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Instructor’s Solutions Manual, Section 3.1 Exercise 1 Solutions to Exercises, Section 3.1 For Exercises 1–8, evaluate the indicated quantities. Do not use a calculator because otherwise you will not gain the understanding that these exercises should help you attain. 1. 253/2 3 solution 253/2 = (251/2 ) = 53 = 125 Instructor’s Solutions Manual, Section 3.1 2. 85/3 5 solution 85/3 = (81/3 ) = 25 = 32 Exercise 2 Instructor’s Solutions Manual, Section 3.1 3. 323/5 3 solution 323/5 = (321/5 ) = 23 = 8 Exercise 3 Instructor’s Solutions Manual, Section 3.1 4. 813/4 3 solution 813/4 = (811/4 ) = 33 = 27 Exercise 4 Instructor’s Solutions Manual, Section 3.1 Exercise 5 5. 32−4/5 −4 solution 32−4/5 = (321/5 ) = 2−4 = 1 1 = 4 2 16 Instructor’s Solutions Manual, Section 3.1 Exercise 6 6. 8−5/3 −5 solution 8−5/3 = (81/3 ) = 2−5 = 1 1 = 5 2 32 Instructor’s Solutions Manual, Section 3.1 7. (−8)7/3 solution 7 (−8)7/3 = (−8)1/3 = (−2)7 = −128 Exercise 7 Instructor’s Solutions Manual, Section 3.1 8. (−27)4/3 solution 4 (−27)4/3 = (−27)1/3 = (−3)4 = 81 Exercise 8 Instructor’s Solutions Manual, Section 3.1 For Exercises 9–20, expand the indicated expression. √ 9. (2 + 3)2 solution (2 + √ √ √ 2 3)2 = 22 + 2 · 2 · 3 + 3 √ =4+4 3+3 √ =7+4 3 Exercise 9 Instructor’s Solutions Manual, Section 3.1 10. (3 + √ 2)2 solution (3 + √ √ √ 2 2)2 = 32 + 2 · 3 · 2 + 2 √ =9+6 2+2 √ = 11 + 6 2 Exercise 10 Instructor’s Solutions Manual, Section 3.1 √ 11. (2 − 3 5)2 solution √ √ 2 √ (2 − 3 5)2 = 22 − 2 · 2 · 3 · 5 + 32 · 5 √ = 4 − 12 5 + 9 · 5 √ = 49 − 12 5 Exercise 11 Instructor’s Solutions Manual, Section 3.1 √ 12. (3 − 5 2)2 solution √ √ 2 √ (3 − 5 2)2 = 32 − 2 · 3 · 5 · 2 + 52 · 2 √ = 9 − 30 2 + 25 · 2 √ = 59 − 30 2 Exercise 12 Instructor’s Solutions Manual, Section 3.1 13. (2 + √ Exercise 13 3)4 solution Note that (2 + √ √ 2 3)4 = (2 + 3)2 . √ Thus first we need to compute (2 + 3)2 . We already did that in Exercise 9, getting √ √ (2 + 3)2 = 7 + 4 3. Thus (2 + √ √ 2 3)4 = (2 + 3)2 √ = (7 + 4 3)2 √ √ 2 = 72 + 2 · 7 · 4 · 3 + 4 2 · 3 √ = 49 + 56 3 + 16 · 3 √ = 97 + 56 3. Instructor’s Solutions Manual, Section 3.1 14. (3 + √ Exercise 14 2)4 solution Note that (3 + √ √ 2 2)4 = (3 + 2)2 . √ Thus first we need to compute (3 + 2)2 . We already did that in Exercise 10, getting √ √ (3 + 2)2 = 11 + 6 2. Thus (3 + √ √ 2 2)4 = (3 + 2)2 √ = (11 + 6 2)2 √ √ 2 = 112 + 2 · 11 · 6 · 2 + 62 · 2 √ = 121 + 132 2 + 36 · 2 √ = 193 + 132 2. Instructor’s Solutions Manual, Section 3.1 15. (3 + √ x)2 solution (3 + √ √ √ 2 x)2 = 32 + 2 · 3 · x + x √ =9+6 x+x Exercise 15 Instructor’s Solutions Manual, Section 3.1 16. (5 + √ x)2 solution (5 + √ √ √ 2 x)2 = 52 + 2 · 5 · x + x √ = 25 + 10 x + x Exercise 16 Instructor’s Solutions Manual, Section 3.1 17. (3 − √ 2x)2 solution (3 − √ √ √ 2 2x)2 = 32 − 2 · 3 · 2x + 2x √ = 9 − 6 2x + 2x Exercise 17 Instructor’s Solutions Manual, Section 3.1 18. (5 − √ 3x)2 solution (5 − √ √ √ 2 3x)2 = 52 − 2 · 5 · 3x + 3x √ = 25 − 10 3x + 3x Exercise 18 Instructor’s Solutions Manual, Section 3.1 √ 19. (1 + 2 3x)2 solution √ √ 2 √ (1 + 2 3x)2 = 12 + 2 · 2 · 3x + 22 · 3x √ = 1 + 4 3x + 4 · 3x √ = 1 + 4 3x + 12x Exercise 19 Instructor’s Solutions Manual, Section 3.1 √ 20. (3 + 2 5x)2 solution √ √ 2 √ (3 + 2 5x)2 = 32 + 2 · 3 · 2 · 5x + 22 · 5x √ = 9 + 12 5x + 4 · 5x √ = 9 + 12 5x + 20x Exercise 20 Instructor’s Solutions Manual, Section 3.1 Exercise 21 For Exercises 21–32, find a formula for the inverse function f −1 of the indicated function f . 