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culus 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 give...
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). ...
< 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 (...
. 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...
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 ag...
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 spen...
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 ...
: (−∞, ∞) 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 ...
−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...
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 th...
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 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...
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? Kee...
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...
, 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 grap...
(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 The...
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 horizont...
1(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 +...
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...
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 Transformati...
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...
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, ∞). √...
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 a...
, 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...
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-coordinat...
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 ou...
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...
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...
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 ...
−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 reﬂectio...
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 funct...
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...
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 ...
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 vert...
�ect 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 ge...
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 tr...
= 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 − ...
hing 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 ...
−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 ...
(−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....
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 ...
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)...
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...
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 ...
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 Interme...
�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...
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 = −...
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). No...
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. R...
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) ...
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...
− 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 ...
−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...
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...
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 m...
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, ...
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...
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 d...
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 o...
, 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 ...
“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 = m...
(x − √ y = −5x + 7 3 11. y = − 5 3 x 13. y = 8 5 x − 8 15. y = 5 17. y = − 5 4 x + 11 8 19. y = −x 21. f (x) = 2x − 1 slope: m = 2 y-intercept: (0, −1) 2, 0 x-intercept: 1 22. f (x) = 3 − x slope: m = −1 y-intercept: (0, 3) x-intercept: (3, 0) √ 3) 10. y + 12 = 678(x + 1) y = 678x + 666 12. y = −2 6x + 1) 7 x + 29 7 √ ...
. 32. W (x) = 200 +.05x, x ≥ 0 She must make $5500 in weekly sales. 33. C(p) = 0.035p + 1.5 The slope 0.035 means it costs 3.5¢ per page. C(0) = 1.5 means there is a ﬁxed, or start-up, cost of $1.50 to make each book. 34. F (m) = 2.25m + 2.05 The slope 2.25 means it costs an additional $2.25 for each mile beyond the ﬁr...
1 ≤ c ≤ 100 c > 100 D(d) =    − 1 8 2 d + 31 2 2 if if if 0 ≤ d ≤ 15 15 ≤ d ≤ 27 27 ≤ d ≤ 37 D(s) =    2 1 2 s − 3 8 if if if 0 ≤ s ≤ 10 10 ≤ s ≤ 22 22 ≤ s ≤ 37 8 2 8 2 15 27 37 10 22 37 y = D(d) y = D(s) 172 44. 46. 48 Linear and Quadratic Functions 45. 47 32 − (−3)2 3 − (−3) = 0 49. (3(2)2 + 2(2) − 7) − (3...
the temperature increases, on average, at a rate of 2◦F per hour. (c) The average rate of change is T (12)−T (8) 12−8 = −2. Between 2 PM and 6 PM, the temperature decreases, on average, at a rate of 2◦F per hour. (d) The average rate of change is T (12)−T (4) ture, on average, remains constant. 12−4 = 0. Between 10 AM...
if x ≥ 0 In Deﬁnition 2.4, we deﬁne \|x\| using a piecewise-deﬁned function. (See page 62 in Section 1.4.) To check that this deﬁnition agrees with what we previously understood as absolute value, note that since 5 ≥ 0, to ﬁnd \|5\| we use the rule \|x\| = x, so \|5\| = 5. Similarly, since −5 < 0, we use the rule \|x\| = −x, so...
\| = −a, \|b\| = −b and \|ab\| = ab. The equation \|ab\| = \|a\|\|b\| becomes ab = (−a)(−b), which is true. Suppose a is positive and b is negative. Then ab is negative, and we have \|ab\| = −ab, \|a\| = a and \|b\| = −b. The equation \|ab\| = \|a\|\|b\| reduces to −ab = a(−b) which is true. A symmetric argument shows the equation \|ab\| = \|a\|...
to 3x − 1 = 6 or 3x − 1 = −6. Solving the former, we arrive at x = 7 3, and solving the latter, we get x = − 5 3. We may check both of these solutions by substituting them into the original equation and showing that the arithmetic works out. 2. To use the Equality Properties to solve 3 − \|x + 5\| = 1, we ﬁrst isolate t...
