ALGEBRA AND TRIGONOMETRY 5TH EDITION BY JAMES STEWART, LOTHAR REDLIN, SALEEM WATSON TEST BANK

Page 1


Corrections: p. 1, 3, 5, 36 NOTE THAT THESE CORRECTIONS SHOULD BE MADE IN CA-8 ALSO.

CHAPTER P

PREREQUISITES

P.1

Modeling the Real World with Algebra 1

P.2 P.3

Real Numbers 2 Integer Exponents and Scientific Notation 7

P.4

Rational Exponents and Radicals 12

P.5

Algebraic Expressions 16

P.6

Factoring 19

P.7

Rational Expressions 24

P.8

Solving Basic Equations 31

P.9

Modeling with Equations 36 Chapter P Review 42 Chapter P Test 48

¥

FOCUS ON MODELING: Making Optimal Decisions 51

1


P

PREREQUISITES

P.1

MODELING THE REAL WORLD WITH ALGEBRA

1. Using this model, we find that 15 cars have W  4 15  60 wheels. To find the number of cars that have a total of W .If the cars in a parking lot have a total of 124 wheels, we find that there are W wheels, we write W  4X  X  4 X  124 4  31 cars in the lot.

2. If each gallon of gas costs $350, then x gallons of gas costs $35x. Thus, C  35x. We find that 12 gallons of gas would cost C  35 12  $42. 3. If x  $120 and T  006x, then T  006 120  72. The sales tax is $720. 4. If x  62,000 and T  0005x, then T  0005 62,000  310. The wage tax is $310. 5. If   70, t  35, and d  t, then d  70  35  245. The car has traveled 245 miles.   6. V  r 2 h   32 5  45  1414 in3 240 N   30 miles/gallon G 8 175 175 G  7 gallons (b) 25  G  25 

8. (a) T  70  0003h  70  0003 1500  655 F

7. (a) M 

(b) 64  70  0003h  0003h  6  h  2000 ft

  10. (a) P  006s 3  006 123  1037 hp

9. (a) V  95S  95 4 km3  38 km3 (b) 19 km3  95S  S  2 km3 11. (a)

(b) 75  006s 3  s 3  125 so s  5 knots

Depth (ft)

Pressure (lb/in2 )

0

045 0  147  147

10

045 10  147  192

20

045 20  147  237

30

Insert solution for part (c)

(b) We know that P  30 and we want to find d, so we solve the equation 30  147  045d  153  045d 

153  340. Thus, if the pressure is 30 lb/in2 , the depth 045 is 34 ft. d

045 30  147  282

40

045 40  147  327

50

045 50  147  372

60

045 60  147  417 (b) We solve the equation 40x  120,000 

12. (a) Population

Water use (gal)

0

0

1000

40 1000  40,000

2000 3000 4000 5000

x

120,000  3000. Thus, the population is about 3000. 40

40 2000  80,000

40 3000  120,000 40 4000  160,000 40 5000  200,000

13. The number N of cents in q quarters is N  25q. ab 14. The average A of two numbers, a and b, is A  . 2

1


2

CHAPTER P Prerequisites

15. The cost C of purchasing x gallons of gas at $350 a gallon is C  35x.

16. The amount T of a 15% tip on a restaurant bill of x dollars is T  015x. 17. The distance d in miles that a car travels in t hours at 60 mi/h is d  60t. d 18. The speed r of a boat that travels d miles in 3 hours is r  . 3 19. (a) $12  3 $1  $12  $3  $15 (b) The cost C, in dollars, of a pizza with n toppings is C  12  n.

(c) Using the model C  12  n with C  16, we get 16  12  n  n  4. So the pizza has four toppings.

20. (a) 3 30  280 010  90  28  $118         daily days cost miles (b) The cost is    , so C  30n  01m. rental rented per mile driven (c) We have C  140 and n  3. Substituting, we get 140  30 3  01m  140  90  01m  50  01m  m  500. So the rental was driven 500 miles.

21. (a) (i) For an all­electric car, the energy cost of driving x miles is Ce  004x.

(ii) For an average gasoline powered car, the energy cost of driving x miles is C g  012x.

(b) (i) The cost of driving 10,000 miles with an all­electric car is Ce  004 10,000  $400.

(ii) The cost of driving 10,000 miles with a gasoline powered car is C g  012 10,000  $1200.

22. (a) If the width is 20, then the length is 40, so the volume is 20  20  40  16,000 in3 .

(b) In terms of width, V  x  x  2x  2x 3 . 4a  3b  2c  d 4a  3b  2c  1d  0 f  . 23. (a) The GPA is abcd  f abcd  f (b) Using a  2  3  6, b  4, c  3  3  9, and d  f  0 in the formula from part (a), we find the GPA to be 54 463429   284. 649 19

P.2

THE REAL NUMBERS

1. (a) The natural numbers are 1 2 3   .

(b) The numbers     3 2 1 0 are integers but not natural numbers. p 5 , 1729 . (c) Any irreducible fraction with q  1 is rational but is not an integer. Examples: 32 ,  12 23 q   p (d) Any number which cannot be expressed as a ratio of two integers is irrational. Examples are 2, 3, , and e. q

2. (a) ab  ba; Commutative Property of Multiplication

(b) a  b  c  a  b  c; Associative Property of Addition (c) a b  c  ab  ac; Distributive Property

3. (a) In set­builder notation: x  3  x  5 (b) In interval notation: 3 5

(c) As a graph: _3

5

4. The symbol x stands for the absolute value of the number x. If x is not 0, then the sign of x is always positive.

5. The distance between a and b on the real line is d a b  b  a. So the distance between 5 and 2 is 2  5  7. 6. (a) If a  b, then any interval between a and b (whether or not it contains either endpoint) contains infinitely many ba numbers—including, for example a  n for every positive n. (If an interval extends to infinity in either or both 2 directions, then it obviously contains infinitely many numbers.)


SECTION P.2 The Real Numbers

3

(b) No, because 5 6 does not include 5. 7. (a) No: a  b   b  a  b  a in general.

(b) No; by the Distributive Property, 2 a  5  2a  2 5  2a  10  2a  10.

8. (a) Yes, absolute values (such as the distance between two different numbers) are always positive. (b) Yes, b  a  a  b.

 9  3, 10  (b) Integers: 2,  100 2  50, 9  3, 10     (c) Rational numbers: 45  92 , 13 , 16666     53 ,  2,  100 2 , 9  3, 10   (d) Irrational numbers: 2, 314

10. (a) Natural numbers: 2,

9. (a) Natural number: 100 (b) Integers: 0, 100, 8

(c) Rational numbers: 15, 0, 52 , 271, 314, 100, 8  (d) Irrational numbers: 7, 

11. Commutative Property of addition

12. Commutative Property of multiplication

13. Associative Property of addition

14. Distributive Property

15. Distributive Property

16. Distributive Property

17. Commutative Property of multiplication

18. Distributive Property

19. x  3  3  x

20. 7 3x  7  3 x

21. 4 A  B  4A  4B

22. 5x  5y  5 x  y

23. 2 x  y  2x  2y

24. a  b 5  5a  5b   26. 43 6y  43 6 y  8y

25. 5 2x y  5  2 x y  10x y 27.  52 2x  4y   52 2x  52 4y  5x  10y

28. 3a b  c  2d  3ab  3ac  6ad

15 29 29. (a) 23  57  14 21  21  21

15 1 30. (a) 25  38  16 40  40  40

  31. (a) 23 6  32  23  6  23  32  4  1  3       1 5  4  13  1  13 (b) 3  14 1  45  12 4  4 5 5 4 5 20

2 32. (a) 2  3  2  32  23  12  3  13  93  13  83 2

33. (a) 2  3  6 and 2  72  7, so 3  72

34. (a) 3  23  2 and 3  067  201, so 23  067

5  3  10  9  1 (b) 12 8 24 24 24

(b) 6  7

wrong problem

(c) 35  72

15 4 25 (b) 32  58  16  36 24  24  24  24 2

3 2  1 2  1 2  1 45 9 (b) 15 23  51 21  51 21  10 10  12  3  3    10 15 10 5 10 5

(b) 23  067

(c) 06  06

35. (a) False

36. (a) False:

(b) True

(b) False

37. (a) True

(b) False

38. (a) True

 3  173205  17325.

(b) True


4

CHAPTER P Prerequisites

39. (a) x  0

(b) t  4

40. (a) y  0

(d) 5  x  13

(c) a   (e) 3  p  5

(b) z  3

(c) b  8

(d) 0    17

(e) y    2

41. (a) A  B  1 2 3 4 5 6 7 8

42. (a) B  C  2 4 6 7 8 9 10

(b) A  B  2 4 6

(b) B  C  8

43. (a) A  C  1 2 3 4 5 6 7 8 9 10 (b) A  C  7

44. (a) A  B  C  1 2 3 4 5 6 7 8 9 10 (b) A  B  C   46. (a) A  C  x  1  x  5

45. (a) B  C  x  x  5

(b) A  B  x  2  x  4

(b) B  C  x  1  x  4

48. 2 8]  x  2  x  8

47. 3 0  x  3  x  0 _3

0

2

    50. 6  12  x  6  x   12

49. [2 8  x  2  x  8 2

8

_6

51. [2   x  x  2

1

53. x  1  x   1]

54. 1  x  2  x  [1 2] 1

1

55. 2  x  1  x  2 1]

_5

1

58. 5  x  2  x  5 2

_1

_5

(b) 3 5]

(c) 3 

60. (a) [0 2

2

(b) 2 0]

(c)  0]

62. 2 0  1   1 0

61. 2 0  1 1  2 1 _2

2

56. x  5  x  [5 

57. x  1  x  1 

59. (a) [3 5]

1

_ _2

52.  1  x  x  1

2

_2

8

1

_1

0


SECTION P.2 The Real Numbers

63. [4 6]  [0 8  [0 6] 0

64. [4 6]  [0 8  [4 8 6

65.  4  4  _4

_4

8

66.  6]  2 10  2 6] 4

67. (a) 50  50 (b) 13  13 69. (a) 6  4  6  4  2  2 1  1  1 (b) 1 1

71. (a) 2  6  12  12      (b)   13 15  5  5

73. 2  3  5  5 75. (a) 17  2  15

(b) 21  3  21  3  24  24        3   12 55   67  67 (c)  10  11 8    40  40    40   40

2

6

68. (a) 2  8  6  6 (b) 8  2  8  2  6  6 70. (a) 2  12  2  12  10  10 (b) 1  1  1  1  1  1  1  0  1        1 1 72. (a)  6 24    4   4        5 (b)  712 127    5   1  1

74. 25  15  4  4          7 1    49  5    54    18   18 76. (a)  15   21   105 105   105   35  35 (b) 38  57  38  57  19  19.

(c) 26  18  26  18  08  08.

77. (a) Let x  0777   . So 10x  77777     x  07777     9x  7. Thus, x  79 .

13 (b) Let x  02888   . So 100x  288888     10x  28888     90x  26. Thus, x  26 90  45 . 19 (c) Let x  0575757   . So 100x  575757     x  05757     99x  57. Thus, x  57 99  33 .

78. (a) Let x  52323   . So 100x  5232323     1x  52323     99x  518. Thus, x  518 99 .

62 (b) Let x  13777   . So 100x  1377777     10x  137777     90x  124. Thus, x  124 90  45 .

1057 (c) Let x  213535   . So 1000x  21353535     10x  213535     990x  2114. Thus, x  2114 990  495 .        79.   3, so   3    3. 80. 2  1, so 1  2  2  1.

81. a  b, so a  b   a  b  b  a.

82. a  b  a  b  a  b  b  a  2b

83. (a) a is negative because a is positive.

(b) bc is positive because the product of two negative numbers is positive. (c) a  ba  b is positive because it is the sum of two positive numbers.

wrong problem

(d) ab  ac is negative: each summand is the product of a positive number and a negative number, and the sum of two negative numbers is negative. 84. (a) b is positive because b is negative.

(b) a  bc is positive because it is the sum of two positive numbers.

(c) c  a  c  a is negative because c and a are both negative. (d) ab2 is positive because both a and b2 are positive.

85. Distributive Property

5


6

CHAPTER P Prerequisites

86. (a) When L  60, x  8, and y  6, we have L  2 x  y  60  2 8  6  60  28  88. Because 88  108 the post office will accept this package. When L  48, x  24, and y  24, we have L  2 x  y  48  2 24  24  48  96  144, and since 144  108, the post office will not accept this package. (b) If x  y  9, then L  2 9  9  108  L  36  108  L  72. So the length can be as long as 72 in.  6 ft.

m1 m2 m1 m m1n2  m2n1 and y  be rational numbers. Then x  y   2  , n1 n2 n1 n2 n1 n2 m m n  m 2 n1 m m m m m , and x  y  1  2  1 2 . This shows that the sum, difference, and product xy 1  2  1 2 n1 n2 n1 n2 n1 n2 n1n2 of two rational numbers are again rational numbers. However the product of two irrational numbers is not necessarily   irrational; for example, 2  2  2, which is rational. Also, the sum of two irrational numbers is not necessarily irrational;     for example, 2   2  0 which is rational.

87. Let x 

88. 12 

       2 is irrational. If it were rational, then by Exercise 6(a), the sum 12  2   12  2 would be rational, but

this is not the case.  Similarly, 12  2 is irrational. (a) Following the hint, suppose that r  t  q, a rational number. Then by Exercise 6(a), the sum of the two rational numbers r  t and r is rational. But r  t  r  t, which we know to be irrational. This is a contradiction, and hence our original premise—that r  t is rational—was false. a (b) r is a nonzero rational number, so r  for some nonzero integers a and b. Let us assume that rt  q, a rational b a c bc c , implying number. Then by definition, q  for some integers c and d. But then r t  q  t  , whence t  d b d ad that t is rational. Once again we have arrived at a contradiction, and we conclude that the product of a rational number and an irrational number is irrational.

89. x 1 x

1

2

10

100

1000

1

1 2

1 10

1 100

1 1000

As x gets large, the fraction 1x gets small. Mathematically, we say that 1x goes to zero. x

1

05

01

001

0001

1 x

1

1 05  2

1 01  10

1 001  100

1 0001  1000

As x gets small, the fraction 1x gets large. Mathematically, we say that 1x goes to infinity.  90. We can construct the number 2 on the number line by Ï2

transferring the length of the hypotenuse of a right triangle with legs of length 1 and 1.  Similarly, to locate 5, we construct a right triangle with legs of length 1 and 2. By the Pythagorean Theorem, the length   of the hypotenuse is 12  22  5. Then transfer the

length of the hypotenuse to the number line.

_1

0

1 1 Ï5

_1

0

1

Ï2

2

3

2 Ï5

3

1

The square root of any rational number can be located on a number line in this fashion. The circle in the second figure in the text has circumference , so if we roll it along a number line one full rotation, we have found  on the number line. Similarly, any rational multiple of  can be found this way.


SECTION P.3 Integer Exponents and Scientific Notation

7

abab a  b  a  b   a. 2 2 On the other hand, if b  a, then max a b  b and a  b   a  b  b  a. In this case, abba a  b  a  b   b. 2 2 If a  b, then a  b  0 and the result is trivial. a  b  b  a a  b  a  b   a. (b) If a  b, then min a b  a and a  b  b  a. In this case 2 2 a  b  a  b Similarly, if b  a, then  b; and if a  b, the result is trivial. 2 92. Answers will vary.

91. (a) Suppose that a  b, so max a b  a and a  b  a  b. Then

93. (a) Subtraction is not commutative. For example, 5  1  1  5. (b) Division is not commutative. For example, 5  1  1  5.

(c) Putting on your socks and putting on your shoes are not commutative. If you put on your socks first, then your shoes, the result is not the same as if you proceed the other way around. (d) Putting on your hat and putting on your coat are commutative. They can be done in either order, with the same result. (e) Washing laundry and drying it are not commutative. 94. (a) If x  2 and y  3, then x  y  2  3  5  5 and x  y  2  3  5. If x  2 and y  3, then x  y  5  5 and x  y  5. If x  2 and y  3, then x  y  2  3  1 and x  y  5. In each case, x  y  x  y and the Triangle Inequality is satisfied. (b) Case 0: If either x or y is 0, the result is equality, trivially.   xy Case 1: If x and y have the same sign, then x  y    x  y

 if x and y are positive   x  y. if x and y are negative 

Case 2: If x and y have opposite signs, then suppose without loss of generality that x  0 and y  0. Then x  y  x  y  x  y.

P.3

INTEGER EXPONENTS AND SCIENTIFIC NOTATION

1. Using exponential notation we can write the product 5  5  5  5  5  5 as 56 .

2. Yes, there is a difference: 54  5 5 5 5  625, while 54   5  5  5  5  625. 3. In the expression 34 , the number 3 is called the base and the number 4 is called the exponent.

4. (a) When we multiply two powers with the same base, we add the exponents. So 34  35  39 . 35 (b) When we divide two powers with the same base, we subtract the exponents. So 2  33 . 3  2 5. When we raise a power to a new power, we multiply the exponents. So 34  38 .

6. (a) 21 

1 2

(b) 23 

1 8

(c)

 1 1 2 2

1 (d) 3  23  8 2

7. To move a number raised to a power from numerator to denominator or from denominator to numerator change the sign of 1 1 a 3 1 6a 2 the exponent. So a 2  2 , 2  b2 , 2  3 2 , and 3  6a 2 b3 . a b b a b b

8. Scientists express very large or very small numbers using scientific notation. In scientific notation, 8,300,000 is 83  106 and 00000327 is 327  105 .


8

CHAPTER P Prerequisites

9. (a) No,

 2  2 9 2 3   . 3 2 4

  (b) No, 53  125 and 53   53  125.

 3 10. (a) No, x 2  x 23  x 6 .

3   3 (b) No, 2x 4  23  x 4  8x 12 .

  11. (a) 26   26  64

12. (a) 53  125   (b) 53   53  125  2 (c) 52  25  4

(b) 26  64  2 27 33 (c) 15  33   25 52  0 1 5  21  13. (a) 3 2

  14. (a) 23  20   23  1  8 1 1 (b) 23  20   3  1   8 2   53 125 3 3 (c)   5 27 33

23 1 1  3  8 30 2  2  2 9 3 2   (c) 3 2 4

(b)

15. (a) 53  5  531  625

16. (a) 38  35  385  313

107  1074  1000 104  4 (c) 35  354  320

(b) 54  52  542  25  3 (c) 22  223  64 710  7108  72  49 78 62 (b) 2  622  64  1296 6

(b)

83 83  1  831  82  64 8 8 81 81 1 (b)  1  811  82  8 64 8 1 3 1 31 4 5  (c) 5  5  5 625

17. (a)

(c) 41  42  412  43 

18. (a)

1 64

19. (a) t 5 t 2  t 52  t 7  2  2 (b) 4z 3  42 z 3  16z 32  16z 6

20. (a) a 4 a 6  a 46  a 10 3  3  8 (b) 2b3  23 b3  8b33   9 b 2 (c) 2y 10 y 11  2y 1011  2y 1   y

1 21. (a) x 5  x 3  x 53  x 2  2 x

1 22. (a) y 2  y 5  y 25  y 3  3 y

(c) x 3 x 5  x 35  x 2

(b) 2 4 5  245  1  (c)

1 

x 16  x 1610  x 6 x 10

a 9 a 2  a 921  a 6 a 3  3  3  (b) a 2 a 4  a 24  a 6  a 63  a 18

23. (a)

 x 3  5x 6  x 3  5x 9 5x 6   3 2 8 2    25. (a) 3x 3 y 2 2y 3  3  2x 3 y 2 y 3  6x 3 y 5 (c)

1 (b) z 5 z 3 z 4  z 534  z 2  2 z y7 y0 1 (c) 10  y 7010  y 3  3 y y z2 z4 24. (a) 3 1  z 2431  z 4 z z 4  4   4 (b) 2a 3 a 2  24 a 32  16 a 5  16a 20 3    2z 3  33  z 23  2z 3  54z 9 (c) 3z 2


SECTION P.3 Integer Exponents and Scientific Notation

2    z3  (b) 52 z 2

52 z2

2

9

 2   52 2 z 3 254 z3    2 z z2

   4n 2 8m 2 n 4 12 n 2  8  12 m 2 n 42  2 m 3       27a 14 a 2 b1  33 a 43 b23 a 2 b1  33 a 122 b61  (b) 3a 4 b2 b7

26. (a)

7 x 2 y 1 25 y 1  x  x y x 5 3  a 33 a9 a3  (b)  3  6 2 2b 8b 23 b2

27. (a)

y 2 z 3 y 1  2 3  3 y 1 y z yz 2  2  2  y8 x 3 y 2 33 y 22 6 y 4 (b)  x  x  x 3 y 2 x 12 3  5  19 a 3 b2 a2 2533 b523 c33  a b 29. (a)  a b c3 c9  2 u 1  2 10 1233  2223   (b)  3  u u 11 u 3  2 28. (a)

 2 22 432 522 232 x 10 2x 3 y 2 x 4 z2 x 30. (a)  y z  4 5 3 4 4y z yz  3 rs 2 332 s 2322  r 9 s 2 (b)  2  r r 3 s 2 

31. (a)

8a 3 b4 4a 8  4a 35 b45  9 5 5 2a b b

32. (a)

5x y 2  5x 11 y 23  5x 2 y x 1 y 3

33. (a) 69,300,000  693  107

3 125 y  513 x 23 y 3  6 3 5x 2 x y 3  9 2a 1 b 3 a 1323 b333  a (b)  2 a 2 b3 8b12 (b)

34. (a) 129,540,000  12954  108

(b) 7,200,000,000,000  72  1012

(b) 7,259,000,000  7259  109

(c) 0000028536  28536  105

(c) 0000 000 001 4  14  109

(d) 00001213  1213  104

(d) 00007029  7029  104

35. (a) 319  105  319,000

36. (a) 71  1014  710,000,000,000,000

(b) 2721  108  272,100,000

(b) 6  1012  6,000,000,000,000

(c) 2670  108  0000 000 026 70

(c) 855  103  000855

(d) 9999  109  0000 000 009 999

(d) 6257  1010  0000 000 000 625 7

37. (a) 5,900,000,000,000 mi 59  1012 mi


10

CHAPTER P Prerequisites

(b) 0000 000 000 000 4 cm  4  1013 cm

(c) 33 billion billion molecules  33  109  109  33  1019 molecules

38. (a) 93,000,000 mi  93  107 mi

(b) 0000 000 000 000 000 000 000 053 g  53  1023 g

(c) 5,970,000,000,000,000,000,000,000 kg  597  1024 kg    39. 72  109 1806  1012  72  1806  109  1012  130  1021  13  1020    40. 1062  1024 861  1019  1062  861  1024  1019  914  1043

1295643 1295643  109     109176  01429  1019  1429  1019 41.  3610  2511 3610  1017 2511  106     731  10 16341  1028 731 16341  1028 731  16341   101289  63  1038 42.  9 00000000019 19 19  10    5 2 1582  10 162  10 162  1582 00000162 001582     105283  0074  1012 43.   594621  58 594621000 00058 594621  108 58  103  74  1014

 9 3542  106 8774796 35429  1054  105448  319  104  10102  319  10106 44.   12  12 48 27510376710 4 505  10 505  10         45. 1050  1010   1050 , whereas 10101  10100   10100 10  1  9  10100  1050 . So 1010 is closer to 1050 than 10100 is to 10101 .

46. (a) b5 is negative since a negative number raised to an odd power is negative.

(b) b10 is positive since a negative number raised to an even power is positive. (c) ab2 c3 we have positive negative2 negative3  positive positive negative which is negative.

(d) Since b  a is negative, b  a3  negative3 which is negative. (e) Since b  a is negative, b  a4  negative4 which is positive. (f)

a 3 c3 negative positive3 negative3 positive negative which is negative.    6 6 positive positive positive b c negative6 negative6

47. Since one light year is 59  1012 miles, Centauri is about 43  59  1012  254  1013 , or 25,400,000,000,000 miles away. 93  107 mi t st  s  500 s  8 13 min. s 186 000    103 liters   3 14 2  133  1021 liters 49. Volume  Average depth Area  37  10 m 36  10 m m3

48. 93  107 mi  186 000

50. Each person’s share is equal to

2670  1013 National debt   80555  104  $80,555. Population 33145  108

51. First, we estimate the total mass of the stars in the observable universe:     Number of stars Total star mass  Number of galaxies Mass of typical star  15  2  177 1012  1011  1030 Galaxy  531  1053 kg


SECTION P.3 Integer Exponents and Scientific Notation

11

Thus, the number of hydrogen atoms in the observable universe is 531  1053 Total star mass   318  1080 Mass of a single hydrogen atom 167  1027

52. (a)

Weight

Height

A

295 lb

4232

obese

B

105 lb

5 ft 10 in.  70 in.

1695

underweight

C

220 lb

5 ft 6 in.  66 in.

overweight

110 lb

6 ft 4 in.  76 in.

2678

D

5 ft 2 in.  62 in.

2012

normal

(b) Answers will vary.   53. I  15,000 10037512n  1

BMI  703

W H2

Patient

Year

Total interest

1

$68910

2

140985

3

216372

4

295222

5

377694

Result

54. Since 106  103  103 it would take 1000 days  274 years to spend the million dollars.

Since 109  103  106 it would take 106  1,000,000 days  273972 years to spend the billion dollars.  5 18 185  25  32 (b) 206  056  20  056  106  1,000,000 55. (a) 5  9 9 m factors

   am a  a    a 56. (a) n  . Because m  n, we can cancel n factors of a from numerator and denominator and are left with a a  a       a n factors

m  n factors of a in the numerator. Thus, n factors

(b)

57. (a)

 a n b

n factors

      a a a a  a    a an        n b b b b b   b     b

 a n b

am  a mn . an

n factors

1 1 bn  n n  n n a b a ab

a n 1a n 1 bm (b) m   n  bm  n m b 1b a a


12

CHAPTER P Prerequisites

P.4

RATIONAL EXPONENTS AND RADICALS

 1. (a) Using exponential notation we can write 3 5 as 513 .  (b) Using radicals we can write 512 as 5. 2  12  2    5212  5 and 5  512  5122  5. (c) No. 52  52  3  12 2. 412  23  8; 43  6412  8 3. Because the denominator is of the form

    a, we multiply numerator and denominator by a: 1  1  3  33 . 3

3

3

4. 513  523  51  5

 4a 2  2a.    6. No. For example, if a  2, then a 2  4  8  2 2, but a  2  0.     7. (a) 144  122  12 8. (a) 49  72  7    5  3 (b) 5 32  25  2 (b) 3 125  53  5       3 3 3 1 4 1 4 4 1 1 (c)  27   13   13   (c) 5. No. If a is negative, then

81

3

3

     9. (a) 3 3 16  3 3 8  2  3 3 8 3 2  6 3 2     18 18 2 2 (b)     81 9 3 81     3 3 3 3 27 93 (c)     4 2 2 4      11. (a) 3 15  45  9  5  3 5   48 48  (b)    16  4 3 3      3 (c) 3 24 3 18  3 24  18  63  2  6 3 2   72 72   93 13. (a)   8 8     (b) 3 9 3 24  3 9  24  3 216  6     1 4 1 1 1 1 1  4   4  (c) 4 32 8 32 8 256 4  4 4 15. (a) x  x   4 4 (b) 16x 8  24 x 8  2x 2

   3 3 274  33 3    3 3     (b) 4 64y 6  4 24 y 4  4y 2  2 y 4 4y 2     3 3 19. (a) 3 8x 9 y 3  23 x 3 y 3  2x 3 y     (b) 4 8x 6 y 2 4 2x 2 y 2  4 8x 6 y 2  2x 2 y 2  4 16x 8 y 4    4 4  24 x 2 y 4  2x 2 y 17. (a)

     10. (a) 2 3 81  2 3 27  3  2 3 27 3 3  6 3 3    18 92 3 2 (b)    5 5 25    2 3 12 43 (c)    49 7 49      12. (a) 10 32  320  64  5  8 5   54 54 (b)   3 6 6      3 (c) 3 15 3 75  3 15  75  53  9  5 3 9     3 27 3 3 3 9 3 3 9 14. (a)    3  4 2 4 2 8 2     5 5 1 5 5 1 (b) 4 128  4 128  32  2    3 1 12 3 1 3 12  (c)    3 96 8 2 96  15  5 16. (a) x 10  x 10  x2 13   (b) 3 x 3 y 6  x 3 y 6  x y2    4 4 4 18. (a) x 2 z 5  z 4  x 2 z  z x 2 z, z  0   2 (b) x 6 y 6  x 3 y 3  x3 y3       2 3 2 2 20. (a) 16x 4 y 6 z 2  42 x 2 y z  4x 2 y3 z  19  3  3 512x 9  512x 9  2x (b)


SECTION P.4 Rational Exponents and Radicals

           32  18  16  2  9  2  16 2  9 2  4 2  3 2  7 2            (b) 75  48  25  3  16  3  25 3  16 3  5 3  4 3  9 3        22. (a) 125  45  25  5  9  5  5 5  3 3  8 5          (b) 3 54  3 16  3 27  2  3 8  2  3 27 3 2  3 8 3 2  3 2         23. (a) 9a 3  a  9 a 2  a  a  3a a  a  3a  1 a         x (b) 16x  x 5  16 x  x 4 x  x 2  4      3 3 3 24. (a) x 4  3 8x  x 3  x  23 x  x  2 3 x              (b) 4 18rt 3  5 32r 3 t 5  4 32  2 r t 2  t  5 42  2 r 2  r t 4  t  4 3 t 2rt  5 4 r t 2 2rt    20rt 2  12t 2rt           25. (a) 36x 2  36x 4  62 x 2 1  x 2  6x 1  x 2 (b) 81x 2  81y 2  92 x 2  y 2  9 x 2  y 2       26. (a) 27a 3  63a 2  32 a 2 3a  7  3a 3a  7 (b) 25t 2  100t 2  125t 2  5 5t, t  0   28. 5 6  615 27. 10  1012  15  1 1 1 5 29. 735  73  73 30. 652  52   12   5 6 65 6 21. (a)

1 1 31.   12  512 5 5 1 1 1 33. 534  34     4 3 4 5 125 5

1 1 32.   13  713 3 7 7 1 34. y 15  y 32   y3 1 1 36.   34  x 34 4 3 x x

1 1 35.   23  x 23 3 2 x x  37. (a) 1614  4 16  2

 (b) 813  3 8  2

 38. (a) 2713  3 27  3

 (b) 813  3 8  2

39. (a) 3225 

  2 5 32  22  4

(b)

  2 3 125  52  25  32   3  3 125 25 5 25     (b) 64 8 512 64

40. (a) 12523 

1 1 1 (c) 2743    4  4  81 3 3 27

41. (a) 523  513  52313  51  5

 335  33525  315  5 3 325 3   3  (c) 3 4  413  4133  41  4 (b)

42. (a) 327  3127  327127  32  9

  12 3 1 9 4    9 4 2 49

1 1 (c) 912    3 9  13 1 1 1  (c)    3 8 2 8 3  34   4 8 16 16 (c)    4 81 27 81

13


14

CHAPTER P Prerequisites

723 1  72353  71  7 753   10  10 1 1 (c) 5 6  615  61510  62  2  36 6     43. When x  3, y  4, z  1 we have x 2  y 2  32  42  9  16  25  5.      4 44. When x  3, y  4, z  1 we have 4 x 3  14y  2z  4 33  14 4  2 1  4 27  56  2  4 81  34  3. (b)

45. When x  3, y  4, z  1 we have

 23  23  23  113 9x23  2y23  z 23  9  323  2  423  123  33  32  22  1  9  4  1  14.

1 . 46. When x  3, y  4, z  1 we have x y2z  3  42  1  122  144   48. (a) 4b12 8b14  412  8b1214  16b34 47. (a) x 34 x 54  x 3454  x 2 2    (b) y 23 y 43  y 2343  y 2 5a 12  32  5a 34212  45a 2 (b) 3a 34

43 23  432313  53 13 6  (b) 3x 12 y 13  36 x 126 y 136  36  729x 3 y 2

49. (a)

51. (a)

23  8a 6 b32  823 a 623 b3223  4a 4 b 

35

23  1 50. (a) 8y 3  823 y 323  2 4y 13  1 4 6 413 613 (b) u  u   43 2 u  32  8b9 (b) 4a 4 b6  432 a 432 b632  8a 6 b9  6 a

x3  x 535 y 1335  15 y 13  12  16u 6  23 (b) u 3  2  u 313  213 1612 u 612  2312  4u 23 u 3  13  4u 4 

52. (a)

x 5 y 13

12 x 1 32 x 142 x 112 9x 12 x 12 9    4 53. (a) 4 2 412 y y y y2 y2 y 4  2  44323 34 134 22 132 43 423 42 213 313  (b)    z 1 z 12 z 14 z 122 81z 4 z 81z 5 81z 5  3 3    y 23 y 2 x 1 x 1 x 1 y 23 2 54. (a)   3  1  3 2  x 1 2 x y x y x 1 y 2  2  2  13  13 2  13  2 y3 23 3 4 3 23 3 6 z x y6 16y 6 z 43 x 1 y 2 8y 8 x y 4y z (b)    2  2  13 12 4 43 x x 8z 2z x 813 z 4 x 12       12 12 55. (a) x 3  x 3  x 312  x 32 56. (a) x 5  x 5  x 512  x 52  15  14   5 4 (b) x 6  x 6  x 615  x 65 (b) x 6  x 6  x 614  x 32     4 57. (a) 6 y 5 3 y 2  y 56  y 23  y 5623  y 32 58. (a) b3 b  b3412  b54           3 2 (b) 5 3 x 2 4 x  5  2x 1314  10x 712 (b) 2 a a  2a 1223  2a 76     6 60. (a) 5 x 3 y 2 10 x 4 y 16  x 35410 y 251610  x y 2 59. (a) 4st 3 s 3 t 2  412 s 1236 t 3226  2st 116   4 7 3 x 8x 2 7434 (b)   x  x (b)   813 x 2312  2x 16 4 3 x x 

3x 14 y

2 


SECTION P.4 Rational Exponents and Radicals

15

13 12       61. (a) 3 y y  y 112  y 3213  y 12  s 54 62. (a) s s 3  s 132     2 4 3 3y 3u 18u 5  9u 2 3 27y 3 54x y (b)   (b)   5 3 3 3 2 x  2x y x 2u        12 1 6 6 3 1 12 12 3 64. (a)       4 3 63. (a)       6 3 6 6 6 3 3 3         2 15 3 3 2 6 12 43 5 (b) (b)     2 2 2 2 5 5 5 5   9 234 948 9 8 8 513 835  14  34  (c)  (c)   23  13  4 3 2 2 5 2 2 5 5 2 5       1 5x 5x s s 3t 3st 1 65. (a)         66. (a) 5x 3t 3t 3t 3t 5x 5x 5x       3 2 3 x x 5 5x b a a a b2 (b)   (b)      6 3 3 5 5 5 5 b b b2 b2    5 2 5 2 25 25 1 x 1 x 1 c 1 c (c) 5 3  35  25  (c) 35  35  25  x x x x c c c c         1 1 1 1 5 5 1 2 2 67. (a) 4 4  4 (b)       4 64 4 40 8 16 4 4  43 40  4 x 32 x 324 x6 68. (a)    x 6 y2 y 12 y 124 y 2 12  5 u 5u (b) 25u 2  4  2512 u 212  412  5 u  2  2 . Since u  0, this is equivalent to  2 .     4 2 12 24 y 69. (a) y y  y y   12 4 (b) 81 8 z 8  8112 8412 z 8412  9  z. Since   0 and z  0, this is equivalent to 9z. 70. (a)

(b)

x 32

y 12

4 

4  3 3 z 6

12

4

y3

12 z 212

x 2



x 32

y 12

412

4 

x 2 x 6  x 2 x4  2 3  3 y y y y

32 2   z z

71. (a) Since 12  13 , 212  213 .  12  13  12  13 (b) 12  212 and 12  213 . Since  12   13 , we have 12  12 .

 112  112  343112 ; 413  4412  44  256112 . So 714  413 . 72. (a) We find a common root: 714  7312  73  16  16   (b) We find a common root: 3 5  513  526  52  2516 ; 3  312  336  33  2716 . So   3 5  3. 1 mile 73. First convert 1135 feet to miles. This gives 1135 ft  1135   0215 mi. Thus the distance you can see is given 5280 feet    by D  2r h  h 2  2 3960 0215  02152  17028  413 miles.   74. (a) Using f  04 and substituting d  65, we obtain s  30 f d  30  04  65  28 mi/h.    (b) Using f  05 and substituting s  50, we find d. This gives s  30 f d  50  30  05 d  50  15d  2500  15d  d  500 3  167 feet.


16

CHAPTER P Prerequisites

75. Since 1 day  86,400 s, 36525 days  31,557,600 s. Substituting, we obtain 13  23  667  1011  199  1030 d  315576  107  15  1011 m  15  108 km. 2 4 7523  005012  17707 ft/s. 24123  0040 (b) Since the volume of the flow is V  A, the canal discharge is 17707  75  13280 ft3 s.

76. (a) Substituting the given values we get V  1486  77. (a)

n

1

2

5

10

100

21n

211  2

212  1414

215  1149

2110  1072

21100  1007

So when n gets large, 21n decreases to 1. (b) n  1n 1 2

1  11 1 2

2  05

 12 1 2

 0707

 1n So when n gets large, 12 increases to 1.

P.5

5  15 1 2

 0871

10  110 1  0933 2

100  1100 1  0993 2

ALGEBRAIC EXPRESSIONS

 1. (a) 2x 3  12 x  3 is a polynomial. (The constant term is not an integer, but all exponents are integers.)  (b) x 2  12  3 x  x 2  12  3x 12 is not a polynomial because the exponent 12 is not an integer. (c)

1

x 2  4x  7

is not a polynomial. (It is the reciprocal of the polynomial x 2  4x  7.)

(d) x 5  7x 2  x  100 is a polynomial.  3 (e) 8x 6  5x 3  7x  3 is not a polynomial. (It is the cube root of the polynomial 8x 6  5x 3  7x  3.)  4  2 (f) 3x  5x  15x is a polynomial. (Some coefficients are not integers, but all exponents are integers.)

2. To add polynomials we add like terms. So     3x 2  2x  4  8x 2  x  1  3  8 x 2  2  1 x  4  1  11x 2  x  5.

3. To subtract polynomials we subtract like terms. So     2x 3  9x 2  x  10  x 3  x 2  6x  8  2  1 x 3  9  1 x 2  1  6 x  10  8  x 3  8x 2  5x  2.

4. We use FOIL to multiply two polynomials:x  2 x  3  x  x  x  3  2  x  2  3  x 2  5x  6. 5. The Special Product Formula for the “square of a sum” is A  B2  A2  2AB  B 2 . So 2x  32  2x2  2 2x 3  32  4x 2  12x  9.

6. The Special Product Formula for the “product of the sum and difference of terms” is A  B A  B  A2  B 2 . So 6  x 6  x  62  x 2  36  x 2 .

7. (a) No; x  52  x 2  2 5x  25  x 2  25. (b) Yes; x  a2  x 2  2xa  a 2 .

8. (a) Yes; by a Special Product Formula, x  5 x  5  x 2  25. (b) No, x  a x  a  x 2  a 2 , by a Special Product Formula.

9. Type: binomial. Terms: 5x 3 and 6. Degree: 3.

10. Type: trinomial. Terms: 2x 2 , 5x, and 3. Degree: 2.


SECTION P.5 Algebraic Expressions

11. Type: monomial. Term: 8. Degree: 0.

12. Type: monomial. Terms: 12 x 7 . Degree: 7. 13. Type: quadrinomial. Terms: x, x 2 , x 3 , and x 4 . Degree: 4.   14. Type: binomial. Terms: 2x and  3. Degree: 1. 15. 12x  7  5x  12  12x  7  5x  12  7x  5

16. 5  3x  2x  8  x  3     17. 2x 2  3x  1  3x 2  5x  4  2x 2  3x  1  3x 2  5x  4  x 2  2x  3     18. 3x 2  x  1  2x 2  3x  5  3x 2  x  1  2x 2  3x  5  x 2  4x  6     19. 5x 3  4x 2  3x  x 2  7x  2  5x 3  4x 2  3x  x 2  7x  2  5x 3  3x 2  10x  2     20. 5x 2  3  1  4x  3x 2  5x 2  3  1  4x  3x 2  8x 2  4x  4 21. 3 x  1  4 x  2  3x  3  4x  8  7x  5

22. 8 2x  5  7 x  9  16x  40  7x  63  9x  103     23. 4 x 2  3x  5  3 x 2  2x  1  4x 2  12x  20  3x 2  6x  3  x 2  6x  17       24. 5x 3  3x 2  2x  2 3  x  4x 3  5x 3  3x 2  2x  2 3  2 x  2 4x 3  13x 3  3x 2  6

25. 4   5  42  20

26. 10x 1  2x  10x  20x 2  20x 2  10x   27. y 3 y 2  y  y 3  y 2  y 3  y  y 5  y 4   28. 4z 2  z 2  4z 3  8z     29. x 3 x 2  3x  2x x 4  3x 2  x 5  3x 4  2x 5  6x 3  x 5  3x 4  6x 3     30. 4x 1  x 3  3x 3 x 3  x  4x  4x 4  3x 6  3x 4  3x 6  7x 4  4x   31. 4x x  2  2 x 2  4x  2x 2 x  1  4x 2  8x  2x 2  8x  2x 3  2x 2  2x 3     32. 6x 3 x 2  1  2x 2  3x 2  2 2x  4  6x 5  6x 3  4x  6x 3  4x  8  6x 5  12x 3  8 33. x  4 x  10  x 2  10x  4x  4  10  x 2  14x  40 34. y  1 y  7  y 2  7y  y  1 7  y 2  8y  7

35. z  1 4z  5  4z 2  5z  4z  1 5  4z 2  z  5

36. 3x  2 x  5  3x 2  3  5x  2x  2 5  3x 2  13x  10

37. 3t  2 7t  4  21t 2  12t  14t  8  21t 2  26t  8 38. 4s  1 2s  5  8s 2  18s  5 39. 3x  5 2x  1  6x 2  10x  3x  5  6x 2  7x  5 40. 7y  3 2y  1  14y 2  13y  3 41. x  3y 2x  y  2x 2  5x y  3y 2

42. 4x  5y 3x  y  12x 2  19x y  5y 2

43. 3u  4 4u  3  12u 2  25u  12 2

44. 2r  s r  2s  2r 2  3rs  2s 2

45. 4x  32  16x 2  24x  9

46. 2  7y2  49y 2  28y  4

47. 1  3z2  9z 2  6z  1

48. 2u  32  4u 2  12u  9

49. y  3x2  y 2  6x y  9x 2

50. 5x  y2  25x 2  10x y  y 2

17


18

CHAPTER P Prerequisites

51. 2x  3y2  4x 2  12x y  9y 2 52. r  2s2  r 2  4rs  4s 2 2 2      2 53. 5  y 3  52  y 3 2 5 y 3  y 6 10y 3 25 54. x 4  3  x 8  6x 4  9 55.   7   7  2  49

56. 5  y 5  y  25  y 2

57. 3x  4 3x  4  3x2  42  9x 2  16 58. 2y  5 2y  5  4y 2  25           x 2 x 2  x 4 60. y 2 y  2  y 2 59.   61. y  23  y 3  3y 2 2  3y 22  23  y 3  6y 2  12y  8 62. x  33  x 3  9x 2  27x  27

63. 2  5u3  125u 3  150u 2  60u  8

64. 3x  23  27x 3  54x 2  36x  8   65. x  2 x 2  2x  3  x 3  2x 2  3x  2x 2  4x  6  x 3  4x 2  7x  6   66. x  1 2x 2  x  1  2x 3  x 2  1   67. 2x  5 x 2  x  1  2x 3  2x 2  2x  5x 2  5x  5  2x 3  7x 2  7x  5   68. 1  2x x 2  3x  1  2x 3  5x 2  x  1  2       x x x x x x x xx   71. y 13 y 23  y 53  y 1323  y 1353  y 2  y 69.

73. 75.

x 12  y 12 x 2  a2



2

  x  2 xy  y

 x 2  a2  x 4  a4



  x  1 x  x 2  x    72. x 14 2x 34  x 14  2x  x    1 2 1 74. u u2 u u    76. x 12  y 12 x 12  y 12  x  y     78. h2  1  1 h2  1  1  h2 70. x 32

    ab a  b  a  b2     2 79. x  1  x 2 x  1  x 2  x  12  x 2  x 2  2x  1  x 4  x 4  x 2  2x  1       80. x  2  x 2 x  2  x 2  x 4  3x 2  4

77.

81. 2x  y  3 2x  y  3  2x  y2  32  4x 2  4x y  y 2  9

82. x  y  z x  y  z  x 2  y 2  z 2  2yz      83. (a) 12 a  b2  a 2  b2  12 a 2  2ab  b2  a 2  b2  12 2ab  ab. 2  2          a 2  b2  a 2  b2 (b) a 2  b2  a 2  b2  a 2  b2  a 2  b2        a 2  b2  a 2  b2 a 2  b2  a 2  b2  2b2 2a 2  4a 2 b2    84. LHS  a 2  b2 c2  d 2  a 2 c2  a 2 d 2  b2 c2  b2 d 2 .

RHS  ac  bd2  ad  bc2  a 2 c2  2abcd  b2 d 2  a 2 d 2  2abcd  b2 c2  a 2 c2  a 2 d 2  b2 c2  b2 d 2 .    So LHS  RHS, that is, a 2  b2 c2  d 2  ac  bd2  ad  bc2 .

85. (a) The height of the box is x, its width is 6  2x, and its length is 10  2x. Since Volume  height  width  length, we have V  x 6  2x 10  2x.   (b) V  x 60  32x  4x 2  60x  32x 2  4x 3 , degree 3.


SECTION P.6 Factoring

19

    (c) When x  1, the volume is V  60 1  32 12  4 13  32, and when x  2, the volume is     V  60 2  32 22  4 23  24.

86. (a) The width is the width of the lot minus the setbacks of 10 feet each. Thus width  x  20 and length  y  20. Since Area  width  length, we get A  x  20 y  20. (b) A  x  20 y  20  x y  20x  20y  400

(c) For the 100  400 lot, the building envelope has A  100  20 400  20  80 380  30,400. For the 200  200, lot the building envelope has A  200  20 200  20  180 180  32,400. The 200  200 lot has a larger building envelope.   87. (a) A  4000 1  r3  2000 1  3r  3r 2  r 3  2000  6000r  6000r 2  2000r 3 , degree 3. (b) Remember that % means divide by 100, so 2%  002. Interest rate r

2%

35%

5%

6%

10%

Amount A

$424483

$443487

$463050

$476406

$532400

88. (a) When x  1, x  52  1  52  36 and x 2  25  12  25  26. (b) x  52  x 2  10x  25

89. (a) The degree of the product is the sum of the degrees of the original polynomials. (b) The degree of the sum could be lower than either of the degrees of the original polynomials, but is at most the largest of the degrees of the original polynomials.    (c) Product: 2x 3  x  3 2x 3  x  7  4x 6  2x 4  14x 3  2x 4  x 2  7x  6x 3  3x  21  4x 6  4x 4  20x 3  x 2  10x  21     Sum: 2x 3  x  3  2x 3  x  7  4.

P.6

FACTORING

1. (a) The polynomial 2x 3  3x 2  10x has three terms: 2x 3 , 3x 2 , and 10x.   (b) The factor x is common to each term, so 2x 3  3x 2  10x  x 2x 2  3x  10 .

  2. The greatest common factor in the expression 18x 3  30x is 6x, and the expression factors as 6x 3x 2  5 .

3. To factor the trinomial x 2  8x  12 we look for two integers whose product is 12 and whose sum is 8. These integers are 6 and 2, so the trinomial factors as x  6 x  2. 4. The Special Factoring Formula for the “difference of squares” is A2  B 2  A  B A  B. So 49x 2  9  7x  3 7x  3.

5. The Special Factoring Formula for a “perfect square” is A2  2AB  B 2  A  B2 . So x 2  10x  25  x  52 .

6. The greatest common factor in the expression 4 x  12  x x  12 is x  12 , and the expression factors as 4 x  12  x x  12  x  12 4  x.

7. 10x  15  5 2x  3   9. 4x 4  x 2  x 2 4x 2  1  x 2 2x  1 2x  1   11. 4x 3 y 2  6x y 3  8x 2 y 4  2x y 2 2x 2  3y  4x y 2

13. y y  6  9 y  6  y  6 y  9

8. 12  3y  3 4  y

  10. 3x 4  6x 3  x 2  x 2 3x 2  6x  1   12. 7x 4 y 2  14x y 3  21x y 4  7x y 2 x 3  2y  3y 2

14. z  22  5 z  2  z  2 [z  2  5]  z  2 z  3


20

CHAPTER P Prerequisites

15. z 2  11z  18  z  9 z  2

16. x 2  4x  5  x  5 x  1

17. 10x 2  19x  6  5x  2 2x  3

18. 6y 2  11y  21  y  3 6y  7

19. 3x 2  16x  5  3x  1 x  5

20. 5x 2  7x  6  5x  3 x  2

21. 3x  22  8 3x  2  12  [3x  2  2] [3x  2  6]  3x  4 3x  8

22. 2 a  b2  5 a  b  3  [a  b  3] [2 a  b  1]  a  b  3 2a  2b  1 23. x 2  400  x  20 x  20

24. 64  y 2  8  y 8  y

25. 36a 2  49  6a  7 6a  7

26. 4z 2  81  2z  9 2z  9

27. x 2  49y 2  x  7y x  7y

28. 4u 2  25 2  2u  5 2u  5

29. z  52  92  z  5  3 z  5  3

30. y 2  x  12  y  x  1 y  x  1

31. x 2  12x  36  x 2  2 6x  62  x  62

32. 16  8y  y 2  y 2  2 4y  42  y  42

33. 100  20z  z 2  z 2  2 10z  102  z  102

34. 2  14  49  2  2 7  72    72

35. 25  30t  9t 2  3t2  2 3t 5  52  3t  52

36. 16z 2  24z  9  4z2  2 4z 3  32  4z  32

37. 4u 2  36u  81 2  2u2  2 2u 9  92  2u  92

38. 4y 2  20yz  25z 2  2y2  2 2y 5z  5z2  2y  5z2     39. x 3  125  x 3  53  x  5 x 2  5x  25 40. y 3  64  y 3  43  y  4 y 2  4y  16     41. 83  27  23  33  2  3 42  6  9 42. 1  27a 3  13  3a3  1  3a 1  3a  9a 2     43. 27x 3  y 3  3x3  y 3  3x  y 3x2  3x y  y 2  3x  y 9x 2  3x y  y 2   44. 8x 3  y 3  2x3  y 3  2x  y 4x 2  2x y  y 2     3   2   45. a 3  b6  a 3  b2  a  b2 a 2  ab2  b2  a  b2 a 2  ab2  b4   46. 64x 3 y 3  27  4x y3  33  4x y  3 16x 2 y 2  12x y  9   47. x 3  4x 2  x  4  x 2 x  4  1 x  4  x  4 x 2  1   48. 3x 3  x 2  6x  2  x 2 3x  1  2 3x  1  3x  1 x 2  2   49. 5x 3  x 2  5x  1  x 2 5x  1  5x  1  x 2  1 5x  1   50. 18x 3  9x 2  2x  1  9x 2 2x  1  2x  1  9x 2  1 2x  1   51. x 3  x 2  x  1  x 2 x  1  1 x  1  x  1 x 2  1   52. x 5  x 4  x  1  x 4 x  1  1 x  1  x  1 x 4  1

53. Start by factoring out the power of x with the smallest exponent, that is, x 23 . Thus, x 23  3x 53  x 23 1  3x

54. Start by factoring out the power of x with the smallest exponent, that is, x 14 . Thus, x 34  5x 14  x 14 x  5

55. Start by factoring out the power of x with the smallest exponent, that is, x 32 .     Thus, x 32  x 12  x 12  x 32 1  x  x 2  x 32 x 2  x  1


SECTION P.6 Factoring

56. Start by factoring out the power of x with the smallest exponent, that is, x 13 .   Thus, x 53  x 23  2x 13  x 13 x 2  x  2  12 57. Start by factoring out the power of x 2  1 with the smallest exponent, that is, x 2  1 .

12  12  12     12    x2  3 x2  1  2  x2  1 x2  3    2 x2  1  x2  1 . Thus, x 2  1 x2  1 2x  1 58. x 12 x  112  x 12 x  112  x 12 x  112 [x  1  x]  x 12 x  112 2x  1    x x 1   3 2 2 3 2 60. 12x  3x  3x 4  x 59. 2x  12x  2x 1  6x 61. 15x 3  10x 2  5x 2 3x  2

62. 6x yz  3xz  3xz 2y  1

63. x 2  2x  8  x  4 x  2

64. x 2  14x  48  x  8 x  6

65. y 2  4y  21  y  7 y  3

66. t 2  10t  24  t  4 t  6

67. 2x 2  5x  3  2x  3 x  1   69. 9x 2  36x  45  9 x 2  4x  5  9 x  5 x  1

68. 2x 2  7x  4  2x  1 x  4

71. 12x 2  x  20  4x  5 3x  4

72. 15  26t  8t 2  4t  3 2t  5

73. x 2  121  x  11 x  11

74. 4t 2  900  2t  30 2t  30

75. 49  4y 2  7  2y 7  2y

76. 4t 2  9s 2  2t  3s 2t  3s

77. t 2  6t  9  t  32

78. x 2  10x  25  x  52

70. 8x 2  10x  3  4x  3 2x  1

79. y 2  10yz  25z 2  y  5z2   81. 1  8t 3  1  2t 1  2t  4t 2

80. r 2  6rs  9s 2  r  3s2   82. 27x 3  1000  3x  10 9x 2  30x  100     83. 8x 3  125  2x3  53  2x  5 2x2  2x 5  52  2x  5 4x 2  10x  25    3    2     x 2  4 x 2  42  x 2  4 x 4  4x 2  16 84. x 6  64  x 6  26  x 2  43  x 2  4   85. x 3  2x 2  x  x x 2  2x  1  x x  12   86. 3x 3  27x  3x x 2  9  3x x  3 x  3   87. x 6  x 5  42x 4  x 4 x 2  x  42  x 4 x  7 x  6   88. 3t 4  2t 3  5t 2  t 2 3t 2  2t  5  t 2 t  1 3t  5   89. x 4 y 3  x 2 y 5  x 2 y 3 x 2  y 2  x 2 y 3 x  y x  y

90. 18y 3 x 2  2x y 4  2x y 3 9x  y 3     91. 27x 6  y 3  3x 2  y 3  3x 2  y 9x 4  3x 2 y  y 2    92. a 3  64b6  a  4b2 a 2  4ab2  16b4     93. y 3  5y 2  9y  45  y y 2  9  5 y 2  9  y  5 y  3 y  3   94. y 3  y 2  y  1  y 2 y  1  y  1  y 2  1 y  1  y  1 y  1 y  1  y  12 y  1

21


22

CHAPTER P Prerequisites

      95. 3x 3  x 2  12x  4  3x x 2  4  x 2  4  3x  1 x 2  4  3x  1 x  2 x  2   96. 9x 3  18x 2  x  2  9x 2 x  2  x  2  9x 2  1 x  2  3x  1 3x  1 x  2

97. a  b2  a  b2  [a  b  a  b] [a  b  a  b]  2b 2a  4ab             1 2 1 1 1 1 2 1 98. 1   1 1  1  1  1 x x x x x x      1 1 1 1 2 4 1 1   1 1 2  x x x x x x        99. x 2 x 2  1  9 x 2  1  x 2  1 x 2  9  x  1 x  1 x  3 x  3        100. a 2  1 b  22  4 a 2  1  a 2  1 b  22  4  a  1 a  1 [b  2  2] [b  2  2]  a  1 a  1 b b  4

101. Start by factoring out the power of x with the smallest exponent, that is, x 32 . So   1  x2 x 32  2x 12  x 12  x 32 1  2x  x 2  x 32 1  x2  . x 32   102. x  172  x  132  x  132 x  12  1  x  132 [x  1  1] [x  1  1]  x  132 x  2 x

103. x  1 x  22  x  12 x  2  x  1 x  2 [x  2  x  1]  3 x  1 x  2     104. y 4 y  23  y 5 y  24  y 4 y  23 1  y y  2  y 4 y  23 y 2  2y  1  y 4 y  23 y  12

105. Start by factoring y 2  7y  10, and then substitute a 2  1 for y. This gives  2            a 2  1  7 a 2  1  10  a 2  1  2 a 2  1  5  a 2  1 a 2  4  a  1 a  1 a  2 a  2 2             a 2  2a  1  a 2  2a  3 a 2  2a  1 106. a 2  2a  2 a 2  2a  3  a 2  2a  3

 a  1 a  3 a  12    2 2 107. x 2  3 x  13  4x  12 x  13  x  13 x 2  3  4x  12          x 2  3  4x  1  x  13 x 2  4x  4 x 2  4x  2  x  13 x 2  3  4x  1    x  13 x  22 x 2  4x  2   108. x  12 x  2  6 x  1 x  2  9 x  2  x  2 x  12  6 x  1  9

 x  2 [x  1  3]2  x  2 x  22    4  5  4   109. 5 x 2  4 2x x  24  x 2  4 4 x  23  2 x 2  4 x  23 5 x x  2  x 2  4 2   4   4    2 x 2  4 x  23 5x 2  10x  2x 2  8  2 x 2  4 x  23 7x 2  10x  8      110. 3 2x  12 2 x  312  2x  13 12 x  312  2x  12 x  312 6 x  3  2x  1 12      2x  12 x  312 6x  18  x  12  2x  12 x  312 7x  35 2 111.

1 2    13 43  43   43   1 x2  3  3 x  3 x2  3 x 2  3  23 x 2  x 2  3  23 x 2 x 2  3  x2  3  43 3 x2  3

112. 12 x 12 3x  412  32 x 12 3x  412  12 x 12 3x  412 [3x  4  3x]  12 x 12 3x  412 4  2x 12 3x  412


SECTION P.6 Factoring

23

2    113. 4a 2 c2  c2  b2  a 2  2ac2  c2  b2  a 2        2ac  c2  b2  a 2 2ac  c2  b2  a 2 (difference of squares)     2ac  c2  b2  a 2 2ac  c2  b2  a 2       c2  2ac  a 2  b2 (regrouping)  b2  c2  2ac  a 2     b2  c  a2 c  a2  b2 (perfect squares)

 [b  c  a] [b  c  a] [c  a  b] [c  a  b] (each factor is a difference of squares)  b  c  a b  c  a c  a  b c  a  b  a  b  c a  b  c a  b  c a  b  c

114. (a) Mowed portion  field  habitat

(b) Using the difference of squares, we get b2  b  2x2  [b  b  2x] [b  b  x]  2x 2b  2x  4x b  x.

115. The volume of the shell is the difference between the volumes of the outside cylinder (with radius R) and the inside cylinder   R r (with radius r). Thus V   R 2 h  r 2 h   R 2  r 2 h   R  r R  r  h  2   h  R  r. The 2 R r R r and 2  is the average circumference (length of the rectangular box), h is the height, and average radius is 2 2 R r R  r is the thickness of the rectangular box. Thus V   R 2 h  r 2 h  2   h  R  r  2  average radius  2 height  thickness R

length

rl

h

h

thickness

116. (a) 5282  5272  528  527 528  527  1 1055  1055

(b) 10202  10102  1020  1010 1020  1010  10 2030  20,300

(c) 501  499  500  1 500  1  5002  12  250,000  1  249,999

(d) 1002  998  1000  2 1000  2  10002  22  1,000,000  4  999,996 117. (a)

A2  A  1

A  1 

A  1

A  1

A2  A A2

 1

A  1

A2  A  1

A3  A2  A

A3

A3  A2  A  1

 1

A  1

A3  A2  A  1

A 4  A3  A2  A A4

 1  (b) Based on the pattern in part (a), we suspect that A5  1  A  1 A4  A3  A2  A  1 . Check: 

A 4  A 3  A2  A  1

A  1

A4  A3  A2  A  1

A 5  A 4  A 3  A2  A

A5  1   The general pattern is An  1  A  1 An1  An2      A2  A  1 , where n is a positive integer.


24

CHAPTER P Prerequisites

  118. (a) A  B A2  AB  B 2  A3  A2 B  AB 2  A2 B  AB 2  B 3  A3  B 3   (b) A  B A2  AB  B 2  A3  A2 B  AB 2  A2 B  AB 2  B 3  A3  B 3

P.7

RATIONAL EXPRESSIONS

P x , where P and Q are polynomials. 1. A rational expression has the form Q x 3x (a) 2 is a rational expression. x 1  x 1 (b) is not a rational expression. A rational expression must be a polynomial divided by a polynomial, and the 2x  3  numerator of the expression is x  1, which is not a polynomial.

x3  x xx 2  1  is a rational expression. x 3 x 3 2. To simplify a rational expression we cancel factors that are common to the numerator and denominator. So, the expression x  1x  2 x 1 simplifies to . x  3x  2 x 3 3. To multiply two rational expressions we multiply their numerators together and multiply their denominators together. So 2 x 2x 2x  is the same as .  2 x 1 x 3 x  1  x  3 x  4x  3 (c)

4. (a)

2 x 1   has three terms. x x  1 x  12

(b) The least common denominator of all the terms is x x  12 . (c)

2 x 2x x  1 1 x x x  12 x  12  2x x  1  x 2       2 2 2 x x  1 x  1 x  1 x x  1 x x  12 x  1 

x 2  2x  1  2x 2  2x  x 2

5. (a) Yes. Cancelling x  1, we have

x x  12

x x  1 x  12

x . x 1

(b) No; x  52  x 2  10x  25  x 2  25, so x  5 

2x 2  1

x x  12

x 2  10x  25 

x 2  25.

3 a a 3a   1 . 3 3 3 3 (b) No. We cannot “separate” the denominator in this way; only the numerator, as in part (a). (See also Exercise 101.)

6. (a) Yes,

7. The domain of 4x 2  10x  3 is all real numbers.

8. The domain of x 4  x 3  9x is all real numbers.

9. Since x  3  0 we have x  3. Domain: x  x  3

10. Since 3t  6  0 we have t  2. Domain: t  t  2

11. Since x  3  0, x  3. Domain; x  x  3

12. Since x  1  0, x  1. Domain; x  x  1

13. x 2  x  2  x  1 x  2  0  x  1 or 2, so the domain is x  x  1 2. x x  14. 2 . x  2 x  2  0  x  2, so the domain is x  x  2. x  2 x  2 x 4  x 2 15. is defined for x  x  2 x  3, so its domain is x  x  2. x 3  x 2 16. 2 is defined for x  x  2 x  3, so its domain is x  x  2 x  3. x 9


SECTION P.7 Rational Expressions

17. 19. 21. 22.

x  5 x  5 x  5 x  5   12 x  5 2x  10 2 x  5 x 2 1 x 2   2 x  2  2  2 x x x 4

x 1 x 2  7x  8 x  8 x  1   x 2 x  8 x  2 x 2  10x  16

18. 20.

 x x2  2

x2  2 x 3  2x   2 x x  1 x 1 x x

x2  x  2 x 2 x  2 x  1   2 x 1  1  1 x x x 1

x 2  x  12 x 4 x  4 x  3   x 2 x  2 x  3 x 2  5x  6

y y  1 y y2  y   y1 y  1 y  1 y2  1   y y 2  5y  24 y8 y y  8 y  3 y 3  5y 2  24y      24. y y  3 y  3 y3 y 3  9y y y2  9   2x 6 x 2x 3 2 x 2x  3 x  2 x 2x  3 2x  x  6x    25. 2x  3 2x  3 x  2 2x  3 x  2 2x 2  7x  6 23.

26. 27. 28. 29.

 x  1 1  x  x  1 1  x2 1  x 1  x      3 2 2 x 1 x2  x  1 x  1 x  x  1 x  1 x  x  1 4x

x2  4

4x 1 x 2 x 2    16x 4 x  2 x  2 x  2 16x

x 5 x 2  25 x  4 x  5 x  5 x  4     x 4 x  4 x  4 x  5 x 2  16 x  5

x 3 x 2  2x  15 x  5 x  5 x  3 x  5    x 2 x 2 x  5 x  5 x  2 x 2  25

 x  1 x  1 1x x 2  2x  3 3  x x  3 x  1  x  3       x 3 x 1 x 1 1x x  3 x  1 x 2  2x  3 3  x 2t  3 2t  3 1 2t  3 2t  3  31. 2  2  2 t  9 2t  3 2t  3 t  9 4t 2  9 t 9 30.

32.

33. 34. 35. 36.

3y 2  9y y2  9 3y y  3 y  3 y  3 3   2   2  3 2y  3 y  9y 2y  3y  9 y y  9 y  3 2y  3

x 3  2x 2  8x x 2  2x  24 x 4 x x  2 x  4 x  4 x  6    x 4 x  6 x  2 x x  4 x  4 x 2  8x  12 x 3  16x

xy 2x 2  x y  y 2 x 2  2x y  y 2 2x  y x  y x  y2  2   2 2 2 x  2y x  y x  2y 2x  y x  y x  x y  2y 2x  3x y  y

x 3 x 2  7x  12 x  3 2x 2  7x  15 x 3 x 5 x  5 2x  3  2  2     2 2x  3 2x  3 x  3 x  4 2x  3 x  4 4x  9 2x  7x  15 4x  9 x 2  7x  12 2x  1

2x 2  x  15

2x  1 6x 2  x  2 x 3 1    x 3 x  3 2x  5 2x  1 3x  2 2x  5 3x  2

x3 x 2  2x  1 x 3 x  1 x  1 x3 x 1    37.  x 2 x  1 x x 1 x x  1 x x 2  2x  1

2x 2  3x  2 x 2 x2  x  2 x  2 2x  1 x  1 x  2 2x 2  3x  2 x2  1      38. x 1 x  1 x  1 x  2 2x  1 x2  1 2x 2  5x  2 2x 2  5x  2 x2  x  2

25


26

CHAPTER P Prerequisites

x 1 x xy    z y z yz y x z xz x x    40. yz z 1 y y 39.

1 x 3 1 x 4    x 3 x 3 x 3 x 3 3x  2 3x  2 2 x  1 3x  2  2x  2 x 4 42. 2    x 1 x 1 x 1 x 1 x 1 1 2 x 3 2 x  5 x  3  2x  10 3x  7 43.      x 5 x 3 x  5 x  3 x  5 x  3 x  5 x  3 x  5 x  3 41. 1 

44. 45. 46. 47. 48.

1 1 x 1 x 1 x 1x 1 2x      x 1 x 1 x  1 x  1 x  1 x  1 x  1 x  1 x  1 x  1

3 1 3 x  2 1 x  1 3x  6  x  1 2x  5      x 1 x 2 x  1 x  2 x  1 x  2 x  1 x  2 x  1 x  2

3 x x  6 x 3 x  4 x 2  6x  3x  12 x 2  3x  12      x 4 x 6 x  4 x  6 x  4 x  6 x  4 x  6 x  4 x  6

5 3 5 2x  3 3 10x  15  3 10x  18 2 5x  9       2 2 2 2 2x  3 2x  32 2x  3 2x  3 2x  3 2x  3 2x  32 x

x  12

49. u  1 

x 2 2 x  1 3x  2 x  2x  2     2 2 x 1  1  1 x x x  1 x  1 x  12

u u u 2  2u  1  u u 2  3u  1 u  1 u  1     u1 u1 u1 u1 u1

3 2b2 3ab 4a 2 2b2  3ab  4a 2 4 2   2  2 2  2 2  2 2  2 ab b a a b a b a b a 2 b2 x 1 1 1 1 1 x 2x  1  2 51. 2  2  2   2  2 x x  1 x x x x x x  1 x x  1 x x  1 50.

1 1 x2 x 1 x2  x  1 1  2  3  3  3  3  x x x x x x x3 1 1 2 2 2 x  4 1    53.   x  3 x 2  7x  12 x  3 x  3 x  4 x  3 x  4 x  3 x  4 2x  8  1 2x  7   x  3 x  4 x  3 x  4 52.

54.

55. 56.

57. 58.

x

x2  4

x x 1 1 x 2     x 2 x  2 x  2 x  2 x  2 x  2 x  2 x  2 2x  2 2 x  1   x  2 x  2 x  2 x  2

1 1 1 1 x 3 1 x 2       x  3 x2  9 x  3 x  3 x  3 x  3 x  3 x  3 x  3 x  3 x  3

x 2 x 2    x  1 x  2 x  1 x  4 x 2  x  2 x 2  5x  4 2 x  2 x 2  4x  2x  4 x 2  6x  4 x x  4     x  1 x  2 x  4 x  1 x  2 x  4 x  1 x  2 x  4 x  1 x  2 x  4 3 4 3 4 2 x  1 3x 4 2x  2  3x  4 5x  6 2 2           x x  1 x2  x x x  1 x x  1 x x  1 x x  1 x x  1 x x  1 x x  1 1 1 1 z2 x2 y2 x 2  y2  z2       x 3 y3 z x y3 z3 x 3 yz 3 x 3 y3 z3 x 3 y3 z3 x 3 y3 z3 x 3 y3 z3


SECTION P.7 Rational Expressions

59.

1 1 x x  1 x 1 x 2  2x  1 1       x 2 x  1 x 2 x  12 x 3 x  12 x 3 x  12 x 3 x  12 x 3 x  12 x 3 x  12 

60.

27

x  12

x 3 x  12

1  3 x

2 2 1 3 3 1    2   x  1 x  12 x  1 x  12 x  1 x  1 x 1 2 x  1 3 x  1 x  1 x  1    x  1 x  12 x  1 x  12 x  1 x  12 

x2  1

2x  2 3x  3 x 2  1  2x  2  3x  3 x2  x  4     x  1 x  12 x  1 x  12 x  1 x  12 x  1 x  12 x  1 x  12

  x 1  1x x 1     61. 1 1 1  2x x x 2 x 2 1  1x

  y 1  2y y 2   62. 3   3 3y y y 1 y 1 1  2y

  1 1 1   2 x 1 x 3 x  2  1 x 2 x 2     63. 1 1 x 1 x  2  1 1 x  2 1  x 2 x 2 1 c  1  c11  c 64. 1 c11 c2 1 c1 1

  1 1 1 1  x  1 x  3  2 x  1 x  3  x  1 x 1 x 3 65. x  1 x  3    x 1 x  1 x  3 x  1 x  1 x  1 x  3 x  1 x  1 x  3 2  x  1 x  3   x 3 x 2  x 2  2x  3  x 2  2x  8 5  3  1   2  4 x x x x 66. x  4 x  1    x 3 x  4 x  3 x  1 x  4 x  3 x  1 x  4 x  3 x  1   x x xy x  x 2 y  1 x2 y  x2 y y  67. y  y   x y 2  y 2  y 2 x  1 y xy y  x x x

   y y xy x  y y  x2 2 y  y2 x x x      68. x  x x y2  x 2 x x  y2 y x y y  y y x


28

CHAPTER P Prerequisites

y x x 2  y2  x 2 y2 x 2  y2 xy y x xy   x y. An alternative method is to multiply the 69.  2   2 2 2 1 1 x y 1 y  x y x  x2 y2 x 2 y2 numerator and denominator by the common denominator of both the numerator and denominator, in this case x 2 y 2 :     y x y x 2  y2   x y x 2 2 3 3 x y  xy x y y x y x    2   x y. 1 1 1 1 x 2 y2 y  x2 y2  x 2   x2 y2 x2 y2   x x 2  y2 y xy x y2 y x y2 x 3  x y2  x y2 x3 70. x  x  2 y x x y  x y  x  x 2  y2  x 2  y2  x 2  y2  2 2 x y x  y2   y x y x 1 x2 1 y2   y2  x 2 xy yx y  x y  x x y x2 y2 x 2 y2 x 2 y2   71. 1     y x 2 y2 2 y 2 y  x 1 1 y  x xy x  y 1 x x   xy xy x y   1 1  x 2 y2 yx y2  x 2 x 2  y 2 y  x y  x x2 y2    .    2  Alternatively, 1 1 2 2 2 1 1 x y y  x xy x y x y xy  x y  x y 1 1 u 1 72.   1  1 1u 1u u1 1 u 1 x 1x 1 x 73. 1    1 1 x 1 x 1 1x 1 x x 2x 1 2x  3 x 1 1x 1   1 1 74. 1  1 x 2 x 2 x 2 1  x  1 1 1x 1 1  1 1  x  h 1  x  1  x  1  x  h   75. h h 1  x 1  x  h 1  x 1  x  h 76. In calculus it is necessary to eliminate the h in the denominator, and we do this by rationalizing the numerator: 1 1       x  x  h 1 x  x h x  x h x x h                 . h h x x h x  x h h x x h x  x h x x h x  x h   1 1 2  x 2  2xh  h 2  2 2 2 x 2 x  x  h 2x  h x  h x    77. h hx 2 x  h2 hx 2 x  h2 x 2 x  h2 x 2  y 2

x 2  h 2  2xh  3x  3h  x 2  3x h h  2x  3 x  h2  3 x  h  x 2  3x    2x  h  3 h h h   2      x2 x2 1 1  x2 1 x  79. 1    1     2 2 2 2 1x 1x 1x 1x 1  x2 1  x2       1 2 2x 3 1 1 1 80. 1  x 3  3  1  x6  3   1  x 6  12   x 6  12  6 6 6 4x 4x 16x 16x 16x   2   1 1  x3  3   x 3  3  4x 4x 78.


SECTION P.7 Rational Expressions

81.

82.

2 x  3 x  53  3 x  52 x  32 x  56

4x 3 1  x3  3 1  x2 1 x 4 1  x6

x  3 x  52 [2 x  5  3 x  3] x  56

x 3 1  x2 [4 1  x  3x] 1  x6

x  3 19  x x  54

x 3 4  x 1  x4

x 2 2 1  x12  x 1  x12 1  x12 [2 1  x  x]   1x 1x 1  x32   12  12 12   1  x2 1  x2 1  x2  x2  x2 1  x2 1 84.   32 2 2 1x 1x 1  x2 83.

85.

3 1  x13  x 1  x23 1  x23

1  x23 [3 1  x  x] 1  x23

2x  3

1  x43   7  3x12 7  3x  32 x 7  32 x 7  3x12  32 x 7  3x12   86. 7  3x 7  3x 7  3x32   1 3  10 3  10  87.        9  10  10  3 3  10 3  10 3  10     2  5 3  3 63 5    88.   6  3 5     45 2 5 2 5 2 5

      2 5 3 5 3    5 3 89.           53 5 3 5 3 5 3   x 1 x 1 1 1    90.  x 1 x 1 x 1 x 1        3 y y y 3y y 3 y y y 91.         3 y 3y 3 y 3 y 3 y        2 x  y x  y x y  2 x  y 2 x  y    2 x  y 2 x 2 y 92.        xy x y x y x y       2  5 2  5 2 5 45 1   93.           5 5 2 5 5 2 5 5 2 5       3 5 3 5 3 5 35 2 1       94.        2 2 3 5 3 5 2 3 5 2 3 5       r 2 r 2 r 2 r 2   95.      5 5 r 2 5 r 2       x  x h x  x h x  x h x  x  h 96.              h x x h h x x h x  x h h x x h x  x h h 1            h x x h x  x h x x h x  x h    x2  1  x x2  1  x x2  1  x2 1 97. x 2  1  x     2 2 2 1 x 1x x 1x x 1x       x 1x 1 x 1 x x 1 x  98. x  1  x       1 x 1 x x 1 x x 1 x 2

2

29


30

CHAPTER P Prerequisites

99. (a) R 

1 1 R R R1 R2   1 2  1 1 1 1 R1 R2 R2  R1   R1 R2 R1 R2

(b) Substituting R1  10 ohms and R2  20 ohms gives R  100. (a) The average cost A 

200 10 20  67 ohms.  30 20  10

Cost 500  6x  001x 2  . number of shirts x

(b) x

10

20

50

100

200

500

1000

Average cost

$5610

$3120

$1650

$1200

$1050

$1200

$1650

101. x x2  9 x 3

280

290

295

299

2999

3

3001

301

305

310

320

580

590

595

599

5999

?

6001

601

605

610

620

From the table, we see that the expression

x2  9 approaches 6 as x approaches 3. We simplify the expression: x 3

x2  9 x  3 x  3   x  3, x  3. Clearly as x approaches 3, x  3 approaches 6. This explains the result in the x 3 x 3 table. 2 2 102. No, squaring  changes its value by a factor of  . x x 103. Answers will vary. Algebraic Error 1 1 1   a b ab

a  b2  a 2  b2  a 2  b2  a  b ab  b a 1 a  ab b am  a mn an 104. (a)

Counterexample 1 1 1   2 2 22

1  32  12  32  52  122  5  12 26  6 2 1  1 11 5 3  352 32

5 a a 5a    1  , so the statement is true. 5 5 5 5

(b) This statement is false. For example, take x  5 and y  2. Then LHS  RHS 

5 5 x  , and 2  . y 2 2

(c) This statement is false. For example, take x  0 and y  1. Then LHS  RHS 

1 1 1 1   , and 0  . 1y 11 2 2

x 1 51 6    2, while y1 21 3 0 x   0, while xy 01

(d) This statement is false. For example, take x  1 and y  1. Then LHS  2 2a 2 RHS    1, and 2  1. 2b 2

a  b

2

  1  2, while 1


SECTION P.8 Solving Basic Equations

31

    a  a a 1 1  a  1 a  1  . (e) This statement is true: b b b b b (f) This statement is true:

1 x x2 1 1  x  x2      1  x. x x x x x

105. (a) x x

1 x

1

3

1 2

9 10

99 100

999 1000

9999 10,000

2

3333

25

2011

20001

2000001

200000001

It appears that the smallest possible value of x 

1 is 2. x

(b) Because x  0, we can multiply both sides by x and preserve the inequality: x 

  1 1 2x x  2x  x x

x 2  1  2x  x 2  2x  1  0  x  12  0. The last statement is true for all x  0, and because each step is 1 reversible, we have shown that x   2 for all x  0. x

P.8

SOLVING BASIC EQUATIONS

1. Substituting x  3 in the equation 4x  2  10 makes the equation true, so the number 3 is a solution of the equation.

2. Subtracting 4 from both sides of the given equation, 3x  4  10, we obtain 3x  4  4  10  4  3x  6. Multiplying by 13 , we have 13 3x  13 6  x  2, so the solution is x  2.

x  2x  10 is equivalent to 52 x  10  0, so it is a linear equation. 2 2 2 (b)  2x  1 is not linear because it contains the term , a multiple of the reciprocal of the variable. x x (c) x  7  5  3x  4x  2  0, so it is linear.

3. (a)

4. (a) x x  1  6  x 2  x  6 is not linear because it contains the square of the variable.  (b) x  2  x is not linear because it contains the square root of x  2.

(c) 3x 2  2x  1  0 is not linear because it contains a multiple of the square of the variable.

5. (a) This is true: If a  b, then a  x  b  x.

(b) This is false, because the number could be zero. However, it is true that multiplying each side of an equation by a nonzero number always gives an equivalent equation.

(c) This is false. For example, 5  5 is false, but 52  52 is true.

 6. To solve the equation x 3  125 we take the cube root of each side. So the solution is x  3 125  5.

7. (a) When x  2, LHS  4 2  7  8  7  1 and RHS  9 2  3  18  3  21. Since LHS  RHS, x  2 is not a solution. (b) When x  2, LHS  4 2  7  8  7  15 and RHS  9 2  3  18  3  15. Since LHS  RHS, x  2 is a solution.

8. (a) When x  1, LHS  2  5 1  2  5  7 and RHS  8  1  7. Since LHS  RHS, x  1 is a solution. (b) When x  1, LHS  2  5 1  2  5  3 and RHS  8  1  9. Since LHS  RHS, x  1 is not a solution.

9. (a) When x  2, LHS  1  [2  3  2]  1  [2  1]  1  1  0 and RHS  4 2  6  2  8  8  0. Since LHS  RHS, x  2 is a solution. (b) When x  4 LHS  1  [2  3  4]  1  [2  1]  1  3  2 and RHS  4 4  6  4  16  10  6. Since LHS  RHS, x  4 is not a solution.


32

CHAPTER P Prerequisites

1 1  12   12  12  1 and RHS  1. Since LHS  RHS, x  2 is a solution. 24 2 1 is not defined, so x  4 is not a solution. (b) When x  4 the expression 44

10. (a) When x  2, LHS  12 

11. (a) When x  1, LHS  2 113  3  2 1  3  2  3  5. Since LHS  1, x  1 is not a solution. (b) When x  8 LHS  2 813  3  2 2  3  4  3  1  RHS. So x  8 is a solution.

432 23 8    4 and RHS  4  8  4. Since LHS  RHS, x  4 is a solution. 46 2 2  32 23 832 292 (b) When x  8, LHS     272 and RHS  8  8  0. Since LHS  RHS, x  8 is not a 86 2 2 solution. 0a a a 13. (a) When x  0, LHS     RHS. So x  0 is a solution. 0b b b ba ba  is not defined, so x  b is not a solution. (b) When x  b, LHS  bb 0    2 b b b2 b b2 b2 b 14. (a) When x  , LHS  b  14 b2     0  RHS. So x  is a solution. 2 2 2 4 2 4 2  2   2 1 1 b 1 1 1  14 b2  2  1  , so x  is not a solution. b (b) When x  , LHS  b b b 4 b b 12. (a) When x  4, LHS 

15. 5x  6  14  5x  20  x  4

16. 3x  4  7  3x  3  x  1

17. 7  2x  15  2x  8  x  4

18. 4x  95  1  4x  96  x  24

19. 12 x  7  3  12 x  4  x  8

20. 2  13 x  4  13 x  6  x  18

21. 3x  3  5x  3  0  8x  x  0

22. 2x  3  5  2x  4x  2  x  12

23. 7x  1  4  2x  9x  3  x  13

24. 1  x  x  4  3  2x  x   32

25. x  3  4x  3  5x  x  35

26. 2x  3  7  3x  5x  4  x  45

27. x3  1  53 x  7  x  3  5x  21  4x  24  x  6

3 x  3  4x  10  3x  30  x  40 28. 25 x  1  10

29. 2 1  x  3 1  2x  5  2  2x  3  6x  5  2  2x  8  6x  6  8x  x   34

30. 5 x  3  9  2 x  2  1  5x  15  9  2x  4  1  5x  24  2x  3  7x  21  x  3   31. 4 y  12  y  6 5  y  4y  2  y  30  6y  3y  2  30  6y  9y  32  y  32 9

32. r  2 [1  3 2r  4]  61  r  2 1  6r  12  61  r  2 6r  11  61  r  12r  22  61  13r  39  r 3 33. x  13 x  12 x  5  0  6x  2x  3x  30  0 (multiply both sides by 6)  x  30

34. 23 y  12 y  3   y  21 11

y1  8y  6 y  3  3 y  1  8y  6y  18  3y  3  14y  18  3y  3  11y  21 4

x 1 x 1   6x  8x  2x  x  1  24x  7x  1  24x  1  17x  x  17 2 4 x 1 1 5x    18x  15x  2 x  1  1  3x  2x  1  x  1 36. 3x  2 3 6 35. 2x 

37. x  1 x  2  x  2 x  3  x 2  x  2  x 2  5x  6  x  2  5x  6  6x  8  x  43


SECTION P.8 Solving Basic Equations

33

38. x x  1  x  32  x 2  x  x 2  6x  9  x  6x  9  5x  9  x   95

39. x  1 4x  5  2x  32  4x 2  x  5  4x 2  12x  9  x  5  12x  9  13x  14  x  14 13

40. t  42  t  42  32  t 2  8t  16  t 2  8t  16  32  16t  32  t  2 1 4 41.   1  3  4  3x (multiply both sides by the LCD, 3x)  1  3x  x   13 x 3x 6 2 42.  5   4  2  5x  6  4x  4  9x   49  x x x 4 2x  1   5 2x  1  4 x  2  10x  5  4x  8  6x  13  x  13 43. 6 x 2 5 2x  7  23  2x  7 3  2 2x  4 (cross multiply)  6x  21  4x  8  2x  29  x  29 44. 2 2x  4 3 2   2 t  1  3 t  6 [multiply both sides by the LCD, t  1 t  6]  2t  2  3t  18  20  t 45. t 6 t 1 5 6   6 x  4  5 x  3  6x  24  5x  15  x  39 46. x 3 x 4 3 1 47. 1   3 6  3x  3  2 [multiply both sides by 6 x  1]  18  3x  3  2  3x  15  2  x 1 2 3x  3 3x  13  x  13 3 48.

49.

5 12x  5  2   12x  5 x  2x 6x  3  5 6x  3  12x 2  5x  12x 2  6x  30x  15  6x  3 x 12x 2  5x  12x 2  24x  15  19x  15  x   15 19

1 1 1 10     10 z  1  5 z  1  2 z  1  10 10z [multiply both sides by 10z z  1]  z 2z 5z z1 3 z 3 z  1  100z  3z  3  100z  3  97z  97

1 4 15    0  3  t  4 3  t  15  0  3  t  12  4t  15  0  3t  30  0  3t  30 3  t 3  t 9  t2  t  10 x 1 51. 2   x  2 2x  4  2 [multiply both sides by 2 x  2]  x  4x  8  2  3x  6  x  2. 2x  4 x 2 But substituting x  2 into the original equation does not work, since we cannot divide by 0. Thus there is no solution. 5 2 1   x  3  5  2 x  3  x  2  2x  6  x  4  52. x  3 x2  9 x 3 3 1 6x  12 53.   2  3 x  x  4  6x  12 (multiply both sides by x x  4]  3x  7x  16  4x  16 x 4 x x  4x  x  4. But substituting x  4 into the original equation does not work, since we cannot divide by 0. Thus, there is no solution. 1 2 1 54.   2x  1  2 x  1  1  1. This is an identity for x  0 and x   12 , so the solutions are  2 x 2x  1 2x  x all real numbers except 0 and  12 . 50.

55. x 2  25  x  5

56. 3x 2  48  x 2  16  x  4  57. 5x 2  15  x 2  3  x   3   58. x 2  1000  x   1000  10 10

  59. 8x 2  64  0  x 2  8  0  x 2  8  x   8  2 2   60. 5x 2  125  0  5 x 2  25  0  x 2  25  x  5 61. x 2  16  0  x 2  16 which has no real solution.


34

CHAPTER P Prerequisites

62. 6x 2  100  0  6x 2  100  x 2   50 3 , which has no real solution.   2 63. x  3  5  x  3   5  x  3  5    4 7 2 64. 3x  4  7  3x  4   7  3x  4  7  x  3 65. x 3  27  x  2713  3

66. x 5  32  0  x 5  32  x  3215  2      67. 0  x 4  16  x 2  4 x 2  4  x 2  4 x  2 x  2 x 2  4  0 has no real solution. If x  2  0, then x  2. If x  2  0, then x  2. The solutions are 2.   16 27 2716 3 27 6 6 x    16   68. 64x  27  x  64 64 2 64

69. x 4  64  0  x 4  64 which has no real solution.

70. x  13  8  0  x  13  8  x  1  813  2  x  1. 14   8114  x  2  3. So x  2  3, then x  1. If 71. x  24  81  0  x  24  81  x  24 x  2  3, then x  5. The solutions are 5 and 1.

72. x  14  16  0  x  14  16, which has no real solution.

73. 3 x  33  375  x  33  125  x  3  12513  5  x  3  5  8   74. 4 x  25  1  x  25  14  x  2  5 14  x  2  5 14  75. 3 x  5  x  53  125 3  3   14 76. x 43  16  0  x 43  16  24  x 43  24  212  x 4  212  x   212  23  8 15   23  8 77. 2x 53  64  0  2x 53  64  x 53  32  x  3235  25

32  32   62  x  63  216 78. 6x 23  216  0  6x 23  216  x 23  36  62  x 23

944  313 302 161 80. 836  095x  997  095x  161  x   169 095 582  506 81. 215x  463  x  119  115x  582  x  119 195  059 82. 395  x  232x  200  195  332x  x  332

79. 302x  148  1092  302x  944  x 

4497  4366 103 84. 214 x  406  227  011x  214x  86684  227  011x  225x  109584  x  48704  487 026x  194 85.  176  026x  194  176 303  244x  026x  194  533  429x  455x  727  303  244x 727  160 x 455 173x 320 86.  151  173x  151 212  x  173x  320  151x  022x  320  x   1455 212  x 022 12 d 12 M 88. d  r T H  T  87. r  M r rH 83. 316 x  463  419 x  724  316x  1463  419x  3034  4497  103x  x 

89. P V  n RT  R 

PV nT

90. F  G

mM Fr 2 m  2 GM r


SECTION P.8 Solving Basic Equations

91. P  2l  2  2  P  2l   

35

P  2l 2

1 1 1    R1 R2  R R2  R R1 (multiply both sides by the LCD, R R1 R2 ). Thus R1 R2  R R1  R R2  R R1 R2 R R2 . R1 R2  R  R R2  R1  R2  R  3V 3V r  93. V  13 r 2 h  r 2  h h  mM mM mM 2 r  G 94. F  G 2  r  G F F r  3V 3V r  3 95. V  43 r 3  r 3  4 4  2 2 2 2 2 2 96. a  b  c  b  c  a  b   c2  a 2        i 2 i i 2 A i A A A  1    1   i  100  100 97. A  P 1   1 100 P 100 100 P 100 P P     98. a 2 x  a  1  a  1 x  a 2 x  a  1 x   a  1  a 2  a  1 x  a  1  a 2  a  1 x  a  1 92.

a  1 x  2 a a1 ax  b 2d  b 99.  2  ax  b  2 cx  d  ax  b  2cx  2d  ax  2cx  2d  b  a  2c x  2d  b  x  cx  d a  2c a1 a1 b1 100.    a a  1  a a  1  b b  1  a 2  a  a 2  a  b2  b  2a  b2  b  b  b  a a  12 b2  b

8  25 0032 250  25   000055. So the beam shrinks 10,000 10,000 000055  12025  0007 m, so when it dries it will be 12025  0007  12018 m long. 0032  25  5  0032  25  75  0032  (b) Substituting S  000050 we get 000050  10,000 75   234375. So the water content should be 234375 kg/m3 . 0032 3150  840. So the toy manufacturer can 102. Substituting C  3600 we get 3600  450  375x  3150  375x  x  375 manufacture 840 toy trucks. 101. (a) The shrinkage factor when   250 is S 

103. (a) Solving for  when P  10,000 we get 10,000  156 3   3  64102    86 km/h.

(b) Solving for  when P  50,000 we get 50,000  156 3   3  320513    147 km/h.

104. Substituting F  300 we get 300  03x 34  1000  103  x 34  x 14  10  x  104  10,000 lb.   105. (a) C  12 1000 22  1000 981 1  200,000  211,810 (b) 12  2  gh  P  C  P  C  12  2  gh.

  If the height increases to 5 m, then P  211,810  12 1000 22  1000 981 5  160,760 Pa.   If h  1 m and   4 ms, then P  211,810  12 1000 42  1000 981 1  194,000 Pa.

106. (a) 3 0  k  5  k 0  k  1  k  5  k  1  2k  6  k  3

(b) 3 1  k  5  k 1  k  1  3  k  5  k  k  1  k  2  1  k  3

(c) 3 2  k  5  k 2  k  1  6  k  5  2k  k  1  k  1  k  1. x  2 is a solution for every value of k. That is, x  2 is a solution to every member of this family of equations.


36

CHAPTER P Prerequisites

107. When we multiplied by x, we introduced x  0 as a solution. When we divided by x  1, we are really dividing by 0, since x  1  x  1  0.

P.9

MODELING WITH EQUATIONS

1. An equation modeling a real­world situation can be used to help us understand a real­world problem using mathematical methods. We translate real­world ideas into the language of algebra to construct our model, and translate our mathematical results back into real­world ideas in order to interpret our findings. 2. In the formula I  Pr t for simple interest, P stands for principal, r for interest rate, and t for time (in years). 3. (a) A square of side x has area A  x 2 .

(b) A rectangle of length l and width  has area A  l. (c) A circle of radius r has area A  r 2 .

5  16 ounces 4. Balsamic vinegar contains 5% acetic acid, so a 32 ounce bottle of balsamic vinegar contains 32  5%  32  100

of acetic acid.

1 1 wall  . x hours x d rt d d rt d 6. Solving d  rt for r, we find   r  . Solving d  rt for t, we find  t  . t t t r r r 7. If n is the first integer, then n  1 is the middle integer, and n  2 is the third integer. So the sum of the three consecutive integers is n  n  1  n  2  3n  3. 5. A painter paints a wall in x hours, so the fraction of the wall she paints in one hour is

8. If n is the middle integer, then n  1 is the first integer, and n  1 is the third integer. So the sum of the three consecutive integers is n  1  n  n  1  3n. 9. If n is the first even integer, then n  2 is the second even integer and n  4 is the third. So the sum of three consecutive even integers is n  n  2  n  4  3n  6.

10. If n is the first integer, then the next integer is n  1. The sum of their squares is   n 2  n  12  n 2  n 2  2n  1  2n 2  2n  1.

11. If s is the third test score, then since the other test scores are 78 and 82, the average of the three test scores is 160  s 78  82  s  . 3 3 12. If q is the fourth quiz score, then since the other quiz scores are 8, 8, and 8, the average of the four quiz scores is 888q 24  q  . 4 4 13. If x dollars are invested at 2 12 % simple interest, then the first year you will receive 0025x dollars in interest. 14. If n is the number of months the apartment is rented, and each month the rent is $945, then the total rent paid is 945n. 15. Since  is the width of the rectangle, the length is three times the width, or 4. Then area  length  width  4    42 ft2 .

four

16. Since  is the width of the rectangle, the length is   6. The perimeter is 2  length  2  width  2   6  2   4  12 d distance  . 17. If d is the given distance, in miles, and distance  rate  time, we have time  rate 55 1h  34 s mi. 18. Since distance  rate time we have distance  s  45 min 60 min 19. If x is the quantity of pure water added, the mixture will contain 25 oz of salt and 3  x gallons of water. Thus the 25 concentration is . 3x


SECTION P.9 Modeling with Equations

37

20. If p is the number of pennies in the purse, then the number of nickels is 2p, the number of dimes is 4  2 p, and the number of quarters is 2 p  4  2 p  4 p  4. Thus the value (in cents) of the change in the purse is 1  p  5  2 p  10  4  2 p  25  4 p  4  p  10 p  40  20 p  100 p  100  131 p  140. 21. If d is the number of days and m the number of miles, then the cost of a rental is C  65d  040m. In this case, d  3 and 88 C  283, so we solve for m: 283  65  3  040m  283  195  04m  04m  88  m   220. Thus, the 04 truck was driven 220 miles.   100 if m  250 22. The cost of speaking for m minutes on this plan is C  In this case m  250  100  025 m  250 if m  250

and C  12050, so we solve 100  025 m  250  1205  025 m  250  205  m  250  4 205  82  m  250  82  332. Therefore, the tourist used 332 minutes of talk in the month of June. 23. If x is the student’s score on their final exam, then because the final counts twice as much as each midterm, the average 82  75  71  2x 228  2x 114  x 114  x score is   . For the average to be 80%, we must have  80%  08  3 100  200 500 250 250 114  x  250 08  200  x  86. So the student scored 86% on their final exam. 24. Six students scored 100 and three students scored 60. Let x be the average score of the remaining 25  6  3  16 students. 6 100  3 60  16x  084  780  16x  084 2500  2100 Because the overall average is 84%  084, we have 25 100  16x  1320  x  1320 16  825. Thus, the remaining 16 students’ average score was 825%.

25. Let m be the amount invested at 2 12 %. Then 12,000  m is the amount invested at 3%. Since the total interest is equal to the interest earned at 25% plus the interest earned at 3%, we have 318  0025m  003 12,000  m  318  0025m  360  003m  0005m  42  m 

42  8400. Thus, 0005

$8400 is invested at 2 12 % and 12,000  8400  $3600 is invested at 3%.

26. Let m be the amount invested at 5%. Then 8000  m is the total amount invested. Thus 4% of the total investment  interest earned at 3 12 %  interest earned at 5% 40  4000. Thus, $4000 must be So 004 8000  m  0035 8000  005m  320  004m  280  005m  m  001 invested at 5%. 2625  0075 or 75%. 27. Using the formula I  Prt and solving for r, we get 26250  3500  r  1  r  3500   28. If $3000 is invested at an interest rate a%, then $5000 is invested at a  12 %, so, remembering that a is expressed as a 1

a 2 a  1  5000   1  30a  50a  25  80a  25. Since the total interest 100 100 is $265, we have 265  80a  25  80a  240  a  3. Thus, $3000 is invested at 3% interest. 29. Let x be the monthly salary. Since annual salary  12  monthly salary  Christmas bonus, we have 180,100  12x  7300  172,800  12x  x  14,400. The monthly salary is $14,400. 30. Let s be the assistant’s annual salary. Then the foreman’s annual salary is 115s. Their total income is the sum of their salaries, so s  115s  113,305  215s  113,305  s  52,700. Thus, the assistant’s annual salary is $52,700. 31. Let x be the overtime hours worked. Since gross pay  regular salary  overtime pay, we obtain the equation 16650 814  1850  35  1850  15  x  814  64750  2775x  16650  2775x  x   6. Thus, the lab 2775 technician worked 6 hours of overtime. 32. Let x be the number of hours worked by the plumber. Then the cabinet maker works for 9x hours. The total labor charge is percentage, the total interest is I  3000 

the sum of their charges, so 2610  150x  80 9x  2610  150x  720x  x  2610 870  3. Thus, the plumber works for 3 hours and the cabinet maker works for 3  9  27 hours.


38

CHAPTER P Prerequisites

33. All ages are in terms of the daughter’s age 7 years ago. Let y be age of the daughter 7 years ago. Then 11y is the age of the movie star 7 years ago. Today, the daughter is y  7, and the movie star is 11y  7. But the movie star is also 4 times his daughter’s age today. So 4 y  7  11y  7  4y  28  11y  7  21  7y  y  3. Thus, today the movie star is 11 3  7  40 years old. 34. Let h be number of home runs Babe Ruth hit. Then h  41 is the number of home runs that Hank Aaron hit. So 1469  h  h  41  1428  2h  h  714. Thus Babe Ruth hit 714 home runs. 35. Let n be the number of nickels. Then there are also n dimes and n quarters. The total value of the coins in the purse is the sum of the values of nickels, dimes, and quarters, so 280  005n  010n  025n  28  04 p  p  28 04  7. So the

purse contains 7 nickels, 7 dimes, and 7 quarters. 36. Let q be the number of quarters. Then 2q is the number of dimes, and 2q  5 is the number of nickels. Thus 300  value of nickels value of dimes value of quarters, so 300  005 2q  5  010 2q  025q  300  010q  025  020q  025q  275  055q  q  275 055  5. Thus you have 5 quarters, 2 5  10 dimes, and 2 5  5  15 nickels.

37. Let l be the length of the garden. Since area  width  length, we obtain the equation 1125  25l  l  1125 25  45 ft. So

the garden is 45 feet long. 38. Let  be the width of the pasture. Then the length of the pasture is 3. Since area  length  width we have 132,300   3  32  2  1323,300  44,100    210. Thus the width of the pasture is 210 feet.

39. Let  be the width of the building lot. Then the length of the building lot is 5. Since a half­acre is 12  43,560  21,780 and area is length times width, we have 21,780   5  52  2  4,356    66. Thus the width of the building lot is 66 feet and the length of the building lot is 5 66  330 feet.

40. Let x be the length of a side of the square plot. As shown in the figure,

x

area of the plot  area of the building  area of the parking lot. Thus,

x 2  60 40  12,000  2,400  12,000  14,400  x  120. So the plot of

x

land measures 120 feet by 120 feet.

60 40

41. The shaded area is the sum of the area of a square and the area of a triangle. So A  y 2  12 y y  32 y 2 . We are given    that the area is 120 in2 , so 120  32 y 2  y 2  80  y   80  4 5. y is positive, so y  4 5  894 in. x

42. First we write a formula for the area of the figure in terms of x. Region A has dimensions 10 cm and x cm and region B has dimensions 6 cm and x cm. So the shaded region has area 10  x  6  x  16x cm2 . We are given that this is equal to 144 cm2 , so 144  16x  x  144 16  9 cm.

10 cm

A

6 cm B

x

43. Let x be the width of the strip. Then the length of the mat is 20  2x, and the width of the mat is 15  2x. Now the perimeter is twice the length plus twice the width, so 102  2 20  2x  2 15  2x  102  40  4x  30  4x  102  70  8x  32  8x  x  4. Thus the strip of mat is 4 inches wide. 44. Let x be the width of the strip. Then the width of the poster is 100  2x and its length is 140  2x. The perimeter of the printed area is 2 100  2 140  480, and the perimeter of the poster is 2 100  2x  2 140  2x. Now we use the

fact that the perimeter of the poster is 1 12 times the perimeter of the printed area: 2 100  2x  2 140  2x  32  480 

480  8x  720  8x  240  x  30. The blank strip is thus 30 cm wide.

45. Let x be the length of the person’s shadow, in meters. Using similar triangles,  x  5. Thus the person’s shadow is 5 meters long.

x 10  x   20  2x  6x  4x  20 6 2


SECTION P.9 Modeling with Equations

39

46. Let x be the height of the tall tree. Here we use the property that corresponding sides in similar triangles are proportional. The base of the similar triangles starts at x 5 150   15 25 25 x  5  15 150  25x  125  2250  25x  2375  x  95. Thus the

x-5

eye level of the woodcutter, 5 feet. Thus we obtain the proportion

5

tree is 95 feet tall.

25

15

125

47. Let x be the amount (in mL) of 60% acid solution to be used. Then 300  x mL of 30% solution would have to be used to yield a total of 300 mL of solution. 60% acid

30% acid

Mixture

mL

x

300

Rate (% acid)

060

300  x

Value

060x

030

050

030 300  x

050 300

60  200. 03 So 200 mL of 60% acid solution must be mixed with 100 mL of 30% solution to get 300 mL of 50% acid solution.

Thus the total amount of pure acid used is 060x  030 300  x  050 300  03x  90  150  x 

48. The amount of pure acid in the original solution is 300 50%  150. Let x be the number of mL of pure acid added. Then 150  x  60%  06  the final volume of solution is 300  x. Because its concentration is to be 60%, we must have 300  x 30 150  x  06 300  x  150  x  180  06x  04x  30  x   75. Thus, 75 mL of pure acid must be 04 added.

49. Let x be the number of grams of silver added. The weight of the rings is 5  18 g  90 g. 5 rings

Pure silver

Mixture

Grams

90

x

Rate (% gold)

090

0

90  x

090 90

0x

Value

075

075 90  x

So 090 90  0x  075 90  x  81  675  075x  075x  135  x  135 075  18. Thus 18 grams of silver

must be added to get the required mixture.

50. Let x be the number of liters of water to be boiled off. The result will contain 6  x liters. Original

Water

Final

Liters

6

Concentration

120

x

6x

Amount

120 6

0

200 6  x

0

200

So 120 6  0  200 6  x  720  1200  200x  200x  480  x  24. Thus 24 liters need to be boiled off.


40

CHAPTER P Prerequisites

51. Let x be the number of liters of coolant removed and replaced by water. 60% antifreeze

60% antifreeze (removed)

Water

Mixture

Liters

36

x

x

36

Rate (% antifreeze)

060

060

0

050

060 36

060x

0x

050 36

Value

036  06. Thus 06 liters 06

So 060 36  060x  0x  050 36  216  06x  18  06x  036  x  must be removed and replaced by water.

52. Let x be the number of gallons of 2% bleach removed from the tank. This is also the number of gallons of pure bleach added to make the 5% mixture. Original 2%

Pure bleach

5% mixture

100  x

x

100

002

1

005

002 100  x

1x

005  100

Gallons Concentration Bleach

So 002 100  x  x  005  100  2  002x  x  5  098x  3  x  306. Thus 306 gallons need to removed and replaced with pure bleach. 53. Let c be the concentration of fruit juice in the cheaper brand. The new mixture consists of 650 mL of the original fruit punch and 100 mL of the cheaper fruit punch. Original Fruit Punch

Cheaper Fruit Punch

mL

650

100

750

Concentration

050

c

048

050  650

100c

048  750

Juice

Mixture

So 050  650  100c  048  750  325  100c  360  100c  35  c  035. Thus the cheaper brand is only 35% fruit juice. 54. Let x be the number of ounces of $300oz tea Then 80  x is the number of ounces of $275oz tea. $300 tea

$275 tea

Ounces

x

Rate (cost per ounce)

300

80  x

Value

300x

Mixture 80

275

290

275 80  x

290 80

So 300x  275 80  x  290 80  300x  220  275x  232  025x  12  x  48. The mixture uses 48 ounces of $300oz tea and 80  48  32 ounces of $275oz tea.

1 of the car per 55. Let t be the time in minutes it would take to wash the car if the friends worked together. Friend 1 washes 25 1 of the car per minute. The sum of these fractions is equal to the fraction of the job they minute, while Friend 2 washes 35

1 1 875t 875t 875t 1       875  35t  25t  60t  t  875 60  t 25 35 t 25 35 t  14 35 60 minutes, or 14 minutes 35 seconds can do working together, so we have

56. Let t be the time, in minutes, it takes the landscaper to mow the lawn. Since the assistant is half as fast, it would take them 1 1 10t 10t 10t 1       t  10  5  15. Thus, it would the 2t minutes to mow the lawn alone. Thus, 10 t 2t 10 t 2t assistant 2 15  30 minutes to mow the lawn alone.


SECTION P.9 Modeling with Equations

41

57. Let t be the number of hours it would take your friend to paint a house alone. Then working together, it takes 23 t hours. Because it takes you 7 hours, we have

3 14t 1 1 1 14t 14t 14t   2    2  2  2t  14  21  t  72 . Thus, it 7 t 7 t t t t 3

3

would take your friend 35 h to paint a house alone. 58. Let h be the time, in hours, to fill the swimming pool using the smaller hose alone. Since the larger hose takes 20% less 1 1  1  16  08  16  08h  128  16  08h  time, it takes 08h to fill the pool alone. Thus 16   16  h 08h h  288 08  36. Thus, the smaller hose takes 36 hours to fill the pool alone, and the larger hose takes 08 36  288 hours. 59. Let t be the time in hours that the commuter spent on the train. Then 11 2  t is the time in hours that commuter spent on the bus. We construct a table: Rate By train By bus

Time

40

t

60

11  t 2

Distance 

40t

60 11 2 t

  The total distance traveled is the sum of the distances traveled by bus and by train, so 300  40t  60 11 2 t 

300  40t  330  60t  30  20t  t  30 20  15 hours. So the time spent on the train is 55  15  4 hours.

60. Let r be the speed of the slower cyclist, in mi/h. Then the speed of the faster cyclist is 2r. Rate

Time

Distance

Slower cyclist

r

2

2r

Faster cyclist

2r

2

4r

When they meet, they will have traveled a combined total of 90 miles, so 2r  4r  90  6r  90  r  15. The speed of the slower cyclist is 15 mi/h, while the speed of the faster cyclist is 2 15  30 mi/h. 61. Let r be the speed of the plane from Montreal to Los Angeles. Then r  020r  120r is the speed of the plane from Los Angeles to Montreal. Rate Montreal to L.A.

r

L.A. to Montreal

12r

Time 2500 r 2500 12r

Distance 2500 2500

2500 55 2500 2500 2500      r 12r 6 r 12r 33,000 55  12r  2500  6  12  2500  6  66r  18,000  15,000  66r  33,000  r  66  500. Thus the plane flew The total time is the sum of the times each way, so 9 16 

at a speed of 500 mi/h on the trip from Montreal to Los Angeles.

ft 22 62. Let x be the speed of the car in mi/h. Since a mile contains 5280 ft and an hour contains 3600 s, 1 mi/h  5280 3600 s  15 ft/s. 220 220 The truck is traveling at 50  22 15  3 ft/s. So in 6 seconds, the truck travels 6  3  440 feet. Thus the back end

of the car must travel the length of the car, the length of the truck, and the 440 feet in 6 seconds, so its speed must be 1430440  242 ft/s. Converting to mi/h, we have that the speed of the car is 242  15  55 mi/h. 6 3 3 22

63. Let x be the distance from the fulcrum to where the 125­pound friend sits. Then substituting the known values into the formula given, we have 100 8  125x  800  125x  x  64. So the 125­pound friend should sit 64 feet from the fulcrum.


42

CHAPTER P Prerequisites

64. Let  be the largest weight that can be hung. In this exercise, the edge of the building acts as the fulcrum, so the 240 lb man is sitting 25 feet from the fulcrum. Then substituting the known values into the formula given in Exercise 63, we have 240 25  5  6000  5    1200. Therefore, 1200 pounds is the largest weight that can be hung. l

65. Let l be the length of the lot in feet. Then the length of the diagonal is l  10. We apply the Pythagorean Theorem with the hypotenuse as the diagonal. So

l 2  502  l  102  l 2  2500  l 2  20l  100  20l  2400  l  120.

50

l+10

Thus the length of the lot is 120 feet.

66. Let r be the radius of the running track. The running track consists of two semicircles and two straight sections 110 yards long, so we get the equation 2r  220  440  2r  220  r  110   3503. Thus the radius of the semicircle is about 35 yards. 67. Let h be the height in feet of the structure. The structure is composed of a right cylinder with radius 10 and height 23 h and a cone with base radius 10 and height 13 h. Using the formulas for the volume of a cylinder and that of a cone, we obtain the      100 equation 1400   102 23 h  13  102 13 h  1400  200 3 h  9 h  126  6h  h (multiply both sides

by

9 )  126  7h  h  18. Thus the height of the structure is 18 feet. 100

68. Let h be the height of the break, in feet. Then the portion of the bamboo above the break is 10  h. Applying the Pythagorean Theorem, we obtain

h 2  32  10  h2  h 2  9  100  20h  h 2  91  20h  h  91 20  455. Thus the break is 455 ft above the ground.

10-h

h

3

69. Answers will vary.

CHAPTER P REVIEW 1. (a) Since there are initially 250 tablets and the patient takes 2 tablets per day, the number of tablets T that are left in the bottle after x days is T  250  2x. (b) After 30 days, there are 250  2 30  190 tablets left.

(c) We set T  0 and solve: T  250  2x  0  250  2x  x  125. The patient will run out after 125 days.

2. (a) The total cost is $11 per calzone plus the $7 delivery charge, so C  11x  7. (b) Four calzones would cost 11 4  7  $51.

(c) We solve C  11x  7  40  11x  33  x  3. You can order three calzones.

3. (a) 16 is rational. It is an integer, and more precisely, a natural number.

(b) 16 is rational. It is an integer, but because it is negative, it is not a natural number.  16  4 is rational. It is an integer, and more precisely, a natural number.  (d) 2 is irrational. (c)

(e) 83 is rational, but is neither a natural number nor an integer. (f)  82  4 is rational. It is an integer, but because it is negative, it is not a natural number.

4. (a) 5 is rational. It is an integer, but not a natural number.

(b)  25 6 is rational, but is neither an integer nor a natural number.  (c) 25  5 is rational, a natural number, and an integer. (d) 3 is irrational.


CHAPTER P 3 (e) 24 16  2 is rational, but is neither a natural number nor an integer.

(f) 1020 is rational, a natural number, and an integer. 5. Commutative Property for addition.

6. Commutative Property for multiplication.

7. Distributive Property.

8. Distributive Property.

5 4 9 3 5 2      6 3 6 6 6 2 5 4 1 5 2 (b)     6 3 6 6 6 15  12 33 9 15 12     11. (a) 8 5 85 21 2 15  5 55 25 15 12     (b) 8 5 8  12 84 32

11 21 22 1 7     10 15 30 30 30 11 21 22 43 7     (b) 10 15 30 30 30 30  35 55 25 30 12     12. (a) 7 35 7  12 12 2 30  12 6  12 72 30 12     (b) 7 35 7  35 77 49

9. (a)

10. (a)

13. x  [2 6  2  x  6

14. x  0 10]  0  x  10

_2

6

15. x   4]  x  4

0

16. x  [2   2  x 4

17. x  5  x  [5 

_2

18. x  3  x   3

5

_3

  20. 0  x  12  x  0 12

19. 1  x  5  x  1 5] _1

5

0

  21. (a) A  B  1 0 12  1 2 3 4

22. (a) C  D  1 2]

(b) A  B  1

(b) C  D  0 1]

23. (a) A  C  1 2   (b) B  D  12  1

24. (a) A  D  0 1   (b) B  C  12  1

25. 1  4  1  4  3  3    27. 212 812  2  8  16  4 29. 21613  31.

1 1  16   3 13 216 216

  242    242 2  121  11 2

33. (a) 5  3  2  2

(b) 5  3  8  8

10

1 2

26. 5  10  4  5  10  4  5  6  1

9  8  1 28. 23  32  18  19  72 72 72  23 30. 6423  43  42  16

32.

   2 50  100  10

34. (a) 4  0  4  4 (b) 4  4  8  8

Review

43


44

CHAPTER P Prerequisites

 35. (a) 3 7  713  5 (b) 74  745  6

x 5  x 56  9  12 9 (b) x  x  x 92

37. (a)

 3 7 5  573 3   3  (b) 4 5  514  534  12  38. (a) y 3  y 3  y 32 2   2  (b) 8 y  y 18  y 14 36. (a)

 3  2  4 a3 b b3  a 6  a 6 b2  b12  a 66 b212  b14 a2 3    2 x 1 y 2  27x 3 y 6  4 x 2 y 2  27  4 x 32 y 62  12x y 8 (b) 3x y 2 3 9 9

39. (a)

x 4 3x2 x 4  9x 2   9x 423  9x 3 3 x x3  6 r 2 s 43 r 12 s 8 (b)   r 122 s 86  r 10 s 2 r 13 s r 2s6    2 3 x 3 y y4  3 x 6 y4 y2  3 x 6 y6  x 2 y2 41. (a)    (b) 8 z 10  8  z 10  4 z5 40. (a)

8r 12 s 3 4r 52  4r 122 s 34  4r 52 s 7  2 4 s7 2r s 2  4 6 a 2 b4 c6 ab2 c3 2 a 26 b48 c6  4a 4 b12 c6  4a c (b)   2 2a 3 b4 22 a 6 b8 b12

42. (a)

43. 78,250,000,000  7825  1010

44. 208  108  0000 000 020 8      293  106 1582  1014 000000293 1582  1014 293  1582 ab   1061412   45. 12 12 c 28064 28064  10 28064  10  165  1032 46. 80

times 60 minutes 24 hours 365 days     90 years  38  109 times minute hour day year

47. x 2  5x  14  x  7 x  2   48. 12x 2  10x  8  2 6x 2  5x  4  2 3x  4 2x  1 49. 2x 2 y  6x y 2  2x y x  3y

  50. 4y 2 z 3  10y 3 z  12y 5 z 2  2y 2 z 2z 2  5y  6y 3 z

51. 3x 2  2x  1  3x  1 x  1  2      52. x 4  x 2  2  x 2  x 2  2  x 2  2 x 2  1  x 2  2 x  1 x  1

53. 4t 2  13t  12  4t  3 t  4  2 54. x 4  2x 2  1  x 2  1  [x  1 x  1]2  x  12 x  12   55. 16  4t 2  4 4  t 2  4 2  t 2  t  4 t  2 t  2        56. 2y 6  32y 2  2y 2 y 4  16  2y 2 y 2  4 y 2  4  2y 2 y 2  4 y  2 y  2        57. x 6  1  x 3  1 x 3  1  x  1 x 2  x  1 x  1 x 2  x  1


CHAPTER P

Review

    58. 16a 4 b2  2ab5  2ab2 8a 3  b3  2ab2 2a  b 4a 2  2ab  b2   59. x 3  27  x  3 x 2  3x  9     60. 3y 3  81x 3  3 y 3  27x 3  3 y  3x y 2  3x y  9x 2   61. 3x 12  2x 12  5x 32  x 12 5x 3  2x 2  3x  x 12 5x  3 x  1   62. 7x 32  8x 12  x 12  x 32 x 2  8x 2  7  x 32 x  1 x  7   63. 5x 3  15x 2  x  3  5x 2 x  3  x  3  5x 2  1 x  3   64. 3  32  4  12  2   3  4   3  2  4   3    3   2   2 65. a  b2  3 a  b  10  a  b  5 a  b  2

66. 3x  22  3x  2  6  [3x  2  3] [3x  2  2]  3x  1 3x  4

67. 2y  7 2y  7  4y 2  49

  68. 1  x 2  x  3  x 3  x  2  x  x 2  9  x 2  2  x  x 2  9  x 2  7  x   69. x 2 x  2  x x  22  x 3  2x 2  x x 2  4x  4  x 3  2x 2  x 3  4x 2  4x  2x 3  6x 2  4x   x x 2  2x  3 x 3  2x 2  3x 70.   x 2  2x  3 x x             71. x x  1 2 x  1  x  x 2 x  1  2x x  x  2x  x  2x 32  x  x 12

72. 2x  13  2x3  3 2x2 1  3 2x 12  13  8x 3  12x 2  6x  1 x 2  2x  3 x 3 x  3 x  1   2 2x  3 x  1 2x  3 2x  5x  3   2 t  1 t  t  1 t2  t  1 t3  1   74. 2 t 1 t  1 t  1 t 1 73.

75. 76.

x 2  4x  5 x 2  12x  36 x 6 x 1 x 6 x  5 x  1 x  62    2    2 x 5 x 1 x 5 x  5 x  5 x  6 x  1 x  25 x  7x  6

x 2  2x  15 x 2  x  12 x 1 x  5 x  3 x  1 x  1     x 4 x  5 x  1 x  4 x  3 x 2  6x  5 x2  1

77. x  78.

x x  1 1 x2  x  1 1    x 1 x 1 x 1 x 1

1 x x2  1 x x  1 x2  1  x2  x x 1            2  2 2 2 x 1 x 1 x  1 x  1 x  1 x  1 x  1 x  1 x  1 x 2  1

1 3 2 x  22 x x  2 3x 2      2 x x  2 x  2 x x  22 x x  22 x x  22   2 x 2  4x  4  x 2  2x  3x 2x 2  8x  8  x 2  2x  3x 3x 2  7x  8    x x  22 x x  22 x x  22 1 1 1 2 1 2   80.    x  2 x2  4 x2  x  2 x  2 x  2 x  2 x  2 x  1 x 1 2 x  2 x  2 x  1    x  2 x  1 x  2 x  2 x  1 x  2 x  2 x  1 x  2 79.

x 2  2x  5 x 2  x  2  x  1  2x  4  x  2 x  1 x  2 x  2 x  1 x  2

45


46

CHAPTER P Prerequisites

x 2 1 1   2  2x 2x  2  x  1  1 x  2  1  1 81. x x 2 x 2 2x x 2 2x x 2 2x 1 1 1 1   1 x 1  x x  1  x x  1  x  1  x  82. x 1 1 1 1 x x  1 2x  1 x  1  x   x x 1 x x 1     x 44 x x 2 x 2 x 2    83.  x 4 x 2 x 2 x 2       h 1 x h x x h x x h x x  h  x   84.            h h x h x h x h x h x h x x h x   1 11 11 1 85.       11 11 11 11    3 6 6 3 3 6 86.        6 2 6 6 6    10 10  10 2 10 21    87.   10  10 2 2  1 21 21 21     14 3 2 42  14 2 14 42  14 2 88. 62 2        2  7 3 2 3 2 3 2 32  2     x x 2 x 2 x 2x  x x x  89.        2 x 4 2 x 2 x 2 x 22  x     x 4 x 4 x 2 x 2 x 2    90.  x 4 x 2 x 2 x 2

x 5 is defined whenever x  10  0  x  10, so its domain is x  x  10. x  10 2x 92. 2 is defined whenever x 2  9  0  x 2  9  x  3, so its domain is x  x  3 and x  3. x 9   x is defined whenever x  0 (so that x is defined) and x 2  3x  4  x  1 x  4  0  x  1 and 93. 2 x  3x  4 x  4. Thus, its domain is x  x  0 and x  4.  x 3 is defined whenever x  3  0  x  3 and x 2  4x  4  x  22  0  x  2. Thus, its domain is 94. 2 x  4x  4 x  x  3. 91.

95. This statement is false. For example, take x  1 and y  1. Then LHS  x  y3  1  13  23  8, while RHS  x 3  y 3  13  13  1  1  2, and 8  2.    1a 1 a 1 a 1 a 1     .   96. This statement is true for a  1:    1a 1a 1 a 1 a 1  a 1  a

97. This statement is true:

12 y 12 12  y     1. y y y y

   98. This statement is false. For example, take a  1 and b  1. Then LHS  3 a  b  3 1  1  3 2, while      RHS  3 a  3 b  3 1  3 1  1  1  2, and 3 2  2.    99. This statement is false. For example, take a  1. Then LHS  a 2  12  1  1, which does not equal  a  1. The true statement is a 2  a.


CHAPTER P

100. This statement is false. For example, take x  1. Then LHS  and

1 5  . 5 4

Review

47

1 1 1 1 1 1 5 1   , while RHS      , x 4 14 5 x 4 1 4 4

101. 3x  12  24  3x  12  x  4

102. 5x  7  42  5x  49  x  49 5

103. 7x  6  4x  9  3x  15  x  5

104. 8  2x  14  x  3x  6  x  2

105. 13 x  12  2  2x  3  12  2x  15  x  15 2 6 3 106. 23 x  35  15  2x  10x  9  3  30x  40x  6  x   40 20

107. 2 x  3  4 x  5  8  5x  2x  6  4x  20  8  5x  2x  26  8  5x  3x  18  x  6 5 x  5 2x  5    3 x  5  2 2x  5  5  3x  15  4x  10  5  x  30  x  30 108. 2 3 6 x 1 2x  1 109.   x  1 2x  1  2x  1 x  1  2x 2  3x  1  2x 2  3x  1  6x  0  x  0 x 1 2x  1 x 1 7 110. 3   x  3 x  2  1  x  3x  6  1  2x  7  x   x 2 x 2 2 3x 3x x x 1 2     x  1 x  2  x x  1  x  x  2  x 2  x  2  0. Since this 111. x 1 3x  6 3 x  2 x 2 last equation is never true, there is no real solution to the original equation. 112. x  22  x  42  x  22  x  42  0  [x  2  x  4] [x  2  x  4]  0  [x  2  x  4] [x  2  x  4]  6 2x  2  0  2x  2  0  x  1. 113. x 2  144  x  12

7 114. 4x 2  49  x 2  49 4  x  2

115. x 3  27  0  x 3  27  x  3.

116. 6x 4  15  0  6x 4  15  x 4   52 . Since x 4 must be nonnegative, there is no real solution.

117. x  13  64  x  1  4  x  1  4  5.   118. x  22  2  0  x  22  2  x  2   2  x  2  2.  119. 3 x  3  x  33  27. 2  120. x 23  4  0  x 13  4  x 13  2  x  8.

121. 4x 34  500  0  4x 34  500  x 34  125  x  12543  54  625. 122. x  215  2  x  2  25  32  x  2  32  34. xy  2A  x  y  x  2A  y. 123. A  2 V  xz . 124. V  x y  yz  xz  V  y x  z  xz  V  xz  y x  z  y  x z 1 1 1 1 1 11 11 125. Multiply through by t: J     tJ  1   t  , J  0. t 2t 3t 2 3 6 6J  q q q q q q 126. F  k 1 2 2  r 2  k 1 2  r   k 1 2 . (In real­world applications, r represents distance, so we would take the F F r positive root.) 127. Let x be the number of pounds of raisins. Then the number of pounds of nuts is 50  x. Raisins

Nuts

Mixture

Pounds

x

50

Rate (cost per pound)

320

50  x 240

272

So 320x  240 50  x  272 50  320x  120  240x  136  08x  16  x  20. Thus the mixture uses 20 pounds of raisins and 50  20  30 pounds of nuts.


48

CHAPTER P Prerequisites

128. Let t be the number of hours that the district supervisor drives. Then the store manager drives for t  14 hours. Supervisor Manager

Rate

Time

Distance

45

t

40

t  14

45t   40 t  14

  When they pass each other, they will have traveled a total of 160 miles. So 45t  40 t  14  160  45t  40t  10  160  85t  170  t  2. Since the supervisor leaves at 2:00 P. M . and travels for 2 hours, they pass each other at 4:00 P. M .

129. Let x be the amount invested in the account earning 15% interest. Then the amount invested in the account earning 25% is 7000  x. 15% Account

25% Account

Total

Amount invested

x

7000

Interest earned

0015x

7000  x

0025 7000  x

12025

From the table, we see that 0015x  0025 7000  x  12025  0015x  175  0025x  12025  5475  001x  x  5475. Thus, $5475 is invested in the account earning 15% interest and $1525 is invested in the account earning 25% interest. 130. The amount of interest currently earned is 6000 003  $180 per year. If a total of $300 is desired, another $120 in interest must be earned at a rate of 125% per year. If the additional amount invested is x, we have 00125x  120  x  9600. Thus, an additional $9600 must be invested at 125% simple interest to earn a total of $300 interest per year. 131. Let t be the time it would take the interior decorator to paint a living room if they work alone. It would take the assistant 1 2t hours alone, and it would take the apprentice 3t hours alone. Thus, the decorator does of the job per hour, the assistant t 1 1 1 1 1 of the job per hour, and the apprentice does of the job per hour. So    1  6  3  2  6t  does 2t 3t t 2t 3t 6t  11  t  11 6 . Thus, it would take the decorator 1 hour 50 minutes to paint the living room alone.

132. Let  be width of the pool. Then the length of the pool is 2, and its volume is 8  2  8464  162  8464  2  529    23. Since   0, we reject the negative value. The pool is 23 feet wide, 2 23  46 feet long, and 8 feet deep.

CHAPTER P TEST 1. (a) The cost is C  9  15x.

(b) There are four extra toppings, so x  4 and C  9  15 4  $15.

(c) We have C  195  9  15x  15x  105  x  7. Thus, a $1950 pizza has 7 toppings.

2. (a) 5 is rational. It is an integer, and more precisely, a natural number.  (b) 5 is irrational. (c)  93  3 is rational, and it is an integer.

(d) 1,000,000 is rational, and it is an integer.

3. (a) A  B  0 1 5   (b) A  B  2 0 12  1 3 5 7 4. (a)

_4

2

[4 2

0

3

[0 3]


CHAPTER P

(b) 0

(c) 2  4  6  6 5. (a)

2

_4

[4 2  [0 3]  [0 2

_5

3

3

[4 2  [0 3]  [4 3]

2

5 3] (b) x  3  x   3]; 1  x  4  x  [1 4

2 

6. (a) 34  81

1 1 (c) 34  4  81 3  2 32 9 2  2  (e) 3 4 2  8 32 9 4 3   (g) 16 216 24 7. (a) 186,000,000,000  186  1011

(b) 34  81

375 (d) 72  37572  33  27 3  5 2 32 1 (f)    4 2 16  34 (h) 1634  24  23  18

(b) 0000 000 396 5  3965  107

a 3 b2 a2  b ab3      (b) 200  32  10 2  4 2  6 2  2  (c) 2x 12 y 2 3x 14 y 1  2  32 x 12214 y 221  18x  2  (d) 3a 3 b3 4ab2  3a 3 b3  42 a 2 b4  48a 5 b7   2 (e) 9x 2 y 4  3x y 2  3 x y 2 12  12  12  y2 4x 9 y 3 4x 8 y4 (f)    4 7 4 8 xy y 4x 2x

8. (a)

9. (a) z 4z  3  2z 3  2z  4z 2  3z  6z  4z 2  3z

(b) x  3 4x  5  4x 2  5x  12x  15  4x 2  7x  15        2  2 a b a b  a  b ab (c) (d) 2x  32  2x2  2 2x 3  32  4x 2  12x  9

(e) x  23  x3  3 x2 2  3 x 22  23  x 3  6x 2  12x  8   (f) x 2 x  3 x  3  x 2 x 2  9  x 4  9x 2

10. (a) 4x 2  25  2x  5 2x  5

(b) 2x 2  5x  12  2x  3 x  4

  (c) x 3  3x 2  4x  12  x 2 x  3  4 x  3  x  3 x 2  4  x  3 x  2 x  2     (d) x 4  27x  x x 3  27  x x  3 x 2  3x  9   (e) 2x 32  8x 12  10x 12  2x 12 x 2  4x  5  2x 12 x  5 x  1

Test

49


50

CHAPTER P Prerequisites

  (f) x 4 y 2  9x 2 y 2  x 2 y 2 x 2  9  x 2 y 2 x  3 x  3 11. (a) (b)

2  4  3 3   3   1   3   3   1 2  2  3

x 1 2x 2  x  1 x  3 2x  1 x  1 x  3    2x  1 x 3 x  3 x  3 2x  1 x2  9

x2 x2 x 1 x 1  x  1 x  2     x 2 x  2 x  2 x  2 x  2 x  2 x  2 x  2   2 2 x  x x 2 x 2 1    x 2 x  2 x  2 x  2 x  2 y x x y   y2  x 2  x  y y  x y  x y  x x y x y xy (d)      x  y   1 1 xy 1 1 xy xy xy   y x y x   3  2 6 6 6 632 12. (a)      332    3 3 2 3 2 3 2 4 2 2 2       10 52     10 10 52    (b)   50  2 10  5 2  2 10 5  4 52 52 52   1 1 x 1 x 1 (c)       1x 1 x 1 x 1 x (c)

x2

x2  4

13. (a) 4x  3  2x  7  4x  2x  7  3  2x  10  x  5.   3 (b) 8x 3  125  8x 3  3 125  2x  5  x   52 . 32   6432  x  83  512. (c) x 23  64  0  x 23  64  x 23

x 3 x   x 2x  1  x  3 2x  5  2x 2  x  2x 2  5x  6x  15  x  x  15  2x  15 2x  5 2x  1  x  15 2 .   (e) 3 x  12  18  0  3 x  12  18  x  12  6  x  1   6  x   6  1  E E 14. E  mc2   c2  c  . (We take the positive root because c represents the speed of light, which is positive.) m m 15. Let t be the time (in hours) it took the trucker to drive from Amity to Belleville. Then 44  t is the time it took the trucker to drive from Belleville to Amity. Since the distance from Amity to Belleville equals the distance from Belleville to Amity, we have 50t  60 44  t  50t  264  60t  110t  264  t  24 hours. Thus the distance is 50 24  120 mi. (d)


Making Optimal Decisions

51

FOCUS ON MODELING Making Optimal Decisions 

1. (a) The total cost is 

cost of printer



maintenance cost

 

number of months



printing cost

 

number of months

. Each

month the printing cost is 8000  003  240. Thus we get C1  5800  25n  240n  5800  265n.       rental number printing number    . Each month the printing cost (b) In this case the cost is  cost of months cost of months

(c)

is 8000  006  480. Thus we get C2  95n  480n  575n. Years

n

Purchase

Rental

1

12

8,980

6,900

2

24

12,160

13,800

3

36

15,340

20,700

4

48

18,520

27,600

5

60

21,700

34,500

6

72

24,880

41,400

(d) The cost is the same when C1  C2 are equal. So 5800  265n  575n  5800  310n  n  1871 months.

daily

daily

2. (a) The cost of Plan 1 is 3  

cost

The cost of Plan 2 is 3  

cost



cost per mile

 

number of miles

  3  65  015x  195  015x.

  3  90  270.

(b) When x  400, Plan 1 costs 195  015 400  $255 and Plan 2 costs $270, so Plan 1 is cheaper. When x  800, Plan 1 costs 195  015 800  $315 and Plan 2 costs $270, so Plan 2 is cheaper.

(c) The cost is the same when 195  015x  270  015  75x  x  500. So both plans cost $270 when you drive 500 miles.

setup

cost per

price per



number

3. (a) The total cost is  (b) The revenue is 

cost

tire

 

tire

of tires

 

number of tires

. So C  8000  22x.

. So R  49x.

(c) Profit  Revenue  Cost. So P  R  C  49x  8000  22x  27x  8000.

(d) Break even is when profit is zero. Thus 27x  8000  0  27x  8000  x  2963. So they need to sell at least 297 tires to break even.


52

FOCUS ON MODELING

4. (a) Option 1: In this option the width is constant at 100. Let x be the increase in length. Then the additional area is   increase   100x. The cost is the sum of the costs of moving the old fence, and of installing the width   in length

new one. The cost of moving is $6  100  $600 and the cost of installation is 2  10  x  20x, so the total cost is C  600 . Substituting in the area C  20x  600. Solving for x, we get C  20x  600  20x  C  600  x  20   C  600 we have A1  100  5 C  600  5C  3,000. 20 Option 2: In this option the length is constant at 180. Let y be the increase in the width. Then the additional area is   increase   180y. The cost of moving the old fence is 6  180  $1080 and the cost of installing the new length   in width

(b)

one is 2  10  y  20x, so the total cost is C  20y  1080. Solving for y, we get C  20y  1080  20y  C  1080   C  1080 C  1080 . Substituting in the area we have A2  180  9 C  1080  9C  9,720. y 20 20 Cost C

Area gain A1 from Option 1

Area gain A2 from Option 2

$1100

2,500 ft2

180 ft2

$1200

3,000 ft2

1,080 ft2

$1500

4,500 ft2

3,780 ft2

$2000

7,000 ft2

8,280 ft2

$2500

9,500 ft2

12,780 ft2

$3000

12,000 ft2

17,280 ft2

(c) If the farmer has only $1200, Option 1 gives the greatest gain. If the farmer has only $2000, Option 2 gives the greatest gain. 5. (a) Design 1 is a square and the perimeter of a square is four times the length of a side. 24  4x, so each side is x  6 feet long. Thus the area is 62  36 ft2 .

12 . Thus, the area is Design 2 is a circle with perimeter 2r and area r 2 . Thus we must solve 2r  24  r    2 12 144  458 ft2 . Design 2 gives the largest area.    

(b) In Design 1, the cost is $3 times the perimeter p, so 120  3 p and the perimeter is 40 feet. By part (a), each side is then 40  10 feet long. So the area is 102  100 ft2 . 4

In Design 2, the cost is $4 times the perimeter p. Because the perimeter is 2r, we get 120  4 2r so  2 120 15 225 15 r  . The area is r 2    716 ft2 . Design 1 gives the largest area.  8   

6. (a) Plan 1: Tomatoes every year. Profit  acres  Revenue  cost  100 1600  300  130,000. Then for n years the profit is P1  130,000n. (b) Plan 2: Soybeans followed by tomatoes. The profit for two years is Profit  acres      soybean tomato     100 1200  1600  280,000. Remember that no fertilizer is revenue revenue needed in this plan. Then for 2k years, the profit is P2  280,000k.

(c) When n  10, P1  130,000 10  1,300,000. Since 2k  10 when k  5, P2  280,000 5  1,400,000. So Plan B is more profitable.


Making Optimal Decisions

53

7. (a) Data (GB) 1 15 2 25 3 35 4

Plan A

Plan B $25

25  5 200  $35

Plan C $40

40  5 150  $4750

$60 60  5 100  $65

25  10 200  $45

40  10 150  $55

60  10 100  $70

25  20 200  $65

40  20 150  $70

60  20 100  $80

25  15 200  $55 25  25 200  $75 25  30 200  $85

40  15 150  $6250 40  25 150  $7750

40  30 150  $85

60  15 100  $75 60  25 100  $85 60  30 100  $90

(b) For Plan A: CA  25  2 10x  10  20x  5. For Plan B: CB  40  15 10x  10  15x  25. For Plan C: CC  60  1 10x  10  10x  50. Note that these equations are valid only for x  1.

(c) If 22 GB are used, Plan A costs 25  12 2  $49, Plan B costs 40  12 15  $58, and Plan C costs 60  12 1  $72. If 37 GB are used, Plan A costs 25  27 2  $79, Plan B costs 40  27 15  $8050, and Plan C costs 60  27 1  $87. If 49 GB are used, Plan A costs 25  39 2  $103, Plan B costs 40  39 15  $9850, and Plan C costs 60  39 1  $99.

(d) (i) We set CA  CB  20x  5  15x  25  5x  20  x  4. Plans A and B cost the same when 4 GB are used.

(ii) We set CA  CC  20x  5  10x  50  10x  45  x  45. Plans A and C cost the same when 45 GB are used.