21. f (x) = x 9 solution By the definition of roots, the inverse of f is the function f −1 defined by f −1 (y) = y 1/9 . Instructor’s Solutions Manual, Section 3.1 Exercise 22 22. f (x) = x 12 solution By the definition of roots, the inverse of f is the function f −1 defined by f −1 (y) = y 1/12 . Instructor’s Solutions Manual, Section 3.1 Exercise 23 23. f (x) = x 1/7 solution By the definition of roots, f = g −1 , where g is the function −1 defined by g(y) = y 7 . Thus f −1 = (g −1 ) = g. In other words, f −1 (y) = y 7 . Instructor’s Solutions Manual, Section 3.1 Exercise 24 24. f (x) = x 1/11 solution By the definition of roots, f = g −1 , where g is the function −1 defined by g(y) = y 11 . Thus f −1 = (g −1 ) = g. In other words, f −1 (y) = y 11 . Instructor’s Solutions Manual, Section 3.1 Exercise 25 25. f (x) = x −2/5 solution To find a formula for f −1 , we solve the equation x −2/5 = y 5 for x. Raising both sides of this equation to the power − 2 , we get −5/2 . Hence x=y f −1 (y) = y −5/2 . Instructor’s Solutions Manual, Section 3.1 Exercise 26 26. f (x) = x −17/7 solution To find a formula for f −1 , we solve the equation x −17/7 = y 7 for x. Raising both sides of this equation to the power − 17 , we get −7/17 . Hence x=y f −1 (y) = y −7/17 . Instructor’s Solutions Manual, Section 3.1 27. f (x) = Exercise 27 x4 81 4 solution To find a formula for f −1 , we solve the equation x81 = y for x. Multiplying both sides by 81 and then raising both sides of this 1 equation to the power 4 , we get x = (81y)1/4 = 811/4 y 1/4 = 3y 1/4 . Hence f −1 (y) = 3y 1/4 . Instructor’s Solutions Manual, Section 3.1 Exercise 28 28. f (x) = 32x 5 solution To find a formula for f −1 , we solve the equation 32x 5 = y for x. Dividing both sides by 32 and then raising both sides of this 1 equation to the power 5 , we get x = (y/32)1/5 = y 1/5 /321/5 = y 1/5 /2. Hence y 1/5 . f −1 (y) = 2 Instructor’s Solutions Manual, Section 3.1 Exercise 29 29. f (x) = 6 + x 3 solution To find a formula for f −1 , we solve the equation 6 + x 3 = y for x. Subtracting 6 from both sides and then raising both sides of this 1 equation to the power 3 , we get x = (y − 6)1/3 . Hence f −1 (y) = (y − 6)1/3 . Instructor’s Solutions Manual, Section 3.1 Exercise 30 30. f (x) = x 6 − 5 solution To find a formula for f −1 , we solve the equation x 6 − 5 = y for x. Adding 5 to both sides and then raising both sides of this 1 equation to the power 6 , we get x = (y + 5)1/6 . Hence f −1 (y) = (y + 5)1/6 . Instructor’s Solutions Manual, Section 3.1 Exercise 31 31. f (x) = 4x 3/7 − 1 solution To find a formula for f −1 , we solve the equation 4x 3/7 − 1 = y for x. Adding 1 to both sides, then dividing both sides by 7 4, and then raising both sides of this equation to the power 3 , we get y+1 7/3 . Hence x= 4 y + 1 7/3 . f −1 (y) = 4 Instructor’s Solutions Manual, Section 3.1 Exercise 32 32. f (x) = 7 + 8x 5/9 solution To find a formula for f −1 , we solve the equation 7 + 8x 5/9 = y for x. Subtracting 7 from both sides, then dividing both 9 sides by 8, and then raising both sides of this equation to the power 5 , y−7 9/5 . Hence we get x = 8 f −1 (y) = y − 7 9/5 . 8 Instructor’s Solutions Manual, Section 3.1 Exercise 33 For Exercises 33–38, find a formula for (f ◦ g)(x) assuming that f and g are the indicated functions. 33. f (x) = x 1/2 and g(x) = x 3/7 solution 1/2 (f ◦ g)(x) = f g(x) = f (x 3/7 ) = (x 3/7 ) = x 3/14 Instructor’s Solutions Manual, Section 3.1 34. f (x) = x 5/3 Exercise 34 and g(x) = x 4/9 solution 5/3 (f ◦ g)(x) = f g(x) = f (x 4/9 ) = (x 4/9 ) = x 20/27 Instructor’s Solutions Manual, Section 3.1 35. f (x) = 3 + x 5/4 and Exercise 35 g(x) = x 2/7 solution 5/4 (f ◦ g)(x) = f g(x) = f (x 2/7 ) = 3 + (x 2/7 ) = 3 + x 5/14 Instructor’s Solutions Manual, Section 3.1 36. f (x) = x 2/3 − 7 and Exercise 36 g(x) = x 9/16 solution 2/3 −7 (f ◦ g)(x) = f g(x) = f (x 9/16 ) = (x 9/16 ) = x 3/8 − 7 Instructor’s Solutions Manual, Section 3.1 37. f (x) = 5x √ 2 and g(x) = x Exercise 37 √ 8 solution √ (f ◦ g)(x) = f g(x) = f (x 8 ) = 5(x √ 8 √ 2 ) = 5x √ 16 = 5x 4 Instructor’s Solutions Manual, Section 3.1 38. f (x) = 7x √ 12 and g(x) = x Exercise 38 √ 3 solution √ (f ◦ g)(x) = f g(x) = f (x 3 ) = 7(x √ 3 √ ) 12 = 7x √ 36 = 7x 6 Instructor’s Solutions Manual, Section 3.1 Exercise 39 For Exercises 39–46, find all real numbers x that satisfy the indicated equation. √ 39. x − 5 x + 6 = 0 √ solution This equation involves x; thus we make the substitution √ √ x = y. Squaring both sides of the equation x = y gives x = y 2 . With these substitutions, the equation above becomes y 2 − 5y + 6 = 0. This new equation can now be solved either by factoring the left side or by using the quadratic formula (see Section 2.2). Let’s factor the left side, getting (y − 2)(y − 3) = 0. Thus y = 2 or y = 3 (the same result could have been obtained by using the quadratic formula). √ √ √ Substituting x for y now shows that x = 2 or x = 3. Thus x = 4 or x = 9. Instructor’s Solutions Manual, Section 3.1 Exercise 40 √ 40. x − 7 x + 12 = 0 √ solution This equation involves x; thus we make the substitution √ √ x = y. Squaring both sides of the equation x = y gives x = y 2 . With these substitutions, the equation above becomes y 2 − 7y + 12 = 0. This new equation can now be solved either by factoring the left side or by using the quadratic formula (see Section 2.2). Let’s factor the left side, getting (y − 3)(y − 4) = 0. Thus y = 3 or y = 4 (the same result could have been obtained by using the quadratic formula). √ √ √ Substituting x for y now shows that x = 3 or x = 4. Thus x = 9 or x = 16. Instructor’s Solutions Manual, Section 3.1 41. x − √ Exercise 41 x=6 √ solution This equation involves x; thus we make the substitution √ √ x = y. Squaring both sides of the equation x = y gives x = y 2 . Making these substitutions and subtracting 6 from both sides, we have y 2 − y − 6 = 0. This new equation can now be solved either by factoring the left side or by using the quadratic formula. Let’s use the quadratic formula, getting y= 1± √ 1 + 24 1±5 = . 2 2 Thus y = 3 or y = −2 (the same result could have been obtained by factoring). √ √ √ Substituting x for y now shows that x = 3 or x = −2. The first possibility corresponds to the solution x = 9. There are no real √ numbers x such that x = −2. Thus x = 9 is the only solution to this equation. Instructor’s Solutions Manual, Section 3.1 42. x − √ Exercise 42 x = 12 √ solution This equation involves x; thus we make the substitution √ √ x = y. Squaring both sides of the equation x = y gives x = y 2 . Making these substitutions and subtracting 12 from both sides, we have y 2 − y − 12 = 0. This new equation can now be solved either by factoring the left side or by using the quadratic formula. Let’s use the quadratic formula, getting y= 1± √ 1 + 48 1±7 = . 2 2 Thus y = 4 or y = −3 (the same result could have been obtained by factoring). √ √ √ Substituting x for y now shows that x = 4 or x = −3. The first possibility corresponds to the solution x = 16. There are no real √ numbers x such that x = −3. Thus x = 16 is the only solution to this equation. Instructor’s Solutions Manual, Section 3.1 Exercise 43 43. x 2/3 − 6x 1/3 = −8 solution This equation involves x 1/3 and x 2/3 ; thus we make the substitution x 1/3 = y. Squaring both sides of the equation x 1/3 = y gives x 2/3 = y 2 . Making these substitutions and adding 8 to both sides, we have y 2 − 6y + 8 = 0. This new equation can now be solved either by factoring the left side or by using the quadratic formula. Let’s factor the left side, getting (y − 2)(y − 4) = 0. Thus y = 2 or y = 4 (the same result could have been obtained by using the quadratic formula). Substituting x 1/3 for y now shows that x 1/3 = 2 or x 1/3 = 4. Thus x = 23 or x = 43 . In other words, x = 8 or x = 64. Instructor’s Solutions Manual, Section 3.1 Exercise 44 44. x 2/3 + 3x 1/3 = 10 solution This equation involves x 1/3 and x 2/3 ; thus we make the substitution x 1/3 = y. Squaring both sides of the equation x 1/3 = y gives x 2/3 = y 2 . Making these substitutions and subtracting 10 from both sides, we have y 2 + 3y − 10 = 0. This new equation can now be solved either by factoring the left side or by using the quadratic formula. Let’s factor the left side, getting (y − 2)(y + 5) = 0. Thus y = 2 or y = −5 (the same result could have been obtained by using the quadratic formula). Substituting x 1/3 for y now shows that x 1/3 = 2 or x 1/3 = −5. Thus x = 23 or x = (−5)3 . In other words, x = 8 or x = −125. Instructor’s