Exam (elaborations) TEST BANK FOR Calculus 8th Editon (3 Volume Set) By Ron Larson, Robert P. Hostetler and Bruce H. Edwards (Study And Solutions Guide)

Exam (elaborations) TEST BANK FOR Calculus 8th Editon (3 Volume Set) By Ron Larson, Robert P. Hostetler and Bruce H. Edwards (Study And Solutions Guide) 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 P A R T I C H A P T E R P Preparation for Calculus Section P.1 Graphs and Models Solutions to Odd-Numbered Exercises 2 1. x-intercept: y-intercept: Matches graph (b) 0, 2 4, 0 y  12 x  2 3. x-intercepts: y-intercept: Matches graph (a) 0, 4 2, 0, 2, 0 y  4  x2 5. x 2 2 −4 −6 −8 4 6 8 −8 −6 −4 4 6 8 (−4, −5) (−2, −2) (0, 1) (2, 4) (4, 7) y y  32 x  1 7. x 2 −4 −2 −6 6 −6 −4 4 6 (−3, −5) (3, −5) (−2, 0) (0, 4) (2, 0) y y  4  x2 x 0 2 4 y 5 2 1 4 7 4 2 x 0 2 3 y 5 0 4 0 5 3 2 9. x 2 −2 4 6 −6 −4 2 (−3, 1) (−1, 1) (−4, 2) (−2, 0) (0, 2) (1, 3) (−5, 3) y y  x  2 11. x 4 6 8 2 −6 −8 −4 −10 10 − 18 (0, −4) (1, −3) (4, −2) (16, 0) y (9, −1) y  x  4 x 0 1 y 3 2 1 0 1 2 3 5 4 3 2 1 x 0 1 4 9 16 y 4 3 2 1 0 Section P.1 Graphs and Models 3 13. Note that y  4 when x  0. Xmin = -3 Xmax = 5 Xscl = 1 Ymin = -3 Ymax = 5 Yscl = 1 15. (a) (b) x, 3  4, 3 3  5  4 2, y  2, 1.73 y  5  2  3  1.73 −6 6 −3 5 (−4.00, 3) (2, 1.73) 17. y-intercept: x-intercepts: x  2, 1; 2, 0, 1, 0 0  x  2x  1 0  x2  x  2 y  2; 0, 2 y  02  0  2 y  x2  x  2 19. y-intercept: x-intercepts: x  0, ±5; 0, 0; ±5, 0 0  x25  x5  x 0  x225  x2 y  0; 0, 0 y  0225  02 y  x225  x2 21. y-intercept: None. x cannot equal 0. x-intercepts: x  4; 4, 0 0  2  x 0  32  x x y  32  x x 23. y-intercept: x-intercept: x  0; 0, 0 x20  x2  40  0 y  0; 0, 0 02y  02  4y  0 x2y  x2  4y  0 25. Symmetric with respect to the y-axis since y  x2  2  x2  2. 27. Symmetric with respect to the x-axis since y2  y2  x3  4x. 29. Symmetric with respect to the origin since xy  xy  4. 31. No symmetry with respect to either axis or the origin. y  4  x  3 33. Symmetric with respect to the origin since y  x x2  1 . y  x x2  1 35. is symmetric with respect to the y-axis since y  x3  x  x3  x  x3  x. y  x3  x 37. Intercepts: Symmetry: none 23 , 0, 0, 2 0 y 2 (0, 2) 1 x 1 2 3 1 , 23 y  3x  2 4 Chapter P Preparation for Calculus 39. Intercepts: Symmetry: none x 8 10 (8, 0) y 2 4 , 4) 2 ( 2 0 2 8 6 10 8, 0, 0, 4 y  x 2  4 41. Intercepts: Symmetry: y-axis x (1, 0) 0, 1) y 2 ( 1, 0) ( 1 2 −2 2 1, 0, 1, 0, 0, 1 y  1  x2 43. Intercepts: Symmetry: none x 2 −2 8 10 12 −10 −8 −6 (−3, 0) 2 4 (0, 9) y 3, 0, 0, 9 y  x  32 45. Intercepts: Symmetry: none y 5 4 3 3 x 2 3 , 1 1 1 , 3 2 ( 2 0) (0 2) 3 2, 0, 0, 2 y  x3  2 47. Intercepts: Symmetry: none Domain: x 1 2 3 4 5 −2 6 −4 −3 −1 1 2 3 4 (−2, 0) (0, 0) y x ≥ 2 0, 0, 2, 0 y  xx  2 49. Intercepts: Symmetry: origin x 1 −2 −3 −4 2 3 4 −4 −3 −2 −1 2 3 4 (0, 0) y 0, 0 x  y3 51. Intercepts: none Symmetry: origin y 3 1 2 x 1 2 3 y  1 x 53. Intercepts: Symmetry: y-axis 0, 6, 6, 0, 6, 0 x 2 2 −4 −2 −6 −8 4 6 8 −8 −4 −2 4 6 8 (−6, 0) (0, 6) (6, 0) y  6  x y 55. Intercepts: Symmetry: x-axis 0, 3, 0, 3, 9, 0 1 −4 −11 4 (−9, 0) (0, −3) (0, 3) y  ±x  9 y2  x  9 y2  x  9 57. Intercepts: Symmetry: x-axis 6, 0, 0, 2, 0, 2 8 −3 −1 3 (6, 0) (0 , 2 ) ( 0 , − 2 ) y  ±2  x 3 3y2  6  x x  3y2  6 Section P.1 Graphs and Models 5 59. y  x  2x  4x  6 (other answers possible) 61. Some possible equations: y  3 x y  3x3  x y  x3 y  x 63. The corresponding y-value is Point of intersection: 1, 1 y  1. 