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we give you for a and b. How many functions did you ﬁnd that failed to satisfy the conditions above? Did f (x) = x2 work? What about f (x) =? Did you ﬁnd an attribute x or f (x) = 3x + 7 or f (x) = common to those functions that did succeed? You should have, because there is only one extremely special family of functions that actually works here. Thus we return to our previous statement, in general, f (a + b) = f (a) + f (b). 1 x √ 1.4 Function Notation 1.4.3 Answers 1. f (x) = 2x+3 4 Domain: (−∞, ∞) 69 2. f (x) = 2(x+3) 4 = x+3 2 Domain: (−∞, ∞) 3. f (x Domain: (−∞, ∞) √ 4. f (x) = 2x + 3 Domain: − 3 2, ∞ 5. f (x) = 2(x + 3) = √ 2x + 6 Domain: [−3, ∞) 7. f (x) = 4√ x−13 Domain: [0, 169) ∪ (169, ∞) 6. f (x) = 2 √ x + 3 Domain: [−3, ∞) 8. f (x) = 4√ x−13 Domain: (13, ∞) 9. f (x) = 4√ x − 13 Domain: (0, ∞) 11. For f (x) = 2x + 1 10. f (x) = 4 x − 13 = 2√ x − 13 Domain: (0, ∞) f (3) = 7 f (−1) = −1 f 3 2 = 4 f (4x) = 8x + 1 4f (x) = 8x + 4 f (−x) = −2x + 1 f (x − 4) = 2x − 7 f (x) − 4 = 2x − 3 f x2 = 2x2 + 1 12. For f (x) = 3 − 4x f (3) = −9 f (−1) = 7 f 3 2 = −3 f (4x) = 3 − 16x 4f (x) = 12 − 16
x f (−x) = 4x + 3 f (x − 4) = 19 − 4x f (x) − 4 = −4x − 1 f x2 = 3 − 4x2 70 Relations and Functions 13. For f (x) = 2 − x2 f (3) = −7 f (−14x) = 2 − 16x2 4f (x) = 8 − 4x2 f (−x) = 2 − x2 f (x − 4) = −x2 + 8x − 14 f (x) − 4 = −x2 − 2 f x2 = 2 − x4 14. For f (x) = x2 − 3x + 2 f (3) = 2 f (−14x) = 16x2 − 12x + 2 4f (x) = 4x2 − 12x + 8 f (−x) = x2 + 3x + 2 f (x − 4) = x2 − 11x + 30 f (x) − 4 = x2 − 3x − 2 f x2 = x4 − 3x2 + 2 15. For f (x) = x x−1 f (3) = 3 2 f (4x) = 4x 4x−1 f (x − 4) = x−4 x−5 16. For f (x) = 2 x3 f (3) = 2 27 f (4x) = 1 32x3 f (x − 4) = 2 (x−4)3 = 2 x3−12x2+48x−64 17. For f (x) = 6 f (3) = 6 f (4x) = 6 f (−1) = 1 2 4f (x) = 4x x−1 f (x) − 4 = x x−1 − 4 = 4−3x x−1 f (−1) = −2 4f (x) = 8 x3 f (x) − 4 = 2 x3 − 4 = 2−4x3 x3 f (−1) = 6 4f (x) = 24 f (x − 4) = 6 f (x (−x) = x x+1 f x2 = x2 x2−1 f 3 2 = 16 27 f (−x) = − 2 x3 f x2 = 2 x6 f 3 2 = 6 f (−x) = 6 f x2 = 6 1.4
Function Notation 71 18. For f (x) = 0 f (3) = 0 f (−1) = 0 f (4x) = 0 4f (x) = 0 f (x − 4) = 0 f (x) − 4 = −4 f 3 2 = 0 f (−x) = 0 f x2 = 0 19. For f (x) = 2x − 5 f (2) = −1 f (−2) = −9 f (2a) = 4a − 5 2f (a) = 4a − 10 f (a + 2) = 2a − 1 f (a) + f (2) = 2a − −5a a 20. For f (x) = 5 − 2x f (a) 2 = 2a−5 2 f (a + h) = 2a + 2h − 5 f (2) = 1 f (−2) = 9 f (2a) = 5 − 4a 2f (a) = 10 − 4a f (a + 2) = 1 − 2a f (a) + f (2) = 6 − 2a f 2 a = 5 − 4 a = 5a−4 a 21. For f (x) = 2x2 − 1 f (a) 2 = 5−2a 2 f (a + h) = 5 − 2a − 2h f (2) = 7 f (−2) = 7 f (2a) = 8a2 − 1 2f (a) = 4a2 − 2 f (a + 2) = 2a2 + 8a + 7 f (a) + f (2) = 2a2 + 6 f 2 a = 8 a2 − 1 = 8−a2 a2 f (a) 2 = 2a2−1 2 f (a + h) = 2a2 + 4ah + 2h2 − 1 72 Relations and Functions 22. For f (x) = 3x2 + 3x − 2 f (2) = 16 f (−2) = 4 f (2a) = 12a2 + 6a − 2 2f (a) = 6a2 + 6a − 4 f (a+2) = 3a2 +15a+16 f (a) + f (2) = 3a2 + 3a + 14 f 2 a = 12 a2 + 6 a − 2 = 12+
6a−2a2 a2 23. For f (x) = √ 2x + 1 f (a) 2 = 3a2+3a−2 2 f (a + h) = 3a2 + 6ah + 3h2 + 3a + 3h − 2 f (2) = √ 5 f (−2) is not real f (2a) = √ 4a + 1 2f (a) = 2 √ 2a + 1 f (a + 2) = √ 2a + 5 f (a)+f (2) = √ 2a + 1+ √ 5 f (a) 2 = √ 2a+1 2 f (a+h) = √ 2a + 2h + +4 a 24. For f (x) = 117 f (2) = 117 f (−2) = 117 f (2a) = 117 2f (a) = 234 f (a + 2) = 117 f (a) + f (2) = 234 f 2 a = 117 25. For f (x) = x 2 f (2) = 1 2f (aa) 2 = 117 2 f (a + h) = 117 f (−2) = −1 f (2a) = a f (a + 2) = a+2 2 f (a) 2 = a 4 f (a) + f (2) = a 2 + 1 = a+2 2 f (a + h) = a+h 2 1.4 Function Notation 73 26. For f (x) = 2 x f (2) = 1 2f (a (−2) = −1 f (a + 2) = 2 a+2 f (a) 2 = 1 a f (2a) = 1 a f (a) + f (2) = 2 a + 1 = a+2 2 f (a + h) = 2 a+h 27. For f (x) = 2x − 1, f (0) = −1 and f (x) = 0 when x = 1 2 28. For f (x) = 3 − 2 5 x, f (0) = 3 and f (x) = 0 when x = 15 2 29. For f (x) = 2x2 − 6, f (0) = −6 and f (x) = 0 when x = ± √ 3 30
. For f (x) = x2 − x − 12, f (0) = −12 and f (x) = 0 when x = −3 or x = 4 31. For f (x) = 32. For f (x) = √ √ x + 4, f (0) = 2 and f (x) = 0 when x = −4 1 − 2x, f (0) = 1 and f (x) = 0 when x = 1 2 33. For f (x) = 3 4−x, f (0) = 3 4 and f (x) is never equal to 0 34. For f (x) = 3x2−12x 4−x2, f (0) = 0 and f (x) = 0 when x = 0 or x = 4 35. (a) f (−4) = 1 (b) f (−3) = 2 (d) f (3.001) = 1.999 (e) f (−3.001) = 1.999 36. (a) f (4) = 4 (d) f (0) = 1 37. (−∞, ∞) 39. (−∞, −1) ∪ (−1, ∞) 41. (−∞, ∞) (b) f (−3) = 9 (e) f (−1) = 1 (c) f (3) = 0 √ (f) f (2) = 5 (c) f (1) = 0 (f) f (−0.999) ≈ 0.0447 38. (−∞, ∞) 40. (−∞, −2) ∪ (−2, 1) ∪ (1, ∞) 42. (−∞, − √ √ 3) ∪ (− √ 3, 3) ∪ ( √ 3, ∞) 43. (−∞, −6) ∪ (−6, 6) ∪ (6, ∞) 44. (−∞, 2) ∪ (2, ∞) 45. (−∞, 3] 46. − 5 2, ∞ 74 47. [−3, ∞) 49. 1 3, ∞ 51. (−∞, ∞) 53. 1 3, 6 ∪ (6, ∞) 55. (−∞, 8) ∪ (8, ∞
) 57. (8, ∞) 59. (−∞, 8) ∪ (8, ∞) 61. [0, 5) ∪ (5, ∞) Relations and Functions 48. (−∞, 7] 50. 1 523, ∞) 54. (−∞, ∞) 56. [0, 8) ∪ (8, ∞) 58. [7, 9] 60. −∞, −, 1 2 ∪ 1 2, ∞ 62. [0, 25) ∪ (25, ∞) 63. A(3) = 9, so the area enclosed by a square with a side of length 3 inches is 9 square inches. The solutions to A(x) = 36 are x = ±6. Since x is restricted to x > 0, we only keep x = 6. This means for the area enclosed by the square to be 36 square inches, the length of the side needs to be 6 inches. Since x represents a length, x > 0. 64. A(2) = 4π, so the area enclosed by a circle with radius 2 meters is 4π square meters. The solutions to A(r) = 16π are r = ±4. Since r is restricted to r > 0, we only keep r = 4. This means for the area enclosed by the circle to be 16π square meters, the radius needs to be 4 meters. Since r represents a radius (length), r > 0. 65. V (5) = 125, so the volume enclosed by a cube with a side of length 5 centimeters is 125 cubic centimeters. The solution to V (x) = 27 is x = 3. This means for the volume enclosed by the cube to be 27 cubic centimeters, the length of the side needs to 3 centimeters. Since x represents a length, x > 0. 66. V (3) = 36π, so the volume enclosed by a sphere with radius 3 feet is 36π cubic feet. The is r = 2. This means for the volume enclosed by the sphere to be 32π 3 solution to V (r) = 32π 3 cubic feet, the radius needs to 2 feet. Since r represents a radius (length), r > 0. 67. h(0) = 64, so at the moment the object is dropped oﬀ the building, the object is 64 feet oﬀ of the ground. The solutions to h(
t) = 0 are t = ±2. Since we restrict 0 ≤ t ≤ 2, we only keep t = 2. This means 2 seconds after the object is dropped oﬀ the building, it is 0 feet oﬀ the ground. Said diﬀerently, the object hits the ground after 2 seconds. The restriction 0 ≤ t ≤ 2 restricts the time to be between the moment the object is released and the moment it hits the ground. 68. T (0) = 3, so at 6 AM (0 hours after 6 AM), it is 3◦ Fahrenheit. T (6) = 33, so at noon (6 hours after 6 AM), the temperature is 33◦ Fahrenheit. T (12) = 27, so at 6 PM (12 hours after 6 AM), it is 27◦ Fahrenheit. 1.4 Function Notation 75 69. C(0) = 27, so to make 0 pens, it costs15 $2700. C(2) = 11, so to make 2000 pens, it costs $1100. C(5) = 2, so to make 5000 pens, it costs $2000. 70. F (0) = 16.00, so in 1980 (0 years after 1980), the average fuel economy of passenger cars in the US was 16.00 miles per gallon. F (14) = 20.81, so in 1994 (14 years after 1980), the average fuel economy of passenger cars in the US was 20.81 miles per gallon. F (28) = 22.64, so in 2008 (28 years after 1980), the average fuel economy of passenger cars in the US was 22.64 miles per gallon. 71. P (0) = 0 which means in 1803 (0 years after 1803), there are no Sasquatch in Portage County. 22 ≈ 139.77, so in 2008 (205 years after 1803), there were between 139 and 140 P (205) = 3075 Sasquatch in Portage County. 72. (a) C(20) = 300. It costs $300 for 20 copies of the book. (b) C(50) = 675, so it costs $675 for 50 copies of the book. C(51) = 612, so it costs $612 for 51 copies of the book. (c) 56 books. 73. (a) S(10) = 17.5, so it costs $17.50
to ship 10 comic books. (b) There is free shipping on orders of 15 or more comic books. 74. (a) C(750) = 25, so it costs $25 to talk 750 minutes per month with this plan. (b) Since 20 hours = 1200 minutes, we substitute m = 1200 and get C(1200) = 45. It costs $45 to talk 20 hours per month with this plan. (c) It costs $25 for up to 1000 minutes and 10 cents per minute for each minute over 1000 minutes. 75. (a) 0.785 = 0, 117 = 117, −2.001 = −3, and π + 6 = 9 15This is called the ‘ﬁxed’ or ‘start-up’ cost. We’ll revisit this concept on page 82. 76 Relations and Functions 1.5 Function Arithmetic In the previous section we used the newly deﬁned function notation to make sense of expressions such as ‘f (x) + 2’ and ‘2f (x)’ for a given function f. It would seem natural, then, that functions should have their own arithmetic which is consistent with the arithmetic of real numbers. The following deﬁnitions allow us to add, subtract, multiply and divide functions using the arithmetic we already know for real numbers. Function Arithmetic Suppose f and g are functions and x is in both the domain of f and the domain of g.a The sum of f and g, denoted f + g, is the function deﬁned by the formula (f + g)(x) = f (x) + g(x) The diﬀerence of f and g, denoted f − g, is the function deﬁned by the formula (f − g)(x) = f (x) − g(x) The product of f and g, denoted f g, is the function deﬁned by the formula (f g)(x) = f (x)g(x) The quotient of f and g, denoted, is the function deﬁned by the formula f g f g (x) = f (x) g(x), provided g(x) = 0. aThus x is an element of the intersection of the two domains. In other words, to add two functions, we add their outputs;
to subtract two functions, we subtract their outputs, and so on. Note that while the formula (f + g)(x) = f (x) + g(x) looks suspiciously like some kind of distributive property, it is nothing of the sort; the addition on the left hand side of the equation is function addition, and we are using this equation to deﬁne the output of the new function f + g as the sum of the real number outputs from f and g. Example 1.5.1. Let f (x) = 6x2 − 2x and g(x) = 3 − 1 x. 1. Find (f + g)(−1) 2. Find (f g)(2) 3. Find the domain of g − f then ﬁnd and simplify a formula for (g − f )(x). 1.5 Function Arithmetic 77 4. Find the domain of g f then ﬁnd and simplify a formula for g f (x). Solution. 1. To ﬁnd (f + g)(−1) we ﬁrst ﬁnd f (−1) = 8 and g(−1) = 4. By deﬁnition, we have that (f + g)(−1) = f (−1) + g(−1) = 8 + 4 = 12. 2. To ﬁnd (f g)(2), we ﬁrst need f (2) and g(2). Since f (2) = 20 and g(2) = 5 2, our formula yields (f g)(2) = f (2)g(2) = (20) 5 2 = 50. 3. One method to ﬁnd the domain of g−f is to ﬁnd the domain of g and of f separately, then ﬁnd the intersection of these two sets. Owing to the denominator in the expression g(x) = 3 − 1 x, we get that the domain of g is (−∞, 0) ∪ (0, ∞). Since f (x) = 6x2 − 2x is valid for all real numbers, we have no further restrictions. Thus the domain of g − f matches the domain of g, namely, (−∞, 0) ∪ (0, ∞). A second method is to analyze the formula for (g − f )
(x) before simplifying and look for the usual domain issues. In this case, (g − f )(x) = g(x) − f (x) = 3 − 1 x − 6x2 − 2x, so we ﬁnd, as before, the domain is (−∞, 0) ∪ (0, ∞). Moving along, we need to simplify a formula for (g − f )(x). In this case, we get common denominators and attempt to reduce the resulting fraction. Doing so, we get (g − f )(x) = g(x) − f (x) = 3 − 1 x − 6x2 − 2x = 3 − 1 x − 6x2 + 2x get common denominators = = = + − − 1 x 6x3 x 2x2 3x x x 3x − 1 − 6x3 − 2x2 x −6x3 − 2x2 + 3x − 1 x 4. As in the previous example, we have two ways to approach ﬁnding the domain of g f. First, we can ﬁnd the domain of g and f separately, and ﬁnd the intersection of these two sets. In f (x), we are introducing a new denominator, namely f (x), so we addition, since need to guard against this being 0 as well. Our previous work tells us that the domain of g is (−∞, 0) ∪ (0, ∞) and the domain of f is (−∞, ∞). Setting f (x) = 0 gives 6x2 − 2x = 0 (x) = g(x) g f 78 Relations and Functions 3. As a result, the domain of g f is all real numbers except x = 0 and x = 1 3, or or x = 0, 1 (−∞, 0) ∪ 0, 1 3 ∪ 1 3, ∞. Alternatively, we may proceed as above and analyze the expression simplifying. In this case, g f (x) = g(x) f (x) before g f (x) = g(x) f (x) = 3 − 1 x 6x2 − 2x We see immediately from the ‘little’ denominator that x = 0. To keep the ‘big’ denominator away from 0, we solve 6x2 − 2
x = 0 and get x = 0 or x = 1 3. Hence, as before, we ﬁnd the domain of g f to be (−∞, 0) ∪ 0, 1 ∪ 1 3 Next, we ﬁnd and simplify a formula for (x). 3, ∞. g f g f (x(x) f (x) 3 − 1 x 6x2 − 2x 3 − 1 x · x x x 3 − 6x2 − 2x 1 x (6x2 − 2x) x 3x − 1 (6x2 − 2x) x 3x − 1 2x2(3x − 1) 1 (3x − 1) 2x2 (3x − 1) 1 2x2 simplify compound fractions factor cancel Please note the importance of ﬁnding the domain of a function before simplifying its expression. In number 4 in Example 1.5.1 above, had we waited to ﬁnd the domain of g f until after simplifying, we’d just have the formula 1 2x2 to go by, and we would (incorrectly!) state the domain as (−∞, 0)∪(0, ∞), since the other troublesome number, x = 1 3, was canceled away.1 1We’ll see what this means geometrically in Chapter 4. 1.5 Function Arithmetic 79 Next, we turn our attention to the diﬀerence quotient of a function. Deﬁnition 1.8. Given a function f, the diﬀerence quotient of f is the expression f (x + h) − f (x) h We will revisit this concept in Section 2.1, but for now, we use it as a way to practice function notation and function arithmetic. For reasons which will become clear in Calculus, ‘simplifying’ a diﬀerence quotient means rewriting it in a form where the ‘h’ in the deﬁnition of the diﬀerence quotient cancels from the denominator. Once that happens, we consider our work to be done. Example 1.5.2. Find and simplify the diﬀerence quotients for the following functions 1. f (x) = x2 − x − 2 Solution. 2. g(x) = 3 2x +
1 3. r(x) = √ x 1. To ﬁnd f (x + h), we replace every occurrence of x in the formula f (x) = x2 − x − 2 with the quantity (x + h) to get f (x + h) = (x + h)2 − (x + h) − 2 = x2 + 2xh + h2 − x − h − 2. So the diﬀerence quotient is f (x + h) − f (x) h = = = = = x2 + 2xh + h2 − x − h − 2 − x2 − x − 2 h x2 + 2xh + h2 − x − h − 2 − x2 + x + 2 h 2xh + h2 − h h h (2x + h − 1) h h (2x + h − 1) h factor cancel = 2x + h − 1. 80 Relations and Functions 2. To ﬁnd g(x + h), we replace every occurrence of x in the formula g(x) = 3 2x+1 with the quantity (x + h) to get which yields g(x + h) − g(x) h g(x + h) = = 3 2(x + h) + 1 3 2x + 2h + 1, 3 2x + 2h + 1 h 3 2x + 2h + 1 h − − 3 2x + 1 3 2x + 1 · (2x + 2h + 1)(2x + 1) (2x + 2h + 1)(2x + 1) 3(2x + 1) − 3(2x + 2h + 1) h(2x + 2h + 1)(2x + 1) 6x + 3 − 6x − 6h − 3 h(2x + 2h + 1)(2x + 1) −6h h(2x + 2h + 1)(2x + 1) −6h h(2x + 2h + 1)(2x + 1) −6 (2x + 2h + 1)(2x + 1). = = = = = = = Since we have managed to cancel the original ‘h’ from the denominator, we are done. 3. For r(x) = √ x, we get r(x + h)
= √ x + h so the diﬀerence quotient is r(x + h) − r(x In order to cancel the ‘h’ from the denominator, we rationalize the numerator by multiplying by its conjugate.2 2Rationalizing the numerator!? How’s that for a twist! 1.5 Function Arithmetic 81 r(x + h) − r(x Multiply by the conjugate. Diﬀerence of Squares. √ x + h2 )2 x (x + h Since we have removed the original ‘h’ from the denominator, we are done. As mentioned before, we will revisit diﬀerence quotients in Section 2.1 where we will explain them geometrically. For now, we want to move on to some classic applications of function arithmetic from Economics and for that, we need to think like an entrepreneur.3 Suppose you are a manufacturer making a certain product.4 Let x be the production level, that is, the number of items produced in a given time period. It is customary to let C(x) denote the function which calculates the total cost of producing the x items. The quantity C(0), which represents the cost of producing no items, is called the ﬁxed cost, and represents the amount of money required to begin production. Associated with the total cost C(x) is cost per item, or average cost, denoted C(x) and read ‘C-bar’ of x. To compute C(x), we take the total cost C(x) and divide by the number of items produced x to get C(x) = C(x) x On the retail end, we have the price p charged per item. To simplify the dialog and computations in this text, we assume that the number of items sold equals the number of items produced. From a 3Not really, but “entrepreneur” is the buzzword of the day and we’re trying to be trendy. 4Poorly designed resin Sasquatch statues, for example. Feel free to choose your own entrepreneurial fantasy. 82 Relations and Functions retail perspective, it seems natural to think of the number of items sold, x, as a function of the price charged, p. After all, the retailer can easily adjust the price to sell more product. In the language of functions, x would
be the dependent variable and p would be the independent variable or, using function notation, we have a function x(p). While we will adopt this convention later in the text,5 we will hold with tradition at this point and consider the price p as a function of the number of items sold, x. That is, we regard x as the independent variable and p as the dependent variable and speak of the price-demand function, p(x). Hence, p(x) returns the price charged per item when x items are produced and sold. Our next function to consider is the revenue function, R(x). The function R(x) computes the amount of money collected as a result of selling x items. Since p(x) is the price charged per item, we have R(x) = xp(x). Finally, the proﬁt function, P (x) calculates how much money is earned after the costs are paid. That is, P (x) = (R − C)(x) = R(x) − C(x). We summarize all of these functions below. Summary of Common Economic Functions Suppose x represents the quantity of items produced and sold. The price-demand function p(x) calculates the price per item. The revenue function R(x) calculates the total money collected by selling x items at a price p(x), R(x) = x p(x). The cost function C(x) calculates the cost to produce x items. The value C(0) is called the ﬁxed cost or start-up cost. The average cost function C(x) = C(x) x Here, we necessarily assume x > 0. calculates the cost per item when making x items. The proﬁt function P (x) calculates the money earned after costs are paid when x items are produced and sold, P (x) = (R − C)(x) = R(x) − C(x). It is high time for an example. Example 1.5.3. Let x represent the number of dOpi media players (‘dOpis’6) produced and sold in a typical week. Suppose the cost, in dollars, to produce x dOpis is given by C(x) = 100x + 2000, for x ≥ 0, and the price, in dollars per dOpi, is given by p(x) = 450 − 15x for 0 ≤