Solutions Manual, Section 3.1 Exercise 45 45. x 4 − 3x 2 = 10 solution This equation involves x 2 and x 4 ; thus we make the substitution x 2 = y. Squaring both sides of the equation x 2 = y gives x 4 = y 2 . Making these substitutions and subtracting 10 from both sides, we have y 2 − 3y − 10 = 0. This new equation can now be solved either by factoring the left side or by using the quadratic formula. Let’s factor the left side, getting (y − 5)(y + 2) = 0. Thus y = 5 or y = −2 (the same result could have been obtained by using the quadratic formula). Substituting x 2 for y now shows that x 2 = 5 or x 2 = −2. The first of √ √ these equations implies that x = 5 or x = − 5; the second equation is not satisfied by any real value of x. In other words, the original √ √ equation implies that x = 5 or x = − 5. Instructor’s Solutions Manual, Section 3.1 Exercise 46 46. x 4 − 8x 2 = −15 solution This equation involves x 2 and x 4 ; thus we make the substitution x 2 = y. Squaring both sides of the equation x 2 = y gives x 4 = y 2 . Making these substitutions and adding 15 to both sides, we have y 2 − 8y + 15 = 0. This new equation can now be solved either by factoring the left side or by using the quadratic formula. Let’s factor the left side, getting (y − 3)(y − 5) = 0. Thus y = 3 or y = 5 (the same result could have been obtained by using the quadratic formula). Substituting x 2 for y now shows that x 2 = 3 or x 2 = 5. Thus the √ √ √ original equation has four solutions: x = 3, x = − 3, x = 5, or √ x = − 5. Instructor’s Solutions Manual, Section 3.1 Exercise 47 47. Suppose x is a number such that 3x = 4. Evaluate 3−2x . solution −2 3−2x = (3x ) = 4−2 = 1 42 = 1 16 Instructor’s Solutions Manual, Section 3.1 Exercise 48 48. Suppose x is a number such that 2x = 13 . Evaluate 2−4x . solution −4 2−4x = (2x ) = 1 −4 3 = 34 = 81 Instructor’s Solutions Manual, Section 3.1 Exercise 49 49. Suppose x is a number such that 2x = 5. Evaluate 8x . solution x 8x = (23 ) = 23x 3 = (2x ) = 53 = 125 Instructor’s Solutions Manual, Section 3.1 Exercise 50 50. Suppose x is a number such that 3x = 5. Evaluate 1 x x = (3−2 ) solution 9 = 3−2x −2 = (3x ) = 5−2 = 1 25 1 x 9 . Instructor’s Solutions Manual, Section 3.1 Exercise 51 For Exercises 51–56, evaluate the indicated quantities assuming that f and g are the functions defined by f (x) = 2x and g(x) = x+1 . x+2 51. (f ◦ g)(−1) solution (f ◦ g)(−1) = f g(−1) = f (0) = 20 = 1 Instructor’s Solutions Manual, Section 3.1 Exercise 52 52. (g ◦ f )(0) solution (g ◦ f )(0) = g f (0) = g(20 ) = g(1) = 2 3 Instructor’s Solutions Manual, Section 3.1 53. (f ◦ g)(0) solution 1 (f ◦ g)(0) = f g(0) = f ( 2 ) = 21/2 ≈ 1.414 Exercise 53 Instructor’s Solutions Manual, Section 3.1 54. Exercise 54 (g ◦ f )( 32 ) solution 3 23/2 + 1 3 (g ◦ f )( 2 ) = g f ( 2 ) = g(23/2 ) = 3/2 2 +2 ≈ 0.793 Instructor’s Solutions Manual, Section 3.1 55. Exercise 55 (f ◦ f )( 12 ) solution 1 1 (f ◦ f )( 2 ) = f f ( 2 ) = f (21/2 ) ≈ f (1.41421) = 21.41421 ≈ 2.66514 Instructor’s Solutions Manual, Section 3.1 56. Exercise 56 (f ◦ f )( 35 ) solution 3 3 (f ◦ f )( 5 ) = f f ( 5 ) = f (23/5 ) ≈ f (1.51572) = 21.51572 ≈ 2.85941 Instructor’s Solutions Manual, Section 3.1 Exercise 57 57. Find an integer m such that √ 2 (3 + 2 5)2 − m is an integer. √ solution First we evaluate (3 + 2 5)2 : √ √ 2 √ (3 + 2 5)2 = 32 + 2 · 3 · 2 · 5 + 22 · 5 √ = 9 + 12 5 + 4 · 5 √ = 29 + 12 5. Thus √ √ 2 (3 + 2 5)2 − m = (29 + 12 5 − m)2 . If we choose m = 29, then we have √ √ 2 (3 + 2 5)2 − m = (12 5)2 = 122 · √ 2 5 = 122 · 5, which is an integer. Any choice other than m = 29 will leave a term √ √ involving 5 when (29 + 12 5 − m)2 is expanded. Thus m = 29 is the only solution to this exercise. Instructor’s Solutions Manual, Section 3.1 Exercise 58 58. Find an integer m such that √ 2 2 (5 − 2 3) − m is an integer. √ solution First we evaluate (5 − 2 3)2 : √ √ 2 √ (5 − 2 3)2 = 52 − 2 · 5 · 2 · 3 + 22 · 3 √ = 25 − 20 3 + 4 · 3 √ = 37 − 20 3. Thus √ √ 2 (5 − 2 3)2 − m = (37 − 20 3 − m)2 . If we choose m = 