1  x 3  3x 2  x  2x  1 2x  y  1 ⇒y  2x  1 x  y  2 ⇒y  2  x 65. The corresponding y-value is Point of intersection: 5, 2 y  2. x  5 5x  25 14  2x  3x  11 7  x  3x  11 2 3x  2y  11 ⇒y  3x  11 2 x  y  7 ⇒y  7  x 67. The corresponding y-values are (for ) and (for ). Points of intersection: 2, 2, 1, 5 y  5 x  1 y  2 x  2 x  2, 1 0  x  2x  1 0  x2  x  2 6  x2  4  x x  y  4 ⇒y  4  x x2  y  6 ⇒y  6  x2 69. The corresponding y-values are and Points of intersection: 1, 2, 2, 1 y  2 y  1. x  1 or x  2 0  2x2  2x  4  2x  1x  2 5  x2  x2  2x  1 5  x2  x  12 x  y  1 ⇒y  x  1 x2  y2  5 ⇒y2  5  x2 71. The corresponding y-values are and Points of intersection: 0, 0, 1, 1, 1, 1 y  1. y  0, y  1, x  0, x  1, or x  1 xx  1x  1  0 x3  x  0 x3  x y  x y  x3 73. −4 6 −8 4 y = −x 2 + 3x − 1 y = x 3 − 2x 2 + x − 1 (2, 1) (0, −1) (− 1 , −5) 1, 5, 0, 1, 2, 1 x  1, 0, 2 xx  2x  1  0 x3  x2  2x  0 x3  2x2  x  1  x2  3x  1 y  x2  3x  1 y  x3  2x2  x  1 6 Chapter P Preparation for Calculus 75. The other root, does not satisfy the equation This problem can also be solved by using a graphing utility and finding the intersection of the graphs of C and R. x  2949, R  C. x  3133 units 0  10.8241x2  65,830.25x  100,000,000 Use the Quadratic Formula. 30.25x  10.8241x2  65,800x  100,000,000 5.5x2  3.29x  10,0002 5.5x  10,000  3.29x 77. (a) Using a graphing utility, you obtain (c) For the year 2004, and y  187.2 CPI. t  34 y  0.0153t2  4.9971t  34.9405 (b) 35 −50 −5 250 79. If the diameter is doubled, the resistance is changed by approximately a factor of For instance, and y40  6.36125. 14. y20  26.555 0 0 400 100 85. Distance to the origin Distance to Note: This is the equation of a circle! 1  K2x 2  1  K2y 2  4K2x  4K2  0 x2  y2  K2x2  4x  4  y2 x2  y2  Kx  22  y2, K  1  K  2, 0 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 . Section P.2 Linear Models and Rates of Change Section P.2 Linear Models and Rates of Change 7 1. m  1 3. m  0 5. m  12 7. 2 3 4 5 −1 1 x 1 3 4 5 (2, 3) m = −2 m = 1 32 m = − m is undefined y 9. x y −1 1 2 3 5 6 7 (3, −4) (5, 2) −2 −3 −4 −5 1 2 3  6 2  3 m  2  4 5  3 11. undefined x 1 2 −1 −2 3 4 5 6 −2 −1 1 3 4 5 6 (2, 1) (2, 5) y  4 0 m  5  1 2  2 13.  12 14  2 m  23  16 12  34 x −1 −2 −3 2 3 −3 −2 1 2 3 y ( ) 12 23 , − ( ) 34 16 − , 15. Since the slope is 0, the line is horizontal and its equation is Therefore, three additional points are and 3, 1. y  1. 0, 1, 1, 1, 17. The equation of this line is . Therefore, three additional points are 0, 10, 2, 4, and 3, 1. y  3x  10 y  7  3x  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. 8 Chapter P Preparation for Calculus 21. (a) 270 260 250 2 3 4 5 6 7 8 9 Year (0 ↔ 1990) Population (in millions) 1 (b) The slopes of the line segments are The population increased most rapidly from 1991 to 1992. m  2.9 270.3  267.7 8  7  2.6 267.7  265.2 7  6  2.5 265.2  262.8 6  5  2.4 262.8  260.3 5  4  2.5 260.3  257.7 4  3  2.6 257.7  255.0 3  2  2.7 255.0  252.1 2  1  2.9 23. Therefore, the slope is and the y-intercept is 0, 4. m  15 y  15 x  4 x  5y  20 25. The line is vertical. Therefore, the slope is undefined and there is no y-intercept. x  4 27. x y −4 −3 −2 −1 1 1 2 4 5 (0, 3) 0  3x  4y  12 4y  3x  12 y  34 x  3 29. x y 1 2 3 4 −1 2 3 4 (0, 0) 2x  3y  0 3y  2x y  23 x 31. x 1 2 −3 −1 −2 −4 −5 3 −2 −1 1 2 3 4 5 6 (3, −2) y y  3x  11  0 y  3x  11 y  2  3x  9 y  2  3x  3 33. x 2 2 −8 4 6 8 −8 −6 −4 −2 4 6 8 (2, 6) (0, 0) y y  3x y  0  3x  0 m  6  0 2  0  3 35. x y −2 −1 2 3 4 5 −1 −2 −3 −5 1 2 (2, 1) (0, −3) 0  2x  y  3 y  1  2x  4 y  1  2x  2 m  1  3 2  0  2 37. 