x − 3 + 4. Solution y = + 4 Set up an equation. 2 ____ x − 3 2 _____ x − 3 2 ____ y − 4 2 ____ = Subtract 4 from both sides. Multiply both sides by x − 3 and divide by y − 4. x = + 3 Add 3 to both sides. So f −1 (y) = + 3 or f −1 (x) = 2 _____ y − 4 2 ____ x − 4 + 3. Analysis The domain and range of f exclude the values 3 and 4, respectively. f and f −1 are equal at two points but are not the same function, as we can see by creating Table 5. x f (x) 1 3 2 2 Table 5 5 5 f −1(y) y Example 9 Solving to Find an Inverse with Radicals Find the inverse of the function f (x) = 2 + √ — x − 4. Solution y − 2)2 = x − 4 x = (y − 2)2 + 4 So f −1 (x) = (x − 2)2 + 4. The domain of f is [4, ∞). Notice that the range of f is [2, ∞), so this means that the domain of the inverse function f −1 is also [2, ∞). Analysis The formula we found for f −1(x) looks like it would be valid for all real x. However, f −1 itself must have an inverse (namely, f) so we have to restrict the domain of f −1 to [2, ∞) in order to make f −1 a one-to-one function. This domain of f −1 is exactly the range of f. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 262 CHAPTER 3 FUNCTIONS Try It #8 What is the inverse of the function f (x) = 2 − √ — 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, ∞), on which this function is one-to-one, and graph it as in Figure 7. f (x) 5 4 3 2 1 –1–1 –2 –5 –4 –3 –2 21 3 4 5 x Figure 7 Quadratic function with domain restricted to [0, ∞). Restricting the domain to [0, ∞) 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 −1(x) = √ What happens if we graph both f and f −1 on the same set of axes, using the x-axis for the input to both f and f −1? We notice a distinct relationship: The graph of f −1(x) is the graph of f (x) reflected about the diagonal line y = x, which we will call the identity line, shown in Figure 8. — x. f (x) y = x f –1(x) 21 1–1 –2 –3 –4 –5 –5 –4 –3 –2 Figure 8 Square and square-root functions on the non-negative 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 10 Given the graph of f (x) in Figure 9, sketch a graph of f −1(x). Finding the Inverse of a Function Using Reﬂection about the Identity Line Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.7 INVERSE FUNCTIONS 263 y 5 4 3 2 1 –1–1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x Figure 9 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, ∞) and range of (−∞, ∞
), so the inverse will have a domain of (−∞, ∞) and range of (0, ∞). 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 10. y 5 4 3 2 1 –1–1 –2 –3 –4 –5 –5 –4 –3 –2 f –1(x) y = x f (x) 21 3 4 5 x Figure 10 The function and its inverse, showing reﬂection about the identity line Try It #9 Draw graphs of the functions f and f −1 from Example 8. Q & A… Is there any function that is equal to its own inverse? Yes. If f = f −1, then f (f (x)) = x, and we can think of several functions that have this property. The identity function does, and so does the reciprocal function, because Any function f (x) = c − x, where c is a constant, is also equal to its own inverse. 1 _ = x 1 _ x Access these online resources for additional instruction and practice with inverse functions. • Inverse Functions (http://openstaxcollege.org/l/inversefunction) • One-to-one Functions (http://openstaxcollege.org/l/onetoone) • Inverse Function Values Using Graph (http://openstaxcollege.org/l/inversfuncgraph) • Restricting the Domain and Finding the Inverse (http://openstaxcollege.org/l/restrictdomain) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 264 CHAPTER 3 FUNCTIONS 3.7 SECTION EXERCISES VERBAL 1. n eff 2. fx=x 3. 4. 5. ALGEBRAIC 6. fx=a−xa 7. 8. 9. 10., fi 11. fx=x+ 12. fx=x+ 13. fx=−x 14. fx=x + 15. fx=x + 16. fx=−x 17. fx=−x 21. fx=√x−+ — 25. fx= _____ x− 26
. fx= x_____+ x+ 18. 22. 27. + fx=√x — − fx=+ √ x fx= x_____− x+ 19. fx=√−x — 23. 28. — fx=−√ x x______+ −x fx= 20. 24. 29. — fx= fx=+ √x− _____ x+ x______+ fx= −x 30. fx=x + x−∞ 31. fx=x + x+ −∞ 32. fx=x −x+ ∞ fif. Thn fif 33. fx=x+ 34. fx=x− 35. fx=x − 36. fx= x − gx= x − x a. fgxg fx b. fxgx f xgx 37. fx=√ x− g x=x + — 38. fx=−x+ g x= x− − GRAPHICAL 39. fx=√x — 40. — fx=√ x+ 41. fx=−x+ 42. fx=x − Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 3.7 SECTION EXERCISES 265 44. x – – – – y – – – – – – 45 43. 46 – – – – – – – – – – 47. y 48. y 49. y x x 50. y x x fFigure 11 – – – – 51. f 52. fx= 53. f − 54. f −x Figure 11 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 266 CHAPTER 3 FUNCTIONS Figure 12 55. f − 56. ff − 57. f f 58. f f NUMERIC y –– – – –– – – – – f x Figure 12 f 59. f=f − 61. f −−=−f− 60. f=f − 62. f −−=−f− Table 6 x f (x) Table 6 63. f 65. f − 64. fx= 66. f −x= 67. fTable 7 f −x x f (x) Table 7 TECHNOLOGY, fi. Th fx= 68
. fx=x − 69. _____ x− 70. fx= _ x− REAL-WORLD APPLICATIONS 71. xy __ x+ fx= 72. ThC Cr=πr rCrπ 73. Th tdt=t td t Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 3 REVIEW 267 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 the new function formed by function composition, when the output of one function is used as the input of another 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 a function whose graph is unchanged by horizontal reflection, f (x) = f (−x), and is symmetric about the y-axis function a relation in which each input value yields a unique output value horizontal compression a transformation that compresses a function’s graph horizontally, by multiplying the input by a constant b > 1 horizontal line test a method of testing whether a function is one-to-one by determining whether any horizontal line intersects the graph more than once horizontal reflection a transformation that reflects a function’s graph across the y-axis by multiplying the input by −1 horizontal shift a transformation that shifts a function’s graph left or right by adding a positive or negative constant to the input horizontal stretch a transformation that stretches a function’s graph horizontally by multiplying the input by a constant 0 < b < 1 increasing function a function is increasing in some open interval if f (b) > f (a) for any two input values a and b in the given interval where b > a independent variable an input variable input each object or value in a domain that relates to another object or value by a relationship known as a function interval notation a method of describing a set that includes all numbers between a lower limit and an upper limit; the lower and upper values are listed between brackets or parentheses, a square bracket indicating inclusion in the set, and a parenthesis indicating exclusion inverse function for any one-to-one function
f (x), the inverse is a function f −1(x) such that f −1(f (x)) = x for all x in the domain of f; this also implies that f ( f −1(x)) = x for all x in the domain of f −1 local extrema collectively, all of a function’s local maxima and minima local maximum a value of the input where a function changes from increasing to decreasing as the input value increases. local minimum a value of the input where a function changes from decreasing to increasing as the input value increases. odd function a function whose graph is unchanged by combined horizontal and vertical reflection, f (x) = − f (−x), and is symmetric about the origin one-to-one function a function for which each value of the output is associated with a unique input value output each object or value in the range that is produced when an input value is entered into a function piecewise function a function in which more than one formula is used to define the output range the set of output values that result from the input values in a relation rate of change the change of an output quantity relative to the change of the input quantity relation a set of ordered pairs Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 268 CHAPTER 3 FUNCTIONS set-builder notation a method of describing a set by a rule that all of its members obey; it takes the form {x. statement about x} vertical compression a function transformation that compresses the function’s graph vertically by multiplying the output by a constant 0 < a < 1 vertical line test a method of testing whether a graph represents a function by determining whether a vertical line intersects the graph no more than once vertical reflection a transformation that reflects a function’s graph across the x-axis by multiplying the output by −1 vertical shift a transformation that shifts a function’s graph up or down by adding a positive or negative constant to the output vertical stretch a transformation that stretches a function’s graph vertically by multiplying the output by a constant a > 1 Key Equations Constant function f (x) = c, where c is a constant Identity function f (x) = x Absolute value function f (x) =.x. Reciprocal squared function f (x) = Quadratic function Cubic function Reciprocal function Square root function Cube root function Average rate of change f (x) = x
2 f (x) = x3 1_ f (x) = x 1 _ x2 f (x) = √ — x f (x) = ∆y _ ∆x = — x 3 √ f (x2) − f (x1) _ x2 − x1 Composite function (f ∘#g)(x) = f (g(x)) Vertical shift g(x) = f (x) + k (up for k > 0) Horizontal shift g(x) = f (x − h) (right for h > 0) Vertical reflection g(x) = −f (x) Horizontal reflection g(x) = f (−x) Vertical stretch g(x) = af (x) (a > 0) Vertical compression g(x) = af (x) (0 < a < 1) Horizontal stretch g(x) = f (bx) (0 < b < 1) Horizontal compression g(x) = f (bx) (b > 1) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 3 REVIEW 269 Key Concepts 3.1 Functions and Function Notation • A relation is a set of ordered pairs. A function is a specific type of relation in which each domain value, or input, leads to exactly one range value, or output. See Example 1 and Example 2. • Function notation is a shorthand method for relating the input to the output in the form y = f (x). See Example 3 and Example 4. • In tabular form, a function can be represented by rows or columns that relate to input and output values. See Example 5. • To evaluate a function, we determine an output value for a corresponding input value. Algebraic forms of a function can be evaluated by replacing the input variable with a given value. See Example 6 and Example 7. • To solve for a specific function value, we determine the input values that yield the specific output value. See Example 8. • An algebraic form of a function can be written from an equation. See Example 9 and Example 10. • Input and output values of a function can be identified from a table. See Example 11. • Relating input values to output values on a graph is another way to evaluate a function. See Example 12. • A function is one-to-one if each output value corresponds to only one input