37, then we have √ √ 2 (5 − 2 3)2 − m = (−20 3)2 = (−20)2 · √ 2 3 = (−20)2 · 3, which is an integer. Any choice other than m = 37 will leave a term √ √ involving 3 when (37 − 20 3 − m)2 is expanded. Thus m = 37 is the only solution to this exercise. Instructor’s Solutions Manual, Section 3.1 Problem 59 Solutions to Problems, Section 3.1 √ √ 59. Sketch the graph of the functions x + 1 and x + 1 on the interval [0, 4]. √ √ solution The graphs of x + 1 (blue) and x + 1 (red) on the interval [0, 4] are shown below. y 3 2 1 1 2 3 4 x √ The graph of x + 1 is obtained by shifting up one unit the graph of √ x on the interval [0, 4]. √ The graph of x + 1 is obtained by shifting left one unit the graph of √ x on the interval [1, 5]. Instructor’s Solutions Manual, Section 3.1 Problem 60 60. Sketch the graph of the functions 2x 1/3 and (2x)1/3 on the interval [0, 8]. solution The graphs of 2x 1/3 (blue) and (2x)1/3 (red) on the interval [0, 8] are shown below. y 4 3 2 1 2 4 6 8 x The graph of 2x 1/3 is obtained by vertically stretching the graph of x 1/3 on the interval [0, 8] by a factor of 2. Note that (2x)1/3 = 21/3 x 1/3 . Thus the graph of (2x)1/3 is obtained by vertically stretching the graph of x 1/3 on the interval [0, 8] by a factor of 21/3 (approximately 1.26). Instructor’s Solutions Manual, Section 3.1 Problem 61 61. Sketch the graphs of the functions x 1/4 and x 1/5 on the interval [0, 81]. solution The graphs of x 1/4 (blue) and x 1/5 (red) on the interval [0, 81] are shown below. For x in the interval (1, 81), we have the inequality x 1/4 > x 1/5 . However, for x in the interval (0, 1), the reverse inequality x 1/4 < x 1/5 holds (although the interval (0, 1) is such a small part of the interval [0, 81] that this behavior is difficult to see in the graph below): y 3 8115 1 81 x Instructor’s Solutions Manual, Section 3.1 62. Show that Problem 62 √ 2 + 3 = 32 + 12 . solution We need to verify that the square of Here is that computation: 3 2 + 2 1 2 = 3 2 +2 =2+ √ 3 3 2 · 1 2 + 3 2 1 2 + 1 2 equals 2 + √ 3. Instructor’s Solutions Manual, Section 3.1 63. Show that Problem 63 √ 2 − 3 = 32 − 12 . solution We need to verify that the square of Here is that computation: 3 2 − 2 1 2 = 3 2 −2 =2− √ 3 3 2 · 1 2 + 3 2 1 2 − 1 2 equals 2 − √ 3. Instructor’s Solutions Manual, Section 3.1 64. Show that Problem 64 √ √ 9 − 4 5 = 5 − 2. solution We need to verify that the square of Here is that computation: √ √ √ ( 5 − 2)2 = 5 − 2 5 · 2 + 4 √ =9−4 5 √ 5 − 2 equals 9 − 4 5. Instructor’s Solutions Manual, Section 3.1 Problem 65 √ √ 65. Show that (23 − 8 7)1/2 = 4 − 7. √ √ solution We need to verify that the square of 4 − 7 equals 23 − 8 7. Here is that computation: (4 − √ √ 7)2 = 16 − 2 · 4 7 + 7 √ = 23 − 8 7 Instructor’s Solutions Manual, Section 3.1 Problem 66 66. Make up a problem similar in form to the problem above, without duplicating anything in this book. solution In general there is no way to simplify an expression of the √ form m + n k. Thus the way to make up a problem similar to the previous one, start with an expression of the desired form and square √ it. For example, if we start by squaring 5 − 3, then we have (5 − √ √ 3)2 = 25 − 2 · 5 3 + 3 √ = 28 − 10 3. Thus we can now make up the following problem: √ √ 1 Show that 28 − 10 3 2 = 5 − 3. Instructor’s Solutions Manual, Section 3.1 Problem 67 √ √ 67. Show that (99 + 70 2)1/3 = 3 + 2 2. √ solution We need to verify that the cube of 3 + 2 2 equals √ 99 + 70 2. Here is that computation: √ √ √ (3 + 2 2)3 = (3 + 2 2)2 (3 + 2 2) √ √ = (9 + 2 · 3 · 2 2 + 8)(3 + 2 2) √ √ = (17 + 12 2)(3 + 2 2) √ √ = 51 + 34 2 + 36 2 + 24 · 2 √ = 99 + 70 2 Instructor’s Solutions Manual, Section 3.1 Problem 68 √ √ 68. Show that (−37 + 30 3)1/3 = −1 + 2 3. √ solution We need to verify that the cube of −1 + 2 3 equals √ −37 + 30 3. Here is that computation: √ √ √ (−1 + 2 3)3 = (−1 + 2 3)2 (−1 + 2 3) √ √ = (1 − 2 · 2 3 + 12)(−1 + 2 3) √ √ = (13 − 4 3)(−1 + 2 3) √ √ = −13 + 26 3 + 4 3 − 8 · 3 √ = −37 + 30 3 Instructor’s Solutions Manual, Section 