1 2 3 4 5 −2 6 7 8 9 x −1 1 2 3 4 6 7 8 9 (5, 0) (2, 8) y 3y  8x  40  0 y   8 3 x  40 3 y  0   8 3 x  5 m  8  0 2  5   8 3 Section P.2 Linear Models and Rates of Change 9 39. Undefined. Vertical line 1 2 3 4 5 −2 6 7 8 9 x −1 1 2 3 4 6 7 8 9 (5, 1) (5, 8) y x  5 m  8  1 5  5 41. x 1 2 1 3 4 −4 −3 −2 −1 2 3 4 ( ) 12 72 , ( ) 34 0, y 22x  4y  3  0 y  11 2 x  3 4 y  3 4  11 2 x  0 m  72  34 12  0  114 12  11 2 43. 1 2 4 −1 −2 1 2 (3, 0) y x x  3  0 x  3 45. 3x  2y  6  0 x 2  y 3  1 47. x  y  3  0 a  3 ⇒x  y  3 3 a  1 1 a  2 a  1 x a  y a  1 49. x y −3 −2 −1 1 2 3 4 5 −2 −4 −5 −6 1 2 y  3  0 y  3 51. x y −2 −1 1 2 −1 3 y  2x  1 53. x 1 2 1 3 4 −2 −2 −3 −4 −4 −3 2 3 4 y 2y  3x  1  0 y  32 x  12 y  2  32 x  1 55. y 1 x 2 3 1 2 3 2 1 y  2x  3 2x  y  3  0 10 Chapter P Preparation for Calculus 57. The lines do not appear perpendicular. 10 −10 −10 10 The lines appear perpendicular. 15 −10 −15 10 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. 1 59. (a) (b) x  2y  4  0 2y  2  x  2 y  1  12 x  2 2x  y  3  0 y  1  2x  4 y  1  2x  2 m  2 y  2x  32 4x  2y  3 61. (a) (b) 40y  24x  53  0 40y  35  24x  18 y  78  35 x  34  24y  40x  9  0 24y  21  40x  30 y  78  53 x  34  m  53 y  53 x 5x  3y  0 63. (a) (b) y  5 ⇒y  5  0 x  2 ⇒x  2  0 65. The slope is 125. Hence,  125t  2415 V  125t  1  2540 67. The slope is Hence,  2000t  22,400 2000. V  2000t  1  20,400 69. You can use the graphing utility to determine that the points of intersection are and Analytically, The slope of the line joining and is Hence, an equation of the line is y  2x. y  0  2x  0 0, 0 2, 4 m  4  02  0  2. x  2 ⇒y  4 ⇒2, 4. x  0 ⇒y  0 ⇒0, 0 2xx  2  0 2x2  4x  0 x2  4x  x2 0, 0 2, 4. −1 −3 6 (0, 0) (2, 4) 5 Section P.2 Linear Models and Rates of Change 11 71. The points are not collinear. m1  m2 m2  2  0 2  1   2 3 m1  1  0 2  1  1 73. Equations of perpendicular bisectors: Letting in either equation gives the point of intersection: This point lies on the third perpendicular bisector, x  0. 0, a2  b2  c2 2c . x  0 y  c 2  a  b c x  b  a 2  y  c 2  a  b c x  a  b 2  ( b , c) (−a, 0) ( a , 0) a + b 2 b − a 2 c 2 c ( 2 ) ( ) , , y x 75. Equations of altitudes: Solving simultaneously, the point of intersection is b, a2  b2 c . y   a  b c x  a x  b y  a  b c x  a ( b , c) ( a , 0) (−a, 0) y x 77. Find the equation of the line through the points and For F  72, C  22.2. 5F  9C  160  0 F  95 C  32 F  32  95 C  0 m  180 100  95 0, 32 100, 212. 79. (a) (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. W2  1.30x  9.20 W1  0.75x  12.50 (b) Using a graphing utility, the point of intersection is approximately Analytically, y  0.756  12.50  17. 3.3  0.55x⇒x  6 0.75x  12.50  1.30x  9.20 6, 17. 0 0 30 50 (6, 17) 12 Chapter P Preparation for Calculus 81. (a) Two points are and The slope is or x  1 15 1330  p p  15x  750  580  15x  1330 p  580  15x  50 m  625  580 47  50  15. 50, 580 47, 625. (b) If (c) If p  595, x  1 15 1330  595  49 units. p  655, x  1 15 1330  655  45 units. 