(iii) We set CB  CC  15x  25  10x  50  5x  25  x  5. Plans B and C cost the same when 5 GB are used. 8. (a) In this plan, Company A gets $32 million and Company B gets $32 million. Company A’s investment is $14 million, so they make a profit of 32  14  $18 million. Company B’s investment is $26 million, so they make a profit of 32  26  $06 million. So Company A makes three times the profit that Company B does, which is not fair.

(b) The original investment is 14  26  $4 million. So after giving the original investment back, they then share the profit of $24 million. So each gets an additional $12 million. So Company A gets a total of 14  12  $26 million and Company B gets 26  12  $38 million. So even though Company B invests more, they make the same profit as Company A, which is not fair.

(c) The original investment is $4 million, so Company A gets 14 4  64  $224 million and Company B gets 26  64  $416 million. This seems the fairest. 4


Corrections: p. 2 NOTE THAT THESE CORRECTIONS SHOULD BE MADE IN CA-8 ALSO.

CHAPTER 1

EQUATIONS AND GRAPHS

1.1 1.2

The Coordinate Plane 1 Graphs of Equations in Two Variables 8

1.3 1.4 1.5

Circles 19 Lines 24 Solving Quadratic Equations 34

1.6

Complex Numbers 43

1.7

Solving Other Types of Equations 46

1.8

Solving Inequalities 54

1.9

Solving Absolute Value Equations and Inequalities 73

1.10

Solving Equations and Inequalities Graphically 76

1.11

Modeling Variation 86 Chapter 1 Review 90 Chapter 1 Test 109

¥

FOCUS ON MODELING: Fitting Lines to Data 113

1


1

EQUATIONS AND GRAPHS

1.1

THE COORDINATE PLANE

1. (a) The point that is 3 units to the right of the y­axis and 5 units below the x­axis has coordinates 3 5.

(b) The point 2 7 is 2 units to the right of the y­axis and 7 units above the x­axis, so it is closer to the y­axis.

2. If x is positive and y is negative, then the point x y is in Quadrant IV.  3. The distance between the points a b and c d is c  a2  d  b2 . So the distance between 1 2 and 7 10 is     7  12  10  22  62  82  36  64  100  10. 4. The point midway between a b and c d is     8 12 1  7 2  10     4 6. 2 2 2 2

 ac bd  . So the point midway between 1 2 and 7 10 is 2 2

5. The points have coordinates A 5 1, B 1 2, C 2 6, D 6 2, E 4 1, F 2 0, G 1 3, and H 2 2. 6. A and B lie in Quadrant I and E and G lie in Quadrant III. y

7.

5l (0, 5)

(_1, 0)

y

8. (_2.5, 3.5)

(21 , 23) 0

x

5

(_5, 0)

(_1, _2)

9. x y  x  2

5l

10. x y  y  2

y 5l

0

0

(2, 0)

x 5 (2.6, _1.3)

y 5l

5

x

0

5

x

1


2

CHAPTER 1 Equations and Graphs

11. x y  y  1

12. x y  x  3

y 5l

0

13. x y  2  x  4

5l

5

x

0

14. x y  0  y  2

y 5l

0

15. x y  x  1 and y  3

5

x

x

5

x

5

x

y

0

16. x y  x  2 and y  1

5l

y 5l

5

x

17. x y  1  x  1 and y  4

0

18. x y  1  x  1 and  2  y  2 y

y

5l

5l

0

5

5l

y

0

y

5

x

0

5

x


SECTION 1.1 The Coordinate Plane

19. x y  x y  0

20. x y  x y  0

y

y

5l

5l

0

5

x

0

5

x

21. The two points are 0 2 and 3 0.     (a) d  3  02  0  22  32  22  9  4  13     30 02 (b) Midpoint:   32  1 2 2 22. The two points are 2 1 and 2 2.     (a) d  2  22  1  22  42  32  16  9  25  5     2  2 1  2   0 12 (b) Midpoint: 2 2 23. The two points are 3 3 and 5 3.     (a) d  3  52  3  32  82  62  64  36  100  10   3  5 3  3   1 0 (b) Midpoint: 2 2 24. The two points are 2 3 and 4 1.      (a) d  2  42  3  12  62  22  36  4  50  2 10   2  4 3  1   1 2 (b) Midpoint: 2 2 25. (a)

26. (a)

y

y (_2, 5)

(6, 16) 1l

(10, 0) 2

(0, 8) 2l 1

 (b) d  0  62  8  162    62  82  100  10   0  6 8  16 (c) Midpoint:   3 12 2 2

x

 2  102  5  02    122  52  169  13     2  10 5  0 (c) Midpoint:   4 52 2 2

(b) d 

x

3


4

CHAPTER 1 Equations and Graphs y

27. (a)

y

28. (a)

5l

(_4,5)

(_1, 1)

0

5

1

(3,_2)

x

(_6, _3)

 3  42  2  52      72  72  49  49  98  7 2     4  3 5  2 (c) Midpoint:    12  32 2 2

(b) d 

y

29. (a)

1l

x

 1  62  1  32    52  42  41     6  1 3  1 (c) Midpoint:    72  1 2 2

(b) d 

y

30. (a)

(_6, 2) 1l

1l 1

(6, _2)

(5, 0)

1

x

x

(0, _6)

  (b) d  6  62  2  22  122  42     144  16  160  4 10   6  6 2  2   0 0 (c) Midpoint: 2 2

  1  52  3  32  42  4.   d A C  1  12  3  32  62  6. So

31. d A B 

the area is 4  6  24.

y

A

B

 (b) d  0  52  6  02     52  62  25  36  61     0  5 6  0   52  3 (c) Midpoint: 2 2 32. The area of a parallelogram is its base times its height. Since two sides are parallel to the x­axis, we use the length of one of these as the base. Thus, the base is   d A B  1  52  2  22  42  4. The height is the change in the y coordinates, thus, the height is 6  2  4. So the area of the parallelogram is base  height  4  4  16. y

C

1l 1 C

D

x D

1l

A 1

B x


SECTION 1.1 The Coordinate Plane

33. From the graph, the quadrilateral ABC D has a pair of parallel sides, so ABC D is a trapezoid. The area is   b1  b2 h. From the graph we see that 2   b1  d A B  1  52  0  02  42  4;   b2  d C D  4  22  3  32  22  2; and h is the difference in y­coordinates is 3  0  3. Thus   42 the area of the trapezoid is 3  9. 2

34. The point S must be located at 0 4. To find the area, we find the length of one side and square it. This gives  d Q R  5  02  1  62   52  52    25  25  50  2 50  50. So the area is y

y

D 1l A

1

C

R B

5

Q

P

1l 1

x

x

    6  02  7  02  62  72  36  49  85.     d 0 B  5  02  8  02  52  82  25  64  89.

35. d 0 A 

Thus point A 6 7 is closer to the origin.     36. d E C  6  22  3  12  42  22  16  4  20.     d E D  3  22  0  12  52  12  25  1  26.

Thus point C is closer to point E.      37. d P R  1  32  1  12  42  22  16  4  20  2 5.    d Q R  1  12  1  32  0  42  16  4. Thus point Q 1 3 is closer to point R.

    38. (a) The distance from 7 3 to the origin is 7  02  3  02  72  32  49  9  58. The distance from     3 7 to the origin is 3  02  7  02  32  72  9  49  58. So the points are the same distance from the origin.

  (b) The distance from a b to the origin is a  02  b  02  a 2  b2 . The distance from b a to the origin is    b  02  a  02  b2  a 2  a 2  b2 . So the points are the same distance from the origin.

39. Since we do not know which pair are isosceles, we find the length of all three sides.      d A B  3  02  1  22  32  32  9  9  18  3 2.     d C B  3  42  1  32  12  42  1  16  17.     d A C  0  42  2  32  42  12  16  1  17. So sides AC and C B have the same length.


6

CHAPTER 1 Equations and Graphs

40. Since the side AB is parallel to the x­axis, we use this as the base in the formula area  12 base  height. The height is the change in the y­coordinates. Thus, the base is 2  4  6 and the height is 4  1  3. So the area is 12 6  3  9.

41. (a) Here we have A  2 2, B  3 1, and C  3 3. So     d A B  3  22  1  22  12  32  1  9  10;      d C B  3  32  1  32  62  22  36  4  40  2 10;      d A C  3  22  3  22  52  52  25  25  50  5 2.

Since [d A B]2  [d C B]2  [d A C]2 , we conclude that the triangle is a right triangle.   (b) The area of the triangle is 12  d C B  d A B  12  10  2 10  10.

    52  42  25  16  41; 11  62  3  72      16  25  41; d A C  2  62  2  72  42  52      d B C  2  112  2  32  92  12  81  1  82.

42. d A B 

Since [d A B]2  [d A C]2  [d B C]2 , we conclude that the triangle is a right triangle. The area is    41 1 41  41  2 . 2

43. We show that all sides are the same length (its a rhombus) and then show that the diagonals are equal. Here we have A  2 9, B  4 6, C  1 0, and D  5 3. So     d A B  4  22  6  92  62  32  36  9  45;     d B C  1  42  0  62  32  62  9  36  45;     d C D  5  12  3  02  62  32  36  9  45;     d D A  2  52  9  32  32  62  9  36  45. So the points form a      rhombus. Also d A C  1  22  0  92  32  92  9  81  90  3 10,      and d B D  5  42  3  62  92  32  81  9  90  3 10. Since the diagonals are equal, the rhombus is a square.

     42  82  16  64  80  4 5. 3  12  11  32       22  42  4  16  20  2 5. d B C  5  32  15  112       d A C  5  12  15  32  62  122  36  144  180  6 5. So d A B  d B C  d A C,

44. d A B 

and the points are collinear.

45. Let P  0 y be such a point. Setting the distances equal we get   0  52  y  52  0  12  y  12    25  y 2  10y  25  1  y 2  2y  1  y 2  10y  50  y 2  2y  2  12y  48  y  4. Thus, the point is P  0 4. Check:     0  52  4  52  52  12  25  1  26;     0  12  4  12  12  52  25  1  26.


7

SECTION 1.1 The Coordinate Plane

13 06  2 2

   2 3. So the length of the median CC  is d C C         18 02  2 2   92  1 . So the length of the median B B  2  8  3  2  37. The midpoint of AC is B  2 2          9  3 2  1  62  109 . The midpoint of BC is A  3  8  6  2  11  4 . So the length is d B B   2 2 2 2 2      11  1 2  4  02  145 . of the median A A is d A A  2 2

46. The midpoint of AB is C  

47. As indicated by Example 3, we must find a point S x1  y1  such that the midpoints

y

of P R and of QS are the same. Thus     x1  1 y1  1 4  1 2  4    . Setting the x­coordinates equal, 2 2 2 2

x 1 4  1  1  4  1  x1  1  x1  2Setting the we get 2 2 y 1 2  4  1  2  4  y1  1  y1  3. y­coordinates equal, we get 2 2 Thus S  2 3. 48. We solve the equation 6  8

y

C D

A

1

P

x S

2  7 1  7  2 2     41 24   52  3 . of B D is 2 2

(b) The midpoint of AC is

   52  3 , the midpoint

(c) Since the they have the same midpoint, we conclude that the

B

1l

1

2x 2x to find the x coordinate of B. This gives 6   12  2  x  x  10. Likewise, 2 2

3y  16  3  y  y  13. Thus, B  10 13. 2

49. (a)

1l

R

Q

x

diagonals bisect each other.

   a b a0 b0    . Thus, 2 2 2 2    2 a 2  b b2 a2 a 2  b2 d C M  0  0    ; 2 2 4 4 2     2 a  a 2  b 2 2  b b2 a2 a 2  b2 d A M   a  0    ;   2 2 2 2 4 4 2     2 2  b a  a 2  b 2 b2 a2 a 2  b2 d B M  0  b    .    2 2 2 2 4 4 2

50. We have M 

51. (a) The point 5 3 is shifted to 5  3 3  2  8 5. (b) The point a b is shifted to a  3 b  2.

(c) Let x y be the point that is shifted to 3 4. Then x  3 y  2  3 4. Setting the x­coordinates equal, we get x  3  3  x  0. Setting the y­coordinates equal, we get y  2  4  y  2So the point is 0 2.


8

CHAPTER 1 Equations and Graphs

(d) A  5 1, so A  5  3 1  2  2 1; B  3 2, so B   3  3 2  2  0 4; and C  2 1, so C   2  3 1  2  5 3.

52. (a) The point 3 7 is reflected to the point 3 7.

(b) The point a b is reflected to the point a b. (c) Since the point a b is the reflection of a b, the point 4 1 is the reflection of 4 1.

(d) A  3 3, so A  3 3; B  6 1, so B   6 1; and C  1 4, so C   1 4.   53. (a) d A B  32  42  25  5. (b) We want the distances from C  4 2 to D  11 26. The walking distance is 4  11  2  26  7  24  31 blocks. Straight­line distance is    4  112  2  262  72  242  625  25 blocks.

(c) The two points are on the same avenue or the same street.   3  27 7  17   15 12, which is at the intersection of 15th Street and 12th Avenue. 54. (a) The midpoint is at 2 2 (b) They each must walk 15  3  12  7  12  5  17 blocks.

55. The midpoint of the line segment is 66 45. The pressure experienced by an ocean diver at a depth of 66 feet is 45 lb/in2 .

56. Let the area of the triangle be A. As hinted, we draw a rectangle

y

5  1  4 units high, so its area is A R  6  4  24. The areas of the three

right triangles inside the rectangle but outside the original triangle are A1  12 7  1 5  2  9, A2  12 4  1 2  1  32 , and

A3  12 7  4 5  1  6. Since the area of the rectangle is the sum of

(7, 5)

(1, 5)

circumscribing the triangle. This rectangle is 7  1  6 units long and

AÁ (1, 2)

A

1

Aª (1, 1)

0

1

(4, 1)

A£ (7, 1) x

the areas of the right triangles and the original triangle, we have A  A R  A1  A2  A3  24  9  32  6  15 2 .

57. Following the hint, we consider an equilateral triangle with side length 1 anywhere in the plane. Each of its vertices is either red or blue, and each is 1 unit away from the other two. So at least two have the same color and are exactly 1 unit apart.

1.2

GRAPHS OF EQUATIONS IN TWO VARIABLES

1. If the point 2 3 is on the graph of an equation in x and y, then the equation is satisfied when we replace x by 2 and y by 3. ?

?

We check whether 2 3  2  1  6  3. This is false, so the point 2 3 is not on the graph of the equation 2y  x  1.

To complete the table, we express y in terms of x: 2y  x  1  y  12 x  1  12 x  12 . x

y

2

 12

1

0

x y

0

1 2

1

1

2

3 2

  2  12 1 0   0 12 1 1   2 32

y 1

0

1

x

2. (a) The x­coordinates of the points where the graph of an equation intersects the x­axis are called x­intercepts. Similarly, the y­coordinates of the points where the graph of an equation intersects the x­axis are called y­intercepts.


SECTION 1.2 Graphs of Equations in Two Variables

9

(b) To find the x­intercept(s) of the graph of an equation we set y equal to 0 in the equation and solve for x: 2 0  x  1  x  1, so the x­intercept of 2y  x  1 is 1.

(c) To find the y­intercept(s) of the graph of an equation we set x equal to 0 in the equation and solve for y: 2y  0  1  y  12 , so the y­intercept of 2y  x  1 is 12 .

3. (a) If a graph is symmetric with respect to the x­axis and a b is on the graph, then a b is also on the graph. (b) If a graph is symmetric with respect to the y­axis and a b is on the graph, then a b is also on the graph. (c) If a graph is symmetric about the origin and a b is on the graph, then a b is also on the graph. 4. (a) The x­intercepts are 5 and 3, and the y­intercepts are 2. (b) The graph is symmetric about the x­axis. ?

?

?

?

5. 0 5: 3 0  5  5  0  0  5  5  0. Yes. 2 1: 3 2  1  5  0  6  1  5  0. No. ?

?

2 1: 3 2  1  5  0  6  1  5  0. Yes. So 0 5 and 2 1 lie points on the graph of this equation. ?  ?  6. 4 3: 3  1  2 4  3  9. Yes. ?  ?  1 0: 0  1  2 1  0  1. No. ?  ?  0 1: 1  1  2 0  1  1. Yes. So 4 3 and 0 1 lie on the graph of the equation. ?

?

7. 1 1: 5 1  2 1  7  5  2  7. Yes. ?

?

?

?

2 1: 5 2  2 1  7  10  2  7. No. 1 6: 5 1  2 6  7  5  12  7. Yes. So 1 1 and 1 6 lie on the graph of the equation.

  ? ? 8. 1 1: 1 12  1  1  1 2  1. No.      ? ? 1 12 : 12 12  1  1  12 2  1. Yes.      ? ? 1 12 : 12 12  1  1  12 2  1. Yes.     So both 1 12 and 1 12 lie on the graph of this equation. ?

?

?

?

?

?

9. 0 2: 02  0 2  22  4  0  0  4  4. Yes. 1 2: 12  1 2  22  4  1  2  4  4. No.

2 2: 22  2 2  22  4  4  4  4  4. Yes. So 0 2 and 2 2 lie on the graph of this equation. ?

?

10. 0 1: 02  12  1  0  0  1  1  0. Yes.   2  2  ? ? 1  1 : 1  1  1  0  12  12  1  0. Yes. 2 2 2 2     2  2 ? ? 3 1 3  12  1  0  34  14  1  0. Yes. 2 2 : 2     So 0 1, 1  1 , and 23  12 all lie on the graph of this equation. 2

2


10

CHAPTER 1 Equations and Graphs

11. y  3x x

y

3

9

12. y  5x

y 5l

6

2 1

3

0

0

1

3

2

0

5

x

y

3

15

2

10

1

5

1

5

2

10

3

15

0

6

3

9

13. y  2x  3 y

3

4

2

1

5l

y

5

1

3

1

1

3

1

0

3

1

5

1

2

7

3

3

9

5

0

5

x

0

15. 4x  5y  40  y  8  45 x y

2

96

0

8

2

64

4

48

6

32

8

16

10

0

12

16

5

x

5

x

y 5l

0

4 5 7 9

16. 2x  3y  12  y  23 x  4

y

y

x

y

3

6

1

47

0

5l

1 3

0

0

0

x

1

x

5l

14. y  x  4

y

x

y

x

5

x

6

4 33 2

0

9

2

12

4

5

0

_5

5

x


11

SECTION 1.2 Graphs of Equations in Two Variables

17. y  x 2  3

y

x

y

3

6

2

1

1

2

0 1 2

18. y  3  x 2

5l

y

3

6

2

1

1

2

0

3

1

2

1

2

6

3

1

3

0

x

1

2

3

19. y  x 2  1 x

y

3

8

2

3

5l

0

0

1

5

x

x

y

3

15

2

5

1

1

0

0

1

2

3

2

y

5

4

3 1

0 1

y 5l

0 0

5

x

y 5l

0

5

x

5

15

x

y

5

4

3

x

1

1

22. y  x  1

2

1

3

3

8

x

0

6

1

21. y  x  1

5l

20. y  2x 2  3

y

0

1

3

y

x

y 5l

2

1

0

0

1

0

1

2

3

2

3

4

5

4

5

6

0

5

x


12

CHAPTER 1 Equations and Graphs

23. y  x 3

24. x  y 3 . Since x  y 3 is solved for x in terms of y, we

insert values for y and find the corresponding values of x

y

x

y

3

27

in the table below.

5l

x

y

27

3

8

2

1

1

1

1

8

2

27

3

8

2 1

1

0

0

1

1

2

0

1

x

0

8

3

27

25. x  y 4 . Since x  y 3 is solved for x in terms of y, we

insert values for y and find the corresponding values of x in the table below. x

y

81

3

16

y

2

1

1

1

0

0

0

1

1

16

2

81

3

5

x

2

10

1

8

0 1 2 3 4

6 4 2

0 2

0

0

y

3

80

2

15

1

0

0

1

1

0

2

15

2 0

1

x

80

y  0  0  2x  6  x  3, so the x­intercept is 3, and

y

x­axis symmetry: 2x  y  6, which is not the same as

1l

x  0  y  2 0  6  6, so the y­intercept is 6.

2x  y  6, so the graph is not symmetric with respect to the x­axis. y­axis symmetry: 2x  y  6, which is not the same as

2x  y  6, so the graph is not symmetric with respect to the y­axis.

Origin symmetry: 2x  y  6, which is not the same as

2x  y  6, so the graph is not symmetric with respect to the origin.

x

5

y

x

27. (a) Solve for y: 2x  y  6  y  2x  6. y

5l

26. y  1  x 4

3

x

y

0

1

x


SECTION 1.2 Graphs of Equations in Two Variables

(b) y  2 x  12 x

y

2

18

1

8

0

2

1

0

2

2

3

8

13

y

y  0  0  2 x  12  x  1, so the x­intercept is 1,

and x  0  y  2 0  12  2, so the y­intercept is 2.

x­axis symmetry: y  2 x  12 , which is not the same as

y  2 x  12 , so the graph is not symmetric with respect to

the x­axis.

y­axis symmetry: y  2 x  12 , which is not the same as

1

y  2 x  12 , so the graph is not symmetric with respect to

1

the y­axis.

x

Origin symmetry: y  2 x  12  y  2 x  12 ,

which is not the same as y  2 x  12 , so the graph is not

symmetric with respect to the origin.

28. (a) Solve for y: x  4y  8  y  14 x  2. x

y

2

 52

0 2 4 8

2

 32

1

0

12

1

y  0  0  14 x  2  x  8, so the x­intercept is 8, and

y

x­axis symmetry: y  2x  6, which is not the same as

1l

x  0  y  14 0  2  2, so the y­intercept is 2.

y  2x  6, so the graph is not symmetric with respect to the

x­axis. y­axis symmetry: y  2 x  6  y  2x  6, which is

0

x

1

not the same as y  2x  6, so the graph is not symmetric with respect to the y­axis.

Origin symmetry: y  2 x  6  y  2x  6, which is not the same as y  2x  6, so the graph is not symmetric

with respect to the origin. (b) y  x 2  4 x

y

3

5

2

0

1

3

0

4

1

3

2

0

3

5

y  0  0  x 2  4  x  2, so the x­intercepts are

y

x­axis symmetry: y  x 2  4, which is not the same as

1l

2, and x  0  y  02  4  4, so the y­intercept is 4. y  x 2  4, so the graph is not symmetric with respect to

the x­axis.

y­axis symmetry: y   x2  4  x 2  4, so the graph

is symmetric with respect to the y­axis.

Origin symmetry: y  x 2  4, which is not the same as y  x 2  4, so the graph is not symmetric with respect to

the origin.

0

1

x


14

CHAPTER 1 Equations and Graphs

29. (a) y 

x 2

x

y

0

2

1

3  1 2  1 3

2 3 4

3  6

6

1

9

5

 y  0  0  x  2, so there is no x­intercept, and x  0   y  0  2  2, so the y­intercept is 2.  x­axis symmetry: y  x  2, which is not the same as  y  x  2, so the graph is not symmetric with respect to

y

the x­axis.

 y­axis symmetry: y  x  2, which is not the same as  y  x  2, so the graph is not symmetric with respect to

the y­axis.

  x  2  y   x  2,  which is not the same as y  x  2, so the graph is not Origin symmetry: y 

1l 0

x

1

symmetric with respect to the origin. (b) y   x x

y

5

5

3

3

1

1

1

1

0 3 5

30. (a) y 

0

x  0  y   0  0, so the y­intercept is 0.

same as y   x, so the graph is not symmetric with

y­axis symmetry: y   x   x, so the graph is

respect to the origin.

4

0

5 6

1  2

8

2

13

3

5

x

symmetric with respect to the y­axis.

5

y

0

respect to the x­axis.

Origin symmetry: y   x  y  x, which is not the

x

5l

x­axis symmetry: y   x  y  x, which is not the

3

x 4

y

y  0   x  0  x  0, so the x­intercept is 0, and

same as y   x, so the graph is not symmetric with

y 00

x  4  x  4, so the x­intercept is 4, and  x  0  y  0  4, so there is no y­intercept.  x­axis symmetry: y  x  4, which is not the same as  y  x  4, so the graph is not symmetric with respect to

the x­axis.

 y­axis symmetry: y  x  4, which is not the same as  y  x  4, so the graph is not symmetric with respect to

the y­axis.

  Origin symmetry: y  x  4  y   x  4,  which is not the same as y  x  4, so the graph is not

symmetric with respect to the origin.

y

1l

0

1

x


SECTION 1.2 Graphs of Equations in Two Variables

(b) x  y. Here y is not a function of x. y

x

3

3 2

2

31. (a) y 

y  0  x  0  0, so the x­intercept is 0, and x  0 

x  y, so the graph is symmetric with respect to the x­axis.

0

0

graph is not symmetric with respect to the y­axis.

1

1

2

2

Origin symmetry: x  y  y, which is not the same,

3

3

y

2

0  3

1

0 1

2  3

2

0

5l

x­axis symmetry: x  y  y, which is the same as

1

x

y

y  0  y  0, so the y­intercept is 0.

1

 4  x2

15

y­axis symmetry: x  y, which is not the same, so the

0

5

x

so the graph is not symmetric with respect to the origin.

 4  x 2  0  x  2, so the x­intercepts are  2, and x  0  y  4  02  2, so the y­intercept is 2.  x­axis symmetry: y  4  x 2 , which is not the same as  y  4  x 2 , so the graph is not symmetric with respect to y 0

y

2l

0

the x­axis.

  y­axis symmetry: y  4  x2  4  x 2 , so the graph

2

x

is symmetric with respect to the y­axis.   Origin symmetry: y  4  x2  4  x 2    y   4  x 2 , which is not the same as y  4  x 2 , so

the graph is not symmetric with respect to the origin. (b) y  x 3  4x y

x

3

15

2

0

1

3

0

0

1

3

2

0

3

15

  y  0  0  x 3  4x  x x 2  4  x x  2 x  2, so

y

the x­intercepts are 0 and 2, and x  0  y  03  4 0  0, so the y­intercept is 0.

x­axis symmetry: y  x 3  4x  y  x 3  4x, which

is not the same as y  x 3  4x, so the graph is not symmetric with respect to the x­axis.

y­axis symmetry: y  x3  4 x  x 3  4x, which

is not the same as x  x 3  4x, so the graph is not symmetric

with respect to the y­axis.

Origin symmetry: y  x3  4 x  y  x 3  4x, so

the graph is symmetric with respect to the origin.

10l 0

1

x


16

CHAPTER 1 Equations and Graphs

 32. (a) y   4  x 2 x

y

2

0   3

1

0 1 2

 y  0   4  x 2  0  x  2, so the x­intercepts are  2, and x  0  y   4  2, so the y­intercept is 2.   x­axis symmetry: y   4  x 2  y  4  x 2 , which  is not the same as y   4  x 2 , so the graph is not

2   3

symmetric with respect to the x­axis.   y­axis symmetry: y   4  x2   4  x 2 , so the

0

y

2l

0

2

x

graph is symmetric with respect to the y­axis.   Origin symmetry: y   4  x2  y  4  x 2 ,  which is not the same as y   4  x 2 , so the graph is not symmetric with respect to the origin.

(b) x  y 3  y 

 3

y  0  x  03  0, so the x­intercept is 0, and x  0 

y

x

8 4

2  34

1

1

0 1

x.

0

4

1  3 4

8

2

y

0  y 3  y  0, so the y­intercept is 0.

x­axis symmetry: x  y3  y 3 , which is not the same

as x  y 3 , so the graph is not symmetric with respect to the x­axis.

1l 0

y­axis symmetry: x  y 3  x  y 3 , which is not the

10

x

same as x  y 3 , so the graph is not symmetric with respect to the y­axis.

Origin symmetry: x  y3  x  y 3 , so the graph is

symmetric with respect to the origin.

33. (a) To find x­intercepts, set y  0. This gives 0  x  6  x  6, so the x­intercept is 6. To find y­intercepts, set x  0. This gives y  0  6  6, so the y­intercept is 6.   (b) To find x­intercepts, set y  0. This gives 0  x 2  5  x 2  5  x   5, so the x­intercepts are  5. To find y­intercepts, set x  0. This gives y  02  5  5, so the y­intercept is 5.

34. (a) To find x­intercepts, set y  0. This gives 4x 2  25 02  100  x 2  25  x  5, so the x­intercepts are 5. To find y­intercepts, set x  0. This gives 4 02  25y 2  100  y 2  4  y  2, so the y­intercepts are 2.

(b) To find x­intercepts, set y  0. This gives x 2  x 0  3 0  1  x 2  1  x  1, so the x­intercepts are 1. To find y­intercepts, set x  0. This gives 02  0y  3y  1  y  13 , so the y­intercept is 13 .

35. (a) To find x­intercepts, set y  0. This gives 9x 2  4 02  36  9x 2  36  x 2  4  x  2, so the x­intercepts are 2. To find y­intercepts, set x  0. This gives 9 02  4y 2  36  y 2  9, so there is no y­intercept.

(b) To find x­intercepts, set y  0. This gives 0  2x 0  4x  1  x  14 , so the x­intercept is 14 . To find y­intercepts,

set x  0. This gives y  2 0 y  4 0  1, so the y­intercept is 1.  36. (a) To find x­intercepts, set y  0. This gives 0  x 2  16  x 2  16  x  4, so the x­intercepts are 4. To find  y­intercepts, set x  0. This gives y  02  16, so there is no y­intercept.  (b) To find x­intercepts, set y  0. This gives 0  64  x 3  x 3  64  x  4, so the x­intercept is 4. To find  y­intercepts, set x  0. This gives y  64  03  y  8, so the y­intercept is 8.


SECTION 1.2 Graphs of Equations in Two Variables

17

37. To find x­intercepts, set y  0. This gives 0  4x  x 2  0  x 4  x  0  x or x  4, so the x­intercepts are 0 and 4. To find y­intercepts, set x  0. This gives y  4 0  02  y  0, so the y­intercept is 0.

x 2 02 x2  1  1  x 2  9  x  3, so the x­intercepts are 3 and 3. 9 4 9 y2 02 y 2  1  1  y 2  4  x  2, so the y­intercepts are 2 and 2. To find y­intercepts, set x  0. This gives 9 4 4 39. To find x­intercepts, set y  0. This gives x 4  02  x 0  16  x 4  16  x  2. So the x­intercepts are 2 and 2.

38. To find x­intercepts, set y  0. This gives

To find y­intercepts, set x  0. This gives 04  y 2  0 y  16  y 2  16  y  4. So the y­intercepts are 4 and 4.

40. To find x­intercepts, set y  0. This gives x 2  03  x 2 02  64  x 2  64  x  8. So the x­intercepts are 8 and 8. To find y­intercepts, set x  0. This gives 02  y 3  02 y 2  64  y 3  64  y  4. So the y­intercept is 4.

41. x­axis symmetry: y  x 4  x 2  y  x 4  x 2 , which is not the same as y  x 4  x 2 , so the graph is not symmetric with respect to the x­axis. y­axis symmetry: y  x4  x2  x 4  x 2 , so the graph is symmetric with respect to the y­axis.

Origin symmetry: y  x4  x2  y  x 4  x 2 , which is not the same as y  x 4  x 2 , so the graph is not symmetric with respect to the origin. 42. x­axis symmetry: x  y4  y2  y 4  y 2 , so the graph is symmetric with respect to the x­axis.

y­axis symmetry: x  y 4  y 2 , which is not the same as x  y 4  y 2 , so the graph is not symmetric with respect to the y­axis.

Origin symmetry: x  y4  y2  x  y 4  y 2 , which is not the same as x  y 4  y 2 , so the graph is not symmetric with respect to the origin. 43. x­axis symmetry: y  x 3  10x  y  x 3  10x, which is not the same as y  x 3  10x, so the graph is not symmetric with respect to the x­axis. y­axis symmetry: y  x3  10 x  y  x 3  10x, which is not the same as y  x 3  10x, so the graph is not symmetric with respect to the y­axis. Origin symmetry: y  x3  10 x  y  x 3  10x  y  x 3  10x, so the graph is symmetric with respect to the origin. 44. x­axis symmetry: y  x 2  x  y  x 2  x, which is not the same as y  x 2  x, so the graph is not symmetric with respect to the x­axis. y­axis symmetry: y  x2  x  y  x 2  x, so the graph is symmetric with respect to the y­axis. Note that x  x.

Origin symmetry: y  x2  x  y  x 2  x  y  x 2  x, which is not the same as y  x 2  x, so the graph is not symmetric with respect to the origin. 45. x­axis symmetry: x 2 y4  x 4 y2  2  x 2 y 4  x 4 y 2  2, so the graph is symmetric with respect to the x­axis. y­axis symmetry: x2 y 4  x4 y 2  2  x 2 y 4  x 4 y 2  2, so the graph is symmetric with respect to the y­axis.

Origin symmetry: x2 y4  x4 y2  2  x 2 y 4  x 4 y 2  2, so the graph is symmetric with respect to the origin. (Note that if a graph is symmetric with respect to each coordinate axis, it is symmetric with respect to the origin. The converse is not true, as shown in the next exercise.) 46. x­axis symmetry: x 3 y  x y3  1  x 3 y  x y 3  1  x 3 y  x y 3  1  1, so the graph is not symmetric with respect to the x­axis. y­axis symmetry: x3 y  x y 3  1  x 3 y  x y 3  1  x 3 y  x y 3  1  1, so the graph is not symmetric with respect to the y­axis. Origin symmetry: x3 y  x y3  1  x 3 y  x y 3  1, so the graph is symmetric with respect to the origin.


18

CHAPTER 1 Equations and Graphs

47. Symmetric with respect to the y­axis.

48. Symmetric with respect to the x­axis.

y

y

x

0

0

49. Symmetric with respect to the origin.

x

50. Symmetric with respect to the origin. y

y

x

0

0

51. (a) Symmetric about the x­axis:

(b) Symmetric about the y­axis:

x

(c) Symmetric about the origin:

y

y

y

1

1

1

x

1

52. (a) Symmetric about the x­axis.

x

1

(b) Symmetric about the y­axis.

(c) Symmetric about the origin.

y

y

y

1

1

1

1

x

0

1

x

1

x

1

x

53. (a) From the graph, it appears that I 10  900 lm, I 30  100 lm, and I 40  55 lm. (b) It appears that I x  100 lm for 10 cm  x  30 cm.

54. (a) From the graph, it appears that the closest the satellite gets to the center of the moon is approximately 5 Mm and farthest approximately 70 Mm. x  3252 Setting y  0 in the equation (which obviously corresponds to the desired distances), we have 1 1420  x  3252  1420  x  325   1420  377  x  52 or 702. Thus, the smallest and largest distances are 52 Mm and 702 Mm.


SECTION 1.3 Circles

19

  102 100 x  3252 x  3252  1  1  x  3252  1420 113 (b) Setting y  10, we have 213 1420 213 1420 213   x  325   75333  x  505 or 5995. The distances of these points from the center of the moon are   d1  5052  102  112 Mm and d2  59952  102  608 Mm.

55. True. If a b is on the graph, then by symmetry about the x­axis, the point a b is on the graph. Then by symmetry about the y­axis, the point a b is on the graph. Thus, the graph is symmetric with respect to the origin. 56. False. For example, the graph of y  x 3 is symmetric with respect to the origin, but not with respect to either axis.

1.3

CIRCLES

1. The equation of a circle with center C h k and radius r is x  h2  y  k2  r 2 . So the equation of the circle with

center 1 3 and radius 2 is x  12  y  32  4. The equation of the circle in the figure is [x  1]2  y  22  32  x  12  y  22  9.

 9  3. The graph of  2 2 x  5  y  8  16 is a circle with center  5  8  5 8 and radius 16  4.  2 3. To complete the square of the expression x 2  bx, we add b2  14 b2 . To complete the square of x 2  12x, we add   1 122  36 to get x 2  12x  36  x  62 . 4

2. The graph of the equation x 2  y 2  9 is a circle with center 0 0 and radius

4. To determine if the graph of x 2  2x  y 2  4y  31 is a circle, we complete the squares of the x­terms and the y­terms: x 2  2x  1  y 2  4y  4  31  1  4  x  12  y  22  36. So the graph of this equation is a circle with center 1 2 and radius 6.

5. x 2  y 2  9 has center 0 0 and radius 3.

6. x 2  y 2  5 has center 0 0 and radius

y

 5.

y

1l

1l 1

1

x

7. x  22  y 2  9 has center 2 0 and radius 3. y

x

8. x 2  y  12  4 has center 0 1 and radius 2. y

1l 1l 0

0 1

x

1

x


20

CHAPTER 1 Equations and Graphs

9. x  32  y  42  25 has center 3 4 and radius 5. 10. x  12  y  22  36 has center 1 2 and radius 6.

y

y

1l 1

x

1l 1

x

11. Using h  3, k  1, and r  2, we get x  32  y  12  22  x  32  y  12  4. 12. Using h  2, k  5, and r  3, we get x  22  y  52  32  x  22  y  52  9.

13. The equation of a circle centered at the origin is x 2  y 2  r 2 . Using the point 4 7 we solve for r 2 . This gives 42  72  r 2  16  49  65  r 2 . Thus, the equation of the circle is x 2  y 2  65.

14. Using h  1 and k  5, we get x  12  y  52  r 2  x  12  y  52  r 2 . Next, using the point 4 6, we solve for r 2 . This gives 4  12  6  52  r 2  130  r 2 . Thus, an equation of the circle is

x  12  y  52  130.

  1  5 1  9   2 5. The radius is one half the 15. The center is at the midpoint of the line segment, which is 2 2    diameter, so r  12 1  52  1  92  12 36  64  12 100  5. Thus, the equation of the circle is x  22  y  52  52 or x  22  y  52  25.

  1  7 3  5 16. The center is at the midpoint of the line segment, which is   3 1. The radius is one half the 2 2   diameter, so r  12 1  72  3  52  4 2. Thus, an equation of the circle is x  32  y  12  32.

17. Since the circle is tangent to the x­axis, it must contain the point 7 0, so the radius is the change in the y­coordinates. That is, r  3  0  3. So the equation of the circle is x  72  y  32  32 , which is x  72  y  32  9.

18. Since the circle with r  5 lies in the first quadrant and is tangent to both the x­axis and the y­axis, the center of the circle is at 5 5. Therefore, the equation of the circle is x  52  y  52  25.

19. From the figure, the center of the circle is 2 2. The radius is distance from the center to 0 2, so r  2. Thus, an equation of the circle is x  22  y  22  22  x  22  y  22  4.

20. From the figure, the center of the circle is 1 1. The radius is the distance from the center to the point 2 0. Thus,    r  1  22  1  02  9  1  10, and an equation of the circle is x  12  y  12  10.   21. From the figure, the center of the circle is 0 12 . The radius is the distance from the center to 0 1, so r  12 . Thus, an 2  2 2   equation of the circle is x  02  y  12  12  x 2  y  12  14 .   22. From the figure, the center of the circle is 12  0 . The radius is the distance from the center to 1 0, so r  12 . Thus, an 2 2  2   equation of the circle is x  12  y  02  12  x  12  y 2  14 .  2 23. To complete the square, we add 82  16: x 2  8x  16  x  42 .


SECTION 1.3 Circles

21

2  24. To complete the square, we add 12  36: y 2  12y  36  x  62 . 2

2   2  94 : y 2  3x  94  y  32 . 25. To complete the square, we add 3 2 26. To complete the square, we add

  2 3 2

 34 : x 2 

 2   3x  34  x  23 .

 2  2  2  2 27. Completing the square gives x 2  y 2  4x  6y  12  0  x 2  4x  42  y 2  6y  62  12  42  62

 x 2  4x  4  y 2  6y  9  12  4  9  x  22  y  32  1. Thus, the circle has center 2 3 and radius 1.

 2  2 28. Completing the square gives x 2  y 2  8x  5  0  x 2  8x  82  y 2  5  82  x 2  8x  16  y 2  5  16   x  42  y 2  11. Thus, the circle has center 4 0 and radius 11. 2 2  2   2   y 2  12 y  12  18  12  12 29. Completing the square gives x 2  y 2  12 x  12 y  18  x 2  12 x  12 2 2 2 2 2  2  1  y2  1 y  1  1  1  1  2  1  x  1  x 2  12 x  16  y  14  14 . Thus, the circle has center 2 16 8 16 16 8 4 4   1   1 and radius 1 . 4 4 2         1  0  x 2  1 x  12 2  y 2  2y  2 2   1  12 2  2 2 30. Completing the square gives x 2  y 2  12 x 2y  16 2 2 2 16 2 2 2   x  14  y  12  1.   Thus, the circle has center  14  1 and radius 1. 2   2 9. 31. Completing the square gives 2x 2  2y 2  3x  0  x 2  y 2  32 x  0  x  34  y 2  34  16   Thus, the circle has center 34  0 and radius 34 .

2  1  37 . 32. Completing the square gives 3x 2  3y 2  6x  y  0  x 2  y 2  2x  13 y  0  x  12  y  16  1  36 36    37 1 Thus, the circle has center 1 6 and radius 6 .

33. x 2  8x  y 2  6y  25  0  x  42  y  32  25  16  9  x  42  y  32  25  25  0. The graph is the point 4 3. 2    34. x 2  x  y 2  10y  25  0  x  12  y  52  14 . The graph is a circle with center 12  5 and radius 12 .

35. x 2  6x  y 2  2y  8  0  x  32  y  12  8  9  1  x  32  y  12  18. The graph is a circle with  center 3 1 and radius 3 2. 2 2   36. x 2  x  y 2  4y  9  0  x  12  y  22  9  14  4  x  12  y  22   19 4 . The graph is empty.

2    7 2   7 . The graph is empty. 2 37. x 2  8x  y 2  7y  30  0  x  42  y  72  30  16  49 4  x  4  y  2 4

38. x 2  12x  y 2  4y  40  0  x  62  y  22  40  36  4  x  62  y  22  0. The graph is the point 6 2.


22

CHAPTER 1 Equations and Graphs

39. x  32  y  22  4. Setting y  0, we have x  32  4  4  x  3  0  x  3, so the

x­intercept is 3. Setting x  0, we have 9  y  22  4

 y  22  5, so there is no y­intercept.

40. x 2  6x  y 2  6y  9  0  x  32  y  32  9.

Setting y  0, we have x  32  9  9  x  3, so the x­intercept is 3. Setting x  0, we have 9  y  32  9

 y  3, so the y­intercept is 3. y

y

1l

1l 1

1

x

41. x  32  y  12  10. Setting y  0, we have

x  32  9  x  0 or x  6, so the x­intercepts are 0

and 6. Setting x  0, we have y  12  1  y  0 or y  2, so the y­intercepts are 0 and 2.

x

42. x 2  2x  y 2  7y  8  0  2  x  12  y  72  85 4 . Setting y  0 in the original

equation, we have x 2  2x  8  x  2 x  4  0, so the x­intercepts are 2 and 4. Setting x  0, we have

y

y 2  7y  8  y  1 y  8  0, so the y­intercepts

are 1 and 8.

y

1l 1

x

1l 1

  43. x y  x 2  y 2  1 . This is the set of points on or inside the circle x 2  y 2  1. y

1l

x

  44. x y  x 2  y 2  4 . This is the set of points outside the circle x 2  y 2  4.

y

2l

1

x

2

x


SECTION 1.3 Circles

45. Completing the square gives x 2  y 2  4y  12  0     4 2 4 2  12    x 2  y 2  4y  2 2

23

  46. x y  x 2  y 2  9 and y  x . This is the top

x 2  y  22  16. Thus, the center is 0 2, and the

radius is 4. So the circle x 2  y 2  4, with center 0 0

quarter of the interior of the circle of radius 3. Thus, the   area is 14   32  94 . y

and radius 2 sits completely inside the larger circle. Thus,

the area is   42    22  16  4  12.

3l 3

47. We find the distance between the transmitter at 10 5 and the person at 50 60: d1 

x

 50  102  60  52  68 mi.

This is less than 75 mi, so this person would be able to receive the signal.  For the person at 50 60, d2  50  102  60  52  885 mi, so this person would not be able to receive the signal.

y

48. By definition, if the epicenter is 30 mi from A 10 50, then it lies on the circle with radius 30 centered at A, with equation

50

x  102  y  502  302 . Similarly, it lies on the circle with radius 50

centered at B 30 10, with equation x  302  y  102  502 , and the circle with radius 40 centered at C 50 20, with equation

A E

B

C 50

x  502  y  202  402 .

Graphing these three circles, we see that the epicenter lies at E 10 20. Check: 10  102  20  502  302 , 10  302  20  102  502 , and 10  502  20  202  402 .

49. (a) x  22  y  12  9 has center 2 1 and radius 3.

x  62  y  42  16 has center 6 4 and radius 4.     The distance between centers is 2  62  1  42  42  32  16  9  25  5.

Because 5  3  4, these circles intersect.

(b) x 2  y  22  4 has center 0 2 and radius 2.

x  52  y  142  9 has center 5 14 and radius 3.     The distance between centers is 0  52  2  142  52  122  25  144  169  13.

Because 13  2  3, these circles do not intersect.

(c) x  32  y  12  1 has center 3 1 and radius 1.

x  22  y  22  25 has center 2 2 and radius 5.     The distance between centers is 3  22  1  22  12  32  1  9  10.  Since 10  1  5, these circles intersect.

x


24

CHAPTER 1 Equations and Graphs

50. If the distance d between the centers of the circles is greater than the sum r1  r2 of their radii, then the circles do not intersect, as shown in the first diagram. If d  r1  r2 , then the circles intersect at a single point, as shown in the second diagram. If d  r1  r2 , then the circles intersect at two points, as shown in the third diagram.

rª CÁ rÁ

CÁ rÁ

d

d

d

Case 1 d  r1  r2

Case 2 d  r1  r2

51. Completing the square gives x 2  y 2  ax  by  c  0  x 2  ax  

 x

a 2 2

 y

b 2

2

 c 

a 2  b2 4

 a 2 2

Case 3 d  r1  r2  y 2  by 

 2  a 2  b 2 b  c   2 2 2

.

a 2  b2 a 2  b2  0, this is the equation of a circle; if c   0, it is the equation of a single point, and if 4 4 a 2  b2  0, its graph is empty. c  4     a b a 2  b2 If the equation represents a circle, its center is    and its radius is c   12 a 2  b2  4ac. 2 2 4 If c 

1.4

LINES

1. We find the “steepness” or slope of a line passing through two points by dividing the difference in the y­coordinates of these 51  2. points by the difference in the x­coordinates. So the line passing through the points 0 1 and 2 5 has slope 20 2. (a) The line with equation y  3x  2 has slope 3. (b) Any line parallel to this line has slope 3.

(c) Any line perpendicular to this line has slope  13 . 3. The point­slope form of the equation of the line with slope 3 passing through the point 1 2 is y  2  3 x  1. 4. For the linear equation 2x  3y  12  0, the x­intercept is 6 and the y­intercept is 4. The equation in slope­intercept form is y   23 x  4. The slope of the graph of this equation is  23 .

5. The slope of a horizontal line is 0. The equation of the horizontal line passing through 2 3 is y  3. 6. The slope of a vertical line is undefined. The equation of the vertical line passing through 2 3 is x  2. 7. (a) Yes, the graph of y  3 is a horizontal line 3 units below the x­axis.

(b) Yes, the graph of x  3 is a vertical line 3 units to the left of the y­axis.

(c) No, a line perpendicular to a horizontal line is vertical and has undefined slope. (d) Yes, a line perpendicular to a vertical line is horizontal and has slope 0.


SECTION 1.4 Lines y

8.

25

Yes, the graphs of y  3 and x  3 are perpendicular lines.

5l x=_3

0

x

5 y=_3

y  y1 1 1 1  0 10. m  2    x2  x1 30 3 3 y2  y1 3 3 2  1 12. m     x2  x1 3  5 8 8 y2  y1 2 1 3  1 14. m     x2  x1 2  4 6 3 2  2 y2  y1  16. m  0 x2  x1 63 y  y1 20  2. 17. For 1 , we find two points, 1 2 and 0 0 that lie on the line. Thus the slope of 1 is m  2  x2  x1 1  0 y  y1 32 For 2 , we find two points 0 2 and 2 3. Thus, the slope of 2 is m  2   12 . For 3 we find the points x2  x1 20 y  y1 1  2  3. For 4 , we find the points 2 1 and  2 2 and 3 1. Thus, the slope of 3 is m  2 x 2  x1 32 1 1 y  y1 2  1 2 2. Thus, the slope of 4 is m  2   .  x2  x1 2  2 4 4

2 02 y  y1   2  9. m  2 x2  x1 0  1 1 y  y1 5 3  2 11. m  2   x2  x1 3  3 6 y2  y1 44 13. m  0  x2  x1 05 y  y1 53 2 15. m  2    1 x2  x1 68 2

18. (a)

y

m=2

m=1

(b)

y

m=3

1 m=_ 2

1 m=_ 2

m=0

1 m=_ 3

1l

1l 1

1

x

x 1 m=_ _ 3

m=_1

04  1. Since the y­intercept is 4, 40 the equation of the line is y  mx  b  1x  4. So y  x  4, or x  y  4  0. 04 20. We find two points on the graph, 0 4 and 2 0. So the slope is m   2. Since the y­intercept is 4, the 2  0 equation of the line is y  mx  b  2x  4, so y  2x  4  2x  y  4  0. 0  3 21. We choose the two intercepts as points, 0 3 and 2 0. So the slope is m   32 . Since the y­intercept is 3, 20 the equation of the line is y  mx  b  32 x  3, or 3x  2y  6  0. 19. First we find two points 0 4 and 4 0 that lie on the line. So the slope is m 

22. We choose the two intercepts, 0 4 and 3 0. So the slope is m  equation of the line is y  mx  b   43 x  4  4x  3y  12  0.

0  4   43 . Since the y­intercept is 4, the 3  0


26

CHAPTER 1 Equations and Graphs

23. Using y  mx  b, we have y  3x  2 or 3x  y  2  0. 24. Using y  mx  b, we have y  25 x  4  2x  5y  20  0.

25. Using the equation y  y1  m x  x1 , we get y  1  3 x  4  y  1  3x  12  3x  y  11  0.

26. Using the equation y  y1  m x  x1 , we get y  4  1 x  2  y  4  x  2  x  y  2  0.

27. Using the equation y  y1  m x  x1 , we get y  7  23 x  1  3y  21  2x  2  2x  3y  19  2x  3y  19  0.

28. Using the equation y  y1  m x  x1 , we get y  5   72 x  3  2y  10  7x  21  7x  2y  31  0.

5 61 y  y1   5. Substituting into y  y1  m x  x1 , we get  29. First we find the slope, which is m  2 x2  x1 12 1 y  6  5 x  1  y  6  5x  5  5x  y  11  0. 3  2 y  y1  55  1. Substituting into y  y1  m x  x1 , we get  30. First we find the slope, which is m  2 x2  x1 4  1 y  3  1 x  4  y  3  x  4  x  y  1  0. y  y1 6 1  5 31. We are given two points, 2 5 and 5 1. Thus, the slope is m  2   2. Substituting into  x2  x1 52 3 y  y1  m x  x1 , we get y  5  2 x  2  y  2x  9 or 2x  y  9  0. 77 y  y1  0. Substituting into  32. We are given two points, 1 7 and 4 7. Thus, the slope is m  2 x2  x1 41 y  y1  m x  x1 , we get y  7  0 x  1  y  7 or y  7  0. 3 3  0 y  y1   3. Using the y­intercept,  33. We are given two points, 1 0 and 0 3. Thus, the slope is m  2 x2  x1 01 1 we have y  3x  3  y  3x  3 or 3x  y  3  0. y  y1 60  68  34 . Using the y­intercept 34. We are given two points, 8 0 and 0 6. Thus, the slope is m  2  x2  x1 0  8 we have y  34 x  6  3x  4y  24  0.

35. Since the equation of a line with slope 0 passing through a b is y  b, the equation of this line is y  3.

36. Since the equation of a line with undefined slope passing through a b is x  a, the equation of this line is x  3.

37. Since the equation of a line with undefined slope passing through a b is x  a, the equation of this line is x  2. 38. Since the equation of a line with slope 0 passing through a b is y  b, the equation of this line is y  1.

39. Any line parallel to y  2x 8 has slope 2. The desired line passes through 1 4, so substituting into y y1  m x  x1 , we get y  4  2 [x  1]  y  2x  6 or 2x  y  6  0. 1  2. The desired line passes through 3 2, so substituting 40. Any line perpendicular to y   12 x  7 has slope  12 into y  y1  m x  x1 , we get y  2  2 [x  3]  y  2x  8 or 2x  y  8  0. 41. Since the equation of a horizontal line passing through a b is y  b, the equation of the horizontal line passing through 4 5 is y  5. 42. Any line parallel to the y­axis has undefined slope and an equation of the form x  a. Since the graph of the line passes through the point 4 5, the equation of the line is x  4. 43. Since 3x  2y  4  2y  3x  4  y   32 x  2, the slope of this line is  32 . Thus, the line we seek is given by y  4   32 x  3  y  4   32 x  92  y   32 x  17 2  3x  2y  17  0.

44. Since 2x  3y  4  0  3y  2x  4  y   23 x  43 , the slope of this line is m   23 . Substituting m   23 and b  6 into the slope­intercept formula, the line we seek is given by y   23 x  6  2x  3y  18  0.

45. Any line parallel to x  5 has undefined slope and an equation of the form x  a. Thus, an equation of the line is x  1.

46. Any line perpendicular to y  1 has undefined slope and an equation of the form x  a. Since the graph of the line passes through the point 2 6, an equation of the line is x  2.


SECTION 1.4 Lines

27

47. First find the slope of 3x  4y  7  0  4y  3x  7  y   34 x  74 . Thus, the slope of any line perpendicular 1 to 3x  4y  7  0 is m   34  43 . Since it passes through 2 1, the equation of the line we seek is

y  1  43 x  2  3y  3  4x  8  4x  3y  11  0.

48. First find the slope of the line 4x  8y  1. This gives 4x  8y  1  8y  4x  1  y  12 x  18 . So the slope of the     1  2. The equation of the line we seek is y   23  2 x  12 line that is perpendicular to 4x  8y  1 is m   12  y  23  2x  1  6x  3y  1  0.

49. First find the slope of the line passing through 2 5 and 2 1. This gives m  of the line we seek is y  7  1 x  1  x  y  6  0. 50. First find the slope of the line passing through 3 5 and 6 2: m 

4 15   1, and so the equation 2  2 4

25 3 1    , and so the slope of the line 6  3 9 3

1  3. Thus an equation of the line we seek is y  10  3 x  4  3x  y  22  0. that is perpendicular is m   13

51. (a)

52. (a)

y

(_2, 1)

y

1l 1

x 1l 1

(b) y  1  32 x  2  2y  2  3 x  2 

b=_1 b=_3

8

b=_6

4 _4

_2

0

2x  y  7  0.

54.

_4

b=6

_8

b=3 b=1

b=0

y  2x  b, b  0, 1, 3, 6. They have the same

slope, so they are parallel.

_2

0 _4

4

3

m= 2

4

_4 2

x

(b) y  1  2 x  4  y  1  2x  8 

2y  2  3x  6  3x  2y  8  0.

53.

(4, _1)

_8

2

4

m= 34 m= 41

m=0 1 m=_ 4 3

m=_ 4 3

m=_ 2

y  mx  3, m  0,  14 ,  34 ,  32 . Each of the lines

contains the point 0 3 because the point 0 3

satisfies each equation y  mx  3. Since 0 3 is on the y­axis, they all have the same y­intercept.


28

CHAPTER 1 Equations and Graphs 3

m= 2

55. 4 2 _2

0 _2

m=6 y

56. m= 34

4

6

8

m=0

m=1

10

m= 41 2

m=2 1

m= 2

1

m=_ 4

_12

_8

_4

x m=0 1 4 m=_ 2

0

m=_1

3

m=_ 4

_4

_10 m=_6

3

m=_ 2

y  m x  3, m  0,  14 ,  34 ,  32 . Each of the lines

m=_2

y  2  m x  3, m  0,  12 , 1, 2, 6. Each of the

contains the point 3 0 because the point 3 0 satisfies

lines contains the point 3 2 because the point 3 2

each equation y  m x  3. Since 3 0 is on the x­axis,

satisfies each equation y  2  m x  3.

we could also say that they all have the same x­intercept.

57. y  x  4. So the slope is 1 and the y­intercept is 4. y

58. y   12 x  1. So the slope is  12 and the y­intercept is 1.

y

1 1

x

1 1

59. 2x  y  7  y  2x  7. So the slope is 2 and the y­intercept is 7.

y

x

60. 2x  5y  0  5y  2x  y  25 x. So the slope is 2 and the y­intercept is 0. 5

y

1l 1 1 1

x

x


SECTION 1.4 Lines

61. 4x  5y  10  5y  4x  10  y   45 x  2. So

62. 3x  4y  12  4y  3x  12  y  34 x  3. So

y

y

the slope is  45 and the y­intercept is 2.

29

the slope is 34 and the y­intercept is 3.

1l

1l 1

1

x

x

63. y  4 can also be expressed as y  0x  4. So the slope is 64. x  5 cannot be expressed in the form y  mx  b. So 0 and the y­intercept is 4.

the slope is undefined, and there is no y­intercept. This is a vertical line.

y

y

1l 1

x

1l 1

x

65. x  3 cannot be expressed in the form y  mx  b. So the 66. y  2 can also be expressed as y  0x  2. So the slope slope is undefined, and there is no y­intercept. This is a vertical line.

is 0 and the y­intercept is 2. y

y

1l 1

1l 1

x

x


30

CHAPTER 1 Equations and Graphs

67. 3x  2y  6  0. To find x­intercepts, we set y  0 and solve for x: 3x  2 0  6  0  3x  6  x  2, so the x­intercept is 2. To find y­intercepts, we set x  0 and solve for y: 3 0  2y  6  0  2y  6  y  3, so the y­intercept is 3. y

68. 6x  7y  42  0. To find x­intercepts, we set y  0 and solve for x: 6x  7 0  42  0  6x  42  x  7, so the x­intercept is 7. To find y­intercepts, we set x  0 and solve for y: 6 0  7y  42  0  7y  42  y  6, so the y­intercept is 6. y

1

2 1

2

x

69. 12 x  13 y  1  0. To find x­intercepts, we set y  0 and

solve for x: 12 x  13 0  1  0  12 x  1  x  2,

x

70. 13 x  15 y  2  0. To find x­intercepts, we set y  0 and solve for x: 13 x  15 0  2  0  13 x  2  x  6, so

so the x­intercept is 2. To find y­intercepts, we set x  0 and solve for y:

the x­intercept is 6. To find y­intercepts, we set x  0 and solve for y:

y­intercept is 3.

y­intercept is 10.

1 0  1 y  1  0  1 y  1  y  3, so the 2 3 3 y

1 0  1 y  2  0  1 y  2  y  10, so the 3 5 5

1

y

2 1

2

x

71. y  6x  4. To find x­intercepts, we set y  0 and solve for x: 0  6x  4  6x  4  x   23 , so the x­intercept is  23 . To find y­intercepts, we set x  0 and solve for y: y  6 0  4  4, so the y­intercept is 4. y

x

72. y  4x  10. To find x­intercepts, we set y  0 and

solve for x: 0  4x  10  4x  10  x   52 , so the x­intercept is  52 .

To find y­intercepts, we set x  0 and solve for y: y  4 0  10  10, so the y­intercept is 10. y

2 1

1 1

x

x


31

SECTION 1.4 Lines

73. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation y  2x  3 has slope 2. The line with equation 2y  4x  5  0  2y  4x  5  y  2x  52 also has slope 2, and so the lines are parallel.

74. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation y  12 x  4 has slope 12 .

1 , and so the lines are neither The line with equation 2x  4y  1  4y  2x  1  y   12 x  14 has slope  12   12

parallel nor perpendicular.

75. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation 2x  5y  8 

5y  2x  8  y  25 x  85 has slope 25 . The line with equation 10x  4y  1  4y  10x  1  y   52 x  14 has 1 , and so the lines are perpendicular. slope  52   25

76. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation 15x  9y  2 

9y  15x  2  y  53 x  29 has slope 53 . The line with equation 3y  5x  5  3y  5x  5  y  53 x  53 has slope 5 , and so the lines are parallel. 3

77. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation 7x  3y  2 

3y  7x  2  y  73 x  23 has slope 73 . The line with equation 9y  21x  1  9y  21x  1  y   73  19 has 1 , and so the lines are neither parallel nor perpendicular. slope  73   73

78. To determine if the lines are parallel or perpendicular, we find their slopes. The line with equation 6y  2x  5 

6y  2x  5  y  13 x  56 has slope 13 . The line with equation 2y  6x  1  2y  6x  1  y  3x  12 has 1 , and so the lines are perpendicular. slope 3   13

79. We first plot the points to find the pairs of points that determine each side. Next we find the slopes of opposite sides. The slope of AB is slope of DC is

y

41 3 1   , and the 71 6 2

C

10  7 3 1   . Since these slope are equal, these two sides 5  1 6 2

are parallel. The slope of AD is

D

6 71   3, and the slope of BC is 1  1 2

B

6 10  4   3. Since these slope are equal, these two sides are parallel. 57 2 Hence ABC D is a parallelogram.

1l

A

x

1

80. We first plot the points to determine the perpendicular sides. Next find the slopes of 4 2 3  1   , and the slope of AC is the sides. The slope of AB is 3  3 6 3

y

C

9 3 8  1    . Since 9  3 6 2

   slope of AB  slope of AC  23  32  1 the sides are perpendicular, and ABC is a right triangle.

B

1l

A

1

x


32

CHAPTER 1 Equations and Graphs

81. We first plot the points to find the pairs of points that determine each side. Next we 2 1 31   and the find the slopes of opposite sides. The slope of AB is 11  1 10 5 2 1 68   . Since these slope are equal, these two sides slope of DC is 0  10 10 5 61 5 are parallel. Slope of AD is   5, and the slope of BC is 01 1 5 38   5. Since these slope are equal, these two sides are parallel. 11  10 1

y

C D

1l

B A

1

x

Since slope of AB  slope of AD  15  5  1, the first two sides are each perpendicular to the second two sides. So the sides form a rectangle.

82. (a) The slope of the line passing through 1 1 and 3 9 is

8 91   4. The slope of the line passing through 1 1 31 2

21  1 20   4. Since the slopes are equal, the points are collinear. 61 5 4 73   2. The slope of the line passing through (b) The slope of the line passing through 1 3 and 1 7 is 1  1 2 12 15  3  . Since the slopes are not equal, the points are not collinear. 1 3 and 4 15 is 4  1 5 and 6 21 is

83. We need the slope and the midpoint of the line AB. The midpoint of AB is

17 42  2 2

 4 1, and the slope of

2  4 6 1 1   1. The slope of the perpendicular bisector will have slope   1. Using the 71 6 m 1 point­slope form, the equation of the perpendicular bisector is y  1  1 x  4 or x  y  3  0. AB is m 

84. We find the intercepts (the length of the sides). When x  0, we have 2y  3 0  6  0  2y  6  y  3, and when y  0, we have 2 0  3x  6  0  3x  6  x  2. Thus, the area of the triangle is 12 3 2  3.

85. (a) We start with the two points a 0 and 0 b. The slope of the line that contains them is

b b0   . So the equation 0a a

b of the line containing them is y   x  b (using the slope­intercept form). Dividing by b (since b  0) gives a y x x y    1    1. b a a b x y (b) Setting a  6 and b  8, we get   1  4x  3y  24  4x  3y  24  0. 6 8

86. (a) The line tangent at 3 4 will be perpendicular to the line passing through the points 0 0 and 3 4. The slope of 4  0 4 1 3 this line is   . Thus, the slope of the tangent line will be   . Then the equation of the tangent 30 3 4 43 line is y  4  34 x  3  4 y  4  3 x  3  3x  4y  25  0.

(b) Since diametrically opposite points on the circle have parallel tangent lines, the other point is 3 4.


SECTION 1.4 Lines

33

87. Using the diagram provided, we calculate the coordinates of the midpoints: m 1 has coordinates         ca 0b 0a 0b    12 a 12 b and m 2 has coordinates  12 a  c  12 b . 2 2 2 2

1b  1b 2  0, and so it is parallel to the third side (a segment of the The line joining the midpoints thus has slope 1 2  c  12 a a 2 x­axis).      1 a  c  1 a 2  1 b  1 b 2  1 c2  1 c, half the length of the third The line joining the midpoints has length 2 2 2 2 4 2

side of the triangle; and so we have verified both conclusions of the theorem.

88. (a) The slope represents the increase in the average surface temperature in  C per year. The T ­intercept is the average surface temperature in 1950, or 15 C. (b) In 2050, t  2050  1950  100, so T  002100  15  17 degrees Celsius.

89. (a) The slope is 00417D  00417 200  834. It represents the increase in dosage for each one­year increase in the child’s age. (b) When a  0, c  834 0  1  834 mg. 90. (a)

(b) The slope, 4, represents the decline in number of spaces sold for

y

each $1 increase in rent. The y­intercept is the number of spaces at the flea market, 200, and the x­intercept is the cost per space when the

200l

manager rents no spaces, $50. 100l

20

91. (a)

40

60

80 100

x

(b) The slope is the cost per toaster oven, $6. The y­intercept, $3000, is

y

the monthly fixed cost—the cost that is incurred no matter how many toaster ovens are produced.

10,000l

5000l

500

1000

20

10

x

(b) Substituting a for both F and C, we have

92. (a) C F

30 22

4

14

0

10

20

30

32

50

68

86

a  95 a  32   45 a  32 

a  40 . Thus both scales agree at

40 .

t  t1 80  70 10 5 93. (a) Using n in place of x and t in place of y, we find that the slope is 2    . So the linear n2  n1 168  120 48 24 5 n  168  t  80  5 n  35  t  5 n  45. equation is t  80  24 24 24

5 150  45  7625 F  76 F. (b) When n  150, the temperature is approximately given by t  24


34

CHAPTER 1 Equations and Graphs

94. (a) Using t in place of x and V in place of y, we find the slope of the line

(b)

y

using the points 0 4000 and 4 200. Thus, the slope is

4000l

3800 200  4000   950. Using the V ­intercept, the m 40 4 linear equation is V  950t  4000.

3000l 2000l

(c) The slope represents a decrease in the value of the computer of $950 each year. The V ­intercept is the original price of the computer.

1000l

(d) When t  3, the value of the computer is given by

1

V  950 3  4000  1150.

95. (a) We are given

434 change in pressure   0434. Using P for 10 feet change in depth 10

pressure and d for depth, and using the point P  15 when d  0, we have P  15  0434 d  0  P  0434d  15.

(c) The slope represents the increase in pressure per foot of descent. The y­intercept represents the pressure at the surface. (d) When P  100, then 100  0434d  15  0434d  85  d  1959 ft. Thus the pressure is 100 lb/in3 at a depth of

approximately 196 ft.

(b)

2

3

4

x

y 60l 50l 40l 30l 20l 10l 10

20

30

40

50

60 x

96. Slope is the rate of change of one variable per unit change in another variable. So if the slope is positive, then the temperature is rising. Likewise, if the slope is negative then the temperature is decreasing. If the slope is 0, then the temperature is not changing. 97. We label the three points A, B, and C. If the slope of the line segment AB is equal to the slope of the line segment BC, then the points A, B, and C are collinear. Using the distance formula, we find the distance between A and B, between B and C, and between A and C. If the sum of the two smaller distances equals the largest distance, the points A, B, and C are collinear. Another method: Find an equation for the line through A and B. Then check if C satisfies the equation. If so, the points are collinear.

1.5

SOLVING QUADRATIC EQUATIONS

1. (a) The Quadratic Formula states that x 

b 

 b2  4ac . 2a

(b) In the equation 12 x 2  x  4  0, a  12 , b  1, and c  4. So, the solution of the equation is     1  12  4 12 4 13   x  2 or 4.  1 1 2 2

2. (a) To solve the equation x 2  4x  5  0 by factoring, we write x 2  4x  5  x  5 x  1  0 and use the Zero­Product Property to get x  5 or x  1.  2 (b) To solve by completing the square, we add 5 to both sides to get x 2  4x  5, and then add  42 to both sides to get x 2  4x  4  5  4  x  22  9  x  2  3  x  5 or x  1.


SECTION 1.5 Solving Quadratic Equations

35

(c) To solve using the Quadratic Formula, we substitute a  1, b  4, and c  5, obtaining    4  42  4 1 5 4  36   2  3  x  5 or x  1. x 2 1 2 3. For the quadratic equation ax 2  bx  c  0 the discriminant is D  b2  4ac. If D  0, the equation has two real solutions; if D  0, the equation has one real solution; and if D  0, the equation has no real solution. 4. There are many possibilities. For example, x 2  1 has two solutions, x 2  0 has one solution, and x 2  1 has no solution.

5. x 2  13x  30  0  x  15 x  2  0  x  15  0 or x  2  0. Thus, x  15 or x  2. 6. x 2  13x  30  0  x  10 x  3  0  x  10  0 or x  3  0. Thus, x  3 or x  10.

7. x 2  x  6  x 2  x  6  0  x  2 x  3  0  x  2  0 or x  3  0. Thus, x  2 or x  3.

8. x 2  4x  21  x 2  4x  21  0  x  3 x  7  0  x  3  0 or x  7  0. Thus, x  3 or x  7.

9. 5x 2  9x  2  0  5x  1 x  2  0  5x  1  0 or x  2  0. Thus, x   15 or x  2.

10. 6x 2  x  12  0  3x  4 2x  3  0  3x  4  0 or 2x  3  0. Thus, x   43 or x  32 .

11. 2s 2  5s  3  2s 2  5s  3  0  2s  1 s  3  0  2s  1  0 or s  3  0. Thus, s   12 or s  3.

12. 4y 2  9y  28  4y 2  9y  28  0  4y  7 y  4  0  4y  7  0 or y  4  0. Thus, y   74 or y  4.

13. 12z 2  44z  45  12z 2  44z  45  0  6z  5 2z  9  0  6z  5  0 or 2z  9  0. Thus, z   56 or z  92 .

14. 42  4  3  42  4  3  0  2  1 2  3  0  2  1  0 or 2  3  0. If 2  1  0, then    12 ; if 2  3  0, then   32 .

15. x 2  5 x  100  x 2  5x  500  x 2  5x  500  0  x  25 x  20  0  x  25  0 or x  20  0. Thus, x  25 or x  20.

16. 6x x  1  21  x  6x 2  6x  21  x  6x 2  5x  21  0  2x  3 3x  7  0  2x  3  0 or 3x  7  0. If 2x  3  0, then x   32 ; if 3x  7  0, then x  73 .

  17. x 2  10x  2  0  x 2  10x  2  x 2  10x  25  2  25  x  52  23  x  5   23  x  5  23   18. x 2  6x  2  0  x 2  6x  2  x 2  6x  9  2  9  x  32  11  x  3   11  x  3  11.   19. x 2  6x  11  0  x 2  6x  11  x 2  6x  9  11  9  x  32  20  x  3  2 5  x  3  2 5. 2  3 3 20. x 2  3x  74  0  x 2  3x  74  x 2  3x  94  74  94  x  32  16 4  4  x  2  2  x   2  2  x  12 or x   72 .

2  3  x 2  x  14  34  14  x  12  1  x  12  1  x   12  1. So 4 x   12  1   32 or x   12  1  12 .     25  x  5 2  21  x  5   21   21  22. x 2  5x  1  0  x 2  5x  1  x 2  5x  25  1  4 4 2 4 2 4 2 21. x 2  x  34  0  x 2  x 

x  52  221 . 23. x 2  22x  21  0  x 2  22x  21  x 2  22x  112  21  112  21  121  x  112  100  x  11  10  x  11  10. Thus, x  1 or x  21.

24. x 2  18x  19  x 2  18x  92  19  92  19  81  x  92  100  x  9  10  x  9  10, so x  1 or x  19.       25. 5x 2  10x  1  5 x 2  2x  1  5 x 2  2x  1  1  5  x  12  65  x  1   65  x  1  530 26. 2x 2  16x  5  0  x 2  8x  52  0  x 2  8x   52  x 2  8x  16   52  16  x  42  27 2    3 6 x  4   27 2  x  4  2 .


36

CHAPTER 1 Equations and Graphs

  49 7 2  17  27. 2x 2  7x  4  0  x 2  72 x  2  0  x 2  72 x  2  x 2  72 x  49 16  2  16  x  4 16   17 7 x  74   17 16  x   4  4 .    25  x  5 2  153  x  5   153 28. 4x 2  5x  8  0  x 2  54 x  2  0  x 2  54 x  2  x 2  54 x  25  2  64 64 8 64 8 64 

 x   58  3 817 .

29. x 2  8x  12  0  x  2 x  6  0  x  2 or x  6.

30. x 2  3x  18  0  x  3 x  6  0  x  3 or x  6.

31. x 2  8x  20  0  x  10 x  2  0  x  10 or x  2.

32. 10x 2  9x  7  0  5x  7 2x  1  0  x   75 or x  12 .

33. 2x 2  x  3  0  x  1 2x  3  0  x  1  0 or 2x  3  0. If x  1  0, then x  1; if 2x  3  0, then x   32 .

34. 3x 2  7x  4  0  3x  4 x  1  0  3x  4  0 or x  1  0. Thus, x   43 or x  1.

 35. 3x 2  6x  5  0  x 2  2x  53  0  x 2  2x  53  x 2  2x  1  53  1  x  12  83  x  1   83  

x  1  2 3 6 .

36. x 2  6x  1  0       2   6 62  4 1 1  b  b  4ac 6  36  4 6  32 64 2 x      3  2 2. 2a 2 1 2 2 2 37. x 2  43 x  49  0  9x 2  12x  4  0  3x  22  0  x  23 .     b  b2  4ac 6  62  4 4 1 6  52 3  13 1 2 2    . 38. 2x 3x  2  0  4x 6x 1  0  x  2a 2 4 8 4 39. 4x 2  16x  9  4x 2  16x  9  0  2x  1 2x  9  0  2x  1  0 or 2x  9  0. If 2x  1  0, then x  12 ; if 2x  9  0, then x   92 .

40. 0  x 2  4x  1  0       2   4 42  4 1 1  b  b  4ac 4  16  4 4  12 42 3 x      2  3. 2a 2 1 2 2 2 41. 4x  x 2  1  x 2  4x  1  0        4  42  4 1 1  b  b2  4ac 4  16  4 4  12 42 3 x     2 3 2a 2 1 2 2 2     2   5 52  4 1 3 5  25  12 5  13 b  b  4ac    . 42. 3  5z  z 2  0  z  2a 2 1 2 2    2  4ac  2  22  4 7 4 2  4  112 b  b 2 2   , so 43. 7x  2x  4  7x  2x  4  0  x  2a 2 7 4 there is no real solution.    3  9  4 1 5 3  29 b  b2  4ac 44. z z  3  5  z 2  3z  5  0  z    . 2a 2 2     2  4ac  2  22  4 3 2 2  4  24 2  20 b  b 45. 3x 2  2x  2  0  x     . Since the 2a 2 3 6 6 discriminant is less than 0, the equation has no real solution.


SECTION 1.5 Solving Quadratic Equations

  7  72  4 5 5

 b2  4ac  2a 2 5 Since the discriminant is less than 0, the equation has no real solution.

b  46. 5x 2  7x  5  x 

7

37

  7  51 49  100  . 10 10

47. x 2  0011x  0064  0     0011  00112  4 1 0064 0011  0000121  0256 0011  0506 x  .  2 1 2 2 0011  0506 0011  0506 Thus, x   0259 or x   0248. 2 2 48. x 2  2450x  1500  0      2450  24502  4 1 1500 2450  60025  6 2450  00025 2450  0050 x    . Thus, 2 1 2 2 2 2450  0050 2450  0050 x  1250 or x   1200. 2 2 49. x 2  2450x  1501  0      2450  24502  4 1 1501 2450  60025  6004 2450  00015 x   . 2 1 2 2 Thus, there is no real solution. 50. x 2  1800x  0810  0      1800  18002  4 1 0810 1800  324  324 1800  0 x     0900. Thus the only 2 1 2 2 solution is x  0900. 51. h 

1 gt 2   t  1 gt 2   t  h 0 0 2 2

 0. Using the Quadratic Formula,       0    0 2  4 12 g h  0   02  2gh   t .  g 2 12 g

n n  1  2S  n 2  n  n 2  n  2S  0. 2   1  12  4 1 2S 1  1  8S n  . 2 1 2

52. S 

Using the Quadratic Formula,

53. A  2x 2  4xh  2x 2  4xh  A  0. Using the Quadratic Formula,       4h  4 4h 2  2A  4h  4h2  4 2 A 4h  16h 2  8A 4h  2 4h 2  2A    x  2 2 4 4 4     2 2h  4h 2  2A 2h  4h 2  2A   4 2 54. A  2r 2  2r h  2r 2  2r h  A  0. Using the Quadratic Formula,     2h  2h2  4 2 A 2h  4 2 h 2  8 A h   2 h 2  2 A   . r 2 2 4 2


38

CHAPTER 1 Equations and Graphs

1 1 1    c s  b  c s  a  s  a s  b  cs  bc  cs  ac  s 2  as  bs  ab  sa s b c s 2  a  b  2c s  ab  ac  bc  0. Using the Quadratic Formula,   a  b  2c  a  b  2c2  4 1 ab  ac  bc s  2 1   a  b  2c  a 2  b2  4c2  2ab  4ac  4bc  4ab  4ac  4bc  2  2 2 2  a  b  2c  a  b  4c  2ab  2     2 4 2 1 1 4  2  r 2 1  r   r 2 1  r  2  r 1  r  2r 2  4 1  r  r  r 2  2r 2  4  4r 56.  r 1r r 1r r r     5  52  4 1 4 5  25  16 5  41   .  r 2  5r  4  0. Using the Quadratic Formula, r  2 1 2 2 55.

57. D  b2  4ac  62  4 1 1  32. Since D is positive, this equation has two real solutions.

58. x 2  6x  9  x 2  6x  9, so D  b2  4ac  62  4 1 9  36  36  0. Since D  0, this equation has one real solution.

59. D  b2  4ac  2202  4 1 121  484  484  0. Since D  0, this equation has one real solution.

60. D  b2  4ac  2212  4 1 121  48841  484  00441. Since D  0, this equation has two real solutions.   61. D  b2  4ac  52  4 4 13 8  25  26  1. Since D is negative, this equation has no real solution.

62. D  b2  4ac  r2  4 1 s  r 2  4s. Since D is positive, this equation has two real solutions. 1 63. a 2 x 2  2ax  1  0  ax  12  0  ax  1  0. So ax  1  0 then ax  1  x   . a 64. ax 2  2a  1 x  a  1  0  [ax  a  1] x  1  0  ax  a  1  0 or x  1  0. If ax  a  1  0, a1 ; if x  1  0, then x  1. then x  a 65. We want to find the values of k that make the discriminant 0. Thus k 2  4 4 25  0  k 2  400  k  20

66. We want to find the values of k that make the discriminant 0. Thus D  362  4 k k  0  4k 2  362  2k  36  k  18.

67. Let n be one number. Then the other number must be 55  nsince n  55  n  55. Because

the product is 684, we have n 55  n  684  55n  n 2  684  n 2  55n  684  0 

   55 552 41684 5517  72  36 or n   55 30252736  552 289  5517 21 2 2 . So n  2 2 5517 38 n  2  2  19. In either case, the two numbers are 19 and 36. 68. Let n be one even number. Then the next even number is n  2. Thus we get the equation n 2  n  22  1252 

  n 2  n 2  4n  4  1252  0  2n 2  4n  1248  2 n 2  2n  624  2 n  24 n  26. So n  24 or n  26. Thus the consecutive even integers are 24 and 26 or 26 and 24.

69. Let  be the width of the garden in feet. Then the length is   30. Thus 2800     30  2  30  2800  0    70   40  0. If   70  0, then   70, which is impossible. Therefore   40  0, and so   40. The garden is 40 feet wide and 70 feet long. 70. Let  be the width of the bedroom. Then its length is   5. Since area is length times width, we have

234    5   2  5  2  5  234  0    18   13  0    18 or   13. Since the width must be positive, the width is 13 feet.


SECTION 1.5 Solving Quadratic Equations

39

71. Let  be the width of the garden in feet. We use the perimeter to express the length l of the garden in terms of width. Since the perimeter is twice the width plus twice the length, we have 200  2  2l  2l  200  2  l  100  . Using the formula for area, we have 2400   100    100  2  2  100  2400  0    40   60  0. So   40  0    40, or   60  0    60. If   40, then l  100  40  60. And if   60, then l  100  60  40. So the length is 60 feet and the width is 40 feet.

72. Let  be the width of the lot in feet. Then the length is   8. Using the Pythagorean Theorem, we have

2    82  2322  2  2  16  64  53824  22  16  53670  0  2  8  26880  0    168   160  0. Since widths must be positive, the width is 160 feet and the length is 168 feet.

73. First we write a formula for the area of the figure in terms of x. Region A has dimensions 14 in. and x in. and region B has dimensions 13  x in. and x in. So

the area of the figure is 14  x  [13  x x]  14x  13x  x 2  x 2  27x. We

are given that this is equal to 160 in2 , so 160  x 2  27x  x 2  27x  160  0  x  32 x  5  x  32 or x  5. But x must be positive, so x  5 in.

x A

14 in. 13 in. x

B

74. The shaded area is the sum of the area of a rectangle and the area of a triangle. So A  y 1  12 y y  12 y 2  y. We

are given that the area is 1200 cm2 , so 1200  12 y 2  y  y 2  2y  2400  0  y  50 y  48  0. y is positive, so y  48 cm.

 2  2 75. Let h be the height the ladder reaches (in feet). Using the Pythagorean Theorem we have 7 12  h 2  19 12   2  2  2  2  15 225 1296  h 2  39  h 2  39  15  1521 2 4 4 2 4  4  4  324. So h  324  18 feet. 76. Let h be the height of the flagpole, in feet. Then the length of each guy wire is h  5. Since the distance between the points where the wires are fixed to the ground is equal to one guy wire, the triangle is equilateral, and the flagpole is the perpendicular bisector of the base. Thus from the Pythagorean Theorem, we get   1 h  5 2  h 2  h  52  h 2  10h  25  4h 2  4h 2  40h  100  h 2  30h  75  0  2

     30 302 4175 3 . Since h  3020 3  0, we reject it. Thus  30 900300  30 2 1200  3020 h 2 2 2 21   3  15  10 3  3232 ft  32 ft 4 in. the height is h  3020 2

77. Let t be the time in hours that it takes you to wash all the windows. Then it takes your roommate t  32 hours to wash all the windows, and the sum of the fractions of the job you can do individually per hour equals the fraction 1 1 1 4 9  9  of the job you can do together. Since 1 hour 48 minutes  1  48 60  1  5  5 , we have t  t  3 2

5

2 1   59  9 2t  3  2 9t  5t 2t  3  18t  27  18t  10t 2  15t  10t 2  21t  27  0 t 2t  3    21  212  4 10 27 21  441  1080 21  39 21  39 9 t    . So t    2 10 20 20 20 10 21  39 or t   3. Since t  0 is impossible, you can wash the windows alone in 3 hours, and it takes your roommate 20 3  32  4 12 hours.


40

CHAPTER 1 Equations and Graphs

78. Let t be the time, in hours, it takes your manager to deliver all the flyers alone. Then it takes the assistant t  1 1 1 1 1  hours to deliver all the flyers alone, and it takes the group 04t hours to do it together. Thus   4 t t 1 04t 1 1 1 4t  04t  04t   10  t t  1  4 t  1  4t  10 t  1  04t  1  t  4  4 t t 1 t 1 t 2  t  4t  4  4t  10t  10  t 2  t  6  0  t  3 t  2  0. So t  3 or t  2. Since t  2 is impossible, it takes your manager 3 hours to deliver all the flyers alone.

79. Let x be the rate, in mi/h, at which the salesman drove between Ajax and Barrington. Cities

Distance

Rate

Ajax  Barrington

120

x

Barrington  Collins

150

x  10

Time 120 x 150 x  10

distance to fill in the “Time” column of the table. Since the second part of the trip rate 1 hour) more than the first, we can use the time column to get the equation 120  1  150  took 6 minutes (or 10 x 10 x  10 120 10 x  10  x x  10  150 10x  1200x  12,000  x 2  10x  1500x  x 2  290x  12,000  0  We have used the equation time 

   290 2902 4112,000 290 84,10048,000 290 36,100    290190  145  95. Hence, the salesman x 2 2 2 2

drove either 50 mi/h or 240 mi/h between Ajax and Barrington. (The first choice seems more likely!)

80. Let x be the rate, in mi/h, at which the trucker drove from Tortula to Cactus. Cities

Distance

Rate

Tortula  Cactus

250

x

Cactus  Dry Junction

360

x  10

Time 250 x 360 x  10

distance to fill in the time column of the table. We are given that the sum of rate 360 250   11  250 x  10  360x  the times is 11 hours. Thus we get the equation x x  10 11x x  10  250x  2500  360x  11x 2  110x  11x 2  500x  2500  0      500  5002  4 11 2500 500  250,000  110,000 500  360,000 500  600    . Hence, x 2 11 22 22 22 the trucker drove either 454 mi/h (impossible) or 50 mi/h between Tortula and Cactus. We have used time 


SECTION 1.5 Solving Quadratic Equations

41

81. Let r be the rowing rate in km/h of the crew in still water. Then their rate upstream was r  3 km/h, and their rate downstream was r  3 km/h. Distance

Rate

Upstream

6

r 3

Downstream

6

r 3

Time 6 r 3 6 r 3

Since the time to row upstream plus the time to row downstream was 2 hours 40 minutes  83 hour, we get the equation

6 8 6    6 3 r  3  6 3 r  3  8 r  3 r  3  18r  54  18r  54  8r 2  72  r 3 r 3 3   0  8r 2  36r  72  4 2r 2  9r  18  4 2r  3 r  6. Since 2r  3  0  r   32 is impossible, the solution is

r  6  0  r  6. So the rate of the rowing crew in still water is 6 km/h.

82. Let r be the speed of the southbound boat. Then r  3 is the speed of the eastbound boat. In two hours the southbound boat has traveled 2r miles and the eastbound boat has traveled 2 r  3  2r  6 miles. Since they are traveling is directions with are 90 apart, we can use the Pythagorean Theorem to get 2r 2  2r  62  302  4r 2  4r 2  24r  36  900    8r 2  24r  864  0  8 r 2  3r  108  0  8 r  12 r  9  0. So r  12 or r  9. Since speed is positive, the speed of the southbound boat is 9 mi/h.

83. Let x be the length of one side of the cardboard, so we start with a piece of cardboard x by x. When 4 inches are removed from each side, the base of the box is x  8 by x  8. Since the volume is 100 in3 , we get 4 x  82  100 

x 2  16x  64  25  x 2  16x  39  0  x  3 x  13  0So x  3 or x  13. But x  3 is not possible, since then the length of the base would be 3  8  5 and all lengths must be positive. Thus x  13, and the piece of cardboard is 13 inches by 13 inches. 84. Let r be the radius of the can. Now using the formula V  r 2 h with V  40 cm3 and h  10, we solve for r. Thus 40  r 2 10  4  r 2  r  2. Since r represents radius, r  0. Thus r  2, and the diameter is 4 cm.

85. Using h 0  288, we solve 0  16t 2  288, for t  0. So 0  16t 2  288  16t 2  288  t 2  18     t   18  3 2. Thus it takes 3 2  424 seconds for the ball the hit the ground.

86. (a) Using h 0  96, half the distance is 48, so we solve the equation 48  16t 2  96  48  16t 2  3  t 2    t   3. Since t  0, it takes 3  1732 s.  (b) The ball hits the ground when h  0, so we solve the equation 0  16t 2  96  16t 2  96  t 2  6  t   6.  Since t  0, it takes 6  2449 s. 87. We are given  o  40 ft/s.

  (a) Setting h  24, we have 24  16t 2  40t  16t 2  40t  24  0  8 2t 2  5t  3  0  8 2t  3 t  1  0

 t  1 or t  1 12 Therefore, the ball reaches 24 feet in 1 second (ascending) and again after 1 12 seconds (descending).

(b) Setting h  48, we have 48  16t 2  40t  16t 2  40t  48  0  2t 2  5t  6  0    5  23 5  25  48  . However, since the discriminant D  0, there are no real solutions, and hence the t 4 4 ball never reaches a height of 48 feet.

(c) The greatest height h is reached only once. So h  16t 2  40t  16t 2  40t  h  0 has only one solution. Thus D  402  4 16 h  0  1600  64h  0  h  25. So the greatest height reached by the ball is 25 feet.

(d) Setting h  25, we have 25  16t 2  40t  16t 2  40t  25  0  4t  52  0  t  1 14 . Thus the ball reaches the highest point of its path after 1 14 seconds.


42

CHAPTER 1 Equations and Graphs

(e) Setting h  0 (ground level), we have 0  16t 2  40t  2t 2  5t  0  t 2t  5  0  t  0 (start) or t  2 12 . So the ball hits the ground in 2 12 s.

88. If the maximum height is 100 feet, then the discriminant of the equation 16t 2   0 t  100  0 must equal zero. So

0  b2  4ac   0 2  4 16 100   02  6400   0  80. Since  0  80 does not make sense, we must have  0  80 ft/s.

89. Let x be the distance from the center of the earth to the dead spot (in thousands of miles). Now setting F  0, we have 0012K K 0012K K  2   K 239  x2  0012K x 2  57121  478x  x 2  0012x 2  0 2  x x 239  x2 239  x2

0988x 2  478x  57121  0. Using the Quadratic Formula, we obtain    478  4782  4 0988 57,121 478  228,484  225,742192  x  2 0988 1976  478  52362 478  2741808   241903  26499.  1976 1976 So either x  241903  26499  268 or x  241903  26499  215. Since 268,000 is greater than the distance from the earth to the moon, we reject the first root; thus x  215,000 miles.

90. Let y be the circumference of the circle, so 360  y is the perimeter of the square. Use the circumference to find the  2 radius, r, in terms of y: y  2r  r  y 2. Thus the area of the circle is  y 2  y 2  4. Now if the  2 perimeter of the square is 360  y, the length of each side is 14 360  y  and the area of the square is 14 360  y .  2      Setting these areas equal, we obtain y 2  4  14 360  y  y 2   14 360  y  2y  360   y          2   y  360 . Therefore, y  360  2    1691. Thus one wire is 1691 in. long and the other is 1909 in. long. 91. Let x equal the original length of the reed in cubits. Then x  1 is the piece that fits 60 times along the length of the field, that is, the length is 60 x  1. The width is 30x. Then converting cubits to ninda, we have

2 2 375  60 x  1  30x  12  25 2 x x  1  30  x  x  x  x  30  0  x  6 x  5  0. So x  6 or 12 x  5. Since x must be positive, the original length of the reed is 6 cubits.

92. (a) x 2  9x  20  0  x  4 x  5  0  x  4 or x  5. The product of the solutions is 4  5  20, the constant term in the original equation; and their sum is 4  5  9, the negative of the coefficient of x in the original equation.   b  b2  4c b  b2  4c 2 and r2  . (b) In general, the equation x  bx  c  0 has solutions r1  2 2      2   1 b  b2  4c b  b2  4c  14 b2  b2  4c  c and   b2  4c Thus, r1r2  b2  2 2 4   2b b  b2  4c b  b2  4c    b. r1  r2  2 2 2     b b 2 b 93. (a) We make the substitution x  u  : x 2  bx  c  0  u  c  0 b u 2 2 2   b2 b2 b2  bu   c  0  u2   c  0. u 2  bu  4 2 4   49 2 (b) Here b  5 and c  6, so the substitution x  u  52 results in the equation u 2  25 4 6  0u  4  u   72  x   72  52  6 or x  72  52  1.


SECTION 1.6 Complex Numbers

1.6

43

COMPLEX NUMBERS

1. The imaginary number i has the property that i 2  1.

2. For the complex number 3  4i the real part is 3 and the imaginary part is 4.

3. (a) The complex conjugate of 3  4i is 3  4i  3  4i.   (b) 3  4i 3  4i  32  42  25

4. If 3  4i is a solution of a quadratic equation with real coefficients, then 3  4i  3  4i is also a solution of the equation. 5. Yes, every real number a is a complex number of the form a  0i.

6. Yes. For any complex number z, z  z  a  bi  a  bi  a  bi  a  bi  2a, which is a real number. 7. 3  8i: real part 3, imaginary part 8.

8.  5  i  5  i: real part 5, imaginary part 1.

9.

10.

2  5i   23  53 i: real part  23 , imaginary part  53 . 3

4  7i  2  72 i: real part 2, imaginary part 72 . 2

13.  23 i: real part 0, imaginary part  23 .     15. 3  4  3  2i: real part 3, imaginary part 2.

12.  12 : real part  12 , imaginary part 0.   14. 3i: real part 0, imaginary part 3.    16. 2  5  2  i 5: real part 2, imaginary part  5.

17. 3  2i  5i  3  2  5 i  3  7i

18. 3i  2  3i  2  [3  3] i  2  6i

19. 5  3i  4  7i  5  4  3  7 i  1  10i

20. 3  4i2  5i  3  2[4  5] i  59i     22. 3  2i 5  13 i  3  5 2  13 i  2 73 i

11. 3: real part 3, imaginary part 0.

21. 6  6i  9  i  6  9  6  1 i  3  5i       23. 7  12 i  5  32 i  7  5   12  32 i  2  2i

24. 4  i  2  5i  4  i  2  5i  4  2  1  5 i  6  6i

25. 12  8i  7  4i  12  8i  7  4i  12  7  8  4 i  19  4i 26. 6i  4  i  6i  4  i  4  6  1 i  4  7i

27. 4 1  2i  4  8i

28. 2 3  4i  6  8i

29. 3  4i 2  5i  6  15i  8i  20i 2  6  7i  20 1  26  7i

30. 5  i 6  2i  30  10i  6i  2i 2  30  16i  2 1  28  16i

31. 6  5i 2  3i  12  18i  10i  15i 2  12  15  18  10 i  27  8i 32. 2  i 3  7i  6  14i  3i  7i 2  6  7  14  3 i  1  17i

33. 3  2i 3  2i  32  2i2  13

34. 10  i 10  i  102  i 2  101

35. 3  2i2  9  12i  4 1  5  12i

36. 10  i2  100  20i  1  99  20i 1 i i i 1  i 37.    2  i i i 1 i 1 1 1i 1i 1i 1 1 1i 38.      2  2i  1i 1i 1i 11 2 1  i2 1  3i 5  5i 1  3i 1  2i 39.   1  i  1  2i 5 1  2i 1  2i 5  5i 2i 2  i 1  3i   12  12 i  40. 1  3i 10 1  3i 1  3i


44

CHAPTER 1 Equations and Graphs

10i 1  2i 10i  20i 2 5 4  2i 10i 20  10i      4  2i  1  2i 1  2i 1  2i 14 5 1  4i 2 1 2  3i 2  3i 2  3i 1 2  3i 2  3i      13  42. 2  3i1  13 2 2  3i 2  3i 2  3i 4  9 13 4  9i

41.

43.

4  6i 4  6i 3i 12i  18i 2 18 12 18  12i      i  2  43 i  3i 3i 3i 9 9 9 9i 2

3  5i 3  5i 15i 45i  75i 2 75 45 75  45i      i  13  15 i  15i 15i 15i 225 225 225 225i 2 1 1 1i 1 1i 1i 1  i 1i 1i 1         i   45. 1i 1i 1i 1i 1i 1i 2 2 1  i2 1  i2 44.

3  i  6i  2i 2 5  5i 2  i 10  5i  10i  5i 2 10  5  5  10 i 1  2i 3  i      2i 2i 2i 2i 5 4  i2 15  5i 15 5   5  5i  3  i 5  5 48. i 10  i 2  15  1 47. i 3  i 2 i  i  2 49. 3i5  35 i 2 i  243 12 i  243i 50. 2i4  24 i 4  16 1  16  250  250 51. i 1000  i 4  1250  1 52. i 1002  i 4 i 2  1i 2  1 46.

   54. 8  4 1  2i 2       1 55. 4 9  2i 3i  6 56. 2 32  16 1  4i            57. 2  1 3  3  6  2 3i  3i  3  6  3  3  2 3 i                58. 3  4 6  8  3  2i 6  2 2i  18  2 6i  2 6i  4 2i 2            3 2  4 2  2 6  2 6 i   2  4 6i      2 1  2i 2  2 2i 2  8  2  59.    1  2 1  2i 1  2i      2 2 2i 6i 36 2i 2i    60.     2i   2 1 2i 2 9 2i  3i 2i 2i

53.

   25  25 1  5i

61. x 2  25  0  x 2  25  x  5i 

62. 2x 2  5  0  x 2   52  x   210 i   6  36  4 13  3  12 16  3  2i 63. x 2  6x  13  0  x  2 Or: x 2  6x  13  0  x 2  6x  9  13  9  x  32  4  x  3  2i  x  3  2i    2  22  4 1 2 2  4  8 2  4 2  2i 2 64. x  2x  2  0  x      1  i 2 1 2 2 2   2  36 2  4  4 2 5 65. 2x 2  2x  5  0  x    12  32 i 2 2 4   12  122  4 8 5 12  16 2   34  14 i 66. 8x  12x  5  0  x  2 8 16       1  12  4 1 1 1  1  4 1  3 1  3i 2      12  23 i 67. x  x  1  0  x  2 1 2 2 2


SECTION 1.6 Complex Numbers

68. x 2  3x  3  0  x 

  3  32  4 1 3 2 1  4  42  4 9 4

3

45

    3  3 3  3i 9  12    32  23 i 2 2 2

  128  29  4 9 2 i 2 9 18    6 2  20 2  4  4 1 6 70. t  2   0  t 2  2t  6  0  t    1  5i t 2 2 71. 6x 2  12x  7  0  69. 9x 2  4x  4  0  x 

4

      12 122 467 144168 6i  12  2 6i  1  6 i  12 12  1212 24  122 x 12 12 12 6 26

72. x 2  12 x  1  0     2    1  12   4 1 1   12  14  4  12   15  1  1 15i 2 4 x    2 2   14  415 i 2 1 2 2 2 73. z    3  4i  5  2i  3  4i  5  2i  8  2i 74. z    3  4i  5  2i  8  2i  8  2i

75. z  z  3  4i 3  4i  32  42  25

76. z    3  4i 5  2i  15  6i  20i  8i 2  23  14i

77. LHS  z    a  bi  c  di  a  bi  c  di  a  c  b  d i  a  c  b  d i. RHS  z    a  bi  c  di  a  c  b  d i  a  c  b  d i. Since LHS  RHS, this proves the statement.

78. LHS  z  a  bi c  di  ac  adi  bci  bdi 2  ac  bd  ad  bc i  ac  bd  ad  bc i.

RHS  z    a  bi  c  di  a  bi c  di  ac  adi  bci  bdi 2  ac  bd  ad  bc i. Since LHS  RHS, this proves the statement.    2 79. LHS  z2  a  bi  a  bi2  a 2  2abi  b2 i 2  a 2  b2  2abi.     RHS  z 2  a  bi2  a 2  2abi  b2 i 2  a 2  b2  2abi  a 2  b2  2abi. Since LHS  RHS, this proves the statement.

80. z  a  bi  a  bi  a  bi  z.

81. z  z  a  bi  a  bi  a  bi  a  bi  2a, which is a real number.

82. z  z  a  bi  a  bi  a  bi  a  bi  a  bi  a  bi  2bi, which is a pure imaginary number.

83. z  z  a  bi  a  bi  a  bi  a  bi  a 2  b2 i 2  a 2  b2 , which is a real number.

84. Suppose z  z. Then we have a  bi  a  bi  a  bi  a  bi  0  2bi  b  0, so z is real. Now if z is real, then z  a  0i(where a is real). Since z  a  0i, we have z  z.  b  b2  4ac 85. Using the Quadratic Formula, the solutions to the equation are x  . Since both solutions are nonreal, we 2a   b 4ac  b2 have b2  4ac  0  4ac  b2  0, so the solutions are x   i, where 4ac  b2 is a real number. 2a 2a Thus the solutions are complex conjugates of each other. 86. i  i, i 5  i 4  i  i, i 9  i 8  i  i;

i 3  i, i 7  i 4  i 3  i, i 11  i 8  i 3  i;

i 2  1, i 6  i 4  i 2  1, i 10  i 8  i 2  1; i 4  1, i 8  i 4  i 4  1, i 12  i 8  i 4  1.

Because i 4  1, we have i n  i r , where r is the remainder when n is divided by 4, that is, n  4  k  r , where k is an

integer and 0  r  4. Since 4446  4  1111  2, we must have i 4446  i 2  1.


46

CHAPTER 1 Equations and Graphs

1.7

SOLVING OTHER TYPES OF EQUATIONS

Note: In cases where both sides of an equation are squared, the implication symbol  is sometimes used loosely. For example,  2  x  x  1 “” x  x  12 is valid only for positive x. In these cases, inadmissible solutions are identified later in the

solution.

1. (a) To solve the equation x 3  4x 2  0 we factor the left­hand side: x 2 x  4  0, as above. (b) The solutions of the equation x 2 x  4  0 are x  0 and x  4.   2. (a) Isolating the radical in 2x  x  0, we obtain 2x  x.  2 2x  x2  2x  x 2 . (b) Now square both sides:

(c) Solving the resulting quadratic equation, we find 2x  x 2  x 2  2x  x x  2  0, so the solutions are x  0 and x  2.  (d) We substitute these possible solutions into the original equation: 2  0  0  0, so x  0 is a solution, but  2  2  2  4  0, so x  2 is not a solution. The only real solution is x  0.

3. The equation x  12  5 x  1  6  0 is of quadratic type. To solve the equation we set W  x  1. The resulting

quadratic equation is W 2  5W  6  0. Solving this equation we get W  3 W  2  0  W  2 or W  3. Thus the solution to the original equation is x  1  2 or x  1  3  x  1 or x  2. You can verify that these are both solutions to the original equation.

4. The equation x 6  7x 3  8  0 is of quadratic type. To solve the equation we set W  x 3 . The resulting quadratic equation is W 2  7W  8  0. Solving this equation we get W 2  7W  8  W  8 W  1  W  8 or W  1. Thus the   solution to the given equation is x  3 8  2 or x  3 1  1. You can verify that these are both solutions to the original equation.

5. x 2  x  0  x x  1  0  x  0 or x  1  0. Thus, the two real solutions are 0 and 1.

6. 3x 3  6x 2  0  3x 2 x  2  0  x  0 or x  2  0. Thus, the two real solutions are 0 and 2.   7. x 3  25x  x 3  25x  0  x x 2  25  0  x x  5 x  5  0  x  0 or x  5  0 or x  5  0. The three real solutions are 5, 0, and 5.

   8. x 5  5x 3  x 5  5x 3  0  x 3 x 2  5  0  x  0 or x 2  5  0. The solutions are 0 and  5.    9. x 5  3x 2  0  x 2 x 3  3  0  x  0 or x 3  3  0. The solutions are 0 and 3 3.      10. 6x 5  24x  0  6x x 4  4  0  6x x 2  2 x 2  2  0. Thus, x  0, or x 2  2  0 (which has no solution), or  x 2  2  0. The solutions are 0 and  2.    11. 0  4z 5  10z 2  2z 2 2z 3  5 . If 2z 2  0, then z  0. If 2z 3  5  0, then 2z 3  5  z  3 52 . The solutions are 0  and 3 52 .     3 2  2 . The solutions are 0 12. 0  125t 10  2t 7  t 7 125t 3  2 . If t 7  0, then t  0. If 125t 3  2  0, then t  3 125 5  3

and 52 .

    13. 0  x 5  8x 2  x 2 x 3  8  x 2 x  2 x 2  2x  4  x 2  0, x  2  0, or x 2  2x  4  0. If x 2  0, then x  0; if x  2  0, then x  2, and x 2  2x  4  0 has no real solution. Thus the solutions are x  0 and x  2.


SECTION 1.7 Solving Other Types of Equations

47

 14. 0  x 4  64x  x x 3  64  x  0 or x 3  64  0. If x 3  64  0, then x 3  64  x  4. The solutions are 0 and 4.

  15. 0  x 3  5x 2  6x  x x 2  5x  6  x x  2 x  3  x  0, x  2  0, or x  3  0. Thus x  0, or x  2, or x  3. The solutions are x  0, x  2, and x  3.   16. 0  x 4  x 3  6x 2  x 2 x 2  x  6  x 2 x  3 x  2. Thus either x 2  0, so x  0,or x  3, or x  2. The solutions are 0, 3, and 2.

  17. 0  x 4  4x 3  2x 2  x 2 x 2  4x  2 . So either x 2  0  x  0, or using the Quadratic Formula on x 2  4x  2  0,       42 412 8  42 2  2  2. The solutions are 0, 2  2, and  4 2 168  4 we have x  4 21 2 2  2  2.   18. 0  y 5  8y 4  4y 3  y 3 y 2  8y  4 . If y 3  0, then y  0. If y 2  8y  4  0, then using the Quadratic Formula, we    8  82  4 1 4    8  48 have y    4  2 3. Thus, the three solutions are 0, 4  2 3, and 4  2 3. 2 1 2 19. 3x  54  3x  53  0. Let y  3x  5. The equation becomes y 4  y 3  0      y y 3  1  y y  1 y 2  y  1  0. If y  0, then 3x  5  0  x   53 . If y  1  0, then 3x  5  1  0  x   43 . If y 2  y  1  0, then 3x  52  3x  5  1  0  9x 2  33x  31  0. The discriminant is b2  4ac  332  4 9 31  27  0, so this case gives no real solution. The solutions are x   53 and x   43 .

20. x  54  16 x  52  0. Let y  x  5. The equation becomes y 4  16y 2  y 2 y  4 y  4  0. If y 2  0, then x  5  0 and x  5. If y  4  0, then x  5  4  0 and x  1. If y  4  0, then x  5  4  0 and x  9. Thus, the solutions are 9, 5, and 1.   21. 0  x 3  5x 2  2x  10  x 2 x  5  2 x  5  x  5 x 2  2 . If x  5  0, then x  5. If x 2  2  0, then   x 2  2  x   2. The solutions are 5 and  2.   22. 0  2x 3  x 2  18x  9  x 2 2x  1  9 2x  1  2x  1 x 2  9  2x  1 x  3 x  3. The solutions are  12 , 3, and 3.

  23. x 3  x 2  x  1  x 2  1  0  x 3  2x 2  x  2  x 2 x  2  x  2  x  2 x 2  1 . Since x 2  1  0 has no real solution, the only solution comes from x  2  0  x  2.

  24. 7x 3  x  1  x 3  3x 2  x  0  6x 3  3x 2  2x  1  3x 2 2x  1  2x  1  2x  1 3x 2  1  2x  1  0  or 3x 2  1  0. If 2x  1  0, then x  12 . If 3x 2  1  0, then 3x 2  1  x 2  13  x   13 . The solutions are 12  and  13 .

x2  50  x 2  50 x  100  50x  5000  x 2  50x  5000  0  x  100 x  50  0  x  100  0 x  100 or x  50  0. Thus x  100 or x  50. The solutions are 100 and 50. 1 2 26.   0  x 2  2 x  1  0  x 2  2x  2  0  x  1 x2     2  22  4 1 2 2 48 2  4 x   . Since the radicand is negative, there is no real solution. 2 1 2 2   2 y 1   y  1 y  2  y 2  1 2  y 2  3y  2  2y 2  2  y 2  3y  0  y y  3  0  27. 2 y2 y 1 y  0 or 3. 25.


48

CHAPTER 1 Equations and Graphs

3 3  1   3  1   1     3  32  2  1  2  3  22    1  0   1   1 2  1  0    1 or 12 .     1 1 1 1   54  4 x  1 x  2   4 x  1 x  2 54  29. x 1 x 2 x 1 x 2 28.

4 x  2  4 x  1  5 x  1 x  2  4x  8  4x  4  5x 2  5x  10  5x 2  3x  14  0 

5x  7 x  2  0. If 5x  7  0, then x   75 ; if x  2  0, then x  2. The solutions are  75 and 2.

x 5 5 28   2  x  2 x  5  5 x  2  28  x 2  7x  10  5x  10  28  x 2  2x  8  0  x 2 x 2 x 4 x  2 x  4  0  x  2  0 or x  4  0  x  2 or x  4. However, x  2 is inadmissible since we can’t divide by 0 in the original equation, so the only solution is 4.   x  x2 x  2x x x2  2    5x  x 2  2  5x 3x  4  x 2  2  15x 2  20x  0  14x 2  20x  2  5x  31. x 3x  4 3  4x 3  4x       20  202  4 14 2 20  400  112 20  512 20  16 2 5  4 2     . The x  2 14 28 28 28 7  5  4 2 . solutions are 7     3 1   3  1x x 4 4 x  3x  1  2x 2  4x  2x 2  7x  1  0. Using the Quadratic 32.  x 2   x  x 2  x x 2  4x 2  4x     7  72  4 2 1 7  57 7  57  . Both are admissible, so the solutions are . Formula, we find x  2 2 4 4    2 4x  3  25  4x  3  4x  28  x  7 33. 5  4x  3  52   34. 8x  1  3  8x  1  9  8x  10  x  54   35. 2x  1  3x  5  2x  1  3x  5  x  4   36. 3  x  x 2  1  3  x  x 2  1  x 2  x  2  0  x  1 x  2  0  x  1 or x  2.   37. 2x  1  1  x  2x  1  x  1  2x  1  x  12  2x  1  x 2  2x  1  0  x 2  4x  x x  4. Potential solutions are x  0 and x  4  x  4. These are only potential solutions since squaring is not a reversible  ? operation. We must check each potential solution in the original equation. Checking x  0: 2 0  1  1  0     ? ? ? ? 1  1  0 is false. Checking x  4: 2 4  1  1  4, 9  1  4, 3  1  4 is true. Thus, the only solution is x  4.   38. 2x  x  1  8  x  1  8  2x  x  1  8  2x2  x  1  64  32x  4x 2  30.

0  4x 2  33x  63  4x  21 x  3. Potential solutions are x  21 4 and x  3. Substituting each of these solutions

into the original equation, we see that x  3 is a solution, but x  21 4 is not. Thus 3 is the only solution.     2 x  1  x 2  6x  9  x  1  x 2  7x  10  0  39. x  x  1  3  x  3  x  1  x  32  x  2 x  5  0. Potential solutions are x  2 and x  5. We must check each potential solution in the original   equation. Checking x  2: 2  2  1  3, which is false, so x  2 is not a solution. Checking x  5: 5  5  1  3  5  2  3, which is true, so x  5 is the only solution.    2 40. 3  x  2  1  x  3  x  1  x  3  x  1  x2  3  x  x 2  2x  1  x 2  3x  2  0. Using   3  17 3  32  4 1 2  . Substituting the Quadratic Formula to find the potential solutions, we have x  2 1 2 

each of these solutions into the original equation, we see that x  32 17 is a solution, but x  32 17 is not. Thus 

x  32 17 is the only solution.


SECTION 1.7 Solving Other Types of Equations

49

41. x 4  4x 2  3  0. Let y  x 2 . Then the equation becomes y 2  4y  3  0  y  1 y  3  0, so y  1 or y  3. If  y  1, then x 2  1  x  1, and if y  3, then x 2  3  x   3.

42. 2x 4  4x 2  1  0. The LHS is the sum of two nonnegative numbers and a positive number, so 2x 4  4x 2  1  1  0. This equation has no real solution.    43. 0  x 6  26x 3  27  x 3  27 x 3  1 . If x 3  27  0  x 3  27, so x  3. If x 3  1  0  x 3  1, so x  1. The solutions are 3 and 1.

   44. x 8  15x 4  16  0  x 8  15x 4  16  x 4  16 x 4  1 . If x 4  16  0, then x 4  16 which is impossible (for real numbers). If x 4  1  0  x 4  1, so x  1. The solutions are 1 and 1.

45. 0  x  52  3 x  5  10  [x  5  5] [x  5  2]  x x  7  x  0 or x  7. The solutions are 0 and 7.     x 1 x 1 2 x 1 46. Let   . Then 0   3 becomes 0  2  4  3    1   3. Now if   1  0, 4 x x x x 1 x 1 x 1 x 1 then 10  1  x  1  x  x   12 , and if   3  0, then 3  0  3 x x x x 1 1 1  x  1  3x  x   4 . The solutions are  2 and  4 .  2   1 1 1 . Then  8  0 becomes 2  2  8  0    4   2  0. So   4  0 2 47. Let   x 1 x 1 x 1 1    4, and   2  0    2. When   4, we have  4  1  4x  4  3  4x  x   34 . When x 1 1  2  1  2x  2  3  2x  x   32 . Solutions are  34 and  32 .   2, we have x 1  2 x x 4x 48. Let   . Then  4 becomes 2  4  4  0  2  4  4    22 . Now if  x 2 x 2 x 2 x x 20  2  x  2x  4  x  4. The solution is 4.   2  0, then x 2 x 2 49. Let u  x 23 . Then 0  x 43  5x 23  6 becomes u 2  5u  6  0  u  3 u  2  0  u  3  0 or u  2  0.  If u  3  0, then x 23  3  0  x 23  3  x  332  3 3. If u  2  0, then x 23  2  0  x 23  2     x  232  2 2. The solutions are 3 3 and 2 2.    50. 4 x  112  5 x  132  x  152  0  x  1 4  5 x  1  x  12  0         x  1 4  5x  5  x 2  2x  1  0  x  1 x 2  3x  0  x  1  x x  3  0  x  1 or x  0 or x  3. The solutions are 1, 0, and 3.

51. Let u  x  4; then 0  2 x  473  x  443  x  413  2u 73  u 43  u 13  u 13 2u  1 u  1. So

u  x  4  0  x  4, or 2u  1  2 x  4  1  2x  7  0  2x  7  x  72 , or u  1  x  4  1  x  5  0

 x  5. The solutions are 4, 72 ,and 5.

  52. x 32  10x 12  25x 12  0  x 12 x 2  10x  25  0  x 12 x  52  0. Now x 12  0, so the only solution is x  5.

  53. x 12  x 12  6x 32  0  x 32 x 2  x  6  0  x 32 x  2 x  3  0. Now x 12  0, and furthermore

the original equation cannot have a negative solution. Thus, the only solution is x  3.    54. Let u  x. Then 0  x  5 x  6 becomes u 2  5u  6  u  3 u  2  0. If u  3  0, then x  3  0     x  3  x  9. If u  2  0, then x  2  0  x  2  x  4. The solutions are 9 and 4.     55. 2x  12  14 x  3  4 2x  12  4 14 x  3  8x  2  x  12  7x  14  x  2


50

CHAPTER 1 Equations and Graphs

56. x 2  7x  12  0  x  4 x  3  0  x  4  0 or x  3  0. Thus, x  4 or x  3. 57.

4a  a  4  a  a 3a  2  4  a  3a 2  2a  3a 2  a  4  0  a  1 3a  4  0. Thus, a  1 or 3a  2 a  43 .

3x  5  8  3x  5  8 x  5  3x  5  8x  40  5x  45  x  9 x 5 1 1 2  60   3  10  15   1  35  165  x   33 59.   3    7 5 3 2 58.

60. 6x x  1  21  x  6x 2  6x  21  x  6x 2  5x  21  0  2x  3 3x  7  0  2x  3  0 or 3x  7  0.

If 2x  3  0, then x   32 ; if 3x  7  0, then x  73 .   61. x  9  3x  0  x  9  3x  x 2  9  3x  x 2  3x  9  0       3  32  4 1 9 3  9  36 3  3 5 3  3 5 b  b2  4ac    . However, is not a x  2a 2 1 2 2 2       3  3 5  3  3 5   93 is the sum of two negative nonzero numbers and hence is nonzero. solution because 2 2  3  3 5 Thus, the only solution is x  . 2    x 5 62. 3x  12    3x  6  x  5 (multiply both sides by 3)  2x  1  x   12 3      4 63. Let u  x. Then 0  x  3 4 x  4  u 2  3u  4  u  4 u  1. So u  4  4 x  4  0  4 x  4     x  44  256, or u  1  4 x  1  0  4 x  1. However, 4 x is the positive fourth root, so this cannot equal 1. The only solution is 256.      64. x 6  x 3  6  0  x 3  2 x 3  3  0  x 3  2 or x 3  3  x   3 2 or 3 3 65. z  12  8 z  1  15  0  [z  1  5] [z  1  3]  0  z  6 z  4  0  z  4 or 6

66. Let u  x 16 . (We choose the exponent 16 because the LCD of 2, 3, and 6 is 6.) Then x 12  3x 13  3x 16  9    x 36  3x 26  3x 16  9  u 3  3u 2  3u  9  0  u 3  3u 2  3u  9  u 2 u  3  3 u  3  u  3 u 2  3 . So u  3  0 or u 2  3  0. If u  3  0, then x 16  3  0  x 16  3  x  36  729. If u 2  3  0, then x 13  3  0  x 13  3  x  33  27. The solutions are 729 and 27.

67. 2t  52  2t  32  8  4t 2  20t  25  4t 2  12t  9  8  32t  8  t   14 2    x x x  6  0. Let   , as in Example 6. Then 2  7  6    1   6  0    1 68. 7 x 1 x 1 x 1 x x 6 or   6  either  1, which has no solution; or  6  x  . The solution is 65 . x 1 x 1 5 1 4 4 69. 3  2   0  1  4x  4x 2  0  1  2x2  0  1  2x  0  2x  1  x   12 . The solution is  12 . x x x x4 we get, 0  1  4x 2  x 4 . Substituting u  x 2 , we get 0  1  4u  u 2 , and 4      4 42 411  4 2164  42 12  422 3  2  3. Substituting using the Quadratic Formula, we get u  21        back, we have x 2  2  3, and since 2  3 and 2  3 are both positive we have x   2  3 or x   2  3.         Thus the solutions are  2  3, 2  3,  2  3, and 2  3.

70. 0  4x 4  16x 2  4. Multiplying by


SECTION 1.7 Solving Other Types of Equations

71.



x  5  x  5. Squaring both sides, we get

x  5  x  25 

51

x  5  25  x. Squaring both sides again, we

get x  5  25  x2  x  5  625  50x  x 2  0  x 2  51x  620  x  20 x  31. Potential solutions are x  20 and x  31. We must check each potential solution in the original equation.    Checking x  20: 20  5  20  5  25  20  5  5  20  5, which is true, and hence x  20 is a solution.    Checking x  31: 36  31  5  37  5, which is false, and hence x  31 is not a 31  5  31  5  solution. The only real solution is x  20.

72.

   3 4x 2  4x  x  4x 2  4x  x 3  0  x 3  4x 2  4x  x x 2  4x  4  x x  22 . So x  0 or x  2. The

solutions are 0 and 2.

         73. x 2 x  3  x  332  0  x 2 x  3  x  332  0  x  3 x 2  x  3  0  x  3 x 2  x  3 . 

If x  312  0, then x  3  0  x  3. If x 2  x  3  0, then using the Quadratic Formula x  12 13 . The 

solutions are 3 and 12 13 .

74. Let u 

  2 2 11  x 2 . By definition of u we require it to be nonnegative. Now 11  x 2    1  u   1. 2 u 11  x

Multiplying both sides by u we obtain u 2  2  u  0  u 2  u  2  u  2 u  1. So u  2 or u  1. But since u   must be nonnegative, we only have u  2  11  x 2  2  11  x 2  4  x 2  7  x   7. The solutions are   7.     75. x  x  2  2. Squaring both sides, we get x  x  2  4  x  2  4  x. Squaring both sides again, we get x  2  4  x2  16  8x  x 2  0  x 2  9x  14  0  x  7 x  2. If x  7  0, then x  7. If x  2  0, then x  2. So x  2 is a solution but x  7 is not, since it does not satisfy the original equation.

76.

        1  x  2x  1  5  x. We square both sides to get 1  x  2x  1  5  x      2   x  2x  1  4  x  16  8 x  x  2x  1  16  8 x. Again, squaring both sides, we obtain   2   2x  1  16  8 x  256  256 x  64x  62x  255  256 x. We could continue squaring both sides until  we found possible solutions; however, consider the last equation. Since we are working with real numbers, for x to be  defined, we must have x  0. Then 62x  255  0 while 256 x  0, so there is no solution.

    77. 0  x 4  5ax 2  4a 2  a  x 2 4a  x 2 . Since a is positive, a  x 2  0  x 2  a  x  a. Again, since a is    positive, 4a  x 2  0  x 2  4a  x  2 a. Thus the four solutions are  a and 2 a.   b 78. 0  a 3 x 3  b3  ax  b a 2 x 2  abx  b2 . So ax  b  0  ax  b  x   or a       ab  ab2  4 a 2 b2 ab  3a 2 b2 b    x , but this gives no real solution. Thus, the solution is x   . 2 2 a 2 a 2a 79.

   x  a  x  a  2 x  6. Squaring both sides, we have             x  a  x  a  2 x  6  2x  2 x  a x  a  2x  12  2 x  a x  a  12 x a 2 x  a     x a x  a  6. Squaring both sides again we have x  a x  a  36  x 2  a 2  36  x 2  a 2  36     x   a 2  36. Checking these answers, we see that x   a 2  36 is not a solution (for example, try substituting  a  8), but x  a 2  36 is a solution. 


52

CHAPTER 1 Equations and Graphs

80. Let   x 16 . Then x 13  2 and x 12  3 , and so

       0  3  a2  b  ab  2   a  b   a  2  b   a  3 x  b 6 x  a . So 6 x  a  0     a  6 x  x  a 6 is one solution. Setting the first factor equal to zero, we have 3 x  b  0  3 x  b  x  b3 .  However, the original equation includes the term b 6 x, and we cannot take the sixth root of a negative number, so this is not a solution. The only solution is x  a 6 .

81. The volume is 180 ft3 , so x x  4 x  9  180  x 3  5x 2  36x  180  x 3  5x 2  36x  180  0    x 2 x  5  36 x  5  0  x  5 x 2  36  0  x  5 x  6 x  6  0  x  6 is the only positive solution. So the box is 2 feet by 6 feet by 15 feet.

  82. Let r be the radius of the larger sphere, in millimeters. Equating the volumes, we have 43 r 3  43  23  33  43   r 3  23  33  44  r 3  99  r  3 99  463. Therefore, the radius of the larger sphere is about 463 mm. 83. Let r be the radius of the tank, in feet. The volume of the spherical tank is 43 r 3 and is also 750  01337  100275. So 4 r 3  100275  r 3  23938  r  288 feet. 3

84. Let x be the length of the hypotenuse of the triangle, in feet. Then one of the other sides has length x  7 feet, and since the perimeter is 392 feet, the remaining side

must have length 392  x  x  7  399  2x. From the Pythagorean Theorem,

x-7

x

we get x  72  399  2x2  x 2  4x 2  1610x  159250  0. Using the

Quadratic Formula, we get   2 44159250 x  1610 161024  16108 44100  1610210 , and so x  2275 or x  175. But if x  2275, then the 8

side of length x  7 combined with the hypotenuse already exceeds the perimeter of 392 feet, and so we must have x  175. Thus the other sides have length 175  7  168 and 399  2 175  49. The lot has sides of length 49 feet, 168 feet, and 175 feet.

900 . After 5 people 85. Let x be the number of people originally intended to take the trip. Then originally, the cost of the trip is x   900 900 4500 cancel, there are now x  5 people, each paying  2. Thus 900  x  5  2  900  900  2x   10 x x x 4500  0  2x  10   0  2x 2  10x  4500  2x  100 x  45. Thus either 2x  100  0, so x  50, or x x  45  0, x  45. Since the number of people on the trip must be positive, originally 50 people intended to take the trip. 120,000 . If one person joins the group, then there would n   120,000 120,000 be n  1 members in the group, and each person would pay  6000. So n  1  6000  120,000 n n     n  n  120,000  6000 n  1  120,000  20  n n  1  20n  n 2  19n  20  20n   6000 n 6000

86. Let n be the number of people in the group, so each person now pays

0  n 2  n  20  n  4 n  5. Thus n  4 or n  5. Since n must be positive, there are now 4 friends in the group.

87. Let x be the height of the pile in feet. Then the diameter is 3x and the radius is 32 x feet. Since the volume of the cone is    4000  3x 2 3x 3 4000 3 3 1000 ft , we have x  1000   1000  x  x  3  752 feet. 3 2 4 3 3


SECTION 1.7 Solving Other Types of Equations

53

88. Let h be the height of the screens in inches. The width of the smaller screen is h  7 inches, and the width of the bigger  screen is 18h inches. The diagonal measure of the smaller screen is h 2  h  72 , and the diagonal measure of the     larger screen is h 2  18h2  424h 2  206h. Thus h 2  h  72  3  206h  h 2  h  72  206h  3. Squaring both sides gives h 2  h 2  14h  49  424h 2  1236h  9  0  224h 2  2636h  40. Applying

  2636 26362 422440 10532496  2636  3245 . So the Quadratic Formula, we obtain h   2636 448 2224 448

h

2636  3245  1313. Thus, the screens are approximately 131 inches high. 448

89. Let x be the length, in miles, of the abandoned road to be used. Then the length of the abandoned road not used  is 40  x, and the length of the new road is 102  40  x2 miles, by the Pythagorean Theorem. Since the  cost of the road is cost per mile  number of miles, we have 100,000x  200,000 x 2  80x  1700  6,800,000   2 x 2  80x  1700  68  xSquaring both sides, we get 4x 2  320x  6800  4624  136x  x 2  

1  18488  x  136 3x 2  184x  2176  0  x  184 3385626112 6 6 3 or x  16. Since 45 3 is longer than the existing  road, 16 miles of the abandoned road should be used. A completely new road would have length 102  402 (let x  0)  and would cost 1700  200,000  83 million dollars. So no, it would not be cheaper.

90. Let x be the distance, in feet, that you go on the boardwalk before veering off onto the sand. The distance along the  boardwalk from where you started to the point on the boardwalk closest to the umbrella is 7502  2102  720 ft. Thus the    distance you walk on the sand is 720  x2  2102  518,400  1440x  x 2  44,100  x 2  1440x  562,500. Along boardwalk Across sand

Distance

Rate

x

4

x 2  1440x  562,500

2

Time x 4

x 2  1440x  562,500 2

Since 4 minutes 45 seconds  285 seconds, we equate the time it takes to walk along the boardwalk and across the sand   x x 2  1440x  562,500  1140  x  2 x 2  1440x  562,500. Squaring both to the total time to get 285   4 2  sides, we get 1140  x2  4 x 2  1440x  562,500  1,299,600  2280x  x 2  4x 2  5760x  2,250,000    0  3x 2  3480x  950,400  3 x 2  1160x  316,800  3 x  720 x  440. So x  720  0  x  720, and x  440  0  x  440. Checking x  720, the distance across the sand is

210 210 feet. So 720 4  2  180  105  285 seconds. Checking x  440, the distance across the sand is  350 720  4402  2102  350 feet. So 440 4  2  110  175  285 seconds. Since both solutions are less than or equal

to 720 feet, we have two solutions: Either you walk 440 feet down the boardwalk and then head towards the umbrella, or you walk 720 feet down the boardwalk and then head towards the umbrella.

91. Let d be the distance from the lens to the object. Then the distance from the lens to the image is d  4. So substituting 1 1 1   . Now we multiply by the F  48, x  d, and y  d  4, and then solving for x, we have 48 d d 4 LCD, 48d d  4, to get d d  4  48 d  4  48d  d 2  4d  96d  192  0  d 2  136d  192  136  104 . So d  16 or d  12. Since d  4 must also be positive, the object is 12 cm from the lens. d 2


54

CHAPTER 1 Equations and Graphs

  d d 1 2  1 2  1 3  0  . Letting   d, we have 3  14  1090 92. Since the total time is 3 s, we have 3  1090 4 4 1090  545  591054 . Since   0, we have d    1151, so d  13256. The well  22  545  6540  0    4 is 1326 ft deep.   93. (a) Method 1: Let u  x, so u 2  x. Thus x  x  2  0 becomes u 2  u  2  0  u  2 u  1  0. So u  2   or u  1. If u  2, then x  2  x  4. If u  1, then x  1  x  1. So the possible solutions are 4 and   1. Checking x  4 we have 4  4  2  4  2  2  0. Checking x  1 we have 1  1  2  1  1  2  0. The only solution is 4.   Method 2: x  x  2  0  x  2  x  x 2  4x  4  x  x 2  5x  4  0  x  4 x  1  0. So the possible solutions are 4 and 1. Checking will result in the same solution. 1 1 12 10 (b) Method 1: Let u  , so u 2   1  0 becomes 12u 2  10u  1  0. Using . Thus  2 2 x 3 x 3  3  3 x x      102 4121 52  102 13  5 13 . If u  5 13 , then the quadratic formula, we have u  10 212  10 24 24 12 12      12 5 13   5  13 1 12  513   x 3  5  13. So x  2  13. 12 5 13 5 13 x 3 12       12 5 13  5  13 1 5 13 5 13 12     x 3  , then   5  13. So If u  12 12 5 13 5 13 x 3 12  x  2  13.  The solutions are 2  13.   10 12  1  0  x  32   Method 2: Multiplying by the LCD, x  32 , we get x  32 x 3 x  32 12  10 x  3  x  32  0  12  10x  30  x 2  6x  9  0  x 2  4x  9  0. Using the Quadratic    2 4 52  42 13  2  13. The solutions are 2  13. Formula, we have u  4 4 419  2 2 22

1.8

SOLVING INEQUALITIES

1. (a) If x  5, then x  3  5  3  x  3  2. (b) If x  5, then 3  x  3  5  3x  15.

(c) If x  2, then 3  x  3  2  3x  6.

(d) If x  2, then x  2.

2. To solve the nonlinear inequality

x 1  0 we x 2

first observe that the numbers 1 and 2 are zeros of the numerator and denominator. These numbers divide the real line into the three intervals  1, 1 2, and 2 . The endpoint 1 satisfies the inequality, because

Sign of x  1 Sign of x  2

Sign of x  1  x  2

 1

1 2

2 

21 1  1  0  0, but 2 fails to satisfy the inequality because is not 1  2 22

defined. Thus, referring to the table, we see that the solution of the inequality is [1 2.

3. (a) No. For example, if x  2, then x x  1  2 1  2  0, but x  0. (b) No. For example, if x  2, then x x  1  2 3  6  5, but x  5.


SECTION 1.8 Solving Inequalities

4. (a) To solve 3x  7, start by dividing both sides of the inequality by 3.

(b) To solve 5x  2  1, start by adding 2 to both sides of the inequality.

5. x

2  3x  13

6. x

5 17  13 ; no 1 5  13 ; no 0 2 3 5 6

1  5 3

1  2x  5x

11  25; yes

5

2  0; no

1

3  5; yes

2 3 5 6

 13  10 3 ; no  23  25 6 ; no

0

0  13 ; no 1  1 ; yes 2 3 1  13 ; yes 47  13 ; yes 7  13 ; yes 13  13 ; yes

1  0; yes

1  5

1  5; no

347  1118; no

3

5  The elements 56 , 1, 5, 3, and 5 satisfy the inequality.

5  15; no

5

9  25; no

The elements 5, 1, and 0 satisfy the inequality. 8.

7. x 5

1  2x  4  7

1  14  7; no

1

1  6  7; no

2 3 5 6

1   83  7; no 1   73  7; no

0

1  5 3 5

x

x

1  4  7; no

5 1 0 2 3 5 6

1  5 3 5

1

2  4  2; no

2 3 5 6

2  73  2; no

0

1  2  7; no

1  5

1  2  7; yes

3

1  047  7; no

1  6  7; yes

1 1 x  2  15  12 ; yes 1  12 ; yes

1 is undefined; no 0 3  1 ; no 2 2 6  1 ; no 5 2 1  12 ; no 045  12 ; yes 1  1 ; yes 3 2 1  1 ; yes 5 2

 The elements 5, 1, 5, 3, and 5 satisfy the inequality.

2  8  2; no

5

2  3  2; no

2  13 6  2; no 2  2  2; no

2  076  2; yes

2  0  2; yes

5 2  2  2; yes  The elements 5, 3, and 5 satisfy the inequality.

The elements 3 and 5 satisfy the inequality. 9.

2  3  x  2

10. x 5

x2  2  4 27  4; no

1

3  4; yes

0

2  4; yes

2 3 5 6

22  4; yes 9 97  4; yes 36

1  5

3  4; yes

3

11  4; no

5

27  4; no

7  4; no

The elements 1, 0, 23 , 56 , and 1 satisfy the inequality.

55


56

CHAPTER 1 Equations and Graphs

  11. 2x  7  x  72 . Interval:  72 Graph:

  12. 4x  10  x   52 . Interval:   52 Graph:

7 2

13. 2x  5  3  2x  8  x  4 Interval: 4 . Graph:

14. 3x  11  5  3x  6  x  2 Interval:  2. Graph:

4

15. 7  x  5  x  2  x  2

Interval: [7 . Graph:

2

17. 2x  1  0  2x  1  x   12 1

__ 2

12 __ 11

4 1 22. 25 x  1  15  2x  12 5 x  5  x  3

  Interval:   13 . Graph:

3

23. 13 x  2  16 x  1  16 x  3  x  18  18. Graph:

1

__ 3

24. 23  12 x  16  x (multiply both sides by 6) 

Interval:

4  3x  1  6x  3  9x  13  x   Interval:  13 . Graph:

_18

25. 4  3x   1  8x  4  3x  1  8x  5x  5  x  1

Interval:  1]. Graph:

_1

27. 2  x  5  4  3  x  1

1 _ 3

26. 2 7x  3  12x  16  14x  6  12x  16  2x  22  x  11

Interval:  11]. Graph:

11

28. 8  x  3  12  5  x  15 _3

_1

29. 1  2x  5  7  4  2x  12  2  x  6 2

2

  Interval:  12 11 . Graph:

_3

21. x  32  12 x  12 x  32  x  3

Interval: 2 6. Graph:

Interval: 2 . Graph:

20. 5  3x  8x  7  11x  12  x  12 11

19. 4x  7  8  9x  5x  15  x  3.

Interval: [3 1. Graph:

7

18. 0  4x  8  0  x  2  x  2

  Interval:   12 . Graph:

Interval: 3 . Graph:

_2

16. 5  3x  16  3x  21  x  7

Interval:  2]. Graph:

Interval: 3 . Graph:

_ 52

6

Interval: [5 15]. Graph:

_5

15

30. 1  3x  4  16  3  3x  12  1  x  4 Interval: 1 4]. Graph:

_1

4


57

SECTION 1.8 Solving Inequalities

31. 2  8  2x  1  10  2x  9  5  x  92  92  x  5

  Interval: 92  5 . Graph:

33.

9 _ 2

32. 3  3x  7  12  10  3x   13 2  13  10 3 x 6

  13 Interval:  10 3   6 . Graph:

5

2x  13 2 1    2  2x  13  8 (multiply each 6 12 3

 21 Interval: 15 2  2 . Graph:

15 __ 2

13

_ _6_

4  3x 1 1    (multiply each expression by 20) 2 5 4 10  4 4  3x  5  10  16  12x  5 

34. 

21 expression by 12)  15  2x  21  15 2 x  2.

10

_ _3_

11 11 13 26  12x  11  13 6  x  12  12  x  6 .

21 __ 2

  13 . Graph:  Interval: 11 12 6

11 __ 12

13 __ 6

35. x  2 x  3  0. The expression on the left of the inequality changes sign where x  2 and where x  3. Thus we must check the intervals in the following table. Interval Sign of x  2

 2

2 3

3 

Sign of x  3

Sign of x  2 x  3

From the table, the solution set is x  2  x  3. Interval: 2 3. Graph:

_2

3

36. x  5 x  4  0. The expression on the left of the inequality changes sign when x  5 and x  4. Thus we must check the intervals in the following table. Interval Sign of x  5

 4

4 5

5 

Sign of x  4

Sign of x  5 x  4

From the table, the solution set is x  x  4 or 5  x. Interval:  4]  [5 . Graph:

_4

5

37. x 2x  7  0. The expression on the left of the inequality changes sign where x  0 and where x   72 . Thus we must check the intervals in the following table. Interval Sign of x Sign of 2x  7

Sign of x 2x  7

    72

   72  0

0 

From the table, the solution set is   x  x   72 or 0  x .   Interval:   72  [0 . Graph:

7

__ 2

0


58

CHAPTER 1 Equations and Graphs

38. x 2  3x  0. The expression on the left of the inequality changes sign when x  0 and x  23 . Thus we must check the intervals in the following table. Interval

 0

Sign of x

Sign of 2  3x

Sign of x 2  3x

  0 23 

From the table, the solution set is   x  x  0 or 23  x .   Interval:  0]  23   .