3.1 69. Show that if x and y are positive numbers with x = y, then √ x−y √ √ = x + y. x− y solution √ √ x+ y x−y x−y √ √ = √ √ ·√ √ x− y x− y x+ y = = √ √ (x − y)( x + y) x−y √ x+ y Problem 69 Instructor’s Solutions Manual, Section 3.1 70. Explain why Problem 70 10100 ( 10200 + 1 − 10100 ) is approximately equal to 1 2. solution 10100 ( 10200 + 1 − 10100 ) √ ( 10200 + 1 + 10100 ) = 10100 ( 10200 + 1 − 10100 ) · √ ( 10200 + 1 + 10100 ) = 10100 (10200 + 1 − 10200 ) √ 10200 + 1 + 10100 10100 = √ 200 10 + 1 + 10100 ≈ = 10100 + 10100 10100 1 2 Instructor’s Solutions Manual, Section 3.1 Problem 71 √ 71. Explain why the equation x 2 = x is not valid for all real numbers x √ and should be replaced by the equation x 2 = |x|. √ solution Note that (−3)2 = 9 = 3. Thus for x = −3, it is not true √ that x 2 = x. More generally, if t is a a positive number, then there are two real numbers whose square equals t; the positive number whose square √ equals t is defined to be t. Now for any real number x = 0, the number x 2 is positive, and we can apply the remark above to t = x 2 . Specifically, there exist two real numbers whose square equals x 2 ; these two numbers are x and −x. By √ the remark above, x 2 equals x if x > 0 and equals −x if x < 0. In √ other words, we have x 2 = |x| (derived here under the assumption that x = 0, but the desired equation also obviously holds when x = 0). Instructor’s Solutions Manual, Section 3.1 Problem 72 √ 72. Explain why the equation x 8 = x 4 is valid for all real numbers x, with no necessity for using absolute value. solution Suppose x is a real number with x = 0. Then x 8 is a positive number. The two numbers whose square equals x 8 are x 4 and −x 4 . Regardless of whether x is positive or negative, x 4 is positive. √ Thus x 8 = x 4 (derived here under the assumption that x = 0, but the desired equation also obviously holds when x = 0). Instructor’s Solutions Manual, Section 3.1 Problem 73 73. Show that if x and y are positive numbers, then x+y < √ x + y. [In particular, if x and y are positive numbers, then √ √ √ x + y = x + y.] solution Suppose x and y are positive numbers. Then √ x+y <x+2 x y +y √ = ( x + y)2 . Because everything in sight is positive, we can take square roots of both sides of the inequality above, preserving the direction of the inequality, getting √ x + y < x + y. Instructor’s Solutions Manual, Section 3.1 Problem 74 74. Show that if 0 < x < y, then √ y − x < y − x. solution Because x < y, we know that y − x > 0. Applying the result from the previous problem, we have Subtracting √ √ y = x + (y − x) < x + y − x. x from both sides of the inequality above shows that √ y − x < y − x. Instructor’s Solutions Manual, Section 3.1 Problem 75 75. Explain why the spoken phrase “the square root of x plus one” could be interpreted in two different ways that would not give the same result. solution The phrase “the square root of x plus one” could mean “(the square root of x) plus one”, which in symbols is √ x + 1. Or the phrase “the square root of x plus one” could mean “the square root of (x plus one)”, which in symbols is √ x + 1. These two interpretations of the spoken phrase do not give the same result. For example, if x = 4, the first interpretation gives 3 and the √ second interpretation gives 5. Instructor’s Solutions Manual, Section 3.1 Problem 76 76. One of the graphs in this section suggests that √ √ x< 3x if 0<x<1 √ √ x> 3x if x > 1. and Explain why each of these inequalities holds. solution First suppose 0 < x < 1. Then x 1/6 < 1 because if x 1/6 were greater than or equal to 1, then multiplying x 1/6 by itself six times (which produces x) would give a number greater than or equal to 1. Multiplying both sides of the inequality above by x 1/3 and using the identity x 1/6 · x 1/3 = x 1/2 gives x 1/2 < x 1/3 , as desired. Now suppose x > 1. Then x 1/6 > 1 because if x 1/6 were less than or equal to 1, then multiplying x 1/6 by itself six times (which produces x) would give a number less than or Instructor’s