1500 0 0 50 83. 4x  3y  10  0 ⇒d  40  30  10 42  32  10 5  2 85. x  y  2  0 ⇒d  12  11  2 12  12  5 2  5 2 2 87. A point on the line is The distance from the point to is d  10  11  5 12  12  1  5 2  4 2  2 2. x  y  1 0, 1. 0, 1 x  y  5  0 89. If then is the horizontal line The distance to is If then is the vertical line The distance to is (Note that A and B cannot both be zero.) The slope of the line is The equation of the line through perpendicular to is: The point of intersection of these two lines is: (1) (2) (By adding equations (1) and (2)) (3) (4) (By adding equations (3) and (4)) —CONTINUED— y  BC  ABx1  A2y1 A2  B2 A2  B2y  BC  ABx1  A2y1 Bx  Ay  Bx1  Ay1⇒ABx  A2 y  ABx1  A2 y1 Ax  By  C ⇒ ABx  B2y  BC x  AC  B2x1  ABy1 A2  B2 A2  B2x  AC  B2x1  ABy1 Bx  Ay  Bx1  Ay1⇒ B2x  ABy  B2x1  ABy1 Ax  By  C ⇒ A2x  ABy  AC Bx1  Ay1  Bx  Ay Ay  Ay1  Bx  Bx1 y  y1  B A x  x1 Ax  By  C  0 x1, y1Ax  By  C  0 AB.  d  x1  C A  Ax1  C A  Ax1  By1  C A2  B2 . x1, y1B  0, Ax  C  0 x  CA.  d  y1  C B  By1  C B  Ax1  By1  C A2  B2 . x1, y1A  0, By  C  0 y  CB.  Section P.2 Linear Models and Rates of Change 13 89. —CONTINUED— point of intersection The distance between and this point gives us the distance between and the line  Ax1  By1  C A2  B2  A2  B2C  Ax1  By12 A2  B22  AC  By1  Ax1 A2  B2 2  BC  Ax1  By1 A2  B2 2  AC  ABy1  A2x1 A2  B2 2  BC  ABx1  B2y1 A2  B2 2 d  AC  B2x1  ABy1 A2  B2  x12  BC  ABx1  A2y1 A2  B2  y12 x1 Ax  By  C  0. , y1x1  , y1 AC  B2x1  ABy1 A2  B2 , BC  ABx1  A2y1 A2  B2  91. For simplicity, let the vertices of the rhombus be and as shown in the figure. The slopes of the diagonals are then and Since the sides of the Rhombus are equal, and we have Therefore, the diagonals are perpendicular. m1m2  c a  b  c b  a  c2 b2  a2  c2 c2  1. a2  b2  c2, m2  c b  a m1  . c a  b a, 0, b, c, a  b, c, 0, 0, ( a , 0) ( b , c) (a + b , c) (0, 0) y x 93. Consider the figure below in which the four points are collinear. Since the triangles are similar, the result immediately follows. (x , y ) 1 1 (x , y ) 1 * 1 * (x , y ) 2 2 (x , y ) 2 * 2 * y x y2   y1  x2   x1   y2  y1 x2  x1 95. True. m2   1 m1 bx  ay  c2⇒y  b a x  c2 a ⇒m2  b a ax  by  c1⇒y   a b x  c1 b ⇒m1   a b Section P.3 Functions and Their Graphs 14 Chapter P Preparation for Calculus 1. (a) (b) (c) (d) f x  1  2x  1  3  2x  5 f b  2b  3 f 3  23  3  9 f 0  20  3  3 3. (a) (b) (c) (d) gt  1  3  t  12  t2  2t  2 g2  3  22  3  4  1 g3  3  32  3  3  0 g0  3  02  3 5. (a) (c) f  3   cos2 3   cos 2 3   1 2 f 0  cos20  cos 0  1 (b) f   4   cos2  4   cos  2   0 7. f x  x  f x x  x  x3  x3 x  x3  3x2x  3xx2  x3  x3 x  3x2  3xx  x2, x  0 9.  1  x  1 x  2x  1  1  x  1 1  x  1  2  x x  2x  11  x  1  1 x  11  x  1 , x  2 f x  f 2 x  2  1x  1  1 x  2 11. Domain: Range:  , 0 x  3 ≥ 0 ⇒ 3,  hx  x  3 13. Domain: all k an integer Range:  , 1, 1,  t  4k  2, t 4  2k  1 2 ⇒t  4k  2 f t  sec t 4 15. Domain: Range:  , 0, 0,   , 0, 0,  f x  1 x 17. (a) (b) (c) (d) (Note: for all t) Domain: Range:  , 1, 2,   ,  t2  1 ≥ 0 f t2  1  2t2  1  2t2  4 f 2  22  2  6 f 0  20  2  2 f 1  21  1  1 f x  2x  1, 2x  2, x 0 x ≥ 0 19. (a) (b) (c) (d) Domain: Range:  , 0  1,   ,  f b2  1  b2  1  1  b2 f 3  3  1  2 f 1  1  1  0 f 3  3  1  4 f x  x  1, x  1, x 1 x ≥ 1 Section P.3 Functions and Their Graphs 15 