Math Chapter 2 Properties of Functions Key

Chapter 2 NONLINEAR FUNCTIONS 9. y = 2x + 3 2.1 Properties of Functions 1. The x-value of 82 corresponds to two y-values, 93 and 14. In a function, each value of x must correspond to exactly one value of y: The rule is not a function. x y ¡2 ¡1 ¡1 1 0 3 1 5 2 7 3 9 2 3 4 Pairs: (¡2; ¡1); (¡1; 1); (0; 3); (1; 5); (2; 7); (3; 9) Range: f¡1; 1; 3; 5; 7; 9g 2. Each x-value corresponds to exactly one y-value. The rule is a function. 3. Each x-value corresponds to exactly one y-value. The rule is a function. 4. 9 corresponds to 3 and ¡3; 4 corresponds to 2 and ¡2; and 1 corresponds to ¡1 and 1. The rule is not a function. 5. y = x3 + 2 Each x-value corresponds to exactly one y-value. The rule is a function. 6. y = p x Each x-value corresponds to exactly one y-value. The rule is a function. 10. y = ¡3x + 9 x y ¡2 15 ¡1 12 0 9 1 6 2 3 3 0 Pairs: (¡2; 15); (¡1; 12); (0; 9); (1; 6); (2; 3); (3; 0) Range: f0; 3; 6; 9; 12; 15g 7. x = jyj Each value of x (except 0) corresponds to two yvalues. The rule is not a function. 8. x = y 2 + 4 Solve the rule for y: p y 2 = x ¡ 4 or y = § x ¡ 4 11. 2y ¡ x = 5 2y = 5 + x y= Each value of x (greater than 4) corresponds to two y-values p p y = x ¡ 4 and y = ¡ x ¡ 4: The rule is not a function. 5 1 x+ 2 2 x y ¡2 3 2 ¡1 2 0 5 2 1 3 7 2 Pairs:(¡2; 32 ); (¡1; 2); (0; 52 ); (1; 3); (2; 72 ); (3; 4) Section 2.1 Properties of Functions Range: f 32 ; 2; 5 2 3; 7 2; 4g 14. y = (x ¡ 2)(x + 2) x y ¡2 0 ¡1 ¡3 0 ¡4 1 ¡3 2 0 3 5 Pairs: (¡2; 0); (¡1; ¡3); (0; ¡4); (1; ¡3); (2; 0); (3; 5) Range: f¡4; ¡3; 0; 5g 12. 6x ¡ y = ¡1 ¡y = ¡6x ¡ 1 y = 6x + 1 x y ¡2 ¡11 ¡1 ¡5 0 1 1 7 2 13 3 19 15. y = x2 Pairs: (¡2; ¡11); (¡1; ¡5); (0; 1); (1; 7); (2; 13); (3; 19) Range: f¡11; ¡5; 1; 7; 13; 19g x y ¡2 4 ¡1 1 0 0 1 1 2 4 3 9 Pairs: (¡2; 4); (¡1; 1); (0; 0); (1; 1); (2; 4); (3; 9) Range: f0; 1; 4; 9g 13. y = x(x + 2) x y ¡2 0 ¡1 ¡1 0 0 1 3 Pairs: (¡2; 0); (¡1; ¡1); (0; 0); (1; 3); (2; 8); (3; 15) Range: f¡1; 0; 3; 8; 15g 2 8 3 15 16. y = ¡4x2 x y ¡2 ¡16 ¡1 ¡4 0 0 1 ¡4 2 ¡16 3 ¡36 Pairs: (¡2; ¡16); (¡1; ¡4); (0; 0); (1; ¡4); (2; ¡16); (3; ¡36) Range: f¡36; ¡16; ¡4; 0g Chapter 2 17. y = 1 x+3 x y 20. y = ¡2 1 ¡1 0 1 2 3 1 2 1 3 1 4 1 5 1 6 x y Pairs: (¡2; 1); (¡1; 12 ); (0; 13 ); (1; 14 ); (2; 15 ); (3; 16 ) Range: f1; 1 1 1 1 1 2; 3; 4; 5; 6g NONLINEAR FUNCTIONS 2x + 1 x+3 ¡2 ¡3 ¡1 ¡ 12 0 1 1 3 3 4 2 1 3 7 6 Pairs: (¡2; ¡3); (¡1; ¡ 12 ); (0; 13 ); (1; 34 ); (2; 1); (3; 76 ) Range: f¡3; ¡ 12 ; 13 ; 34 ; 1; 76 g 18. y = x y ¡2 x+4 ¡2 ¡1 ¡1 ¡ 23 0 ¡ 12 1 ¡ 25 2 ¡ 13 3 ¡ 27 Pairs: (¡2; ¡1); (¡1; ¡ 23 ); (0; ¡ 12 ); 21. f(x) = 2x (1; ¡ 25 ); (2; ¡ 13 ); (3; ¡ 27 ) x can take on any value, so the domain is the set of real numbers, (¡1; 1): Range: f¡1; ¡ 23 ; ¡ 12 ; ¡ 25 ; ¡ 13 ; ¡ 27 g 22. f(x) = 2x + 3 x can take on any value, so the domain is the set of real numbers, which is written (¡1; 1): 23. f(x) = x4 19. y = x can take on any value, so the domain is the set of real numbers, (¡1; 1): 2x ¡ 2 x+4 x y ¡2 ¡3 ¡1 ¡ 43 0 ¡ 12 1 0 2 3 1 3 4 7 Pairs:(¡2; ¡3); (¡1; ¡ 43 ); (0; ¡ 12 ); (1; 0); (2; 13 ); (3; 47 ) Range: f¡3; ¡ 43 ; ¡ 12 ; 0; 1 4 3; 7g 24. f(x) = (x + 3)2 x can take on any value, so the domain is the set of real numbers, (¡1; 1): 25. f(x) = p 4 ¡ x2 For f(x) to be a real number, 4 ¡ x2 ¸ 0: Solve 4 ¡ x2 = 0: (2 ¡ x)(2 + x) = 0 x = 2 or x = ¡2 The numbers form the intervals (¡1; ¡2); (¡2; 2); and (2; 1): Only values in the interval (¡2; 2) satisfy the inequality. The domain is [¡2; 2]: Section 2.1 Properties of Functions r 31. f(x) = ¡ 26. f (x) = j3x ¡ 6j x can take on any value, so the domain is the set of real numbers, (¡1; 1): p 27. f (x) = (x ¡ 3)1=2 = x ¡ 3 For f (x) to be a real number, Use the values ¡4 and 4 to divide the number line into 3 intervals, (¡1; ¡4); (¡4; 4) and (4; 1). Only the values in the intervals (¡1; ¡4) and (4; 1) satisfy the inequality. The domain is The domain is [3; 1): p 28. f (x) = (3x + 5)1=2 = 3x + 5 For f (x) to be a real number, (¡1; ¡4) [ (4; 1): 3x + 5 ¸ 0 3x ¸ ¡5 r 32. f(x) = ¡ 1 1 (3x) ¸ (¡5) 3 3 5 x¸¡ : 3 In interval notation, the domain is [¡ 53 ; 1): 2 2 = 2 1¡x (1 ¡ x)(1 + x) or or (x ¡ 5)(x + 1) ¸ 0 when x ¸ 5 and when x · ¡1: The domain is (¡1; ¡1] [ [5; 1): 1+x =0 x = ¡1: (¡1; ¡1) [ (¡1; 1) [ (1; 1): 34. f(x) = p 15x2 + x ¡ 2 The expression under the radical must be nonnegative. 15x2 + x ¡ 2 ¸ 0 (5x + 2)(3x ¡ 1) ¸ 0 ¡8 x2 ¡ 36 Solve (5x + 2)(3x ¡ 1) = 0: 2 In order for f (x) to be a real number, x ¡ 36 cannot be equal to 0. When x2 ¡ 36 = 0; x2 = 36 x = 6 or x = ¡6: Thus, the domain is any real number except 6 or ¡6: In interval notation, the domain is (¡1; ¡6) [ (¡6; 6) [ (6; 1): p p x2 ¡ 4x ¡ 5 = (x ¡ 5)(x + 1) See the method used in Exercise 25. Thus, x can be any real number except §1: The domain is 30. f (x) = 5 x2 + 36 x can take on any value. No choice