Solutions Manual, Section 3.1 Problem 76 equal to 1. Multiplying both sides of the inequality above by x 1/3 and using the identity x 1/6 · x 1/3 = x 1/2 gives x 1/2 > x 1/3 , as desired. Instructor’s Solutions Manual, Section 3.1 Problem 77 77. What is the domain of the function (3 + x)1/4 ? solution For (3 + x)1/4 to be defined as a real number, we must have 3 + x ≥ 0, which is equivalent to the inequality x ≥ −3. Thus the domain of the function (3 + x)1/4 is the interval [−3, ∞). Instructor’s Solutions Manual, Section 3.1 Problem 78 78. What is the domain of the function (1 + x 2 )1/8 ? solution Because x 2 ≥ 0 for every real number x, we have 1 + x 2 ≥ 1 for every real number x. In particular, 1 + x 2 is positive for every real number x, which means that (1 + x 2 )1/8 is defined as a real number for every real number x. Thus the domain of the function (1 + x 2 )1/8 is the set of real numbers. Instructor’s Solutions Manual, Section 3.1 Problem 79 79. Suppose x is a positive number. Using only the definitions of roots and integer powers, explain why 3 6 (x 1/2 ) = (x 1/4 ) . solution We have (x 1/4 )2 2 = (x 1/4 )2 · (x 1/4 )2 = x 1/4 · x 1/4 · x 1/4 · x 1/4 = x, where the last equality above holds because of the definition of x 1/4 . 2 The equation (x 1/4 )2 = x states that (x 1/4 )2 is a number whose square equals x. Thus (x 1/4 )2 must equal x 1/2 (because everything in sight is positive). Now if we cube both sides of the equality x 1/2 = (x 1/4 )2 , we get 3 (x 1/2 ) = (x 1/4 )2 · (x 1/4 )2 · (x 1/4 )2 = x 1/4 · x 1/4 · x 1/4 · x 1/4 · x 1/4 · x 1/4 6 = (x 1/4 ) , as desired. Instructor’s Solutions Manual, Section 3.1 Problem 80 80. Suppose x is a positive number and n is a positive integer. Using only the definitions of roots and integer powers, explain why n 2n (x 1/2 ) = (x 1/4 ) . solution In the solution to the previous problem we showed that x 1/2 = (x 1/4 )2 . Raise both sides of this equality to the nth power, getting n n (x 1/2 ) = (x 1/4 )2 = (x 1/4 )2 · (x 1/4 )2 · · · · · (x 1/4 )2 n times = x 1/4 ·x 1/4 1/4 · · · · · x 2n times = (x as desired. 1/4 2n ) , Instructor’s Solutions Manual, Section 3.1 Problem 81 81. Suppose x is a positive number and n and p are positive integers. Using only the definitions of roots and integer powers, explain why np n (x 1/2 ) = (x 1/(2p) ) . solution We have (x 1/(2p) )p 2 = (x 1/(2p) )p · (x 1/(2p) )p 1/(2p) = x 1/(2p) · x 1/(2p) · · · · · x 2p times = x, where the last equality above holds because of the definition of x 1/(2p) . 2 The equation (x 1/(2p) )p = x states that (x 1/(2p) )p is a number whose square equals x. Thus (x 1/(2p) )p must equal x 1/2 (because everything in sight is positive). Raise both sides of the equality x 1/2 = (x 1/(2p) )p to the nth power, getting n n (x 1/2 ) = (x 1/(2p) )p = (x 1/(2p) )p · (x 1/(2p) )p · · · · · (x 1/(2p) )p n times = x 1/(2p) ·x 1/(2p) · · · · · x np times np = (x 1/(2p) ) . 1/(2p) Instructor’s Solutions Manual, Section 3.1 Problem 82 82. Suppose x is a positive number and m, n, and p are positive integers. Using only the definitions of roots and integer powers, explain why np n (x 1/m ) = (x 1/(mp) ) . solution We have (x 1/(mp) )p m = (x 1/(mp) )p · (x 1/(mp) )p · · · · · (x 1/(mp) )p m times 1/(mp) = x 1/(mp) · x 1/(mp) · · · · · x mp times = x, where the last equality above holds because of the definition of x 1/(mp) . m = x states that (x 1/(mp) )p is a number The equation (x 1/(mp) )p th whose m power equals x. Thus (x 1/(mp) )p must equal x 1/m (because everything in sight is positive). Raise both sides of the equality x 1/m = (x 1/(mp) )p to the nth power, getting Instructor’s Solutions Manual, Section 3.1 Problem 82 n n (x 1/m ) = (x 1/(mp) )p = (x 1/(mp) )p · (x 1/(mp) )p · · · · · (x 1/(mp) )p n times 1/(mp) = x 1/(mp) · x 1/(mp) · · · · · x np times np = (x 1/(mp) ) as desired. , Instructor’s Solutions Manual, Section 3.1 Problem 83 83. Using the result from the problem above, explain why the definition of exponentiation of a positive number