21. Domain: Range:  ,   ,  8 y x 2 4 6 2 4 2 f x  4  x 23. Domain: Range: 0,  1,  y x 1 2 3 2 1 hx  x  1 25. Domain: Range: 0, 3 3, 3 y x 2 4 2 4 2 4 2 f x  9  x2 27. Domain: Range: 2, 2  ,  t y 2 3 −1 1 2 gt  2 sin t 29. y is not a function of x. Some vertical lines intersect the graph twice. x  y 2  0 ⇒y  ±x 31. y is a function of x. Vertical lines intersect the graph at most once. 33. y is not a function of x since there are two values of y for some x. x2  y2  4 ⇒y  ±4  x2 35. y is not a function of x since there are two values of y for some x. y2  x2  1 ⇒y  ±x2  1 37. If then If then If then Thus, f x  21  x, 2, 2x  1, x 0 0 ≤ x 2 . x ≥ 2. x ≥ 2, f x  x  x  2  2x  2  2x  1. 0 ≤ x 2, f x  x  x  2  2. x 0, f x  x  x  2  2x  2  21  x. f x  x  x  2 39. The function is Since satisfies the equation, c  2. Thus, gx  2x2. gx  cx2. 1, 2 41. The function is since it must be undefined at Since satisfies the equation, Thus, rx  32x. x  0. 1, 32 c  32. rx  cx, 43. (a) For each time t, there corresponds a depth d. (b) Domain: Range: (c) t d 5 10 15 20 25 30 1 2 3 4 5 6 0 ≤ d ≤ 30 0 ≤ t ≤ 5 45. t 27 t t t 1 2 3 9 18 d 16 Chapter P Preparation for Calculus 47. (a) The graph is shifted 3 units to the left. x −6 −2 −4 4 −6 −4 −2 2 4 y (b) The graph is shifted 1 unit to the right. x −6 −2 −4 4 2 −2 2 4 6 8 y (c) The graph is shifted 2 units upward. x −2 4 6 2 −4 −2 2 4 6 y (d) The graph is shifted 4 units downward. x −2 −8 −6 −4 −4 −2 2 4 6 y (e) The graph is stretched vertically by a factor of 3. x −2 −8 −10 −6 −4 −4 −2 4 6 y (f) The graph is stretched vertically by a factor of 14 . x −6 4 2 −4 −2 2 4 6 y 49. (a) Vertical shift 2 units upward 4 3 2 1 1 2 3 4 y x y  x  2 (b) Reflection about the x-axis 1 2 3 4 1 2 1 3 x y y  x (c) Horizontal shift 2 units to the right 3 4 6 3 2 1 1 −1 −2 2 4 5 x y y  x  2 51. (a) (b) If 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. Ht  Tt  1, T4  16 , T15  23 53. Domain: Domain: No. Their domains are different.  f g  g f  for x ≥ 0.  ,  g f x  g f x  gx2  x2  x 0,   f gx  f gx  f x  x2  x, x ≥ 0 f x  x2, gx  x 55. Domain: all Domain: all No, f g  g f. x  0 g f x  g f x  g 3 x  3 x2  1  9 x2  1  9  x2 x2 x  ±1  f gx  f gx  f x2  1  3 x2  1 f x  3 x , gx  x2  1 Section P.3 Functions and Their Graphs 17 57. A rt represents the area of the circle at time t. A rt  Art  A0.6t  0.6t2  0.36t 2 59. Even f x  x24  x2  x24  x2  f x 61. Odd f x  x cosx  x cos x  f x 63. (a) If f is even, then is on the graph. 32 , 4 (b) If f is odd, then is on the graph. 32 , 4 65. Odd  f x   a2n1x2n1  . . .  a3x3  a1x f x  a2n1x2n1  . . .  a3x3  a1x 67. Let where f and g are even. Then . Thus, is even. Let where f and g are odd. Then Thus, F x is even. F x  f xgx  f x gx  f xgx  F x. F x F x  f xgx F x  f xgx  f xgx  F x F x  f xgx 69. and are even. is even. −4 4 −1 5 f xgx  x2  1x4  x6  x4 f x  x2  1 gx  x4 is odd and is even. is odd. −6 6 −4 4 f xgx  x3  xx2  x5  x3 f x  x3  x gx  x2 71. (a) The maximum volume appears to be 1024 cm3. (b) Yes, V is a function of x. 0 0 7 1200 (c) Domain: 0 x 12 V  x24  2x2  4x12  x2 (d) Maximum volume is for box having dimensions 4  16  16 cm. V  1024 cm3 12 −100 −1 1100 x length and width volume V 1 484 2 800 3 972 4 1024 5 980 6 24  26 864 24  25 24  24 24  23 24  22 24  21 73. False; let Then f 3  f 3  9, but 3  3. f x  x2. 