2 3

Graph:

0

2 _ 3

39. x 2  3x  18  0  x  3 x  6  0. The expression on the left of the inequality changes sign where x  6 and where x  3. Thus we must check the intervals in the following table. Interval Sign of x  3 Sign of x  6

Sign of x  3 x  6

 3

3 6

6 

From the table, the solution set is x  3  x  6. Interval: [3 6]. Graph:

_3

6

40. x 2  8x  7  0  x  7 x  1  0. The expression on the left of the inequality changes sign where x  7 and where x  1. Thus we must check the intervals in the following table. Interval Sign of x  7 Sign of x  1

Sign of x  7 x  1

 1

1 7

7 

From the table, the solution set is x  x  1 or x  7. Interval:  1  7 . Graph:

1

7

41. 3x 2  5x  2  3x 2  5x  2  0  x  2 3x  1  0. The expression on the left of the inequality changes sign where x  2 and where x  13 . Thus we must check the intervals in the following table. Interval Sign of x  2

Sign of 3x  1

Sign of x  2 3x  1

 2 

  2 13 

1 3

From the table, the solution set is   x  x  2 or x  13 .   Interval:  2]  13   . Graph:

_2

1 _ 3

42. x 2  x  2  x 2  x  2  0  x  1 x  2  0. The expression on the left of the inequality changes sign when x  1 and x  2. Thus we must check the intervals in the following table. Interval Sign of x  1 Sign of x  2

Sign of x  1 x  2

 1

1 2

2 

From the table, the solution set is x  1  x  2. Interval: 1 2. Graph:

_1

2


SECTION 1.8 Solving Inequalities

59

43. 3x 2  3x  2x 2  4  x 2  3x  4  0  x  1 x  4  0. The expression on the left of the inequality changes sign where x  1 and where x  4. Thus we must check the intervals in the following table. Interval Sign of x  1 Sign of x  4

Sign of x  1 x  4

 1

1 4

4 

From the table, the solution set is x  1  x  4. Interval: 1 4. Graph:

_1

4

44. 5x 2  3x  3x 2  2  2x 2  3x  2  0  2x  1 x  2  0. The expression on the left of the inequality changes sign when x  12 and x  2. Thus we must check the intervals in the following table. Interval Sign of 2x  1 Sign of x  2

Sign of 2x  1 x  2

 2 

  2 12 

1 2

From the table, the solution set is   x  x  2 or 12  x .   Interval:  2]  12   . Graph:

_2

1 _ 2

45. x 2  3 x  6  x 2  3x  18  0  x  3 x  6  0. The expression on the left of the inequality changes sign where x  6 and where x  3. Thus we must check the intervals in the following table. Interval Sign of x  3 Sign of x  6

Sign of x  3 x  6

 3

3 6

6 

From the table, the solution set is x  x  3 or 6  x. Interval:  3  6 . Graph:

_3

6

46. x 2  2x  3  x 2  2x  3  0  x  3 x  1  0. The expression on the left of the inequality changes sign when x  3 and x  1. Thus we must check the intervals in the following table. Interval Sign of x  3 Sign of x  1

Sign of x  3 x  1

 3

3 1

1 

From the table, the solution set is x  x  3 or 1  x. Interval:  3  1 . Graph:

_3

1

47. x 2  4  x 2  4  0  x  2 x  2  0. The expression on the left of the inequality changes sign where x  2 and where x  2. Thus we must check the intervals in the following table. Interval Sign of x  2 Sign of x  2

Sign of x  2 x  2

 2

2 2

2 

From the table, the solution set is x  2  x  2. Interval: 2 2. Graph:

_2

2


60

CHAPTER 1 Equations and Graphs

48. x 2  9  x 2  9  0  x  3 x  3  0. The expression on the left of the inequality changes sign when x  3 and x  3. Thus we must check the intervals in the following table. Interval Sign of x  3

 3

3 3

3 

Sign of x  3

Sign of x  3 x  3

From the table, the solution set is x  x  3 or 3  x. Interval:  3]  [3 . Graph:

_3

3

49. x  2 x  1 x  3  0. The expression on the left of the inequality changes sign when x  2, x  1, and x  3. Thus we must check the intervals in the following table. Interval Sign of x  2 Sign of x  1 Sign of x  3

Sign of x  2 x  1 x  3

 2

2 1

1 3

3 

From the table, the solution set is x  x  2 or 1  x  3. Graph:

_2

1

Interval:  2]  [1 3].

3

50. x  5 x  2 x  1  0. The expression on the left of the inequality changes sign when x  5, x  2, and x  1. Thus we must check the intervals in the following table. Interval Sign of x  5 Sign of x  2 Sign of x  1

Sign of x  5 x  2 x  1

 1

1 2

2 5

5 

From the table, the solution set is x  1  x  2 or 5  x. Interval: 1 25 . Graph:

_1

2

5

51. x  4 x  22  0. Note that x  22  0 for all x  2, so the expression on the left of the original inequality changes sign only when x  4. We check the intervals in the following table. Interval Sign of x  4

Sign of x  22

Sign of x  4 x  22

 2

2 4

4 

From the table, the solution set is x  x  2 and x  4. We exclude the

endpoint 2 since the original expression cannot be 0. Interval:  2  2 4. Graph:

_2

4


SECTION 1.8 Solving Inequalities

61

52. x  4 x  22  0. Note that x  22  0 for all x  2, so the expression on the left of the original inequality changes sign only at x  4. We check the intervals in the following table. Interval Sign of x  4

 2

2 4

4 

Sign of x  22

Sign of x  32 x  1

From the table, the solution set is x  x  4.

(The point 2 is already excluded.) Interval: 4 . Graph:

4

53. x  32 x  2 x  5  0. The left­hand side is 0 when x  5, 3, or 2. We check the intervals in the following table. Interval Sign of x  32

 5

5 3

3 2

2 

_3

2

Sign of x  5 Sign of x  2

Sign of x  22 x  3 x  1

When x  3, the left­hand side is equal to 0 and the inequality is satisfied. Thus, the solution set is x  x  5, x  3, or x  2. Interval:  5]  3  [2 . Graph:

_5

  54. 4x 2 x 2  9  0  4x 2 x  3 x  3  0. The expression on the left of the inequality changes sign when x  3 and x  0. Thus we must check the intervals in the following table. Interval Sign of 4x 2 Sign of x  3

Sign of x  3   Sign of 4x 2 x 2  9

 3

3 0

0 3

3 

From the table, the solution set is x  3  x  3. (The endpoint 0 is included since the original expression is allowed to be 0.) Interval: [3 3]. Graph:

_3

3

  55. x 3  4x  0  x x 2  4  0  x x  2 x  2  0. The expression on the left of the inequality changes sign where x  0, x  2 and where x  4. Thus we must check the intervals in the following table. Interval Sign of x Sign of x  2 Sign of x  2

Sign of x x  2 x  2

 2

2 0

0 2

2 

From the table, the solution set is x  2  x  0 or x  2. Interval: 2 02 . Graph:

_2

0

2


62

CHAPTER 1 Equations and Graphs

  56. 9x  x 3  0  x 3  9x  x x 2  9  x x  3 x  3. The expression on the left of the inequality changes sign when x  3, x  0, and x  3. Thus we must check the intervals in the following table. Interval Sign of x  3 Sign of x

Sign of x  3

Sign of x x  3 x  3

 3

3 0

0 3

3 

From the table, the solution set is x  3  x  0 or 3  x. Interval: [3 0][3 . Graph:

_3

0

3

  57. x 4  x 2  x 4  x 2  0  x 2 x 2  1  0  x 2 x  1 x  1  0. The expression on the left of the inequality changes sign where x  0, where x  1, and where x  1. Thus we must check the intervals in the following table. Interval Sign of x 2 Sign of x  1 Sign of x  1

Sign of x 2 x  1 x  1

 1

1 0

0 1

1 

From the table, the solution set is x  x  1 or 1  x. Interval:  1  1 . Graph:

_1

1

    58. x 5  x 2  x 5  x 2  0  x 2 x 3  1  0  x 2 x  1 x 2  x  1  0. The expression on the left of the inequality   1  12  4 1 1 1  3 2 changes sign when x  0 and x  1. But the solution of x  x  1  0 are x   . 2 1 2 Since these are not real solutions. The expression x 2  x  1 does not changes signs, so we must check the intervals in the following table. Interval Sign of x 2 Sign of x  1

Sign of x 2  x  1   Sign of x 2 x  1 x 2  x  1

 0

0 1

1 

From the table, the solution set is x  1  x. Interval: 1 . Graph:

1


SECTION 1.8 Solving Inequalities

59.

x 3  0. The expression on the left of the inequality changes sign where x  1 and where x  3. Thus we must check x 1 the intervals in the following table. Interval Sign of x  1

Sign of x  3 x 3 Sign of x 1

60.

 1

1 3

3 

Sign of 2x  6

cannot equal 0 we must have x  1. Interval:  1  [3 . Graph:

 3

3 2

2 

Sign of x  2 2x  6 Sign of x 2

_1

3

From the table, the solution set is x  3  x  2. Interval: 3 2. Graph:

_3

2

x x 5 x  2 4x  10 2x  5 x 5 50  0 0  0. The expression on the left x 2 x 2 x 2 x 2 x 2 x 2 of the inequality changes sign when x  2 and x  52 . Thus we must check the intervals in the following table. Interval

 2

Sign of 2x  5

Sign of x  2 2x  5 Sign of x 2

62.

From the table, the solution set is x  x  1 or x  3. Since the denominator

2x  6  0. The expression on the left of the inequality changes sign when x  3 and x  2. Thus we must check the x 2 intervals in the following table. Interval

61.

63

  2 52 

5 2

From the table, the solution set is     x  2  x  52 . Interval: 2 52 . Graph:

2

5 _ 2

x 4 x  4  5 2x  1 9x  9 x 4 5 50 0  0. The expression on the left of the inequality 2x  1 2x  1 2x  1 2x  1 changes sign when x  1 and x   12 . Thus we must check the intervals in the following table. Interval Sign of 9x  9

Sign of 2x  1 9x  9 Sign of 2x  1

  1  12

   12  

 1 

From the table, the solution set is   x  x  1 or  12  x   .   Interval:  1   12   . Graph:

_1

1

__ 2


64

63.

CHAPTER 1 Equations and Graphs

2x  1 2x  1 3 x  5 x  16 2x  1 3 3 0  0  0. The expression on the left of the inequality x 5 x 5 x 5 x 5 x 5 changes sign where x  16 and where x  5. Thus we must check the intervals in the following table. Interval

 5

5 16

16 

Sign of x  16

Sign of x  5 x  16 Sign of x 5

64.

From the table, the solution set is x  x  5 or x  16. Since the denominator cannot equal 0, we must have x  5. Interval:  5  [16 . Graph:

5

16

3x 3x 3x 2x 3x 1 1 0  0  0. The expression on the left of the inequality changes 3x 3x 3x 3x 3x sign when x  0 and x  3. Thus we must check the intervals in the following table. Since the denominator cannot equal 0, we must Interval Sign of 3  x

Sign of 2x 2x Sign of 3x

65.

 0

0 3

3 

have x  3. The solution set is x  0  x  3. Interval: [0 3. Graph:

0

3

4 4 xx 4  x2 4 2  x 2  x  x  x 0  0 0  0. The expression on the left of the x x x x x x inequality changes sign where x  0, where x  2, and where x  2. Thus we must check the intervals in the following table. Interval Sign of 2  x Sign of x

Sign of 2  x 2  x 2  x Sign of x

 2

2 0

0 2

2 

From the table, the solution set is x  2  x  0 or 2  x. Interval: 2 02 . Graph:

_2

0

2


SECTION 1.8 Solving Inequalities

66.

65

x x 3x x  1 2x  3x 2 x 2  3x x  3x   3x  0   0 0  0. The expression on x 1 x 1 x 1 x 1 x 1 x 1 the left of the inequality changes sign whenx  0, x   23 , and x  1. Thus we must check the intervals in the following table. Interval Sign of x

Sign of x  1 2  x 2  x Sign of x

Graph:

67. 1 

_1

2

__ 3

   23  0

0 

Sign of 2  3x

From the table, the solution set is

  1  23

 1

   x  x  1 or  23  x  0 . Interval:  1   23  0 .

0

2 2 2 x x  1 2x 2 x  1 x 2  x  2x  2x  2 2  1  0   0 0 x 1 x x 1 x x x  1 x x  1 x x  1 x x  1

x2  x  2 x  2 x  1 0  0. The expression on the left of the inequality changes sign where x  2, where x x  1 x x  1 x  1, where x  0, and where x  1. Thus we must check the intervals in the following table. Interval Sign of x  2 Sign of x  1 Sign of x

Sign of x  1 x  2 x  1 Sign of x x  1

 2

2 1

1 0

0 1

1 

Since x  1 and x  0 yield undefined expressions, we cannot include them in the solution. From the table, the solution set is x  2  x  1 or 0  x  1. Interval: [2 1  0 1]. Graph:

_2

_1

0

1


66

68.

CHAPTER 1 Equations and Graphs

4 3 4 3x 4 x  1 x x  1 3x  4x  4  x 2  x 3  1  1  0   0 0 x 1 x x 1 x x x  1 x x  1 x x  1 x x  1 4  x2 2  x 2  x 0  0 The expression on the left of the inequality changes sign when x  2, x  2, x x  1 x x  1 x  0, and x  1. Thus we must check the intervals in the following table. Interval Sign of 2  x Sign of 2  x Sign of x

Sign of x  1 2  x 2  x Sign of x x  1

 2

2 0

0 1

1 2

2 

Since x  0 and x  1 give undefined expressions, we cannot include them in the solution. From the table, the solution set is x  2  x  0 or 1  x  2. Interval: [2 0  1 2]. Graph:

69.

_2

0

1

2

x 1 x 2 x 1 x 2 x  2 x  2 x  1 x  3    0  0 x 3 x 2 x 3 x 2 x  3 x  2 x  2 x  3 x 2  4  x 2  2x  3 2x  1 0  0. The expression on the left of the inequality x  3 x  2 x  3 x  2 changes sign where x   12 , where x  3, and where x  2. Thus we must check the intervals in the following table. Interval Sign of 2x  1 Sign of x  3

Sign of x  2 2x  1 Sign of x  3 x  2 From the table, the solution set is

Graph:

_3

1

__ 2

2

  3  12

   12  2

2 

 3 

   x  3  x   12 or 2  x . Interval: 3  12  2 .


SECTION 1.8 Solving Inequalities

70.

67

1 x 2 1 x 1 x 2x 1 2x  3  0  0 0  0. The x 1 x 2 x  1 x  2 x  1 x  2 x  1 x  2 x  1 x  2 expression on the left of the inequality changes sign when x   32 , x  1, and x  2. Thus we must check the intervals in the following table. Interval Sign of 2x  3 Sign of x  1

Sign of x  2 2x  3 Sign of x  1 x  2

  2  32

   32  1

1 

 2 

  From the table, the solution set is x  x  2 or  32  x  1 The points x  2 and x  1 are

  excluded from the solution because the expression is undefined at those values. Interval:  2   32  1 . Graph:

_2

3

__ 2

_1

  71. 0  5  2x  2x  5  x  52 . Interval:  52 . Graph:

72.

5 _ 2

6 6 6 6x 6 x  1 x x  1 6  1  10   0 x 1 x x 1 x x x  1 x x  1 x x  1 x 2  x  6 6x  6x  6  x 2  x x  3 x  2 0 0  0. The x x  1 x x  1 x x  1 expression on the left of the inequality changes sign where x  3, where x  2, where x  0, and where x  1. Thus we must check the intervals in the following table. Interval Sign of x  3 Sign of x  2 Sign of x

Sign of x  1 x  3 x  2 Sign of x x  1

 2

2 0

0 1

1 3

3 

From the table, the solution set is x  2  x  0 or 1  x  3. The points x  0 and x  1 are excluded from the solution set because they make the denominator zero. Interval: [2 0  1 3]. Graph:

_2

0

1

3


68

CHAPTER 1 Equations and Graphs

  73. 16x  x 3  0  x 3  16x  x x 2  16  x x  4 x  4. The expression on the left of the inequality changes sign when x  4, x  0, and x  4. Thus we must check the intervals in the following table. Interval Sign of x  4

 4

4 0

0 4

4 

Sign of x

Sign of x  4

Sign of x x  4 x  4

From the table, the solution set is x  4  x  0 or 4  x. Interval: [4 0]  [4 . Graph:

74. 5  3x  4  14  9  3x  18  3  x  6. Interval: [3 6]. Graph:

3

_4

0

4

6

75. x 2  5x  6  0  x  3 x  2  0. The expression on the left of the inequality changes sign when x  3 and x  2. Thus we must check the intervals in the following table. Interval Sign of x  3

 3

3 2

2 

Sign of x  2

Sign of x  3 x  2 76.

From the table, the solution set is x  x  3 or  2  x.

Interval:  3  2 . Graph:

_3

_2

x x 3 x  1 2x  3 x 3 3  0  0  0. The expression on the left of the inequality x 1 x 1 x 1 x 1 x 1 changes sign when x   32 and x  1. Thus we must check the intervals in the following table. Interval Sign of 2x  3

    32

   32  1

1 

From the table, the solution set is     x   32  x  1 . Interval:  32  1 .

Graph:

Sign of x  1 2x  3 Sign of x 1

3

_1

__ 2

77. x  32 x  1  0. Note that x  32  0 for all x  3, so the expression on the left of the original inequality changes sign only when x  1. We check the intervals in the following table. Interval Sign of x  32 Sign of x  1

Sign of x  32 x  1

 3

3 1

1 

From the table, the solution set is x  x  1. (The endpoint 3 is already excluded.) Interval: 1 . Graph:

_1


SECTION 1.8 Solving Inequalities

78. 1  3  2x  5  2  2x  8  1  x  4. Interval: [1 4].

79. For

Graph:

1

4

x 2  9 to be defined as a real number we must have x 2  9  0  x  3 x  3  0. The expression in the

inequality changes sign at x  3 and x  3. Interval Sign of x  3 Sign of x  3

Sign of x  3 x  3

 3

3 3

3 

Thus x  3 or x  3.

80. For

x 2  5x  50 to be defined as a real number we must have x 2  5x  50  0  x  5 x  10  0. The

expression on the left of the inequality changes sign when x  5 and x  10. Thus we must check the intervals in the following table. Interval Sign of x  5

Sign of x  10

Sign of x  5 x  10

 5

5 10

10 

Thus x  5 or x  10.

81. For

12 1 to be defined as a real number we must have x 2  3x  10  0  x  2 x  5  0. The x 2  3x  10

expression in the last inequality changes sign at x  2 and x  5. Interval Sign of x  2 Sign of x  5

Sign of x  2 x  5

 2

2 5

5 

Thus x  2 or x  5, and the solution set is  2  5 .

69


70

CHAPTER 1 Equations and Graphs

82. For 4

1x 1x to be defined as a real number we must have  0The expression on the left of the inequality changes 2x 2x

sign when x  1 and x  2. Thus we must check the intervals in the following table. Interval Sign of 1  x

Sign of 2  x 1x Sign of 2x

 2

2 1

1 

Thus 2  x  1 and the solution set is 2 1]. Note that x  2 has been excluded from the solution set because the expression is undefined at that value.

83. a bx  c  bc (where a, b, c  0)  bx  c  x

bc 1 bc  bx  cx  a a b

 c c c c bc c   x    a a b a b

a  b c cb  ca  . ab ab

84. We have a  bx  c  2a, where a, b, c  0  a  c  bx  2a  c 

ac 2a  c x . b b

85. Inserting the relationship C  59 F  32, we have 20  C  30  20  59 F  32  30  36  F  32  54  68  F  86. 86. Inserting the relationship F  95 C  32, we have 50  F  95  50  95 C  32  95  18  95 C  63  10  C  35.

87. Let x be the average number of miles driven per day. Each day the cost of Plan A is 95  040x, and the cost of Plan B is 135. Plan B saves money when 135  95  040x  40  04x  100  x. So Plan B saves money when you average more than 100 miles a day.

88. Let m be the number of minutes of international calls placed per month. Then under Plan A, the cost will be 25  005m, and under Plan B, the cost will be 5012m. To determine when Plan B is advantageous, we must solve 25005m  5012m  20  007m  2857  m. So Plan B is advantageous if a person places fewer than 286 minutes of international calls per month.

89. We need to solve 6400  035m  2200  7100 for m. So 6400  035m  2200  7100  4200  035m  4900  12,000  m  14,000. She plans on driving between 12,000 and 14,000 miles.


SECTION 1.8 Solving Inequalities

71

h , where T is the temperature in  C, and h is the height in meters. 100 (b) Solving the expression in part (a) for h, we get h  100 20  T . So 0  h  5000  0  100 20  T   5000  0  20  T  50  20  T  30  20  T  30. Thus the range of temperature is from 20 C down to 30 C.

90. (a) T  20 

91. (a) Let x be the number of $3 increases. Then the number of seats sold is 120  x. So P  200  3x

 3x  P  200  x  13 P  200. Substituting for x we have that the number of seats sold is

120  x  120  13 P  200   13 P  560 3 .

(b) 90   13 P  560 3  115  270  360  P  200  345  270  P  560  345  290  P  215 

290  P  215. Putting this into standard order, we have 215  P  290. So the ticket prices are between $215 and $290.

92. If the customer buys x pounds of coffee at $650 per pound, then his cost c will be 650x. Thus x 

c . Since the 65

scale’s accuracy is 003 lb, and the scale shows 3 lb, we have 3  003  x  3  003  297 

c  303  65

650 297  c  650 303  19305  c  19695. Since the customer paid $1950, he could have been over­ or undercharged by as much as 195 cents. 93. 00004 

4,000,000  001. Since d 2  0 and d  0, we can multiply each expression by d 2 to obtain d2

00004d 2  4,000,000  001d 2 . Solving 00004d 2  4,000,000, we have d 2  10,000,000,000  d  100,000; solving 4,000,000  001d 2 , we have 400,000,000  d 2  20,000  d. Putting these together, we have 20,000  d  100,000. 94.

  600,000  500  600,000  500 x 2  300 (Note that x 2  300  300  0, so we can multiply both sides by the 2 x  300

denominator and not worry that we might be multiplying both sides by a negative number or by zero.) 1200  x 2  300  0  x 2  900  0  x  30 x  30. The expression in the inequality changes sign at x  30 and x  30. However, since x represents distance, we must have x  0. Interval Sign of x  30 Sign of x  30

Sign of x  30 x  30

0 30

30 

So x  30 and you must stand at least 30 meters from the center of the fire. 95. (a) We substitute d  1320 ft and t  10 s into the given formula and find the value of a that results in a quarter­mile time 2 of exactly 10 s: 1320  12 a 102  1320  50a  a  1320 50  264 fts . Thus, the quarter­mile time will be less

than 10 s if a  264 fts2 .

2 1320 . Taking the positive root, we find 32  that the quarter­mile time for a (downward) quarter­mile under Earth’s gravity is 21320  91 s. 32

(b) We substitute a  g  32 fts2 and solve for t: 1320  12 32 t 2  t 2 


72

CHAPTER 1 Equations and Graphs

96. Solve 30  10  09  001 2 for 10    75. We have 30  10  09  001 2  001 2  09  20  0  01  4 01  5  0. The possible endpoints are 01  4  0  01  4    40 and 01  5  0  01  5    50. Interval

10 40

40 50

50 75

Sign of 01  4 Sign of 01  5

Sign of 01  4 01  5 Thus he must drive between 40 and 50 mi/h. 97. 240   

  2 1  2    240  0  1   3   80  0. The expression in the inequality changes sign at  20 20 20

  60 and   80. However, since  represents the speed, we must have   0. Interval

0 60

60 

1  3 Sign of 20

Sign of   80   1   3   80 Sign of 20

So Kerry must drive between 0 and 60 mi/h.

  98. Solve 2400  20x  2000  8x  00025x 2  2400  20x  2000  8x  00025x 2  00025x 2  12x  4400  0  00025x  1 x  4400  0. The expression on the left of the inequality changes sign whenx  400 and x  4400.

Since the manufacturer can only sell positive units, we check the intervals in the following table. Interval Sign of 00025x  1 Sign of x  4400

Sign of 00025x  1 x  4400

0 400

400 4400

4400 

So the manufacturer must sell between 400 and 4400 units to enjoy a profit of at least $2400. 99. Let x be the length of the garden and  its width. Using the fact that the perimeter is 120 ft, we must have 2x  2  120    60  x. Now since the area must be at least 800 ft2 , we have 800  x 60  x  800  60x  x 2  x 2  60x  800  0  x  20 x  40  0. The expression in the inequality changes sign at x  20 and x  40. However, since x represents length, we must have x  0. Interval Sign of x  20 Sign of x  40

Sign of x  20 x  40 The length of the garden should be between 20 and 40 feet.

0 20

20 40

40 


SECTION 1.9 Solving Absolute Value Equations and Inequalities

100. Case 1: a  b  0

73

We have a  a  a  b, since a  0, and b  a  b  b, since b  0. So a 2  a  b  b2 , that

is a  b  0  a 2  b2 . Continuing, we have a  a 2  a  b2 , since a  0 and b2  a  b2  b, since b2  0. So a 3  ab2  b3 . Thus a  b  0  a 3  b3 . So a  b  0  a n  bn , if n is even, and a n  b, if n is odd. Case 2: 0  a  b

We have a  a  a  b, since a  0, and b  a  b  b, since b  0. So a 2  a  b  b2 . Thus 0 

a  b  a 2  b2 . Likewise, a 2  a  a 2  b and b  a 2  b  b2 , thus a 3  b3 . So 0  a  b  a n  bn , for all positive integers n. Case 3: a  0  b

If n is odd, then a n  bn , because a n is negative and bn is positive. If n is even, then we could have

either a n  bn or a n  bn . For example, 1  2 and 12  22 , but 3  2 and 32  22 . 101. The rule we want to apply here is “a  b  ac  bc if c  0 and a  b  ac  bc if c  0 ”. Thus we cannot simply multiply by x, since we don’t yet know if x is positive or negative, so in solving 1 

3 , we must consider two cases. x

Case 1: x  0

Multiplying both sides by x, we have x  3. Together with our initial condition, we have 0  x  3.

Case 2: x  0

Multiplying both sides by x, we have x  3. But x  0 and x  3 have no elements in common, so this

gives no additional solution. Hence, the only solutions are 0  x  3. 102.

c a c ad ad a  , so by Rule 3, d  d   c. Adding a to both sides, we have  a  c  a. Rewriting the left­hand b d b d b b side as

ad ab a b  d a ac   and dividing both sides by b  d gives  . b b b b bd

Similarly, a  c 

c b  d ac c cb c  , so  . d d bd d

103. (a) Because x is nonnegative, x  y  x 2  x y, and because y is nonnegative, x  y  x y  y 2 . Thus, by transitivity, x 2  y2.

xy x  y2  xy  . Expanding, this becomes 4x y  x  y2  x 2  y 2  2x y  2 4 x 2  y 2  2x y  0  x  y2  0. This is true for any x and y, so the original inequality is true for all nonnegative x and y.

(b) By part (a),

1.9

xy 

SOLVING ABSOLUTE VALUE EQUATIONS AND INEQUALITIES

1. The equation x  3 has the two solutions 3 and 3.

2. (a) The solution of the inequality x  3 is the interval [3 3].

(b) The solution of the inequality x  3 is a union of two intervals  3]  [3 .

3. (a) The set of all points on the real line whose distance from zero is less than 3 can be described by the absolute value inequality x  3.

(b) The set of all points on the real line whose distance from zero is greater than 3 can be described by the absolute value inequality x  3.

4. (a) 2x  1  5 is equivalent to the two equations 2x  1  5 and 2x  1  5. (b) 3x  2  8 is equivalent to 8  3x  2  8.


74

CHAPTER 1 Equations and Graphs

5. 5x  20  5x  20  x  4.

6. 3x  10  3x  10  x   10 3.

7. 5 x  3  28  5 x  25  x  5  x  5.

8. 12 x  7  2  12 x  9  x  18  x  18.

9. x  3  2 is equivalent to x  3  2  x  3  2  x  1 or x  5.

10. 2x  3  7 is equivalent to either 2x  3  7  2x  10  x  5; or 2x  3  7  2x  4  x  2. The two solutions are x  5 and x  2. 11. x  4  05 is equivalent to x  4  05  x  4  05  x  45 or x  35.

12. x  4  3. Since absolute value is always nonnegative, there is no solution.

13. 2x  3  11 is equivalent to either 2x  3  11  2x  14  x  7; or 2x  3  11  2x  8  x  4. The two solutions are x  7 and x  4.

14. 2  x  11 is equivalent to either 2  x  11  x  9; or 2  x  11  x  13. The two solutions are x  9 and x  13.

15. 4  3x  6  1   3x  6  3  3x  6  3, which is equivalent to either 3x  6  3  3x  3  x  1; or 3x  6  3  3x  9  x  3. The two solutions are x  1 and x  3.

16. 5  2x  6  14  5  2x  8 which is equivalent to either 5  2x  8  2x  3  x   32 ; or 5  2x  8  3 13 2x  13  x  13 2 . The two solutions are x   2 and x  2 .

17. 3 x  5  6  15  3 x  5  9  x  5  3, which is equivalent to either x  5  3  x  2; or x  5  3  x  8. The two solutions are x  2 and x  8.

18. 20  2x  4  15  2x  4  5. Since the absolute value is always nonnegative, there is no solution.             19. 8  5  13 x  56   33  5  13 x  56   25   13 x  56   5, which is equivalent to either 13 x  56  5  13 x  35 6 

1 5 1 25 25 25 35 x  35 2 ; or 3 x  6  5  3 x   6  x   2 . The two solutions are x   2 and x  2 .         3 9 20.  35 x  2  12  4   35 x  2  92 which is equivalent to either 35 x  2  92  35 x  52  x  25 6 ; or 5 x  2   2  3 x   13  x   65 . The two solutions are x  25 and x   65 . 5 2 6 6 6

21. x  1  3x  2, which is equivalent to either x  1  3x  2  2x  3  x   32 ; or x  1   3x  2  x  1  3x  2  4x  1  x   14 . The two solutions are x   32 and x   14 .

22. x  3  2x  1 is equivalent to either x  3  2x  1  x  2  x  2; or x  3   2x  1  x  3  2x  1  3x  4  x   43 . The two solutions are x  2 and x   43 .

23. x  5  5  x  5. Interval: [5 5].

24. 2x  20  20  2x  20  10  x  10. Interval: [10 10].

    25. 2x  7 is equivalent to 2x  7  x  72 ; or 2x  7  x   72 . Interval:   72  72   .

26. 12 x  1  x  2 is equivalent to x  2 or x  2. Interval:  2]  [2 .

27. x  4  10 is equivalent to 10  x  4  10  6  x  14. Interval: [6 14].

28. x  3  9 is equivalent to x  3  9  x  6; or x  3  9  x  12. Interval:  6  12 . 29. x  1  1 is equivalent to x  1  1  x  0; or x  1  1  x  2. Interval:  2]  [0 .

30. x  4  0 is equivalent to x  4  0  x  4  0  x  4. The only solution is x  4.

31. 2x  1  3 is equivalent to 2x  1  3  2x  4  x  2; or 2x  1  3  2x  2  x  1. Interval:  2]  [1 .

32. 3x  2  7 is equivalent to 3x  2  7  3x  5  x   53 ; or 3x  2  7  3x  9  x  3. Interval:     53  3 .


SECTION 1.9 Solving Absolute Value Equations and Inequalities

75

33. 2x  3  04  04  2x  3  04  26  2x  34  13  x  17. Interval: [13 17].   34. 5x  2  6  6  5x  2  6  4  5x  8   45  x  85 . Interval:  45  85 .   x  2    2  2  x  2  2  6  x  2  6  4  x  8. Interval: 4 8. 35.  3  3     x  1   4   1 x  1  4  1 x  1  4  x  1  8 which is equivalent to either x  1  8  x  7; or 36.   2 2 2 x  1  8  x  9. Interval:  9]  [7 .

37. x  6  0001  0001  x  6  0001  6001  x  5999. Interval: 6001 5999. 38. x  a  d  d  x  a  d  a  d  x  a  d. Interval: a  d a  d.

39. 4 x  2  3  13  4 x  2  16  x  2  4  4  x  2  4  6  x  2. Interval: 6 2.

40. 3  2x  4  1   2x  4  2  2x  4  2 which is equivalent to either 2x  4  2  2x  2  x  1; or 2x  4  2  2x  6  x  3. Interval:  3]  [1 .

41. 8  2x  1  6   2x  1  2  2x  1  2  2  2x  1  2  1  2x  3   12  x  32 .   Interval:  12  32 .

42. 7 x  2  5  4  7 x  2  1  x  2   17 . Since the absolute value is always nonnegative, the inequality is true for all real numbers. In interval notation, we have  .     1     43. 12 4x  13   56  4x  13   53 , which is equivalent to either 4x  13  53  4x  43  x  13 ; or 4x    53  3    

4x  2  x   12 . Interval:   12  13   .             44. 2  12 x  3  3  51  2  12 x  3  48   12 x  3  24  24  12 x  3  24  27  12 x  21  54  x  42. Interval: [54 42].

45. 1  x  4. If x  0, then this is equivalent to 1  x  4. If x  0, then this is equivalent to 1  x  4  1  x  4  4  x  1. Interval: [4 1]  [1 4].

46. 0  x  5  12 . For x  5, this is equivalent to  12  x  5  12  92  x  11 2 . Since x  5 is excluded, the solution is     9  5  5 11 . 2 2

47.

48.

1 13  2  1  2 x  7 (x  7)  x  7  12   12  x  7  12   15 2  x   2 and x  7. x  7     13 . Interval:  15  7  7  2 2 1  5  15  2x  3, since 2x  3  0, provided 2x  3  0  x  32 . Now for x  32 , we have 2x  3

1  2x  3 is equivalent to either 1  2x  3  16  2x  8  x; or 2x  3   1  2x  14  x  7 . 5 5 5 5 5 5 5     7 8 Interval:  5  5   .

49. x  3

50. x  2

51. x  7  5

52. x  2  4

53. x  2

54. x  1

55. x  3

56. x  4

57. (a) Let x be the thickness of the laminate. Then x  0020  0003.

(b) x  0020  0003  0003  x  0020  0003  0017  x  0023.    h  682    2  2  h  682  2  58  h  682  58  624  h  740. Thus 95% of the adult males are 58.  29  29 between 624 in and 740 in.


76

CHAPTER 1 Equations and Graphs

59. x  1 is the distance between x and 1; x  3 is the distance between x and 3. So x  1  x  3 represents those points closer to 1 than to 3, and the solution is x  2, since 2 is the point halfway between 1 and 3. If a  b, then the ab solution to x  a  x  b is x  . 2

1.10 SOLVING EQUATIONS AND INEQUALITIES GRAPHICALLY 1. The solutions of the equation x 2  2x  3  0 are the x­intercepts of the graph of y  x 2  2x  3. 2. The solutions of the inequality x 2  2x  3  0 are the x­coordinates of the points on the graph of y  x 2  2x  3 that lie above the x­axis. 3. (a) From the graph, it appears that the graph of y  x 4  3x 3  x 2  3x has x­intercepts 1, 0, 1, and 3, so the solutions to the equation x 4  3x 3  x 2  3x  0 are x  1, x  0, x  1, and x  3.

(b) From the graph, we see that where 1  x  0 or 1  x  3, the graph lies below the x­axis. Thus, the inequality x 4  3x 3  x 2  3x  0 is satisfied for x  1  x  0 or 1  x  3  [1 0]  [1 3].

4. (a) The graphs of y  5x  x 2 and y  4 intersect at x  1 and at x  4, so the equation 5x  x 2  4 has solutions x  1 and x  4. (b) The graph of y  5x  x 2 lies strictly above the graph of y  4 when 1  x  4, so the inequality 5x  x 2  4 is satisfied for those values of x, that is, for x  1  x  4  1 4.

5. (a)

(b) x­intercepts 0, 1; y­intercept 0. y  x 3  x 2  x 2 x  1, so y  0 

1

x  0 or x  1 and x  0  y  0.

(c) The graph is not symmetric with respect to either axis or the origin. ­2

­1

1

2

­1

y  x 3  x 2 ; [2 2] by [1 1] (b) x­intercepts 0, 2; y­intercept 0. y  x 4  2x 3  x 3 x  2, so y  0 

6. (a)

x  0 or x  2; x  0  y  0.

2

(c) The graph is not symmetric with respect to either axis or the origin. ­2

­1

1

2

3

­2

y  x 4  2x 3 ; [2 3] by [3 3]


SECTION 1.10 Solving Equations and Inequalities Graphically

7. (a)

(b) No x­intercept; y­intercept 2. y  0 has no solution; x  0 

1 ­4

77

­2

2

y

4

2  2. 1

(c) The graph is symmetric with respect to the y­axis:

­1

­2 ­3

2 x2  1

2 .  2 x 1

2 y 2 ; [5 5] by [3 1] x 1 (b) x­intercepts 1, y­intercept 1. y  0   3 x  1; x  0  y  1  02  1.

8. (a) 2

­4

­2

2

(c) The graph is symmetric with respect to the y­axis:   3 3 1  x2  1  x 2 .

4

­2 ­4

y

 3 1  x2  0  1  x2  0 

 3 1  x 2 ; [5 5] by [5 3]

9. Although the graphs of y  3x 2  6x  12 and  7 x 2 appear to intersect in the viewing y  7  12 rectangle [4 4] by [1 3], there is no point of intersection. You can verify this by zooming in.

10. Although the graphs of y 

y  15 41  3x appear to intersect in the viewing

rectangle [8 8] by [1 8], there is no point of intersection. You can verify this by zooming in. 8

3

6

2

4

1 ­4

­2

 49  x 2 and

2 2

­1

4

­8 ­6 ­4 ­2

2 4 6 8

11. The graphs of y  6  4x  x 2 and y  3x  18 appear to 12. The graphs of y  x 3  4x and y  x  5 appear to have have two points of intersection in the viewing rectangle

one point of intersection in the viewing rectangle [4 4]

[6 2] by [5 20]. You can verify that x  4 and

by [15 15]. The solution is x  2627.

x  3 are exact solutions.

10

20 ­4

10

­2

2 ­10

­6

­4

­2

2

4


78

CHAPTER 1 Equations and Graphs

13. Algebraically: 3x  2  5x  4  6  2x  x  3.

Graphically: We graph the two equations y1  3x  2 and y2  5x  4 in the viewing rectangle [1 4] by [1 13].

Zooming in, we see that the solution is x  3.

14. Algebraically: 23 x  11  1  x  53 x  10  x  6.

Graphically: We graph the two equations y1  23 x  11

and y2  1  x in the viewing rectangle [12 2] by

[2 8]. Zooming in, we see that the solution is x  6.

10 5

5

­1

1

2

3

4 ­12 ­10 ­8 ­6 ­4 ­2

15. Algebraically:

1 2   7  2x x 2x

5.  4  1  14x  x  14

2 1  x 2x

 2x 7

1 2  x 2x and y2  7 in the viewing rectangle [2 2] by [2 8]. Zooming in, we see that the solution is x  036.

Graphically: We graph the two equations y1 

5

2

6 5 4    16. Algebraically: x  2 2x 2x  4     6 5 4   2x x  2  2x x  2 x  2 2x 2x  4 2x 4  x  2 6  x 5  8x  6x  12  5x  12  3x  4  x. Graphically: We graph the two equations 6 4 5  and y2  in the viewing y1  x  2 2x 2x  4 rectangle [5 5] by [10 10]. Zooming in, we see that there is only one solution at x  4. 10

­2

­1

1

2 ­4

­2

2

4

­10

 17. Algebraically: 4x 2  8  0  x 2  2  x   2.

Graphically: We graph the equation y1  4x 2  8 and determine where this curve intersects the x­axis. We use the viewing rectangle [2 2] by [4 4]. Zooming in, we see that solutions are x  141 and x  141. 4

18. Algebraically: x 3  10x 2  0  x 2 x  10  0  x  0 or x  10. Graphically: We graph the equation y  x 3  10x 2 and determine where this curve intersects the x­axis. We use the viewing rectangle [12 2] by [5 5]. We see that the solutions are x  0 and x  10. 4

2

2

­2

­1

­2 ­4

1

2 ­12 ­10 ­8 ­6 ­4 ­2 ­2 ­4

2


SECTION 1.10 Solving Equations and Inequalities Graphically

19. Algebraically: x 2  9  0  x 2  9, which has no real solution. Graphically: We graph the equation y  x 2  9 and see that this curve does not intersect the x­axis. We use the viewing rectangle [5 5] by [5 30]. 30 20 10

79

20. Algebraically: x 2  3  2x  x 2  2x  3  0     2  22  4 1 3 2  8  . x 2 1 2 1 Because the discriminant is negative, there is no real solution. Graphically: We graph the two equations y1  x 2  3 and y2  2x in the viewing rectangle [4 6] by [6 12], and see that the two curves do not intersect. 10

­4

­2

2

4

5 ­4

­2

2

4

6

­5

4 21. Algebraically: 81x 4  256  x 4  256 81  x   3 .

Graphically: We graph the two equations y1  81x 4 and

y2  256 in the viewing rectangle [2 2] by [250 260].

Zooming in, we see that solutions are x  133. 260

22. Algebraically: 2x 5  243  0  2x 5  243  x 5  243 2   5 5 243 3 x  2  2 16. Graphically: We graph the equation y  2x 5  243 and

determine where this curve intersects the x­axis. We use the viewing rectangle [5 10] by [5 5]. Zooming in, we see that the solution is x  261.

255

4 2 ­2

­1

0

1

2 ­5

5

­2

10

­4

23. Algebraically: x  54  80  0  x  54  80     x  5   4 80  2 4 5  x  5  2 4 5.

Graphically: We graph the equation y1  x  54  80

and determine where this curve intersects the x­axis. We use the viewing rectangle [1 9] by [5 5]. Zooming in, we see that solutions are x  201 and x  799.

24. Algebraically: 3 x  53  72  x  53  72 3  24    x  5  3 24  x  5  3 24.

Graphically: We graph the two equations y1  3 x  53 and y2  72 in the viewing rectangle [3 1] by [65 78]. Zooming in, we see that the solution is x  212

4 75

2 ­2 ­4

2

4

6

70

8 ­3

­2

­1

65


80

CHAPTER 1 Equations and Graphs

25. We graph y  x 2  11x  30 in the viewing rectangle

26. We graph y  x 2  075x  0125 in the viewing

[2 8] by [01 01]. The solutions appear to be exactly

rectangle [2 2] by [01 01]. The solutions are

x  5 and x  6. [In fact

x  025 and x  050.

x 2  11x  30  x  5 x  6.]

0.1

0.1

­2

0.0

4

6

­1

8

1

2

­0.1

­0.1

27. We graph y  x 3  6x 2  11x  6 in the viewing

28. Since 16x 3  16x 2  x  1  16x 3  16x 2  x  1  0, we graph y  16x 3  16x 2  x  1 in the viewing

rectangle [1 4] by [01 01]. The solutions are x  100, x  200,and x  300.

rectangle [2 2] by [01 01]. The solutions are: x  100, x  025, and x  025.

0.1

0.1 ­1

1

2

3

4 ­2

­0.1

­1

1

2

­0.1

29. We first graph y  x 

x  1 in the viewing rectangle [1 5] by [01 01] and

find that the solution is near 16. Zooming in, we see that solutions is x  162.

0.1

­1

1

2

3

4

­0.1

 1  x2     1  x  1  x 2  0Since x is only defined

30. 1 

x

for x  0, we start with the viewing rectangle [1 5] by [1 1]. In this rectangle, there

appears to be an exact solution at x  0 and

another solution between x  2 and x  25. We then use the viewing rectangle [23 235] by [001 001], and isolate the second solution as x  2314. Thus the solutions are x  0 and x  231.

1

­1 ­1

0.01

1

2

3

4

5

0.00

­0.01

2.32

2.34

5


SECTION 1.10 Solving Equations and Inequalities Graphically

31. We graph y  x 13  x in the viewing rectangle [3 3] by [1 1]. The solutions

1

are x  1, x  0, and x  1, as can be verified by substitution.

­3

­2

­1

1

2

3

­1

32. Since x 12 is defined only for x  0, we start by

1

graphing y  x 12  x 13  x in the viewing

0.01

rectangle [1 5] by [1 1] We see a solution at x  0 and another one between x  3 and x  35. We then use the viewing rectangle

[33 34] by [001 001], and isolate the second

­1

1

2

3

4

0.00

5

­1

3.35

3.40

­0.01

solution as x  331. Thus, the solutions are x  0 and x  331.   2x  1 and y  3x  5 in the viewing rectangle [0 9] by [0 5]   and see that the only solution to the equation 2x  1  3x  5 is x  4, which

33. We graph y 

can be verified by substitution.

4 2 0

0

2

  3  x and y  x 2  1 in the viewing rectangle [4 4] by [0 4]   and see that the equation 3  x  x 2  1 has solutions x  1 and x  2,

34. We graph y 

4

6

8

4 3

which can be verified by substitution.

2 1 ­4

 2x  1  1 and y  x in the viewing rectangle [1 6] by [0 6]  and see that the only solution to the equation 2x  1  1  x is x  4, which can

35. We graph y 

be verified by substitution.

­2

0

2

4

6 4 2 0

4

6

x  1 and y  8 in the viewing rectangle [2 6] by  [2 12] and see that the only solution to the equation 2x  x  1  8 is x  3,

36. We graph y  2x 

2

10

which can be verified by substitution.

5

­2

2

4

6

81


82

CHAPTER 1 Equations and Graphs

37. x 3  2x 2  x  1  0, so we start by graphing

100

the function y  x 3  2x 2  x  1 in the viewing

1

rectangle [10 10] by [100 100]. There

appear to be two solutions, one near x  0 and

­10

­5

another one between x  2 and x  3. We then use the viewing rectangle [1 5] by [1 1] and

5

10

­1

­100

1

2

3

4

5

­1

zoom in on the only solution, x  255. 38. x 4  8x 2  2  0. We start by graphing the

10

function y  x 4  8x 2  2 in the viewing

1

rectangle [10 10] by [10 10]. There appear to be four solutions between x  3 and x  3. We then use the viewing rectangle [5 5] by

­10

­5

5

[1 1], and zoom to find the four solutions x  278, x  051, x  051, and x  278. 39. x x  1 x  2  16 x 

x x  1 x  2  16 x  0. We start by graphing

the function y  x x  1 x  2  16 x in the

­4

viewing rectangle [5 5] by [10 10]. There

­4

­2

­10

­1

10

1

­2

appear to be three solutions. We then use the

10

2

4

­2

­1

­10

viewing rectangle [25 25] by [1 1] and

2

4

1

2

­1

zoom into the solutions at x  205, x  000, and x  105. 40. x 4  16  x 3 . We start by graphing the functions y1  x 4 and y2  16  x 3 in the viewing rectangle [10 10] by [5 40]. There appears to be two solutions, one near x  2 and another one near x  2. We then use the viewing

rectangle [24 22] by [27 29], and zoom in to find the solution at x  231. We then use the viewing rectangle [17 18] by [95 105], and zoom in to find the solution at x  179. 40

29

30 28

20 10 ­10

­5

5

10

­2.4

­2.3

27 ­2.2

10.4 10.2 10.0 9.8 9.6 1.70

1.75

41. We graph y  x 2 and y  3x  10 in the viewing rectangle [4 7] by [5 30].

1.80

30

The solution to the inequality is [2 5].

20 10 ­4

­2

2

4

6


SECTION 1.10 Solving Equations and Inequalities Graphically

42. Since 05x 2  0875x  025  05x 2  0875x  025  0, we graph

83

4

y  05x 2  0875x  025 in the viewing rectangle [3 1] by [5 5]. Thus the

2

solution to the inequality is [2 025].

­3

­2

­1

1

­2 ­4

43. Since x 3  11x  6x 2  6  x 3  6x 2  11x  6  0, we graph

4

y  x 3  6x 2  11x  6 in the viewing rectangle [0 5] by [5 5]. The solution

2

set is  100]  [200 300].

0

2

­2

4

­4

44. Since 16x 3  24x 2  9x  1 

0.01

4

16x 3  24x 2  9x  1  0, we graph

2

y  16x 3  24x 2  9x  1 in the viewing

rectangle [3 1] by [5 5]. From this rectangle,

­3

­2

­1

we see that x  1 is an x­intercept, but it is

­2

1

­1.0

0.00

­0.5

­4

unclear what is occurring between x  05 and

­0.01

x  0. We then use the viewing rectangle [1 0] by [001 001]. It shows y  0 at x  025. Thus in interval notation,

the solution is 1 025  025 .

45. Since x 13  x  x 13  x  0, we graph y  x 13  x

in the viewing rectangle [3 3] by [1 1]. From this, we

  05x 2  1  2 x  05x 2  1  2 x  0, we  graph y  05x 2  1  2 x in the viewing rectangle

46. Since

find that the solution set is 1 0  1 .

[1 1] by [1 1]. We locate the x­intercepts at

and y2  x in the same viewing rectangle, and see that

x  0535. Thus in interval notation, the solution is approximately  0535]  [0535 .

Another Method: As in Example 7, we graph y1  x 13 x 13  x for 1  x  0 and for 1  x. 1

1

2 1

­1.0 ­2

2 ­1

­2

­1 ­2

2

­0.5

0.5 ­1

1.0


84

CHAPTER 1 Equations and Graphs

47. Since x  12  x  12  x  12  x  12  0, we graph y  x  12  x  12 in the viewing

rectangle [2 2] by [5 5]. The solution set is  0.

48. Since x  12  x 3  x  12  x 3  0, we graph

y  x  12  x 3 in the viewing rectangle [4 4] by

[1 1]. The x­intercept is close to x  2. Using a trace

function, we obtain x  2148. Thus the solution is [2148 .

4 2

1 ­2

­1

­2

1

2

­4

­4

­2

2

4

­1

49. We graph the equations y  3x 2  3x and y  2x 2  4 in the viewing rectangle

[2 6] by [10 50]. We see that the two curves intersect at x  1 and at x  4,

40

the inequality 3x 2  3x  2x 2  4 has solution set 1 4.

20

and that the first curve is lower than the second for 1  x  4 . Thus, we see that

­2

2

50. We graph the equations y  5x 2  3x and y  3x 2  2 in the viewing rectangle

[3 2] by [5 25]. We see that the two curves intersect at x  2 and at x  12 ,

6

20

which can be verified by substitution. The first curve is higher than the second for

x  2 and for x  12 , so the solution set of the inequality 5x 2  3x  3x 2  2 is    2]  12   .

4

10

­3

­2

­1

1

2

51. We graph the equation y  x  32 x  2 x  5 in the viewing rectangle

[7 3] by [120 20] and see that the inequality x  32 x  2 x  5  0 has

the solution set  5]  3  [2 .

­6

­4

­2

2

­50 ­100

  52. We graph the equation y  4x 2 x 2  9 in the viewing rectangle [5 5] by   [100 100] and see that the inequality 4x 2 x 2  9  0 has the solution set [3 3].

100

­4

­2 ­100

2

4


SECTION 1.10 Solving Equations and Inequalities Graphically

53. To solve 5  3x  8x  20 by drawing the graph of a single equation, we isolate

85

1

all terms on the left­hand side: 5  3x  8x  20  5  3x  8x  20  8x  20  8x  20  11x  25  0 or 11x  25  0. We graph y  11x  25, and see that the solution is x  227, as in Example 2.

­1

1

2

3

­1

54. Graphing y  x 3  6x 2  9x and y 

x in the viewing rectangle [001 002]

0.2

by [005 02], we see that x  0 and x  001 are solutions of the equation  x 3  6x 2  9x  x.

0.1

­0.01

0.01

0.02

(c) We graph the equations y  15,000 and

55. (a) We graph the equation y  10x  05x 2  0001x 3  5000 in the viewing

y  10x  05x 2  0001x 3  5000 in the viewing

rectangle [0 450] by [5000 20000].

rectangle [250 450] by [11000 17000]. We use a zoom or trace function on a graphing calculator, and find that

20000

the company’s profits are greater than $15,000 for 279  x  400.

10000

16000

0

100 200 300 400

14000

(b) From the graph it appears that

12000

0  10x  005x 2  0001x 3  5000 for

300

400

100  x  500, and so 101 cooktops must be produced

to begin to make a profit. 56. (a)

(b) Using a zoom or trace function, we find that y  10 for x  667. We  x 2 could estimate this since if x  100, then 5280  000036. So for    x 2   15x. Solving 15x  10 we x  100 we have 15x  5280

15 10 5 0

0

50

100

get 15  100 or x  100 15  667 mi.

57. Answers will vary. 58. Calculators perform operations in the following order: exponents are applied before division and division is applied before addition. Therefore, Y_1=x^1/3 is interpreted as y  interpreted as y 

x x1  , which is the equation of a line. Likewise, Y_2=x/x+4 is 3 3

x  4  1  4  5. Instead, enter the following: Y_1=x^(1/3), Y_2=x/(x+4). x


86

CHAPTER 1 Equations and Graphs

1.11 MODELING VARIATION 1. If the quantities x and y are related by the equation y  5x then we say that y is directly proportional to x, and the constant of proportionality is 5. 5 2. If the quantities x and y are related by the equation y  then we say that y is inversely proportional to x, and the constant x of proportionality is 5. x 3. If the quantities x, y, and z are related by the equation z  5 then we say that z is directly proportional to x and inversely y proportional to y. 4. Because z is jointly proportional to x and y, we must have z  kx y. Substituting the given values, we get 10  k 4 5  20k  k  12 . Thus, x, y, and z are related by the equation z  12 x y.

5. (a) In the equation y  3x, y is directly proportional to x. (b) In the equation y  3x  1, y is not proportional to x.

3 , y is not proportional to x. x 1 3 (b) In the equation y  , y is inversely proportional to x. x

6. (a) In the equation y 

7. T  kx, where k is constant.

8. P  k, where k is constant.

9.  

10.   kmn, where k is constant.

k , where k is constant. z ks 11. y  , where k is constant. t

12. P 

k , where k is constant. T

 13. z  k y, where k is constant.

14. A 

kx 2 , where k is constant. t3

15. V  klh, where k is constant.

16. S  kr 2 2 , where k is constant.

k P2t 2  , where k is constant. 18. A  k x y, where k is constant. 3 b 19. Since y is directly proportional to x, y  kx. Since y  32 when x  8, we have 32  k 8  k  4. So y  4x. k k 24 20.  is inversely proportional to t, so   . Since   3 when t  8, we have 3   k  24, so   . t 8 t k k 75 21. A varies inversely as r, so A  . Since A  15 when r  5, we have 15   k  75. So A  . r 5 r 17. R 

22. P is directly proportional to T , so P  kT . Since P  60 when T  72, we have 60  k 72  k  56 . So P  56 T . 23. Since A is directly proportional to x and inversely proportional to t, A  have 42 

k 7 18x  k  18. Therefore, A  . 3 t

kx . Since A  42 when x  7 and t  3, we t

24. S  kpq. Since S  350 when p  7 and q  20, we have 350  k 7 20  350  140k  k  52 . So S  52 pq.

k k  k  216. 25. Since W is inversely proportional to the square of r, W  2 . Since W  24 when r  3, we have 24  r 32 216 So W  2 . r 10  15 xy xy  k  10. So t  10 . 26. t  k . Since t  125 when x  10, y  15, and r  12, we have 125  k r 12 r


SECTION 1.11 Modeling Variation

87

27. Since C is jointly proportional to l, , and h, we have C  klh. Since C  128 when l    h  2, we have 128  k 2 2 2  128  8k  k  16. Therefore, C  16lh.  2 28. H  kl 2 2 . Since H  36 when l  2 and   13 , we have 36  k 22 13  36  49 k  k  81. So H  81l 2 2 .

k k 275 k  29. R   . Since R  25 when x  121, 25    k  275. Thus, R   . 11 x x 121 a 2 2 abc abc . Since M  128 when a  d and b  c  2, we have 128  k  4k  k  32. So M  32 . 30. M  k d a d x3 31. (a) z  k 2 y (b) If we replace x with 3x and y with 2y, then z  k

  x3 27 , so z changes by a factor of 27 k  4. 4 y2 2y2

3x3

x2 32. (a) z  k 4 y

  x2 9 9. k 4 , so z changes by a factor of 16 (b) If we replace x with 3x and y with 2y, then z  k  16 y 2y4 3x2

33. (a) z  kx 3 y 5

(b) If we replace x with 3x and y with 2y, then z  k 3x3 2y5  864kx 3 y 5 , so z changes by a factor of 864.

k 34. (a) z  2 3 x y

(b) If we replace x with 3x and y with 2y, then z 

k 3x2 2y3

1 k 1 . , so z changes by a factor of 72 72 x 2 y 3

35. (a) The force F needed is F  kx.

(b) Since F  30 N when x  9 cm and the spring’s natural length is 5 cm, we have 30  k 9  5  k  75 Ncm. (c) From part (b), we have F  75x. Substituting x  11  5  6 into F  75x gives F  75 6  45 N.

36. (a) C  kpm

(b) Since C  60,000 when p  120 and m  4000, we get 60,000  k 120 4000  k  18 . So C  18 pm.

(c) Substituting p  92 and m  5000, we get C  18 92 5000  $57,500.

37. (a) P  ks 3 .

(b) Since P  96 when s  20, we get 96  k  203  k  0012 W(mih)3 . So P  0012s 3 .

(c) Substituting x  30, we get P  0012  303  324 watts.

38. Let the amount of soluble CO2 be x and the temperature of the water be T . Because these quantities are inversely k k proportional, x  . We are given that when T  273 K, x  3 g, so 3   k  819. Thus, if T  298, then T 273 819  275 g. x 298 39. D  ks 2 . Since D  150 when s  40, we have 150  k 402 , so k  009375. Thus, D  009375s 2 . If D  200, then 200  009375s 2  s 2  21333, so s  46 mi/h (for safety reasons we round down).   40. L  ks 2 A. Since L  1700 when s  50 and A  500, we have 1700  k 502 500  k  000136. Thus   L  000136s 2 A. When A  600 and s  40 we get the lift is L  000136 402 600  13056 lb.

41. F  k As 2 . Since F  220 when A  40 and s  5. Solving for k we have 220  k 40 52  220  1000k   k  022. Now when A  28 and F  175 we get 175  0220 28 s 2  284090  s 2 so s  284090  533 mi/h.


88

CHAPTER 1 Equations and Graphs

42. (a) T 2  kd 3

3  (b) Substituting T  365 and d  93  106 , we get 3652  k  93  106  k  166  1019 .  3 (c) T 2  166  1019 279  109  360  109  T  600  104 . Hence the period of Neptune is 6.00104 days 164 years.

43. (a) P 

kT . V

(b) Substituting P  332, T  400, and V  100, we get 332  P

83T . V

(c) Substituting T  500 and V  80, we have P  about 519 kPa.

k 400  k  83. Thus k  83 and the equation is 100

83 500  51875 kPa. Hence the pressure of the sample of gas is 80

s 2 r (b) For the first car we have 1  1600 and s1  60 and for the second car we have 2  2500. Since the forces are equal

44. (a) F  k

we have k

2500  s22 16  602 1600  602 k   s22 , so s2  48 mi/h. r r 25

k 45. (a) The loudness L is inversely proportional to the square of the distance d, so L  2 . d k (b) Substituting d  10 and L  70, we have 70  2  k  7000. 10   1 k k , so the loudness is changed by a factor of 14 .  (c) Substituting 2d for d, we have L  4 d2 2d2   k k , so the loudness is changed by a factor of 4. (d) Substituting 12 d for d, we have L   2  4 d2 1d 2

46. (a) The power P is jointly proportional to the area A and the cube of the velocity , so P  k A 3 .   (b) Substituting 2 for  and 12 A for A, we have P  k 12 A 23  4k A 3 , so the power is changed by a factor of 4.  3 (c) Substituting 12  for  and 3A for A, we have P  k 3A 12   38 Ak 3 , so the power is changed by a factor of 38 . 47. (a) R 

kL d2

k 12 7  0002916.  k  2400 00052 7 3 4375 (c) Substituting L  3 and d  0008, we have R    137 ohms.  2400 00082 32 (b) Since R  140 when L  12 and d  0005, we get 140 

(d) If we substitute 2d for d and 3L for L, then R 

k 3L 2d2

3 kL , so the resistance is changed by a factor of 34 . 4 d2

48. Let S be the final size of the cabbage, in pounds, let N be the amount of nutrients it receives, in ounces, and let c be the N number of other cabbages around it. Then S  k . When N  20 and c  12, we have S  30, so substituting, we have c   N 20 30  k 12  k  18. Thus S  18 . When N  10 and c  5, the final size is S  18 10 5  36 lb. c


SECTION 1.11 Modeling Variation

89

  ES k60004 6000 4  204  160,000. So the sun   300 EE k3004 produces 160,000 times the radiation energy per unit area than the Earth.

49. (a) For the sun, E S  k60004 and for earth E E  k3004 . Thus

(b) The surface area of the sun is 4 435,0002 and the surface area of the Earth is 4 3,9602 . So the sun has   4 435,0002 435,000 2  times the surface area of the Earth. Thus the total radiation emitted by the sun is 3,960 4 3,9602   435,000 2 160,000   1,930,670,340 times the total radiation emitted by the Earth. 3,960  50. (a) Let T and l be the period and the length of the pendulum, respectively. Then T  k l.

 T2 T2 2T 2 (b) T  k l  T 2  k 2 l  l  2 . If the period is doubled, the new length is  4 2  4l. So we would 2 k k k quadruple the length l to double the period T .

51. (a) Since f is inversely proportional to L, we have f  (b) If we replace L by 2L we have

k , where k is a positive constant. L

k k  12   12 f . So the frequency of the vibration is cut in half. 2L L

52. (a) Since r is jointly proportional to x and P  x, we have r  kx P  x, where k is a positive constant.

(b) When 10 people are infected the rate is r  k10 5000  10  49,900k. When 1000 people are infected the rate is r  k  1000  5000  1000  4,000,000k. So the rate is much higher when 1000 people are infected. Comparing 1000 people infected 4,000,000k these rates, we find that   80. So the infection rate when 1000 people are infected 10 people infected 49,900k is about 80 times as large as when 10 people are infected. (c) When the entire population is infected the rate is r  k 5000 5000  5000  0. This makes sense since there are no more people who can be infected.

02  2105 . 1002 Now, because the range is R  500 km for these values of e and , we can use the first equation to calculate the value of C: C 500   C  100, so a full charge for this vehicle is 100 kWh. 02 C Substituting e  k 2 into the first given equation, we have R  2 . k 100 100  296 km; and at 80 kmh, its range is R   781 km. So at 130 kmh, its range is R  2 5 2  10 130 2  105 802

53. We substitute e  02 kWhkm and   100 km/h into the second given equation: 02  k 1002  k 

54. (a) We are given that the mass flow rate is m  A0  0 . Because m is constant, we can equate the flow rates for the wider A  and narrower sections of pipe: A0  0   A    0 0 . The velocity of the fluid is indeed inversely proportional A to the cross­sectional area of the pipe. r02  0 A0  0 06 m2 5 ms   45 ms. Thus,  A r 2 02 m2 the fluid flows through the constricted section of pipe at 45 m/s.

(b) We substitute the given values into our expression for :  

L 25  1026 14 . 55. Using B  k 2 with k  0080, L  25  1026 , and d  24  1019 , we have B  0080  2  347  10 d 24  1019 The star’s apparent brightness is about 347  1014 Wm2 .


90

CHAPTER 1 Equations and Graphs

 L L L 2 56. First, we solve B  k 2 for d: d  k  d  k because d is positive. Substituting k  0080, L  58  1030 , and B B d  58  1030 16 B  82  10 , we find d  0080  238  1022 , so the star is approximately 238  1022 m from earth. 82  1016 57. Examples include radioactive decay and exponential growth in biology.

CHAPTER 1 REVIEW 1. (a)

y

Q

(b) The distance from P to Q is  d P Q  5  22  12  02    49  144  193     3 5  2 12  0 (c) The midpoint is    6 . 2 2 2

1 1

(d) The line has slope m 

P

x

12  0   12 7 , and has 5  2

equation y  0   12 7 x  2 

24 y   12 7 x  7  12x  7y  24  0. y

Q

(e) The radius of this circle was found in part (b). It is  r  d P Q  193. So an equation is  2 193  x  22  y 2  193. x  22  y  02  Q

y

2

1 1

2. (a)

x

1

Q

x

P

P

y 1

P 2

P

x

(b) The distance from P to Q is  d P Q  2  72  11  12     25  100  125  5 5     2  7 11  1 9 (c) The midpoint is    6 . 2 2 2


CHAPTER 1

10 11  1   2, 27 5 and its equation is y  11  2 x  2 

(d) The line has slope m 

y  11  2x  4  y  2x  15. y

Review

(e) The radius of this circle was found in part (b). It is  r  d P, Q  5 5. So an equation is   2 x  72  y  12  5 5  x  72  y  12  125.

1 1

y

x

P

2 2

x

P

Q

Q

y

3. (a)

4

P

x

4 Q

(d) The line has slope m 

(c) The midpoint is

16 8 2  14   6  4 10 5

and equation y  2   85 x  6 

8 38 y  2   85 x  48 5  y  5x  5 . y

6  4 2  14  2 2

x

4

 1 6.

(e) The radius of this circle was found in part (b). It is  r  d P Q  2 89. So an equation is   2 [x  6]2  y  22  2 89  x  62  y  22  356.

4

P

(b) The distance from P to Q is  d P Q  6  42  [2  14]2     100  256  356  2 89

P

Q

y

4 4

x

Q

y

4. (a)

2 Q

2

P

x

(b) The distance from P to Q is  d P Q  [5  3]2  [2  6]2     64  16  80  4 5. (c) The midpoint is

5  3 2  6  2 2

 1 4.

91


92

CHAPTER 1 Equations and Graphs

(d) The line has slope m 

2  6  48  12 , and 5  3

has equation y  2  12 x  5  y  2  12 x  52  y  12 x  92 .

(e) The radius of this circle was found in part (b). It is  r  d P Q  4 5. So an equation is  2   2 x  52  y  2  4 5  x  52  y  22  80.

y

y

2 2

Q

P

x

2 Q

5. x y  x  4 or y  2

2

x

P

6. x y  x  1 and y  4

y

y

1

1 1

x

1

x

   7. d A C  74 and 4  12  4  32  4  12  4  32     d B C  5  12  3  32  5  12  3  32  72. Therefore, B is closer to C. 8. The circle with center 3 4 and radius

 2  6 has equation x  32  y  42  6  x  32  y  42  6.

9. The center is C  5 1, and the point P  0 0 is on the circle. The radius of the circle is    r  d P C  0  52  0  12 = 0  52  0  12  26. Thus, the equation of the circle is x  52  y  12  26.

    1 11 21 38    , and the radius is 12 of the distance from P to Q, or 10. The midpoint of segment P Q is 2 2 2 2    r  12  d P, Q  12 2  12  3  82  12 2  12  3  82  r  12 34. Thus the equation is 2  2  x  12  y  11  17 2 2 .


CHAPTER 1

11. (a) x 2  y 2  8x  2y  13  0  x 2  8x  y 2  2y  13      x 2  8x  16  y 2  2y  1  13  16  1 

Review

93

(b) The circle has center 4 1 and radius 2.

y

x  42  y  12  4, an equation of a circle.

1 x

1

12. (a) 2x 2  2y 2  2x  8y  12  x 2  x  y 2  4y  14      x 2  x  14  y 2  4y  4  14  14  4  

x  12

2

  (b) The circle has center 12  2 

and radius 3 2 2 . y

 y  22  92 , an equation of a circle.

1l 1

x

    13. (a) x 2  y 2  72  12x  x 2  12x  y 2  72  x 2  12x  36  y 2  72  36  x  62  y 2  36. Since the left side of this equation must be greater than or equal to zero, this equation has no graph.

14. (a) x 2  y 2  6x  10y  34  0  x 2  6x  y 2  10y  34      x 2  6x  9  y 2  10y  25  34  9  25  x  32  y  52  0, an equation of a point.

(b) This is the equation of the point 3 5. y

1l 1

x


94

CHAPTER 1 Equations and Graphs

15. y  2  3x x

y

2

8

0

2 3

17.

16. 2x  y  1  0  y  2x  1

y

2 0

1

x

18. y

x

y

2

14 7

2

19. y  16  x 2 x

y

3

7

1

15

0

16

1

15

3

7

y

0

0

1

1

2

4

3

9

3

 12

0

1 1 1

x

1

x

1

x

y

x

y

4

5

0

0

4

5

20. 8x  y 2  0  y 2  8x

y

y

y x   0  5x  4y  0 4 5

x

x

y

8

8

2

4

0

0

1

y

1

2 1

x

y x

2

2

0

2

21. x 

y 0

1

y x   1  y  72 x  7 2 7

0

x

y

 22. y   1  x 2

y

0

1

x

1 1 2

1

0 1

x

1

y

  23

1

0

1

x


CHAPTER 1

Review

95

23. x  16  y 2 (a) x­axis symmetry: replacing y by y gives x  16  y2  16  y 2 , which is the same as the original equation, so the graph is symmetric with respect to the x­axis. y­axis symmetry: replacing x by x gives x  16  y 2  x  y 2  16, which is not the same as the original equation, so the graph is not symmetric with respect to the y­axis. Origin symmetry: replacing x by x and y by y gives x  16  y2  x  16  y 2 , which is not the same as the original equation, so the graph is not symmetric with respect to the origin. (b) To find x­intercepts, we set y  0 and solve for x: x  16  02  16, so the x­intercept is 16.

To find y­intercepts, we set x  0 and solve for y: 0  16  y 2  y  4, so the y­intercept are 4 and 4.

24. x 2  4y 2  9 (a) x­axis symmetry: replacing y by y gives x 2  4 y2  9  x 2  4y 2  9, which is the same as the original equation, so the graph is symmetric with respect to the x­axis. y­axis symmetry: replacing x by x gives x2  4 y2  9  x 2  4y 2  9, which is the same as the original equation, so the graph is symmetric with respect to the y­axis. Origin symmetry: replacing x by x and y by y gives x2  4 y2  9  x 2  4y 2  9, the same as the original equation, so the graph is symmetric with respect to the origin. (b) To find x­intercepts, we set y  0 and solve for x: x 2  4 02  9  x  3, so the x­intercepts are 3 and 3. To find y­intercepts, we set x  0 and solve for y: 02  4y 2  9  y   32 , so the y­intercepts are 32 and  32 .

25. x 2  9y  9 (a) x­axis symmetry: replacing y by y gives x 2  9 y  9  x 2  9y  9, so the graph is not symmetric with respect to the x­axis. y­axis symmetry: replacing x by x gives x2  9y  9  x 2  9y  9, so the graph is symmetric with respect to the y­axis. Origin symmetry: replacing x by x and y by y gives x2  9 y  9  x 2  9y  9, so the graph is not symmetric with respect to the origin. (b) To find x­intercepts, we set y  0 and solve for x: x 2  9 0  9  x  3, so the x­intercepts are 3 and 3. To find y­intercepts, we set x  0 and solve for y: 02  9y  9  y  1, so the y­intercept is 1.

26. x  12  y 2  4 (a) x­axis symmetry: replacing y by y gives x  12  y2  4  x  12  y 2  4, so the graph is symmetric with respect to the x­axis. y­axis symmetry: replacing x by x gives x  12  y 2  4  1  x2  y 2  4, so the graph is not symmetric with respect to the y­axis. Origin symmetry: replacing x by x and y by y gives x  12  y2  4  1  x2  y 2  4, so the graph is not symmetric with respect to the origin. (b) To find x­intercepts, we set y  0 and solve for x: x  12  02  4  x  12  4  x  3 or 1, so the x­intercepts are 3 and 1.  To find y­intercepts, we set x  0 and solve for y: 0  12  y 2  4  y 2  3  y   3, so the y­intercepts are   3 and  3.


96

CHAPTER 1 Equations and Graphs

27. 9x 2  16y 2  144 (a) x­axis symmetry: replacing y by y gives 9x 2  16 y2  144  9x 2  16y 2  144, so the graph is symmetric with respect to the x­axis. y­axis symmetry: replacing x by x gives 9 x2  16y 2  144  9x 2  16y 2  144, so the graph is symmetric with respect to the y­axis. Origin symmetry: replacing x by x and y by y gives 9 x2  16 y2  144  9x 2  16y 2  144, so the graph is symmetric with respect to the origin. (b) To find x­intercepts, we set y  0 and solve for x: 9x 2  16 02  144  9x 2  144  x  4, so the x­intercepts are 4 and 4. To find y­intercepts, we set x  0 and solve for y: 9 02  16y 2  144  16y 2  144, so there is no y­intercept. 4 x 4 (a) x­axis symmetry: replacing y by y gives y  , which is different from the original equation, so the graph is not x symmetric with respect to the x­axis. 4 , which is different from the original equation, so the graph is not y­axis symmetry: replacing x by x gives y  x symmetric with respect to the y­axis. 4 4 Origin symmetry: replacing x by x and y by y gives y   y  , so the graph is symmetric with respect x x to the origin. 4 (b) To find x­intercepts, we set y  0 and solve for x: 0  has no solution, so there is no x­intercept. x To find y­intercepts, we set x  0 and solve for y. But we cannot substitute x  0, so there is no y­intercept.

28. y 

29. x 2  4x y  y 2  1 (a) x­axis symmetry: replacing y by y gives x 2  4x y  y2  1, which is different from the original equation, so the graph is not symmetric with respect to the x­axis. y­axis symmetry: replacing x by x gives x2  4 x y  y 2  1, which is different from the original equation, so the graph is not symmetric with respect to the y­axis. Origin symmetry: replacing x by x and y by y gives x2  4 x y  y2  1  x 2  4x y  y 2  1, so the graph is symmetric with respect to the origin. (b) To find x­intercepts, we set y  0 and solve for x: x 2  4x 0  02  1  x 2  1  x  1, so the x­intercepts are 1 and 1. To find y­intercepts, we set x  0 and solve for y: 02  4 0 y  y 2  1  y 2  1  y  1, so the y­intercepts are 1 and 1. 30. x 3  x y 2  5 (a) x­axis symmetry: replacing y by y gives x 3  x y2  5  x 3  x y 2  5, so the graph is symmetric with respect to the x­axis. y­axis symmetry: replacing x by x gives x3  x y 2  5, which is different from the original equation, so the graph is not symmetric with respect to the y­axis. Origin symmetry: replacing x by x and y by y gives x3  x y2  5, which is different from the original equation, so the graph is not symmetric with respect to the origin.   (b) To find x­intercepts, we set y  0 and solve for x: x 3  x 02  5  x 3  5  x  3 5, so the x­intercept is 3 5. To find y­intercepts, we set x  0 and solve for y: 03  0y 2  5 has no solution, so there is no y­intercept.


CHAPTER 1

31. (a) We graph y  x 2  6x in the viewing rectangle [10 10] by [10 10].

32. (a) We graph y 

97

Review

 5  x in the viewing rectangle

[10 6] by [1 5].

10

4 2

­10

­5

5

10 ­10

­10

(b) From the graph, we see that the x­intercepts are 0

­5

5

(b) From the graph, we see that the x­intercept is 5 and

and 6 and the y­intercept is 0.

the y­intercept is approximately 224.

33. (a) We graph y  x 3  4x 2  5x in the viewing rectangle [4 8] by [30 20]. 20

x2 x2  y2  1  y2  1   34. (a) We graph 4 4  x2 in the viewing rectangle [3 3] by y  1 4 [2 2].

­4

­2

2

4

6

8

2 1

­20 ­3

(b) From the graph, we see that the x­intercepts are 1,

­2

­1 ­1

1

2

3

­2

0, and 5 and the y­intercept is 0.

(b) From the graph, we see that the x­intercepts are 2 and 2 and the y­intercepts are 1 and 1.

35. (a) The line that has slope 2 and y­intercept 6 has the slope­intercept equation

(c)

y

y  2x  6. (b) An equation of the line in general form is 2x  y  6  0.

1 1

x


98

CHAPTER 1 Equations and Graphs

36. (a) The line that has slope  12 and passes through the point 6 3 has

(c)

y

equation y  3   12 x  6  y  3   12 x  6  y   12 x.

(b)  12 x  3  y  3  x  6  2y  6  x  2y  0.

1

37. (a) The line that passes through the points 3 2 and 1 4 has slope m

(c)

1

x

1

x

1

x

1

x

y

3 4  2  , so, using the second point for convenience, an 1  3 2

equation of the line is y  4  32 x  1  y  32 x  52 .

1

(b) y  32 x  52  2y  3x  5  3x  2y  5  0.

38. (a) The line that has x­intercept 4 and y­intercept 12 passes through the points

(c)

y

12  0  3 and the equation is 4 0 and 0 12, so m  04 y  0  3 x  4  y  3x  12. (b) y  3x  12  3x  y  12  0. 2

39. (a) The vertical line that passes through the point 3 2 has equation x  3.

(c)

y

(b) x  3  x  3  0. 1


CHAPTER 1

40. (a) The horizontal line with y­intercept 5 has equation y  5.

(c)

99

Review

y

(b) y  5  y  5  0. 1 1

41. (a) The line containing 2 4 and 4 4 has slope

(c)

x

y

4  4 8 m   4, and the line passing through the origin with 42 2 this slope has equation y  4x. 1

(b) y  4x  4x  y  0.

42. (a) 2x  5y  10  5y  2x  10  y  25 x  2, so the given line has slope (c)

1

x

1

x

y

m  25 . Thus, an equation of the line passing through 1 1 parallel to this line is y  1  25 x  1  y  25 x  35 .

1

(b) y  25 x  35  5y  2x  3  2x  5y  3  0.

43. (a) The line y  12 x  10 has slope 12 , so a line perpendicular to this one has

(c)

y

1  2. In particular, the line passing through the origin slope  12

perpendicular to the given line has equation y  2x. (b) y  2x  2x  y  0.

1 1

x


100

CHAPTER 1 Equations and Graphs

44. (a) The line with equation x  4y  7  0  4y  x  7  y  14 x  74 has (c)

y

1  4. Since the slope 14 , so any line perpendicular to it has slope  14

desired line passes through 2 3, it has equation y  3  4 x  2  y  4x  11.

(b) y  4x  11  4x  y  11  0. 1 1

x

45. The line with equation y   13 x  1 has slope  13 . The line with equation 9y  3x  3  0  9y  3x  3  y   13 x  13 also has slope  13 , so the lines are parallel.

46. The line with equation 5x  8y  3  8y  5x  3  y  58 x  38 has slope 58 . The line with equation 10y  16x  1  1 has slope  8   1 , so the lines are perpendicular. 10y  16x  1  y   85 x  10 5 58

47. (a) The slope, 03, represents the increase in length of the spring for each unit increase in weight . The s­intercept is the resting or natural length of the spring. (b) When   5, s  03 5  25  15  25  40 inches.

48. (a) We use the information to find two points, 0 60000 and 3 70500. Then the slope is 70,500  60,000 10,500 m   3,500. So S  3,500t  60,000. 30 3 (b) The slope represents the accountant’s annual increase in salary, $3500, and the S­intercept represents the accountant’s initial salary, $60,000. (c) When t  12, the accountant’s salary will be S  3500 12  60,000  42,000  60,000  $102,000.

49. x 2  9x  14  0  x  7 x  2  0  x  7 or x  2.

50. x 2  24x  144  0  x  122  0  x  12  0  x  12.

51. 2x 2  x  1  2x 2  x  1  0  2x  1 x  1  0. So either 2x  1  0  2x  1  x  12 ; or x  1  0  x  1.   52. x  1  x 2  3  x  1  x 2  3  x 2  x  2  0  x  1 x  2  0  x  1 or x  2. However, x  1   fails to satisfy the original equation because 12  3  2, which is undefined. Thus, the only solution is x  2.   53. 0  4x 3  25x  x 4x 2  25  x 2x  5 2x  5  0. So either x  0; or 2x  5  0  2x  5  x  52 ; or 2x  5  0  2x  5  x   52 .

   54. x 3  2x 2  5x  10  0  x 2 x  2  5 x  2  0  x  2 x 2  5  0  x  2 or x   5.

55. 3x 2  4x  1  0           2 2 7 4 42 431 b b2 4ac 4 1612 42 7 7 4 28    6    2 . x 2a 6 6 6 3 23

    3 32 419 2 4ac b b   3 2936  3 227 , which are not real numbers. 56. x 2  3x  9  0  x  2a 21

There is no real solution. 1 2 57.   3  x  1  2 x  3 x x  1  x  1  2x  3x 2  3x  0  3x 2  6x  1  x x 1          2 3 6 6 62 431 b b2 4ac 6 3612 6 24 62 6 3 6 . x       2a 6 6 6 6 3 23


CHAPTER 1

58.

Review

101

1 8 x   2  x x  2  x  2  8  x 2  2x  x  2  8  x 2  3x  10  0  x  2 x  5  0 x 2 x 2 x 4  x  2 or x  5. However, since x  2 makes the expression undefined, we reject this solution. Hence the only solution is x  5.

     59. x 4  8x 2  9  0  x 2  9 x 2  1  0  x  3 x  3 x 2  1  0  x  3  0  x  3, or x  3  0  x  3, however x 2  1  0 has no real solution. The solutions are x  3.

  60. x  4 x  32. Let u  x. Then u 2  4u  32  u 2  4u  32  0  u  8 u  4  0 So either u  8  0 or   u  4  0. If u  8  0, then u  8  x  8  x  64. If u  4  0, then u  4  x  4, which has no real solution. So the only solution is x  64. 61. x  32  3 x  3  4  0  [x  3  4] [x  3  1]  0  x  7 x  2  0  x  2 or x  7.   62. 9x 12  6x 12  x 32  0  x 12 9  6x  x 2  0  x 12 x  32  0  x  3.

63. x  7  4  x  7  4  x  7  4, so x  11 or x  3.

64. 2x  5  9 is equivalent to 2x  5  9  2x  5  9  x 

59 . So x  2 or x  7. 2

65. (a) 2  3i  1  4i  2  1  3  4 i  3  i

(b) 2  i 3  2i  6  4i  3i  2i 2  6  i  2  8  i

66. (a) 3  6i  6  4i  3  6  [6  4] i  3  2i   (b) 4i 2  12 i  8i  2i 2  8i  2  2  8i 4  2i 4  2i 2  i 8  8i  2i 2 6  8i 8  8i  2      65  85 i  2i 2i 2i 41 5 4  i2      (b) 1  1 1  1  1  i 1  i  1  i  i  i 2  1  1  2

67. (a)

2  3i 1  5i 2  5i  3i 2 2  3i 1  i    12  52 i   2 1i 2 1  i 1  i 1i     (b) 10  40  i 10  2i 10  20i 2  20

68. (a)

69. x 2  16  0  x 2  16  x  4i   70. x 2  12  x   12  2 3i b  71. x 2  6x  10  0  x  72. 2x 2  3x  2  0  x 

   6  62  4 1 10 6  36  40 b2  4ac    3  i 2a 2 1 2

 3 

 32  4 2 2 2 2

3

  3 7 7   i 4 4 4

   73. x 4  256  0  x 2  16 x 2  16  0  x  4 or x  4i   74. x 3  2x 2  4x  8  0  x  2 x 2  4  x  2 or x  2i


102

CHAPTER 1 Equations and Graphs

75. Let r be the athlete’s running speed, in mi/h. Then they cycle at r  8 mi/h. Rate

Cycle

r 8

Run

r

Time

Distance

4 r 8 25 r

4 25

25 4   1. Multiplying by 2r r  8, we r 8 r get 4 2r   25 2 r  8  2r r  8  8r  5r  40  2r 2  16r  0  2r 2  3r  40  Since the total time of the workout is 1 hour, we have

   3 32 4240 3 9320 3 329 . Since r  0, we reject the negative value. The athlete’s running r    22 4 4  3 329  378 mi/h. speed is r  4 x2  1500  20x  x 2  x 2  20x  1500  0  x  30 x  50  0. So 76. Substituting 75 for d, we have 75  x 

20 x  30 or x  50. The speed of the car was 30 mi/h.

77. Let x be the length of one side in cm. Then 28  x is the length of the other side. Using the Pythagorean Theorem, we   have x 2  28  x2  202  x 2  784  56x  x 2  400  2x 2  56x  384  0  2 x 2  28x  192  0 

2 x  12 x  16  0. So x  12 or x  16. If x  12, then the other side is 28  12  16. Similarly, if x  16, then the other side is 12. The sides are 12 cm and 16 cm. 80 and the total amount of fencing material is 78. Let l be length of each garden plot. The width of each plot is then l     480 80 4 l  6  88. Thus 4l   88  4l 2  480  88l  4l 2  88l  480  0  4 l 2  22l  120  0  l l 4 l  10 l  12  0. So l  10 or l  12. If l  10 ft, then the width of each plot is 80 10  8 ft. If l  12 ft, then the

width of each plot is 80 12  667 ft. Both solutions are possible.

80. 12  x  7x  12  8x  32  x.   Interval:  32

79. 3x  2  11  3x  9  x  3. Interval: 3 . Graph:

-3

Graph:

81. 3  x  2x  7  10  3x  10 3 x   Interval: 10 3  Graph:

3 2

82. 7  3x  1  1  6  3x  0  2  x  0 Interval: [2 0]. Graph:

10 3

_2

0

83. x 2  7x  8  0  x  8 x  1  0. The expression on the left of the inequality changes sign where x  8 and where x  1. Thus we must check the intervals in the following table. Interval Sign of x  8 Sign of x  1

Sign of x  8 x  1

 1

1 8

8 

Interval:  1  8   Graph:

_1

8


CHAPTER 1

Review

103

84. x 2  1  x 2  1  0  x  1 x  1  0. The expression on the left of the inequality changes sign when x  1 and x  1. Thus we must check the intervals in the following table. Interval: [1 1]

Interval Sign of x  1

 1

1 1

1 

Sign of x  1

Sign of x  1 x  1

85.

Graph:

_1

1

2x  5 2x  5 x 1 x 4 2x  5 1 1  0  0  0. The expression on the left of the inequality x 1 x 1 x 1 x 1 x 1 changes sign where x  1 and where x  4. Thus we must check the intervals in the following table. Interval Sign of x  4

4 1

1 

defined at this value. Thus the solution is [4 1.

Graph:

Sign of x  1 Sign of

We exclude x  1, since the expression is not

 4

x 4 x 1

_4

_1

86. 2x 2  x  3  2x 2  x  3  0  2x  3 x  1  0. The expression on the left of the inequality changes sign when 1 and 32 . Thus we must check the intervals in the following table. Interval

 1

Sign of 2x  3

Sign of x  1

Sign of 2x  3 x  1

87.

  1 32 

 3 2 

  Interval:  1]  32   Graph:

3 2

_1

x 4 x 4 0  0. The expression on the left of the inequality changes sign where x  2, where x  2, 2  2 x  2 x x 4 and where x  4. Thus we must check the intervals in the following table. Interval

Sign of x  4 Sign of x  2

 2

2 2

2 4

4 

Sign of x  2     x 4 Sign of     x  2 x  2 Since the expression is not defined when x  2we exclude these values and the solution is  2  2 4]. Graph:

_2

2

4


104

88.

CHAPTER 1 Equations and Graphs

5 5 5  0  0 2  0. The 0 2 x  1 x  2 x  2 x x  1  4 x  1 x  1 x  4 expression on the left of the inequality changes sign when 2 1and 2. Thus we must check the intervals in the following table. 5

x 3  x 2  4x  4

Interval Sign of x  1

 2

2 1

1 2

2 

Sign of x  2 Sign of x  2

5 Sign of x  1 x  2 x  2 Interval:  2  1 2 Graph:

_2

1

2

89. x  5  3  3  x  5  3  2  x  8. Interval: [2 8] Graph:

2

90. x  4  002  002  x  4  002  398  x  402 Interval: 398 402

8

Graph:

3.98

4.02

91. 2x  1  1 is equivalent to 2x  1  1 or 2x  1  1. Case 1: 2x  1  1  2x  0  x  0. Case 2: 2x  1  1  2x  2  x  1. Interval:  1]  [0 . Graph:

_1

0

92. x  1 is the distance between x and 1 on the number line, and x  3 is the distance between x and 3. We want those points that are closer to 1 than to 3. Since 2 is midway between 1 and 3, we get x   2 as the solution. Graph: 2

93. (a) For

 24  x  3x 2 to define a real number, we must have 24  x  3x 2  0  8  3x 3  x  0. The expression

on the left of the inequality changes sign where 8  3x  0  3x  8  x  83 ; or where x  3. Thus we must

check the intervals in the following table. Interval Sign of 8  3x Sign of 3  x

Sign of 8  3x 3  x

 3 

  3 83 

 8 3

  Interval: 3 83 . Graph:

_3

8 3


CHAPTER 1

(b) For  4

1 x  x4

Review

105

    to define a real number we must have x  x 4  0  x 1  x 3  0  x 1  x 1  x  x 2  0.

The expression on the left of the inequality changes sign where x  0; or where x  1; or where 1  x  x 2  0    12 411 x  1 21  1 214 which is imaginary. We check the intervals in the following table. Interval: 0 1. Interval

 0

0 1

Sign of x

1 

Sign of 1  x

Sign of 1  x  x 2

  Sign of x 1  x 1  x  x 2 

6 9 6 9  r3   3  r  3 . Thus r  94. We have 8  43 r 3  12     

Graph: 0

 3

1

  6 3 9  .  

95. From the graph, we see that the graphs of y  x 2  4x and y  x  6 intersect at x  1 and x  6, so these are the solutions of the equation x 2  4x  x  6.

96. From the graph, we see that the graph of y  x 2  4x crosses the x­axis at x  0 and x  4, so these are the solutions of the equation x 2  4x  0.

97. From the graph, we see that the graph of y  x 2  4x lies below the graph of y  x  6 for 1  x  6, so the inequality x 2  4x  x  6 is satisfied on the interval [1 6].

98. From the graph, we see that the graph of y  x 2  4x lies above the graph of y  x  6 for   x  1 and 6  x  , so the inequality x 2  4x  x  6 is satisfied on the intervals  1] and [6 .

99. From the graph, we see that the graph of y  x 2  4x lies above the x­axis for x  0 and for x  4, so the inequality x 2  4x  0 is satisfied on the intervals  0] and [4 .

100. From the graph, we see that the graph of y  x 2  4x lies below the x­axis for 0  x  4, so the inequality x 2  4x  0 is satisfied on the interval [0 4]. 101. x 2  4x  2x  7. We graph the equations y1  x 2  4x 102.

x  4  x 2  5. We graph the equations y1 

x 4

and y2  2x  7 in the viewing rectangle [10 10] by

and y2  x 2  5 in the viewing rectangle [4 5] by

solutions x  1 and x  7.

solutions x  250 and x  276.

[5 25]. Using a zoom or trace function, we get the

­10

­5

[0 10]. Using a zoom or trace function, we get the

20

10

10

5

5

10

­4

­2

0

2

4


106

CHAPTER 1 Equations and Graphs

103. x 4  9x 2  x  9. We graph the equations y1  x 4  9x 2 104. x  3  5  2. We graph the equations and y2  x  9 in the viewing rectangle [5 5] by

y1  x  3  5 and y2  2 in the viewing rectangle

[25 10]. Using a zoom or trace function, we get the

[20 20] by [0 10]. Using Zoom and/or Trace, we get the

solutions x  272, x  115, x  100, and x  287.

solutions x  10, x  6, x  0, and x  4.

10 ­4

10

­2 ­10

2

5

4

­20

­20

105. x 2  12  4x. We graph the equations y1  x 2 and

y2  12  4x in the viewing rectangle [8 4] by [0 40].

­10

0

10

20

106. x 3  4x 2  5x  2. We graph the equations

y1  x 3  4x 2  5x and y2  2 in the viewing rectangle

Using a zoom or trace function, we find the points of

[10 10] by [5 5]. We find that the point of intersection

intersection are at x  6 and x  2. Since we want

is at x  507. Since we want x 3  4x 2  5x  2, the solution is the interval 507 .

x 2  12  4x, the solution is the union of intervals

 6  2 .

4

40

2

30 20

­10

­5

10

­2

5

10

­4 ­8

­6

­4

­2

0

2

4

    108. x 2  16  10  0. We graph the equation     y1  x 4  4x 2 and y2  12 x  1 in the viewing rectangle y  x 2  16  10 in the viewing rectangle [10 10] by [5 5] by [5 5]. We find the points of intersection are [10 10]. Using a zoom or trace function, we find that the at x  185, x  060, x  045, and x  200. Since x­intercepts are x  510 and x  245. Since we   we want x 4  4x 2  12 x  1, the solution is   want x 2  16  10  0, the solution is approximately 185 060  045 200.  510]  [245 245]  [510 .

107. x 4  4x 2  12 x  1. We graph the equations

4

10

2 ­4

­2

­2 ­4

2

4 ­10

­5

5 ­10

10


CHAPTER 1

Review

107

109. Here the center is at 0 0, and the circle passes through the point 5 12, so the radius is    r  5  02  12  02  25  144  169  13. The equation of the circle is x 2  y 2  132 

x 2  y 2  169. The line shown is the tangent that passes through the point 5 12, so it is perpendicular to the line 12 12  0   . The slope of the line we seek is through the points 0 0 and 5 12. This line has slope m 1  5  0 5 5 1 1 5 x  5  y  12  5 x  25  m2    . Thus, an equation of the tangent line is y  12  12  12 12 m1 125 12 5 x  169  5x  12y  169  0. y  12 12

110. Because the circle is tangent to the x­axis at the point 5 0 and tangent to the y­axis at the point 0 5, the center is at 5 5 and the radius is 5. Thus an equation is x  52  y  52  52  x  52  y  52  25. The slope of 4 4 51    , so an equation of the line we seek is the line passing through the points 8 1 and 5 5 is m  58 3 3 y  1   43 x  8  4x  3y  35  0.

111. Since M varies directly as z we have M  kz. Substituting M  120 when z  15, we find 120  k 15  k  8. Therefore, M  8z. k k 112. Since z is inversely proportional to y, we have z  . Substituting z  12 when y  16, we find 12   k  192. y 16 192 Therefore z  . y k 113. (a) The intensity I varies inversely as the square of the distance d, so I  2 . d k (b) Substituting I  1000 when d  8, we get 1000   k  64,000. 82 64,000 64,000 (c) From parts (a) and (b), we have I  . Substituting d  20, we get I   160 candles. d2 202 114. Let f be the frequency of the string and l be the length of the string. Since the frequency is inversely proportional to the k 5280 k  k  5280. Therefore f  . For length, we have f  . Substituting l  12 when k  440, we find 440  l 12 l 5280  l  5280 f  660, we must have 660  660  8. So the string needs to be shortened to 8 inches. l 115. Let  be the terminal velocity of the parachutist in mi/h and  be his weight in pounds. Since the terminal velocity is  directly proportional to the square root of the weight, we have   k . Substituting   9 when   160, we solve   9 for k. This gives 9  k 160  k    0712. Thus   0712 . When   240, the terminal velocity is 160    0712 240  11 mi/h. 116. Let r be the maximum range of the baseball and  be the velocity of the baseball. Since the maximum range is directly proportional to the square of the velocity, we have r  l 2 . Substituting   60 and r  242, we find 242  k 602 

k  00672. If   70, then we have a maximum range of r  00672 702  3294 feet. k 117. The speed  is inversely proportional to the square root of the density d, so    . In fresh water with density d k 3 d1  1 gcm , the speed is  1  1480 ms, so 1480    k  1480. Thus, in seawater with density 10273 gcm3 , we 1 1480 have     1460 ms. 10273  118. (a) Because   122 , for fixed D the angular distance  is directly proportional to the wavelength . Thus, the angular D distance is smaller for shorter wavelengths.


108

CHAPTER 1 Equations and Graphs

 (b) For fixed , if we substitute D1  2D in the formula   122 then the angular distance becomes D     1  1  122  122  . Thus, for a fixed wavelength, if the diameter of the mirror is doubled, 1  122 D1 2D 2 D 2 the angular distance is halved.  c c  1  z2   c   1  z2  c    119. We solve the first equation for : 1  z  c c       c 1  z2  1 . c 1  z2   1  z2  c     1  z2  1  c 1  z2  1    1  z2  1     3  105 1  22  1 8 5  24  105 kms.  3  10 Substituting z  2 and c  3  105 , we find   2 10 1  2  1  , so substituting  from above and H0  208 Mlykms, we have By Hubble’s Law,   H0 D  D  H0 24  105  11,538 megalight­years. 208 120. (a) The first compression wave is traveling at about 6 kms, so it reaches the seismograph station 120 km away in about D

120  20 seconds. 6

The first transverse wave is traveling at about 4 kms, so it reaches the seismograph station 120 km away in about 120  30 seconds. 4

Thus, the time difference is about 10 seconds. (b) If the epicenter is d km from the seismograph station, then the time difference x between the detection of compression d d d and transverse waves is x    seconds. Thus, d  12x. This agrees with our result in part (a). 4 6 12 121. (a) y  2 x  3 is the graph of the absolute value function y  x, stretched vertically by a factor of 2 and shifted downward 3 units. It has graph III (b) 2y  3x  2  y  32 x  1 represents a line with slope 32 , x­intercept 23 , and y­intercept 1. It has graph V.

(c) y  x 4 has y 0  0 and y  0 elsewhere. It is an even function, and has graph II.

(d) x  12  y  12  9 is the equation of a circle with center 1 1 and radius 3. It has graph IV.

(e) 6y  x  3  y   16 x  12 represents a line with slope  16 , x­intercept 3, and y­intercept 12 . It has graph I.

(f) y 

6x is an odd function with y 0  0 and horizontal asymptote the x­axis. It has graph VII. 1  x4

(g) x  y 3 is an odd function whose graph contains the points 1 1 and 1 1. It has graph VIII.

(h) x 2  2x  y 2  4y  1  0  x  12  y  22  1  1  4  4 is the equation of a circle with center 1 2 and radius 2. It has graph VI.


CHAPTER 1

Test

109

CHAPTER 1 TEST 1. (a)

y

There are several ways to determine the coordinates of S. The diagonals of a

S

P

square have equal length and are perpendicular. The diagonal P R is horizontal and has length is 6 units, so the diagonal QS is vertical and also has length 6.

R

Thus, the coordinates of S are 3 6.    (b) The length of P Q is 0  32  3  02  18  3 2. So the area of   2 P Q RS is 3 2  18.

1 1

Q

x

y

2. (a)

(b) The x­intercepts occur when y  0, so 0  4  x 2  x 2  4  x  2. The y­intercept occurs when x  0, so y  4.

(c) x­axis symmetry: y  4  x 2  y  x 2  4, which is not the same as the original equation, so the graph is not symmetric with respect to the x­axis.

1 x

1

y­axis symmetry: y  4  x2  4  x 2 , which is the same as the original

equation, so the graph is symmetric with respect to the y­axis.

Origin symmetry: y  4  x2  y  4  x 2 , which is not the same

as the original equation, so the graph is not symmetric with respect to the origin. 3. (a)

(b) The distance between P and Q is    d P Q  3  52  1  62  64  25  89.     3  5 1  6 (c) The midpoint is   1 72 . 2 2   (d) The center of the circle is the midpoint, 1 72 , and the length of the radius is  1 89 . Thus the equation of the circle whose diameter is P Q is 2 2   2 2   x  12  y  72  12 89  x  12  y  72  89 4.

y Q

P

1 1

4.

x

(a) x 2  y 2  5 has center 0 0 and  radius 5.

(b) x  12  y  32  4  22 has

(c) x 2  y 2  10x  16  0 

y

y

center 5 0 and radius 3.

center 1 3 and radius 2.

x  52  y 2  16  25  9 has y

1

1 1

x

1

x

1 1

x


110

CHAPTER 1 Equations and Graphs

5. (a) x  4  y 2 . To test for symmetry with respect to the x­axis, we replace y

y

with y: x  4  y2  x  4  y 2 , so the graph is symmetric with

respect to the x­axis.

To test for symmetry with respect to the y­axis, we replace x with x:

1

x  4  y 2 is different from the original equation, so the graph is not

1

x

1

x

symmetric with respect to the y­axis.

For symmetry with respect to the origin, we replace x with x and y with

y: x  4  y2  x  4  y 2 , which is different from the

original equation, so the graph is not symmetric with respect to the origin. To find x­intercepts, we set y  0 and solve for x: x  4  02  4, so the x­intercept is 4.

To find y­intercepts, we set x  0 and solve for y:: 0  4  y 2  y 2  4  y  2, so the y­intercepts are 2 and 2.

(b) y  x  2. To test for symmetry with respect to the x­axis, we replace y

y

with y: y  x  2 is different from the original equation, so the

graph is not symmetric with respect to the x­axis.

To test for symmetry with respect to the y­axis, we replace x with x:

1

y  x  2  x  2 is different from the original equation, so the

graph is not symmetric with respect to the y­axis.

To test for symmetry with respect to the origin, we replace x with x and

y with y: y  x  2  y   x  2, which is different from the

original equation, so the graph is not symmetric with respect to the origin. To find x­intercepts, we set y  0 and solve for x: 0  x  2  x  2  0  x  2, so the x­intercept is 2.

To find y­intercepts, we set x  0 and solve for y: y  0  2  2  2, so the y­intercept is 2.

6. (a) To find the x­intercept, we set y  0 and solve for x: 3x  5 0  15

(b)

y

 3x  15  x  5, so the x­intercept is 5.

To find the y­intercept, we set x  0 and solve for y: 3 0  5y  15  5y  15  y  3, so the y­intercept is 3.

(c) 3x  5y  15  5y  3x  15  y  35 x  3.

1 1

x

(d) From part (c), the slope is 35 .

(e) The slope of any line perpendicular to the given line is the negative 1  5. reciprocal of its slope, that is,  35 3

7. (a) The line through 6 7 and 1 3 has slope 2x  y  5  0.

3  7  2, so an equation is y  7  2 x  6  y  2x  5  16

(b) 3x  y  10  0  y  3x  10, so the slope of the line we seek is 3. Using the point­slope form, y  6  3 x  3  y  6  3x  9  3x  y  3  0. x y (c) Using the intercept form we get   1  2x  3y  12  2x  3y  12  0. 6 4


CHAPTER 1

8. (a) When x  100 we have T  008 100  4  8  4  4, so the

(b)

Test

111

T

temperature at one meter is 4 C.

(c) The slope represents the raise in temperature as the depth increase.

5

The T ­intercept is the surface temperature of the soil and the x­intercept represents the depth of the “frost line”, where the soil 20

below is not frozen.

40

60

80

100 120 x

_5

9. (a) x  5  14  12 x  2x  10  28  x  3x  18  x  6 (b)

2x 2x  1   2x x  2x  1 x  1 (x  1, x  0)  2x 2  2x 2  x  1  0  x  1  x  1 x 1 x

(c) x 2  x  12  0  x  4 x  3  0. So x  4 or x  3.      4  42  4 2 1 4  16  8 4  8 4  2 2 2  2     . (d) 2x 2  4x  1  0  x  2 2 4 4 4 2     (e) 3  x  5  2  3  x  5  4  1  x  5. (Note that this is impossible, so there can be no solution.) Squaring both sides again, we get 1  x  5  x  4. But this does not satisfy the original equation, so there is no solution. (You must always check your final answers if you have squared both sides when solving an equation, since extraneous answers may be introduced, as here.)     (f) x 4  3x 2  2  0  x 2  1 x 2  2  0. So x 2  1  0  x  1 or x 2  2  0  x   2. Thus the   solutions are x  1, x  1, x   2, and x  2. 10 10 10 2 (g) 3 x  4  10  0  3 x  4  10  x  4  10 3  x  4   3  x  4  3 . So x  4  3  3 or 22 2 22 x  4  10 3  3 . Thus the solutions are x  3 and x  3 .

10. (a) 3  2i  4  3i  3  4  2i  3i  7  i

(b) 3  2i  4  3i  3  4  2i  3i  1  5i

(c) 3  2i 4  3i  3  4  3  3i  2i  4  2i  3i  12  9i  8i  6i 2  12  i  6 1  18  i

3  2i 4  3i 12  17i  6i 2 6 17 12  17i  6 3  2i      i  4  3i 4  3i 4  3i 16  9 25 25 16  9i 2  24  124  1 (e) i 48  i 2         2   (f) 2  2 8  2  2  8  2 2  2 8  2  4  2i  4i  2  6  2i (d)

4  11. Using the Quadratic Formula, 2x 2  4x  3  0  x 

   4  8 42  4 2 3   1  22 i. 2 2 4

12. Let  be the width of the parcel of land. Then   70 is the length of the parcel of land. Then 2    702  1302 

2  2  140  4900  16,900  22  140  12,000  0  2  70  6000  0    50   120  0. So   50 or   120. Since   0, the width is   50 ft and the length is   70  120 ft.

13. (a) 4  5  3x  17  9  3x  12  3  x  4. Expressing in standard form we have: 4  x  3. Interval: [4 3. Graph:

_4

3


112

CHAPTER 1 Equations and Graphs

(b) x x  1 x  2  0. The expression on the left of the inequality changes sign when x  0, x  1, and x  2. Thus we must check the intervals in the following table. Interval Sign of x Sign of x  1 Sign of x  2

Sign of x x  1 x  2

 2

2 0

0 1

1 

From the table, the solution set is x  2  x  0 or 1  x. Interval: 2 0  1 . Graph:

_2

0

1

(c) x  4  3 is equivalent to 3  x  4  3  1  x  7. Interval: 1 7. Graph: (d)

1

7

3  2x x 2  x 3  2x 2x  x 2  3  2x 3  2x  x x  0   0  0 2x 2x 2x 2x 2x x 2  4x  3 x  1 x  3   0. The expression on the left of the inequality changes sign at x  1, x  2, and x 2 x 2 x  3. Thus we must check the intervals in the following table. Interval

 1

1 2

2 3

3 

Sign of x  1 Sign of x  3

Sign of x  2 x  1 x  3 Sign of x 2 Intervals: 1 2  3 

Graph:

1

2

3

14. 5  59 F  32  10  9  F  32  18  41  F  50. Thus the medicine is to be stored at a temperature between 41 F and 50 F.

 15. For 6x  x 2 to be defined as a real number 6x  x 2  0  x 6  x  0. The expression on the left of the inequality changes sign when x  0 and x  6. Thus we must check the intervals in the following table. Interval

Sign of x Sign of 6  x

 0

0 6

6 

Sign of x 6  x   From the table, we see that 6x  x 2 is defined when 0  x  6.


Fitting Lines to Data

16. (a) x 3  9x  1  0. We graph the equation

113

(b) x 2  1  x  1. We graph the equations

y  x 3  9x  1 in the viewing rectangle [5 5]

y1  x 2  1 and y2  x  1 in the viewing

by [10 10]. We find that the points of

rectangle [5 5] by [5 10]. We find that the

intersection occur at x  294, 011, 305.

points of intersection occur at x  1 and x  2. Since we want x 2  1  x  1, the solution is

10

the interval [1 2].

10 ­4

­2

2

4 5

­10 ­4

­2

2

4

­5

17. (a) M  k

h 2 L

  4 62

h 2 (b) Substituting   4, h  6, L  12, and M  4800, we have 4800  k  k  400. Thus M  400 . 12 L   3 102  12,000. So the beam can support 12,000 pounds. (c) Now if L  10,   3, and h  10, then M  400 10

FOCUS ON MODELING Fitting Lines to Data

1. (a) Using a graphing calculator, we obtain the regression line y  18807x  8265.

we get y  18807 58  8265  1917 cm.

y

180

160

140 40

(b) Using x  58 in the equation y  18807x  8265,

50

Femur length (cm)

x


114

FOCUS ON MODELING

2. (a) Using a graphing calculator,

(b) For a per capita GDP of $80,000, the model predicts

we obtain the regression line y  017565x  18265.

y

carbon emissions of y  017565 80  18265  159 tons per capita. For $32,000, it predicts

14 12 10 8 6 4 2

y  017565 32  18265  74 tons per capita. (c) A linear model is reasonable for these data. One limitation is that there are no data points beyond per­capita income above $51,000, so the model cannot reliably make predictions for GDPs much larger than $51,000.

0

10

60 x

20 30 40 50 GDP per capita ($000)

3. (a) Using a graphing calculator, we obtain the regression line y  6451x  01523.

(b) Using x  18 in the equation y  6451x  01523, we get y  6451 18  01523  116 years.

y 100

80 60 40 20 0

2

4

6

8

10

12

14

16

18

20 x

Diameter (in.)

4. (a) Using a graphing calculator, we obtain the regression

y  4857x  22097, we get y  265 chirps per

line y  4857x  22097.

minute.

y

200

100

0

50

60

(b) Using x  100 F in the equation

70

80

Temperature (°F)

90

x


Fitting Lines to Data

5. (a) Using a graphing calculator, we obtain the regression line y  013198x  72514

115

(b) Using x  25 in the regression line equation, we get y  013198 25  72514  395 million km2 .

y

This is roughly 10% less than the actual figure of

8

44 million km2 . (c) Despite fluctuations over brief periods, the model

6

seems fairly accurate. If external circumstances 4

change (reduced or increased CO2 emissions, for example), it may become less reliable. It is unlikely

2

to be accurate far into the future.

0

10

x

20

Years since 1994 6. (a) Using a graphing calculator, we obtain the regression line y  0168x  1989.

(b) Using the regression line equation y  0168x  1989, we get y  813% when

x  70%.

y

20

10

0

20

40

60

80

100 x

Flow rate (%) 7. (a) Using a graphing calculator, we obtain

(b) The correlation coefficient is r  098, so linear model is appropriate for x between 80 dB and

y  39018x  4197.

104 dB.

y

(c) Substituting x  94 into the regression equation, we

100

get y  39018 94  4197  53. So the intelligibility is about 53%.

50

0

80

90

100

110 x

Noise level (dB) 8. Students should find a fairly strong correlation between shoe size and height. 9. Results will depend on student surveys in each class.


Corrections: 19, 20, 33, 42, 70, 77, 79, 88, 101 NOTE THAT THESE CORRECTIONS SHOULD BE MADE IN ALL THREE BOOKS.

CHAPTER 2

FUNCTIONS

2.1 2.2

Functions 1 Graphs of Functions 10

2.3

Getting Information from the Graph of a Function 22

2.4

Average Rate of Change of a Function 33

2.5 2.6 2.7

Linear Functions and Models 38 Transformations of Functions 44 Combining Functions 59

2.8

One-to-One Functions and Their Inverses 69 Chapter 2 Review 81 Chapter 2 Test 95

¥

FOCUS ON MODELING: Modeling with Functions 99

1


2

FUNCTIONS

2.1

FUNCTIONS

1. If f x  x 3  1, then (a) the value of f at x  1 is f 1  13  1  0. (b) the value of f at x  2 is f 2  23  1  9.

(c) the net change in the value of f between x  1 and x  2 is f 2  f 1  9  0  9.

2. For a function f , the set of all possible inputs is called the domain of f , and the set of all possible outputs is called the range of f . x 5 have 5 in their domain because they are defined when x  5. However, 3. (a) f x  x 2  3x and g x  x    h x  x  10 is undefined when x  5 because 5  10  5, so 5 is not in the domain of h. 0 55 (b) f 5  52  3 5  25  15  10 and g 5    0. 5 5 4. (a) Verbal: “Subtract 4, then square and add 3.” (b) Numerical: x

f x

0

19

2

7

4

3

6

7

5. A function f is a rule that assigns to each element x in a set A exactly one element called f x in a set B. Diagram (a) does not represent a function because the input value 5 in set A has two output values in set B. Diagram (b) does represent a function because each number in set A has exactly one output value in set B. 6. (a) From the table, f 1  4 and f 2  4.

(b) Yes, a function can have the same output for two different inputs. (However, no input can have more than one possible output.)

7. Yes, it is possible that f 1  f 2  5. [For instance, let f x  5 for all x.]

8. No, it is not possible to have f 1  5 and f 1  6. A function assigns each value of x in its domain exactly one value of f x. 9. Multiplying x by 3 gives 3x, then subtracting 5 gives f x  3x  5.

10. Adding 2 gives x  2, then multiplying by 5 gives f x  5 x  2.

11. Squaring gives x 2 , adding 1 gives x 2  1, then taking the square root gives f x 

x 2  1.   x 1 . 12. Adding 1 gives x  1, taking the square root gives x  1, then dividing by 6 gives f x  6 13. 5x  1: Multiply by 5, then add 1.   14. 4 x 2  2 : Square, subtract 2, then multiply by 4.  x 4 : Take the square root, subtract 4, then divide by 3. 15. 3

1


2

CHAPTER 2 Functions

16.

x2  9 : Square, then add 9, then take the square root, then divide by 2. 2

17. Machine diagram for f x 

x  1.

18. Machine diagram for f x 

1

subtract 1, then take square root

0

2

subtract 1, then take square root

1

5

subtract 1, then take square root

2

19. f x  2 x  12

3 . x 2

3

subtract 2, take reciprocal, multiply by 3

3

_1

subtract 2, take reciprocal, multiply by 3

_1

1

subtract 2, take reciprocal, multiply by 3

_3

20. g x  2x  3

x

f x

x

g x

1

2 1  12  8

3

2 3  3  3

0 1 2 3

2 12  2

2 1  12  0 2 2  12  2 2 3  12  8

2

2 2  3  1

1

2 1  3  5

0 3

2 0  3  3 2 3  3  9

21. f x  3x 2  1; f 2  3 22  1  12  1  13; f 2  3 22  1  12  1  13; f 0  3 02  1  1;  2      2   5 3 5  1  15  1  16. f 13  3 13  1  3 19  1  43 ; f

22. f x  4x  x 3 ; f 2  4 2  23  8  8  0; f 0  4 0  03  0; f 2  4 2  23  8  8  0;      3 f 1  4 1  13  3; f 12  4 12  12  2  18  15 8 .

1  2 3 10 1 12 1 1  a a1 1x ; g 2   ; g 0   ; g 2    ; g a   ; 5 5 5 5 5 5 5 5 5   1  x2 1  a  2 3a g x2  ; g a  2   . 5 5 5        x 3 1  3 2 03 3 13 a3 ; h 1   ; h 0   ; h 1   1; h a  ; 24. h x  2 2 2 2 2 2 2       a2  2  3 x 1  2 a2  1 x  2  3 h x  2   ;h a 2   . 2 2 2 2 23. g x 

25. f x  x 2  2x; f 0  02  2 0  0; f 3  32  2 3  9  6  15; f 3  32  2 3  9  6  3;    2   1 2 1 1 1   2  . 2 f a  a 2  2 a  a 2  2a; f x  x2  2 x  x 2  2x; f a a a a a   1 1  1  1  2; h 2  2  1  5 ; h 1  1  1  1  2  5 ; ; h 1  1  1 2 2 2 1 t 2 2 2 2   1 1 1 1 1 ;h   h x  1  x  1    x. 1 x 1 x x x x

26. h t  t 


SECTION 2.1 Functions

3

 

  1 1 1 1  2 1 1 1  1 1x 1 1 2    2  ; ; g 2     ; g 1  , which is undefined; g  3 1 1x 1  2 3 3 1  1 2 3 1 2 2    1  x2  1 2  x2 1  a 1a 1  a  1 1a1 2a  2    g a   ; g a  1    ;g x 1  . 1  a 1a 1  a  1 1a1 a 1  x2  1 x2 t 2 2  2 22 02 a2 28. g t  ; g 2   0; g 2  , which is undefined; g 0   1; g a  ; t 2 2  2 22 02 a2  a2  2  2  a3 a2 a12  .  2 ; g a  1  g a2  2  2 a  1  2 a 1 a 22 a 4 29. k x  3x 2  x  1; k 1  3 12  1  1  5; k 0  3 02  0  1  1; k 2  3 22  2  1  11;    2   k 5 3 5  5  1  16  5; k a  1  3 a  12  a  1  1  3a 2  6a  3  a  1  1  3a 2  7a  5;  2   k x 2  3 x 2  x 2  1  3x 4  x 2  1. 27. g x 

30. k x  x 4  x 3 ; k 2  24  23  16  8  24; k 1  14  13  1  1  2;  4  3  a 4  a 3 a  a3 a 4  3a 3  2  a4    ; k a  a2  a2  a8  a6 ; k 1  14  13  0; k   3 3 3 81 27 81    4  3 1 1 1 1 1t 1 k   4  3  4 .  t t t t t t 31. f x  2 x  1; f 2  2 2  1  2 3  6; f 0  2 0  1  2 1  2;         f 12  2  12  1  2 12  1; f 2  2 2  1  2 1  2; f x  1  2 x  1  1  2 x;            f x 2  2  2  x 2  2  1  2 x 2  1  2x 2  2 (since x 2  1  0 ). 2 1 x 2 1 ; f 2    1; f 1    1; f x is not defined at x  0; x 2 2 1 1         x 2  1x 5 5 x x2 1 .   1; f x 2  2  2  1 since x 2  0, x  0; f   f 5  x 5 5 x 1x x x

32. f x 

33. Since 5  5, we have f 5  3 5  1  14. Since 0  5, we have f 0  3 0  1  1. Since 13  5, we have     f 13  3 13  1  2. Since 5  5, we have f 5  52  1  24. Since 6  5, we have f 6  62  1  35. 34. Since 3  2, we have f 3  5. Since 0  2, we have f 0  5. Since 2  2, we have f 2  5. Since 3  2, we have f 3  2 3  3  3. Since 5  2, we have f 5  2 5  3  7.

35. Since 4  1, we have f 4  42  2 4  16  8  8. Since  32  1, we have    2   f  32   32  2  32  94  3   34 . Since 1  1, we have f 1  12  2 1  1  2  1. Since

1  0  1, we have f 0  0. Since 25  1, we have f 25  1. 36. Since 5  0, we have f 5  3 5  15. Since 0  0  2, we have f 0  0  1  1. Since 0  1  2, we have f 1  1  1  2. Since 0  2  2, we have f 2  2  1  3. Since 5  2, we have f 5  5  22  9.

37. f x  2  x  22  1  x 2  4x  4  1  x 2  4x  5; f x  f 2  x 2  1  22  1  x 2  1  4  1  x 2  6. 38. f 2x  3 2x  1  6x  1; 2 f x  2 3x  1  6x  2.    2 39. f x 2  x 2  4; f x  [x  4]2  x 2  8x  16. x  x  f x 6x  18 3 2x  6 6  18  2x  18;    2x  6 40. f 3 3 3 3 3 41. f x  3x  2, so f 1  3 1  2  1 and f 5  3 5  2  13. Thus, the net change is f 5  f 1  13  1  12. 42. f x  4  5x, so f 3  4  5 3  11 and f 5  4  5 5  21. Thus, the net change is f 5  f 3  21  11  10.


4

CHAPTER 2 Functions

43. g t  1  t 2 , so g 2  1  22  1  4  3 and g 5  1  52  24. Thus, the net change is g 5  g 2  24  3  21. 44. h t  t 2 5, so h 3  32 5  14 and h 6  62 5  41. Thus, the net change is h 6h 3  4114  27. 45. f a  3  a; f a  h  3  a  h  3  a  h;

h f a  h  f a 3  a  h  3  a    1. h h h

46. f a  a 2  4a; f a  h  a  h2  4 a  h  a 2  2ah  h 2  4a  4h  a 2  4a  2ah  4h  h 2 ;     a 2  2ah  h 2  4a  4h  a 2  4a f a  h  f a 2ah  h 2  4h    2a  4  h h h h 47. f a  5; f a  h  5;

55 f a  h  f a   0. h h

1 1 ; f a  h  ; a1 ah1 a1 ah1 1 1   f a  h  f a a  1 a  h  1 a  1 a  h  1  ah1 a1  h h h h 1 a  1 a  h  1   . h a  1 a  h  1

48. f a 

ah a ; f a  h  ; a1 ah1 a a  h  1 a  h a  1 a ah   f a  h  f a a  h  1 a  1 a  h  1 a  1 a  h  1 a  1   h h h   a  h a  1  a a  h  1 2 a  a  ah  h  a 2  ah  a a  h  1 a  1   h h a  h  1 a  1 1  a  h  1 a  1

49. f a 

ah1 a1 ; f a  h  ; a ah a  h  1 a  a  1 a  h ah1 a1  a 2  ah  a  a 2  ah  a  h f a  h  f a a a  h a  h a    h h h ah a  h h 1   ah a  h a a  h

50. f a 

51. f a  3  5a  4a 2 ;

  f a  h  3  5 a  h  4 a  h2  3  5a  5h  4 a 2  2ah  h 2

 3  5a  5h  4a 2  8ah  4h 2 ;     2  8ah  4h 2  3  5a  4a 2 3  5a  5h  4a f a  h  f a  h h 3  5a  5h  4a 2  8ah  4h 2  3  5a  4a 2 5h  8ah  4h 2   h h h 5  8a  4h  5  8a  4h.  h


SECTION 2.1 Functions

5

52. f a  a 3 ; f a  h  a  h3  a 3  3a 2 h  3ah 2  h 3 ;     a 3  3a 2 h  3ah 2  h 3  a 3 f a  h  f a 3a 2 h  3ah 2  h 3   h h h   2 2 h 3a  3ah  h  3a 2  3ah  h 2 .  h 53. f x  3x. Since there is no restriction, the domain is all real numbers,  . Since every real number y is three times the real number 13 y, the range is all real numbers  .

54. f x  5x 2  4. Since there is no restriction, the domain is all real numbers,  . Since 5x 2  0 for all x, 5x 2  4  4 for all x, so the range is [4 .

55. f x  x  3. Since there is no restriction, the domain is all real numbers,  . Since x  0 for all x, x  3  3 for all x, so the range is [3 .   56. f x  2  x  1. We must have x  1  0  x  1, so the domain is [1 . Since x  1  0 for all x,  2  x  1  2, so the range is [2 .

57. f x  3x, 2  x  6. The domain is [2 6], f 2  3 2  6, and f 6  3 6  18, so the range is [6 18].

58. f x  5x 2  4, 0  x  2. The domain is [0 2], f 0  5 02  4  4, and f 2  5 22  4  24, so the range is [4 24]. 2 59. f x  . Since the denominator cannot equal 0, we have 3  x  0  x  3. Thus the domain is x  x  3. In 3x interval notation, the domain is  3  3 . x . Since the denominator cannot equal 0, we have 4  x  0  x  4. Thus the domain is x  x  4. In 60. f x  4x interval notation, the domain is  4  4 . x 2 . Since the denominator cannot equal 0, we have x 2  1  0  x 2  1  x  1. Thus the domain is 61. f x  2 x 1 x  x  1. In interval notation, the domain is  1  1 1  1 . x4 62. f x  2 . Since the denominator cannot equal 0, x 2  x  6  0  x  3 x  2  0  x  3 or x  2. x x 6 In interval notation, the domain is  3  3 2  2 .  63. f t  2  t. We must have 2  t  0  t  2. Thus, the domain is  2].  64. g t  t 2  9. The argument of the square root is positive for all t, so the domain is  .  65. f t  3 2t  5. Since the odd root is defined for all real numbers, the domain is the set of real numbers,  .  66. g x  7  3x. For the square root to be defined, we must have 7  3x  0  7  3x  73  x. Thus the domain is    73 .  67. f t  t 2  25. Since the square root is defined as a real number only for nonnegative numbers, we require that

t 2  25  0  t  5. So the domain is t  t  5 or t  5. In interval notation, the domain is  5]  [5 .  68. g t  36  t 2 . We must have 36  t 2  0  t  6. Thus the domain is [6 6].  2x 69. g x  . We require 2  x  0, and the denominator cannot equal 0. Now 2  x  0  x  2, and 3  x  0 3x  x  3. Thus the domain is x  x  2 and x  3, which can be expressed in interval notation as [2 3  3 .  x 70. g x  2 . We must have x  0 for the numerator and 2x 2  x  1  0 for the denominator. So 2x 2  x  1  0 2x  x  1      2x  1 x  1  0  2x  1  0 or x  1  0  x  12 or x  1. Thus the domain is 0 12  12   .


6

CHAPTER 2 Functions

71. g x  a table:

 4

x 2  6x. Since the input to an even root must be nonnegative, we have x 2  6x  0  x x  6  0. We make

Sign of x Sign of x  6

Sign of x x  6

 0

0 6

6 

Thus the domain is  0]  [6 .  72. g x  x 2  2x  8. We must have x 2  2x  8  0  x  4 x  2  0. We make a table: Sign of x  4 Sign of x  2

Sign of x  4 x  2

 2

2 4

4 

Thus the domain is  2]  [4 . 4 . Since the input to an even root must be nonnegative and the denominator cannot equal 0, we have 73. f x   2x 2  x  0  x  2. Thus the domain is  2. 3x . Since the input to an even root must be nonnegative and the denominator cannot equal 0, we have 74. f x   x 2 x  2  0  x  2. Thus the domain is 2 .

x  12 . Since the input to an even root must be nonnegative and the denominator cannot equal 0, we have 75. f x   2x  1   2x  1  0  x  12 . Thus the domain is 12   .

x 76. f x   . Since the input to an even root must be nonnegative and the denominator cannot equal 0, we have 4 9  x2 9  x 2  0  3  x 3  x  0. We make a table: Interval Sign of 3  x Sign of 3  x Thus the domain is 3 3.

Sign of x  4 x  2

 3

3 3

3 

77. To evaluate f x, square the input and add 1 to the result. (a) f x  x 2  1

(c)

y

(b) x

f x

2

5

1

2

0

1

1

2

2

5

1 1

x


SECTION 2.1 Functions

78. To evaluate f x, add 2 to the input and square the result. (a) f x  x  22

y

(c)

(b) x

g x

5

9 1

3

0

2

1

1

1

1

1

9

x

79. Let T x be the amount of sales tax charged in Lemon County on a purchase of x dollars. To find the tax, take 8% of the purchase price. (a) T x  008x

(c)

y

(b) x

T x

2

016

4

032

6

048

8

064

1l

1

x

80. Let V d be the volume of a sphere of diameter d. To find the volume, take the cube of the diameter, then multiply by  and divide by 6. 3 (a) V d  d 3  6   6d

(c)

y

(b) x

f x

2

4  42 3 32  335 3

4 6 8

81. f x 

36  113

256  268 3

10l 1

x

  1 if x is rational

The domain of f is all real numbers, since every real number is either rational or

 1

The domain of f is all real numbers, since every real number is either rational or

 5 if x is irrational

irrational; and the range of f is 1 5. 82. f x 

if x is rational  5x if x is irrational

irrational. If x is irrational, then 5x is also irrational [see Exercise P.2.88(b).] Also, if x is irrational, then f x5  x, so every irrational number is in the range of f . Thus, the range of f is x  x  1 or x is irrational.

7


8

CHAPTER 2 Functions

    0 2  50 and V 20  50 1  20 2  0. 83. (a) V 0  50 1  20 20

(c)

(b) V 0  50 represents the volume of the full tank at time t  0, and

V 20  0 represents the volume of the empty tank twenty minutes

later.

(d) The net change in V as t changes from 0 minutes to 20 minutes is V 20  V 0  0  50  50 gallons.

x

V x

0

50

5

28125

10

125

15

3125

20

0

84. (a) S 2  4 22  16  5027, S 3  4 32  36  11310.

(b) S 2 represents the surface area of a sphere of radius 2, and S 3 represents the surface area of a sphere of radius 3.   05c2 075c2  866 m, L 075c  10 1   661 m, and 85. (a) L 05c  10 1  2 c c2  09c2  436 m. L 09c  10 1  c2 (b) It will appear to get shorter. a 4 567  108 W W T , where a  567  108 3004  1462 2 , , so L 300  4 2   m m K 567  108 W 567  108 W L 350  3504  2708 2 , and L 1000  10004  18,050 2 .   m m (b) The radiance increases dramatically (to the fourth power) as temperature increases.

86. (a) L T  

(c) The sun’s radiance is approximately L 5778  87. (a) R 1 

R 10 

13  7 104

 1  4 104  13  7 1004

R 100 

1  4 1004

567  108 W 57784  201  107 2 .  m

20  2 mm, 5

(b)

 166 mm, and

13  7 10004

 148 mm. 1  4 10004

(c) The net change in R as x changes from 10 to 100 is R 100  R 10  148  166  018 mm.   88. (a)  01  18500 025  012  4440,    04  18500 025  042  1665.

(b) They tell us that the blood flows much faster (about 275 times faster) 01 cm from the center than 01 cm from the edge.

(d) The net change in V as r changes from 01 cm to 05 cm is V 05  V 01  0  4440  4440 cms.   89. (a) D 01  2 3960 01  012  79201  281 miles   D 02  2 3960 02  022  158404  398 miles

x

R x

1

2

10

166

100

148

200

144

500

141

1000

139

r

 r

0

4625

01

4440

02

3885

03

2960

04

1665

05

0

(c)


SECTION 2.1 Functions

9

  (b) 1135 feet  1135 miles  0215 miles. D  2 3960 0215  02152  1702846  413 miles 0215 5280   (c) D 7  2 3960 7  72  55489  2356 miles

(d) The net change in D as h changes from 1135 ft (or 0215 mi) to 7 mi is D 7 D 0215  2356413  1943 miles.

90. (a) From the table, P 1960  13 million, P 1980  27 million, and P 2020  72 million.

(b) The net change in population from 1960 to 1980 was P 1980  P 1960  27  13  14 million and the net change from 1980 to 2020 was P 2020  P 1980  72  27  45 million.

91. (a) Since 0  5000  10,000 we have T 5000  0. Since 10,000  12,000  20,000 we have T 12,000  008 12,000  10,000  160. Since 20,000  25,000 we have T 25,000  800  015 25,000  20,000  1550. (b) There is no tax on $5000, a tax of $160 on $12,000 income, and a tax of $1550 on $25,000.

92. (a) C 25  25  9  $34; C 45  45  9  $54; C 50  $50; and C 65  $65. (b) The answers represent the total prices of the goods purchased, including shipping.   114x if 0  x  2 93. (a) T x   228  99 x  2 if x  2

(b) T 2  114 2  $228; T 3  228  99 3  2  $327; and T 5  228  99 5  2  $525.

(c) The answers represent the total costs of the lodgings.     15 40  x if 0  x  40 94. (a) F x  0 if 40  x  65    15 x  65 if x  65

(b) F 30  15 40  10  15  10  $150; F 50  $0; and F 75  15 75  65 15  10  $150. (c) The answers represent fines for violating speed limits on the freeway.

T (¡F) 400

96.

95. We assume the grass grows linearly. h (in.) 2

300 200 100

1

0

97.

0

W

W

W

T (¡F)

W

1

t (h)

t (d)

98.

V ($000) 20

50

10

0

noon

midnight Time of day

0

10

20

30

t (yr)

If the car becomes a collectible antique, the value begins to rise again. 99. Answers will vary. 100. Answers will vary. 101. Answers will vary.


10

CHAPTER 2 Functions

2.2

GRAPHS OF FUNCTIONS

1. To graph the function f we plot the points x f x in a coordinate plane. To graph f x  x 2  2, we plot the     points x x 2  2 . So, the point 3 32  2  3 7 is

on the graph of f . The height of the graph of f above the

x

f x

x y

2

2

2 2

1

1

1 1

0

x­axis when x  3 is 7.

0

1

x

1 1

1

2

1

0 2

2

1

y

2

2 2

2. If f 4  10 then the point 4 10 is on the graph of f . 3. If the point 3 7 is on the graph of f , then f 3  7.

4. (a) f x  x 2 is a power function with an even exponent. It has graph IV. (b) f x  x 3 is a power function with an odd exponent. It has graph II.  (c) f x  x is a root function. It has graph I. (d) f x  x is an absolute value function. It has graph III.

5. Because the input 1 has two different outputs (1 and 2), this set of ordered pairs does not define y as a function of x.  6. For any positive x, there are two possible values of y, since x  4y 2  14 x  y 2  y   12 x. Thus, the equation does not define y as a function of x.

7. Because the input x  10 corresponds to outputs y  10 and y  15, this table does not define y as a function of x. 8. This graph fails the Vertical Line Test, and so does not define y as a function of x. 9.

10.

y

x

f x  x  2

6

4

2

4

2

1

6

0

4

1

2

2

0 2

2

0

x

1

0

2

2

4

4

6

3

6

8

4

11. x 3 2

0

f x  x  3, 3  x  3

1

x

12.

y

x

6 5 3

1

2

2

1

3

0

0 1

1 1

x

2 3

y

f x  4  2x 8

1 x

1

4 x 3 , 2 0x 5

f x 

15 1 05 0

4

05

5

1

y

1 1

x


11

SECTION 2.2 Graphs of Functions

13.

14.

y

x

f x  x 2

4

16

5

3

9

4

2

4

1

1

0

x

23 14 7

3

1 x

1

0

2

2

1

1

0

15.

g x  x 2  6x  9

x

16.

y

16

1

1

4

2

1

3

0

4

1

5

4

7

16

17. x 3 2 1

r x  3x 4

10

0

x

1

3

0

0

1

3

2

48

3

243

g x  x 3  8

3

19

1

16

5

4

4

1

3

0

2

1

1

4

48

x

2

7

243

0 7

0

8

1

9

2

16

3

35

x

r x  20  x 4

3

61

0

x

1

20.

4 1

x

19

0

20

1

19

2

4

3

61

x

g x  x  23

5

27

4

8

3

1

2

x

y 0

1

x

_10

y 20 10

4

1

100

1

16

2

y

19.

g x   x  32

18.

y

1

2

x

1

y

f x  x 2  2

0

1

1

0

8

1

27

_1

0

x

1

_10

y

0

4 1

x


12

CHAPTER 2 Functions

21.

3

27 1

1

0

0

1

1

8

0

f x  2 

y

x

4

0

9

1

_1

2

_2

3

1

 12  14

0

1 C t  2 t

2

4

6

1 4

5

3

12

4

21

5

32

6

x

C t  

 32

4 16 

1 _1 0

1

1

1

1

2

2

1 4

x

H x  2x

0

y

27.

2 1

0

2

4 2 0

1 1

x

y

2 0

2

t

 2 1

 12  13

y

x

8

x

4

1 t 1

28.

10 6

1 _4 0

2

1

 12

t

x 4

1

2

4

3

1 2

3

1

y

1 2

16

4

0

1

1 4 1 2

5

f x 

26.

y

x

10

3

3

x

8

0 _1

2

24.

1

0

2

1

x

4

x

0

1

2

1

25.

0

27

2

25

1

8

1

16

1

3

23. x

x

10

1

2

8

2

27

3

27

0 _1

y

 k x   3 x

x 1

2

8

22.

y

 k x  3 x

x

H x  x  2 4

1

3

0

2

1

1

2

0

3

1

4

2

2 0

2

x


13

SECTION 2.2 Graphs of Functions

29. x

x

0

5 0

0

1

2

2

4

5

10

10 4

2 1 1

31.

x

1

2

0

0

1

0

3

0

x

f x 

3

1

2

1

1

1

32.

y

f x  2x  2 12

5

8

2

0

2

1

0

2

2

5

8

1 1

33. f x  8x  x 2

(a) [5 5] by [5 5]

x

y

G x  x  x

5

0

2

x

30.

y

G x  x  x

1 x

1

y

x x

1

0

undefined

1

1

2

1

3

1

x

1

(b) [10 10] by [10 10] 10

4 2 ­4

­2

2

­2

4

­10

­5

­4

5

10

5

10

­10

(c) [2 10] by [5 20]

(d) [10 10] by [100 100]

20

100

10 ­10 ­2

2

4

6

8

10

The viewing rectangle in part (c) produces the most appropriate graph of the equation.