x ≤ 30. 1. Find and interpret C(0). 2. Find and interpret C(10). 3. Find and interpret p(0) and p(20). 4. Solve p(x) = 0 and interpret the result. 5. Find and simplify expressions for the revenue function R(x) and the proﬁt function P (x). 6. Find and interpret R(0) and P (0). 7. Solve P (x) = 0 and interpret the result. 5See Example 5.2.4 in Section 5.2. 6Pronounced ‘dopeys’... 1.5 Function Arithmetic 83 Solution. 1. We substitute x = 0 into the formula for C(x) and get C(0) = 100(0) + 2000 = 2000. This means to produce 0 dOpis, it costs $2000. In other words, the ﬁxed (or start-up) costs are $2000. The reader is encouraged to contemplate what sorts of expenses these might be. 2. Since C(x) = C(x) x, C(10) = C(10) 10 = 3000 10 = 300. This means when 10 dOpis are produced, the cost to manufacture them amounts to $300 per dOpi. 3. Plugging x = 0 into the expression for p(x) gives p(0) = 450 − 15(0) = 450. This means no dOpis are sold if the price is $450 per dOpi. On the other hand, p(20) = 450 − 15(20) = 150 which means to sell 20 dOpis in a typical week, the price should be set at $150 per dOpi. 4. Setting p(x) = 0 gives 450 − 15x = 0. Solving gives x = 30. This means in order to sell 30 dOpis in a typical week, the price needs to be set to $0. What’s more, this means that even if dOpis were given away for free, the retailer would only be able to move 30 of them.7 5. To ﬁnd the revenue, we compute R(x) = xp(x) = x(450 − 15x) = 450x − 15x2. Since the formula for p(x) is valid only for 0 ≤ x ≤ 30, our formula R(
x) is also restricted to 0 ≤ x ≤ 30. For the proﬁt, P (x) = (R − C)(x) = R(x) − C(x). Using the given formula for C(x) and the derived formula for R(x), we get P (x) = 450x − 15x2−(100x+2000) = −15x2+350x−2000. As before, the validity of this formula is for 0 ≤ x ≤ 30 only. 6. We ﬁnd R(0) = 0 which means if no dOpis are sold, we have no revenue, which makes sense. Turning to proﬁt, P (0) = −2000 since P (x) = R(x)−C(x) and P (0) = R(0)−C(0) = −2000. This means that if no dOpis are sold, more money ($2000 to be exact!) was put into producing the dOpis than was recouped in sales. In number 1, we found the ﬁxed costs to be $2000, so it makes sense that if we sell no dOpis, we are out those start-up costs. 7. Setting P (x) = 0 gives −15x2 + 350x − 2000 = 0. Factoring gives −5(x − 10)(3x − 40) = 0 so x = 10 or x = 40 3. What do these values mean in the context of the problem? Since P (x) = R(x) − C(x), solving P (x) = 0 is the same as solving R(x) = C(x). This means that the solutions to P (x) = 0 are the production (and sales) ﬁgures for which the sales revenue exactly balances the total production costs. These are the so-called ‘break even’ points. The solution x = 10 means 10 dOpis should be produced (and sold) during the week to recoup the cost of production. For x = 40 3 = 13.3, things are a bit more complicated. Even though x = 13.3 satisﬁes 0 ≤ x ≤ 30, and hence is in the domain of P, it doesn’t make sense in the context of this problem to produce a fractional part of a dOpi.8 Eval
uating P (13) = 15 and P (14) = −40, we see that producing and selling 13 dOpis per week makes a (slight) proﬁt, whereas producing just one more puts us back into the red. While breaking even is nice, we ultimately would like to ﬁnd what production level (and price) will result in the largest proﬁt, and we’ll do just that... in Section 2.3. 7Imagine that! Giving something away for free and hardly anyone taking advantage of it... 8We’ve seen this sort of thing before in Section 1.4.1. 84 1.5.1 Exercises Relations and Functions In Exercises 1 - 10, use the pair of functions f and g to ﬁnd the following values if they exist. (f + g)(2) (f g) 1 2 (f − g)(−1) f g (0) (g − f )(1) g f (−2) 1. f (x) = 3x + 1 and g(x) = 4 − x 2. f (x) = x2 and g(x) = −2x + 1 3. f (x) = x2 − x and g(x) = 12 − x2 4. f (x) = 2x3 and g(x) = −x2 − 2x − 3 √ 5. f (x) = x + 3 and g(x) = 2x − 1 √ 6. f (x) = 4 − x and g(x) = √ x + 2 7. f (x) = 2x and g(x) = 9. f (x) = x2 and g(x) = 1 2x + 1 1 x2 8. f (x) = x2 and g(x) = 3 2x − 3 10. f (x) = x2 + 1 and g(x) = 1 x2 + 1 In Exercises 11 - 20, use the pair of functions f and g to ﬁnd the domain of the indicated function then ﬁnd and simplify an expression for it. (f + g)(x) (f − g)(x) (f g)(x) f g (x) 11. f (x) = 2x + 1 and g(x) = x
− 2 12. f (x) = 1 − 4x and g(x) = 2x − 1 13. f (x) = x2 and g(x) = 3x − 1 14. f (x) = x2 − x and g(x) = 7x 15. f (x) = x2 − 4 and g(x) = 3x + 6 16. f (x) = −x2 + x + 6 and g(x) = x2 − 9 17. f (x) = x 2 and g(x) = 19. f (x) = x and g(x) = 2 x √ x + 1 18. f (x) = x − 1 and g(x) = 1 x − 1 20. f (x) = √ x − 5 and g(x) = f (x) = √ x − 5 In Exercises 21 - 45, ﬁnd and simplify the diﬀerence quotient f (x + h) − f (x) h for the given function. 21. f (x) = 2x − 5 23. f (x) = 6 25. f (x) = −x2 + 2x − 1 22. f (x) = −3x + 5 24. f (x) = 3x2 − x 26. f (x) = 4x2 1.5 Function Arithmetic 85 27. f (x) = x − x2 28. f (x) = x3 + 1 29. f (x) = mx + b where m = 0 30. f (x) = ax2 + bx + c where a = 0 31. f (x) = 33. f (x) = 2 x 1 x2 35. f (x) = 37. f (x) = 1 4x − 3 x x − 9 √ √ √ 39. f (x) = 41. f (x) = 43. f (x) = x − 9 −4x + 5 ax + b, where a = 0. 32. f (x) = 34. f (x) = 36. f (x 3x x + 1 38. f (x) = 40. f (x) = 42. f (x) = x2 2x + 1 √ 2x + 1 √ 4 − x 44. f
(x) = x √ x √ 45. f (x) = 3 x. HINT: (a − b) a2 + ab + b2 = a3 − b3 In Exercises 46 - 50, C(x) denotes the cost to produce x items and p(x) denotes the price-demand function in the given economic scenario. In each Exercise, do the following: Find and interpret C(0). Find and interpret C(10). Find and interpret p(5) Find and simplify R(x). Find and simplify P (x). Solve P (x) = 0 and interpret. 46. The cost, in dollars, to produce x “I’d rather be a Sasquatch” T-Shirts is C(x) = 2x + 26, x ≥ 0 and the price-demand function, in dollars per shirt, is p(x) = 30 − 2x, 0 ≤ x ≤ 15. 47. The cost, in dollars, to produce x bottles of 100% All-Natural Certiﬁed Free-Trade Organic Sasquatch Tonic is C(x) = 10x + 100, x ≥ 0 and the price-demand function, in dollars per bottle, is p(x) = 35 − x, 0 ≤ x ≤ 35. 48. The cost, in cents, to produce x cups of Mountain Thunder Lemonade at Junior’s Lemonade Stand is C(x) = 18x + 240, x ≥ 0 and the price-demand function, in cents per cup, is p(x) = 90 − 3x, 0 ≤ x ≤ 30. 49. The daily cost, in dollars, to produce x Sasquatch Berry Pies C(x) = 3x + 36, x ≥ 0 and the price-demand function, in dollars per pie, is p(x) = 12 − 0.5x, 0 ≤ x ≤ 24. 86 Relations and Functions 50. The monthly cost, in hundreds of dollars, to produce x custom built electric scooters is C(x) = 20x + 1000, x ≥ 0 and the price-demand function, in hundreds of dollars per scooter, is p(x) = 140 − 2x, 0 ≤ x ≤ 70. In Exercises 51 - 62, let f be the function deﬁned by f = {(−3, 4), (−2, 2), (−
1, 0), (0, 1), (1, 3), (2, 4), (3, −1)} and let g be the function deﬁned g = {(−3, −2), (−2, 0), (−1, −4), (0, 0), (1, −3), (2, 1), (3, 2)}. Compute the indicated value if it exists. 51. (f + g)(−3) 54. (g + f )(1) 57. 60. f g g f (−2) (−1) 52. (f − g)(2) 55. (g − f )(3) 58. 61. f g g f (−1) (3) 53. (f g)(−1) 56. (gf )(−3) 59. 62. f g g f (2) (−3) 1.5 Function Arithmetic 87 1.5.2 Answers 1. For f (x) = 3x + 1 and g(x) = 4 − x (f + g)(2) = 9 (f g) 1 2 = 35 4 (f − g)(−1) = −7 (g − f )(1) = −1 f g (0) = 1 4 g f (−2) = − 6 5 2. For f (x) = x2 and g(x) = −2x + 1 (f + g)(2) = 1 (f − g)(−1) = −2 (g − f )(1) = −2 (f g) 1 2 = 0 f g (0) = 0 g f (−2) = 5 4 3. For f (x) = x2 − x and g(x) = 12 − x2 (f + g)(2) = 10 (f − g)(−1) = −9 (g − f )(1) = 11 (f g) 1 2 = − 47 16 f g (0) = 0 g f (−2) = 4 3 4. For f (x) = 2x3 and g(x) = −x2 − 2x − 3 (f + g)(2) = 5 (f g) 1 2 = − 17 16 (f − g)(−1) = 0 (g − f )(1) = −8 f g (0) = 0 g f (−2) = 3 16
5. For f (x) = √ x + 3 and g(x) = 2x − 1 (f + g)(2) = 3 + √ 5 (f − g)(−1) = 3 + √ 2 (f g) 1 2 = 0 f g √ (0) = − 3 (g − f )(1) = −1 g f (−2) = −5 6. For f (x) = √ 4 − x and g(x) = √ (f + g)(2f − g)(−1) = −1 + √ 5 (f g) 1 2 = √ 35 2 f g √ 2 (0) = (g − f )(1) = 0 g f (−2) = 0 88 Relations and Functions 7. For f (x) = 2x and g(x) = 1 2x+1 (f + g)(2) = 21 5 (f g) 1 2 = 1 2 (f − g)(−1) = −1 f g (0) = 0 (g − f )(1) = − 5 3 g f (−2) = 1 12 8. For f (x) = x2 and g(x) = 3 2x−3 (f + g)(2) = 7 (f g) 1 2 = − 3 8 (f − g)(−1) = 8 5 f g (0) = 0 (g − f )(1) = −4 g f (−2) = − 3 28 9. For f (x) = x2 and g(x) = 1 x2 (f + g)(2) = 17 4 (f g) 1 2 = 1 (f − g)(−1) = 0 f g (0) is undeﬁned. (g − f )(1) = 0 g f (−2) = 1 16 10. For f (x) = x2 + 1 and g(x) = 1 x2+1 (f + g)(2) = 26 5 (f g) 1 2 = 1 (f − g)(−1) = 3 2 f g (0) = 1 (g − f )(1) = − 3 2 g f (−2) = 1 25 11. For f (x) = 2x + 1 and g(x) = x − 2 (f + g)(x) = 3
x − 1 Domain: (−∞, ∞) (f g)(x) = 2x2 − 3x − 2 Domain: (−∞, ∞) 12. For f (x) = 1 − 4x and g(x) = 2x − 1 (f + g)(x) = −2x Domain: (−∞, ∞) (f g)(x) = −8x2 + 6x − 1 Domain: (−∞, ∞) (f − g)(x) = x + 3 Domain: (−∞, ∞) f g (x) = 2x+1 x−2 Domain: (−∞, 2) ∪ (2, ∞) (f − g)(x) = 2 − 6x Domain: (−∞, ∞) f (x) = 1−4x g 2x−1 Domain: −∞, 1 2 ∪ 1 2, ∞ 1.5 Function Arithmetic 89 13. For f (x) = x2 and g(x) = 3x − 1 (f + g)(x) = x2 + 3x − 1 Domain: (−∞, ∞) (f g)(x) = 3x3 − x2 Domain: (−∞, ∞) 14. For f (x) = x2 − x and g(x) = 7x (f + g)(x) = x2 + 6x Domain: (−∞, ∞) (f g)(x) = 7x3 − 7x2 Domain: (−∞, ∞) (f − g)(x) = x2 − 3x + 1 Domain: (−∞, ∞) f (x) = x2 3x−1 g Domain: −∞, 1 3 ∪ 1 3, ∞ (f − g)(x) = x2 − 8x Domain: (−∞, ∞) f g (x) = x−1 7 Domain: (−∞, 0) ∪ (0, ∞) 15. For f (x) = x2 − 4 and g(x) = 3x + 6 (f + g)(x) = x2 + 3x + 2 Domain: (−∞, ∞) (f g)(x) = 3x3 + 6x2 − 12x − 24 Domain: (−∞, �
�) 16. For f (x) = −x2 + x + 6 and g(x) = x2 − 9 (f − g)(x) = x2 − 3x − 10 Domain: (−∞, ∞) f g (x) = x−2 3 Domain: (−∞, −2) ∪ (−2, ∞) (f + g)(x) = x − 3 Domain: (−∞, ∞) (f g)(x) = −x4 + x3 + 15x2 − 9x − 54 Domain: (−∞, ∞) (f − g)(x) = −2x2 + x + 15 Domain: (−∞, ∞) f g (x) = − x+2 x+3 Domain: (−∞, −3) ∪ (−3, 3) ∪ (3, ∞) 17. For f (x) = x 2 and g(x) = 2 x (f + g)(x) = x2+4 2x Domain: (−∞, 0) ∪ (0, ∞) (f g)(x) = 1 Domain: (−∞, 0) ∪ (0, ∞) (f − g)(x) = x2−4 2x Domain: (−∞, 0) ∪ (0, ∞) f g (x) = x2 4 Domain: (−∞, 0) ∪ (0, ∞) 90 Relations and Functions 18. For f (x) = x − 1 and g(x) = 1 x−1 (f + g)(x) = x2−2x+2 x−1 Domain: (−∞, 1) ∪ (1, ∞) (f − g)(x) = x2−2x x−1 Domain: (−∞, 1) ∪ (1, ∞) (f g)(x) = 1 Domain: (−∞, 1) ∪ (1, ∞) f g (x) = x2 − 2x + 1 Domain: (−∞, 1) ∪ (1, ∞) √ x + 1 19. For f (x) = x and g(x) = √ (f + g)(x) = x + Domain: [−1, ∞) x + 1 (f − g)(
x) = x − Domain: [−1, ∞) √ x + 1 (f g)(x) = x √ x + 1 Domain: [−1, ∞) 20. For f (x) = √ x − 5 and g(x) = f (x) = √ x − 5 f g (x) = x√ Domain: (−1, ∞) x+1 √ (f + g)(x) = 2 Domain: [5, ∞) x − 5 (f g)(x) = x − 5 Domain: [5, ∞) 21. 2 23. 0 25. −2x − h + 2 27. −2x − h + 1 29. m 31. 33. 35. −2 x(x + h) −(2x + h) x2(x + h)2 −4 (4x − 3)(4x + 4h − 3) (f − g)(x) = 0 Domain: [5, ∞) f g (x) = 1 Domain: (5, ∞) 22. −3 24. 6x + 3h − 1 26. 8x + 4h 28. 3x2 + 3xh + h2 30. 2ax + ah + b 32. 34. 36. 3 (1 − x − h)(1 − x) −2 (x + 5)(x + h + 5) 3 (x + 1)(x + h + 1) 1.5 Function Arithmetic 91 37. −9 (x − 9)(x + h − 9) 39. √ 41. √ 434 −4x − 4h + 5 + √ −4x + 5 a ax + ah + b + √ ax + b 38. 2x2 + 2xh + 2x + h (2x + 1)(2x + 2h + 1) 40. √ 42. √ 2 2x + 2h + 1 + √ 2x + 1 − 44. 3x2 + 3xh + h2 (x + h)3/2 + x3/2 45. 46. 47. 48. 1 (x + h)2/3 + (x + h)1/3x1/3 + x2/3 C(0) = 26, so the ﬁxed costs are $26. C(10
) = 4.6, so when 10 shirts are produced, the cost per shirt is $4.60. p(5) = 20, so to sell 5 shirts, set the price at $20 per shirt. R(x) = −2x2 + 30x, 0 ≤ x ≤ 15 P (x) = −2x2 + 28x − 26, 0 ≤ x ≤ 15 P (x) = 0 when x = 1 and x = 13. These are the ‘break even’ points, so selling 1 shirt or 13 shirts will guarantee the revenue earned exactly recoups the cost of production. C(0) = 100, so the ﬁxed costs are $100. C(10) = 20, so when 10 bottles of tonic are produced, the cost per bottle is $20. p(5) = 30, so to sell 5 bottles of tonic, set the price at $30 per bottle. R(x) = −x2 + 35x, 0 ≤ x ≤ 35 P (x) = −x2 + 25x − 100, 0 ≤ x ≤ 35 P (x) = 0 when x = 5 and x = 20. These are the ‘break even’ points, so selling 5 bottles of tonic or 20 bottles of tonic will guarantee the revenue earned exactly recoups the cost of production. C(0) = 240, so the ﬁxed costs are 240¢ or $2.40. C(10) = 42, so when 10 cups of lemonade are made, the cost per cup is 42¢. p(5) = 75, so to sell 5 cups of lemonade, set the price at 75¢ per cup. R(x) = −3x2 + 90x, 0 ≤ x ≤ 30 P (x) = −3x2 + 72x − 240, 0 ≤ x ≤ 30 P (x) = 0 when x = 4 and x = 20. These are the ‘break even’ points, so selling 4 cups of lemonade or 20 cups of lemonade will guarantee the revenue earned exactly recoups the cost of production. 92 49. Relations and Functions C(0) = 36, so the daily ﬁxed costs are $36. C(10) = 6.6, so when 10 pies are made, the cost per pie is $6.60. p(5) = 9
.5, so to sell 5 pies a day, set the price at $9.50 per pie. R(x) = −0.5x2 + 12x, 0 ≤ x ≤ 24 P (x) = −0.5x2 + 9x − 36, 0 ≤ x ≤ 24 P (x) = 0 when x = 6 and x = 12. These are the ‘break even’ points, so selling 6 pies or 12 pies a day will guarantee the revenue earned exactly recoups the cost of production. 50. C(0) = 1000, so the monthly ﬁxed costs are 1000 hundred dollars, or $100,000. C(10) = 120, so when 10 scooters are made, the cost per scooter is 120 hundred dollars, or $12,000. p(5) = 130, so to sell 5 scooters a month, set the price at 130 hundred dollars, or $13,000 per scooter. R(x) = −2x2 + 140x, 0 ≤ x ≤ 70 P (x) = −2x2 + 120x − 1000, 0 ≤ x ≤ 70 P (x) = 0 when x = 10 and x = 50. These are the ‘break even’ points, so selling 10 scooters or 50 scooters a month will guarantee the revenue earned exactly recoups the cost of production. 51. (f + g)(−3) = 2 52. (f − g)(2) = 3 53. (f g)(−1) = 0 54. (g + f )(1) = 0 55. (g − f )(3) = 3 56. (gf )(−3) = −8 57. 60. f g g f (−2) does not exist (−1) does not exist 58. 61. f g g f (−1) = 0 (3) = −2 59. 62. f g g f (2) = 4 (−3) = − 1 2 1.6 Graphs of Functions 93 1.6 Graphs of Functions In Section 1.3 we deﬁned a function as a special type of relation; one in which each x-coordinate was matched with only one y-coordinate. We spent most of our time in that section looking at functions graphically because they were, after all, just sets of points in the plane. Then in Section 1
.4 we described a function as a process and deﬁned the notation necessary to work with functions algebraically. So now it’s time to look at functions graphically again, only this time we’ll do so with the notation deﬁned in Section 1.4. We start with what should not be a surprising connection. The Fundamental Graphing Principle for Functions The graph of a function f is the set of points which satisfy the equation y = f (x). That is, the point (x, y) is on the graph of f if and only if y = f (x). Example 1.6.1. Graph f (x) = x2 − x − 6. Solution. To graph f, we graph the equation y = f (x). To this end, we use the techniques outlined in Section 1.2.1. Speciﬁcally, we check for intercepts, test for symmetry, and plot additional points as needed. To ﬁnd the x-intercepts, we set y = 0. Since y = f (x), this means f (x) = 0. f (x) = x2 − x − 6 0 = x2 − x − 6 0 = (x − 3)(x + 2) factor x − 3 = 0 or x + 2 = 0 x = −2, 3 So we get (−2, 0) and (3, 0) as x-intercepts. To ﬁnd the y-intercept, we set x = 0. Using function notation, this is the same as ﬁnding f (0) and f (0) = 02 − 0 − 6 = −6. Thus the y-intercept is (0, −6). As far as symmetry is concerned, we can tell from the intercepts that the graph possesses none of the three symmetries discussed thus far. (You should verify this.) We can make a table analogous to the ones we made in Section 1.2.1, plot the points and connect the dots in a somewhat pleasing fashion to get the graph below on the right. x −3 −2 −1 0 1 2 3 4 f (x) 6 0 (x, f (x)) (−3, 6) (−2, 0) −4 (−1, −4) (0, −6) −6 (1, −6) −6 (2,
−4) −4 (3, 0) 0 (4, 63−2−1 −1 1 2 3 4 x −2 −3 −4 −5 −6 94 Relations and Functions Graphing piecewise-deﬁned functions is a bit more of a challenge. Example 1.6.2. Graph: f (x) = 4 − x2 x − 3, if x < 1 if x ≥ 1 Solution. We proceed as before – ﬁnding intercepts, testing for symmetry and then plotting additional points as needed. To ﬁnd the x-intercepts, as before, we set f (x) = 0. The twist is that we have two formulas for f (x). For x < 1, we use the formula f (x) = 4 − x2. Setting f (x) = 0 gives 0 = 4 − x2, so that x = ±2. However, of these two answers, only x = −2 ﬁts in the domain x < 1 for this piece. This means the only x-intercept for the x < 1 region of the x-axis is (−2, 0). For x ≥ 1, f (x) = x − 3. Setting f (x) = 0 gives 0 = x − 3, or x = 3. Since x = 3 satisﬁes the inequality x ≥ 1, we get (3, 0) as another x-intercept. Next, we seek the y-intercept. Notice that x = 0 falls in the domain x < 1. Thus f (0) = 4 − 02 = 4 yields the y-intercept (0, 4). As far as symmetry is concerned, you can check that the equation y = 4 − x2 is symmetric about the y-axis; unfortunately, this equation (and its symmetry) is valid only for x < 1. You can also verify y = x − 3 possesses none of the symmetries discussed in the Section 1.2.1. When plotting additional points, it is important to keep in mind the restrictions on x for each piece of the function. The sticking point for this function is x = 1, since this is where the equations change. When x = 1, we use the formula f (x) = x − 3, so the point on the graph (1, f (1)) is (1, −2). However, for all values less than
