study guide for Calculus by Larson(9th) - PA R T I C H A P T E R P Preparation for Calculus Section P.1 Graphs and Models 2 Section P.2 Linear Models

study guide for Calculus by Larson(9th) - PA R T I C H A P...

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Unformatted text preview: PA R T 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 C H A P T E R P Preparation for Calculus Section P.1 Graphs and Models Solutions to Odd-Numbered Exercises 1 1. y   2 x  2 3. y  4  x2 x-intercept: 4, 0 x-intercepts: 2, 0, 2, 0 y-intercept: 0, 2 y-intercept: 0, 4 Matches graph (b) Matches graph (a) 7. y  4  x2 5. y  32x  1 x 4 2 0 2 4 x 3 2 0 2 3 y 5 2 1 4 7 y 5 0 4 0 5 y y 8 (4, 7) 6 2 (−2, 0) (0, 1) −8 −6 −4 2 −4 (0, 4) (2, 4) 4 (−4, − 5) 6 x 4 6 −6 8 4 (− 3, − 5) (3, − 5) −4 −6  11. y  x  4 9. y  x  2 x 5 4 3 2 1 0 1 x 0 1 4 9 16 y 3 2 1 0 1 2 3 y 4 3 2 1 0 y y 10 8 6 4 2 6 4 (−5, 3) (−4, 2) 2 −6 −4 (1, 3) (0, 2) −2 (−1, 1) (−3, 1) x (− 2, 0) −2 2 6 −2 −8  (2, 0) x −4 (− 2, −2) −6 2 2 −4 −6 −8 − 10 (4, − 2) (16, 0) x 2 (1, − 3) (0, − 4) 12 14 16 18 (9, − 1) Section P.1 13. 15. Xmin = -3 Xmax = 5 Xscl = 1 Ymin = -3 Ymax = 5 Yscl = 1 3 5 (− 4.00, 3) (2, 1.73) −6 6 −3 (a) 2, y  2, 1.73 Note that y  4 when x  0. (b) x, 3  4, 3 y  02  0  2 y-intercept: y  0225  02 y-intercept: y  0; 0, 0 y  2; 0, 2 0  x2  x  2 x-intercepts:  y  5  2  3  1.73  3  5  4  19. y  x225  x2 17. y  x2  x  2 21. y  Graphs and Models x-intercepts: 0  x225  x2 0  x  2x  1 0  x25  x5  x x  2, 1; 2, 0, 1, 0 x  0, ± 5; 0, 0; ± 5, 0 32  x  x 23. x2y  x2  4y  0 y-intercept: None. x cannot equal 0. x-intercepts: 32  x 0 x y-intercept: 02y  02  4y  0 y  0; 0, 0 0  2  x x-intercept: x  4; 4, 0 x20  x2  40  0 x  0; 0, 0 25. Symmetric with respect to the y-axis since 27. Symmetric with respect to the x-axis since y  x  2  x  2. 2 y2  y2  x3  4x. 2 31. y  4  x  3 29. Symmetric with respect to the origin since xy  xy  4. No symmetry with respect to either axis or the origin.  x y  x2  1 y  35. y  x3  x is symmetric with respect to the y-axis 33. Symmetric with respect to the origin since       since y  x3  x   x3  x  x3  x . x . x2  1 37. y  3x  2 y Intercepts:  23 , 0, 0, 2 2 (0, 2) 1 2 3, Symmetry: none 0 x 1 2 1 3 4 Chapter P 39. y  Preparation for Calculus x 4 2 43. y  x  32 41. y  1  x2 Intercepts: Intercepts: Intercepts: 1, 0, 1, 0, 0, 1 8, 0, 0, 4 3, 0, 0, 9 Symmetry: y-axis 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 −2 4) 2 6 1 8 2 2 − 10 − 8 − 6 10 47. y  xx  2 45. y  x3  2 3 2, 0 , 0, 2    −2 Symmetry: origin Symmetry: none y 4 Domain: x ≥ 2 y 3 2 5 y 4 2, 0) (0, 2) 2 1 2 1 (− 2, 0) −4 −3 −1 3 4 −4 2 3 2 −3 3 1 1 −2 4 x x −4 −3 −2 −1 5 1 3 (0, 0) 6 3 3 4 Intercepts: 0, 0 0, 0, 2, 0 Symmetry: none x 2 (− 3, 0) 49. x  y3 Intercepts: Intercepts: ( (0, 9) 8 (1, 0) (0, 0) x 1 2 3 4 −2 1 x Intercepts: none 51. y   