value. See Example 13. • A graph represents a function if any vertical line drawn on the graph intersects the graph at no more than one point. See Example 14. • The graph of a one-to-one function passes the horizontal line test. See Example 15. 3.2 Domain and Range • The domain of a function includes all real input values that would not cause us to attempt an undefined mathematical operation, such as dividing by zero or taking the square root of a negative number. • The domain of a function can be determined by listing the input values of a set of ordered pairs. See Example 1. • The domain of a function can also be determined by identifying the input values of a function written as an equation. See Example 2, Example 3, and Example 4. • Interval values represented on a number line can be described using inequality notation, set-builder notation, and interval notation. See Example 5. • For many functions, the domain and range can be determined from a graph. See Example 6 and Example 7. • An understanding of toolkit functions can be used to find the domain and range of related functions. See Example 8, Example 9, and Example 10. • A piecewise function is described by more than one formula. See Example 11 and Example 12. • A piecewise function can be graphed using each algebraic formula on its assigned subdomain. See Example 13. 3.3 Rates of Change and Behavior of Graphs • A rate of change relates a change in an output quantity to a change in an input quantity. The average rate of change is determined using only the beginning and ending data. See Example 1. • Identifying points that mark the interval on a graph can be used to find the average rate of change. See Example 2. • Comparing pairs of input and output values in a table can also be used to find the average rate of change. See Example 3. • An average rate of change can also be computed by determining the function values at the endpoints of an interval described by a formula. See Example 4 and Example 5. • The average rate of change can sometimes be determined as an expression. See Example 6. • A function is increasing where its rate of change is positive and decreasing where its rate of change is negative. See Example 7. • A local maximum is where a function changes from increasing to decreasing and has an output value larger (more positive or less negative) than output values at neighboring input values. Download the OpenStax text for free at
http://cnx.org/content/col11759/latest. 270 CHAPTER 3 FUNCTIONS • A local minimum is where the function changes from decreasing to increasing (as the input increases) and has an output value smaller (more negative or less positive) than output values at neighboring input values. • Minima and maxima are also called extrema. • We can find local extrema from a graph. See Example 8 and Example 9. • The highest and lowest points on a graph indicate the maxima and minima. See Example 10. 3.4 Composition of Functions • We can perform algebraic operations on functions. See Example 1. • When functions are combined, the output of the first (inner) function becomes the input of the second (outer) function. • The function produced by combining two functions is a composite function. See Example 2 and Example 3. • The order of function composition must be considered when interpreting the meaning of composite functions. See Example 4. • A composite function can be evaluated by evaluating the inner function using the given input value and then evaluating the outer function taking as its input the output of the inner function. • A composite function can be evaluated from a table. See Example 5. • A composite function can be evaluated from a graph. See Example 6. • A composite function can be evaluated from a formula. See Example 7. • The domain of a composite function consists of those inputs in the domain of the inner function that correspond to outputs of the inner function that are in the domain of the outer function. See Example 8 and Example 9. • Just as functions can be combined to form a composite function, composite functions can be decomposed into simpler functions. • Functions can often be decomposed in more than one way. See Example 10. 3.5 Transformation of Functions • A function can be shifted vertically by adding a constant to the output. See Example 1 and Example 2. • A function can be shifted horizontally by adding a constant to the input. See Example 3, Example 4, and Example 5. • Relating the shift to the context of a problem makes it possible to compare and interpret vertical and horizontal shifts. See Example 6. • Vertical and horizontal shifts are often combined. See Example 7 and Example 8. • A vertical reflection reflects a graph about the x-axis. A graph can be reflected vertically by multiplying the output by –1. • A horizontal reflection reflects a graph about the y-axis. A graph can be reflected horizontally by multiplying the
input by –1. • A graph can be reflected both vertically and horizontally. The order in which the reflections are applied does not affect the final graph. See Example 9. • A function presented in tabular form can also be reflected by multiplying the values in the input and output rows or columns accordingly. See Example 10. • A function presented as an equation can be reflected by applying transformations one at a time. See Example 11. • Even functions are symmetric about the y-axis, whereas odd functions are symmetric about the origin. • Even functions satisfy the condition f (x) = f (−x). • Odd functions satisfy the condition f (x) = −f (−x). • A function can be odd, even, or neither. See Example 12. • A function can be compressed or stretched vertically by multiplying the output by a constant. See Example 13, Example 14, and Example 15. • A function can be compressed or stretched horizontally by multiplying the input by a constant. See Example 16, Example 17, and Example 18. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 3 REVIEW 271 • The order in which different transformations are applied does affect the final function. Both vertical and horizontal transformations must be applied in the order given. However, a vertical transformation may be combined with a horizontal transformation in any order. See Example 19 and Example 20. 3.6 Absolute Value Functions • Applied problems, such as ranges of possible values, can also be solved using the absolute value function. See Example 1. • The graph of the absolute value function resembles a letter V. It has a corner point at which the graph changes direction. See Example 2. • In an absolute value equation, an unknown variable is the input of an absolute value function. • If the absolute value of an expression is set equal to a positive number, expect two solutions for the unknown variable. See Example 3. 3.7 Inverse Functions • If g(x) is the inverse of f (x), then g(f (x)) = f (g(x)) = x. See Example 1, Example 2, and Example 3. • Each of the toolkit functions has an inverse. See Example 4. • For a function to have an inverse, it must be one-to-one (pass the horizontal line test). • A function that is not one-to-one over its entire domain may be one-to-one on part
of its domain. • For a tabular function, exchange the input and output rows to obtain the inverse. See Example 5. • The inverse of a function can be determined at specific points on its graph. See Example 6. • To find the inverse of a formula, solve the equation y = f (x) for x as a function of y. Then exchange the labels x and y. See Example 7, Example 8, and Example 9. • The graph of an inverse function is the reflection of the graph of the original function across the line y = x. See Example 10. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 272 CHAPTER 3 FUNCTIONS CHAPTER 3 REVIEW EXERCISES FUNCTIONS AND FUNCTION NOTATION For the following exercises, determine whether the relation is a function. 1. {(a, b), (c, d), (e, d)} 2. {(5, 2), (6, 1), (6, 2), (4, 8)} 3. y 2 + 4 = x, for x the independent variable and y the dependent variable 4. Is the graph in Figure 1 a function? y 25 20 15 10 5 –5 –5 –10 –15 –20 –25 –25 –20 –15 –10 f 5 10 15 20 25 x Figure 1 For the following exercises, evaluate the function at the indicated values: f (−3); f (2); f (−a); −f (a); f (a + h). 5. f (x) = −2x 2 + 3x 6. f (x) = 2.3x − 1. For the following exercises, determine whether the functions are one-to-one. 7. f (x) = −3x + 5 8. f (x) =.x − 3. For the following exercises, use the vertical line test to determine if the relation whose graph is provided is a function. 9. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 10. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 11. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4
5 x For the following exercises, graph the functions. 12. f (x) =.x + 1. 13. f (x) = x 2 − 2 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 3 REVIEW 273 For the following exercises, use Figure 2 to approximate the values. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 14. f (2) 15. f (−2) 21 3 4 5 x 16. If f (x) = −2, then solve for x. 17. If f (x) = 1, then solve for x. Figure 2 For the following exercises, use the function h(t) = −16t 2 + 80t to find the values. 18. h(2) − h(1) __ 2 − 1 19. h(a) − h(1) __ a − 1 DOMAIN AND RANGE For the following exercises, find the domain of each function, expressing answers using interval notation. 20. f (x) = 2 _ 3x + 2 21. f (x) = x − 3 ___________ x 2 − 4x − 12 23. Graph this piecewise function: f (x) = { x + 1 −2x − 3 x < −2 x ≥ −2 — x − 6 _ 22. f (x) = # √ x − 4 √ — RATES OF CHANGE AND BEHAVIOR OF GRAPHS For the following exercises, find the average rate of change of the functions from x = 1 to x = 2. 24. f (x) = 4x − 3 25. f (x) = 10x 2 + x 26. f (x) = − #2 _ x 2 For the following exercises, use the graphs to determine the intervals on which the functions are increasing, decreasing, or constant. 27. y 28. –5 –4 –3 –2 10 8 6 4 2 –1 –2 –4 –6 –8 –10 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 29. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x
30. Find the local minimum of the function graphed in Exercise 27. 31. Find the local extrema for the function graphed in Exercise 28. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 274 CHAPTER 3 FUNCTIONS 32. For the graph in Figure 3, the domain of the function is [−3, 3]. The range is [−10, 10]. Find the absolute minimum of the function on this interval. 33. Find the absolute maximum of the function graphed in Figure 3. –5 –4 –3 –2 y 10 8 6 4 2 –1 –2 –4 –6 –8 –10 21 3 4 5 x Figure 3 COMPOSITION OF FUNCTIONS For the following exercises, find (f ∘#g)(x) and (g ∘ f)(x) for each pair of functions. 34. f (x) = 4 − x, g(x) = −4x 35. f (x) = 3x + 2, g(x) = 5 − 6x 36. f (x) = x 2 + 2x, g(x) = 5x + 1 37. f (x(x) = x 38. f (x) =, g(x _____ 2 For the following exercises, find (f ∘#g) and the domain for (f ∘#g)(x) for each pair of functions. 39. f (x __, g(x) = x 42. f (x) = 1 _ x2 − 1, g(x) = √ — x + 1 40. f (x) =, g(x(x) = √ 41. f (x) = — x For the following exercises, express each function H as a composition of two functions f and g where H(x) = (f ∘#g)(x). 43. H(x) = √ _______ 2x − 1 ______ 3x + 4 44. H(x) = 1 _ (3x2 − 4)−3 TRANSFORMATION OF FUNCTIONS For the following exercises, sketch a graph of the given function. 45. f (x) = (x − 3)2 46. f (x) = (x + 4)3 48. f (x) = −x3 49. f (x) = 3 √ — −x 51
. f (x) = 4[.x − 2. − 6] 52. f (x) = −(x + 2)2 − 1 47. f (x) = √ — x + 5 50. f (x) = 5 √ — −x − 4 For the following exercises, sketch the graph of the function g if the graph of the function f is shown in Figure 4. y 5 4 3 2 1 –1 –1 –2 –5 –4 –3 –2 21 3 4 5 x Figure 4 53. g(x) = f (x − 1) 54. g(x) = 3f (x) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 3 REVIEW 275 For the following exercises, write the equation for the standard function represented by each of the graphs below. 55. y 56. y –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x –5 –4 –3 –2 5 4 3 2 1 –1 –1 –2 –3 –4 –5 21 3 4 5 x For the following exercises, determine whether each function below is even, odd, or neither. 57. f (x) = 3x4 58. g(x) = √ — x 1 _ x + 3x 59. h(x) = For the following exercises, analyze the graph and determine whether the graphed function is even, odd, or neither. 60. y 61. y 62. y –25 –20 –15 –10 25 20 15 10 5 –5 –5 –10 –15 –20 –25 5 10 15 20 25 x –10 –8 –6 –4 25 20 15 10 5 –2 –5 –10 –15 –20 –25 42 6 8 10 x –10 –8 –6 –4 ABSOLUTE VALUE FUNCTIONS For the following exercises, write an equation for the transformation of f (x) = \| x \|. 63. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 64. 21 3 4 5 x –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 65. 21 3 4 5 x –5 –4 –3 –2