by a positive rational number gives the same result even if the positive rational number is not expressed in reduced form. solution Suppose x is a positive number and m and n are positive m integers such that n is in reduced form (meaning that m and n have no n common factors other than 1). Then x n/m is defined to equal (x 1/m ) . However, suppose p is a positive integer. Then np mp np mp equals n m, although is not in reduced form. If we apply the definition of exponentiation by a rational number to x (np)/(mp) , we see that x (np)/(mp) is defined to np equal (x 1/(mp) ) , which the previous problem tells us is equal to n (x 1/m ) . Thus we get the same result even if we use a fraction not in reduced form. Instructor’s Solutions Manual, Section 3.1 84. Using the result that 25/2 is irrational. Problem 84 √ 2 is irrational (proved in Section 0.1), show that solution Note that √ 2= 25/2 25/2 25/2 . = = 24/2 22 4 √ Thus if 25/2 were a rational number, then we would have written 2 as √ the quotient of two rational numbers, which would mean that 2 would be a rational number (which is not true). Thus 25/2 is not a rational number. Instructor’s Solutions Manual, Section 3.1 85. Using the result that solution Note that Problem 85 √ 2 is irrational, explain why 21/6 is irrational. √ 3 2 = 21/6 . √ Thus if 21/6 were a rational number, then we would have written 2 as √ the cube of a rational number, which would mean that 2 would be a rational number (which is not true). Thus 21/6 is not a rational number. Instructor’s Solutions Manual, Section 3.1 Problem 86 86. Suppose you have a calculator that can only compute square roots. Explain how you could use this calculator to compute 71/8 . solution Note that 1/2 1/2 71/8 = (71/2 ) , which can be written as 7 1/8 # √ = 7. Thus to calculate 71/8 on a calculator that can only compute square roots, enter 7 and then press the square-root button three times. Instructor’s Solutions Manual, Section 3.1 Problem 87 87. Suppose you have a calculator that can only compute square roots and can multiply. Explain how you could use this calculator to compute 73/4 . solution Note that 1/2 73/4 = 71/2 · 71/4 = 71/2 · (71/2 ) , which can be written as 73/4 = √ √ 7· 7. Thus to calculate 73/4 on a calculator that can only compute square roots and can multiply, enter 7 and then press the square-root button, then multiply that result by the number obtained by entering 7 and pressing the square-root button twice. Instructor’s Solutions Manual, Section 3.1 Problem 88 88. Give an example of three irrational numbers x, y, and z such that xyz is a rational number. solution Note that √ 2 2 = 21/4 . √ Thus if 21/4 were a rational number, then we would have written 2 as √ the square of a rational number, which would mean that 2 would be a rational number (which is not true). Thus 21/4 is not a rational number. Now take x = 21/4 , y = 21/4 , and z = 21/2 . Then x, y, and z are irrational numbers and xyz = 21/4 · 21/4 · 21/2 = 2. Instructor’s Solutions Manual, Section 3.1 Problem 89 89. Give an example of three irrational numbers x, y, and z such that z (x y ) is a rational number. solution Let x = y = z = z √ 2. Then (x y ) = x yz = √ √2·√2 √ 2 2 = 2 = 2. Instructor’s Solutions Manual, Section 3.1 Problem 90 90. Is the function f defined by f (x) = 2x for every real number x an even function, an odd function, or neither? solution Note that f (−1) = 12 and f (1) = 2. Thus f (−1) = f (1) and f (−1) = −f (1). Hence f is neither an even function nor an odd function. Instructor’s Solutions Manual, Section 3.1 Problem 91 91. What is wrong with the following string of equalities, which seems to show that −1 = 1? −1 = i · i = −1 −1 = (−1)(−1) = 1 = 1 solution The error in the string of equalities above is with the “equality” −1 −1 = (−1)(−1), which would follow from the identity a b = ab by taking a = −1 and b = −1. However, as stated in the text, this identity assumes that a and b are positive. Thus this identity cannot be used with a = −1 and b = −1.