for x will produce a zero in the denominator. Also, no choice for x will produce a negative number under the radical. The domain is (¡1; 1): 33. f (x) = Since division by zero is not de…ned, (1 ¡ x) ¢ (1 + x) 6= 0: When (1 ¡ x)(1 + x) = 0; 1¡x=0 x=1 (x¡4)¢(x+4) > 0, since (x¡4)¢(x+4) < 0 would produce a negative radicand and (x¡4)¢(x+4) = 0 would lead to division by zero. Solve (x ¡ 4) ¢ (x + 4) = 0: x ¡ 4 = 0 or x + 4 = 0 x = 4 or x = ¡4 x¡3¸0 x ¸ 3: 29. f(x) = r 2 2 = ¡ : x2 ¡ 16 (x ¡ 4)(x + 4) 5x + 2 = 0 5x = ¡2 x=¡ 2 5 or 3x ¡ 1 = 0 3x = 1 or x= 1 3 Use these numbers to¢divide the ¡ ¢ number ¡ line¢ into ¡ 3 intervals, ¡1; ¡ 25 ; ¡ 25 ; 13 ; and 13 ; 1 : ¢ ¢ ¡ ¡ Only values in the intervals ¡1; ¡ 25 ¡and 13 ; 1 ¤ satisfy the inequality. The domain is ¡1; ¡ 25 [ £1 ¢ 3; 1 : Chapter 2 1 1 =p 35. f(x) = p 2 3x + 2x ¡ 1 (3x ¡ 1)(x + 1) (3 ¡ 2)(x + 1) > 0; since the radicand cannot be negative and the denominator of the function cannot be zero. Solve (3 ¡ 1)(x + 1) = 0: 3¡1 =0 x = 13 or or x+1 =0 x = ¡1 Use the values ¡1 and 13 to divide the number line into 3 intervals, (¡1; ¡1); (¡1; 4) and ( 13 ; 1): Only the values in the intervals (¡1; ¡ 13 ) and ( 13 ; 1) satisfy the inequality. The domain is (¡1; ¡1) [ ( 13 ; 1): r 36. f (x) = μ ¶ μ ¶2 μ ¶ 1 1 1 ¡4 ¡ =3 ¡ +1 (b) f ¡ 2 2 2 3 = +2+1 4 15 = 4 (c) f (a) = 3(a)2 ¡ 4(a) + 1 = 3a2 ¡ 4a + 1 μ ¶ μ ¶ μ ¶2 2 2 2 ¡4 (d) f =3 +1 m m m 8 12 +1 = 2¡ m m 12 ¡ 8m + m2 or m2 (e) x2 3¡x For f (x) to be a real number, x2 ¸ 0 and 3 ¡ x > 0 ¡1 < x < 1 and x<3 x < 3: The domain is (¡1; 3): 37. By reading the graph, the domain is all numbers greater than or equal to ¡5 and less than 4. The range is all numbers greater than or equal to ¡2 and less than or equal to 6. Domain: [¡5; 4); range: [¡2; 6] 38. By reading the graph, the domain is all numbers greater than or equal to ¡5: The range is all numbers greater than or equal to 0. Domain: [¡5; 1) Range: [0; 1) 39. By reading the graph, x can take on any value, but y is less than or equal to 12. Domain: (¡1; 1); range: (¡1; 12] 40. By reading the graph, both x and y can take on any values. Domain: (¡1; 1) Range: (¡1; 1) 41. f (x) = 3x2 ¡ 4x + 1 (a) f (4) = 3(4)2 ¡ 4(4) + 1 = 48 ¡ 16 + 1 = 33 NONLINEAR FUNCTIONS f (x) = 1 3x2 ¡ 4x + 1 = 1 3x2 ¡ 4x = 0 x(3x ¡ 4) = 0 x = 0 or x = 4 3 42. f(x) = (x + 3)(x ¡ 4) (a) f(4) = (4 + 3)(4 ¡ 4) = (7)(0) = 0 ¶μ ¶ μ ¶ μ 1 1 1 ¡ ¡4 (b) f ¡ = ¡ +3 2 2 2 μ ¶μ ¶ 5 9 