1, we use the formula f (x) = 4 − x2. As we have discussed earlier in Section 1.2, there is no real number which immediately precedes x = 1 on the number line. Thus for the values x = 0.9, x = 0.99, x = 0.999, and so on, we ﬁnd the corresponding y values using the formula f (x) = 4 − x2. Making a table as before, we see that as the x values sneak up to x = 1 in this fashion, the f (x) values inch closer and closer1 to 4 − 12 = 3. To indicate this graphically, we use an open circle at the point (1, 3). Putting all of this information together and plotting additional points, we get (x, f (x)) x f (x) (0.9, 3.19) 0.9 3.19 0.99 ≈ 3.02 (0.99, 3.02) 0.999 ≈ 3.002 (0.999, 3.002) y 4 3 2 1 −3 −2 −1 −1 1 2 3 x −2 −3 −4 1We’ve just stepped into Calculus here! 1.6 Graphs of Functions 95 In the previous two examples, the x-coordinates of the x-intercepts of the graph of y = f (x) were found by solving f (x) = 0. For this reason, they are called the zeros of f. Deﬁnition 1.9. The zeros of a function f are the solutions to the equation f (x) = 0. In other words, x is a zero of f if and only if (x, 0) is an x-intercept of the graph of y = f (x). Of the three symmetries discussed in Section 1.2.1, only two are of signiﬁcance to functions: symmetry about the y-axis and symmetry about the origin.2 Recall that we can test whether the graph of an equation is symmetric about the y-axis by replacing x with −x and checking to see if an equivalent equation results. If we are graphing the equation y = f (x), substituting −x for x results in the equation y = f (−x). In order for this equation to be equivalent to the original equation y = f (x) we
need f (−x) = f (x). In a similar fashion, we recall that to test an equation’s graph for symmetry about the origin, we replace x and y with −x and −y, respectively. Doing this substitution in the equation y = f (x) results in −y = f (−x). Solving the latter equation for y gives y = −f (−x). In order for this equation to be equivalent to the original equation y = f (x) we need −f (−x) = f (x), or, equivalently, f (−x) = −f (x). These results are summarized below. Testing the Graph of a Function for Symmetry The graph of a function f is symmetric about the y-axis if and only if f (−x) = f (x) for all x in the domain of f. about the origin if and only if f (−x) = −f (x) for all x in the domain of f. For reasons which won’t become clear until we study polynomials, we call a function even if its graph is symmetric about the y-axis or odd if its graph is symmetric about the origin. Apart from a very specialized family of functions which are both even and odd,3 functions fall into one of three distinct categories: even, odd, or neither even nor odd. Example 1.6.3. Determine analytically if the following functions are even, odd, or neither even nor odd. Verify your result with a graphing calculator. 1. f (x) = 5 2 − x2 2. g(x) = 5x 2 − x2 3. h(x) = 5. j(x) = x2 − 5x 2 − x3 x 100 − 1 4. i(x) = 6. p(x) = 5x 2x − x3 x + 3 if x < 0 if x ≥ 0 −x + 3, Solution. The ﬁrst step in all of these problems is to replace x with −x and simplify. 2Why are we so dismissive about symmetry about the x-axis for graphs of functions? 3Any ideas? 96 1. 2. Relations and Functions f (x) = f (−x) = f (−x) = 5 2 − x2 5 2 − (−x)2 5 2 − x2 f (−x) = f (x) Hence, f is even.
The graphing calculator furnishes the following. This suggests4 that the graph of f is symmetric about the y-axis, as expected. g(x) = g(−x) = g(−x) = 5x 2 − x2 5(−x) 2 − (−x)2 −5x 2 − x2 It doesn’t appear that g(−x) is equivalent to g(x). To prove this, we check with an x value. After some trial and error, we see that g(1) = 5 whereas g(−1) = −5. This proves that g is not even, but it doesn’t rule out the possibility that g is odd. (Why not?) To check if g is odd, we compare g(−x) with −g(x) −g(x) = − 5x 2 − x2 −5x 2 − x2 = −g(x) = g(−x) Hence, g is odd. Graphically, 4‘Suggests’ is about the extent of what it can do. 1.6 Graphs of Functions 97 3. 4. The calculator indicates the graph of g is symmetric about the origin, as expected. h(x) = h(−x) = h(−x) = 5x 2 − x3 5(−x) 2 − (−x)3 −5x 2 + x3 Once again, h(−x) doesn’t appear to be equivalent to h(x). We check with an x value, for example, h(1) = 5 but h(−1) = − 5 3. This proves that h is not even and it also shows h is not odd. (Why?) Graphically, The graph of h appears to be neither symmetric about the y-axis nor the origin. i(x) = i(−x) = i(−x) = 5x 2x − x3 5(−x) 2(−x) − (−x)3 −5x −2x + x3 The expression i(−x) doesn’t appear to be equivalent to i(x). However, after checking some x values, for example x = 1 yields i(1) = 5 and i(−1) = 5, it appears that i(−x) does, in fact, equal i(x). However, while this suggests i is
even, it doesn’t prove it. (It does, however, prove 98 Relations and Functions i is not odd.) To prove i(−x) = i(x), we need to manipulate our expressions for i(x) and i(−x) and show that they are equivalent. A clue as to how to proceed is in the numerators: in the formula for i(x), the numerator is 5x and in i(−x) the numerator is −5x. To re-write i(x) with a numerator of −5x, we need to multiply its numerator by −1. To keep the value of the fraction the same, we need to multiply the denominator by −1 as well. Thus i(x) = = = 5x 2x − x3 (−1)5x (−1) (2x − x3) −5x −2x + x3 Hence, i(x) = i(−x), so i is even. The calculator supports our conclusion. 5. j(x) = x2 − x 100 j(−x) = (−x)2 − j(−x) = x2 + x 100 − 1 −x 100 − 1 − 1 The expression for j(−x) doesn’t seem to be equivalent to j(x), so we check using x = 1 to get j(1) = − 1 100. This rules out j being even. However, it doesn’t rule out j being odd. Examining −j(x) gives 100 and j(−1) = 1 j(x) = x2 − −j(x) = − x 100 x2 − −j(x) = −x2 + + 1 − 1 − 1 x 100 x 100 The expression −j(x) doesn’t seem to match j(−x) either. Testing x = 2 gives j(2) = 149 50 and j(−2) = 151 50, so j is not odd, either. The calculator gives: 1.6 Graphs of Functions 99 The calculator suggests that the graph of j is symmetric about the y-axis which would imply that j is even. However, we have proven that is not the case. 6. Testing the graph of y = p(x) for symmetry is complicated by the fact p(x) is a piecewisedeﬁned function. As always
, we handle this by checking the condition for symmetry by checking it on each piece of the domain. We ﬁrst consider the case when x < 0 and set about ﬁnding the correct expression for p(−x). Even though p(x) = x + 3 for x < 0, p(−x) = −x + 3 here. The reason for this is that since x < 0, −x > 0 which means to ﬁnd p(−x), we need to use the other formula for p(x), namely p(x) = −x+3. Hence, for x < 0, p(−x) = −(−x)+3 = x + 3 = p(x). For x ≥ 0, p(x) = −x + 3 and we have two cases. If x > 0, then −x < 0 so p(−x) = (−x) + 3 = −x + 3 = p(x). If x = 0, then p(0) = 3 = p(−0). Hence, in all cases, p(−x) = p(x), so p is even. Since p(0) = 3 but p(−0) = p(0) = 3 = −3, we also have p is not odd. While graphing y = p(x) is not onerous to do by hand, it is instructive to see how to enter this into our calculator. By using some of the logical commands,5 we have: The calculator bears shows that the graph appears to be symmetric about the y-axis. There are two lessons to be learned from the last example. The ﬁrst is that sampling function values at particular x values is not enough to prove that a function is even or odd − despite the fact that j(−1) = −j(1), j turned out not to be odd. Secondly, while the calculator may suggest mathematical truths, it is the Algebra which proves mathematical truths.6 5Consult your owner’s manual, instructor, or favorite video site! 6Or, in other words, don’t rely too heavily on the machine! 100 Relations and Functions 1.6.1 General Function Behavior The last topic we wish to address in this section is general function behavior. As you shall see in the next several chapters, each family of functions has its own unique attributes and we will study them
all in great detail. The purpose of this section’s discussion, then, is to lay the foundation for that further study by investigating aspects of function behavior which apply to all functions. To start, we will examine the concepts of increasing, decreasing and constant. Before deﬁning the concepts algebraically, it is instructive to ﬁrst look at them graphically. Consider the graph of the function f below. (6, 5.5) (−2, 4.54 −3 −2 −1 −1 1 2 3 4 5 6 7 x (−4, −3) −2 −3 −4 −5 −6 −7 −8 −9 (4, −6) (3, −8) (5, −6) The graph of y = f (x) Reading from left to right, the graph ‘starts’ at the point (−4, −3) and ‘ends’ at the point (6, 5.5). If we imagine walking from left to right on the graph, between (−4, −3) and (−2, 4.5), we are walking ‘uphill’; then between (−2, 4.5) and (3, −8), we are walking ‘downhill’; and between (3, −8) and (4, −6), we are walking ‘uphill’ once more. From (4, −6) to (5, −6), we ‘level oﬀ’, and then resume walking ‘uphill’ from (5, −6) to (6, 5.5). In other words, for the x values between −4 and −2 (inclusive), the y-coordinates on the graph are getting larger, or increasing, as we move from left to right. Since y = f (x), the y values on the graph are the function values, and we say that the function f is increasing on the interval [−4, −2]. Analogously, we say that f is decreasing on the interval [−2, 3] increasing once more on the interval [3, 4], constant on [4, 5], and ﬁnally increasing once again on [5, 6]. It is extremely important to notice that the behavior (increasing, decreasing or constant) occurs on an interval on the x-axis. When we say that the function f is increasing 1
.6 Graphs of Functions 101 on [−4, −2] we do not mention the actual y values that f attains along the way. Thus, we report where the behavior occurs, not to what extent the behavior occurs.7 Also notice that we do not say that a function is increasing, decreasing or constant at a single x value. In fact, we would run into serious trouble in our previous example if we tried to do so because x = −2 is contained in an interval on which f was increasing and one on which it is decreasing. (There’s more on this issue – and many others – in the Exercises.) We’re now ready for the more formal algebraic deﬁnitions of what it means for a function to be increasing, decreasing or constant. Deﬁnition 1.10. Suppose f is a function deﬁned on an interval I. We say f is: increasing on I if and only if f (a) < f (b) for all real numbers a, b in I with a < b. decreasing on I if and only if f (a) > f (b) for all real numbers a, b in I with a < b. constant on I if and only if f (a) = f (b) for all real numbers a, b in I. It is worth taking some time to see that the algebraic descriptions of increasing, decreasing and constant as stated in Deﬁnition 1.10 agree with our graphical descriptions given earlier. You should look back through the examples and exercise sets in previous sections where graphs were given to see if you can determine the intervals on which the functions are increasing, decreasing or constant. Can you ﬁnd an example of a function for which none of the concepts in Deﬁnition 1.10 apply? Now let’s turn our attention to a few of the points on the graph. Clearly the point (−2, 4.5) does not have the largest y value of all of the points on the graph of f − indeed that honor goes to (6, 5.5) − but (−2, 4.5) should get some sort of consolation prize for being ‘the top of the hill’ between x = −4 and x = 3. We say that the function f has a local maximum8 at the point (−2, 4.5), because the y-coordinate 4.5
is the largest y-value (hence, function value) on the curve ‘near’9 x = −2. Similarly, we say that the function f has a local minimum10 at the point (3, −8), since the y-coordinate −8 is the smallest function value near x = 3. Although it is tempting to say that local extrema11 occur when the function changes from increasing to decreasing or vice versa, it is not a precise enough way to deﬁne the concepts for the needs of Calculus. At the risk of being pedantic, we will present the traditional deﬁnitions and thoroughly vet the pathologies they induce in the Exercises. We have one last observation to make before we proceed to the algebraic deﬁnitions and look at a fairly tame, yet helpful, example. If we look at the entire graph, we see that the largest y value (the largest function value) is 5.5 at x = 6. In this case, we say the maximum12 of f is 5.5; similarly, the minimum13 of f is −8. 7The notions of how quickly or how slowly a function increases or decreases are explored in Calculus. 8Also called ‘relative maximum’. 9We will make this more precise in a moment. 10Also called a ‘relative minimum’. 11‘Maxima’ is the plural of ‘maximum’ and ‘mimima’ is the plural of ‘minimum’. ‘Extrema’ is the plural of ‘extremum’ which combines maximum and minimum. 12Sometimes called the ‘absolute’ or ‘global’ maximum. 13Again, ‘absolute’ or ‘global’ minimum can be used. 102 Relations and Functions We formalize these concepts in the following deﬁnitions. Deﬁnition 1.11. Suppose f is a function with f (a) = b. We say f has a local maximum at the point (a, b) if and only if there is an open interval I containing a for which f (a) ≥ f (x) for all x in I. The value f (a) = b is called ‘a local maximum value of f ’ in this case. We say f has a local minimum at the point (a, b) if and only if
there is an open interval I containing a for which f (a) ≤ f (x) for all x in I. The value f (a) = b is called ‘a local minimum value of f ’ in this case. The value b is called the maximum of f if b ≥ f (x) for all x in the domain of f. The value b is called the minimum of f if b ≤ f (x) for all x in the domain of f. It’s important to note that not every function will have all of these features. Indeed, it is possible to have a function with no local or absolute extrema at all! (Any ideas of what such a function’s graph would have to look like?) We shall see examples of functions in the Exercises which have one or two, but not all, of these features, some that have instances of each type of extremum and some functions that seem to defy common sense. In all cases, though, we shall adhere to the algebraic deﬁnitions above as we explore the wonderful diversity of graphs that functions provide us. Here is the ‘tame’ example which was promised earlier. It summarizes all of the concepts presented in this section as well as some from previous sections so you should spend some time thinking deeply about it before proceeding to the Exercises. Example 1.6.4. Given the graph of y = f (x) below, answer all of the following questions. y (0, 3) 4 3 2 1 (−2, 0) (2, 0) −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 (−4, −3) (4, −3) 1.6 Graphs of Functions 103 1. Find the domain of f. 2. Find the range of f. 3. List the x-intercepts, if any exist. 4. List the y-intercepts, if any exist. 5. Find the zeros of f. 7. Determine f (2). 6. Solve f (x) < 0. 8. Solve f (x) = −3. 9. Find the number of solutions to f (x) = 1. 10. Does f appear to be even, odd, or neither? 11. List the intervals on which f is increasing. 12. List the intervals on which f is decreasing. 13. List the local maximum
s, if any exist. 14. List the local minimums, if any exist. 15. Find the maximum, if it exists. 16. Find the minimum, if it exists. Solution. 1. To ﬁnd the domain of f, we proceed as in Section 1.3. By projecting the graph to the x-axis, we see that the portion of the x-axis which corresponds to a point on the graph is everything from −4 to 4, inclusive. Hence, the domain is [−4, 4]. 2. To ﬁnd the range, we project the graph to the y-axis. We see that the y values from −3 to 3, inclusive, constitute the range of f. Hence, our answer is [−3, 3]. 3. The x-intercepts are the points on the graph with y-coordinate 0, namely (−2, 0) and (2, 0). 4. The y-intercept is the point on the graph with x-coordinate 0, namely (0, 3). 5. The zeros of f are the x-coordinates of the x-intercepts of the graph of y = f (x) which are x = −2, 2. 6. To solve f (x) < 0, we look for the x values of the points on the graph where the y-coordinate is less than 0. Graphically, we are looking for where the graph is below the x-axis. This happens for the x values from −4 to −2 and again from 2 to 4. So our answer is [−4, −2) ∪ (2, 4]. 7. Since the graph of f is the graph of the equation y = f (x), f (2) is the y-coordinate of the point which corresponds to x = 2. Since the point (2, 0) is on the graph, we have f (2) = 0. 8. To solve f (x) = −3, we look where y = f (x) = −3. We ﬁnd two points with a y-coordinate of −3, namely (−4, −3) and (4, −3). Hence, the solutions to f (x) = −3 are x = ±4. 9. As in the previous problem, to solve f (x) = 1, we look for points on the graph where the y-coord
inate is 1. Even though these points aren’t speciﬁed, we see that the curve has two points with a y value of 1, as seen in the graph below. That means there are two solutions to f (x) = 1. 104 Relations and Functions y 4 3 2 1 −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 −4 10. The graph appears to be symmetric about the y-axis. This suggests14 that f is even. 11. As we move from left to right, the graph rises from (−4, −3) to (0, 3). This means f is increasing on the interval [−4, 0]. (Remember, the answer here is an interval on the x-axis.) 12. As we move from left to right, the graph falls from (0, 3) to (4, −3). This means f is decreasing on the interval [0, 4]. (Remember, the answer here is an interval on the x-axis.) 13. The function has its only local maximum at (0, 3) so f (0) = 3 is the local minimum value. 14. There are no local minimums. Why don’t (−4, −3) and (4, −3) count? Let’s consider the point (−4, −3) for a moment. Recall that, in the deﬁnition of local minimum, there needs to be an open interval I which contains x = −4 such that f (−4) < f (x) for all x in I diﬀerent from −4. But if we put an open interval around x = −4 a portion of that interval will lie outside of the domain of f. Because we are unable to fulﬁll the requirements of the deﬁnition for a local minimum, we cannot claim that f has one at (−4, −3). The point (4, −3) fails for the same reason − no open interval around x = 4 stays within the domain of f. 15. The maximum value of f is the largest y-coordinate which is 3. 16. The minimum value of f is the smallest y-coordinate which is −3. With few exceptions, we will not develop techniques in College Algebra which allow us to determine the intervals on which a function is increasing, decreasing or constant or to ﬁnd