­5 ­100


14

CHAPTER 2 Functions

34. f x  x 2  4x  32

(b) [10 10] by [10 10]

(a) [3 3] by [5 5]

10

4 2 ­3

­2

­1 ­2

1

2

3

­10

­5

5

­4

10

­10

(c) [7 7] by [30 5]

(d) [6 10] by [40 5] ­5

­6 ­4 ­2 ­10

2

4

5

10

2

4

6 ­10 ­20

­20

­30

­30

­40

The viewing rectangle in part (d) produces the most appropriate graph of the equation. 35. f x  3x 3  9x  20

(b) [5 5] by [10 20]

(a) [2 2] by [10 10] 10

20 10

­2

­1

1

2 ­4

­2 ­10

­10

(c) [5 5] by [20 20]

(d) [3 3] by [40 20] 20

20

10 ­3 ­4

­2 ­10

2

4

­2

­1 ­20

­20

1

2

3

­40

The viewing rectangle in part (d) produces the most appropriate graph of the equation. 36. f x  x 4  10x 2  5x

(b) [5 5] by [10 10]

(a) [1 1] by [10 10] 10

­1.0

­0.5 ­10

10

0.5

1.0

­4

­2 ­10

2

4


SECTION 2.2 Graphs of Functions

(c) [5 5] by [40 20]

(d) [10 10] by [20 20] 20

20 10

­4

­2 ­20

2

4 ­10

­5

­40

­10

5

10

­20

The viewing rectangle in part (c) produces the most appropriate graph of the equation.   1  2x if x  0 if x  1 38. f x  37. f x   x  1 if x  1  2 if x  0 y

y

1

2 x

1

39. f x 

 x

if x  0

40. f x 

 x  1 if x  0

y

x

1

  2x  3 if x  1  3x

if x  1 y

1

1 x

1

1

  3  x if x  1 41. f x   2x 2 if x  1

x

  4  x 2 if x  1 42. f x   x  5 if x  1

y

y

4 2 0 _2

1 1

x

_4 _6

1

x

15


16

CHAPTER 2 Functions

43. f x 

  0 if x  2

 3 if x  2

  x 2 if x  1 44. f x   1 if x  1

y

y

1

1 x

1

  if x  2  4 2 45. f x  x if 2  x  2    x  6 if x  2

    1 if x  1 46. f x  x if 1  x  1    1 if x  1

y

y

1

1

  x 2  6x  12 if x  2  6x

x

1

x

1

47. f x 

x

1

if x  2

20 10

­2

  2x  x 2 if x  1 48. f x   x  13 if x  1

0

2

4

6

The first graph shows the output of a typical graphing device. However, the actual graph

of this function is also shown, and its difference from the graphing device’s version should be noted. y 1

2 ­3 ­2 ­1 ­2

    2 if x  2 49. f x  x if 2  x  2    2 if x  2

1

1

2

3

  if x  1  1 50. f x  1  x if 1  x  2    2 if x  2

x


SECTION 2.2 Graphs of Functions

17

51. The curves in parts (a) and (c) are graphs of a function of x, by the Vertical Line Test. 52. The curves in parts (b) and (c) are graphs of functions of x, by the Vertical Line Test.   53. Solving for y in terms of x gives x 2  3y  7  3y  x 2  7  y  13 x 2  7 . This defines y as a function of x.

54. Solving for y in terms of x gives 10x  y  5  y  10x  5. This defines y as a function of x.  55. Solving for y in terms of x gives y 3  x  5  y 3  x  5  y  3 x  5. This defines y as a function of x.  3 56. Solving for y in terms of x gives x 2  y 13  1  y 13  x 2  1  y  x 2  1 . This defines y as a function of x.

 57. Solving for y in terms of x gives x  y 2  y   x. The last equation gives two values of y for a given value of x. Thus, this equation does not define y as a function of x.   58. Solving for y in terms of x gives x 2  y  12  4  y  12  4  x 2  y  1   4  x 2  y  1  4  x 2 . The last equation gives two values of y for a given value of x. Thus, this equation does not define y as a function of x.  59. Solving for y in terms of x gives 2x  4y 2  3  4y 2  2x  3  y   12 2x  3. The last equation gives two values of y for a given value of x. Thus, this equation does not define y as a function of x.  60. Solving for y in terms of x gives 2x 2  4y 2  3  4y 2  2x 2  3  y   12 2x 2  3. The last equation gives two values of y for a given value of x. Thus, this equation does not define y as a function of x.

61. Solving for y in terms of x using the Quadratic Formula gives 2x y  5y 2  4  5y 2  2x y  4  0      2x  2x2  4 5 4 2x  4x 2  80 x  x 2  20   . The last equation gives two values of y for a y 2 5 10 5 given value of x. Thus, this equation does not define y as a function of x.   62. Solving for y in terms of x gives y  x  5  y  x  5  y  x  52 . This defines y as a function of x. 63. Solving for y in terms of x gives 2 x  y  0  y  2 x. This defines y as a function of x.

64. Solving for y in terms of x gives 2x  y  0  y  2x. Since a  a, the last equation gives two values of y for a given value of x. Thus, this equation does not define y as a function of x.  65. Solving for y in terms of x gives x  y 3  y  3 x. This defines y as a function of x.  66. Solving for y in terms of x gives x  y 4  y   4 x. The last equation gives two values of y for any positive value of x. Thus, this equation does not define y as a function of x. 67.

y

68.

3

2

10

1 0

y 20

1

2

3

4

5

6 x

0

1

2

3

4

5 x

The input value 1 has two output values, so the relation

Each input value has a single output value, so this relation

does not define y as a function of x. Its domain is

defines y as a function of x. Its domain is 1 2 3 4 5

0 1 4 5 6 and its range is 1 2 3.

and its range is 5 10 15 20.

  69. y  13  x  y  1  3 x  y  3 x  1. This defines y as a function of x since every real number x has one and only one cube root. 70. The set x y  yx   does not define y as a function of x. For example, 2 4 and 2 6 both belong to the set.


18

CHAPTER 2 Functions

71. (a) f x  x 2  c, for c  0, 2, 4, and 6. c=4

10 8

c=2

(b) f x  x 2  c, for c  0, 2, 4 , and 6. c=_2

6

4

0 ­2

c=_6

4

c=0

2 ­2

c=0

8

6

­4

c=_4

10

c=6

2

2

­4

4

0 ­2

­2

­4

­4

­6

­6

­8

­8

­10

­10

2

4

(c) The graphs in part (a) are obtained by shifting the graph of f x  x 2 upward c units, c  0. The graphs in part (b) are obtained by shifting the graph of f x  x 2 downward c units.

72. (a) f x  x  c2 , for c  0, 1, 2, and 3.

(b) f x  x  c2 , for c  0, 1, 2, and 3.

c=0 c=1

10 8

c=_1 c=0

c=2

c=_2

c=3

c=_3

10 8

6

6

4 2 ­4

0 ­2

­2

2

4

4 2 ­4

0 ­2

­2

­4

­4

­6

­6

­8

­8

­10

­10

2

4

(c) The graphs in part (a) are obtained by shifting the graph of y  x 2 to the right 1, 2, and 3 units, while the graphs in part (b) are obtained by shifting the graph of y  x 2 to the left 1, 2, and 3 units.

73. (a) f x  cx 2 , for c  1, 12 , 2, and 4. c=1

c=4

10

(b) f x  cx 2 , for c  1, 1,  12 , and 2.

c=2

10

8 4

8

c=1

1 c=_ 2

6

6 4 2

2 ­4

­2

0 ­2

2

4

­4

­2

­4

­4

­6

­6

2

4 1

c=__ 2

­8

­8 ­10

0 ­2

c=_4

­10

c=_2

(c) As c increases, the graph of f x  cx 2 is stretched vertically. As c decreases, the graph of f is flattened. When c  0, the graph is reflected about the x­axis.


SECTION 2.2 Graphs of Functions

74. (a) f x  x 1n , for n  2, 4, and 6.

(b) f x  x 1n , for n  3, 5, and 7. 2

3

n=2

2

n=3 n=5 n=7

1

n=4 n=6

1

0

­1

1

2

19

3

­3

­2

­1

0

1

2

3

­1

4

­2

­1

(c) Graphs of even roots are similar to y 

 x, graphs of odd roots are similar to y  3 x. As n increases, the graph of

y  x 1n becomes steeper near x  0 and flatter for x  1.

75. The slope of the line segment joining the points 2 1 and 4 6 is m 

6  1   76 . Using the point­slope form, 4  2

we have y  1   76 x  2  y   76 x  73  1  y   76 x  43 . Thus the function is f x   76 x  43 for 2  x  4.

76. The slope of the line containing the points 3 2 and 6 3 is m  of the line, we have y  3  59 x  6  y  59 x  10 3 3 

3  x  6.

2  3 5   59 . Using the point­slope equation 3  6 9

5 x  13 . Thus the function is f x  59 x  13 , for 9

 77. First solve the circle for y: x 2  y 2  9  y 2  9  x 2  y   9  x 2 . Since we seek the top half of the circle, we   choose y  9  x 2 . So the function is f x  9  x 2 , 3  x  3.  78. First solve the circle for y: x 2  y 2  9  y 2  9  x 2  y   9  x 2 . Since we seek the bottom half of the circle,   we choose y   9  x 2 . So the function is f x   9  x 2 , 3  x  3. 05 for 10  r  100. As the balloon r2 is inflated, the skin gets thinner, as we would expect.

79. We graph T r 

80. We graph P   141 3 for 1    10. As wind speed increases, so does power output, as expected. P

T

20,000

0.004 0.003 10,000

0.002 0.001 0

50

r

0

label horizontal axis "v"

5

10


Match label on graph V(t)

20

Match label on graph:

CHAPTER 2 Functions

  t 2 81. y  50 1  , 0  t  20 20

82. y 

L(T)

567  108 

V

20000

20

10000

0

20 t

10

0

0

1000 T

500

Match label on graph

  84. y  18,500 025  r 2 , 0  r  05

2 83. L   10 1  , 0    300,000 300,0002

L

v

v(r)

10

4000

5

0

T 4 , 0  T  1000

L

40

0

2000

0

1e+5

2e+5

0

3e+5 v

0.0

0.1

0.2

0.3

0.4

0.5

r

300

400

500

600

x

P

85.     060    084 P x   108      132

1.40

if 0  x  1 if 1  x  2 if 2  x  3

if 3  x  35

1.20 1.00 0.80 0.60 0.40 0.20 0

86. (a) E x 

  1000  012x

if 0  x  300  4600  017 x  300 if 300  x

1

2

4

3

x

(b) E

100 80 60 40 20 0

100

200


SECTION 2.2 Graphs of Functions

21

87. The graph of x  y 2 is not the graph of a function because both 1 1 and 1 1 satisfy the equation x  y 2 . The graph of x  y 3 is the graph of a function because x  y 3  x 13  y. If n is even, then both 1 1 and 1 1 satisfies the equation x  y n , so the graph of x  y n is not the graph of a function. When n is odd, y  x 1n is defined for all real

numbers, and since y  x 1n  x  y n , the graph of x  y n is the graph of a function.

    88. (a) The graphs of f x  x 2  x  6 and g x  x 2  x  6 are shown in the viewing rectangle [10 10] by [10 10].

10

­10

10

­5

5

10

­10

­5

5

­10

10

­10

For those values of x where f x  0 , the graphs of f and g coincide, and for those values of x where f x  0, the graph of g is obtained from that of f by reflecting the part below the x­axis about the x­axis.     (b) The graphs of f x  x 4  6x 2 and g x  x 4  6x 2  are shown in the viewing rectangle [5 5] by [10 15]. 10

­4

­2

10

2

4

­4

­10

­2

2

4

­10

For those values of x where f x  0 , the graphs of f and g coincide, and for those values of x where f x  0, the graph of g is obtained from that of f by reflecting the part below the x­axis above the x­axis. (c) In general, if g x   f x, then for those values of x where f x  0, the graphs of f and g coincide, and for those values of x where f x  0, the graph of g is obtained from that of f by reflecting the part below the x­axis above the x­axis. y

y

x

y  f x

x

y  g x

89. (a) The relation “x is the sibling of y” is not a function, in general, because a person x might have more than one sibling. (b) The relation “x is the birth mother of y” is not a function, in general, because x might have given birth to more than one child. (c) The relation “x is a student in your school and y is their ID number” is a function because every student x has a unique ID number. (d) The relation “x is the age of a student and y is their shoe size” is not a function because two students might well be the same age and have different shoe sizes. For example, it is possible that 21 10 and 21 8 are both in the relation.


22

CHAPTER 2 Functions

90. The relation in Exercise 5 is 1 1  1 2  2 3  3 4. Because each y­value has one corresponding x­value, the relation defines x as a function of y. The relation in Exercise 6 is defined by x  4y 2 . Because every value of y has one corresponding x­value, the relation defines x as a function of y. The relation in Exercise 7 is 8 11  10 15  2 11  10 10. Because the y­value 11 has two corresponding x­values (8 and 2), the relation does not define x as a function of y. There exist horizontal lines that intersect the graph in Exercise 8 more than once. Thus, some y­values have multiple corresponding x­values, and so the relation does not define x as a function of y.

2.3

GETTING INFORMATION FROM THE GRAPH OF A FUNCTION

1. To find a function value f a from the graph of f we find the height of the graph above the x­axis at x  a. From the graph of f we see that f 3  4 and f 5  6. The net change in f between x  3 and x  5 is f 5  f 3  6  4  2. 2. The domain of the function f is all the x­values of the points on the graph, and the range is all the corresponding y­values. From the graph of f we see that the domain of f is the interval [1 7] and the range of f is the interval [0 7]. 3. (a) If f is increasing on an interval, then the y­values of the points on the graph rise as the x­values increase. From the graph of f we see that f is increasing on the intervals 1 2 and 4 5. (b) If f is decreasing on an interval, then y­values of the points on the graph fall as the x­values increase. From the graph of f we see that f is decreasing on the intervals 2 4 and 5 7. 4. (a) A function value f a is a local maximum value of f if f a is the largest value of f on some interval containing a. From the graph of f we see that there are two local maximum values of f : one maximum is 7, and it occurs when x  2; the other maximum is 6, and it occurs when x  5.

(b) A function value f a is a local minimum value of f if f a is the smallest value of f on some interval containing a. From the graph of f we see that there is one local minimum value of f . The minimum value is 2, and it occurs when x  4.

5. The solutions of the equation f x  0 are the x­intercepts of the graph of f . The solution of the inequality f x  0 is the set of x­values at which the graph of f is on or above the x­axis. From the graph of f we find that the solutions of the equation f x  0 are x  1 and x  7, and the solution of the inequality f x  0 is the interval [1 7]. 6. (a) To solve the equation 2x  1  x  4 graphically we graph the

y

functions f x  2x  1 and g x  x  4 on the same set of axes and determine the values of x at which the graphs of f and g intersect. From the graph, we see that the solution is x  1.

1 0

1

x

(b) To solve the inequality 2x  1  x  4 graphically we graph the functions f x  2x  1 and g x  x  4 on the same set of axes and find the values of x at which the graph of g is higher than the graph of f . From the graphs in part (a) we see that the solution of the inequality is  1. 7. (a) h 2  1, h 0  1, h 2  3, and h 3  4. (b) Domain: [3 4]. Range: [1 4].

(c) h 3  3, h 2  3, and h 4  3, so h x  3 when x  3, x  2, or x  4.

(d) The graph of h lies below or on the horizontal line y  3 when 3  x  2 or x  4, so h x  3 for those values of x. (e) The net change in h between x  3 and x  3 is h 3  h 3  4  3  1.

8. (a) g 4  3, g 2  2, g 0  2, g 2  1, and g 4  0. (b) Domain: [4 4]. Range: [2 3].


SECTION 2.3 Getting Information from the Graph of a Function

(c) g 4  3. [Note that g 2  1 not 3.]

(d) It appears that g x  0 for 1  x  18 and for x  4; that is, for x  1  x  18  4.

(e) g 1  0 and g 2  1, so the net change between x  1 and x  2 is 1  0  1. 9. (a) f 0  3  15  g 0, so f 0 is larger.

(b) f 1  25  1  g 1, so f 1 is larger.

(c) f x  g x for x  2 and x  2.

(d) f x  g x for 4  x  2 and 2  x  4; that is, on the intervals [4 2] and [2 4]. (e) f x  g x for 2  x  2; that is, on the interval 2 2.

10. (a) The graph of g is higher than the graph of f at x  6, so g 6 is larger.

(b) The graph of f is higher than the graph of g at x  3, so f 3 is larger.

(c) The graphs of f and g intersect at x  2, x  5, and x  7, so f x  g x for these values of x. (d) f x  g x for 1  x  2 and approximately 5  x  7; that is, on [1 2] and [5 7].

(e) f x  g x for 2  x  5 and approximately 7  x  8; that is, on 2 5 and 7 8].

11. From the graph, the domain of f is 3 3] and its range is [2 3]. 12. From the graph, the domain of f is 3 3 and its range is 3 2]. 13. From the graph, the domain of f is [3 3] and its range is 3 2 3. 14. From the graph, the domain of f is 3 3] and its range is [3 3]. y

15. (a)

y

16. (a)

2 0

x

1

2 0

1

(b) Domain:  ; Range:   17. (a)

y

x

(b) Domain:  ; Range:   18. (a)

y

1 0

1

x

(b) Domain:  ; Range: [3 

4

0

1

x

(b) Domain:  ; Range: [2 

23


24

CHAPTER 2 Functions

19. (a)

20. (a)

y

y

1 0

1 0

1

x

1

x

(b) Domain: 1 4; Range: 4 2 (b) Domain: [2 5]; Range: [4 3]

21. (a)

22. (a)

y

y

10 0 2 0

x

1

(b) Domain: [3 3]; Range: [1 8]

(b) Domain: [3 3]; Range: [28 26]

23. (a)

24. (a)

­6

­4

10

2

5

1

­2

4

6

­1

(b) Domain: [1 ; Range: [0 

26. (a)

2 ­6 ­4 ­2 ­2

2

2

(b) Domain:  ; Range: [1 

25. (a)

x

1

2 2

­4 ­6

(b) Domain: [6 6]; Range: [6 0]

4

6

1 ­6 ­4 ­2 ­1

2

4

(b) Domain:  ; Range: 0 2]

6


SECTION 2.3 Getting Information from the Graph of a Function

27. (a)

28. (a)

10

1 ­6 ­4 ­2 ­1

2

4

6

­4

­2

2

4

­10

(b) Domain:  ; Range: [1 1]

(b) Domain:  ; Range: [9 

y

29.

y

30.

y=8x-9

y=4x-5 1l y=5-x

1l 1

1

x y=3-4x

x

(a) From the graph, we see that 4x  5  5  x when

(a) From the graph, we see that 3  4x  8x  9 when

(b) From the graph, we see that 4x  5  5  x when

(b) From the graph, we see that 3  4x  8x  9 when

x  1.

x  2.

x  1.

x  2.

y

31.

y

32.

1l

1

x y=_x@l

y=x@l y=3-4x 1l

y=2-xl 1

x

(a) From the graph, we see that x 2  2  x when

(a) From the graph, we see that x 2  3  4x when

(b) From the graph, we see that x 2  2  x when

(b) From the graph, we see that x 2  3  4x when

x  2 or x  1.

2  x  1.

x  1 or x  3.

1  x  3.

25


26

CHAPTER 2 Functions

34.

33.

20

10

10 ­4

­2

2

­10 ­20

­10

­30

­20

(a) We graph y  x 3  3x 2 (black) and

y  x 2  3x  7 (gray). From the graph, we see

4

6

(a) We graph y  5x 2  x 3 (black) and

y  x 2  3x  4 (gray). From the graph, we see

that the graphs intersect at x  432, x  112,

that the graphs intersect at x  058, x  129, and

and x  144.

(b) From the graph, we see that

2

x  529.

(b) From the graph, we see that

x 3  3x 2  x 2  3x  7 on approximately

5x 2  x 3  x 2  3x  4 on approximately

[432 112] and [144 .

[058 129] and [529 .

(a) We graph y  16x 3  16x 2 (black) and y  x  1 (gray). From the

35.

graph, we see that the graphs intersect at x  1, x  025, and

2

x  025.

­2

­1

(b) From the graph, we see that 16x 3  16x 2  x  1 on [1 025]

1

and [025 .

­2

36.

(a) We graph y  1 

4

x (black) and y 

x 2  1 (gray). From the

graph, we see that the solutions are x  0 and x  231.   (b) From the graph, we see that 1  x  x 2  1 on approximately

3 2

0 231.

1 ­1

0

1

2

3

4

5

37. (a) The domain is [1 4] and the range is [1 3]. (b) The function is increasing on 1 1 and 2 4 and decreasing on 1 2. 38. (a) The domain is [2 3] and the range is [2 3]. (b) The function is increasing on 0 1 and decreasing on 2 0 and 1 3. 39. (a) The domain is [3 3] and the range is [2 2]. (b) The function is increasing on 2 1 and 1 2 and decreasing on 3 2, 1 1, and 2 3. 40. (a) The domain is [2 2] and the range is [2 2]. (b) The function is increasing on 1 1 and decreasing on 2 1 and 1 2.


SECTION 2.3 Getting Information from the Graph of a Function

41. (a) f x  x 2  5x is graphed in the viewing rectangle [2 7] by [10 10].

27

42. (a) f x  x 3  4x is graphed in the viewing rectangle [10 10] by [10 10].

10

10

­2

2

4

6

­10

­5

­10

5

10

­10

(b) The domain is   and the range is [625 .

(c) The function is increasing on 25 . It is decreasing on  25.

43. (a) f x  2x 3  3x 2  12x is graphed in the viewing rectangle [3 5] by [25 20].

(b) The domain and range are  . (c) The function is increasing on  115 and 115 . It is decreasing on 115 115.

44. (a) f x  x 4  16x 2 is graphed in the viewing rectangle [10 10] by [70 10].

20 ­10

10 ­2

­5

5

10

­20 2

­10

4

­40 ­60

­20

(b) The domain and range are  .

(b) The domain is   and the range is [64 .

(c) The function is increasing on  1 and 2 .

(c) The function is increasing on 283 0 and

It is decreasing on 1 2.

283 . It is decreasing on  283 and 0 283.

45. (a) f x  x 3  2x 2  x  2 is graphed in the viewing rectangle [5 5] by [3 3].

46. (a) f x  x 4  4x 3  2x 2  4x  3 is graphed in the viewing rectangle [3 5] by [5 5]. 4

2

2 ­4

­2

2

4

­2

­2

­2

2

4

­4

(b) The domain and range are  .

(b) The domain is   and the range is [4 .

(c) The function is increasing on  155 and

(c) The function is increasing on 04 1 and 24 .

022 . It is decreasing on 155 022.

It is decreasing on  04 and 1 24.


28

CHAPTER 2 Functions

47. (a) f x  x 25 is graphed in the viewing rectangle [10 10] by [5 5].

48. (a) f x  4  x 23 is graphed in the viewing rectangle [10 10] by [10 10].

10

4 2 ­10

­5

5

­2

10

­10

­5

5

10

­4

­10

(b) The domain is   and the range is [0 .

(b) The domain is   and the range is  4].

(c) The function is increasing on 0 . It is decreasing

(c) The function is increasing on  0. It is

on  0.

49. (a) f x  2 

x  3 is graphed in the viewing

rectangle [4 10] by [0 6].

decreasing on 0 .  50. (a) f x  25  x 2 is graphed in the viewing rectangle [6 6] by [1 6].

6

6

4

4

2

2

­4 ­2 0

2

4

6

­6

8 10

­4

­2

2

4

6

(b) The domain is [3  and the range is [2 .

(b) The domain is [5 5] and the range is [0 5].

(c) The function is increasing on 3 .

(c) The function is increasing on 5 0 and decreasing on 0 5.

51. (a) Local maxima: 3 at x  1 and 4 at x  3. Local minimum: 3 at x  1.

(b) The function is increasing on  1 and 1 3 and decreasing on 1 1 and 3 .

52. (a) Local maximum: 2 at x  0. Local minima: 1 at x  2 and 0 at x  2.

(b) The function is increasing on 2 0 and 2  and decreasing on  2 and 0 2.

53. (a) Local maximum: 3 at x  0. Local minima: 1 at x  2 and 1 at x  1.

(b) The function is increasing on 2 0 and 1  and decreasing on  2 and 0 1.

54. (a) Local maxima: 3 at x  2 and 2 at x  1. Local minima: 0 at x  1 and 1 at x  2.

(b) The function is increasing on  2, 1 1, and 2  and decreasing on 2 1 and 1 2.

55. (a) In the first graph, we see that f x  x 3  x has a local minimum and a local maximum. Smaller x­ and y­ranges show that f x has a local maximum of about 038 when x  058 and a local minimum of about 038 when x  058. 5

0.5

0.50 ­0.3

0.4 ­5

­0.4

5 ­5

­0.60

­0.55

0.3 ­0.50

­0.5

(b) The function is increasing on  058 and 058  and decreasing on 058 058.

0.55

0.60


SECTION 2.3 Getting Information from the Graph of a Function

29

56. (a) In the first graph, we see that f x  3  x  x 2  x 3 has a local minimum and a local maximum. Smaller x­ and y­ranges show that f x has a local maximum of about 400 when x  100 and a local minimum of about 281 when x  033. 4 2

­2

0

2

­0.40

­0.35

2.9

4.1

2.8

4.0

2.7 ­0.30

3.9

0.9

1.0

1.1

(b) The function is increasing on 033 100 and decreasing on  033 and 100 .

57. (a) In the first graph, we see that g x  x 4  2x 3  11x 2 has two local minimums and a local maximum. The local maximum is g x  0 when x  0. Smaller x­ and y­ranges show that local minima are g x  1361 when x  171 and g x  7332 when x  321. ­5

5

­1.75

­1.70

­1.65 ­13.4

­73.0

3.1

­50

­13.6

­73.5

­100

­13.8

­74.0

3.2

3.3

(b) The function is increasing on 171 0 and 321  and decreasing on  171 and 0 321.

58. (a) In the first graph, we see that g x  x 5  8x 3  20x has two local minimums and two local maximums. The local maximums are g x  787 when x  193 and g x  1302 when x  104. Smaller x­ and y­ranges show that local minimums are g x  1302 when x  104 and g x  787 when x  193. Notice that since g x is odd, the local maxima and minima are related. 20

­2.0

­1.8 ­7.8

13.1 13.0

­5

­7.9

5

12.9 ­20

­8.0 ­1.2

­1.0

1.0

1.2

7.90 ­12.8 7.85 ­13.0 ­13.2

7.80 1.90

1.95

2.00

(b) The function is increasing on  193, 104 104, and 193  and decreasing on 193 104 and 104 193.


30

CHAPTER 2 Functions

 59. (a) In the first graph, we see that U x  x 6  x has only a local maximum. Smaller x­ and y­ranges show that U x has a local maximum of about 566 when x  400. 10

5.70

5

5.65 5.60 5

3.9

4.0

4.1

(b) The function is increasing on  400 and decreasing on 400 600.  60. (a) In the first viewing rectangle below, we see that U x  x x  x 2 has only a local maximum. Smaller x­ and y­ranges show that U x has a local maximum of about 032 when x  075. 1.0

0.40

0.5

0.35

0.0

0.0

0.5

0.30

1.0

0.7

0.8

0.9

(b) The function is increasing on 0 075 and decreasing on 075 1. 1  x2 has a local minimum and a local maximum. Smaller x­ and y­ranges x3 show that V x has a local maximum of about 038 when x  173 and a local minimum of about 038 when x  173.

61. (a) In the first graph, we see that V x 

2

0.40

1.6

1.7

1.8

­0.30 0.35 ­5

­0.35

5 ­1.8

­2

­1.7

0.30 ­1.6

­0.40

(b) The function is increasing on  173 and 173  and decreasing on 173 0 and 0 173.

1 has only a local maximum. Smaller x­ and x2  x  1 y­ranges show that V x has a local maximum of about 133 when x  050.

62. (a) In the first viewing rectangle below, we see that V x  2

1.40 1.35

­5

5

­0.6

­0.5

1.30 ­0.4

(b) The function is increasing on  050 and decreasing on 050 .

63. (a) At 5 A . M . the graph shows that the power consumption is about 11 gigawatts. Since t  22 represents 10 P. M ., the graph shows that the power consumption at 10 P. M . is about 14 gigawatts. (b) The power consumption is lowest between 3 A . M . and 4:30 A . M . The power consumption is highest at about 7 P. M .


31

SECTION 2.3 Getting Information from the Graph of a Function

(c) The net change in power consumption from 5 A . M . to 10 P. M . is P 22  P 5  14  11  3 gigawatts. 64. (a) The first noticeable movements occurred at time t  5 seconds. (b) It seemed to end at time t  30 seconds.

(c) Maximum intensity was reached at t  17 seconds.

65. (a) This person appears to be gaining weight steadily until the age of 21 when this person’s weight gain slows down. The person continues to gain weight until the age of 30, at which point this person experiences a sudden weight loss. Weight gain resumes around the age of 32, and the person dies at about age 68. Thus, the person’s weight W is increasing on 0 30 and 32 68 and decreasing on 30 32. (b) The sudden weight loss could be due to a number of reasons, among them major illness, a weight loss program, etc. (c) The net change in the person’s weight from age 10 to age 20 is W 20  W 10  150  50  100 lb. 66. (a) Measuring in hours since midnight, the representative’s distance from home D is increasing on 8 9, 10 12, and 15 17, constant on 9 10, 12 13, and 17 18, and decreasing on 13 15 and 18 19. (b) The representative travels away from home and stops to make a sales call between 9 A . M . and 10 A . M ., and then travels further from home for a sales call between 12 noon and 1 P. M . Next, the representative travels along a route that veers closer to home before going further away again. A final sales call is made between 5 P. M . and 6 P. M ., then the representative returns home. (c) The net change in the distance D from noon to 1 P. M . is D 1 P. M .  D noon  0. 67. (a) The function W is increasing on 0 150 and 300 365 and decreasing on 150 300. (b) W has a local maximum at x  150 and a local minimum at x  300.

(c) The net change in the depth W from 100 days to 300 days is W 300  W 100  25  75  50 ft.

68. (a) The function P is increasing on 0 25 and decreasing on 25 50. (b) The maximum population was 50,000, and it was attained at x  25 years, which represents the year 1995. (c) The net change in the population P from 1990 to 2010 is P 40  P 20  40  40  0.

69. Runner A won the race. All runners finished the race. Runner B fell, but got up and finished the race.

70. (a)

a (m/s²)

71. (a)

2

E 400 300

1

200 100

0

2000

4000 x (km)

(b) As the distance x increases, the acceleration a due to gravity decreases. The rate of decrease is rapid at first, and slows as the distance increases.

0

50

100 150 200 250 300

(b) As the temperature T increases, the energy E increases. The rate of increase gets larger as the temperature increases.

T


32

CHAPTER 2 Functions

72. In the first graph, we see the general location of the minimum of V  99987  006426T  00085043T 2  00000679T 3 is around T  4. In the second graph, we isolate the minimum, and from this graph, we see that the minimum volume of 1 kg of water occurs at T  396 C. 1005

999.76

1000

999.75

995

0

999.74

20

3.5

4.0

4.5

10 . In the second graph, we isolate the  5 minimum, and from this graph, we see that energy is minimized when   75 mi/h.

73. In the first graph, we see the general location of the minimum of E   273 3 10000

4700 4650

5000 6

8

10

4600

7.4

7.5

7.6

74. In the first graph, we see the general location of the maximum of  r   32 1  r  r 2 is around r  07 cm. In the second graph, we isolate the maximum, and from this graph we see that at the maximum velocity is approximately 047 when r  067 cm. 1.0

0.50 0.48

0.5

0.46 0.0

0.6

0.8

1.0

75. (a) f x is always increasing, and f x  0 for all x. y

0

0.60

0.65

0.70

(b) f x is always decreasing, and f x  0 for all x. y

x

0

x


SECTION 2.4 Average Rate of Change of a Function

(c) f x is always increasing, and f x  0 for all x.

33

(d) f x is always decreasing, and f x  0 for all x.

y

y

x 0

0

x

76. To find the fixed points of a function algebraically, set f x  x and solve. To find the fixed points graphically, find the points of intersection with the graphs of y  f x and y  x. (a) 5x  x 2  x  x 2  4x  0  x x  4  0  x  0 or 4. make bold

(b) x 3  x  1  x  x 3  1  0  x  1. (c)

(d)

y

y

1 0

2.4

1 1

0

x

1

x

From the graph, the fixed points of the function are

From the graph, we see that this function has no

4, 0, and 3.

fixed point.

AVERAGE RATE OF CHANGE OF A FUNCTION

100 miles  50 mi/h. 2 hours f b  f a 2. The average rate of change of a function f between x  a and x  b is average rate of change  . ba

1. If you travel 100 miles in two hours then your average speed for the trip is average speed 

3. The average rate of change of the function f x  x 2 between x  1 and x  5 is f 1 average rate of change  f 5  51

25  1 24 52  12    6. 4 4 4

4. (a) The average rate of change of a function f between x  a and x  b is the slope of the secant line between a f a and b f b. (b) The average rate of change of the linear function f x  3x  5 between any two points is 3.

5. (a) Yes, the average rate of change of a function between x  a and x  b is the slope of the secant line through a f a f b  f a and b f b; that is, . ba (b) Yes, the average rate of change of a linear function y  mx  b is the same (namely m) for all intervals.

6. (a) No, the average rate of change of an increasing function is positive over any interval.


34

CHAPTER 2 Functions

(b) No, just because the average rate of change of a function between x  a and x  b is negative, it does not follow

that the function is decreasing on that interval. For example, f x  x 2 has negative average rate of change between x  2 and x  1, but f is increasing for 0  x  1.

7. (a) The net change is f 4  f 1  5  3  2.

(b) We use the points 1 3 and 4 5, so the average rate of change is 8. (a) The net change is f 5  f 1  2  4  2. (b) We use the points 1 4 and 5 2, so the average rate of change is 9. (a) The net change is f 5  f 0  2  6  4. (b) We use the points 0 6 and 5 2, so the average rate of change is 10. (a) The net change is f 5  f 1  4  0  4.

53 2  . 41 3 24 2 1   . 51 4 2 4 26  . 50 5

(b) We use the points 1 0 and 5 4, so the average rate of change is 11. (a) The net change is f 7  f 4  [5 7  3]  [5 4  3]  15. f 7  f 4 15 (b) The average rate of change is   5. 74 3

40 4 2   . 5  1 6 3

12. (a) The net change is s 5  s 1  [4  2 5]  [4  2 1]  8. 8 s 5  s 1   2. (b) The average rate of change is 51 4     13. (a) The net change is g 10  g 6  2  12 10  2  12 6  8. (b) The average rate of change is

8 1 g 10  g 6   . 10  6 16 2

14. (a) The net change is h 5  h 2 

9 3 5  4 3 2  4   . 5 5 5 9

h 5  h 2 3  5  . 52 3 5     15. (a) The net change is f 3  f 1  3 32  3  3 12  1  30  4  26. (b) The average rate of change is

26 f 3  f 1   13. 31 2     16. (a) The net change is f 0  f 3  3 0  02  3 3  32  0  18  18. (b) The average rate of change is

f 0  f 3 18   6. 0  3 3       17. (a) The net change is f 10  f 0  103  4 102  03  4 02  600  0  600. (b) The average rate of change is

f 10  f 0 600   60. 10  0 10     18. (a) The net change is g 2  g 2  24  23  22  24  23  22  12  28  16. (b) The average rate of change is

(b) The average rate of change is

19. The difference quotient is

g 2  g 2 16   4. 2  2 4

4a 2  8ah  4h 2  4a 2 4 a  h2  4a 2 f a  h  f a   8a  4h.  h h a  h  a


SECTION 2.4 Average Rate of Change of a Function

35

20. The difference quotient is

  3  10 a  h2  3  10a 2 10a 2  20ah  10h 2  10a 2 f a  h  f a   20a  10h.  h h a  h  a 1 1  f a  h  f a a  a  h 1 a  h a 21. The difference quotient is   .  h ah a  h a a  h a  h  a 22. The difference quotient is 2  2 f a  h  f a 2 a  1  2 a  h  1 2h 2    ah1 a1  . h h a  1 a  h  1 h a  1 a  h  1 a  h  a a  1 a  h  1 23. The difference quotient is     f a  h  f a h 1 ah a ah a a  h  a             . h a  h  a ah a h ah a h ah a ah a

24. The difference quotient is f a  h  f a  a  h  a

2  22 a ah2

h

2a 2  2 a  h2 a 2 a  h2 h

  2a 2  2 a 2  2ah  h 2 ha 2 a  h2

25. (a) The average rate of change is     1b  3  1a  3 1b  3  1a  3 1 b  a 1 f b  f a 2 2 2   2  2  . ba ba ba ba 2

2h 2a  h ha 2 a  h2



2 2a  h

a 2 a  h2

.

(b) The slope of the line f x  12 x  3 is 12 , which is also the average rate of change.

26. (a) The average rate of change is g b  g a 4 b  a 4b  2  4a  2    4. ba ba ba (b) The slope of the line g x  4x  2 is 4, which is also the average rate of change.

27. The function f has a greater average rate of change between x  0 and x  1. The function g has a greater average rate of change between x  1 and x  2. The functions f and g have the same average rate of change between x  0 and x  15. 28. The average rate of change of f is constant, that of g increases, and that of h decreases. 29. (a) The average rate of change is that the depth decreased.

50  75 25 1 W 200  W 100     ft/day. The negative sign indicates 200  100 200  100 100 4

(b) Answers will vary. One interval on which the average rate of change is 0 is [200 350]. 50  40 10 P 25  P 20    2. Thus, the average rate of change over this time 25  20 25  20 5 period is 2000 people per year.

30. (a) The average rate of change is

(b) On average, the population grew at a rate of 2000 per year between 1990 and 1995. (c) There are many, including 1990–2010 and 1970–2016 (approximately). 6375  4869  3765 persons/yr. 2012  2008 4921  6288  34175 persons/yr. (b) The average rate of change of population between 2014 and 2018 is 2018  2014 (c) The population was increasing from 2002 to 2012.

31. (a) The average rate of change of population between 2008 and 2012 is

(d) The population was decreasing from 2012 to 2020. 800  400 400 100    476 m/s. 152  68 84 21 400 1,600  1,200   268 m/s. (b) The average speed is 412  263 149

32. (a) The average speed is


36

CHAPTER 2 Functions

(c) Lap

Length of time to run lap

Average speed of lap

1

32

625 m/s

2

36

556 m/s

3

40

500 m/s

4

44

455 m/s

5

51

392 m/s

6

60

333 m/s

7

72

278 m/s

8

77

260 m/s

The runner is slowing down throughout the run.

3629  1146  2483 cakesyr. 2020  2010 638  1042  404 cakesyr. (b) The average rate of change between 2011 and 2012 was 2012  2011 1145  638  507 cakesyr. (c) The average rate of change between 2012 and 2013 was 2013  2012 (d) Year Snack cakes sold Change in sales from previous year

33. (a) The average rate of change of sales between 2010 and 2020 was

2010

1146

2011

1042

2012

638

104

2013

1145

2014

1738

593

2015

1804

66

2016

1121

2017

1987

683

2018

2533

546

2019

2983

450

2020

3629

646

404

507

866

Sales increased most quickly between 2016 and 2017, and decreased most quickly between 2015 and 2016.


SECTION 2.4 Average Rate of Change of a Function

37

34. Year

Number of books

2000

530

2001

590

2002

650

2006

890

2010

1130

2012

1250

2015

1430

2017

1550

2018

1610

2019

1670

2020

1730

35. The average rate of change of the temperature of the soup over the first 20 minutes is 119  200 81 T 20  T 0    405 F/min. Over the next 20 minutes, it is 20  0 20  0 20 T 40  T 20 89  119 30    15 F/min. The first 20 minutes had a higher average rate of change of 40  20 40  20 20 temperature (in absolute value).

36. (a) (i) Between 1860 and 1890, the average rate of change was about 83,300 farms per year. (ii) Between 1950 and 1980, the average rate of change was 100,000 farms per year.

45  20 y 1890  y 1860   00833, a gain of 1890  1860 30 25  55 y 1980  y 1950   01, a loss of about 1980  1950 30

(b) From the graph, it appears that the steepest rate of decline was during the period from 1950 to 1960.

d 10  d 0 100   10 ms. 10  0 10 (b) Skier A gets a great start, but slows at the end of the race. Skier B maintains a steady pace. Runner C is slow at the beginning, but accelerates down the hill.

37. (a) For all three skiers, the average rate of change is

38. (a) Skater B won the race, because he travels 500 meters before Skater A. A 10  A 0 200  0 (b) Skater A’s average speed during the first 10 seconds is   20 ms. 10  0 10 B 10  B 0 100  0 Skater B’s average speed during the first 10 seconds is   10 ms. 10  0 10 500  400 A 40  A 25   7 ms. (c) Skater A’s average speed during his last 15 seconds is 40  25 15 B 35  B 20 500  200 Skater B’s average speed during his last 15 seconds is   20 ms. 35  20 15


38

CHAPTER 2 Functions

39. t a

t b

3

35

3

31

3

301

3

3001

3

30001

Average Speed 

f b  f a ba

16 352  16 32  104 35  3

16 312  16 32  976 31  3

16 3012  16 32  9616 301  3

16 30012  16 32  96016 3001  3

16 300012  16 32  960016 30001  3

From the table it appears that the average speed approaches 96 fts as the time intervals get smaller and smaller. It seems reasonable to say that the speed of the object is 96 fts at the instant t  3.

2.5

LINEAR FUNCTIONS AND MODELS

1. If f is a function with constant rate of change, then (a) f is a linear function of the form f x  ax  b. (b) The graph of f is a line.

2. If f x  5x  7, then (a) The rate of change of f is 5.

(b) The graph of f is a line with slope 5 and y­intercept 7.

3. From the graph, we see that y 2  50 and y 0  20, so the slope of the graph is 50  20 y 2  y 0   15 galmin. m 20 2

4. From Exercise 3, we see that the pool is being filled at the rate of 15 gallons per minute. 5. If a linear function has positive rate of change, its graph slopes upward.

6. f x  3 is a linear function because it is of the form f x  ax  b, with a  0 and b  3. Its slope (and hence its rate of change) is 0.   7. f x  5  2x  2x  5 is linear with a  2 and  b  5. 9. f x 

20  x   15 x  4 is linear with a   15 and 5

8. f x  12 x  4  12 x  2 is linear with a  12 and b  2.

10. f x 

4 2x  4  2  is not linear. x x

b  4. 11. f x  x 2  3x  3x 2  2x is not linear. 13. f x 

x  1 is not linear.

12. f x  3 6  5x  15x  18 is linear with a  15 and b  18.

14. f x  2x  52  4x 2  20x  25 is not linear.


SECTION 2.5 Linear Functions and Models y

15. x

f x  2x  3 7

2

5

1

3

1

1

1

0

2

1

0

3

x

1

3

The slope of the graph of f x  2x  3 is 2. y

16. x

g x  3x  1

2

7

1

4

0 1

1

1

0

2

2

5

3

8

1

x

The slope of the graph of g x  3x  1 is 3. 17.

r t   23 t  2

t 1

267

0

2

1

133

2

067

3

0

4

067

y

2 0

2

t

The slope of the graph of r t   23 t  2 is  23 . 18.

h t  12  34 t

t 2 1

2

125

0

05

1

025

2 3

y

1 175

The slope of the graph of h t  12  34 t is  34 .

1 0

1

t

39


40

CHAPTER 2 Functions y

19. (a)

y

20. (a)

1 _1

1 0

0

1

z

x

1

_10

(b) The graph of f x  2x  6 has slope 2.

(b) The graph of g z  3z  9 has slope 3.

(c) f x  2x  6 has rate of change 2.

(c) g z  3z  9 has rate of change 3.

y

21. (a)

y

22. (a)

1

1 0

x

1

(b) The graph of f x  2  3x has slope 3. (c) f x  2  3x has rate of change 3.

z

1

(b) The graph of g z   z  3  z  3 has slope 1. (c) g z  z  3 has rate of change 1.

y

23. (a)

0

24. (a)

y

1 1 0

(b) The graph of h t   15 .

1

t

5  2t   15 t  12 has slope 10

(c) h t   15 t  12 has rate of change  15 .

0

1

w

(b) The graph of s   05  2 has slope 05. (c) s   05  2 has rate of change 05.


SECTION 2.5 Linear Functions and Models y

25. (a)

26. (a)

41

y 1 0

x

1

1 _1

0 _1

t

1

(b) The graph of f t   32 t  2 has slope  32 . (c) f t   32 t  2 has rate of change  32 .

(b) The graph of g x  54 x  10 has slope 54 . (c) g x  54 x  10 has rate of change 54 .

27. The linear function f with rate of change 5 and initial value 10 has equation f x  5x  10.

28. The linear function f with rate of change 3 and initial value 1 has equation f x  3x  1. 29. The linear function f with slope 12 and y­intercept 3 has equation f x  12 x  3.

30. The linear function f with slope  45 and y­intercept 2 has equation f x   45 x  2.

31. (a) From the table, we see that for every increase of 2 in the value of x, f x increases by 3. Thus, the rate of change of f is 32 . (b) When x  0, f x  7, so b  7. From part (a), a  32 , and so f x  32 x  7.

32. (a) From the table, we see that f 3  11 and f 0  2. Thus, when x increases by 3, f x decreases by 9, and so the rate of change of f is 3. (b) When x  0, f x  2, so b  2. From part (a), a  3, and so f x  3x  2.

33. (a) From the graph, we see that f 0  3 and f 1  4, so the rate of change of f is (b) From part (a), a  1, and f 0  b  3, so f x  x  3. 34. (a) From the graph, we see that f 0  4 and f 2  0, so the rate of change of f is (b) From part (a), a  2, and f 0  b  4, so f x  2x  4. 35. (a) From the graph, we see that f 0  2 and f 4  0, so the rate of change of f is (b) From part (a), a   12 , and f 0  b  2, so f x   12 x  2.

43  1. 10 04  2. 20 02 1  . 40 2

36. (a) From the graph, we see that f 0  1 and f 2  0, so the rate of change of f is (b) From part (a), a  12 , and f 0  b  1, so f x  12 x  1. 37.

f

Increasing the value of a makes the graph of f steeper. In other words, it

a=2

increases the rate of change of f .

a=1 a= 21

1 01

1 0  1  . 20 2

t


42

CHAPTER 2 Functions

38.

f

Increasing the value of b moves the graph of f upward, but does not affect

b=2

the rate of change of f . b=1 b= 21

1

0

39. (a)

t

1

T 38

(b) The slope of T x  150x  32 is the value of a, 150. (c) The amount of trash is changing at a rate equal to the slope of the graph,

36

150 thousand tons per year.

34

The graph is incorrect (it does not match the function--for instance T(10)=1532). The revsied graph should use 0 < x < 3 and 0 < T < 500 labeling tics every 100 on the Taxis.

32

0

40. (a)

10

20

30

x

f 900

(b) The slope of the graph of f x  200  32x is 32.

700

(c) Ore is being produced at a rate equal to the slope of the graph, 32 thousand tons per year.

800 600 500 400 300 200 100

0

5

10

15

20

25 x

41. (a) Let V t  at  b represent the volume of hydrogen. The balloon is being filled at the rate of 05 ft3 s, so a  05, and initially it contains 2 ft3 , so b  2. Thus, V t  05t  2.

(b) We solve V t  15  05t  2  15  05t  13  t  26. Thus, it takes 26 seconds to fill the balloon. 42. (a) Let V t  at  b represent the volume of water. The pool is being filled at the rate of 10 galmin, so a  10, and initially it contains 300 gal, so b  300. Thus, V t  10t  300. (b) We solve V t  1300  10t  300  1300  10t  1000  t  100. Thus, it takes 100 minutes to fill the pool.

1 . The ramp 43. (a) Let H x  ax  b represent the height of the ramp. The maximum rise is 1 inch per 12 inches, so a  12 1 x. starts on the ground, so b  0. Thus, H x  12

1 150  125. Thus, the ramp reaches a height of 125 inches. (b) We find H 150  12

1200  0075, or 75%. 15,000 500 The road biker descends 500 vertical feet over 10,000 feet, so the grade of the road is  005, or 5%. 10,000

44. The mountain biker descends 1200 vertical feet over 15,000 feet, so the grade of the road is


43

SECTION 2.5 Linear Functions and Models

45. (a) From the graph, we see that the slope of the engineer’s trip is steeper than that of the manager. Thus, the engineer traveled faster. 70 7 (b) The points 0 0 and 6 7 are on the engineer’s graph, so their speed is  miles per minute or 60 6   7  70 mih. 60 6 16  10 The points 0 10 and 6 16 are on the manager’s graph, so their speed is 60   60 mih. 60 1 hmin  1 mi/min and the engineer’s speed is (c) t is measured in minutes, so the manager’s speed is 60 mih  60

1 hmin  7 mi/min. Thus, the manager’s distance is modeled by f t  1 t  0  10  t  10 and the 70 mih  60 6

engineer’s distance is modeled by g t  76 t  0  0  76 t.

46. (a) Let d t represent the distance traveled. When t  0, d  0, and when

(b)

40  0 t  50, d  40. Thus, the slope of the graph is  08. The 50  0 y­intercept is 0, so d t  08t.

(c) The speed of the bus is equal to the slope of the graph of d, that is, 08 mimin or 08 60  48 mih.

d 110 100 90 80 70 60 50 40 30 20 10 0

20

40

60

80 100 120 t

6 , so if we take 0 0 as the starting point, the 47. Let x be the horizontal distance and y the elevation. The slope is  100

6 x. We have descended 1000 ft, so we substitute y  1000 and solve for x: 1000   6 x  elevation is y   100 100 1 16,667  316 mi. x  16,667 ft. Converting to miles, the horizontal distance is 5280

48. (a) Taking x  0 to correspond to the year 1980, we have a  20 and b  024, so D x  20  024x.

(b)

D 30

(c) The rate of sedimentation is equal to the slope of the graph, 024 cmyr or 24 mmyr.

20 10

0

10

20

30

40

50

x

49. The atmospheric pressure is 100 kPa at sea level (x  0 km), and it decreases by 12 kPa for each kilometer increase in elevation, so a linear model is f x  100  12x. At the peak of Mt. Rainier, the atmospheric pressure is estimated to be f 44  100  12 44  472 kPa. 50. The boiling point of water is 100 C at 100 kPa and it decreases by 375 C for each 10 kPa drop in pressure. In 375 this case the linear model g x has g 100  100 and its graph has slope  0375, so an equation is 10 g x  100  0375 x  100  g x  0375 x  625. We estimate the boiling point of water at 88 kPa to be g 88  0375 88  625  955 C.


44

CHAPTER 2 Functions

51. (a) Let C x  ax  b be the cost of driving x miles. In May, driving

(b)

480 miles cost $380, and in June, driving 800 miles cost $460. Thus, the

C 600

points 480 380 and 800 460 are on the graph, so the slope is

500

a

1 460  380  . We use the point 480 380 to find the value of b: 800  480 4

380  14 480  b  b  260. Thus, C x  14 x  260.

(c) The rate at which the cost increases is equal to the slope of the line, that is 1 . So the cost increases by $025 for every additional mile driven. 4

400 300 200 100 200 400 600 800 1000 1200 1400 x

0

The slope of the graph of C x  14 x  260 is the value of a, 14 . 52. (a) Let C x  ax  b be the cost of producing x chairs in one day. The first (b) day, it cost $2200 to produce 100 chairs, and the other day it cost $4800 to produce 300 chairs.. Thus, the points 100 2200 and 300 4800 are on 4800  2200  13. We use the point 300  100 100 2200 to find the value of b: 2200  13 100  b  b  900. Thus,

the graph, so the slope is a  C x  13x  900.

(c) The rate at which the factory’s cost increases is equal to the slope of the line, that is $13chair.

C 10,000 9000 8000 7000 6000 5000 4000 3000 2000 1000 0

100 200 300 400 500 600 x

The slope of the graph of C x  13x  900 is the value of a, 13. f x2   f x1  ax  ax1 ax2  b  ax1  b   2 . x2  x1 x2  x1 x 2  x1 a x2  x1  ax  ax1   a. (b) Factoring the numerator and cancelling, the average rate of change is 2 x2  x1 x2  x1

53. (a) By definition, the average rate of change between x1 and x2 is

f x  f a  c. x a (b) Multiplying the equation in part (a) by x  a, we obtain f x  f a  c x  a. Rearranging and adding f a to both sides, we have f x  cx   f a  ca, as desired. Because this equation is of the form f x  Ax  B with constants A  c and B  f a  ca, it represents a linear function with slope c and y­intercept f a  ca.

54. (a) The rate of change between any two points is c. In particular, between a and x, the rate of change is

2.6

TRANSFORMATIONS OF FUNCTIONS

1. (a) The graph of y  f x  3 is obtained from the graph of y  f x by shifting upward 3 units. (b) The graph of y  f x  3 is obtained from the graph of y  f x by shifting left 3 units.

2. (a) The graph of y  f x  3 is obtained from the graph of y  f x by shifting downward 3 units. (b) The graph of y  f x  3 is obtained from the graph of y  f x by shifting right 3 units.

3. (a) The graph of y   f x is obtained from the graph of y  f x by reflecting about the x­axis.

(b) The graph of y  f x is obtained from the graph of y  f x by reflecting about the y­axis.


SECTION 2.6 Transformations of Functions

45

4. (a) The graph of f x  2 is obtained from that of y  f x by shifting upward 2 units, so it has graph II.

(b) The graph of f x  3 is obtained from that of y  f x by shifting to the left 3 units, so it has graph I.

(c) The graph of f x  2 is obtained from that of y  f x by shifting to the right 2 units, so it has graph III.

(d) The graph of f x  4 is obtained from that of y  f x by shifting downward 4 units, so it has graph IV. 5. If f is an even function, then f x  f x and the graph of f is symmetric with respect to the y­axis. 6. If f is an odd function, then f x   f x and the graph of f is symmetric with respect to the origin. 7. (a) The graph of y  f x  11 can be obtained by shifting the graph of y  f x upward 11 units. (b) The graph of y  f x  8 can be obtained by shifting the graph of y  f x to the left 8 units.

8. (a) The graph of y  f x  7 can be obtained by shifting the graph of y  f x to the right 7 units.

(b) The graph of y  f x  10 can be obtained by shifting the graph of y  f x downward 10 units.

9. (a) The graph of y  14 f x can be obtained by reflecting the graph of y  f x about the y­axis, then shrinking the resulting graph vertically by a factor of 14 .

(b) The graph of y  5 f x can be obtained by reflecting the graph of y  f x about the x­axis, then stretching the resulting graph vertically by a factor of 5. 10. (a) The graph of y  6 f x can be obtained by reflecting the graph of y  f x about the x­axis, then stretching the resulting graph vertically by a factor of 6. (b) The graph of y  23 f x can be obtained by reflecting the graph of y  f x about the y­axis, then shrinking the resulting graph vertically by a factor of 23 .

11. (a) The graph of y  f x  1  5 can be obtained by shifting the graph of y  f x to the right 1 unit and downward 5 units. (b) The graph of y  f x  2  4 can be obtained by shifting the graph of y  f x to the left 2 units and downward 4 units. 12. (a) The graph of y  f x  4  6 can be obtained by shifting the graph of y  f x to the right 4 units and upward 6 units. (b) The graph of y  f x  2  9 can be obtained by shifting the graph of y  f x to the left 2 units and upward 9 units. 13. (a) The graph of y  5  f x can be obtained by reflecting the graph of y  f x about the y­axis, then shifting the resulting graph upward 5 units. (b) The graph of y  3  12 f x  2 can be obtained by shifting the graph of y  f x to the left 2 units, then shrinking vertically by a factor of 12 , then reflecting about the x­axis, then shifting upward 3 units.

14. (a) The graph of y  10  f x  1 can be obtained by shifting the graph of y  f x to the left 1 unit, then reflecting about the x­axis, then shifting upward 10 units. (b) The graph of y  4 f x  5  8 can be obtained by reflecting the graph of y  f x about the y­axis, shifting to the right 5 units, stretching vertically by a factor of 4, and shifting down 8 units. 15. (a) The graph of y  2  f 5x can be obtained by shrinking the graph of y  f x horizontally by a factor of 15 ,

reflecting about the x­axis, and shifting up 2 units.   (b) The graph of y  1  f 12 x  1 can be obtained by shifting the graph of y  f x to the left 1 unit, stretching

horizontally by a factor of 2, then shifting upward 1 unit.   16. (a) The graph of y  f 13 x  2 can be obtained by stretching the graph of y  f x horizontally by a factor of 3 and shifting downward 2 units.

(b) The graph of y  f 2 x  3  1 can be obtained by shifting the graph of y  f x to the right 3 units, shrinking horizontally by a factor of 12 , then shifting downward 1 unit.


46

CHAPTER 2 Functions

17. (a) The graph of g x  x  22 is obtained by shifting the graph of f x to the left 2 units. (b) The graph of g x  x 2  2 is obtained by shifting the graph of f x upward 2 units.

18. (a) The graph of g x  x  43 is obtained by shifting the graph of f x to the right 4 units. (b) The graph of g x  x 3  4 is obtained by shifting the graph of f x downward 4 units.

19. (a) The graph of g x  x  2  2 is obtained by shifting the graph of f x to the left 2 units and downward 2 units.

(b) The graph of g x  g x  x  2  2 is obtained from by shifting the graph of f x to the right 2 units and upward 2 units.

 20. (a) The graph of g x   x  1 is obtained by reflecting the graph of f x about the x­axis, then shifting the resulting graph upward 1 unit.  (b) The graph of g x  x  1 is obtained by reflecting the graph of f x about the y­axis, then shifting the resulting graph upward 1 unit. 21. (a)

y

y

(b) y=x@

y=x@

1 0

1

x

y=2(x+3)@

y=x@-4

2 0

(c)

y

1

x

y

(d)

y=x@

y=x@

1

1

0

0

1

x

1

x

y=1-x@ y=(x+1)@-3

22. (a)

y

(b)

y

y=Ïx+1

y=Ïx y=Ïx-2

1 1

x

y=Ïx

1 1

x


SECTION 2.6 Transformations of Functions y

(c)

y

(d)

y=Ïx+2+2

y=Ïx

1

y=Ïx

1

1

47

1

x

x y=_Ïx+1

23. The graph of y  x  1 is obtained from that of y  x by shifting to the left 1 unit, so it has graph II. Its range is [0 . 24. y  x  1 is obtained from that of y  x by shifting to the right 1 unit, so it has graph IV. Its range is [0 . 25. The graph of y  x  1 is obtained from that of y  x by shifting downward 1 unit, so it has graph I. Its range is [1 . 26. The graph of y   x is obtained from that of y  x by reflecting about the x­axis, so it has graph III. Its range is  0]. 27. f x  x 2  5. Shift the graph of y  x 2 downward 5 units.

28. f x  x 2  2. Shift the graph of y  x 2 upward 2 units. y

y

y=x@+2 y=x@

y=x@ 1 0

1

x

1

0

29. f x  3 units.

x  3. Shift the graph of y  y

x upward

x

1

y=x@-5

30. f x  x  5. Shift the graph of y  x downward 5 units.

y

y=Ïx+3

y=| x |

1 1

y=Ïx

1 1

x

x y=| x |-5


48

CHAPTER 2 Functions

31. f x  x  52 . Shift the graph of y  x 2 to the right 5 units.

32. f x  x  12 . Shift the graph of y  x 2 to the left 1 unit.

y

y y=x@ y=(x-5)@

y=x@

y=(x+1)@

5 1

x

1

33. f x  x  2. Shift the graph of y  x to the left 2 units.

y

1

34. f x  4 units.

x  4. Shift the graph of y 

x to the right

y

y=| x+2 |

y=Ïx

y=| x |

y=Ïx-4

1

1 1

x

1

x

35. f x  x 3 . Reflect the graph of y  x 3 about the x­axis.

y

x

36. f x   x. Reflect the graph of y  x about the x­axis.

y

y=| x |

y=x#

2 1

1

x y=_x#

1

x y=_| x |


SECTION 2.6 Transformations of Functions

  37. f x  4 x. Reflect the graph of y  4 x about the y­axis.

49

  38. f x  3 x. Reflect the graph of y  3 x about the y­axis.

y

y

2 y=Îx 4 y=Ï_x

4 y=Ïx

2

1

x y=Î_x

10

x

39. f x  5x 2 . Stretch the graph of y  x 2 vertically by a factor of 5.

y

y=5x@

40. f x  13 x. Shrink the graph of y  x vertically by a factor of 13 .

y

y=|x| y=x@ 1

1

2 1

x

  41. f x   15 x. Reflect the graph of y  x about the y­axis, then shrink vertically by a factor of 15 .

0

y=3|x| x

1

  42. f x  2 x. Reflect the graph of y  x about the y­axis, then stretch vertically by a factor of 2. y

y

y=2Ï_x y=Ïx

1

y=Ïx

1 1 1 y=_5 Ïx

x

1

x


50

CHAPTER 2 Functions

43. f x  x  4  2. Shift the graph of y  x to the right 44. f x  x  12  3. Shift the graph of y  x 2 to the left 4 units, then shift upward 2 units.

1 unit, then shift downward 3 units. y

y

y=(x+1)@-3 y=|x|

y=x@ 1

y=|x-4|+2

0

1 0

1

x

x

1

  45. f x  2 x  4  3. Shift the graph of y  x to the left 4 units, stretch vertically by a factor of 2, reflect about the x­axis, then shift upward 3 units.

46. f x  1  12 x  2. Shift the graph of y  x to the right 2 units, shrink vertically by a factor of 12 , reflect

about the x­axis, then shift upward 1 unit.

y

y y=|x|

y=Ïx

1

1 x

1

x

1

y=_2Ïx+4+3

1

y=1- 2 |x-2|

47. f x  12 x  22  3. Shift the graph of y  x 2 to the left 2 units, shrink vertically by a factor of 12 , then shift

downward 3 units.

  48. f x  2 x  1  3. Shift the graph of y  x to the right 1 unit, stretch vertically by a factor of 2, then shift upward 3 units. y

y y=x@

y=2Ïx-1+3

1

y=Ïx 1

1

y= 2 (x+2)@-3

x

1 1

x


51

SECTION 2.6 Transformations of Functions

  49. f x  12 x  4  3. Shrink the graph of y  x

vertically by a factor of 12 , then shift the result to the left

50. f x  3  2 x  12 . Stretch the graph of y  x 2 vertically by a factor of 2, reflect the result about the

x­axis, then shift the result to the right 1 unit and upward

4 units and downward 3 units.

3 units.

y

y y=x@

y=Ïx 1

1 4

1

x 1

x

y=3-2(x-1)@

y=2Ïx+4-3

51. y  f x  10. When f x  x 2 , y  x 2  10.

52. y  f x  4. When f x 

53. y  f x  3. When f x  x 4 , y  x  34 .

54. y  f x  8. When f x  x 3 , y  x  83 .

55. y  f x  2  5. When f x  x, y  x  2  5.

56. y   f x  4  3. When f x  x,

  57. y  f x  1. When f x  4 x, y  4 x  1.

58. y   f x  2. When f x  x 2 , y   x  22 .

59. y  2 f x  3  2. When f x  x 2 ,

60. y  12 f x  1  3. When f x  x,

61. g x  f x  2  x  22  x 2  4x  4

62. g x  f x  3  x 3  3

63. g x  f x  1  2  x  1  2  65. g x   f x  2   x  2

64. g x  2 f x  2 x 66. g x   f x  2  1   x  22  1  x 2  4x  3

67. (a) y  f x  4 is graph #3.

68. (a) y  13 f x is graph #2.

y  2 x  32  2.

x, y 

y   x  4  3.

y  12 x  1  3.

(b) y  f x  3 is graph #1.

(b) y   f x  4 is graph #3.

(c) y  2 f x  6 is graph #2.

(c) y  f x  4  3 is graph #1.

(d) y   f 2x is graph #4.

(d) y  f x is graph #4.

69. (a) y  f x  2

(b) y  f x  2

y

1

(c) y  2 f x

y

1 1

x

x  4.

y

1 1

x

1

x


52

CHAPTER 2 Functions

(d) y   f x  3

(e) y  f x

y

y

1 x

(b) y  f x

y

1

1

1

x

(c) y  f x  2

y

1 1

(d) y  f x  2

y

1 1

70. (a) y  f x  1

(f) y  12 f x  1

(e) y   f x

y

1

(f) y  2 f x

y

71. (a) y  f 2x

x

(b) y  f

y

1x 2

y=f(2x)

y=f(x)

x

1

x

y

y

1 1

x

1 1

x

1

y

x

1 1

x

1 1

x

1

1 x) y=f(_ 2

1 1

y=f(x)

x


SECTION 2.6 Transformations of Functions

(c) y  2 f 2x

(d) y  2 f

y

1x 2

y y=2f(2x)

1

y=f(3x)

1x 3

y

y=f(x)

1 y=f(x)

1 y=f( _ 3 x)

x

1

3

(d) y  1  f

y

y=f(3(x+1))

x 1 x) y=_2f(_ 2

(b) y  f

y

(c) y  f 3 x  1

1

x

y=f(x)

72. (a) y  f 3x

y=f(x)

1

1

1

1x 3

y=f(x)

1 y=1-f( _ 3 x)

1

y=f(x)

73. y  [[2x]]

  74. y  14 x

y

1

8

(d)

4

­8

­6

­4

2

4

6

x

2

x

y

x

For part (b), shift the graph in (a) to the left 5 units; for part (c), shift the graph in (a) to the left 5 units, and stretch it vertically by a factor of 2; for part (d), shift

(c)

the graph in (a) to the left 5 units, stretch it vertically by a factor of 2, and then shift

(b)

it upward 4 units.

(a) ­2 0

3

1 1

75.

x

y

x

1

53

8


54

CHAPTER 2 Functions

76. (a)

­8

­6

­4

­2

(b)

(c)

6

For (b), reflect the graph in (a) about the x­axis; for (c), stretch the graph in (a)

4

vertically by a factor of 3 and reflect about the x­axis; for (d), shift the graph in (a)

2

to the right 5 units, stretch it vertically by a factor of 3, and reflect it about the 2

4

6

8

­2

4

(c)

(d)

For part (b), shrink the graph in (a) vertically by a factor of 13 ; for part (c), shrink

(a) (b)

the graph in (a) vertically by a factor of 13 , and reflect it about the x­axis; for

2

­4

­2

2 ­2

important to obtain the graphs in part (c) and (d).

­4 ­6

77.

x­axis. The order in which each operation is applied to the graph in (a) is not

4

(c)

6 (d)

part (d), shift the graph in (a) to the right 4 units, shrink vertically by a factor of 13 , and then reflect it about the x­axis.

­4

78. (b)

­6

­4

4

For (b), shift the graph in (a) to the left 3 units; for (c), shift the graph in (a) to the

2 (a)

left 3 units and shrink it vertically by a factor of 12 ; for (d), shift the graph in (a) to the left 3 units, shrink it vertically by a factor of 12 , and then shift it downward

(c)

2

4

­2

6

3 units. The order in which each operation is applied to the graph in (a) is not important to sketch (c), while it is important in (d).

(d) ­4

 (b) y  f 2x  2 2x  2x2   4x  4x 2

 79. (a) y  f x  2x  x 2

­4

­2

1x 2

(c) y  f    x  14 x 2

 

  2 2 12 x  12 x

4

4

4

2

2

2

­2

2

4

­4

­2

­4

­2

2

4

­4

­2

­4

­2

2

4

­4

The graph in part (b) is obtained by horizontally shrinking the graph in part (a) by a factor of 12 (so the graph is half as wide). The graph in part (c) is obtained by horizontally stretching the graph in part (a) by a factor of 2 (so the graph is twice as wide).    80. (a) y  f x  2x  x 2 (b) y  f x  2 x  x2 (c) y   f x   2 x  x2    2x  x 2   2x  x 2

­4

­2

4

4

4

2

2

2

­2 ­4

2

4

­4

­2

­2 ­4

2

4

­4

­2

­2 ­4

2

4


SECTION 2.6 Transformations of Functions

 (d) y  f 2x  2 2x  2x2     2x  x 2  4x  4x 2

  (e) y  f  12 x    x  14 x 2

4

  2 2  12 x   12 x

4

2 ­4

­2

55

 

2 2

­2

4

­4

­4

­2

2

­2

4

­4

The graph in part (b) is obtained by reflecting the graph in part (a) about the y­axis. The graph in part (c) is obtained by rotating the graph in part (a) through 180 about the origin [or by reflecting the graph in part (a) first about the x­axis and then about the y­axis]. The graph in part (d) is obtained by reflecting the graph in part (a) about the y­axis and then horizontally shrinking the graph by a factor of 12 (so the graph is half as wide). The graph in part (e) is obtained by reflecting the graph in part (a) about the y­axis and then horizontally stretching the graph by a factor of 2 (so the graph is twice as wide).

81. f x  x 4 . f x  x4  x 4  f x. Thus f x

82. f x  x 3 . f x  x3  x 3   f x. Thus

y

y

is even.

f x is odd.

2 1

x

2 1

x

83. f x  x 2  x. f x  x2  x  x 2  x. Thus 84. f x  x 4  4x 2 . f x  f x. Also, f x   f x, so f x is

neither odd nor even.

y

f x  x4  4 x2  x 4  4x 2  f x. Thus f x is even.

y

1 1

x

1 1

x


56

CHAPTER 2 Functions

85. f x  x 3  x.

86. f x  3x 3  2x 2  1.

f x  3 x3  2 x2  1  3x 3  2x 2  1.

f x  x3  x  x 3  x     x 3  x   f x.

Thus f x  f x. Also f x   f x, so f x is neither odd nor even.

Thus f x is odd.

y

y

1

1

   87. f x  1  3 x. f x  1  3 x  1  3 x. Thus f x  f x. Also f x   f x, so f x is

neither odd nor even.

x

1

x

1

88. f x  x  1x. f x  x  1 x  x  1x   x  1x   f x.

y

Thus f x is odd.

y

1 x

1

1 x

1

89. (a) Even

(b) Odd y

y

1

1 1

x

1

x


57

SECTION 2.6 Transformations of Functions

90. (a) Even

(b) Odd y

y

1

1 1

1

x

x

91. (a) Since f x  x 2  4  0, for 2  x  2, the graph of y  g x is found by sketching the graph of y  f x for x  2 and x  2, then reflecting about the x­axis the part of the graph of y  f x for 2  x  2. y

(b)

y

1

1 1

x

    g x  4x  x 2  y

92. (a)

x

1

    h x  x 3  y

(b)

y=f(|x|) y=|f(x)| 0 y=f(x)

0

x

x

y=f(x)

93. (a) The bungee jumper drops to a height of 200 feet, bounces up and down, then settles at 350 feet.

(b)

y (ft) 500

(c) To obtain the graph of H from that of h, we shift downward 100 feet. Thus, H t  h t  100.

0

4

t (s)


58

CHAPTER 2 Functions

(b)

94. (a) The swimmer swims two and a half laps, slowing down with each

80

successive lap. In the first 30 seconds they swim 50 meters, so their

60

average speed is 50 30  167 ms.

40 20

(c) Here the swimmer swims 60 meters in 30 seconds, so the average speed is 60  2 ms. 30

0

100

200

t

This graph is obtained by stretching the original graph vertically by a factor of 12.

95. (a) The trip to the park corresponds to the first piece of the graph. The class travels 800 feet in 10 minutes, so their average speed is 800 10  80 ftmin. The second (horizontal) piece of the graph stretches from t  10 to t  30, so the class spends 20 minutes at the park. The park is 800 feet from the school.

(b)

(c)

y 600

y 1000 800

400

600 400

200 0

200 20

40

0

60 t

20

40

60 t

The new graph is obtained by shrinking the original

This graph is obtained by shifting the original graph

graph vertically by a factor of 050. The new average

to the right 10 minutes. The class leaves ten minutes

speed is 40 ftmin, and the new park is 400 ft from

later than it did in the original scenario.

the school.

96. (a) For z  11, the graph of g x  f 1  z x  f 12x is obtained from that of f x by shrinking horizontally by a 1. factor of 12

(b) g x  f 12x, so in this case 12x  480  x  40. The dip occurs at 40 nm at the source. (c) The observed dip at 410 nm corresponds to a wavelength of 410 12  34 nm at the source. Similarly,

the observed dips at 430, 480, and 660 nm

480 correspond to dips at 430 12  36, 12  40, and 660  55 nm at the source. 12

g (lm) 2 1 0

30

40

50

60

x (nm)

97. To obtain the graph of g x  x  22  5 from that of f x  x  22 , we shift to the right 4 units and upward 5 units.

98. To obtain the graph of g x from that of f x, we reflect the graph about the y­axis, then reflect about the x­axis, then shift upward 6 units.


SECTION 2.7 Combining Functions

99.

y

y

1

y

1 1

x

59

1 1

x

1

x

f x  [[x]]

g x  [[2x]] h x  [[3x]]   1 The graph of k x  [[nx]] is that of a step function with steps units wide, and the graph of k x  n1 x is that of a n step function with steps n units wide. 100. f even implies f x  f x; g even implies g x  g x; f odd implies f x   f x; and g odd implies g x  g x. If f and g are both even, then  f  g x  f x  g x  f x  g x   f  g x and f  g is even. If f and g are both odd, then  f  g x  f x  g x   f x  g x    f  g x and f  g is odd. If f is odd and g is even, then  f  g x  f x  g x   f x  g x, which is neither odd nor even.

101. f even implies f x  f x; g even implies g x  g x; f odd implies f x   f x; and g odd implies g x  g x. If f and g are both even, then  f g x  f x  g x  f x  g x   f g x. Thus f g is even. If f and g are both odd, then  f g x  f x  g x   f x  g x  f x  g x   f g x. Thus f g is even If f if odd and g is even, then  f g x  f x  g x  f x  g x   f x  g x    f g x. Thus f g is odd.

102. f x  x n is even when n is an even integer and f x  x n is odd when n is an odd integer. These names were chosen because polynomials with only terms with odd powers are odd functions, and polynomials with only terms with even powers are even functions. 103. Any function that is periodic with period 1 has this property. So if we draw any function on the interval [0 1 and repeat its graph on every other interval [n n  1, n an integer, we will have such a function. One example is f x  x  [[x]].

2.7

COMBINING FUNCTIONS

1. From the graphs of f and g in the figure, we find  f  g 2  f 2  g 2  3  5  8,   f 2 3 f  . 2   f  g 2  f 2  g 2  3  5  2,  f g 2  f 2 g 2  3  5  15, and g g 2 5

2. By definition, f  g x  f g x. So, if g 2  5 and f 5  12, then f  g 2  f g 2  f 5  12.

3. If the rule of the function f is “add one” and the rule of the function g is “multiply by 2” then the rule of f  g is “multiply by 2, then add one” and the rule of g  f is “add one, then multiply by 2.” 4. We can express the functions in Exercise 3 algebraically as f x  x  1, g x  2x,  f  g x  2x  1, and g  f  x  2 x  1.

5. (a) The function  f  g x is defined for all values of x that are in the domains of both f and g. (b) The function  f g x is defined for all values of x that are in the domains of both f and g.

(c) The function  f g x is defined for all values of x that are in the domains of both f and g, and g x is not equal to 0. 6. The composition  f  g x is defined for all values of x for which x is in the domain of g and g x is in the domain of f .


60

CHAPTER 2 Functions

7. Both f x  3x and g x  1  x have domain  , and the intersection of their domains is  .  f  g x  3x  1  x  2x  1 with domain  ;  f  g x  3x  1  x  4x  1 with domain  ;   3x f with domain  1  1 . x   f g x  3x 1  x  3x  3x 2 with domain  ; and g 1x 8. Both f x  3  2x and g x  2x  1 have domain  , and the intersection of their domains is  .  f  g x  3  2x  2x  1  4 with domain  ;  f  g x  3  2x  2x  1  2  4x with domain   3  2x f with domain x   ;  f g x  3  2x 2x  1  4x 2  4x  3 with domain  ; and g 2x  1       12   12   . 9. Both f x  x 3  x 2 and g x  x 2 have domain  , and the intersection of their domains is  .

 f  g x  x 3  2x 2 with domain  ;  f  g x  x 3 with domain  ;  f g x  x 5  x 4 with   x3  x2 f  x  1 with domain  0  0 . domain  ; and x  g x2

10. Both f x  x 2  1 and g x  x 2  2 have domain  , and the intersection of their domains is  .

 f  g x  2x 2  3 with domain  ;  f  g x  1 with domain  ;      x2  1 f with domain  .  f g x  x 2  1 x 2  2  x 4  3x 2  2 with domain  ; and x  2 g x 2

11. Both f x  5  x and g x  x 2  3x have domain  , and the intersection of their domains is  .    f  g x  5  x  x 2  3x  x 2  4x  5 with domain  ;    f  g x  5  x  x 2  3x  x 2  2x  5 with domain  ;    f g x  5  x x 2  3x  x 3  8x 2  15x with domain  ; and   5x f 5x with domain  0  0 3  3 .  x  2 g x x  3 x  3x 12. f x  x 2  2x has domain  ; g x  3x 2  1 has domain  ; and the intersection of their domains is  .    f  g x  x 2  2x  3x 2  1  4x 2  2x  1 with domain  ;    f  g x  x 2  2x  3x 2  1  2x 2  2x  1 with domain  ;     f g x  x 2  2x 3x 2  1  3x 4  6x 3  x 2  2x with domain  ; and       x 2  2x f , 3x 2  1  0  x   33 with domain x  x   33 . x  2 g 3x  1 13. f x  [3 5].

  25  x 2 has domain [5 5]; g x  x  3 has domain [3 ; and the intersection of their domains is

     f  g x  25  x 2  x  3 with domain [3 5];  f  g x  25  x 2  x  3 with domain [3 5];      25  x 2 f 2 with domain 3 5]. x   f g x  25  x x  3 with domain [3 5]; and g x 3


61

SECTION 2.7 Combining Functions

  14. f x  16  x 2 has domain [4 4]; g x  x 2  1 has domain  1]  [1 ; and the intersection of their domains is [4 1]  [1 4].      f  g x  16  x 2  x 2  1 with domain [4 1]  [1 4];  f  g x  16  x 2  x 2  1 with domain       16  x 2 f 2 2 [4 1]  [1 4];  f g x  with 16  x x  1 with domain [4 1]  [1 4]; and x  g x2  1 domain [4 1  1 4]. 1 3 15. f x  has domain x  x  1; g x  has domain x  x  2; the intersection of their domains is x 1 x 2 x  x  1 2 or, in interval notation,  1  1 2  2 . 1 3 4x  1 x  2  3 x  1    with domain  1  1 2  2 ;  f  g x  x 1 x 2 x  1 x  2 x  1 x  2 3 2x  5 1 x  2  3 x  1    with domain  1  1 2  2 ;  f  g x  x 1 x 2 x  1 x  2 x  1 x  2 3 3 1   with domain  1  1 2  2 ; and  f g x  x 1 x 2 x  1 x  2 1   f x  1  x  2 with domain  1  1 2  2 . x  3 g 3 x  1 x 2 3 x 16. f x  has domain x  x  3; g x  has domain x  x  3; and the intersection of their domains is x 3 x 3 x  x  3 or, in interval notation,  3  3 3  3 .

3 x x2  9 x x 2  6x  9 3   2   with domain x  x  3;  f  g x  x 3 x 3 x 3 x 3 x 9 x2  9 3 x 3x with domain x  x  3;  f g x    2 with domain x  x  3; and x 3 x 3 x 9 3   3 x  3 3x  9 f x   2 with domain x  x  0 3. x  x 3  g x x  3 x  3x x 3     17. f x  x  3  x. The domain of x is [0 , and the domain of 3  x is  3]. Thus, the domain of f is  3]  [0   [0 3].     1x 1x . The domain of x  4 is [4 , and the domain of is  0  0 1]. Thus, the 18. f x  x  4  x x domain of f is [4    0  0 1]  [4 0  0 1]. 1 19. h x  x  314  . Since the fourth root is an even root and the denominator cannot equal 0, x  3  0  x  314 x  3 So the domain is 3 .   1 x 3 . The domain of x  3 is [3 , and the domain of is x  1. Since x  1 corresponds to the 20. k x  x 1 x 1 union of intervals  1  1 , the domain is [3    1  1   [3 1  1 .  f  g x 

21.

22.

y

y

23.

f

g 0

f+g 0

4

f+g

f

g

2

f

x

f+g

g

x ­4

­2

0

2

4


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CHAPTER 2 Functions

24. 5

25.

4 3

­4

2

0

­2

26.

3

f

2

4

f+g f

2

2

1

­20

g 1

g

20 f

1

f+g

40

f+g

g

­40

­4

­2

00

In Solutions 27–32, f x  4x  5 and g x  x 2  2.   27. (a) f g 1  f 12  2  f 3  4 3  5  17 (b) g  f 1  g 4 1  5  g 9  92  2  83

28. (a) f  f 0  f 4 0  5  f 5  4 5  5  25   (b) g g 1  g 12  2  g 3  32  2  11   29. (a)  f  g 2  f g 2  f 22  2  f 6  4 6  5  29 (b) g  f  1  g  f 1  g 4 1  5  g 1  12  2  3

30. (a)  f  f  1  f  f 1  f 4 1  5  f 9  4 9  5  41   (b) g  g 0  g g 0  g 02  2  g 2  22  2  6     31. (a)  f  g x  f g x  f x 2  2  4 x 2  2  5  4x 2  13

(b) g  f  x  g  f x  g 4x  5  4x  52  2  16x 2  40x  27

32. (a)  f  f  x  f  f x  f 4x  5  4 4x  5  5  16x  25    2 (b) g  g x  g g x  g x 2  2  x 2  2  2  x 4  4x 2  6

33. f g 2  f 5  4

34. f 0  0, so g  f 0  g 0  3.

35. g  f  4  g  f 4  g 2  5

36. g 0  3, so  f  g 0  f 3  0.

37. g  g 2  g g 2  g 1  4

38. f 4  2, so  f  f  4  f 2  2.

39. From the table, g 2  5 and f 5  6, so f g 2  6.

40. From the table, f 2  3 and g 3  6, so g  f 2  6.

41. From the table, f 1  2 and f 2  3, so f  f 1  3. 42. From the table, g 2  5 and g 5  1, so g g 2  1.

43. From the table, g 6  4 and f 4  1, so  f  g 6  1.

44. From the table, f 2  3 and g 3  6, so g  f  2  6.

45. From the table, f 5  6 and f 6  3, so  f  f  5  3. 46. From the table, g 3  6 and g 6  4, so g  g 3  4.

47. f x  2x  3 has domain  ; g x  4x  1 has domain  .  f  g x  f 4x  1  2 4x  1  3  8x  1 with domain  . g  f  x  g 2x  3  4 2x  3  1  8x  11 with domain  .  f  f  x  f 2x  3  2 2x  3  3  4x  9 with domain  . g  g x  g 4x  1  4 4x  1  1  16x  5 with domain  .

2

4


SECTION 2.7 Combining Functions

x 48. f x  6x  5 has domain  . g x  has domain  . 2 x  x  6  5  3x  5 with domain  .  f  g x  f 2 2 6x  5  3x  52 with domain  . g  f  x  g 6x  5  2  f  f  x  f 6x  5  6 6x  5  5  36x  35 with domain  . x x  x  2  with domain  . g  g x  g 2 2 4 49. f x  x 2 has domain  ; g x  x  1 has domain  .

 f  g x  f x  1  x  12  x 2  2x  1 with domain  .     g  f  x  g x 2  x 2  1  x 2  1 with domain  .    2  f  f  x  f x 2  x 2  x 4 with domain  . g  g x  g x  1  x  1  1  x  2 with domain  .

 1 50. f x  3 x has domain  . g x  3 has domain x  x  0   0  0 . x    1 1 1  3 3  with domain x  x  0   0  0 .  f  g x  f x x3 x   1 1 g  f  x  g 3 x    3  with domain x  x  0   0  0 . 3 x x        f  f  x  f 3 x 3 3 x  9 x with domain  .   1 1   x 3 with domain x  x  0   0  0 . g  g x  g x3 1x 3 1 51. f x  x 2  1 has domain  . g x   has domain 0 . x     1 1 1 2    1   1 with domain 0 .  f  g x  f  x x x   1 with domain  . g  f  x  g x 2  1   x2  1   2   f  f  x  f x 2  1  x 2  1  1  x 4  2x 2  2 with domain  .    1 1     4 x with domain 0 . g  g x  g  x 1 x 52. f x  x  4 has domain  . g x  x  4 has domain  .  f  g x  f x  4  x  4  4 with domain  . g  f  x  g x  4  x  4  4  x with domain  .  f  f  x  f x  4  x  4  4  x  8 with domain  . g  g x  g x  4  x  4  4  x  4  4 (x  4  4 is always positive). The domain is  .

63


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CHAPTER 2 Functions

x has domain x  x  1; g x  2x  1 has domain   x 1 2x  1 2x  1 with domain x  x  0   0  0 .   f  g x  f 2x  1  2x 2x  1  1     2x x x 2 1   1 with domain x  x  1   1  1  g  f  x  g x 1 x 1 x 1 x   x x x 1 x x   x 1    .  f  f  x is defined whenever both f x and  f  f  x  f x 1 x  1 x  x  1 2x 1 1 x 1     f  f x are defined; that is, whenever x  1 and 2x  1  0  x   12 , which is  1  1  12   12   .

53. f x 

g  g x  g 2x  1  2 2x  1  1  4x  2  1  4x  3 with domain  . x 1 has domain x  x  1; g x  has domain x  x  0. x 1 x   1 1 1 1      1 x .  f  g x is defined whenever both g x and f g x are  f  g x  f x x 1 1 x 1 1

54. f x 

x

x

defined, so the domain is x  x  1 0.   1 x 1 x  x  . g  f  x is defined whenever both f x and g  f x are defined, so the g  f  x  g x 1 x x1

domain is x  x  1 0.   x x x x     xx1  .  f  f  x is defined whenever both f x and  f  f  x  f x x 1  1 2x 1 x  1 x1  1 x1   f  f x are defined, so the domain is x  x  1  12 .   1 1  1  x. g  g x is defined whenever both g x and g g x are defined, so the domain is g  g x  g x x

x  x  0.

2 x has domain x  x  0; g x  has domain x  x  2. x x 2   2x  4 2 x  x  .  f  g x is defined whenever both g x and f g x are defined; that  f  g x  f x 2 x x 2 is, whenever x  0 and x  2. So the domain is x  x  0 2. 2   1 2 2 x   . g  f  x is defined whenever both f x and g  f x are defined;  g  f  x  g 2 x 2  2x 1x 2 x that is, whenever x  0 and x  1. So the domain is x  x  0 1.   2 2   x.  f  f  x is defined whenever both f x and f  f x are defined; that is, whenever  f  f  x  f 2 x x x  0. So the domain is x  x  0. x   x x x  xx  2   . g  g x is defined whenever both g x and g  g x  g x 2 x  2  2 3x 4 x 2 x 2   g g x are defined; that is whenever x  2 and x   43 . So the domain is x  x  2  43 .

55. f x 


SECTION 2.7 Combining Functions

65

1 1 has domain x  x  1; g x  2 has domain  . x 1 x 1   1 1 x2  1 1      1 with domain x  x  0.   f  g x  f 1 x2  1 1  x2  1 x2  1 x2  1   1 x 2  2x  1 1 x  12   2  with domain x  x  1. g  f  x  g  2 2 x 1 x  2x  2 1  x  1 1 1 x 1   1 1x x 1 1   , with domain x  x  1 2.   f  f  x  f 1 x 1 1  x  1 x 2 1 x 1  2 21   x x 4  2x 2  1 1 1  with domain  .  g  g x  g 2  2  4 2 x 1 x  2x 2  2 1 1  x2  1 1 x2  1

56. f x 

1 57. f x   has domain x  x  0; g x  x 2  4x has domain  . x   1 .  f  g x is defined whenever 0  x 2  4x  x x  4. The product of two  f  g x  f x 2  4x   x 2  4x numbers is positive either when both numbers are negative or when both numbers are positive. So the domain of f  g is x  x  0 and x  4  x  x  0 and x  4 which is  0  4 .       1 4 1 2 1 1      . g  f  x is defined whenever both f x and g  f x are 4  g  f  x  g  x x x x x defined, that is, whenever x  0. So the domain of g  f is 0 .   1 1    x 14 .  f  f  x is defined whenever both f x and f  f x are defined, that is,  f  f  x  f  x 1  x whenever x  0. So the domain of f  f is 0 .    2   g  g x  g x 2  4x  x 2  4x  4 x 2  4x  x 4  8x 3  16x 2  4x 2  16x  x 4  8x 3  12x 2  16x

with domain  .

 58. f x  x 2 has domain  . g x  x  3 has domain [3 .    2 x 3  x  3  x  3 with domain [3 .  f  g x  f      g  f  x  g x 2  x 2  3. For the domain we must have x 2  3  x   3 or x  3. Thus the domain is        3  3  .    2  f  f  x  f x 2  x 2  x 4 with domain  .     x  3  3. For the domain we must have x  3  3  x  3  9  x  12, so the g  g x  g x  3  domain is [12 .


66

CHAPTER 2 Functions

 x has domain [0 ; g x  3 x has domain  .       f  g x  f 3 x  1  3 x  1  6 x. For 6 x to be defined we must have x  0, so the domain of f  g is [0 .      g  f  x  g 1  x  3 1  x with domain [0 .          f  f  x  f 1  x  1  1  x. As in Example 4, for x to be defined we must have x  0, and for 1  x   to be defined we must have 1  x  0  x  1  x  1. Thus, to satisfy both conditions we must have 0  x  1, so the domain of f  f is [0 1].     g  g x  g 3 x  3 3 x  9 x with domain  .   60. f x  x 2  1 has domain  1]  [1 ; g x  1  x has domain  1].    2    1x  1  x  1  1  x  1  x with domain  0].  f  g x  f      x 2  1  1  x 2  1. For x 2  1 to be defined we must have x 2  1  0  x 2  1  x  1 g  f  x  g     or x  1, and for 1  x 2  1 to be defined we must have 1  x 2  1  0  x 2  1  1  x 2  1  1  x 2  2           x  2   2  x  2. Thus, the domain of g  f is  2 1  1 2 .    2   2 x 1  x 2  1  1  x 2  2. For x 2  2 to be defined we must have x 2  2  0   f  f  x  f         x  2  x   2 or x  2. These values also satisfy x 2 1  0, so the domain of f  f is   2  2  .      g  g x  g 1  x  1  1  x. For 1  x to be defined we must have 1  x  0  x  1, and for     1  1  x to be defined we must have 1  1  x  0  1  x  1  1  x  1  x  0. Thus, the domain of g  g is [0 1].    61.  f  g  h x  f g h x  f g x  1  f x 1  x 11   3  62. g  h x  g x 2  2  x 2  2  x 6  6x 4  12x 2  8.   1 .  f  g  h x  f x 6  6x 4  12x 2  8  6 x  6x 4  12x 2  8    4    x 5  x 5 1 63.  f  g  h x  f g h x  f g x  f       3 3 3   x x x 3 .  f  g  h x  f    . 64. g  h x  g x   3 3 3 x 1 x 1 x 1 59. f x  1 

For Exercises 65–78, many answers are possible.

65. F x  x  95 . Let f x  x 5 and g x  x  9, then F x   f  g x.   66. F x  x  1. If f x  x  1 and g x  x, then F x   f  g x.

x x2 and g x  x 2 , then F x   f  g x. . Let f x  67. F x  2 x  4 x 4 1 1 68. F x  . If f x  and g x  x  3, then F x   f  g x. x 3 x     3 69. F x  1  x . Let f x  x and g x  1  x 3 , then F x   f  g x.     70. F x  1  x. If f x  1  x and g x  x, then F x   f  g x.   71. F x  1  x 3  1. If f x  1  x and g x  x 3  1, then F x   f  g x. 1 1 and g x  x 2  x  1, then F x   f  g x. 72. F x   . If f x   3 3 2 x x x 1

1 1 73. F x  2 . Let f x  , g x  x  1, and h x  x 2 , then F x   f  g  h x. x x 1


SECTION 2.7 Combining Functions

67

    74. F x  3 x  1. If g x  x  1 and h x  x, then g  h x  x  1, and if f x  3 x, then F x   f  g  h x.   9  75. F x  4  3 x . Let f x  x 9 , g x  4  x, and h x  3 x. Then F x   f  g  h x.   2 2 76. F x    2 . If g x  3  x and h x  x, then g  h x  3  x, and if f x  x 2 , then 3 x F x   f  g  h x.   3  x x 77. F x   . If h x  x, g x  , and f x  x 3 , then F x   f  g  h x. x 1 x 1  1 1 1 . If h x  x 2  1, g x  1  , and f x  , then F x   f  g  h x. 78. F x  1 x x 1  x2  1 79. Yes. If f x  m 1 x  b1 and g x  m 2 x  b2 , then  f  g x  f m 2 x  b2   m 1 m 2 x  b2   b1  m 1 m 2 x  m 1 b2  b1, which is a linear function, because it is of the form y  mx  b. The slope is m 1 m 2 . 80. (a) From the graphs we can see that  f  g 1  f 1  g 1  2  1  3. The only graph that satisfies this condition is graph VI. (b) The graphs of f and g intersect at x  23 and f x  g x for x & 23, so  f  g x  0 for x & 23, corresponding to graph I. (c) Because f and g are both negative for x  1, have opposite signs for 1  x  0, and are both positive for x  0, the graph of f g is negative only on 1 0, corresponding to graph III. (d) g 2 x is everywhere nonnegative, corresponding to graph II. (e) Because f 1  0, the graph of g f has a vertical asymptote at x  1, corresponding to graph V.

(f) f x  x  1, so the graph of f  g is the graph of g translated upward 1 unit, corresponding to graph IV. 81. The price per sticker is 015  0000002x and the number sold is x, so the revenue is R x  015  0000002x x  015x  0000002x 2 .

82. As found in Exercise 81, the revenue is R x  015x  0000002x 2 , and the cost is 0095x  00000005x 2 , so the profit   is P x  015x  0000002x 2  0095x  00000005x 2  0055x  00000015x 2 .

83. (a) Because the ripple travels at a speed of 60 cm/s, the distance traveled in t seconds is the radius, so g t  60t. (b) The area of a circle is r 2 , so f r  r 2 .

(c)  f  g t   g t2   60t2  3600t 2 cm2 . This function represents the area of the ripple as a function of time.

84. (a) Let f t be the radius of the spherical balloon in centimeters. Since the radius is increasing at a rate of 2 cm/s, the radius is r  f t  2t after t seconds. (b) The volume of the balloon can be written as g r  43 r 3 .

3 (c) g  f  t  43  2t3  32 3 t . g  f represents the volume as a function of time.

85. Let r be the radius of the spherical balloon in centimeters. Since the radius is increasing at a rate of 2 cm/s, the radius is r  2t   after t seconds. Therefore, the surface area of the balloon can be written as S  4r 2  4 2t2  4 4t 2  16t 2 .

86. (a) From Exercises 2.5.49–50, f x  100  12x and g x  0375 x  625. Thus, h x  g  f  x  g 100  12x  0375 100  12x  625  100  45x. The inputs of h are the inputs of f (elevations), and its outputs are the outputs of g (boiling points). The function h models the boiling point of water as a function of elevation. (b) Using the formula for h, we estimate the boiling point of water at the peak of Mt. Rainier to be h 44  100  45 44  802 C.


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(c) We solve h x  91  100  45x  91  x  2. Water boils at 91 C approximately 2 km above sea level. 87. (a) f x  080x

(b) g x  x  50

(c)  f  g x  f x  50  080 x  50  080x  40. f  g represents applying the $50 coupon, then the 20% discount. g  f  x  g 080x  080x  50. g  f represents applying the 20% discount, then the $50 coupon. So applying the 20% discount, then the $50 coupon gives the lower price.

88. (a) f x  090x

(b) g x  x  100

(c)  f  g x  f x  100  090 x  100  090x  90. f  g represents applying the $100 coupon, then the 10% discount. g  f  x  g 090x  090x  100. g  f represents applying the 10% discount, then the $100 coupon. So applying the 10% discount, then the $100 coupon gives the lower price.

89. Let t be the time since the plane flew over the radar station. (a) Let s be the distance in miles between the plane and the radar station, and let d be the horizontal distance that the plane  has flown. Using the Pythagorean theorem, s  f d  1  d 2 . (b) Since distance  rate  time, we have d  g t  350t.   (c) s t   f  g t  f 350t  1  350t2  1  122,500t 2 .

90. A x  105x A  A x  A A x  A 105x  105 105x  1052 x.     A  A  A x  A A  A x  A 1052 x  105 1052 x  1053 x.     A  A  A  A x  A A  A  A x  A 1053 x  105 1053 x  1054 x. A represents the amount in

the account after 1 year; A  A represents the amount in the account after 2 years; A  A  A represents the amount in the account after 3 years; and A  A  A  A represents the amount in the account after 4 years. We can see that if we compose n copies of A, we get 105n x.

91. g x  2x  1 and h x  4x 2  4x  7.

Method 1: Notice that 2x  12  4x 2  4x  1. We see that adding 6 to this quantity gives

2x  12  6  4x 2  4x  1  6  4x 2  4x  7, which is h x. So let f x  x 2  6, and we have

 f  g x  2x  12  6  h x. Method 2: Since g x is linear and h x is a second degree polynomial, f x must be a second degree polynomial, that is, f x  ax 2  bx  c for some a, b, and c. Thus f g x  f 2x  1  a 2x  12  b 2x  1  c 

4ax 2  4ax  a  2bx  b  c  4ax 2  4a  2b x  a  b  c  4x 2  4x  7. Comparing this with f g x, we

have 4a  4 (the x 2 coefficients), 4a  2b  4 (the x coefficients), and a  b  c  7 (the constant terms)  a  1 and

2a  b  2 and a  b  c  7  a  1, b  0 c  6. Thus f x  x 2  6.

f x  3x  5 and h x  3x 2  3x  2. Note since f x is linear and h x is quadratic, g x must also be quadratic. We can then use trial and error to find g x.

Another method is the following: We wish to find g so that  f  g x  h x. Thus f g x  3x 2  3x  2  3 g x  5  3x 2  3x  2  3 g x  3x 2  3x  3  g x  x 2  x  1.


SECTION 2.8 One-to-One Functions and Their Inverses

69

92. If g x is even, then h x  f g x  f g x  h x. So yes, h is always an even function.

If g x is odd, then h is not necessarily an odd function. For example, if we let f x  x  1 and g x  x 3 , g is an odd   function, but h x   f  g x  f x 3  x 3  1 is not an odd function.

If g x is odd and f is also odd, then h x   f  g x  f g x  f g x   f g x    f  g x  h x. So in this case, h is also an odd function. If g x is odd and f is even, then h x   f  g x  f g x  f g x  f g x   f  g x  h x, so in this case, h is an even function.

93.

y

y

y

y

y

1l

1l

1l

1l

1l

1

0

x

0

1

x

0

1

x

0

x

1

0

1

y  u x y  u x  3 y  xu x y  u x  1  u x  2 y  2u x In parts (a)–(d), many answers are possible. Note that for any function f x, the graph of y  u x f x is 0 for x  0 and identical to the graph of f x for x  0. (a) This is the graph of y  u x  2.

(b) This is the graph of y  3u x  1.

(c) This is similar to the fourth graph above. One possible equation is y  3u x  2  3u x  4.

(d) This is the fifth graph above shifted one unit to the right. One possible equation is y  x  1 u x  1.

94. f 0 x 

1 , f 1 x   f 0  f 0  x  1x

1

1 1 1x

1x  1x 1

x 1 , x

1 1 x  x, and f 3 x   f 0  f 2  x   f 0 x.  x 1 x  x  1 1x 1 x This cycle continues, so f 3n  f 0 for any natural number n. Since 1000  3 333  1, we conclude that x 1 . f 1000 x  f 1 x  x f 2 x   f 0  f 1  x 

2.8

ONE-TO-ONE FUNCTIONS AND THEIR INVERSES

1. A function f is one­to­one if different inputs produce different outputs. You can tell from the graph that a function is one­to­one by using the Horizontal Line Test. 2. (a) For a function to have an inverse, it must be one­to­one. f x  x 2 is not one­to­one, so it does not have an inverse. However g x  x 3 is one­to­one, so it has an inverse.  (b) The inverse of g x  x 3 is g1 x  3 x.

3. (a) Proceeding backward through the description of f , we can describe f 1 as follows: “Take the third root, subtract 5, then divide by 3.”  3 x 5 . (b) f x  3x  53 and f 1 x  3 4. Yes, the graph of f is one­to­one, so f has an inverse. Because f 4  1, f 1 1  4, and because f 5  3, f 1 3  5.

x


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CHAPTER 2 Functions

5. If the point 3 4 is on the graph of f , then the point 4 3 is on the graph of f 1 . [This is another way of saying that f 3  4  f 1 4  3.] 6. (a) This is false in general. For instance, if f x  x, then f 1 x  x, but

1 1   f 1 x. f x x

(b) This is true, by definition. 7. By the Horizontal Line Test, f is not one­to­one.

8. By the Horizontal Line Test, f is one­to­one.

9. By the Horizontal Line Test, f is one­to­one.

10. By the Horizontal Line Test, f is not one­to­one.

11. By the Horizontal Line Test, f is not one­to­one.

12. By the Horizontal Line Test, f is one­to­one.

13. f x  2x  4. If x1  x2 , then 2x1  2x2 and 2x1  4  2x2  4. So f is a one­to­one function.

14. f x  3x  2. If x1  x2 , then 3x1  3x2 and 3x1  2  3x2  2. So f is a one­to­one function.    15. g x  x. If x1  x2 , then x1  x2 because two different numbers cannot have the same square root. Therefore, g is a one­to­one function. 16. g x  x. Because every number and its negative have the same absolute value (for example, 1  1  1), g is not a one­to­one function. 17. h x  x 2  2x. Because h 0  0 and h 2  2  2 2  0 we have h 0  h 2. So f is not a one­to­one function. 18. h x  x 3  8. If x1  x2 , then x13  x23 and x13  8  x23  8. So f is a one­to­one function.

19. f x  x 4  5. Every nonzero number and its negative have the same fourth power. For example, 14  1  14 , so f 1  f 1. Thus f is not a one­to­one function.

20. f x  x 4  5, 0  x  2. If x1  x2 , then x14  x24 because two different positive numbers cannot have the same fourth power. Thus, x14  5  x24  5. So f is a one­to­one function.

21. r t  t 6  3, 0  t  5. If t1  t2 , then t16  t26 because two different positive numbers cannot have the same sixth power. Thus, t16  3  t26  3. So r is a one­to­one function.

22. r t  t 4  1. Every nonzero number and its negative have the same fourth power. For example, 14  1  14 , so r 1  r 1. Thus r is not a one­to­one function. 1 1 1 23. f x  2 . Every nonzero number and its negative have the same square. For example, 1 , so 2 x 1 12 f 1  f 1. Thus f is not a one­to­one function. 1 1 1 24. f x  . If x1  x2 , then  . So f is a one­to­one function. x x1 x2 25. (a) f 5  9. Since f is one­to­one, f 1 9  5. (b) f 1 10  0. Since f is one­to­one, f 0  10.

26. (a) f 3  6. Since f is one­to­one, f 1 6  3. (b) f 1 12  8. Since f is one­to­one, f 8  12.

27. f x  5  2x. Since f is one­to­one and f 1  5  2 1  3, then f 1 3  1. (Find 1 by solving the equation 5  2x  3.)

28. To find g1 5, we find the x value such that g x  5; that is, we solve the equation g x  x 2  4x  5. Now

x 2  4x  5  x 2  4x  5  0  x  1 x  5  0  x  1 or x  5. Since the domain of g is [2 , x  1 is

the only value where g x  5. Therefore, g1 5  1. 29. (a) Because f 6  2, f 1 2  6.

(b) Because f 2  5, f 1 5  2.

(c) Because f 0  6, f 1 6  0.

30. (a) Because g 4  2, g1 2  4.

(b) Because g 7  5, g1 5  7.

(c) Because g 8  6, g1 6  8.

In #30, replace "g" with "f"


SECTION 2.8 One-to-One Functions and Their Inverses

31. From the table, f 4  5, so f 1 5  4. 33. f 1  f 1  1

71

32. From the table, f 5  0, so f 1 0  5.   34. f f 1 6  6

  35. From the table, f 6  1, so f 1 1  6. Also, f 2  6, so f 1 6  2. Thus, f 1 f 1 1  f 1 6  2.

  36. From the table, f 5  0, so f 1 0  5. Also, f 4  5, so f 1 5  4. Thus, f 1 f 1 0  f 1 5  4. 37. f g x  f 4 x  5  14 [4 x  5  5]  x  5  5  x for all x.     g  f x  g 14 x  5  4 14 x  5  5  x for all x. Thus f and g are inverses of each other.

3x  3  1  x for all x. 38. f g x  f 3x  3  x   3x  g  f x  g 1 3  1  3  x for all x. Thus f and g are inverses of each other. 3 3     39. f g x  f 32 x  9  23 32 x  9  6  x for all x.     g  f x  g 23 x  6  32 23 x  6  9  x for all x. Thus f and g are inverses of each other. 

 x 7  7  x for all x. 4 4x  7  7 g  f x  g 4x  7   x for all x. Thus f and g are inverses of each other. 4   1 1 41. f g x  f   x for all x  0. Since f x  g x, we also have g  f x  x for all x  0. Thus f and x 1x g are inverses of each other.      5 42. f g x  f 5 x  5 x  x for all x.    5 g  f x  g x 5  x 5  x for all x. Thus f and g are inverses of each other. 40. f g x  f

x 7 4

4



  2 x 9  x  9  9  x  9  9  x for all x  9.       g  f x  g x 2  9  x 2  9  9  x 2  x for all x  0. Thus f and g are inverses of each other.

43. f g x  f

  3  44. f g x  f x  113  x  113  1  x  1  1  x for all x. 3    g  f x  g x 3  1  x  113  1  x  1  1  x for all x. Thus f and g are inverses of each other. 

 1 1   x for all x  0. 1   1 x 1 1 x   1 1   1  x  1  1  x for all x  1. Thus f and g are inverses of each other.  g  f x  g 1 x 1 x 1   2     4  x2  4  4  x 2  4  4  x 2  x 2  x, for all 0  x  2. (Note that the last 46. f g x  f

45. f g x  f

equality is possible since x  0.)    2    2 g  f x  g 4x  4 4  x 2  4  4  x 2  x 2  x, for all 0  x  2. (Again, the last equality is possible since x  0.) Thus f and g are inverses of each other.


72

CHAPTER 2 Functions

2x2  2

4x 2x  2  2 x  1   x for all x  1.  2x  2  2 x  1 4  2 x1    2 x2  2  2 x  2  2 x  2 4x x 2 x2    x for all x  2. Thus f and g are inverses of  x2 g  f x  g x 2 x  2  1 x  2 4 1

47. f g x  f

2x  2 x 1

x1  2x2

x2

each other.

54x  5 5  4x  5 1  3x 19x 13x     x for all x  13 . 54x 3 5  4x  4 1  3x 19 3 13x  4    5  4 x5  19x 5 3x  4  4 x  5 x 5 3x4      for all x   43 . Thus f and g are inverses g  f x  g x5 3x  4 3x  4  3  5 19 x 13

48. f g x  f

5  4x 1  3x

3x4

of each other.

49. f x  3x  15. y  3x  15  3x  y  15  x  13 y  5. So f 1 x  13 x  5.     Check: f f 1 x  3 13 x  5  15  x and f 1  f x  13 3x  15  5  x  5  5  x.

50. f x  8  3x. y  8  3x  3x  8  y  x  83  13 y. So f 1 x   13 x  83 .     Check: f f 1 x  8  3  13 x  83  x and f 1  f x   13 8  3x  83   83  x  83  x.

51. f x  34 x  12. y  34 x  12  34 x  y  12  x  43 y  16. So f 1 x  43 x  16.       Check: f f 1 x  34 43 x  16  12  x and f 1  f x  43 34 x  12  16  x  16  16  x.

3x 3x .y  10y  3  x  x  3  10y. So f 1 x  10x  3. 10 10    3  10x  3  3x  x and f 1  f x  10  3  3  x  3  x. Check: f f 1 x  10 10   53. f x  5  4x 3 . y  5  4x 3  4x 3  5  y  x 3  14 5  y  x  3 14 5  y. So f 1 x  3 14 5  x.  3        3 3 1 1  x and f 1  f x  3 14 5  5  4x 3  x 3  x. Check: f f x  5  4 4 5  x 52. f x 

  54. f x  3x 3  8. y  3x 3  8  3x 3  y  8  x 3  13 y  83  x  3 13 y  83 . So f 1 x  3 13 x  8.  3        3 Check: f f 1 x  3 3 13 x  8  8  x and f 1  f x  3 13 3x 3  8  8  x 3  x. 1 1 1 1 1 .y  x  2   x   2. So f 1 x   2. x 2 x 2 y y x   1 1   x and f 1  f x  1  2  x  2  2  x. Check: f f 1 x   1 2 2 x2 x

55. f x 

x 2 x 2 .y  y x  2  x  2  x y  2y  x  2  x y  x  2  2y  x y  1  2 y  1 x 2 x 2 2 y  1 2 x  1 x  . So f 1 x  . y1 x 1 2 x  1 2   4x 2 x  1  2 x  1 x 1 1   x and Check: f f x   2 x  1 2 x  1  2 x  1 4 2 x 1   x2 2 x2  1 2 [x  2  x  2] 2 2x   x.  f 1  f x  x2  1 4 x  2  x  2

56. f x 

x2


SECTION 2.8 One-to-One Functions and Their Inverses

73

x 2y x 2x .y  y 2  x  x  x 1  y  2y  x  . So f 1 x  . 2x 2x y 1 x 1   2x x   2 2x 2x 2x 2x  Check: f f 1 x  x  1    x and f 1  f x     x. 2x x 2 x  1  2x x  2  x 2 2 2x  1 x 1

57. f x 

4x 5y 5x 4x .y  y x  5  4x  x 4  y  5y  x  . So f 1 x  . x 5 x 5 4 y 4x   5x 4   20x 4 x Check: f f 1 x  x and   5x 5x  5 4  x 5 4x   4x 5 x5 5 4x 20x    f 1  f x    x. 4 x  5  4x 20 4  4x

58. f x 

x5

2x  5 2x  5 .y  y x  7  2x  5  x y  7y  2x  5  x y  2x  7y  5  x y  2  7y  5 x 7 x 7 7y  5 7x  5 x  . So f 1 x  . y2 x 2   7x  5 5 2   2 7x  5  5 x  2 x 2 Check: f f 1 x   x and  7x  5 7x  5  7 x  2 7 x 2   2x5 7 x7  5 7 2x  5  5 x  7 19x   f 1  f x     x. 2x5  2 2x  5  2 x  7 19

59. f x 

x7

4x  2 4x  2 .y  y 3x  1  4x  2  3x y  y  4x  2  4x  3x y  y  2  x 4  3y  y  2 3x  1 3x  1 y2 x 2 x  . So f 1 x  . 4  3y 4  3x   x 2 2 4   4 x  2  2 4  3x 4  3x    Check: f f 1 x   x and x 2 3 x  2  4  3x 1 3 4  3x   4x2  2 4x  2  2 3x  1 10x 3x1    f 1  f x    x. 4x2 4  1  3  2 10 3x 4x 43

60. f x 

3x1

2x  3 2x  3 .y  y 1  5x  2x  3  y  5x y  2x  3  2x  5x y  y  3  x 2  5y  y  3 1  5x 1  5x y3 x 3 x  . So f 1 x  . 5y  2 5x  2   2x3  3   2x  3  3 1  5x 17x 15x   Check: f f 1 x  x and f 1  f x     x. 5 2x  3  2 1  5x 17 5 2x3  2

61. f x 

15x


74

CHAPTER 2 Functions

3  4x y3 3  4x .y  y 8x  1  3  4x  8x y  y  3  4x  4x 2y  1  y  3  x  . 8x  1 8x  1 4 2y  1 x 3 So f 1 x  . 4 2x  1   34x  3   3  4x  3 8x  1 20x 8x1    Check: f f 1 x  x and f 1  f x      x. 34x 8  4x  4  1 20 3 8x 1 4 2

62. f x 

8x1

x3  1

x3  1

   x 3  3y  1  x  3 3y  1. So f 1 x  3 3x  1. 3    3  3      x3  1 3x  1  1 3 3 1 1  x and f  f x  3  1  x 3  1  1  x. Check: f f x  3 3

63. f x 

3

.y

7  7      64. f x  x 5  6 . y  x 5  6  7 y  x 5  6  x 5  7 y  6  x  5 7 y  6. Thus, f 1 x  5 7 x  6. 7   5    7 5  7 7 x 6 6  x 66  x and Check: f f 1 x    7 5 7  5 5 f 1  f x  x  6  6  x 5  6  6  x.    65. f x  2  3 x. y  2  3 x  y  2  3 x  x  y  23 . Thus, f 1 x  x  23 .    3   3    3 Check: f f 1 x  2  x  23  x and f 1  f x  2  3 x  2  3 x  x.       66. f x  3 6x  5. y  3 6x  5  y 3  6x  5  x  16 y 3  5 . Thus, f 1 x  16 x 3  5 .       3 3 x  5  5  x and  3 6  16 x 3  5  5  Check: f f 1 x     3 f 1  f x  16 3 6x  5  5  16 6x  5  5  x. 67. f x  x 32  1. y  x 32  1  y  1  x 32  x  y  123 . Thus, f 1 x  x  123 . 32 23  23       1  x and f 1  f x  x 32  1  1  x 32  x. Check: f f 1 x  x  123

68. f x  x  235 . y  x  235  y 53  x  2  x  y 53  2. Thus, f 1 x  x 53  2. 35 53      x and f 1  f x  x  235  2  x  2  2  x. Check: f f 1 x  x 53  2  2

69. (a), (b) f x  3x  6

70. (a), (b) f x  16  x 2 , x  0

y

y

f f Ð!

1 1

f

x

f Ð!

2 2

(c) f x  3x  6. y  3x  6  3x  y  6  x  13 y  6. So f 1 x  13 x  6.

x

(c) f x  16  x 2 , x  0. y  16  x 2   x 2  16  y  x  16  y. So  f 1 x  16  x, x  16. (Note: x  0  f x  16  x 2  16.)


SECTION 2.8 One-to-One Functions and Their Inverses

71. (a), (b) f x  x 3  1

72. (a), (b) f x 

y

x 1

y

f f Ð!

f Ð!

1 x

1

f 1 x

1

(c) f x  x 3  1  y  x 3  1  x 3  y  1    x  3 y  1. So f 1 x  3 x  1.

73. (a), (b) f x  3 

y

(c) f x 

x  1, x  1. y 

f 1 x  x 2  1, x  0.

x  1, x  1

f f Ð!

f Ð!

1

  (c) f x  3  x  1, x  1. y  3  x  1   y  3  x  1  y  32  x  1  x  y  32  1 and y  3. So

x

1

x

1

  (c) f x  2  x  1, x  1. y  2  x  1   y  2  x  1  y  22  x  1  x  y  22  1 and y  2. So f 1 x  x  22  1, x  2.

f 1 x  x  32  1, x  3.

75. f x  x 3  x. Using a graphing device and the

Horizontal Line Test, we see that f is not a one­to­one function. For example, f 0  0  f 1.

­2

y

f 1

x  1, y  0

 y 2  x  1  x  y 2  1 and y  0. So

74. (a), (b) f x  2 

x 1

2

76. f x  x 3  x. Using a graphing device and the

Horizontal Line Test, we see that f is a one­to­one function.

­2

2

75


76

CHAPTER 2 Functions

x  12 . Using a graphing device and the x 6 Horizontal Line Test, we see that f is a one­to­one

77. f x 

function.

78. f x 

x 3  4x  1. Using a graphing device and the

Horizontal Line Test, we see that f is not a one­to­one function. For example, f 0  1  f 2.

10 2 ­20

20 ­2

­10

79. f x  x  x  6. Using a graphing device and the Horizontal Line Test, we see that f is not a one­to­one function. For example f 0  6  f 2.

0

2

80. f x  x  x. Using a graphing device and the

Horizontal Line Test, we see that f is a one­to­one function. 20

10

­10

­5

10

5 ­20

­10

82. (a) y  f x  2  12 x  12 x  2  y  x  4  2y.

81. (a) y  f x  2  x  x  y  2. So f 1 x  x  2.

So f 1 x  4  2x.

(b)

(b)

10

5

­5

5

­10

10

­5 ­10

83. (a) y  g x 

x  3, y  0  x  3  y 2 , y  0

 x  y 2  3, y  0. So g1 x  x 2  3, x  0.

(b)

84. (a) y  g x  x 2  1, x  0  x 2  y  1, x  0    x  y  1. So g1 x  x  1. (b)

10

5

10

5

5

10 5

10


f

SECTION 2.8 One-to-One Functions and Their Inverses

77

 4  y (since x  0, we take the

85. If we restrict the domain of f x to [0 , then y  4  x 2  x 2  4  y  x   positive square root). So f 1 x  4  x.

 If we restrict the domain of f x to  0], then y  4  x 2  x 2  4  y  x   4  y (since x  0, we take the  negative square root). So f 1 x   4  x. g

86. If we restrict the domain of g x to [1 , then y  x  12  x  1    root)  x  1  y. So g 1 x  1  x.

y (since x  1 we take the positive square

 If we restrict the domain of g x to  1], then y  x  12  x  1   y (since x  1 we take the negative square   root)  x  1  y. So g 1 x  1  x. h

87. If we restrict the domain of h x to [2 , then y  x  22  x  2    root)  x  2  y. So h 1 x  2  x.

y (since x  2, we take the positive square

 If we restrict the domain of h x to  2], then y  x  22  x  2   y (since x  2, we take the negative   square root)  x  2  y. So h 1 x  2  x.

88. k x  x  3  k

   x  3 if x  3  0  x  3  x 3

if x  3  0  x  3

If we restrict the domain of k x to [3 , then y  x  3  x  3  y. So k 1 x  3  x.

If we restrict the domain of k x to  3], then y   x  3  y  x  3  x  3  y. So k 1 x  3  x. y

89. f Ð!

90. y

2 f

2

f

x 1

f Ð! 1

x

91. f x  x 2  9, x  0. The range is [9 . y  x 2  9  y  9  x 2  x  x  9.

y  9. Thus, f 1 x 

92. f x  x 2  2x  1  x  12 , x  1. The range is [0 . y  x  12  x  x  0.

y  1. Thus, f 1 x 

x  9, x  1,

1 1 1 1 , x  0. . Thus, f 1 x   93. f x  4 , x  0. The range is 0 . y  4  x   4 4 y x x x 1 1 94. f x  2 , x  0. The range is 0 1]. y  2  x2 y  1  y  x  x 1 x 1 0  x  1. 95. f x 

x, 0  x  9. The range is [0 3]. y 

1y . Thus, f 1 x  y

1x , x

x  x  y 2 . Thus, the inverse function is f 1 x  x 2 , 0  x  3.

96. f x  x 2  6x, x  3.The range is [9 . y  x 2  6x  x 2  6x  9  y  9  x  32  y  9    x  y  9  3. Thus, the inverse function is f 1 x  x  9  3, x  9.


78

CHAPTER 2 Functions

97. (a)

(b) Yes, the graph is unchanged upon reflection about

y

the line y  x. (c) y 

1 0

98. (a)

1

1 1 1  x  , so f 1 x  . x y x

x

(b) Yes, the graph is unchanged upon reflection about

y

the line y  x.

x 3  y x  1  x  3  x 1 y3 x y  1  y  3  x  . Thus, y1

(c) y  1 0

1

x

f 1 x 

x 3 . x 1

99. (a) The price of a pizza with no toppings (corresponding to the y­intercept) is $16, and the cost of each additional topping (the rate of change of cost with respect to number of toppings) is $150. Thus, f n  16  15n. (b) p  f n  16  15n  p  16  15n  n  23  p  16. Thus, n  f 1  p  23  p  16. This function represents the number of toppings on a pizza that costs x dollars.

(c) f 1 25  23 25  16  23 9  6. Thus, a $25 pizza has 6 toppings. 100. (a) f x  500  80x.

p  500 1 p  500 . So x  f 1  p  . f 80 80 represents the number of hours the investigator spends on a case for x dollars. 720 1220  500 (c) f 1 1220    9. If the investigator charges $1220, they spent 9 hours investigating the case. 80 80        V t 2 t 2 t 2 V t 101. (a) V  f t  100 1   1   , 0  t  40. V  100 1   1 40 40 100 40 40 100    t V 1  t  40  4 V . Since t  40, we must have t  f 1 V   40  4 V . f 1 represents time that 40 10 has elapsed since the tank started to leak.  (b) f 1 15  40  4 15  245 minutes. In 245 minutes the tank has drained to just 15 gallons of water.     102. (a)   g r  18,500 025  r 2 .   18,500 025  r 2    4625  18,500r 2  18,500r 2  4625      4625   4625   4625   1 2 1 r  r  . Since r represents a distance, r  0, so g   . g  18,500 18,500 18,500 represents the radial distance from the center of the vein at which the blood has velocity .  4625  30 (b) g 1 30   0498 cm. The velocity is 30 cms at a distance of 0498 cm from the center of the artery 18,500 or vein. (b) p  f x  500  80x. p  500  80x  80x  p  500  x 


SECTION 2.8 One-to-One Functions and Their Inverses

79

103. (a) D  f  p  3 p  150. D  3 p  150  3 p  150  D  p  50  13 D. So f 1 D  50  13 D. f 1 D represents the price that is associated with demand D.

(b) f 1 30  50  13 30  40. So when the demand is 30 units, the price per unit is $40. g

104. (a) F  g C  95 C  32. F  95 C  32  95 C  F  32  C  59 F  32. So g1 F  59 F  32. g1 F represents the Celsius temperature that corresponds to the Fahrenheit temperature of F.

(b) F 1 86  59 86  32  59 54  30. So 86 Fahrenheit is the same as 30 Celsius. 105. (a) U  f x  079x

(b) U  079x  x  U079  1265823U , so f 1 U   1265823U . f 1 U  represents the value of U US dollars in Canadian dollars.

(c) f 1 12,250  1265823 12,250  15,50633. So $12,250 in US currency is worth $15,50633 in Canadian currency.   01x, if 0  x  20,000 106. (a) f x   2000  02 x  20,000 if x  20,000 (b) We find the inverse of each piece of the function f :

f 1 x  01x. T  01x  x  10T . So f 11 T   10T .

f 2 x  2000  02 x  20,000  02x  2000. T  02x  2000  02x  T  2000  x  5T  10,000. So f 21 T   5T  10,000. Since f 0  0 and f 20,000  2000 we have f 1 T   taxpayer’s income.

  10T ,

if 0  T  2000

 5T  10,000 if T  2000

This represents the

(c) f 1 10,000  5 10,000  10,000  60,000. The required income is 60,000. 107. (a) f x  085x.

(b) g x  x  1000.

(c) H x   f  g x  f x  1000  085 x  1000  085x  850.

(d) P  H x  085x  850. P  085x  850  085x  P  850  x  1176P  1000. So

H 1 P  1176P  1000. The function H 1 represents the original sticker price for a given discounted price P.

(e) H 1 13,000  1176 13,000  1000  16,288. So the original price of the car is $16,288 when the discounted price ($1000 rebate, then 15% off) is $13,000. 108. f x  mx  b. Notice that f x1   f x2   mx1  b  mx2  b  mx1  mx2 . We can conclude that x1  x2 if and only if m  0. Therefore f is one­to­one if and only if m  0. If m  0, f x  mx  b  y  mx  b  mx  y  b x 

yb x b . So, f 1 x  . m m

2x  1 is “multiply by 2, add 1, and then divide by 5 ”. So the reverse is “multiply by 5, subtract 1, and then 5   5x  1  2  1 5x  1  1 5x 5x  1 5x  1 2 . Check: f  f 1 x  f    x divide by 2 ” or f 1 x  2 2 5 5 5   2x  1   5 1 2x  1 2x  1  1 2x 5 and f 1  f x  f 1     x. 5 2 2 2

109. (a) f x 


80

CHAPTER 2 Functions

1 1   3 is “take the negative reciprocal and add 3 ”. Since the reverse of “take the negative x x reciprocal” is “take the negative reciprocal ”, f 1 x is “subtract 3 and take the negative reciprocal ”, that is,     1 x 3 1 1 1 1 . Check: f  f x  f  3  3  x  3  x and f x   3 1 1 x 3 x 3 1 x 3   1 1 1 x     x. f 1  f x  f 1 3   1  1 1 x 1  3 3 x x  (c) f x  x 3  2 is “cube, add 2, and then take the square root”. So the reverse is “square, subtract 2, then take     3 the cube root ” or f 1 x  x 2  2. Domain for f x is  3 2  ; domain for f 1 x is [0 . Check:    3    3 2 3 2 1 x 2  x  2  2  x 2  2  2  x 2  x (on the appropriate domain) and f  f x  f    2   3 3 3 1 1 3 f  f x  f x 2  x 3  2  2  x 3  2  2  x 3  x (on the appropriate domain). (b) f x  3 

(d) f x  2x  53 is “double, subtract 5, and then cube”. So the reverse is “take the cube root, add  3 x 5 Domain for both f x and f 1 x is  . Check: 5, and divide by 2” or f 1 x  2     3  3 3   3   3 x 5 x 5 3 f  f 1 x  f  2  5  3 x  5  5  3 x  x 3  x and 2 2  x   2x  53  5 2x 2x  5  5 1 1 3 f    x.  f x  f 2x  5  2 2 2 In a function like f x  3x  2, the variable occurs only once and it easy to see how to reverse the operations step by step. But in f x  x 3  2x  6, you apply two different operations to the variable x (cubing and multiplying by 2) and then add 6, so it is not possible to reverse the operations step by step.

110. f I x  f x; therefore f  I  f . I  f x  f x; therefore I  f  f .

By definition, f  f 1 x  x  I x; therefore f  f 1  I . Similarly, f 1  f x  x  I x; therefore f 1  f  I .

111. (a) We find g 1 x: y  2x  1  2x  y  1  x  12 y  1. So g1 x  12 x  1. Thus     2  f x  h  g1 x  h 12 x  1  4 12 x  1  4 12 x  1  7  x 2  2x  1  2x  2  7  x 2  6.

(b) f  g  h  f 1  f  g  f 1  h  I  g  f 1  h  g  f 1  h. Note that we compose with f 1 on the left

on each side of the equation. We find f 1 : y  3x  5  3x  y  5  x  13 y  5. So f 1 x  13 x  5.        Thus g x  f 1  h x  f 1 3x 2  3x  2  13 3x 2  3x  2  5  13 3x 2  3x  3  x 2  x  1.            112.  f  g  g 1  f 1 x  f  g  g1  f 1 x  f  g  g1  f 1 x  f  f 1 x  x and            g 1  f 1   f  g x  g 1  f 1  f  g x  g 1  f 1  f  g x  g1  g x  x.


CHAPTER 2

Review

81

CHAPTER 2 REVIEW 1. “Square, then subtract 5” can be represented by the function f x  x 2  5. x 2. “Divide by 2, then add 9” can be represented by the function g x   9. 2 3. f x  3 x  10: “Add 10, then multiply by 3.”  4. f x  6x  10: “Multiply by 6, then subtract 10, then take the square root.”

5. g x  x 2  4x

6. h x  3x 2  2x  5

x

g x

x

1

5

2

3

1

4

0

0

1

3

2 3

4 3

0 1 2

h x

5

0

11

7. C x  5000  30x  0001x 2 (a) C 1000  5000  30 1000  0001 10002  $34,000 and

C 10,000  5000  30 10,000  0001 10,0002  $205,000.

(b) From part (a), we see that the total cost of printing 1000 copies of the book is $34,000 and the total cost of printing 10,000 copies is $205,000. (c) C 0  5000  30 0  0001 02  $5000. This represents the fixed costs associated with getting the print run ready.

(d) The net change in C as x changes from 1000 to 10,000 is C 10,000  C 1000  205,000  34,000  $171,000, and 171,000 C 10,000  C 1000   $19copy. the average rate of change is 10,000  1000 9000 8. E x  400  003x (a) E 2000  400  003 2000  $460 and E 15 000  400  003 15,000  $850.

(b) From part (a), we see that if the salesperson sells $2000 worth of goods, they make $460, and if they sell $15,000 worth of goods, they make $850. (c) E 0  400  003 0  $400 is the salesperson’s base weekly salary.

(d) The net change in E as x changes from 2000 to 15,000 is E 15,000  E 2000  850  460  $390, and the average 390 E 15,000  E 2000   $003 per dollar. rate of change is 15,000  2000 13,000 (e) Because the value of goods sold x is multiplied by 003 or 3%, we see that the salesperson earns a percentage of 3% on the goods that they sell. 9. f x  x 2  4x  6; f 0  02  4 0  6  6; f 2  22  4 2  6  2;

f 2  22  4 2  6  18; f a  a2  4 a  6  a 2  4a  6; f a  a2  4 a  6  a 2  4a  6;

f x  1  x  12 4 x  16  x 2 2x 14x 46  x 2 2x 3; f 2x  2x2 4 2x6  4x 2 8x 6.     10. f x  4  3x  6; f 5  4  15  6  1; f 9  4  27  6  4  21;        f a  2  4  3a  6  6  4  3a; f x  4  3 x  6  4  3x  6; f x 2  4  3x 2  6.

11. f x  x 2  8  f a  a 2  8, f a  h  a  h2  8  a 2  h 2  2ah  8, and   2  h 2  2ah  8  a 2  8 a 2ah  h 2 f a  h  f a    2a  h. h h h


82

CHAPTER 2 Functions

1 1 1  f a  , f a  h  , and x 2 a2 ah2 1 1  h 1 f a  h  f a a  2  a  h  2  ah2 a2    . h h h a  2 a  h  2 h a  2 a  h  2 a  2 a  2  h

12. f x

13. By the Vertical Line Test, figures (b) and (c) are graphs of functions. By the Horizontal Line Test, figure (c) is the graph of a one­to­one function. 14. (a) f 2  1 and f 2  2.

(b) The net change in f from 2 to 2 is f 2  f 2  2  1  3, and the average rate of change is 3 f 2  f 2  . 2  2 4 (c) The domain of f is [4 5] and the range of f is [4 4]. (d) f is increasing on 4 2 and 1 4; f is decreasing on 2 1 and 4 5. (e) f has local maximum values of 1 (at x  2) and 4 (at x  4).

(f) f is not a one­to­one, for example, f 2  1  f 0. There are many more examples. 15. Domain: We must have x  5  0  x  5. In interval notation, the domain is [5 .  Range: For x in the domain of f , we have x  5  x  5  0  x  5  0  f x  0. So the range is [0 .

1 has domain x  x  2. For x  2, f takes on all negative real numbers, and for x  2, it takes on all x 2 positive real numbers. Thus, the range of f is  0  0 .

16. f x 

17. f x  7x  15. The domain is all real numbers,  .

  2x  1 . Then 2x  1  0  x  12 . So the domain of f is x  x  12 . 2x  1  19. f x  x 2  4. x 2  4  0 for all x, so the domain is all real numbers,  . 18. f x 

20. f x  3x  

21. f x 

2 . The domain of f is the set of x where x  1  0  x  1. So the domain is 1 . x 1

1 1 1   . The denominators cannot equal 0, therefore the domain is x  x  0 1 2. x x 1 x 2

2x 2  5x  3 2x 2  5x  3  . The domain of g is the set of all x where the denominator is not 0. So the 2 2x  1 x  3 2x  5x  3   domain is x  2x  1  0 and x  3  0  x  x   12 and x  3 .

22. g x 

23. h x 

  4  x  x 2  1. We require the expression inside the radicals be nonnegative. So 4  x  0  4  x; also

x 2  1  0  x  1 x  1  0. We make a table: Interval Sign of x  1 Sign of x  1

Sign of x  1 x  1

 1

1 1

1 

Thus the domain is  4]   1]  [1    1]  [1 4].  3 2x  1 . Since the roots are both odd, the domain is the set of all x where the denominator is not 0. Now 24. f x   3 2x  2   3 2x  2  0  3 2x  2  2x  8  x  4. Thus the domain of f is x  x  4.


CHAPTER 2

25. f x  2  34 x

26. f x  3 1  2x, 2  x  2 y

y

1

2

x

1

27. f x  3x 2  4

Review

1

28. f x  12 x 2  8

y

x

y 1 1

1 1

29. f x 

x

x

x 5

30. f x 

y

 3 x  1 y

1 1 1

x

1

x

83


84

CHAPTER 2 Functions

31. f x  13 x  32  2

 32. f x  2 x  4  3

y

y

1 1

x

1

x

1 1

 33. f x  4 x  2

x

34. f x   12 x  12  2

y

y

1

1 1

x

 36. f x  3 x

35. f x  12 x 3 y

1

2 0

37. f x  5  x

y

1

0

x

38. f x  3  x  2

y

1

x

2

y

1 1

x

1

x


CHAPTER 2

1 39. f x   2 x

40. f x  y

Review

85

1 x  13

y

1 0

1

x 5 0

1

x

    x if x  0 42. f x  x 2 if 0  x  2    1 if x  2

  1  x if x  0 41. f x  1 if x  0

y

y

1 1

x

1 1

x

 43. x  y 2  14  y 2  14  x  y   14  x, so the original equation does not define y as a function of x.

44. 3x 

y 8

y  3x  8  y  3x  82 , so the original equation defines y as a function of x.

 13 45. x 3  y 3  27  y 3  x 3  27  y  x 3  27 , so the original equation defines y as a function of x (since the cube root function is one­to­one).

 46. 2x  y 4  16  y 4  2x  16  y   4 2x  16, so the original equation does not define y as a function of x.


86

CHAPTER 2 Functions

47. (a) The ordered pairs in this relation are 3 3, 2 0, 0 1, 2 3, and 3 3.

(b) The ordered pairs in this relation are 3 3, 2 1, 0 2, 2 5, and 3 3. y

y

1 x

1

1 x

1

Each input in the domain has exactly one output in the range, so this relation defines y as a function of

The input 2 has two different outputs (1 and 5), so

x. The domain is 3 2 0 2 3 and the range is

this relation does not define y as a function of x. The

3 1 0 3.

domain is 3 0 2 3 and the range is 2 1 3 5.

48. (a) This relation defines y as a function of x because each value x in the domain has exactly one corresponding value of y. (b) This relation does not define y as a function of x because the x­value 68 corresponds to two different y­values. (c) This relation does not define y as a function of x because every integer x­value corresponds to an infinite number of different integer y­values. (d) This relation does not define y as a function of x because every positive number x has two fourth roots.

49. We graph f x 

 9  x 2 in the viewing rectangle

[4 4] by [1 4].

50. We graph f x 

[5 5] by [1 6].

4

x 2  3 in the viewing rectangle

5

2

­4

­2

2

4

(a) From the graph, the domain of f is [3 3] and the range of f is [0 3]. (b) f x  0 at x  3. (c) f x  1 on approximately 283 283.

­5

5

(a) From the graph, the domain of f is  173]  [173  and the range of f is [0 .

(b) f x  0 at x  173. (c) f x  1 on  2 and 2 .


CHAPTER 2

51. We graph f x 

[5 5] by [1 5].

x 3  4x  1 in the viewing rectangle

Review

87

52. We graph f x  x 4  x 3  x 2  3x  6 in the viewing rectangle [3 4] by [20 100]. 100

4

50

2 ­5

5

(a) From the graph, the domain of f is approximately [211 025]  [186  and the range of f is

[0 .

(b) f x  0 at x  211, x  025, and x  186. (c) f x  1 on 2 0 and 2 .

­2

2

4

(a) From the graph, the domain of f is   and the range of f is approximately [710 .

(b) f x  0 at x  149 and x  127. (c) f x  1 on approximately  154 and 138 .

53. We graph f x  2x 2  4x  5 in the viewing rectangle [2 4] by [1 10].

54. We graph f x  1  x  x 2 in the viewing rectangle [4 3] by [10 2].

10 ­4

­2

5

­5

­2

2

4

(a) The local minimum value is f 1  3. There is no local maximum.

(b) f is decreasing on  1 and increasing on 1 . 55. We graph f x  33  16x  25x 3 in the viewing rectangle [2 2] by [10 10].

­10

  (a) The local maximum value is f  12  54 . There is

no local minimum.   (b) f is increasing on   12 and decreasing on    12   .

56. We graph f x  x 3  4x 2 in the viewing rectangle [2 5] by [15 10].

10

­2

10

2 ­10

(a) The local minimum value is approximately f 046  281 and the local maximum value is

approximately f 046  379.

(b) f is decreasing on approximately  046 and 046  and increasing on approximately 046 046.

2

­2

2

4

­10

(a) The local maximum value is f 0  0 and the local

minimum value is approximately f 267  948.

