5 1 05 1 x 1 1 1 2 2 3 x 1 2 3 e i f ii 27 14 chapter

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Unformatted text preview: intercept is the original value. y x 2 4 8 6 10 10. A line through (6, −1) has the form y +1 = m(x − 6). The intercepts are x = 6+1/m and y = −6m − 1. Set −(6+1/m)(6m +1) = 3, or 36m2 +15m +1 = (12m +1)(3m +1) = 0 with roots m = −1/12, −1/3; thus y + 1 = −(1/3)(x − 6) and y + 1 = −(1/12)(x − 6). 11. (a) VI 12. In all cases k must be positive, or negative values would appear in the chart. Only kx−3 decreases, so that must be f (x). Next, kx2 grows faster than kx3/2 , so that would be g (x), which grows faster than h(x) (to see this, consider ratios of successive values of the functions). Finally, experimentation (a spreadsheet is handy) for values of k yields (approximately) f (x) = 10x−3 , g (x) = x2 /2, h(x) = 2x1.5 . 13. (a) (b) IV (c) III (d) V y -2 y -1 1 2 x 30 -10 20 -20 10 -30 x -2 -1 1 2 -10 -40 -20 -50 -30 -60 y (b) y 10 4 8 2 6 x -4 -2 2 4 4 -2 2 -4 x -4 -2 2 4 y (c) y 2 1.5 1 0.5 1 x -1 1 -1 -2 2 3 x 1 2 3 (e) I (f ) II 27 14. Chapter 1 y (a) y 80 40 40 -2 x x 2 -2 2 -40 y (b) y 4 x -2 2 -2 x -2 2 -4 -4 y (c) y 4 x -1 x -3 15. -2 -1 -2 (a) (b) y 10 10 6 5 4 x -2 y (c) y -1 1 5 2 2 x -2 -1 x -5 -2 -1 1 1 2 -5 2 -2 -10 -10 16. (a) (b) y (c) y y 3 2 2 x 2 x 2 x 2 17. y (a) 80 60 40 20 6 x 4 2 x -3 -2 -1 y (b) 8 1 -1 -20 -40 -60 -80 1 2 3 4 5 Exercise Set 1.6 28 (c) (d) y -5 -4 -3 -2 -1 123 x y 40 -10 20 -20 x 1 -30 2 3 4 5 -20 -40 -40 -50 18. (a) (b) y y 1 1 x 1 x -2 y (c) 1 y (d) 4 4 x -2 -6 2 x -2 -4 19. y (a) (b) 1.5 y 1 1 0.5 0.5 -3 -2 -1 1 x x 1 -0.5 -1.5 30 20 20 10 10 x x 1 -10 2 -2 3 -1 1 -10 -20 2 2 3 -20 -30 -30 y (a) 6 5 y (d) 30 20. 4 -1 y -1 3 -0.5 -1 (c) 2 y (b) 30 20 10 10 x -10 1 3 4 x -1 1 -10 -20 -30 29 Chapter 1 y (c) y (d) 106 15 10 x -1 1 5 x -106 -4 -2 2 y 21. 15 10 5 x -4 22. -2 2 y (a) y (b) 1.5 1.5 1 1 0.5 0.5 -2 23. 2 x x -2 2 √ t = 0.445 d 2.3 0 25 0 24. (a) t = 0.373r1.5 25. (a) (c) (b) 238,000 km N·m V (L) 0.25 P (N/m2 ) 80 × 103 (c) 1.89 days (b) 20 N·m 0.5 40 × 103 1.0 20 × 103 1.5 13.3 × 103 2.0 10 × 103 26. If the side of the square base is x and the height of the container is y then V = x2 y = 100; minimize A = 2x2 + 4xy = 2x2 + 400/x. A graphing utility with a zoom feature suggests that the solution is a 1 cube of side 100 3 cm. 27. (a) F = k/x2 so 0.0005 = k/(0.3)2 and k = 0.000045 N·m2 . (b) 0.000005 N Exercise Set 1.6 (c) 30 (d) When they approach one another, the force becomes infinite; when they get far apart it tends to zero. (b) W = C/50002 = (3.2 × 1010 )/(25 × 106 ) = 1280 lb. (d) F No, but W is very small when x is large. 10-5 5 28. (a) 10 d 2000 = C/(4000)2 , so C = 3.2 × 1010 lb·mi2 (c) W 20000 15000 10000 5000 x 2000 4000 6000 8000 10000 29. (a) II; y = 1, x = −1, 2 (c) IV; y = 2 30. The denominator has roots x = ±1, so x2 − 1 is the denominator. To determine k use the point (0, −1) to get k = 1, y = 1/(x2 − 1). 31. Order the six trigonometric functions as sin, cos, tan, cot, sec, (a) pos, pos, pos, pos, pos, pos (b) neg, (c) pos, neg, neg, neg, neg, pos (d) neg, (e) neg, neg, pos, pos, neg, neg (f ) neg, 32. (a) (c) (e) neg, zero, undef, zero, undef, neg zero, neg, zero, undef, neg, undef neg, neg, pos, pos, neg, neg (b) (d) (f ) 33. (a) (c) sin(π − x) = sin x; 0.588 sin(2π + x) = sin x; 0.588 (b) cos(−x) = cos x; 0.924 (d) cos(π − x) = − cos x; −0.924 (e) sin 2x = ±2 sin x 1 − sin2 x; use the + sign for x small and positive; 0.951 (f ) cos2 x = 1 − sin2 x; 0.654 (a) sin(3π + x) = − sin x; −0.588 (b) cos(−x − 2π ) = cos x; 0.924 (c) sin(8π + x) = sin x ; 0.588 (e) cos(3π + 3x) = −4 cos3 x + 3 cos x; −0.384 (d) sin(x/2) = ± (1 − cos x)/2; use the negative sign for x small and negative; −0.195 1 − cos2 x ; 0.172 (f ) tan2 x = cos2 x (a) −a (c) (e) −c −b 34. 35. √ (g) ±2b 1 − b2 (i) 1/b (k) 1/c (b) I; y = 0, x = −2, 3 (d) III; y = 0, x = −2 csc: zero, undef, zero, undef, neg pos, neg, neg, pos, neg pos, neg, neg, pos, neg pos, neg, neg, neg, neg, pos pos, zero, undef, zero, undef, pos neg, neg, pos, pos, neg, neg (b) b (h) (j) (l) 2b2 − 1 −1/a (1 − b)/2 √ (d) ± 1 − a2 (f ) −a 31 Chapter 1 36. (a) The distance is 36/360 = 1/10th of a great circle, so it is (1/10)2πr = 2, 513.27 mi. 37. If the arc length is 1, then solve the ratio 38. The distance travelled is equal to the length of that portion of the circumference of the wheel which touches the road, and that is the fraction 225/360 of a circumference, so a distance of (225/360)(2π )3 = 11.78 ft 39. The second quarter revolves twice (720◦ ) about its own center. 40. Add r to itself until you exceed 2πr; since 6r < 2πr < 7r, you can cut off 6 pieces of pie, but there’s not enough for a full seventh piece. We conclude that there is no exact solution of the equation ‘One pie = 2πr’. 41. (a) y = 3 sin(x/2) (b) y = 4 cos 2x (c) y = −5 sin 4x 42. (a) y = 1 + cos πx (b) y = 1 + 2 sin x (c) y = −5 cos 4x 43. (a) y = sin(x + π/2) (b) y = 3 + 3 sin(2x/9) (c) y = 1 + 2 sin(2(...
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