the local maximums and local minimums analytically; this is the business of Calculus.15 When we have need to ﬁnd such beasts, we will resort to the calculator. Most graphing calculators have ‘Minimum’ and ‘Maximum’ features which can be used to approximate these values, as we now demonstrate. 14but does not prove 15Although, truth be told, there is only one step of Calculus involved, followed by several pages of algebra. 1.6 Graphs of Functions 105 Example 1.6.5. Let f (x) =. Use a graphing calculator to approximate the intervals on which f is increasing and those on which it is decreasing. Approximate all extrema. 15x x2 + 3 Solution. Entering this function into the calculator gives Using the Minimum and Maximum features, we get To two decimal places, f appears to have its only local minimum at (−1.73, −4.33) and its only local maximum at (1.73, 4.33). Given the symmetry about the origin suggested by the graph, the relation between these points shouldn’t be too surprising. The function appears to be increasing on [−1.73, 1.73] and decreasing on (−∞, −1.73]∪[1.73, ∞). This makes −4.33 the (absolute) minimum and 4.33 the (absolute) maximum. Example 1.6.6. Find the points on the graph of y = (x − 3)2 which are closest to the origin. Round your answers to two decimal places. Solution. Suppose a point (x, y) is on the graph of y = (x − 3)2. Its distance to the origin (0, 0) is given by d = (x − 0)2 + (y − 0)2 = = = x2 + y2 x2 + [(x − 3)2]2 x2 + (x − 3)4 Since y = (x − 3)2 x2 + (x − 3)4 is the distance from (0, 0) to the point (x, y) Given a value for x, the formula d = on the curve y = (x − 3)2. What we have deﬁned, then, is a function d(x) which we wish to 106 Relations and Functions minimize over all values of x. To accomplish this
task analytically would require Calculus so as we’ve mentioned before, we can use a graphing calculator to ﬁnd an approximate solution. Using the calculator, we enter the function d(x) as shown below and graph. Using the Minimum feature, we see above on the right that the (absolute) minimum occurs near x = 2. Rounding to two decimal places, we get that the minimum distance occurs when x = 2.00. To ﬁnd the y value on the parabola associated with x = 2.00, we substitute 2.00 into the equation to get y = (x − 3)2 = (2.00 − 3)2 = 1.00. So, our ﬁnal answer is (2.00, 1.00).16 (What does the y value listed on the calculator screen mean in this problem?) 16It seems silly to list a ﬁnal answer as (2.00, 1.00). Indeed, Calculus conﬁrms that the exact answer to this problem is, in fact, (2, 1). As you are well aware by now, the authors are overly pedantic, and as such, use the decimal places to remind the reader that any result garnered from a calculator in this fashion is an approximation, and should be treated as such. 1.6 Graphs of Functions 107 1.6.2 Exercises In Exercises 1 - 12, sketch the graph of the given function. State the domain of the function, identify any intercepts and test for symmetry. 1. f (x) = 2 − x 4. f (x) = 4 − x2 7. f (x) = x(x − 1)(x + 2) 2. f (x) = x − 2 3 5. f (x) = 2 √ 8. f (x) = x − 2 10. f (x) = 3 − 2 √ x + 2 √ 11. f (x) = 3 x 3. f (x) = x2 + 1 6. f (x) = x3 √ 9. f (x) = 5 − x 12. f (x) = 1 x2 + 1 In Exercises 13 - 20, sketch the graph of the given piecewise-deﬁned function. 13. f (x) = 15. f (x) = 17. f (
x) = 19. f (x) =       4 − x 2 if if x ≤ 3 x > 3 −3 2x − 3 3 if if if 2x − 4 3x if if x < 0 x ≥ 0 x2 3 − x 4 x ≤ −2 if if −2 < x < 2 x ≥ 2 if 14. f (x) = 16. f (x) =    x2 2x if if x ≤ 0 x > 0 x2 − 4 4 − x2 x2 − 4 x ≤ −2 if if −2 < x < 2 x ≥ 2 if 18. f (x if −4 ≤ x < 5 x ≥ 5 if 20. f (x) =    1 x x x √ if −6 < x < −1 if − if In Exercises 21 - 41, determine analytically if the following functions are even, odd or neither. 21. f (x) = 7x 22. f (x) = 7x + 2 23. f (x) = 7 24. f (x) = 3x2 − 4 25. f (x) = 4 − x2 26. f (x) = x2 − x − 6 27. f (x) = 2x3 − x 30. f (x) = x3 + x2 + x + 1 28. f (x) = −x5 + 2x3 − x √ 31. f (x) = 1 − x 33. f (x) = 0 √ 34. f (x) = 3 x 29. f (x) = x6 − x4 + x2 + 9 √ 1 − x2 32. f (x) = 35. f (x) = 3√ x2 108 Relations and Functions 36. f (x) = 3 x2 39. f (x) = x2 − 3 x − 4x3 37. f (x) = 2x − 1 x + 1 40. f (x) = √ 9 4 − x2 38. f (x) = 41. f (x) = 3x x2 + 1 3
√ x3 + x 5x In Exercises 42 - 57, use the graph of y = f (x) given below to answer the question. y 5 4 3 2 1 −5 −4 −3 −2 −1 −1 1 2 3 4 5 x −2 −3 −4 −5 42. Find the domain of f. 43. Find the range of f. 44. Determine f (−2). 45. Solve f (x) = 4. 46. List the x-intercepts, if any exist. 47. List the y-intercepts, if any exist. 48. Find the zeros of f. 49. Solve f (x) ≥ 0. 50. Find the number of solutions to f (x) = 1. 51. Does f appear to be even, odd, or neither? 52. List the intervals where f is increasing. 53. List the intervals where f is decreasing. 54. List the local maximums, if any exist. 55. List the local minimums, if any exist. 56. Find the maximum, if it exists. 57. Find the minimum, if it exists. 1.6 Graphs of Functions 109 In Exercises 58 - 73, use the graph of y = f (x) given below to answer the question. y 5 4 3 2 1 −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 −5 58. Find the domain of f. 59. Find the range of f. 60. Determine f (2). 61. Solve f (x) = −5. 62. List the x-intercepts, if any exist. 63. List the y-intercepts, if any exist. 64. Find the zeros of f. 65. Solve f (x) ≤ 0. 66. Find the number of solutions to f (x) = 3. 67. Does f appear to be even, odd, or neither? 68. List the intervals where f is increasing. 69. List the intervals where f is decreasing. 70. List the local maximums, if any exist. 71. List the local minimums, if any exist. 72. Find the maximum, if it exists. 73. Find the minimum, if it exists. In Exercises 74 - 77, use your graphing calculator to approximate the local and absolute extrema of the given function. Approximate the
intervals on which the function is increasing and those on which it is decreasing. Round your answers to two decimal places. 74. f (x) = x4 − 3x3 − 24x2 + 28x + 48 75. f (x) = x2/3(x − 4) 76. f (x) = √ 9 − x2 77. f (x) = x √ 9 − x2 110 Relations and Functions In Exercises 78 - 85, use the graphs of y = f (x) and y = g(x) below to ﬁnd the function valuex) y = g(x) 78. (f + g)(0) 79. (f + g)(1) 80. (f − g)(1) 81. (g − f )(2) 82. (f g)(2) 83. (f g)(1) 84. f g (4) 85. g f (2) The graph below represents the height h of a Sasquatch (in feet) as a function of its age N in years. Use it to answer the questions in Exercises 86 - 90. y 8 6 4 2 15 30 45 60 N y = h(N ) 86. Find and interpret h(0). 87. How tall is the Sasquatch when she is 15 years old? 88. Solve h(N ) = 6 and interpret. 89. List the interval over which h is constant and interpret your answer. 90. List the interval over which h is decreasing and interpret your answer. 1.6 Graphs of Functions 111 For Exercises 91 - 93, let f (x) = x be the greatest integer function as deﬁned in Exercise 75 in Section 1.4. 91. Graph y = f (x). Be careful to correctly describe the behavior of the graph near the integers. 92. Is f even, odd, or neither? Explain. 93. Discuss with your classmates which points on the graph are local minimums, local maximums or both. Is f ever increasing? Decreasing? Constant? In Exercises 94 - 95, use your graphing calculator to show that the given function does not have any extrema, neither local nor absolute. 94. f (x) = x3 + x − 12 95. f (x) = −5x + 2 96. In Exercise 71 in Section 1.4, we saw that the population of Sas
quatch in Portage County could be modeled by the function P (t) = 150t t + 15 your graphing calculator to analyze the general function behavior of P. Will there ever be a time when 200 Sasquatch roam Portage County?, where t = 0 represents the year 1803. Use 97. Suppose f and g are both even functions. What can be said about the functions f + g, f − g, f g and f g? What if f and g are both odd? What if f is even but g is odd? 98. One of the most important aspects of the Cartesian Coordinate Plane is its ability to put Algebra into geometric terms and Geometry into algebraic terms. We’ve spent most of this chapter looking at this very phenomenon and now you should spend some time with your classmates reviewing what we’ve done. What major results do we have that tie Algebra and Geometry together? What concepts from Geometry have we not yet described algebraically? What topics from Intermediate Algebra have we not yet discussed geometrically? It’s now time to “thoroughly vet the pathologies induced” by the precise deﬁnitions of local maximum and local minimum. We’ll do this by providing you and your classmates a series of Exercises to discuss. You will need to refer back to Deﬁnition 1.10 (Increasing, Decreasing and Constant) and Deﬁnition 1.11 (Maximum and Minimum) during the discussion. 99. Consider the graph of the function f given below. y 3 2 1 −2 −1 −1 1 2 x −2 −3 112 Relations and Functions (a) Show that f has a local maximum but not a local minimum at the point (−1, 1). (b) Show that f has a local minimum but not a local maximum at the point (1, 1). (c) Show that f has a local maximum AND a local minimum at the point (0, 1). (d) Show that f is constant on the interval [−1, 1] and thus has both a local maximum AND a local minimum at every point (x, f (x)) where −1 < x < 1. 100. Using Example 1.6.4 as a guide, show that the function g whose graph is given below does not have a local maximum at (−3, 5) nor does it have a
local minimum at (3, −3). Find its extrema, both local and absolute. What’s unique about the point (0, −4) on this graph? Also ﬁnd the intervals on which g is increasing and those on which g is decreasing. y 5 4 3 2 1 −3 −2 −1 −1 1 2 3 x −2 −3 −4 101. We said earlier in the section that it is not good enough to say local extrema exist where a function changes from increasing to decreasing or vice versa. As a previous exercise showed, we could have local extrema when a function is constant so now we need to examine some functions whose graphs do indeed change direction. Consider the functions graphed below. Notice that all four of them change direction at an open circle on the graph. Examine each for local extrema. What is the eﬀect of placing the “dot” on the y-axis above or below the open circle? What could you say if no function value were assigned to x = 02 −1 −1 1 2 x −2 −1 −1 1 2 x (a) Function I (b) Function II 1.6 Graphs of Functions 113 y 4 3 2 1 −2 −1 −2 −1 1 2 x (c) Function III (d) Function IV 114 1.6.3 Answers 1. f (x) = 2 − x Domain: (−∞, ∞) x-intercept: (2, 0) y-intercept: (0, 2) No symmetry 2. f (x) = x − 2 3 Domain: (−∞, ∞) x-intercept: (2, 0) y-intercept: 0, − 2 3 No symmetry 3. f (x) = x2 + 1 Domain: (−∞, ∞) x-intercept: None y-intercept: (0, 1) Even 4. f (x) = 4 − x2 Domain: (−∞, ∞) x-intercepts: (−2, 0), (2, 0) y-intercept: (0, 4) Even 5. f (x) = 2 Domain: (−∞, ∞) x-intercept: None y-intercept: (0, 2) Even Relations and Functions y 3 2 1 −2 −1 1 2 3 x −1 y 1 −1
1 2 3 4 x −2 −1 −2 −1 1 2 x 1 2 x −2 −1 1 2 x 1.6 Graphs of Functions 115 6. f (x) = x3 Domain: (−∞, ∞) x-intercept: (0, 0) y-intercept: (0, 0) Odd 7. f (x) = x(x − 1)(x + 2) Domain: (−∞, ∞) x-intercepts: (−2, 0), (0, 0), (1, 0) y-intercept: (0, 0) No symmetry √ 8. f (x) = x − 2 Domain: [2, ∞) x-intercept: (2, 0) y-intercept: None No symmetry 9. f (x) = √ 5 − x Domain: (−∞, 5] x-intercept: (5, 0) √ y-intercept: (0, 5) No symmetry 2−1 −1 1 2 x −2 −3 −4 −5 −6 −7 −8 y 4 3 2 1 −2 −4 −3 −2 −1 1 2 3 4 5 x 116 Relations and Functions 10. f (x) = 3 − 2 √ x + 2 Domain: [−2, ∞) 4, 0 x-intercept: 1 y-intercept: (0, 3 − 2 √ 2) No symmetry √ 11. f (x) = 3 x Domain: (−∞, ∞) x-intercept: (0, 0) y-intercept: (0, 0) Odd 12. f (x) = 1 x2 + 1 Domain: (−∞, ∞) x-intercept: None y-intercept: (0, 1) Even 13. 15. y 5 4 3 2 1 −4−3−2−1 −1 −2 −1 1 2 x y 2 1 −8−7−6−5−4−3−2−1 −2 y 1 −2 −3−2−1 1 2 3 x 14. 16. y 5 4 3 2 1 −2−1 1 2 x 1.6 Graphs of Functions 117 17. 19. y 3 2 1 −2 −1 1 x −1 −2 −3 −4 6 5 4 3 2 1 y −2 −1 1
2 3 x 21. odd 24. even 27. odd 22. neither 25. even 28. odd 30. neither 31. neither 33. even and odd 34. odd 36. even 39. odd 42. [−5, 3] 45. x = −3 37. neither 40. even 43. [−5, 4] 18. 20. y 3 2 1 −4 −3 −2 −6−5−4−3−2−1 23. even 26. neither 29. even 32. even 35. even 38. odd 41. even 44. f (−2) = 2 48. −4, −1, 1 49. [−4, −1] ∪ [1, 3] 50. 4 46. (−4, 0), (−1, 0), (1, 0) 47. (0, −1) 118 Relations and Functions 51. neither 52. [−5, −3], [0, 2] 53. [−3, 0], [2, 3] 54. f (−3) = 4, f (2) = 3 56. f (−3) = 4 55. f (0) = −1 57. f (−5) = −5 58. [−4, 4] 61. x = −2 64. −4, 0, 4 67. neither 70. none 72. none 74. No absolute maximum Absolute minimum f (4.55) ≈ −175.46 Local minimum at (−2.84, −91.32) Local maximum at (0.54, 55.73) Local minimum at (4.55, −175.46) Increasing on [−2.84, 0.54], [4.55, ∞) Decreasing on (−∞, −2.84], [0.54, 4.55] 59. [−5, 5) 60. f (2) = 3 62. (−4, 0), (0, 0), (4, 0) 63. (0, 0) 65. [−4, 0] ∪ {4} 66. 3 68. [−2, 2) 69. [−4, −2], (2, 4] 71. f (−2) = −5, f (2) = 3 73. f (−2) = −5 75. No absolute maximum No absolute minimum Local maximum at (0, 0) Local minimum at (1.60, −3.
28) Increasing on (−∞, 0], [1.60, ∞) Decreasing on [0, 1.60] 76. Absolute maximum f (0) = 3 77. Absolute maximum f (2.12) ≈ 4.50 Absolute minimum f (±3) = 0 Local maximum at (0, 3) No local minimum Increasing on [−3, 0] Decreasing on [0, 3] Absolute minimum f (−2.12) ≈ −4.50 Local maximum (2.12, 4.50) Local minimum (−2.12, −4.50) Increasing on [−2.12, 2.12] Decreasing on [−3, −2.12], [2.12, 3] 78. (f + g)(0) = 4 79. (f + g)(1) = 5 80. (f − g)(1) = −1 81. (g − f )(2) = 0 82. (f g)(2) = 9 83. (f g)(1) = 6 84. f g (4) = 0 85. g f (2) = 1 86. h(0) = 2, so the Sasquatch is 2 feet tall at birth. 87. h(15) = 6, so the Saquatch is 6 feet tall when she is 15 years old. 88. h(N ) = 6 when N = 15 and N = 60. This means the Sasquatch is 6 feet tall when she is 15 and 60 years old. 89. h is constant on [30, 45]. This means the Sasquatch’s height is constant (at 8 feet) for these years. 1.6 Graphs of Functions 119 90. h is decreasing on [45, 60]. This means the Sasquatch is getting shorter from the age of 45 to the age of 60. (Sasquatchteoporosis, perhaps?) 91. 92. Note that f (1.1) = 1, but f (−1.1) = −2, so f is neither even nor odd.... y 6 5 4 3 2 1 −6 −5 −4 −3 −2 −1 1 2 3 4 5 6 x −2 −3 −4 −5 −6... The graph of f (x) = x. 120 Relations and Functions 1.7 Transformations In this section, we study how the graphs of functions change, or
transform, when certain specialized modiﬁcations are made to their formulas. The transformations we will study fall into three broad categories: shifts, reﬂections and scalings, and we will present them in that order. Suppose the graph below is the complete graph of a function f. y 5 4 3 2 (0, 1) (5, 5) (2, 3) (4, 3x) The Fundamental Graphing Principle for Functions says that for a point (a, b) to be on the graph, f (a) = b. In particular, we know f (0) = 1, f (2) = 3, f (4) = 3 and f (5) = 5. Suppose we wanted to graph the function deﬁned by the formula g(x) = f (x) + 2. Let’s take a minute to remind ourselves of what g is doing. We start with an input x to the function f and we obtain the output f (x). The function g takes the output f (x) and adds 2 to it. In order to graph g, we need to graph the points (x, g(x)). How are we to ﬁnd the values for g(x) without a formula for f (x)? The answer is that we don’t need a formula for f (x), we just need the values of f (x). The values of f (x) are the y values on the graph of y = f (x). For example, using the points indicated on the graph of f, we can make the following table. x 0 2 4 5 (x, f (x)) f (x) g(x) = f (x) + 2 (x, g(x)) (0, 1) (2, 3) (4, 3) (5, 50, 3) (2, 5) (4, 5) (5, 7) In general, if (a, b) is on the graph of y = f (x), then f (a) = b, so g(a) = f (a) + 2 = b + 2. Hence, (a, b+2) is on the graph of g. In other words, to obtain the graph of g, we add 2 to the y-coordinate of each point on the graph of f. Geometrically, adding 2 to the
y-coordinate of a point moves the point 2 units above its previous location. Adding 2 to every y-coordinate on a graph en masse is usually described as ‘shifting the graph up 2 units’. Notice that the graph retains the same basic shape as before, it is just 2 units above its original location. In other words, we connect the four points we moved in the same manner in which they were connected before. We have the results side-by-side at the top of the next page. 1.7 Transformations 121 y 7 6 5 4 3 2 (0, 1) (5, 5) (2, 3) (4, 3) (5, 7) (2, 5) (4, 5) y 7 6 5 4 (0, 3x) shift up 2 units −−−−−−−−−−−−→ add 2 to each y-coordinate 1 2 3 4 5 x y = g(x) = f (x) + 2 You’ll note that the domain of f and the domain of g are the same, namely [0, 5], but that the range of f is [1, 5] while the range of g is [3, 7]. In general, shifting a function vertically like this will leave the domain unchanged, but could very well aﬀect the range. You can easily imagine what would happen if we wanted to graph the function j(x) = f (x) − 2. Instead of adding 2 to each of the y-coordinates on the graph of f, we’d be subtracting 2. Geometrically, we would be moving the graph down 2 units. We leave it to the reader to verify that the domain of j is the same as f, but the range of j is [−1, 3]. What we have discussed is generalized in the following theorem. Theorem 1.2. Vertical Shifts. Suppose f is a function and k is a positive number. To graph y = f (x) + k, shift the graph of y = f (x) up k units by adding k to the y-coordinates of the points on the graph of f. To graph y = f (x) − k, shift the graph of y = f (x) down k units by subtracting k from the y-coordinates of the points on the graph of f. The key to understanding Theorem 1.2
and, indeed, all of the theorems in this section comes from an understanding of the Fundamental Graphing Principle for Functions. If (a, b) is on the graph of f, then f (a) = b. Substituting x = a into the equation y = f (x) + k gives y = f (a) + k = b + k. Hence, (a, b + k) is on the graph of y = f (x) + k, and we have the result. In the language of ‘inputs’ and ‘outputs’, Theorem 1.2 can be paraphrased as “Adding to, or subtracting from, the output of a function causes the graph to shift up or down, respectively.” So what happens if we add to or subtract from the input of the function? Keeping with the graph of y = f (x) above, suppose we wanted to graph g(x) = f (x + 2). In other words, we are looking to see what happens when we add 2 to the input of the function.1 Let’s try to generate a table of values of g based on those we know for f. We quickly ﬁnd that we run into some diﬃculties. 1We have spent a lot of time in this text showing you that f (x + 2) and f (x) + 2 are, in general, wildly diﬀerent algebraic animals. We will see momentarily that their geometry is also dramatically diﬀerent. 122 Relations and Functions x 0 2 4 5 (x, f (x)) f (x) (0, 1) (2, 3) (4, 3) (5, 5) 1 3 3 5 g(x) = f (x + 2) f (0 + 2) = f (2) = 3 f (2 + 2) = f (4) = 3 f (4 + 2) = f (6) =? f (5 + 2) = f (7) =? (x, g(x)) (0, 3) (2, 3) When we substitute x = 4 into the formula g(x) = f (x + 2), we are asked to ﬁnd f (4 + 2) = f (6) which doesn’t exist because the domain of f is only [0, 5].