(b) f is decreasing on approximately 0 267 and increasing on  0 and approximately 267 .


88

CHAPTER 2 Functions

57. We graph f x  x 23 6  x13 in the viewing rectangle [5 10] by [8 8].

    58. We graph f x  x 4  16 in the viewing rectangle [3 3] by [5 30].

5

20

­5

5

10

­5 ­2

(a) The local maximum value is f 4  317 and the local minimum value is f 0  0.

(b) f is decreasing on  0 and 4  and increasing on 0 4.

2

(a) The local maximum value is f 0  16 and the local minimum values are f 2  0.

(b) f is decreasing on  2 and 0 2 and increasing on 2 0 and 2 .

f 8  f 4 4   1. 84 4 35 7 g 30  g 10   . 60. The net change is g 30  g 10  30  5  35 and the average rate of change is 30  10 20 4 4 f 2  f 1  . 61. The net change is f 2  f 1  6  2  4 and the average rate of change is 2  1 3 Replace "f" with "g" f 3  f 1 6 62. The net change is f 3  f 1  1  5  6 and the average rate of change is   3. 31 2     63. The net change is f 4  f 1  42  2 4  12  2 1  8  1  9 and the average rate of change is 59. The net change is f 8  f 4  8  12  4 and the average rate of change is

9 f 4  f 1   3. 41 3

64. The net change is g a  h  g a  a  h  12  a  12  2ah  2h  h 2 and the average rate of change is 2ah  2h  h 2 g a  h  g a   2a  2  h. aha h

65. f x  2  3x2  9x 2  12x  4 is not linear. It cannot be expressed in the form f x  ax  b with constant a and b.     2x  10 66. g x    2 5 5 x  2 5 is linear with a  2 5 5 and b  2 5. 5 67. (a)

68. (a)

y

1 0

1

x

y

1 0

(b) The slope of the graph is the value of a in the equation f x  ax  b  3x  2; that is, 3. (c) The rate of change is the slope of the graph, 3.

1

x

(b) The slope of the graph is the value of a in the equation f x  ax  b   12 x  3; that is,  12 .

(c) The rate of change is the slope of the graph,  12 .

69. The linear function with rate of change 2 and initial value 3 has a  2 and b  3, so f x  2x  3.


CHAPTER 2

Review

89

70. The linear function whose graph has slope 12 and y­intercept 1 has a  12 and b  1, so f x  12 x  1. 71. Between x  0 and x  1, the rate of change is f x  2x  3. 72. Between x  0 and x  2, the rate of change is is f x   14 x  6.

f 1  f 0 53   2. At x  0, f x  3. Thus, an equation is 10 1

55  6 f 2  f 0    14 . At x  0, f x  6. Thus, an equation 20 2

73. The points 0 4 and 8 0 lie on the graph, so the rate of change is y   12 x  4.

04 1   . At x  0, y  4. Thus, an equation is 80 2

74. The points 0 4 and 2 0 lie on the graph, so the rate of change is is y  2x  4.

0  4  2. At x  0, y  4. Thus, an equation 20

75. f x  12 x  6 (a) The average rate of change of f between x  0 and x  2 is     1 2  6  1 0  6 f 2  f 0 5  6 1 2 2    , and the average rate of change of f between x  15 20 2 2 2 and x  50 is     1 50  6  1 15  6 19  32 1 f 50  f 15 2 2    . 50  15 35 35 2 (b) The rates of change are the same.

(c) Yes, f is a linear function with rate of change 12 . 76. f x  8  3x

f 2  f 0 [8  3 2]  [8  3 0] 28    3, 20 2 2 and the average rate of change of f between x  15 and x  50 is [8  3 50]  [8  3 15] f 50  f 15 142  37    3. 50  15 35 35 (b) The rates of change are the same. (a) The average rate of change of f between x  0 and x  2 is

(c) Yes, f is a linear function with rate of change 3.

77. f x  x  12

(a) The average rate of change of f between x  0 and x  2 is average rate of change of f between x  15 and x  50 is

f 2  f 0 2  12  0  12   0, and the 20 20

f 50  f 15 2401  196 50  12  15  12  .50  12   63. 50  15 35 35 (b) The rates of change are not the same. (c) No, f is not a linear function. 1 78. f x  x 3

1

1

 f 2  f 0 1  23 03   , and the average 20 2 15 1  1 1 f 50  f 15  503 153   . rate of change of f between x  15 and x  50 is 50  15 35 954 (b) The rates of change are not the same.

(a) The average rate of change of f between x  0 and x  2 is

(c) No, f is not a linear function.


90

CHAPTER 2 Functions

79. (a) (i) y  f x  8. Shift the graph of f x upward 8 units. (ii) g x  x 3  8

(b) (i) y  f x  8. Shift the graph of f x to the left 8 units. (ii) g x  x  83

(c) (i) y  1  2 f x. Stretch the graph of f x vertically by a factor of 2, then shift it upward 1 unit. (ii) g x  1  2x 3

(d) (i) y  f x  2  2. Shift the graph of f x to the right 2 units, then downward 2 units. (ii) g x  x  23  2

(e) (i) y  f x. Reflect the graph of f x about the y­axis. (ii) g x  x3  x 3

(f) (i) y   f x. Reflect the graph of f x first about the y­axis, then reflect about the x­axis. (ii) g x   x3  x 3

(g) (i) y   f x. Reflect the graph of f x about the x­axis. (ii) g x  x 3

(h) (i) y  f 1 x. Reflect the graph of f x about the line y  x.  (ii) g x  3 x 80. (a) y  f x  2

(b) y   f x

y

1

(c) y  3  f x

y

1 1

(d) y  12 f x  1

(e) y  f 1 x

y

1 1

x

1

x

(f) y  f x

y

1

1 1

x

y

x

y

1 1

x

1

x

81. (a) f x  2x 5  3x 2  2. f x  2 x5  3 x2  2  2x 5  3x 2  2. Since f x  f x, f is not even.  f x  2x 5  3x 2  2. Since  f x  f x, f is not odd.   (b) f x  x 3  x 7 . f x  x3  x7   x 3  x 7   f x, hence f is odd.

(c) f x 

1  x2 1  x2 1  x2 . f    f x. Since f x  f x, f is even. x 1  x2 1  x2 1  x2


CHAPTER 2

Review

91

1 1 1 1 . f x  .  f x   . Since f x  f x , f is not even, and since  x 2 2x x 2 x  2 f x   f x, f is not odd.

(d) f x 

82. (a) This function is odd. (b) This function is neither even nor odd. (c) This function is even. (d) This function is neither even nor odd. 2. 83. (a) The average rate of change is the same between any two points if the function is linear, as is graph  5 is increasing on  2 and decreasing on 2 . (b) Graph  4 has domain [1 . (c) Graph 

3 has f x  f x for all x in its domain, so it is even. (d) Graph  1 is decreasing on  . (e) Graph  3 has two local minima. (f) Graph 

1, 2 , and  4 all pass the Horizontal Line Test and thus have inverse functions. (g) Graphs 

    84. h t  16t 2  48t  32  16 t 2  3t  32  16 t 2  3t  94  32  36   2   16 t 2  3t  94  68  16 t  32  68 The stone reaches a maximum height of 68 feet.

85. (a) If the rate is doubled, the volume is given by Vc  2V t  16t. (b) The volume is given by Vb  V t  4  08 t  4, t  4.

(c) In this case, the volume is given by Va  V t  50  08t  5.

 3V 3V  r V   3 86. (a) Solving V  43 r 3 for r, we have r 3  4 4   3 08t 3t  3 . This function models the radius of the balloon as a function (b) r  V  t  r V t  r 08t  3 4 5 of t.   3 50 30 (c) r  V  50  3  3  212 ft 5     2 12  3960 12 3960   3960  h  3960    h  3960   1 . Thus,  87. (a)   144 3960  h 144 3960  h     12 1 x  3960   1 . This represents the astronaut’s height above the surface of the earth as a function of their x weight.   12 1 (b)  64  3960   1  1980. This means that if the astronaut weighs 64 lb, then they are 1980 mi above the 64 surface of the earth. 

88. The higher the value of k, the higher the maximum yield. (In fact, it appears that the maximum yield is 12 k.) A nitrogen level of 1 ppm seems to deliver the maximum yield, regardless of the value of k. 5x . The maximum yield is Y 1  52 tons per acre. For k  5, Y x  1  x2


92

CHAPTER 2 Functions

89. f x  x  2, g x  x 2

90. f x  x 2  1, g x  3  x 2 5

5

­4

­2

2

­2

2

91. f x  x 2  3x 2 and g x 4  3x.

(a)  f  g x  x 2  3x  2  4  3x  x 2  6x  6   (b)  f  g x  x 2  3x  2  4  3x  x 2  2   (c)  f g x  x 2  3x  2 4  3x  4x 2  12x  8  3x 3  9x 2  6x  3x 3  13x 2  18x  8   x 2  3x  2 f , x  43 (d) x  g 4  3x

(e)  f  g x  f 4  3x  4  3x2  3 4  3x  2  16  24x  9x 2  12  9x  2  9x 2  15x  6     (f) g  f  x  g x 2  3x  2  4  3 x 2  3x  2  3x 2  9x  2

 92. f x  1  x 2 and g x  x  1. (Remember that the proper domains must apply.)   2  x 1 1 x 1 1x 1 x (a)  f  g x  f      (b) g  f  x  g 1  x 2  1  x 2  1  x 2  x   (c)  f  g 2  f g 2  f 2  1  f 1  1  12  2.   (d)  f  f  2  f  f 2  f 1  22  f 5  1  52  26.

(e)  f  g  f  x  f g  f  x  f x  1  x2  1  x 2 . Note that g  f  x  x by part (b).  (f) g  f  g x  g  f  g x  g x  x  1. Note that  f  g x  x by part (a).

x  1 and g x  x  x 2 .     f  g x  f x  x 2  x  x 2  1, and the domain is [0 1].    2     g  f  x  g x  1  x  1  x  1  x  1  x  2 x  1  x  x, and the domain is [0 .    x 1  x  1  1, and the domain is [0 .  f  f  x  f     2    g  g x  g x  x 2  x  x 2  x  x 2  x  x 2  x 2  2x 3  x 4  x 4  2x 3  2x 2  x, and the

93. f x 

domain is  .


CHAPTER 2

Review

93

2 , has domain x  x  4. x, has domain x  x  0. g x  x 4    2 2 .  f  g x is defined whenever both g x and f g x are defined; that is,   f  g x  f x 4 x 4 2 2 whenever x  4 and  0. Now  0  x  4  0  x  4. So the domain of f  g is 4 . x 4 x 4   2 . g  f  x is defined whenever both f x and g  f x are defined; that is, whenever g  f  x  g x   x 4   x  0 and x  4  0. Now x  4  0  x  16. So the domain of g  f is [0 16  16 .    x  x  x 14 .  f  f  x is defined whenever both f x and f  f x are defined; that is,  f  f  x  f whenever x  0. So the domain of f  f is [0 .   2 x 4 2 x  4 2    g  g x is defined whenever both g x and  g  g x  g 2 x 4 2  4 x  4 9  2x 4 x 4 g g x are defined; that is, whenever x  4 and 9  2x  0. Now 9  2x  0  2x  9  x  92 . So the domain of   g  g is x  x  92  4 .   95. f x  1  x, g x  1  x 2 and h x  1  x.          2     f 1 12 x x  f  g  h x  f g h x  f g 1  x  f 1  1  x       2         f x  2 x  1  x  2 x  1  2 x  x  1  x  1  x

94. f x 

96. If h x 

x and g x  1  x , then g  h x  g

   1  f  g  h x  f 1  x     T x. 1 x

   1 x  1  x. If f x   , then x

97. f x  3  x 3 . If x1  x2 , then x13  x23 (unequal numbers have unequal cubes), and therefore 3  x13  3  x23 . Thus f is a one­to­one function.   98. g x  2  2x  x 2  x 2  2x  1  1  x  12  1. Since g 0  2  g 2 , as is true for all pairs of numbers

equidistant from 1, g is not a one­to­one function. 1 99. h x  4 . Since the fourth powers of a number and its negative are equal, h is not one­to­one. For example, x 1 1  1 and h 1   1, so h 1  h 1. h 1  14 14      100. r x  2  x  3. If x1  x2 , then x1  3  x2  3, so x1  3  x2  3 and 2  x1  3  2  x2  3. Thus r is one­to­one. 101. p x  33  16x  25x 3 . Using a graphing device and 102. q x  33  16x  25x 3 . Using a graphing device and the Horizontal Line Test, we see that p is not a one­to­one

the Horizontal Line Test, we see that q is a one­to­one

function.

function. 10

­5

10

5 ­10

­5

5 ­10

103. f x  3x  2  y  3x  2  3x  y  2  x  13 y  2. So f 1 x  13 x  2.


94

CHAPTER 2 Functions

2x  1 2x  1 y  2x  1  3y  2x  3y  1  x  12 3y  1  So f 1 x  12 3x  1. 3 3    105. f x  x  13  y  x  13  x  1  3 y  x  3 y  1. So f 1 x  3 x  1.    106. f x  1  5 x  2. y  1  5 x  2  y  1  5 x  2  x  2  y  15  x  2  y  15 . So

104. f x 

f 1 x  2  x  15 .

107. The graph passes the Horizontal Line Test, so f has an inverse. Because f 1  0, f 1 0  1, and because f 3  4, f 1 4  3.

108. The graph fails the Horizontal Line Test, so f does not have an inverse. 109. (a), (b) f x  x 2  4, x  0

110.

y

f

  (a) If x1  x2 , then 4 x1  4 x2 , and so   1  4 x1  1  4 x2 . Therefore, f is a one­to­one function.

f Ð!

(b), (c)

y

1 1

f Ð!

x

f 1

(c) f x  x 2  4, x  0  y  x 2  4, y  4   x 2  y  4  x  y  4. So  f 1 x  x  4, x  4.

1

x

   (d) f x  1  4 x. y  1  4 x  4 x  y  1  x  y  14 . So f 1 x  x  14 ,

x  1. Note that the domain of f is [0  , so  y  1  4 x  1. Hence, the domain of f 1 is

[1 .

111. (a) y  x  2 is the graph of the absolute value function y  x, shifted downward 2 units. It has Graph VI, and defines y as a function of x. (b) y  x  12  2 is the graph of y  x 2 , shifted to the left 1 unit and downward 2 units. It has Graph IV, and defines y as a function of x. (c) x  22  y  12  4 is the graph of a circle centered at 2 1 with radius 2. It has Graph V, and does not define y as a function of x. (d) y  x  23 is the graph of y  x 3 , shifted to the right 2 units. It has Graph I, and defines y as a function of x.

(e) x  y 2  3 is the graph of y  x 2 , reflected about the line y  x and shifted to the left 3 units. It has Graph VIII, and does not define y as a function of x. 1 (f) y  is undefined at x  0 and has Graph III. It defines y as a function of x. x (g) y  3 x  32  3 is the graph of y  x 2 , shifted to the left 3 units, reflected about the x­axis, stretched vertically by a factor of 3, and shifted 3 units upward. It has Graph VII, and defines y as a function of x.   (h) y  2  x is the graph of of y  x, reflected about the y­axis and shifted to the right 2 units. It has Graph II, and defines y as a function of x.


CHAPTER 2

95

Test

CHAPTER 2 TEST

1. By the Vertical Line Test, figures (a) and (b) are graphs of functions. By the Horizontal Line Test, only figure (a) is the graph of a one­to­one function.

     0 2 2 a2 a2 2. (a) f 0   0; f 2   ; f a  2   . 01 21 3 a21 a3  x (b) f x  . Our restrictions are that the input to the radical is nonnegative and that the denominator must not be 0. x 1 Thus, x  0 and x  1  0  x  1. (The second restriction is made irrelevant by the first.) In interval notation, the domain is [0 .   10 2    f 10  f 2 3 10  11 2 10  1 2  1 (c) The average rate of change is   . 10  2 10  2 264

3. (a) “Subtract 2, then cube the result” can be expressed

(c)

y

algebraically as f x  x  23 . (b) x

f x

1

27

0 1 2 3

2

1 x

8 1

0 1

4 8 (d) We know that f has an inverse because it passes the Horizontal Line Test. A verbal description for f 1 is, “Take the cube root, then add 2.”    (e) y  x  23  3 y  x  2  x  3 y  2. Thus, a formula for f 1 is f 1 x  3 x  2.

4. (a) f 3  2 and f 2  3.

(b) The net change between x  3 and x  2 is f 2  f 3  3  2  5. The average rate of change over this 5 f 2  f 3   1. interval is 2  3 5 (c) f has domain [5 5] and range [4 4]. (d) f is increasing on 5 4 and 1 3 and decreasing on 4 1 and 3 5.

(e) f has a local minimum value of 4 at x  1 and local maximum values of 1 at x  4 and 4 at x  3. (f) No, f fails the Horizontal Line Test and so is not one­to­one.


96

CHAPTER 2 Functions

5. R x  500x 2  3000x

(a) R 2  500 22  3000 2  $4000 represents their total

(b) We can see from the graph that the

sales revenue when their price is $2 per bar and

revenue increases until the price reaches

R 4  500 42  3000 4  $4000 represents their total

$3, then decreases. R

sales revenue when their price is $4 per bar

5000

(c) The maximum revenue is $4500, and it is achieved at a price

4000

of x  $3.

3000 2000 1000 0

1

2

3

4

x

5

      6. The net change is f 2  h  f 2  2  h2  2 2  h  22  2 2  4  h 2  4h  4  2h  0  2h  h 2 and the average rate of change is

2h  h 2 f 2  h  f 2   2  h. 2h2 h

7. (a) f x  x  52  x 2  10x  25 is not linear because it cannot be

(b)

y

expressed in the form f x  ax  b for constants a and b. g x  1  5x is linear.

y=g(x)

y=f(x)

(c) g x has rate of change 5. 10 0

8. (a) f x  x 2

1

x

(b) g x  x  42  1. To obtain the graph of g, shift the graph of f to the left 4 units and

y

downward 1 unit. y

y=x@

2 0

1

x

y=x@

2 y=(x+4)@-1

0

1

x

9. (a) The graph of y  f x  3  2 can be obtained by shifting the graph of f x to the right 3 units, then upward 2 units.   (b) If f x  x, then y  f x  3  2  x  3  2.

10. (a) The graph of y  f x can be obtained by reflecting the graph of f x about the y­axis.   (b) If f x  x, then y  f x  x.


CHAPTER 2

11. (a) f 2  1  2  1  2  3 (since 2  1 ).

97

Test

y

(b)

f 1  1  1  0 (since 1  1 ).

1 x

1

12. f x  x 2  x  1; g x  x  3.

 (a)  f  g x  f x  g x  x 2  x  1  x  3  x 2  2x  2   (b)  f  g x  f x  g x  x 2  x  1  x  3  x 2  4

(c)  f  g x  f g x  f x  3  x  32  x  3  1  x 2  6x  9  x  3  1  x 2  5x  7     (d) g  f  x  g  f x  g x 2  x  1  x 2  x  1  3  x 2  x  2

(e) f g 2  f 1  12  1  1  1. [We have used the fact that g 2  2  3  1.] (f) g  f 2  g 7  7  3  4. [We have used the fact that f 2  22  2  1  7.]

(g) g  g  g x  g g g x  g g x  3  g x  6  x  6  3  x  9. [We have used the fact that g x  3  x  3  3  x  6.]

13. (a) f x  x 3  1 is one­to­one because each real number has a unique cube.

(b) g x  x  1 is not one­to­one because, for example, g 2  g 0  1.

1 1 1    x for all x  0, and g  f x   2  x  2  2  x for all x  2. Thus, by 1 1 1 2 2 x x 2 x the Inverse Function Property, f and g are inverse functions.

14. f g x  

x 3 5y  3 x 3 . y   2x  5 y  x  3  x 2y  1  5x  3  x   . Thus, 2x  5 2x  5 2y  1 5x  3 f 1 x   . 2x  1    (b) f x  3  x, x  3 and f 1 x  3  x 2 , 16. (a) f x  3  x, x  3  y  3  x 

15. f x 

y 2  3  x  x  3  y 2 . Thus f 1 x  3  x 2 , x  0.

x 0

y

f 1 x

1

f Ð!

17. The domain of f is [0 6], and the range of f is [1 7]. 18. The graph passes through the points 0 1 and 4 3, so f 0  1 and f 4  3.


98

CHAPTER 2 Functions

19. The graph of f x  2 can be obtained by shifting the graph of f x to the right

y

2 units. The graph of f x  2 can be obtained by shifting the graph of f x

y=f(x)+2

upward 2 units.

y=f(x-2) 1

f 1

x

20. The net change of f between x  2 and x  6 is f 6  f 2  7  2  5 and the average rate of change is 5 f 6  f 2  . 62 4 21. Because f 0  1, f 1 1  0. Because f 4  3, f 1 3  4. 22. y

f fÐ! 1

1

x

23. (a) f x  3x 4  14x 2  5x  3. The graph is shown in the viewing rectangle [10 10] by [30 10].

­10

10 ­20

(b) No, by the Horizontal Line Test. (c) The local maximum is approximately 255 when x  018, as shown in the first viewing rectangle [015 025] by [26 25]. One local minimum is approximately 2718 when x  161, as shown in the second viewing rectangle [165 155] by [275 27]. The other local minimum is approximately 1193 when x  143, as shown is the viewing rectangle [14 15] by [12 119]. 0.15 ­2.50 ­2.55 ­2.60

0.20

0.25

­1.65

­1.60

­1.55 ­27.0 ­27.2 ­27.4

1.40 ­11.90

1.45

1.50

­11.95 ­12.00

(d) Using the graph in part (a) and the local minimum, 2718, found in part (c), we see that the range is [2718 .

(e) Using the information from part (c) and the graph in part (a), f x is increasing on the intervals 161 018 and 143  and decreasing on the intervals  161 and 018 143.


Modeling with Functions

99

FOCUS ON MODELING Modeling with Functions 1. Let  be the width of the building lot. Then the length of the lot is 3. So the area of the building lot is A   32 ,   0. 2. Let  be the width of the poster. Then the length of the poster is   10. So the area of the poster is A      10  2  10.

3. Let  be the width of the base of the rectangle. Then the height of the rectangle is 12 . Thus the volume of the box is given by the function V   12 3 ,   0. 4. Let r be the radius of the cylinder. Then the height of the cylinder is 4r. Since for a cylinder V  r 2 h, the volume of the cylinder is given by the function V r   r 2 4r  4r 3 .

5. Let P be the perimeter of the rectangle and y be the length of the other side. Since P  2x  2y and the perimeter is 20, we have 2x  2y  20  x  y  10  y  10  x. Since area is A  x y, substituting gives A x  x 10  x  10x  x 2 , and since A must be positive, the domain is 0  x  10.

6. Let A be the area and y be the length of the other side. Then A  x y  16  y  P  2x  2 

32 16  2x  , where x  0. x x

7.

16 . Substituting into P  2x  2y gives x

Let h be the height of an altitude of the equilateral triangle whose side has length x, x

x

h 1 _x 2

as shown in the diagram. Thus the area is given by A  12 xh. By the Pythagorean   2 Theorem, h 2  12 x  x 2  h 2  14 x 2  x 2  h 2  34 x 2  h  23 x.

Substituting into the area of a triangle, we get    A x  12 xh  12 x 23 x  43 x 2 , x  0.

 8. Let d represent the length of any side of a cube. Then the surface area is S  6d 2 , and the volume is V  d 3  d  3 V .   2 Substituting for d gives S V   6 3 V  6V 23 , V  0. A r  9. We solve for r in the formula for the area of a circle. This gives A  r 2  r 2  r A 

A , A  0. 

A , so the model is 

C 10. Let r be the radius of a circle. Then the area is A  r 2 , and the circumference is C  2r  r  . Substituting for r 2  2 C2 C , C  0. gives A C    2 4 60 . x2 The surface area, S, of the box is the sum of the area of the 4 sides and the area of the base and top. Thus   240 240 60  2x 2  S  4xh  2x 2  4x  2x 2 , so the model is S x   2x 2 , x  0. 2 x x x

11. Let h be the height of the box in feet. The volume of the box is V  60. Then x 2 h  60  h 

12. By similar triangles,

12 5d 5   5 L  d  12L  5d  7L  L  . The model is L d  57 d. L L d 7


100

FOCUS ON MODELING

13.

Let d1 be the distance traveled south by the first ship and d2 be the distance

traveled east by the second ship. The first ship travels south for t hours at 5 mi/h, so dÁ

d1  15t and, similarly, d2  20t. Since the ships are traveling at right angles to

D

each other, we can apply the Pythagorean Theorem to get    D t  d12  d22  15t2  20t2  225t 2  400t 2  25t.

14. Let x be one of the numbers. Then the other number is 60  x, so the product is given by the function P x  x 60  x  60x  x 2 .

Let b be the length of the base, l be the length of the equal sides, and h be the

15. l

h

b

l

height in centimeters. Since the perimeter is 8, 2l  b  8  2l  8  b   2 l  12 8  b. By the Pythagorean Theorem, h 2  12 b  l 2   h  l 2  14 b2 . Therefore the area of the triangle is  b1 1 2 2 A  12  b  h  12  b l 2  14 b2  4 8  b  4 b 2  b b b   64  16b  b2  b2  64  16b   4 4  b  b 4  b 4 4 4  so the model is A b  b 4  b, 0  b  4.

16. Let x be the length of the shorter leg of the right triangle. Then the length of the other triangle is 2x. Since it is a right    triangle, the length of the hypotenuse is x 2  2x2  5x 2  5 x (since x  0 ). Thus the perimeter of the triangle is     P x  x  2x  5 x  3  5 x.

 2 2 17. Let  be the length of the rectangle. By the Pythagorean Theorem, 12   h 2  102   h 2  102  4     2  4 100  h 2    2 100  h 2 (since   0 ). Therefore, the area of the rectangle is A  h  2h 100  h 2 ,  so the model is A h  2h 100  h 2 , 0  h  10.

18. Using the formula for the volume of a cone, V  13 r 2 h, we substitute V  100 and solve for h. Thus 100  13 r 2 h  h r 

300 . r 2


Modeling with Functions

(b) Let x be one number: then 19  x is the other number, and so the product, p, is

19. (a) We complete the table. First number

Second number

Product

1 2 3 4 5 6 7 8 9 10 11

18 17 16 15 14 13 12 11 10 9 8

18 34 48 60 70 78 84 88 90 90 88

101

p x  x 19  x  19x  x 2 .   (c) p x  19x  x 2   x 2  19x   2   2  19   x 2  19x  19 2 2   x  952  9025

So the product is maximized when the numbers are both 95.

From the table we conclude that the numbers is still increasing, the numbers whose product is a maximum should both be 95. 20. Let the positive numbers be x and y. Since their sum is 100, we have x  y  100  y  100  x. We wish to minimize

the sum of squares, which is S  x 2  y 2  x 2  100  x2 . So S x  x 2  100  x2  x 2  10,000  200x      x 2  2x 2  200x  10,000  2 x 2  100x  10,000  2 x 2  100x  2500  10,000  5000  2 x  502  5000.

Thus the minimum sum of squares occurs when x  50. Then y  100  50  50. Therefore both numbers are 50.

21. (a) Let x be the width of the field (in feet) and l be the length of the field (in feet). Since the farmer has 2400 ft of fencing we must have 2x  l  2400. 2000

200

400

200

Area=2000(200)=400,000 1000 700

1000

1000

700 Area=400(1000)=400,000

Area=1000(700)=700,000

Width

Length

Area

200

2000

400,000

300

1800

540,000

400

1600

640,000

500

1400

700,000

600

1200

720,000

700

1000

700,000

800

800

640,000

It appears that the field of largest area is about 600 ft  1200 ft.

(b) Let x be the width of the field (in feet) and l be the length of the field (in feet). Since the farmer has 2400 ft of fencing we must have 2x  l  2400  l  2400  2x. The area of the fenced­in field is given by   A x  l  x  2400  2x x  2x 2  2400x  2 x 2  1200x .     (c) The area is A x  2 x 2  1200x  6002  2 6002  2 x  6002  720,000. So the maximum area occurs when x  600 ft and l  2400  2 600  1200 ft.

rancher

22. (a) Let  be the width of the rectangular area (in feet) and l be the length of the field (in feet). Since the farmer has 750 feet of fencing, we must have 5  2l  750  2l  750  5  l  52 150  . Thus the total area of the four   pens is A   l    52  150     52 2  150 .


102

FOCUS ON MODELING

(b) We complete the square to get       A    52 2  150   52 2  150  752  52  752   52   752  140625. Therefore, the largest possible total area of the four pens is 14,0625 ft2 .

23. (a) Let x be the length of the fence along the road. If the area is 1200, we have 1200  x width, so the width of the garden   1200 7200 1200 . Then the cost of the fence is given by the function C x  5 x  3 x  2   8x  . is x x x (b) We graph the function y  C x in the viewing

(c) We graph the function y  C x and y  600 in

cost is minimized when x  30 ft. Then the

From this we get that the cost is at most $600

rectangle [0 75]  [0 800]. From this we get the width is 1200 30  40 ft. So the length is 30 ft and

the width is 40 ft.

the viewing rectangle [10 65]  [450 650].

when 15  x  60. So the range of lengths he can fence along the road is 15 feet to 60 feet. 600

500

500 0

0

50

20

40

24. (a) Let x be the length of wire in cm that is bent into a square. So 10  x is the length of wire in 10  x x and , and the area cm that is bent into the second square. The width of each square is 4 4    x 2 x2 100  20x  x 2 10  x 2 and . Thus the sum of the areas is of each square is   4 16 4 16

x2 100  20x  x 2 100  20x  2x 2    18 x 2  54 x  25 4 . 16 16 16 (b) We complete the square.     1 x 2  10x  25  1 x 2  10x  25  25  25  1 x  52  25 . A x  18 x 2  54 x  25  4 8 4 8 8 8 8 4 25 2 So the minimum area is 8 cm when each piece is 5 cm long. A x 

25. (a) Let h be the height in feet of the straight portion of the window. The circumference of the semicircle is C  12 x. Since the perimeter of the window is 30 feet, we have x  2h  12 x  30.

Solving for h, we get 2h  30  x  12 x  h  15  12 x  14 x. The area of the window is  2   A x  xh  12  12 x  x 15  12 x  14 x  18 x 2  15x  12 x 2  18 x 2 .   120 (b) A x  15x  18   4 x 2   18   4 x 2  x 4    2  2 60 450 120 450 60 1 2  x   18   4 x     8   4 x  4 4 4 4 4 The area is maximized when x 

60  840, and hence h  15  12 840  14  840  420. 4

60


Modeling with Functions

26. (a) The height of the box is x, the width of

103

(b) We graph the function y  V x in the viewing rectangle

the box is 12  2x, and the length of the

[0 6]  [200 270].

box is 20  2x. Therefore, the volume of the box is

250

V x  x 12  2x 20  2x  4x 3  64x 2  240x, 0  x  6

200

(c) From the graph, the volume of the box

0

5

From the calculator we get that the volume of the box is

with the largest volume is 262682 in3

greater than 200 in3 for 1174  x  3898 (accurate to 3

when x  2427.

decimal places).

27. (a) Let x be the length of one side of the base and let h be the height of the box in feet. Since the volume of 12 the box is V  x 2 h  12, we have x 2 h  12  h  2 . The surface area, A, of the box is sum of the x area of the four sides and the area of the base. Thus the surface area of the box is given by the formula   12 48 A x  4xh  x 2  4x  x2   x 2 , x  0. x x2 (b) The function y  A x is shown in the first viewing rectangle below. In the second viewing rectangle, we isolate the minimum, and we see that the amount of material is minimized when x (the length and width) is 288 ft. Then the 12 height is h  2  144 ft. x 26

50

25 0

0

2

24

4

3.0

28. Let A, B, C, and D be the vertices of a rectangle with base AB on the x­axis and its other two vertices C and D above the x­axis and lying on the parabola y  8  x 2 . Let C have the coordinates x y, x  0. By symmetry, the coordinates of D must be x y. So the width of the rectangle is 2x, and the length is y  8  x 2 . Thus the area of the rectangle is   A x  length  width  2x 8  x 2  16x  2x 3 . The graphs of A x below show that the area is maximized when

x  163. Hence the maximum area occurs when the width is 326 and the length is 533. y

y=8-x@

D

A

C

x

B

x

20

18

10

17

0

0

2

4

16

1.5

2.0


104

FOCUS ON MODELING

29. (a) Let x be the width of the pen and l be the length in meters. We use the area to establish a relationship between 100 . So the amount of fencing used is x and l. Since the area is 100 m2 , we have l  x  100  l  x   2 200  2x 100 F  2l  2x  2  2x  . x x (b) Using a graphing device, we first graph F in the viewing rectangle [0 40] by [0 100], and locate the approximate location of the minimum value. In the second viewing rectangle, [8 12] by [39 41], we see that the minimum value of F occurs when x  10. Therefore the pen should be a square with side 10 m. 100

41

50

40

0

0

20

40

39

8

10

12

  30. (a) The distance from B to R is d1  0  32  x  02  x 2  9 and the distance   from R to A is d2  6  02  6  x2  x  62  36, so the total distance is   d x  d1 x  d2 x  x 2  9  x  62  36. (b) Using a graphing device, we find that the distance is minimized when x  2. 11.0 10

10.8 10.6

0

0

5

31. (a) Let t1 represent the time, in hours, spent walking, and let t2 represent the time spent rowing. Since the distance walked is x and the walking speed is 5 mi/h, the time spent walking is t1  15 x. By the

Pythagorean Theorem, the distance rowed is   d  22  7  x2  x 2  14x  53, and so the time spent  rowing is t2  12  x 2  14x  53. Thus the total time is  T x  12 x 2  14x  53  15 x.

1.8

2.0

2.2

(b) We graph y  T x. Using the zoom function, we see that T is minimized when x  613. You should land at a point 613 miles from point B. 4 2 0

0

2

4

6


Modeling with Functions

32. (a) Let x be the distance from point B to C, in miles. Then the distance from A to C is  flying from A to C then C to D is f x  14 x 2  25  10 12  x.

105

x 2  25, and the energy used in

(b) Using a graphing device, we find that the energy expenditure is minimized when the distance from B to C is about 51 miles. 200

169.1

100

169.0

0

0

5

10

168.9

5.0

5.1

5.2

 33. (a) Using the Pythagorean Theorem, we have that the height of the upper triangles is 25  x 2 and the height of the lower   triangles is 144  x 2 . So the area of the each of the upper triangles is 12 x 25  x 2 , and the area of the each of the  lower triangles is 12 x 144  x 2 . Since there are two upper triangles and two lower triangles, we get that the total area          25  x 2  144  x 2 . is A x  2  12 x 25  x 2  2  12 x 144  x 2  x    (b) The function y  A x  x 25  x 2  144  x 2 is shown in the first viewing rectangle below. In the second viewing rectangle, we isolate the maximum, and we see that the area of the kite is maximized when x  4615. So the length of the horizontal crosspiece must be 2  4615  923. The length of the vertical crosspiece is   52  46152  122  46152  1300. 100

60.1

50

60.0

0

0

2

4

59.9 4.60

4.62

4.64


Corrections: 1, 12, 13, 16, 45, 85, 90, 91, 94, 95, 96, 97, 99, 113, 116, 120, 122, 123, 124, 130, 134, 138

CHAPTER 3

NOTE THAT THESE CORRECTIONS SHOULD BE MADE IN ALL THREE BOOKS.

POLYNOMIAL AND RATIONAL FUNCTIONS

3.1

Quadratic Functions and Models 1

3.2

Polynomial Functions and Their Graphs 11

3.3

Dividing Polynomials 28

3.4

Real Zeros of Polynomials 37

3.5

Complex Zeros and the Fundamental Theorem of Algebra 70

3.6 3.7

Rational Functions 80 Polynomial and Rational Inequalities 101 Chapter 3 Review 115 Chapter 3 Test 135

¥

FOCUS ON MODELING: Fitting Polynomial Curves to Data 139

1


3

POLYNOMIAL AND RATIONAL FUNCTIONS

3.1

QUADRATIC FUNCTIONS AND MODELS

1. To put the quadratic function f x  ax 2  bx  c in vertex form we complete the square. 2. The quadratic function f x  a x  h2  k is in vertex form. (a) The graph of f is a parabola with vertex h k. (b) If a  0 the graph of f opens upward. In this case f h  k is the minimum value of f .

(c) If a  0 the graph of f opens downward. In this case f h  k is the maximum value of f .

3. The graph of f x  3 x  22  6 is a parabola that opens upward, with its vertex at 2 6, and f 2  6 is the minimum value of f . 4. The graph of f x  3 x  22  6 is a parabola that opens downward, with its vertex at 2 6, and f 2  6 is the maximum value of f . 6. f x   12 x 2  2x  6

5. f x  x 2  6x  5 (a) The vertex is 3 4, the x­intercepts are 1 and 5, and

(a) The vertex is 2 8, the x­intercepts are 6 and 2,

it appears that the y­intercept is approximately 5.

and the y­intercept is 6.

(b) Maximum value of f : 4

(b) Maximum value of f : 8

(c) Domain  , range:  4]

(c) Domain:  , range:  8]

2x^2 - 4x -1

7. f x  x 2  6x  5

8. f x  3x 2  6x  1

(a) The vertex is 1 3, the x­intercepts are

(a) The vertex is 1 4, the x­intercepts are

(b) Minimum value of f : 3

(b) Minimum value of f : 4

(c) Domain:  , range: [3 

(c) Domain:  , range: [4 

 2 6  02 and 22, and the y­intercept is 1. 2

 32 3  22 and 02, and the y­intercept is 1. 3

9. (a) f x  x 2  4x  9  x  22  4  9  x  22  5

(c)

y

(b) The vertex is at 2 5. x­intercepts: y  0  0  x  22  5  x  22  5. This has

no real solution, so there is no x­intercept.

y­intercept: x  0  y  0  22  5  9. The y­intercept is 9. (d) Domain:  , range: [5  1 0

1

x

1


2

CHAPTER 3 Polynomial and Rational Functions

10. (a) f x  x 2  6x  8  x  32  9  8  x  32  1

y

(c)

(b) The vertex is at 3 1.

x­intercepts: y  0  0  x  32  1  x  3  1  x  4

or 2. The x­intercepts are 4 and 2.

y­intercept: x  0  y  0  32  1  8. The y­intercept is 8.

(d) Domain:  , range: [1 .

11. (a) f x  x 2  6x  x 2  6x  x 2  6x  9  9  x  32  9

1 1x

(c)

(b) The vertex is at 3 9.

y

1

x­intercepts: y  0  0  x 2  6x  x x  6. So x  0 or x  6.

x

1

The x­intercepts are 0 and 6.

y­intercept: x  0  y  0. The y­intercept is 0. (d) Domain:  , range: [9 

12. (a) f x  x 2  8x  x 2  8x  16  16  x  42  16

(c)

y

(b) The vertex is at 4 16.

x­intercepts: y  0  0  x 2  8x  x x  8. So x  0 or

x  8. The x­intercepts are 0 and 8.

3

y­intercept: x  0  y  0. The y­intercept is 0.

1

x

1

x

(d) Domain:  , range: [16 

    13. (a) f x  3x 2  6x  3 x 2  2x  3 x 2  2x  1  3

(c)

y

 3 x  12  3

(b) The vertex is at 1 3.

x­intercepts: y  0  0  3 x  12  3  x  12  1 

x  2 or 0. The x­intercepts are 2 and 0.

y­intercept: x  0  y  3 02  6 0  0. The y­intercept is 0.

(d) Domain:  , range: [3 

1


3

SECTION 3.1 Quadratic Functions and Models

 14. (a) f x  x 2  10x   x 2  10x   x 2  10x  25  25   x  52  25

(c)

y

(b) The vertex is at 5 25. 10

x­intercepts: y  0  0   x  52  25  x  52  25 

x  0 or 10. The x­intercepts are 0 and 10.

x

1

y­intercept: x  0  y  02  10 0  0. The y­intercept is 0.

(d) Domain:  , range:  25]

15. (a) f x  x 2  4x  3  x 2  4x  4  4  3  x  22  1

(c)

y

(b) The vertex is at 2 1.

x­intercepts: y  0  0  x 2  4x  3  x  1 x  3. So x  1 or x  3. The x­intercepts are 1 and 3.

1

y­intercept: x  0  y  3. The y­intercept is 3.

1

x

1

x

(d) Domain:  , range: [1 

16. (a) f x  x 2  2x  2  x 2  2x  1  1  2  x  12  1

(c)

y

(b) The vertex is at 1 1. x­intercepts: y  0  x  12  1  0  x  12  1. Since

this last equation has no real solution, there is no x­intercept.

1

y­intercept: x  0  y  2. The y­intercept is 2. (d) Domain:  , range: [1 

  17. (a) f x  x 2  10x  15   x 2  10x  25  25  15

(c)

y

  x  52  10

(b) The vertex is at 5 10. x­intercepts: y  0  0   x  52  10  x  52  10 

  x  5   10  x  5  10. y­intercept: x  0  y  15. The y­intercept is 15. (d) Domain:  , range:  10]

2 1

x


4

CHAPTER 3 Polynomial and Rational Functions

  18. (a) f x  x 2  12x  11   x 2  12x  36  36  11

(c)

y

  x  62  25

(b) The vertex is at 6 25. x­intercepts: y  0  0   x  62  25  x  62  25 

x  1 or 11. y­intercept: x  0  y  11. The y­intercept is 11. (d) Domain:  , range:  25]

  19. (a) f x  3x 2  6x  7  3 x 2  2x  1  3  7  3 x  12  4

2

(c)

1

x

y

(b) The vertex is at 1 4.

x­intercepts: y  0  0  3 x  12  4  3 x  12  4.

Since this equation has no real solution, there is no x­intercept. y­intercept: x  0  y  7. The y­intercept is 7. (d) Domain:  , range: [4 

2 x

1

  20. (a) f x  3x 2  6x  2  3 x 2  2x  1  3  2

(c)

y

 3 x  12  1

(b) The vertex is at 1 1. x­intercepts: y  0  0  3 x  12  1  0  x  12  13     x  1   13  x  1  13 . The x­intercepts are 1  13 and  1  13 .

2 x

1

y­intercept: x  0  y  2. The y­intercept is 2.

(d) Domain:  , range:  1]   21. (a) f x  05x 2  6x  16  12 x 2  12x  36  18  16

(c)

y

 12 x  62  2

(b) The vertex is at 6 2.

x­intercepts: y  0  0  12 x  62  2  x  62  4 

x  8 or 4. y­intercept: x  0  y  16. The y­intercept is 16. (d) Domain:  , range: [2 

2 1

x


5

SECTION 3.1 Quadratic Functions and Models

 22. (a) f x  2x 2  12x  10  2 x 2  6x  9  18  10  2 x  32  8

(c)

y

(b) The vertex is at 3 8. The x­intercepts are 5 and 1 and the y­intercept is 10.

2 0

(d) Domain:  , range: [8 

  23. (a) f x  4x 2  12x  1  4 x 2  3  1 2 2    4 x  32  9  1  4 x  32  10   (b) The vertex is at  32  10 .

x

y

(c)

10

 2 2  x­intercepts: y  0  0  4 x  32  10  x  32  52     x  32   52  x   32  52   32  210 . The x­intercepts are

 32 

1

5

1

  10 10 3 2 and  2  2 .

_3

_2

x

_1

y­intercept: x  0  y  4 02  12 0  1  1. The y­intercept

is 1.

(d) Domain:  , range:  10]

  24. (a) f x  3x 2  2x  2  3 x 2  23 x  2 2 2    3 x  13  13  2  3 x  13  73   (b) The vertex is at  13   73 . 

x­intercepts: y  0  0  3 x  13 

2

 73  x  13

(c)

2

10

 79  

x  13   37  x   13  37 . The x­intercepts are  13  37 and

  13  37 .

y­intercept: x  0  y  3 02  2 0  2  2. The y­intercept

is 2.

  (d) Domain:  , range:  73  

y 20

1

x


6

CHAPTER 3 Polynomial and Rational Functions

  25. (a) f x  x 2  2x  1  x 2  2x  1    x 2  2x  1  1  1  x  12  2 (b)

y

  26. (a) f x  x 2  8x  8  x 2  8x  16  8  16  x  42  8

(b)

y

2

1 1

(c) The minimum value is f 1  2.   27. (a) f x  4x 2  8x  1  4 x 2  2x  1    4 x 2  2x  1  4  1

(c) The minimum value is f 4  8.   28. (a) f x  2x 2  12x  14  2 x 2  6x  14  2 x  32  18  14

 2 x  32  4

 4 x  12  5

(b)

y

x

1

x

(b)

y

1

x

1

(c) The minimum value is f 1  5.   29. (a) f x  x 2  3x  3   x 2  3x  3     x 2  3x  94  3  94 2    x  32  21 4 (b)

x

1

1

y

(c) The minimum value is f 3  4.   30. (a) f x  1  6x  x 2   x 2  6x  1     x 2  6x  9  1  9   x  32  10

(b)

y

2

1 1

  (c) The maximum value is f  32  21 4.

x

1

(c) The maximum value is f 3  10.

x


SECTION 3.1 Quadratic Functions and Models

31. (a) f x  3x 2  12x  13  3 x 2  4x  13    3 x 2  4x  4  13  12  3 x  22  1

(b)

32. (a) f x  2x 2  12x  20  2 x 2  6x  20    2 x 2  6x  9  18  20  2 x  32  2

y

(b)

y

2 x

1

2 1

(c) The minimum value is f 2  1. 

(c) The minimum value is f 3  2.   34. (a) f x  3  4x  4x 2  4 x 2  x  3    4 x 2  x  14  3  1 2   4 x  12  4

33. (a) f x  1  x  x 2   x 2  x  1     x 2  x  14  1  14 2    x  12  54 (b)

x

(b)

y

y

1

1 1

  (c) The maximum value is f  12  54 .

x

1

x

  (c) The maximum value is f  12  4.

35. For f x  7x 2  14x  5 we have a  7, b  14, and c  5. Because a  0, the maximum value is       b  f  14  f  7 12  14 1  5  2. f  2a 1 27

36. For f x  6x 2  48x  1 we have a  6, b  48, and c  1. Because a  0, the minimum value is     b  f  48  f 4  6 42  48 4  1  95. f  2a 26

37. For f t  4t 2  40t  110 we have a  4, b  40, and c  110. Because a  0, the minimum value is     b  f  40  f 5  4 52  40 5  110  10. f  2a 24

38. For g x  5x 2  60x  200 we have a  5, b  60, and c  200. Because a  0, the maximum value is     b  g  60 2 g  2a 25  g 6  5 6  60 6  200  20.

39. For f s  s 2  12s  16 we have a  1, b  12, and c  16. Because a  0, the minimum value is     b  f  12  f 06  062  12 06  16  1564. f  2a 21

7


8

CHAPTER 3 Polynomial and Rational Functions

40. For g x  100x 2  1500x we have a  100, b  1500, and c  0. Because a  0, the minimum value is           b  g  1500  g 15  100 15 2  1500 15  5625. g  2a 2100 2 2 2 41. For h x  12 x 2  2x  6 we have a  12 , b  2, and c  6. Because a  0, the minimum value is     b h  2 1 2 h  2a 212  h 2  2 2  2 2  6  8

x2  2x  7   13 x 2  2x  7 we have a   13 , b  2, and c  7. Because a  0, the maximum value is 42. For f x   3     b  f  2 1 2 f  2a 213  f 3   3 3  2 3  7  10. 43. For f x  3  x  12 x 2   12 x 2  x  3 we have a   12 , b  1, and c  3. Because a  0, the maximum value is     b  f  1 1 7 2 f  2a 212  f 1   2 1  1  3  2 .

44. For g x  2x x  4  7  2x 2  8x  7 we have a  2, b  8, and c  7. Because a  0, the minimum value is     b  g  8  g 2  2 22  8 2  7  1. g  2a 22 45. (a) The graph of f x  x 2  179x  321 is shown. The minimum value is f 090  401. ­1.0

­0.9

­0.8 ­3.9 ­4.0

(b) f x  x 2  179x  321 has a  1, b  179, and c  321. Because a  0, the minimum value is     b  f  179  f 0895 f  2a 21  08952  179 0895  321  4011025

­4.1

46. (a) The graph of f x  1  x 

 2 2x is

shown. The maximum value is f 035  118. 1.180

  (b) f x  1  x  2 x 2   2x 2  x  1 has  a   2, b  1, and c  1. Because a  0, the maximum value is       b 1    f 42 f  2a  f   2  2

    2    2 42  42  1  82  1

1.175 1.170 0.30

0.35

0.40

 1176777

47. The vertex is 2 3, so the parabola has equation y  a x  22  3. Substituting the point 3 1, we have 1  a 3  22  3  a  4, so f x  4 x  22  3.

48. The vertex is 1 5, so the parabola has equation y  a x  12  5. Substituting the point 3 7, we have

7  a 3  12  5  a  3, so f x  3 x  12  5.   49. Substituting t  x 2 , we have f t  3  4t  t 2   t 2  4t  4  4  3   t  22  7. Thus, the maximum value  is 7, when t  2 (or x   2).   50. Substituting t  x 3 , we have f t  2  16t  4t 2  4 t 2  4t  4  16  2  4 t  22  14. Thus, the minimum  value is 14, when t  2 (or x   3 2).


SECTION 3.1 Quadratic Functions and Models

9

  2  2   2    16 54  16 t  54  25. Thus the maximum 51. y  f t  40t  16t 2  16 t 2  52  16 t 2  52 t  54   height attained by the ball is f 54  25 feet.         32 x 2  x  5   2 x 2  25 x  5   2 x 2  25 x  25 2  2 25 2  5 52. (a) We complete the square: y   400 25 2 25 2 4 25 4   2 x  25 2  65 , so the maximum height of the ball is 65  8125 ft.  y   25 4 8 8

2 x 2  x  5. Using the Quadratic (b) The ball hits the ground when its vertical displacement y is 0, that is, when 0   25    2 5  1  1  4  25 25  5 65  . Taking the positive root, we find that x  163 ft. Formula, we find x  4 4  25     53. R x  80x  04x 2  04 x 2  200x  04 x 2  200x  10,000  4,000  04 x  1002  4,000. So

revenue is maximized at $4,000 when 100 units are sold.     54. P x  0001x 2  3x  1800  0001 x 2  3000x  1800  0001 x 2  3000x  2 250 000  1800 

2250  0001 x  15002  450. The maximum profit is $450, and occurs when 1500 cans are sold.     1 n 2   1 n 2  60n   1 n 2  60n  900  10   1 n  302  10. Since the maximum of 55. E n  23 n  90 90 90 90 the function occurs when n  30, the viewer should watch the commercial 30 times for maximum effectiveness.     56. C t  006t  00002t 2  00002 t 2  300t  00002 t 2  300t  22,500  45  00002 t  1502  45. The maximum concentration of 45 mg/L occurs after 150 minutes.

57. A n  n 900  9n  9n 2  900n is a quadratic function with a  9 and b  900, so by the formula, the maximum 900 b   50 trees, and because a  0, this gives a maximum value. or minimum value occurs at n   2a 2 9 58. A n  700  n 10  001n  001n 2  10n  001 700 n  7000  001n 2  3n  7000. This is a quadratic b 3 function with a  001 and b  3, so the maximum (a  0) occurs at x     150. Since n  150 2a 2 001 is the number of additional vines that should be planted, the total number of vines that maximizes grape production is 700  150  850 vines. 59. The area of the fenced­in field is given by A x  2400  2x x  2x 2  2400x. Thus, by the formula in this section, 2400 b   600. The maximum area occurs when x  600 feet the maximum or minimum value occurs at x   2a 2 2 and l  2400  2 600  1200 feet. 60. The total area of the four pens is A   52  150     52 2  375. Thus, by the formula, the maximum or minimum value occurs at   

375 b      75. Therefore, the largest possible total area of the four pens is 2a 2 5 2

  A 75   52 752  375 75  14,0625 square feet.

61. A x  15x  18   4 x 2 , so by the formula, the maximum area occurs when x   and h  15  12 840  14  840  42 ft.

15 b   84 ft   2a 2 1   4

62. A x  18 x 2  54 x  25 4 , so by the formula, the maximum or minimum area occurs where x     25 2 minimum area is 18 52  54 5  25 4  8 cm when each piece is 5 cm long.

8

5 b    4   5. The 2a 2 1 8


10

CHAPTER 3 Polynomial and Rational Functions

63. (a) The area of the corral is A x  x 1200  x  1200x  x 2  x 2  1200x.

(b) A is a quadratic function with a  1 and b  1200, so by the formula, it has a maximum or minimum at 1200 b   600, and because a  0, this gives a maximum value. The desired dimensions are 600 ft by x  2a 2 1 600 ft.

64. (a) The dimensions of the gutter are x inches and 30  x  x  30  2x inches, so the cross sectional area is A  x 30  2x  30x  2x 2 .

(b) Since A is a quadratic function with a  2 and b  30, the maximum occurs at x  

b 30   75 inches. 2a 2 2

(c) The maximum cross section is A 75  2 752  30 75  1125 in2 .

65. (a) To model the revenue, we need to find the total attendance. Let x be the ticket price. Then the amount by which the ticket price is lowered is 10  x, and we are given that for every dollar it is lowered, the attendance increases by 3000; that is, the increase in attendance is 3000 10  x. Thus, the attendance is 27,000  3000 10  x, and since each spectator pays $x, the revenue is R x  x [27,000  3000 10  x]  3000x 2  57,000x.

(b) Since R is a quadratic function with a  3000 and b  57,000, the maximum occurs at 57,000 b   95; that is, when admission is $950. x  2a 2 3000 (c) We solve R x  0 for x: 3000x 2  57,000x  0  3000x x  19  0  x  0 or x  19. Thus, if admission is $19, nobody will attend and no revenue will be generated.

66. (a) Let x be the price per feeder. Then the amount by which the price is increased is x  10, and we are given that for every dollar increase, sales decrease by 2; that is, the change in sales is 2 x  10, so the total number sold is 20  2 x  10  40  2x. The profit per feeder is equal to the sale price minus the cost, that is, x  6. Multiplying the number of feeders sold by the profit per feeder sold, we find the profit to be P x  40  2x x  6  2x 2  52x  240.

(b) Using the formula, profit is maximized when x  

52 b   13; that is, when the society charges $13 per 2a 2 2

feeder. The maximum weekly profit is P 13  2 132  52 13  240  $98.

y

67. Because f x  x  m x  n  0 when x  m or x  n, those are its x­intercepts. By symmetry, we expect that the vertex is halfway between mn . We obtain the graph shown at right. these values; that is, at x  2

Expanding, we see that f x  x 2  m  n x  mn, a quadratic

function with a  1 and b   m  n. Because a  0, the minimum value occurs at x  

m n b  , the x­value of the vertex, as expected. 2a 2

y=(x-a)(x-b)

a

0

a+b 2

b

x


SECTION 3.2 Polynomial Functions and Their Graphs

11

68. The revenue is R x  23,500x  1000x 2 and the attendance is A x  9500  1000 14  x, so at a ticket

price of $1175, the revenue is R 1175  23,500 1175  1000 11752  $138,060 and the attendance is A 1175  9500  1000 14  1175  11,750. The arena’s capacity is 15,000, so we set A x  15,000 and solve for x: 15,000  9500  1000 14  x  1000 14  x  5500  14  x  55  x  $850, the ticket price at which the arena will be filled to capacity. The

revenue in that case is R 85  23,500 85  1000 852  $127,500, which is less than the maximum revenue of $138,060. The model reflects the fact that lower prices generally correspond to more attendance, and vice versa; and revenue is the product of price and attendance. In general, quadratic models are appropriate for modeling profit and revenue due to this fact.

69. The point Q x y lies on the line y  2x  3, so it can be written as Q x 2x  3. The distance from P 3 2 to Q is    thus g x  [2x  3  2]2  x  32  2x  12  x  32  5x 2  10x  10. Note that in general, the  minimum value of g x  f x occurs at the value of x at which f x is minimized. Let f x  5x 2  10x  10. We   find the vertex form of f : f x  5 x 2  2x  1  5  10  5 x  12  5. Thus, f (and g) have minimum values at x  1. The point closest to P on the line y  2x  3 is therefore 1 2 1  3  1 1.

3.2

POLYNOMIAL FUNCTIONS AND THEIR GRAPHS

1. Graph I cannot be that of a polynomial because it is not smooth (it has a cusp.) Graph II could be that of a polynomial function, because it is smooth and continuous. Graph III could not be that of a polynomial function because it has a break. Graph IV could not be that of a polynomial function because it is not smooth.

2. (a) y  x 3  8x 2  2x  15 has odd degree and a positive leading coefficient, so y   as x   and y   as x  .

(b) y  2x 4  12x  100 has even degree and a negative leading coefficient, so y   as x   and y   as x  .

3. (a) If c is a zero of the polynomial P, then P c  0.

(b) If c is a zero of the polynomial P, then x  c is a factor of P x.

(c) If c is a zero of the polynomial P, then c is an x­intercept of the graph of P.

4. (a) This is impossible. If P has degree n  3, it has at most n  1  2 local extrema.

(b) This is possible. For example, y  x 3 has degree 3 and no local maxima or minima.

(c) This is possible. For example, P x  x 4 has one local maximum and no local minima.


Q(x)

12

CHAPTER 3 Polynomial and Rational Functions

5. (a) P x  12 x 2  2

(b) P x  2 x  32 y

y

1 1

(_2, 0)

x

(2, 0)

(0, 18) 10

(0, _2)

1

Domain:  ; range: [2  (c) P x  4x 2  1

R(x)

y

x

(3, 0)

Domain:  ; range: [0  (d) P x   x  22 (_2, 0)

S(x)

y 1 0

1

x

(0, _4)

2 (0, 1)

1

x

Domain:  ; range:  0]

Domain:  ; range: [1 

Q(x)

6. (a) P x  x 4  1

(_1, 0)

(b) P x  x  14

y (0, 1)

(1, 0)

y

x

2

(0, 1)

(_1, 0)

Domain:  ; range:  1]

Domain:  ; range: [0 

1x


R(x)

SECTION 3.2 Polynomial Functions and Their Graphs

(c) P x  6x 4  6

S(x)

(d) P x  19 x  34

y

y (0, 9)

1

(_1, 0)

(1, 0)

0

x 1 0

(0, _6)

x

(3, 0)

Domain:  ; range: [0 

Domain:  ; range: [6 

7. (a) P x  x 3  8

1

(b) Q x  x 3  27

y

y

1 1

(2, 0)

x (0, 27)

(0, _8)

4

(3, 0) 1

Domain:  ; range:   (c) R x   x  23

(_2, 0)

x

Domain:  ; range:   (d) S x  12 x  13  4

y

y

1 1

x

(0, _72) (0, _8)

Domain:  ; range:  

(_1, 0)

1 1

Domain:  ; range:  

x

13


14

CHAPTER 3 Polynomial and Rational Functions

8. (a) P x  x  35

(b) Q x  2 x  35  64

y

y (0, 422)

(0, 243) (_1, 0)

50

(_3, 0)

100 1 x

1 x

Domain:  ; range:  

Domain:  ; range:  

(c) R x   12 x  25

(d) S x   12 x  25  16

y

y (0, 16) (0, 32)

2

(2, 0) 1

x

4 1

Domain:  ; range:  

(4, 0)

x

Domain:  ; range:  

  9. (a) P x  x x 2  4  x 3  4x has odd degree and a positive leading coefficient, so y   as x   and y   as x  .

(b) This corresponds to graph III.   10. (a) Q x  x 2 x 2  4  x 4  4x 2 has even degree and a negative leading coefficient, so y   as x   and y   as x  .

(b) This corresponds to graph I. 11. (a) R x  x 5  5x 3  4x has odd degree and a negative leading coefficient, so y   as x   and y   as x  . (b) This corresponds to graph V.

12. (a) S x  12 x 6  2x 4 has even degree and a positive leading coefficient, so y   as x   and y   as x  . (b) This corresponds to graph II.

13. (a) T x  x 4  2x 3 has even degree and a positive leading coefficient, so y   as x   and y   as x  . (b) This corresponds to graph VI.

14. (a) U x  x 3  2x 2 has odd degree and a negative leading coefficient, so y   as x   and y   as x  . (b) This corresponds to graph IV.


SECTION 3.2 Polynomial Functions and Their Graphs

15. P x  x  2 x  5

16. P x  4  x x  1 y

y

(0, 4) (_1, 0) 2 (_2, 0)

2

(4, 0)

1

x

(5, 0) x

1

(0, _10)

17. P x  x x  2 x  3

18. P x  x  3 x  1 x  4

y

y

1 (_3, 0)

(2, 0)

1 (0, 0)

(12, 0) x

(_3, 0)

4

(1, 0) 2

19. P x   2x  1 x  1 x  3

y

(0, 12)

(0, 3) 1

(_1, 0)

1

(_2, 0)

21. P x  x  2 x  1 x  2 x  3 y

(0, 12) 0 (_1, 0)

(2, 0) 1

(3, 0) x

( _23 , 0) 1

(3, 0)

x

22. P x  x x  1 x  1 2  x y

(_1, 0)

(_2, 0)

4

(1/2, 0) x

5

x

20. P x  x  3 x  2 3x  2

y

(_3, 0)

(4, 0)

1 0

(0, 0) 1 (1, 0)

(2, 0)

x

15


16

CHAPTER 3 Polynomial and Rational Functions

24. P x  15 x x  52

23. P x  2x x  22 y

y

1 (0, 0)

2 (0, 0)

(2, 0)

1

x

1

x

(5, 0)

2

25. P x  x  1 x  12 2x  3

26. P x   x  23 x  1 x  32 y

y

(_2, 0) (_1, 0) 2 0 (_2, 0)

(0, 72)

20

(1, 0)

0

(3, 0)

x

2

(3/2, 0) 1

x

(0, _6)

1 x  22 x  32 27. P x  12

28. P x  x  12 x  23

y

y

(0, 8) 2 (_2, 0)

(0, 3)

1 (1, 0)

x

1 (_2, 0)

1

(3, 0)

x

30. P x  19 x 2 x  32 x  32

29. P x  x x  22 x  23 y

10 (_2, 0)

y

1 (0, 0)

5 (2, 0)

x

(_3, 0)

(0, 0)

1

(3, 0) x


SECTION 3.2 Polynomial Functions and Their Graphs

31. P x  x 3  x 2  6x  x x  2 x  3

32. P x  x 3  2x 2  8x  x x  2 x  4

y

2 (0, 0)

(_2, 0)

1

y

4

(_4, 0) (3, 0)

x

33. P x  x 3  x 2  12x  x x  3 x  4 y

(0, 0)

1

(2, 0)

x

  34. P x  2x 3  x 2  x  x 2x 2  x  1  x 2x  1 x  1 y

4 (0, 0) 1

(_3, 0)

(4, 0)

1 x

( _21 , 0)

(0, 0)

1

(_1, 0)

35. P x  x 4  3x 3  2x 2  x 2 x  1 x  2 y

1

36. P x  x 5  9x 3  x 3 x  3 x  3 y

(_3, 0)

10 (0, 0)

(1, 0) (0, 0)

1

(2, 0)

x

x

(3, 0) 1

x

17


18

CHAPTER 3 Polynomial and Rational Functions

37. P x  x 3  x 2  x  1  x  1 x  12 y

38. P x  x 3  3x 2  4x  12  x  3 x  2 x  2 y

(_2, 0)

2

(_3, 0) (_1, 0)

1

(2, 0) 1

(1, 0)

x

x

1 (0, _1)

(0, _12)

 2 40. P x  18 2x 4  3x 3  16x  24  2  18 x  22 2x  32 x 2  2x  4

39. P x  2x 3  x 2  18x  9  x  3 2x  1 x  3 y

y

40 (0, 9)

(_3, 0)

( _21 , 0)

1

(3, 0)

x (0, 72) 20 1

(_ _32 , 0)

41. P x  x 4  2x 3  8x  16    x  22 x 2  2x  4

y

(_2, 0)

10 (2, 0) 1 (0, _16)

(0, 16) 1 (2, 0)

x

42. P x  x 4  2x 3  8x  16    x  2 x  2 x 2  2x  4

y

10

(2, 0)

x

x


SECTION 3.2 Polynomial Functions and Their Graphs

  43. P x  x 4  3x 2  4  x  2 x  2 x 2  1 y

19

 2 44. P x  x 6  2x 3  1  x 3  1  2  x  12 x 2  x  1 y

1

(_2, 0)

(2, 0)

x

1 (0, _4)

(0, 1)

1

1 (1, 0)

x

45. P x  3x 3  x 2  5x  1; Q x  3x 3 . Since P has odd degree and positive leading coefficient, it has the following end behavior: y   as x   and y   as x  . On a large viewing rectangle, the graphs of P and Q look almost the same. On a small viewing rectangle, we see that the graphs of P and Q have different intercepts. 100

10

50 ­3

­2

­1

P 1

5

Q

­50

2

3

­1.5 ­1

­0.5 ­5

­100

­10

P

Q

0.5

1

1.5

46. P x   18 x 3  14 x 2  12x; Q x   18 x 3 . Since P has odd degree and negative leading coefficient, it has the following end behavior: y   as x   and y   as x  . On a large viewing rectangle, the graphs of P and Q look almost the same. On a small viewing rectangle, the graphs of P and Q look very different and seem (wrongly) to have different end behavior. 60 40

1000 P

Q

Q

500 ­6

­30 ­20 ­10 ­500 ­1000

10

20

30

­4

P

20 ­2

0 ­20 ­40 ­60 ­80

2

4

6


20

CHAPTER 3 Polynomial and Rational Functions

47. P x  x 4  7x 2  5x  5; Q x  x 4 . Since P has even degree and positive leading coefficient, it has the following end behavior: y   as x   and y   as x  . On a large viewing rectangle, the graphs of P and Q look almost the same. On a small viewing rectangle, the graphs of P and Q look very different and we see that they have different intercepts. 600

10

500 Q

400

P

P

300

Q ­3

200

­2

­1

­6

­4

1

2

3

­10

100 ­2 0 ­100

0

2

4

6

­20

48. P x  x 5  2x 2  x; Q x  x 5 . Since P has odd degree and negative leading coefficient, it has the following end behavior: y   as x   and y   as x  . On a large viewing rectangle, the graphs of P and Q look almost the same. On a small viewing rectangle, the graphs of P and Q look very different. 3

200 Q

P

100

­3

­2

­1 0 ­100

Q 1

2

P ­1

3

2 1 0 ­1

1

­2

­200

­3

49. P x  x 11  9x 9 ; Q x  x 11 . Since P has odd degree and positive leading coefficient, it has the following end behavior: y   as x   and y   as x  . On a large viewing rectangle, the graphs of P and Q look like they have the same end behavior. On a small viewing rectangle, the graphs of P and Q look very different and seem (wrongly) to have different end behavior. 600000 400000

Q

P

2

4

100

Q

­6

­4

­2 0 ­200000

6

­1

0

1

­50

P

­400000

P

50

200000

Q

­100

­600000

50. P x  2x 2  x 12 ; Q x  x 12 . Since P has even degree and negative leading coefficient, it has the following end behavior: y   as x   and y   as x  . On a large viewing rectangle, the graphs of P and Q look almost the same. On a small viewing rectangle, the graphs of P and Q look very different. 200 ­3

­2

­1 0 ­200

1

2

2

3

­400 ­600 P Q

­800

P

­1

0

Q

­2

­1000 ­1200

­4

1


SECTION 3.2 Polynomial Functions and Their Graphs

51. (a) x­intercepts are 0 and 4, y­intercept is 0.

52. (a) x­intercepts are 0 and 45, y­intercept is 0.

(b) Local maximum at 2 4, no local minimum.

(b) Local maximum at 0 0, local minimum at 3 3.

(c) Domain  , range  4].

(c) Domain  , range  .

53. (a) x­intercepts are 2 and 1, y­intercept is 1. (b) Local maximum at 1 0, local minimum at 1 2.

54. (a) x­intercepts are 0 and 4, y­intercept is 0. (b) No local maximum, local minimum at 3 3. (c) Domain  , range [3 .

(c) Domain  , range  . 55. y  10x  x 2 , [2 12] by [15 30]

No local minimum. Local maximum at 5 25. Domain:  , range:  25].

56. y  x 3  3x 2 , [2 5] by [10 10]

Local minimum at 2 4. Local maximum at 0 0. Domain:  , range:  .

30

10

20 10 ­2 ­2 ­10

2

4

6

2

4

8 10 12 ­10

57. y  x 3  12x  9, [5 5] by [30 30]

Local maximum at 2 25. Local minimum at 2 7. Domain:  , range:  .

58. y  2x 3  3x 2  12x  32, [5 5] by [60 30] Local minimum at 2 52. Local maximum at 1 25.

Domain:  , range:  .

20

20 ­4

­2

2

4

­4

­20

­2 ­20

2

4

­40 ­60

59. y  x 5  9x 3 , [4 4] by [50 50]

Local minimum at approximately 232 4517, local maximum at approximately 232 4517.

60. y  x 4  18x 2  32, [5 5] by [100 100]

Local minima at 3 49 and 3 49. Local maximum at 0 32.

Domain:  , range: [49 .

Domain:  , range:  .

100

40 20 ­4

­2

­20 ­40

2

4

­4

­2 ­100

2

4

21


22

CHAPTER 3 Polynomial and Rational Functions

61. y  3x 5  5x 3  3, [3 3] by [5 10]

62. y  x 5  5x 2  6, [3 3] by [5 10]

Domain:  , range:  .

Domain:  , range:  .

Local maximum at 1 5. Local minimum at 1 1.

Local minimum at 126 124. Local maximum at 0 6.

10

10

5 ­3

­2

­1

1

2

3

­3

­2

­1

1

2

3

­5

63. y  2x 2  3x  5 has one local maximum at 075 613.

64. y  x 3  12x has no local maximum or minimum. 20

10

10

5 ­4

­2

­4 2

­2 ­10

2

4

­20

4

­5

65. y  x 3  x 2  x has one local maximum at 033 019 66. y  6x 3  3x  1 has no local maximum or minimum. and one local minimum at 100 100.

20 10

2 ­3 ­3

­2

­1

1

2

3

­2

­1 ­10

67. y  x 4  5x 2  4 has one local maximum at 0 4 and two local minima at 158 225 and 158 225. 10

­1 ­5

3

68. y  12x 5  375x 4  7x 3  15x 2  18x has two local

maxima at 050 465 and 297 1210 and two local minima at 140 2744 and 140 254. 20

5 ­2

2

­20

­2

­3

1

1

2

3

­4

­2

2 ­20


SECTION 3.2 Polynomial Functions and Their Graphs

69. y  x  25  32 has no maximum or minimum.

23

 3 70. y  x 2  2 has one local minimum at 0 8.

60

10

40 20

­3

­2

2

4

­2

6

­1

1

2

3

­10

71. y  x 8  3x 4  x has one local maximum at 044 033 and two local minima at 109 115 and 112 336.

72. y  13 x 7  17x 2  7 has one local maximum at 0 7 and one local minimum at 171 2846. 20

4 2 ­3

­2

­1 ­2

­3 1

2

­2

3

­1

1

2

3

­20

­4

73. y  cx 3 ; c  1, 2, 5, 12 . Increasing the value of c stretches 74. P x  x  c4 ; c  1, 0, 1, 2. Increasing the value of c shifts the graph to the right.

the graph vertically.

100

c=5

c=0

c=1

c=_1 ­4

­2

2

c=2

1

4

0

­2

1 c=_ 2

2

­1

­100

c=2

c=1

2

75. P x  x 4  c; c  1, 0, 1, and 2. Increasing the value of c moves the graph up.

76. P x  x 3  cx; c  2, 0, 2, 4. Increasing the value of c makes the “bumps” in the graph flatter.

c=2 c=1 c=0 c=_1

6 4

10

c=0

c=_2

c=2

c=_4

2 ­2

­1

0

0

1

­2 ­10

2


24

CHAPTER 3 Polynomial and Rational Functions

77. P x  x 4  cx; c  0, 1, 8, and 27. Increasing the value 78. P x  x c ; c  1, 3, 5, 7. The larger c gets, the flatter the of c causes a deeper dip in the graph, in the fourth

graph is near the origin, and the steeper it is away from the

quadrant, and moves the positive x­intercept to the right.

origin.

c=1 c=0

­4

10

20

­2

0

c=7

c=5

c=3

c=27

c=1 2 c=8

4

­2

­1

0

1

2

­20

­10 ­40

79. (a)

y

y=x#-2x@-x+2 y=_x@+5x+2

5 1

x

(b) The two graphs appear to intersect at 3 points. (c) x 3  2x 2  x  2  x 2  5x  2  x 3  x 2  6x  0    x x 2  x  6  0  x x  3 x  2  0. Then either x  0, x  3, or x  2. If x  0, then y  2; if x  3 then

y  8 if x  2, then y  12. Hence the points where the

two graphs intersect are 0 2, 3 8, and 2 12.

80. Graph 1 belongs to y  x 4 . Graph 2 belongs to y  x 2 . Graph 3 belongs to y  x 6 . Graph 4 belongs to y  x 3 . Graph 5 belongs to y  x 5 .

81. (a) Let P x be a polynomial containing only odd powers of x. Then each term of P x can be written as C x 2n1 , for some constant C and integer n. Since C x2n1  C x 2n1 , each term of P x is an odd function. Thus by part (a), P x is an odd function. (b) Let P x be a polynomial containing only even powers of x. Then each term of P x can be written as C x 2n , for some constant C and integer n. Since C x2n  C x 2n , each term of P x is an even function. Thus by part (b), P x is an even function. (c) Since P x contains both even and odd powers of x, we can write it in the form P x  R x  Q x, where R x contains all the even­powered terms in P x and Q x contains all the odd­powered terms. By part (d), Q x is an odd function, and by part (e), R x is an even function. Thus, since neither Q x nor R x are constantly 0 (by assumption), by part (c), P x  R x  Q x is neither even nor odd.     (d) P x  x 5  6x 3  x 2  2x  5  x 5  6x 3  2x  x 2  5  PO x  PE x where PO x  x 5  6x 3  2x and PE x  x 2  5. Since PO x contains only odd powers of x, it is an odd function, and since PE x contains only even powers of x, it is an even function.


SECTION 3.2 Polynomial Functions and Their Graphs

82. (a) From the graph, P x  x 3  4x  x x  2 x  2 has three x­intercepts, one local maximum, and one local minimum.

25

 (b) From the graph, Q x  x 3  4x  x x 2  4 has one x­intercept and no local maximum or minimum.

10

­5

10

5

­5

5

­10

­10

  (c) For the x­intercepts of P x  x 3  ax, we solve x 3  ax  0. Then we have x x 2  a  0  x  0 or x 2  a. If         x 2  a, then x   a. So P has 3 x­intercepts. Since P x  x x 2  a  x x  a x  a , by part (c) of problem 67, P has 2 local extrema. For the x­intercepts of Q x  x 3  ax, we solve x 3  ax  0. Then we   have x x 2  a  0  x  0 or x 2  a. The equation x 2  a has no real solutions because a  0. So Q has 1

x­intercept. We now show that Q is always increasing and hence has no extrema. Ifx1  x2 , then ax1  ax2 (because

a  0) and x13  x23 . So we have x13  ax1  x23  ax2 , and hence Q x1   Q x2 . Thus Q is increasing, that is, its graph always rises, and so it has no local extrema. 83. (a) P x  x  1 x  3 x  4. Local maximum at 18 21.

Local minimum at 36 06.

(b) Since Q x  P x  5, each point on the graph of Q has y­coordinate 5 units more than the corresponding point on the graph of P. Thus Q has a local maximum at 18 71 and a local minimum at 35 44.

10 10 5 5

­10

84. (a) P x  x  2 x  4 x  5 has one local maximum and one local minimum.

(b) Since P a  P b  0, and P x  0 for a  x  b (see the table

below), the graph of P must first rise and then fall on the interval a b,

and so P must have at least one local maximum between a and b. Using similar reasoning, the fact that P b  P c  0 and P x  0 for

10

b  x  c shows that P must have at least one local minimum between b and c. Thus P has at least two local extrema.

5 ­10

Interval Sign of x  a Sign of x  b Sign of x  c

Sign of x  a x  b x  c

 a a b b c c  

85. Since the polynomial shown has five distinct zeros, it has at least five factors, and so its degree is greater than or equal to 5. (Alternately, we can use the theorem on Local Extrema of Polynomials: Because the function has four local extrema, its degree is at least 5.)


26

CHAPTER 3 Polynomial and Rational Functions c=2 c=1

y 2

86. If c  0, the polynomial P x  cx 4  2x 2 has only a maximum at x  0. For c  0, P x has a

c=3

c=21

1

maximum at x  0 and minima at q, where q

decreases as c increases. Because the graph of P is

_1

_2

fundamentally different for positive and negative values of c, c  0 is called a transitional value.

0

1

_1

x

2

_2 _3 _4

87. P x  8x  03x 2  00013x 3  372 4000

c=_2 c=_1

c=0

c=41

(a) For the firm to break even, P x  0. From the graph, we see that P x  0 when x  252. But unless the firm also manufactures

meta­blenders, it cannot produce fractions of a blender, so it must produce at least 26 blenders a year.

2000

(b) No, the profit does not increase indefinitely. The largest profit is 0

0

approximately $327622, which occurs when the firm produces

200

166 blenders per year.

88. P t  120t  04t 4  1000

(a) A maximum population of approximately 1380 is attained after 422 months. (b) The rabbit population disappears after approximately 842 months.

1000 500 0

0

5

10

89. (a) The length of the bottom is 40  2x, the width of the bottom is 20  2x, and the height is x, so the volume of the box is V  x 20  2x 40  2x  4x 3  120x 2  800x.

(c) Using the domain from part (b), we graph V in the viewing rectangle [0 10] by [0 1600]. The maximum volume is V  15396 when x  423.

(b) Since the height and width must be positive, we must have x  0 and 20  2x  0, and so the domain of V is 0  x  10.

1000

0

0

5

10


SECTION 3.2 Polynomial Functions and Their Graphs

90. (a) Let h be the height of the box. Then the total length of all

(c) Using the domain from part (b), we graph V in

12 edges is 8x  4h  144 in. Thus, 8x  4h  144 

the viewing rectangle [0 18] by [0 2000]. The

2x  h  36  h  36  2x. The volume of the box is equal to   area of baseheight  x 2 36  2x  2x 3 36x 2 .

27

maximum volume is V  1728 in3 when x  12 in. 2000

Therefore, the volume of the box is

1000

V x  2x 3  36x 2  2x 2 18  x. (b) Since the length of the base is x, we must have x  0. Likewise, the height must be positive so 36  2x  0 

0

0

10

x  18. Putting these together, we get that the domain of V

is 0  x  18. 91.

The graph of y  x 100 is close to the x­axis for x  1, but passes through the

y (_1, 1)

points 1 1 and 1 1. The graph of y  x 101 behaves similarly except that the

(1, 1)

1

y­values are negative for negative values of x, and it passes through 1 1

y=x@

y=x$

instead of 1 1. x

1

y=x%

y=x#

(1, _1)

92. No, it not possible for a third­degree polynomial to have exactly one local extremum. The end behavior of such a polynomial is the same as that of y  kx 3 , and for this function, the values of y go off in opposite directions as x   and x  . But for a function with just one extremum, the values of y go in the same direction (either both positive or both negative) on both sides of the extremum. Neither is it possible for any polynomial to have two local maxima and no local minimum. All polynomials are continuous and defined everywhere, so any two local maxima must have a local minimum between them. An example of a polynomial with six local extrema is P x  x  1 x  2 x  3 x  4 x  5 x  6 x  7. 93.

y

1

Recall that if f has a fixed point at x, then the graph of f intersects the (1, 1)

functions f i with domain [0 1] and range contained in [0 1]. It quickly

becomes clear that all such functions have a fixed point.

y=x fª 0 (0, 0)

line y  x. We draw the line y  x and sketch a few graphs of continuous

1

x


28

CHAPTER 3 Polynomial and Rational Functions

3.3

DIVIDING POLYNOMIALS

1. If we divide the polynomial P by the factor x  c, and we obtain the equation P x  x  c Q x  R x, then we say that x  c is the divisor, Q x is the quotient, and R x is the remainder.

2. (a) If we divide the polynomial P x by the factor x  c, and we obtain a remainder of 0, then we know that c is a zero of P. (b) If we divide the polynomial P x by the factor x  c, and we obtain a remainder of k, then we know that P c  k.

3.

3

3 3

4.

4

6

9

9

3

13

P x  D x

x 3

5. 3x  2

 3x  3 

4x 2 

12x 3  12x 3 

8x

16x 2  8x 2

24x 2 

x  1 x

24x 2  16x

15x  1 15x  10 11

Thus, the quotient is 4x 2  8x  5 and the remainder is 11, and

12x 3  16x 2  x  1 P x  D x 3x  2   11 2  4x  8x  5  3x  2

10

0

7

0

7

2x 2  10x  7 7 P x   2x  . D x x 5 x 5

2x 2

6.

5

10

Thus, the quotient is 2x and the remainder is 7, and

13 . x 3

2 2

Thus, the quotient is 3x  3 and the remainder is 13, and 3x 2  6x  4

5

5x  1

 45

10x 3  2x 2  4x  1 10x 3  2x 2

 4x  1  4x  45 9 5

Thus, the quotient is 2x 2  45 and the remainder is 95 , and 9

10x 3  2x 2  4x  1  2 4  P x 5   2x  5  . D x 5x  1 5x  1


SECTION 3.3 Dividing Polynomials

2x 2  x 

7. x2  4

1

8.

2x 4  x 3  9x 2 2x 4

x 2  3x  1

 8x 2

x 3  x 2 x 3

2x 3  6x 2  17x 

2x 5  0x 4 

x2

x 3  2x 2

6x 4  18x 3  6x 2

17x 3  8x 2 

43x 2  14x  5

43x 2  129x  43

Thus, the quotient is 2x 2  x  1 and the remainder is P x  D x

2x 4  x 3  9x 2 x2  4

3x

17x 3  51x 2  17x

 4

4x  4

4x  4, and

3x  5

2x 5  6x 4  2x 3

 4x

x 2  4x

43

x 3  2x 2 

6x 4 

 4x  4   2x 2  x  1  2 . x 4

115x  48

Thus, the quotient is 2x 3  6x 2  17x  43 and the remainder is 115x  48 and

2x 5  x 3  2x 2  3x  5 P x  D x x 2  3x  1  

115x  48  2x 3  6x 2  17x  43  2 x  3x  1

9. Long division:

x 5

10. Long division: 3x 2 

10x

3x 3 

15x 2

3x 3 

 50

5x 2 

0x 

5

x 2

10x 2  0x

5x 3  10x 2  10x  23 5x 4

5x 4  10x 3

 10x 2  3x  2

10x 3  10x 2

10x 2  50x

50x  50x 

10x 3  20x 2

10x 2  3x

5

10x 2  20x

250

23x  2

245

23x  46

Thus, the quotient is 3x 2  10x  50 and the remainder is

245, so

  P x  3x 3 5x 2 5  x  5 3x 2  10x  50 245.

Synthetic division:

5

3 3

5

0

5

15

50

250

10

50

245

Thus, the quotient is 3x 2  10x  50 and the remainder is

245, as above.

29

48

P x  5x 4  10x 2  3x  2    x  2  5x 3  10x 2  10x  23  48

Synthetic division: 2

5 5

0

10

3

2

10

20

20

46

10

10

23

48

Thus, the quotient is 5x 3  10x 2  10x  23 and the

remainder is 48, as above.


30

CHAPTER 3 Polynomial and Rational Functions

x2

11.

 1

12.

2x 3  3x 2  2x

2x  3

2x  1

2x 3  3x 2

2x 2 

4x 3

x  4

4x 3  2x 2

 7x  9

2x 2  7x

2x

2x 2  x

2x  3

8x  9

3

8x  4

Thus, the quotient is x 2  1 and the remainder is 3, and   P x  2x 3  3x 2  2x  2x  3 x 2  1  3.

13. 2x 2  1

4x 2  2x  1

5

Thus

  P x  4x 3  7x  9  2x  1  2x 2  x  4  5.

14.

8x 4  4x 3  6x 2 8x 4

3x 2  3x  1

 4x 2

4x 3  2x 2 4x 3

2x 2  2x 2

9x 3  6x 2 

3x  2

27x 5  9x 4  0x 3  3x 2  0x  3 27x 5  27x 4  9x 3

18x 4  9x 3  3x 2 18x 4  18x 3  6x 2

2x

9x 3  3x 2

2x

9x 3  9x 2  3x

 1

6x 2  3x  3

2x  1

6x 2  6x  2

Thus, the quotient is 4x 2  2x  1 and the remainder is 2x  1, and

P x  27x 5  9x 4  3x 2  3      3x 2  3x  1  9x 3  6x 2  3x  2  3x  5

     2x 2  1  4x 2  2x  1  2x  1

15. x 2

3x  5

Thus

P x  8x 4  4x 3  6x 2

x  1

16.

x 2  3x  7

x 2  2x

x  7 x  2

5

Thus, the quotient is x  1 and the remainder is 5.

x 3

x2 

x  2

x 3  2x 2  x 3  3x 2

x  1

x 2  x x 2  3x

2x  1 2x  6 5

Thus, the quotient is x 2  x  2 and the remainder is 5.


SECTION 3.3 Dividing Polynomials

17. 3x  1

3x 2 

2x 2  2x  1

18.

x

9x 3  6x 2  x  1

8x 3  2x 2  2x  3

4x  3

9x 3  3x 2

31

8x 3  6x 2

3x 2  x

8x 2  2x

3x 2  x

8x 2  6x

1

4x  3 4x  3

Thus, the quotient is 3x 2  x and the remainder is 1.

0

Thus, the quotient is 2x 2  2x  1 and the remainder is 0. 19.

4x x2  x  1

 2

20.

4x 3  2x 2

4x 3  4x 2  4x

x 2  3x  2

 3

2x 2  4x

x2  x  5

x 4  4x 3  0x 2 

x  3

x 4  3x 3  2x 2

x 3  2x 2 

2x 2  2x  2

x

x 3  3x 2 

2x

5x 2 

6x  5

5x 2 

Thus, the quotient is 4x  2 and the remainder is 6x  5.

x  3

15x  10

14x  13

Thus, the quotient is x 2  x  5 and the remainder is 14x  13.

21. 2x 2  0x  5

3x 

1

22.

6x 3  2x 2  22x  0 6x 3

 15x

2x 2 

2x 2

3 3x 2  7x

7x  0

9x 2 

x  5

9x 2  21x

20x  5

 5

Thus, the quotient is 3 and the remainder is 20x  5.

7x  5

Thus, the quotient is 3x  1 and the remainder is 7x  5. x4

23. x2  1

24.

 1

x 6  0x 5  x 4  0x 3  x 2  0x  1

x6

 x4

0

 x2 x2

 1

4x 2  6x  8

1 x3  x2  5 x  7 2 2 4 2x 5  7x 4  0x 3  0x 2  0x  13

2x 5  3x 4  4x 3

4x 4  4x 3  0x 2 4x 4  6x 3  8x 2

 1

10x 3  8x 2  0x

0

10x 3  15x 2  20x

Thus, the quotient is x 4  1 and the remainder is 0.

7x 2  20x  13

7x 2  21 2 x  14

19 x  1 2

Thus, the quotient is 12 x 3  x 2  52 x  74 and the

remainder is 19 2 x  1.


32

CHAPTER 3 Polynomial and Rational Functions

25. The synthetic division table for this problem takes the following form.

26. The synthetic division table for this problem takes the following form.

3

2

3

5

2

6

3

1

6

1

Thus, the quotient is 2x  1 and the remainder is 6. 27. The synthetic division table for this problem takes the

1

1

4

1

2

2

1

6

Thus, the quotient is x  2 and the remainder is 6. 28. The synthetic division table for this problem takes the following form.

following form. 3

1

3

1

0

3

2

2

0 8

2

2

4 4

8

3

16 13

Thus, the quotient is 3x  2 and the remainder is 2.

Thus, the quotient is 4x  8 and the remainder is 13.

29. The synthetic division table for this problem takes the

30. Since x  5  x  5, the synthetic division table for this problem takes the following form.

following form. 2

3 3

2

1

6

8

4

9

18 13

5

Thus, the quotient is 3x 2  4x  9 and the remainder is 13. 31. Since x  4  x  4 and

x 3  10x  13  x 3  0x 2  10x  13, the synthetic

division table for this problem takes the following form. 1

4

0

16

1

11

Thus, the quotient is x 2  4x  6 and the remainder is 11. 33. Since x 5  3x 3  6  x 5  0x 4  3x 3  0x 2  0x  6, the synthetic division table for this problem takes the following form. 1

1

0

3

0

0

1

1

4

4

1

4

4

4

6

4

2

Thus, the quotient is x 4  x 3  4x 2  4x  4 and the remainder is 2.

10

25

25

25

5

5

35

following form. 3

1

0

1

1

3

6

2

6

10

0

18

24

8

24

Thus, the quotient is x 3  2x 2  6x  8 and the remainder is 24.

34. The synthetic division table for this problem takes the following form. 3

1

30

32. Since x 4  x 3  10x  x 4  x 3  0x 2  10x  0, the synthetic division table for this problem takes the

24

6

4

20

Thus, the quotient is 5x 2  5x  5 and the remainder is 35.

13

10

4

5

5

5

1

9

3

1

6

27 18 9

27

27 0

Thus, the quotient is x 2  6x  9 and the remainder is 0.


SECTION 3.3 Dividing Polynomials

35. The synthetic division table for this problem takes the following form.

36. The synthetic division table for this problem takes the following form.

1 2

2

3 1

2

2

1

2

0

0

1

4

 23

37. Since x 3  27  x 3  0x 2  0x  27, the synthetic

division table for this problem takes the following form. 3

1 1

0

0

3

9

3

9

0

4

3

3

0

5

6

12

6

3

12

5

5

4

8

4

16 16

0

8

is 0.

0

2

0 0

40

2 2

9

1

1

5

10

6

  Therefore, by the Remainder Theorem, P 12  6. 1

1 1 2

1

5

2

3

3

2

Therefore, by the Remainder Theorem, P 1  2. 44. P x  2x 3  21x 2  9x  200, c  11

2 0

1 2

1

11

7

2

0 2

7

45. P x  5x 4  30x 3  40x 2  36x  14, c  7 30

2 2

1

Therefore, by the Remainder Theorem, P 2  7.

5

0

Thus, the quotient is x 3  2x 2  4x  8 and the remainder

6

10

43. P x  x 3  2x 2  7, c  2

1

0

42. P x  x 3  x 2  x  5, c  1

Therefore, by the Remainder Theorem, P 2  12.

1

1

0

17

6

7

2

2

1 1

41. P x  x 3  3x 2  7x  6, c  2

1

4

following form.

0

1

2

1

1  29 7 9

38. The synthetic division table for this problem takes the

27

Therefore, by the Remainder Theorem, P 2  17.

2

6

1  23 1 3

40. P x  2x 2  9x  1, c  12

2

1

4

5

remainder is 79 .

2

39. P x  x 4  x 2  5, c  2 1

10

Thus, the quotient is 6x 3  6x 2  x  13 and the

27

Thus, the quotient is x 2  3x  9 and the remainder is 0.

2

6 6

Thus, the quotient is 2x 2  4x and the remainder is 1.

7

33

36

14 497

35

35

35

5

5

71

483

Therefore, by the Remainder Theorem, P 7  483.

21

9

22

11

1

20

200

220 20

Therefore, by the Remainder Theorem, P 11  20. 46. P x  6x 5  10x 3  x  1, c  2 2

6 6

0

10

0

1

1

12

24

68

136

274

12

34

68

137

273

Therefore, by the Remainder Theorem, P 2  273.


34

CHAPTER 3 Polynomial and Rational Functions

47. P x  x 7  3x 2  1 c3

3

48. P x  2x 6  7x 5  40x 4  7x 2  10x  112, c  3

 x 7  0x 6  0x 5  0x 4  0x 3  3x 2  0x  1 1 1

0

0

0

0

0

3

9

27

81

243

720

2160

3

9

27

81

240

720

2159

3

3

2

40

0

6

39

3

13

2

1

7

1

3

7

9

2

10

112

6

12

4

100

Therefore, by the Remainder Theorem, P 3  100.

Therefore by the Remainder Theorem, P 3  2159.

49. P x  3x 3  4x 2  2x  1, c  23 2 3

3

4 2

3

6

2

4 2

50. P x  x 3  x  1, c  14 1 4

1 4 3 7 3

01

1 1

2

3

021

21

0279

279

8279

52. (a) P x  6x 7  40x 6  16x 5  200x 4  60x 3  69x 2  13x  139, c  7 6 6

40

16

42

14

2

30

1

1

1 16  15 16

1  15 64

49 64

8

01 Therefore, by the Remainder Theorem, P 01  8279.

7

0 1 4 1 4

  Therefore, by the Remainder Theorem, P 14  49 64 .

  Therefore, by the Remainder Theorem, P 23  73 . 51. P x  x 3  2x 2  3x  8, c  01

1

200

60

69

13

70

7

10

10

1

20

210

70

139

140 1

Therefore, by the Remainder Theorem, P 7  1.

(b) P 7  6 77  40 76  16 75  200 74  60 73  69 72  13 7  139  6 823,543  40 117,649  16 16,807  200 2401  60 343  69 49  13 7  139  1

which agrees with the value obtained by synthetic division, but requires more work.

53. P x  x 3  3x 2  3x  1, c  1 1

1

3

1

1

2

3 2 1

54. P x  x 3  2x 2  3x  10, c  2 1

2

1

1

0

Since the remainder is 0, x  1 is a factor.

2 2

1

4

3

10

5

0

8

10

Since the remainder is 0, x  2 is a factor.


SECTION 3.3 Dividing Polynomials

55. P x  2x 3  7x 2  6x  5, c  12 1 2

2 2

7

6

1

4

8

10

35

56. P x  x 4  3x 3  16x 2  27x  63, c  3, 3 3

5

1

3

16

3

5

1

0

18

6

63

27

6

2

63

0

21

Since the remainder is 0, x  3 is a factor. We next show that x  3 is also a factor by using synthetic division on

Since the remainder is 0, x  12 is a factor.

the quotient of the above synthetic division, x 3  6x 2  2x  21. 3

1 1

6

2

3

9

3

21 21

0

7

Since the remainder is 0, x  3 is a factor.

57. P x  x 3  3x 2  18x  40, c  4 4

1 1

3

18 28

4

10

7

58. P x  x 3  5x 2  2x  10, c  5

1

1 3

1

2

11 6

5

15 15

0

Since the remainder is 0, we know that 3 is a zero and   P x  x  3 x 2  2x  5 . Now x 2  2x  5  0   2  22  4 1 5  1  6. Hence, when x  2 1  the zeros are 3 and 1  6.

5

1

10

2

5

0

Hence, the zeros are 4, 2, and 5.

3

1

40

Since the remainder is 0, we know that 4 is a zero and   P x  x  4 x 2  7x  10  x  4 x  2 x  5.

59. x 3  x 2  11x  15, c  3

5

40

0

0

10

0

2

Since the remainder is 0, we know that 5 is a zero and        P x  x  5 x 2  2  x  5 x  2 x  2 .  Hence, the zeros are 5 and  2.

60. P x  3x 4  x 3  21x 2  11x  6, c  2, 13 2

3 3

1

21

6

14

7

7

6

11 14

6

3

0

Since the remainder is 0, we know that 2 is a zero and   P x  x  2 3x 3  7x 2  7x  3 . 1 3

3

7

1

3

6

7

3

2

3

9

0

Since the remainder is 0, we know that 13 is a zero and    P x  x  2 x  13 3x 2  6x  9    3 x  2 x  13 x  1 x  3.

Hence, the zeros are 2, 13 , 1, and 3.


36

CHAPTER 3 Polynomial and Rational Functions

61. P x  3x 4  8x 3  14x 2  31x  6, c  2, 3 3

2

8 6

3

31

6

28

28

14

3

6

14

14

3

14

3

2

1

0

Since the remainder is 0, we know that 2 is a zero and   P x  x  2 3x 3  14x 2  14x  3 . 3

62. P x  2x 4  13x 3  7x 2  37x  15, c  1, 3

14

3

9

15

3

5

1

3x 2  5x  1  0 when    5  52  4 3 1 5  37 x  , and 2 3 6  5  37 . hence, the zeros are 2, 3, and 6

2

2

7

37

15

15

22

15

22

15

15

0

Since the remainder is 0, we know that 1 is a zero and   P x  x  1 2x 3  15x 2  22x  15 . 3

0

Since the remainder is 0, we know that 3 is a zero and   P x  x  2 x  3 3x 2  5x  1 . Now

13

2

15

22

15

2

6

27

15

9

5

0

The remainder is 0, so 3 is a zero and   P x  x  1 x  3 2x 2  9x  5

 x  1 x  3 2x  1 x  5.

Hence, the zeros are 1, 3,  12 , and 5.

63. Since the zeros are x  1, x  1, and x  3, the factors are x  1, x  1, and x  3. Thus P x  x  1 x  1 x  3  x 3  3x 2  x  3.

64. Since the zeros are x  2, x  0, x  2, and x  4, the factors are x  2, x, x  2, and x  4.

Thus P x  c x  2 x x  2 x  4. If we let c  1, then P x  x 4  4x 3  4x 2  16x.

65. Since the zeros are x  1, x  1, x  3, and x  5, the factors are x  1, x  1, x  3, and x  5. Thus P x  x  1 x  1 x  3 x  5  x 4  8x 3  14x 2  8x  15.

66. Since the zeros are x  2, x  1, x  0, x  1, and x  2, the factors are x  2, x  1, x, x  1, and x  2. Thus P x  c x  2 x  1 x x  1 x  2. If we let c  1, then P x  x 5  5x 3  4x.

67. Since the zeros of the polynomial are 2, 0, 1, and 3, it follows that P x  C x  2 x x  1 x  3  C x 4 2C x 3 

5C x 2  6C x. Since the coefficient of x 3 is to be 4, 2C  4, so C  2. Therefore, P x  2x 4  4x 3  10x 2  12x is the polynomial.   68. Since the zeros of the polynomial are 1, 0, 2, and 12 , it follows that P x  C x  1 x x  2 x  12  C x 4 

3 C x 3  3 C x 2 C x. Since the coefficient of x 3 is to be 3,  3 C  3, so C  2. Therefore, P x  2x 4 3x 3 3x 2 2x 2 2 2

is the polynomial.

  2 and integer coefficients, the fourth zero must be  2, otherwise the      constant term would be irrational. Thus, P x  C x  1 x  1 x  2 x  2  C x 4  3C x 2  2C. Requiring

69. Since the polynomial degree 4 and zeros 1, 1, and

that the constant term be 6 gives C  3, so P x  3x 4  9x 2  6.   70. Since the polynomial degree 5 and zeros 2, 1, 2, and 5 and integer coefficients, the fourth zero must be  5,       otherwise the constant term would be irrational. Thus, P x  C x  2 x  1 x  2 x  5 x  5  C x 5  C x 4  9C x 3  9C x 2  20C x  20C. Requiring that the constant term be 40 gives C  2, so P x  2x 5  2x 4  18x 3  18x 2  40x  40.


SECTION 3.4 Real Zeros of Polynomials

37

71. The y­intercept is 2 and the zeros of the polynomial are 1, 1, and 2.   It follows that P x  C x  1 x  1 x  2  C x 3  2x 2  x  2 . Since P 0  2 we have   2  C 03  2 02  0  2  2  2C  C  1 and P x  x  1 x  1 x  2  x 3  2x 2  x  2.

72. The y­intercept is 4 and the zeros of the polynomial are 1 and 2 with 2 being degree two.     It follows that P x  C x  1 x  22  C x 3  3x 2  4 . Since P 0  4 we have 4  C 03  3 02  4  4  4C  C  1 and P x  x 3  3x 2  4.

73. The y­intercept is 4 and the zeros of the polynomial are 2 and 1 both being degree two.   It follows that P x  C x  22 x  12  C x 4  2x 3  3x 2  4x  4 . Since P 0  4 we have   4  C 04  2 03  3 02  4 0  4  4  4C  C  1. Thus P x  x  22 x  12  x 4  2x 3  3x 2  4x  4.

74. The y­intercept is 2 and the zeros of the polynomial are 2, 1, and 1 with 1 being degree two.   It follows that P x  C x  2 x  1 x  12  C x 4  x 3  3x 2  x  2 . Since P 0  2 we have   4  C 04  03  3 02  0  2  2  2C so C  1 and P x  x 4  x 3  3x 2  x  2.

75. (a) By the Remainder Theorem, the remainder when P x  6x 1000  17x 562  12x  26 is divided by x  1 is P 1  6 11000  17 1562  12 1  26  6  17  12  26  3.

(b) If x  1 is a factor of Q x  x 567  3x 400  x 9  1, then Q 1 must equal 0. By the Remainder Theorem,

Q 1  1567  3 1400  19  1  1  3  1  1  0, so x  1 is a factor.      76. R x  x 5  2x 4  3x 3  2x 2  3x  4  x 4  2x 3  3x 2  2x  3 x  4  x 3  2x 2  3x  2 x  3 x  4      x 2  2x  3 x  2 x  3 x  4  [x  2 x  3] x  2 x  3 x  4

So to calculate R 3, we start with 3, then subtract 2, multiply by 3, add 3, multiply by 3, subtract 2, multiply by 3, add 3, multiply by 3, and add 4, to get 157.

3.4

REAL ZEROS OF POLYNOMIALS

1. If the polynomial function P x  an x n  an1 x n1     a1 x  a0 has integer coefficients, then the only numbers that p could possibly be rational zeros of P are all of the form , where p is a factor of the constant coefficient a0 and q is a factor q of the leading coefficient an . The possible rational zeros of P x  6x 3  5x 2  19x  10 are 1,  12 ,  13 ,  16 , 2,  23 , 5,  52 ,  53 ,  56 , 10, and  10 3.

2. Using Descartes’ Rule of Signs, we can tell that the polynomial P x  x 5  3x 4  2x 3  x 2  8x  8 has 1, 3 or 5 positive real zeros and no negative real zero. 3. This is true. If c is a real zero of the polynomial P, then P x  x  c Q x, and any other zero of P x is also a zero of Q x  P x  x  c. 4. False. Consider the polynomial P x  x  1 x  2 x  4  x 3  x 2  10x  8. An upper bound is 3, but 3 is not a lower bound.

5. P x  x 3  4x  6. The constant term is 6, with factors 1, 2, 3, 6; and since the leading term is 1, these are the possible rational zeros of P. 6. Q x  x 5  3x 2  5x  10 has possible rational zeros 1, 2, 5, 10.


38

CHAPTER 3 Polynomial and Rational Functions

7. R x  3x 4  2x 3  8x 2  9. The constant term is 9, with factors 1, 3, 9; and the leading term is 3, with factors 1, 3. Thus, the possible rational zeros of R are  11 ,  31 ,  91 ,  13 ,  33 ,  93 . Eliminating duplicates, they are 1, 3, 9,  13 .

8. S x  5x 6  3x 4  20x 2  15. The constant term is 15, with factors 1, 3, 5, 15; and the leading term is 5, with 1 3 5 15 factors 1, 5. Thus, the possible rational zeros of S are  11 ,  31 ,  51 ,  15 1 ,  5 ,  5 ,  5 ,  5 . Eliminating duplicates,

they are 1, 3, 5, 15,  15 ,  35 .

9. T x  6x 5  8x 3  5. The constant term is 5, with factors 1, 5; and the leading term is 6, with factors 1, 2, 3, 6. Thus, the possible rational zeros of T are  11 ,  51 ,  12 ,  52 ,  13 ,  53 ,  16 ,  56 ; that is, 1, 5,  12 ,  52 ,  13 ,  53 ,  16 ,  56 .

10. U x  12x 5  6x 3  2x  8. The constant term is 8, with factors 1, 2, 4, 8; and the leading term is 12, with

factors 1, 2, 3, 4, 6, 12. Thus, the possible rational zeros of U are  11 ,  21 ,  41 ,  81 ,  12 ,  22 ,  42 ,  82 ,  13 ,

1 ,  2 ,  4 ,  8 . Eliminating duplicates, they are 1, 2, 4,  23 ,  43 ,  83 ,  14 ,  24 ,  44 ,  84 ,  16 ,  26 ,  46 ,  86 ,  12 12 12 12 1. 8,  12 ,  13 ,  14 ,  23 ,  43 ,  83 ,  16 ,  12

11. (a) P x  5x 3  x 2  5x  1 has possible rational zeros 1,  15 . (b) From the graph, the actual zeros are 1, 15 , and 1.

12. (a) P x  3x 3  4x 2  x  2 has possible rational zeros 1, 2,  13 ,  23 . (b) From the graph, the actual zeros are 1 and 23 .

13. (a) P x  2x 4  9x 3  9x 2  x  3 has possible rational zeros 1, 3,  12 ,  32 . (b) From the graph, the actual zeros are  12 , 1, and 3.

14. (a) P x  4x 4  x 3  4x  1 has possible rational zeros 1,  12 ,  14 . (b) From the graph, the actual zeros are 14 and 1.

15. P x  x 3  5x 2  8x  12. The possible rational zeros are 1, 2, 3, 4, 6, 12. P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x  x 3  5x 2  8x  12 has 1 variation in sign and hence P has 1 negative real zero. 1

1

5

1

8

12

4

12

1 4 12 0  x  1 is a zero.   P x  x 3  5x 2  8x  12  x  1 x 2  4x  12  x  1 x  6 x  2. Therefore, the zeros are 2, 1, and 6.

16. P x  x 3  4x 2  19x  14. The possible rational zeros are 1, 7, 14. P x has 1 variation in sign and hence 1 positive real zero. P x  x 3  4x 2  19x  14 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1

1

4 1

1

19

14

3

22

22

36

1

1

4 1

19 5

14 14

 x  1 is not a zero. 1 5 14 0  x  1 is a zero.   P x  x 3  4x 2  19x  14  x  1 x 2  5x  14  x  2 x  1 x  7. Thus, the zeros are 2, 1, and 7. 3


SECTION 3.4 Real Zeros of Polynomials

39

17. P x  x 3  5x 2  3x  9. The possible rational zeros are 1, 3, 9. P x has 2 variation in sign and hence 0 or 2 positive real zero.P x  x 3  5x 2  3x  9 has 1 variation in sign and hence P has 1 negative real zero. 1

1

3

9

4

1

5

1

1

1

1

5 1

9

9

 x  1 is not a zero. 1 6 9 0  x  1 is a zero.   P x  x 3  5x 2  3x  9  x  1 x 2  6x  9  x  1 x  32 . Therefore, the zeros are 1 and 3. 4

1

8

3

6

18. P x  x 3  3x  2. The possible rational zeros are 1, 2. P x has 1 variation in sign and hence 1 positive real zero.P x  x 3  3x  2 has 2 variations in sign and P has hence 0 or 2 negative real zeros. 1

1

0

3

1 1

1

1

1

2

2

0

3

2

2

2

4

2

 x  1 is not a zero. 1 2 1 0  x  2 is a zero.  P x  x 3  3x  2  x  2 x 2  2x  1  x  2 x  12 . Therefore, the zeros are 2 and 1. 2

4

19. P x  x 3  6x 2  12x  8. The possible rational zeros are 1, 2, 4, 8. P x has 3 variations in sign and hence 1 or 3 positive real zeros. P x  x 3  6x 2  12x  8 has no variations in sign and hence P has no negative real zero. 1

1

6

1

1

12

5 7

1

2

8

12

6

7

2

8 8

8

 x  1 is not a zero. 1 4 4 0  x  2 is a zero.   P x  x 3  6x 2  12x  8  x  2 x 2  4x  4  x  23 . Therefore, the only zero is x  2. 5

1

20. P x  x 3  12x 2  48x  64. The possible rational zeros are 1, 2, 4, 8, 16, 32, 64. P x has 0 variations in sign and hence no positive real zero. P x  x 3  12x 2  48x  64 has 3 variations in sign and hence P has 1 or 3 negative real zeros. 1

1

1

12

48

64

1

11

37

11

37

27

2  x  1 is not a zero. 4

1 1

1

12

48

64

4

32

64

8

16

1

0

12

48

64

2

20

56

10

28

6

 x  2 is not a zero.

 x  4 is a zero.

P x  x 3  12x 2  48x  64  x  4 x 2  8x  16  x  43 . Therefore, the only zero is x  4. 21. P x  x 3  27x  54. The possible rational zeros are 1, 2, 3, 6, 9, 18, 27. P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x  x 3  27x  30 has 1 variation in sign and hence P has 1 negative real zero. 1

1

0

1 1

1

27

54

1

26

26

80

2

1

0

2  x  1 is not a zero. 3

1

0 3

1

9

2

4

23

54

46

8

 x  2 is not a zero.

54 54

18 0  x  3 is a zero.  P x  x 3  27x  54  x  3 x 2  3x  18  x  3 x  6 x  3. Therefore, the zeros are 6 and 3. 

3

27

1

27


40

CHAPTER 3 Polynomial and Rational Functions

22. P x  x 3  5x 2  9x  45. The possible rational zeros are 1, 3, 9, 15, 45. P x has 1 variation in sign and

hence 1 positive real zero. P x  x 3  5x 2  9x  45 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1

1

5

9

1 1

6

6

2

45 3

3

48

1

5 2

 x  1 is not a zero. 3

1

5

9

3 1

1

7

9

45

5

35

14

10  x  2 is not a zero.

45

24

45

8

15 0  x  3 is a zero.  P x  x 3  5x 2  9x  45  x  3 x 2  8x  15  x  3 x  5 x  3. Therefore, the zeros are 5, 3, and 3. 

23. P x  x 3  3x 2  x  3. The possible rational zeros are 1, 3. P x has 1 variation in sign and hence 1 positive real zero. P x  x 3  3x 2  x  4 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1

1

1

3

1

1

2

2

3 3

3 0  x  1 is a zero.  So P x  x 3  3x 2  x  3  x  1 x 2  2x  3  x  1 x  3 x  1. Therefore, the zeros are 1, 3, and 1. 

24. P x  x 3  4x 2  11x  30. The possible rational zeros are 1, 2, 3, 5, 10, 15, 30. P x has 2 variations

in sign and hence 0 or 2 positive real zeros. P x  x 3  4x 2  11x  30 has 1 variation in sign and hence P has 1 negative real zero. 1

1

4 1

1

11

30

3

14

14

16

1

2

4

2

11

30

4

30

1 2 15 0  x  2 is a zero.  So P x  x 3  4x 2  11x  30  x  2 x 2  2x  15  x  2 x  5 x  3. Thus, the zeros are 3, 2, and 5. 3

25. Method 1: P x  x 4  5x 2  4The possible rational zeros are 1, 2, 4. P x has 1 variation in sign and hence 1 positive real zero. P x  x 4  5x 2  4 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1

1

0

1

0

4

4

4

5

1

1 1 4 4 0  x  1 is a zero.   Thus P x  x 4  5x 2  4  x  1 x 3  x 2  4x  4 . Continuing with the quotient we have: 1

1

1

1

4 0

4 4

1 0 4 0  x  1 is a zero.   P x  x 4  5x 2  4  x  1 x  1 x 2  4  x  1 x  1 x  2 x  2. Therefore, the zeros are 1, 2.

Method 2: Substituting u  x 2 , the polynomial becomes P u  u 2  5u  4, which factors:    u 2  5u  4  u  1 u  4  x 2  1 x 2  4 , so either x 2  1 or x 2  4. If x 2  1, then x  1; if x 2  4, then x  2. Therefore, the zeros are 1 and 2.


SECTION 3.4 Real Zeros of Polynomials

41

26. P x  x 4  2x 3  3x 2  8x  4. Using synthetic division, we see that x  1 is a factor of P x: 1

1

8

1

3 1

4

1

4

2

1

4 4

4

0

 x  1 is a zero.

We continue by factoring the quotient, and we see that x  1 is again a factor: 1

1

1

1

1

4

4

0

4

0

4 0  x  1 is a zero.   P x  x 4  2x 3  3x 2  8x  4  x  1 x  1 x 2  4  x  12 x  2 x  2

Therefore, the zeros are 1 and 2.

27. P x  x 4  6x 3  7x 2  6x  8. The possible rational zeros are 1, 2, 4, 8. P x has 1 variation in sign and hence 1 positive real zero. P x  x 4  6x 3  7x 2  6x  8 has 3 variations in sign and hence P has 1 or 3 negative real zeros. 1

1

6

1

7

1

7

7

14

6

8

8

0

14

8

 x  1 is a zero

  and there are no other positive zeros. Thus P x  x 4  6x 3  7x 2  6x  8  x  1 x 3  7x 2  14x  8 . Continuing

by factoring the quotient, we have:

1

1

1

7

14

8

1

6

8

6

8

0

 x  1 is a zero.  So P x  x 4  6x 3  7x 2  6x  8  x  1 x  1 x 2  6x  8  x  1 x  1 x  2 x  4. Therefore, the 

zeros are 4, 2, and 1.

28. P x  x 4  x 3  23x 2  3x  90. The possible rational zeros are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90. Since P x has 2 variations in sign, P has 0 or 2 positive real zeros. Since P x  x 4  x 3  23x 2  3x  90 has 2 variations in sign, P has 0 or 2 negative real zeros. 1

1

3

90

0

0

23

26

23

26

1

1

1

1

1 3

23

3

21

12

1

2

27

1

42

90

21

45

1

21

45

2

64

45

3

90

2

1

1 5

1

5

23

30

0

 x  2 is a zero.

45

1 6 9 0  x  5 is a zero. 4 9 72   P x  x  2 x  5 x 2  6x  9  x  2 x  5 x  32 . Therefore, the zeros are 3, 2, and 5. 1


42

CHAPTER 3 Polynomial and Rational Functions

29. P x  9x 4  82x 2  9 has possible rational zeros 1, 3, 9,  13 ,  19 . Since P x has 2 variations in sign, there are 0 or 2 positive real zeros, and since P x  9x 4  9x 2  36 has 2 variations in sign, P has 0 or 2 negative real zeros. 1

9

0

9 9

9

0

9

9

73

73

73

73

64

82

9

3

0

9

0

9

81

3

9

1

3

82

27 27

0

 x  3 is a zero

Because P is even, we know that x  3 is also a zero. Thus, P x  9x 4  82x 2     9  x 2  9 9x 2  1  x  3 x  3 3x  1 3x  1, and the zeros are 3 and  13 . Note: Since P x contains only even powers of x, factoring by substitution also works. Let x 2  u; then    P u  9u 2  82u  9  u  9 9u  1  x 2  9 9x 2  1 , which gives the same result.

30. P x  6x 4 23x 3 13x 2 32x 16. The possible rational zeros are 1, 2, 4, 8, 16,  12 ,  13 ,  23 ,  43 ,  83 ,  16 3 ,  16 . Since P x has 2 variations in sign, P has 0 or 2 positive real zeros. Since P x  6x 4  23x 3  13x 2  32x  16 has 2 variations in sign, P has 0 or 2 negative real zeros. 1

6 6

32

16

6

13 17

30

2

17

30

23

4

6

2

18

23

13

6

1

23

12

2

24

6 6

32

16

4

36

16

9 

4

0

11

13

32

16

22

70

76

35

38

60

 x  4 is a zero 

P x  6x 4  23x 3  13x 2  32x  16  x  4 6x 3  x 2  9x  4 . We continue:  12

6

1

9

4

1

3

4

2 8 0  x   12 is a zero   Thus, P x  x  4 2x  1 3x 2  x  4  x  1 2x  1 3x  4 x  4 with zeros 1,  12 , 43 , and 4. 6

31. P x  6x 4  7x 3  9x 2  7x  3. The possible rational zeros are 1, 2, 3,  12 ,  13 ,  16 ,  32 . P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x  6x 4  7x 3  9x 2  7x  3 has 2 variations in sign and hence P has

0 or 2 negative real zeros.

1 6

7 9 7 6

13

3

4 3

6 13

4 3 0  x  1 is a zero.   Thus P x  6x 4  7x 3  9x 2  7x  3  x  1 6x 3  13x 2  4x  3 . Continuing by factoring the quotient, we have 1

6

13

4

6

7

3 3

6 7 3 0  x  1 is a zero.   Thus P x  x  1 x  1 6x 2  7x  3  x  1 x  1 2x  3 3x  1 with zeros  32 , 1, 13 , and 1.


SECTION 3.4 Real Zeros of Polynomials

43

32. P x  6x 3  37x 2  5x  6. The possible rational zeros are 1, 2, 3, 6,  12 ,  13 ,  16 ,  23 ,  32 . Since P x has 1 variation in sign, P has 1 positive real zero. Since P x  6x 3  37x 2  5x  6 has 2 variations in sign, P has 0 or 2

negative real zeros. 1

6 2

37

5

2

39

39

44

1 2

6

6

37

5

3

20

40

25

44 38

 x  1 is an upper bound. 1 3

6

37

5

2

13

6

6 25 2 13 2

 x  12 is an upper bound.

6

6

6 39 18 0  x  13 is a zero.   P x  3x  1 2x 2  13x  6  3x  1 2x  1 x  6. Therefore, the zeros are 6,  12 , and 13 . 33. Factoring by grouping can be applied to this exercise.

  4x 3  4x 2  x  1  4x 2 x  1  x  1  x  1 4x 2  1  x  1 2x  1 2x  1. Therefore, the zeros are 1 and  12 .

  34. We use factoring by grouping: P x  2x 3  3x 2  2x  3  2x x 2  1      3 x 2  1  x 2  1 2x  3  x  1 x  1 2x  3. Therefore, the zeros are 32 and 1. 35. P x  4x 3  7x  3. The possible rational zeros are 1, 3,  12 ,  32 ,  14 ,  34 . Since P x has 2 variations in sign, there are 0 or 2 positive zeros. Since P x  4x 3  7x  3 has 1 variation in sign, there is 1 negative zero. 1 2

4

0 2

3

7

1

3

2 6 0  x  12 is a zero.      P x  x  12 4x 2  2x  6  2x  1 2x 2  x  3  2x  1 x  1 2x  3  0. Thus, the zeros are  32 , 4

1 , and 1. 2

1 . Since P x has 2 36. P x  12x 3  25x 2  x  2. The possible rational zeros are 1, 2,  12 ,  13 ,  23 ,  14 ,  16 ,  12

variations in sign, P has 0 or 2 positive real zeros, and since P x  12x 3  25x 2  x  2 has 1 variation in sign, P has 1 negative real zero. 1

12

25 12

12

1

2

13

12

2

12

25

24

1

2

2

2

10 12 1 1 0  x  2 is a zero.   P x  12x 3  25x 2  x  2  x  2 12x 2  x  1  4x  1 3x  1 x  2. Therefore, the zeros are  14 , 13 ,

and 2.

13

12


44

CHAPTER 3 Polynomial and Rational Functions

37. P x  24x 3  10x 2  13x  6. The possible rational zeros are 1, 2, 3, 6,  12 ,  32 ,  13 ,  23 ,  14 ,  34 ,  16 ,  18 , 1 ,  1 . P x has 1 variation in sign and hence 1 positive real zero. P x  24x 3  10x 2  13x  6 has 2  38 ,  12 24

variations in sign, so P has 0 or 2 negative real zeros. 1 24

10 13 6 24

24 14 3 24

2 24

14 1

1 7  x  1 is not a zero.

10 13

24 38 6 24

6

72 186 519

24 62 173 525  x  3 is not a zero.  12

24

10

6

76 126

63 132  x  2 is not a zero.

10 13

6

144 804 4746

24 134 791 4752  x  6 is not a zero.

13

6

1

12

10 13 48

6

2 12 0  x   12 is a zero.   Thus, P x  24x 3  10x 2  13x  6  2x  1 12x 2  x  6  3x  2 2x  1 4x  3 has zeros  23 ,  12 , and 34 . 24

1 ,  3 ,  3 . P x has 2 38. P x  12x 3  20x 2  x  3. The possible rational zeros are 1, 2, 3,  12 ,  13 ,  14 ,  16 ,  12 2 4

variations in sign and hence 0 or 2 positive real zeros. P x  12x 3  20x 2  x  3 has 1 variation in sign and hence P has 1 negative real zero. 1

3

12

1

3

12

12

8

7

8

7

4

12

20

1

3

36

48

147

16

49

150

20

12

2  x  1 is not a zero.

12 1 2

1

3

24

8

18

4

9

21

20

1

 x  3 is not a zero. 12 14    Thus, P x  12x 3  20x 2  x  3  x  12 12x 2  14x  6 . Continuing:

6

12

12

20

6

3 2

12

14

6

4

0

18

12

6  x  32 is a zero.

Thus, P x  2x  1 2x  3 3x  1 has zeros 12 , 32 , and  13 .

7

 x  2 is not a zero. 3 3 0

 x  12 is a zero.


SECTION 3.4 Real Zeros of Polynomials

45

39. P x  6x 4  13x 3  32x 2  45x  18. The possible rational zeros are 1, 2, 3, 6, 9, 18,  12 ,  13 ,  16 ,  23 . P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x  6x 4  13x 3  32x 2  45x  18 has 2

variations in sign and hence P has 0 or 2 negative real zeros. 1

6

13

6

19

32

6

19

45

18

13

58

2

6

13 12

13 58 40  x  1 is not a zero.   P x  x  2 6x 3  25x 2  18x  9 . Continuing: 6

3

25

18

18

21

6

32

45

18

9

50

25

18

36

18

0

 x  2 is a zero.

9

9

6 7 3 0  x  3 is a zero.   P x  x  2 x  3 3x 2  8x  3  x  2 x  3 3x  1 2x  3, so the zeros are 3,  32 , 13 , and 2.

40. P x  2x 4  11x 3  11x 2  15x  9. The possible rational zeros are 1, 3, 9,  12 ,  32 ,  92 . Since P x has 1

variation in sign, P has 1 or 3 positive real zeros. Since P x  2x 4  11x 3  11x 2  15x  9 has 3 variations in sign, P has 1 or 3 negative real zeros.

1

2

11

11

2

13

15

24

9

9

2 13 24 9 0  x  1 is a zero.   P x  x  1 2x 3  13x 2  24x  9 . We continue by factoring the quotient: 3

2

13

24

9

6

57

243

 12

2

13

24

9

1

6

9

252  x  2 is an upper bound. 2 12 18 0  x   12 is a root.   P x  x  1 2x  1 x 2  6x  9  x  1 2x  1 x  32 . Therefore, the zeros are 3,  12 , and 1. 2

19

81

zero.


46

CHAPTER 3 Polynomial and Rational Functions

41. P x  x 5  3x 4  9x 3  31x 2  36. The possible rational zeros are 1, 2, 3, 4, 6, 8, 9, 12, 18. P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x  x 5  3x 4  9x 3  31x 2  36 has 3 variations in sign and hence P has 1 or 3 negative real zeros. 1

1

3

9

1

4

1

4

5

36

31

0

36

5

36

36

36 36 0  x  1 is a zero.   So P x  x 5  3x 4  9x 3  31x 2  36  x  1 x 4  4x 3  5x 2  36x  36 . Continuing by factoring the quotient, 5

we have:

1

1

4 1

1

5

1

0

0

36

36

72

3

1

5

4 3

1

7

2

36

21

1

4 2

1 36

48

5

12

6

7

36

14

22

36 44 80

36

36

16

12 0  x  3 is a zero.  So P x  x  1 x  3 x 3  7x 2  16x  12 . Since we have 2 positive zeros, there are no more positive zeros, so 

we continue by factoring the quotient with possible negative zeros. 1

1 1

7

16

12

1

6

10

10

2

1

1

7

16

12

2

10

12

0  x  2 is a zero.  Then P x  x  1 x  3 x  2 x 2  5x  6  x  1 x  3 x  22 x  3. Thus, the zeros are 1, 3, 2,

and 3.

6

2

5

6


SECTION 3.4 Real Zeros of Polynomials

47

42. P x  x 5  4x 4  3x 3  22x 2  4x  24 has possible rational zeros 1, 2, 3, 4, 6, 8, 12, 24. Since P x has 3 variations in sign, there are 1 or 3 positive real zeros. Since P x  x 5  4x 4  3x 3  22x 2  4x  24 has 2 variations in sign, P has 0 or 2 negative real zeros. 1

1

4

1

1

3

22

3

6

4

2

7

1

0

2

  P x  x  22 x 3  7x  6 2

1

0

7

2

1

1

4

14

2

7

8

12

0

14

12

7

6

0

4

24

12

0

16

8

24  x  2 is a zero.

 x  2 is a zero again.

1

3

6

4

3

22

2

4

12

3 6 16 12 12   P x  x  2 x 4  2x 3  7x 2  8x  12 2

1

2

24

16

0

7

3

6

9

6

6

1 3 2 0  x  3 is a zero. 2 3 12   P x  x  22 x  3 x 2  3x  2  x  22 x  3 x  1 x  2  0. Therefore, the zeros are 1, 2, and 3. 1

43. P x  3x 5  14x 4  14x 3  36x 2  43x  10 has possible rational zeros 1, 2, 5, 10,  13 ,  23 ,  53 ,  10 3 . Since P x has 2 variations in sign, there are 0 or 2 positive real zeros. Since P x  3x 5  14x 4  14x 3  36x 2  43x  10

has 3 variations in sign, P has 1 or 3 negative real zeros. 1

3

14 3

3

14

36

43

10

11

25

11

54

2

3

8

3

2

6

30

24

5

4

68

184

14 6

11 25 11 54 64   P x  x  2 3x 4  8x 3  30x 2  24x  5 2

3 3

8

5

3

14

36

43

10

16

60

48

10

30

24

5

8 15

30 35

24

25

0

 x  2 is a zero.

5

5

34 92 189 3 7 5 1 0  x  5 is a zero.   P x  x  2 x  5 3x 3  7x 2  5x  1 . Since 3x 3  7x 2  5x  1 has no variation in sign, there are no more

positive zeros.

1

3

7

5

1

3

4

1

3 4 1 0  x  1 is a zero.   P x  x  2 x  5 x  1 3x 2  4x  1  x  2 x  5 x  1 x  1 3x  1. Therefore, the zeros are 1,

 13 , 2, and 5.


48

CHAPTER 3 Polynomial and Rational Functions

44. P x  2x 6  3x 5  13x 4  29x 3  27x 2  32x  12 has possible rational zeros 1, 2, 3, 4,

6, 12,  12 ,  32 . Since P x has 5 variations in sign, there are 1 or 3 or 5 positive real zeros. Since

P x  2x 6  3x 5  13x 4  29x 3  27x 2  32x  12 has 3 variations in sign, P has 1 or 3 negative real zeros. 1

2

3

13

29

2

1

14

1

14

2

2

2

13

1

11

1

11

4

2

29

3

2

15

15

32

14

2

4

10

13

7

13

26

20

20

8

12

12

0  x  2 is a zero.   P x  x  2 2x 5  x 4  11x 3  7x 2  13x  6 . We continue with the quotient: 2

7

12

12

12

27

22

32

27

6

6

10

2

6

2 5 1 5 3 0  x  2 is a zero again.   P x  x  22 2x 4  5x 3  x 2  5x  3 . We continue with the quotient, first noting 2 is no longer a possible rational

solution:

3

2

5 6

2

11

5

1

22

42

21

47

3

94 91

 x  3 is an upper bound.

We know that there is at least 1 more positive zero. 1 2

2

5 1

1

2

5 1

3

3

2 6 2 6 0  x  12 is a zero.   P x  x  22 x  12 2x 3  6x 2  2x  6 . We can factor 2x 3  6x 2  2x  6 by grouping;         2x 3  6x 2  2x  6  2x 3  6x 2  2x  6  2x  6 x 2  1 . So P x  2 x  22 x  12 x  3 x 2  1 . 

Since x 2  1 has no real zeros, the zeros of P are 3, 2, and 12 .

45. P x  3x 3  5x 2  2x  4. The possible rational zeros are 1, 2, 4,  13 ,  23 ,  43 . P x has 1 variation in sign and hence 1 positive real zero. P x  3x 3  5x 2  2x  4 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1 3 5 2 4 5 2 4 1 3 3

3

2

0

3

2

4

2 0 4 3 2 4 0  x  1 is a zero.   So P x  x  1 3x 2  2x  4 . Using the Quadratic Formula on the second factor, we have    2 434  13 13 . Therefore, the zeros of P are 1 and 13 13 . x  2 223


SECTION 3.4 Real Zeros of Polynomials

49

46. P x  3x 4  5x 3  16x 2  7x  15. The possible rational zeros are 1, 3, 5, 15,  13 ,  53 . P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x  3x 4  5x 3  16x 2  7x  15 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1

3

3

3

5

16

2 18 11 4   So P x  x  3 3x 3  4x 2  4x  5 . Continuing:

3

4

5

3

16

7

15

2

18

11

3

1

9

3

4

4

3

1

7

15

12

12

15

4

5

0

 x  3 is a zero.

5 5

3 1 5 0  x  1 is a zero.     Thus, P x  x  3 3x 3  4x 2  4x  5  x  1 x  3 3x 2  x  5 . Using the Quadratic Formula on the last    2 435 factor, we have x  1 123  16 61 . Therefore, the zeros of P are 1, 3, and 16 61 .

47. P x  x 4  6x 3  4x 2  15x  4. The possible rational zeros are 1, 2, 4. P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x  x 4  6x 3  4x 2  15x  4 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1 1 6 4 15 4 4 15 4 2 1 6 1

1

5

5

1 4

14

1 14

1

18

6

4

4

15

4

8

16

4

8

4

4

2 4 1 0  x  4 is a zero.  So P x  x  4 x 3  2x 2  4x  1 . Continuing by factoring the quotient, we have: 4

1

2

4

1

2

1

1

2

4

8

4

1

1

16

15

 x  4 is an upper bound. 

1 1

2

8

7

4

1

3

1

3

1

0

14

18

1

 x  1 is a zero.

So P x  x  4 x  1 x 2  3x  1 . Using the Quadratic Formula on the third factor, we have:    3 32 411 3 13 . Therefore, the zeros are 4, 1, and 3 13 .  x 2 2 21


50

CHAPTER 3 Polynomial and Rational Functions

48. P x  x 4  2x 3  2x 2  3x  2. The possible rational zeros are 1, 2. P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x  x 4  2x 3  2x 2  3x  2 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1

1

2

1 

1

3

2

3 1

2

3

1

2

2

0

3

1 2

P x  x  1 x 3  3x 2  x  2 . Continuing with the quotient: 1 1 3 1 2 1 4

1 1

5

2 1

1

1 2

1 4 5

 x  1 is a zero. 3

1 2

2 2

2

3  x  1 is an 1 2 1 1 1 1 1 0  x  2 is a zero. upper bound.   So P x  x  1 x  2 x 2  x  1 . Using the Quadratic Formula on the third factor, we have    1  12  4 1 1 1  5 5. x  . Therefore, the zeros are 1, 2, and 1 2 2 1 2

49. P x  x 4  7x 3  14x 2  3x  9. The possible rational zeros are 1, 3, 9. P x has 3 variations in sign and hence 1 or 3 positive real zeros. P x  x 4  7x 3  14x 2  3x  4 has 1 variation in sign and hence P has 1 negative real zero. 1

1

7

1

1

14

6

3

8

9

1

3

14

7

5

3

3

12

6

2

3

3

3

3

1

1

9 9

6 8 5 4 1 4 2 3 0  x  3 is a zero.   So P x  x  3 x 3  4x 2  2x  3 . Since the constant term of the second term is 3, 9 are no longer possible zeros.

Continuing by factoring the quotient, we have:

3

1 1

4

0

 x  3 is a zero again.

So P x  x  32 x 2  x  1 . Using the Quadratic Formula on the second factor, we have:    1 12 411 1 5 . Therefore, the zeros are 3 and 1 5 .  x 2 2 21


SECTION 3.4 Real Zeros of Polynomials

51

50. P x  x 5  4x 4  x 3  10x 2  2x  4. The possible rational zeros are 1, 2, 4. P x has 3 variations in sign and hence 1 or 3 positive real zeros. P x  x 5  4x 4  x 3  10x 2  2x  4 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1

1

4 1

1

1

10

2

3

4

6

2

4

8 1

3

is 2, 4 are no longer possible zeros. 1

2

1

0

2

0

2

1

10

2

4

10

0

0

5

4

4

2

0

 x  2 is a zero.

Since the constant term of the second factor

Continuing by factoring the quotient, we have:

0

2

10

20

1

1

5

4

2

4 6 8 4   So P x  x  2 x 4  2x 3  5x 2  2 . 2

1

1

1

2 1

5 3

0

2

2

2

5 10 18 1 3 2 2 0  x  1 is a zero.   So P x  x  2 x  1 x 3  3x 2  2x  2 . Continuing by factoring the quotient, we have:

1

3

1

2

4

2

2

4 2 0  x  1 is a zero again.  So P x  x  2 x  12 x 2  4x  2 . Using the Quadratic Formula on the second factor, we have: 

     4 42 412 4 8  42 2  2  2. Therefore, the zeros are 1, 2, and 2  2.  x 21 2 2

51. P x  4x 3  6x 2  1. The possible rational zeros are 1,  12 ,  14 . P x has 2 variations in sign and hence 0 or 2 positive real zeros. P x  4x 3  6x 2  1 has 1 variation in sign and hence P has 1 negative real zero. 1

4

6 4

So P x 

0

1

2

2

1 2

4

6

2

0

1

2

1

4 2 2 1 4 4 2 0  x  12 is a zero.   x  12 4x 2  4x  2 . Using the Quadratic Formula on the second factor, we have:

     4 42 442 4 48  44 3  1 3 . Therefore, the zeros are 1 and 1 3 .  x 8 8 2 2 2 24


52

CHAPTER 3 Polynomial and Rational Functions

52. P x  3x 3  5x 2  8x  2. The possible rational zeros are 1, 2,  13 ,  23 . P x has 1 variation in sign and hence 1 positive real zero. P x  3x 3  5x 2  8x  2 has 2 variations in sign and hence P has 0 or 2 negative real zeros. 1

1 3

3

5 3

8

2

3

2

10

2

10

12

3

5

8

2

 43

1

3

2

 28 3

5

8

2

0

2

5

8 2

3

1

12

6

14

3

5

6

2 3

 28 9

4

3

 46 9

3

8

3

3

8  13

2

0

3

5

8

3

2

3

10

 26 3

3

5

8

2

3

11

6

22

14

28 30

2

2

1

2

 20 3

2

Thus we have tried all the positive rational zeros, so we try the negative zeros. 1

8

2

2

3 6 6 0  x   13 is a zero.       So P x  x  13 3x 2  6x  6  3 x  13 x 2  2x  2 . Using the Quadratic Formula on the second factor, we      2 22 412  22 12  222 3  1  3. Therefore, the zeros are  13 and 1  3. have: x  21

53. P x  2x 4  15x 3  17x 2  3x  1. The possible rational zeros are 1,  12 . P x has 1 variation in sign and hence 1 positive real zero. P x  2x 4  15x 3  17x 2  3x  1 has 3 variations in sign and hence P has 1 or 3 negative real zeros. 1 2

2 2  12

15

17

3

1

8

16

25

25 2 31 2

2

15

17

3

1

7

5

1 31 4 27 4

 x  12 is an upper bound. 1

1

14 10 2 0  x   12 is a zero.       So P x  x  12 2x 3  14x 2  10x  2  2 x  12 x 3  7x 2  5x  1 . 2

1

1

7

5

1

6

1 1

1 6 1 0  x  1 is a zero.       So P x  x  12 2x 3  14x 2  10x  2  2 x  12 x  1 x 2  6x  1 Using the Quadratic Formula on the 

   6 62 411 10  3  10. Therefore, the zeros are 1,  1 , third factor, we have x   62 40  62 21 2 2

and 3 

 10.