75. True, the function is even. Section P.4 Fitting Models to Data 18 Chapter P Preparation for Calculus 1. Quadratic function 3. Linear function 5. (a), (b) Yes. The cancer mortality increases linearly with increased exposure to the carcinogenic substance. (c) If x  3, then y  136. x y 3 6 9 12 15 50 100 150 200 250 7. (a) (b) The model fits well. (c) If F  55, then d  0.06655  3.63 cm. 0 0 10 125 F = 15.13 d + 0.10 d  0.066F or F  15.1d  0.1 9. (a) Let per capita energy usage (in millions of Btu) per capita gross national product (in thousands) (b) (c) Denmark, Japan, and Canada (d) Deleting the data for the three countries above, (r  0.9202 is much closer to 1.) y  0.0959x  1.0539 420 0 0 y = 0.08x + 5.0 40 r  0.7052 y  0.0764x  4.9985  0.08x  5.0 y  x  11. (a) (b) For 2002, and y1  y2  y3 t  12  31.06 centsmile 8 0 0 y1 y2 y3 y1 + y2 + y3 15 y3  0.0917t  0.7917 y2  0.1095t  2.0667 y1  0.0343t3  0.3451t2  0.8837t  5.6061 13. (a) (b) (c) The cubic model is better. 8 25 0 y2 = −0.01t3 + 0.55t2 + 0.24t + 33.14 y1 = 4.04t + 28.96 70 y2  0.0099t3  0.5488t2  0.2399t  33.1414 y1  4.0367t  28.9644 (d) (e) The slope represents the average increase per year in the number of people (in millions) in HMOs. (f) For 2000, and million. (linear) y2 million (cubic)  80.5 y1 t  10,  69.3 8 25 0 70 y3  0.4297t2  0.5994t  32.9745 Review Exercises for Chapter P Review Exercises for Chapter P 19 15. (a) (b) (c) If x  4.5, y  214 horsepower. 0 0 7 300 y  1.81x3  14.58x2  16.39x  10 17. (a) Yes, y is a function of t. At each time t, there is one and only one displacement y. (b) The amplitude is approximately The period is approximately (c) One model is . (d) 0 0 4 0.9 y  0.35 sin4t  2 20.375  0.125  0.5. 2.35  1.652  0.35. 19. Answers will vary. 1. y-intercept y  0 ⇒ 0  2x  3 ⇒x  x-intercept 32 ⇒32 , 0 x  0 ⇒y  20  3  3 ⇒0, 3 y  2x  3 3. y-intercept y  0 ⇒ 0  x-intercept x  1 x  2 ⇒x  1 ⇒1, 0 x  0 ⇒y  0  1 0  2  1 2 ⇒0, 1 2 y  x  1 x  2 5. Symmetric with respect to y-axis since x2y  x2  4y  0. x2y  x2  4y  0 7. 3 y x 1 2 3 1 2 1 2 y  12 x  32 9. Slope: y-intercept: 2 3 1 3 2 1 1 x y 65 2 5 y  25 x  65 25 x  y  65 13 x  56 y  1 11. y x 5 5 10 5 y  7  6x  x2 20 Chapter P Preparation for Calculus 13. Domain: 5 4 3 2 1 2 3 4 5 1 y x , 5 y  5  x 15. Xmin = -5 Xmax = 5 Xscl = 1 Ymin = -30 Ymax = 10 Yscl = 5 y  4x2  25 17. Point: 4, 1 y  1 x  4 7x  28 4x  4y  20 3x  4y  8 19. You need factors and Multiply by x to obtain origin symmetry .  x3  4x. y  xx  2x  2 x  2 x  2. 21. Slope  52  1 5  32  32 72  3 7 x y 1 2 3 4 5 1 2 3 4 5 3 , 1 2 5 2 (5, ) ( ) 23. t  7 3 1  t   4 3 1  t 1  0  1  5 1  2 25. y x (0, −5) −4 −4 −8 −2 −2 2 2 4 4 6 8 2y  3x  10  0 y  32 x  5 y  5  32 x  0 27. y x −8 −8 −6 −4 −2 2 4 −6 −4 2 4 (−3, 0) 3y  2x  6  0 y  23 x  2 y  0  23 x  3 Review Exercises for Chapter P 21 29. (a) (c) 2x  y  0 y  2x m  4  0 2  0  2 0  7x  16y  78 16y  64  7x  14 y  4  7 16 x  2 (b) Slope of line is (d) x  2  0 x  2 0  5x  3y  22 3y  12  5x  10 y  4  5 3 x  2 5 3 . 