The same thing happens when we attempt to ﬁnd g(5). What we need here is a new strategy. We know, for instance, f (0) = 1. To determine the corresponding point on the graph of g, we need to ﬁgure out what value of x we must substitute into g(x) = f (x + 2) so that the quantity x + 2, works out to be 0. Solving x + 2 = 0 gives x = −2, and g(−2) = f ((−2) + 2) = f (0) = 1 so (−2, 1) is on the graph of g. To use the fact f (2) = 3, we set x + 2 = 2 to get x = 0. Substituting gives g(0) = f (0 + 2) = f (2) = 3. Continuing in this fashion, we get x −(x) = f (x + 2) g(−2) = f (0) = 1 g(0) = f (2) = 3 g(2) = f (4) = 3 g(3) = f (5) = 5 (x, g(x)) (−2, 1) (0, 3) (2, 3) (3, 5) In summary, the points (0, 1), (2, 3), (4, 3) and (5, 5) on the graph of y = f (x) give rise to the points (−2, 1), (0, 3), (2, 3) and (3, 5) on the graph of y = g(x), respectively. In general, if (a, b) is on the graph of y = f (x), then f (a) = b. Solving x + 2 = a gives x = a − 2 so that g(a − 2) = f ((a − 2) + 2) = f (a) = b. As such, (a − 2, b) is on the graph of y = g(x). The point (a − 2, b) is exactly 2 units to the left of the point (a, b) so the graph of y = g(x) is obtained by shifting the graph y = f (x) to the left 2 units, as pictured below. y 5 4 3 2 (0, 1) (5, 5) (
2, 3) (4, 3) y (3, 5) (2, 3) 5 4 (0, 3) 2 1 (−2, 1) −2 −x) shift left 2 units −−−−−−−−−−−−→ subtract 2 from each x-coordinate −2 −(x) = f (x + 2) Note that while the ranges of f and g are the same, the domain of g is [−2, 3] whereas the domain of f is [0, 5]. In general, when we shift the graph horizontally, the range will remain the same, but the domain could change. If we set out to graph j(x) = f (x − 2), we would ﬁnd ourselves adding 1.7 Transformations 123 2 to all of the x values of the points on the graph of y = f (x) to eﬀect a shift to the right 2 units. Generalizing these notions produces the following result. Theorem 1.3. Horizontal Shifts. Suppose f is a function and h is a positive number. To graph y = f (x + h), shift the graph of y = f (x) left h units by subtracting h from the x-coordinates of the points on the graph of f. To graph y = f (x − h), shift the graph of y = f (x) right h units by adding h to the x-coordinates of the points on the graph of f. In other words, Theorem 1.3 says that adding to or subtracting from the input to a function amounts to shifting the graph left or right, respectively. Theorems 1.2 and 1.3 present a theme which will run common throughout the section: changes to the outputs from a function aﬀect the y-coordinates of the graph, resulting in some kind of vertical change; changes to the inputs to a function aﬀect the x-coordinates of the graph, resulting in some kind of horizontal change. Example 1.7.1. 1. Graph f (x) = √ x. Plot at least three points. 2. Use your graph in 1 to graph g(x) = √ x − 1. 3. Use your graph in 1 to graph j(x) = 4. Use your graph in 1 to graph m(x) = Solution. √ x − 1. √ x
+ 3 − 2. 1. Owing to the square root, the domain of f is x ≥ 0, or [0, ∞). We choose perfect squares to build our table and graph below. From the graph we verify the domain of f is [0, ∞) and the range of f is also [0, ∞). x 0 1 4 f (x) 0 1 2 (x, f (x)) (0, 0) (1, 1) (4, 2) (4, 2) y 2 (1, 1) 1 (0, 0x) = √ x 2. The domain of g is the same as the domain of f, since the only condition on both functions is that x ≥ 0. If we compare the formula for g(x) with f (x), we see that g(x) = f (x) − 1. In other words, we have subtracted 1 from the output of the function f. By Theorem 1.2, we know that in order to graph g, we shift the graph of f down one unit by subtracting 1 from each of the y-coordinates of the points on the graph of f. Applying this to the three points we have speciﬁed on the graph, we move (0, 0) to (0, −1), (1, 1) to (1, 0), and (4, 2) to (4, 1). 124 Relations and Functions The rest of the points follow suit, and we connect them with the same basic shape as before. We conﬁrm the domain of g is [0, ∞) and ﬁnd the range of g to be [−1, ∞). (4, 2) y 2 (1, 1) 1 (0, 01, 0) (4, 1) 1 2 3 4 x (0, −1) y = f (x) = √ x shift down 1 unit −−−−−−−−−−−−→ subtract 1 from each y-coordinate y = g(x) = √ x − 1 3. Solving x − 1 ≥ 0 gives x ≥ 1, so the domain of j is [1, ∞). To graph j, we note that j(x) = f (x − 1). In other words, we are subtracting 1 from the input of f. According to Theorem 1.
3, this induces a shift to the right of the graph of f. We add 1 to the x-coordinates of the points on the graph of f and get the result below. The graph reaﬃrms that the domain of j is [1, ∞) and tells us that the range of j is [0, ∞). (4, 2) y 2 (1, 1) 1 (0, 0) y 2 1 (2, 1) (5, 2x) = √ x shift right 1 unit −−−−−−−−−−−−→ add 1 to each x-coordinate (1, 0) 2 3 4 5 x y = j(x) = √ x − 1 4. To ﬁnd the domain of m, we solve x + 3 ≥ 0 and get [−3, ∞). Comparing the formulas of f (x) and m(x), we have m(x) = f (x + 3) − 2. We have 3 being added to an input, indicating a horizontal shift, and 2 being subtracted from an output, indicating a vertical shift. We leave it to the reader to verify that, in this particular case, the order in which we perform these transformations is immaterial; we will arrive at the same graph regardless as to which transformation we apply ﬁrst.2 We follow the convention ‘inputs ﬁrst’,3 and to that end we ﬁrst tackle the horizontal shift. Letting m1(x) = f (x + 3) denote this intermediate step, Theorem 1.3 tells us that the graph of y = m1(x) is the graph of f shifted to the left 3 units. Hence, we subtract 3 from each of the x-coordinates of the points on the graph of f. 2We shall see in the next example that order is generally important when applying more than one transformation to a graph. 3We could equally have chosen the convention ‘outputs ﬁrst’. 1.7 Transformations 125 (4, 2) y 2 (1, 1) 1 (0, 0) −3 −2 −1 1 2 3 4 x −1 −2 y = f (x) = √ x y (1, 2) (−2, 1) 2 1 (−3, 0) −3 −2 −1 1
2 3 4 x −1 −2 shift left 3 units −−−−−−−−−−−−→ subtract 3 from each x-coordinate y = m1(x) = f (x + 3) = √ x + 3 Since m(x) = f (x + 3) − 2 and f (x + 3) = m1(x), we have m(x) = m1(x) − 2. We can apply Theorem 1.2 and obtain the graph of m by subtracting 2 from the y-coordinates of each of the points on the graph of m1(x). The graph veriﬁes that the domain of m is [−3, ∞) and we ﬁnd the range of m to be [−2, ∞). y (1, 2) (−2, 1) 2 1 (−3, 0) −3 −2 −1 1 2 3 4 x −1 −2 y = m1(x) = f (x + 3) = √ x + 3 shift down 2 units −−−−−−−−−−−−→ subtract 2 from each y-coordinate y 2 1 (1, 0) −3 −2 −1 1 2 3 4 (−2, −1) (−3, −2) −1 −2 x y = m(x) = m1(x Keep in mind that we can check our answer to any of these kinds of problems by showing that any of the points we’ve moved lie on the graph of our ﬁnal answer. For example, we can check that (−3, −2) is on the graph of m by computing m(−3) = 0 − 2 = −2 (−3) + 3 − 2 = √ We now turn our attention to reﬂections. We know from Section 1.1 that to reﬂect a point (x, y) across the x-axis, we replace y with −y. If (x, y) is on the graph of f, then y = f (x), so replacing y with −y is the same as replacing f (x) with −f (x). Hence, the graph of y = −f (x) is the graph of f reﬂected across the x-axis. Similarly, the graph of y = f (−x) is the graph
of f reﬂected across the y-axis. Returning to the language of inputs and outputs, multiplying the output from a function by −1 reﬂects its graph across the x-axis, while multiplying the input to a function by −1 reﬂects the graph across the y-axis.4 4The expressions −f (x) and f (−x) should look familiar - they are the quantities we used in Section 1.6 to test if a function was even, odd or neither. The interested reader is invited to explore the role of reﬂections and symmetry of functions. What happens if you reﬂect an even function across the y-axis? What happens if you reﬂect an odd function across the y-axis? What about the x-axis? 126 Relations and Functions Theorem 1.4. Reﬂections. Suppose f is a function. To graph y = −f (x), reﬂect the graph of y = f (x) across the x-axis by multiplying the y-coordinates of the points on the graph of f by −1. To graph y = f (−x), reﬂect the graph of y = f (x) across the y-axis by multiplying the x-coordinates of the points on the graph of f by −1. Applying Theorem 1.4 to the graph of y = f (x) given at the beginning of the section, we can graph y = −f (x) by reﬂecting the graph of f about the x-axis y 5 4 3 2 (5, 5) (2, 3) (4, 3) 1 2 3 4 5 x (0, 1) −1 −2 −3 −4 −5 y = f (x) reﬂect across x-axis −−−−−−−−−−−−→ multiply each y-coordinate by −0, −1) −2 −3 −4 −5 (4, −3) (2, −3) (5, −5) y = −f (x) By reﬂecting the graph of f across the y-axis, we obtain the graph of y = f (−x). y 5 4 3 2 (0, 1) (5, 5) (−5, 5) (2, 3) (4
, 3) (−2, 3) (−4, 3) y 5 4 3 2 (0, 1) −5 −4 −3 −2 −x) reﬂect across y-axis −−−−−−−−−−−−→ multiply each x-coordinate by −1 −5 −4 −3 −2 − (−x) With the addition of reﬂections, it is now more important than ever to consider the order of transformations, as the next example illustrates. √ Example 1.7.2. Let f (x) = functions. Also, state their domains and ranges. √ √ 1. g(x) = −x 2. j(x) = 3 − x 3. m(x) = 3 − √ x x. Use the graph of f from Example 1.7.1 to graph the following 1.7 Transformations 127 Solution. √ 1. The mere sight of √ −x usually causes alarm, if not panic. When we discussed domains in Section 1.4, we clearly banished negatives from the radicands of even roots. However, we must remember that x is a variable, and as such, the quantity −x isn’t always negative. For −(−4) = 2 is perfectly well-deﬁned. To ﬁnd the example, if x = −4, −x = 4, thus domain analytically, we set −x ≥ 0 which gives x ≤ 0, so that the domain of g is (−∞, 0]. Since g(x) = f (−x), Theorem 1.4 tells us that the graph of g is the reﬂection of the graph of f across the y-axis. We accomplish this by multiplying each x-coordinate on the graph of f by −1, so that the points (0, 0), (1, 1), and (4, 2) move to (0, 0), (−1, 1), and (−4, 2), respectively. Graphically, we see that the domain of g is (−∞, 0] and the range of g is the same as the range of f, namely [0, ∞). −x = y (1, 1) 2 1 (0, 0) (4, 2) (−4, 2) y 2 1 (−1, 1) (0, 0) −4
−3 −2 −x) = √ x reﬂect across y-axis −−−−−−−−−−−−→ multiply each x-coordinate by −1 −4 −3 −2 −1 1 2 3 4 x y = g(x) = f (−x) = √ −x √ 2. To determine the domain of j(x) = √ √ −x + 3 = f (−x + 3). Comparing this formula with f (x) = 3 − x, we solve 3 − x ≥ 0 and get x ≤ 3, or (−∞, 3]. To determine which transformations we need to apply to the graph of f to obtain the graph of j, we rewrite j(x) = x, we see that not only are we multiplying the input x by −1, which results in a reﬂection across the y-axis, but also we are adding 3, which indicates a horizontal shift to the left. Does it matter in which order we do the transformations? If so, which order is the correct order? Let’s consider the point (4, 2) on the graph of f. We refer to the discussion leading up to Theorem 1.3. We know f (4) = 2 and wish to ﬁnd the point on y = j(x) = f (−x + 3) which corresponds to (4, 2). We set −x + 3 = 4 and solve. Our ﬁrst step is to subtract 3 from both sides to get −x = 1. Subtracting 3 from the x-coordinate 4 is shifting the point (4, 2) to the left. From −x = 1, we then multiply5 both sides by −1 to get x = −1. Multiplying the x-coordinate by −1 corresponds to reﬂecting the point about the y-axis. Hence, we perform the horizontal shift ﬁrst, then follow it with the reﬂection about the y-axis. Starting with f (x) = x, we let j1(x) be the intermediate function which shifts the graph of f 3 units to the left, j1(x) = f (x + 3). √ y (1, 1) 2 1 (0, 0) (4, 2) y (1, 2) (−2, 1)
2 1 (−3, 0) −4 −3 −2 −x) = √ x shift left 3 units −−−−−−−−−−−−→ subtract 3 from each x-coordinate −4 −3 −2 −1 1 2 3 4 x y = j1(x) = f (x + 3) = √ x + 3 5Or divide - it amounts to the same thing. 128 Relations and Functions To obtain the function j, we reﬂect the graph of j1 about y-axis. Theorem 1.4 tells us we −x + 3, have j(x) = j1(−x). Putting it all together, we have j(x) = j1(−x) = f (−x + 3) = which is what we want.6 From the graph, we conﬁrm the domain of j is (−∞, 3] and we get that the range is [0, ∞). √ y (1, 2) (−2, 1) 2 1 (−3, 0) y 2 (−1, 2) (2, 1) (3, 0) −4 −3 −2 −1 1 2 3 4 x y = j1(x) = √ x + 3 reﬂect across y-axis −−−−−−−−−−−−→ multiply each x-coordinate by −1 −4 −3 −2 −1 1 2 3 4 x y = j(x) = j1(−x) = √ −x + 3 √ 3. The domain of m works out to be the domain of f, [0, ∞). Rewriting m(x) = − x + 3, we see m(x) = −f (x) + 3. Since we are multiplying the output of f by −1 and then adding 3, we once again have two transformations to deal with: a reﬂection across the x-axis and a vertical shift. To determine the correct order in which to apply the transformations, we imagine trying to determine the point on the graph of m which corresponds to (4, 2) on the graph of f. Since in the formula for m(x), the input to f is just x, we substitute to ﬁnd m(4) = −f (4) + 3 = −2 + 3 = 1. Hence,
(4, 1) is the corresponding point on the graph of m. If we closely examine the arithmetic, we see that we ﬁrst multiply f (4) by −1, which corresponds to the reﬂection across the x-axis, and then we add 3, which corresponds to the vertical shift. If we deﬁne an intermediate function m1(x) = −f (x) to take care of the reﬂection, we get y (1, 1) (4, 20, 0) −1 −2 y 3 2 1 (0, 0) −1 1 2 3 4 x (1, −1) −2 y = f (x) = √ x reﬂect across x-axis −−−−−−−−−−−−→ multiply each y-coordinate by −1 y = m1(x) = −f (x) = − (4, −2) √ x To shift the graph of m1 up 3 units, we set m(x) = m1(x) + 3. Since m1(x) = −f (x), when we put it all together, we get m(x) = m1(x) + 3 = −f (x) + 3 = − x + 3. We see from the graph that the range of m is (−∞, 3]. √ 6If we had done the reﬂection ﬁrst, then j1(x) = f (−x). Following this by a shift left would give us j(x) = j1(x + 3) = f (−(x + 3)) = f (−x − 3) = −x − 3 which isn’t what we want. However, if we did the reﬂection ﬁrst and followed it by a shift to the right 3 units, we would have arrived at the function j(x). We leave it to the reader to verify the details. √ 1.7 Transformations 129 y 3 2 1 (0, 0) −1 1 2 3 4 x (1, −1) −2 (4, −2) shift up 3 units −−−−−−−−−−−−→ add 3 to each y-coordinate (0, 3) y (1, 2) (4, 1) 1 2 3 4
x 2 1 −1 −2 y = m1(x) = − √ x y = m(x) = m1(x) + 3 = − √ x + 3 We now turn our attention to our last class of transformations known as scalings. A thorough discussion of scalings can get complicated because they are not as straight-forward as the previous transformations. A quick review of what we’ve covered so far, namely vertical shifts, horizontal shifts and reﬂections, will show you why those transformations are known as rigid transformations. Simply put, they do not change the shape of the graph, only its position and orientation in the plane. If, however, we wanted to make a new graph twice as tall as a given graph, or one-third as wide, we would be changing the shape of the graph. This type of transformation is called nonrigid for obvious reasons. Not only will it be important for us to diﬀerentiate between modifying inputs versus outputs, we must also pay close attention to the magnitude of the changes we make. As you will see shortly, the Mathematics turns out to be easier than the associated grammar. Suppose we wish to graph the function g(x) = 2f (x) where f (x) is the function whose graph is given at the beginning of the section. From its graph, we can build a table of values for g as before. y 5 4 3 2 (0, 1) (5, 5) (2, 3) (4, 3x) x 0 2 4 5 (x, f (x)) f (x) g(x) = 2f (x) (0, 1) (2, 3) (4, 3) (5, 5) 1 3 3 5 2 6 6 10 (x, g(x)) (0, 2) (2, 6) (4, 6) (5, 10) In general, if (a, b) is on the graph of f, then f (a) = b so that g(a) = 2f (a) = 2b puts (a, 2b) on the graph of g. In other words, to obtain the graph of g, we multiply all of the y-coordinates of the points on the graph of f by 2. Multiplying all of the y-coordinates of all of the points on the graph of f by 2 causes what is known as a
‘vertical scaling7 by a factor of 2’, and the results are given on the next page. 7Also called a ‘vertical stretching’, ‘vertical expansion’ or ‘vertical dilation’ by a factor of 2. 130 Relations and Functions y 10 9 8 7 6 5 4 3 2 (0, 1) (5, 5) (2, 3) (4, 3) y 10 (5, 10) (2, 6) (4, 6) 9 8 7 6 5 4 3 (0, 2x) vertical scaling by a factor of 2 −−−−−−−−−−−−−−−−−−−−→ multiply each y-coordinate by 2 1 2 3 4 5 x y = 2f (x) If we wish to graph y = 1 of f by 1 2. This creates a ‘vertical scaling8 by a factor of 1 2 ’ as seen below. 2 f (x), we multiply the all of the y-coordinates of the points on the graph y 5 4 3 2 (0, 1) (5, 5) (2, 3) (4, 3x) vertical scaling by a factor of 1 2 −−−−−−−−−−−−−−−−−−−−→ multiply each y-coordinate by 1 2 These results are generalized in the following theorem. y 5 4 3 2 1 0, 1 2 5, 5 2 2, 3 2 4x) Theorem 1.5. Vertical Scalings. Suppose f is a function and a > 0. To graph y = af (x), multiply all of the y-coordinates of the points on the graph of f by a. We say the graph of f has been vertically scaled by a factor of a. If a > 1, we say the graph of f has undergone a vertical stretching (expansion, dilation) by a factor of a. If 0 < a < 1, we say the graph of f has undergone a vertical shrinking (compression, contraction) by a factor of 1 a. 8Also called ‘vertical shrinking’, ‘vertical compression’ or ‘vertical contraction’ by a factor of 2. 1.7 Transformations 131 A few remarks about Theorem 1.5 are in order. First, a note about the verbiage. To the authors, the words
‘stretching’, ‘expansion’, and ‘dilation’ all indicate something getting bigger. Hence, ‘stretched by a factor of 2’ makes sense if we are scaling something by multiplying it by 2. Similarly, we believe words like ‘shrinking’, ‘compression’ and ‘contraction’ all indicate something getting smaller, so if we scale something by a factor of 1 2, we would say it ‘shrinks by a factor of 2’ - not ‘shrinks by a factor of 1 2 ’. This is why we have written the descriptions ‘stretching by a factor of a’ and ‘shrinking by a factor of 1 a ’ in the statement of the theorem. Second, in terms of inputs and outputs, Theorem 1.5 says multiplying the outputs from a function by positive number a causes the graph to be vertically scaled by a factor of a. It is natural to ask what would happen if we multiply the inputs of a function by a positive number. This leads us to our last transformation of the section. Referring to the graph of f given at the beginning of this section, suppose we want to graph g(x) = f (2x). In other words, we are looking to see what eﬀect multiplying the inputs to f by 2 has on its graph. If we attempt to build a table directly, we quickly run into the same problem we had in our discussion leading up to Theorem 1.3, as seen in the table on the left below. We solve this problem in the same way we solved this problem before. For example, if we want to determine the point on g which corresponds to the point (2, 3) on the graph of f, we set 2x = 2 so that x = 1. Substituting x = 1 into g(x), we obtain g(1) = f (2 · 1) = f (2) = 3, so that (1, 3) is on the graph of g. Continuing in this fashion, we obtain the table on the lower rightx, f (x)) f (x) (0, 1) (2, 3) (4, 3) (5, 5) (x, g(x)) (0, 1) (2, 3) g(x) = f (2x) f (2 · 0) =
f (0) = 1 f (2 · 2) = f (4) = 3 f (2 · 4) = f (8) =? f (2 · 5) = f (10) =? In general, if (a, b) is on the graph of f, then f (a) = b. Hence g a = f (a) = b so that 2 2, b is on the graph of g. In other words, to graph g we divide the x-coordinates of the points on a the graph of f by 2. This results in a horizontal scaling9 by a factor of 1 2. g(x) = f (2x) g(0) = f (0) = 1 g(1) = f (2) = 3 g(2) = f (4) = 3 g 5 = f (5x, g(x)) (0, 0) (1, 3) (2, 3) 2, 5 5 2x 0, 1) (5, 5) (2, 3) (4, 3) y 2, 5 5 (1, 3) (2, 3) 5 4 3 2 (0, 1x) horizontal scaling by a factor of 1 2 −−−−−−−−−−−−−−−−−−−−→ multiply each x-coordinate by (x) = f (2x) 9Also called ‘horizontal shrinking’, ‘horizontal compression’ or ‘horizontal contraction’ by a factor of 2. 132 Relations and Functions If, on the other hand, we wish to graph y = f 1 2 x, we end up multiplying the x-coordinates of the points on the graph of f by 2 which results in a horizontal scaling10 by a factor of 2, as demonstrated below. y (5, 5) (2, 3) (4, 3) 5 4 3 2 (0, 1) y 5 4 3 2 (0, 1) (10, 5) (4, 3) (8, 3 10 x y = f (x) horizontal scaling by a factor of 2 −−−−−−−−−−−−−−−−−−−−→ multiply each x-coordinate by 10 x y = g(x) = f 1 2 x We have the following theorem. Theorem 1.6. Horizontal Scalings.