10 8 6 4 2 –2 –2 –4 –6 –8 –10 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 42 6 8 10 x 21 3 4 5 x For the following exercises, graph the absolute value function. 66. f (x) = \| x − 5 \| 67. f (x) = −\| x − 3 \| 68. f (x) = \| 2x − 4 \| Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 276 CHAPTER 3 FUNCTIONS INVERSE FUNCTIONS For the following exercises, find f −1(x) for each function. 69. f (x) = 9 + 10x 70. f (x) = x _ x + 2 For the following exercise, find a domain on which the function f is one-to-one and non-decreasing. Write the domain in interval notation. Then find the inverse of f restricted to that domain. 71. f (x) = x 2 + 1 72. Given f (x) = x 3 − 5 and g(x. Find f (g(x)) and g(f (x)). b. What does the answer tell us about the relationship between f (x) and g(x)? For the following exercises, use a graphing utility to determine whether each function is one-to-one. 1 _ x 73. f (x) = 76. If f (1) = 4, find f −1(4). 74. f (x) = −3x 2 + x 75. If f (5) = 2, find f −1(2). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. CHAPTER 3 PRACTICE TEST 277 CHAPTER 3 PRACTICE TEST For the following exercises, determine whether each of the following relations is a function. 1. y = 2x + 8 2. {(2, 1), (3, 2), (−1, 1), (0, −2)} For the following exercises, evaluate the function f (x) = −3x 2 + 2x at the given input. 3. f (−2) 4. f (a) 5. Show that the function f (x) = −2(x − 1)2 + 3 is not one-to-one. 6.
Write the domain of the function f (x) = √ — 3 − x in interval notation. 7. Given f (x) = 2x 2 − 5x, find f (a + 1) − f (1). 8. Graph the function f (x) = { x + 1 if −2 < x < 3 x ≥ 3 −x if 9. Find the average rate of change of the function f (x) = 3 − 2x 2 + x by finding f (b) − f (a) _ b − a. For the following exercises, use the functions f (x) = 3 − 2x 2 + x and g(x) = √ — x to find the composite functions. 10. ( g ∘ f )(x) 11. ( g ∘ f )(1) 12. Express H(x) = 3 √ — 5x 2 − 3x as a composition of two functions, f and g, where ( f ∘#g )(x) = H(x). For the following exercises, graph the functions by translating, stretching, and/or compressing a toolkit function. 13. f (x) = √ — x + 6 − 1 14. f (x) = 1 _ x + 2 − 1 For the following exercises, determine whether the functions are even, odd, or neither. 5 _ 15. f (x) = − x2 + 9x 6 1_ 17. f (x) = x 5 _ 16. f (x) = − x 3 + 9x 5 18. Graph the absolute value function f (x) = −2\| x − 1 \| + 3. For the following exercises, find the inverse of the function. 19. f (x) = 3x − 5 20. f (x) = 4 _____ x + 7 For the following exercises, use the graph of g shown in Figure 1. y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –5 –4 –3 –2 21. On what intervals is the function increasing? 22. On what intervals is the function decreasing? 21 3 4 5 x 23. Approximate the local minimum of the function. Express the answer as an ordered pair. 24. Approximate the local maximum of the function. Express the answer as an ordered pair. Figure 1 Download the OpenStax text for free at http://cnx.
org/content/col11759/latest. 278 CHAPTER 3 FUNCTIONS For the following exercises, use the graph of the piecewise function shown in Figure 2. 25. Find f (2). 26. Find f (−2). 27. Write an equation for the piecewise function. y 5 4 3 2 1 f 21 3 4 5 x –5 –4 –3 –2 –1 –1 –2 –3 –4 –5 Figure 2 For the following exercises, use the values listed in Table 1. x F (x 11 6 13 7 15 8 17 Table 1 28. Find F (6). 29. Solve the equation F (x) = 5. 30. Is the graph increasing or decreasing on its domain? 31. Is the function represented by the graph one-to-one? 32. Find F −1(15). 33. Given f (x) = −2x + 11, find f −1(x). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. Linear Functions 4 Figure 1 A bamboo forest in China (credit: “JFXie”/Flickr) CHAPTER OUTLINE 4.1 Linear Functions 4.2 Modeling with Linear Functions 4.3 Fitting Linear Models to Data Introduction Imagine placing a plant in the ground one day and finding that it has doubled its height just a few days later. Although it may seem incredible, this can happen with certain types of bamboo species. These members of the grass family are the fastest-growing plants in the world. One species of bamboo has been observed to grow nearly 1.5 inches every hour.[6] In a twenty-four hour period, this bamboo plant grows about 36 inches, or an incredible 3 feet! A constant rate of change, such as the growth cycle of this bamboo plant, is a linear function. Recall from Functions and Function Notation that a function is a relation that assigns to every element in the domain exactly one element in the range. Linear functions are a specific type of function that can be used to model many real-world applications, such as plant growth over time. In this chapter, we will explore linear functions, their graphs, and how to relate them to data. 6 http://www.guinnessworldrecords.com/records-3000/fastest-growing-plant/ 279 Download the OpenStax text for free at http://cnx.org/content/col11759
/latest. 280 CHAPTER 4 LINEAR FUNCTIONS LEARNING OBJECTIVES In this section, you will: • Represent a linear function. • Determine whether a linear function is increasing, decreasing, or constant. • Interpret slope as a rate of change. • Write and interpret an equation for a linear function. • Graph linear functions. • Determine whether lines are parallel or perpendicular. • Write the equation of a line parallel or perpendicular to a given line. 4.1 LINEAR FUNCTIONS Figure 1 Shanghai MagLev Train (credit: “kanegen”/Flickr) Just as with the growth of a bamboo plant, there are many situations that involve constant change over time. Consider, for example, the first commercial maglev train in the world, the Shanghai MagLev Train (Figure 1). It carries passengers comfortably for a 30-kilometer trip from the airport to the subway station in only eight minutes.[7] Suppose a maglev train travels a long distance, and that the train maintains a constant speed of 83 meters per second for a period of time once it is 250 meters from the station. How can we analyze the train’s distance from the station as a function of time? In this section, we will investigate a kind of function that is useful for this purpose, and use it to investigate real-world situations such as the train’s distance from the station at a given point in time. Representing Linear Functions The function describing the train’s motion is a linear function, which is defined as a function with a constant rate of change, that is, a polynomial of degree 1. There are several ways to represent a linear function, including word form, function notation, tabular form, and graphical form. We will describe the train’s motion as a function using each method. Representing a Linear Function in Word Form Let’s begin by describing the linear function in words. For the train problem we just considered, the following word sentence may be used to describe the function relationship. • The train’s distance from the station is a function of the time during which the train moves at a constant speed plus its original distance from the station when it began moving at constant speed. The speed is the rate of change. Recall that a rate of change is a measure of how quickly the dependent variable changes with respect to the independent variable. The rate of change for this example is constant, which means that it is the same for each input value
. As the time (input) increases by 1 second, the corresponding distance (output) increases by 83 meters. The train began moving at this constant speed at a distance of 250 meters from the station. 7 http://www.chinahighlights.com/shanghai/transportation/maglev-train.htm Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 281 Representing a Linear Function in Function Notation Another approach to representing linear functions is by using function notation. One example of function notation is an equation written in the slope-intercept form of a line, where x is the input value, m is the rate of change, and b is the initial value of the dependent variable. Equation form y = mx + b Function notation f (x) = mx + b In the example of the train, we might use the notation D(t) in which the total distance D is a function of the time t. The rate, m, is 83 meters per second. The initial value of the dependent variable b is the original distance from the station, 250 meters. We can write a generalized equation to represent the motion of the train. Representing a Linear Function in Tabular Form D(t) = 83t + 250 A third method of representing a linear function is through the use of a table. The relationship between the distance from the station and the time is represented in Figure 2. From the table, we can see that the distance changes by 83 meters for every 1 second increase in time. 1 second 1 second 1 second t D(t) 0 250 1 333 2 416 3 499 83 meters 83 meters 83 meters Figure 2 Tabular representation of the function D showing selected input and output values Q & A… Can the input in the previous example be any real number? No. The input represents time, so while nonnegative rational and irrational numbers are possible, negative real numbers are not possible for this example. The input consists of non-negative real numbers. Representing a Linear Function in Graphical Form Another way to represent linear functions is visually, using a graph. We can use the function relationship from above, D(t) = 83t + 250, to draw a graph, represented in Figure 3. Notice the graph is a line. When we plot a linear function, the graph is always a line. The rate of change, which is constant, determines the sl
ant, or slope of the line. The point at which the input value is zero is the vertical intercept, or y-intercept, of the line. We can see from the graph that the y-intercept in the train example we just saw is (0, 250) and represents the distance of the train from the station when it began moving at a constant speed 500 400 300 200 100 0 1 2 3 4 Time (s) 5 Figure 3 The graph of D(t) = 83t + 250. Graphs of linear functions are lines because the rate of change is constant. Notice that the graph of the train example is restricted, but this is not always the case. Consider the graph of the line f (x) = 2x + 1. Ask yourself what numbers can be input to the function. In other words, what is the domain of the function? The domain is comprised of all real numbers because any number may be doubled, and then have one added to the product. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 282 CHAPTER 4 LINEAR FUNCTIONS linear function A linear function is a function whose graph is a line. Linear functions can be written in the slope-intercept form of a line f (x) = mx + b where b is the initial or starting value of the function (when input, x = 0), and m is the constant rate of change, or slope of the function. The y-intercept is at (0, b). Example 1 Using a Linear Function to Find the Pressure on a Diver The pressure, P, in pounds per square inch (PSI) on the diver in Figure 4 depends upon her depth below the water surface, d, in feet. This relationship may be modeled by the equation, P(d) = 0.434d + 14.696. Restate this function in words. Figure 4 (credit: Ilse Reijs and Jan-Noud Hutten) Solution To restate the function in words, we need to describe each part of the equation. The pressure as a function of depth equals four hundred thirty-four thousandths times depth plus fourteen and six hundred ninety-six thousandths. Analysis The initial value, 14.696, is the pressure in PSI on the diver at a depth of 0 feet, which is the surface of the water. The rate of change, or slope, is 0.434 PSI per foot.