us x2 + x − 6 = 0, or (x + 3)(x − 2) = 0. We get x = −3 or x = 2. Since only x = −3 satisﬁes x < 0, this is the answer we keep. For x ≥ 0, \|x\| = x, so the equation \|x\| = x2 − 6 becomes x = x2 − 6. From this, we get x2 − x − 6 = 0 or (x − 3)(x + 2) = 0. Our solutions are x = 3 or x = −2, and since only x = 3 satisﬁes x...
hing linear functions (from Section 2.1) and piecewise-deﬁned functions (from Section 1.4). Example 2.2.2. Graph each of the following functions. 1. f (x) = \|x\| 2. g(x) = \|x − 3\| 3. h(x) = \|x\| − 3 4. i(x) = 4 − 2\|3x + 1\| 176 Linear and Quadratic Functions Find the zeros of each function and the x- and y-intercepts of e...
notice that when x ≥ 0, we get to ﬁll in the point at (0, 0), which eﬀectively ‘plugs’ the hole indicated by the open circle. Thus we get, y 4 3 2 1 −3 −2 −1 1 2 3 x f (x) = \|x\| 1Actually, since functions can have at most one y-intercept (Do you know why?), as soon as we found (0, 0) as the x-intercept, we knew this w...
∞). The function g is increasing on [3, ∞) and decreasing on (−∞, 3]. The relative and absolute minimum value of g is 0 which occurs at (3, 0). As before, there is no relative or absolute maximum value of g. 3. Setting h(x) = 0 to look for zeros gives \|x\| − 3 = 0. As in Example 2.2.1, we isolate the absolute value to ...
intercepts are 1 for a y-intercept of (0, 2). Rewriting the formula for i(x) without absolute values gives i(x) = 4 − 2(−(3x + 1)), 4 − 2(3x + 1), if if (3x + 1) < 0 (3x + 1) ≥ 0 = 6x + 6, −6x + 2, if x < − 1 3 if x ≥ − 1 3 The usual analysis near the trouble spot x = − 1 and we get the distinctive ‘∨’ shape: 3 gives t...
− 3), Theorem 1.7 tells us to add 3 to each of the x-values of the points on the graph of y = f (x) to obtain the graph of y = g(x). This shifts the graph of y = f (x) to the right 3 units and moves the point (−1, 1) to (2, 1), (0, 0) to (3, 0) and (1, 1) to (4, 1). Connecting these points in the classic ‘∨’ fashion p...
= 4 − 2f (3x + 1) = −2f (3x + 1) + 4 and apply Theorem 1.7. First, we take care of the changes on the ‘inside’ of the absolute value. Instead of \|x\|, we have \|3x + 1\|, so, in accordance with Theorem 1.7, we ﬁrst subtract 1 from each of the x-values of points on the graph of y = f (x), then divide each of those new val...
of each graph, if any exist. From the graph, determine the domain and range of each function, list the intervals on which the function is increasing, decreasing or constant, and ﬁnd the relative and absolute extrema, if they exist. 1. f (x) = \|x\| x Solution. 2. g(x) = \|x + 2\| − \|x − 3\| + 1 1. We ﬁrst note that, due to...
.) 2. To ﬁnd the zeros of g, we set g(x) = 0. The result is \|x + 2\| − \|x − 3\| + 1 = 0. Attempting to isolate the absolute value term is complicated by the fact that there are two terms with absolute values. In this case, it easier to proceed using cases by re-writing the function g with two separate applications of Deﬁ...
also (0, 0). To graph g, we start with x < −2 and graph the horizontal line y = −4 with an open circle at (−2, −4). For −2 ≤ x < 3, we graph the line y = 2x and the point (−2, −4) patches the hole left by the previous piece. An open circle at (3, 6) completes the graph of this part. Finally, we graph the horizontal li...