Suppose f is a function and b > 0. To graph y = f (bx), divide all of the x-coordinates of the points on the graph of f by b. We say the graph of f has been horizontally scaled by a factor of 1 b. If 0 < b < 1, we say the graph of f has undergone a horizontal stretching (expansion, dilation) by a factor of 1 b. If b > 1, we say the graph of f has undergone a horizontal shrinking (compression, con- traction) by a factor of b. Theorem 1.6 tells us that if we multiply the input to a function by b, the resulting graph is scaled horizontally by a factor of 1 b since the x-values are divided by b to produce corresponding points on the graph of y = f (bx). The next example explores how vertical and horizontal scalings sometimes interact with each other and with the other transformations introduced in this section. Example 1.7.3. Let f (x) = functions. Also, state their domains and ranges. √ x. Use the graph of f from Example 1.7.1 to graph the following 1. g(x) = 3 √ x 2. j(x) = √ 9x 3. m(x) = 1 − x+3 2 Solution. 1. First we note that the domain of g is [0, ∞) for the usual reason. Next, we have g(x) = 3f (x) so by Theorem 1.5, we obtain the graph of g by multiplying all of the y-coordinates of the points on the graph of f by 3. The result is a vertical scaling of the graph of f by a factor of 3. We ﬁnd the range of g is also [0, ∞). 10Also called ‘horizontal stretching’, ‘horizontal expansion’ or ‘horizontal dilation’ by a factor of 2. 1.7 Transformations 133 y (4, 2) (1, 1) 6 5 4 3 2 1 (0, 0) y (4, 6) 6 5 4 3 2 1 (0, 0) (1, 3x) = √ x vertical scale by a factor of 3 −−−−−−−−−−−−−−−−−−−−−→ multiply each y-coordinate by 3 1 2 3 4 x
y = g(x) = 3f (x) = 3 √ x 2. To determine the domain of j, we solve 9x ≥ 0 to ﬁnd x ≥ 0. Our domain is once again [0, ∞). We recognize j(x) = f (9x) and by Theorem 1.6, we obtain the graph of j by dividing the x-coordinates of the points on the graph of f by 9. From the graph, we see the range of j is also [0, ∞). (4, 2) y (1, 1) 2 1 (0, 00, 0x) = √ x horizontal scale by a factor of 1 9 −−−−−−−−−−−−−−−−−−−−−→ multiply each x-coordinate by (x) = f (9x) = √ 9x 3. Solving x+3 2 x + 3 2 + 1, or m(x) = −f 1 2 corresponds to a shift to the left by 3 2 ≥ 0 gives x ≥ −3, so the domain of m is [−3, ∞). To take advantage of what 1 + 1. 2 x + 3 we know of transformations, we rewrite m(x) = − 2 2 x + 3 Focusing on the inputs ﬁrst, we note that the input to f in the formula for m(x) is 1 2. Multiplying the x by 1 2 corresponds to a horizontal stretching by a factor of 2, and adding the 3 2. As before, we resolve which to perform ﬁrst by thinking about how we would ﬁnd the point on m corresponding to a point on f, in this case, (4, 2). To use f (4) = 2, we solve 1 2 = 4. Our ﬁrst step is to subtract the 3 2 x + 3 2 (the horizontal shift) to obtain 1 2. Next, we multiply by 2 (the horizontal stretching) and obtain x = 5. We deﬁne two intermediate functions to handle ﬁrst the shift, then the 2 will shift the In accordance with Theorem 1.3, m1(x stretching. graph of f to the left 3 2 units. 134 Relations and Functions 1, 1) (4, 2) −3 −2 1 (0, 0) −1 −
1 −2 1 2 3 4 5 x −3 −2 − 3 −x) = √ x shift left 3 2 units −−−−−−−−−−−−→ subtract 3 2 from each x-coordinate −2 y = m1(x Next, m2(x) = m1 graph of m1 by a factor of 2 will, according to Theorem 1.6, horizontally stretch the 3 −2 − 3 −2 y = m1(x) = x + 3 2 y 2 (−1, 1) (5, 2) −2 (−3, 0) −1 1 2 3 4 5 x −1 −2 horizontal scale by a factor of 2 −−−−−−−−−−−−−−−−−−→ multiply each x-coordinate by 2 y = m2(x) = m1 We now examine what’s happening to the outputs. From m(x) = −f 1 + 1, we see that the output from f is being multiplied by −1 (a reﬂection about the x-axis) and then a 1 is added (a vertical shift up 1). As before, we can determine the correct order by looking at how the point (4, 2) is moved. We already know that to make use of the equation f (4) = 2, we need to substitute x = 5. We get m(5) = −f 1 + 1 = −f (4) + 1 = −2 + 1 = −1. We see that f (4) (the output from f ) is ﬁrst multiplied by −1 then the 1 is added meaning we ﬁrst reﬂect the graph about the x-axis then shift up 1. Theorem 1.4 tells us m3(x) = −m2(x) will handle the reﬂection. 2 (5 (−1, 1) (5, 2) y 2 1 (−3, 0) −2 (−3, 0) −1 1 2 3 4 5 x −2 −1 1 2 3 4 5 x −1 −2 y = m2(x) = 1 2 x + 3 2 (−1, −1) −2 (5, −2) reﬂect across x-axis −−−−−−−−−−−−→ multiply each y-coordinate by −
1 y = m3(x) = −m2(x.7 Transformations 135 Finally, to handle the vertical shift, Theorem 1.2 gives m(x) = m3(x) + 1, and we see that the range of m is (−∞, 1]. y 2 1 (−3, 0) y (−3, 1) 2 (−1, 0) −2 −1 1 2 3 4 5 x −2 −1 1 2 3 4 5 x (−1, −1) −2 (5, −2) −2 (5, −1) y = m3(x) = −m2(x) = − 1 2 x + 3 2 shift up 1 unit −−−−−−−−−−−−→ add 1 to each y-coordinate y = m(x) = m3(x √ √ √ √ 9x = x = 3 Some comments about Example 1.7.3 are in order. First, recalling the properties of radicals from Intermediate Algebra, we know that the functions g and j are the same, since j and g have the same domains and j(x) = x = g(x). (We invite the reader to verify that all of the points we plotted on the graph of g lie on the graph of j and vice-versa.) Hence, for f (x) = x, a vertical stretch by a factor of 3 and a horizontal shrinking by a factor of 9 result in the same transformation. While this kind of phenomenon is not universal, it happens commonly enough with some of the families of functions studied in College Algebra that it is worthy of note. Secondly, to graph the function m, we applied a series of four transformations. While it would have been easier on the authors to simply inform the reader of which steps to take, we have strived to explain why the order in which the transformations were applied made sense. We generalize the procedure in the theorem below. √ Theorem 1.7. Transformations. Suppose f is a function. If A = 0 and B = 0, then to graph g(x) = Af (Bx + H) + K 1. Subtract H from each of the x-coordinates of the points on the graph of f. This results in a horizontal shift to the left if H > 0 or right if H < 0. 2. Divide the x-coordinates of the points on the
graph obtained in Step 1 by B. This results in a horizontal scaling, but may also include a reﬂection about the y-axis if B < 0. 3. Multiply the y-coordinates of the points on the graph obtained in Step 2 by A. This results in a vertical scaling, but may also include a reﬂection about the x-axis if A < 0. 4. Add K to each of the y-coordinates of the points on the graph obtained in Step 3. This results in a vertical shift up if K > 0 or down if K < 0. Theorem 1.7 can be established by generalizing the techniques developed in this section. Suppose (a, b) is on the graph of f. Then f (a) = b, and to make good use of this fact, we set Bx + H = a If B and solve. We ﬁrst subtract the H (causing the horizontal shift) and then divide by B. 136 Relations and Functions = Af B · a−H is a positive number, this induces only a horizontal scaling by a factor of 1 If B < 0, then B. we have a factor of −1 in play, and dividing by it induces a reﬂection about the y-axis. So we have x = a−H B as the input to g which corresponds to the input x = a to f. We now evaluate g a−H B + H + K = Af (a) + K = Ab + K. We notice that the output from f is B ﬁrst multiplied by A. As with the constant B, if A > 0, this induces only a vertical scaling. If A < 0, then the −1 induces a reﬂection across the x-axis. Finally, we add K to the result, which is our vertical shift. A less precise, but more intuitive way to paraphrase Theorem 1.7 is to think of the quantity Bx + H is the ‘inside’ of the function f. What’s happening inside f aﬀects the inputs or x-coordinates of the points on the graph of f. To ﬁnd the x-coordinates of the corresponding points on g, we undo what has been done to x in the same way we would solve an equation. What’s happening to the output can be thought of as things happening �
�outside’ the function, f. Things happening outside aﬀect the outputs or y-coordinates of the points on the graph of f. Here, we follow the usual order of operations agreement: we ﬁrst multiply by A then add K to ﬁnd the corresponding y-coordinates on the graph of g. Example 1.7.4. Below is the complete graph of y = f (x). Use it to graph g(x) = 4−3f (1−2x) 2. y (0, 3) 3 2 1 (−2, 0) (2, 0) −4 −3 −2 −1 1 2 3 4 x −1 −2 −3 (−4, −3) (4, −3) Solution. We use Theorem 1.7 to track the ﬁve ‘key points’ (−4, −3), (−2, 0), (0, 3), (2, 0) and (4, −3) indicated on the graph of f to their new locations. We ﬁrst rewrite g(x) in the form presented in Theorem 1.7, g(x) = − 3 2 f (−2x + 1) + 2. We set −2x + 1 equal to the x-coordinates of the key points and solve. For example, solving −2x + 1 = −4, we ﬁrst subtract 1 to get −2x = −5 then divide by −2 to get x = 5 2. Subtracting the 1 is a horizontal shift to the left 1 unit. Dividing by −2 can be thought of as a two step process: dividing by 2 which compresses the graph horizontally by a factor of 2 followed by dividing (multiplying) by −1 which causes a reﬂection across the y-axis. We summarize the results in the table on the next page. 1.7 Transformations 137 a −2x + 1 = a x (a, f (a)) x = 5 (−4, −3) −4 −2x + 1 = −4 2 x = 3 (−2, 0) −2 −2x + 1 = −2 2 x = 1 −2x + 1 = 0 2 −2x + 1 = 2 x = − 1 2 −2x + 1 = 4 x = − 3 2 (2, 0) (
4, −3) (0, 3) 2 0 4 Next, we take each of the x values and substitute them into g(x) = − 3 corresponding y-values. Substituting x = 5 2, and using the fact that f (−4) = −3, we get 2 f (−2x + 1) + 2 to get the 5 2 g = − 3 2 f − (−4) + 2 = − 3 2 (−3) + 2 = 9 2 + 2 = 13 2 We see that the output from f is ﬁrst multiplied by − 3 2. Thinking of this as a two step process, multiplying by 3 2 followed by a reﬂection across the x-axis. Adding 2 results in a vertical shift up 2 units. Continuing in this manner, we get the table below. 2 then by −1, we have a vertical stretching by a factor of (x) 13 2 2 − 5 2 2 13 2 2 (x, g(x)) 5 2, 13, 13 2 2 To graph g, we plot each of the points in the table above and connect them in the same order and fashion as the points to which they correspond. Plotting f and g side-by-side gives y (0, 3) 6 5 4 3 2 1 (2, 0) (−2, 0) − 3 2, 13 2 y 5 2, 13 4 −3 −2 −1 −1 1 2 3 4 x −4 −3 −2 −1 −1 1 2 3 4 x (−4, −3) −2 −3 −4 (4, −3) −2 −3 −4 1 2, − 5 2 138 Relations and Functions The reader is strongly encouraged11 to graph the series of functions which shows the gradual transformation of the graph of f into the graph of g. We have outlined the sequence of transformations in the above exposition; all that remains is to plot the ﬁve intermediate stages. Our last example turns the tables and asks for the formula of a function given a desired sequence of transformations. If nothing else, it is a good review of function notation. Example 1.7.5. Let f (x) = x2. Find and simplify the formula of the function g(x) whose graph is the result of f undergoing the following sequence of transformations. Check your answer using a graphing calculator. 1. Vertical shift up 2 units 2. Reﬂection across the x-
axis 3. Horizontal shift right 1 unit 4. Horizontal stretching by a factor of 2 Solution. We build up to a formula for g(x) using intermediate functions as we’ve seen in previous examples. We let g1 take care of our ﬁrst step. Theorem 1.2 tells us g1(x) = f (x)+2 = x2 +2. Next, we reﬂect the graph of g1 about the x-axis using Theorem 1.4: g2(x) = −g1(x) = − x2 + 2 = −x2 − 2. We shift the graph to the right 1 unit, according to Theorem 1.3, by setting g3(x) = g2(x − 1) = −(x − 1)2 − 2 = −x2 + 2x − 3. Finally, we induce a horizontal stretch by a factor of 2 2 x2 4 x2 + x − 3. using Theorem 1.6 to get g(x) = g3 We use the calculator to graph the stages below to conﬁrm our result. 2 x − 3 which yields g(x shift up 2 units −−−−−−−−−−−−→ add 2 to each y-coordinate y = f (x) = x2 y = g1(x) = f (x) + 2 = x2 + 2 reﬂect across x-axis −−−−−−−−−−−−→ multiply each y-coordinate by −1 y = g1(x) = x2 + 2 y = g2(x) = −g1(x) = −x2 − 2 11You really should do this once in your life. 1.7 Transformations 139 y = g2(x) = −x2 − 2 y = g3(x) = g2(x − 1) = −x2 + 2x − 3 shift right 1 unit −−−−−−−−−−−−→ add 1 to each x-coordinate y = g3(x) = −x2 + 2x − 3 y = g(x) = g3 2 x = − 1 1 4 x2 + x − 3 horizontal stretch by a factor of 2 −−−−−−−−−−−−−−−−−−→
multiply each x-coordinate by 2 We have kept the viewing window the same in all of the graphs above. This had the undesirable consequence of making the last graph look ‘incomplete’ in that we cannot see the original shape of f (x) = x2. Altering the viewing window results in a more complete graph of the transformed function as seen below. This example brings our ﬁrst chapter to a close. In the chapters which lie ahead, be on the lookout for the concepts developed here to resurface as we study diﬀerent families of functions. y = g(x) 140 1.7.1 Exercises Relations and Functions Suppose (2, −3) is on the graph of y = f (x). In Exercises 1 - 18, use Theorem 1.7 to ﬁnd a point on the graph of the given transformed function. 1. y = f (x) + 3 2. y = f (x + 3) 3. y = f (x) − 1 4. y = f (x − 1) 5. y = 3f (x) 6. y = f (3x) 7. y = −f (x) 8. y = f (−x) 9. y = f (x − 3) + 1 10. y = 2f (x + 1) 11. y = 10 − f (x) 12. y = 3f (2x) − 1 13. y = 1 2 f (4 − x) 16. y = f 7 − 2x 4 14. y = 5f (2x + 1) + 3 15. y = 2f (1 − x) − 1 17. y = f (3x) − 1 2 18. y = 4 − f (3x − 1) 7 The complete graph of y = f (x) is given below. In Exercises 19 - 27, use it and Theorem 1.7 to graph the given transformed function. y 4 3 2 1 (2, 2) (−2, 2) −4 −3 −2 −1 (0, 0) 2 3 4 x The graph for Ex. 19 - 27 19. y = f (x) + 1 20. y = f (x) − 2 21. y = f (x + 1) 22. y = f (x − 2) 23. y = 2f (x) 24. y
= f (2x) 25. y = 2 − f (x) 26. y = f (2 − x) 27. y = 2 − f (2 − x) 28. Some of the answers to Exercises 19 - 27 above should be the same. Which ones match up? What properties of the graph of y = f (x) contribute to the duplication? 1.7 Transformations 141 The complete graph of y = f (x) is given below. In Exercises 29 - 37, use it and Theorem 1.7 to graph the given transformed function. y (0, 42, 0) (4, −2) −4 −3 −1 (−2, 0) −1 −2 −3 −4 The graph for Ex. 29 - 37 29. y = f (x) − 1 30. y = f (x + 1) 32. y = f (2x) 33. y = −f (x) 31. y = 1 2 f (x) 34. y = f (−x) 35. y = f (x + 1) − 1 36. y = 1 − f (x) 37. y = 1 2 f (x + 1) − 1 The complete graph of y = f (x) is given below. In Exercises 38 - 49, use it and Theorem 1.7 to graph the given transformed function. y (0, 3) 3 2 1 −3 −2 −1 1 2 (−3, 0) −1 x 3 (3, 0) The graph for Ex. 38 - 49 38. g(x) = f (x) + 3 39. h(x) = f (x) − 1 2 41. a(x) = f (x + 4) 42. b(x) = f (x + 1) − 1 44. d(x) = −2f (x) 45. k(x) = f 2 3 x 47. n(x) = 4f (x − 3) − 6 48. p(x) = 4 + f (1 − 2x) 40. j(x) = f x − 2 3 43. c(x) = 3 5 f (x) 46. m(x) = − 1 49. q(x) = − 1 4 f (3x) 2 f x+4 2 − 3 142 Relations and Functions The complete graph of y = S(
x) is given below. y (1, 3) (0, 0) 1 (2, 0) x 3 2 1 −1 −2 (−2, 0) −2 −1 −3 (−1, −3) The graph of y = S(x) The purpose of Exercises 50 - 53 is to graph y = 1 one step at a time. 2 S(−x + 1) + 1 by graphing each transformation, 50. y = S1(x) = S(x + 1) 51. y = S2(x) = S1(−x) = S(−x + 1) 52. y = S3(x) = 1 2 S2(x) = 1 2 S(−x + 1) 53. y = S4(x) = S3(x) + 1 = 1 2 S(−x + 1) + 1 √ Let f (x) = sequence of transformations. x. Find a formula for a function g whose graph is obtained from f from the given 54. (1) shift right 2 units; (2) shift down 3 units 55. (1) shift down 3 units; (2) shift right 2 units 56. (1) reﬂect across the x-axis; (2) shift up 1 unit 57. (1) shift up 1 unit; (2) reﬂect across the x-axis 58. (1) shift left 1 unit; (2) reﬂect across the y-axis; (3) shift up 2 units 59. (1) reﬂect across the y-axis; (2) shift left 1 unit; (3) shift up 2 units 60. (1) shift left 3 units; (2) vertical stretch by a factor of 2; (3) shift down 4 units 61. (1) shift left 3 units; (2) shift down 4 units; (3) vertical stretch by a factor of 2 62. (1) shift right 3 units; (2) horizontal shrink by a factor of 2; (3) shift up 1 unit 63. (1) horizontal shrink by a factor of 2; (2) shift right 3 units; (3) shift up 1 unit 1.7 Transformations 143 √ 64. The graph of y = f (x) = 3 x is given below on the left and the graph of y = g(x
) is given on the right. Find a formula for g based on transformations of the graph of f. Check your answer by conﬁrming that the points shown on the graph of g satisfy the equation y = g(x). 11−10−9−8−7−6−5−4−3−2−1 −11−10−9−8−7−6−5−4−3−2−1 −2 −3 −4 −5 y = 3√ x −2 −3 −4 −5 y = g(x) √ 65. For many common functions, the properties of Algebra make a horizontal scaling the same as a vertical scaling by (possibly) a diﬀerent factor. For example, we stated earlier that √ x. With the help of your classmates, ﬁnd the equivalent vertical scaling produced. What about √ by the horizontal scalings y = (2x)3, y = \|5x\|, y = 3 √ 2 x2 y = (−2x)3, y = \| − 5x\|, y = 3? −27x and y = − 1 27x and y = 1 2 x2 9x = 3 66. We mentioned earlier in the section that, in general, the order in which transformations are applied matters, yet in our ﬁrst example with two transformations the order did not matter. (You could perform the shift to the left followed by the shift down or you could shift down and then left to achieve the same result.) With the help of your classmates, determine the situations in which order does matter and those in which it does not. 67. What happens if you reﬂect an even function across the y-axis? 68. What happens if you reﬂect an odd function across the y-axis? 69. What happens if you reﬂect an even function across the x-axis? 70. What happens if you reﬂect an odd function across the x-axis? 71. How would you describe symmetry about the origin in terms of reﬂections? 72. As we saw in Example 1.7.5, the viewing window on the graphing calculator aﬀects how we see the transformations done to a graph. Using two diﬀerent calculators, ﬁnd viewing windows so that f (x) = x2 on