53. y  6  x y 3 y 8 Intercepts: 6 2 0, 6, 6, 0, 6, 0 1 Symmetry: origin x 1 2 Symmetry: y-axis 3 (0, 6) 4 2 (− 6, 0) −8 −4 −2 −2 (6, 0) x 2 4 6 8 −4 −6 −8 57. x  3y2  6 55. y2  x  9 y2  x  9 3y2  6  x y  ± x  9 4 Intercepts: 0, 3, 0, 3, 9, 0 Symmetry: x-axis y± (0, 3) (−9, 0) − 11 1 (0, − 3) −4 2  3x Intercepts: 3 (0, 2 ) (6, 0) −1 8 6, 0, 0, 2, 0,  2 Symmetry: x-axis (0, − 2 ) −3 Section P.1 59. y  x  2x  4x  6 (other answers possible) Graphs and Models 61. Some possible equations: yx y  x3 y  3x3  x 3 x y  63. xy2⇒y2x xy7⇒y7x 65. 2x  y  1 ⇒ y  2x  1 3x  2y  11 ⇒ y  2  x  2x  1 7x 3  3x 1x 3x  11 2 3x  11 2 14  2x  3x  11 The corresponding y-value is y  1. 5x  25 Point of intersection: 1, 1 x5 The corresponding y-value is y  2. Point of intersection: 5, 2 67. x2  y  6 ⇒ y  6  x2 69. x2  y 2  5 ⇒ y 2  5  x 2 xy1⇒yx1 xy4⇒y4x 5  x2  x  12 6  x2  4  x 0  x2  x  2 5  x2  x2  2x  1 0  x  2x  1 0  2x2  2x  4  2x  1x  2 x  2, 1 x  1 or x  2 The corresponding y-values are y  2 and y  1. The corresponding y-values are y  2 (for x  2) and y  5 (for x  1). Points of intersection: 1, 2, 2, 1 Points of intersection: 2, 2, 1, 5 71. y  x3 y  x3  2x2  x  1 73. yx y  x2  3x  1 x3  x x3  2x2  x  1  x2  3x  1 x3  x  0 xx  1x  1  0 x3  x2  2x  0 xx  2x  1  0 x  1, 0, 2 x  0, x  1, or x  1 The corresponding y-values are y  0, y  1, and y  1. 1, 5, 0, 1, 2, 1 4 Points of intersection: 0, 0, 1, 1, 1, 1 −4 y = x 3 − 2x 2 + x − 1 (2, 1) (0, −1) 6 (−1, −5) −8 y = −x 2 + 3 x − 1 5 6 Chapter P Preparation for Calculus 75. 5.5x  10,000  3.29x  5.5x  2  3.29x  10,0002 30.25x  10.8241x2  65,800x  100,000,000 0  10.8241x2  65,830.25x  100,000,000 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 by using a graphing utility and finding the intersection of the graphs of C and R. 77. (a) Using a graphing utility, you obtain (b) 250 y  0.0153t2  4.9971t  34.9405 (c) For the year 2004, t  34 and y  187.2 CPI. −5 35 − 50 79. 400 0 100 0 If the diameter is doubled, the resistance is changed by approximately a factor of 14. For instance, y20  26.555 and y40  6.36125. 81. False; x-axis symmetry means that if 1, 2 is on the graph, then 1, 2 is also on the graph. 83. True; the x-intercepts are b ± b2  4ac 2a ,0 . 85. Distance to the origin  K  Distance to 2, 0 x2  y2  Kx  22  y2, K  1 x2  y 2  K 2x2  4x  4  y2 1  K 2 x 2  1  K 2y 2  4K 2x  4K 2  0 Note: This is the equation of a circle! Section P.2 Section P.2 Linear Models and Rates of Change 1. m  1 3. m  0 7. 9. m  y m=1 5 4 2 1 5. m  12 2  4 53 11. m  6 3 2  y m = − 32 m is undefined 1 −1  (2, 3) 3 Linear Models and Rates of Change 51 22 4 0 undefined 3 x 3 4 2 5 y (5, 2) 1 m = −2 −1 6 x 1 2 3 5 6 4 −2 3 −3 −4 (2, 5) 5 7 2 (3, − 4) (2, 1) 1 −5 −2 −1 −1 x 1 3 4 5 6 −2 13. m  23  16 12  34 y 3 2 12  2 14 (− 12 , 23 ) −3 −2 (− 34 , 16 ) x 1 2 3 −1 −2 −3 15. Since the slope is 0, the line is horizontal and its equation is y  1. Therefore, three additional points are 0, 1, 1, 1, and 3, 1. 