This tells us that the pressure on the diver increases 0.434 PSI for each foot her depth increases. Determining Whether a Linear Function Is Increasing, Decreasing, or Constant The linear functions we used in the two previous examples increased over time, but not every linear function does. A linear function may be increasing, decreasing, or constant. For an increasing function, as with the train example, the output values increase as the input values increase. The graph of an increasing function has a positive slope. A line with a positive slope slants upward from left to right as in Figure 5(a). For a decreasing function, the slope is negative. The output values decrease as the input values increase. A line with a negative slope slants downward from left to right as in Figure 5(b). If the function is constant, the output values are the same for all input values so the slope is zero. A line with a slope of zero is horizontal as in Figure 5(c). Increasing function Decreasing function Constant function f (x) f (x) f (x) f f f x x x (a) (b) Figure 5 (c) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 283 increasing and decreasing functions The slope determines if the function is an increasing linear function, a decreasing linear function, or a constant function. • f (x) = mx + b is an increasing function if m > 0. • f (x) = mx + b is an decreasing function if m < 0. • f (x) = mx + b is a constant function if m = 0. Example 2 Deciding Whether a Function Is Increasing, Decreasing, or Constant Some recent studies suggest that a teenager sends an average of 60 texts per day.[8] For each of the following scenarios, find the linear function that describes the relationship between the input value and the output value. Then, determine whether the graph of the function is increasing, decreasing, or constant. a. The total number of texts a teen sends is considered a function of time in days. The input is the number of days, and output is the total number of texts sent. b. A teen has a limit of 500 texts per month in his or her data plan. The input is the number of days, and output is the total number of texts remaining for the month. c. A teen has an
unlimited number of texts in his or her data plan for a cost of $50 per month. The input is the number of days, and output is the total cost of texting each month. Solution Analyze each function. a. The function can be represented as f (x) = 60x where x is the number of days. The slope, 60, is positive so the function is increasing. This makes sense because the total number of texts increases with each day. b. The function can be represented as f (x) = 500 − 60x where x is the number of days. In this case, the slope is negative so the function is decreasing. This makes sense because the number of texts remaining decreases each day and this function represents the number of texts remaining in the data plan after x days. c. The cost function can be represented as f (x) = 50 because the number of days does not affect the total cost. The slope is 0 so the function is constant. Interpreting Slope as a Rate of Change In the examples we have seen so far, we have had the slope provided for us. However, we often need to calculate the slope given input and output values. Recall that given two values for the input, x1 and x2, and two corresponding values for the output, y1 and y2—which can be represented by a set of points, (x1, y1) and (x2, y2)—we can calculate the slope m. m = change in output (rise) __ change in input (run) = ∆y _ ∆x = y2 − y1 _ x2 − x1 Note in function notation two corresponding values for the output y1 and y2 for the function f, y1 = f (x1) and y2 = f (x2), so we could equivalently write Figure 6 indicates how the slope of the line between the points, (x1, y1) and (x2, y2), is calculated. Recall that the slope measures steepness, or slant. The greater the absolute value of the slope, the steeper the line is. m = f (x2) − f (x1) _ x2 − x1 8 http://www.cbsnews.com/8301-501465_162-57400228-501465/teens-are-sending-60-texts-a-day-study-says/ Download the
OpenStax text for free at http://cnx.org/content/col11759/latest. 284 1. CHAPTER 4 LINEAR FUNCTIONS x2 – x1 y2 – y1 y 10 x2, y2) (x1, y1) 0 –1 321 4 ∆y ∆x = y2 – y1 x2 – x1 m = x Figure 6 The slope of a function is calculated by the change in y divided by the change in x. It does not matter which coordinate is used as the (x2, y2) and which is the (x1, y1), as long as each calculation is started with the elements from the same coordinate pair. Q & A… Are the units for slope always units for the output __? units for the input Yes. Think of the units as the change of output value for each unit of change in input value. An example of slope could be miles per hour or dollars per day. Notice the units appear as a ratio of units for the output per units for the input. calculate slope The slope, or rate of change, of a function m can be calculated according to the following: m = change in output (rise) __ change in input (run) = ∆y _ ∆x = y2 − y1 _ x2 − x1 where x1 and x2 are input values, y1 and y2 are output values. How To… Given two points from a linear function, calculate and interpret the slope. 1. Determine the units for output and input values. 2. Calculate the change of output values and change of input values. 3. Interpret the slope as the change in output values per unit of the input value. Example 3 Finding the Slope of a Linear Function If f (x) is a linear function, and (3, −2) and (8, 1) are points on the line, find the slope. Is this function increasing or decreasing? Solution The coordinate pairs are (3, −2) and (8, 1). To find the rate of change, we divide the change in output by the change in input. m = change in output __ = change in input 1 − (−2) ________ 8 − 3 3 __ = 5 We could also write the slope as m = 0.6. The function is increasing because m > 0. Analysis As noted earlier, the order in which we write the points does not matter when we compute the
slope of the line as long as the first output value, or y-coordinate, used corresponds with the first input value, or x-coordinate, used. Note that if we had reversed them, we would have obtained the same slope. m = (−2) − (1) _________ 3 − 8 = −3 ___ −5 3 _ = 5 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 285 Try It #1 If f (x) is a linear function, and (2, 3) and (0, 4) are points on the line, find the slope. Is this function increasing or decreasing? Example 4 Finding the Population Change from a Linear Function The population of a city increased from 23,400 to 27,800 between 2008 and 2012. Find the change of population per year if we assume the change was constant from 2008 to 2012. Solution The rate of change relates the change in population to the change in time. The population increased by 27,800 − 23,400 = 4,400 people over the four-year time interval. To find the rate of change, divide the change in the number of people by the number of years. 4,400 people __ 4 years = 1,100 people__ year So the population increased by 1,100 people per year. Analysis Because we are told that the population increased, we would expect the slope to be positive. This positive slope we calculated is therefore reasonable. Try It #2 The population of a small town increased from 1,442 to 1,868 between 2009 and 2012. Find the change of population per year if we assume the change was constant from 2009 to 2012. Writing and Interpreting an Equation for a Linear Function Recall from Equations and Inequalities that we wrote equations in both the slope-intercept form and the point-slope form. Now we can choose which method to use to write equations for linear functions based on the information we are given. That information may be provided in the form of a graph, a point and a slope, two points, and so on. Look at the graph of the function f in Figure 7. y f 10 8 6 4 2 –2 –2 –4 (0, 7) (4, 4) 42 6 8 10 x Figure 7 –10 –8 –6 –4 We are not given the slope of the line, but we can choose any
two points on the line to find the slope. Let’s choose (0, 7) and (4, 4). We can use these points to calculate the slope. m = y2 − y1 _ x2 − x1 = 4 − 7 _____ 4 − 0 = − #3 _ 4 Now we can substitute the slope and the coordinates of one of the points into the point-slope form. y − y1 = m(x − x1) y − 4 = − #3 _ (x − 4) 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 286 CHAPTER 4 LINEAR FUNCTIONS If we want to rewrite the equation in the slope-intercept form, we would find y − 4 = − #3 _ (x − 4) 4 y − 4 = − #3 _ x + 7 4 If we wanted to find the slope-intercept form without first writing the point-slope form, we could have recognized that the line crosses the y-axis when the output value is 7. Therefore, b = 7. We now have the initial value b and the slope m so we can substitute m and b into the slope-intercept form of a line. f (x) = mx + b ↑ ↑ − #3 _ 7 4 f (x) = − #3 _ x + 7 4 So the function is f (x) = − #3 x + 7, and the linear equation would be y = − #3 _ _ x + 7. 4 4 How To… Given the graph of a linear function, write an equation to represent the function. 1. Identify two points on the line. 2. Use the two points to calculate the slope. 3. Determine where the line crosses the y-axis to identify the y-intercept by visual inspection. 4. Substitute the slope and y-intercept into the slope-intercept form of a line equation. Example 5 Writing an Equation for a Linear Function Write an equation for a linear function given a graph of f shown in Figure 8. y f 642 8 10 x –10 –8 –6 –4 10 8 6 4 2 –2 –2 –4 –6 –8 –10 Figure 8 Solution Identify two points on the line, such as (0, 2) and (−2, −4). Use the points to calculate the slope. m = y2