45 = ¡ =¡ 2 2 4 (c) f(a) = [(a) + 3][(a) ¡ 4] = (a + 3)(a ¡ 4) ¶μ ¶ μ ¶ μ 2 2 2 +3 ¡4 = (d) f m m m μ ¶μ ¶ 2 + 3m 2 ¡ 4m = m m (2 + 3m)(2 ¡ 4m) m2 2(2 + 3m)(1 ¡ 2m) or m2 = (e) f (x) = 1 (x + 3)(x ¡ 4) = 1 x2 ¡ x ¡ 12 = 1 x2 ¡ x ¡ 13 = 0 p 1 § 1 + 52 x= 2 p 1 § 53 x= 2 x ¼ ¡3:140 or x ¼ 4:140 Section 2.1 Properties of Functions 2x + 1 x¡2 9 2(4) + 1 = (a) f(4) = 4¡2 2 μ ¶ 2(¡ 12 ) + 1 1 (b) f ¡ = 2 ¡ 12 ¡ 2 (e) 43. f (x) = = = 3x + 2 =1 2x ¡ 4 3x + 2 = 2x ¡ 4 x = ¡6 45. The domain is all real numbers between the endpoints of the curve, or [¡2; 4]. The range is all real numbers between the minimum and maximum values of the function or [0; 4]. ¡1 + 1 5 2 0 5 2 =0 2a + 1 2(a) + 1 = (a) ¡ 2 a¡2 ¡2¢ μ ¶ +1 2 2 (d) f = 2m m ¡ 2 m (c) f(a) = = = 4 m m + m 2 2m m ¡ m 4+m m 2¡2m m m 4+m ¢ m 2 ¡ 2m 4+m = 2 ¡ 2m f (x) = 1 = (e) (a) f (¡2) = 0 (b) f(0) = 4 μ ¶ 1 =3 (c) f 2 (d) From the graph, f (x) = 1 when x = ¡1:5; 1:5; or 2.5. 46. The domain is all real numbers between the end points of the curve, or [¡2; 4]. The range is all real numbers between the minimum and maximum values of the function or [0; 5]. (a) f (¡2) = 5 μ ¶ 1 =1 (c) f 2 (b) f (0) = 0 (d) From the graph, if f (x) = 1; x = ¡0:2; 0:5; 1:2; or 2:8: 2x + 1 =1 x¡2 2x + 1 = x ¡ 2 x = ¡3 3x + 2 2x ¡ 4 12 + 2 14 7 3(4) + 2 = = = (a) f(4) = 2(4) ¡ 4 8¡4 4 2 ¡ 1¢ μ ¶ 3 3 ¡ +2 ¡ +2 1 (b) f ¡ = 2 = ¡ 21 ¢ 2 ¡1 ¡ 4 2 ¡2 ¡ 4 44. f (x) = 1 ¡ 32 + 42 1 = 2 =¡ ¡5 ¡5 10 3a + 2 3(a) + 2 = (c) f (a) = 2(a) ¡ 4 2a ¡ 4 ¡2¢ μ ¶ +2 3 m 2 (d) f = ¡2¢ m 2 m ¡4 = = f(x) = 1 6 m 4 m + ¡ 2m m 4m m = 6+2m m 4¡4m m m 6 + 2m ¢ = m 4 ¡ 4m 3+m 6 + 2m = = 4 ¡ 4m 2 ¡ 2m 47. The domain is all real numbers between the endpoints of the curve, or [¡2; 4]. The range is all real numbers between the minimum and maximum values of the function or [¡3; 2]. (a) f (¡2) = ¡3 (b) f(0) = ¡2 μ ¶ 1 = ¡1 (c) f 2 (d) From the graph, f (x) = 1 when x = 2:5: 48. The domain is all real numbers between the end points of the curve, or [¡2; 4]. The range is all real numbers between the minimum and maximum values of the function, or in this case, {3}. (a) f (¡2) = 3 (b) f(0) = 3 μ ¶ 1 =3 (c) f 2 (d) From the graph, f (x) is 1 nowhere. Chapter 2 49. f (x) = 6x2 ¡ 2 f(t + 1) = 6(t + 1)2 ¡ 2 = 6(t2 + 2t + 1) ¡ 2 = 6t2 + 12t + 6 ¡ 2 = 6t2 + 12t + 4 50. f (x) = 6x2 ¡ 2 f(2 ¡ r) = 6(2 ¡ r)2 ¡ 2 = 6(4 ¡ 4r + r2 ) ¡ 2 = 24 ¡ 24r + 6r2 ¡ 2 = 6r2 ¡ 24r + 22 51. g(r + h) = (r + h)2 ¡ 2(r + h) + 5 = r2 + 2hr + h2 ¡ 2r ¡ 2h + 5 52. g(z ¡ p) = (z ¡ p)2 ¡ 2(z ¡ p) + 5 = z 2 ¡ 2zp + p2 ¡ 2z + 2p + 5 μ ¶ μ ¶2 μ ¶ 3 3 3 53. g ¡2 = +5 q q q 6 9 = 2 ¡ +5 q q or 59. A vertical line drawn anywhere through the graph will intersect the graph in only one place. The graph represents a function. 