17. The equation of this line is y  7  3x  1 y  3x  10 . Therefore, three additional points are 0, 10, 2, 4, and 3, 1. 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 y-values to the change in x-values will always be the same. See Section P.2 Exercise 93 for a proof. 7 8 Chapter P Population (in millions) 21. (a) Preparation for Calculus (b) The slopes of the line segments are 270 255.0  252.1  2.9 21 260 257.7  255.0  2.7 32 250 260.3  257.7  2.6 43 1 2 3 4 5 6 7 8 9 Year (0 ↔ 1990) 262.8  260.3  2.5 54 265.2  262.8  2.4 65 267.7  265.2  2.5 76 270.3  267.7  2.6 87 The population increased most rapidly from 1991 to 1992. m  2.9 23. x  5y  20 y 25. x  4  15 x 4 Therefore, the slope is m  0, 4. 27. y  34 x  3 1 5 The line is vertical. Therefore, the slope is undefined and there is no y-intercept. and the y-intercept is y  23 x 29. 4y  3x  12 y  2  3x  3 31. y  2  3x  9 3y  2x 0  3x  4y  12 2x  3y  0 y  3x  11 y  3x  11  0 y y 5 4 4 3 y 3 (0, 3) 2 2 2 1 1 (0, 0) x x −4 −3 −2 −1 1 1 2 3 4 −2 −1 −1 x 1 2 4 5 6 (3, − 2) −2 −1 3 −3 −4 −5 33. m  60 3 20 35. m  y  0  3x  0 1  3 2 20 y  1  2x  4 8 3y  8x  40  0 y 6 (2, 6) 4 2 y (2, 1) 1 (0, 0) x 2 4 6 8 −2 −1 −1 x 2 −2 −8 40 8 y x 3 3 0  2x  y  3 y −8 −6 −4 −2 80 8  25 3 8 y  0   x  5 3 y  1  2x  2 y  3x 2 37. m  −3 −5 (0, −3) 3 4 5 9 8 7 6 5 4 3 2 1 −1 −2 (2, 8) (5, 0) x 1 2 3 4 6 7 8 9 Section P.2 81 55 39. m  Undefined. Vertical line x  5 41. m  Linear Models and Rates of Change 72  34 114 11   12  0 12 2 y x3 43. x30 3 11  x  0 4 2 y y 9 8 7 6 5 4 3 2 1 −1 y (5, 8) 11 3 x 2 4 2 1 22x  4y  3  0 −1 y (5, 1) x 1 2 3 4 (3, 0) 1 4 6 7 8 9 3 −2 −2 ( 12 , 72 ) 2 1 −4 −3 −2 −1 y x  1 2 3 45. ( 0, 34 ) x 1 2 3 4 47. 3x  2y  6  0 y x  1 a a 1 2  1 a a 3 1 a a3⇒xy3 xy30 49. 51. y  2x  1 y  3 y30 y 3 y 2 1 x −3 −2 −1 1 2 3 4 5 −2 −2 x −1 1 2 −1 −4 −5 −6 y  2  32x  1 53. y 3 2x  55. 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 −4 −3 −2 1 −2 −3 −4 2 3 4 3 3 2 4 x 9 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 1 are negative reciprocals of each other. You must use a square setting in order for perpendicular lines to appear perpendicular. 61. 5x  3y  0 59. 4x  2y  3 y  2x  2 y  53x m2 m  53 3 y  1  2x  2 (a) (a) 24y  21  40x  30 y  1  2x  4 24y  40x  9  0 2x  y  3  0 y1 (b) y  78  53x  34  1  2 x  2 (b) y  78   35x  34  40y  35  24x  18 2y  2  x  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  125t  1  2540  125t  2415 67. The slope is 2000. Hence, V  2000t  1  20,400  2000t  22,400 69. 