− y1 _ x2 − x1 −4 − 2 _______ −2 − 0 −6 ___ −2 = = Substitute the slope and the coordinates of one of the points into the point-slope form. = 3 y − y1 = m(x − x1) y − (−4) = 3(x − (−2)) y + 4 = 3(x + 2) Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 287 We can use algebra to rewrite the equation in the slope-intercept form. y + 4 = 3(x + 2) y + 4 = 3x + 6 y = 3x + 2 Analysis This makes sense because we can see from Figure 9 that the line crosses the y-axis at the point (0, 2), which is the y-intercept, so b = 2. y 10 8 6 4 2 –10 –2 –4 –6 –8 (−2, −4) –2 –4 –6 –8 –10 (0, 2) 642 8 10 x Example 6 Writing an Equation for a Linear Cost Function Figure 9 Suppose Ben starts a company in which he incurs a fixed cost of $1,250 per month for the overhead, which includes his office rent. His production costs are $37.50 per item. Write a linear function C where C(x) is the cost for x items produced in a given month. Solution The fixed cost is present every month, $1,250. The costs that can vary include the cost to produce each item, which is $37.50 for Ben. The variable cost, called the marginal cost, is represented by 37.5. The cost Ben incurs is the sum of these two costs, represented by C(x) = 1250 + 37.5x. Analysis If Ben produces 100 items in a month, his monthly cost is found by substitution 100 for x. So his monthly cost would be $5,000. C(100) = 1,250 + 37.5(100) = 5,000 Example 7 Writing an Equation for a Linear Function Given Two Points If f is a linear function, with f (3) = −2, and f (8) = 1, find an equation for the function in slope-intercept form. Solution We can write the given points using coordinates. We can then use
the points to calculate the slope. f (3) = −2 → (3, −2) f (8) = 1 → (8, 1) m = y2 − y1 _ x2 − x1 1 − (−2) ________ 8 − 3 = Substitute the slope and the coordinates of one of the points into the point-slope form. = #3 _ 5 y − y1 = m(x − x1) y − (−2) =# #3 _ (x − 3) 5 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 288 CHAPTER 4 LINEAR FUNCTIONS We can use algebra to rewrite the equation in the slope-intercept form. y + 2 =# #3 _ (x − 3) 5 x −# #9 y + 2 =# #3 _ _ 5 5 19_ y = #3 _ x − 5 5 Try It #3 If f (x) is a linear function, with f (2) = −11, and f (4) = −25, find an equation for the function in slope-intercept form. Modeling Real-World Problems with Linear Functions In the real world, problems are not always explicitly stated in terms of a function or represented with a graph. Fortunately, we can analyze the problem by first representing it as a linear function and then interpreting the components of the function. As long as we know, or can figure out, the initial value and the rate of change of a linear function, we can solve many different kinds of real-world problems. How To… Given a linear function f and the initial value and rate of change, evaluate f (c). 1. Determine the initial value and the rate of change (slope). 2. Substitute the values into f (x) = mx + b. 3. Evaluate the function at x = c. Example 8 Using a Linear Function to Determine the Number of Songs in a Music Collection Marcus currently has 200 songs in his music collection. Every month, he adds 15 new songs. Write a formula for the number of songs, N, in his collection as a function of time, t, the number of months. How many songs will he own in a year? Solution The initial value for this function is 200 because he currently owns 200 songs, so N(0) = 200, which means that b = 200. The number of songs increases
by 15 songs per month, so the rate of change is 15 songs per month. Therefore we know that m = 15. We can substitute the initial value and the rate of change into the slope-intercept form of a line. f (x) = mx + b ↑ ↑ 15 200 N(t) = 15t + 200 Figure 10 We can write the formula N(t) = 15t + 200. With this formula, we can then predict how many songs Marcus will have in 1 year (12 months). In other words, we can evaluate the function at t = 12. N(12) = 15(12) + 200 = 180 + 200 = 380 Marcus will have 380 songs in 12 months. Analysis Notice that N is an increasing linear function. As the input (the number of months) increases, the output (number of songs) increases as well. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 289 Example 9 Using a Linear Function to Calculate Salary Based on Commission Working as an insurance salesperson, Ilya earns a base salary plus a commission on each new policy. Therefore, Ilya’s weekly income, I, depends on the number of new policies, n, he sells during the week. Last week he sold 3 new policies, and earned $760 for the week. The week before, he sold 5 new policies and earned $920. Find an equation for I(n), and interpret the meaning of the components of the equation. Solution The given information gives us two input-output pairs: (3,760) and (5,920). We start by finding the rate of change. m = 920 − 760 ________ 5 − 3 $160 _______ 2 policies = $80 per policy = Keeping track of units can help us interpret this quantity. Income increased by $160 when the number of policies increased by 2, so the rate of change is $80 per policy. Therefore, Ilya earns a commission of $80 for each policy sold during the week. We can then solve for the initial value. I(n) = 80n + b 760 = 80(3) + b when n = 3, I(3) = 760 760 − 80(3) = b 520 = b The value of b is the starting value for the function and represents Ilya’s income when n = 0, or when no new policies are sold
. We can interpret this as Ilya’s base salary for the week, which does not depend upon the number of policies sold. We can now write the final equation. I(n) = 80n + 520 Our final interpretation is that Ilya’s base salary is $520 per week and he earns an additional $80 commission for each policy sold. Example 10 Using Tabular Form to Write an Equation for a Linear Function Table 1 relates the number of rats in a population to time, in weeks. Use the table to write a linear equation. Number of weeks, w 0 2 4 6 Number of rats, P(w) 1,000 1,080 1,160 1,240 Table 1 Solution We can see from the table that the initial value for the number of rats is 1,000, so b = 1,000. Rather than solving for m, we can tell from looking at the table that the population increases by 80 for every 2 weeks that pass. This means that the rate of change is 80 rats per 2 weeks, which can be simplified to 40 rats per week. P(w) = 40w + 1000 If we did not notice the rate of change from the table we could still solve for the slope using any two points from the table. For example, using (2, 1080) and (6, 1240) m = 1240 − 1080 __________ 6 − 2 = 160___ 4 = 40 Q & A… Is the initial value always provided in a table of values like Table 1? No. Sometimes the initial value is provided in a table of values, but sometimes it is not. If you see an input of 0, then the initial value would be the corresponding output. If the initial value is not provided because there is no value of input on the table equal to 0, find the slope, substitute one coordinate pair and the slope into f (x) = mx + b, and solve for b. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 290 CHAPTER 4 LINEAR FUNCTIONS Try It #4 A new plant food was introduced to a young tree to test its effect on the height of the tree. Table 2 shows the height of the tree, in feet, x months since the measurements began. Write a linear function, H(x), where x is the number of months since the start of the experiment. x 0 2 4 8 12 H(x) 12
.5 13.5 14.5 16.5 18.5 Table 2 Graphing Linear Functions Now that we’ve seen and interpreted graphs of linear functions, let’s take a look at how to create the graphs. There are three basic methods of graphing linear functions. The first is by plotting points and then drawing a line through the points. The second is by using the y-intercept and slope. And the third method is by using transformations of the identity function f (x) = x. Graphing a Function by Plotting Points To fi nd points of a function, we can choose input values, evaluate the function at these input values, and calculate output values. The input values and corresponding output values form coordinate pairs.We then plot the coordinate pairs on a grid. In general, we should evaluate the function at a minimum of two inputs in order to find at least two points on the graph. For example, given the function, f (x) = 2x, we might use the input values 1 and 2. Evaluating the function for an input value of 1 yields an output value of 2, which is represented by the point (1, 2). Evaluating the function for an input value of 2 yields an output value of 4, which is represented by the point (2, 4). Choosing three points is often advisable because if all three points do not fall on the same line, we know we made an error. How To… Given a linear function, graph by plotting points. 1. Choose a minimum of two input values. 2. Evaluate the function at each input value. 3. Use the resulting output values to identify coordinate pairs. 4. Plot the coordinate pairs on a grid. 5. Draw a line through the points. Example 11 Graphing by Plotting Points Graph f (x) = − #2 __ x + 5 by plotting points. 3 Solution Begin by choosing input values. This function includes a fraction with a denominator of 3, so let’s choose multiples of 3 as input values. We will choose 0, 3, and 6. Evaluate the function at each input value, and use the output value to identify coordinate pairs0) = − #2 __ (0) + 5 = 5 ⇒ (0, 5) 3 f (3) = − #2 __ (3) + 5 = 3 ⇒ (3, 3) 3 f (6) = − #2 __ (6) +
5 = 1 ⇒ (6, 1) 3 Plot the coordinate pairs and draw a line through the points. Figure 11 represents the graph of the function 2 __ x + 5. f (x) = − 3 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 291 f(x) (0, 5) f 6 5 4 3 2 1 (3, 3) (6, 1) – – __ x + 5. Figure 11 The graph of the linear function f (x) = − # 3 Analysis The graph of the function is a line as expected for a linear function. In addition, the graph has a downward slant, which indicates a negative slope. This is also expected from the negative, constant rate of change in the equation for the function. Try It #5 Graph f (x) = − #3 _ x + 6 by plotting points. 4 Graphing a Function Using y-intercept and Slope Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. The first characteristic is its y-intercept, which is the point at which the input value is zero. To find the y-intercept, we can set x = 0 in the equation. The other characteristic of the linear function is its slope. Let’s consider the following function. f (x) =# #1 _ x + 1 2 1 _ The slope is. Because the slope is positive, we know the graph will slant upward from left to right. The y-intercept is 2 the point on the graph when x = 0. The graph crosses the y-axis at (0, 1). Now we know the slope and the y-intercept. rise _ We can begin graphing by plotting the point (0, 1). We know that the slope is rise over run, m = run. From our example, we have m =# #1 _, which means that the rise is 1 and the run is 2. So starting from our y-intercept (0, 1), we can 2 rise 1 and then run 2, or run 2 and then rise 1. We repeat until we have a few points, and then we draw a line through the points as shown in Figure 12. y 5 4 3 2 (0, 1) –2 –1 f y-intercept ←Rise = 1
