Preview text I C H A P T E R P Preparation for Calculus Section P.1 Graphs and Models. 2 Section P.2 Linear Models and Rates of Change. 7 Section P.3 Functions and Their Graphs. 14 Section P.4 Fitting Models to Data. 18 Review Exercises. 19 Problem Solving. 23 PA R T Section P.1 15.
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LIBRO Y SOLUCIONARIO DE: CALCULO (DE UNA VARIABLE y DE VARIAS VARIABLES).
Xmin Xmax 5 Xscl 1 Ymin Ymax 5 Yscl 1 5 4.00, 3) (2, 1.73) 6 (a) Note that y 4 when x 0. (b) y 02 0 2 y 02 y y 0 x2 x 2 y 2 3 19. Y 3 0 x2 0 0 x x 0, 5, x 13. Graphs and Models 23. X2y x2 4y 0 None.
X cannot equal 0. X 02 4y 0 y 0 2 x x2 0 x 25. Symmetric with respect to the since 27. Symmetric with respect to the since y 2 x 2. Symmetric with respect to the origin since xy 4.
No symmetry with respect to either axis or the origin. Y x3 x is symmetric with respect to the 33. Symmetric with respect to the origin since since y x3 x. Y 2 y Intercepts: 23, 2 (0, 2) 1 2 3, Symmetry: none 0 x 1 2 1 3 4 Chapter P 39. Y Preparation for Calculus x 2 43. Y 1 x2 Intercepts: Intercepts: Intercepts: Symmetry: Symmetry: none Symmetry: none y y y 12 2 2 (8, 0) 2 2 8 4 10 ( 1, 0) 2 (0, 10 (0, 1) x x 4) 2 6 1 8 2 2 10 8 6 10 47. Y 2 3 2, 0, Symmetry: origin Symmetry: none y 4 Domain: x y 3 2 5 y 4 2, 0) (0, 2) 2 1 2 1 2, 0) 3 4 2 3 2 3 1 1 4 x x 5 1 3 (0, 0) 6 3 3 4 Intercepts: Symmetry: none 49.
X y3 Intercepts: Intercepts: x 2 3, 0) 45. Y x3 2 ( (0, 9) 8 (1, 0) (0, 0) x 1 2 3 4 1 x Intercepts: none 51. Y 6 x y 3 y 8 Intercepts: 6 2 1 Symmetry: origin x 1 2 Symmetry: 3 (0, 6) 4 2 6, 0) (6, 0) x 2 4 6 8 57. Y2 x 9 y2 x 9 3y2 6 x y 9 4 Intercepts: Symmetry: (0, 3) 0) 11 1 (0, 3) 3x Intercepts: 3 (0, 2 ) (6, 0) 8 Symmetry: (0, 2 ) 6 Chapter P Preparation for Calculus 75. 3.29x 2 30.25x 10.8241x2 65,800x 0 10.8241x2 65,830.25x Use the Quadratic Formula. X 3133 units The other root, x 2949, does not satisfy the equation R C.
This problem can also be solved using a graphing utility and finding the intersection of the graphs of C and R. (a) Using a graphing utility, you obtain (b) 250 y 4.9971t 34.9405 (c) For the year 2004, t 34 and y 187.2 CPI. 400 0 100 0 If the diameter is doubled, the resistance is changed approximately a factor of For instance, and 6.36125. Symmetry means that if is on the graph, then is also on the graph. The are 4ac 2a,0. Distance to the origin K Distance to y2 y2, K 1 x2 y 2 K 4x 4 K x 2 K 2 4K 2x 4K 2 0 Note: This is the equation of a circle!
50 Section P.2 Section P.2 3. M y 5 4 (2, 3) 2 2 11. M 6 2 4 0 undefined 3 x 3 4 2 5 y (5, 2) 1 m 6 x 1 2 3 5 6 4 3 (2, 5) 5 7 2 (3, 4) (2, 1) 1 x 1 3 4 5 6 13.
M y 3 2 12, 23 ) 34, 16 ) x 1 2 3 15. Since the slope is 0, the line is horizontal and its equation is y 1. Therefore, three additional points are and 17.
The equation of this line is y 7 y 10. Therefore, three additional points are and 19. Given a line L, you can use any two distinct points to calculate its slope. Since a line is straight, the ratio of the change in to the change in will always be the same. See Section P.2 Exercise 93 for a proof. M y m 32 m is undefined 1 7 Linear Models and Rates of Change 1.
M 1 3 Linear Models and Rates of Change Section P.2 39. Vertical line x 5 41.