60. A vertical line is not a function since the one xvalue in the domain corresponds to more than one, in fact, in…nitely many y-values. The graph does not represent a function. 61. f(x) = 2x + 1 (a) f (x + h) = 2(x + h) + 1 = 2x + 2h + 1 (b) f(x + h) ¡ f (x) = 55. A vertical line drawn anywhere through the graph will intersect the graph in only one place. The graph represents a function. 56. A vertical line drawn anywhere through the graph will intersect the graph in only one place. The graph represents a function. 57. A vertical line drawn through the graph may intersect the graph in two places. The graph does not represent a function. 58. A vertical line drawn through the graph may intersect the graph in two or more places. The graph does not represent a function. 2x + 2h + 1 2x + 1 = 2x + 2h + 1 ¡ 2x ¡ 1 = 2h (c) 9 ¡ 6q + 5q 2 q2 μ μ ¶ μ ¶2 ¶ 5 5 5 54. g ¡ ¡2 ¡ = ¡ +5 z z z 25 10 +5 = 2+ z z 25 10z 5z 2 = 2+ 2 + 2 z z z 25 + 10z + 5z 2 = z2 NONLINEAR FUNCTIONS f (x + h) ¡ f (x) h = 2x+2h+1 2x+1 h 2x + 2h + 1 ¡ 2x ¡ 1 = h 2h = h =2 62. f(x) = x2 ¡ 3 (a) f(x + h) = (x + h)2 ¡ 3 = (x2 + 2xh + h2 ) ¡ 3 = x2 + 2xh + h2 ¡ 3 (b) f (x + h) ¡ f(x) = (x2 + 2xh + h2 ¡ 3) ¡ (x2 ¡ 3) = x2 + 2xh + h2 ¡ 3 ¡ x2 + 3 = 2xh + h2 (c) 2xh + h2 f(x + h) ¡ f (x) = h h 2x + h = h = 2x + h Section 2.1 Properties of Functions 63. f (x) = 2x2 ¡ 4x ¡ 5 65. f(x) = (a) f (x + h) = 2(x + h)2 ¡ 4(x + h) ¡ 5 = 2(x2 + 2hx + h2 ) ¡ 4x ¡ 4h ¡ 5 = 2x2 + 4hx + 2h2 ¡ 4x ¡ 4h ¡ 5 (a) f (x + h) = 1 1 ¡ x+h x μ ¶ ³x´ 1 1 x+h ¡ = x x+h x x+h = (c) f (x + h) ¡ f(x) h 4hx + 2h2 ¡ 4h = h h(4x + 2h ¡ 4) = h 1 x+h (b) f(x + h) ¡ f (x) (b) f (x + h) ¡ f (x) = 2x2 + 4hx + 2h2 ¡ 4x ¡ 4h ¡ 5 ¡ (2x2 ¡ 4x ¡ 5) = 2x2 + 4hx + 2h2 ¡ 4x ¡ 4h ¡ 5 ¡ 2x2 + 4x + 5 = 4hx + 2h2 ¡ 4h (c) 1 x = x ¡ (x + h) x(x + h) = ¡h x(x + h) f (x + h) ¡ f (x) h 1 ¡1 = x+h x h ¡ x ¢ 1 ³ x+h ´ 1 x+h x ¡ x x+h = h = x¡(x+h) (x+h)x h ¸ 1 x¡x¡h = h (x + h)x · ¸ 1 ¡h = h (x + h)x ¡1 = x(x + h) · = 4x + 2h ¡ 4 64. f (x) = ¡4x2 + 3x + 2 (a) f (x + h) = ¡4(x + h)2 + 3(x + h) + 2 = ¡4(x2 + 2hx + h2 ) + 3x + 3h + 2 = ¡4x2 ¡ 8hx ¡ 4h2 + 3x + 3h + 2 (b) f(x + h) ¡ f (x) = ¡4x2 ¡ 8hx ¡ 4h2 + 3x + 3h + 2 ¡ (¡4x2 + 3x + 2) = ¡4x2 ¡ 8hx ¡ 4h2 + 3x + 3h + 2 + 4x2 ¡ 3x ¡ 2 = ¡8hx ¡ 4h2 + 3h 66. f(x) = ¡ 1 x2 1 (x + h)2 1 =¡ 2 x + 2xh + h2 μ ¶ 1 1 (b) f(x + h) ¡ f (x) = ¡ 2 ¡ ¡ x + 2xh + h2 x2 1 1 =¡ 2 + 2 2 x + 2xh + h x x2 =¡ 2 2 x (x + 2xh + h2 ) (a) f(x + h) = ¡ + 2 (c) ¡8hx ¡ 4h + 3h f (x + h) ¡ f (x) = h h h(¡8x ¡ 4h + 3) = h = ¡8x ¡ 4h + 3 (x2 + 2xh + h2 ) x2 (x2 + 2xh + h2 ) = ¡x2 + x2 + 2xh + h2 x2 (x2 + 2xh + h2 ) = 2xh + h2 x2 (x2 + 2xh + h2 ) Chapter 2 (c) 67. f (x + h) ¡ f (x) = h f (x) = 3x f(¡x) = 3(¡x) = ¡(3x) = ¡f (x) f (x) = 5x f(¡x) = 5(¡x) = ¡(5x) = ¡f (x) The function is odd. 69. f (x) = 2x2 f(¡x) = 2(¡x)2 = 2x2 = f (x) The function is even. 70. f (x) = x2 ¡ 3 f(¡x) = (¡x)2 ¡ 3 = x2 ¡ 3 = f (x) The function is even. 1 71. f (x) = 2 x +4 1 f(¡x) = (¡x)2 + 4 1 = 2 x +4 = f (x) The function is even. 72. 73. h 2x + h = 2 2 x (x + 2xh + h2 ) The function is odd. 68. h(2x+h) x2 (x2 +2xh+h2 ) f (x) = x3 + x f(¡x) = (¡x)3 + (¡x) = ¡x3 ¡ x = ¡(x3 + x) = ¡f (x) The function is odd. NONLINEAR FUNCTIONS x x2 ¡ 9 ¡x f (¡x) = (¡x)2 ¡ 9 x =¡ 2 x ¡9 f(x) = = ¡f(x) The function is odd. 74. f(x) = jx ¡ 2j f (¡x) = j¡x ¡ 2j = j¡(x + 2)j = jx + 2j f(¡x) does not equal either f(x) or f(¡x). The function is neither even nor odd. 75. (a) The independent variable is the years. (b) The dependent variable is the number of Internet users. (c) f (2003) = 719 million users. (d) The domain is 1995 · x · 2006: The range is 16,000,000 · y · 1,043,000,000: 76. If x is a whole number of days, the cost of renting a saw in dollars is S(x) = 28x + 8: For x in whole days and a fraction of a day, substitute the next whole number for x in 28x + 8; because a fraction of a day is charged as a whole day. ¡ ¢ (a) S 21 = S(1) = 28(1) + 8 = 36 The cost is $36. (b) S(1) = 28(1) + 8 = 36 The cost is $36. ¡ ¢ (c) S 1 14 = S(2) = 28(2) + 8 = 56 + 8 = 64 The cost is $64. ¡ ¢ (d) S 3 12 = S(4) = 28(4) + 8 = 112 + 8 = 120 The cost is $120. (e) S(4) = 28(4) + 8 = 112 + 8 = 120 The cost is $120. ¡ 1¢ = S(5) = 28(5) + 8 = 140 + 8 = 148 (f ) S 4 10 The cost is $148. ¡ 9¢ = S(5) = 28(5) + 8 = 140 + 8 = 148 (g) S 4 10 The cost is $148. Section 2.1 Properties of Functions (h) To continue the graph, continue the horizontal bars up and to the right. (i) The independent variable is x; the number of full and partial days. (j) The dependent variable is S; the cost of renting a saw. 77. If x is a