↑Run = 2 1 2 4 3 Figure 12 x 5 6 7 graphical interpretation of a linear function In the equation f (x) = mx + b • b is the y-intercept of the graph and indicates the point (0, b) at which the graph crosses the y-axis. • m is the slope of the line and indicates the vertical displacement (rise) and horizontal displacement (run) between each successive pair of points. Recall the formula for the slope: m = change in output (rise) ___ = change in input (run) ∆y _ ∆x = y2 − y1 _ x2 − x1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 292 CHAPTER 4 LINEAR FUNCTIONS Q & A… Do all linear functions have y-intercepts? Yes. All linear functions cross the y-axis and therefore have y-intercepts. (Note: A vertical line parallel to the y-axis does not have a y-intercept, but it is not a function.) How To… Given the equation for a linear function, graph the function using the y-intercept and slope. 1. Evaluate the function at an input value of zero to find the y-intercept. 2. Identify the slope as the rate of change of the input value. 3. Plot the point represented by the y-intercept. 4. Use rise _ run to determine at least two more points on the line. 5. Sketch the line that passes through the points. Example 12 Graphing by Using the y-intercept and Slope Graph f (x) = − #2 _ x + 5 using the y-intercept and slope. 3 Solution Evaluate the function at x = 0 to find the y-intercept. The output value when x = 0 is 5, so the graph will cross the y-axis at (0, 5). According to the equation for the function, the slope of the line is − #2 _. This tells us that for each vertical decrease in 3 the “rise” of −2 units, the “run” increases by 3 units in the horizontal direction. We can now graph the function by first plotting the y-intercept on the graph in Figure 13. From the initial value (0, 5) we move down 2 units and to the right 3 units. We can extend the line to
the left and right by repeating, and then draw a line through the points. f f(x) 6 5 4 3 2 1 – – __ x + 5 and shows how to calculate the rise over run for the slope. Figure 13 Graph of f (x ) = − 3 Analysis The graph slants downward from left to right, which means it has a negative slope as expected. Try It #6 Find a point on the graph we drew in Example 12 that has a negative x-value. Graphing a Function Using Transformations Another option for graphing is to use a transformation of the identity function f (x) = x. A function may be transformed by a shift up, down, left, or right. A function may also be transformed using a reflection, stretch, or compression. Vertical Stretch or Compression In the equation f (x) = mx, the m is acting as the vertical stretch or compression of the identity function. When m is negative, there is also a vertical reflection of the graph. Notice in Figure 14 that multiplying the equation of f (x) = x by m stretches the graph of f by a factor of m units if m > 1 and compresses the graph of f by a factor of m units if 0 < m < 1. This means the larger the absolute value of m, the steeper the slope. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 293 y 6 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –6 –5 –4 –3 –2 1 2 3 4 5 6 x Figure 14 Vertical stretches and compressions and reﬂections on the function f (x) = x. Vertical Shift In f (x) = mx + b, the b acts as the vertical shift, moving the graph up and down without affecting the slope of the line. Notice in Figure 15 that adding a value of b to the equation of f (x) = x shifts the graph of f a total of b units up if b is positive and.b. units down if b is negative(x) = f(x) = f(x) = f(x) = f(x) = x 42 6 8 10 y 10 8 6 4 2 –2 –2 –4 –6 –8 –10 –10 –8 –6 –4 Figure 15 This graph
illustrates vertical shifts of the function f (x) = x. Using vertical stretches or compressions along with vertical shifts is another way to look at identifying different types of linear functions. Although this may not be the easiest way to graph this type of function, it is still important to practice each method. How To… Given the equation of a linear function, use transformations to graph the linear function in the form f (x) = mx + b. 1. Graph f (x) = x. 2. Vertically stretch or compress the graph by a factor m. 3. Shift the graph up or down b units. Example 13 Graphing by Using Transformations Graph f (x) = #1 __ x − 3 using transformations. 2 1 Solution The equation for the function shows that m =# #1 __ __. The so the identity function is vertically compressed by 2 2 equation for the function also shows that b = −3 so the identity function is vertically shifted down 3 units. First, graph the identity function, and show the vertical compression as in Figure 16. Then show the vertical shift as in Figure 17. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 294 1. CHAPTER 4 LINEAR FUNCTIONS y = x y = x1 2 x 5 6 7 21 3 4 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –7 –6 –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 –7 –6 –5 –4 –3 –2 y = x1 2 y = x − 3 1 2 x 21 3 4 5 6 7 Figure 16 The function, y = x, compressed by a factor of 1 __. 2 1 __ x, shifted down 3 units. Figure 17 The function y =# 2 Try It #7 Graph f (x) = 4 + 2x, using transformations. Q & A… In Example 15, could we have sketched the graph by reversing the order of the transformations? No. The order of the transformations follows the order of operations. When the function is evaluated at a given input, the corresponding output is calculated by following the order of operations. This is why we performed the compression first. For example, following the order: Let the input be 2. f (2) =# #1 __ (22 Writing the Equation for a Function from the Graph
of a Line Earlier, we wrote the equation for a linear function from a graph. Now we can extend what we know about graphing linear functions to analyze graphs a little more closely. Begin by taking a look at Figure 18. We can see right away that the graph crosses the y-axis at the point (0, 4) so this is the y-intercept. y f 42 6 8 10 x –10 –8 –6 –4 10 8 6 4 2 –2 –2 –4 –6 –8 –10 Figure 18 Then we can calculate the slope by finding the rise and run. We can choose any two points, but let’s look at the point (−2, 0). To get from this point to the y-intercept, we must move up 4 units (rise) and to the right 2 units (run). So the slope must be m = rise run =# #4 _ _ = 2 2 Substituting the slope and y-intercept into the slope-intercept form of a line gives y = 2x + 4 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 295 How To… Given a graph of linear function, find the equation to describe the function. 1. Identify the y-intercept of an equation. 2. Choose two points to determine the slope. 3. Substitute the y-intercept and slope into the slope-intercept form of a line. Example 14 Matching Linear Functions to Their Graphs Match each equation of the linear functions with one of the lines in Figure 19. a. f (x) = 2x + 3 b. g(x) = 2x − 3 c. h(x) = −2x + 3 d. j(x) =# #1 __ 2 x + 3 –7 –6 –5 –4 –3 –2 y 5 4 3 2 1 –1 –1 –2 –3 –4 –5 I II III 21 3 4 5 6 7 x IV Figure 19 Solution Analyze the information for each function. a. This function has a slope of 2 and a y-intercept of 3. It must pass through the point (0, 3) and slant upward from left to right. We can use two points to find the slope, or we can compare it with the other functions listed. Function g has the same slope, but a
different y-intercept. Lines I and III have the same slant because they have the same slope. Line III does not pass through (0, 3) so f must be represented by line I. b. This function also has a slope of 2, but a y-intercept of −3. It must pass through the point (0, −3) and slant upward from left to right. It must be represented by line III. c. This function has a slope of −2 and a y-intercept of 3. This is the only function listed with a negative slope, so it must be represented by line IV because it slants downward from left to right. 1 _ and a y-intercept of 3. It must pass through the point (0, 3) and slant upward d. This function has a slope of 2 from left to right. Lines I and II pass through (0, 3), but the slope of j is less than the slope of f so the line for j must be flatter. This function is represented by Line II. Now we can re-label the lines as in Figure 20. j(x) = x + 3 1 2 –7 –6 –5 –4 –3 –2 g(x) = 2 x − 3 21 1 –1 –2 –3 –4 –5 f (x) = 2 x + 3 h(x) = −2 x + 3 Figure 20 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 296 CHAPTER 4 LINEAR FUNCTIONS Finding the x-intercept of a Line So far, we have been finding the y-intercept of a function: the point at which the graph of the function crosses the y-axis. Recall that a function may also have an x-intercept, which is the x-coordinate of the point where the graph of the function crosses the x-axis. In other words, it is the input value when the output value is zero. To fi nd the x-intercept, set a function f (x) equal to zero and solve for the value of x. For example, consider the function shown. Set the function equal to 0 and solve for x. f (x) = 3x − 6 0 = 3x − 6 6 = 3x 2 = x x = 2 The graph of the function crosses the x-axis at the point (2, 0