5 (2, 4) −3 (0, 0) 6 −1 You can use the graphing utility to determine that the points of intersection are 0, 0 and 2, 4. Analytically, x2  4x  x2 2x2  4x  0 2xx  2  0 x  0 ⇒ y  0 ⇒ 0, 0 x  2 ⇒ y  4 ⇒ 2, 4. The slope of the line joining 0, 0 and 2, 4 is m  4  02  0  2. Hence, an equation of the line is y  0  2x  0 y  2x. Section P.2 71. m1  m2  Linear Models and Rates of Change 10  1 2  1 2  0 2  2  1 3 m1  m2 The points are not collinear. y 73. Equations of perpendicular bisectors: y y c ab ab x  2 c 2  c ba ab x  2 c 2    Letting x  0 in either equation gives the point of intersection: 0, a 2 (b, c) ( b −2 a , 2c ) ( a +2 b , 2c ) (− a, 0) x (a, 0)  b2  c2 . 2c  This point lies on the third perpendicular bisector, x  0. 75. Equations of altitudes: y y ab x  a c (b, c) xb y ab x  a c (a, 0) x (− a, 0) Solving simultaneously, the point of intersection is b, a 2  b2 . c  77. Find the equation of the line through the points 0, 32 and 100, 212. 9 m  180 100  5 9 F  32  5 C  0 F  95 C  32 5F  9C  160  0 For F  72, C  22.2. 79. (a) W1  0.75x  12.50 (b) 50 W2  1.30x  9.20 (c) Both jobs pay $17 per hour if 6 units are produced. For someone who can produce more than 6 units per hour, the second offer would pay more. For a worker who produces less than 6 units per hour, the first offer pays more. (6, 17) 0 30 0 Using a graphing utility, the point of intersection is approximately 6, 17. Analytically, 0.75x  12.50  1.30x  9.20 3.3  0.55x ⇒ x  6 y  0.756  12.50  17. 11 12 Chapter P Preparation for Calculus 81. (a) Two points are 50, 580 and 47, 625. The slope is m (b) 50 625  580  15. 47  50 p  580  15x  50 0 p  15x  750  580  15x  1330 1 If p  655, x  15 1330  655  45 units. 1 or x  15 1330  p 83. 4x  3y  10  0 ⇒ d  85. x  y  2  0 ⇒ d  1500 0 1 (c) If p  595, x  15 1330  595  49 units. 40  30  10  10  2 42  32 5 12  11  2  1  1 2 2 5 5 2  2 2 87. A point on the line x  y  1 is 0, 1. The distance from the point 0, 1 to x  y  5  0 is d 10  11  5  1  5  12  12 2 4 2  2 2. 89. If A  0, then By  C  0 is the horizontal line y  CB. The distance to x1, y1 is  By  C  Ax  By  C.  C B  A  B B  Ax  C  Ax  By  C.  C A  A  B A d  y1   1 1 1 2 2 If B  0, then Ax  C  0 is the vertical line x  CA. The distance to x1, y1 is d  x1   1 1 1 2 2 (Note that A and B cannot both be zero.) The slope of the line Ax  By  C  0 is AB. The equation of the line through x1, y1 perpendicular to Ax  By  C  0 is: y  y1  B x  x1 A Ay  Ay1  Bx  Bx1 Bx1  Ay1  Bx  Ay The point of intersection of these two lines is: Ax  By  C ⇒ Bx  Ay  Bx1  Ay1 ⇒ A2x  ABy  AC B 2x  ABy  B2x 1 (1)  ABy1 (2) A2  B2x  AC  B2x1  ABy1 (By adding equations (1) and (2)) x Ax  By  C ⇒ AC  B2x1  ABy1 A2  B2 ABx  B2y  BC Bx  Ay  Bx1  Ay1⇒ ABx  A2 y  ABx1  (3) A2 y1 (4) A2  B2y  BC  ABx1  A2y1 (By adding equations (3) and (4)) y —CONTINUED— BC  ABx1  A2y1 A2  B2 Section P.2 Linear Models and Rates of Change 89. —CONTINUED— Ay AC A Bx B ABy , BC A ABx  point of intersection B 2 1 2 1 2 1 2 2 1 2 The distance between x1, y1 and this point gives us the distance between x1, y1 and the line Ax  By  C  0. Ay