whole number of days, the cost of renting a car is given by C(x) = 54x + 44: For x in whole days plus a fraction of a day, substitute the next whole number for x in 54x + 44 because a fraction of a day is charged as a whole day. ¡ ¢ (a) C 34 = C(1) = 54(1) + 44 = $98 ¡9¢ (b) C 10 = C(1) = $98 (c) C(1) = $98 ¡ ¢ (d) C 1 58 = C(2) = 54(2) + 44 = $152 (e) C(2:4) = C(3) = 54(3) + 44 = $206 (f ) 79. (a) (i) (ii) y = f(5) = 19:7(5)0:753 ¼ 66 kcal/day y = f(25) = 19:7(25)0:753 ¼ 222 kcal/day (b) Since 1 pound equals 0.454 kg, then x = g(z) = 0:454z is the number of kilograms equal in weight to z pounds. 80. (a)(i) By the given function f, a muskrat weighing 800 g expends f(800) = 0:01(800)0:88 ¼ 3:6, or approximately 3.6 kcal/km when swimming at the surface of the water. (ii) A sea otter weighing 20,000 g expends f (20,000) = 0:01(20,000)0:88 ¼ 61, or approximately 61 kcal/km when swimming at the surface of the water. (b) If z is the number of kilogram of an amimal’s weight, then x = g(z) = 1000z is the number of grams since 1 kilogram equals 1000 grams. 81. (a) In the graph, the curves representing wood and coal intersect approximately at the point (1880; 50). So, in 1880 use of wood and coal were both about 50% of the global energy consumption. (b) In the graph, the curves representing oil and coal intersect approximately at the point (1965; 35). So, in 1965 use of oil and coal were both about 35% of the global energy consumption. 82. (a) P = 2L + 2W (g) Yes, C is a function. (h) No, C is not a linear function. 78. (a) The curve in the graph crosses the point with x-coordinate 17:37 and y-coordinate of approximately 140. So, at time 17 hours, 37 minutes the whale reaches a depth of about 140 m. (b) The curve in the graph crosses the point with x-coordinate 17:39 and y-coordinate of approximately 240. So, at time 17 hours, 39 minutes the whale reaches a depth of about 240 m. 500 : W μ ¶ 500 P (W ) = 2 + 2W W 1000 + 2W P (W ) = W However, LW = 500; so L = (b) Since L = 500 W ; W 6= 0 but W could be any positive value. Therefore, the domain of P is 0 < W < 1: Chapter 2 (c) 83. (a) Let w = the width of the …eld; l = the length. The perimeter of the …eld is 6000 ft, so 2l + 2w = 6000 l + w = 3000 l = 3000 ¡ w: Thus, the area of the …eld is given by A = lw A = (3000 ¡ w)w: (b) Since l = 3000 ¡ w and w cannot be negative, 0 · w · 3000: The domain of A is 0 · w · 3000: (c) NONLINEAR FUNCTIONS