the one calculator looks like g(x) = 3x2 on the other. 144 1.7.2 Answers 1. (2, 0) 4. (3, −3) 7. (2, 3) 10. (1, −6) 13. 2, − 3 2 16. − 1 2, −3 Relations and Functions 2. (−1, −3) 5. (2, −9) 8. (−2, −3) 3. (2, −4) 6. 2 3, −3 9. (5, −2) 11. (2, 13) 12. y = (1, −10) 14. 1 17. 2 2, −12 3, −2 19. y = f (x) + 1 y (−2, 3) 4 3 2 1 (2, 3) (0, 1) −4 −3 −2 −1 1 2 3 4 21. y = f (x + 1) y 4 3 2 1 (1, 2) (−3, 2) −5 −4 −3 (−1, 0) 1 2 3 23. y = 2f (x) (−2, 4) (2, 4) y 4 3 2 1 −4 −3 −2 −1 (0, 0) 2 3 4 x x x 15. (−1, −7) 18. (1, 1) 20. y = f (x) − 2 y 2 1 −4 (−2, 0) −1 −1 1 (2, 0) x 4 −2 (0, −2) 22. y = f (x − 2) y 4 2 1 (0, 2) (4, 2) −2 −1 (2, 0) 3 4 5 6 24. y = f (2x) y 4 3 2 1 (1, 2) (−1, 2) −4 −3 −2 −1 (0, 0) 2 3 4 x x 1.7 Transformations 25. y = 2 − f (x) y (0, 2) 2 1 −4 −3 (−2, 0) (2, 0) 3 4 x 145 26. y = f (2 − x) y 4 2 1 (0, 2) (4, 2) −2 −1 (2, 0) 3 4 5 6 x 27. y = 2 − f (2 − x) 29. y = f (x) − 1 y 4
3 2 1 (2, 2) y (0, 3) 4 3 2 1 −2 −1 (0, 0) 2 (4, 0) 5 6 x −4 −3 −2 −1 (−2, −1) 1 2 3 4 x (2, −1) (4, −3) −1 −2 −3 −4 30. y = f (x + 1) (−1, 4) y 4 3 2 1 −4 −3 −2 (−3, 0) −1 −1 1 (1, 0) 2 3 4 x −2 −3 −4 (3, −2) 31. y = 1 2 f (x) −4 −3 y (0, 2) 4 3 2 1 −1 −2 −3 −4 −1 (−2, 0) 1 3 4 x (2, 0) (4, −1) 146 Relations and Functions 32. y = f (2x) 33. y = −f (x) y (0, 4) (1, 0) 2 3 4 x (2, −2) y (0, 4) x 4 1 3 (2, 0) 4 3 2 1 −4 −3 −2 (−1, 0) 34. y = f (−x) −2 −3 −4 4 3 2 1 −4 −3 (−4, −2) −1 (−2, 0) −1 −2 −3 −4 36. y = 1 − f (x) y 4 3 2 1 (−2, 1) (4, 3) (2, 1) −4 −3 −2 −1 −1 1 2 3 4 x −2 −3 −4 (0, −3) y 4 3 2 1 (−2, 0) −4 −3 −2 −1 −1 (4, 2) (2, 0) 1 2 3 4 x −2 −3 −4 (0, −4) 35. y = f (x + 1) − 1 y (−1, 3) 4 3 2 1 −4 −3 −2 −1 −1 1 3 2 (1, −1) 4 x (−3, −1) −2 −3 −4 (3, −3) 37. y = 1 2 f (x + 1) − 1 y 4 3 2 1 (−1, 1) −4 −3 −2 (−3, −1) −1 −1 1 3 2
(1, −1) 4 x −2 −3 −4 (3, −2) 1.7 Transformations 147 38. g(x) = f (x) + 3 y 39. h(x) = f (x) − 1 2 y (0, 6) 6 5 4 3 2 1 (3, 3) (−3, 3) −3 −2 −1 1 2 3 x −1 0, 5 2 3 2 1 1 −1 2 3 3, − 1 2 x −3 −2 −1 −3, − 1 2 40. j(x 41. a(x) = f (x + 4) (−4, 3) y 3 2 1 3 2 1 −3 −2 −1 3, 0 − 7 −1 1 2 3 11 3, 0 x −7 −6 −5 −4 −3 −2 −1 (−7, 0) (−1, 0) x 42. b(x) = f (x + 1) − 1 y (−1, 2) 2 1 43. c(x) = 3 5 f (x) 2 1 y 0, 9 5 −4 −3 −2 −1 1 2 x −1 (−4, −1) (2, −1) −3 −2 −1 1 2 (−3, 0) −1 x 3 (3, 0) 44. d(x) = −2f (x) y (−3, 0) (3, 0) −3 −2 −1 1 2 3 x −1 −2 −3 −4 −5 −6 (0, −6) 45. k(x) = f 2 3 x y (0, 3) 3 2 1 −4 −3 −2 −1 2, 0 − 9 148 Relations and Functions 46. m(x) = − 1 4 f (3x) y 47. n(x) = 4f (x − 3) − 6 y (3, 6) (−1, 0) −1 (1, 0) 1 x −1 01 −2 −3 −4 −5 −6 1 2 3 4 5 6 x (0, −6) (6, −6) 48. p(x) = 4 + f (1 − 2x) = f (−2x + 1) + 4 y 1 2, 7 (2, 4) 7 6 5 4 (−1, 4) 3 2 1 −1 −1 1 2
x 49. q(x) = − 1 2 f x+10−9−8−7−6−5−4−3−2−1 −1 (−10, −3) −2 −3 −4 −4, − 9 2 1 2 x (2, −3) 1.7 Transformations 149 50. y = S1(x) = S(x + 1) 51. y = S2(x) = S1(−x) = S(−x + 1) y (0, 3) 3 2 1 (−3, 0) (−1, 0) −3 −2 −1 x (1, 0) −1 −2 −3 y (0, 3) 3 2 1 (1, 0) (3, 0) 1 2 3 x (−1, 0) −1 −2 −3 (−2, −3) (2, −3) 52. y = S3(x) = 1 2 S2(x) = 1 2 S(−x + 1) y 2 1 0, 3 2 (1, 0) (3, 0) 1 2 3 x 2, − 3 2 (−1, 0) −1 −2 54. g(x 56. g(x) = − √ 58. g(x) = 60. g(x) = 2 √ 53. y = S4(x) = S3(x) + 1 = 1 2 S(−x + 1) + 1 y 3 2 0, 5 2 (1, 1) (3, 1) 1 (−1, 1) −1 1 −1 2, − 1 2 x 3 55. g(x) = √ x − 2 − 3 57. g(x) = −( √ x + 1) = − √ −(x + 1) + 2 = 59. g(x) = 61. g(x) = 2 √ x − 1 √ −(x − 3) + 1 = 2x − 6 + 1 62. g(x) = 2x − 3 + 1 √ 64. g(x) = −2 3 √ x + 3 − 1 or g(x) = 2 3 63. g(x) = −x − 3 − 1 150 Relations and Functions Chapter 2 Linear and Quadratic Functions 2.1 Linear Functions We now begin the study of families of functions. Our �
�rst family, linear functions, are old friends as we shall soon see. Recall from Geometry that two distinct points in the plane determine a unique line containing those points, as indicated below. P (x0, y0) Q (x1, y1) To give a sense of the ‘steepness’ of the line, we recall that we can compute the slope of the line using the formula below. Equation 2.1. The slope m of the line containing the points P (x0, y0) and Q (x1, y1) is: provided x1 = x0. m = y1 − y0 x1 − x0, A couple of notes about Equation 2.1 are in order. First, don’t ask why we use the letter ‘m’ to represent slope. There are many explanations out there, but apparently no one really knows for sure.1 Secondly, the stipulation x1 = x0 ensures that we aren’t trying to divide by zero. The reader is invited to pause to think about what is happening geometrically; the anxious reader can skip along to the next example. Example 2.1.1. Find the slope of the line containing the following pairs of points, if it exists. Plot each pair of points and the line containing them. 1See www.mathforum.org or www.mathworld.wolfram.com for discussions on this topic. 152 Linear and Quadratic Functions 1. P (0, 0), Q(2, 4) 2. P (−1, 2), Q(3, 4) 3. P (−2, 3), Q(2, −3) 4. P (−3, 2), Q(4, 2) 5. P (2, 3), Q(2, −1) 6. P (2, 3), Q(2.1, −1) Solution. In each of these examples, we apply the slope formula, Equation 2.1. 1. m = 4 − 2 3 − (−1) = 2 4 = 1 2 3. m = −3 − 3 2 − (−2) = −6 4 = − 3 2 4. m = 2 − 2 4 − (−33 −2 −1 1 2 3 x −1 −2 −3 −4 Q y 3 2 1 P Q −4 −3 −2 −1 1 2 3 4 x 2.1 Linear
Functions 153 5. m = −1 − 3 2 − 2 = −4 0, which is undeﬁned 6. m = −1 − 3 2.1 − 2 = −4 0.1 = −40 1 −2 −3 3 2 1 −1 −2 −3 A few comments about Example 2.1.1 are in order. First, for reasons which will be made clear soon, if the slope is positive then the resulting line is said to be increasing. If it is negative, we say the line is decreasing. A slope of 0 results in a horizontal line which we say is constant, and an undeﬁned slope results in a vertical line.2 Second, the larger the slope is in absolute value, the steeper the line. You may recall from Intermediate Algebra that slope can be described as the ratio ‘ rise run ’. For example, in the second part of Example 2.1.1, we found the slope to be 1 2. We can interpret this as a rise of 1 unit upward for every 2 units to the right we travel along the line, as shown below. y 4 3 2 1 ‘up 1’ ‘over 2’ −1 1 2 3 x 2Some authors use the unfortunate moniker ‘no slope’ when a slope is undeﬁned. It’s easy to confuse the notions of ‘no slope’ with ‘slope of 0’. For this reason, we will describe slopes of vertical lines as ‘undeﬁned’. 154 Linear and Quadratic Functions Using more formal notation, given points (x0, y0) and (x1, y1), we use the Greek letter delta ‘∆’ to write ∆y = y1 − y0 and ∆x = x1 − x0. In most scientiﬁc circles, the symbol ∆ means ‘change in’. Hence, we may write m = ∆y ∆x, which describes the slope as the rate of change of y with respect to x. Rates of change abound in the ‘real world’, as the next example illustrates. Example 2.1.2. Suppose that two separate temperature readings were taken at the ranger station on the top of Mt. Sasquatch: at 6 AM the temperature was 24◦F and at 10 AM it was 32◦
F. 1. Find the slope of the line containing the points (6, 24) and (10, 32). 2. Interpret your answer to the ﬁrst part in terms of temperature and time. 3. Predict the temperature at noon. Solution. 1. For the slope, we have m = 32−24 10−6 = 8 4 = 2. 2. Since the values in the numerator correspond to the temperatures in ◦F, and the values in 2◦ F the denominator correspond to time in hours, we can interpret the slope as 2 =, 1 hour or 2◦F per hour. Since the slope is positive, we know this corresponds to an increasing line. Hence, the temperature is increasing at a rate of 2◦F per hour. 2 1 = 3. Noon is two hours after 10 AM. Assuming a temperature increase of 2◦F per hour, in two hours the temperature should rise 4◦F. Since the temperature at 10 AM is 32◦F, we would expect the temperature at noon to be 32 + 4 = 36◦F. Now it may well happen that in the previous scenario, at noon the temperature is only 33◦F. This doesn’t mean our calculations are incorrect, rather, it means that the temperature change throughout the day isn’t a constant 2◦F per hour. As discussed in Section 1.4.1, mathematical models are just that: models. The predictions we get out of the models may be mathematically accurate, but may not resemble what happens in the real world. In Section 1.2, we discussed the equations of vertical and horizontal lines. Using the concept of slope, we can develop equations for the other varieties of lines. Suppose a line has a slope of m and contains the point (x0, y0). Suppose (x, y) is another point on the line, as indicated below. (x, y) (x0, y0) 2.1 Linear Functions Equation 2.1 yields 155 y − y0 x − x0 m (x − x0) = y − y0 m = y − y0 = m (x − x0) We have just derived the point-slope form of a line.3 Equation 2.2. The point-slope form of the line with slope m containing the point (x0, y0) is the equation y − y0 = m (x −
x0). Example 2.1.3. Write the equation of the line containing the points (−1, 3) and (2, 1). Solution. In order to use Equation 2.2 we need to ﬁnd the slope of the line in question so we use Equation 2.1 to get m = ∆y 3. We are spoiled for choice for a point (x0, y0). We’ll use (−1, 3) and leave it to the reader to check that using (2, 1) results in the same equation. Substituting into the point-slope form of the line, we get 2−(−1) = − 2 ∆x = 1−3 y − y0 = m (x − x0x − (−1)) (x + 1) x − x + 2 3 7 3. We can check our answer by showing that both (−1, 3) and (2, 1) are on the graph of y = − 2 algebraically, as we did in Section 1.2.1. 3 x + 7 3 In simplifying the equation of the line in the previous example, we produced another form of a line, the slope-intercept form. This is the familiar y = mx + b form you have probably seen in Intermediate Algebra. The ‘intercept’ in ‘slope-intercept’ comes from the fact that if we set x = 0, we get y = b. In other words, the y-intercept of the line y = mx + b is (0, b). Equation 2.3. The slope-intercept form of the line with slope m and y-intercept (0, b) is the equation y = mx + b. Note that if we have slope m = 0, we get the equation y = b which matches our formula for a horizontal line given in Section 1.2. The formula given in Equation 2.3 can be used to describe all lines except vertical lines. All lines except vertical lines are functions (Why is this?) so we have ﬁnally reached a good point to introduce linear functions. 3We can also understand this equation in terms of applying transformations to the function I(x) = x. See the Exercises. 156 Linear and Quadratic Functions Deﬁnition 2.1. A linear function is a function of the form f (x
) = mx + b, where m and b are real numbers with m = 0. The domain of a linear function is (−∞, ∞). For the case m = 0, we get f (x) = b. These are given their own classiﬁcation. Deﬁnition 2.2. A constant function is a function of the form f (x) = b, where b is real number. The domain of a constant function is (−∞, ∞). Recall that to graph a function, f, we graph the equation y = f (x). Hence, the graph of a linear function is a line with slope m and y-intercept (0, b); the graph of a constant function is a horizontal line (a line with slope m = 0) and a y-intercept of (0, b). Now think back to Section 1.6.1, speciﬁcally Deﬁnition 1.10 concerning increasing, decreasing and constant functions. A line with positive slope was called an increasing line because a linear function with m > 0 is an increasing function. Similarly, a line with a negative slope was called a decreasing line because a linear function with m < 0 is a decreasing function. And horizontal lines were called constant because, well, we hope you’ve already made the connection. Example 2.1.4. Graph the following functions. Identify the slope and y-intercept. 1. f (x) = 3 2. f (x) = 3x − 1 Solution. 3. f (x) = 4. f (x) = 3 − 2x 4 x2 − 4 x − 2 1. To graph f (x) = 3, we graph y = 3. This is a horizontal line (m = 0) through (0, 3). 2. The graph of f (x) = 3x − 1 is the graph of the line y = 3x − 1. Comparison of this equation with Equation 2.3 yields m = 3 and b = −1. Hence, our slope is 3 and our y-intercept is (0, −1). To get another point on the line, we can plot (1, f (1)) = (1, 2). 2.1 Linear Functions 157 y 4 3 2 1 −3 −2 −1 1 2 3 x f (x) = 3 y 4 3 2 1 −
2−1 −1 1 2 x f (x) = 3x − 1 3. At ﬁrst glance, the function f (x) = 3−2x 4 − 2x rearranging we get f (x) = 3−2x our graph is a line with a slope of − 1 point, we can choose (1, f (1)) to get 1, 1 4 4 = 3. 4 does not ﬁt the form in Deﬁnition 2.1 but after some 2 and b = 3 4 = − 1 4. Hence,. Plotting an additional 2 and a y-intercept of 0, 3 4. We identify. If we simplify the expression for f, we get f (x) = x2 − 4 x − 2 = (x − 2)(x + 2) (x − 2) = x + 2. If we were to state f (x) = x + 2, we would be committing a sin of omission. Remember, to ﬁnd the domain of a function, we do so before we simplify! In this case, f has big problems when x = 2, and as such, the domain of f is (−∞, 2) ∪ (2, ∞). To indicate this, we write f (x) = x + 2, x = 2. So, except at x = 2, we graph the line y = x + 2. The slope m = 1 and the y-intercept is (0, 2). A second point on the graph is (1, f (1)) = (1, 3). Since our function f is not deﬁned at x = 2, we put an open circle at the point that would be on the line y = x + 2 when x = 2, namely (2, 4). y 2 1 −3 −2 −1 1 2 3 x f (x) = 3 − 2x 4 y 4 3 2 1 −1 1 2 3 x f (x) = x2 − 4 x − 2 158 Linear and Quadratic Functions The last two functions in the previous example showcase some of the diﬃculty in deﬁning a linear function using the phrase ‘of the form’ as in Deﬁnition 2.1, since some algebraic manipulations may be needed to rewrite a given function to match ‘the form’. Keep
in mind that the domains of linear and constant functions are all real numbers (−∞, ∞), so while f (x) = x2−4 x−2 simpliﬁed to a formula f (x) = x + 2, f is not considered a linear function since its domain excludes x = 2. However, we would consider f (x) = 2x2 + 2 x2 + 1 to be a constant function since its domain is all real numbers (Can you tell us why?) and f (x) = 2x2 + 2 x2 + 1 = 2 x2 + 1 x2 + 1 = 2 The following example uses linear functions to model some basic economic relationships. Example 2.1.5. The cost C, in dollars, to produce x PortaBoy4 game systems for a local retailer is given by C(x) = 80x + 150 for x ≥ 0. 1. Find and interpret C(10). 2. How many PortaBoys can be produced for $15,000? 3. Explain the signiﬁcance of the restriction on the domain, x ≥ 0. 4. Find and interpret C(0). 5. Find and interpret the slope of the graph of y = C(x). Solution. 1. To ﬁnd C(10), we replace every occurrence of x with 10 in the formula for C(x) to get C(10) = 80(10) + 150 = 950. Since x represents the number of PortaBoys produced, and C(x) represents the cost, in dollars, C(10) = 950 means it costs $950 to produce 10 PortaBoys for the local retailer. 2. To ﬁnd how many PortaBoys can be produced for $15,000, we solve C(x) = 15000, or 80x + 80 = 185.625. Since we can only produce a whole 150 = 15000. Solving, we get x = 14850 number amount of PortaBoys, we can produce 185 PortaBoys for $15,000. 3. The restriction x ≥ 0 is the applied domain, as discussed in Section 1.4.1. In this context, x represents the number of PortaBoys produced. It makes no sense to produce a negative quantity of game systems.5 4The similarity of this name to PortaJohn is deliberate. 5Actually, it makes no
sense to produce a fractional part of a game system, either, as we saw in the previous part of this example. This absurdity, however, seems quite forgivable in some textbooks but not to us. 2.1 Linear Functions 159 4. We ﬁnd C(0) = 80(0) + 150 = 150. This means it costs $150 to produce 0 PortaBoys. As mentioned on page 82, this is the ﬁxed, or start-up cost of this venture. 5. If we were to graph y = C(x), we would be graphing the portion of the line y = 80x + 150 for x ≥ 0. We recognize the slope, m = 80. Like any slope, we can interpret this as a rate of change. Here, C(x) is the cost in dollars, while x measures the number of PortaBoys so m = ∆y ∆x = ∆C ∆x = 80 = 80 1 = $80 1 PortaBoy. In other words, the cost is increasing at a rate of $80 per PortaBoy produced. This is often called the variable cost for this venture. The next example asks us to ﬁnd a linear function to model a related economic problem. Example 2.1.6. The local retailer in Example 2.1.5 has determined that the number x of PortaBoy game systems sold in a week is related to the price p in dollars of each system. When the price was $220, 20 game systems were sold in a week. When the systems went on sale the following week, 40 systems were sold at $190 a piece. 1. Find a linear function which ﬁts this data. Use the weekly sales x as the independent variable and the price p as the dependent variable. 2. Find a suitable applied domain. 3. Interpret the slope. 4. If the retailer wants to sell 150 PortaBoys next week, what should the price be? 5. What would the weekly sales be if the price were set at $150 per system? Solution. 1. We recall from Section 1.4 the meaning of ‘independent’ and ‘dependent’ variable. Since x is to be the independent variable, and p the dependent variable, we treat x as the input variable and p as the output variable. Hence, we are looking for a function of the form p(x) = mx + b.
