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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 functi... |
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) = ... |
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) = 2... |
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 ... |
. 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 whe... |
) 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 ar... |
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 ... |
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... |
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 ... |
(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 frac... |
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 compou... |
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 ... |
= √ 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 + h... |
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 in... |
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 resu... |
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 ar... |
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 t... |
− 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 √ 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) Fi... |
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 )... |
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 ... |
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: ... |
�) 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, ∞)... |
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 −... |
) = 4.6, 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 r... |
.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) = 10... |
.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 ... |
−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 ... |
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 s... |
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... |
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 tha... |
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) th... |
, 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... |
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 i... |
.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 i... |
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 inc... |
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. ... |
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... |
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. Thi... |
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 15Al... |
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... |
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 ... |
√ 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 ... |
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... |
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... |
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 lo... |
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-in... |
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. [... |
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... |
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)... |
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 conne... |
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 ha... |
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 =... |
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 ... |
+ 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 dom... |
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 −−−−−−−−−−... |
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 m... |
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 th... |
, 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) = functio... |
−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 ... |
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... |
(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 ... |
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, n... |
‘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 sc... |
‘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 somet... |
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 s... |
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 >... |
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,... |
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 sca... |
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 −−−−−−−−... |
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 ... |
�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 ... |
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 ... |
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-axi... |
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 fu... |
= 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... |
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(... |
) 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 ... |
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, 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 Func... |
(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... |
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) = ... |
�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 u... |
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... |
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 i... |
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 ch... |
) = 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 (−∞, ∞). ... |
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 ... |
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 ... |
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 pag... |
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... |
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 ... |
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 cen... |
��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 ge... |
) 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.... |
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 t... |
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 ... |
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 ... |
(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 h... |
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, ... |
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