AC A Bx B ABy  x  BC A ABx B AC  ABy  A x By   BC A ABx A B B 2 d 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  1 2 1 2 2 1 2 2 1  y1 2 2 2 AC A ByB Ax   BC A AxB By  A  B C  Ax  By   A  B   1 1 2 2 2 Ax1  By1  C A2  B2 91. For simplicity, let the vertices of the rhombus be 0, 0, a, 0, b, c, and a  b, c, as shown in the figure. The slopes of the diagonals are then m1  y (b, c) (a + b , c ) c c . and m2  ab ba x (0, 0) Since the sides of the Rhombus are equal, a2  b2  c2, and we have m1m2  c ab c c2 (a, 0) c2  b  a  b2  a2  c2  1. Therefore, the diagonals are perpendicular. 93. Consider the figure below in which the four points are collinear. Since the triangles are similar, the result immediately follows. y2  y1 y2  y1  x2  x1 x2  x1 95. True. a c a ax  by  c1 ⇒ y   x  1 ⇒ m1   b b b b c bx  ay  c2 ⇒ y  x  2 a a y m2   (x 2 , y2 ) (x *2 , y*2 ) (x1, y1 ) (x *1, y*1 ) x 1 m1 ⇒ m2  b a 13 14 Chapter P Preparation for Calculus Section P.3 Functions and Their Graphs 3. (a) g0  3  02  3 1. (a) f 0  20  3  3 (b) f 3  23  3  9 (b) g3  3  3  3  3  0 (c) f b  2b  3 (c) g2  3  22  3  4  1 (d) f x  1  2x  1  3  2x  5 (d) gt  1  3  t  12  t2  2t  2 5. (a) f 0  cos20  cos 0  1 (c) f 2  4   cos2 4   cos 2   0 (b) f  3   cos23   cos23   12 7. f x  x  f x x  x3  x3 x3  3x2x  3xx2  x3  x3    3x2  3xx  x2, x  0 x x x 9. f x  f 2 1x  1  1  x2 x2  1  x  1 1  x  1 2x 1   ,x2  x  2x  1 1  x  1 x  2x  11  x  1 x  11  x  1 11. hx   x  3 13. f t  sec Domain: x  3 ≥ 0 ⇒ 3,  Range:  , 0 t 4 t 2k  1  ⇒ t  4k  2 4 2 Domain: all t  4k  2, k an integer Range:  , 1, 1,  15. f x  1 x Domain:  , 0, 0,  Range:  , 0, 0,  17. f x  2x2x  1,2, xx <≥ 00 19. f x  x  1, x < 1 x  1, x ≥ 1 (a) f 1  21  1  1 (a) f 3  3  1  4 (b) f 0  20  2  2 (b) f 1  1  1  0 (c) f 2  22  2  6 (c) f 3  3  1  2 (d) f t  1  2t  1  2t  4 2 2 2 (d) f b  1   b  1  1  b (Note: t2  1 ≥ 0 for all t) Domain:  ,  Domain:  ,  Range:  , 0  1,  2 2 Range:  , 1, 2,  2 Section P.3 21. f x  4  x 23. hx  x  1 y Domain:  ,  8 6 Range:  ,  Functions and Their Graphs 15 y Domain: 1,  2 Range: 0,  1 x 2 1 2 3 x 4 2 2 25. f x  9  x2 4 27. gt  2 sin  t y 4 Domain: 3, 3 Domain:  ,  2 Range: 0, 3 2 2 2 1 Range: 2, 2 x 4 y t 4 2 −1 2 29. x  y 2  0 ⇒ y  ± x y is not a function of x. Some vertical lines intersect the graph twice. 33. x2  y2  4 ⇒ y  ± 4  x2 31. y is a function of x. Vertical lines intersect the graph at most once. 35. y2  x2  1 ⇒ y  ± x2  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. 37. f x  x  x  2 If x < 0, then f x  x  x  2  2x  2  21  x. If 0 ≤ x < 2, then f x  x  x  2  2. If x ≥ 2, then f x  x  x  2  2x  2  2x  1. Thus, 21  x, f x  2, 2x  1, x < 0 0 ≤ x < 2. x ≥ 2. 39. The function is gx  cx2. Since 1, 2 satisfies the equation, c  2. Thus, gx  2x2. 