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  0225  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  x225  x2 17. y  x2  x  2 21. y  Graphs and Models x-intercepts: 0  x225  x2 0  x  2x  1 0  x25  x5  x x  2, 1; 2, 0, 1, 0 x  0, ± 5; 0, 0; ± 5, 0 32  x  x 23. x2y  x2  4y  0 y-intercept: None. x cannot equal 0. x-intercepts: 32  x 0 x y-intercept: 02y  02  4y  0 y  0; 0, 0 0  2  x x-intercept: x  4; 4, 0 x20  x2  40  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 y2  y2  x3  4x. 2 31. y  4  x  3 29. Symmetric with respect to the origin since xy  xy  4. No symmetry with respect to either axis or the origin.  x y  x2  1 y  35. y  x3  x is symmetric with respect to the y-axis 33. Symmetric with respect to the origin since       since y  x3  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  32 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  xx  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  2x  4x  6 (other answers possible) Graphs and Models 61. Some possible equations: yx y  x3 y  3x3  x 3 x y  63. xy2⇒y2x xy7⇒y7x 65. 2x  y  1 ⇒ y  2x  1 3x  2y  11 ⇒ y  2  x  2x  1 7x 3  3x 1x 3x  11 2 3x  11 2 14  2x  3x  11 The corresponding y-value is y  1. 5x  25 Point of intersection: 1, 1 x5 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 xy1⇒yx1 xy4⇒y4x 5  x2  x  12 6  x2  4  x 0  x2  x  2 5  x2  x2  2x  1 0  x  2x  1 0  2x2  2x  4  2x  1x  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. yx y  x2  3x  1 x3  x x3  2x2  x  1  x2  3x  1 x3  x  0 xx  1x  1  0 x3  x2  2x  0 xx  2x  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. 5.5x  10,000  3.29x  5.5x  2  3.29x  10,0002 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 14. For instance, y20  26.555 and y40  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  Kx  22  y2, K  1 x2  y 2  K 2x2  4x  4  y2 1  K 2 x 2  1  K 2y 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 53 11. m  6 3 2  y m = − 32 m is undefined 1 −1  (2, 3) 3 Linear Models and Rates of Change 51 22 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  23  16 12  34 y 3 2 12  2 14 (− 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  3x  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 21 260 257.7  255.0  2.7 32 250 260.3  257.7  2.6 43 1 2 3 4 5 6 7 8 9 Year (0 ↔ 1990) 262.8  260.3  2.5 54 265.2  262.8  2.4 65 267.7  265.2  2.5 76 270.3  267.7  2.6 87 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  3x  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  60 3 20 35. m  y  0  3x  0 1  3 2 20 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 80 8  25 3 8 y  0   x  5 3 y  1  2x  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 81 55 39. m  Undefined. Vertical line x  5 41. m  Linear Models and Rates of Change 72  34 114 11   12  0 12 2 y x3 43. x30 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 a3⇒xy3 xy30 49. 51. y  2x  1 y  3 y30 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  32x  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 m2 m  53 3 y  1  2x  2 (a) (a) 24y  21  40x  30 y  1  2x  4 24y  40x  9  0 2x  y  3  0 y1 (b) y  78  53x  34  1  2 x  2 (b) y  78   35x  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  125t  1  2540  125t  2415 67. The slope is 2000. Hence, V  2000t  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 2xx  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  02  0  2. Hence, an equation of the line is y  0  2x  0 y  2x. Section P.2 71. m1  m2  Linear Models and Rates of Change 10  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 ab ab x  2 c 2  c ba ab 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 ab x  a c (b, c) xb y ab 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  0.756  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  15x  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. 40  30  10  10  2 42  32 5 12  11  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 10  11  5  1  5  12  12 2 4 2  2 2. 89. If A  0, then By  C  0 is the horizontal line y  CB. 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  CA. 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 AB. 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  B2x  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  B2y  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 Bx 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 Bx 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 AC A ByB Ax   BC 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  ab ba x (0, 0) Since the sides of the Rhombus are equal, a2  b2  c2, and we have m1m2  c ab 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) g0  3  02  3 1. (a) f 0  20  3  3 (b) f 3  23  3  9 (b) g3  3  3  3  3  0 (c) f b  2b  3 (c) g2  3  22  3  4  1 (d) f x  1  2x  1  3  2x  5 (d) gt  1  3  t  12  t2  2t  2 5. (a) f 0  cos20  cos 0  1 (c) f 2  4   cos2 4   cos 2   0 (b) f  3   cos23   cos23   12 7. f x  x  f x x  x3  x3 x3  3x2x  3xx2  x3  x3    3x2  3xx  x2, x  0 x x x 9. f x  f 2 1x  1  1  x2 x2  1  x  1 1  x  1 2x 1   ,x2  x  2x  1 1  x  1 x  2x  11  x  1 x  11  x  1 11. hx   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  21  1  1 (a) f 3  3  1  4 (b) f 0  20  2  2 (b) f 1  1  1  0 (c) f 2  22  2  6 (c) f 3  3  1  2 (d) f t  1  2t  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. hx  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. gt  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  21  x. If 0 ≤ x < 2, then f x  x  x  2  2. If x ≥ 2, then f x  x  x  2  2x  2  2x  1. Thus, 21  x, f x  2, 2x  1, x < 0 0 ≤ x < 2. x ≥ 2. 39. The function is gx  cx2. Since 1, 2 satisfies the equation, c  2. Thus, gx  2x2. 41. The function is rx  cx, since it must be undefined at x  0. Since 1, 32 satisfies the equation, c  32. Thus, rx  32x. 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) T4  16 , T15 23 (b) If Ht  Tt  1, then the program would turn on (and off) one hour later. (c) If Ht  Tt  1, then the overall temperature would be reduced 1 degree. 53. f x  x2, gx  x  f gx  f gx  f  x    x   x, x ≥ 0 2 Domain: 0,  g f x  g f x  gx2  x2  x Domain:  ,  No. Their domains are different.  f g  g f  for x ≥ 0. 3 55. f x  , gx  x2  1 x  f gx  f gx  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 rt  Art  A0.6t  0.6t2  0.36t 2 Functions and Their Graphs 59. f x  x24  x2  x24  x2  f x A rt represents the area of the circle at time t. Even 61. f x  x cosx  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  a2n1x2n1  . . .  a3x3  a1x   a2n1x2n1  . . .  a3x3  a1x  f x Odd 67. Let F x  f xgx where f and g are even. Then F x  f xgx  f xgx  F x. Thus, F x is even. Let F x  f xgx where f and g are odd. Then F x  f xgx  f x gx  f xgx  F x. Thus, F x is even. 69. f x  x2  1 and gx  x4 are even. f xgx  x2  1x4  x6  x4 is even. f x  x3  x is odd and gx  x2 is even. f xgx  x3  xx2  x5  x3 is odd. 5 4 −6 −4 6 4 −4 −1 71. (a) x length and width volume V 1 24  21 484 2 24  22 800 3 24  23 972 4 24  24 1024 5 24  25 980 6 24  26 864 (b) 1200 0 The maximum volume appears to be 1024 cm3. (c) V  x24  2x2  4x12  x2 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  0.06655  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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