M Linear Models and Rates of Change 11 0 2 9 43. 3 11 4 2 y y 9 8 7 6 5 4 3 2 1 11 3 2 4 2 1 22x 4y 3 0 x 4 6 7 8 9 3 x 4 ( 12, 72 ) 2 1 y x 2 3 45. ( 0, 34 ) x 1 2 3 4 47.
3x 2y 6 0 y x a a 1 2 a a 3 a 49. 2 y (5, 1) 1 2 3 4 (3, 0) 1 51. Y 1 y y 3 y 2 1 x 1 2 3 4 5 x 1 2 y 2 53.
2x y 3 0 y 2x 3 1 2 2y 3x 1 0 y 1 y x 4 2 3 1 2 1 2 2 1 x 1 2 3 4 3 3 (5, 8) 10 Chapter P 57. Preparation for Calculus 10 10 10 15 10 10 15 10 The lines do not appear perpendicular. The lines appear perpendicular. The lines are perpendicular because their slopes 1 and are negative reciprocals of each other. You must use a square setting in order for perpendicular lines to appear perpendicular. 4x 2y 3 y 2x 2 y 53x m 53 3 y 1 (a) 24y 21 40x 30 y 1 2x 4 24y 40x 9 0 2x y 3 0 (b) y 78 34 1 2 (b) y 78 34 40y 35 18 2y 2 2 40y 24x 53 0 x 2y 4 0 63. (a) x 2 x 2 0 (b) y 5 y 5 0 65.
The slope is 125. Hence, V 2540 125t 2415 67. The slope is Hence, V 20,400 22,400 69. 5 (2, 4) (0, 0) 6 You can use the graphing utility to determine that the points of intersection are and Analytically, x2 4x x2 2x2 4x 0 0 x 0 y 0 x 2 y 4 The slope of the line joining and is m 2. Hence, an equation of the line is y 0 y 2x. (a) 12 Chapter P Preparation for Calculus 81. (a) Two points are and The slope is (b) 50 625 580 47 50 p 580 0 p 750 580 1330 1 If p 655, x 15 45 units.
1 or x 15 83. 4x 3y 10 0 d 85. X y 2 0 d 1500 0 1 (c) If p 595, x 15 49 units. 10 2 42 32 5 1 1 2 2 5 5 2 2 2 12 12 2 4 2 2 2. If A 0, then C 0 is the horizontal line y The distance to is B A B A A B d y1 1 1 1 2 2 If B 0, then Ax C 0 is the vertical line x The distance to is d x1 1 1 1 2 2 (Note that A and B cannot both be zero.) The slope of the line Ax C 0 is The equation of the line through perpendicular to Ax C 0 is: y y1 B A Ay Ay1 Bx Bx1 Bx1 Ay1 Bx Ay The point of intersection of these two lines is: Ax Bx Ay Bx1 Ay1 A2x A B 2x A B2x 1 (1) A1 (2) B2x1 A1 ( adding equations (1) and (2)) Ax B2x1 A1 A2 B2 ABx B2y Bx Ay Bx1 A2 y (3) A2 y1 (4) ABx1 A2y1 ( adding equations (3) and (4)) ABx1 A2y1 A2 B2 87.
A point on the line x y 1 is The distance from the point to x y 5 0 is Section P.2 Linear Models and Rates of Change 13 89. A, ABx point of intersection 2 1 2 1 2 1 2 2 1 2 The distance between and this point gives us the distance between and the line Ax C 0.
A x ABx A A x ABx A 2 1 2 1 2 2 1 2 1 2 1 1 2 2 2 2 1 2 2 2 1 1 2 2 2 2 2 1 2 1 2 2 1 2 2 1 2 2 2 1 A2 B2 91. For simplicity, let the vertices of the rhombus be and b, as shown in the figure.
The slopes of the diagonals are then m1 y (b, c) (a b, c ) c c. And m2 x (0, 0) Since the sides of the Rhombus are equal, a2 b2 c2, and we have m1m2 c c c2 (a, 0) c2 b a b2 a2 Therefore, the diagonals are perpendicular. Consider the figure below in which the four points are collinear. Since the triangles are similar, the result immediately follows.
Y2 y1 x2 x1 95. A c a ax c1 y x 1 m1 b b b b c bx ay c2 y x 2 a a y m2 (x 2, y2 ) (x, ) (x1, y1 ) (x ) x 1 m1 m2 b a 1 2 y1 2 Ax B Ax B 1 Section P.3 21. 1 y Domain:, 8 6 Range:, Functions and Their Graphs 15 y Domain: 1, 2 Range: 0, 1 x 2 1 2 3 x 2 2 25. 2 sin t y 4 Domain: Domain:, 2 Range: 0, 2 2 2 1 Range: x 4 y t 4 2 2 29.