41. The function is rx  cx, since it must be undefined at x  0. Since 1, 32 satisfies the equation, c  32. Thus, rx  32x. 43. (a) For each time t, there corresponds a depth d. 45. (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 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 −6 −4 x −2 2 2 −2 −4 −4 −6 −6 y (c) The graph is shifted 2 units upward. x −2 4 −2 −4 4 6 2 4 6 −4 −6 x −2 2 4 6 −2 −8 y −4 2 −2 2 (e) The graph is stretched vertically by a factor of 3. 8 x −2 4 −4 6 y (d) The graph is shifted 4 units downward. 6 4 −2 4 y (f) The graph is stretched vertically by a factor of 14. x 6 −2 4 2 −4 −4 x −2 −6 −8 −6 − 10 49. (a) y  x  2 (b) y   x y y (c) y  x  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 −2 3 x 1 −1 4 Reflection about the x-axis Vertical shift 2 units upward Horizontal shift 2 units to the right 51. (a) T4  16 , T15 23 (b) If Ht  Tt  1, then the program would turn on (and off) one hour later. (c) If Ht  Tt  1, then the overall temperature would be reduced 1 degree. 53. f x  x2, gx  x  f gx  f gx  f  x    x   x, x ≥ 0 2 Domain: 0,  g f x  g f x  gx2  x2  x Domain:  ,  No. Their domains are different.  f g  g f  for x ≥ 0. 3 55. f x  , gx  x2  1 x  f gx  f gx  f x2  1  3 x2  1 Domain: all x  ± 1 g f x  g f x  g Domain: all x  0 No, f g  g f. 3x   3x  2 1 9 9  x2 1 2 x x2 Section P.3 57. A rt  Art  A0.6t  0.6t2  0.36t 2 Functions and Their Graphs 59. f x  x24  x2  x24  x2  f x A rt represents the area of the circle at time t. Even 61. f x  x cosx  x cos x  f x Odd 63. (a) If f is even, then  2 , 4 is on the graph. (b) If f is odd, then  2 , 4 is on the graph. 3 3 65. f x  a2n1x2n1  . . .  a3x3  a1x   a2n1x2n1  . . .  a3x3  a1x  f x Odd 67. Let F x  f xgx where f and g are even. Then F x  f xgx  f xgx  F x. Thus, F x is even. Let F x  f xgx where f and g are odd. Then F x  f xgx  f x gx  f xgx  F x. Thus, F x is even. 69. f x  x2  1 and gx  x4 are even. f xgx  x2  1x4  x6  x4 is even. f x  x3  x is odd and gx  x2 is even. f xgx  x3  xx2  x5  x3 is odd. 5 4 −6 −4 6 4 −4 −1 71. (a) x length and width volume V 1 24  21 484 2 24  22 800 3 24  23 972 4 24  24 1024 5 24  25 980 6 24  26 864 (b) 1200 0 The maximum volume appears to be 1024 cm3. (c) V  x24  2x2  4x12  x2 Domain: 0 < x < 12 73. False; let f x  x2. Then f 3  f 3  9, but 3  3. 7 0 Yes, V is a function of x. (d) 1100 −1 12 − 100 Maximum volume is V  1024 cm3 for box having dimensions 4 16 16 cm. 75. True, the function is even. 17 18 Chapter P Preparation for Calculus Section P.4 Fitting Models to Data 1. Quadratic function 3. Linear function 5. (a), (b) 7. (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. Yes. The cancer mortality increases linearly with increased exposure to the carcinogenic substance. (c) If F  55, then d  0.06655  3.63 cm. (c) If x  3, then y  136. 9. (a) Let x  per capita energy usage (in millions of Btu) 11. (a...
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