). Q & A… Do all linear functions have x-intercepts? No. However, linear functions of the form y = c, where c is a nonzero real number, are the only examples of linear functions with no x-intercept. For example, y = 5 is a horizontal line 5 units above the x-axis. This function has no x-intercepts, as shown in Figure 21. y y = 5 21 3 4 5 x –5 –4 –3 –2 5 4 3 2 1 –1–1 –2 –3 –4 –5 Figure 21 x-intercept The x-intercept of the function is value of x when f (x) = 0. It can be solved by the equation 0 = mx + b. Example 15 Finding an x-intercept Find the x-intercept of f (x) =# #1 _ x − 3. 2 Solution Set the function equal to zero to solve for x. 0 =# #1 _ x − 3 2 3 =# #1 __ x 2 6 = x x = 6 The graph crosses the x-axis at the point (6, 0). Analysis A graph of the function is shown in Figure 22. We can see that the x-intercept is (6, 0) as we expected. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 297 –10 –8 –6 –4 y 4 2 –2 –2 –4 –6 –8 –10 642 8 10 x Figure 22 Try It #8 Find the x-intercept of f (x) = #1 _ x − 4. 4 Describing Horizontal and Vertical Lines There are two special cases of lines on a graph—horizontal and vertical lines. A horizontal line indicates a constant output, or y-value. In Figure 23, we see that the output has a value of 2 for every input value. The change in outputs between any two points, therefore, is 0. In the slope formula, the numerator is 0, so the slope is 0. If we use m = 0 in the equation f (x) = mx + b, the equation simplifies to f (x) = b. In other words, the value of the function is a constant. This graph represents the function f (x) = 24 −5 –4 –3 –2 –1
21 3 4 5 x Figure 23 A horizontal line representing the function f (x) = 2. A vertical line indicates a constant input, or x-value. We can see that the input value for every point on the line is 2, but the output value varies. Because this input value is mapped to more than one output value, a vertical line does not represent a function. Notice that between any two points, the change in the input values is zero. In the slope formula, the denominator will be zero, so the slope of a vertical line is undefined. m = change of output __ change of input ←#Non-zero real number ← 0 Figure 24 Example of how a line has a vertical slope. 0 in the denominator of the slope. Notice that a vertical line, such as the one in Figure 25, has an x-intercept, but no y-intercept unless it’s the line x = 0. This graph represents the line x = 2. y 5 4 3 2 1 –1–1 –2 –3 –4 –5 –5 –4 –3 –2 21 3 4 5 x x 2 2 y −4 −2 2 0 2 2 2 4 Figure 25 The vertical line, x = 2, which does not represent a function. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 298 CHAPTER 4 LINEAR FUNCTIONS horizontal and vertical lines Lines can be horizontal or vertical. A horizontal line is a line defined by an equation in the form f (x) = b. A vertical line is a line defined by an equation in the form x = a. Example 16 Writing the Equation of a Horizontal Line Write the equation of the line graphed in Figure 26. –10 –8 –6 –4 y –2 –2 –4 –6 –8 –10 642 8 10 x f Figure 26 Solution For any x-value, the y-value is −4, so the equation is y = −4. Example 17 Writing the Equation of a Vertical Line Write the equation of the line graphed in Figure 27. y 10 8 6 4 2 –2 –2 –4 –6 –8 –10 –10 –8 –6 –4 f 42 6 8 10 x Figure 27 Solution The constant x-value is 7, so the equation is x = 7. Determining Whether Lines are Parallel or Perpendicular The two lines in Figure
28 are parallel lines: they will never intersect. They have exactly the same steepness, which means their slopes are identical. The only difference between the two lines is the y-intercept. If we shifted one line vertically toward the y-intercept of the other, they would become coincident1 –1 –2 –3 –6 –5 –4 –3 –2 21 3 4 5 6 x Figure 28 Parallel lines We can determine from their equations whether two lines are parallel by comparing their slopes. If the slopes are the same and the y-intercepts are different, the lines are parallel. If the slopes are different, the lines are not parallel. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 299 f (x) = −2x + 6} f (x) = −2x − 4 parallel f (x) = 3x + 2} f (x) = 2x + 2 not parallel Unlike parallel lines, perpendicular lines do intersect. Their intersection forms a right, or 90-degree, angle. The two lines in Figure 29 are perpendicular. –6 –5 –4 –3 –2 y 4 3 2 1 –1 –1 –2 –3 –4 –5 –6 –7 –8 21 3 4 5 6 x Figure 29 Perpendicular lines Perpendicular lines do not have the same slope. The slopes of perpendicular lines are different from one another in a specific way. The slope of one line is the negative reciprocal of the slope of the other line. The product of a number and its reciprocal is 1. So, if m1 and m2 are negative reciprocals of one another, they can be multiplied together to yield −1. m1m2 = −1 1 1 _ _ To fi nd the reciprocal of a number, divide 1 by the number. So the reciprocal of 8 is is 8. To, and the reciprocal of 8 8 find the negative reciprocal, first find the reciprocal and then change the sign. As with parallel lines, we can determine whether two lines are perpendicular by comparing their slopes, assuming that the lines are neither horizontal nor vertical. The slope of each line below is the negative reciprocal of the other so the lines are perpendicular. The product of the slopes is −1. f (x) =# #1 1 _ _ is −4 x + 2 negative reciprocal of 4 4 1 _ f (x
) = −4x + 3 negative reciprocal of −4 is 4 1 −4 ( ) = −1 _ 4 parallel and perpendicular lines Two lines are parallel lines if they do not intersect. The slopes of the lines are the same. f (x) = m1x + b1 and g(x) = m2x + b2 are parallel if m1 = m2. If and only if b1 = b2 and m1 = m2, we say the lines coincide. Coincident lines are the same line. Two lines are perpendicular lines if they intersect at right angles. f (x) = m1x + b1 and g(x) = m2x + b2 are perpendicular if and only if m1m2 = −1, and so m2 = − # 1 ___ m1 Example 18 Identifying Parallel and Perpendicular Lines Given the functions below, identify the functions whose graphs are a pair of parallel lines and a pair of perpendicular lines. f (x) = 2x + 3 g(x) =# #1 _ x − 4 2 h(x) = −2x + 2 j(x) = 2x − 6 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 30 0 CHAPTER 4 LINEAR FUNCTIONS Solution Parallel lines have the same slope. Because the functions f (x) = 2x + 3 and j(x) = 2x − 6 each have a slope 1 _ of 2, they represent parallel lines. Perpendicular lines have negative reciprocal slopes. Because −2 and are negative 2 reciprocals, the equations, g(x) =# #1 _ x − 4 and h(x) = −2x + 2 represent perpendicular lines. 2 Analysis A graph of the lines is shown in Figure 30. h(x) = −2x + 2 y f (x) = 2x + 3 j(x) = 2x − 6 g(x) = x − 4 1 2 42 6 8 10 x –10 –8 –6 –4 10 8 6 4 2 –2 –2 –4 –6 –8 –10 Figure 30 The graph shows that the lines f (x) = 2x + 3 and j(x) = 2x − 6 are parallel, and the lines g(x) =# #1 _ 2 h(x) = −2x + 2 are perpendicular. x
− 4 and Writing the Equation of a Line Parallel or Perpendicular to a Given Line If we know the equation of a line, we can use what we know about slope to write the equation of a line that is either parallel or perpendicular to the given line. Writing Equations of Parallel Lines Suppose for example, we are given the equation shown. f (x) = 3x + 1 We know that the slope of the line formed by the function is 3. We also know that the y-intercept is (0, 1). Any other line with a slope of 3 will be parallel to f (x). So the lines formed by all of the following functions will be parallel to f (x). g(x) = 3x + 6 h(x) = 3x + 1 p(x) = 3x +# #2 __ 3 Suppose then we want to write the equation of a line that is parallel to f and passes through the point (1, 7). This type of problem is often described as a point-slope problem because we have a point and a slope. In our example, we know that the slope is 3. We need to determine which value for b will give the correct line. We can begin with the point-slope form of an equation for a line, and then rewrite it in the slope-intercept form. y − y1 = m(x − x1) y − 7 = 3(x − 1) y − 7 = 3x − 3 y = 3x + 4 So g(x) = 3x + 4 is parallel to f (x) = 3x + 1 and passes through the point (1, 7). How To… Given the equation of a function and a point through which its graph passes, write the equation of a line parallel to the given line that passes through the given point. 1. Find the slope of the function. 2. Substitute the given values into either the general point-slope equation or the slope-intercept equation for a line. 3. Simplify. Example 19 Finding a Line Parallel to a Given Line Find a line parallel to the graph of f (x) = 3x + 6 that passes through the point (3, 0). Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 301 Solution The slope of the given line is 3. If we choose the slope-inter
cept form, we can substitute m = 3, x = 3, and f (x) = 0 into the slope-intercept form to find the y-intercept. g(x) = 3x + b 0 = 3(3) + b b = −9 The line parallel to f (x) that passes through (3, 0) is g(x) = 3x − 9. Analysis We can confirm that the two lines are parallel by graphing them. Figure 31 shows that the two lines will never intersect. y 5 4 3 2 1 Right 1 –6 –3 –4 –5 y = 3x + 6 –2 –1 –1 –2 –3 –4 –5 Up 3 Up 3 1 2 4 5 6 3 Right 1 x y = 3x − 9 Figure 31 Writing Equations of Perpendicular Lines We can use a very similar process to write the equation for a line perpendicular to a given line. Instead of using the same slope, however, we use the negative reciprocal of the given slope. Suppose we are given the following function: f (x) = 2x + 4. Any function with a slope of − #1 The slope of the line is 2, and its negative reciprocal is − #1 __ __ will be perpendicular to 2 2 f (x). So the lines formed by all of the following functions will be perpendicular to f (x). g(x) = − #1 __ x + 4 2 h(x) = − #1 __ x + 2 2 x − #1 p(x) = − #1 __ __ 2 2 As before, we can narrow down our choices for a particular perpendicular line if we know that it passes through a given point. Suppose then we want to write the equation of a line that is perpendicular to f (x) and passes through the point (4, 0). We already know that the slope is − #1 __. Now we can use the point to find the y-intercept by substituting 2 the given values into the slope-intercept form of a line and solving for b. g(x) = mx + b 0 = − #1 __ (4) + b 2 0 = −2 + b 2 = b b = 2 The equation for the function with a slope of − #1 __ and a y-intercept of 2 is 2 g(x) = − #1 __ x + 2. 2 So g(x) = − #1