\|2x − 1\| = x + 1 11. 4 − \|x\| = 2x + 1 12. \|x − 4\| = x − 5 13. \|x\| = x2 14. \|x\| = 12 − x2 15. \|x2 − 1\| = 3 Prove that if \|f (x)\| = \|g(x)\| then either f (x) = g(x) or f (x) = −g(x). Use that result to solve the equations in Exercises 16 - 21. 16. \|3x − 2\| = \|2x + 7\| 17. \|3x + 1\| = \|4x\| 18. \|1 − 2x\| = \|x + 1\| 19. \|4 − x\|...
th parts of Theorem 2.1. 36. Prove The Triangle Inequality: For all real numbers a and b, \|a + b\| ≤ \|a\| + \|b\|. 184 Linear and Quadratic Functions 2.2.2 Answers 1. x = −6 or x = 6 4. x = −1 or x = 1 7. x = −3 or x = 3 2. x = −3 or x = 11 3 3. x = −3 or x = 11 5. x = − 1 2 or x = 1 10 8. x = − 13 8 or x = 53 8 6. no solu...
) Relative and absolute minimum at (0, 0) No relative or absolute maximum 25. f (x) = −3\|x\| f (0) = 0 x-intercept (0, 0) y-intercept (0, 0) Domain (−∞, ∞) Range (−∞, 0] Increasing on (−∞, 0] Decreasing on [0, ∞) Relative and absolute maximum at (0, 0) No relative or absolute minimum 26. f (x) = 3\|x + 4\| − 4 = 0, − 8 f ...
x\| − 3 as x < 0 x ≥ 0 −3 2x − 3 f (x) = if if = 0 f 3 2 x-intercept 3 2, 0 y-intercept (0, −3) Domain (−∞, ∞) Range [−3, ∞) Increasing on [0, ∞) Constant on (−∞, 0] Absolute minimum at every point (x, −3) where x ≤ 0 No absolute maximum Linear and Quadratic Functions where x < −4 Absolute maximum at every point (x, 1) ...
) = \|x + 4\| + \|x − 2\| as −2x − 2 6 2x + 2 x < −4 if if −4 ≤ x < 2 x ≥ 2 if f (x) =    No zeros No x-intercept y-intercept (0, 6) Domain (−∞, ∞) Range [6, ∞) Decreasing on (−∞, −4] Constant on [−4, 2] Increasing on [2, ∞) Absolute minimum at every point (x, 6) where −4 ≤ x ≤ 2 No absolute maximum Relative minimum a...
found. y 4 3 2 1 (2, 4) (1, 1) (−2, 4) (−1, 1) −2 −1 1 (0, 0) f (x) = x2 2 x Much like many of the absolute value functions in Section 2.2, knowing the graph of f (x) = x2 enables us to graph an entire family of quadratic functions using transformations. Example 2.3.1. Graph the following functions starting with the g...
(x) = f (x + 2) − 3 = (x + 2)2 − 3 From the graph, we see that the vertex has moved from (0, 0) on the graph of y = f (x) to (−2, −3) on the graph of y = g(x). This sets [−3, ∞) as the range of g. We see that the graph of y = g(x) crosses the x-axis twice, so we expect two x-intercepts. To ﬁnd these, we set y = g(x) = ...
) to (5, −7). y (3, 1) 2 1 (2, −1) 3 4 5 (4, −1) x 1 −1 −2 −3 −4 −5 −6 (2, 4) (1, 1) (1, −7) (5, −7) y 4 3 2 1 (−2, 4) (−1, 1) −2 −1 (0, 0) 1 2 x f (x) = x2 −−−−−−−−−−−−→ h(x) = −2f (x − 3) + 1 = −2(x − 3)2 + 1 The vertex is (3, 1) which makes the range of h (−∞, 1]. From our graph, we know that there are two x-interce...
’ formulas for g(x) and h(x) satisfy Deﬁnition 2.5, they do not lend themselves to graphing easily. For that reason, the form of g and h presented in Example 2.3.2 is given a special name, which we list below, along with the form presented in Deﬁnition 2.5. Deﬁnition 2.6. Standard and General Form of Quadratic Function...
y = k, so (h, k) is on the graph. If x = h, then x − h = 0 so (x − h)2 is a positive number. If a > 0, then a(x − h)2 is positive, thus y = a(x − h)2 + k is always a number larger than k. This means that when a > 0, (h, k) is the lowest point on the graph and thus the parabola must open upwards, making (h, k) the vert...
y = a(x−h)2+k is a parabola ‘opening upwards’ if a > 0, and ‘opening downwards’ if a < 0. Moreover, the symmetry enjoyed by the graph of y = x2 about the y-axis is translated to a symmetry about the vertical line x = h which is the vertical line through the vertex.6 This line is called the axis of symmetry of the para...
must also subtract it. We collect the terms which create the perfect square and gather the remaining constant terms. Putting it all together, we get f (x) = x2 − 4x + 3 2 (−4) = −2.) = x2 − 4x + 4 − 4 + 3 (Add and subtract (−2)2 = 4 to (x2 + 4x).) = x2 − 4x + 4 − 4 + 3 (Group the perfect square trinomial.) = (x − 2)2 ...