53

SECTION 3.4 Real Zeros of Polynomials

54. P x  4x 5  18x 4  6x 3  91x 2  60x  9. The possible rational zeros are 1, 3, 9,  12 ,  32 ,  92 ,  14 ,  34 ,  94 .

P x has 4 variations in sign and hence 0 or 2 or 4 positive real zeros. P x  4x 5  18x 4  6x 3  91x 2  60x  9 has 1 variation in sign and hence P has 1 negative real zero. 1

4

18

4

14

4

6

91

14

20

9

60

71

3

4

6

91

18

72

18

1

12

9

60 57

9

20 71 11 10 4 6 24 19 3 0  x  3 is a zero.   So P x  x  3 4x 4  6x 3  24x 2  19x  3 . Continuing by factoring the quotient, we have: 4

3

6

24

12

18

19

3

18

3

4 6 6 1 0  x  3 is a zero again.   So P x  x  32 4x 3  6x 2  6x  1 . Continuing by factoring the quotient, we have: 3

4

6

12

1 2

1

6

54

144

4

6

2

6

4

1

1

48 1445  x  3 is an upper bound. 4 8 2 0  x  12 is a zero.       So P x  x  32 x  12 4x 2  8x  2  2 x  32 x  12 2x 2  4x  1 . Using the Quadratic Formula on     2 421 the second factor, we have x  4 422  44 24  1  26 . Therefore, the zeros are 12 , 3, and 1  26 . 4

18

55. (a) P x  x 3  3x 2  x  3 has possible rational zeros 1, 3. 1

1

3

1

4

3

4

3

1

3

1

0

(b)

y

1

x  1 is a zero.

  So P x  x  1 x 2  4x  3   x  1 x  3 x  1.

1

x

1

x

The real zeros of P are 3, 1, and 1.

56. (a) P x  x 3  3x 2  6x  8 has possible rational zeros 1, 2, 4,

(b)

y

8. 1

1

3 1

1 2

1

4

8

2

10

3

6

8

4

0

5

10

2

2

4

2 1

6

8  x  2 is a zero.

  So P x  x  2 x 2  5x  4  x  2 x  4 x  1. The real zeros of P are 4, 1, and 2.


54

CHAPTER 3 Polynomial and Rational Functions

57. (a) P x  2x 3  7x 2  4x  4 has possible rational zeros 1, 2, 4,

(b)

y

 12 .

1 2 7

4

4

2 2 7

2 5 1

4

4

4 6 4

2 5 1 3

2 3 2 0  x  2 is a zero.   So P x  x  2 2x 2  3x  2 . Continuing: 2

2

3

2

1

0

4

2

1 x

1

2  x  2 is a zero again.

Thus P x  x  22 2x  1. The real zeros of P are 2 and  12 . 58. (a) P x  3x 3  17x 2  21x  9 has possible rational zeros 1, 3,

(b)

y 2

9,  13 ,  23 . 3

1

1

17

21

3

20

3

20

41

1 3

3

17 1

6

3

18

27

x

9

41 32

21

 x  1 is an upper bound. 9

9  x  13 is a zero.

0

        So P x  x  13 3x 2  18x  27  3 x  13 x 2  6x  9  3 x  13 x  32 . The real zeros of P are 3 and 13 .

59. (a) P x  x 4  5x 3  6x 2  4x  8 has possible rational zeros 1,

(b)

y

2, 4, 8. 1

1

5

1

2

6

4

4

2

8

6

1

4

2

6

2

1

5

6

4

6

0

8

2

1

3

0

2 1

x

8

4

0

 x  2 is a zero.

  So P x  x  2 x 3  3x 2  4 and the possible rational zeros are restricted to 1, 2, 4. 2

1

3

2

0

4

2

4

1 1 2 0  x  2 is a zero again.  P x  x  22 x 2  x  2  x  22 x  2 x  1  x  23 x  1. So the real zeros of P are 1 and 2. 


55

SECTION 3.4 Real Zeros of Polynomials

60. (a) P x  x 4  10x 2  8x  8 has possible rational zeros 1, 2,

(b)

y

4, 8. 1 1

0 10 1 1

1 1 4 1

0

8 8

9 17

9 17

10

9 8

0 10

2 1

8 8

2 4 12 40

6 20 32

1 2

5

8

1

4 16 24 64

6 16 72  x  4 is an upper bound.

1 4 1

x

1

0

10

8

1

1

9

8

1

2

1

0

10

8

2

4

12

1 9 1 7 1 2   So P x  x  2 x 3  2x 2  6x  4 . Continuing, we have: 1

2

1

2

6

2

4

8

0

 x  2 is a zero.

4

4

8

4

6

8

 x  2 is a zero again.   P x  x  22 x 2  4x  2 . Using the Quadratic Formula on the second factor, we have      2 412 8  42 2  2  2. So the real zeros of P are 2 and 2  2.  4 x  4 421 2 2 1

0

2

61. (a) P x  x 5  x 4  5x 3  x 2  8x  4 has possible rational zeros

(b)

y

1, 2, 4. 1

1

1

1

2

1

1

2

1

1

8

4

0

5

4

4

5

4

5

1

0

5

1

8

4

6

10

4

2

1

4

8 1 1

3 5 2 0  x  2 is a zero.   So P x  x  2 x 4  x 3  3x 2  5x  2 , and the possible rational zeros are restricted to 1, 2. 2

1

1

1

1

3

5

2

3

3

1

0

1

3

3

1

1

2

1

2

6

6

2

 x  2 is a zero again.  So P x  x  22 x 3  3x 2  3x  1 , and the possible rational zeros are restricted to 1. 

1

2 1 0  x  1 is a zero.  So P x  x  22 x  1 x 2  2x  1  x  22 x  13 ., and the real zeros of P are 1 and 2. 

x


56

CHAPTER 3 Polynomial and Rational Functions

62. (a) P x  x 5  2x 4  8x 3  16x 2  16x  32 has possible rational

(b)

y

zeros 1, 2, 4, 8, 16. 1

1

2

8

1 1 2

1

2 2

1

4

8

3

3

5

0

0

16

32

5

21

5

21

5

16

32

32

32

16

8

16

10 1

27

x

 x  2 is a zero.  So P x  x  2 x 4  4x 3  16x  16 . Continuing: 

16

0

16

2

1 1

4

0

2

12

6

12

16

24

16

16

8

0  x  2 is a zero again.  Thus, P x  x  22 x 3  6x 2  12x  8  x  22 x  23 , and the real zeros of P are 2. 

63. P x  x 3  x 2  x  3. Since P x has 1 variation in sign, P has 1 positive real zero. Since P x  x 3  x 2  x  3 has 2 variations in sign, P has 2 or 0 negative real zeros. Thus, P has 1 or 3 real zeros. 64. P x  2x 3  x 2  4x  7. Since P x has 3 variations in signs, P has 3 or 1 positive real zeros. Since P x  2x 3  x 2  4x  7 has no variation in sign, there is no negative real zero. Thus, P has 1 or 3 real zeros. 65. P x  2x 6  5x 4  x 3  5x  1. Since P x has 1 variation in sign, P has 1 positive real zero. Since P x  2x 6  5x 4  x 3  5x  1 has 1 variation in sign, P has 1 negative real zero. Therefore, P has 2 real zeros. 66. P x  x 4  x 3  x 2  x  12. Since P x has no variations in sign, P has no positive real zero. Since P x  x 4  x 3  x 2  x  12 has 4 variations in sign, P has 4, 2, or 0 negative real zeros. Therefore, P x has 0, 2, or 4 real zeros. 67. P x  x 5  4x 3  x 2  6x. Since P x has 2 variations in sign, P has 2 or 0 positive real zeros. Since P x  x 5  4x 3  x 2  6x has no variation in sign, P has no negative real zero. Therefore, P has a total of 1 or 3 real zeros (since x  0 is a zero, but is neither positive nor negative). 68. P x  x 8  x 5  x 4  x 3  x 2  x  1. Since P x has 6 variations in sign, the polynomial has 6, 4, 2, or 0 positive real zeros. Since P x has no variation in sign, the polynomial has no negative real zeros. Therefore, P has 6, 4, 2, or 0 real zeros. 69. P x  2x 3  5x 2  x  2; a  3, b  1 3

2 2

1

2 2

5

1

6

3

1

4

5

1

2

7

7

8

2 12 14

alternating signs  lower bound.

2

8 6

Therefore a  3 and b  1 are lower and upper bounds.

all nonnegative  upper bound.


SECTION 3.4 Real Zeros of Polynomials

70. P x  x 4  2x 3  9x 2  2x  8; a  3, b  5 1

3

2

1 5

1

2

8

18

48

9 15

3 5

6

2

9

2

8

15

30

160

3

6

32

168

5

1

56

16

Alternating signs  lower bound.

All nonnegative  upper bound.

Therefore a  3 and b  5 are lower and upper bounds.

71. P x  8x 3  10x 2  39x  9; a  3, b  2 8

3

10

8 2

42

14

3

10

39

8

16 8

9

39

24

26

57

9

0

alternating signs  lower bound.

9

52

26

13

35

all nonnegative  upper bound.

Therefore a  3 and b  2 are lower and upper bounds. Note that x  3 is also a zero.

72. P x  3x 4  17x 3  24x 2  9x  1; a  0, b  6 0

3

3

24

0

0

17

9

1

0

0

24

9

1

17

24

18

6

9

180

1026

1

30

171

1027

3 6

17

3

Alternating signs  lower bound. 1 All nonnegative  upper bound.

Therefore a  0 and b  6 are lower and upper bounds. Note that because P x alternates in sign, by Descartes’ Rule of Signs, 0 is automatically a lower bound.

73. P x  x 4  2x 3  3x 2  5x  1; a  2, b  1 2

1 1

1

1 1

2

3

5

2

0

6

0

3

2

3

1

3

6

3

6

11

1 2

1

1 5

Alternating signs  lower bound.

1

11 10

Therefore a  2 and b  1 are lower and upper bounds.

All nonnegative  upper bound.


58

CHAPTER 3 Polynomial and Rational Functions

74. P x  x 4  3x 3  4x 2  2x  7; a  4, b  2 1

4

3

1 2

4

2

7

0

2

1

4

4 1

1

3 2

1

0

8

4

2

7

6

10

13

10

5

12

20 All nonnegative  upper bound.

Therefore a  4 and b  2 are lower and upper bounds.

75. P x  2x 4  6x 3  x 2  2x  3; a  1, b  3 2

1

3

2

8

2

2 3

1

6

11

9

8

9

11

14

6

1

2

3

3

3

1

6

2 2

6

0

0

1

Alternating signs  lower bound.

All nonnegative  upper bound.

Therefore a  1 and b  3 are lower and upper bounds.

76. P x  3x 4  5x 3  2x 2  x  1; a  1, b  2 3

1

5

8

3

3 2

2 6

5

2 2

0

1

0

1

3

6

3

1

1 5

6

8

Alternating signs  lower bound.

4

5 1

Alternating signs  lower bound.

1

2 1

Therefore a  1 and b  2 are lower and upper bounds.

All nonnegative  upper bound.

77. P x  x 3  3x 2  4 and use the Upper and Lower Bounds Theorem: 1

1

1 3

1

3 1 4

4

4

4

0

0

4

3

0

0

0

0

4

3

1

0

4

alternating signs  lower bound.

all nonnegative  upper bound.

Therefore 1 is a lower bound (and a zero) and 3 is an upper bound. (There are many possible solutions.)

78. P x  2x 3  3x 2  8x  12 and using the Upper and Lower Bounds Theorem: 2

2

3 4

2 3

8 14

12

12

7

6

2

3

8

12

9

3

3

1

15

6

2 (There are many possible solutions.)

0

Alternating signs  x  2 is a lower bound (and a zero).

All nonnegative  x  3 is an upper bound.


SECTION 3.4 Real Zeros of Polynomials

79. P x  x 4  2x 3  x 2  9x  2. 1

1

1

2

1

1

2

9

1

0

9

1

0

9

7

3

1

2

1

3

3

1

4

1 1

1

1

2

1

9

2

12

9

3

11

2

1

2

0

0

1

2

9

2

14

7

12

all positive  upper bound.

2 13

4

4

3

1 1

9

3

1

2

15

13

alternating signs  lower bound.

Therefore 1 is a lower bound and 3 is an upper bound. (There are many possible solutions.)

80. Set P x  x 5  x 4  1. 1

1 1

1

1 1

0

0

0

1

1

0

0

0

0

0

0

0

0

1

0

0

0

1

2

2

2

2

1

1 1 2

2

(There are many possible solutions.)

81. P x  2x 4  3x 3  4x 2  3x  2. 1

2

2

2

3 2

  P x  x  1 2x 3  5x 2  x  2

1

2

Alternating signs  x  1 is a lower bound.

1

4

5

5

2

All nonnegative  x  1 is an upper bound.

1

2

3

1

2 0

2

5

1

2

3

 x  1 is a zero.

2 2

2 3 2 0  x  1 is a zero.   P x  x  1 x  1 2x 2  3x  2  x  1 x  1 2x  1 x  2. Therefore, the zeros are 2, 12 , 1.

59


60

CHAPTER 3 Polynomial and Rational Functions

82. P x  2x 4  15x 3  31x 2  20x  4. The possible rational zeros are 1, 2, 4,  12 . Since all of the coefficients are positive, there are no positive zeros. Since P x  2x 4  15x 3  31x 2  20x  4 has 4 variations in sign, there are 0, 2, or 4 negative real zeros. 1 2 15

31

20

4

2 2 15

2 13 18 2

2 13

18

2

2

9

2 11

2

2

4 2 11

4 14 10 7

20

4

4 22 18 4

  P x  x  2 2x 3  11x 2  9x  2 : 2 2 11

31

3

2

 12 2 11

2

8 12 12

2

5 12

9

9

0  x  2 is a zero.

9

2

1 5 2

3 14

2 10

4

0 x   12 is a zero.

    P x  x  2 2x  1 x 2  5x  2 . Now if x 2  5x  2  0, then x  5 25412  52 17 . Thus, the zeros 2 

are 2,  12 , and 52 17 .

83. Method 1: P x  4x 4  21x 2  5 has 2 variations in sign, so by Descartes’ rule of signs there are either 2 or 0 positive zeros. If we replace x with x, the function does not change, so there are either 2 or 0 negative zeros. Possible rational zeros are 1,  12 ,  14 , 5,  52 ,  54 By inspection, 1 and 5 are not zeros, so we must look for non­integer solutions: 1 2

4

0 2

4

2

21

0

5

1

10

5

20

10

0

 x  12 is a zero.

   P x  x  12 4x 3  2x 2  20x  10 , continuing with the quotient, we have:  12

4 4

P x  x  12



2 2

0

20

10

20

0

0

10

 x   12 is a zero.

    1  2 4x  20  0. If 4x 2  20  0, then x   5. Thus the zeros are  12   5. x 2

Method 2: Substituting u  x 2 , the equation becomes 4u 2  21u  5  0, which factors:     4u 2  21u  5  4u  1 u  5  4x 2  1 x 2  5 . Then either we have x 2  5, so that x   5, or we have   x 2  14 , so that x   14   12 . Thus the zeros are  12   5.


SECTION 3.4 Real Zeros of Polynomials

61

  84. P x  6x 4  7x 3  8x 2  5x  x 6x 3  7x 2  8x  5 . So x  0 is a zero. Continuing with the quotient,

1 Q x  6x 3  7x 2  8x  5. The possible rational zeros are 1, 5,  12 ,  52 ,  ,  53 ,  16 ,  56 . Since Q x has 2 3

variations in sign, there are 0 or 2 positive real zeros. Since Q x  6x 4  7x 3  8x 2  5x has 1 variation in sign, there is 1 negative real zero. 6

1

6

5

6

8 1

9

1

9

4

7

1 2

6

5

7

8

5

30

115

535

6

23

107

540

6

7

5

3

8

6

2

5

4

10

0

All positive  upper bound.

 x  12 is a zero.

  P x  x 2x  1 3x 2  2x  5  x 2x  1 3x  5 x  1. Therefore, the zeros are 0, 1, 12 and 53 .

85. P x  x 5  7x 4  9x 3  23x 2  50x  24. The possible rational zeros are 1, 2, 3, 4, 6, 8, 12, 24. P x has 4 variations in sign and hence 0, 2, or 4 positive real zeros. P x  x 5  7x 4  9x 3  23x 2  50x  24 has 1 variation in sign, and hence P has 1 negative real zero. 1

1

7 1

1

6

9

23

6

3

3

26

50

26

24

24 24

0

 x  1 is a zero.

  P x  x  1 x 4  6x 3  3x 2  26x  24 ; continuing with the quotient, we try 1 again. 1

1 1

3

26

1

5

2

5

2

24

1

5

2

24

6

16

6

24

24 0

 x  1 is a zero again.

  P x  x  12 x 3  5x 2  2x  24 ; continuing with the quotient, we start by trying 1 again. 1

1 1

2

24

1

4

6

4

6

5

18

2

2

1

3

8

8

3

1 1

2

24

3

6

24

2

8

5

0

 x  3 is a zero.

  P x  x  12 x  3 x 2  2x  8  x  12 x  3 x  4 x  2. Therefore, the zeros are 2, 1, 3 4.


62

CHAPTER 3 Polynomial and Rational Functions

1 86. P x  8x 5  14x 4  22x 3  57x 2  35x  6. The possible rational zeros are 1, 2, 3, 6,  , 2 1  32 ,  14 ,  34 ,  ,  38 . Since P x has 4 variations in sign, there are 0, 2, or 4 positive real zeros. Since 8 P x  8x 5  14x 4  22x 3  57x 2  35x  6 has 1 variation in sign, there is 1 negative real zero. 1

8

14

8

22

22

57

22

0

0

57

8

35 57

6

2

92

92

98

8

30

8

14

8

30

22

57

38

19

60

16

6

35 38

76

6

3

0

 x  2 is a zero.

  P x  x  2 8x 4  30x 3  38x 2  19x  3 . All the other real zeros are positive. 1

38

8

8

16

22

16

22

8

16

22 8

8 Since f

  1 2

14

2

14

3

0

3

  P x  x  2 x  1 8x 3  22x 2  16x  3 . 1

3

19

 x  1 is a zero.

1 2

3

8

2

22 4

8

1

16

3 7 2 1 2

9

7

18

 0  f 1, there must be a zero between 12 and 1. We try 34 : 3 4

8

6

8

16

22

12 4

16

3

3 0

 x  34 is a zero.

    P x  x  2 x  1 4x  3 2x 2  4x  1 . Now, 2x 2  4x  1  0 when x  4 16421  22 2 . Thus, the 22 

zeros are 1, 34 , 2, and 22 2 .

87. P x  x 3  x  2. The only possible rational zeros of P x are 1 and 2. 1

1

0 1

1

1

1

2

0

2

1

2

1

0

0 2

1

2

1

2

3

4

4

6

1

1 1

0 1 1

1

2

0

2

1

0

Since the row that contains 1 alternates between nonnegative and nonpositive, 1 is a lower bound and there is no need to try 2. Therefore, P x does not have any rational zeros.


SECTION 3.4 Real Zeros of Polynomials

63

88. P x  2x 4  x 3  x  2. The only possible rational zeros of P x are 1, 2,  12 . 1 2

2

1

2 2

1

1

2 1 2 5 2

1

0

0

0

0

1

1 2

2

0

3

0

1

2

3

3

2

3

2  12

All nonnegative  x  12 is an upper bound.

4

Alternating signs  x  1 is a lower bound. 2 2

Therefore, there is no rational zero.

1 1 2

0

1

2

1

 12 1 2

 14 7 4

1

89. P x  3x 3  x 2  6x  12 has possible rational zeros 1, 2, 3, 4, 6, 12,  13 ,  23 ,  43 . 3 1

3

1 2

2

3

5

1

3

4

2

3

7

12

6

3 1 3 2 3 4 3  13  23  43

8

4 4

20

2

14

8

4

all positive  x  2 is an upper bound alternating signs  x  2 is a lower bound

Therefore, there is no rational zero.

3

1 0

6 6

 16 3

3

1

3

3

3

2

 16 3

3

4

3 3

5

2

2 3

12 10 76 9 28 3 124 9 44 3 100 9

90. P x  x 50 5x 25 x 2 1. The only possible rational zeros of P x are 1. Since P 1  150 5 125 12 1  4 and P 1  150  5 125  12  1  6, P x does not have a rational zero.

91. P x  x 3  3x 2  4x  12, [4 4] by [15 15]. The

possible rational zeros are 1, 2, 3, 4, 6, 12. By

observing the graph of P, the rational zeros are x  2, 2, 3.

92. P x  x 4  5x 2  4, [4 4] by [30 30].

The possible rational solutions are 1, 2, 4.

By observing the graph of the equation, the solutions of the given equation are x  1, 2.

10

­4

­2

20

2 ­10

4

­4

­2

2 ­20

4


64

CHAPTER 3 Polynomial and Rational Functions

93. P x  2x 4  5x 3  14x 2  5x  12, [2 5] by

94. P x  3x 3  8x 2  5x  2, [3 3] by [10 10]

[40 40]. The possible rational zeros are 1, 2, 3,

The possible rational solutions are 1, 2,  13 ,  23 . By

4, 6, 12,  12 ,  32 . By observing the graph of P, the

observing the graph of the equation, the only real solution of the given equation is x  2.

zeros are  32 , 1, 1, 4.

10

40 20 ­2

2

­20

­2

4

2 ­10

­40

95. x 4  x  4  0. Possible rational solutions are 1, 2, 4. 1

1 1

1

0

0

1

1

1

1

1

0

1

1

1

1

4

0

4

1

0 1

1

0

0

2

4

1

2

1

0

1

2

0

1

1

2

4

1

2

2

2

2

2

1

4

4

7

10

0

1

4

8

4

14

18

8

4

 x  2 is an upper bound.

14

9

 x  2 is a lower bound.

Therefore, we graph the function P x  x 4  x  4 in the viewing rectangle [2 2] by [5 20] and see there are two

solutions. In the viewing rectangle [13 125] by [01 01], we find the solution x  128. In the viewing rectangle

[1516] by [01 01], we find the solution x  153. Thus the solutions are x  128, 153. 20 10 ­1.30 ­2

­1

1

­1.28

­1.26

2

0.1

0.1

0.0

0.0

­0.1

­0.1

1.55

1.60

96. 2x 3  8x 2  9x  9  0. Possible rational solutions are 1, 3, 9,  12 ,  32 ,  92 . 1

2

8

2

2

6

9 6 3

9

3

2

3

6

6

2

9

8

6

2

3

9 9 0

 x  3 is a zero.

We graph P x  2x 3  8x 2  9x  9 in the viewing rectangle [4 6] by

[40 40]. It appears that the equation has no other real solution. We can factor   2x 3  8x 2  9x  9  x  3 2x 2  2x  3 . Since the quotient is a quadratic expression, we can use the Quadratic Formula to locate the other possible  2 423 , which are not real. So the only solution is x  3. solutions: x  2 222

40 20 ­4

­2 ­20 ­40

2

4

6


SECTION 3.4 Real Zeros of Polynomials

65

97. 400x 4  400x 3  1096x 2  588x  909  0. 1 4 4 1096 588 4

4 8 2 4

909

8 296 884

296 884

025

4 1096 588

909

8

8

4 4

2 4

8

4 12 3 4

592 008

004

296

4 1096 588

901

4

24 2608 4040

1304

4 1096 12 8

909

202 4949  x  2 is an upper bound. 588

24 3912

1304

909 135

45 14409  x  3 is a lower bound.

Therefore, we graph the function P x  400x 4  400x 3  1096x 2  588x  909 in the viewing rectangle [3 2] by [10 40]. There appear to be two solutions. In the viewing rectangle [16 14] by [01 01], we find the solution x  150. In the viewing rectangle [08 12] by [0 1], we see that the graph comes close but does not go through the x­axis. Thus there is no solution here. Therefore, the only solution is x  150. 40

0.1

20 ­1.6 ­3

­2

­1

1

1.0 0.5

0.0

­1.5

0.0

2

­0.1

0.8

0.9

1.0

1.1

1.2

98. x 5  2x 4  096x 3  5x 2  10x  48  0. Since all the coefficients are positive, there is no positive solution. So x  0 is an upper bound. 2 1 1

2 096 2

5

10

48

0 192 616 768

0 096

308

3 1

384 288

2 096 3

5

10

48

3 1188 2064 9192

1 1 396

688 3064 8712  x  3 is a lower bound.

Therefore, we graph P x  x 5  2x 4  096x 3  5x 2  10x  48 in the viewing rectangle [3 0] by [10 5] and see that there are three possible solutions. In the viewing rectangle [175 17] by [01 01], we find the solution x  171. In the viewing rectangle [125 115] by [01 01], we find the solution x  120. In the viewing

rectangle [085 075] by [01 01], we find the solution x  080. So the solutions are x  171, 120, 080. 0.1 ­2

0.1

0.1

0 ­1.74 ­10

­1.72

0.0 ­0.1

­1.25

­1.20

0.0 ­0.1

­0.85

­0.80

0.0 ­0.1


66

CHAPTER 3 Polynomial and Rational Functions

  99. Let r be the radius of the silo. The volume of the hemispherical roof is 12 43 r 3  23 r 3 . The volume of the cylindrical   section is  r 2 30  30r 2 . Because the total volume of the silo is 15,000 ft3 , we get the following equation: 2 r 3  30r 2  15000  2 r 3  30r 2  15000  0  r 3  45r 2  22500  0. Using a graphing device, we 3 3

first graph the polynomial in the viewing rectangle [0 15] by [10000 10000]. The solution, r  1128 ft., is shown in the viewing rectangle [112 114] by [1 1]. 1

10000

0

11.3

0

11.4

­1

10

­10000

100. The volume of the box is V  1500  x 20  2x 40  2x  4x 3  120x 2  800x    4x 3  120x 2  800x  1500  4 x 3  30x 2  200x  375  0. Clearly, we must have 20  2x  0, and so 0  x  10. 5

1

30 5

200

375

125

375

1 25 75 0  x  5 is a zero.   x 3  30x 2  200x  375  x  5 x 2  25x  75  0. Using the Quadratic Formula, we find the other zeros: 

 252 325  2552 13 . Since 2552 13  10, the two answers are: x  height  5 cm, x  25 6254175 2 

width  20  2 5  10 cm, and length  40  2 5  30 cm; and x  height  2552 13  349 cm,         width  20  25  5 13  5 13  5  1303 cm, and length  40  25  5 13  15  5 13  3303 cm. 101. Let r be the radius of the cone and cylinder and let h be the height of the cone. Since the height and diameter are equal, we get h  2r. So the volume of the

cylinder is V1  r 2  cylinder height  20r 2 , and the volume of the cone is V2  13 r 2 h  13 r 2 2r  23 r 3 . Since the total volume is

500 , it follows 3

500  r 3  30r 2  250  0. By Descartes’ Rule of that 23 r 3  20r 2  3 Signs, there is 1 positive zero. Since r is between 276 and 2765 (see the table), the radius should be 276 m (correct to two decimals).

r 1 2 3 27 276 277

r 3  30r 2  250 219 122

47

1162 233

144

2765

144

28

715

 102. (a) Let x be the length, in ft, of each side of the base and let h be the height. The volume of the box is V  2 2  hx 2 ,    and so hx 2  2 2. The length of the diagonal on the base is x 2  x 2  2x 2 , and hence the length of the diagonal  between opposite corners is 2x 2  h 2  x  1. Squaring both sides of the equation, we have 2x 2  h 2  x 2  2x  1      h 2  x 2  2x  1  h  x 2  2x  1. Therefore, 2 2  hx 2  x 2  2x  1 x 2    x 2  2x  1 x 4  8  x 6  2x 5  x 4  8  0.


SECTION 3.4 Real Zeros of Polynomials

67

(b) We graph y  x 6  2x 5  x 4  8 in the viewing rectangle [0 5] by [10 10], and we see that there are two solutions. In the second viewing rectangle, [14 15] by [1 1], we see the solution x  145. The third viewing rectangle, [225 235] by [1 1], shows the solution x  231. 10

1

0

2

1

0

4

­10

1.45

0

1.50

­1

2.30

2.35

­1

 If x  width  length  145 ft, then height  x 2  2x  1  134 ft, and if x  width  length  231 ft, then  height  x 2  2x  1  053 ft.

103. Let b be the width of the base, and let l be the length of the box. Then the length plus girth is l  4b  108, and the volume

is V  lb2  2200. Solving the first equation for l and substituting this value into the second equation yields l  108  4b    V  108  4b b2  2200  4b3  108b2  2200  0  4 b3  27b2  550  0. Now P b  b3  27b2  550

has two variations in sign, so there are 0 or 2 positive real zeros. We also observe that since l  0 b  27, so b  27 is an upper bound. Thus the possible positive rational real zeros are 1 2 3 10 11 22 25. 1

1

27

0

550

1

1

26

26

26

26

524

1

27

5

5

2

0

550

110

550

1

27

0

550

1

2

50

100

25

50

450

1 22 110 0  b  5 is a zero.     P b  b  5 b2  22b  110 . The other zeros are b  22 48441110  222 924  2230397 . The positive 2 2

answer from this factor is b  2620Thus we have two possible solutions, b  5 or b  2620. If b  5, then

l  108  4 5  88; if b  2620, then l  108  4 2620  320. Thus the length of the box is either 88 in. or 320 in.

104. Given that x is the length of a side of the rectangle, we have that the length of the diagonal is x  10, and the length of   the other side of the rectangle is x  102  x 2 . Hence x x  102  x 2  5000  x 2 20x  100  25,000,000  2x 3  10x 2  2,500,000  0  x 3  5x 2  1,250,000  0. The first viewing rectangle, [0 120] by [100 500],

shows there is one solution. The second viewing rectangle, [106 1061] by [01 01], shows the solution is x  10608.

Therefore, the dimensions of the rectangle are 47 ft by 106 ft.

0.1

400 200 0

0.0 50

100

106.05

106.10

­0.1

105. (a) An odd­degree polynomial must have a real zero. The end behavior of such a polynomial requires that the graph of the polynomial heads off in opposite directions as x   and x  . Thus the graph must cross the x­axis. (b) There are many possibilities one of which is P x  x 4  1.


68

CHAPTER 3 Polynomial and Rational Functions

     (c) P x  x x  2 x  2  x 3  2x.          (d) P x  x  2 x  2 x  3 x  3  x 4  5x 2  6. If a polynomial with integer coefficients has no real zeros, then the polynomial must have even degree.

a for x we have 3    a 3 a 2 a x 3  ax 2  bx  c  u  c a u b u 3 3 3   a3 a2 ab a2 2a 3 2 2  u  au  u  a u  u  bu  c 3 27 3 9 3

106. (a) Substituting u 

a3 a3 ab a2 2a 2 u  au 2  u  bu  c 3 27 3 9 3     2a 2 a3 a3 ab a2 3 2  b u    c  u  a  a u  3 3 27 9 3     a2 2a 3 ab 3  u  b u  c 3 27 3  u 3  au 2 

(b) x 3  6x 2  9x  4  0. Setting a  6, b  9, and c  4, we have: u 3 

62 9 3

u  16  18  4  u 3  3u  2.

        2 3   3 3 2 2 2 22 3 33         3 1  3 1  1  1  2 107. (a) Using the cubic formula, x  2 4 27 2 4 27 2

1

0

2

3 4

2

2

1 2 1 0  So x  2 x 2  2x  1  x  2 x  12  0  x  2, 1. Using the methods from this section, we have 

1

1

0

1

3

1

2

2

1 1 2 0  So x 3  3x  2  x  1 x 2  x  2  x  12 x  2  0  x  2, 1. 

Since this factors easily, the factoring method was easier.

(b) Using the cubic formula,         2 3 3 3 4 3 33 4 42  4       x  2 4 27 2 4 27         3 3 3 3 2  4  1  2  4  1  2  5  2  5 


SECTION 3.4 Real Zeros of Polynomials

Using a graphing calculator, we see that P x  x 3  3x  4 has one zero. Using methods from this section, P x has possible rational zeros 1, 2, 4. 1

1

1

0

3

4

1

1

4

1

4

4

1

1

0

3

4

1

1

4

69

 1 is an upper bound. 1 1 4 0  x  1 is a zero.   P x  x 3  3x  4  x  1 x 2  x  4 . Using the Quadratic Formula we have

  1 12 414  1 215 which is not a real number. Since it is not easy to see that x  2

    3 3 2  5  2  5  1, we see that the factoring method was much easier.

108. (a) Since z  b, we have z  b  0. Since all the coefficients of Q x are nonnegative, and since z  0, we have Q z  0 (being a sum of positive terms). Thus, P z  z  b  Q z  r  0, since the sum of a positive number and a nonnegative number. (b) In part (a), we showed that if b satisfies the conditions of the first part of the Upper and Lower Bounds Theorem and z  b, then P z  0. This means that no real zero of P can be larger than b, so b is an upper bound for the real zeros. (c) Suppose b is a negative lower bound for the real zeros of P x. Then clearly b is an upper bound for P1 x  P x. Thus, as in Part (a), we can write P1 x  x  b  Q x  r, where r  0 and the coefficients of Q are all nonnegative, and P x  P1 x  x  b  Q x  r  x  b  [Q x]  r . Since the coefficients of Q x are all nonnegative, the coefficients of Q x will be alternately nonpositive and nonnegative, which proves the second part of the Upper and Lower Bounds Theorem. 109. P x  x 5  x 4  x 3  5x 2  12x  6 has possible rational zeros 1, 2, 3, 6. Since P x has 1 variation in sign,

there is 1 positive real zero. Since P x  x 5  x 4  x 3  5x 2  12x  6 has 4 variations in sign, there are 0, 2, or 4 negative real zeros. 1

1

1

1

1

0

1

0

1

5

12

6

1

6

18

6

18

24

2

1

6 15

1 2

1

3 1 1 1 5 12 6 3

1

1

1

1

5 2

3

12

6

6

36

18

42

1 1 1 1 5 12 6

30 54

1

2

2 1

5 10 18 48  3 is an upper bound. 1 2 1 6   P x  x  1 x 4  2x 3  x 2  6x  6 , continuing with the quotient we have 1

2

1 1

2 1 3

1

3 4

6 4 10

6

6

6

0  x  1 is a zero.

6 10

4

 1 is a lower bound.

Therefore, there is 1 rational zero, namely 1. Since there are 1, 3 or 5 real zeros, and we found 1 rational zero, there must

be 0, 2 or 4 irrational zeros. However, since 1 zero must be positive, there must be at least one irrational zero. Therefore, there is exactly 1 rational zero and there are either 2 or 4 irrational zeros.


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3.5

COMPLEX ZEROS AND THE FUNDAMENTAL THEOREM OF ALGEBRA

1. The polynomial P x  5x 2 x  43 x  7 has degree 6. It has zeros 0, 4, and 7. The zero 0 has multiplicity 2, and the zero 4 has multiplicity 3. 2. (a) If a is a zero of the polynomial P, then x  a must be a factor of P x.

(b) If a is a zero of multiplicity m of the polynomial P, then x  am must be a factor of P x when we factor P completely.

3. A polynomial of degree n  1 has exactly n zeros, if a zero of multiplicity m is counted m times.

4. If the polynomial function P has real coefficients and if a  bi is a zero of P, then a  bi is also a zero of P. So if 3  i is a zero of P, then 3  i is also a zero of P.

5. P x  x 4  1 (a) True, P has degree 4, so by the Zeros Theorem it has four (not necessarily distinct) complex zeros. (b) True, by the Complete Factorization Theorem, this is true.

(c) False, the fourth power of any real number is nonnegative, so P x  1 for all real x and P has no real zeros.

6. P x  x 3  x

  (a) False. P x  x x 2  1 , and x 2  1 has no real zeros. (b) True, P 0  0.

(c) False, P x cannot be factored into linear factors with real coefficients because x 2  1 has no real zeros.   7. (a) x 4  4x 2  0  x 2 x 2  4  0. So x  0 or x 2  4  0. If x 2  4  0 then x 2  4  x  2i. Therefore, the solutions are x  0 and 2i.

(b) To get the complete factorization, we factor the remaining quadratic factor P x  x 2 x  4  x 2 x  2i x  2i.   8. (a) x 5  9x 3  0  x 3 x 2  9  0. So x  0 or x 2  9  0. If x 2  9  0 then x  3i. Therefore, the zeros of P are x  0, 3i.

(b) Since 3i and 3i are the zeros from x 2  9  0, x  3i and x  3i are the factors of x 2  9. Thus the complete   factorization is P x  x 3 x 2  9  x 3 x  3i x  3i.   9. (a) x 3  2x 2  2x  0  x x 2  2x  2  0. So x  0 or x 2  2x  2  0. If x 2  2x  2  0 then   2 22 412 2 4   22i x 2 2 2  1  i. Therefore, the solutions are x  0, 1  i.

(b) Since 1  i and 1  i are zeros, x  1  i  x  1  i and x  1  i  x  1  i are the factors of x 2  2x  2.   Thus the complete factorization is P x  x x 2  2x  2  x x  1  i x  1  i.   10. (a) x 3  x 2  x  0 x x 2  x  1  0. So x  0 or x 2  x  1  0. If x 2  x  1  0 then

    1 12 411  12 3   12  i 23 . Therefore, the zeros of P are x  0,  12  i 23 . x 21   (b) The zeros of x 2  x  1  0 are  12  i 23 and  12  i 23 , so factoring we get x 2  x               x   12  i 23  x  12  i 23 x  12  i 23 . Thus the complete factorization is 1  x   12  i 23        P x  x x 2  x  1  x x  12  i 23 x  12  i 23 .

 2 11. (a) x 4  2x 2  1  0  x 2  1  0  x 2  1  0  x 2  1  x  i. Therefore the zeros of P are x  i.


SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra

71

(b) Since i and i are zeros, x  i and x  i are the factors of x 2  1. Thus the complete factorization is  2 P x  x 2  1  [x  i x  i]2  x  i2 x  i2 .

    12. (a) x 4  x 2  2  0  x 2  2 x 2  1  0. So x 2  2  0 or x 2  1  0. If x 2  2  0 then x 2  2  x   2.  And if x 2  1  0 then x 2  1  x  i. Therefore, the zeros of P are x   2, i. (b) To get the complete factorization, we factor the quadratic factors to get         P x  x 2  2 x 2  1  x  2 x  2 x  i x  i.

     13. (a) x 4  16  0  0  x 2  4 x 2  4  x  2 x  2 x 2  4 . So x  2 or x 2  4  0. If x 2  4  0 then x 2  4  x  2i. Therefore the zeros of P are x  2, 2i.

(b) Since i and i are zeros, x  i and x  i are the factors of x 2  1. Thus the complete factorization is   P x  x  2 x  2 x 2  4  x  2 x  2 x  2i x  2i.  2  14. (a) x 4  6x 2  9  0  x 2  3  0  x 2  3. So x  i 3 are the only zeros of P (each of multiplicity 2). (b) To get the complete factorization, we factor the quadratic factor to get 2    2    2   2  x i 3 .  x i 3 P x  x 2  3  x  i 3 x  i 3

  15. (a) x 3  8  0  x  2 x 2  2x  4  0. So x  2 or x 2  2x  4  0. If x 2  2x  4  0 then

     2 22 414  2 212  22i2 3  1i 3. Therefore, the zeros of P are x  2, 1  i 3. x 2

        (b) Since 1  i 3 and 1  i 3 are the zeros from the x 2  2x  4  0, x  1  i 3 and x  1  i 3 are the factors of x 2  2x  4. Thus the complete factorization is           P x  x  2 x 2  2x  4  x  2 x  1  i 3 x  1i 3       x  2 x  1  i 3 x  1  i 3

  16. (a) x 3  8  0  x  2 x 2  2x  4  0. So x  2 or x 2  2x  4  0. If x 2  2x  4  0 then

   2 22 414 3  1i 3. Therefore, the zeros of P are x  2, 1  i 3.  22 12  22i x 2 2

        (b) Since 1  i 3 and 1  i 3 are the zeros from x 2  2x  4  0, x  1  i 3 and x  1  i 3 are the factors of x 2  2x  4. Thus the complete factorization is           P x  x  2 x 2  2x  4  x  2 x  1  i 3 x  1  i 3       x  2 x  1  i 3 x  1  i 3

       17. (a) x 6  1  0  0  x 3  1 x 3  1  x  1 x 2  x  1 x  1 x 2  x  1 . Clearly, x  1 are solutions. 

If x 2  x  1  0, then x  1 1411  12 3   12  23 so x   12  i 23 . And if x 2  x  1  0, then 2 

 1 2 3  12  x  1 1411 2

    3 1  i 3 . Therefore, the zeros of P are x  1,  1  i 3 , 1  i 3 .  2 2 2 2 2 2 2


72

CHAPTER 3 Polynomial and Rational Functions 

(b) The zeros of x 2  x  1  0 are  12  i 23 and  12  i 23 , so x 2  x  1 factors as              x   12  i 23  x  12  i 23 x  12  i 23 . Similarly,since x   12  i 23 

the zeros of x 2  x  1  0 are 12  i 23 and 12  i 23 , so x 2  x  1 factors as              1 x  12  i 23  x  12  i 23 x  12  i 23 . Thus the complete  i 23 x 2 factorization is     P x  x  1 x 2  x  1 x  1 x 2  x  1

          x  1 x  1 x  12  i 23 x  12  i 23 x  12  i 23 x  12  i 23

       18. (a) x 6  7x 3  8  0  0  x 3  8 x 3  1  x  2 x 2  2x  4 x  1 x 2  x  1 . Clearly, x  1 and     x  2 are solutions. If x 2  2x  4  0, then x  2 4414  22 12   22  2 23 so x  1  i 3. If 2 

 1 2 3  12  23  12  i 23 . Therefore, the zeros of P are x  1, 2, x 2  x  1  0, then x  1 1411 2   1  i 3, 12  i 23 .      (b) From Exercise 10, x 2  2x  4  x  1  i 3 x  1  i 3 and from Exercise 11,      x 2  x  1  x  12  i 23 x  12  i 23 . Thus the complete factorization is     P x  x  2 x 2  2x  4 x  1 x 2  x  1

          x  2 x  1 x  1  i 3 x  1  i 3 x  12  i 23 x  12  i 23

  19. P x  x 4  16x 2  x 2 x 2  16  x 2 x  4i x  4i. The zeros of P are 0 (multiplicity 2) and 4i (multiplicity 1).   20. P x  9x 6  16x 4  x 4 9x 2  16  x 4 3x  4i 3x  4i. The zeros of P are 0 (multiplicity 4) and  43 i (multiplicity 1).

  21. Q x  x 6  2x 5  2x 4  x 4 x 2  2x  2 . Using the Quadratic Formula to solve x 2  2x  2  0, we  22 412  1  i. The zeros of Q are 0 (multiplicity 4) and 1  i (multiplicity 1), and obtain x  2 21

Q x  x 4 x  1  i x  1  i.   22. Q x  x 5  x 4  x 3  x 3 x 2  x  1 . Using the Quadratic Formula to solve x 2  x  1  0, we obtain    12 411   12  i 23 . The zeros of Q are 0 (multiplicity 3) and  12  i 23 (multiplicity 1); and x  1 21      Q x  x 3 x  12  i 23 x  12  i 23 .   23. P x  x 3  4x  x x 2  4  x x  2i x  2i. The zeros of P are 0, 2i, and 2i, all of multiplicity 1.   24. P x  x 3  x 2  x  x x 2  x  1 . Using the Quadratic Formula, we have

     1 12 411  1 2 3  12  i 23 . The zeros of P are 0, 12  i 23 , and 12  i 23 , all of multiplicity 1; and x 21      P x  x x  12  i 23 x  12  i 23 .

     25. Q x  x 4  1  x 2  1 x 2  1  x  1 x  1 x 2  1  x  1 x  1 x  i x  i. The zeros of Q are 1, 1, i, and i, all of multiplicity 1.


SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra

73

    26. Q x  x 4  625  x 2  25 x 2  25  x  5 x  5 x 2  25  x  5 x  5 x  5i x  5i. The zeros

of Q are 5, 5, 5i, and 5i, all of multiplicity 1.    27. P x  16x 4  81  4x 2  9 4x 2  9  2x  3 2x  3 2x  3i 2x  3i. The zeros of P are 32 ,  32 , 32 i, and  32 i, all of multiplicity 1.

  28. P x  x 3  64  x  4 x 2  4x  16 .

Using the Quadratic Formula, we have

   4 48 44i 3  2  2i 3. The zeros of P are 4, 2  2i 3, and 2  2i 3, all of x  4 164116   2 2 2

     multiplicity 1; and P x  x  4 x  2  2i 3 x  2  2i 3 .   29. P x  x 3  x 2  9x  9  x 2 x  1  9 x  1  x  1 x 2  9  x  1 x  3i x  3i. The zeros of P are 1, 3i, and 3i, all of multiplicity 1.        30. P x  x 6  729  x 3  27 x 3  27  x  3 x 2  3x  9 x  3 x 2  3x  9 . Using the 

 27 2 . Using the      3 227  32  27  32  3 2 3 i Quadratic Formula for x 2  3x  9, we obtain x  3 9419 2 2     The zeros of P are 3, 3,  32  3 2 3 i,  32  3 2 3 i, 32  3 2 3 i, and 32  3 2 3 i, all of multiplicity 1; and          P x  x  3 x  3 x  32  3 2 3 i x  32  3 2 3 i x  32  3 2 3 i x  32  3 2 3 i . 

Quadratic Formula for x 2  3x  9, we obtain x  3 9419  32 27   32  2

   2  31. P x  x 6  10x 4  25x 2  x 2 x 4  10x 2  25  x 2 x 2  5 . The zeros of P are 0 (multiplicity 2) and  5i   2   2 x  5i . (multiplicity 2); and P x  x 2 x  5i    2 32. P x  x 5  18x 3  81x  x x 4  18x 2  81  x x 2  9 . The zeros of P are 0 (multiplicity 1) and 3i

(multiplicity 2); and P x  x x  3i2 x  3i2 .    33. P x  x 4  3x 2  4  x 2  1 x 2  4  x  1 x  1 x  2i x  2i. The zeros of P are 1, 1, 2i, and 2i (all of multiplicity 1).

         34. P x  x 5  7x 3  x 3 x 2  7  x 3 x  i 7 x  i 7 . The zeros of P are 0 (multiplicity 3), i 7, and i 7, both of multiplicity 1.

   2   2  2  x  i 3 . The zeros of P are 0 35. P x  x 5  6x 3  9x  x x 4  6x 2  9  x x 2  3  x x  i 3   (multiplicity 1), i 3 (multiplicity 2), and i 3 (multiplicity 2).   2 2 36. P x  x 6  16x 3  64  x 3  8  x  22 x 2  2x  4 . Using the Quadratic Formula, on x 2  2x  4 we

      2 22 414  2 212  22i2 3  1  i 3. The zeros of P are 2, 1  i 3, and 1  i 3, all of have x  2   2   2 multiplicity 2; and P x  x  22 x  1  i 3 x 1i 3 .

37. Since 1  i and 1  i are conjugates, the factorization of the polynomial must be   P x  a x  [1  i] x  [1  i]  a x 2  2x  2 . If we let a  1, we get P x  x 2  2x  2.

  38. Since 1  i 2 and 1  i 2 are conjugates, the factorization of the polynomial must be           x  1  i 2  c x 2  2x  3 . If we let c  1, we get P x  x 2  2x  3. P x  c x  1  i 2


74

CHAPTER 3 Polynomial and Rational Functions

39. Since 2i and 2i are conjugates, the factorization of the polynomial must be     Q x  b x  3 x  2i x  2i]  b x  3 x 2  4  b x 3  3x 2  4x  12 . If we let b  1, we get Q x  x 3  3x 2  4x  12.

40. Since i is a zero, by the Conjugate Roots Theorem, i is also a zero. So the factorization of the polynomial must be     Q x  b x  0 x  i x  i  bx x 2  1  b x 3  x . If we let b  1, we get Q x  x 3  x.

41. Since i is a zero, by the Conjugate Roots Theorem, i is also a zero. So the factorization of the polynomial must be   P x  a x  2 x  i x  i  a x 3  2x 2  x  2 . If we let a  1, we get P x  x 3  2x 2  x  2.

42. Since 1  i is a zero, by the Conjugate Roots Theorem, 1  i is also a zero. So the factorization of the polynomial must be     Q x  a x  3 x  [1  i] x  [1  i]  a x  3 x 2  2x  2  a x 3  x 2  4x  6 . If we let a  1, we get Q x  x 3  x 2  4x  6.

43. Since the zeros are 1  2i and 1 (with multiplicity 2), by the Conjugate Roots Theorem, the other zero is 1  2i. So a factorization is R x  c x  [1  2i] x  [1  2i] x  12  c [x  1]  2i [x  1]  2i x  12           c [x  1]2  [2i]2 x 2  2x  1  c x 2  2x  1  4 x 2  2x  1  c x 2  2x  5 x 2  2x  1      c x 4  2x 3  x 2  2x 3  4x 2  2x  5x 2  10x  5  c x 4  4x 3  10x 2  12x  5 If we let c  1 we get R x  x 4  4x 3  10x 2  12x  5.

44. Since S x has zeros 2i and 3i, by the Conjugate Roots Theorem, the other zeros of S x are 2i and 3i. So a factorization of S x is         S x  C x  2i x  2i x  3i x  3i  C x 2  4i 2 x 2  9i 2  C x 2  4 x 2  9  C x 4  13x 2  36 If we let C  1, we get S x  x 4  13x 2  36.

45. Since the zeros are i and 1  i, by the Conjugate Roots Theorem, the other zeros are i and 1  i. So a factorization is T x  C x  i x  i x  [1  i] x  [1  i]          C x 2  i 2 [x  1]  i [x  1]  i  C x 2  1 x 2  2x  1  i 2  C x 2  1 x 2  2x  2      C x 4  2x 3  2x 2  x 2  2x  2  C x 4  2x 3  3x 2  2x  2  C x 4  2C x 3  3C x 2  2C x  2C Since the constant coefficient is 12, it follows that 2C  12  C  6, and so   T x  6 x 4  2x 3  3x 2  2x  2  6x 4  12x 3  18x 2  12x  12.

46. Since U x has zeros 12 , 1 (with multiplicity two), and i, by the Conjugate Roots Theorem, the other zero is i. So a factorization of U x is        U x  c x  12 x  12 x  i x  i  12 c 2x  1 x 2  2x  1 x 2  1  12 c 2x 5  3x 4  2x 3  2x 2  1 Since the leading coefficient is 4, we have 4  12 c 2  c. Thus we have   U x  12 4 2x 5  3x 4  2x 3  2x 2  1  4x 5  6x 4  4x 3  4x 2  2.

  47. P x  x 3  2x  4  x  2 x 2  2x  2 , so x  2 or x 2  2x  2  0. Using the Quadratic Formula, we have  22 412  1  i. Thus the zeros are 2 and 1  i. x  2 21


SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra

48. P x  x 3  7x 2  16x  10. We start by trying the possible rational factors of the polynomial: 1

1

7 1

1 

6

16

10 10

6

10

0

 x  1 is a zero.

So P x  x  1 x 2  6x  10 . Using the Quadratic Formula on the second factor, we have   3  i. Thus the zeros are 1 and 3  i. x  6 364110 2 49. P x  x 3  2x 2  2x  1. By inspection, P 1  1  2  2  1  0, and hence x  1 is a zero. 1

1

2

1

1

1

2

1 1

1

1

0

 Thus P x  x  1 x 2  x  1 . So x  1 or x 2  x  1  0. 

Using the Quadratic Formula, we have x  1 1411  1i2 3 . Hence, the zeros are 1 and 1i2 3 . 2 50. P x  x 3  7x 2  18x  18 has possible rational zeros 1, 2, 3, 6, 9, 18. Since all of the coefficients are positive, there are no positive real zeros. 1 1

7 18

18

2 1

1 6 12

7

18

18

3 1

2 10 16

7

18

18

3 12 18

1 5 8 2 1 4 6 0  x  3 is a zero. 6 12 6   So P x  x  3 x 2  4x  6 . Using the Quadratic Formula on the second factor, we have      2  2  i 2. Thus the zeros are 3, 2  i 2. x  4 16416  42 8  42i 2 2 1

51. P x  x 3  3x 2  3x  2.

2

1

3 2

1 

1

3

2 2

2

1

0

Thus P x  x  2 x 2  x  1 . So x  2 or x 2  x  1  0    Using the Quadratic Formula we have x  1 1411  1i2 3 . Hence, the zeros are 2, and 1i2 3 . 2 52. P x  x 3  x  6 has possible zeros 1, 2, 3. 1

1

0

1

1

1

6

0

2

1

0 2

1

4

6

6

1 2 3 0  x  2 is a zero. 0 6     2 4413 2  1  i 2. Thus the  22i P x  x  2 x 2  2x  3 . Now x 2  2x  3 has zeros x  2 2  zeros are 2, 1  i 2. 1

1

75


76

CHAPTER 3 Polynomial and Rational Functions

53. P x  2x 3  7x 2  12x  9 has possible rational zeros 1, 3, 9,  12 ,  32 ,  92 . Since all coefficients are positive, there are no positive real zeros. 2

1

2

7

12

9

2

5

7

5

7

There is a zero between 1 and 2.

 32

2

2

2

2

2

7

12

9

3

6

9

7

12

9

4

6

12

3

6

3

2 4 6 0  x   32 is a zero.      P x  x  32 2x 2  4x  6  2 x  32 x 2  2x  3 . Now x 2  2x  3 has zeros     22 2 x  2 4431   1  i 2. Hence, the zeros are  32 and 1  i 2. 2 2 

54. Using synthetic division, we see that x  3 is a factor of the polynomial: 1

2

9

8

2

2

3

9

9

8

3

6

6

9 9

6

2 2 3 0  x  3 is a zero.   So P x  2x 3  8x 2  9x  9  x  3 2x 2  2x  3 . Using the Quadratic Formula, we find the other two solutions:     2  4  4 3 2 2  20 x   12  25 i. Thus the zeros are 3, 12  25 i. 2 2 4 2

3

6

6

55. P x  x 4  x 3  7x 2  9x  18. Since P x has one change in sign, we are guaranteed a positive zero, and since P x  x 4  x 3  7x 2  9x  18, there are 1 or 3 negative zeros. 1

1

1

7

9

1

2

9

18

18

1 2 9 18 0  Therefore, P x  x  1 x 3  2x 2  9x  18 . Continuing with the quotient, we try negative zeros. 

1

1

2

9

18

1

1

8

2

1

2

1

8

10

0

9

18

0

18

0  P x  x  1 x  2 x 2  9  x  1 x  2 x  3i x  3i. Therefore,the zeros are 1, 2, and 3i. 1

1

2

9

56. P x  x 4  2x 3  2x 2  2x  3 has possible zeros 1, 3. 1

1

2

1

2

2

3

1

3

5

3

1

2 3

2 3

2 3

3 3

1 1 1 1 0  x  3 is a zero. 5 8  P x  x  3 x 3  x 2  x  1 . If we factor the second factor by grouping, we get   x 3  x 2  x  1  x 2 x  1  1 x  1  x  1 x 2  1 . So we have   P x  x  3 x  1 x 2  1  x  3 x  1 x  i x  i. Thus the zeros are 3, 1, i, and i. 1 

1

3


SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra

77

57. We see a pattern and use it to factor by grouping. This gives   P x  x 5  x 4  7x 3  7x 2  12x  12  x 4 x  1  7x 2 x  1  12 x  1  x  1 x 4  7x 2  12          x  1 x 2  3 x 2  4  x  1 x  i 3 x  i 3 x  2i x  2i

 Therefore,the zeros are 1, i 3, and 2i.

           58. P x  x 5  x 3  8x 2  8  x 3 x 2  1  8 x 2  1  x 2  1 x 3  8  x 2  1 x  2 x 2  2x  4

(factoring a sum of cubes). So x  2, or x 2  1  0. If x 2  1  0, then x 2  1  x  i. If x 2  2x  4  0, then     12 x  2 4414  1   1  i 3. Thus, the zeros are 2, i, 1  i 3. 2 2

59. P x  x 4  6x 3  13x 2  24x  36 has possible rational zeros 1, 2, 3, 4, 6, 9, 12, 18. P x has 4 variations in sign and P x has no variation in sign. 1 1 6 13 24 1 5

1 5

8 16

Continuing:

36

2 1 6 13 24

8 16

2 8

20

1 4

36

3 1 6 13 24

10 28

5 14

3 9

8

1 3

36

12 36

4 12

0  x  3 is a zero.

3 1 3 4 12 3 0

1

12

0 4

0  x  3 is a zero.  P x  x  32 x 2  4  x  32 x  2i x  2i. Therefore,the zeros are 3 (multiplicity 2) and 2i. 

60. P x  x 4  x 2  2x  2 has possible rational zeros 1, 2. 1 1 0 1 2 2 1

1 1

1 0 2

0 1 2

1 1 1 0

1 0 2

1

1 1

2

2

1 2 2

0 2 4 1 is an upper bound. 1 1 0 2 0 1 2 2 0   P x  x  12 x 2  2x  2 . Using the Quadratic Formula on x 2  2x  2, we have x  2 248  22i 2  1  i. 

Thus, the zeros of P x are 1 (multiplicity 2) and 1  i.

61. P x  4x 4  4x 3  5x 2  4x  1 has possible rational zeros 1,  12 ,  14 . Since there is no variation in sign, all real zeros (if there are any) are negative. 1

4

4

5

4

1

4

0

5

1

 12

4

 12

4 0 5 1 2   P x  x  12 4x 3  2x 2  4x  2 . Continuing: 

4

4 4

2

4

2

2

0

2

0

4

0

4

5

4

1

2

1

2

1

2

4

2

0

 x   12 is a zero again.

 2   P x  x  12 4x 2  4 . Thus, the zeros of P x are  12 (multiplicity 2) and i.

 x   12 is a zero.


78

CHAPTER 3 Polynomial and Rational Functions

62. P x  4x 4  2x 3  2x 2  3x  1 has possible rational zeros 1,  12 ,  14 . P has one variation in sign, so P has one positive real zero. 4

1

2

2

3

1

4 6 4   P x  x  1 4x 3  6x 2  4x  1 . Continuing:

1

0

 12

4

4

4

1

6

4

1

4

2

2

6

4

1  1 is a zero. 6

4

1

2

2

1

2 2 3 4 4 2 0   12 is a zero.   P x  x  1 x  12 4x 2  4x  2 . Using the Quadratic Formula on 4x 2  4x  2, we find 4

x  4 81632   12  12 i. Thus, P has zeros 1,  12 ,  12  12 i.

63. P x  x 5  3x 4  12x 3  28x 2  27x  9 has possible rational zeros 1, 3, 9. P x has 4 variations in sign and P x has 1 variation in sign.

1

1

12

3 1

1 1

1

10

2

1

9

10

2

10

18

9

18

9 9

0

 x  1 is a zero.

1

1

9

18

1

1

2

27

28

9

1 1

9

0

9

9

1 9 9 0  x  1 is a zero. 1 0 9 0  x  1 is a zero.   P x  x  13 x 2  9  x  13 x  3i x  3i. Therefore,the zeros are 1 (multiplicity 3) and 3i

64. P x  x 5  2x 4  2x 3  4x 2  x  2 has possible rational zeros 1, 2. 1

1

1

1

2

2

1

1

2

3

2

4

1

2

1

2 2

2

4

0

4

1 0

2

2

1 1 3 2 4 1 0 2 0 1 0  x  2 is a zero.   2  P x  x  2 x 4  2x 2  1  x  2 x 2  1  x  2 x  i2 x  i2 . Thus, the zeros of P x are 2, i.   65. (a) P x  x 3  5x 2  4x  20  x 2 x  5  4 x  5  x  5 x 2  4 (b) P x  x  5 x  2i x  2i

66. (a) P x  x 3  2x  4

1

1

0 1

1

1 

2

1

1

4

2

1 5 

P x  x 3  2x  4  x  2 x 2  2x  2

(b) P x  x  2 x  1  i x  1  i      67. (a) P x  x 4  8x 2  9  x 2  1 x 2  9  x  1 x  1 x 2  9 (b) P x  x  1 x  1 x  3i x  3i

1

0 2

1

2

2

4

2

0

4

4


79

SECTION 3.5 Complex Zeros and the Fundamental Theorem of Algebra

 2 68. (a) P x  x 4  8x 2  16  x 2  4

(b) P x  x  2i2 x  2i2        69. (a) P x  x 6  64  x 3  8 x 3  8  x  2 x 2  2x  4 x  2 x 2  2x  4          (b) P x  x  2 x  2 x  1  i 3 x  1  i 3 x  1  i 3 x  1  i 3        70. (a) P x  x 5  16x  x x 4  16  x x 2  4 x 2  4  x x  2 x  2 x 2  4 (b) P x  x x  2 x  2 x  2i x  2i   71. (a) x 4  2x 3  11x 2  12x  x x 3  2x 2  11x  12  0. We first find the bounds for our viewing rectangle. 1 5

1

4

1

11 4

32

6

13

50

3

5

12

2

­20

 x  5 is an upper bound.

­40

 x  4 is a lower bound.

We graph P x  x 4  2x 3  11x 2  12x in the viewing rectangle [4 5] by [50 10] and see that it has 4 real solutions. Since this matches the degree of P x, P x has no nonreal solution. (b) x 4  2x 3  11x 2  12x  5  0. We use the same bounds for our viewing rectangle, [4 5] by [50 10],

(c) x 4  2x 3  11x 2  12x  40  0. We graph

T x  x 4  2x 3  11x 2  12x  40 in the viewing

and see that R x  x 4  2x 3  11x 2  12x  5 has

rectangle [4 5] by [10 50], and see that T has no

2 real solutions. Since the degree of R x is 4, R x

real solution. Since the degree of T is 4, T must have

must have 2 nonreal solutions.

4 nonreal solutions.

5 ­20

40 20

­40

5

72. (a) 2x  4i  1  2x  1  4i  x  12  2i. (b) x 2  i x  0  x x  i  0  x  0, i.

(c) x 2  2i x  1  0  x  i2  0  x  i.

(d) i x 2  2x  i  0. Using the Quadratic Formula, we get           2 22 4ii 2  1 2  1  2 i  1  2 i. x  22i 8  22 i 2i 2i

73. (a) P x  x 2  1  i x  2  2i. So P 2i  2i2  1  i 2i  2  2i  4  2i  2  2  2i  0, and P 1  i  1  i2  1  i 1  i  2  2i  1  2i  1  1  1  2  2i  0.

Therefore, 2i and 1  i are solutions of the equation x 2  1  i x  2  2i  0. However, P 2i  2i2  1  i 2i  2  2i  4  2i  2  2  2i  4  4i, and

P 1  i  1  i2  1  i 1  i  2  2i  2  2i. Since, P 2i  0 and P 1  i  0, 2i and 1  i are not solutions. (b) This does not violate the Conjugate Roots Theorem because the coefficients of the polynomial P x are not all real.


80

CHAPTER 3 Polynomial and Rational Functions

74. (a) Because i and 1  i are zeros, i and 1  i are also zeros. Thus,

   P x  C x  i x  i x  [1  i] x  [1  i]  C x 2  1 x 2  2x  2      C x 4  2x 3  2x 2  x 2  2x  2  C x 4  2x 3  3x 2  2x  2

Because C  1, the polynomial is P x  x 4  2x 3  3x 2  2x  2.

(b) Because i and 1  i are zeros,

    P x  C x  i x  [i  1]  C x 2  xi  x  xi  1  i  C x 2  1  2i x  1  i

Because C  1, the polynomial is P x  x 2  1  2i x  1  i.

75. Because P has real coefficients, the imaginary zeros come in pairs: a  bi (by the Conjugate Roots Theorem), where b  0. Thus there must be an even number of nonreal zeros. Since P is of odd degree, it has an odd number of zeros (counting multiplicity). It follows that P has at least one real zero. 76. P x  2x 4  3x 3  6x 2  12x  8 50

From the graph, we see that P has real zeros  12 and 2. Thus   P x  2x  1 x  2 x 2  4 has complex zeros x  2i, and so P x  2x  1 x  2 x  2i x  2i.

­2

2

4

­50

3.6

RATIONAL FUNCTIONS

1. If the rational function y  r x has the vertical asymptote x  2, then as x  2 , either y   or y  . 2. If the rational function y  r x has the horizontal asymptote y  2, then y  2 as x  . 3. The function r x 

x  1 x  2 has x­intercepts 1 and 2. x  2 x  3

4. The function r has y­intercept 13 .

5. The function r has vertical asymptotes x  2 and x  3.

6. The function r has horizontal asymptote y  1.

7. The graph of s x  r x  1 can be obtained by shifting the graph of r x to the right 1 unit, so its vertical asymptote is also shifted to the right 1 unit (to x  3). The horizontal asymptote is unchanged at y  4.

8. The graph of t x  r x  5 can be obtained by shifting the graph of r x downward 5 units, so its horizontal asymptote is also shifted downward 5 units (to y  1). Its vertical asymptote is unchanged at x  2. x2  x x x  1 x for x  1.   2 x  2 x  1 2x  4 x  1 2x  4 (a) True, r has vertical asymptote x  2.

9. r x 

  (b) False. r does not have vertical asymptote x  1. It has a “hole” at 1 16 , because r is not defined at x  1.

(c) False, r has horizontal asymptote y  12 but not horizontal asymptote y  1. (d) True, r has horizontal asymptote y  12 .


SECTION 3.6 Rational Functions

x2  x 10. True, the graph of a rational function may cross a horizontal asymptote. For example, r x  2 crosses its x x 1   horizontal asymptote y  1 at the point 12  1 . x 11. r x  x 2 (a)

x

r x

x

r x

x

r x

x

r x

15

3

25

5

10

125

0833

21

21

50

1042

10

201

201

100

1020

2001

2001

1000

1002

19 199 1999

19 199 1999

0962

50 100

0980

1000

0998

(b) r x   as x  2 and r x   as x  2 .

(c) r has horizontal asymptote y  1. 4x  1 12. r x  x 2 (a) x

r x

x

r x

x

r x

x

r x

15

14

25

22

10

5125

325

21

94

50

4188

10

201

904

100

4092

2001

9004

1000

4009

19 199 1999

86 896 8996

50 100

3827 3912

1000

3991

(b) r x   as x  2 and r x   as x  2 .

(c) r has horizontal asymptote y  4. 3x  10 13. r x  x  22 (a) x 15

r x

x

22

19 199 1999

430 40,300 4,003,000

r x

25

10

21 201 2001

370 39,700 3,997,000

x

r x

x

r x

10

03125

10

02778

50

00608

100

00302

1000

50

00592

00030

100

00298

1000

00030

x

r x

x

r x

10

209

(b) r x   as x  2 and r x   as x  2 . (c) r has horizontal asymptote y  0.

14. r x  (a)

3x 2  1

x  22 x

r x

x

r x

15

31

25

79

10

4703

19

1183

21

1423

50

3256

199

128,803

201

131,203

100

3124

1999

12,988,003

2001

13,012,003

1000

3012

(b) r x   as x  2 and r x   as x  2 .

(c) r has horizontal asymptote y  3.

50 100 1000

2774 2884 2988

81


82

CHAPTER 3 Polynomial and Rational Functions

In the solutions to Exercises 15–22, let f x 

1 . x

  4 1 15. r x  4  4 f x  2. From this form we see that the graph x 2 x 2

y

of r is obtained from the graph of f by shifting 2 units to the right and stretching vertically by a factor of 4. Thus r has vertical asymptote x  2 and horizontal

asymptote y  0. The domain of r is  2  2  and its range is

4

1

x

 0  0 .

9  9 f x  3. From this form we see that the graph of r is obtained x 3 from the graph of f by shifting 3 units to the left and stretching vertically by a

y

16. r x 

factor of 9. Thus r has vertical asymptote x  3 and horizontal asymptote y  0.

10

The domain of r is  3  3  and its range is  0  0 .

1x

  2 1  2  2 f x  1. From this form we see that the 17. s x   x 1 x 1

y

graph of s is obtained from the graph of f by shifting 1 unit to the left, stretching vertically by a factor of 2, and reflecting about the x­axis. Thus s has vertical

asymptote x  1 and horizontal asymptote y  0. The domain of s is

2

 1  1  and its range is  0  0 .

18. s x 

  3 1  3  3 f x  4. From this form we see that the x 4 x 4

1

x

y

graph of s is obtained from the graph of f by shifting 4 units to the right, stretching vertically by a factor of 3, and then reflecting about the x­axis. Thus s has vertical asymptote x  4 and horizontal asymptote y  0. The domain of s is  4  4  and its range is  0  0 .

1 1

x


83

SECTION 3.6 Rational Functions

1 2x  3 2  f x  2  2 (see the long x 2 x 2 division at right). From this form we see that the graph of t is

19. t x 

y

2 x 2

obtained from the graph of f by shifting 2 units to the right

2x  3 2x  2 1

and 2 units vertically. Thus t has vertical asymptote x  2

1

and horizontal asymptote y  2. The domain of t is   9 3x  3 1  3 39 x 2 x 2 x 2  9 f x  2  3

20. t x 

x

1

 2  2  and its range is  2  2 .

y

3 x 2

3x  3

From this form we see that the graph of t is obtained from the

3x  6

graph of f by shifting 2 units to the left, stretching vertically

9

5

x

1

by a factor of 9, reflecting about the x­axis, and then shifting 3 units vertically. Thus t has vertical asymptote x  2 and horizontal asymptote y  3. The domain of t is

 2  2  and its range is  3  3 .

1 x 2 1   f x  3  1 (see the long x 3 x 3 division at right). From this form we see that the graph of r is

21. r x 

y

1 x 3

x  2

obtained from the graph of f by shifting 3 units to the left,

x  3

reflect about the x­axis, and then shifting vertically 1 unit.

1

Thus r has vertical asymptote x  3 and horizontal

1 1

x

asymptote y  1. The domain of r is  3  3  and its range is  1  1 .

1 2x  9  2 2 x 4 x 4   f x  4  2

22. t x 

1 x 4

y

2 x 4

2x  9

From this form we see that the graph of t is obtained from the

2x  8

graph of f by shifting 4 units to the right, reflecting about the

1

x­axis, and then shifting 2 units vertically. Thus t has vertical

1 1

x

asymptote x  4 and horizontal asymptote y  2. The domain of r is  4  4  and its range is  2  2 .

x 1 . When x  0, we have r 0   14 , so the y­intercept is  14 . The numerator is 0 when x  1, so the x 4 x­intercept is 1.

23. r x 

3x . When x  0, we have s 0  0, so the y­intercept is 0. The numerator is zero when 3x  0 or x  0, so x 5 the x­intercept is 0.

24. s x 

25. t x 

x2  x  2 2 . When x  0, we have t 0   13 , so the y­intercept is 13 . The numerator is 0 when x 6 6

x 2  x  2  x  2 x  1  0 or when x  2 or x  1, so the x­intercepts are 2 and 1.


84

CHAPTER 3 Polynomial and Rational Functions

2 2   12 , so the y­intercept is  12 . The numerator is never zero, so . When x  0, we have r 0  26. r x  2 4 x  3x  4 there is no x­intercept. 27. r x 

x2  9 . Since 0 is not in the domain of r x, there is no y­intercept. The numerator is 0 when x2

x 2  9  x  3 x  3  0 or when x  3, so the x­intercepts are 3.

x3  8 . When x  0, we have r 0  84  2, so the y­intercept is 2. The x­intercept occurs when x 3  8  0  28. r x  2 x 4    x  2 x 2  2x  4  0  x  2 or x  1  i 3, which has only one real solution, so the x­intercept is 2.

29. From the graph, the x­intercept is 3, the y­intercept is 3, the vertical asymptote is x  2, and the horizontal asymptote is y  2.

30. From the graph, the x­intercept is 0, the y­intercept is 0, the horizontal asymptote is y  0, and the vertical asymptotes are x  1 and x  2.

31. From the graph, the x­intercepts are 1 and 1, the y­intercept is about 14 , the vertical asymptotes are x  2 and x  2, and the horizontal asymptote is y  1.

32. From the graph, the x­intercepts are 2, the y­intercept is 6, the horizontal asymptote is y  2, and there are no vertical asymptotes 5 has a vertical asymptote where x  2  0  x  2, and y  0 is a horizontal asymptote because the degree x 2 of the denominator is greater than that of the numerator.

33. r x 

2x  3 34. r x  2 has are vertical asymptotes where x 2  1  0  x  1 or x  1, and y  0 is a horizontal asymptote x 1 because the degree of the denominator is greater than that of the numerator. 3x  10 has no vertical asymptote since x 2  5  0 for all x. There is a horizontal asymptote at y  0 because the x2  5 degree of the denominator is greater than that of the numerator.

35. r x 

   2x 3  x 2 has vertical asymptotes where x 4  16  x 2  4 x 2  4  0  x  2, and y  0 is a horizontal 4 x  16 asymptote because the degree of the denominator is greater than that of the numerator.

36. r x 

37. s x 

  10x 3  7 has vertical asymptotes where x 3  x  x x 2  1  x x  1 x  1  0  x  1, 0, or 1; and 3 x x

horizontal asymptote y  10 1  10. 38. s x  39. r x 

18x 2  9 has no vertical asymptote since 9x 2  1  0 for all x. It has horizontal asymptote y  18 9  2. 9x 2  1

x  1 2x  3 has vertical asymptotes where x  2 4x  7  0  x   74 or x  2, and horizontal x  2 4x  7

asymptote y  40. r x 

1 12  . 14 2

x  3 x  2 has vertical asymptotes where 5x  1 2x  3  0  x   15 or x  32 , and horizontal 5x  1 2x  3

asymptote y 

1 11  . 52 10


85

SECTION 3.6 Rational Functions

41. r x 

6x 3  2 6x 3  2   . Because the quadratic in the denominator has no real zero, r has vertical  2x 3  5x 2  6x x 2x 2  5x  6

asymptote x  0 and horizontal asymptote y  62  3.

5x 3 5x 3 5x 2    42. r x  3  . Because the denominator has no real zero, r has no vertical x  2x 2  5x x x 2  2x  5 x 2  2x  5 asymptote. r has horizontal asymptote y  51  5.

t(x)

x2  2 . A vertical asymptote occurs when x  1  0  x  1. There is no horizontal asymptote because the degree x 1 of the numerator is greater than the degree of the denominator.

43. y 

x 3  3x 2 x 2 x  3 . Because the degree of the numerator is greater than the degree of the denominator,  x  2 x  2 x2  4 the function has no horizontal asymptote. Two vertical asymptotes occur at x  2 and x  2. By using long division, we 4x  12 so y  x  3 is a slant asymptote. see that r x  x  3  2 x 4 2x  2 y . When x  0, y  2, so the y­intercept is 2. When y  0, 45. r x  x 1 2x  2  0  x  1, so the x­intercept is 1. Since the degree of the numerator

44. r x 

and denominator are the same, the horizontal asymptote is y  21  2. There is a

vertical asymptote at x  1. As x  1 , y 

2x  2  , and as x  2 , x 2

1

  3 x  13 1  3x  . When x  0, y  14 , so the y­intercept is 14 . When 46. r x  2x  4 2 x  2

y

y  0, we have x  13  0  x  13 , so the x­intercept is 13 . There is a vertical

1

asymptote where x  2  0  x  2. Because the degree of the denominator

1

3   3 . The and the numerator are the same, the horizontal asymptote is y  2 2   3 domain is x  x  2 and the range is y  y   2 .

47. r x 

3x 2  12x  13 x 2  4x  4

  3 x 2  4x  4  1 x 2  4x  4

3

1 x  22

13 y  13 4 , so the y­intercept is 4 . There is no x­intercept since

. When x  0,

1

x  22

x

1

2x  2 y  . The domain is x  x  1 and the range is y  y  2. x 2

x

y

is positive

on its domain. There is a vertical asymptote at x  2. The horizontal asymptote is y  3. The domain is x  x  2 and the range is y  y  3.

1 1

x


86

CHAPTER 3 Polynomial and Rational Functions

  2 x 2  4x  4  1 2x 2  8x  9 1 48. r x  . When   2  2 2 x  4x  4 x  4x  4 x  22 x  0, y   94 , so the y­intercept is  94 . There is no x­intercept since

1

x  22

y

is

1 0

positive on its domain. There is a vertical asymptote at x  2. The horizontal

1

x

asymptote is y  2. The domain is x  x  2 and the range is y  y  2.

   x 2  8x  16  2 x 2  8x  18 2 49. r x  2 . When   1  2 x  8x  16 x  8x  16 x  42 x  0, y   98 , and so the y­intercept is  98 . There is no x­intercept since 2

x  42

is positive on its domain. There is a vertical asymptote at x  4. The

y

1 0

1

x

horizontal asymptote is y  1. The domain is x  x  4 and the range is y  y  1.

  1 2x 2  4x  2  2 1 x 2  2x  3 1 2 50. r x  2 . is true. When    2 2 2x  4x  2 2x  4x  2 x  12

y

x  0, we have y  32 , so the y­intercept is 2. There is no x­intercept since 1

x  12

is positive on its domain. There is a vertical asymptote at x  1. The

horizontal asymptote is y  12 . The domain is x  x  1 and the range is   y  y  12 .

51. s x 

4x  8 8 . When x  0, y   2, so the y­intercept is 2. x  4 x  1 4 1

1 0

1

x

y

When y  0, 4x  8  0  x  2, so the x­intercept is 2. The vertical asymptotes are x  1 and x  4, and because the degree of the numerator is less than the

degree of the denominator, the horizontal asymptote is y  0. The domain is x  x  1 4 and the range is .

1 1

x


87

SECTION 3.6 Rational Functions

9 . When x  0, y  94 , so the y­intercept is 94 . Since the 52. s x  2 x  5x  4 numerator is never zero, there is no x­intercept. The vertical asymptotes occur when x 2  5x  4  x  1 x  4  0  x  1 or x  4, and because the

degree of the numerator is less than the degree of the denominator, the horizontal

y

2 x

1

asymptote is y  0. The domain is x  x  1 4 and the range is  4]  0 .

9x  18 9 x  2 53. s x  2 . When x  0, y  9, so the y­intercept is  x  1 x  2 x x 2 9. When y  0, we have x  2  0  x  2, so the x­intercept is 2. There are

y

vertical asymptotes where x  1 x  2  0  x  1 or x  2. Because the degree of the denominator is greater than the degree of the numerator, the

4

horizontal asymptote is y  0. The domain is x  x  2 1 and the range is

1

x

1

x

 1]  [9 .

54. s x 

x 2 2 , so the y­intercept is  23 When . When x  0, y  3 x  3 x  1

y

y  0, we have x  2  0  x  2, so the x­intercept is 2. A vertical

asymptote occurs when x  3 x  1  0  x  3 and x  1. Because the degree of the denominator is greater than the degree of the numerator, the

1

horizontal asymptote is y  0. The domain is x  x  3 1 and the range is .

55. r x 

x  1 x  2 . When x  0, y  23 , so the y­intercept is 23 . When x  1 x  3

y

y  0, x  1 x  2  0  x  2, 1, so, the x­intercepts are 2 and 1. The

vertical asymptotes are x  1 and x  3, and because the degree of the

numerator and denominator are the same the horizontal asymptote is y  11  1.

The domain is x  x  1 3 and the range is .

1 1

x


88

CHAPTER 3 Polynomial and Rational Functions

2x 2  10x  12 2 1 6 2 x  1 x  6 . When x  0, y   2,  x  2 x  3 2 3 x2  x  6 so the y­intercept is 2. When y  0, 2 x  1 x  6  0  x  6, 1, so the

y

56. r x 

x­intercepts are 6 and 1. Vertical asymptotes occur when x  2 x  3  0

 x  3 or x  2. Because the degree of the numerator and denominator are the

2

same the horizontal asymptote is y  21  2. The domain is x  x  3 2 and

x

1

the range is .

x 2  2x  8 x  2 x  4 . The function has vertical asyptotes  2 x x  2 x  2x x  2 and x  0. Since x cannot equal 0, there is no y­intercept. The

y

57. r x 

x­intercepts are 4 and 2. Because the degree of the denominator and numerator

are the same, the horizontal asymptote is y  11  1. The domain is

4

x  x  2 0 and the range is  1  [9 .

3x 2  6

  3 x2  2

. We cannot have x  0 or y  0 so, there is no x­ 58. r x  2  x x  4 x  4x or y­intercept. There are vertical asymptotes at x  0 and x  4. Because the

1

y

degree of the numerator and denominator are the same, the horizontal asymptote   is.y  31  3. The domain is x  x  0 4 and the range is  3]  32   .

59. s x 

x 2  2x  1 x  12  2 . Since x  0 is not in the domain of s x, 3 2 x  3x x x  3

there is no y­intercept. The x­intercept occurs when y  0 

x 2  2x  1  x  12  0  x  1, so the x­intercept is 1. Vertical asymptotes

occur when x  0, 3. Since the degree of the numerator is less than the degree of

the denominator, the horizontal asymptote is y  0. The domain is x  x  0 3 and the range is .

x

2 2

x

1

x

y 1


89

SECTION 3.6 Rational Functions

x2  x  6 x  3 x  2 . The x­intercept occurs when y  0   x x  3 x 2  3x x  3 x  2  0  x  2, 3, so the x­intercepts are 2 and 3. There is no

y

60. y 

y­intercept because y is undefined when x  0. The vertical asymptotes are x  0

and x  3. Because the degree of the numerator and denominator are the same,

2

the horizontal asymptotes is y  11  1. The domain is x  x  3 0 and the

x

1

range is .

x 2  2x  1 x  12 61. r x  2   x  2x  1 x  12

 x 1 2 . When x  0, y  1, so the x 1

y

y­intercept is 1. When y  0, x  1, so the x­intercept is 1. A vertical asymptote

occurs at x  1  0  x  1. Because the degree of the numerator and

denominator are the same the horizontal asymptote is y  11  1. The domain is x  x  1 and the range is y  y  0.

1 x

1

9x 2 9x 2 . When x  0, we have y  0, so the  4 x  2 x  1 4x 2  4x  8 graph passes through the origin. Vertical asymptotes occur at x  2 and x  1.

62. r x 

y

Because the degree of the denominator and numerator are the same, the horizontal asymptote is y  94 . The domain is x  x  2 1 and the range is

1

63. r x 

5 x  12 5x 2  10x  5  . When x  0, y  59 , so the y­intercept is x 2  6x  9 x  32

x

1

 0]  [2 .

y

5 . Since r 1  0, the x­intercept is 1. The vertical asymptote is x  3. 9

Because the degree of the denominator and numerator are the same, the horizontal asymptote occurs at y  51  5. The domain is x  x  3 and the range is

y  y  0.

2 2

x


90

CHAPTER 3 Polynomial and Rational Functions

x3  x2 x 2 x  1 64. r x  3  3 . When x  0, we have y  0, so the x  3x  2 x  3x  2

y

y­intercept is 0. When y  0, we have x 2 x  1  0, so the x­intercepts are 0

and 1. Vertical asymptotes occur when x 3  3x  2  0. Since x 3  3x  2  0

when x  2, we can factor x  2 x  12  0, so the vertical asymptotes occur

1

at x  2 and x  1. Because the degree of the denominator and numerator are

the same, the horizontal asymptote is y  11  1. The domain is x  x  1 2

x

1

and the range is .

65. r x 

x 2  4x  5 x 5 x  5 x  1 for x  1. When x  0, y  52 ,   2 x 2  2  1 x x x x 2

y

so the y­intercept is 52 . When y  0, x  5, so the x­intercept is 5. Vertical asymptotes occur when x  2  0  x  2. Because the degree of the

(1, 2)

denominator and numerator are the same, the horizontal asymptote is y  1. The

1

domain is x  x  2 1 and the range is y  y  1 2.

0

1

x

Move

66. r x 

x 2  3x  10 x 2 x  2 x  5   for x  1 x  3 x  5 x  1 x  3 x  5 x  1 x  3

y

x  5. When x  0, we have y  23 , so the y­intercept is 23 . When y  0, we

have x  2, so the x­intercept is 2. Vertical asymptotes occur when

1

x  1 x  3  0  x  1 or 3, so the vertical asymptotes occur at x  1 and x  3. Because the degree of the denominator is greater than that of the

numerator, the horizontal asymptote is y  0. The domain is x  x  5 1 3

0

(_5, _ 327 )

1

x

and the range is .

x 2  2x  3 x  3 x  1   x  3 for x  1. When x  0, x 1 x 1 y  3, so the y­intercept is 3. When y  0, x  3, so the x­intercept is 3.

y

67. r x 

There are no asymptotes. The domain is x  x  1 and the range is

1

y  y  4.

0 (_1, _4)

1

x


91

SECTION 3.6 Rational Functions

x x  1 x  3 x 3  2x 2  3x   x x  1 for x  3. When x  0, x 3 x 3 we have y  0, so the y­intercept is 0. When y  0, x  1 or 0, so the

y

68. r x 

(3, 12)

x­intercepts are 1 and 0. There are no asymptotes. The domain is x  x  3 and   the range is y  y   14 .

1 0

x 3  5x 2  3x  9 . We use synthetic division to check whether the x 1 denominator divides the numerator:

y

69. r x 

1

1

5 1

3

9

6

9

1 6 9 0  x 2  6x  9 x  1 Thus, r x   x 2  6x  9  x  32 for x  1. x 1 When x  0, y  9, so the y­intercept is 9. When y  0, x  3, so the x­intercept 

x

1

(_1, 16)

2 0

x

1

is 3. There are no asymptotes. The domain is x  x  1 and the range is

y  y  0.

x 1 x 2  4x  5 x  1 x  5  for x  5. x  0 is 70. r x  3  x x  2 x  5 x x  2 x  7x 2  10x not in the domain of r , so there is no y­intercept. When y  0, x  1, so the

y

x­intercept is 1. There are vertical asymptotes at x  2 and x  0. Because the

degree of the denominator is less than the degree of the numerator, y  0 is a

Move 1

horizontal asymptote. The domain is x  x  5 2 0 and the range is y  y  0134 or y  1866.

71. r x 

x2 . When x  0, y  0, so the graph passes through the origin. There x 2

0

(_5, _ 327 )

1

x

y

is a vertical asymptote when x  2  0  x  2, with y   as x  2 , and y   as x  2 . Because the degree of the numerator is greater than the

degree of the denominator, there is no horizontal asymptote. Using long division, we see that y  x  2 

4 , so y  x  2 is a slant asymptote. x 2

2 2

x


92

CHAPTER 3 Polynomial and Rational Functions

x x  2 x 2  2x  . When x  0, we have y  0, so the graph passes x 1 x 1 through the origin. Also, when y  0, we have x  0 or 2, so the x­intercepts are

y

72. r x 

2 and 0. The vertical asymptote is x  1. There is no horizontal asymptote, and the line y  x  3 is a slant asymptote because by long division, we have y  x 3

2 . x 1

2 x

1

x 2  2x  8 x  4 x  2  . The vertical asymptote is x  0, thus, x x there is no y­intercept. If y  0, then x  4 x  2  0  x  2, 4, so the

y

73. r x 

x­intercepts are 2 and 4. Because the degree of the numerator is greater than the

degree of the denominator, there are no horizontal asymptotes. By using long

2

8 division, we see that y  x  2  , so y  x  2 is a slant asymptote. x

3x  x 2 x 3  x  . When x  0, we have y  0, so the graph passes 2x  2 2 x  1 through the origin. Also, when y  0, we have x  0 or x  3, so the x­intercepts are 0 and 3. The vertical asymptote is x  1. There is no horizontal asymptote, and the line y   12 x  1 is a slant asymptote because by long division we have

1

x

1

1 y   12 x  1  . x 1

x 2  5x  4 x  4 x  1  . When x  0, y   43 , so the y­intercept x 3 x 3

x

y

74. r x 

75. r x 

2

y

is  43 . When y  0, x  4 x  1  0  x  4, 1, so the two x­intercepts are 4 and 1. A vertical asymptote occurs when x  3, with y   as x  3 , and y   as x  3 . Using long division, we see that 28 , so y  x  8 is a slant asymptote. y  x 8 x 3

5 2

x


93

SECTION 3.6 Rational Functions

x3  4 x3  4 . When x  0, we have  2x  1 x  1 2x 2  x  1  04  4, so the y­intercept is 4. Since x 3  4  0  x   3 4, y 001  the x­intercept is x   3 4. There are vertical asymptotes where

y

76. r x 

1 1

2x  1 x  1  0  x  12 or x  1. Since the degree of the numerator is

x

greater than the degree of the denominator, there is no horizontal asymptote. By long division, we have y  12 x  14  slant asymptote.

77. r x 

3 x  15 4 4 , so the line y  1 x  1 is a 2 4 2x 2  x  1

x3  x2 x 2 x  1  . When x  0, y  0, so the graph passes x  2 x  2 x2  4

y

through the origin. Moreover, when y  0, we have x 2 x  1  0  x  0, 1, so the x­intercepts are 0 and 1. Vertical asymptotes occur when x  2; as

x  2 , y   and as x  2 , y  . Because the degree of the numerator is greater than the degree of the denominator, there is no horizontal 4x  4 , so y  x  1 is a asymptote. Using long division, we see that y  x  1  2 x 4 slant asymptote.

78. r x 

2x 3  2x x2  1

  2x x 2  1

x  1 x  1

2

50 ­5

5 ­20 ­20

1

x

2

2x 2  6x  6 , g x  2x. f has vertical asymptote x  3. x 3

­10

x

y

. When x  0, we have y  0, so the graph

passes through the origin. Also, note that x 2  1  0, for all real x, so the only x­intercept is 0. There are two vertical asymptotes at x  1 and x  1. There is no horizontal asymptote, and the line y  2x is a slant asymptote because by long 4x . division, we have y  2x  2 x 1

79. f x 

1

20 ­50


94

CHAPTER 3 Polynomial and Rational Functions

80. f x 

x 3  6x 2  5 , g x  x  4. f has vertical asymptotes x  0 and x  2. x 2  2x 20

20

­5

5

­20

20

­20

81. f x 

­20

x 3  2x 2  16 , g x  x 2 . f has vertical asymptote x  2. x 2 50

50

­10

10

5 ­50

82. f x 

x 4  2x 3  2x x  12

­4

, g x  1  x 2 . f has vertical asymptote x  1.

­2

2

4

­4

­2

2

­5

­5

­10

­10

4

2x 2  5x has vertical asymptote x  15, x­intercepts 0 and 25, y­intercept 0, local maximum 39 104, 2x  3 12 . From the graph, we see that the end and local minimum 09 06. Using long division, we get f x  x  4  2x  3 behavior of f x is like the end behavior of g x  x  4.

83. f x 

f

2x  3

x  4

2x 2  5x

50

20

2x 2  3x  8x

­10

 8x  12

12

10 ­20

­20

20 ­50


SECTION 3.6 Rational Functions

95

x 4  3x 3  x 2  3x  3 has vertical asymptotes are x  0, x  3, x­intercept 082, and no y­intercept. The x 2  3x local minima are 080 263 and 338 1476. The local maximum is 256 488. By using long division, we see that 3 f x  x 2  1  2 . From the second graph, we see that the end behavior of f x is the same as the end behavior x  3x f 2 of g x  x  1.

84. f x 

x2

x 2  3x

 1

20

x 4  3x 3  x 2  3x  3

50

x 4  3x 3

0x 3  x 2  3x

­5

x 2  3x

5

­5 ­50

­20

3

5

x5 85. f x  3 has vertical asymptote x  1, x­intercept 0, y­intercept 0, and local minimum 14 31. f x 1 x2 Thus y  x 2  3 . From the graph we see that the end behavior of f x is like the end behavior of g x  x 2 . x 1 x2 x3  1

10

x5

10

x5  x2 x2

­5

5

­5

5

­10

­10

Graph of f 86. f x 

f

x4

Graph of f and g

has vertical asymptotes x  141, x­intercept 0, and y­intercept 0. The local maximum is 0 0. The

x2  2

4 . From the second graph, local minima are 2 8 and 2 8. By using long division, we see that f x  x 2  2  2 x 2 2 we see that the end behavior of f x is the same as the end behavior of g x  x  2. x2

x2  2

2

20

x 4  0x 3  0x 2  0x  0

x4

50

 2x 2 2x 2 2x 2

r(x)

 4

­5

5

­5

4

5

x 4  3x 3  6 has vertical asymptote x  3, x­intercepts 16 and 27, y­intercept 2, local maxima 04 18 x 3 6 and 24 38, and local minima 06 23 and 34 543. Thus y  x 3  . From the graphs, we see that the end x 3 behavior of f x is like the end behavior of g x  x 3 .

87. f x 

r

x3

x 3

x 4  3x 3  6

x 4  3x 3

6

100

­5

100

5 ­100

­5

5 ­100


96

CHAPTER 3 Polynomial and Rational Functions

4  x2  x4 x 4  x 2  4 has vertical asymptotes x  1, x­intercepts 16, and y­intercept 4. The local  x  1 x  1 x2  1 maximum is 0 4 and there is no local minimum. 6 Thus y  x 2  2 . From the graphs, we see that the end behavior of f x is like the end behavior of g x  x 2 . x 1 x 2

88. r x 

x2  1

89.

20

x 4  0x 3  x 2  0x  4 x 4

 x2

0

2

 4

20

­5

5

­5

­20

5 ­20

90.

c=4 c=3 c=2

5

c=1 c=5

­5

c=4

5

­10

c=3

c=2

­5

­2

cx has the same basic shape for The graph of r x  2 x 1 all values of c. Increasing the value of c stretches the graph vertically.

x 1 has y  1 as a horizontal x c asymptote and x  c as a vertical asymptote for all values of c. The location of the vertical asymptote changes as c changes.

The graph of r x 

r


SECTION 3.6 Rational Functions

92.

91.

c=1 c=2 c=3 c=4

c=4

5

97

20 4 9 16

c=3

­10

c=2 c=1

­2

2

­20

­5

­40

cx 2 has the same basic shape, x 1 local maximum r 0  0, vertical asymptotes x  1, and horizontal asymptote y  c for all values of c. The horizontal asymptote moves upward as c increases. The graph of r x 

x2  c has the same basic shape and x 5 vertical asymptote x  5 for all values of c. The local maximum and minimum values get closer together as c increases.

The graph of r x 

93. (a) The total cost of producing x purses is C x, so the average cost per purse is (b)

y 100

 158, the average cost per purse decreases, presumably due to For x 

 158, the average cost steadily increases, economies of scale. For x 

80

perhaps because of supply issues or overtime costs. The local

60

minimum of approximately A 158  545 tells us that the lowest

40

average cost is approximately $5450 per purse, and is achieved when

20 0

Total cost C x  . Number of purses x

158 purses are produced. 100 200 300 400 500 600 700 x

3000 3000t  3000  . So as t  , we have t 1 t 1 p t  3000.

94. (a)

(b) p t  2000 0

0

20

40

Please don't use this symbol. I don't think students don't know what it means.


98

CHAPTER 3 Polynomial and Rational Functions

5t 95. c t  2 t 1

(a) The highest concentration of drug is 250 mg/L, and it is reached 1 hour after the drug is administered. (b) The concentration of the drug in the bloodstream goes to 0.

2

0

(c) From the first viewing rectangle, we see that an approximate solution

0

10

5t is near t  15. Thus we graph y  2 and y  03 in the t 1 viewing rectangle [14 18] by [0 05]. So it takes about 1661 hours for the concentration to drop below 03 mg/L.

20

0.4 0.2 0.0

14

16

  64  106  2   96. Substituting for R and g, we have h   . The vertical 2 98 64  106   2

2e+7

asymptote is   11,000, and it represents the escape velocity from the earth’s

1e+7

gravitational pull: 11,000 m/s  1900 mi/h.

0

97. P   P0

s0 s0  

 P   440

332 332  

18

0

10000

4000

If the speed of the train approaches the speed of sound, the pitch of the whistle becomes very loud. This would be experienced as a “sonic boom”—an effect

2000

seldom heard from trains. 0

98. (a)

1 1 1 1 1 1 xF xF 1        y . Using x y F y F x y xF xF

55x . Since y  0, we use the viewing F  55, we get y  x  55 rectangle [0 1000] by [0 250]. (b) y approaches 55 millimeters. (c) y approaches .

0

200

400

200 100 0

0

500

1000


99

SECTION 3.6 Rational Functions

99. (a) The initial amount of salt is 4 lb and salt is being poured in at 5 lb per C (lb/gal) minute. Thus, at time t there is S t  4  5t pounds of salt in the

0.1

tank. Similarly, there is W t  100  50t gallons of water in the tank, so the concentration at time t is C t 

4  5t . 100  50t

0

10

20

t (minutes)

4  5 10  009 lbgal. The tank contains 1000 gal of 100  50 10 water when W t  100  50t  1000  50t  900  t  18 minutes, at which time the concentration is 4  5 18 C 18   0094 lbgal. 100  50 18 (c) As more salt and water is added, the concentration in the tank approaches the ratio of added salt to added water, that is, 4  5t 5  01 lbgal. We can also see this from the graph. As t becomes large, the concentration  50 C t  100  50t C t approaches the horizontal asymptote y  01, regardless of the initial concentration.

(b) After 10 minutes, the concentration is C 10 

2x 1 . Vertical asymptote x  3 and horizontal asymptote y  2: r x  . x 3 x 3 x 4 Vertical asymptotes x  1 and x  1, horizontal asymptote 0, and x­intercept 4: q x  . Of course, x  1 x  1

100. Vertical asymptote x  3: p x 

other answers are possible.

zeros

zeros

x 6  10

130  49x 2

 x2  8  4 has no x­intercept since the numerator has no real roots. Likewise, x 4  8x 2  15 x  8x 2  15 r x has no vertical asymptotes, since the denominator has no real roots. Since the degree of the numerator is two greater

101. r x 

than the degree of the denominator, r x has no horizontal or slant asymptotes. From the division below, we see that the graph of r is close to that of y  x 2  8 for large x. x 4  8x 2  15

x2 

8

x 6  0x 5  0x 4  0x 3  0x 2  0x  10 x6

 8x 4

 15x 2

8x 4

 64x 2

8x 4

15x 2

 10

49x 2

 130

1 1  f x  2. From this form we see that 102. (a) Let f x  2 . Then r x  x x  22

 120

y

the graph of r is obtained from the graph of f by shifting 2 units to the right. Thus

r has vertical asymptote x  2 and horizontal asymptote y  0.

1 1

x


100

CHAPTER 3 Polynomial and Rational Functions

2x 2  4x  5 3  3 f x  1  2. From this form we see that the graph of s is obtained from 2 x 2  2x  1 x  12 the graph of f by shifting 1 unit to the left, stretching vertically by a factor of 3, and shifting 2 units vertically. Thus r has vertical asymptote x  1 and horizontal asymptote y  2.

(b) s x 

y

2 x 2  2x  1

2x 2

2x 2

4x 4x

 

5 2 3 1 x

1

2  3x 2 12x  14 (c) Using long division, we see that p x  2  3  2 which cannot be graphed by transforming x  4x  4 x  4x  4 1 f x  2 . Using long division on q we have: x x 2  4x  4

3

3x 2

3x 2

 

12x  3x 2

0x 12x 12x

  

2 12

3

x 2  4x  4

3x 2 3x 2

14

 

12x 12x

 

0 12 12

12 So q x  2  12 f x  2  3. From this form we see that the graph of q is obtained  3  x  4x  4 x  22 from the graph of f by shifting 2 units to the right, stretching vertically by a factor of 12, and then shifting 3 units vertically down. Thus the vertical asymptote is x  2 and the horizontal asymptote is y  3. We show y  p x just 1 to verify that we cannot obtain p x from y  2 . x

y 1 1

y

1

1

y  q x

x

y  p x

x


SECTION 3.7 Polynomial and Rational Inequalities

3.7

101

POLYNOMIAL AND RATIONAL INEQUALITIES

1. To solve a polynomial inequality, we factor the polynomial into irreducible factors and find all the real zeros of the polynomial. Then we find the intervals determined by the real zeros and use test points in each interval to find the sign of the polynomial on that interval. Sign of x

 2

2 0

0 1

1 

x 2 x 1

P x  x x  2 x  1

From the table, we see that P x  0 on the intervals [2 0] and [1 .

2. To solve a rational inequality, we factor the numerator and the denominator into irreducible factors. The cut points are the real zeros of the numerator and the real zeros denominator. Then we find the intervals determined by the cut points, and we use test points to find the sign of the rational function on each interval. Sign of x 2 x 1 x 3 x 4

 4

4 2

2 1

1 3

3 

   x  2 x  1 r x     x  3 x  4 From the table, we see that r x  0 on the intervals  4, [2 1], and 3 .

3. The inequality x  3 x  5 2x  5  0 already has all terms on one side and the polynomial is factored. The intervals     determined by the zeros 3, 5, and  52 are  5, 5  52 ,  52  3 , and 3 . We make a sign diagram:     5  52  52  3 Sign of  5 3  x 3 x 5

2x  5

P x  x  3 x  5 2x  5

   None of the endpoints satisfies the inequality. The solution is  5   52  3 .

4. The inequality x  1 x  2 x  3 x  4  0 already has all terms on one side and the polynomial is factored. The intervals determined by the zeros 1, 2, 3, and 4 are  4, 4 2, 2 1, 1 3, and 3 . We make a sign diagram: Sign of x 1 x 2 x 3 x 4

P x  x  1 x  2 x  3 x  4

 4

4 2

2 1

1 3

3 

All of the endpoints satisfy the inequality. The solution is  4]  [2 1]  [3 .


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CHAPTER 3 Polynomial and Rational Functions

5. The inequality x  52 x  3 x  1  0 already has all terms on one side and the polynomial is factored. The intervals determined by the zeros 5, 3, and 1 are  5, 5 3, 3 1, and 1 . We make a sign diagram: Sign of

x  52

 5

5 3

3 1

1 

x 3 x 1

P x  x  52 x  3 x  1

None of the endpoints satisfies the inequality. The solution is  5  5 3  1 .

6. The inequality 2x  74 x  13 x  1  0 already has all terms on one side and the polynomial is factored. The     intervals determined by the zeros 72 , 1, and 1 are  1, 1 1, 1 72 , and 72   . We make a sign diagram:     7 1 72 Sign of  1 1 1 2 2x  74 x  13 x 1

P x  2x  74 x  13 x  1

 

  All of the endpoints satisfy the inequality [note that P 72  0]. The solution is [1 1]  72 .

7. We start by moving all terms to one side and factoring: x 3  4x 2  4x  16  x 3  4x 2  4x  16  x  4 x  2 x  2  0. The intervals determined by the zeros 4, 2, and 2 are  4, 4 2, 2 2, and 2 . We make a sign diagram: Sign of x 4 x 2 x 2

P x  x  4 x  2 x  2

 4

4 2

2 2

2 

All of the endpoints satisfy the inequality. The solution is [4 2]  [2 .

8. We start by moving all terms to one side and factoring: 2x 3  18x  x 2  9  2x 3  x 2  18x      9  x  3 2x  1 x  3  0. The intervals determined by the zeros 3, 12 , and 3 are  3, 3 12 , 12  3 , and 3 . We make a sign diagram: Sign of x 3

2x  1 x 3

P x  x  3 2x  1 x  3

 3 

  3 12 

13 2

3 

 

 None of the endpoints satisfies the inequality. The solution is  3  12  3 .

  9. We start by moving all terms to one side and factoring: 2x 3 x 2  918x  2x 3 x 2 18x 9  2x  1 x 2  9  0.   Note that x 2  9  0 for all x, so the sign of P x  2x  1 x 2  9 is negative where 2x  1 is negative and positive   where 2x  1 is positive. The endpoint x  12 does not satisfy the inequality, so the solution is  12 .


SECTION 3.7 Polynomial and Rational Inequalities

103

10. We start by moving all terms to one side and factoring: x 4  3x 3  x  3  x 4  3x 3  x  3  0. The possible rational zeros of P x  x 4  3x 3  x  3 are 1 and 3.

1 1 3 1 3 1

4

3

1 4 3 0    Thus, P x  x  1 x 3  4x 2  4x  3  x  1 x  3 x 2  x  1 . The last factor is positive everywhere, so 

we test the intervals  3, 3 1, and 1 . Sign of x 3 x 1

 3

3 1

1 

P x  x  1 x  3 x 2  x  1

Neither of the endpoints satisfies the inequality. The solution is 3 1.

11. All the terms are on the left size. We factor, using the substitution t  x 2 : x 4  7x 2  18  0       t 2  7t  18  t  2 t  9  0  x 2  2 x 2  9  0  x 2  2 x  3 x  3  0. The first factor is positive everywhere, so we test  3, 3 3, and 3 : Sign of x 3 x 3

P x  x 2  2 x  3 x  3

 3

3 3

3 

Neither of the endpoints satisfies the inequality. The solution is 3 3.

12. All the terms are on the left size. We factor, using the substitution t  x 2 : 4x 4  25x 2  36  0     4t 2  25t  36  4t  9 t  4  0  4x 2  9 x 2  4  2x  3 2x  3 x  2 x  2  0. The zeros are        32 and 2, so we test  2, 2  32 ,  32  32 , 32  2 , and 2 :       32 2  32  32  32 Sign of  2 2  2 x 2

x  32 x  32 x 2 P x

      All of the endpoints satisfy the inequality. 2  32  32  2 .


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CHAPTER 3 Polynomial and Rational Functions

13. All the terms are on the left size. To factor, note that the possible rational zeros of P x  x 3  x 2  17x  15 are 1, 3, 5, 15. 1 1 1 17 1

15

2 15

1 2 15 0  Thus, P x  x  1 x 2  2x  15  x  5 x  1 x  3. The zeros are 5, 1, and 3, so we test  5, 

5 1, 1 3, and 3 :

Sign of x 5

 5

5 1

1 3

3 

x 1 x 3 P x

All of the endpoints satisfy the inequality P x  0, so the solution is [5 1]  [3 .

14. All the terms are on the left size. To factor, note that the possible rational zeros of P x  x 4  3x 3  3x 2  3x  4 are 1, 2, 4. 1 1 3 3 3 4 1

4 1

4

1 4

1 4 

0

Thus, P x  x  1 x 3  4x 2  x  4  x  1 x  4 x 2  1 . The last factor is positive everywhere, so we test  4, 4 1, and 1 :

Sign of x 4

 4

4 1

1 

x 1 P x

None of the endpoints satisfies P x  0, so the solution is 4 1.

3 3 3    15. We start by moving all terms to one side and factoring: x 1  x 2  7 1  x 2  x  7 1  x 2  0 

P x  x  7 1  x3 1  x3  0. The intervals determined by the zeros 1, 1, and 7 are  1, 1 1, 1 7, and 7 . We make a sign diagram: Sign of x 7

1  x3 1  x3 P x

 1

1 1

1 7

7 

None of the endpoints satisfies the inequality. The solution is  1  1 7.


SECTION 3.7 Polynomial and Rational Inequalities

105

16. We start by moving all terms to one side and factoring: x 2 7  6x  1  6x 3  7x 2  1  0. The possible rational zeros of P x  6x 3  7x 2  1 are 1,  12 ,  13 ,  16 .

1 6

7 0 1

6 1

1

6 1 1 0   Thus, P x  x  1 6x 2  x  1  1  x 2x  1 3x  1. The intervals determined by the zeros  13 , 12 , and 1         13 ,  13  12 , 12  1 , and 1 . We make a sign diagram:       11   13  13  12 Sign of 1  2 3x  1 2x  1 1x

P x

    All of the endpoints satisfy the inequality P x  0. The solution is  13  12  [1 .

x 1  0. Since all nonzero terms are already on one side of the inequality symbol and there is no factoring x  10 needed, we find the intervals determined by the cut points 1 and 10. These are  1, 1 10, and 10 . We make a sign diagram:

17. r x 

Sign of

 1

1 10

x 1

10 

x  10

r x

The cut point 1 does not satisfy the inequality, and the cut point 10 is not in the domain of r. Thus, the solution is 1 10.

3x  7  0. Since all nonzero terms are already on one side of the inequality symbol and there is no factoring x 2     needed, we find the intervals determined by the cut points 73 and 2. These are  2, 2 73 , and 73   . We

18. r x 

make a sign diagram:

Sign of 3x  7 x 2

r x

 2 

  2 73 

7 3

  The cut point 73 satisfies equality, but the cut point 2 is not in the domain of r. Thus, the solution is 2 73 .


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CHAPTER 3 Polynomial and Rational Functions

x 3 x  3  2x  5 x 3 1 1  0 0 2x  5 2x  5 2x  5      x  8 r x   0. The intervals determined by the cut points are  8, 8  52 , and  52   . 2x  5     8  52  52   Sign of  8

19. We start by moving all terms to one side and simplifying:

 x  8 2x  5

r x

  The cut point 8 satisfies equality, but  52 is not in the domain of r. Thus, the solution is 8  52 .

20. We start by moving all terms to one side and simplifying:

x 4 x 4 x  4  4 x  5 4 4  0 0 x 5 x 5 x 5

3 x  8 3x  24  0  r x   0. The intervals determined by the cut points are  5, 5 8, and 8 . x 5 x 5 Sign of

 5

5 8

x 8

8 

x 5

r x

The cut point 8 satisfies equality, but 5 is not in the domain of r . Thus, the solution is  5  [8 .

5x  7 5x  7  4x  10 5x  7 1 1  0 0 4x  10 4x  10 4x  10     x 3  r x   0. The intervals determined by the cut points are   52 ,  52  3 , and 3 . 2 2x  5       52  52  3 Sign of 3 

21. We start by moving all terms to one side and simplifying:

x 3

2x  5

r x

  The cut point 3 satisfies equality, but  52 is not in the domain of r. Thus, the solution is  52  3 .

22. We start by moving all terms to one side and simplifying:

4x  6 4x  6 4x  6  2 x  7 2 2 0 0 x 7 x 7 x 7

2 x  10 2x  20 0  0. The intervals determined by the cut points are  7, 7 10, and 10 . x 7 x 7 Sign of x  10 x 7

r x

 7

7 10

10 

The cut point 10 fails to satisfy the strict inequality and 7 is not in the domain of r, so the solution is  7  10 .


SECTION 3.7 Polynomial and Rational Inequalities

23. r x 

2x  5

x 2  2x  35 2x  5

 0.

107

Since all nonzero terms are already on one side, we factor:

2x  5 . Thus, the cut points are 7,  52 , and 5. The intervals determined by these x  7 x  5     points are  7, 7  52 ,  52  5 , and 5 . We make a sign diagram:     7  52  52  5 Sign of  7 5 

r x 

x 2  2x  35

2x  5 x 7 x 5

r x

The cut point  52 satisfies equality, but the cut points 7 and 5 are not in the domain of r. Thus, the solution is   7  52  5 .

4x 2  25 4x 2  25 2x  5 2x  5  0. Since all nonzero terms are already on one side, we factor: r   . x x  3 x  3 x2  9 x2  9       Thus, the cut points are  52 and 3. The intervals determined by these points are  3, 3  52 ,  52  52 , 52  3 ,

24. r x 

and 3 . We make a sign diagram: Sign of 2x  5

  3  52

   52  52

 3 

2x  5 x 3 x 3

r x

53 2

3 

    The cut points  52 satisfy equality, but the cut points 3 are not in the domain of r. Thus, the solution is 3  52  52  3 .

x2 x2 25. r x  2  0. Since all nonzero terms are already on one side, we factor: r x  . The cut x  5 x  2 x  3x  10 points are 5, 0, and 2. We make a sign diagram: Sign of x 5

 5

5 0

0 2

2 

x 2

r x

The cut points 5 and 2 are not in the domain of r, so the solution is 5 2.

x 3 x 3  0. Since all nonzero terms are already on one side, we factor: r x  . The cut points are 26. r x  2 x  6x  9 x  32 3. We make a sign diagram: Sign of

x  32 x 3

r x

 3

3 3

3 

The cut point 3 is not in the domain of r, and the cut point 3 satisfies the inequality, so the solution is  3  3 3].


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CHAPTER 3 Polynomial and Rational Functions

27. We start by moving all terms to one side and factoring: r x 

x 2  3  2 x  1 x2  3  2   0  x 1 x 1

x 2  2x  1 x  12   0. The cut points are 1. We make a sign diagram: x 1 x 1 Sign of

x  12 x 1

r x

 1

1 1

1 

The cut point 1 is not in the domain of r and the cut point 1 does not satisfy the original equation, so the solution is 1 1  1 .   4x  3  x 2  1 x 2  4x  4 4x  3 1 0 0 28. We start by moving all terms to one side and factoring: 2 2 x 1 x 1 x2  1 x  22  0, which is true for all x. The solution is  . r x  2 x 1   x 2  2x  3 x  3 x  1   0. The intervals determined by the cut points are  3, 3  23 , 29. r x  2 3x  2 x  3 3x  7x  6   2  3  1 , 1 3, and 3 .     3  23  23  1 Sign of  3 1 3 3  x 3

3x  2 x 1 x 3

r x

    None of the cut points satisfies the strict inequality, so the solution is  3   23  1  3 .

x 1 x 1   0. The second factor in the denominator is positive for all x, so the intervals   30. r x  3 x 1 x  1 x 2  x  1 determined by the cut points are  1, 1 1, and 1 . Sign of x 1 x 1

r x

 1

1 1

1 

 4

4 3

3 

The cut point 1 satisfies equality, but 1 is not in the domain of r. Thus, the solution is  1  [1 .     x x 2  6x  9  3 x 2  6x  9 x  3 x  32 x 3  3x 2  9x  27    0. The second factor in the 31. r x  x 4 x 4 x 4 numerator is positive for all x, so the intervals determined by the cut points are  4, 4 3, and 3 . Sign of x 4 x 3

r x

The cut point 3 satisfies equality, but 4 is not in the domain of r. Thus, the solution is 4 3].


SECTION 3.7 Polynomial and Rational Inequalities

109

x 2  16 x  4 x  4 x  4 x  4      0. The last factor in the denominator is positive for  32. r x  4   2 x  4 x2  4 x  16 x  2 x  2 x 2  4 all x, so the intervals determined by the cut points are  4, 4 2, 2 2, 2 4, and 4 . Sign of x 4

 4

4 2

2 2

2 4

4 

x 2 x 2 x 4

r x

None of the cut points satisfy the strict inequality. Thus, the solution is 4 2  2 4.

x  12  0. The numerator is nonnegative for all x, but note that x  1 fails to satisfy the strict inequality. x  1 x  2 The intervals determined by the cut points are  2, 2 1, and 1 .

33. r x 

Sign of

x  12

 2

2 1

1 

x 2 x 1

r x

 (except at x  1)

The cut points 2 and 1 are not in the domain of r, so the solution is  2  1 1  1 .

34. r x  1 .

x 2  2x  1

x 3  3x 2  3x  1

x  12

x  13

 0. The intervals determined by the cut points are  1, 1 1, and

Sign of x  13

 1

1 1

1 

x  12

r x

The cut point 1 is not in the domain of r and the cut point 1 satisfies the inequality. Thus, the solution is  1]  1. 5 x 5 x  4  4  0 2 x 1 2 x 1 x 2  7x  18 x x  1  5 2  4 2 x  1 x  2 x  9 0 0  0. The intervals determined by the cut 2 x  1 2 x  1 2 x  1 points are  2, 2 1, 1 9, and 9 .

35. We start by moving all terms to one side and factoring:

Sign of x 2 x 1 x 9

r x

 2

2 1

1 9

9 

The cut point 1 is inadmissible in the original inequality, so the solution is [2 1  [9 .


110

CHAPTER 3 Polynomial and Rational Functions

x 1 x 2 x 1 x 2     0  36. We start by moving all terms to one side and factoring: x 3 x 2 x 3 x 2   2 x  12 x  2 x  2  x  1 x  3  0  r x   0. The intervals determined by the cut points are x  3 x  2 x  3 x  2      3, 3  12 ,  12  2 , and 2 .     3  12  12  2 Sign of  3 2  x 3

    The cut points fail to satisfy the strict inequality, so the solution is 3  12  2 .

x  12 x 2

r x (note negative sign)

6 6 6 6   1  1  0 x 1 x x 1 x x2  x  6 6x  6 x  1  x x  1 x  2 x  3 0  0  r x    0. The intervals determined by the x x  1 x x  1 x x  1 cut points are  2, 2 0, 0 1, 1 3, and 3 .

37. We start by moving all terms to one side and factoring:

Sign of x 2 x

x 1 x 3

r x (note negative sign)

 2

2 0

0 1

1 3

3 

The cut points 0 and 1 are inadmissible in the original inequality, so the solution is [2 0  1 3]. 1 2x 1   2  x 3 x 2 x x 2 1 1 2x x  2 x  1  x  3 x  1  2x x  3    0   0  x 3 x 2 x  2 x  1 x  2 x  1 x  3     3x  1  0. The intervals determined by the cut points are  2, 2  13 ,  13  1 , 1 3, and x  2 x  1 x  3 3 .     2  13  13  1 Sign of  2 1 3 3 

38. We start by moving all terms to one side and factoring:

x 2

3x  1 x 1 x 3

r x

The cut point  13 satisfies equality, but 2, 1, and 3 are not in the domain of r, so the solution is    2   13  1  3 .


SECTION 3.7 Polynomial and Rational Inequalities

111

39. We start by moving all terms to one side and factoring: 1 1 1 x  22 x 2     2 x 1 x 2 x  2 x  1 x  22 x  22 r x 

x  22  x  2  x  1 x  1 x  22

and 2. We make a sign diagram:

x 2  2x  1

x  1 x  22

Sign of x 1

x 1

x  12

x  1 x  22

2 1

1 

r x

x 1

x  22

 2

x  22

x  1 x  22

 0. The cut points are 1

Neither cut point satisfies the original inequality, so the solution is  2  2 1.

1 2 1 1 2 1       0 x x 1 x 2 x x 1 x 2 x 2  3x  2  x 2  2x  2x 2  2x x  1 x  2  x x  2  2x x  1  0   0   x x  1 x  2 x x  1 x  2     3x  2 r x   0. The intervals determined by the cut points are  2, 2 1, 1  23 ,  23  0 , x x  1 x  2 and 0 .     1  23  23  0 Sign of  2 2 1 0 

40. We start by moving all terms to one side and simplifying:

x 2 x 1

3x  2 x

r x

   None of the cut points satisfies the strict inequality. Thus, the solution is 2 1   23  0 .

41. The graph of f lies above that of g where f x  g x; that is, where x 2  3x  10  x 2  3x  10  0  x  2 x  5  0. We make a sign diagram: Sign of x 2 x 5

x  2 x  5

 2

2 5

5 

Thus, the graph of f lies above the graph of g on  2 and 5 . 42. The graph of f lies above that of g where f x  g x; that is, where 

1 . We make a sign diagram: x x  1

Sign of x

1 1 1 1 x  1  x    0 0 x x 1 x x 1 x x  1

 0

0 1

1 

x 1  1  x x  1 Thus, the graph of f lies above the graph of g on 0 1.


112

CHAPTER 3 Polynomial and Rational Functions

43. The graph of f lies above that of g where f x  g x; that is, where 4x  r x 

2x  1 2x  1  0. We make a sign diagram: x       12  12  0 Sign of 2x  1

1 4x 2  1 1  4x   0  0 x x x

  0 12

       Thus, the graph of f lies above the graph of g on  12  0 and 12   .

x

2x  1

r x

1 2

x3  x2  2 2 2 0 44. The graph of f lies above that of g where f x  g x; that is, where x 2  x   x 2  x   0  x x x   x  1 x 2  2x  2  0. The second factor in the numerator is positive for all x. We make a sign diagram:  r x  x Sign of

 0

0 1

1 

x x 1

r x

Thus, the graph of f lies above the graph of g on  0 and 1 .  45. f x  6  x  x 2 is defined where 6  x  x 2   x  2 x  3  0. We make a sign diagram: Sign of x 2

 2

2 3

3 

x 3

 x  2 x  3

Thus, the domain of f is [2 3].  5x 5x is defined where  0 and 5  x  0. We make a sign diagram: 46. g x  5x 5x Sign of 5x

 5

5 5

5 

 1

1 1

1 

5x    5x    5x The cut point 5 is permissible, and so the domain of g is [5 5.    4 47. h x  x 4  1 is defined where x 4  1  x  1 x  1 x 2  1  0. The last factor is positive for all x. We make a sign diagram:

Sign of x 1 x 1

x2  1

x  1 x  1 x 2  1

Thus, the domain of h is  1]  [1 .


SECTION 3.7 Polynomial and Rational Inequalities

48. f x  

1

x 4  5x 2  4 make a sign diagram:

   is defined where x 4  5x 2  4  x 2  4 x 2  1  x  2 x  2 x  1 x  1  0. We Sign of x 2 x 1 x 1 x 2

x 4  5x 2  4

 2

2 1

1 1

1 2

2 

Thus, the domain of h is  2  1 1  2 . 49.

50.

20

20 10

10 ­4

­2

2

­4

4

­2

­10

2

­10 ­20

From the graph, we see that x 3  2x 2  5x  6  0 on

From the graph, we see that 2x 3  x 2  8x  4  0 on    2]   12  2 .

[2 1]  [3 .

51.

52.

­2

­1

4

4

2

2 1

­2

2

­2

­4

­1 ­2

2

3

4

From the graph, we see that x 4  4x 3  8x  0 on

approximately  137  037 1.

approximately  124  0 2  324 .

54.

53.

20

40

10

20 1.6

­1

1

­4

From the graph, we see that 2x 3  3x  1  0 on

0

1

2

  From the graph, we see that 5x 4  8x 3 on 0 85 . 55.

113

­2

­1

1

2

­10

From the graph, we see that x 5  x 3  x 2  6x on

approximately [131 0]  [151 .

    1  x2  x 4 x x  1   1  x2 1  x2  4 x x  1   4 x x  1  0   0     x x x

3x 2  2x  1 1  2x  x 2  4x 2  4x 1  x 3x  1 0  0  r x   0. The domain of r is 0 , and    x x x  both 3x  1 and x are positive there. 1  x  0 for x  1, so the solution is 0 1].


114

CHAPTER 3 Polynomial and Rational Functions

7x  8 56. 23 x 13 x  212  12 x 23 x  212  0  16 x  212 x 13 [4 x  2  3x]  0  r x    0.  3 6 x x 2 Note that the domain of r is 2 . We make a sign diagram with cut points  87 and 0:     2  87  87  0 Sign of 0  7x  8  3 x  x 2

   Neither cut point is a solution. The solution is  87  0 .

r x

57. We want to solve P x  x  a x  b x  c x  d  0, where a  b  c  d. We make a sign diagram with cut points a, b, c, and d: Sign of

 a

a b

b c

c d

x a

d 

x b x c

x d P x

Each cut points satisfies equality, so the solution is  a]  [b c]  [d .

[x  a] x  b x 2  a  b x  ab x  a x  b    0. Note that x c x c x  c 0  a  c, so c  a  0. We make a sign diagram with cut points c, a, and b:

58. Factoring the numerator, we have r x  Sign of

x  c x b

x  a P x

 c

c a

a b

b 

The cut points a and b satisfy equality, but c is not in the domain of P. Thus, the solution is  c  [a b].   2500  275 x 2  2 2500 2500  300  2  275  0  0 59. We want to solve the inequality T x  300  25  2 x 2 x 2 x2  2   25 11x 2  78   0. Because x represents distance, it is positive, and the denominator is positive for all x. Thus, x2  2   the inequality holds where 11x 2  78  x  78 11  266. The temperature is below 300 C at distances greater than 266 meters from the center of the fire.

2  175  0   2  25  4375  0. Using the Quadratic Formula, we find 60. We want to solve d   175    25    25  252  4 1 4375 25  18,125 25  18,125   . Because  is positive, we have    548, so 2 1 2 2 Kerry can travel at up to 548 mih.


CHAPTER 3

61.

Review

88x  x 2 and y  40 in the viewing rectangle 17  17 20

60

We graph N x 

40

[0 100] by [0 60], and see that N x  40 for approximately 95  x  423.

20

Thus, cars can travel at between 95 and 423 mih.

0

0

50

115

100

62. (a) F1 is inversely proportional to the square of the distance between the balls with 5 3 constant of proportionality 3, so F1   2 . F2 is inversely proportional to the x 0 1 cube of the distance with constant of proportionality 1, so F2  3 . Thus, 1 2 3 x ­5 1 3 F x  F1  F2  3  2 . x x     1 3 (b) F x  0  3  2  0  x 2  3x 3  x  13 . F x  0 on 0 13 and F x  0 on 13   . From the graph x x in part (a), the greatest repulsive force occurs when x  12 .

(c) As the distance gets very small, the attractive force increases without bound, and as the distance gets very large, the repulsive force diminishes, becoming arbitrarily close to 0.

CHAPTER 3 REVIEW

  1. (a) f x  x 2  6x  2  x 2  6x  2    x 2  6x  9  2  9

  2. (a) f x  2x 2  8x  4  2 x 2  4x  4    2 x 2  4x  4  4  8  2 x  22  4

 x  32  7

(b)

(b)

y

y

1 _1

0 x

1 0

1

x


116

g(x)

CHAPTER 3 Polynomial and Rational Functions

  3. (a) f x  1  10x  x 2   x 2  10x  1     x 2  10x  25  1  25   x  52  26

(b)

  4. (a) f x  2x 2  12x  2 x 2  6x    2 x 2  6x  9  18  2 x  32  18

(b)

y

y

10 0

1 x

5 0

1

x

 2     5. f x  x 2  3x  1   x 2  3x  1   x 2  3x  94  1  94   x  32  54 has the maximum value 54 when x  32 .

    6. f x  3x 2  18x  5  3 x 2  6x  5  3 x 2  6x  9  5  27  3 x  32  22 has the minimum value 22 when x  3.

    7. We write the height function in vertex form: h t  16t 2  48t  32  16 t 2  3t  32  16 t 2  3t  94  2  32  36  16 t  32  68. The stone reaches a maximum height of 68 ft. 8. We write the profit function in vertex form:

  P x  1500  12x  0004x 2  0004 x 2  3000x  1500    0004 x 2  3000x  15002  1500  0004 15002  0004 x  15002  7500

Thus, the maximum profit of $7500 is achieved when 1500 units are sold.

9. P x  x 3  64

  10. P x  2x 3  16  2 x  2 x 2  2x  4

y

y

5

(0, 64)

(0, _16)

10 1

The domain and range are .

(2, 0) 1

(4, 0)

x

The domain and range are .

x


CHAPTER 3

11. P x  2 x  14  32

117

12. P x  81  x  34 y

y

5

25

(1, 0) x

(_3, 0)

Review

(0, 0)

(6, 0) 1

x

(0, _30)

The domain is  and the range is [32 .

13. P x  32  x  15

The domain is  and the range is  81].

14. P x  3 x  25  96

y

y

(0, 31) 10

(_1, 0)

1

x

20 (0, 0)

1

The domain and range are .

The domain and range are .

15. (a) P x  x  3 x  1 x  5  x 3 7x 2 7x15 has odd degree and a positive leading coefficient, so

  16. (a) P x   x  5 x 2  9 x  2

 x 4  3x 3  19x 2  27x  90

y   as x   and y   as x  . (b)

x

has even degree and a negative leading coefficient, so

y

y   as x   and y   as x  . (b)

y 20

10 1

1 x

x


118

CHAPTER 3 Polynomial and Rational Functions

17. (a) P x   x  12 x  4 x  22  x 5  2x 4  11x 3  8x 2  20x  16

   18. (a) P x  x 2 x 2  4 x 2  9  x 6  13x 4  36x 2 has even degree and a positive leading coefficient, so y   as x   and y   as x  .

has odd degree and a negative leading coefficient, so y   as x   and y   as x  .

(b)

y

y

(b)

100

20 x

1 _5

5

x

_100

19. (a) P x  x 3 x  22 . The zeros of P are 0 and 2,

20. (a) P x  x x  13 x  12 . The zeros of P are 1,

(b) We sketch the graph using the guidelines from

(b) We sketch the graph using the guidelines from

with multiplicities 3 and 2, respectively.

Section 3.2.

0, and 1, with multiplicities 3, 1, and 2, respectively.

Section 3.2. y

y

1

1

x

0.1 1

x

21. y  x 2  8x, [4 12] by [50 30]. x­intercepts: 0 and 22. P x  x 3  4x  1. x­intercepts: 21, 03, and 19. 8. y­intercept: 0. No local minimum. Local maximum

y­intercept: 1. Local maximum 12 41. Local

4 16. End behavior: y   as x  .

minimum 12 21. End behavior: y   as

20

x   and y   as x  . 4

­4 ­2 ­20 ­40

2 4 6 8 10 12

2 ­3

­2

­1 ­2 ­4

1

2

3


CHAPTER 3

23. P x  2x 3  6x 2  2, [2 5] by [6 8].

x­intercepts: 05, 07, and 29. y­intercept: 2. Local

Review

119

24. y  x 4  4x 3 , [5 5] by [30 30]. x­intercepts: 4 and 0. y­intercept: 0. Local minimum at 3 27. No local

maximum is 2 6. Local minimum is 0 2. End

maximum. End behavior: y   as x  .

behavior: y   as x   and y   as x  .

20

5 ­4

­2

2

­2

2

4

­20

4

­5

25. P x  3x 4  4x 3  10x  1. x­intercepts: 01 and 21. 26. P x  x 5  x 4  7x 3  x 2  6x  3. x­intercepts: y­intercept: 1. Local minimum is 14 145. There is

30, 13, and 19. y­intercept: 3. Local maxima are

no local maximum. End behavior: y   as x  .

24 332 and 05 50. Local minima are 06 06 and 16 16. End behavior: y   as x  

20

and y   as x  .

10 ­1

1

­10

2

40

3

20

­20 ­4

­2

2 ­20

27. (a) Use the Pythagorean Theorem and solving for y 2 we have, x 2  y 2  102  y 2  100  x 2 . Substituting   we get S  138x 100  x 2  1380x  138x 3 .

28. (a) The area of the four sides is 2x 2  2x y  1200 

600  x 2 2x y  1200  2x 2  y  . Substituting x   600  x 2  600x  x 3 . we get V  x 2 y  x 2 x

(b) Domain is [0 10]. (c)

(b) 4000

5000

2000 0

0

5

0

10

(d) The strongest beam has width 58 inches. 29.

x 2  5x  2 3

1 1

5

6

2

4

Using synthetic division, we see that Q x  x  2 and R x  4.

20

3x 2  x  5 x 2

2

3

10

(c) V is maximized when x  1414, y  2828. 30.

x 3

0

2

3 3

1 6 5

5 10

5

Using synthetic division, we see that Q x  3x  5 and R x  5.


120

31.

CHAPTER 3 Polynomial and Rational Functions

2x 3  x 2  3x  4 x 5

32. 2

5

3

1

2

Using synthetic division, we see that

294

1

34.

1

0

25

5

1

2

7

85

415

8 17

5

7

49

329

7

47

325

2x 4  3x 3  12 x 4 4

422

83

4

Q x  x 2  7x  47 and R x  325.

x 4  8x 2  2x  7 x 5 5

2

Using synthetic division, we see that

Q x  2x 2  11x  58 and R x  294. 33.

0

1

290

58

11

7

4

55

10

x 3  2x  4 x 7

2 2

3

0

0

8

20

80

5

20

80

12 320

308

Using synthetic division, we see that

Using synthetic division, we see that

Q x  x 3  5x 2  17x  83 and R x  422.

Q x  2x 3  5x 2  20x  80 and R x  308.

35. [AU: Erroneously numbered 31 in ms]

36.

2x 3  x 2  8x  15 x 2  2x  1 x 2  2x  1

2x 

x 4  2x 2  7x x2  x  3 x2  x  3

3

x2 

x 

x4 

x 3  3x 3

2x 3  x 2  8x  15

2x 3  4x 2  2x

x 3  5x 2  7x 4x 2  4x  0

3x 2  6x  3

4x 2  4x  12

12

37. P x  2x 3  9x 2  7x  13; find P 5. 5

2

9

10

2

1

Therefore, P 5  3.

7

5

2

13 10

3

0

x 3  x 2  3x

3x 2  6x  15

Therefore, Q x  2x  3, and R x  12.

4

x 4  0x 3  2x 2  7x 

12

Therefore, Q x  x 2  x  4, and R x  12. 38. Q x  x 4  4x 3  7x 2  10x  15; find Q 3 3

1 1

4

7

10

15

3

3

12

6

1

4

2

21

By the Remainder Theorem, we have Q 3  21.

39. The remainder when dividing P x  x 500  6x 101  x 2  2x  4 by x  1 is

P 1  1500  6 1201  12  2 1  4  8.

40. The remainder when dividing P x  x 101  x 4  2 by x  1 is P 1  1101  14  2  0. Q(x) Q


CHAPTER 3

41. 12 is a zero of P x  2x 4  x 3  5x 2  10x  4 if   P 12  0. 1 2

2

1

5

1 2 Since P

  1 2

1

2

4

10

2 8

4

121

42. x  4 is a factor of

P x  x 5  4x 4  7x 3  23x 2  23x  12 if P 4  0. 4

4

0

1 1

 0, 12 is a zero of the polynomial.

Review

4 4

0

7 0

7

23

12

20

12

23 28

5

3

0

Since P 4  0, x  4 is a factor of the polynomial.

43. (a) P x  x 5  6x 3  x 2  2x  18 has possible rational zeros 1, 2, 3, 6, 9, 18.

(b) Since P x has 2 variations in sign, there are either 0 or 2 positive real zeros. Since P x  x 5 6x 3  x 2 2x 18 has 3 variations in sign, there are 1 or 3 negative real zeros.

44. (a) P x  6x 4  3x 3  x 2  3x  4 has possible rational zeros 1, 2, 4,  12 ,  13 ,  23 ,  43 ,  16 .

(b) Since P x has no variations in sign, there are no positive real zeros. Since P x  6x 4  3x 3  x 2  3x  4 has 4 variations in sign, there are 0, 2, or 4 negative real zeros.

45. (a) P x  3x 7  x 5  5x 4  x 3  8 has possible rational zeros 1, 2, 4, 8,  13 ,  23 ,  43 ,  83 .

(b) Since P x has 2 variations in sign, there are either 0 or 2 positive real zeros. Since P x  3x 7  x 5 5x 4  x 3 8 has 3 variations in sign, there are 1 or 3 negative real zeros.

46. (a) P x  6x 10  2x 8  5x 3  2x 2  12 has possible rational zeros 1, 2, 3, 4, 6, 12,  12 ,  13 ,  16 ,  23 ,  32 ,  43 .

(b) Since P x has 2 variations in sign, there are either 0 or 2 positive real zeros. Since P x  6x 10  2x 8  5x 3  2x 2  12 has 2 variations in sign, there are 0 or 2 negative real zeros.

  47. (a) P x  x 3  16x  x x 2  16

48. (a) P x  x 3  3x 2  4x

 x x  1 x  4

 x x  4 x  4

has zeros 0, 1, and 4 (all of multiplicity 1).

has zeros 4, 0, 4 (all of multiplicity 1). (b)

(b)

y

y

2

5 1

x

1

x


122

CHAPTER 3 Polynomial and Rational Functions

  49. (a) P x  x 4  x 3  2x 2  x 2 x 2  x  2

   50. (a) P x  x 4  5x 2  4  x 2  4 x 2  1  x  2 x  2 x  1 x  1

 x 2 x  2 x  1

Thus, the zeros are 1, 1, 2, 2 (all of multiplicity

The zeros are 0 (multiplicity 2), 2 (multiplicity 1),

1).

and 1 (multiplicity 1).

(b)

(b)

y

y

5

1

x

1

x

1

51. (a) P x  x 4  2x 3  7x 2  8x  12. The possible rational zeros are 1, 2, 3, 4, 6, 12. P has 2 variations in sign, so it has either 2 or 0 positive real zeros. 1

1

2

8

1

7

1

1

8

1

8

2 1 0 6

6

0

12

2

1

2

7

2

0

0

2

12

12

0

14

12

7

6

 P x  x 4  2x 3  7x 2  8x  12  x  2 x 3  7x  6 . Continuing: 2

zero

4

4

1 2 2 10

3 1 0 7 6 3

9

6

1 3

2

0

8

(b)

zero.

0

 x  2 is a root.

y

so x  3 is a root and

  P x  x  2 x  3 x 2  3x  2

5

 x  2 x  3 x  1 x  2

Therefore the real roots are 2, 1, 2, and 3 (all of multiplicity 1).

x

1

x

zeros

52. (a) P x  x 4  2x 3  2x 2  8x  8      x 2 x 2  2x  2  4 x 2  2x  2     x 2  4 x 2  2x  2    x  2 x  2 x 2  2x  2

(b)

y

5

The quadratic is irreducible, so the real roots are 2 (each of multiplicity 1).

1

zeros


zeros

CHAPTER 3

Review

123

53. (a) P x  2x 4  x 3  2x 2  3x  2. The possible rational roots are 1, 2,  12 . P has one variation in sign, and hence 1 positive real root. P x has 3 variations in sign and hence either 3 or 1 negative real roots. 1

zero.