To determine m and b, we use the fact that 20 PortaBoys were sold during the week when the price was 220 dollars and 40 units were sold when the price was 190 dollars. Using function notation, these two facts can be translated as p(20) = 220 and p(40) = 190. Since m represents the rate of change of p with respect to x, we have m = ∆p ∆x = 190 − 220 40 − 20 = −30 20 = −1.5. We now have determined p(x) = −1.5x + b. To determine b, we can use our given data again. Using p(20) = 220, we substitute x = 20 into p(x) = 1.5x + b and set the result equal to 220: −1.5(20) + b = 220. Solving, we get b = 250. Hence, we get p(x) = −1.5x + 250. We can check our formula by computing p(20) and p(40) to see if we get 220 and 190, respectively. You may recall from page 82 that the function p(x) is called the price-demand (or simply demand) function for this venture. 160 Linear and Quadratic Functions 2. To determine the applied domain, we look at the physical constraints of the problem. Certainly, we can’t sell a negative number of PortaBoys, so x ≥ 0. However, we also note that the slope of this linear function is negative, and as such, the price is decreasing as more units are sold. Thus another constraint on the price is p(x) ≥ 0. Solving −1.5x + 250 ≥ 0 results in −1.5x ≥ −250 or x ≤ week, we round down to 166. As a result, a reasonable applied domain for p is [0, 166]. = 166.6. Since x represents the number of PortaBoys sold in a 500 3 3. The slope m = −1.5, once again, represents the rate of change of the price of a system with respect to weekly sales of PortaBoys. Since the slope is negative, we have that the price is decreasing at a rate of $1.50 per PortaBoy sold. (Said diﬀerently, you can sell one more PortaBoy for every $1.50 drop in price.) 4.
To determine the price which will move 150 PortaBoys, we ﬁnd p(150) = −1.5(150)+250 = 25. That is, the price would have to be $25. 5. If the price of a PortaBoy were set at $150, we have p(x) = 150, or, −1.5x+250 = 150. Solving, we get −1.5x = −100 or x = 66.6. This means you would be able to sell 66 PortaBoys a week if the price were $150 per system. Not all real-world phenomena can be modeled using linear functions. Nevertheless, it is possible to use the concept of slope to help analyze non-linear functions using the following. Deﬁnition 2.3. Let f be a function deﬁned on the interval [a, b]. The average rate of change of f over [a, b] is deﬁned as: ∆f ∆x = f (b) − f (a) b − a Geometrically, if we have the graph of y = f (x), the average rate of change over [a, b] is the slope of the line which connects (a, f (a)) and (b, f (b)). This is called the secant line through these points. For that reason, some textbooks use the notation msec for the average rate of change of a function. Note that for a linear function m = msec, or in other words, its rate of change over an interval is the same as its average rate of change. y = f (x) (b, f (b)) (a, f (a)) The graph of y = f (x) and its secant line through (a, f (a)) and (b, f (b)) The interested reader may question the adjective ‘average’ in the phrase ‘average rate of change’. In the ﬁgure above, we can see that the function changes wildly on [a, b], yet the slope of the secant line only captures a snapshot of the action at a and b. This situation is entirely analogous to the 2.1 Linear Functions 161 average speed on a trip. Suppose it takes you 2 hours to travel 100 miles. Your average speed is 100 miles 2 hours = 50 miles per hour. However, it is entirely possible
that at the start of your journey, you traveled 25 miles per hour, then sped up to 65 miles per hour, and so forth. The average rate of change is akin to your average speed on the trip. Your speedometer measures your speed at any one instant along the trip, your instantaneous rate of change, and this is one of the central themes of Calculus.6 When interpreting rates of change, we interpret them the same way we did slopes. In the context of functions, it may be helpful to think of the average rate of change as: change in outputs change in inputs Example 2.1.7. Recall from page 82, the revenue from selling x units at a price p per unit is given by the formula R = xp. Suppose we are in the scenario of Examples 2.1.5 and 2.1.6. 1. Find and simplify an expression for the weekly revenue R(x) as a function of weekly sales x. 2. Find and interpret the average rate of change of R(x) over the interval [0, 50]. 3. Find and interpret the average rate of change of R(x) as x changes from 50 to 100 and compare that to your result in part 2. 4. Find and interpret the average rate of change of weekly revenue as weekly sales increase from 100 PortaBoys to 150 PortaBoys. Solution. 1. Since R = xp, we substitute p(x) = −1.5x + 250 from Example 2.1.6 to get R(x) = x(−1.5x + 250) = −1.5x2 + 250x. Since we determined the price-demand function p(x) is restricted to 0 ≤ x ≤ 166, R(x) is restricted to these values of x as well. 2. Using Deﬁnition 2.3, we get that the average rate of change is ∆R ∆x = R(50) − R(0) 50 − 0 = 8750 − 0 50 − 0 = 175. Interpreting this slope as we have in similar situations, we conclude that for every additional PortaBoy sold during a given week, the weekly revenue increases $175. 3. The wording of this part is slightly diﬀerent than that in Deﬁnition 2.3, but its meaning is to ﬁnd the average rate of change of R over the interval [50, 100]. To �
��nd this rate of change, we compute ∆R ∆x = R(100) − R(50) 100 − 50 = 10000 − 8750 50 = 25. 6Here we go again... 162 Linear and Quadratic Functions In other words, for each additional PortaBoy sold, the revenue increases by $25. Note that while the revenue is still increasing by selling more game systems, we aren’t getting as much of an increase as we did in part 2 of this example. (Can you think of why this would happen?) 4. Translating the English to the mathematics, we are being asked to ﬁnd the average rate of change of R over the interval [100, 150]. We ﬁnd ∆R ∆x = R(150) − R(100) 150 − 100 = 3750 − 10000 50 = −125. This means that we are losing $125 dollars of weekly revenue for each additional PortaBoy sold. (Can you think why this is possible?) We close this section with a new look at diﬀerence quotients which were ﬁrst introduced in Section 1.4. If we wish to compute the average rate of change of a function f over the interval [x, x + h], then we would have ∆f ∆x = f (x + h) − f (x) (x + h) − x = f (x + h) − f (x) h As we have indicated, the rate of change of a function (average or otherwise) is of great importance in Calculus.7 Also, we have the geometric interpretation of diﬀerence quotients which was promised to you back on page 81 – a diﬀerence quotient yields the slope of a secant line. 7So we are not torturing you with these for nothing. 2.1 Linear Functions 2.1.1 Exercises 163 In Exercises 1 - 10, ﬁnd both the point-slope form and the slope-intercept form of the line with the given slope which passes through the given point. 1. m = 3, P (3, −1) 3. m = −1, P (−7, −1) 5. m = − 1 5, P (10, 4) 7. m = 0, P (3, 117) 9. m = −5, P ( √ √ 3
) 3, 2 2. m = −2, P (−5, 8) 4. m = 2 3, P (−2, 1) 6. m = 1 7, P (−1, 4) √ 8. m = − 2, P (0, −3) 10. m = 678, P (−1, −12) In Exercises 11 - 20, ﬁnd the slope-intercept form of the line which passes through the given points. 11. P (0, 0), Q(−3, 5) 13. P (5, 0), Q(0, −8) 15. P (−1, 5), Q(7, 5) 12. P (−1, −2), Q(3, −2) 14. P (3, −5), Q(7, 4) 16. P (4, −8), Q(5, −8) 17. P 1 2, 3 19, − √ √ 2 2, 18. P 2 20, −1, Q √ 2 3, 1 In Exercises 21 - 26, graph the function. Find the slope, y-intercept and x-intercept, if any exist. 21. f (x) = 2x − 1 23. f (x) = 3 25. f (x) = 2 3 x + 1 3 22. f (x) = 3 − x 24. f (x) = 0 26. f (x) = 1 − x 2 27. Find all of the points on the line y = 2x + 1 which are 4 units from the point (−1, 3). 28. Jeﬀ can walk comfortably at 3 miles per hour. Find a linear function d that represents the total distance Jeﬀ can walk in t hours, assuming he doesn’t take any breaks. 29. Carl can stuﬀ 6 envelopes per minute. Find a linear function E that represents the total number of envelopes Carl can stuﬀ after t hours, assuming he doesn’t take any breaks. 30. A landscaping company charges $45 per cubic yard of mulch plus a delivery charge of $20. Find a linear function which computes the total cost C (in dollars) to deliver x cubic yards of mulch. 164 Linear and Quadratic Functions 31. A plumber charges $50 for a service call plus $80 per hour. If
she spends no longer than 8 hours a day at any one site, ﬁnd a linear function that represents her total daily charges C (in dollars) as a function of time t (in hours) spent at any one given location. 32. A salesperson is paid $200 per week plus 5% commission on her weekly sales of x dollars. Find a linear function that represents her total weekly pay, W (in dollars) in terms of x. What must her weekly sales be in order for her to earn $475.00 for the week? 33. An on-demand publisher charges $22.50 to print a 600 page book and $15.50 to print a 400 page book. Find a linear function which models the cost of a book C as a function of the number of pages p. Interpret the slope of the linear function and ﬁnd and interpret C(0). 34. The Topology Taxi Company charges $2.50 for the ﬁrst ﬁfth of a mile and $0.45 for each additional ﬁfth of a mile. Find a linear function which models the taxi fare F as a function of the number of miles driven, m. Interpret the slope of the linear function and ﬁnd and interpret F (0). 35. Water freezes at 0◦ Celsius and 32◦ Fahrenheit and it boils at 100◦C and 212◦F. (a) Find a linear function F that expresses temperature in the Fahrenheit scale in terms of degrees Celsius. Use this function to convert 20◦C into Fahrenheit. (b) Find a linear function C that expresses temperature in the Celsius scale in terms of degrees Fahrenheit. Use this function to convert 110◦F into Celsius. (c) Is there a temperature n such that F (n) = C(n)? 36. Legend has it that a bull Sasquatch in rut will howl approximately 9 times per hour when it is 40◦F outside and only 5 times per hour if it’s 70◦F. Assuming that the number of howls per hour, N, can be represented by a linear function of temperature Fahrenheit, ﬁnd the number of howls per hour he’ll make when it’s only 20◦F outside. What is the applied domain of this function? Why? 37. Economic forces beyond anyone’s control have changed the cost function for PortaBoys to C(
x) = 105x + 175. Rework Example 2.1.5 with this new cost function. 38. In response to the economic forces in Exercise 37 above, the local retailer sets the selling price of a PortaBoy at $250. Remarkably, 30 units were sold each week. When the systems went on sale for $220, 40 units per week were sold. Rework Examples 2.1.6 and 2.1.7 with this new data. What diﬃculties do you encounter? 39. A local pizza store oﬀers medium two-topping pizzas delivered for $6.00 per pizza plus a $1.50 delivery charge per order. On weekends, the store runs a ‘game day’ special: if six or more medium two-topping pizzas are ordered, they are $5.50 each with no delivery charge. Write a piecewise-deﬁned linear function which calculates the cost C (in dollars) of p medium two-topping pizzas delivered during a weekend. 2.1 Linear Functions 165 40. A restaurant oﬀers a buﬀet which costs $15 per person. For parties of 10 or more people, a group discount applies, and the cost is $12.50 per person. Write a piecewise-deﬁned linear function which calculates the total bill T of a party of n people who all choose the buﬀet. 41. A mobile plan charges a base monthly rate of $10 for the ﬁrst 500 minutes of air time plus a charge of 15¢ for each additional minute. Write a piecewise-deﬁned linear function which calculates the monthly cost C (in dollars) for using m minutes of air time. HINT: You may want to revisit Exercise 74 in Section 1.4 42. The local pet shop charges 12¢ per cricket up to 100 crickets, and 10¢ per cricket thereafter. Write a piecewise-deﬁned linear function which calculates the price P, in dollars, of purchasing c crickets. 43. The cross-section of a swimming pool is below. Write a piecewise-deﬁned linear function which describes the depth of the pool, D (in feet) as a function of: (a) the distance (in feet) from the edge of the shallow end of the pool, d. (b) the distance (
in feet) from the edge of the deep end of the pool, s. (c) Graph each of the functions in (a) and (b). Discuss with your classmates how to trans- form one into the other and how they relate to the diagram of the pool. d ft. 37 ft. 8 ft. 15 ft. s ft. 10 ft. 2 ft. In Exercises 44 - 49, compute the average rate of change of the function over the speciﬁed interval. 44. f (x) = x3, [−1, 2] 46. f (x) = √ x, [0, 16] 48. f (x5, 7] 45. f (x) = 1 x, [1, 5] 47. f (x) = x2, [−3, 3] 49. f (x) = 3x2 + 2x − 7, [−4, 2] 166 Linear and Quadratic Functions In Exercises 50 - 53, compute the average rate of change of the given function over the interval [x, x + h]. Here we assume [x, x + h] is in the domain of the function. 50. f (x) = x3 52. f (x) = x + 4 x − 3 51. f (x) = 1 x 53. f (x) = 3x2 + 2x − 7 54. The height of an object dropped from the roof of an eight story building is modeled by: h(t) = −16t2 + 64, 0 ≤ t ≤ 2. Here, h is the height of the object oﬀ the ground in feet, t seconds after the object is dropped. Find and interpret the average rate of change of h over the interval [0, 2]. 55. Using data from Bureau of Transportation Statistics, the average fuel economy F in miles per gallon for passenger cars in the US can be modeled by F (t) = −0.0076t2 + 0.45t + 16, 0 ≤ t ≤ 28, where t is the number of years since 1980. Find and interpret the average rate of change of F over the interval [0, 28]. 56. The temperature T in degrees Fahrenheit t hours after 6 AM is given by: T (t) = − 1 2 t2 + 8t + 32, 0 ≤ t ≤ 12 (a) Find and interpret T (4), T
(8) and T (12). (b) Find and interpret the average rate of change of T over the interval [4, 8]. (c) Find and interpret the average rate of change of T from t = 8 to t = 12. (d) Find and interpret the average rate of temperature change between 10 AM and 6 PM. 57. Suppose C(x) = x2 − 10x + 27 represents the costs, in hundreds, to produce x thousand pens. Find and interpret the average rate of change as production is increased from making 3000 to 5000 pens. 58. With the help of your classmates ﬁnd several other “real-world” examples of rates of change that are used to describe non-linear phenomena. (Parallel Lines) Recall from Intermediate Algebra that parallel lines have the same slope. (Please note that two vertical lines are also parallel to one another even though they have an undeﬁned slope.) In Exercises 59 - 64, you are given a line and a point which is not on that line. Find the line parallel to the given line which passes through the given point. 59. y = 3x + 2, P (0, 0) 60. y = −6x + 5, P (3, 2) 2.1 Linear Functions 61. y = 2 3 x − 7, P (6, 0) 63. y = 6, P (3, −2) 167 621, −1) 64. x = 1, P (−5, 0) (Perpendicular Lines) Recall from Intermediate Algebra that two non-vertical lines are perpendicular if and only if they have negative reciprocal slopes. That is to say, if one line has slope m1 and the other has slope m2 then m1 · m2 = −1. (You will be guided through a proof of this result in Exercise 71.) Please note that a horizontal line is perpendicular to a vertical line and vice versa, so we assume m1 = 0 and m2 = 0. In Exercises 65 - 70, you are given a line and a point which is not on that line. Find the line perpendicular to the given line which passes through the given point. 65. y = 1 3 x + 2, P (0, 0) 67. y = 2 3 x − 7, P (6, 0) 69. y = 6, P (3, −2) 66. y = −6
x + 5, P (3, 2) 681, −1) 70. x = 1, P (−5, 0) 71. We shall now prove that y = m1x + b1 is perpendicular to y = m2x + b2 if and only if m1 · m2 = −1. To make our lives easier we shall assume that m1 > 0 and m2 < 0. We can also “move” the lines so that their point of intersection is the origin without messing things up, so we’ll assume b1 = b2 = 0. (Take a moment with your classmates to discuss why this is okay.) Graphing the lines and plotting the points O(0, 0), P (1, m1) and Q(1, m2) gives us the following set up. y O P Q x The line y = m1x will be perpendicular to the line y = m2x if and only if OP Q is a right triangle. Let d1 be the distance from O to P, let d2 be the distance from O to Q and let d3 be the distance from P to Q. Use the Pythagorean Theorem to show that OP Q is a right triangle if and only if m1 · m2 = −1 by showing d2 3 if and only if m1 · m2 = −1. 2 = d2 1 + d2 168 Linear and Quadratic Functions 72. Show that if a = b, the line containing the points (a, b) and (b, a) is perpendicular to the line y = x. (Coupled with the result from Example 1.1.7 on page 13, we have now shown that the line y = x is a perpendicular bisector of the line segment connecting (a, b) and (b, a). This means the points (a, b) and (b, a) are symmetric about the line y = x. We will revisit this symmetry in section 5.2.) 73. The function deﬁned by I(x) = x is called the Identity Function. (a) Discuss with your classmates why this name makes sense. (b) Show that the point-slope form of a line (Equation 2.2) can be obtained from I using a sequence of the transformations deﬁned in Section 1.7. 2.1 Linear Functions 2.1.2

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