33. Not a function of x since there are two values of y for some x. x y 1 2 3 4 5 6 −1 −2 −3 1 2 3 y  ± x x  y2  0 35. Function of x since there is one value of y for each x. x y −2 −1 3 4 −2 3 4 y  x2  2x 37. (a) does not exist. (b)  1 1  x , x  1, 0 f 1  x  f 1 x  1 1  x  1 1 x  1  1  x 1  xx f 0 f x  1 x 39. (a) Domain: Range: (b) Domain: all Range: all (c) Domain: all x or Range: all y or ,  ,  y  0 or , 0, 0,  x  5 or , 5, 5,  0, 6 36  x2 ≥ 0 ⇒6 ≤ x ≤ 6 or 6, 6 41. (a) —CONTINUED— 3 3 1 3 2 2 x y 2 c c 0 c 2 f x  x3  c, c  2, 0, 2 (b) 3 1 2 2 2 3 x y c 0 c 2 c 2 f x  x  c3, c  2, 0, 2 31. The slope is V3  8503  12,500  $9950 850. V  850t  12,500. 41. —CONTINUED— (c) 2 4 1 1 2 2 x y c 2 c 0 c 2 f x  x  23  c, c  2, 0, 2 (d) 3 3 2 1 3 2 1 1 2 x y 0 2 c c c 2 f x  cx3, c  2, 0, 2 22 Chapter P Preparation for Calculus 43. (a) Odd powers: The graphs of f, g, and h all rise to the right and fall to the left. As the degree increases, the graph rises and falls more steeply. All three graphs pass through the points 0, 0, 1, 1, and 1, 1. 3 −2 −3 f h 2 g f x  x, gx  x3, hx  x5 Even powers: The graphs of f, g, and h all rise to the left and to the right. As the degree increases, the graph rises more steeply. All three graphs pass through the points 1, 1, and 1, 1. 0, 0, 3 0 −3 h g f 4 f x  x2, gx  x4, hx  x6 (b) will look like but rise and fall even more steeply. y  x8 will look like hx  x6, but rise even more steeply. y  x7 hx  x5, 45. (a) A  xy  x12  x  12x  x2 y  12  x 2x  2y  24 x x y y (b) Domain: (c) Maximum area is In general, the maximum area is attained when the rectangle is a square. In this case, x  6. A  36. 0 0 12 40 0 x 12 47. (a) 3 (cubic), negative leading coefficient (b) 4 (quartic), positive leading coefficient (c) 2 (quadratic), negative leading coefficient (d) 5, positive leading coefficient 49. (a) Yes, y is a function of t. At each time t, there is one and only one displacement y. (b) The amplitude is approximately . The period is approximately 1.1. (c) One model is (d) 0 2.2 −0.5 0.5 y  1 4 cos2 1.1 t  1 4 cos5.7t 0.25  0.252  0.25 Problem Solving for Chapter P Problem Solving for Chapter P 23 1. (a) Center: Radius: 5 (c) Slope of line from to is Slope of tangent line is Hence, y  0  Tangent line 3 4 x  6 ⇒ y  3 4 x  9 2 3 4 . 4  0 3  6   4 3 6, 0 3, 4 . 3, 4 x  32  y  42  25 x2  6x  9  y2  8y  16  9  16 x2  6x  y2  8y  0 (b) Slope of line from to is Slope of tangent line is Hence, Tangent line (d) Intersection: 3,  9 4 x  3 3 2 x  9 2  3 4 x  3 4 x  9 2 y  0   3 4 x  0 ⇒ y   3 4 x  3 4 4 3 0, 0 3, 4 3. (a) (c) x 1 2 1 3 4 −2 −1 −1 −2 −3 −4 −4 −3 2 3 4 Hx y x 1 2 1 3 4 −2 −1 −1 −3 −4 −4 −3 2 3 4 Hx  2 y x 1 2 1 3 4 −2 −1 −1 −2 −3 −4 −4 −3 2 3 4 Hx   y 1 0 x ≥ 0 x 0 (b) (d) x 1 2 3 4 −2 −1 −1 −2 −3 −4 −4 −3 2 3 4 Hx y x 1 2 1 3 4 −2 −1 −1 −2 −3 −4 −4 −3 2 3 4 Hx  2 y (e) x 1 2 1 3 4 −2 −1 −1 −2 −3 −4 −4 −3 2 3 4 y 12 Hx (f ) x 1 1 3 4 −2 −1 −1 −2 −3 −4 −4 −3 2 3 4 Hx  2  2 y 24 Chapter P Preparation for Calculus 5. (a) Domain: (b) Maximum of at (c) A50  1250 m2 is the maximum. x  50 m, y  25 m.   1 2 x  502  1250   1 2 x2  100x  2500  1250 Ax   1 2 x2  100x 1250 m2 x  50 m, y  25 m. 110 0 0 1600 0 x 100 Ax  xy  x100  x 2    x2 2  50x x  2y  100 ⇒ y  100  x 2 7. The length of the trip in the water is and the length of the trip over land is Hence, the total time is T  4  x2 2  1  3  x2 4 hours. 1  3  x2. 22  x2, 9. (a) Slope of tangent line is less than 5. (b) Slope of tangent line is greater than 3. (c) Slope of tangent line is less than 4.1. (d) (e) Letting h get closer and closer to 0, the slope approaches 4. Hence, the slope at 2, 4 is 4.  