2 2

1

2

2

3

3

2

5

zeros.

2

3

5 2 0  x  1 is a zero.   P x  2x 4  x 3  2x 2  3x  2  x  1 2x 3  3x 2  5x  2 . (b)

Continuing: 1 2 2

3

2

2 2

2 1 4

 12 2 2

5

1

4 2

3

5

3 5

2

4 2 14

2 1 7 12

2

1 1 2 2

4

y

5

0  x   12 is a zero.

1

x

  P x  x  1 x  12 2x 2  2x  4 . The quadratic is irreducible, so the real zeros are 1 and  12 (each of

multiplicity 1).

54. (a) P x  9x 5  21x 4  10x 3  6x 2  3x  1. The possible rational zeros are 1,  13 ,  19 . P has 3 variations in sign, hence 3 or 1 positive real roots. P x has 2 variations in sign, hence 2 or 0 real negative roots. 1 9 21

zeros.

10

6 3 1

9 12 2

4

1

zeros.

9 12 2 4 1 0  x  1 is a zero.   P x  x  1 9x 4  12x 3  2x 2  4x  1 . Continuing: 1 9 12 2

4

1

9 3 5 1

9 3 5 1   P x  x  12 9x 3  3x 2  5x  1 .

0  x  1 is a zero again. (b)

y

Continuing:

1 9 3 5 1 9

9

6

6

1

4

1

0  x  1 is a zero yet again.   P x  x  13 9x 2  6x  1  x  13 3x  12

So the real zeros of P are 1 (multiplicity 3) and  13 (multiplicity 2).

1

x


124

CHAPTER 3 Polynomial and Rational Functions

  55. Because it has degree 3 and zeros  12 , 2, and 3, we can write P x  C x  12 x  2 x  3 

C x 3  92 C x 2  72 C x  3C. In order that the constant coefficient be 12, we must have 3C  12  C  4, so P x  4x 3  18x 2  14x  12.

56. Because it has degree 4 and zeros 4, 4, and 3i, 3i must also be a zero. Thus,   P x  C x  42 x 2  9  C x 4  8C x 3  25C x 2  72C x  144C.

The coefficient of x 2 is 25  25C, so C  1 and hence P x  x 4  8x 3  25x 2  72x  144.

57. No, there is no polynomial of degree 4 with integer coefficients that has zeros i, 2i, 3i and 4i. Since the imaginary zeros of polynomial equations with real coefficients come in complex conjugate pairs, there would have to be 8 zeros, which is impossible for a polynomial of degree 4.    58. P x  3x 4  5x 2  2  3x 2  2 x 2  1 . Since 3x 2  2  0 and x 2  1  0 have no real zeros, it follows that 3x 4  5x 2  2 has no real zeros.

solutions,

59. P x  x 3  x 2  x  1 has possible rational zeros 1. 1

1

1

1

1

0

1

0

1

1

1

0

 So P x  x  1 x 2  1 . Therefore, the zeros are 1 and i.

 x  1 is a zero.

  60. P x  x 3  8  x  2 x 2  2x  4  0, so 2 is a zero. Using the Quadratic Formula, the other zeros are    2  22  4 1 4  1  12 x  1  3i. 2 2 61. P x  x 3  3x 2  13x  15 has possible rational zeros 1, 3, 5, 15. 1

1

3

1

1

13

15

2

15

2 15 0  x  1 is a zero.   So P x  x 3  3x 2  13x  15  x  1 x 2  2x  15  x  1 x  5 x  3. Therefore, the zeros are 3, 1, and 5.

62. P x  2x 3  5x 2  6x  9 has possible rational zeros 1, 3, 9,  12 ,  32 ,  92 . Since there is one variation in sign, there is a positive real zero. 1 2 5 6 9 2

7

1

3 2

5 6 9

6 33 81

3 2 5 6 9 2

3 12

9

2 11 27 72  x  3 is an upper bound 2 8 6 0  x  32 is a zero.   So P x  2x 3  5x 2  6x  9  2x  3 x 2  4x  3  2x  3 x  3 x  1. Therefore, the zeros are 3, 1 2 7

and 32 .

1 8


CHAPTER 3

Review

125

63. P x  x 4  6x 3  17x 2  28x  20 has possible rational zeros 1, 2, 4, 5, 10, 20. Since all of the coefficients are positive, there are no positive real zeros. 1

1

6

17

28

20

1

5

12

16

2

1

17

28

20

2

8

18

20

9 10 0  x  2 is a zero.  P x  x 4  6x 3  17x 2  28x  20  x  2 x 3  4x 2  9x  10 . Continuing with the quotient, we have 1

5

12

16

1

6

4

2

1

4

2

1

2

9

10

4

10

5

4

0  x  2 is a zero.   Thus P x  x 4  6x 3  17x 2  28x  20  x  22 x 2  2x  5 . Now x 2  2x  5  0 when 

x  2 4451  24i  1  2i. Thus, the zeros are 2 (multiplicity 2) and 1  2i. 2 2

64. P x  x 4  7x 3  9x 2  17x  20 has possible rational zeros 1, 2, 4, 5, 10, 20. 1 1 7

1

9 17 20 2 1 7 8

17

0

9 17 20

2 18

54

1 9 27

37

1 1

74

7

9 17 20

1 6

3

20

54  x  2 is an 1 6 3 20 0  x  1 is a zero. upper bound.   So P x  x 4  7x 3  9x 2  17x  20  x  1 x 3  6x 2  3x  20 . Continuing with the quotient, we have 1 8 17

0 20

1 1

6

3 20

1 5

2

2 1

3 20

4 1

10

2 8

5 2 18

4 5 10

6

3 20

4 8

20

1 2 5 0  x  4 is a zero.  So P x  x 4  7x 3  9x 2  17x  20  x  1 x  4 x 2  2x  5 . Now using the Quadratic Formula on     2  2 6 2  4  4 1 5   1  6. Thus, the zeros are 4, 1, and 1  6. x 2  2x  5 we have: x  2 2 1

1

6

65. P x  x 5  3x 4  x 3  11x 2  12x  4 has possible rational zeros 1, 2, 4. 1 1 3 1 11 12 1 2 3

4

8 4

1 2 3

8 4 0  x  1 is a zero.   P x  x 5  3x 4  x 3  11x 2  12x  4  x  1 x 4  2x 3  3x 2  8x  4 . Continuing with the quotient, we have 1 1 2 3

8 4

1 1 4

4

1 1 4 4 0  x  1 is a zero.     x 5  3x 4  x 3  11x 2  12x  4  x  12 x 3  x 2  4x  4  x  13 x 2  4  x  13 x  2 x  2

Therefore, the zeros are 1 (multiplicity 3), 2, and 2.      66. P x  x 4  81  x 2  9 x 2  9  x  3 x  3 x 2  9  x  3 x  3 x  3i x  3i. Thus, the zeros are 3, 3i.


126

CHAPTER 3 Polynomial and Rational Functions

       67. P x  x 6  64  x 3  8 x 3  8  x  2 x 2  2x  4 x  2 x 2  2x  4 . Now using the Quadratic

Formula to find the zeros of x 2  2x  4, we have   3  1  i 3, and using the Quadratic Formula to find the zeros of x 2  2x  4, we have x  2 4441  22i 2 2       22i2 3  1  i 3. Therefore, the zeros are 2, 2, 1  i 3, and 1  i 3. x  2 4441 2

1 . 68. P x  18x 3  3x 2  4x  1 has possible rational zeros 1,  12 ,  13 ,  16 ,  19 ,  18

1 18

1 18 2

3 4 1

18 21 17

3 4 1

9

6

1

18 21 17 16  x  1 is an upper bound. 18 12 2 0  x  12 is a zero.   So P x  18x 3  3x 2  4x  1  2x  1 9x 2  6x  1  2x  1 3x  12 . Thus the zeros are 12 and  13

(multiplicity 2).

  69. P x  6x 4  18x 3  6x 2  30x  36  6 x 4  3x 3  x 2  5x  6 has possible rational zeros 1, 2, 3, 6. 1 6 18

6 30

6 12

6 12

36

6 36

6 36 

0  x  1 is a zero.     6 x  1 x 3  2x 2  x  6 .

So P x  6x 4  18x 3  6x 2  30x  36  x  1 6x 3  12x 2  6x  36 Continuing with the quotient we have 1 1 2 1 6

2 1 2 1 6

1 1 2

2

1

3 1 2 1 6

0 2

3

6

1 2 0  x  3 is a zero.  So P x  6x 4  18x 3  6x 2  30x  36  6 x  1 x  3 x 2  x  2 . Now x 2  x  2  0 when 1 1 2 8

0 1 8

3

1

7 7 x  1 1412  1i , and so the zeros are 1, 3, and 1i . 2 2 2

    70. P x  x 4  15x 2  54  x 2  9 x 2  6 . If x 2  9, then x  3i. If x 2  6, then x  i 6. Therefore, the  zeros are 3i and i 6. 71. 2x 2  5x  3  2x 2  5x  3  0. The solutions are x  05, 3.

2

4

72. Let P x  x 3  x 2  14x  24. The solutions to P x  0 are x  3, 2, and 4. ­5

5 ­20

­5 ­40


CHAPTER 3

73. x 4  3x 3  3x 2  9x  2  0 has solutions x  024, 424.

Review

127

74. x 5  x  3  x 5  x  3  0. We graph

P x  x 5  x  3. The only real solution is 134. 10

5

­2

2

­50 ­10

75. P x  x 3  2x  4

1

1

0 1

1

1 

2

1

1

4 1 5 

2

1

0 2

1

2

4

2

0

4

2

4

P x  x 3  2x  4  x  2 x 2  2x  2 . Since x 2  2x  2  0 has no real solution, the only real zero of P is

x  2.

     76. P x  x 4  3x 2  4  x 2  1 x 2  4  x  1 x  1 x 2  4 . Since x 2  4  0 has no real solution, the only real zeros of P are x  1 and x  1.

3 . The vertical asymptote is x  4. Because the x 4 denominator has higher degree than the numerator, the horizontal

77. (a) r x 

y

asymptote is y  0. When x  0, y  34 , so the y­intercept is 34 .

There is no x­intercept because the numerator is never 0. The domain

1

of r is  4  4  and its range is  0  0 .   3 1 1 3  3 f x  4, so we (b) If f x  , then r x  x x 4 x 4

1

x

obtain the graph of r by shifting the graph of f to the left 4 units and stretching vertically by a factor of 3.

78. (a) r x  

1 . The vertical asymptote is x  5 and the horizontal x 5

y

asymptote is y  0. When x  0, y  15 , so the y­intercept is 15 .

There is no x­intercept. The domain of r is  5  5  and its range is  0  0 .

  1 1 1    f x  5. (b) If f x  , then r x   x x 5 x 5

Thus, we obtain the graph of r by shifting the graph of f to the right 5 units and reflecting in the x­axis.

1 1

x


128

CHAPTER 3 Polynomial and Rational Functions

3x  4 . The vertical asymptote is x  1. Because the x 1 denominator has the same degree as the numerator, the horizontal

79. (a) r x 

y

asymptote is y  31  3. When x  0, y  4, so the y­intercept is 4. When y  0, 3x  4  0  x  43 , so the x­intercept is 43 . The

domain of r is  1  1  and its range is  3  3 .

1 , then x 3 x  1  1 1 r x  3  3  f x  1. Thus, we x 1 x 1 obtain the graph of r by shifting the graph of f to the right 1 unit,

(b) If f x 

1 x

1

reflecting in the x­axis, and shifting upward 3 units. 80. (a) r x 

2x  5 . The vertical asymptote is x  2 and the horizontal x 2

y

asymptote is y  2. When x  0, y  52 , so the y­intercept is 52 .

When y  0, x   52 , so the x­intercept is  52 . The domain of r is  2  2  and its range is  2  2 .

1

1 , then x 2 x  2  1 1 r x  2  2  f x  2. Thus, we x 2 x 2 obtain the graph of r by shifting the graph of f to the left 2 units and

(b) If f x 

1x

_1

upward 2 units. 12 3x  12 . When x  0, we have r 0   12, so the y­intercept is x 1 1 12. Since y  0, when 3x  12  0  x  4, the x­intercept is 4. The vertical

81. r x 

y

asymptote is x  1. Because the denominator has the same degree as the

numerator, the horizontal asymptote is y  31  3. The domain of r is

5

 1  1  and its range is  3  3 .

82. r x 

1 x  22

1 1 . When x  0, we have r 0  2  , so the y­intercept is 14 . 4 2

1

x

y

Since the numerator is 1, y never equals zero and there is no x­intercept. There is a vertical asymptote at x  2. The horizontal asymptote is y  0 because the

degree of the denominator is greater than the degree of the numerator. The domain of r is  2  2  and its range is 0 . 1 1

x


CHAPTER 3

x 2 x 2 1  83. r x  2 . When x  0, we have r 0  2 8  4 , x  2 x  4 x  2x  8

y

x­intercept is 2. There are vertical asymptotes at x  2 and x  4. The domain

1

129

Review

so the y­intercept is 14 . When y  0, we have x  2  0  x  2, so the of r is  2  2 4  4  and its range is  .

84. r x 

x

1

x 3  27 27 . When x  0, we have r 0  27 4 , so the y­intercept is y  4 . x 4

y

When y  0, we have x 3  27  0  x 3  27  x  3. Thus the x­intercept

is x  3. The vertical asymptote is x  4. Because the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. By long division, we have r x  x 2  4x  16 

37 . So the end behavior of y is x 4

20

like the end behavior of g x  x 2  4x  16. The domain of r is

x

2

 4  4  and its range is  .

85. r x 

x2  9 x  3 x  3  . When x  0, we have r 0  9 1 , so the 2x 2  1 2x 2  1

y 1

y­intercept is 9. When y  0, we have x 2  9  0  x  3 so the x­intercepts

1

x

are 3 and 3. Since 2x 2  1  0, the denominator is never zero so there are no

vertical asymptotes. The horizontal asymptote is at y  12 because the degree of

the denominator and numerator are the same. The domain of r is   and its   range is 9 12 .

86. r x 

2x 2  6x  7 7 . When x  0, we have r 0  7 4  4 , so the y­intercept x 4

y

is y  74 . We use the Quadratic Formula to find the x­intercepts:

   6 62 427 6 92  3 23 . Thus the x­intercepts are x  4 2 22

x  39 and x  09. The vertical asymptote is x  4. Because the degree of the

numerator is greater than the degree of the denominator, there is no horizontal

1 asymptote. By long division, we have r x  2x  2  , so the slant x 4 asymptote is r x  2x  2. The domain of r is  4  4  and its range is approximately  717]  [1283 .

2 1

x


130

CHAPTER 3 Polynomial and Rational Functions

x 2  5x  14 x  7 x  2   x  7 for x  2. The x­intercept is x 2 x 2 7 and the y­intercept is 7, there are no asymptotes, the domain is x  x  2, and

87. r x 

y

(2, 9)

the range is y  y  9.

1 0

 x x 2  3x  10

x x  5 x  2 x 3  3x 2  10x    x x  5 x 2 x 2 x 2 for x  2. The x­intercepts are 0 and 5 and the y­intercept is 0, there are no   asymptotes, the domain is x  x  2, and the range is y  y   25 4 .

88. r x 

1

x

y (_2, 14)

2 0

x 2  3x  18 x 6 x  6 x  3 for x  3. The x­intercept is  89. r x  2  x 5 x  5 x  3 x  8x  15

6 and the y­intercept is  65 , the vertical asymptote is x  5, the horizontal   asymptote is y  1, the domain is x  x  3 5, and the range is y  y  1  92 .

1

x

y

x-5

2 2

x 2  2x  15 x 3 x  3 x  5  90. r x  3  for 2 x  1 x  2 x  5 x  1 x  2 x  4x  7x  10

y

x  5. The x­intercept is 3 and the y­intercept is 32 , the vertical asymptotes are

x  1 and x  2, the horizontal asymptote is y  0, the domain is   x  x  5 1 2, and the range is y  y  19 or y  1 (You can use a graphing calculator to find the range.)

x 3 . From the graph we see that the x­intercept is 3, the y­intercept is 2x  6 05, there is a vertical asymptote at x  3 and a horizontal asymptote at

91. r x 

y  05, and there is no local extremum.

x

(3, _ 92 )

1

(_5, _ 27 )

0

1

x

5

­10 ­8 ­6 ­4 ­2 ­5

2

4


CHAPTER 3

2x  7 . From the graph we see that the x­intercept is 35, the y­intercept 92. r x  2 x 9 is 078, there is a horizontal asymptote at y  0 and no vertical asymptote, the

­20

local minimum is 111 090, and the local maximum is 811 012.

Review

131

20

­1

x3  8 . From the graph we see that the x­intercept is 2, the 93. r x  2 x x 2 y­intercept is 4, there are vertical asymptotes at x  1 and x  2, there is no

10

horizontal asymptote, the local maximum is 0425 3599, and the local

­5

minimum is 4216 7175. By using long division, we see that

10  x f x  x  1  2 , so f has a slant asymptote of y  x  1. x x 2

5 ­10

2x 3  x 2 . From the graph we see that the x­intercepts are 0 and 12 , the x 1 y­intercept is 0, there is a vertical asymptote at x  1, the local maximum is

94. r x 

50

0 0, and the local minima are 157 1790 and 032 003. Using long division, we see that r x  2x 2  3x  3 

3 . So the end x 1

­5

behavior of r is the same as the end behavior of g x  2x 2  3x  3.

5

95. 2x 2  x  3  2x 2  x  3  0  P x  x  1 2x  3  0. The cut points occur where x  1  0 and where 2x  3  0; that is, at x  1 and x  32 . We make a sign diagram:   1 32 Sign of  1 x 1

2x  3 P x

 

3 2

 

Both endpoints satisfy the inequality. The solution is  1]  32   . 96. x 3  3x 2  4x  12  0  x 2 x  3  4 x  3  0  P x  x  2 x  2 x  3  0. The cut points are 2, 2, and 3. We make a sign diagram: Sign of x 2 x 2 x 3 P x

 2

2 2

2 3

3 

All endpoints satisfy the inequality. The solution is  2]  [2 3].


132

CHAPTER 3 Polynomial and Rational Functions

     97. x 4  7x 2  18  0  x 2  9 x 2  2  0  P x  x  3 x  3 x 2  2  0. The last factor is positive for all x. We make a sign diagram:

Sign of x 3

 3

3 3

3 

x 3 P x

Neither endpoint satisfies the strict inequality. The solution is 3 3.

        98. x 8  17x 4  16  0  x 4  16 x 4  1  0  x 2  4 x 2  4 x 2  1 x 2  1  0     P x  x 2  4 x 2  1 x  4 x  4 x  1 x  1  0. The first two factors are positive for all x. We make a sign diagram:

Sign of x 4

 4

4 1

1 1

1 4

4 

x 1 x 1 x 4 P x

None of the endpoints satisfies the strict inequality. The solution is  2  1 1  2 .

99.

5 5 5     0  r x   0. We make a sign diagram: 0  2 x  1 x  2 x  2 x 3  x 2  4x  4 x x  4  x2  4 Sign of x 2 x 1 x 2

r x

 2

2 1

1 2

2 

None of the endpoints is in the domain of r . The solution is  2  1 2.

100.

2 3x  1 2 3 3x  1  2 x  2 9x  3  2x  4 7x  1 3x  1    0 0  0  r x   0. We x 2 3 x 2 3 3 x  2 3 x  2 3 x  2 make a sign diagram:     1 2 17 Sign of  2 7 x 2

7x  1

r x

  The cut point 17 satisfies equality, but 2 is not in the domain of r. The solution is 2 17 .


CHAPTER 3

101.

Review

133

2 3 1 2 3 x x  3  2x x  2  3 x  2 x  3 1       0  0 x 2 x 3 x x 2 x 3 x x x  2 x  3 2 9  2x 4x  18  0  r x   0. We make a sign diagram: x x  2 x  3 x x  2 x  3     9 2 92 Sign of  3 3 0 0 2 2 x 3 x

x 2

9  2x

r x

  The cut point 92 satisfies equality, but the other cut points are not in the domain of r . The solution is 3 0  2 92 . 102.

3 4 1 3 4 x x  3  3x x  2  4 x  2 x  3 1       0  0 x 2 x 3 x x 2 x 3 x x x  2 x  3 7x  24 r x   0. We make a sign diagram: x x  2 x  3       24  24 Sign of 2 0 0 3 3  7 7  2 7x  24 x 2 x

x 3

r x

24 The cut point  24 7 satisfies equality, but the other cut points are not in the domain of r. The solution is  7  2  0 3.

103. f x 

  24  x  3x 2  x  3 8  3x is defined where r x  x  3 8  3x  0. We make a sign diagram:     8 3 83 Sign of  3 3 x 3

8  3x

r x

   Both cut points satisfy equality, so the domain of f is 3 83 . 104. g x   4

  1 1   is defined where r x  x 1  x x 2  x  1  0.       4 4 x  x4 x 1  x3 x 1  x x 2  x  1 1

(Equality is excluded because the denominator cannot be 0.) The last factor of r x is positive for all x. We make a sign diagram: Sign of x 1x

x2  x  1

r x

 0

0 1

1 

Neither cut point satisfies the strict inequality, so the domain of g is 0 1.


134

CHAPTER 3 Polynomial and Rational Functions

105.

106.

40

20

20 ­2 ­3

­2

­1

1

2

­1

3

1

2

3

>

­20

From the graph, we see that x 4  x 3  5x 2  4x  5 on

From the graph, we see that x 5  4x 4  7x 3  12x  2  0

approximately [074 195].

on approximately 106 017  191 .

107. (a) We use synthetic division to show that 1 is a zero of P x  2x 4  5x 3  x  4: 2

1

2

5

0

1

4

2

3

3

4

3 3 4 0   Thus, 1 is a zero and P x  x  1 2x 3  3x 2  3x  4 .

(b) P x has no change of sign, and hence no positive real zeros. But P x  x  1 Q x, so Q cannot have a positive real zero either. 108. We want to find the solutions of x 4  x 2  24x  6x 3  20  P x  x 4  6x 3  x 2  24x  20  0. The possible rational zeros of P are 1, 2, 4, 5, 10, 20. P x has 3 variations in sign and hence 1 or 3 positive real zeros. P x  x 4  6x 3  x 2  24x  20 has 1 variation in sign, and so P has 1 negative real zeros. 1

1

6

1

1

1

24

5

4

2

1

20

5

2

 x  1 is a zero. 1 3  P x  x  1 x  2 x 2  3x  10  x  1 x  2 x  2 x  5, and 5

4

20

20 0

4

20

6

20

10

0

y

 x  2 is a zero.

so the zeros of P (and hence the x­coordinates of the points of intersection) are 2, 1, 2, and 5. From the original equation, the coordinates of the points of intersection are 2 28, 1 26, 2 68, 5 770.

100 1

109. (a) y 

x

x2  4   has vertical asymptotes at x  0 and x  2, so it has graph VII. x x2  4

1 x  22 x  23 has x­intercepts 2 and y­intercept 1 0  22 0  23  2, so it has graph V. (b) y  16 16

(c) y  x  1 x  22 has x­intercepts 1 and 2 and y­intercept 4, so it has graph III.

(d) y  2x  x 2  x 2  x is a parabola that opens downward with x­intercepts 0 and 2 and y­intercept 0. It has graph I. (e) y  x 2 2x  3 x  1 has x­intercepts  32 , 0, and 1, so it has graph IV.

(f) x  2y  y 2  y 2  y is a parabola that opens leftward, so it has graph VIII


CHAPTER 3

135

Test

x2 x2 has vertical asymptotes x  1 and horizontal asymptote y  11  1, so it has  x  1 x  1 x2  1 graph VI.  (h) y  9  x 2 is a semicircle with domain [3 3], so it has graph II.

(g) y 

CHAPTER 3 TEST y

1. f x  x 2  x  6    x2  x  6    x 2  x  14  6  14 2   x  12  25 4

2. g x  2x 2  6x  3 is a quadratic

function with a  2 and b  6, so it has a maximum or minimum value where b 6 3 x     . Because 2a 2 2 2

1 1

x

a  0, this gives the minimum value  2     g  32  2  32 6  32 3   32 .

  Domain:  , range:  25   . 4

3. (a) We write the function in vertex form:       h x  10x  001x 2  001 x 2  1000x  001 x 2  1000x  5002  001 5002  001 x  5002  2500

Thus, the cannonball reaches a maximum height of 2500 feet. (b) By the symmetry of the parabola, we see that the cannonball’s height will be 0 again (and thus it will splash into the water) when x  1000 ft. y

4. f x   x  23  27 has y­intercept y  23  27  19 and x­intercept where  x  23  27  x  1.

10

(0, 19) (1, 0)

x

1

5. (a)

2

1

0 2

1

4

2

4 0

2

5

0

4

2

9

(b) 2x 2  1

x 3  2x 2

2x 5  4x 4  x 3  2x 5

Therefore, the quotient is

4x 4

Q x  x 3  2x 2  2, and the remainder is

4x 4

 x3

1 2 x 2  0x 

 x2  2x 2 x2

R x  9.

7

x2

7

 12 15 2

Therefore, the quotient is Q x  x 3  2x 2  12 and the remainder is R x  15 2 .


136

CHAPTER 3 Polynomial and Rational Functions

6. (a) Possible rational zeros are: 1, 3,  12 ,  32 . (b)

1

2

5

4 7

2

y

3 3

2

2 7 3 0  x  1 is a zero.   P x  x  1 2x 2  7x  3  x  1 2x  1 x  3    2 x  1 x  12 x  3

1

x

(c) The zeros of P are x  1, 3, 12 .

7. P x  x 3  x 2  4x  6. Possible rational zeros are: 1, 2, 3, 6. 1 1 1 4 1

0

1

6

2 1 1 4 2

4

6

2

3 1 1 4 6 3

4

6

6

0 4 10 1 1 2 10 1 2 2 0  x  3 is a zero.   So P x  x  3 x 2  2x  2 . Using the Quadratic Formula on the second factor, we have    2  4 2  2 1 2  22  4 1 2    1  i. So zeros of P x are 3, 1  i, and 1  i. x 2 1 2 2 8. P x  x 4  2x 3  5x 2  8x  4. The possible rational zeros of P are: 1, 2, and 4. Since there are four changes in sign, P has 4, 2, or 0 positive real zeros. 1

1

5

2

1

4

8

4

1

4

1 1 4 4 0  So P x  x  1 x 3  x 2  4x  4 . Factoring the second factor by grouping, we have     P x  x  1 x 2 x  1  4 x  1  x  1 x 2  4 x  1  x  12 x  2i x  2i. 

9. Since 3i is a zero of P x, 3i is also a zero of P x. And since 1 is a zero of multiplicity 2,    P x  x  12 x  3i x  3i  x 2  2x  1 x 2  9  x 4  2x 3  10x 2  18x  9. 10. P x  2x 4  7x 3  x 2  18x  3. (a) Since P x has 4 variations in sign, P x can have 4, 2, or 0 positive real zeros. Since P x  2x 4  7x 3  x 2  18x  3 has no variations in sign, there are no negative real zeros. (b)

4

2 2

7

1

8

4

1

5

18

3

20

8

2

11

Since the last row contains no negative entry, 4 is an upper bound for the real zeros of P x. 1

2 2

7 1 9

1 9 10

18 10 28

3 28 31

Since the last row alternates in sign, 1 is a lower bound for the real zeros of P x.


CHAPTER 3

(c) Using the upper and lower limit from part (b), we graph P x in the viewing rectangle [1 4] by

Test

137

(d) Local minimum 282 7031.

[1 1]. The two real zeros are 017 and 393.

50

1 ­2

2

4

­50 2

4

­1

11. (a) P 1  13  2 12  1  4, Q 1  13  2 12  4 1  8  3, R 1  13  4 1  5  0,

S 1  13  2 12  4 1  8  5, and T 1  13  9 12  27 1  27  8. Thus, only R has value 0 at x  1.   (b) P x  x 3  2x 2  x  x x 2  2x  1  x x  12 , Q x  x 3  2x 2  4x      8  x 2 x  2  4 x  2  x 2  4 x  2  x  2 x  22 , R x  x 3  4x  5  x  1 x  x 2  5 ,   S x  x 3 2x 2 4x 8  x 2 x  24 x  2  x  2 x 2  4 , and T x  x 3 9x 2 27x 27  x  33 , so only the polynomial P has remainder 0 when divided by x  1.

(c) From part (b), only Q x has x  2 as a factor.

(d) From part (b), only T x has 3 as a zero of multiplicity 3. (e) A polynomial with 2 and 2i as zeros must have x  2 and x 2  4 as factors. From part (b), this describes only S x. x2  x  6 2x  1 x 3  27 x 3  9x x 3  6x 2  9x , u x  . 12. r x  2 , s x  2 , t x  , and  x  2 x 2 x 3 x x 2 x 4 x  25 (a) r has horizontal asymptote y  0 because the degree of the denominator is greater than the degree of the numerator. u has horizontal asymptote y  11  1 because the degrees of the numerator and the denominator are the same.

(b) The degree of the numerator of s x is one more than the degree of the denominator, so s has a slant asymptote.   x x 2  6x  9 x x  32 (c) The denominator of s x is never 0, so s has no vertical asymptote.  x    x x  3 x 3 x 3 for x  3, so  has no vertical asymptote.

(d) From part (c),  has a “hole” at 3 0. 2x  1 2x  1  , so r has vertical asymptotes at x  1 (e) r x  2 x  1 x  2 x x 2

y

and x  2. y  0 is a horizontal asymptote because the degree of the numerator is less than the degree of the denominator.

(f) u x 

x2  x  6 x  3 x  2  . When x  0, we have 2 x  5 x  5 x  25

6  6 , so the y­intercept is y  6 . When y  0, we have x  3 or u x  25 25 25

x  2, so the x­intercepts are 3 and 2. The vertical asymptotes are x  5 and

x  5. The horizontal asymptote occurs at y  11  1 because the degree of the

denominator and numerator are the same.

1 2

x


138

CHAPTER 3 Polynomial and Rational Functions

x 2  2x  5

(g) x 2

0 x 3  2x 2 2x 2  9x 2x 2  4x 5x  0 5x  10 10

Thus P x  x 2  2x  5 and t x  13. x 

y

x 3  0x 2  9x 

10 2

x

x 3  9x have the same end behavior. x 2

6x 6  x  x 2x  5 2x 2  4x  6 6x x  1 3  x 0 x 0 0   r x. We 2x  5 2x  5 2x  5 2x  5 x5 2

make a sign diagram:

Sign of x 1

3x

x  52

r x

 1 

  1 52 

53 2

3 

  The cut point 52 is excluded so that the denominator is not 0. The solution is  1]  52  3 .

1 is defined where 4  2x  x 2  0. Using the Quadratic Formula to solve x 2  2x  4  0, we 14. f x   4  2x  x 2   2  22  4 1 4   1  5. The radicand is positive between these two roots, so the domain of have x  2 1     f is 1  5 1  5 .

15. (a)

P x  x 4  4x 3  8x. From the graph, the x­intercepts are approximately

10

124, 0, 2, and 324, P has local maximum P 1  5, and P has local minima P 073  P 273  4.

­2

2

4

(b) From part (a), P x  0 on approximately  124]  [0 2]  [324 .

solutions,


139

Fitting Polynomial Curves to Data

FOCUS ON MODELING Fitting Polynomial Curves to Data

1. (a) Using a graphing calculator, we obtain the quadratic

2. (a) Using a graphing calculator, we obtain the quadratic

polynomial

polynomial

y  0275428x 2  197485x  2735523 (where

y  02783333x 2  184655x  166732 (where

miles are measured in thousands). (b)

y

plants/acre are measured in thousands). y

(b)

100

120

80 60

80

40

40

20 0

25

30

35

40

45

50 x

0

20

30

40

60 x

50

Density (thousand plants/acre)

Pressure (lb/in@)

(c) Moving the cursor along the path of the polynomial,

(c) Moving the cursor along the path of the polynomial, we find that yield when 37,000 plants are planted per

we find that 3585 lb/in2 gives the longest tire life.

3. (a) Using a graphing calculator, we obtain the cubic polynomial

acre is about 135 bushels/acre.

4. (a)

y 60 50

y  000203709x 3  0104522x 2

 1966206x  145576.

(b)

10

y

40 30 20 10

20

0 10

1

2

3

x

Time (s) A quadratic model seems appropriate. 35 x

(b) Using a graphing calculator, we obtain the quadratic

(c) Moving the cursor along the path of the polynomial,

(c) Moving the cursor along the path of the polynomial,

0

5

10

15

20

25

30

Seconds

polynomial y  160x 2  518429x  420714.

we find that the subjects could name about

we find that the ball is 20 ft. above the ground 03

43 vegetables in 40 seconds.

seconds and 29 seconds after it is thrown upward.

(d) Moving the cursor along the path of the polynomial,

(d) Again, moving the cursor along the path of the

we find that the subjects could name 5 vegetables in

polynomial, we find that the maximum height is

about 20 seconds.

462 ft.


140

FOCUS ON MODELING

5. (a) Using a graphing calculator, we obtain the quadratic polynomial y  00120536x 2  0490357x  496571. (c) Moving the cursor along the path of the polynomial, we find that the tank should drain in 190 minutes.

(b)

y 6 5 4 3 2 1 0

10

Time (min)

20 x


COMMENTS: 1,2,9,11,15,36,39,45,51,54,55

CHAPTER 4

EXPONENTIAL AND LOGARITHMIC FUNCTIONS

4.1

Exponential Functions 1

4.2

The Natural Exponential Function 9

4.3

Logarithmic Functions 16

4.4

Laws of Logarithms 24

4.5

Exponential and Logarithmic Equations 28

4.6

Modeling with Exponential Functions 36

4.7

Logarithmic Scales 42 Chapter 4 Review 44 Chapter 4 Test 52

¥

FOCUS ON MODELING: Fitting Exponential and Power Curves to Data 55

1


4

EXPONENTIAL AND LOGARITHMIC FUNCTIONS

4.1

EXPONENTIAL FUNCTIONS

1 , f 0  50  1, f 2  52  25, 1. The function f x  5x is an exponential function with base 5; f 2  52  25

and f 6  56  15,625.

2. (a) f x  2x is an exponential function with base 2. It has graph III.  x (b) f x  2x  12 is an exponential function with base 12 . It has graph I.

(c) The graph of f x  2x is obtained from the graph of 2x by reflecting about the x­axis. Thus, it has graph II.

(d) The graph of f x  2x is obtained from the graph of 2x by reflecting about the x­ and y­axes. Thus, it has graph IV. 3. (a) To obtain the graph of g x  2x  1 we start with the graph of f x  2x and shift it downward 1 unit.

(b) To obtain the graph of h x  2x1 we start with the graph of f x  2x and shift it to the right 1 unit. nt  for compound interest, the letters P, r, n, and t stand for principal, interest rate per 4. In the formula A t  P 1  nr year, number of times interest is compounded per year, and number of years, respectively, and A t stands for the amount accumulated after t years. So if $100 is invested at an interest rate of 6% compounded quarterly, then the amount after  4  2 2 years is 100 1  006  11265. 4  x 5. The exponential function f x  12 has the horizontal asymptote y  0. This means that as x  , we have  x 1  0. 2  x 6. The exponential function f x  12  3 has the horizontal asymptote y  3. This means that as x  , we have  x 1  3  3. 2      5  22195, f 2  0063, f 03  1516. 7. f x  4x ; f 12  412  4  2, f     8. f x  3x1 ; f 12  0577, f 25  5196, f 1  0111, f 14  0439.    x1   ; g 12  0192, g 2  0070, g 35  15588, g 14  1552. 9. g x  13

 3x       10. f x  43 ; f  12  0650, f 6  8281, f 3  0075, f 43  3160.

11.

f x  6x

y

12.

f x  01x

x

y

x

y

2

1 36 1 6

2

100

1

1

0 1

6

2

36

1 1

x

1

10

0

1

1

1 10 1 100

2

Replace "f" with "g" (5x) y

1 1

x

1


2

13.

 x

y

g x  23 x

y

2

9 4 3 2

1

0

1

1

2 3 4 9

2 15.

Replace with "g"

CHAPTER 4 Exponential and Logarithmic Functions

y

14. h x  4x x

y

2

1 16 1 4

1

1

x

1

1

1

4

2

16

2 1

 x 16. h x  4 58

y

h x  5 22x

0

x

y

x

y

2

1775

3

1638

1

2308

0

30

1

39

2

507

3 4

2

64

0

4

1

25

6591

2

156

8568

3

098

x

1

y

1024

0

1

1 x

1

 x 18. f x  8x and g x  18  8x

17. f x  4x and g x  4x y

y

f=g g

f

2

1

x

1

x

1

 x 20. f x  34 and g x  15x .

19. f x  4x and g x  7x . y

y

g

f

f

g

2

5 1

x

Something is wrong here. Both graphs should pass through the point (0, 1).

x

0

1

x


SECTION 4.1 Exponential Functions

3

21. From the graph, f 2  a 2  9, so a  3. Thus f x  3x .

22. From the graph, f 1  a 1  15 , so a  5. Thus f x  5x .   1 , so a  1 . Thus f x  1 x . 23. From the graph, f 2  a 2  16 4 4

 x 24. From the graph, f 3  a 3  8, so a  12 . Thus f x  12 .

25. The graph of f x  5x1 is obtained from that of y  5x by shifting 1 unit to the left, so it has graph II. 26. The graph of f x  5x  1 is obtained from that of y  5x by shifting 1 unit upward, so it has graph I.  x 27. f x  3x  1. The graph of f is obtained by shifting the 28. f x  13  2. The graph of f is obtained by shifting  x graph of y  3x upward 1 unit. The y­intercept is the graph of y  13 downward 2 units. y  f 0  30  1  2. Domain:  .  0 Range: 1 . Horizontal asymptote: y  1. y­intercept: y  f 0  13  2  1. y

Domain:  . Range: 2 . Horizontal symptote: y  2.

y

(1, 4)

(0, 2) 1

x

1

1 x

1

29. The graph of g x  5x is obtained by reflecting the graph of y  5x about the x­axis.

y­intercept: y  g 0  50  1. Domain:  .

Range:  0. Horizontal asymptote: y  0. y 1

(0, _1)

30. The graph of g x  5x is obtained by reflecting the graph of y  5x about the y­axis.

y­intercept: y  g 0  50  1. Domain:  .

Range: 0 . Horizontal asymptote: y  0. y

x

1

(1, _5) 2

(0, 1) 1

x


4

CHAPTER 4 Exponential and Logarithmic Functions

31. h x  3x2 . The graph of h is obtained by shifting the graph of y  3x to the right 2 units.

y­intercept: y  h 0  302  19 . Domain:  .

Range: 0 . Horizontal asymptote: y  0. y

1

32. h x  10x1 . The graph of h is obtained by shifting the graph of y  10x to the left 1 unit.

y­intercept: y  h 0  1001  10.

Domain:  . Range: 0 . Horizontal asymptote: y  0.

y

(2, 1)

(0, 1/9)

20

x

1

(0, 10) 1

33. y  2x  3. The graph is obtained by reflecting the

graph of y  2x about the x­axis and then shifting upward 3 units. y­intercept: y  20  3  4.

Domain:  . Range: 3 . Horizontal asymptote: y  3.

y

x

34. y  2x  3. The graph is obtained by reflecting the

graph of y  2x about the x­axis and shifting upward 3 units. y­intercept: y  20  3  2.

Domain:  . Range:  3. Horizontal asymptote: y  3.

y

(0, 2)

(1, 1)

1 (0, 4)

1

x

1 x

1

35. y  10x  1. The graph is obtained by reflecting the

graph of y  10x about the x­axis and shifting downward 1 unit. y­intercept: y  100  1  2.

Domain:  . Range:  1. Horizontal asymptote: y  1.

y 1

 x 36. y  12  4. The graph is obtained by shifting the  x graph of y  12 downward 4 units.

 0 y­intercept: y  12  4  3. Domain:  .

Range: 4 . Horizontal asymptote: y  4. 1

y

x

(0, _2) 1 1 (0, _3)

x


SECTION 4.1 Exponential Functions

5

37. h x  2x4  1. The graph of h is obtained by shifting

38. y  10x1  5. The graph of y is obtained by shifting the

y­intercept: y  204  1  17 16 . Domain:  .

y­intercept: y  1001  5  5. Domain:  .

the graph of y  2x to the right 4 units and upward 1 unit.

Range: 1 . Horizontal asymptote: y  1.

graph of y  10x to the left 1 unit and downward 5 units. Range: 5 . Horizontal asymptote: y  5. y

y

5

2

(4, 2)

(0, 17/16)

1 4

1

x

x

 x 39. g x  1  3x  3x  1. The graph of g is obtained 40. y  3  15  5x  3. The graph of y is obtained by reflecting the graph of y  3x about the x­ and y­axes

and then shifting upward 1 unit.

by reflecting the graph of y  5x about the x­ and y­axes

y­intercept: y  30  1  0. Domain:  .

y­intercept: y  50  3  2. Domain:  .

and then shifting upward 3 unit.

Range:  1. Horizontal asymptote: y  1. y

Range:  3. Horizontal asymptote: y  3. y

1 (_1, _2)

41. (a)

1

(0, 2)

1

x

x

1

42. (a)

y

y

g f

1

1 1

x

  (b) Since g x  3 2x  3 f x and f x  0, the

height of the graph of g x is always three times the height of the graph of f x  2x , so the graph of g

is steeper than the graph of f .

1

x

 x2 (b) f x  9x2  32  32x2  3x  g x. So f x  g x, and the graphs are the same.


6

CHAPTER 4 Exponential and Logarithmic Functions

43.

f x  x 3

x 0

y

g x  3x

0

1

1

1

3

2

8

9

3

27

27

4

64

81

6

216

729

8

512

6561

10

1000

59,049

x

g(x)=3

f(x)=x#

20 0

x

1

From the graph and the table, we see that the graph of g ultimately increases much more quickly than the graph of f . 44.

f x  x 4

x 0

y

g x  4x

0

1

1

1

4

2

16

16

3

81

64

4

256

256

6

1296

4096

8

4096

65,536

10

10,000

1,048,576

g(x)=4

x

f(x)=x$

100 0

x

1

From the graph and the table, we see that the graph of g ultimately increases much more quickly than the graph of f . 45. (a) From the graphs below, we see that the graph of f ultimately increases much more quickly than the  of g.  graph   (i) [0 5] by [0 20] (ii) [0 25] by 0 107 (iii) [0 50] by 0 108 20

10,000,000

f

8,000,000 6,000,000

10 g

f

1

2

4,000,000

3

4

80,000,000

f

g

40,000,000 20,000,000

0

5

100,000,000

60,000,000

2,000,000

0

g

10

20

0

20

40

(b) From the graphs in parts (a)(i) and (a)(ii), we see that the approximate solutions are x  12 and x  224. 46. (a) (i) [4 4] by [0 20]

  (iii) [0 20] by 0 105

(ii) [0 10] by [0 5000]

20 g

f

4000

f

g

100,000

80,000

10

60,000

2000

f

g

40,000 20,000 ­4

­3

­2

­1 0

1

2

3

4

0

2

4

6

8

10

0

10

(b) From the graphs in parts (i) and (ii), we see that the solutions of 3x  x 4 are x  080, x  152 and x  717.

20


SECTION 4.1 Exponential Functions

47.

c=2 c=1

5

48.

10

c=0.5

4

­2

­1

c=0.25

2

0 ­1

1

2

3

0

­2

The larger the value of c, the more rapidly the graph of shifted horizontally 1 unit. This is because of our choice of c; each c in this exercise is of the form 2k . So f x  2k  2x  2xk . 2

2

4

The larger the value of c, the more rapidly the graph of  x f x  2cx increases. In general, f x  2cx  2c ;  x so, for example, f x  22x  22  4x .

f x  c2x increases. Also notice that the graphs are just

49. y  10xx

c=0.5

4

1 ­3

c=1

6

c=0.25

2

c=2

8

3 c=4

c=4

50. y  x2x 2 1 ­4 ­4

­2

0

2

(a) From the graph, we see that the function is increasing

on 144  and decreasing on  144.

(b) From the graph, we see that the range is

(b) From the graph, we see that the range is

approximately 0 178].

approximately [053 .

10xh  10x f x  h  f x 51. f x  10x , so   10x

52. f x  3x1 , so

2

(a) From the graph, we see that the function is increasing

on  050 and decreasing on 050 .

h

­2

4

h

 10h  1 . h

3xh1  3x1 f x  h  f x   3x1 h h

 3h  1 . h

53. (a) After 1 hour, there are 1500  2  3000 bacteria. After 2 hours, there are 1500  2  2  6000 bacteria. After 3 hours, there are 1500  2  2  2  12,000 bacteria. We see that after t hours, there are N t  1500  2t bacteria. (b) After 24 hours, there are N 24  1500  224  25,165,824,000 bacteria.

54. (a) Because the population doubles every year, there are N t  320  2t mice after t years. (b) After 8 years, there are approximately N 8  320  28  81,920 mice.

7


8

CHAPTER 4 Exponential and Logarithmic Functions

55. Using the formula A t  P 1  ik with P  5000,

004 per month, and k  12  number 12 of years, we fill in the table:

i  4% per year 

56. Using the formula A t  P 1  ik with P  5000, rate per year i per month, and k  12  5  60 months, 12 we fill in the table:

Amount

Rate per year

Amount

1

$520371

1%

$525625

2

$541571

2%

$552539

3

$563636

3%

$580808

4

$586599

4%

$610498

5

$610498

5%

$641679

6

$635371

6%

$674425

Time (years)

2t  57. P  8000, r  00625, and n  2. So A t  8000 1  00625  8000  1031252t . 2

(a) A 5  8000  10312510  10,88252, and so the value of the investment is $10,88252.

(b) A 10  8000  10312520  14,80366, and so the value of the investment is $14,80366. (c) A 15  8000  10312530  20,13765, and so the value of the investment is $20,13765.

365t  . 58. P  3500, r  0035, and n  365. So A t  3500 1  0035 365 365  2  (a) A 2  3500 1  0035  375377, and so the value of the investment is $375377. 365 365  3   388747, and so the value of the investment is $388747. (b) A 3  3500 1  0035 365 365  6   431783, and so the value of the investment is $431783. (c) A 6  3500 1  0035 365

4t  . 59. P  1200, r  00275, and n  4. So A t  1200 1  00275 4 4   123334, and so the value of the investment is $123334. (a) A 1  1200 1  00275 4 8   126761, and so the value of the investment is $126761. (b) A 2  1200 1  00275 4 40   157835, and so the value of the investment is $157835. (c) A 10  1200 1  00275 4 4t  . 60. P  14,000, r  00525, and n  4. So A t  14000 1  00525 4 16   17,24791, and so the amount due is $17,24791. (a) A 4  14000 1  00525 4 24   19,14436, and so the amount due is $19,14436. (b) A 6  14000 1  00525 4 32   21,24932, and so the amount due is $21,24932. (c) A 8  14000 1  00525 4

23   P 10456  10000  13023P  P  767896. 61. We must solve for P in the equation 10000  P 1  009 2 Thus, the present value is $7,67896.

 125 62. We must solve for P in the equation 100000  P 1  008  P 10066760  100000  14898P  12 P  $67,12104.


9

SECTION 4.2 The Natural Exponential Function

   r n 008 12 63. rAPY  1   1. Here r  008 and n  12, so rAPY  1   1  1006666712  1  0083000. n 12 Thus, the annual percentage yield is about 83%.    r n 0055 4  1. Here r  0055 and n  4, so rAPY  1   1  0056145. Thus, the annual 64. rAPY  1  n 4 percentage yield is about 561%. 65. Let T n be the thickness of the paper after n folds. The sheet is 0001 in thick and each fold doubles the thickness, 1 ft  so T n  0001  2n . After 50 folds, the folded paper is T 50  0001  250  1126  1012 in  12 in 1 mi  17,771,460 mi—roughly one­fifth the distance from the earth to the sun! 5280 ft 66. Since f 40  240  1,099,511,627,776, it would take a sheet of paper 4 inches by 1,099,511,627,776 inches. Since there are 12 inches in a foot and 5,280 feet in a mile, 1,099,511,627,776 inches  174 million miles. So the dimensions of the sheet of paper required are 4 inches by about 174 million miles.

4.2

THE NATURAL EXPONENTIAL FUNCTION

1. The function f x  e x is called the natural exponential function. The number e is approximately equal to 271828. 2. In the formula A t  Pert for continuously compound interest, the letters P, r, and t stand for principal, interest rate per year, and number of years respectively, and A t stands for amount accumulated after t years. So, if $100 is invested at an interest rate of 6% compounded continuously, then the amount after 2 years is A 2  100  e0062  $11275. 3. h x  e x ; h 1  2718, h   23141, h 3  0050, h

  2  4113

  4. h x  e3x ; h 13  0368, h 15  0011, h 1  20086, h   12,391648 5.

f x  15e x

y

6. g x  4e13x

x

y

x

y

2

020

3

1087

1 05

055

2

091

0

15

05

247

1

408

2

1108

2 0

1

x

Replace with "f" y

779

1

558

0

4

1

287

2

205

3

147

2 0

1

x


10

CHAPTER 4 Exponential and Logarithmic Functions

7. f x  e x  3. The graph of f is obtained from the graph of y  e x by shifting it upward 3 units. y­intercept: y  f 0  e0  3  4. Domain:  .

Range: 3 . Horizontal asymptote: y  3. y

8. f x  ex  1. The graph of f is obtained from the graph of y  e x by reflecting it about the y­axis and shifting it upward 1 unit. y­intercept:

y  f 0  e0  1  2. Domain:  .

Range: 1 . Horizontal asymptote: y  1. y

1 1

x

1 x

1

9. g x  ex  3. The graph of g is obtained from the graph of y  e x by reflecting it about the y­axis and

shifting it downward 3 units. y­intercept:

y  g 0  e0  3  2. Domain:  .

Range: 3 . Horizontal asymptote: y  3.

10. h x  e x  4. The graph of h is obtained by shifting the graph of y  e x downward 4 units. y­intercept:

y  h 0  e0  4  3. Domain:  .

Range: 4 . Horizontal asymptote: y  4. y

y

1 1

1

x

x

1

11. f x  3  e x . The graph of f is obtained from the graph 12. y  2  ex . The graph is obtained by reflecting the of y  e x by reflecting it about the x­axis, then shifting

graph of y  e x about the y­axis, then reflecting about the

Domain:  . Range:  3. Horizontal

y  2  e0  1. Domain:  . Range:  2.

upward 3 unit. y­intercept: y  f 0  3  e0  2. asymptote: y  3.

y

x­axis, then shifting upward 2 units. y­intercept:

Horizontal asymptote: y  2. y

1

1 1

x

1

x


SECTION 4.2 The Natural Exponential Function

13. y  4  3ex . The graph is obtained by reflecting that of y  e x about the y­axis, then stretching vertically by a factor of 3, then reflecting about the x­axis, then shifting upward 4 units. y­intercept: y  4  3e0  1.

Domain:  . Range:  4. Horizontal asymptote: y  4.

11

14. f x  3e x  1. The graph of f is obtained by stretching that of y  e x vertically by a factor of 3, then shifting 1 unit upward. y­intercept: y  f 0  3e0  1  4. Domain:  . Range: 1 . Horizontal asymptote: y  1. y

y

1 x

1

1

f(x)

1

x

y

f

y

15. y  e x2 . The graph of y  e x2 is obtained from the graph of y  e x by shifting it to the right 2 units.

= f(0)

16. f x  e x3  4. The graph of f x  e x3  4 is obtained by shifting the graph of y  e x to the right 3 units, and then upward 4 units. y­intercept: y­intercept: y  e02  e2  014. Domain:  . Range: 0 . Horizontal asymptote: y  0. y  f 0  e03  4  e3  4  405. Domain:  . Range: 4 . Horizontal y asymptote: y  4. y

1

(3, 5)

(2, 1) x

1

1 x

1

17. h x  e x1  3. The graph of h is obtained from the graph of y  e x by shifting it to the left 1 unit and downward 3 units. y­intercept: y  h 0  e01  3  e1  3  028. Domain:  . Range: 3 . Horizontal asymptote: y  3. y

18. g x  e x1  2. The graph of g is obtained by shifting the graph of y  e x to the right 1 unit, reflecting it about the x­axis, then shifting it downward 2 units. y­intercept: y  g 0  e01  2  e1  2  237. Domain:  . Range:  2. Horizontal asymptote: y  2. y

1 1 1 (0, e-3) (_1, _2)

x (1, _3)

1

x


12

CHAPTER 4 Exponential and Logarithmic Functions 2

2

19. f x  5ex 2

20. f x  10ex5

y

y 6

10

4 5

2 ­4

­2

2

4

No x­intercept. y­intercept: 5. Horizontal asymptote: y  0. Local maximum: f 0  5. No local minimum.

21. l x 

2

x

­2

4

6

8

x

No x­intercept. y­intercept: 10e25  139  1010 .

Horizontal asymptote: y  0. Local

maximum: f 5  10. No local minimum.

20 1  3e15x

22. l x 

100 1  4e05x

y 20

y 100

10

50

­2

2

4

10

x

x

No x­intercept. y­intercept: 5. Horizontal

No x­intercept. y­intercept: 20. Horizontal

asymptotes: y  0 and y  20. No local maximum or

symptotes: y  0 and y  100. No local maximum or

minimum.

minimum.

23. s x  04xe15x

24. s x  25xe35x

y 0.2

y 0.4 0.2

­1

1

2

3

­0.2

4

5

x

­0.5

­0.4

­0.2

­0.6

­0.4

0.5

1.0

1.5

x­intercept: 0. y­intercept: 0. Horizontal

x­intercept: 0. y­intercept: 0. Horizontal

asymptote: y  0. Local maximum:

asymptote: y  0. Local maximum:

approximately s 067  0098. No local minimum.

x

approximately s 029  026. No local minimum.

Answers to Exercises 25–28 will vary. 2

2.0

25. F x  2ex10 can be expressed as  f  g x, where g x  x  102 and f x  2e x . 1 26. F x  e x  ex can be expressed as  f  g x, where g x  e x and f x  x  . x   27. F x  1  e x can be expressed as  f  g x, where g x  1  e x and f x  x.


SECTION 4.2 The Natural Exponential Function

13

3  28. F x  3  e x can be expressed as  f  g x, where g x  3  e x and f x  x 3 . 29. (a)

30. (a)

y

y 1

1 y=_ 2 e¨

y=sinh x

y=cosh x 1 y=_ 2 eШ

1

1 y=_ 2 e¨

1

x

1

ex  e x ex  ex  2 2 e x  ex   cosh x 2

ex  e x ex  ex  2 2 ex  e x e x  ex     sinh x 2 2

(b) cosh x 

31. (a)

a=1

a=1.5

10 a=0.5

a=2

8

4

0

32.

4

0

2 ­2

(b) sinh x 

2

6

­4

x

1

y=_ _ 2 eШ

2

4

 a  xa e (b) As a increases the curve y   exa 2 flattens out and the y intercept increases. 33. g x  x x . Notice that g x is only defined for x  0. The graph of g x is shown in the viewing rectangle

[0 15] by [0 15]. From the graph, we see that there is a local minimum of about 069 when x  037.

0

20

40

 x From the graph, we see that y  1  1x approaches e as x get large.

34. g x  e x  e2x . The graph of g x is shown in the

viewing rectangle [1 2] by [1 6]. From the graph, we see that there is a local minimum of about 189 when x  023. 5

1

0

0

1

­1

1

2

35. D t  50e02t . So when t  3 we have D 3  50e023  274 milligrams. 36. m t  13e0015t (a) m 0  13 kg.

(b) m 45  13e001545  13e0675  6619 kg. Thus the mass of the radioactive substance after 45 days is about 66 kg.


14

CHAPTER 4 Exponential and Logarithmic Functions

37. (a)

(b) From the graph, it appears that the maximum

s 0.2

concentration is reached at t  083 h, or approximately 50 minutes.

0.1

(c) From the graph, it appears that s t decreases to 001 at t  49 h. 1

2

3

4

5

6

t

The drug concentration in the bloodstream increases rapidly until it reaches a peak level, then decreases slowly. 38. (a)

(b) From the graph, the average height appears to be

g 0.004

185 cm.

0.002 0.000

160

180

200

220

x

  39.  t  180 1  e02t

  (a)  0  180 1  e0  180 1  1  0.   (b)  5  180 1  e025  180 0632  11376 ft/s. So the

(c)

y 200

velocity after 5 s is about 1138 ft/s.    10  180 1  e0210  180 08646647  1556 ft/s. So

100

the velocity after 10 s is about 1556 ft/s.

(d) The terminal velocity is 180 ft/s.   40. (a) Q 5  15 1  e0045  15 01813  27345. Thus

x

5

(c)

approximately 27 lb of salt are in the barrel after 5 minutes.   (b) Q 10  15 1  e00410  15 03297  4946Thus

Q

20

10

approximately 49 lb of salt are in the barrel after 10 minutes. 0

(d) The amount of salt approaches 15 lb. This is to be expected, since

0

100

200

t

50 gal  03 lb/gal  15 lb. 41. n t 

5000 1  39e004t

(a) n 0 

5000  125 1  39e0040

(c) From the graph, we see that n t approaches 5000 as t gets large.

(b)

n

6000 4000 2000 0

0

100

200

t


SECTION 4.2 The Natural Exponential Function

42. D t 

15

54 54 . So D 20   1600 ft. 001t 1  29e 1  29e00120

43. Using the formula A t  Pert with P  7000 and r  3%  003, we fill in the table:

44. Using the formula A t  Pert with P  7000 and t  10 years, we fill in the table:

Time (years)

Amount

Rate per year

Amount

1

$721318

1%

$773620

2

$743286

2%

$854982

3

$765922

3%

$944901

4

$789248

4%

$10,44277

5

$813284

5%

$11,54105

6

$838052

6%

$12,75483

45. We use the formula A t  Pert with P  2000 and r  35%  0035. (a) A 2  2000e0035  2  $214502

(b) A 42  2000e0035  4  $230055 (c) A 12  2000e0035  12  $304392

46. We use the formula A t  Pert with P  3500 and r  625%  00625.

(a) A 3  3500e00625  3  $422181 (b) A 6  3500e00625  6  $509247 (c) A 9  3500e00625  9  $614269

47. (a) Using the formula A t  P 1  ik with P  600, i  25% per year  0025, and k  10, we calculate A 10  600 102510  $76805.

  0025 20  $76922. semiannually and k  10  2  20, so A  600 1  (b) Here i  0025 10 2 2   0025 40  $76982. quarterly and k  10  4  40, so A  600 1  (c) Here i  25% per year  0025 10 4 4 (d) Using the formula A t  Pert with P  600, r  25%  0025, and t  10, we have A 10  600e0025  10  $77042.

48. We use the formula A t  Pert with P  8000 and t  12.

(a) If r  2%  002, then A 12  8000e002  12  $10,16999.

(b) If r  3%  003, then A 12  8000e003  12  $11,46664.

(c) If r  45%  0045, then A 12  8000e0045  12  $13,72805. (d) If r  7%  007, then A 12  8000e007  12  $18,53094.

 2 49. Investment 1: After 1 year, a $100 investment grows to A 1  100 1  0025  10252. 2 4   10227. Investment 2: After 1 year, a $100 investment grows to A 1  100 1  00225 4 Investment 3: After 1 year, a $100 investment grows to A 1  100e002  10202. We see that Investment 1 yields the highest return.

2   10519. 50. Investment 1: After 1 year, a $100 investment grows to A 1  100 1  005125 2 Investment 2: After 1 year, a $100 investment grows to A 1  100e005  10512. We see that Investment 1 yields a higher return.

12 12


16

CHAPTER 4 Exponential and Logarithmic Functions

51. (a) A t  Pert  5000e009t (b) 20000

0

0

10

20

(c) A t  25,000 when t  1788 years.

52. We know that e x is increasing and ex is positive, so f x  e x  ex is increasing on 0  and its minimum value on that interval is f 0  e0  e0  2. But f is even, so because it is increasing on 0 , it is decreasing on  0. Therefore, its absolute minimum value is indeed f 0  2.

4.3

LOGARITHMIC FUNCTIONS

1. log x is the exponent to which the base 10 must be raised in order to get x. x

103

102

101

100

101

102

103

1012

log x

3

2

1

0

1

2

3

12

2. The function f x  log9 x is the logarithm function with base 9. So f 9  log9 91  1, f 1  log9 90  0,   f 19  log9 91  1, f 81  log9 92  2, and f 3  log9 912  12 . 3. (a) 53  125, so log5 125  3.

(b) log5 25  2, so 52  25.

4. (a) f x  log2 x is a logarithmic function with base 2. It has graph III.

(b) The graph of f x  log2 x is obtained from that of y  log2 x by reflecting about the y­axis. It has graph II.

(c) f x   log2 x is obtained from that of y  log2 x by reflecting about the x­axis. It has graph I.

(d) f x   log2 x is obtained from that of y  log2 x by reflecting about the x­ and y­axes. It has graph IV.

5. The natural logarithmic function f x  ln x has the vertical asymptote x  0.

6. The natural logarithmic function f x  ln x  1 has the vertical asymptote x  1. 7.

Logarithmic form

Exponential form

log8 8  1

81  8

log4 64  3

log8 4  23

823  4

log4 8  32

log8 81  1

81  18

log8 64  2

log8 512  3

1  2 log8 64

9. (a) 34  81  0 (b) 13  1

13. (a) 3x  5  3 (b) 16  2y

82  64 83  512

1 82  64

10. (a) 51  15  1 (b) 15 5  2 1 z 14. (a) 10 (b) 102t  3

8.

Logarithmic form

Exponential form 43  64

log4 2  12

412  2

1  2 log4 16 log4 21   12 1  5 log4 32 2

1 42  16

11. (a) 813  2 (b) 102  001 15. (a) e2y  10 (b) e2  3x  1

432  8

412  12

1 452  32 1 12. (a) 53  125

(b) 823  4 16. (a) e3  x  2 (b) e1  2x  3


SECTION 4.3 Logarithmic Functions

17. (a) log10 10,000  4   1  2 (b) log5 25 21. (a) log4 70  x

18. (a) log6 36  2   1  1 (b) log10 10

22. (a) log3 10  2x

(b) log12   3

(b) log10 01  4x

25. (a) log2 2  1 (b) log5 1  log5 50  0  1 (c) log12 2  log12 12  1

27. (a) log6 36  log6 62  2 (b) log9 81  log9 92  2

(c) log7 710  10   1  log 33  3 29. (a) log3 27 3  3  3 (b) log13 27  log13 13  (c) log7 7  log7 712  12 31. (a) 3log3 5  5

19. (a) log8 81  1   (b) log2 18  3

23. (a) ln 2  x

 2 (c) log12 025  log12 12  2

28. (a) log2 32  log2 25  5 (b) log5 513  13 (c) log6 1  log6 60  0

30. (a) log5 125  log5 53  3

(b) log49 7  log49 4912  12 (c) log9

 12  3  log9 312  log9 912  14

 32. (a) eln 3  3

(c) 10log 13  13

(b) log10 0001  x  10x  0001  x  3 37. (a) ln x  3  x  e3 (b) ln e2  x  x  2 ln e  2   1  x  4x  1  x  3 39. (a) log4 64 64  3 (b) log12 x  3  12  x  x  18   41. (a) log2 12  x  2x  12  x  1

1 (b) log10 x  3  103  x  x  1000

43. (a) logx 16  4  x 4  16  x  2

(b) logx 8  32  x 32  8  x  823  4

(b) ln t  05x

(b) log4 64  log4 43  3

(c) eln 10  10

35. (a) log6 x  2  x  62  36

24. (a) ln 05  x  1

26. (a) log3 37  7

1 (b) eln1  

(b) ln e4  4   1  ln e1  1 (c) ln e

(b) log12 8  3

(b) ln y  3

(b) 5log5 27  27

33. (a) log8 025  log8 823   23

20. (a) log4 0125   32

 12  2  log4 212  log4 412  14    1 (b) log4 12  log4 21  log4 412   12  3 (c) log4 8  log4 23  log4 412  log4 432  32

34. (a) log4

 0 36. (a) log13 x  0  x  13  1 (b) log4 1  x  4x  1  x  0

38. (a) ln x  1  x  e1  1e   (b) ln 1e  x  x  ln e1   ln e  1   40. (a) log9 13  x  9x  13  x   12 (b) log9 x  05  x  912  3

42. (a) logx 1000  3  x 3  1000  x  10 (b) logx 25  2  x 2  25  x  5 44. (a) logx 6  12  x 12  6  x  36 (b) logx 3  13  x 13  3  x  27

17


18

CHAPTER 4 Exponential and Logarithmic Functions

46. (a) log 50  16990  (b) log 2  01505    (c) log 3 2  06276

45. (a) log 2  03010 (b) log 352  15465   (c) log 23  01761

48. (a) ln 27  32958

47. (a) ln 5  16094

49.

(b) ln 739  20001

(b) ln 253  32308    (c) ln 1  3  10051

(c) ln 546  40000

y

x 1 33 1 32 1 3

f x

y

50.

2

x

1

1

_3

1 43 1 42 1 4

1

0

_4

3

1

32

_5

2

3

0

2

_1

1

2

3

4

5

6

x

_2

f x  log3 x

x 1 103 1 102 1 10

f x 6

_4

4

1

42

_5

2

2

3

1

1

0

2

1

_1

0

_4

10

2

102

_5

4

x

g x

3

4

5

6

x

4 8

1

2 3

f x  2 log x y

54.

2

x

f x

1

1 4 1 2

4

0

0

0

4

1

1

1

1

_2

2

_3

f x

2

_1

1

2

3

4

5

_4

10

2

102

_5

3

1 4

_3

1

x

2

_2

1

6

8

2

3

4

5

6

x

_2

1

52.

1

1

_1

1 16 1 8 1 2

_3

1 103 1 102 1 10

0

2

x

2

53.

1

3

2

0

4

2

y 3

y

51.

g x

0 _1

1

2

3

4

5

6

7

8 x

7

8 x

_2 _3 _4

f x  log12 x y 1 0

3

_1

2

_2

1

_3

1

_4

0

g x  log4 x

1

2

3

4

5

6

_5

g x  log x  1

f x  log2 x  2

55. Since the point 5 1 is on the graph, we have 1  loga 5  a 1  5. Thus the function is y  log5 x.     56. Since the point 12  1 is on the graph, we have 1  loga 12  a 1  12  a  2. Thus the function is y  log2 x.


SECTION 4.3 Logarithmic Functions

19

  57. Since the point 3 12 is on the graph, we have 12  loga 3  a 12  3  a  9. Thus the function is y  log9 x.

58. Since the point 9 2 is on the graph, we have 2  loga 9  a 2  9  a  13 . Thus the function is y  log13 x. 59. The graph of f x  2  ln x is obtained from that of y  ln x by shifting it upward 2 units, as in graph I.

60. The graph of f x  ln x  2 is obtained from that of y  ln x by shifting it to the right 2 units, as in graph II. 61. The graph of y  log4 x is obtained from that of y  4x

62. The graph of y  log3 x is obtained from that of y  3x

y

y

by reflecting it in the line y  x.

by reflecting it in the line y  x.

y=4¨

y=3¨

y=x

y=log£ x

y=log¢ x

1

1 x

1

63. The graph of g x  log5 x is obtained from that of y  log5 x by reflecting it about the y­axis.

Domain:  0. Range:  . Vertical asymptote: x  0.

y=x

1

x

64. The graph of f x   log10 x is obtained from that of y  log10 x by reflecting it about the x­axis.

Domain: 0 . Range:  . Vertical asymptote: x  0.

y

y

1

1 1

x

1

x

65. The graph of f x  log2 x  4 is obtained from that of 66. The graph of g x  ln x  2 is obtained from that of

y  ln x by shifting to the left 2 units. Domain: 2 .

y  log2 x by shifting it to the right 4 units.

Range:  . Vertical asymptote: x  2.

Domain: 4 . Range:  . Vertical asymptote: x  4.

y

y

1 1

1 1

x

x


20

CHAPTER 4 Exponential and Logarithmic Functions

67. The graph of h x  ln x  5 is obtained from that of

y  ln x by shifting to the left 5 units. Domain: 5 .

Range:  . Vertical asymptote: x  5. y

68. The graph of f x  2  log13 x is obtained from that of y  log13 x by reflecting about the x­axis, then shifting

upward 2 units. Domain: 0 . Range:  . Vertical asymptote: x  0. y

1 1

x (1, 2) 1 x

1

69. The graph of y  2  log3 x is obtained from that of

y  log3 x by shifting upward 2 units. Domain: 0 .

Range:  . Vertical asymptote: x  0. y

70. The graph of y  1  log10 x is obtained from that of y  log10 x by reflecting it about the x­axis, and then

shifting it upward 1 unit. Domain: 0 .

Range:  . Vertical asymptote: x  0. y

(1, 2) 1

(1, 1)

1 1

x

x

1

71. The graph of y  log3 x  1  2 is obtained from that of 72. The graph of y  1  ln x is obtained from that of y  log3 x by shifting to the right 1 unit and then

downward 2 units. Domain: 1 . Range:  . Vertical asymptote: x  1. 1

y

y  ln x by reflecting it about the y­axis and then shifting

it upward 1 unit. Domain:  0. Range:  . Vertical asymptote: x  0.

y

x

1

(2, _2)

(_1, 1) 1 1

x


SECTION 4.3 Logarithmic Functions

73. The graph of y  ln x is obtained from that of y  ln x

by reflecting the part of the graph for 0  x  1 about the

74. Note that y  ln x 

  ln x

if x  0  ln x if x  0

21

The graph

of y  ln x is obtained by combining the graph of

x­axis. Domain: 0 . Range: [0 .

y  ln x and its reflection about the y­axis.

Vertical asymptote: x  0.

Domain:  0  0 . Range:  . Vertical

y

asymptote: x  0.

y 1 x

1 1 x

1

75. f x  log10 x  3. We require that x  3  0  x  3, so the domain is 3 . 76. f x  log5 8  2x. Then we must have 8  2x  0  8  2x  4  x, and so the domain is  4.   77. g x  log3 x 2  1 . We require that x 2  1  0  x 2  1  x  1 or x  1, so the domain is  1  1 .   78. g x  ln x  x 2 . Then we must have x  x 2  0 

x 1  x  0. Using the methods from Chapter 1 with the

endpoints 0 and 1, we get the table at right. Thus the domain is 0 1.

Interval

 0

0 1

Sign of x

1 

Sign of 1  x

Sign of x 1  x

79. h x  ln x  ln 2  x. We require that x  0 and 2  x  0  x  0 and x  2  0  x  2, so the domain is 0 2.  80. h x  x  2  log5 10  x . Then we must have x  2  0 and 10  x  0  x  2 and 10  x  2  x  10. So the domain is [2 10.   81. y  log10 1  x 2 has domain 1 1, vertical

asymptotes x  1 and x  1, and local maximum y  0 at x  0.

  82. y  ln x 2  x  ln x x  1 has domain

 0  1 , vertical asymptotes x  0 and x  1, and no local maximum or minimum. 10

­2

2 ­10 ­2

10 ­10


22

CHAPTER 4 Exponential and Logarithmic Functions

83. y  x  ln x has domain 0 , vertical asymptote x  0, and no local maximum or minimum.

84. y  x ln x2 has domain 0 , no vertical asymptote, local minimum y  0 at x  1, and local maximum y  054 at x  014. 10

2 5 ­5 0

ln x has domain 0 , vertical asymptote x  0, x horizontal asymptote y  0, and local maximum y  037

85. y 

at x  272.

2

4

86. y  x log10 x  10 has domain 10 , vertical

asymptote x  10, and local minimum y  362 at x  587.

5 0

10

20 ­10

10

­2 ­5

Answers to Exercises 87–90 will vary.   87. F x  ln x 2  1 can be expressed as  f  g x, where f x  ln x and g x  x 2  1.

88. F x  ln x3 can be expressed as  f  g x, where f x  x 3 and g x  ln x.   89. F x  1  ln x can be expressed as  f  g x, where f x  ln x and g x  1  x.   90. F x  5  log x can be expressed as  f  g x, where f x  5  x and g x  log x.

91. f x  2x and g x  x  1 both have domain  , so  f  g x  f g x  2gx  2x1 with domain   and g  f  x  g  f x  2x  1 with domain  . 2

92. f x  3x and g x  x 2  1 both have domain  , so  f  g x  f g x  3x 1 with domain    2 and g  f  x  g  f x  3x  1  32x  1 with domain  .

93. f x  log2 x has domain 0  and g x  x  2 has domain  , so  f  g x  f g x  log2 x  2 with domain 2  and g  f  x  g  f x  log2 x  2 with domain 0 .   94. f x  log x has domain 0  and g x  x 2 has domain  , so  f  g x  f g x  log x 2 with

domain  0  0  and g  f  x  g  f x  log x2 with domain 0 .  6 95. The graph of g x  x grows faster than the graph of 96. (a) f x  ln x.

4

6 4 2

0

g

10

g

2

f

0

20

30

f

10

20

30

(b) From the graph, we see that the solution to the  equation x  1  ln 1  x is x  1350.


SECTION 4.3 Logarithmic Functions

97. (a)

98. (a)

c=4 c=3 c=2 c=1

2

c=4

6

c=3

4

c=2

1 2

0

23

10

20

30

0

(b) Notice that f x  log cx  log c  log x, so

c=1 10

20

30

(b) As c increases, the graph of f x  c log x

as c increases, the graph of f x  log cx is

stretches vertically by a factor of c.

shifted upward log c units.

  99. (a) f x  log2 log10 x . Since the domain of log2 x is the positive real numbers, we have: log10 x  0  x  100  1. Thus the domain of f x is 1 .   y x (b) y  log2 log10 x  2 y  log10 x  102  x. Thus f 1 x  102 .

100. (a) f x  ln ln ln x. We must have ln ln x  0  ln x  1  x  e. So the domain of f is e . y

(b) y  ln ln ln x  e y  ln ln x  ee  ln x  ee 2x 2x .y  y  y2x  2x x 12 1  2x y  y  2x  y2x  2x 1  y  2x  1 y   y . Thus  x  log2 1y   x f 1 x  log2 . 1x

101. (a) f x 

ey

ex

 x. Thus the inverse function is f 1 x  ee . (b)

x  0. Solving this using the methods from 1x Chapter 1, we start with the endpoints, 0 and 1. Interval

 0

0 1

Sign of x

1 

Sign of 1  x x Sign of 1x

Thus the domain of f 1 x is 0 1.  I  2500 ln 07  89169 moles/liter. 102. Using I  07I0 we have C  2500 ln I0   D  8267 ln 073  2602 years. 103. Using D  073D0 we have A  8267 ln D0 

104. Substituting N  1,000,000 we get t  3 105. When r  6% we have t  t

ln 2  87 years. 008

log N50 log 20,000 3  4286 hours. log 2 log 2

ln 2 ln 2  116 years. When r  7% we have t   99 years. And when r  8% we have 006 007

106. Using k

 025 and substituting C  09C0 we have   C  025 ln 1  09  025 ln 01  058 hours. t  025 ln 1  C0

log 2AW  log 2  1005 log 40    532. Using A  100 and log 2 log 2 log 2 log 2  10010 log 20 523 log 2AW     432. So the smaller icon is  123 times W  10 we find the ID to be log 2 log 2 log 2 432 harder.

107. Using A  100 and W  5 we find the ID to be


24

CHAPTER 4 Exponential and Logarithmic Functions

108. (a) Since 2 feet  24 inches, the height of the graph is 224  1677216 inches. Now, since there are 12 inches per foot and ,216 5280 feet per mile, there are 12 5280  63,360 inches per mile. So the height of the graph is 1,677 63360  2648, or

about 265 miles.   (b) Since log2 224  24, we must be about 224 inches  265 miles to the right of the origin before the height of the

graph of y  log2 x reaches 24 inches or 2 feet.   109. log log 10100  log 100  2      log log log 10googol  log log googol  log log 10100  log 100  2

110. Notice that loga x is increasing for a  1. So we have log4 17  log4 16  log4 42  2. Also, we have log5 24  log5 25  log5 52  2. Thus, log5 24  2  log4 17.

111. The numbers between 1000 and 9999 (inclusive) each have 4 digits, while log 1000  3 and log 10,000  4. Since     log x  3 for all integers x where 1000  x  10,000, the number of digits is log x  1. Likewise, if x is an integer       where 10n1  x  10n , then x has n digits and log x  n  1. Since log x  n  1  n  log x  1, the number   of digits in x is log x  1. 112. We express each side in exponential form: log1a x  y  1a y  x  a y  x. Using the definition of the logarithm on the last equation, we have loga x  y  y   loga x.

Thus, we have shown that log1a x   loga x, and so the graph of

g x  log1a x can be obtained by reflecting the graph of f x  loga x

about the x­axis.

y 4 3 2

f(x)=log4 x

1 0 _1

1

2

3

_2

4 g(x)=log1/4 x

_3 _4

4.4

LAWS OF LOGARITHMS

1. The logarithm of a product of two numbers is the same as the sum of the logarithms of these numbers. So log5 25  125  log5 25  log5 125  2  3  5.

2. The logarithm of a quotient of two numbers is the same as the difference of the logarithms of these numbers. So   25  log 25  log 125  2  3  1. log5 125 5 5

3. The logarithm of a number raised to a power is the same as the power times the logarithm of the number. So   log5 2510  10  log5 25  10  2  20.     x2 y  log x 2 y  log z  log x 2  log y  log z  2 log x  log y  log z 4. log z     x2 y 2 2 . 5. 2 log x  log y  log z  log x  log y  log z  log x y  log z  log z

6. (a) To express log7 12 in terms of common logarithms, we use the Change of Base Formula to write log 12 1079 log7 12    1277. log 7 0845 2485 ln 12 (b) Yes, the result is the same: log7 12    1277. ln 7 1946

5 x


SECTION 4.4 Laws of Logarithms

7. (a) False. log A  B  log A  log B. (b) True. log AB  log A  log B.

A  log A  log B. B log A  log A  log B. (b) False. log B

8. (a) True. log

9. log 50  log 200  log 50  200  4

10. log6 9  log6 24  log6 9  24  3

60  2 11. log2 60  log2 15  log2 15

12. log3 135  log3 45  log3 135 45  1

13. 14 log3 81  14 4  1    15. log5 5  log5 512  12

14.  13 log3 27   13 3  1   16. log5  1  log5 532   32 125   2 6 17. log2 6  log2 15  log2 20  log2 15  log2 20  log2 5  20  log2 8  log2 23  3     100  log3 19  log3 32  2 18. log3 100  log3 18  log3 50  log3 18  50  100 19. log4 16100  log4 42  log4 4200  200  33  log2 299  99 20. log2 833  log2 23   21. log log 1010,000  log 10,000 log 10  log 10,000  1  log 10,000  log 104  4 log 10  4     200  ln e200 ln e  ln e200  200 ln e  200 22. ln ln ee 23. log3 8x  log3 8  log3 x

24. log6 7r  log6 7  log6 r

25. log3 2x y  log3 2  log3 x  log3 y

26. log5 4st  log5 4  log5 s  log5 t  28. log t 5  log t 52  52 log t 27. ln a 3  3 ln a      29. log3 x yz  log3 x yz12  12 log3 x  log3 y  log3 z 30. log5 x y6  6 log5 x y  6 log5 x  log5 y     32. ln x y 2  ln x 12  ln y 2  12 ln x  2 ln y 31. ln a 3 b2  ln a 3  ln b2  3 ln a  2 ln b

4a y  log2 4a  log2 b  2  log2 a  log2 b  log6 y  log6 6z  log6 y  log6 z  1 34. log6 b 6z   a 3 b2 35. log8  log8 a 3  log8 b2  log8 c  3 log8 a  2 log8 b  log8 c c   3x 4 y 2  log7 3  4 log7 x  2 log7 y  log7 2  3 log7 z 36. log7 2z 3  3x 5  12  52 log3 x  log3 y 37. log3 y

33. log2

y3 38. log   3 log y  12 log 2  12 log x 2x     x 3 y4  log x 3 y 4  log z 6  3 log x  4 log y  6 log z 39. log 6 z       x2  loga x 2  loga yz 3  2 loga x  loga y  3 loga z  2 loga x  loga y  3 loga z 40. loga 3 yz

25


26

CHAPTER 4 Exponential and Logarithmic Functions

  x 4  2  12 ln x 4  2    3 42. log x 2  4  13 log x 2  4

41. ln

  x z x  z 12  log  12 log x  z  12 log y y y   4x 2 44. ln 2  ln 4  2 ln x  ln x 2  3 x 3    2 2 3 x  y  13 ln x 2  y 2  13 ln x  y 45. ln xy 43. log

x2  2 ln x  12 ln x  1 x 1        2   x2  4 x2  4 1 log 1 log x 2  4  log x 2  1 3 7 47. log     x   2  2 2 2 x2  1 x3  7 x2  1 x3  7         12 log x 2  4  log x 2  1  2 log x 3  7                48. log x y z  12 log x y z  12 log x  log y z  12 log x  12 log y z     12 log x  12 log y  12 log z  12 log x  14 log y  18 log z   49. log4 6  2 log4 7  log4 6  log4 72  log4 6  72  log4 294 46. ln 

50. 12 log2 5  2 log2 7  log2

  5  log2 49  log2 495

51. 2 log x  3 log x  1  log x 2  log x  13  log 52. 3 ln 2  2 ln x  12 ln x  4  ln 23  ln x 2  ln

x2 x  13

8x 2 x  4  ln  x 4

53. log x  1  log x  1  3 log x  log x  1 x  1  log x 3  log

x2  1 x3

  x 3 1 54. ln x  3  ln x 2  9  ln 2  ln x 3 x 9      x 2 x 2   log5 55. 12 log5 x  2  log5 x 2  4  log5 x  12 log5  2 x 4 x x 3  4x     ac2 a 4 c8 56. 4 log3 a  3 log3 b  2 log3 c  4 log3 a  log3 b3  log3 c2  4 log3 3  log3 12 b b   2  4 x  3  13 log x  2  12 log  57. 13 log x  23  12 log x 4  log x 2  x  6 2 2 x x 6 12  x4 x2 x 2 x  2 x2  log x  2  log  log x  2  log  log  log 2 x 3 x  3 x  2 x  3 x  2 [x  3 x  2]

  bd c 58. loga b  c loga d  r loga s  loga bd c  loga s r  loga r s log 10 log 7  1430677 60. log14 7   0737350 59. log5 10  log 5 log 14 log 4 log 30  0630930 62. log5 30   2113283 61. log9 4  log 9 log 5


SECTION 4.4 Laws of Logarithms

27

log 261 log 532  0493008 64. log6 532   3503061 log 7 log 6 log 125 log 25 65. log4 125   3482892 66. log12 25   0368743 log 4 log 12 ln x 1 1 loge x 67. log3 x    ln x. The graph of y  ln x is loge 3 ln 3 ln 3 ln 3 2

63. log7 261 

shown in the viewing rectangle [1 4] by [3 2]. 2

4

­2

68. Note that logc x 

 1 ln x (by the change of base formula). So ln c

a=2 a=e a=5 a=10

2

the graph of y  logc x is obtained from the graph of y  ln x by

1 depending ln c on whether ln c  1 or ln c  1. All of the graphs pass through 1 0 either shrinking or stretching vertically by a factor of

0

2

4

­2

because logc 1  0 for all c.

1 ln e  ln 10 ln 10     log b log c log d log d 70. loga b logb c logc d   log a log b log c log a 1 1 1 1 log abc 1 1 log a  log b  log c 71.    log x  log x  log x    loga x logb x logc x log x log x

69. log e 

log a

72. loga n x 

log b

log c

1 log x 1 log x log x   loga x  log a n n log a n log a n

73. (a) log P  log c  k log W  log P  log c  log W k  log P  log

c Wk

P

1 log x logabc

1 logabc x

c . Wk

8000 8000  1866 and when W  10 we have P  21  64. 221 10 80 P0 . Substituting P0  80, t  24, and c  03 we have P   305. So the 74. From Example 4(a), P  t  1c 24  103 student should get a score of 30. (b) Using k  21 and c  8000, when W  2 we have P 

75. (a) M  25 log BB0   25 log B  25logB0 .

(b) Suppose B1 and B2 are the brightness of two stars such that B1  B2 and let M1 and M2 be their respective magnitudes. Since log is an increasing function, we have log B1  log B2 . Then log B1  log B2  log B1  log B0  log B2  log B0  log B1 B0   log B2 B0   25 log B1 B0   25 log B2 B0   M1  M2 . Thus the brighter star has less magnitudes.

(c) Let B1 be the brightness of the star Albiero. Then 100B1 is the brightness of Betelgeuse, and its magnitude is     M  25log 100B1 B0   25 log 100  log B1 B0   25 2  log B1 B0   5  25 log B1 B0   5  magnitude of Albiero

  76. (a) log S  log c  k log A  log S  log c  log Ak  log S  log c Ak  S  c Ak .

(b) If A  2A0 when k  3 we get S  c 2A0 3  c  23  A30  8  c A30 . Thus doubling the area increases the number of species eightfold.


28

CHAPTER 4 Exponential and Logarithmic Functions

77. (a) False; log xy  log x  log y  log x  log y.

(b) False; log2 x  log2 y  log2 xy  log2 x  y.   (c) True; the equation is an identity: log5 ab2  log5 a  log5 b2  log5 a  2 log5 b. (d) True; the equation is an identity: log 2z  z log 2.

(e) False; log P  log Q  log P Q  log P log Q. (f) False; log a  log b  log ab  log a  log b.  x (g) False; x log2 7  log2 7x  log2 7 .

(h) True; the equation is an identity. loga a a  a loga a  a  1  a.

(i) False; log x  y  log x  log y. For example, 0  log 3  2  log 3  log 2. (j) True; the equation is an identity:  ln 1A   ln A1  1  ln A  ln A.

78. The error is on the first line: log 01  0, so 2 log 01  log 01.

79. Let f x  x 2 . Then f 2x  2x2  4x 2  4 f x. Now the graph of f 2x is the same as the graph of f shrunk

horizontally by a factor of 12 , whereas the graph of 4 f x is the same as the graph of f x stretched vertically by a factor of 4. Let g x  e x . Then g x  2  e x2  e2 e x  e2 g x. This shows that a horizontal shift of 2 units to the right is the same as a vertical stretch by a factor of e2 .

Let h x  ln x. Then h 2x  ln 2x  ln 2  ln x  ln 2  h x. This shows that a horizontal shrinking by a factor of 12 is the same as a vertical shift upward by ln 2.            80. If the identity is true, then ln x  x 2  1  ln x  x 2  1  0  ln x  x 2  1 x  x 2  1  0     ln x 2  x 2  1  0  ln 1  0. The last statement is true, and all steps are reversible, so the original identity is true.   81. Note that x ln y  x ln Cekx  x kx  ln C, which is on the line Y  k X  ln C.   82. We show that the point ln x ln y  ln x ln ax n satisfies the equation Y  n X  ln a. Substituting, we get Y  n X  ln a  ln ax n  n ln x  ln a  ln ax n  ln ax n , which is true.

4.5

EXPONENTIAL AND LOGARITHMIC EQUATIONS

1. (a) First we isolate e x to get the equivalent equation e x  25.

(b) Next, we take the natural logarithm of each side to get the equivalent equation x  ln 25.

(c) Now we use a calculator to find x  3219.

2. (a) First we combine the logarithms to get the equivalent equation log 3 x  2  log x.

(b) Next, we write each side in exponential form to get the equivalent equation 3 x  2  x. (c) Now we find x  3.

3. Because the function 5x is one­to­one, 5x1  625  5x1  54  x  1  4  x  5. 2

4. Because the function e x is one­to­one, e x  e9  x 2  9  x  3.

5. Because the function 5x is one­to­one, 52x3  1  52x3  50  2x  3  0  x  32 . 1  2x  3  1  x  1. 6. Because the function 10x is one­to­one, 102x3  10

7. Because the function 7x is one­to­one, 72x3  765x  2x  3  6  5x  3x  9  x  3. 8. Because the function e x is one­to­one, e12x  e3x5  1  2x  3x  5  5x  6  x  65 . 2

2

9. Because the function 6x is one­to­one, 6x 1  61x  x 2  1  1  x 2  2x 2  2  x  1.


SECTION 4.5 Exponential and Logarithmic Equations 2

2

10. Because the function 10x is one­to­one, 102x 3  109x  2x 2  3  9  x 2  3x 2  12  x  2.

11. (a) e x  16  ln e x  ln 16  x ln e  ln 16  x  ln 16  4 ln 2 (b) x  2772589

12. (a) e2x  5  ln e2x  ln 5  2x ln e  ln 5  2x  ln 5  x   12 ln 5 (b) x  0804719

13. (a) 10x  6  log 10x  log 6  x log 10  log 6  x   log 6 (b) x  0778151

14. (a) 105x  24  log 105x  log 24  5x log 10  log 24  x  15 log 24 (b) x  0276042

15. (a) 3x5  4  ln 3x5  ln 4  x  5 ln 3  ln 4  x  5 

ln 4 ln 4 x  5 ln 3 ln 3

(b) x  3738140

16. (a) e23x  11  ln e23x  ln 11  2  3x  ln 11  x   13 ln 11  2 (b) x  0132632

17. (a) 3  61x  15  61x  5  ln 61x  ln 5  1  x ln 6  ln 5  x  1  (b) x  0101756

ln 5 ln 6

ln 8 1 18. (a) 5  432x  8  432x  85  3  2x ln 4  ln 85  3  2x  5  x  ln 4 2 (b) x  1330482 15 19. (a) 200 1024t  1500  1024t  15 2  4t ln 102  ln 2  t 

3

ln 85 ln 4

ln 15 2 4 ln 102

(b) t  25437319 12 20. (a) 25 101512t  60  101512t  12 5  12t ln 1015  ln 5  t 

ln 12 5 12 ln 1015

(b) t  4900103

21. (a) 3  e5t  12  5  t  ln 4  t  5  ln 4 (b) t  3613706

22. (a) 5 

 2t3 1 2

 24 

(b) t  0368483

 2t3 1 2

ln 24 5 t  1 1 24  24 5  2t  3 ln 2  ln 5  2t  3  2 ln 12

3

ln 24 5 ln 12

10 ln 03 x ln 2  ln 03  x  10 ln 2 (b) x  17369656

23. (a) 2x10  03 

24. (a) 2x50  06  

50 ln 06 x ln 2  ln 06  x   50 ln 2

(b) x  36848280   25. (a) 4 1  105x  9  1  105x  94  105x  54  5x log 10  log 54  x  15 log 54

(b) x  0019382   ln 45 ln 45 x  1 26. (a) 2 5  3x1  100  5  3x1  50  3x1  45  x  1 ln 3  ln 45  x  1  ln 3 ln 3 (b) x  2464974

29


30

CHAPTER 4 Exponential and Logarithmic Functions

27. (a) 8  e14x  20  e14x  12  1  4x  ln 12  x  (b) x  0371227 28. (a) 1  e4x1  20  e4x1  19  4x  1  ln 19  x 

1  ln 12 4 ln 19  1 4

(b) x  0486110 29. (a) 4x  212x  50  22x  212x  50  22x 1  2  50  22x  50 3  2x ln 2  ln 50  ln 3  x 

ln 50 3 ln 4

(b) x  2029447

30. (a) 125x  53x1  200  53x  53x1  200  53x 1  5  200  53x  100 3  3x log 5  log 100  log 3  2  log 3 3 log 5 (b) x  0726249 x

31. (a) 5t  103t2  t log 5  3t  2  t log 5  3  2  t  (b) t  0869176 32. (a) et  31t  t  1  t ln 3  t 1  ln 3  ln 3  t  (b) t  0523495 33. (a) 5x3  3x1  (b) x  1954364

2 log 5  3

ln 3 ln 3  1

3 ln 3 x ln 5  x  1 ln 3  x ln 5  3 x  1 ln 3  x ln 5  3 ln 3  3 ln 3  x  3 ln 5  3 ln 3

34. (a) 32x1  24x1  2x  1 ln 3  4x  1 ln 2  x 2 ln 3  4 ln 2  ln 3  ln 2  x  (b) x  3114131

ln 6 ln 9  ln 16

50  4  50  4  4ex  46  4ex  115  ex  ln 115  x  x   ln 115 1  ex (b) x  2442347

35. (a)

10  2  10  2  2ex  8  2ex  4  ex  ln 4  x  x   ln 4 1  ex (b) x  1386294    37. e2x  5e x  6  0  e x  6 e x  1  0  e x  6 or e x  1. e x  6 has no solution because e x  0 for all x. 36. (a)

Thus, the only solution occurs where e x  1  x  0.    38. e2x  3e x  10  0  e x  5 e x  2  0  e x  5 or e x  2. The former equation has no solution because e x  0 for all x, so the only solution is x  ln 2  06931.    39. e4x  4e2x  21  0  e2x  7 e2x  3  0  e2x  7 or e2x  3. Now e2x  7 has no solution because

e2x  0 for all x. But we can solve e2x  3  2x  ln 3  x  12 ln 3  05493. So the only solution is x  05493.    40. 34x  32x  6  0  32x  3 32x  2  0  32x  3 or 32x  2. The latter equation has no solution, so we solve

32x  3  2x  1  x  12 .        41. 2x  10 2x  3  0  2x 22x  10  3  2x  0  2x 2x  5 2x  2  0. The first two factors are positive

everywhere, so we solve 2x  2  0  x  1.      42. e x  15ex  8  0  ex e2x  15  8e x  0  ex e x  5 e x  3  0. The first factor is positive everywhere, so the solutions occur where e x  5 or e x  3; that is, x  ln 3  10986 or x  ln 5  16094.


SECTION 4.5 Exponential and Logarithmic Equations

31

 43. x 2 2x  2x  0  2x x 2  1  0  2x  0 (never) or x 2  1  0. If x 2  1  0, then x 2  1  x  1. So the only

solutions are x  1.       44. x 2 10x  x10x  2 10x  x 2 10x  x10x  2 10x  0  10x x 2  x  2  0  10x  0(never) or x 2  x  2  0. If x 2  x  2  0, then x  2 x  1  0  x  2, 1. So the only solutions are x  2, 1.

45. 4x 3 e3x  3x 4 e3x  0  x 3 e3x 4  3x  0  x  0 or e3x  0 (never) or 4  3x  0. If 4  3x  0, then

3x  4  x  43 . So the solutions are x  0 and x  43 .   46. x 2 e x  xe x  e x  0  e x x 2  x  1  0  e x  0 (impossible) or x 2  x  1  0. If x 2  x  1  0, then 

5 . So the solutions are x  1 5 . x  1 2 2

47. log x  2  log x  3  log 4x  log x  2 x  3  log 4x  x  2 x  3  4x  x 2  5x  6  0  x  6 x  1  0  x  1 or x  6. However, x  1 does not satisfy the original equation because log x is defined only for x  0. Thus, the only solution is x  6.

48. log5 x  2  log5 x  5  log5 6x  log5 x  2 x  5  log5 6x  x 2  3x  10  6x  x 2  9x  10  0  x  10 x  1  0  x  1 or x  10. However, only x  10 satisfies the original equation, and so that is the only solution.   49. 2 log x  log 2  log 3x  4  log x 2  log 6x  8  x 2  6x  8  x 2  6x  8  0  x  4 x  2  0  x  4 or x  2. Thus the solutions are x  4 and x  2.       ln x 2  2x  1  x 2  x 2  2x  1  0  x  12  0  x  1. 50. ln x  12  ln 2  2 ln x  ln 2 x  12

Thus the only solution is x  1. 51. log2 3  log2 x  log2 5  log2 x  2  log2 3x  log2 5x  10  3x  5x  10  2x  10  x  5 52. log4 x  2  log4 3  log4 5  log4 2x  3  log4 3 x  2  log4 5 2x  3  3x  6  10x  15  7x  21 x 3 53. log x  9  x  109  1,000,000,000

54. log x  3  4  x  3  104  x  9997 55. ln 4  x  1  4  x  e  x  4  e  12817 56. ln 3x  1  0  3x  1  1  x  0

57. log3 5  x  1  5  x  31  x  14 3  1 58. log12 3x  2  1  3x  2  12  3x  0  x  0

59. 4  log 3  x  3  log 3  x  1  3  x  10  x  7   60. log2 x 2  x  2  2  x 2  x  2  22  4  x 2  x  6  0  x  3 x  2  0  x  3 or x  2. Thus, the solutions are x  3 and x  2. 61. log2 x  log2 x  3  2  log2 [x x  3]  2  x 2  3x  22  x 2  3x  4  0  x  4 x  1  x  1 or x  4. Since log 1  3  log 4 is undefined, the only solution is x  4. 62. log x  log x  3  1  log [x x  3]  1  x 2  3x  10  x 2  3x  10  0  x  2 x  5  0  x  2 or x  5. Since log 2 is undefined, the only solution is x  5.

63. log9 x  5  log9 x  3  1  log9 [x  5 x  3]  1  x  5 x  3  91  x 2  2x  24  0  x  6 x  4  0  x  6 or4. However, x  4 is inadmissible, so x  6 is the only solution.  64. ln x  1  ln x  1  0  ln x  1 x  1  0  x  1 x  1  1  x 2  2  0  x   2. However,   x   2 fails to satisfy the original equation, so x  2 is the only solution.  2 x 1 1 x 1 1 65. log15 x  1  log15 x  1  2  log15 2   25 x  1  x  1  24x  26  x 1 x 1 5 25  x  13 12 .


32

CHAPTER 4 Exponential and Logarithmic Functions

66. log3 x  15  log3 x  1  2  log3 x  15  9x  9  8x  24  x  3

x  15 x  15 x  15 2  32   9  x  15  9 x  1  x 1 x 1 x 1

67. ln x  3  x  ln x  x  3  0. Let f x  ln x  x  3. We need to solve the equation f x  0. From the graph of f , we get x  221.

68. log x  x 2  2  log x  x 2  2  0. Let

f x  log x  x 2  2. We need to solve the equation

f x  0. From the graph of f , we get x  001 or

x  147.

2 0

2

4

0

1

2

­2

69. x 3  x  log10 x  1  x 3  x  log10 x  1  0.

Let f x  x 3  x  log10 x  1. We need to solve the

equation f x  0. From the graph of f , we get x  0 or x  114.

    70. x  ln 4  x 2  x  ln 4  x 2  0. Let   f x  x  ln 4  x 2 . We need to solve the equation f x  0. From the graph of f , we get x  196 or

x  106.

2

­1

1

2

­2

2

­2

71. e x  x  e x  x  0. Let f x  e x  x. We need to

solve the equation f x  0. From the graph of f , we get x  057.

72. 2x  x  1  2x  x  1  0. Let

f x  2x  x  1. We need to solve the equation

f x  0. From the graph of f , we get x  138.

2

­1.0

­0.5

2

0 ­1 ­2

1 ­2

2


SECTION 4.5 Exponential and Logarithmic Equations

73. 4x 

x  4x 

x  0. Let f x  4x 

2

33

2

74. e x  2  x 3  x  e x  2  x 3  x  0. Let

x.

2

We need to solve the equation f x  0. From the graph

f x  e x  2  x 3  x. We need to solve the equation

of f , we get x  036.

f x  0. From the graph of f , we get x  089 or

x  071.

2

2 0

1

2 ­2

­2

2 ­2

  75. log x  2  log 9  x  1  log [x  2 9  x]  1  log x 2  11x  18  1  x 2  11x  18  101  0  x 2  11x  28  0  x  7 x  4. Also, since the domain of a logarithm is positive we must have

0  x 2  11x  18  0  x  2 9  x. Using the methods from Chapter 1 with the endpoints 2, 4, 7, 9 for the intervals, we make the following table: Interval

 2

2 4

4 7

7 9

9 

Sign of x  7 Sign of x  4 Sign of x  2 Sign of 9  x

Sign of x  7 x  4 Sign of x  2 9  x

Thus the solution is 2 4  7 9.

76. 3  log2 x  4  23  x  24  8  x  16.

77. 2  10x  5  log 2  x  log 5  03010  x  06990. Hence the solution to the inequality is approximately the interval 03010 06990.        78. x 2 e x  2e x  0  e x x 2  2  0  e x x  2 x  2  0. We use the methods of Chapter 1 with the endpoints    2 and 2, noting that e x  0 for all x. We make a table: Interval Sign of e x    Sign of x  2    Sign of x  2      Sign of e x x  2 x  2

  Thus  2  x  2.

     2

     2 2

  2 

  ln y 79. To find the inverse of f x  22x , we set y  f x and solve for x. y  22x  ln y  ln 22x  2x ln 2  x  . 2 ln 2 ln x ln x Interchange x and y: y  . Thus, f 1 x  . 2 ln 2 2 ln 2


34

CHAPTER 4 Exponential and Logarithmic Functions

  80. To find the inverse of f x  3x1 , we set y  f x and solve for x. y  3x1  ln y  ln 3x1  x  1 ln 3  x 1

ln y ln y ln x ln x x   1. Interchange x and y: y   1. Thus, f 1 x   1. ln 3 ln 3 ln 3 ln 3

81. To find the inverse of f x  log2 x  1, we set y  f x and solve for x. y  log2 x  1  2 y  2log2 x1  x  1  x  2 y  1. Interchange x and y: y  2x  1. Thus, f 1 x  2x  1.

82. To find the inverse of f x  log 3x, we set y  f x and solve for x. y  log 3x  10 y  3x  x  10x 10x . Thus, f 1 x  . 3 3   1  log 22 log5 x  log 1 83. 22 log5 x  16 2 2 16 

10 y . Interchange 3

x and y: y 

2  4  log5 x   12  x  512  1  04472 5 log5 x

  84. log2 log3 x  4  log3 x  24  16  x  316  43,046,721

1. 85. log x4  log x3  0  log x3 log x  1  log x  0 or log x  1  x  1 or x  10

log x  3  log x  log 3  log x  3  log 3x  x  3  3x  2x  3  x  32   86. log x3  3 log x  log x3  3 log x  0  log x log x2  3  log x  0 or log x2  3  0. Now log x  0     x  1. Also log x2  3  0  log x2  3  log x   3  x  10 3 , so x  10 3  539574 or 

x  10 3  00185. Thus the solutions to the equation are x  1, x  10 3  539574 and x  10 3  00185.     00225 45 87. (a) A 5  5000 1   5000 100562520  559360. Thus the amount after 5 years is $559360. 4     00225 4t  5000 10056254t  2  10056254t  log 2  4t log 1005625  (b) 10000  5000 1  4 log 2 t  3089 years. Thus the investment will double in about 3089 years, or about 30 years 10 months. 4 log 1005625 88. (a) A 2  6500e00454  $778191

  0045t  ln 16  0045t  t  (b) 8000  6500e0045t  16  e 13 13

  1 ln 16 13  461. So the investment will reach 0045

$8000 in about 461 years, or about 4 years 8 months.  4t   89. 8000  5000 1  0035  5000 1008754t  16  1008754t  log 16  4t log 100875  4 t

log 16  1349 years. The investment will increase to $8000 in approximately 13 years and 6 months. 4 log 100875

 2t 90. 5000  3000 1  0065  53  103252t  log 53  2t log 10325  t  2

log 53 2 log 10325

 799. So it takes about

8 years to grow to $5000.

ln 2  815 years. Thus the investment will double in about 815 years. 0085     r 8 r r r 24  8 143577   8 143577  1   143577  1  1 92. 143577  1000 1  2 2 2  2 r  2 8 143577  1  00925. Thus the rate was about 925%. 91. 2  e0085t  ln 2  0085t  t 

  ln 3  126277. So only 5 grams remain after 93. 15e0087t  5  e0087t  13  0087t  ln 13   ln 3  t  0087 approximately 13 days.


SECTION 4.5 Exponential and Logarithmic Equations

35

  94. We want to solve for t in the equation 80 e02t  1  70 (when motion is downwards, the velocity is negative). Then       ln 18 7 1 1 80 e02t  1  70  e02t  1   8  e02t  8  02t  ln 8  t   104 seconds. Thus the 02 velocity is 70 ft/s after about 10 seconds. 10  7337, so there are approximately 7337 fish after 3 years. 1  4e083 10 08t  1  e08t  025  08t  ln 025   5  1  4e08t  10 (b) We solve for t. 5  2  4e 1  4e08t ln 025 t  173. So the population will reach 5000 fish in about 1 year and 9 months. 08 k 96. (a) The length y is inversely proportional to 2x , so y  x  k  2x . To find k, we substitute 200  k  28  2 k  200  28  51200. Thus, we have y  51200  2x .

95. (a) P 3 

(b) y  51200  25  1600 base­pairs

5 2000  5  x ln 2  ln 5  x   ln 128  47 cm (c) 2000  51200  2x  2x  51200 128 128 ln 2   h P P 97. (a) ln    ehk  P  P0 ehk . Substituting k  7 and P0  100 we get P  100eh7 . P0 k P0

(b) When h  4 we have P  100e47  5647 kPa.   T  20 T  20 98. (a) ln  011t   e011t  T  20  200e011t  T  20  200e011t 200 200

(b) When t  20 we have T  20  200e01120  20  200e22  422 F.     13t5  13 I  1  e13t5  e13t5  1  13 I   13 t  ln 1  13 I  99. (a) I  60 13 1  e 60 60 5 60   5 ln 1  13 I . t   13 60   5 ln 1  13 2  0218 seconds. (b) Substituting I  2, we have t   13 60

100. (a) P  M  Cekt  Cekt  M  P  ekt      1 MP MP  t   ln kt  ln C k C

MP  C

(c)

(b) P t  20  14e0024t . Substituting M  20, C  14, k  0024,   1 MP and P  12 into t   ln , we have k C   1 20  12 t  ln  2332. So it takes about 23 months. 0024 14

20 10 0

0

100

200

101. Since 91  9, 92  81, and 93  729, the solution of 9x  20 must be between 1 and 2 (because 20 is between 9 and 81), whereas the solution to 9x  100 must be between 2 and 3 (because 100 is between 81 and 729).   1 102. Notice that log x 1 log x  log x  1, so 1 log x  101 for all x  0. So log x x 1 log x  5 has no solution, and x 1 log x  k has a solution only when k  10.

10

This is verified by the graph of f x  x 1 log x .

0

0

5

10


36

CHAPTER 4 Exponential and Logarithmic Functions

  103. (a) x  1logx1  100 x  1  log x  1logx1  log 100 x  1     2 log x  1 log x  1  log 100  log x  1  log x  1  log x  1  2  0     log x  1  2 log x  1  1  0. Thus either log x  1  2  x  101 or log x  1  1  x  11 10 .         (b) log2 x  log4 x  log8 x  11  log2 x  log2 x  log2 3 x  11  log2 x x 3 x  11  log2 x 116  11 6  11 6 log2 x  11  log2 x  6  x  2  64

  2     ln 3 or 2x  1, which (c) 4x  2x1  3  2x  2 2x  3  0  2x  3 2x  1  0  either 2x  3  x  ln 2 ln 3 has no real solution. So x  is the only real solution. ln 2

4.6

MODELING WITH EXPONENTIAL FUNCTIONS

1. (a) Here n 0  10 and a  15 hours, so n t  10  2t15  10  22t3 .

(b) After 35 hours, there will be n 35  10  22353  106  108 bacteria.   2t 3 ln 1000 (c) n t  10  22t3  10,000  22t3  1000  ln 22t3  ln 1000  ln 2  ln 1000  t   149, 3 2 ln 2 so the bacteria count will reach 10,000 in about 149 hours.

2. (a) Here n 0  25 and a  5 hours, so n t  25  2t5 .

(b) After 18 hours, there will be n 18  25  2185  303 bacteria.   t (c) n t  25  2t5  1,000,000  2t5  40,000  ln 2t5  ln 40,000  ln 2  ln 40,000  5 5 ln 40,000  764, so the bacteria count will reach 1,000,000 in about 764 hours. t ln 2

3. (a) A model for the squirrel population is n t  n 0  2t6 . We are given

(c)

n

1,000,000

that n 30  100,000, so n 0  2306  100,000 

100,000  3125. Initially, there were approximately 25 3125 squirrels.

800,000

(b) In 10 years, we will have t  40, so the population will be

200,000

n0 

600,000 400,000

n 40  3125  2406  317,480 squirrels.

0

10

20

30

50 t (years)

40

Strange notation 4. (a) A model for the bird population is n t  n 0  2t10 . We are given that n 25  13,000, so n 0  22510  13,000 

13,000 n 0  52  2298. Initially, there were approximately 2300 birds. 2

(b) In 5 years, we will have t  30, so the population will be

approximately n 30  2298  23010  18,384  18,400 birds.

(c)

n (’000) 60 50 40 30 20 10 0

10

20

30

40 t


37

SECTION 4.6 Modeling with Exponential Functions

5. (a) Taking t  0 in the year 2005 and measuring n in thousands, we have

(d)

n 0  128 and r  012. Therefore, an exponential model is

n 80 60

n t  128e012t .

40

(b) In 2010, t  5, so the population was approximately n 5  128e0125  233, or 23,300 beavers.

20

(c) Solving n t  50, we get 128e012t  50    ln 128e012t  ln 50  ln 128  012t  ln 50 

0

5

10

15 t

5

10

15

1 ln 50  ln 128  114, so the beaver population will reach t  012

50,000 after about 114 years.

6. (a) Here r  006 and n 0  450 (thousand). Thus, the population at

(d)

n 400

time t modeled by n t  450e006t , where t  0 corresponds to

300

the year 2010.

200

(b) t  2025  2010  15. Then we have

100

n 15  450e00615  183, and we estimate the prairie dog

population in 2025 to be approximately 183,000.

0

(c) Solving n t  300,000, we get 450e006t  300  e006t  23

t

 006t  ln 23  t  68 years.

7. n t  n 0 ert ; n 0  110 million, t  2036  2011  25.

(a) r  003; n 25  110,000,000e00325  110,000,000e075  232,870,000. Thus at a 3% growth rate, the projected population will be approximately 233 million people by the year 2036.

(b) r  002; n 25  110,000,000e00225  110,000,000e050  181,359,340. Thus at a 2% growth rate, the projected population will be approximately 181 million people by the year 2036. 8. (a) In this case, a model for the bacteria population is n t  n 0 ert  22e012t , so after 24 hours the population is approximately n 24  22e01224  392 bacteria.

(b) In this case, a model is n t  n 0 ert  22e005t , so after 24 hours the population is approximately n 24  22e00524  73 bacteria.

9. (a) The doubling time is 18 years and the initial population is 112,000, so a model is n t  112,000  2t18 .

(c)

n (million) 2

(b) We need to find the relative growth rate r. Since the population is 2  112,000  224,000 when t  18, we have 224,000  112,000e18r  2  e18r  ln 2  18r  r  ln182  00385. Thus, a model is

1

0

20

40

60

n t  112,000e00385t . (d) Using the model in part (a), we solve the equation n t  112,000  2t18  500,000  2t18  125 28  t 125 ln 2t18  ln 125 28  18 ln 2  ln 28  t 

population to reach 500,000.

18 ln 125 28 ln 2

 3885. Therefore, it takes about 3885 years for the

t


38

CHAPTER 4 Exponential and Logarithmic Functions

10. (a) The doubling time is 25 years and the initial population is 350,000, so

(c)

a model is n t  350,000  2t25 .

n (million) 2

(b) r  ln252  00277, so a model is n t  350,000e00277t .

1

(d) We solve the equation n t  350,000  2t25  2,000,000  t25  ln 40  t ln 2  ln 40  2t25  40 7  ln 2 7 7 25

0

25 ln 40 7  6286. Therefore, it takes about 629 years for the t ln 2 population to reach 2,000,000.

20

40

60

t

11. (a) The deer population in 2010 was 20,000. (b) Using the model n t  20,000ert and the point 4 31000, we have 31,000  20,000e4r  155  e4r  4r  ln 155  r  14 ln 155  01096. Thus n t  20,000e01096t

(c) n 8  20,000e010968  48,218, so the projected deer population in 2018 is about 48,000. ln 5  1468. Thus, it takes about 147 years (d) 100,000  20,000e01096t  5  e01096t  01096t  ln 5  t  01096 for the deer population to reach 100,000. 12. (a) From the graph, we see that the initial bullfrog population was 100. (b) We use a model of the form n t  n 0 ert with n 0  100. Because we know that the population was 225 at t  2, we 04055t . solve n 2  225  100e2r  225  r  12 ln 225 100  04055. Thus, a model is n t  100e

(c) The estimated population after 15 years is n 15  100e0405515  43,800 frogs.

(d) The population will reach 75,000 when n t  100e04055t  75,000  e04055t  750  t  will take about 163 years for the population to reach 75,000.

ln 750  1632. So it 04055

13. (a) Using the formula n t  n 0 ert with n 0  8600 and n 1  10000, we solve for r, giving 10000  n 1  8600er   50 r 01508t .  50 43  e  r  ln 43  01508. Thus n t  8600e (b) n 2  8600e015082  11627. Thus the number of bacteria after two hours is about 11,600. ln 2 (c) 17200  8600e01508t  2  e01508t  01508t  ln 2  t   4596. Thus the number of bacteria will 01508 double in about 46 hours.

14. (a) Using n t  n 0 ert with n 2  400 and n 6  25,600, we have n 0 e2r  400 and n 0 e6r  25,600. Dividing the

n e6r 25,600 second equation by the first gives 0 2r   64  e4r  64  4r  ln 64  r  14 ln 64  104. Thus the 400 n0e relative rate of growth is about 104%.

(b) Since r  14 ln 64  12 ln 8, we have from part (a) n t  n 0 e

1 2 ln 8 t

. Since n 2  400, we have 400  n 0 eln 8 

400 n 0  ln 8  400 8  50. So the initial size of the culture was 50. e (c) Substituting n 0  50 and r  104, we have n t  n 0 ert  50e104t .

(d) n 45  50e10445  50e468  53885, so the size after 45 hours is approximately 5400. ln 1000 (e) n t  50,000  50e104t  e104t  1000  104t  ln 1000  t   664. Hence the population will 104 reach 50,000 after roughly 6 hours 40 minutes.


SECTION 4.6 Modeling with Exponential Functions

39

15. (a) Using n t  n 0 ert with n 0  49 and n 19  44, we have n 0  49 and n 0 e19r  44  49e19t  44  e19r  44 49 1 44 000566t , we solve 35  49e000566t   19r  ln 44 49  r  19 ln 49  000566. Using the model n t  49e

000566t  ln 35 49  t  594 years, so the population is projected to decline to 35 million in the year 2059.

(b) We solve n t  12 n 0  n 0 e000566t  000566t  ln 12  t  1225 years.

10  t  239 years, so the 16. (a) A model is given by n t  78e00104t . We solve 10  78e00104t  00104t  ln 78

population of the world is projected to reach 10 billion in late 2043.

(b) 2n 0  n 0 e00104t  2  e00104t  ln 2  00104t  t  666. So at the current growth rate, it will take approximately 666 years for the population to double. 17. (a) Setting t  0 in 1950 and substituting M  11, r  00189, and A

M  n0 11  25  34 into the logistic growth model  n0 25

11 M , we have n t  . Solving 1  Aert 1  34e00189t 11 1   34e00189t  10 n t  10, we obtain 10  1  34e00189t

(b)

n (bn) 10

n t 

0

50

100

1 ,  t   1 ln 1  1866. Thus, the world’s e00189t  34 00189 34

population is projected to reach 10 billion in the year 2136.

18. (a) From Exercise 5, n 0  128 and r  012. Therefore, with M  80 and A

M  n0 80  128  5 25, a logistic model is  n0 128

n t 

150

200 t

Strange notation n (’000) 160 120

80 M  . 1  Aert 1  525e012t

80

(b) We see from the graph that the models diverge dramatically as time goes on. The exponential model increases without bound, while the logistic

40 0

10

20

30

5

10

15

40 t

model approaches the carrying capacity. M with n 0  8, M  6000, r  057, and 1  Aert M  n0 6000 6000  8 A  749, we have n t   . n0 8 1  749e057t

19. (a) Using n t 

(b) From the graph, the number of infections reaches 5900 on the 18th or 19th day.

n 6000 5000 4000 3000 2000 1000 0

M with n 0  18, M  10, r  02, and 1  Aert 4 10 10  18 M  n0   , we have n t   . A n0 18 9 1  4 e02t

20. (a) Using n t 

9

(b) The fish population declines rapidly at first, then more gradually, approaching 10,000 as t  .

20 t

Strange notation

n (’000) 20 15 10 5 0

5

10

15

20 t

21. (a) Because the half­life is 1600 years and the sample weighs 22 mg initially, a suitable model is m t  22  2t1600 .


40

CHAPTER 4 Exponential and Logarithmic Functions

(b) From the formula for radioactive decay, we have m t  m 0 ert , where m 0  22 and r 

ln 2 ln 2   0000433. h 1600

Thus, the amount after t years is given by m t  22e0000433t .

(c) m 4000  22e00004334000  389, so the amount after 4000 years is about 4 mg.

9  e0000433t  (d) We have to solve for t in the equation 18  22 e0000433t . This gives 18  22e0000433t  11   9   ln 11 9 t  0000433t  ln 11  4634, so it takes about 463 years. 0000433

22. (a) Because the half­life is 30 years and the sample weighs 10 g initially, a suitable model is m t  10  2t30 . ln 2 ln 2 (b) Using m t  m 0 ert with m 0  10 and h  30, we have r    00231. Thus m t  10e00231t . h 30 (c) m 80  10e0023180  16 grams.    ln 5 (d) 2  10e00231t  15  e00231t  ln 15  00231t  t   70 years. 00231 ln 2 23. By the formula in the text, m t  m 0 ert where r  , so m t  50e[ln 229]t . We need to solve for t in the h     ln 2 32  t   29  ln 32  1867, so it t  ln   equation 32  50e[ln 229]t . This gives e[ln 229]t  32 50 50 50 29 ln 2 takes about 18 years and 8 months. 24. From the formula for radioactive decay, we have m t  m 0 ert , where r 

ln 2 . Since h  30, we have h

ln 2  00231 and m t  m 0 e00231t . In this exercise we have to solve for t in the equation 005m 0  m 0 e00231t 30 ln 005  1297. So it will take about 130 s.  e00231t  005  00231t  ln 005  t  00231 ln 2 25. By the formula for radioactive decay, we have m t  m 0 ert , where r  , in other words m t  m 0 e[ln 2 h ]t . In h ln 2 48 this exercise we have to solve for h in the equation 200  250e[ln 2 h ]48  08  e[ln 2 h ]48  ln 08   h ln 2  48  1491 hours. So the half­life is approximately 149 hours. h  ln 08 ln 2 . In other words, m t  m 0 e[ln 2 h ]t . 26. From the formula for radioactive decay, we have m t  m 0 ert , where r  h (a) Using m 3  058m 0 , we have to solve for h in the equation 058m 0  m 3  m 0 e[ln 2 h ]3 . 3 ln 2 3 ln 2  ln 058  h    38 days. Thus the Then 058m 0  m 0 e[3 ln 2 h ]  e[3 ln 2 h ]  058   h ln 058 half­life of radon­222 is about 38 days. (b) Here we have to solve for t in the equation 02m  m e[ln 2382]t . So we have 02m  m e[ln 2382]t  r

0

0

0

0

382 ln 02 ln 2 t  ln 02  t    887. So it takes roughly 9 days for a sample of 02  e[ln 2382]t  

382 Radon­222 to decay to 20% of its original mass.

ln 2

27. By the formula in the text, m t  m 0 e[ln 2 h ]t , so we have 065  1  e[ln 25730]t  ln 065   t 

5730 ln 065  3561. Thus the artifact is about 3560 years old. ln 2

ln 2 t 5730

ln 2 ln 2 . Since h  5730, r   0000121 h 5730 and m t  m 0 e0000121t . We need to solve for t in the equation 059m 0  m 0 e0000121t  e0000121t  059  ln 059  43606. So the mummy was buried about 4360 years ago. 0000121t  ln 059  t  0000121

28. From the formula for radioactive decay, we have m t  m 0 ert where r 


SECTION 4.6 Modeling with Exponential Functions

29. We use the radioactive decay model m t  m 0 ert with m 0  005 mol and r 

41

ln 2 . Solving m t  004 for t, we get 45

ln 2 t  ln 45  t  145. Thus, the age of the sample is approximately 145 billion years. 45 ln 2 30. We use the radioactive decay model m t  m 0 ert with m 0  00005  00015  0002 mol and r  . Solving 07 ln 2 t  ln 14  t  14. According to this model, the age m t  00005 for t, we get 00005  0002e[ln 207]t   07 of the sample is approximately 14 billion years. This is concordant with the age estimate given by the 238 U­to­206 Pb decay model. ln 2 31. We use the radioactive decay model m t  m 0 ert with m t  08m 0 and r  and solve for t: 08  e[ln 2432]t 432 ln 2 t  ln 08  t  1391 years. (Note that the absolute quantity of 141 Am is irrelevant.)   432 004  005e[ln 245]t  

32. (a) We use Newton’s Law of Cooling: T t  Ts  D0 ekt with k  01947, Ts  60, and D0  986  60  386 . So T t  60  386e01947t .

12  01947t  ln (b) Solve T t  72. So 72  60  386e01947t  386e01947t  12  e01947t    1 12 ln  600, and the time of death was about 6 hours ago. t  01947 386

386

12 386

33. (a) T 0  65  145e0050  65  145  210 F.

(b) T 10  65  145e00510  1529. Thus the temperature after 10 minutes is about 153 F.

(c) 100  65  145e005t  35  145e005t  02414  e005t  ln 02414  005t  t  

ln 02414  284. 005

Thus the temperature will be 100 F in about 28 minutes. kt 34. Using Newton’s Law of Cooling, T t  Ts  D0 ekt with Ts  75 and D0  185  75  110. So T t  75  110e  . 15 (a) Since T 30  150, we have T 30  75  110e30k  150  110e30k  75  e30k  15 22  30k  ln 22   1 ln 15 . Thus we have T 45  75  110e4530 ln1522  1369, and so the temperature of the turkey  k   30 22

after 45 minutes is about 137 F.

25 5  (b) The temperature will be 100 F when 75  110et30 ln1522  100  et30 ln1522   22 110   5       ln 22 t 5  t  30    1161. So the temperature will be 100 F after about 2 hours.  ln ln 15 22 22 30 ln 15 22

35. We use Newton’s Law of Cooling: T t  Ts  D0 ekt , with Ts  20 and

D0  100  20  80. So T t  20  80ekt . Since T 15  75, we have   11 20  80e15k  75  80e15k  55  e15k  11 16  15k  ln 16    1 ln 11 . Thus T 25  20  80e2515ln1116  628, and so the k   15 16 temperature after another 10 min is 63 C. The function

T t  20  80e115ln1116t is shown in the viewing rectangle [0 30] by

[50 100].

100 80 60 0

20


42

CHAPTER 4 Exponential and Logarithmic Functions

4.7

LOGARITHMIC SCALES

    1. (a) pH   log H   log 50  103  23     (b) pH   log H   log 32  104  35     (c) pH   log H   log 50  109  83     2. pH   log H   log 31  108  75 and the substance is basic.

    3. (a) pH   log H  30  H  103 M     (b) pH   log H  65  H  1065  32  107 M     4. (a) pH   log H  46  H  1046 M  25  105 M     (b) pH   log H  73  H  1073 M  50  108 M         5. 40  107  H  16  105  log 40  107  log H  log 16  105       log 40  107  pH   log 16  105  64  pH  48. Therefore the range of pH readings for cheese is approximately 48 to 64.

      6. 158  104  10pH  158  103  log 158  104  log 10pH  log 158  103 

log 158  4  pH  log 158  3  28  pH  38. The pH ranges from 28 to 38.       7. (a) For the California red wine, we have pH   log H  log H  32  H  1032  63  104 M. For       the Italian white wine, pH   log H  log H  29  H  1029  13  103 M. (b) The California red wine has lower hydrogen ion concentration.       8. (a) pH   log H  log H  55  H  1055  32  106 M. (b) The saliva was more acidic when the patient was sick.

(c) As pH increases, hydrogen ion concentration decreases. So as the patient gets better, the pH of their saliva will increase and its hydrogen ion concentration will decrease. 3125 I with S  104 and I  3125, so M  log 4  55. S 10 I I (b) M  log  10 M   I  S  10 M . We have M  48 and S  104 , so I  104  1048  63. S S

9. (a) M  log

10. (a) M  log

I 721 with S  104 and I  721, so M  log10 4  59. S 10

(b) I  104  1058  631.

11. Let I0 be the intensity of the smaller earthquake and I1 the intensity of the larger earthquake. Then I1  20I0 . 20I0 I I  log 20  log I0  log S. Then Notice that M0  log 0  log I0  log S and M1  log 1  log S S S M1  M0  log 20  log I0  log S  log I0  log S  log 20  13. Therefore the magnitude is 13 times larger.

I 12. Let the subscript S represent the San Francisco earthquake and J the Japan earthquake. Then we have M S  log S  83 S 83 I I 10 J S  49  I J  S  1049 . So  49  1034  25119, and so the San Francisco  I S  S  1083 and M J  log S IJ 10 earthquake was 2500 times more intense than the Japan earthquake.


SECTION 4.7 Logarithmic Scales

13. Let the subscript J represent the Japan earthquake and S represent the San Francisco earthquake. Then M J  log

43

IJ  91 S

I I S  1091  I J  S  1091 and M S  log S  83  I S  S  1083 . So J   1008  63, and hence the Japan S IS S  1083 earthquake was about six times more intense than the San Francisco earthquake. 14. Let the subscript N represent the Northridge, California earthquake and K the Kobe, Japan earthquake. Then I I I 1072 M N  log N  68  I N  S  1068 and M K  log K  72  I K  S  1072 . So K  68  1004  251, and S S IN 10 so the Kobe, Japan earthquake was 25 times more intense than the Northridge, California earthquake.     20  105 I  10 log  10 log 2  107  10 log 2  log 107  10 log 2  7  73. Therefore the I0 10  1012 intensity level was 73 dB.

15.   10 log

32  102 I  10 log  10 log 32  1010  105 dB. I0 10  1012 I I  70  10 log  log I  12  7  log I  5, so the intensity was 105 wattsm2 . 17.   10 log I0 10  1012   I 18. 98  10 log 12  log I  1012  98  log I  98  log 1012  22  I  1022  63  103 . So the 10 intensity was 63  103 wattsm2 . 16.   10 log

  I 31  105  10 log  10 log 31  107  75 dB. 12 I0 10  10 I I (b) Here   90 dB, so 90  10 log 12  9  log 12  9  log I  log 1012  log I  12  9  10 10 I  103 Wm2 .

19. (a) The intensity is 31  105 Wm2 , so   10 log

(c) The ratio of the intensities is

103 Ie   323. Is 31  105

  IM  20. Let the subscript M represent the power mower and C the rock concert. Then 106  10 log 1012       IC  log IC  1012  120  log I M  1012  106  I M  1012  10106 . Also 120  10 log 12 10

IC 1012  106  1014  2512, and so the ratio of intensity is roughly 25. IM 10     k I1 k k k  2 log d1  10 log and I1  2  1  10 log 2  10 log  20 log d1 . Similarly, 21. (a) 1  10 log I0 I I d1 d1 I0 0 0 k  20 log d2 . Substituting the expression for 1 gives 2  10 log I0   k d  20 log d1  20 log d1  20 log d2  1  20 log d1  20 log d2  1  20 log 1 . 2  10 log I0 d2 d1 2  120  20 log 02  106, and so the (b) 1  120, d1  2, and d2  10. Then 2  1  20 log  120  20 log 10 d2 intensity level at 10 m is approximately 106 dB. IC  1012  10120 . So


44

CHAPTER 4 Exponential and Logarithmic Functions

CHAPTER 4 REVIEW   2  9739, f 25  55902   7  18775, f 55  135765 2. f x  3  2x ; f 22  0653, f

1. f x  5x ; f 15  0089, f

3. g x  4e x2 ; g 07  0269, g 1  1472, g   12527   3  26888, g 36  174098 4. g x  74 e x1 ; g 2  0644, g 5. f x  3x  1. Domain  , range 1 , horizontal asymptote y  1.

 x 6. f x  12  5. Domain  , range 5 , horizontal asymptote y  5.

y

y

1 1

(0, 2) 1

x

(0, _4) x

1

7. g x  4x1 . Domain  , range 0 , horizontal asymptote y  0. y

8. g x  2x1 . Domain  , range  0, horizontal asymptote y  0. y

1

1

(0, _1/2)

x

1 (0, 1/4) x

1

9. h x  e x2  1. Domain  , range  1, 10. h x  3ex  1. Domain  , range 1 , horizontal asymptote y  1.

horizontal asymptote y  1. y

y 1 1

x

(0, 4) (0, _1-e@)

1 1

x


CHAPTER 4

11. f x  log3 x  2. Domain 2 , range  , vertical asymptote x  2.

Review

12. f x   log4 x  2. Domain 2 ,

range  , vertical asymptote x  2. y

y

1 x

1

1 (0, _1)

x

1

(0, - 1/2)

13. f x  log13 x  1. Domain 0 , range  , vertical asymptote x  0. y

14. f x  1  log12 x  2. Domain 2 , range  , vertical asymptote x  2. y

(0, 2) 1 1

x

1 x

1

15. g x  log2 x  2. Domain  0,

16. g x   log3 x  3  2. Domain 3 ,

range  , vertical asymptote x  0.

range  , vertical asymptote x  3.

y

y

1 1

x 1 2

x

45


46

CHAPTER 4 Exponential and Logarithmic Functions

17. g x  2 ln x. Domain 0 , range  , vertical asymptote x  0.

  18. g x  ln x 2 .

Domain x  x  0   0  0 ,

y

range  , vertical asymptote x  0. y

1 x

1

1 1

x

2

19. f x  10x  log 1  2x. Since log u is defined only for u  0, we require 1  2x  0  2x  1  x  12 , and so   the domain is  12 .   20. g x  log 2  x  x 2 . We must have 2  x  x 2  0 (since log y is defined only for y  0)  x 2  x  2  0  x  2 x  1  0. The endpoints of the intervals are 2 and 1. Interval Sign of x  2 Sign of x  1

Sign of x  2 x  1

 1

1 2

2 

 2

2 2

2 

Thus the domain is 1 2.   21. h x  ln x 2  4 . We must have x 2  4  0 ( since ln y is defined only for y  0)  x 2  4  0  x  2 x  2  0. The endpoints of the intervals are 2 and 2. Interval Sign of x  2 Sign of x  2

Sign of x  2 x  2

Thus the domain is  2  2 .

22. k x  ln x. We must have x  0. So x  0  x  0 or x  0. Since x  0  x  0, the domain is x  0 or x  0 which is equivalent to x  0. In interval notation,  0  0 . 23. log2 1024  10  210  1024

24. log6 37  x  6x  37

25. log x  y  10 y  x

26. ln c  17  e17  c

27. 26  64  log2 64  6

28. 4912  17  log49 17   12

29. 10x  74  log10 74  x  log 74  x   31. log2 128  log2 27  7

30. ek  m  ln m  k   32. log8 1  log8 80  0

33. 10log 45  45   35. ln e6  6

34. log 0000001  log 106  6   36. log4 8  log4 432  32


CHAPTER 4 1  log 33  3 37. log3 27 3

39. log5

 5  log5 512  12

41. log 25  log 4  log 25  4  log 102  2

38. 2log2 13  13 2  40. e2 ln 7  eln 7  72  49    42. log3 243  log3 352  52

Review

   23 3 43. log2 1623  log2 24  log2 292  92 44. log5 250  log5 2  log5 250 2  log5 125  log5 5  3   45. log8 6  log8 3  log8 2  log8 63  2  log8 4  log8 823  23   46. log10 log10 10100  log10 100  log10 102  2   47. log AB 2 C 3  log A  2 log B  3 log C       48. log2 x x 2  1  log2 x  log2 x 2  1  log2 x  12 log2 x 2  1            1 x2  1 x2  1 1 ln x 2  1  ln x 2  1 1 ln x  1 x  1  ln x 2  1   ln  49. ln 2 2 2 x2  1 x2  1     12 ln x  1  ln x  1  ln x 2  1         4x 3  log 4x 3  log y 2 x  15  log 4  3 log x  2 log y  5 log x  1 50. log 5 2 y x  1        x 2 1  5x3/2   log5 x 2 1  5x3/2  log5 x x 2  1  2 log5 x  32 log5 1  5x  12 log5 x 3  x 51. log5 x3  x     2 log5 x  32 log5 1  5x  12 log5 x  log5 x 2  1    2 log5 x  32 log5 1  5x  12 log5 x  log5 x  1  log5 x  1    3 4   x  12  13 ln x 4  12  ln x  16  12 ln x  3 52. ln  x  16 x  3   53. log 6  4 log 2  log 6  log 24  log 6  24  log 96       54. log x  log x 2 y  3 log y  log x  x 2 y  y 3  log x 3 y 4    2 x  y3/2 55. 32 log2 x  y  2 log2 x 2  y 2  log2 x  y3/2  log2 x 2  y 2  log2  2 x 2  y2

2 x  1 56. log5 2  log5 x  1  13 log5 3x  7  log5 [2 x  1]  log5 3x  713  log5  3 3x  7    x2  4 57. log x  2  log x  2  12 log x 2  4  log [x  2 x  2]  log x 2  4  log  x2  4       5  5   ln x  4 x 2  4x 58. 12 ln x  4  5 ln x 2  4x  12 ln x  4 x 2  4x

59. 26x3  8  26x3  23  6x  3  3  x  1  1x  1x  2 60. 13  19  13  13  1  x  2  x  1

61. 53x2  2  ln 53x2  ln 2  3x  2 ln 5  ln 2  3x  2 

ln 2  2 ln 2  x  ln 5  0523 ln 5 3

62. 1043x  5  log 1043x  log 5  4  3x  log 5  x  13 4  log 5  1100

47


48

CHAPTER 4 Exponential and Logarithmic Functions

63. 25x1  34x  ln 25x1  ln 34x  5x  1 ln 2  4  x ln 3  x 5 ln 2  ln 3  4 ln 3  ln 2  x

ln 81 4 ln 3  ln 2 2   0811 5 ln 2  ln 3 5 ln 2  ln 3

  64. 102x5  e3x1  ln 102x5  ln e3x1  25 x ln 10  3x  1  x 3  25 ln 10  1  x 2

1

5 ln 10  3

5  0481 2 ln 10  15

  65. x 3 34x  x 2 34x  6x  34x  34x  x  x 2  x  6  0  34x  0 or x  0 or x 2  x  6  x  2 x  3  0. The first of these equations has no solution, so the solutions are x  0, x  2, and x  3.

 2  2 66. e2x  6e x  9  0  e x  6e x  9  0  e x  3  0  e x  3  0  ln e x  ln 3  x  ln 3  110

67. log x  log x  1  log 12  log x x  1  log 12  x x  1  12  x 2  x  12  0  x  3 x  4  0  x  4 or 3. Since log 4 is undefined, the only solution is x  3. 68. ln x  2  ln 3  ln 5x  7  ln 3 x  2  ln 5x  7  3 x  2  5x  7  2x  1  x  12 , but since ln x  2 is undefined for x  12 , there is no solution.

69. log2 1  x  4  1  x  24  x  1  16  15

  70. ln 2x  3  1  0  ln 2x  3  1  2x  3  e1  x  12 1e  3  168

71. log3 x  8  log3 x  2  log3 x x  8  2  x x  8  9  x 2  8x  9  0  x  9 x  1  0  x  1 or 9. We reject 1 because it does not satisfy the original equation, so the only solution is x  9. 72. log12 x  5  log12 x  2  1  log12

x 5 1 x 5 1   2 x  5  x  2  x  12 x 2 x 2 2

3 log 063 2x log 5  log 063  x    0430618 3 2 log 5   log 7  2602452 74. 23x5  7  3x  5 log 2  log 7  x  13 5  log 2

73. 52x /3  063 

75. 52x1  34x1  2x  1 log 5  4x  1 log 3  2x log 5  log 5  4x log 3  log 3  log 3  log 5  2303600 x 2 log 5  4 log 3   log 3  log 5  x  4 log 3  2 log 5 1 ln 10000  0614023 76. e15k  10000  15k  ln 10000  k   15

77. y  e xx2 . Vertical asymptote x  2, horizontal asymptote y  272, no maximum or minimum. 10

78. y  10x  5x . No vertical asymptote, horizontal

asymptote y  0, local minimum of about 013 at x  052.