(−1) from both the x2 and x terms. We then follow the completing the square recipe as above. g(x) = −x2 − x + 6 = (−1) x2 + x + 6 x2 + x + 1 = (−1) = (−1) x2 + x + 1 4 = − x + 1 + 25 4 2 2 4 4 − 1 + 6 + (−1) − 1 4 (Factor the coeﬃcient of x2 from x2 and x.) + 6 (Group the perfect square trinomial.) 7If you forget why ...
coeﬃcient of x2 from x2 and x.) x2 + = a x2 + = a x2 + = a b a b a b a b2 4a2 + c x + x + b2 4a2 − b2 4a2 − a = a x + 2 b 2a + 4ac − b2 4a b2 4a2 + c (Group the perfect square trinomial.) (Factor and get a common denominator.) so that Comparing this last expression with the standard form, we identify (x − h) with x + ...
= 0, then the solutions to ax2 + bx + c = 0 are −b ± x = √ b2 − 4ac 2a. Assuming the conditions of Equation 2.5, the solutions to ax2 + bx + c = 0 are precisely the zeros of f (x) = ax2 + bx + c. Since f (x) = ax2 + bx + c = a x + 2 b 2a + 4ac − b2 4a the equation ax2 + bx + c = 0 is equivalent to a x + 2 b 2a + 4ac −...
2 + bx + c = 0 has exactly one real solution. If b2 − 4ac > 0, the equation ax2 + bx + c = 0 has exactly two real solutions. The proof of Theorem 2.3 stems from the position of the discriminant in the quadratic equation, and is left as a good mental exercise for the reader. The next example exploits the fruits of all o...
1.5x2 + 170x − 150 = 0. The mere thought of trying to factor the left hand side of this equation could do serious psychological damage, so we resort to the quadratic formula, Equation 2.5. Identifying a = −1.5, b = 170, and c = −150, we obtain x = = = = √ −b ± b2 − 4ac 2a −170 ± 1702 − 4(−1.5)(−150) √ −170 ± 2(−1.5) 28...
0.89 and 112.44. As we saw in Example 1.5.3, these are the ‘break-even’ points of the proﬁt function, where enough product is sold to recover the cost spent to make the product. More importantly, we see from the graph that as long as x is between 0.89 and 112.44, the graph y = P (x) is above the x-axis, meaning y = P ...
feet of grazing area, how many alpaca can Donnie keep in his pasture? 198 Linear and Quadratic Functions Solution. It is always helpful to sketch the problem situation, so we do so below. river w pasture w l We are tasked to ﬁnd the dimensions of the pasture which would give a maximum area. We let w denote the width o...
means that there is a maximum value to be found, and we know it occurs at the vertex. Using the vertex formula, we ﬁnd w = − 200 2(−2) = 50, and A(50) = −2(50)2 + 200(50) = 5000. Since w = 50 lies in the applied domain, 0 < w < 100, we have that the area of the pasture is maximized when the width is 50 feet. To ﬁnd th...
− b 2, and y = 1 4 = −6.25. Plotting, we get the parabola seen below on the left. To obtain 2 points on the graph of y = f (x) = \|x2 − x − 6\|, we can take points on the graph of g(x) = x2 − x − 6 and apply the absolute value to each of the y values on the parabola. We see from the graph of g that for x ≤ −2 or x ≥ 3, ...
x-axis. This is a general template to graph functions of the form f (x) = \|g(x)\|. From this perspective, the graph of f (x) = \|x\| can be obtained by reﬂecting the portion of the line g(x) = x which is below the x-axis back above the x-axis creating the characteristic ‘∨’ shape. 200 Linear and Quadratic Functions 2.3.1...

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