X y 2 0 y y is not a function of x. Some vertical lines intersect the graph twice. X2 y2 4 y x2 31. Y is a function of x. Vertical lines intersect the graph at most once.
Y2 x2 1 y 1 y is not a function of x since there are two values of y for some x. 3 y is not a function of x since there are two values of y for some x. F x x 2 If x 0, then f 2 If 0 x 2, then f x 2. If x 2, then f x 2x 2 Thus, f 2, x 0 0 x 2. The function is cx2.
Since satisfies the equation, c Thus, 41. The function is since it must be undefined at x 0. Since satisfies the equation, c 32. (a) For each time t, there corresponds a depth d. (b) Domain: 0 t 5 27 Range: 0 d 30 (c) d 18 d 9 30 25 t1 20 15 10 5 t 1 2 3 4 5 6 t2 t3 t 4 16 Chapter P Preparation for Calculus y 47. (a) The graph is shifted 3 units to the left. Y (b) The graph is shifted 1 unit to the right.
4 4 2 2 2 y (c) The graph is shifted 2 units upward. X 4 4 6 2 4 6 x 2 4 6 y 2 2 (e) The graph is stretched vertically a factor of 3. 8 x 4 6 y (d) The graph is shifted 4 units downward.
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6 4 4 y (f) The graph is stretched vertically a factor of 14. X 6 4 2 x 10 49.
(a) y 2 (b) y y y (c) y 2 y 4 1 4 3 x 3 1 2 3 4 2 2 1 1 x 2 1 2 3 1 2 3 4 5 6 3 x 1 4 Reflection about the Vertical shift 2 units upward Horizontal shift 2 units to the right 51. (a) 16, 23 (b) If then the program would turn on (and off) one hour later. (c) If 1, then the overall temperature would be reduced 1 degree. F x2, f f f x, x 0 2 Domain: 0, f f x Domain:, No. Their domains are different. F, x2 1 x f f f 3 x2 1 Domain: all x 1 f f g Domain: all x 0 No, f g g f.
2 9 9 x2 2 x x2 x 18 Chapter P Preparation for Calculus Section P.4 Fitting Models to Data 1. Quadratic function 3.
Linear function 5. (a) d 0.066F or F 15.1d 0.1 y 250 (b) 125 200 150 100 F 15.13 d 0.10 50 0 x 3 6 9 12 10 0 15 The model fits well. The cancer mortality increases linearly with increased exposure to the carcinogenic substance. (c) If F 55, then d 3.63 cm. (a) Let x per capita energy usage (in millions of Btu) 11. (a) y1 0.0343t3 0.3451t2 0.8837t 5.6061 y per capita gross national product (in thousands) y2 0.1095t 2.0667 y 0.0764x 4.9985 0.08x 5.0 r 0.7052 (b) y3 0.0917t 0.7917 (b) 15 y1 y2 y3 40 y1 y2 0 8 0 0 420 0 y 0.08x 5.0 y3 For 2002, t 12 and y1 y2 y3 31.06 (c) Denmark, Japan, and Canada (d) Deleting the data for the three countries above, y 0.0959x 1.0539 (r 0.9202 is much closer to 1.) 13.
(a) y1 4.0367t 28.9644 (d) y3 0.4297t2 0.5994t 32.9745 y2 0.5488t2 0.2399t 33.1414 (b) 70 70 y1 4.04t 28.96 0 8 25 0 8 25 y2 3 0.55t 2 0.24t 33.14 (c) The cubic model is better. (e) The slope represents the average increase per year in the number of people (in millions) in HMOs. (f) For t 10, and y1 69.3 million.
(linear) y2 80.5 million (cubic) (c) If x 3, then y 136. Review Exercises for Chapter P 15. (a) y 14.58x2 16.39x 10 (b) 19 17.
(a) Yes, y is a function of t. At each time t, there is one and only one displacement y. 300 (b) The amplitude is approximately 0.35. 0 The period is approximately 7 0 0.5. (c) If x 4.5, y 214 horsepower. (c) One model is y 0.35 2.
(d) 4 0.9 0 0 19. Answers will vary. Review Exercises for Chapter P 1. Y 2x 3 x 0 y 3 3 3 y 0 0 2x 3 x 2 2, 3. Symmetric with respect to since 1 1 0, 2 2 4y 0 x 1 7. Y 12 x 32 x2y x2 4y 0. Y 7 6x x2 1 5 9.
3 x 6 y 1 25 x y 65 y y y 25 x 65 3 2 2 5 Slope: 1 6 5 5 x 1 2 3 y x 1 10 3 2 2 x 3 2 1 1 1 5 5.
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