4  h, h  0  4h  h2 h  2  h2  4 h Slope  f 2  h  f 2 2  h  2 Slope  4.41  4 2.1  2  4.1. Slope  4  1 2  1  3. Slope  9  4 3  2  5. 11. (a) At and the sounds are equal. (b) Circle of radius 2 centered at 1, 0 x  12  y2  4 x2  2x  y2  3 3x2  3y2  6x  9 x  32  y2  4x2  y2 I x2  y2  2I x  32  y2 3 −3 −6 x  1 x  3 3 Problem Solving for Chapter P 25 13. Let Then or Thus, 0, 0, 2, 0 and 2, 0 are on the curve. y  0. x4  2x2 ⇒ x  0 x2  2. x2  y22  2x2  y2 x4  2x2y2  y4  2x2  2y2  0 x4  2x2  1  2x2y2  2y2  y4  1 x2  12  y22x2  2  y4  1 x  12x  12  y2x  12  x  12  y4  1 x  12  y2 x  12  y2  1 (− 2 , 0) ( 2 , 0) (0, 0) x y −2 −2 −1 1 2 2 d1d2  1 C H A P T E R 1 Limits and Their Properties Section 1.1 A Preview of Calculus . . . . . . . . . . . . . . . . . . . . 27 Section 1.2 Finding Limits Graphically and Numerically . . . . . . . . 27 Section 1.3 Evaluating Limits Analytically . . . . . . . . . . . . . . . 31 Section 1.4 Continuity and One-Sided Limits . . . . . . . . . . . . . . 37 Section 1.5 Infinite Limits . . . . . . . . . . . . . . . . . . . . . . . . 42 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 27 C H A P T E R 1 Limits and Their Properties Section 1.1 A Preview of Calculus Solutions to Odd-Numbered Exercises 1. Precalculus: 20 ftsec15 seconds  300 feet 3. Calculus required: slope of tangent line at is rate of change, and equals about 0.16. x  2 5. Precalculus: Area  sq. units 12 bh  12 53  15 2 7. Precalculus: Volume  243  24 cubic units 9. (a) (b) The graphs of are approximations to the tangent line to at (c) The slope is approximately 2. For a better approximation make the list numbers smaller: 0.2, 0.1, 0.01, 0.001 y2 y1 x  1. 8 −2 −4 6 (1, 3) 11. (a) (b) (c) Increase the number of line segments.  2.693  1.302  1.083  1.031  6.11 D2  1  52 2  1  52  53 2  1  53  54 2  1  54  12 D1  5  12  1  52  16  16  5.66 Section 1.2 Finding Limits Graphically and Numerically 1. Actual limit is 13 lim . x→2 x  2 x2  x  2  0.3333 x 1.9 1.99 1.999 2.001 2.01 2.1 f x 0.3448 0.3344 0.3334 0.3332 0.3322 0.3226 3. lim Actual limit is 123. x→0 x  3  3 x  0.2887 x 0.1 0.01 0.001 0.001 0.01 0.1 f x 0.2911 0.2889 0.2887 0.2887 0.2884 0.2863    5. Actual limit is   1 lim  16 . x→3 1x  1  14 x  3  0.0625 x 2.9 2.99 2.999 3.001 3.01 3.1 f x 0.0641 0.0627 0.0625 0.0625 0.0623 0.0610 7. lim (Actual limit is 1.) (Make sure you use radian mode.) x→0 sin x x  1.0000 x 0.1 0.01 0.001 0.001 0.01 0.1 f x 0.9983 0.99998 1.0000 1.0000 0.99998 0.9983    9. lim x→3 4  x  1 11. lim x→2 f x  lim x→2 4  x  2 15. tan x does not exist since the function increases and decreases without bound as x approaches 2. lim x→2 17. does not exist since the function oscillates between 1 and 1 as x approaches 0. lim x→0 cos1x (b) (c) lim does not exist. The values of C jump from 1.75 to 2.25 at t  3. t→3 Ct lim t→3.5 Ct  2.25 t 2 2.5 2.9 3 3.1 3.5 4 C 1.25 1.75 1.75 1.75 2.25 2.25 2.25 19. (a) 0 0 5 3 Ct  0.75  0.50t  1 t 3 3.3 3.4 3.5 3.6 3.7 4 C 1.75 2.25 2.25 2.25 2.25 2.25 2.25 21. You need to find such that implies That is, 1 9 x  1  1 11 . 10 9  1 x  1 10 11  1 10 9 x 10 11 9 10 1 x 11 10 1  0.1 1 x 1  0.1 0.1 1 x  1 0.1 f x  1  1 x  1 0.1. 0 x  1   So take Then implies Using the first series of equivalent inequalities, you obtain f x  1  1 x  1  0.1.  1 11 x  1 1 9 .  1 11 x  1 1 11