__ x + 2 is perpendicular to f (x) = 2x + 4 and passes through the point (4, 0). Be aware that perpendicular 2 lines may not look obviously perpendicular on a graphing calculator unless we use the square zoom feature. Q & A… A horizontal line has a slope of zero and a vertical line has an undefined slope. These two lines are perpendicular, but the product of their slopes is not −1. Doesn’t this fact contradict the definition of perpendicular lines? No. For two perpendicular linear functions, the product of their slopes is −1. However, a vertical line is not a function so the definition is not contradicted. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. 30 2 CHAPTER 4 LINEAR FUNCTIONS How To… Given the equation of a function and a point through which its graph passes, write the equation of a line perpendicular to the given line. 1. Find the slope of the function. 2. Determine the negative reciprocal of the slope. 3. Substitute the new slope and the values for x and y from the coordinate pair provided into g(x) = mx + b. 4. Solve for b. 5. Write the equation for the line. Example 20 Finding the Equation of a Perpendicular Line Find the equation of a line perpendicular to f (x) = 3x + 3 that passes through the point (3, 0). Solution The original line has slope m = 3, so the slope of the perpendicular line will be its negative reciprocal, or − #1 __. Using this slope and the given point, we can find the equation for the line. 3 g(x) = − #1 __ x + b 3 0 = − #1 __ (3 The line perpendicular to f (x) that passes through (3, 0) is g(x) = − #1 __ x + 1. 3 Analysis A graph of the two lines is shown in Figure 321 –1 –2 –3 –6 –5 –4 –3 –2 y f (x) = 3 x + 6 g(x) = − x + 1 1 3 x 21 3 4 5 6 Note that that if we graph perpendicular lines on a graphing calculator using standard zoom, the lines may not appear to be perpendicular. Adjusting the window will make it possible to zoom in further to see the intersection more closely. Figure 32 Try It #9 Given the function
h(x) = 2x − 4, write an equation for the line passing through (0, 0) that is a. parallel to h(x) b. perpendicular to h(x) How To… Given two points on a line and a third point, write the equation of the perpendicular line that passes through the point. 1. Determine the slope of the line passing through the points. 2. Find the negative reciprocal of the slope. 3. Use the slope-intercept form or point-slope form to write the equation by substituting the known values. 4. Simplify. Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 LINEAR FUNCTIONS 303 Example 21 Finding the Equation of a Line Perpendicular to a Given Line Passing through a Point A line passes through the points (−2, 6) and (4, 5). Find the equation of a perpendicular line that passes through the point (4, 5). Solution From the two points of the given line, we can calculate the slope of that line. Find the negative reciprocal of the slope. m1 = 5 − 6 _______ 4 − (−2) = −1___ 6 = − #1 __ 6 m2 = −1_ − #1 __ 6 = −1 ( − #6 __ ) 1 = 6 We can then solve for the y-intercept of the line passing through the point (4, 5). g(x) = 6x + b 5 = 6(4) + b 5 = 24 + b −19 = b b = −19 The equation for the line that is perpendicular to the line passing through the two given points and also passes through point (4, 5) is y = 6x − 19 Try It #10 A line passes through the points, (−2, −15) and (2,−3). Find the equation of a perpendicular line that passes through the point, (6, 4). Access this online resource for additional instruction and practice with linear functions. • Linear Functions (http://openstaxcollege.org/l/linearfunctions) • Finding Input of Function from the Output and Graph (http://openstaxcollege.org/l/ﬁndinginput) • Graphing Functions Using Tables (http://openstaxcollege.org/l/graphwithtable) Download the OpenStax text for free at http://cnx.
org/content/col11759/latest. 30 4 CHAPTER 4 LINEAR FUNCTIONS 4.1 SECTION EXERCISES VERBAL 1. Terry is skiing down a steep hill. Terry’s elevation, E(t), in feet after t seconds is given by E(t) = 3000 − 70t. Write a complete sentence describing Terry’s starting elevation and how it is changing over time. 2. Jessica is walking home from a friend’s house. After 2 minutes she is 1.4 miles from home. Twelve minutes after leaving, she is 0.9 miles from home. What is her rate in miles per hour? 3. A boat is 100 miles away from the marina, sailing directly toward it at 10 miles per hour. Write an equation for the distance of the boat from the marina after t hours. 4. If the graphs of two linear functions are perpendicular, describe the relationship between the slopes and the y-intercepts. 5. If a horizontal line has the equation f (x) = a and a vertical line has the equation x = a, what is the point of intersection? Explain why what you found is the point of intersection. ALGEBRAIC For the following exercises, determine whether the equation of the curve can be written as a linear function. 6. y =# #1 __ x + 6 4 9. 3x + 5y = 15 12. −2x 2 + 3y2 = 6 7. y = 3x − 5 10. 3x 2 + 5y = 15 13. − #x − 3 ______ = 2y 5 8. y = 3x 2 − 2 11. 3x + 5y 2 = 15 For the following exercises, determine whether each function is increasing or decreasing. 14. f (x) = 4x + 3 15. g(x) = 5x + 6 16. a(x) = 5 − 2x 17. b(x) = 8 − 3x 20. j(x) =# #1 __ x − 3 2 23. m(x) = − #3 __ x + 3 8 18. h(x) = −2x + 4 21. p(x) =# #1 __ x − 5 4 19. k(x) = −4x + 1 22. n(x) = − #1 __ x − 2 3 For the following exercises, find the slope of the line that passes through the
two given points. 24. (2, 4) and (4, 10) 27. (8, −2) and (4, 6) 25. (1, 5) and (4, 11) 28. (6, 11) and (−4, 3) 26. (−1, 4) and (5, 2) For the following exercises, given each set of information, find a linear equation satisfying the conditions, if possible. 29. f (−5) = −4, and f (5) = 2 30. f (−1) = 4 and f (5) = 1 31. Passes through (2, 4) and (4, 10) 33. Passes through (−1, 4) and (5, 2) 32. Passes through (1, 5) and (4, 11) 34. Passes through (−2, 8) and (4, 6) 35. x-intercept at (−2, 0) and y-intercept at (0, −3) 36. x-intercept at (−5, 0) and y-intercept at (0, 4) For the following exercises, determine whether the lines given by the equations below are parallel, perpendicular, or neither. 37. 4x − 7y = 10 7x + 4y = 1 39. 3y + 4x = 12 −6y = 8x + 1 40. 6x − 9y = 10 38. 3y + x = 12 −y = 8x + 1 3x + 2y = 1 Download the OpenStax text for free at http://cnx.org/content/col11759/latest. SECTION 4.1 SECTION EXERCISES 305 For the following exercises, find the x- and y-intercepts of each equation 41. f (x) = −x + 2 44. k(x) = −5x + 1 42. g(x) = 2x + 4 45. −2x + 5y = 20 43. h(x) = 3x − 5 46. 7x + 2y = 56 For the following exercises, use the descriptions of each pair of lines given below to find the slopes of Line 1 and Line 2. Is each pair of lines parallel, perpendicular, or neither? 47. Line 1: Passes through (0, 6) and (3, −24) Line 2: Passes through (−1, 19) and (8, −
71) 48. Line 1: Passes through (−8, −55) and (10, 89) Line 2: Passes through (9, −44) and (4, −14) 49. Line 1: Passes through (2, 3) and (4, −1) Line 2: Passes through (6, 3) and (8, 5) 50. Line 1: Passes through (1, 7) and (5, 5) Line 2: Passes through (−1, −3) and (1, 1) 51. Line 1: Passes through (2, 5) and (5, −1) Line 2: Passes through (−3, 7) and (3, −5) For the following exercises, write an equation for the line described. 52. Write an equation for a line parallel to f (x) = −5x − 3 53. Write an equation for a line parallel to g (x) = 3x − 1 and passing through the point (2, −12). and passing through the point (4, 9). 54. Write an equation for a line perpendicular to 55. Write an equation for a line perpendicular to h (t) = −2t + 4 and passing through the point (−4, −1). p (t) = 3t + 4 and passing through the point (3, 1). GRAPHICAL For the following exercises, find the slope of the lines graphed. 57. 321 4 5 6 x –6 –5 –4 –3 –2 56. –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 –6 y 6 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 –6 321 4 5 6 x For the following exercises, write an equation for the lines graphed. 58. –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 –6 59. 321 4 5 6 x –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 –6 60. 321 4 5 6 x –5 –4 –3 –2 y 6 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 –6 321 4 5 x Download the
OpenStax text for free at http://cnx.org/content/col11759/latest. 30 6 61. CHAPTER 4 LINEAR FUNCTIONS y 6 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 –6 –5 –4 –3 –2 62. 321 4 5 x –6 –5 –4 –3 –2 y 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 63. 321 4 5 6 x –6 –5 –4 –3 –2 y 5 4 3 2 1 –1 0 –1 –2 –3 –4 –5 321 4 5 6 x For the following exercises, match the given linear equation with its graph in Figure 33. B 5 A 4 3 2 1 –5 –4 –3 –2 –1–1 –2 –3 –4 –5 21 3 4 5 E F Figure 33 C D 64. f (x) = −x − 1 65. f (x) = −2x − 1 66. f (x) = − #1 __ x − 1 2 67. f (x) = 2 68. f (x) = 2 + x 69. f (x) = 3x + 2 For the following exercises, sketch a line with the given features. 70. An x-intercept of (−4, 0) and y-intercept of (0, −2) 71. An x-intercept of (−2, 0) and y-intercept of (0, 4) 72. A y-intercept of (0, 7) and slope − #3 __ 2 2 __ 73. A y-intercept of (0, 3) and slope 5 74. Passing through the points (−6, −2) and (6, −6) 75. Passing through the points (−3, −4) and (3, 0) For the following exercises, sketch the graph of each equation. 76. f (x) = −2x − 1 77. g(x) = −3x + 2 79. k(x) =# #2 __ x − 3 3 82. x = 3 80. f (t) = 3 + 2t 83. x = −2 For the following exercises, write the equation of the line shown in the graph. 85. –6 –5 –4 –3 –2 y 6 5 4 3 2 1 –1–1 –

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