2 5 1 ­20

0

20 ­2

2


CHAPTER 4

 79. y  log x 3  x . Vertical asymptotes x  1, x  0,

x  1, no horizontal asymptote, local maximum of about

041 when x  058.

Review

49

80. y  2x 2  ln x. Vertical asymptote x  0, no horizontal asymptote, local minimum of about 119 at x  050. 100 50

­1

1

2

0

0

5

10

­2

81. 3 log x  6  2x. We graph y  3 log x and y  6  2x in 82. 4  x 2  e2x . From the graphs, we see that the solutions the same viewing rectangle. The solution occurs where the two graphs intersect. From the graphs, we see that the

are x  064 and x  2.

5

solution is x  242. 10 ­2 5

2

10

­10

83. ln x  x  2We graph the function f x  ln x  x  2, and we see that the graph lies above the x­axis for

016  x  315. So the approximate solution of the given inequality is 016  x  315.

84. e x  4x 2  e x  4x 2  0. We graph the function

f x  e x  4x 2 , and we see that the graph lies below the

x­axis for  041  071 431.

2

5 ­10

0

5

­2

85. f x  e x  3ex  4x. We graph the function f x,

and we see that the function is increasing on  0 and 110  and that it is decreasing on 0 110.

­2

86. The line has x­intercept at x  e0  1. When x  ea ,

y  ln ea  a. Therefore, using the point­slope equation,

a a0 we have y  0  a x  1  y  a x  1. e 1 e 1

2 ­5

87. log4 15 

log 15  1953445 log 4

  log 3 4  0147839 88. log7 34  log 7


50

CHAPTER 4 Exponential and Logarithmic Functions

89. log9 028 

log 028  0579352 log 9

90. log100 250 

log 250  1198970 log 100

91. Notice that log4 258  log4 256  log4 44  4 and so log4 258  4. Also log5 620  log5 625  log5 54  4 and so log5 620  4. Then log4 258  4  log5 620 and so log4 258 is larger.     x x 92. f x  23 . Then y  23  log2 y  3x  log3 log2 y  x, and so the inverse function is f 1 x  log3 log2 x . Since log3 y is defined only when y  0, we have log2 x  0  x  1. Therefore the domain is 1 , and the range is  .  r nt . 93. P  12,000, r  010, and t  3. Then A  P 1     n 23

(a) For n  2, A  12,000 1  010  12,000 1056  $16,08115. 2 123   $16,17818. (b) For n  12, A  12,000 1  010 12 3653   $16,19764. (c) For n  365, A  12,000 1  010 365

(d) For n  , A  Pert  12,000e0103  $16,19831.

94. P  5000, r  0085, and n  2.  215 (a) For t  15, A  5000 1  0085  5000  104253  $566498. 2

7 (b) We want to find t such that A  7000. Then A  5000  104252t  7000  104252t  7000 5000  5        log 75 log 75 7 t   404, and so the investment will amount to $7000 2t log 10425  log 5  2t  log 10425 2 log 10425 after approximately 4 years.

(c) In this case, we solve n t  n 0 ert for t when n 0  5000, r  0085, and n t  7000: 7000  5000e0085t  7 7  e0085t  ln 7  0085t  t  ln 5  396, so the investment will grow to $7000 in just under 4 years. 5 5 0085

 r nt with P  100,000, r  0052, n  365, and A  100,000  10,000  110,000, 95. We use the formula A  P 1  n  365t 365t    and solve for t: 110,000  100,000 1  0052   11  1  0052  log 11  365t log 1  0052 365 365 365 t

log 11   1833. The account will accumulate $10,000 in interest in approximately 18 years.  365 log 1  0052 365

96. We solve n t  n 0 ert for n t  2n 0 and r  0045: 2n 0  n 0 e0045t  ln 2  0045t  t  retirement savings plan will double in about 154 years.

ln 2  15403. The 0045

  00425 365 97. After one year, a principal P will grow to the amount A  P 1   P 104341. The formula for simple 365 interest is A  P 1  r. Comparing, we see that 1  r  104341, so r  004341. Thus the annual percentage yield is 4341%.   0032 12 98. A  P 1   P 103247  P 1  r  r  3247% 12 99. (a) Using the model n t  n 0 ert , with n 0  30 and r  015, we have the formula n t  30e015t . (b) n 4  30e0154  55.

  015t  015t  ln 50  t  (c) 500  30e015t  50 3 e 3 reach 500 in about 19 years.

  1 ln 50 3  1876. So the stray cat population will 015


CHAPTER 4

51

Review

100. Using the model n t  n 0 ert , with n 0  10000 and n 1  25000, we have 25000  n 1  10000er1  er  52  r  ln 52  0916. So n t  10000e0916t .

(a) Here we must solve the equation n t  20000 for t. So n t  10000e0916t  20000  e0916t  2  ln 2  0756. Thus the doubling period is about 45 minutes. 0916t  ln 2  t  0916 (b) n 3  10000e09163  156250, so the population after 3 hours is about 156,250.

101. (a) From the formula for radioactive decay, r 

ln 2 and n t  150  et ln 275,380 or, equivalently, 75,380

n t  150  2t75,380 .

(b) n 1000  150  2100075380  14863 mg (c) We solve n t  50  13  2t75380  t  75,380 119,474 years.

ln 3  119,474. Thus, only 50 mg remains after approximately ln 2

ln 2 . So m t  m 0 e[ln 2 h ]t . h (a) Using m 8  033m 0 , we solve for h. We have 033m 0  m 8  m 0 eln 2 h  033  e8 ln 2 h  8 ln 2 8 ln 2  ln 033  h    5002. So the half­life of this element is roughly 5 days.  h ln 033 (b) m 12  m e[ln 25]12  019m , so about 19% of the original mass remains.

102. From the formula for radioactive decay, we have m t  m 0 ert , where r 

0

0

103. From the formula for radioactive decay, we have m t  m 0 ert , where r 

ln 2 ln 2 . Since h  4, we have r   0173 h 4

and m t  m 0 e0173t . (a) Using m 20  0375, we solve for m 0 . We have 0375  m 20  m 0 e017320  003125m 0  0375  0375  12. So the initial mass of the sample was about 12 g. m0  003125 (b) m t  12e0173t or, equivalently, m t  12  2t4 . (c) m 3  12e01733  7135. So there are about 71 g remaining after 3 days.

(d) Here we solve m t  015 for t: 015  12e0173t  00125  e0173t  0173t  ln 00125  ln 00125 t  253. So it will take about 25 days until only 15% of the substance remains. 0173

104. (a) Using n 0  1500 and n 5  3200 in the formula n t  n 0 ert , we have 3200  n 5  1500e5r  e5r  32 15      32 1 32 01515t . 5r  ln 15  r  5 ln 15  01515. Thus n t  1500  e

(b) We have t  2031  2020  11 so n 11  1500e0151511  7940. Thus in 2031 the bird population should be about 7940.

105. (a) The doubling time is a  15 hr, so the relative growth rate is 2 r  ln 15  0462.

carrying capacity

(b) Since the initial population is 1400, the model we want has the form 1400 n t  , where r  0462 and 1  Aert M  n0 1400  100 A  13. Thus, a model is  n0 100 n t 

1400 . 1  13e0462t

(c)

n 1400 1000 700

0

10

It appears to take about 555 hr for the yeast to reach 700 colonies.

t


52

CHAPTER 4 Exponential and Logarithmic Functions

106. We use Newton’s Law of Cooling: T t  Ts  D0 ekt with k  00341, Ts  60 and D0  190  60  130.

3  So 90  T t  60  130e00341t  90  60  130e00341t  130e00341t  30  e00341t  13   3  t   ln 313  430, so the engine cools to 90 F in about 43 minutes. 00341t  ln 13 00341       8 107. H  13  10 M. Then pH   log H   log 13  108  79, and so fresh egg whites are basic.     108. pH  19   log H . Then H  1019  126  102 M.

109. Let I0 be the intensity of the smaller earthquake and I1 be the intensity of the larger earthquake. Then I1  35I0 . Since     I I M  log , we have M0  log 0  65 and S S       35I0 I I1  log  log 35  log 0  log 35  M0  log 35  65  804. So the magnitude on the M1  log S S S Richter scale of the larger earthquake is approximately 80.     I I 110. Let the subscript J represent the jackhammer and W the whispering:  J  132  10 log J  log J  132 I0 I0 I I I 10132  J  10132 . Similarly W  1028 . So J   10104  251  1010 , and so the ratio of intensities is I0 I0 IW 1028

251  1010 .

111. (a) y  2x has graph VI because it grows without bound as x increases and has asymptote y  0. (b) y   ln x has graph VIII because it has domain 0  and is decreasing.

(c) 2x  3y  6 has graph V because it is linear with x­intercept 3 and y­intercept 2. 1 (d) y  1  3 has graph III because it has vertical asymptote x  0 and horizontal asymptote y  1. x (e) y  log2 x has graph II because it has domain 0  and is increasing. 2

(f) y  4e x 4 has graph VII because it has range 0  and is even.

(g) y  2  2x  x 2 has graph IV because it is a parabola that opens downward.

(h) y  10xex has graph I because it has the characteristic shape of a surge function.

CHAPTER 4 TEST 1. (a)

y

(b) (0, 2)

1 0

(1, 0)

y 1 1

2

x

f x  3  3x has domain  , range  3,

g x  log3 x  3 has domain 3 , range

 , and vertical asymptote x  3.   2. (a) f t  ln 2t  3 is defined where 2t  3  0  2t  3  t  32 , so its domain is 32   .   (b) g x  log x 2  1 is defined where x 2  1  0  x  1, so its domain is  1  1 . and horizontal asymptote y  3.

x


CHAPTER 4

3. (a) 62x  25  log6 62x  log6 25  2x  log6 25

Test

(b) ln A  3  eln A  e3  A  e3

4. (a) 10log 36  36

(b) ln e3  3  12   log3 332  32 (c) log3 27  log3 33   3 (d) log2 80  log2 10  log2 80 10  log2 8  log2 2  3

(e) log8 4  log8 823  23

(f) log6 4  log6 9  log6 4  9  log6 62  2

5. (a) log

x y3 z2

 log x  log y 3  log z 2  log x  3 log y  2 log z

     1 x x 12 x  ln  12 ln x  12 ln y  ln (b) ln y y 2 y      x2  1 1 x2  1 (c) log  log 3  12 log x 2  1  3 log x  log x  1 3 2 x x  1 x x 1  12 log x 2  1  32 log x  12 log x  1 

  6. (a) log a  2 log b  log a  log b2  log ab2

  x 2  25 x  5 x  5  ln  ln x  5 (b) ln x 2  25  ln x  5  ln x 5 x 5

(c) log3 x  2 log3 x  1  3 log3 y  log3 x  log3 x  12  log3 y 3  log3

x y3 x  12

7. (a) 34x  3100  4x  100  x  25 2

(b) e3x2  e x  3x  2  x 2  x 2  3x  2  0  x  1 x  2  0  x  1 or x  2   3x1   3x1  6  23  65  3x  1 ln 23  ln 65  3x ln 23  ln 65  ln 23  ln 65  32  ln 95  (c) 5 23 x

ln 95

3 ln 23

 0483

(d) 10x3  62x  log 10x3  log 62x  x  3  2x log 6  x 2 log 6  1  3  x 

3  539 2 log 6  1

8. (a) log 2x  3  2x  103  x  12 1000  500

(b) log x  1  log 2  log 5x  log 2 x  1  log 5x  2x  2  5x  3x  2  x  23

(c) 5 ln 3  x  4  ln 3  x  45  3  x  e45  x  3  e45  0774 (d) log4 x  3  log4 x  1  2  log4

x 3 x 3 2  42  16  x  3  16x  16  x  19 15 x 1 x 1

9. Using the Change of Base Formula, we have log12 27 

log 27  1326. log 12

53


54

CHAPTER 4 Exponential and Logarithmic Functions

10. (a) From the formula for population growth, we have 8000  1000er1

n (d)

y

 8  er  r  ln 8  207944. Thus n t  1000e207944t .

(b) n 15  1000e20794415  22,600 (c) 15,000  1000e207944t  15  e207944t  ln 15  207944t  ln 15  13. Thus the population will reach 15,000 after t 207944 approximately 13 hours. 12t  11. (a) A t  12,000 1  0056 , where t is in years. 12

50,000 1

x

t

  365t  0056 3653 0056 . So A 3  12,000 1   $14,19506. (b) A t  12,000 1  365 365   (c) A t  12,000e0056t . So 20,000  12,000e0056t  5  3e0056t  ln 5  ln 3e0056t  ln 5  ln 3  0056t  1 ln 5  ln 3  912. Thus, the amount will grow to $20,000 in approximately 912 years. t  0056

12. (a) The initial mass is m 0  3 and the half­life is h  10, so using the formula m t  m 0 2t h , we have m t  3  2t10 .

(b) Using the radioactive decay model with m 0  3 and r  lnh2  ln102 , we have m t  3e[ln 210]t  3e00693t . (c) After 1 minute  60 seconds, the amount remaining is m 60  3e0069360  0047 g.     106 106 106 00693t 6 00693t 00693t  ln e  00693t  ln   10  e   ln (d) We solve 3e 3 3 3   1 106 ln  215, so there is 1 g of 91 Kr remaining after about 215 seconds, or 36 minutes. t  00693 3   IJ IJ  64  1064   13. Let the subscripts J and P represent the two earthquakes. Then we have M J  log S S   I I I 1064 S 1064 S  I J . Similarly, M P  log P  31  1031  P  1031 S  I P . So J  31  1033  19953, S S IP 10 S and so the Japan earthquake was about 1995 times more intense than the Pennsylvania earthquake.


Fitting Exponential and Power Curves to Data

55

FOCUS ON MODELING Fitting Exponential and Power Curves to Data 1. (a)

(b) Let t be the time (in hours) and y be amount of

y 5

iodine­131 (in grams). Using a graphing device, we obtain the exponential model y  abt , where

4.8

a  479246 and b  099642.

4.6

(c) To find the half­life of iodine­131, we must find the time when the sample has decayed to half its original

4.4

mass. Setting y  240 g, we get

4.2

240  479246  099642t 

4 3.8 0

10

20

30

40

ln 240  ln 479246  t ln 099642  ln 240  ln 479246  1928 h, or approximately t ln 099642 8 days.

50 x

Time (h)

t 2. (a)

y

(b) We let t represent the time (in seconds) and y the

5

distance fallen (in meters). Using a graphing device, we obtain the power model: y  49622t 20027 .

4

(c) When t  3 the model predicts that y  44792 m.

3 2 1

0

0.2

0.4

0.6

0.8

1

x

Time (s)

3. (a) Let A be the area of the cave and S the number of species of bat. Using a graphing device, we obtain the power function model S  014A064 . From the graph,

we see that the model fits the data reasonably well. (b) According to the model, there are

S  014 205064  4 species of bat living in the El

Sapo cave.

t

S

7 6 5 4 3 2 1

0

100

200

300

400

Area (m@ )

500 A


56

FOCUS ON MODELING

4. (a)

ln L

ln L

2

2

1

0

1

2

4

6

8

10

12

14

16

18

20 w

_2

0

_1

1

2 ln w

Let  denote weight and L wingspan. From semi­log and log­log plots of the data, it appears that a power function is more appropriate. Using a graphing device, we calculate the power function model L  3056238039520 . L

100

50

0

2

4

6

8

10

12

14

16

18

20 w

(b) According to our model from part (a), the wingspan required for a 300­lb bird to fly is approximately L 300  3056238  300039520  291 in. This is almost three times the actual wingspan of an ostrich.

5. (a) Using a graphing device, we find the model ln I  57267458  002999x 

I  e57267458002999x  3069687e002999x . The “murkiness” constant is k  003.

(b) At the bottom of the twilight zone, the light intensity is approximately

I 1000  307e0031000  29  1011  3  1012 , so these species can thrive in the twilight zone. 6. (a) Using a graphing device, we find the model c , where a  4910976596, y 1  aebx b  04981144989, and c  500855793. The model appears to fit the data very well. (b) From the model, the carrying capacity is c  500.

y 500 400 300 200 100

0

2 4 6 8 10 12 14 16 18 x


COMMENTS: 8, 37

CHAPTER 5

TRIGONOMETRIC FUNCTIONS: RIGHT TRIANGLE APPROACH

5.1

Angle Measure 1

5.2

Trigonometry of Right Triangles 6

5.3

Trigonometric Functions of Angles 11

5.4

Inverse Trigonometric Functions and Right Triangles 15

5.5 5.6

The Law of Sines 20 The Law of Cosines 25 Chapter 5 Review 30 Chapter 5 Test 35

¥

FOCUS ON MODELING: Surveying 37

1


5

TRIGONOMETRIC FUNCTIONS: RIGHT TRIANGLE APPROACH

5.1

ANGLE MEASURE y

1. (a) The radian measure of an angle  is the length of the arc that subtends the angle in a circle of radius 1.

400¡

 . (b) To convert degrees to radians we multiply by 180

2¹/3

(c) To convert radians to degrees we multiply by 180  . (d) An angle is in standard position if it is drawn in the x y­plane with its

210¡

0

x

_¹/4

vertex at the origin and its initial side on the positive x­axis. 2. (a) If a central angle  is drawn in a circle of radius r, the length of the arc subtended by  is s  r . (b) The area of the sector with central angle  is A  12 r 2 .

3. (a) The angular speed of the point is  

 . t

s (b) The linear speed of the point is   . t (c) The linear speed  and the angular speed ome are related by the equation   r.

4. No, if the common angular speed is , Object A has linear speed 2, while Object B has angular speed 5. Object B has greater linear speed.  rad   rad  0349 rad 5. 20  20  180  9

 rad  2  0698 rad 6. 40  40  180  9

 rad  3 rad  0942 rad 7. 54  54  180  10

 rad  5 rad  1309 rad 8. 75  75  180  12

 rad    rad  0785 rad 9. 45  45  180  4

 rad    rad  0524 rad 10. 30  30  180  6

 rad  5 rad  1745 rad 11. 100  100  180  9

 rad  10 rad  3491 rad 12. 200  200  180  9

 rad  50 rad  17453 rad 13. 1000  1000  180  9

 rad  20 rad  62832 rad 14. 3600  3600  180 

 rad   7 rad  1222 rad 15. 70  70  180  18

 rad   5 rad  2618 rad 16. 150  150  180  6

 17. 76  76  180   210

 18. 43  43  180   240

 19. 56  56  180   150

 20.  32   32  180   270

540  21. 3  3  180     1719

360  22. 2  2  180      1146

630  23. 35  35  180      2005

324  24. 18  18  180     1031

    180  18 25. 10  10

  5  180  50 26. 518  18

   2  180  24 27.  215  15

 13 180  28.  13 12   12    195

29. 50 is coterminal with 50  360  410 , 50  720  770 , 50  360  310 , and 50  720  670 . (Other answers are possible.) 30. 135 is coterminal with 135  360  495 , 135  720  855 , 135  360  225 , and 135  720  585 . (Other answers are possible.) 1


2

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

31. 34 is coterminal with 34  2  114 , 34  4  194 , 34  2   54 , and 34  4   134 . (Other answers are possible.) 11 13 32. 116 is coterminal with 116  2  236 , 116  4  356 , 116  2    6 , and 6  4   6 . (Other answers

are possible.)

 7  15  9  17 33.   4 is coterminal with  4  2  4 ,  4  4  4 ,  4  2   4 , and  4  4   4 . (Other answers are possible.)

34. 45 is coterminal with 45  360  315 , 45  720  675 , 45  360  405 , and 45  720  765 . (Other answers are possible.) 35. Since 430  70  360 , the angles are coterminal.

36. Since 330  30   360 , the angles are coterminal.

37. Since 176  56  126  2; the angles are coterminal.

38. Since 323  113  213  7 is not a multiple of 2, the angles are not coterminal.

39. Since 875  155  720  2  360 , the angles are coterminal.

40. Since 340  50  290 is not a multiple of 360 , the angles are not coterminal.

41. Since 400  360  40 , the angles 400 and 40 are coterminal. 42. Since 375  360  15 , the angles 375 and 15 are coterminal.

43. Since 780  2  360  60 , the angles 780 and 60 are coterminal.

44. Since 100  260  360 is a multiple of 360 , the angles 100 and 260 are coterminal. 45. Since 800  3  360  280 , the angles 800 and 280 are coterminal.

46. Since 1270  190  1080  3  360 is a multiple of 360 , the angles 1270 and 190 are coterminal.

47. Since 196  2  76 , the angles 196 and 76 are coterminal.

5  48. Since  53  2   3 , the angles  3 and 3 are coterminal.

49. Since 25  12  2  , the angles 25 and  are coterminal.

50. Since 10  2  3717, the angles 10 and 10  2 are coterminal. 17  51. Since 174  2  2   4 , the angles 4 and 4 are coterminal.

52. Since 512  32  24  12  2, the angles 512 and 32 are coterminal.

53. Using the formula s  r, s  56  9  152 .

 7 35   54. Using the formula s  r, the length of the arc is s  140  5 5  122. 180 9 9 10 180 s  2 rad  2   1146 55.    r 5  8 s 56. Solving for r, we have r  , so the radius of the circle is r   4.  2 57. Solving for s, we have s  r , so the length of the arc is 4  2  8 cm.  8  838 m. 58. Solving for s, we have s  r   12  40   180 3 14 14 180 s rad    891 . 59. Solving for , we have    r 9 9  15 5 5 180 s  rad    955 . 60. Solving for , we have    r 9 3 3  s 15 18 61. r     573 m  56  s 20 72 62. r     2292 cm   50 180  


SECTION 5.1 Angle Measure   32  4  128  4468 63. (a) A  12 r 2   12  82  80  180  9 9

(b) A  12 r 2   12  102  05  25

  2A 2  12 , so the radius is  586. 64. (a) A  12 r 2   r   07   2A 2  12 (b) r     303  150  180    65.   34 rad and r  8 m, so A  12 r 2   12 82 34  24  754 m2 . 66.   145  145 

   29 and r  6 ft, so A  1 r 2   1 62 29  29  456 ft2 .  36 2 2 36 2 180

67. A  90 m2 and   160  160 

 8 1 8 2 rad. Thus, A  12 r 2   90   r r    180 9 2 9

 , so A  1 r 2   20  5 r 2  r  68. A  20 m2 and   512 2 12

2  20 

2  90 

9  80 m. 8

 4 6 12   55 m. 5 

69. r  80 mi and A  1600 mi2 , so A  12 r 2   1600  12 802     12 rad. 70. The area of the circle is r 2  600 m2 , so r  A  12 r 2  

1 600 900  3  2865 m2 . 2  

600 . 

Thus, the area of the sector is

71. Referring to the figure, we have AC  3  1  4, BC  1  2  3, and AB  2  3  5. Since

2

AB 2  AC 2  BC 2 , then by the Pythagorean Theorem, the

triangle is a right triangle. Therefore,    2 and  2 A  12 r 2   12  12   2  4 ft .

3 A

72. The triangle is equilateral, so 1   3 rad. To find  2 , we use the formula  2   1  2   3  1  0047 rad, or approximately 27 .

3

B 2

1 1 C

1 s   1 rad. Thus, r 1

73. (a) Between 1:00 P. M . and 1:45 P. M ., the minute hand traverses three­quarters of a complete revolution, or 3 2  3 rad. 4 2

The hour hand moves three­quarters of the way from 12 to 1, which is itself one­twelfth of a revolution. So the hour   1 2   rad. hand traverses 34 12 8

(b) Between 1:00 P. M . and 6:45 P. M ., the minute hand traverses five complete revolutions plus three­quarters of a revolution; that is, 5 2  34 2  232 rad. The hour hand moves through five­twelfths of a revolution, plus three­quarters of the way from 6 to 7; that is,   5 3 1 23 12 2  4 12 2  24 rad.

3


4

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

    74. The area of the sector is A  B  12 12  3  6 . The height h of the     12 

equilateral triangle is B  12

  3 2

B 1

1 2  3 , and so its area is 2 2

1/2 1/2

h

A

1

 43 . Thus, the area of the region in question is 

3  A  6 B 6  4 .

75. The circumference of each wheel is d  28 in. If the wheels revolve 10,000 times, the distance traveled is 1 ft 1 mi 10,000  28 in.    1388 mi. 12 in. 5280 ft 76. Since the diameter is 30 in., we have r  15 in. In one revolution, the arc length (distance traveled) is s  r  2  1 rev  67227 rev. 15  30 in. The total distance traveled is 1 mi  5280 ft/mi  12 in/ft  63,360 in.  63,360 in.  30 in Therefore the car wheel will make approximately 672 revolutions.  rad   rad. 77. We find the measure of the angle in degrees and then convert to radians.   405 255  15 and 15 180  12   3960  330  1036725 and so the distance between the two cities is Then using the formula s  r , we have s  12 roughly 1037 mi.

 rad   rad. Then using the formula s  r , the length of the arc is 78.   35  30  5  5  180  3

s 3  3960  110  345575. So the distance between the two cities is roughly 346 mi.

1 of its orbit which is 2 rad. Then s  r  2  93,000,000  1,600,9113, so the 79. In one day, the earth travels 365 365 365

distance traveled is approximately 16 million miles.

80. Since the sun is so far away, we can assume that the rays of the sun are parallel when striking the earth. Thus, the angle 500 s 180  500 formed at the center of the earth is also   72 . So r    3980 mi, and the circumference     72  180 72 2  180  500  25,000 mi. 72   1   1   rad   rad. Then s  r    3960  1152, and so a 81. The central angle is 1 minute  60 60 180 10,800 10,800 is c  2r 

nautical mile is approximately 1152 mi.     219,900 ft2 . 82. (a) The area is A  12 r 2   12  3002  280  180 

(b) The nozzle 300 ft from the center has three times as much area to cover as the nozzle 100 ft from the center, so it should spray 3 40  120 gallons per minute.

83. The area is equal to the area of the large sector (with radius 34 in.) minus the area of the small sector (with radius 14 in.)      1131 in.2 . Thus, A  12 r12   12 r22   12 342  142 135  180 

84. The area available to the cow is shown in the diagram. Its area is the sum of four quarter­circles:   A  14  1002  502  402  302  3750

 11,781 ft2

50 ft 30 ft

20 ft 50 ft

100 ft

60 ft 50 ft

40 ft


SECTION 5.1 Angle Measure

5

45  2 rad  90 rad/min. 1 min 45  2  16  1440 in./min  45239 in./min. (b) The linear speed is   1 6000  2 rad  12,000 rad/min. 86. (a) The angular speed is   1 min

85. (a) The angular speed is  

(b) The linear speed is   87.  

5 6000  2  12 250  ft/s  2618 ft/s. 60 3

8  2  2 32   6702 ft/s. 15 15

600  2  11 12 ft  1 mi  60 min  125 mi/h  393 mi/h. 1 min 5280 ft 1 hr 1  2  3960 1 day 89. 23 h 56 min 4 s  239344 hr. So the linear speed is   103957 mi/h. 1 day 239344 hr

88.  

50 mi/h 1h 5280 ft linear speed     2200 rad/min. radius 2 ft 60 min 1 mi angular speed 2200 rad/min (b) The rate of revolution is   350 rev/min. 2 2 2 100  2  020 m   209 m/s. 91.   60 s 3 40  2  4 linear speed of pedal   160 rad/min. 92. (a) The angular speed is   radius of wheel sprocket 2 (b) The linear speed of the bicycle is   angular speed  radius  160 rad/min  13 in  2080 in/min  62 mi/h. 90. (a) The radius is 2 ft, so the angular speed is  

93. (a) The circumference of the opening is the length of the arc subtended by the angle  on the flat piece of paper, that is, C  s  r   6  53  10  314 cm.

C 10   5 cm. 2 2  (c) By the Pythagorean Theorem, h 2  62  52  11, so h  11  33 cm.  (d) The volume of a cone is V  13 r 2 h. In this case V  13   52  11  868 cm3 . (b) Solving for r, we find r 

94. (a) With an arbitrary angle , the circumference of the opening is   92 3 C 2 2  , h  6  r  36  2 , and C  6, r  2      92 9 2 9 V  13 r 2 h   2 36  2  2  2 42  2 . 3   

(b)

100 50 0

0

2

4

6

(c) The volume seems to be maximized for   513 rad, or about 293 . 

95. Refer to Exercise 74. The equilateral triangle has area 43 and each of the three edge regions has area        1  12    3    3 , so the total area of the Reuleaux triangle is 3  3   3    3 . Its perimeter is 2 3 4 6 4 4 6 4 2 2    . 3  3 96. Answers will vary—although of course everybody prefers radians.


6

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

5.2 1. (a)

TRIGONOMETRY OF RIGHT TRIANGLES (b) sin  

opposite

adjacent opposite opposite , cos   , and tan   . hypotenuse hypotenuse adjacent

adjacent

¬

hypotenuse

2. The trigonometric ratios do not depend on the size of the triangle because all right triangles with angle  are similar. 1 1 1 3. The reciprocal identities state that csc   , sec   , and cot   . sin  cos  tan  4. (a) x  r cos  and y  r sin .  (b) If r  6 and   30 then x  6 cos 30  3 3 and y  6 sin 30  3.

5. sin   45 , cos   35 , tan   43 , csc   54 , sec   53 , cot   34

7 , cos   24 , tan   7 , csc   25 , sec   25 , cot   24 6. sin   25 7 7 25 24 24   9 2 7. The remaining side is obtained by the Pythagorean Theorem: 41  402  81  9. Then sin   40 41 , cos   41 , 41 41 9 tan   40 9 , csc   40 , sec   9 , cot   40

8. The hypotenuse is obtained by the Pythagorean Theorem: 17 17 8 tan   15 8 , csc   15 , sec   8 , cot   15

  8 82  152  289  17. Then sin   15 17 , cos   17 ,

9. The remaining side is obtained by the Pythagorean Theorem: 

  32  22  13. Then sin   2

13

 2 1313 ,

cos   3  3 1313 , tan   23 , csc   213 , sec   313 , cot   32 13    10. The remaining side is obtained by the Pythagorean Theorem: 82  72  15. Then sin   78 , cos   815 , 

tan   7  7 1515 , csc   87 , sec   8  8 1515 , cot   715 15 15   2 2 11. c  5  3  34 

(a) sin   cos   3  3 3434 34   12. b  72  42  33 (a) sin   cos   47

(b) tan   cot   35

(b) tan   cot   4

33

(c) sec   csc   534 (c) sec   csc   7

13. (a) sin 22  037461

 (b) cot 38  2  1  041421

14. (a) cos 45  080902

(b) csc 48  134563

15. (a) sec 1  185082

(b) tan 51  123490

16. (a) csc 10  575877 x 17. Since sin 30  , we have x  25 sin 30  25  12  25 2. 25  12 12 12 18. Since sin 45  , we have x   1  12 2. x sin 45 

(b) sin 35  035078

2   x   , we have x  13 sin 60  13  23  132 3 . 19. Since sin 60 

13

33


SECTION 5.2 Trigonometry of Right Triangles

20. Since tan 30 

7

 4 4 4 , we have x   1  4 3.  x tan 30  3

12 12 , we have x   1651658. 21. Since tan 36  x tan 36 25 25 , we have x   3130339. 22. Since sin 53  x sin 53 y x  cos   x  28 cos , and  sin   y  28 sin . 23. 28 28 4 4 x  tan   x  4 tan , and  cos   y   4 sec . 24. 4 y cos    26. cos   12 25. tan   56 . Then the third side is x  52  62  61. 13 . The third side is     y  132  122  25  5. The other five ratios are The other five ratios are sin   5 61 , cos   6 61 , 61   61 61 csc   5 , sec   6 , and cot   65 . Ï61

61

5 , tan   5 , csc   13 , sec   13 , and sin   13 12 5 12

cot   12 5.

13

5

5

¬

¬

12

6

27. cot   1. Then the third side is r 

  12  12  2. 

The other five ratios are sin   1  22 , 2   cos   1  22 , tan   1, csc   2, and 2  sec   2.

Ï2

28. tan  

  3. The third side is r  12  3  2. The other 

five ratios are sin   23 , cos   12 , csc   2 , 3

sec   2, and cot   1 . 3

¬

2

1

Ï3

1

¬ 1

29. csc   11 6 . The third side is x 

    112  62  85. The 30. cot   53 . The third side is x  52  32  34. The 

6 , cos   85 , other five ratios are sin   11 11

other five ratios are sin   3 3434 , cos   5 3434 ,

tan   6 8585 , sec   118585 , and cot   685 .

tan   35 , csc   334 and sec   534 .

11

Ï34

6

¬

¬ Ï85

    1  3  1 3 31. sin   cos 6 6 2 2 2

32. sin 30 csc 30  sin 30 

1 1 sin 30

33. sin 30 cos 60  sin 60 cos 30  12  12  23  23  14  34  1

5

3


8

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

 2 3  12  34  14  1 2   2  2 35. cos 30 2  sin 30 2  23  12  34  14  12 2      2 2    3  1  1  1    2  1 36. sin    31  18 3  1  18 3  2 3  1 3 cos 4  sin 4 cos 3 2 2 2  2    2 2  18 4  2 3  14 2  3  2      2 2  3  2  2  1  12  14  2 12 37. cos  4  sin 6 2  2 2 4 2        2  2   3 3   2  38. sin   12  2  94  2 3 tan 6  csc 4 2  3  2 39. This is an isosceles right triangle, so the other leg has length 16 tan 45  16, the hypotenuse has length 34. sin 60 2  cos 60 2 

  2

 16  16 2  2263, and the other angle is 90  45  45 . sin 45

40. The other leg has length 100 tan 75  2679, the hypotenuse has length 90  75  15 . 41. The other leg has length 35 tan 52  4480, the hypotenuse has length 90  52  38 .

100  10352, and the other angle is sin 75 35  5685, and the other angle is cos 52

42. The adjacent leg has length 1000 cos 68  37461, the opposite leg has length 1000 sin 68  92718, and the other angle is 90  68  22 .

 43. The adjacent leg has length 335 cos  8  3095, the opposite leg has length 335 sin 8  1282, and the other angle is     3 . 2

8

8

44. The opposite leg has length 723 tan  6  4174, the hypotenuse has length     . 2

6

723  8348, and the other angle is cos  6

3

45. The adjacent leg has length

106 106  14590, the hypotenuse has length   18034, and the other angle is tan  sin 5 5

    3 . 2

5

10

46. The adjacent leg has length 425 cos 38  16264, the opposite leg has length 425 sin 38  39265, and the other angle is   3   . 2

8

comma

8

1  045 cos   2  089, tan   1 , csc   224, sec   224  112, cot   200. 47. sin   224 224 2 2

sin 40  064, cos 40  077, tan 40 

48.

064  083, csc 40  156, 077

sec 40  131, cot 40  120. 40¡

49. x 

100 100   2309 tan 60 tan 30

50. Let d be the length of the base of the 60 triangle. Then tan 60  dx 

85 85 x   d  981. tan 30 tan 30

51. Let h be the length of the shared side. Then sin 60 

85 85 85 d   49075, and so tan 30  d tan 60 dx

50 h 50 h h   57735  sin 65   x   637 h sin 60 x sin 65


9

SECTION 5.2 Trigonometry of Right Triangles

52. Let h be the hypotenuse of the top triangle. Then sin 30  x

10 h   58. tan 60 tan 60

53.

From the diagram, sin  

10 ¬

¹ -¬ 2

y

5 5 h h    10, and so tan 60   h sin 30 x

y x and tan   , so x  y sin   10 sin  tan . y 10

x

¬

54. sin  

b 1 d a  a  sin , tan    b  tan , cos    c  sec , cos    d  cos  1 1 c 1

55. Let h be the height, in feet, of the Empire State Building. Then tan 11 

h  h  5280  tan 11  1026 ft. 5280

56. (a) Let r be the distance, in feet, between the plane and the Gateway Arch. Therefore, sin 22 

35,000  r

35,000  93,431 ft. sin 22 (b) Let x be the distance, also in feet, between a point on the ground directly below the plane and the Gateway Arch. Then 35,000 35,000 x   86,628 ft. tan 22  x tan 22 r

57. (a) Let h be the distance, in miles, that the beam has diverged. Then tan 05 

h  240,000

h  240,000  tan 05  2100 mi.

(b) Since the deflection is about 2100 mi whereas the radius of the moon is about 1000 mi, the beam will not strike the moon. 58. Let x be the distance, in feet, of the ship from the base of the lighthouse. Then tan 23 

200 200 x   471 ft. x tan 23

59. Let h represent the height, in feet, that the ladder reaches on the building. Then sin 72 

h  h  20 sin 72  19 ft. 20

60. Let h be the height, in feet, of the communication tower. Then sin 65 

h  h  600 sin 65  544 ft. 600

61. Let h be the height, in feet, of the kite above the ground. Then sin 50 

h  h  450 sin 50  345 ft. 450

62. 18¡ x 14¡

hÁ hª

Let h 1 be the height of the flagpole above elevation and let h 2 be the height below, h as shown in the figure. So tan 18  1  h 1  x tan 18 . Similarly, x h 2  x tan 14 . Since the flagpole is 60 feet tall, we have h 1  h 2  60, so x tan 18  tan 14   60  x 

60  1045 ft. tan 18  tan 14

63. Let h 1 be the height of the window in feet and h 2 be the height from the window to the top of the tower. Then tan 25 

h1 325

h2  h 2  325  tan 39  263 ft. Therefore, the height of the window 325 is approximately 152 ft and the height of the tower is approximately 152  263  415 ft.

 h 1  325  tan 25  152 ft. Also, tan 39 


10

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

64.

52¡ 52¡

car, and d2 be the distance, in feet, between the same point and the other car. Then

5150

d2 

Let d1 be the distance, in feet, between a point directly below the plane and one

35¡

35¡

tan 52 

5150 5150 5150  d1   402362 ft, and tan 35   d1 tan 52 d2

5150  735496 ft. So the distance between the two cars is about d1  d2  402362  735496  11,379 ft. tan 35

65. Let d1 be the distance, in feet, between a point directly below the plane and one car, and d2 be the distance, in feet, between d d the same point and the other car. Then tan 52  1  d1  5150  tan 52  65917 ft. Also, tan 38  2  5150 5150 d2  5150  tan 38  40236 ft. So in this case, the distance between the two cars is about 2570 ft. 66. Let x be the distance, in feet, between a point directly below the balloon and the first mile post. Let h be the height, h h in feet, of the balloon. Then tan 22  and tan 20  . So h  x tan 22  x  5280 tan 20  x x  5280 5280  tan 20 x  47,977 ft. Therefore h  47,9769  tan 22  19,384 ft  37 mi. tan 22  tan 20 67. Let x be the horizontal distance, in feet, between a point on the ground directly below the top of the mountain and h the point on the plain closest to the mountain. Let h be the height, in feet, of the mountain. Then tan 35  x  h 1000  tan 32 and tan 32  . So h  x tan 35  x  1000 tan 32  x   82942. Thus x  1000 tan 35  tan 32  h  82942  tan 35  5808 ft.

68.

Since the angle of elevation from the observer is 45 , the distance from the observer is h, as shown in the figure. Thus, the length of the leg in the smaller right h 75¡ h

ã600ã

h  600  h tan 75  h  600  h 600 tan 75 600 tan 75  h 1  tan 75   h   473 m. 1  tan 75 triangle is 600  h. Then tan 75 

600-h

69. Let d be the distance, in miles, from the earth to the sun.

Then sec 8985 

d  240,000

d  240,000  sec 8985  917 million miles.

 70. (a) s  r    rs  6155 3960  15543 rad  8905

(b) Let d represent the distance, in miles, from the center of the earth to the moon. Since cos   d

3960 , we have d

3960 3960   239,9615. So the distance AC is 239,9615  3960  236,000 mi. cos  cos 8905

r 71. Let r represent the radius, in miles, of the earth. Then sin 60276  r600  r  600 sin 60276  r  sin 60276 600 sin 60276  r 1  sin 60276   r  600 1sin 60276  3960099. So the earth’s radius is about 3960 mi. 

,000 , we have 72. Let d represent the distance, in miles, from the earth to Alpha Centauri. Since sin 0000211  93,000 d 93,000,000 13 d  sin 0000211  25,253,590,022,410. So the distance from the earth to Alpha Centauri is about 253  10 mi.

73. Let d be the distance, in AU, between Venus and the sun. Then sin 463 

d  d, so d  sin 463  0723 AU. 1

74. If two triangles are similar, then their corresponding angles are equal and their corresponding sides are proportional. That       is, if triangle ABC is similar to triangle A B  C  then AB  r  A B  , AC  r  A C  , and BC  r  B  C  . Thus when we express any trigonometric ratio of these lengths as a fraction, the factor r cancels out.


SECTION 5.3 Trigonometric Functions of Angles  75. From the diagram, we see that tan 60  x1  x  cot 60  33 . Thus, the     perimeter of ABC is 3 1  33  3  3.

B

1 A

5.3

11

60¡ x

P

C

1

TRIGONOMETRIC FUNCTIONS OF ANGLES

1. If the angle  is in standard position and P x y is a point on the terminal side of , and r is the distance from the origin to x y y P, then sin   , cos   , and tan   . r r x 2. The sign of a trigonometric function of  depends on the quadrant in which the terminal angle of  lies. For example, if  is in quadrant II, sin  is positive. In quadrant III, cos  is negative. In quadrant IV, sin  is negative. 3. (a) If  is in standard position, then the reference angle  is the acute angle formed by the terminal side of  and the x­axis. So the reference angle for   100 is   80 and that for   190 is   10 .

(b) If  is any angle, the value of a trigonometric function of  is the same, except possibly for sign, as the value of the trigonometric function of . So sin 100  sin 80 and sin 190   sin 10 .

4. The area A of a triangle with sides of lengths a and b and with included angle  is given by the formula A  12 ab sin . So the area of the triangle with sides 4 and 7 and included angle   30 is 12 4 7 sin 30  7.

5. (a) The reference angle for 135 is 180  135  45 .

6. (a) The reference angle for 155 is 180  155  25 .

(b) The reference angle for 195 is 195  180  15 .

(b) The reference angle for 280 is 360  280  80 .

(c) The reference angle for 300 is 360  300  60 .

(c) The reference angle for 390 is 390  360  30 .

7. (a) The reference angle for 250 is 250  180  70 . (b) The reference angle for 485 is 540  485  55 . (c) The reference angle for 100 is

8. (a) The reference angle for 99 is 180  99  81 . (b) The reference angle for 199 is 199  180  19 .

(c) The reference angle for 359 is 360  359  1 .

180  100  80 .

 is   7  3 . 9. (a) The reference angle for 710 10 10

(b) The reference angle for 98 is 98     8.

(c) The reference angle for 103 is 103  3   3.

11. (a) The reference angle for 57 is   57  27 .

10. (a) The reference angle for 56 is   56   6.

(b) The reference angle for 109 is 109     9.

(c) The reference angle for 237 is 237  3  27 .

12. (a) The reference angle for 23 is 23  2  03.

(b) The reference angle for 14 is 14    04.

(b) The reference angle for 23 is   23  084.

(c) The reference angle for 14 is 14 because 14   2.

(c) The reference angle for 10 is 10  10  0.

 13. cos 150   cos 30   23

14. sin 210   sin 30   12

15. tan 315   tan 45  1

16. cot 60    cot 60   tan160   33

17. csc 300   csc 60   sin160   2 3 3 

19. sin 225   sin 45   22

18. sec 330  sec 30  cos130  2 3 3  20. csc 135  csc 45  sin145  2


12

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

 22. tan 120   tan 60   3

21. sec 420   sec 60  cos160  2  23. cot 570  cot 30  tan130  3

24. sin 810  sin 90  1

25. sin 32   sin  2  1    27. tan  43   tan  3  3     29. csc  56   csc  6  2 31. sec 173  sec  3 

1 26. cos 43   cos  3  2    3 28. cos  116  cos  6  2 

2 3 30. sec 76   sec  6  3

1 2 cos  3

32. csc 54   csc  4 

 1  2 sin  4 

  1  33. cot   4   cot 4  tan   1 4

2 1 34. cos 74  cos  4  2  2

35. tan 52  tan  2 which is undefined.

1 36. sin 116   sin  6  2

37. Since sin   0 and cos   0,  is in quadrant III. 38. Since both tan  and sin  are negative,  is in quadrant IV. 39. sec   0  cos   0. Also tan   0  in quadrant IV.

sin   0  sin   0 (since cos   0). Since sin   0 and cos   0,  is cos 

40. Since csc   0  sin   0 and cos   0,  is in quadrant II. 41. tan    24   12

42. sin    23

3 42

 35

43. cos     23

3 32

  22

    2 2 2 2 45. sec   2 33  313 46. csc    1 33   310   47. cos    35 . Then y   52  32   42  4, since  is in quadrant IV. Thus, sin   45 , tan    43 , csc   54 , 44. cot    23

sec    53 , and cot    34 .  48. tan   34 , so r  32  42  5. Because  is in quadrant III, x  4 and y  3, and so sin    35 , cos    45 , csc    53 , sec    54 , and cot   43 .    49. cot   2, so tan    12 and r  22  12  5. Thus, because  is in quadrant IV, sin    1   55 , 5    cos   2 5 5 , tan    12 , csc    5, and sec   25 .    50. csc   2. Then sin   12 and x  22  12  3. Because  is in quadrant I, sin   12 , cos   23 ,    tan   1  33 , sec   2  2 3 3 , and cot   3. 3 3     51. sin    23 , so x  32  22  5. Because  is in quadrant III, cos    35 , tan   2 5 5 , csc    32 , 

sec    3 5 5 , and cot   25 .     130 , cos    7 130 , tan    9 , 52. cot    79 , so r  72  92  130. Because  is in quadrant II, sin   9 130 7 130 csc  

  130 , and sec    130 . 7 9

53. sin   14  0, so because cos   0,  is in quadrant I and x    csc   4, sec   4 1515 , and cot   15.

    42  12  15. Thus, cos   415 , tan   1515 ,

   54. cos   13  0, so because tan   0,  is in quadrant IV and y   32  12  2 2. Thus, sin    2 3 2 ,    tan   2 2, csc    3 4 2 , sec   3, and cot    42 .


SECTION 5.3 Trigonometric Functions of Angles

13

  32  12  10. Because sin   0 and tan   0,  is in quadrant III and we have     sin    3 1010 , cos    1010 , csc    310 , sec    10, and cot   13 .   7 2 2 56. csc    12 7 , so sin    12 and because cos   0 and csc   0,  is in quadrant III. Thus, x   12  7   95 55. tan   3  31 , so r 

7 , cos    95 , tan   7 95 , sec    12 95 , and cot   95 . and sin    12 7 12 95 95

  57. csc   4, so sin    14 and because tan   0 and csc   0,  is in quadrant IV. Thus, x  42  12  15 and     sin    14 , cos   415 , tan    1515 , sec   4 1515 , and cot    15.    3 , so because sec   0 and sin   0,  is in quadrant I and x  102  32  91. Thus, cos   91 , 58. sin   10 10 

10 91 91 tan   3 9191 , csc   10 3 , sec   91 , and cot   3 .   sin  1  cos2  2 59. Since sin  is negative in quadrant III, sin    1  cos  and we have tan    . cos  cos    1  sin2  cos   because cos   0 in quadrant II. 60. cot   sin  sin   61. cos2   sin2   1  cos   1  sin2  because cos   0 in quadrant IV.

1 1  because all trigonometric functions are positive in quadrant I. cos  1  sin2   63. sec2   1  tan2   sec    1  tan2  because sec   0 in quadrant II.  64. csc2   1  cot2   csc    1  cot2  because csc   0 in quadrant III.   3 2  65. If    3 , then sin 2  sin 3  2 and 2 sin   2 sin 3  3.     2   sin  2  3 and sin  2  sin  2  0890. 66. If    , then sin 3 3 4 9 62. sec  

67. The area is 12 7 9 sin 72  300.

68. The area is 12 10 22 sin 10  191.  69. The area is 12 102 sin 60  25 3  433. 

70. The area is 12 132 sin 60  1694 3  732.

 71. Let the length of the other side be x. Then A  16  12 5 x sin 36  x  32 5 csc 36  1089 in.    48  96  4 6  98 cm. 72. Let the lengths of the equal sides be x. Then A  12 x 2 sin   24  12 x 2 sin 56  x  5  sin 6

  4 . For the triangle defined by the two sides, 73. For the sector defined by the two sides, A1  12 r 2   12  22  120  180  3   1 1   A2  2 ab sin   2  2  2  sin 120  2 sin 60  3. Thus the area of the region is A1  A2  43  3  246.

74. The area of the entire circle is r 2    122  144, the area of the sector is 12 r 2   12  122   3  24,  1 1  2 and the area of the triangle is 2 ab sin   2  12  sin 3  36 3, so the area of the shaded region is   A1  A2  A3  144  24  36 3  120  36 3  4393.   75. The height of the equilateral triangle is 22  12  3, and so its ¹/3 ¹/3   1 area is 12 2 3  3. The area of each sector within the triangle is  ¹/3 1  12     , and so the area of the shaded area is 3   . 1 2 3 6 2


14

CHAPTER 5 Trigonometric Functions: Right Triangle Approach A

76. We use the area formula A  12 ab sin . 

a

3 2 2 4 For P Q R, A1  12 a 2 sin  3  4 a  1  a  3 ; for     APC, A2  12 a 2a sin 23  23 a 2  23 4  2.

P 2¹/3 a

3

B

Thus, the area of ABC is A  A1  3A2  7.

Q

¹/3 a

2a R C

77. (a) tan  

5280 ft h , so h  tan   1 mile   5280 tan  ft. 1 mile 1 mile

(b) 

20

60

80

85

h

1922

9145

29,944

60,351

78. (a) Let the depth of the water be h. Then the cross­sectional area of the gutter is a trapezoid whose height is h  10 sin . The bases are 10 and 10  2 10 cos   10  20 cos . Thus, the area is b  b2 10  10  20 cos  A   1 h   10 sin   100 sin   100 sin  cos . 2 2 (b)

(c)

200

130.0 129.5

100 0

0

129.0 1.00

1

1.05

1.10

From the graph, the largest area is achieved when   1047 rad  60 . 79. (a) From the figure in the text, we express depth and width in terms of .

(b)

depth width Since sin   and cos   , we have depth  20 sin  20 20 and width  20 cos . Thus, the cross­section area of the beam is A   depth width  20 cos  20 sin   400 cos  sin . (c) The beam with the largest cross­sectional area is the square beam,   10 2 by 10 2 (about 1414 by 1414).

200

0

0

1

80. Using depth  20 sin  and width  20 cos , from Exercise 79, we have strength  k width depth2 . Thus S   k 20 cos  20 sin 2  8000k cos  sin2 .

81. (a) On Earth, the range is R 

 02 sin 2 g

122 sin  3 32

 9 3   3897 ft and the height is 4

122 sin2   2 sin2  6  9  05625 ft. H 0  2g

2  32

16

2 2  122 sin  3  23982 ft and H  12 sin 6  3462 ft (b) On the moon, R 

82. Substituting, t 

52

 2000  5 10  158 s.  16 sin 30

2  52


SECTION 5.4 Inverse Trigonometric Functions and Right Triangles

83. (a) W  302  038 cot   065 csc 

15

10

(b) From the graph, it appears that W has its minimum value at about   0946  542 .

5 0

84. (a) We label the lengths L 1 and L 2 as shown in the figure. Since sin  

¬ 6 Lª

9 6 and cos   , we L1 L2

have L 1  9 csc  and L 2  6 sec . Thus

40

0

LÁ ¬

(d) The minimum value of L  is the shortest

2

20

¬

L   L 1  L 2  9 csc   6 sec .

(c) The minimum value of L  is 2107.

(b)

0

0.0

0.5

1.0

1.5

9

distance the pipe must pass through.

85. sin 4  00697564737 but sin 4  07568024953. Your partner has found sin 4 instead of sin 4 radians.

O P O P adj   O P. Since QS is tangent to the circle at R, O R Q is a right triangle. Then  O R hyp 1 R Q O Q opp hyp   tan    R Q and sec    O Q. Since  SO Q is a right angle S O Q is a right O R O R adj adj O S S R hyp adj    O S and cot    S R. Summarizing, we have triangle and  O S R  . Then csc   O R O R opp opp sin   P R, cos   O P, tan   R Q, sec   O Q, csc   O S, and cot   S R.   1 1 1  tan2   1  sec2  87. (a) sin2   cos2   1  sin2   cos2   2 cos  cos2    1 1 (b) sin2   cos2   1  sin2   cos2   1  1  cot2   csc2  sin2  sin2 

86. cos  

88. Let x represent the desired real number, in radians. We wish to find

the smallest nonzero solution of the equation sin x  sin 180x . (Since

y 1

180x y=sin ¹

x is in radians, 180x  is in degrees.) Now if a is in degrees with

0  a  180 and sin a  b, then either a  b or 180  a  b. The graph shows that the first positive solution occurs when

y=sin x 0

2

x

_1 180x 180x  180, so we have 180   x.   180 180  3088. You can verify that the value of sin is the same whether Solving this equation, we get x  180   180   your calculator is set to radian mode or degree mode.

0

5.4

INVERSE TRIGONOMETRIC FUNCTIONS AND RIGHT TRIANGLES

1. For a function to have an inverse, it must be one­to­one. To define the inverse sine function we restrict the domain of the   sine function to the interval   2 2 .   2. (a) The function sin1 has domain [1 1] and range   2 2 .


16

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

(b) The function cos1 has domain [1 1] and range [0 ].   (c) The function tan1 has domain  and range   2 2 .

8  sin1 4 3. (a)   sin1 10 5

6  cos1 3 (b)   cos1 10 5

  5 , we let   cos1 5 . So 4. To find tan cos1 13 13

5 . We then complete the right triangle in cos   13

the figure and use the triangle to find that   5  12 . tan cos1 13 5

(c)   tan1 68  tan1 43

13 ¬

12

5

 3  3

5. (a) sin1 1   2

(b) cos1 0   2

(c) tan1

6. (a) sin1 0  0

(b) cos1 1  

(c) tan1 0  0

   7. (a) sin1  22    4    8. (a) sin1  23    3

9. sin1 030  0305  17458

   (b) cos1  22  34   (b) cos1  12  23

  11. cos1 13  1231  70529

(c) tan1 1    4    (c) tan1  3    3

10. cos1 02  1772  101537 12. sin1 56  0985  56443

13. tan1 3  1249  71565

14. tan1 4  132582  75964

15. cos1 3 is undefined.

16. sin1 2 is undefined.

6  3 , so   sin1 3  369 . 17. sin   10 5 5

7 , so   tan1 7  213 . 18. tan   18 18

9 , so   tan1 9  347 . 19. tan   13 13

3 1 3  254 . 20. sin   30 7 70  7 , so   sin

21. sin   47 , so   sin1 47  348 .

22. cos   89 , so   cos1 98  273 .

23. tan   34 , so   tan1 43  369 .

  25. tan    32 , so   tan1  32  563 .

  24. tan    24   12 , so   180  tan1 12  1534

26. tan   3, so   180  tan1 3  2516 .

  27. We use sin1 to find one solution in the interval 90  90 . sin   23    sin1 23  418 . Another solution with  between 0 and 180 is obtained by taking the supplement of the angle: 180  418  1382 . So the solutions of the equation with  between 0 and 180 are approximately   418 and   1382 .   28. One solution is given by   cos1 43  414 . This is the only solution, because cos x is one­to­one on 0 180 .     29. One solution is given by   cos1  25  1136 . This is the only solution, because cos x is one­to­one on 0 180 .

  30. tan1 20  871 , so the only solution in 0 180 is approximately 180  871  929 .   31. tan1 5  787 . This is the only solution on 0 180 .     32. One solution is   sin1 45  531 . Another solution on 0 180 is approximately 180  531  1269 .


SECTION 5.4 Inverse Trigonometric Functions and Right Triangles

  33. To find cos sin1 45 , first let   sin1 54 . Then  is the number in the interval     2  2 whose sine is 45 . We draw a right triangle with  as one of its acute

5

angles, with opposite side 4 and hypotenuse 5. The remaining leg of the triangle is

¬

found by the Pythagorean Theorem to be 3. From the figure we get   cos sin1 54  sin   35 .

17

4

3

    Another method: By the cancellation properties of inverse functions, sin sin1 54 is exactly 45 . To find cos sin1 45 , we

first write the cosine function in terms of the sine function. Let u  sin1 45 . Since 0  u   2 , cos u is positive, and since       2   9 3 cos2 u  sin2 u  1, we can write cos u  1  sin2 u  1  sin2 sin1 45  1  45  1  16 25  25  5 .   Therefore, cos sin1 54  35 .   34. To find cos tan1 34 , we draw a right triangle with angle , opposite side 4, and   adjacent side 3. From the figure we see that cos tan1 34  cos   35 .

  12 , we draw a right triangle with 35. To find sec sin1 13

angle , opposite side 12, and hypotenuse 13. From the   12  sec   13 . figure we see that sec sin1 13 5 13

¬

5

5 ¬

4

3

     7 7 , we draw a  csc cos1 25 36. To find csc cos1  25 right triangle with angle , adjacent side 7, and hypotenuse 25. From the figure we see that    7  csc   25 csc cos1  25 24 . 25

¬

12

7

24

     1 12 , we draw   sin tan 37. To find sin tan1  12 5 5 a right triangle with angle , opposite side 12, and adjacent side 5. From the figure we see that      sin    12 sin tan1  12 5 13 . 13 12

¬

     38. To find sec tan1  23  sec tan1 32 , we draw a

right triangle with angle , opposite side 2, and adjacent side 3. From the figure we see that     sec tan1  23  sec   313 .

5

Ï13 ¬

3

2


18

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

    39. To find sin sec1 4  sin sec1 4 , we draw a right triangle with angle , adjacent side 1, and

hypotenuse 4. From the figure we see that    sin sec1 4  sin   415 . 4

¬

  40. To find csc cot1 43 , we draw a right triangle with angle

, adjacent side 3, and opposite side 4. From the figure we   see that csc cot1 34  csc   54 . 5

1

Ï15

¬

4

3

  41. We want to find cos sin1 x . Let   sin1 x, so sin   x. We sketch a right

1

triangle with an acute angle , opposite side x, and hypotenuse 1. By the  Pythagorean Theorem, the remaining leg is 1  x 2 . From the figure we have    cos sin1 x  cos   1  x 2 .

x

¬ Ï1-x@

Another method: Let u  sin1 x. We need to find cos u in terms of x. To do so, we write cosine in terms of sine. Note     1 x. Now cos u  1  sin2 u is positive because u lies in the interval     . that   2  u  2 because u  sin 2 2      Substituting u  sin1 x and using the cancellation property sin sin1 x  x gives cos sin1 x  1  x 2 .

  42. We want to find cot cos1 x . Let   cos1 x, so cos   x. We sketch a right triangle with an acute angle , adjacent side x, and hypotenuse 1. By the  Pythagorean Theorem, the remaining leg is 1  x 2 . From the figure we have   x cot cos1 x  cot    . 1  x2

1 ¬

x

  43. We want to find sec cos1 x . Let   cos1 x, so cos   x. We sketch a right triangle with an acute angle , opposite side x, and hypotenuse 1. By the  Pythagorean Theorem, the remaining leg is 1  x 2 . From the figure we have   1 sec cos1 x  sec   . x

1 ¬

Ïx@+1 ¬

  45. We want to find sec tan1 x . Let   tan1 x, so tan   x. We sketch a right triangle with an acute angle , opposite side x, and adjacent side 1. By the  Pythagorean Theorem, the hypotenuse is x 2  1. From the figure we have    sec tan1 x  sec   x 2  1.

Ï1-x@

x

  44. We want to find cos tan1 x . Let   tan1 x, so tan   x. We sketch a right triangle with an acute angle , opposite side x, and adjacent side 1. By the  Pythagorean Theorem, the hypotenuse is x 2  1. From the figure we have   1 cos tan1 x  sin    . 2 x 1

Ï1-x@

1

Ïx@+1 ¬

x

1

x


SECTION 5.4 Inverse Trigonometric Functions and Right Triangles

  46. We want to find tan sin1 x . Let   sin1 x, so sin   x. We sketch a right triangle with an acute angle , opposite side x, and hypotenuse 1. By the  Pythagorean Theorem, the remaining leg is 1  x 2 . From the figure we have   x . tan sin1 x  tan    1  x2

1

19

x

¬ Ï1-x@

47. Let  represent the angle of elevation of the ladder. Let h represent the height, in feet, that the ladder reaches on the 6  03    cos1 03  1266 rad  725 . By the Pythagorean Theorem, h 2  62  202  building. Then cos   20   h  400  36  364  19 ft. 96  08    tan1 08  0675  387 . 48. Let  be the angle of elevation of the sun. Then tan   120

49. (a) Solving tan   h2 for h, we have h  2 tan .

(b) Solving tan   h2 for  we have   tan1 h2.

50. (a) Solving tan   50s. Solving for , we have   tan1 50s.

5 1 5  682 . (b) Set s  20 ft to get tan   50 20  2 . Solving for , we have   tan 2

51. (a) Solving sin   h680 for  we have   sin1 h680.    (b) Set h  500 to get   sin1 500 680  0826 rad  473 .

52. The tourist, the base of the low side of the tower, and the top of the low side form a right triangle in which tan   39 56     tan1 39 56  40 .

3960 . Solving for , we get 53. (a) Since the radius of the earth is 3960 miles, we have the relationship cos   h3960   3960 .   cos1 h3960

(b) The arc length is s  radius included angle  3960 2  7920.   3960 (c) s  7920 cos1 h3960     3960 (d) When h  100 we have s  7920 cos1 1003960  7920 cos1 3960 4060  17615 miles.       3960 3960 3960 2450 1 (e) When s  2450 we have 2450  7920 cos1 h3960  2450 7920  cos h3960  h3960  cos 7920      2450 h  3960  3960 sec 2450 7920  h  3960 sec 7920  3960  1973 mi.

   1 1 1  sin  sin1 08102  541 7 tan 10 2  3  1 tan 10     1 1 1  1 (b) For n  2,   sin  483 . For n  3,   sin  322 . For n  4, 5 tan 15 7 tan 15   1 1  245 . n  0 and n  1 are outside of the domain for   15 , because  3732   sin1  9 tan 15 tan 15 1 and  1244, neither of which is in the domain of sin1 . 3 tan 15

54. (a)   sin1

55. We have sin   k sin , where   594 and k  133. Substituting, sin 594  133 sin  sin 594  06472. Using a calculator, we find that   sin1 06472  403 , so  sin   133   4  2  4 403   2 594   424 .


20

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

1 56. Let   sec1 x. Then sec   x, as shown in the figure. Then cos   , so x     1 1   cos1 . Thus, sec1 x  cos1 , x  1. x x In particular, sec1 2  cos1 12   3.

1 Let   csc1 x. Then csc   x, as shown in the figure. Then sin   , so x     1 1 . Thus, csc1 x  sin1 , x  1.   sin1 x x In particular, csc1 3  sin1 31  0340.

1 Let   cot1 x. Then cot   x, as shown in the figure. Then tan   , so x     1 1 1 1 1   tan . Thus, cot x  tan , x  1. x x In particular, cot1 4  tan1 41  0245.

5.5

x ¬ 1

x

1

º

1 

x

THE LAW OF SINES

sin B sin C sin A   . a b c 2. (a) The Law of Sines can be used to solve triangles in cases ASA or SSA.

1. In triangle ABC with sides a, b, and c the Law of Sines states that

(b) The Law of Sines can give ambiguous solutions in case SSA. sin 110 6 sin 110 sin 40  , so x   88. 3. For ABC we have 6 x sin 40    sin  sin 50 4. For P Q R we have  , so   sin1 65 sin 50  668 . 6 5 376 sin 57 5.  C  180  984  246  57 . x   3188. sin 984 17 sin 1144  254. 6.  C  180  375  281  1144 . x  sin 375 267 sin 52 7.  C  180  52  70  58 . x   248. sin 58 563 sin 67 8. sin    0646. Then   sin1 0646  403 . 802 36 sin 120 9. sin C   0693   C  sin1 0693  439 . 45 185 sin 50 10.  C  180  102  28  50 . x   1449. sin 102 65 sin 46 65 sin 20 11.  C  180  46  20  114 . Then a   512 and b   243. sin 114 sin 114   2 sin 100 2 sin 30 12.  B  180  30  100  50 . Then c   257 and a   131. sin 50 sin 50  12 sin 44 13.  B  68 , so  A  180  68  68  44 and a   899. sin 68  34 sin 80 14. sin B   0515, so  B  sin1 0515  310 . Then  C  180  80  31  690 and 65 65 sin 69 c  62. sin 80


SECTION 5.5 The Law of Sines

15.  C  180  50  68  62 . Then 230 sin 68 230 sin 50  1995 and b   2415. a sin 62 sin 62

16.  C  180  110  23  47 . Then 50 sin 110 50 sin 23  267 and b   642. a sin 47 sin 47 C

C

50¡

A

68¡ 230

23¡

A

110¡ 50

B

B

17.  B  180  30  65  85 . Then 10 sin 30 10 sin 65 a  50 and c   91. sin 85 sin 85

18.  C  180  95  22  63 . Then 420 sin 95 420 sin 63 b  11169 and c   9990. sin 22 sin 22 C

C 10

65¡

30¡

A

420 95¡

22¡

A

B

19.  A  180  51  29  100 . Then 44 sin 51 44 sin 100  894 and c   705. a  sin 29 sin 29

C

10¡

100¡

51¡

A

44 29¡

B

20.  A  180  100  10  70 . Then 115 sin 10 115 sin 70  1097 and b   203. a  sin 100 sin 100

C

A

21

115

B

B

15 sin 110  0503   B  sin1 0503  302 . Then 28 28 sin 398  C  180  110  302  398 , and so c   191. Thus  B  302 ,  C  398 , and c  191. sin 110 40 sin 37  0802   C1  sin1 0822  534 or  C2  180  534  1266 . 22. sin C  30 30 sin 896 If  C1  534 , then  B1  180  37  534  896 and b1   498. sin 37 30 sin 164  141. If  C2  1266 , then  B2  180  37  1266  164 and b2  sin 37 Thus, one triangle has  B1  896 ,  C1  534 , and b1  498; the other has  B2  164 ,  C2  1266 , and b2  141.

21. Since  A  90 there is only one triangle. sin B 

23.  A  125 is the largest angle, but since side a is not the longest side, there can be no such triangle.

45 sin 38  0660   B1  sin1 0660  413 or  B2  180  413  1387 . 42 42 sin 1007  67. If  B1  413 , then  A1  180  38  413  1007 and a1  sin 38  42 sin 33 If  B2  1387 , then  A2  180  38  1387  33 and a2   39. sin 38      Thus, one triangle has A1  1007 , B1  413 , and a1  67; the other has A2  33 ,  B2  1387 , and a2  39.

24. sin B 


22

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

30 sin 25  0507   C1  sin1 0507  3047 or  C2  180  3047  14953 . 25 25 sin 12453  4873. If  C1  3047 , then  A1  180  25  3047  12453 and a1  sin 25  25 sin 547 If  C2  14953 , then  A2  180  25  14953  547 and a2   564. sin 25    Thus, one triangle has  A1  125 ,  C1  30 , and a1  49; the other has  A2  5 ,  C2  150 , and a2  56. 100 sin 30 26. sin B   23   B1  sin1 32  418 or  B2  180  418  1382 . 75 75 sin 1082  1425. If  B1  418 , then  C1  180  30  418  1082 and c1  sin 30  75 sin 118 If  B2  1382 , then  C2  180  30  1382  118 and c2   307. sin 30     Thus, one triangle has B1  418 , C1  1082 , and c1  1425; the other has  B2  1382 ,  C2  118 , and c2  307. 100 sin 50 27. sin B   1532. Since sin   1 for all , there can be no such angle B, and thus no such triangle. 50 80 sin 135 28. sin B   0566   B1  sin1 0566  344 or  B2  180  344  1456 . 100 100 sin 106 If  B1  344 , then  C  180  135  344  106 and c   259. sin 135 If  B2  180  344  1456 , then  A   B2  135  1456  180 , so there is no such triangle. Thus, the only possible triangle is  B  344 ,  C  106 , and c  259. 26 sin 29 29. sin A   0840   A  sin1 0840  572 or  A  180  572  1228 . Only the second solution 15 15 sin 281 satisfies  A  90 , so  B  180  29  1228  282 and b   146. sin 29 82 sin 58  0953, so  C  sin1 0953  724 or  C  180  724  1076 . Only the first solution 30. sin C  73 73 sin 496 satisfies  C  90 , so  A  180  58  724  496 and a   656. sin 58   sin 45 sin  31. Here  B  45 , AC  22  32  13, and B D  5. Thus, by the Law of Sines,    5 13   5 sin 45  787 .   sin1  13       sin 45 sin  3 2 sin 45  1 2 2   716 .     sin 32. Here A  45 , BC  1  3  10, and AB  3 2, so    10 10 3 2 25. sin C 

sin B 28 sin 30 sin 30   sin B   07, so 20 28 20  B  sin1 07  44427 . Since BC D is isosceles,  B   B DC  44427 . Thus,  BC D  180  2  B  91146  911 .

33. (a) From ABC and the Law of Sines we get

(b) From ABC we get  BC A  180   A   B  180  30  44427  105573 . Hence  DC A   BC A   BC D  105573  91146  144 . 12 sin 25  525. 34. By symmetry,  DC B  25 , so  A  180  25  50  105 . Then by the Law of Sines, AD  sin 105 35. (a) Let a be the distance from satellite to the tracking station A in miles. Then the subtended angle at the satellite is 50 sin 842  C  180  93  842  28 , and so a   1018 mi. sin 28 (b) Let d be the distance above the ground in miles. Then d  10183 sin 87  1017 mi.


SECTION 5.5 The Law of Sines

36. (a) Let x be the distance from the plane to point A. Then x 5

sin 48  377 mi. sin 100

(b) Let h be the height of the plane. Then sin 32 

200 sin 52  219 ft. sin 46

sin 48 x sin 48       AB sin 180  32  48  sin 100

h  h  377 sin 32  200 mi. x

37.  C  180  82  52  46 , so by the Law of Sines, have AC 

23

AC AB AB sin 52   AC  , so substituting we   sin 52 sin 46 sin 46

312 sin 486  0444   ABC  sin1 0444  264 , and so  BC A  180  486  264  105 . 527 527 sin 105  6785 ft. Then the distance between A and B is AB  sin 486

38. sin  ABC 

39.

We draw a diagram. A is the position of the tourist and C is the top of the tower.

C

 B  90  397  8603 and so  C  180  301  8603  6387 .

Thus, by the Law of Sines, the length of the tower is

3.97¡

B

30.1¡

100

40.

A

BC 

The situation is illustrated in the diagram.

C 165

180

67¡ B

100 sin 301  559 m. sin 6387

A

D

AC  165 sin 67  1519 ft, so using the Pythagorean Theorem  we can calculate B A  1652  15192  644 ft and  AD  1802  15192  966 ft. Thus the anchor points are B A  AD  644  966  161 ft apart.

41. The angle subtended by the top of the tree and the sun’s rays is  A  180  90  52  38 . Thus the height of the tree 215 sin 30 is h   175 ft. sin 38 42.

C

x

Œ

Let x be the length of the wire, as shown in the figure. Since   12 , other angles in ABC are   90  58  148 , and   180  12  148   20 .

Thus,

100 sin 148 x   x  100   155 m.   sin 148 sin 20 sin 20

º B 100

A 58¡

43. Call the balloon’s position R. Then in P Q R, we see that  P  62  32  30 , and  Q  180  71  32  141 . Q R P Q sin 30 Therefore,  R  180  30  141  9 . So by the Law of Sines,   Q R  60   192 m.   sin 30 sin 9 sin 9


24

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

44.

Label the diagram as shown, and let the hill’s angle of elevation be . Then

B º

applying the Law of Sines to ABC,

30

sin   4 sin 8  055669    sin1 055669  338 . But from AB D,

C

 B AD   B    8     90 , so   90  8  338  482 .

120

A

sin  sin 8   120 30

Œ

D

45. Let d be the distance from the earth to Venus, and let  be the angle formed by sun, Venus, and earth. By the Law sin 394 sin    0878, so either   sin1 0878  614 or   180  sin1 0878  1186 . of Sines, 1 0723 0723 d  d  1119 AU; in the second case,  In the first case, sin 180  394  614  sin 394 d 0723  d  0427 AU.  sin 180  394  1186  sin 394 sin 60 b sin 60 sin B  or sin B  . Similarly, applying the Law of b c c sin B sin 120 r sin 120 r b Sines to BC D gives  or sin B  . Since sin 120  sin 60 , we have   r cd cd c cd c sin D sin 60 b sin 60 b  (). Similarly, from ADC and the Law of Sines we have  or sin D  , and r cd b d d a sin 120 b sin 60 a sin 120 b d from B DC we have sin D  . Thus,    . Combining this with cd d cd a cd b c d cd b b ab r a b    1. Solving for r, we find  1      (), we get   r a cd cd cd r a a b ab ab . r ab 43  12 cm. (b) r  43 (c) If a  b, then r is infinite, and so the face is a flat disk.

46. (a) Applying the Law of Sines to ABC, we get

47. By the area formula from Section 5.3, the area of ABC is A  12 ab sin C. Because we are given a and the three sin A a sin B sin B  b  . Thus, angles, we need to find b in terms of these quantities. By the Law of Sines, b a sin A   a 2 sin B sin C a sin B A  12 ab sin C  12 a sin C  . sin A 2 sin A 48. By the area formula from Section 5.3,

1 ab sin C Area of  ABC sin C  12  , because a and b are the same for both    Area of  A B C sin C  ab sin C  2

triangles. 49.

C b

a

B

C b

B

A

a  b: One solution

B

a

C b

a» B

b  a  b sin A: Two solutions

A

C b

a B

a  b sin A: One solution

A

a  b sin A: No solution


SECTION 5.6 The Law of Cosines

25

 A  30 , b  100, sin A  1 . If a  b  100 then there is one triangle. If 100  a  100 sin 30  50, then there are 2

two possible triangles. If a  50, then there is one (right) triangle. And if a  50, then no triangle is possible.

5.6

THE LAW OF COSINES

1. For triangle ABC with sides a, b, and c the Law of Cosines states c2  a 2  b2  2ab cos C.

2. The Law of Cosines is required to solve triangles in cases SSS and SAS.  3. For ABC we have x 2  32  42  2 3 4 cos 35 , so x  25  24 cos 35  23.

13 62  52  32 13  299 .     cos1 15 265 15  5. x 2  212  422  2  21  42  cos 39  441  1764  1764 cos 39  834115 and so x  834115  289.  6. x 2  152  182  2  15  18  cos 108  225  324  540 cos 108  715869 and so x  715869  268.  7. x 2  252  252  2  25  25  cos 140  625  625  1250 cos 140  2207556 and so x  2207556  47.  8. x 2  22  82  2  2  8  cos 88  4  64  32 cos 88  66883 and so x  66883  82.

4. For P Q R we have 32  62  52  2 6 5 cos , so cos  

9. 37832  68012  42152  2  6801  4215  cos . Then cos     cos1 0867  2989 . 10. 15462  6012  12252  2  601  1225  cos . Then cos     cos1 0359  1110 .

37832  68012  42152  0867  2  6801  4215

15462  6012  12252  0359  2  601  1225

11. x 2  242  302  2  24  30  cos 30  576  900  1440 cos 30  228923 and so x 

 228923  151.

156 202  102  122   065    cos1 065  13054 . 2  10  12 240  13. c2  102  182  2  10  18  cos 120  100  324  360 cos 120  604 and so c  604  24576. Then 18 sin 120 sin A   0634295   A  sin1 0634295  394 , and  B  180  120  394  206 . 24576 12. 202  102  122  2  10  12  cos . Then cos  

14. 122  402  442  2  40  44  cos B  cos B 

122  402  442  0964   B  cos1 0964  155 . Then 2  40  44

40 sin 155  0891   A  sin1 0891  630 , and so  C  180  155  630  1015 . 12  4 sin 53 15. c2  32  42  2  3  4  cos 53  9  16  24 cos 53  10556  c  10556  32. Then sin B   0983 325   B  sin1 0983  795 and  A  180  53  795  475 . sin A 

16. a 2  602  302  2  60  30  cos 70  3600  900  3600 cos 70  326873  30 sin 70  0493   a  326873  572. Then sin C  572  C  sin1 0493  295 , and  B  180  70  295  805 . 2

2

2

25 22  0644   A  cos1 0644  499 . Then 17. 202  252  222  2  25  22  cos A  cos A  20 22522

sin B  25 sin20499  0956   B  sin1 0956  729 , and so  C  180  499  729  572 . 

2

2

2

12 16  078125  18. 102  122  162  2  12  16  cos A  cos A  10 21216   A  cos1 078125  386 . Then sin B  12 sin 386  0749  10

 B  sin1 0749  485 , and so  C  180  386  485  929 .


26

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

sin 40  0833   C  sin1 0833  564 or  C  180  564  1236 . 19. sin C  162125 1 2 

sin 836  1932. If  C1  564 , then  A1  180  40  564  836 and a1  125sin 40 

sin 164  550. If  C2  1236 , then  A2  180  40  1236  164 and a2  125sin 40 Thus, one triangle has  A  836 ,  C  564 , and a  1932; the other has  A  164 ,  C  1236 , and a  550. 

52  1024. Since sin   1 for all , there is no such  A, and hence there is no such triangle. 20. sin A  65 sin 50 

55  1065. Since sin   1 for all , there is no such  B, and hence there is no such triangle. 21. sin B  65 sin 50 

sin 61  1094 and c  735 sin 83  1241. 22.  A  180  61  83  36 . Then b  735 sin 36 sin 36 

sin 35  20. 23.  B  180  35  85  60 . Then x  3sin 60 

24. x 2  102  182  2  10  18  cos 40  100  324  360 cos 40  148224 and so x 

sin 30 25. x  50 sin 100  254 

2

2

 148224  122.

2

10 11  205  0932    cos1 0932  213 . 26. 42  102  112  2  10  11  cos . Then cos   4 21011 220

27. b2  1102  1382  2 110 138  cos 38  12,100  19,044  30,360 cos 38  72200 and so b  850. Therefore, 2

2

85 128 using the Law of Cosines again, we have cos   1102110138

2

   8915 .

40  0803    sin1 0803  535 or   180  535  1265 , but 535 doesn’t fit the 28. sin   10 sin 8 

picture, so   1265 .

29. x 2  382  482  2  38  48  cos 30  1444  2304  3648 cos 30  588739 and so x  243. sin 98  11808. 30.  A  180  98  25  57 . Then x  1000 sin 57 

31. The semiperimeter is s  91215  18, so by Heron’s Formula the area is 2   A  18 18  9 18  12 18  15  2916  54. 32. The semiperimeter is s  122  52 , so by Heron’s Formula the area is 2        15 A  52 52  1 52  2 52  2  15 16  4  0968.

 12, so by Heron’s Formula the area is 33. The semiperimeter is s  789 2    A  12 12  7 12  8 12  9  720  12 5  268.

34. The semiperimeter is s  11100101  106, so by Heron’s Formula the area is 2    A  106 106  11 106  100 106  101  302,100  10 3021  5496.

35. The semiperimeter is s  346  13 2 2 , so by Heron’s Formula the area is        13  3 13  4 13  6  455  455  533. A  13 2 2 2 2 16 4

 6, so by Heron’s Formula the area of 36. The smaller triangles have the same area. The semiperimeter of each is s  255 2     each is A  6 6  2 6  5 6  5  24  2 6, so the shaded area is 4 6  980.

37. We draw a diagonal connecting the vertices adjacent to the 100 angle. This forms two triangles. Consider the triangle with sides of length 5 and 6 containing the 100 angle. The area of this triangle is A1  12 5 6 sin 100  1477. To use

Heron’s Formula to find the area of the second triangle, we need to find the length of the diagonal using the Law of Cosines: c2  a 2  b2  2ab cos C  52  62  2  5  6 cos 100  71419  c  845. Thus the second triangle has semiperimeter  8  7  845  117255 and area A2  117255 117255  8 117255  7 117255  845  2600. The area s 2 of the quadrilateral is the sum of the areas of the two triangles: A  A1  A2  1477  2600  4077.


SECTION 5.6 The Law of Cosines

27

38. We draw a line segment with length x bisecting the 60 angle to create two triangles. By the Law of Cosines,    32  42  x 2  2  4x cos 30  x 2  4 3x  7  0. Using the Quadratic Formula, we find x  2 3  5. The minus sign provides the correct length of about 123 (the other solution is about 57, which corresponds to a convex quadrilateral   342 3 5 with the same side lengths), so the semiperimeter of each triangle is s  and the total area of the figure 2      is A  2 s s  3 s  4 s  2 3  5  246. 39.

Label the centers of the circles A, B, and C, as in the figure. By the Law of

C

6

5

6

B 5

4

4 A

Cosines, cos A 

92  102  112 AB 2  AC 2  BC 2   13   A  7053 . 2 AB AC 2 9 10

Now, by the Law of Sines,

sin 7053 sin B sin C   . So 11 AC AB

 1 085710  5899 and sin B  10 11 sin 7053  085710  B  sin

9 sin 7053  077139  C  sin1 077139  5048 . The area of sin C  11

ABC is 12 AB AC sin A  12 9 10 sin 7053   42426.

  2  7053  9848. Similarly, the areas of sectors B and C   4 360 360 are S B  12870 and SC  15859. Thus, the area enclosed between the circles is A  ABC  S A  S B  SC 

The area of sector A is given by S A   R 2 

A  42426  9848  12870  15859  385 cm2 .

 40. By the Law of Cosines we have a 2  62  42  2 6 4 cos 45  52  24 2,     b2  62 x 2 2 6 x cos 30  x 2 6 3x36, and c2  x 2 42 2 x 4 cos 30  45   x 2 2 2  6 x16.        By the Pythagorean Theorem, a 2  b2  c2 , so we have 52  24 2  x 2  6 3x  36  x 2  2 2  6 x  16       12 3  2          26  12 2  3 3x  18  2  6 x  8  36  12 2  3 3  2  6 x, so x     . 3 3 2 6

41. Let c be the distance across the lake, in miles. Then c2  2822  3562  2 282 356  cos 403  5313  c  230 mi. D

42. Suppose ABC D is a parallelogram with AB  DC  5, AD  BC  3, and  A  50 (see the figure). Since opposite angles are equal in a

parallelogram, it follows that  C  50 , and

C

3 50¡l

A B 5 260  130 . 2  By the Law of Cosines, AC 2  32  52  2 3 5  cos 130  AC  9  25  30  cos 130  73. Similarly,  B D  32  52  2 3 5  cos 50  38.

 B   D  360  100  260 . Thus,  B   D 

43. In half an hour, the faster car travels 25 miles while the slower car travels 15 miles. The distance between them is given by the Law of Cosines: d 2  252  152  2 25 15  cos 65    d  252  152  2 25 15  cos 65  5 25  9  30  cos 65  231 mi.

44. Let x be the car’s distance from its original position. Since the car travels

x

at a constant speed of 40 miles per hour, it must have traveled 40 miles east, and then 20 miles northeast (which is 45 east of “due north”). From the diagram, we see that    135 , so   x  202  402  2 20 40  cos 135  10 4  16  16  cos 135  560 mi.

º 45¡ 40

20


28

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

45. The airplane travels a distance of 625  15  9375 miles in its original

direction and 625  2  1250 miles in the new direction. Since it makes a course correction of 10 to the right, the included angle is

10¡

937.5

1250

d

180  10  170 . From the figure, we use the Law of Cosines to get

the expression d 2  93752  12502  2 9375 1250  cos 170  4,749,54942, so d  2179 miles. Thus, the airplane’s distance from its original position is approximately 2179 miles.

46. Let d be the distance between the two boats in miles. After one hour, the boats have traveled distances of 30 miles and 26 miles. Also, the angle subtended by their directions is 180  50  70  60 . Then  d 2  302  262  2  30  26  cos 60  796  d  796  282. Thus the distance between the two boats is about 28 miles. 47. (a) The angle subtended at Egg Island is 100 . Thus using the Law of

Forrest Island

Cosines, the distance from Forrest Island to the boat’s home port is x 2  302  502  2  30  50  cos 100

 900  2500  3000 cos 100  3920945  and so x  3920945  6262 miles.

10¡

(b) Let  be the angle shown in the figure. Using the Law of Sines, sin  

50 sin 100

 07863    sin1 07863  518 . Then

50

x

80¡ Egg Island

20¡

6262   90  20  518  182 . Thus the bearing to the boat’s home port

¬

30

70¡ Home Port

is S 182 E.

48. (a) In 30 minutes the plane flies 100 miles due east, so using the Law of

B

Cosines we have x 2  1002  3002  2  100  300  cos 40

300

 1002 1  9  6 cos 40   1002 5404. Thus, x 

 1002 5404  2325, and so the plane is 2325 miles from

its destination.

(b) Using the Law of Sines, sin  

A

50¡ 40¡

x ¬

100

300 sin 40  0829  2325

  sin1 0829  56 . However, since   90 , the angle we seek is

180  56  124 . Hence the bearing is 124  90  34 , that is, N 34 E.

49. The largest angle is the one opposite the longest side; call this angle . Then by the Law of Cosines, 442  362  222  2 36 22  cos   cos  

362  222  442  009848    cos1 009848  957 . 2 36 22

50. Let  be the angle formed by the cables. The two tugboats and the barge form a triangle: the side opposite  has a length of 120 ft and the other two sides have lengths of 212 and 230 ft. Therefore, 1202  2122  2302  2 212 230  cos   cos  

2122  2302  1202  cos   08557    cos1 08557  31 . 2 212 230

51. Let d be the distance between the kites. Then d 2  3802  4202  2 380 420  cos 30   d  3802  4202  2 380 420  cos 30  211 ft.


SECTION 5.6 The Law of Cosines

29

52. Let x be the length of the wire and  the angle opposite x, as shown in the figure. Since the mountain is inclined 32 , we must have

125

x

  180  90  32   122 . Thus,  x  552  1252  2 55 125  cos 122  161 ft.

¬ 55 32¡

53. Solution 1: From the figure, we see that   106 and sin 74 

3400 b

3400  3537. Thus, x 2  8002  35372  2 800 3537 cos 106 sin 74  x  8002  35372  2 800 3537 cos 106  x  3835 ft.

b

Solution 2: Notice that tan 74 

3400 a

a

Pythagorean Theorem, x 2  a  8002  34002 . So  x  9749  8002  34002  3835 ft.

x

3400  9749. By the tan 74

b

3400

74¡

800

54. Let the person be at point A, the first landmark (at 62 ) be at point B, and the other landmark be at point C. We 1150 1150 1150  AB    2450. Similarly, cos 54  want to find the length BC. Now, cos 62  AB cos 62 AC 1150  1956. Therefore, by the Law of Cosines, BC 2  AB 2  AC 2  2 AB AC  cos 43  AC  cos 54  BC  24502  19562  2 2450 1956  cos 43  BC  1679. Thus, the two landmarks are roughly 1679 feet apart.

 112  148  190 abc   225. Thus, 55. By Heron’s formula, A  s s  a s  b s  c, where s  2 2  A  225 225  112 225  148 225  190  82777 ft2 . Since the land value is $20 per square foot, the value of the lot is approximately 82777  20  $165,554.

56. Having found a  132 using the Law of Cosines, we use the Law of Sines to find  B:

sin 465 sin B   105 132

105 sin 465  0577. Now there are two angles  B between 0 and 180 which have sin B  0577, namely 132  B  352 and  B  1448 . But we must choose  B , since otherwise  A   B  180 . 1 2 1 sin 465 180 sin 465 sin C   sin C   0989, so either  C  815 Using the Law of Sines again, 180 132 132 or  C  985 . In this case we must choose  C  985 so that the sum of the angles in the triangle is  A   B   C  465  352  985  180 . (The fact that the angles do not sum to exactly 180 , and the discrepancies between these results and those of Example 3, are due to roundoff error.) The method in this exercise is slightly easier computationally, but the method in Example 3 is more reliable. sin B 

57. In any ABC, the Law of Cosines gives a 2  b2 c2 2bccos A, b2  a 2 c2 2accos B, and c2  a 2 b2 2abcos C. Adding the second and third equations gives b2  a 2  c2  2ac  cos B

c2  a 2  b2  2ab  cos C

b2  c2  2a 2  b2  c2  2a c cos B  b cos C


30

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

Thus 2a 2  2a c cos B  b cos C  0, and so 2a a  c cos B  b cos C  0. Since a  0 we must have a  c cos B  b cos C  0  a  b cos C  c cos B . The other laws follow from the symmetry of a, b, and c.

CHAPTER 5 REVIEW   7  183 rad 2. (a) 105  105  180 12

    052 rad 1. (a) 30  30  180 6

  5  262 rad (b) 150  150  180 6

  2  126 rad (b) 72  72  180 5

   9  707 rad (c) 405  405  180 4

     035 rad (c) 20  20  180 9

   5  393 rad (d) 225  225  180 4

   7  550 rad (d) 315  315  180 4

 3. (a) 56 rad  56  180   150

 4. (a)  53 rad   53  180   300

 180  (b)   9 rad   9    20

 (b) 109 rad  109  180   200

 (c)  43 rad   43  180   240

900  (c) 5 rad  5  180      2865

720  (d) 4 rad  4  180     2292

 (d) 113 rad  113  180   660

5. r  10 m,   25 rad. Then s  r  10  25  4  126 m. s 7  28 rad  1604 6. s  7 cm, r  25 cm. Then    25 r   5 rad. Then r  s  25  18  90  286 ft. 7. s  25 ft,   50  50  180  18 5 

  13 rad. Then r  s  13  18  18 m. 8. s  13 m,   130  130  180 18 13 

9. Since the diameter is 28 in, r  14 in. In one revolution, the arc length (distance traveled) is s  r  2  14  28 in. The total distance traveled is 60 mi/h  05 h  30 mi  30 mi  5280 ft/mi  12 in./ft  1,900,800 in. The number of 1 rev revolution is 1,900,800 in   216087 rev. Therefore the car wheel will make approximately 21,609 revolutions. 28 in. s 2450   35448 and so the angle is 10. r  3960 miles, s  2450 miles. Then     0619 rad  0619  180 r 3960 approximately 354 . 11. r  5 m,   2 rad. Then A  12 r 2   12  52  2  25 m2 .     18,151 ft2 12. A  12 r 2   12 2002 52  180 

2A 2  125    250 625  04 rad  229 r2 252   2A 100 14. A  50 m2 and   116 rad. Thus, r   11  42 m.  13. A  125 ft2 , r  25 ft. Then  

6

150  2 rad  300 rad/min  9425 rad/min. The linear speed is 15. The angular speed is   1 min 150  2  8   2400 in./min  75398 in./min  6283 ftmin. 1  rad  7000 rad/min  21,9911 rad/min. 16. (a) The angular speed of the engine is e  35002 1 min

e  7000rad/min  09  (b) To find the angular speed  of the wheels, we calculate g    

  77778 rad/min  24,4346 rad/min.

(c) The speed of the car is the angular speed of the wheels times their radius: 77778 rad  11 in  60 min  1 mile  2545 mi/h. min 1 hr 63,360 in.


CHAPTER 5

Review

31

    17. r  52  72  74. Then sin   5 , cos   7 , tan   57 , csc   574 , sec   774 , and cot   75 . 74 74

18. x  19.

    3 , cos   91 , tan   3 , csc   10 , sec   10 , and cot   91 . 102  32  91. Then sin   10 10 3 3 91

x  cos 40 5

20. cos 35  21.

y  5 sin 40  321.

2  x  cos235  244, and tan 35  2y  y  2 tan 35  140. x

1  sin 20 x

22. cos 30 

x  5 cos 40  383, and 5y  sin 40

91

x  sin120  292, and xy  cos 20

2924  311. y  cosx20  09397

x  x  4 cos 30  346, and sin 30  xy  y  x sin 30  346  05  173. 4

23. A  90  20  70 , a  3 cos 20  2819, and b  3 sin 20  1026. A

24. C  90  60  30 , cos 60  20a 

a  20 cos 60   40, and tan 60  b20 

C

b

b  20 tan 60  3464. C

70¡ 3

20¡

a b B A

25. c 

 7  028379  163 , 252  72  24, A  sin1 25

 and C  sin1 24 25  12870  737 . 25 A

27. tan  

c

26. b 

C

30¡

a 60¡ 20

B

 5  03948  226 , 122  52  13, A  sin1 13

 and C  sin1 12 13  11760  674 . b

7 B

A

12

C 5 B

1 1 1 1 a   cot , sin    b   csc  a tan  b sin 

28. Let h be the height of the tower in meters. Then tan 2881 

h  h  1000 tan 2881  550 m. 1000

One side of the hexagon together with radial line segments through its endpoints

29. x

8

forms a triangle with two sides of length 8 m and subtended angle 60 . Let x be the

60¡

length of one such side (in meters). By the Law of Cosines,

8

x 2  82  82  2  8  8  cos 60  64

hexagon is 6x  6  8  48 m.

x  8. Thus the perimeter of the


32

CHAPTER 5 Trigonometric Functions: Right Triangle Approach y

30. As the crankshaft moves in its circular pattern, point Q is

y

determined by the angle , namely it has coordinates Q 2 cos , 2 sin . We split the triangle into two right triangles O Q R and P Q R, as shown in the figure. Let h be the height of the piston. We consider two cases, 0    180 and 180    360 .

If 0    180 , then h is the sum of O R and R P. Using the  Pythagorean Theorem, we find R P  82  2 cos 2 , while

8

h

h

8 R ¬ O 2

Q(2 cos ¬, 2 sin ¬) x

Q(2 cos ¬, 2 sin ¬)

O R is the y­coordinate of the point Q, 2 sin . Thus  h  64  4 cos2   2 sin .

¬

O 2 R

x

 If 180    360 , then h is the difference between R P and R O. Again, R P  64  4 cos2  and O R is the  y­coordinate of the point Q, 2 sin . Thus h  64  4 cos2   2 sin . Since sin   0 for 180    360 , this also  reduces to h  64  4 cos2   2 sin .  Since we get the same result in both cases, the height of the piston in terms of  is h  64  4 cos2   2 sin .

r   ,   0518  r  r  236,900sin 0259 2 r  AB 236,900  sin 0259  r 1  sin 0259   236,900  sin 0259  r   1076 and so the radius of the moon is 1  sin 0259 roughly 1076 miles.

31. Let r represent the radius, in miles, of the moon. Then sin

32. Let d1 represent the horizontal distance from a point directly below the plane to the closer ship in feet, and d2 represent the 35,000 35,000 35,000  d1  , and similarly tan 40   horizontal distance to the other ship in feet. Then tan 52  d1 tan 52 d2 35,000 35,000 35,000 d2  . So the distance between the two ships is d2  d1    14,400 ft. tan 40 tan 40 tan 52   34. csc 94  csc  33. sin 315   sin 45   1   22 4  2 2

35. tan 135   tan 45  1    1   3 37. cot  223  cot 23  cot    3 3 3

39. cos 585  cos 225   cos 45   1   22 2

2 3 2 41. csc 83  csc 23  csc  3  3  3

3 36. cos 56   cos  6  2

38. sin 405  sin 45  1  22 2 40. sec 223  sec 43   sec  3  2 

2 3 42. sec 136  sec  6  3

 43. cot 390   cot 30    cot 30   3 44. tan 234  tan 34   tan  4  1   5 12 13 13 45. r  52  122  169  13. Then sin   12 13 , cos    13 , tan    5 , csc   12 , sec    5 , and 5 . cot    12

46. If  is in standard position, then the terminal point of  on the unit circle is simply cos  sin . Since the terminal point is    given as  23  12 , sin   12 .     47. y  3x  1  0  y  3x  1, so the slope of the line is m  3. Then tan   m  3    60 .   48. 4y  2x  1  0  y  12 x  14 . The slope of the line is m  12 . Then tan   m  12 and r  12  22  5. So   sin    1 , cos    2 , tan   12 , csc    5, sec    25 , and cot   2. 5

5


CHAPTER 5

Review

33

  sin  1  cos2  2  . 49. Since sin  is positive in quadrant II, sin   1  cos  and we have tan   cos  cos  50. sec  

1 1   (because cos   0 in quadrant III). cos   1  sin2 

51. tan2  

sin2  sin2   2 cos  1  sin2 

52. csc2  cos2  

1 sin2 

 cos2  

1  sin2  sin2 

1 sin2 

1

53. tan   37 , sec   43 . Then cos   34 and sin   tan   cos   47 , csc   4  4 7 7 , and cot   3  3 7 7 . 7 7 41 9 40 54. sec   41 40 , csc    9 . Then sin    41 , cos   41 , tan  

9 sin   41 9 , and cot    40 .  40   40 9 cos  41

  55. sin   35 . Since cos   0,  is in quadrant II. Thus, x   52  32   16  4 and so cos    45 , tan    34 , csc   53 , sec    54 , cot    43 .

5 56. sec    13 5 and tan   0. Then cos    13 , and  must be in quadrant III  sin   0. Therefore,   25   12 , tan   sin   12 , csc    13 , and cot   5 . sin    1  cos2    1  169 13 5 12 12 cos 

 4  cos    45   2 since 5 5   sin  cos   0 in quadrant II. But tan     12  sin    12 cos    12  2  1 . Therefore, 5 5 cos     5 1 2 1 sin   cos            5 .

57. tan    12 . sec2   1  tan2   1  14  54  cos2  

5

5

5

 1 1 1 sin   1 sin   1 sin  2 3  2 1     3.  58. sin   2 for  in quadrant I. Then tan   sec    cos  cos  cos  2 1  sin2  1  12 59. By the Pythagorean Theorem, sin2   cos2   1 for any angle . 

3 5 10 5 5  60. cos    23 and  2    . Then   6  2  6  3 . So sin 2  sin 3   sin 3   2 . 

61. sin1 23   3

62. tan1 33   6

63. Let u  sin1 25 and so sin u  25 . Then from the triangle, 64. Let u  cos1 83 then cos u  38 . From the triangle, we      tan sin1 25  tan u  2 . have sin cos1 38  sin u  855 . 21

5

2

u Ï21

8 u

Ï55

3


34

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

65. Let   tan1 x

 tan   x. Then from the   x . triangle, we have sin tan1 x  sin    1  x2

66. Let   sin1 x. Then sin   x. From the triangle, we   1 have csc sin1 x  csc   . x

Ï1+x@

1

¬

x

¬

1

Ï1-x@

x  x    tan1 2 2  10 sin 30 69.  B  180  30  80  70 , and so by the Law of Sines, x   532. sin 70 2 sin 45 70. x   146 sin 105 67. cos  

x  x    cos1 3 3

x

68. tan  

71. x 2  1002  2102  2  100  210  cos 40  21,926133  x  14807 20 sin 60  0247   B  sin1 0247  1433 . Then  C  180  60  1433 10567 , and so 72. sin B  70 70 sin 10567  7782. x sin 60  73. x 2  22  82  2 2 8 cos 120  84  x  84  917 6 sin 3121 4 sin 110  33.  0626   B  3879 . Then  C  180 110 3879  3121 , and so x  74. sin B  6 sin 110      sin 25 23 sin 25 sin  23 sin 25   sin      sin1  541 or 75. By the Law of Sines, 23 12 12 12   180  541  1259 . Because 1259 does not fit the diagram, we must have   541 .   sin 80 4 sin 80 sin  4 sin 80 1   sin      sin  520 . 76. By the Law of Sines, 4 5 5 5 77. By the Law of Cosines, 1202  1002  852  2 100 85 cos , so cos     cos1 016618  804 .

1202  1002  852  016618. Thus, 2 100 85

sin 10 5 sin 10 sin A   sin A   02894, so A  sin1 02894  168 5 3 3 or A  180  168  1632 . Therefore,   180  10  168  1532 or   180  10  1632  68 .

78. We first use the Law of Sines to find  A:

79. After 2 hours the ships have traveled distances d1  40 mi and d2  56 mi. The subtended angle is 180  32  42  106 . Let d be the distance between the two ships in miles. Then by the Law of Cosines, d 2  402  562  2 40 56 cos 106  5970855  d  773 miles.

80. Let h represent the height of the building in feet, and x the horizontal distance from the building to point B. Then h h h tan 241  and tan 302   x  h cot 302 . Substituting for x gives tan 241   x  600 x h cot 302  600  600  tan 241 h  tan 241 h cot 302  600  h   1160 ft. 1  tan 241 cot 302 81. Let d be the distance, in miles, between the points A and B . Then by the Law of Cosines, d 2  322  562  2 32 56 cos 42  14966  d  39 mi. 120 sin 689  12008 miles. Let d be the shortest distance, in miles, to 82.  C  180  423  689  688 . Then b  sin 688  the shore. Then d  b sin A  12008 sin 423  808 miles.


CHAPTER 5

Test

83. A  12 ab sin   12 8 14 sin 35  3212

 568 abc 84. By Heron’s Formula, A  s s  a s  b s  c, where s    95. Thus, 2 2  A  95 95  5 95  6 95  8  1498.   85. (a) csc    2, so sin    22 . The sine function is negative for terminal points below the x­axis, and the reference angle in this case is  4 , so this corresponds to graph VII. 

(b) sin   1  22 has reference angle  4 and its terminal point lies above the x­axis, so this corresponds to graph III. 2   (c)   sin1  35  sin    35 , which corresponds to graph I.

(d) tan   43 corresponds to graph VI.

(e) tan    32 corresponds to graph V, since tangent is negative in quadrant III. (f) sin    2 corresponds to VIII, since 5

4 y 2    . r 5 22  42

  2 x 1    , which lies in the (g)   cos1  1  cos    1 corresponds to graph II since   5 5 r 5 2 2 2  4 domain of the inverse cosine function.

(h) sin   35 corresponds to graph IV since that terminal point has

3 y 3   . 2 2 r 5 3 4

CHAPTER 5 TEST   11 rad. 135  135     3 rad. 1. 330  330  180 6 180 4

180 234   2. 43 rad  43  180   240 . 13 rad  13       745

3. (a) The angular speed is 

120  2 rad  240 rad/min  75398 rad/min. 1 min

(b) The linear speed is 120  2  16  3840 ft/min 1  12,0637 ft/min

 

4. (a) sin 405  sin 45  1  22 2

(b) tan 150   tan 30  1  33 3 (c) sec 53  sec  3 2

(d) csc 52  csc  2 1

 137 mi/h

   2 13  3   2 2 26  6 13 5. r  32  22  13. Then tan   sin      .   3 39 13 3 13 b a  a  24 sin . Also, cos    b  24 cos . 6. sin   24 24   7. cos    13 and  is in quadrant III, so r  3, x  1, and y   32  12  2 2. Then

  1 23  43 2. 3  2   csc   1   4 2 2 2 2 tan  1 sin  1 5 13 13 1 5 , tan    5 . Then sec      tan      . 8. sin   13 12 cos  cos  sin  sin  12 5 12   2 2 2 2 9. sec   1  tan   tan    sec   1. Thus, tan    sec   1 since tan   0 in quadrant II.

tan  cot   csc   tan  

35


36

CHAPTER 5 Trigonometric Functions: Right Triangle Approach

h  h  6 tan 73  196 ft. 6 x  x 11. (a) tan      tan1 4 4   3 3 (b) cos      cos1 x x 10. tan 73 

9 so tan u  9 . From the triangle, r  12. Let u  tan1 40 40   40 9 1 cos tan 40  cos u  41 .

 92  402  41. So

41 u

9

40

13. By the Law of Cosines, x 2  102  122  2 10 12  cos 48  8409  x  91. 14.  C  180  52  69  59 . Then by the Law of Sines, x 

230 sin 69  2505. sin 59

h x h  h  50 tan 20 and tan 28   x  h  50 tan 28 50 50  x  50 tan 28  h  50 tan 28  50 tan 20  84. 15 sin 108 16. Let  A and  X be the other angles in the triangle. Then sin A   0509   A  3063 . Then 28 28 sin 4137  X  180  108  3063  4137 , and so x   195. sin 108 82  62  92  01979, so   cos1 01979  786 . 17. By the Law of Cosines, 92  82 62 2 8 6 cos   cos   2 8 6 15. Let h be the height of the shorter altitude. Then tan 20 

18. We find the length of the third side x using the Law of Cosines: x 2  52  72  2 5 7 cos 75  5588  x  7475. sin 75 5 sin 75 sin    sin    06461, so   sin1 06461  402 . Therefore, by the Law of Sines, 5 7475 7475   50  72 . A triangle  1 r  r sin   1  102 sin 72 . Thus, the area of the 19. (a) A sector  12 r 2   12  102  72  180 180 2  2  72   shaded region is A shaded  A sector  A triangle  50 180  sin 72  153 m2 .

(b) The shaded region is bounded by two pieces: one piece is part of the triangle, the other is part of the circle.   The first part has length l  102  102  2 10 10  cos 72  10 2  2  cos 72 . The second has length   4. Thus, the perimeter of the shaded region is p  l  s  102  2 cos 72  4  243 m. s  10  72  180

20. (a) If  is the angle opposite the longest side, then by the Law of Cosines cos  

92  132  202  06410. Therefore, 2 9 20

  cos1 06410  1299 .

(b) From part (a),   1299 , so the area of the triangle is A  12 9 13 sin 1299  449 units2 . Another way to find  9  13  20 abc   21. Thus, the area is to use Heron’s Formula: A  s s  a s  b s  c, where s  2 2   A  21 21  20 21  13 21  9  2016  449 units2 . 21. Label the figure as shown. Now    85  75  10 , so by the Law of Sines, 100 x  sin 75 sin 10

x  100 

sin 75 h . Now sin 85  sin 10 x

sin 75 sin 85  554 ft. h  x sin 85  100  sin 10

º

x 75¡ 100

85¡

h


Surveying

37

FOCUS ON MODELING Surveying 1. Let x be the distance between the church and City Hall. To apply the Law of Sines to the triangle with vertices at City Hall, the church, and the first bridge, we first need the measure of the angle at the first bridge, which is 180  25  30  125 . x 086 086 sin 125 Then  x   14089. So the distance between the church and City Hall is about   sin 125 sin 30 sin 30 141 miles. 2. To find the distance z between the fire hall and the school, we use the distance found in the text between the bank and the cliff. To find z we first need to find the length of the edges labeled x and y.

bank 50¡

In the bank­second bridge­cliff triangle, the third angle is

1.55

155 sin 50 180  50  60  70 , so x   126. sin 70 In the second bridge­school­cliff triangle, the third angle is

cliff

126 sin 45  109. sin 55 Finally, in the school­fire hall­cliff triangle, the third angle is

80¡

180  80  55  45 , so y 

109 sin 80  131. sin 55 Thus, the fire hall and the school are about 131 miles apart.

180  45  80  55 , so z 

fire hall

55¡

z

70¡

x

60¡

second bridge

45¡

80¡ y 45¡

55¡ school

equal

3. First notice that  D BC  180  20  95  65 and  D AC  180  60  45  75 . AC 20 20 sin 45   AC   146 . From BC D we get From AC D we get   sin 45 sin 75 sin 75 BC 20 20 sin 95 BC     220. By applying the Law of Cosines to ABC we get sin 95 sin 65 sin 65  AB2  AC2 BC2 2 AC BC cos 40  1462 2202 2146220cos 40  205, so AB  205  143 m. Therefore, the distance between A and B is approximately 143 m. 4. Let h represent the height in meters of the cliff, and d the horizontal distance to the cliff. The third horizontal 200 sin 516 h angle is 180  694  516  59 and so d    182857. Then tan 331  sin 59 d   h  d tan 331  182857 tan 331  1192 m. AB BC  sin  sin    sin  d sin  d sin  sin   BC  AB  . Thus, h  BC sin   . sin    sin  cos   sin  cos  cos  sin   cos  sin  d tan  tan  Dividing numerator and denominator by cos  cos , we have h  . tan   tan  800 tan 25 tan 29 d tan  tan    2350 ft (b) h  tan   tan  tan 29  tan 25

5. (a) In ABC,  B  180  , so  C  180    180      . By the Law of Sines,

R R h cos     Rh   R  h  R 1  cos   h cos   R  . With cos  1  cos  sin    2 141 cos 0021 h  141 km and   0021 rad, we have R   6390 km. 1  cos 0021

6. By the Law of Sines,


38

FOCUS ON MODELING

7. Let the surveyor be at point A, the first landmark (with angle of depression 42 ) be at point B, and the other landmark 2430 2430 2430  36316. Similarly, sin 39  be at point C. We want to find BC. Now sin 42   AB  AB AC sin 42 2430  38613. Therefore, by the Law of Cosines, BC2  AB2  AC2  2 AB AC cos 68   AC  sin 39  BC  363162  386132  2 36316 38613 cos 68  4194. Thus, the two landmarks are approximately 4194 ft apart. 8. We start by labeling the edges and calculating the remaining angles, as shown in the first figure. Using the Law of Sines, a 150 150 sin 29 b 150 150 sin 91 we find the following:  a   8397,  b  17318,      sin 29 sin 60 sin 60 sin 91 sin 60 sin 60   17318 17318 sin 32 d 17318 17318 sin 61 c  c   9190,  e   15167, sin 32 sin 87 sin 87 sin 61 sin 87 sin 87   e 15167 15167 sin 41 f 15167 15167 sin 88  e   12804,   f   19504, sin 41 sin 51 sin 51 sin 88 sin 51 sin 51   g 19504 19504 sin 50 h 19504 19504 sin 38  g  14950, and  h   12015. Note that sin 50 sin 92 sin 92 sin 38 sin 92 sin 92 we used two decimal places throughout our calculations. Our results are shown (to one decimal place) in the second figure.

29¡ 150

c

61¡

87¡ 88¡

91¡

60¡ a

32¡ 41¡

128.0

51¡ 38¡

d

b

91.9

e

g

f 50¡ h

92¡

150

173.2

84

151.7

195.0

149.5

120.2

9. Answers will vary. Measurements from the Great Trigonometric Survey were used to calculate the height of Mount Everest to be exactly 29,000 ft, but in order to make it clear that the figure was considered accurate to within a foot, the height was published as 29,002 ft. The accepted figure today is 29,029 ft.


CORRECTIONS: p. 20,21,26,36,56,57

CHAPTER 6

TRIGONOMETRIC FUNCTIONS: UNIT CIRCLE APPROACH

6.1 6.2

The Unit Circle 1 Trigonometric Functions of Real Numbers 6

6.3

Trigonometric Graphs 11

6.4

More Trigonometric Graphs 26

6.5

Inverse Trigonometric Functions and Their Graphs 37

6.6

Modeling Harmonic Motion 40 Chapter 6 Review 46 Chapter 6 Test 54

¥

FOCUS ON MODELING: Fitting Sinusoidal Curves to Data 56

1


6

TRIGONOMETRIC FUNCTIONS: UNIT CIRCLE APPROACH

6.1

THE UNIT CIRCLE

1. (a) The unit circle is the circle centered at 0 0 with radius 1. (b) The equation of the unit circle is x 2  y 2  1.

(c) (i) Since 12  02  1, the point is P 1 0.

(ii) P 0 1

(iii) P 1 0

(iv) P 0 1

2. (a) If we mark off a distance t along the unit circle, starting at 1 0 and moving in a counterclockwise direction, we arrive at the terminal point determined by t.  (b) The terminal points determined by  2 , ,  2 , 2 are 0 1, 1 0, 0 1, and 1 0, respectively.

3. If the terminal point determined by t is P, then the terminal point determined by t  2 is P. The terminal point for t   3       3 3 1 7  1 and so the terminal point for t  . is 2  2 3 is 2  2    2 2 and the y The terminal point determined by t   is 4. 4 2  2 ¹ Ï2 Ï2 ; , 5¹     ; _Ï3 ,1 4 l2l l2l 6 l2l 2 terminal point determined by t  54 is  22   22 .   3 1 The terminal point determined by t    1 x 0 6 is 2   2 and the    Ï3 1 _¹6 ; l2l , _ 2 terminal point determined by t  56 is  23  12 .

(

5¹ ; 4

)

(_Ï2l2l , _Ï2l2l)

(

)

(

)

In general, if the terminal point determined by t is P a b, then by symmetry, the terminal point determined by t   is P  a b.

  2  2  9  16  1, P 3   4 lies on the unit circle. 5. Since 35   45  25 25 5 5 2      7 2  576  49  1, P  24   7 lies on the unit circle.   25 6. Since  24 25 625 25 25 25    2   2  9  7  1, P 3   7 lies on the unit circle. 7. Since 34   47  16 16 4 4     2   2 24  1, P  5   2 6 lies on the unit circle.  8. Since  57   2 7 6  25 7 7 49 49      2  2 9. Since  35  23  59  49  1, P  35  23 lies on the unit circle.   2  2   11 5 25 10. Since 611  56  11 36  36  1, P 6  6 lies on the unit circle.  2 9  y 2  16  y   4 . Since P x y is in quadrant III, y is negative, so the point is 11.  35  y 2  1  y 2  1  25 25 5   3 4 P 5 5 .   7 2  1  x 2  1  49  x 2  576  x   24 . Since P is in quadrant IV, x is positive, so the point is 12. x 2   25 625 625 25   24 7 P 25   25 .   2 13. x 2  13  1  x 2  1  19  x 2  89  x   2 3 2 . Since P is in quadrant II, x is negative, so the point is    P  2 3 2  13 .

1


2

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

14.

 2

4  y 2  21  y   21 . Since P is in quadrant I, y is positive, so the point is  y 2  1  y 2  1  25 25 5    21 2 P 5 5 . 2 5

  2 4  x 2  45  x   3 5 . Since P x y is in quadrant IV, x is positive, so the point is 15. x 2   27  1  x 2  1  49 7 49    P 3 7 5   27 .   2 16.  23  y 2  1  y 2  1  49  y 2  59  y   35 . Since P is in quadrant II, y is positive, so the point is    P  23  35 .

17.

   5 2  y 2  1  y 2  1  25  y 2  144  y   12 . Since its y­coordinate is negative, the point is P 5   12 . 13 169 169 13 13 13

  2  9  x 2  16  x   4 . Since its x­coordinate is positive, the point is P 4   3 . 18. x 2   35  1  x 2  1  25 25 5 5 5    2   19. x 2  23  1  x 2  1  49  x 2  59  x   35 . Since its x­coordinate is negative, the point is P  35  23 .    2 5  x 2  20  x   2 5 . Since its x­coordinate is positive, the point is  1  x 2  1  25 20. x 2   55 25 5     2 5 5 P 5  5 .

   2 21.  32  y 2  1  y 2  1  29  y 2  79  y   37 . Since P lies below the x­axis, its y­coordinate is negative, so     the point is P  32   37 .

  2 4  y 2  21  y   21 . Since P lies above the x­axis, its y­coordinate is positive, so 22.  25  y 2  1  y 2  1  25 25 5    21 2 the point is P  5  5 .

24.

23.

t

Terminal Point

0

1 0    2 2 2  2

 4

2 3 4

t

Terminal Point

t

Terminal Point

1 0      22   22

0 

1 0   3 1  2 2  1 3 2 2

5 4 3 2 7 4

0 1

     22  22

2

1 0

0 1



  2 2 2  2

1 0

6

 3

2 2 3 5 6

 25. t  0, so t  5 corresponds to P x y  1 0. y

P(_1, 0) t=5¹

Q(1, 0)

t=0 x

t

Terminal Point

1 0     23   12     12   23

7 6 4 3 3 2 5 3 11 6

0 1     12  23     23  12

2

1 0

0 1

  1 3 2 2   31 2 2

1 0

26. t  0, so t  3 corresponds to P x y  1 0. y

P(_1, 0)

t=_3¹

Q(1, 0)

t=0

x


SECTION 6.1 The Unit Circle

27. t  0, so t  4 corresponds to P x y  1 0. y

28. t  0, so t  6 corresponds to P x y  1 0. y

t=_4¹ t=0 P(1, 0) x

t=6¹ t=0 P(1, 0) x

3 29. t   2 , so t  2 corresponds to P x y  0 1. y ¹ Q(0, 1) t= 2

5 30. t   2 , so t  2 corresponds to P x y  0 1. y P(0, 1)

t= 2

¹

t= 2 x

t= 2

x

P(0, _1)

5 31. t   2 , so t   2 corresponds to P x y  0 1. y ¹ Q(0, 1) t= 2

3 32. t   2 , so t   2 corresponds to P x y  0 1. y

¹

t=_ 2 t= 2

P(0, 1) Q(0, 1)

x

x

5¹ P(0, _1)

t=_ 2

   3 1 5 33. t   6 and t  6 corresponds to P x y   2  2 . y

P(_ 2 , 2 )

 34. t   3 and t   3 corresponds to    P x y  12   23 . y

Ï3 1

t= 6

x

Q ( 21 , Ï3 2)

¹

t= 3

x ¹ t=_ 3

P ( 21 , _ Ï3 ) 2

3


4

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

3 35. t   4 and t   4 corresponds to     P x y   22   22 . y

5 36. t   4 and t  4 corresponds to     P x y   22   22 .

¹ 4

x

P (_ 2 , _ 2 ) Ï2

Ï2

x

t= 4

P(_ Ï2 , _ Ï2 ) 2 2

t=_ 4

   5 corresponds to P x y  1  3 . 37. t   and t   3 3 2 2 y

y

Q(Ï2 , Ï2 ) 2 2

t=_ 3

P ( 21 , Ï3 2)

¹

t= 3

7 38. t   6 and t   6 corresponds to    P x y   23  12 . y

P(_ 2 , 2 ) Ï3 1

x

x

t=_ 6

   2 corresponds to P x y   1  3 . 40. t   and t  3 3 2 2

7 39. t   4 and t  4 corresponds to    P x y  22   22 .

y

P (_ 21 , Ï3 2)

y

t= 3

t= 4

x x P( Ï2 , _ Ï2 ) 2 2

7 41. t   6 and t  6 corresponds to    P x y   23   12 .

7 42. t   4 and t   4 corresponds to    P x y  22  22 .

y

y

t= 6

Q( 2 , 2) Ï3 1

¹ 6

¹ 4

x

P(_ 2 , _2 ) Ï3

P( Ï2 , Ï2 2 2)

1

t=_ 4

x


SECTION 6.1 The Unit Circle

43. (a) t  43     3

44. (a) t  9  9  0   (b) t     54   4

(b) t  2  53   3   (c) t     76   6

(c) t  256  4   6

(d) t  35    036

(d) t  4    086

45. (a) t    57  27

46. (a) t  115  2   5

(c) t    3  0142

(c) t  2  6  0283

(d) t  2  5  1283

(d) t  7  2  0717

(b) t    79  29

(b) t  97    27

47. (a) t    34   4     2 2 (b) P  2  2

48. (a) t  54     4     2 2 (b) P  2  2

51. (a) t  2  116   6   3 1 (b) P 2   2

52. (a) t   6   (b) P 23   12

49. (a) t   56     6    3 1 (b) P  2   2

50. (a) t  43     3    3 1 (b) P  2   2

53. (a) t  134  3   4     2 2 (b) P  2   2

54. (a) t  136  2   6   3 1 (b) P 2  2

57. (a) t  4  113   3    3 1 (b) P 2  2

58. (a) t  316  5   6    3 1 (b) P  2   2   60. (a) t  10   414   4    2 2 (b) P 2   2

55. (a) t  7  416   6    3 1 (b) P  2  2

56. (a) t  174  4   4    2 2 (b) P 2  2

59. (a) t  163  5   3    3 1 (b) P  2   2

61. t  1  05 08

62. t  25 

63. t  11  05 09   65. Let Q x y  35  45 be the terminal point determined by t. y

¹+t

Q( 35 , 45) ¹-t

2¹+t

t x _t

64. t  42 

P 08 06

P 05 09   (a)   t determines the point P x y   35  45 .   (b) t determines the point P x y  35   45 .   (c)   t determines the point P x y   35   45 .   (d) 2  t determines the point P x y  35  45 .

5


6

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

   66. Let Q x y  34  47 be the terminal point determined by t. y ¹-t 4¹+t t-¹

Q( 34 , Ï7 4) t _t

x

   (a) t determines the point P x y  34   47 .    (b) 4  t determines the point P x y  34  47 .    (c)   t determines the point P x y   34  47 .    (d) t   determines the point P x y   34   47 .

67. The distances P Q and P R are equal because they both subtend arcs of length  3 . Since P x y is  a point on the unit circle, x 2  y 2  1. Now d P Q  x  x2  y  y2  2y and    d R S  x  02  y  12  x 2  y 2  2y  1  2  2y (using the fact that x 2  y 2  1). Setting these equal  gives 2y  2  2y  4y 2  2  2y  4y 2  2y  2  0  2 2y  1 y  1  0. So y  1 or y  12 . Since  2 P is in quadrant I, y  12 is the only viable solution. Again using x 2  y 2  1 we have x 2  12  1  x 2  34     x   23 . Again, since P is in quadrant I the coordinates must be 23  12 .      68. P is the reflection of Q about the line y  x. Since Q is the point Q 23  12 it follows that P is the point P 12  23 .

6.2

TRIGONOMETRIC FUNCTIONS OF REAL NUMBERS

1. If Px y is the terminal point on the unit circle determined by t, then sin t  y, cos t  x, and tan t  yx.

2. If Px y is on the unit circle, then x 2  y 2  1. So for all t we have sin2 t  cos2 t  1. So we can write cosine in terms   of sine as cos t   1  sin2 t and we can write sine in terms of cosine as sin t   1  cos2 t.

3. Because t3 is in quadrant IV, sin t3  0. Also, sin t2  sin t1 because on the interval 0  t  , the value of sin t is larger the closer t is to  2 . Thus, sin t3  sin t1  sin t2 .  4. Because t2 is in quadrant II, cos t2  0. Also, cos t1  cos t3 because on the interval   2  t  2 , the value of cos t is

larger the closer t is to 0. Thus, cos t2  cos t3  cos t1 .

6.

5. t 0

cos t 1

sin t 0

t 0

 2 2

 2 2

0

1

 22

 2 2

1

0

4

2 3 4

 5 4 3 2 7 4

 22

 22

0  2 2

1

  22

2

1

0

6 3

2 2 3 5 6

cos t

sin t

1

 3 2 1 2

t

cos t

sin t

0

1

0

1 2  3 2

0

1

 12

 3 2 1 2

7 6 4 3 3 2 5 3 11 6

1

0

2

 23

  23  12

0

 12 

 23 1 

1 2  3 2

 23

1

0

 12


SECTION 6.2 Trigonometric Functions of Real Numbers    7. (a) sin  23   23 

(b) cos 174  22

(c) tan 176   33

 11. (a) cos 34   22  (b) cos 54   22  (c) cos 74  22

 1  15. (a) cos   3  2   (b) sec   3 2    3   (c) sin   3 2

19. (a) csc 76  2  23  (b) sec   6  3    (c) cot  56  3

23. (a) sin 13  0

(b) cos 14  1 (c) tan 15  0

  8. (a) sin  56   12 

(b) cos 56   23  (c) tan 143   3

 12. (a) sin 34  22  (b) sin 54   22  (c) sin 74   22

  16. (a) tan   4  1    (b) csc   4  2   (c) cot   4  1  20. (a) sec 34   2   (b) cos  23   12    (c) tan  76   33

 9. (a) sin 134   22

   (b) cos  34   22 

(c) tan 76  33

 13. (a) sin 73  23  (b) csc 73  2 3 3  (c) cot 73  33  3  17. (a) cos   6  2

   2 3 (b) csc   3  3    3 (c) tan   6  3 

21. (a) sin 43   23

 10. (a) sin 113   23  (b) cos 116  23

(c) tan 74  1

 14. (a) csc 54   2  (b) sec 54   2

(c) tan 54  1    2 18. (a) sin   4  2    (b) sec  4  2    (c) cot   6  3

25. t  0  sin t  0, cos t  1, tan t  0, sec t  1, csc t and cot t are undefined. 26. t   2  sin t  1, cos t  0, csc t  1, cot t  0, tan t and sec t are undefined. 27. t  

28. t  32

 sin t  0, cos t  1, tan t  0, sec t  1, csc t and cot t are undefined.

 sin t  1, cos t  0, csc t  1, cot t  0, tan t and sec t are undefined.

3  2  2 5  3. 9 3 4 29.  45  35  16 25  25  1. So sin t  5 , cos t   5 , and tan t  4 4 5

30.

 2 3 5

22. (a) csc 23  2 3 3    (b) sec  53  2 (b) sec 116  2 3 3      3 10   1   (c) cot   (c) cos 3 3 3 2   25  24. (a) sin 2  sin 2  24  1   (b) cos 252  cos  2  24  0   (c) cot 252  cot  2  24  0

4  2 9  16  1. So sin t   4 , cos t  3 , and tan t   5   4 .   45  25 25 5 5 3 3 5

1  2   3 . 3  23    2   2   23 3 3 1 1 3 1  4  4  1. So sin t   2 , cos t   2 , and tan t   3. 32.  2   2 1 2  13    2   2 13 7   13 . 13 6 33.  67  713  36 49  49  1. So sin t  7 , cos t   7 , and tan t  6 6 7 9  2  2 9 1600  81  1. So sin t  9 , cos t  40 , and tan t  41  9 .   34. 40 41 41 1681 1681 41 41 40 40 41 12     5 2   12 2  25  144  1. So sin t   12 , cos t   5 , and tan t  13  12 . 35.  13 13 169 169 13 13 5 5  13  2 5     2   2 5  20  1. So sin t  2 5 , cos t  5 , and tan t  5  2. 36. 55  2 5 5  25 25 5 5 5 5    2  2 31.  23  12  34  14  1. So sin t  12 , cos t   23 , and tan t 

7


8

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

21 2  2  21 400  441  1. So sin t  21 , cos t   20 , and tan t  29   21 . 37.  20   29 29 841 841 29 29 20  20 29

38.

24 25

2

7   7 2  576  49  1. So sin t   7 , cos t  24 , and tan t   25   7 .   25 625 625 25 25 24 24

39. (a) 08 (b) 084147 43. (a) 10 (b) 102964

25

40. (a) 07 (b) 069671 44. (a) 36 (b) 360210

41. (a) 09 (b) 093204 45. (a) 06 (b) 057482

42. (a) 03 (b) 028366 46. (a) 09 (b) 088345

47. sin t  cos t. Since sin t is positive in quadrant IV and cos t is negative in quadrant IV, their product is negative. 48. sin t tan t is positive in quadrant IV because both sin t and tan t are negative in that quadrant. 1 tan t  sin t  tan t   sin t  tan t  tan t  sin t  tan2 t  sin t. Since tan2 t is always positive and sin t is negative in 49. cot t cot t quadrant III, the expression is negative in quadrant III. 1  1, provided cos t  0. 50. cos t  sec t is positive in any quadrant, since cos t  sec t  cos t  cos t 51. Quadrant II

52. Quadrant III

53. Quadrant II 54. Quadrant II  55. Because cosine is negative in quadrant III, cos t   1  sin2 t in that quadrant.  56. Because sine is negative in quadrant IV, sin t   1  cos2 t in that quadrant.  57. Because sine is positive in quadrant II, sin t  1  cos2 t in that quadrant.  1  cos2 t in that quadrant. 58. Because tangent is negative and cosine is positive in quadrant IV, tan t   cos t  1  cos2 t 59. Because tangent and cosine are both negative in quadrant II, tan t  in that quadrant. cos t sin t in that quadrant. 60. Because tangent is negative and sine is positive in quadrant II, tan t    1  sin2 t  61. Because tangent is negative in quadrant IV, tan t   sec2 t  1 in that quadrant.  62. Because secant is positive in quadrant IV, sec t  1  tan2 t in that quadrant.  63. Because cosecant is positive in quadrant II, csc t  1  cot2 t in that quadrant.    1 sec2 t  1 in that  64. Because sine and secant are both negative in quadrant III, sin t   1  cos2 t   1  sec t sec2 t quadrant. sin2 t sin2 t  . (Because all trigonometric functions in this expression are squared, the quadrant is 2 cos t 1  sin2 t immaterial.)   1 1  1  cos2 t   1. (Because all trigonometric functions in this expression are squared, the 66. sec2 t  sin2 t  2 cos t cos2 t quadrant is immaterial.)   67. sin t   45 and the terminal point of t is in quadrant IV, so the terminal point determined by t is P x  45 . Since P is    2 9 3 on the unit circle, x 2   45  1. Solving for x gives x   1  16 25   25   5 . Since the terminal point is   in quadrant IV, x  35 . Thus the terminal point is P 35   45 . Thus, cos t  35 , tan t   43 , csc t   54 , sec t  53 , 65. tan2 t 

cot t   34 .


SECTION 6.2 Trigonometric Functions of Real Numbers

9

  7 and the terminal point of t lies in quadrant III, so the terminal point determined by t is P  7  y . Since 68. cos t   25 25   2  7 49   576   24 . Since the terminal  y 2  1. Solving for y gives x   1  625 P is on the unit circle,  25 625 25   7 24 24 24 25 point is in quadrant III, y   24 25 . Thus the terminal point is P  25   25 . Thus, sin t   25 , tan t  7 , csc t   24 , 7 sec t   25 7 , cot t  24 .

  69. sec t  3 and the terminal point of t lies in quadrant IV. Thus, cos t  13 and the terminal point determined by t is P 13  y .     2 Since P is on the unit circle, 13  y 2  1. Solving for y gives y   1  19   89   2 3 2 . Since the terminal      point is in quadrant IV, y   2 3 2 . Thus the terminal point is P 13   2 3 2 . Therefore, sin t   2 3 2 , cos t  13 ,    3   3 2 , cot t    1   2. tan t  2 2, csc t    4 4 2 2

2 2

 2 1 1  17 . 70. tan t  14 and the terminal point of t lies in quadrant III. Since sec2 t  tan2 t 1 we have sec2 t  14 1  16 16    17 17 Thus sec t   17 16   4 . Since sec t  0 in quadrant III we have sec t   4 , so      1 4 1 4 17   1   1717 .       . Since tan t  cos t  sin t we have sin t  14  4 cos t  17 17 sec t 17 17  417        Thus, the terminal point determined by t is P  4 1717   1717 . Therefore, sin t   1717 , cos t   4 1717 , csc t   17, 

sec t   417 , cot t  4.

  12 2  1  169 , so because 2 2 2 71. tan t   12 5 and sin t  0, so t is in quadrant II. Since sec t  tan t  1 we have sec t   5 25    1 13 5 5 secant is negative in quadrant II, sec t   169 25   5 . Thus, cos t  sec t   13 and we have P  13  y . Since      5  12 . Thus, the terminal point determined by t is P  5  12 , and so  13 tan t  cos t  sin t we have sin t   12 5 13 13 13 5 13 13 5 sin t  12 13 , cos t   13 , csc t  12 , sec t   5 , cot t   12 .

  72. csc t  5 and cos t  0, so t is in quadrant II. Thus, sin t  15 and the terminal point determined by t is P x 15 . Since    2 1   2 6 . Since the terminal point is in P is on the unit circle, x 2  15  1. Solving for x gives x   1  25 5       2 6 2 6 1 1 1   6, quadrant II, x   5 and the terminal point is P  5  5 . Thus, sin t  5 , cos t   2 5 6 , tan t    12 2 6   5 6 5 sec t      12 , cot t  2 6. 2 6

  73. sin t   14 , sec t  0, so t is in quadrant III. So the terminal point determined by t is P x  14 . Since P is on the unit     2 1   15   15 . Since the terminal point is in quadrant III, circle, x 2   14  1. Solving for x gives x   1  16 16 4       15 . Thus, the terminal point determined by t is P  415   14 , and so cos t   415 , tan t  1  1515 , x  15 4   csc t  4, sec t   4   4 1515 , cot t  15. 15


10

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

74. tan t  4 and the terminal point of t lies in quadrant II. Since sec2 t  tan2 t 1 we have sec2 t  42 1  161  17.    1 17  1   . Since tan t  cos t  sin t Thus sec t   17. Since sec t  0, we have sec t   17and cos t   17 sec t 17         we have sin t  4  1717  4 1717 . Thus, the terminal point determined by t is P  1717  4 1717 . Thus,     sin t  4 1717 , cos t   1717 , csc t  417 , sec t   17, cot t   14 . For Exercises 75–82, many answers are possible. 75. If f x  x 2 and g x  cos x, then F x   f  g x  cos x2  cos2 x. 76. If f x  e x and g x  sin x, then F x   f  g x  esin x .   77. If f x  x and g x  1  tan x, then F x   f  g x  1  tan x. 78. If f x 

x sin x and g x  sin x, then F x   f  g x  . 1x 1  sin x

  2 79. If f x  e x , g x  x 2 , and h x  sin x, then F x   f  g  h x  f sin2 x  esin x . 80. If f x  sin x, g x 

x, and h x  ln x, then F x   f  g  h x  f

    ln x  sin ln x .

    81. If f x  ln x, g x  x 2 , and h x  cos x, then F x   f  g  h x  f cos2 x  ln cos2 x . 82. If f x  sin x, g x 

x , and h x  e x , then F x   f  g  h x  f 1x

ex 1  ex

 sin

 ex . 1  ex

83. f x  x2 sin x  x 2 sin x   f x, so f is odd. 84. f x  x2 cos 2 x  x 2 cos 2x  f x, so f is even. 85. f x  sin x cos x   sin x cos x   f x, so f is odd. 86. f x  sin x  cos x   sin x  cos x, which is neither f x nor  f x, so f is neither even nor odd. 87. f x  x cos x  x cos x  f x, so f is even. 88. f x  x sin3 x  x [sin x]3  x  sin x3  x sin3 x  f x, so f is even. 89. f x  x3  cos x  x 3  cos x, which is neither f x nor  f x, so f is neither even nor odd. 90. f x  cos sin x  cos  sin x  cos sin x  f x, so f is even. 91. t

0

025

y t 4 283

050 075 100 125 0

92. (a) B 6  80  7 sin  2  87 mmHG

 (b) B 105  80  7 sin 105 12  827 mmHG

283 4 283

(c) B 12  80  7 sin   80 mmHG 

7 3  (d) B 20  80  7 sin 20 12  80  2  739 mmHG

93. (a) I 01  08e03 sin 1  0499 A (b) I 05  08e15 sin 5  0171 A

94. t

0

1

2

4

6

8

12

H t 175 1504 100 386 100 1503 588


SECTION 6.3 Trigonometric Graphs

95. We see from the diagram that every summand in

y

2 100 101 102 200  sin   sin sin sin   sin sin 100 100 100 100 100 100 has a corresponding term whose terminal point is diametrically opposed to its own on the unit circle. But for each of these pairs of terms,  n  n sin n  100  n sin   sin  sin    0, and so the 100 100 100 100 pair sums to 0.

11

t=100

1

k¹ sin 100

sin (k+100)¹ 100

1

0

x

(k+100)¹ 100

t=

Algebraically: 2 100 101 102 200   sin      sin  sin  sin      sin sin 100 100 100 100 100 100        2 2   sin    sin  sin        [sin x  sin 2x]  sin 100 100 100 100  0  0    0  0 96. Notice that if P t  x y, then P t    x y. Thus, (a) sin t    y and sin t  y. Therefore, sin t     sin t.

(b) cos t    x and cos t  x. Therefore, cos t     cos t. y y sin t sin t       tan t. (c) tan t    cos t   x x cos t 97. To prove that AO B  C DO, first note that O B  O D  1 and  O AB   OC D   2 . Now  C O D   AO B     2

 C O D     AO B   AB O. Since we know two angles and one side to be 2

equal, the triangles are (SAA) congruent. Thus AB  OC and O A  C D, so if B has coordinates x y, then D has coordinates y x. Therefore,   (a) sin t   2  x  cos t     (b) cos t   2  y, and sin t  y. Therefore, cos t  2   sin t.   x cos t x (c) tan t   2  y   y   sin t   cot t

6.3

TRIGONOMETRIC GRAPHS

1. If a function f is periodic with period p, then f t  p  f t for every t. The trigonometric functions y  sin x and cos x are periodic, with period 2 and amplitude 1. y

y

1

1

2¹ ¹

_1

2¹ x

¹

x

_1

2. To obtain the graph of y  5  sin x, we start with the graph of y  sin x, then shift it 5 units upward. To obtain the graph of y   cos x, we start with the graph of y  cos x, then reflect it in the x­axis.


12

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

3. (a) The sine and cosine curves y  a sin kx and y  a cos kx, k  0, have amplitude a and period 2k. The sine

(b)

y 3

curve y  3 sin 2x has amplitude 3  3 and period

0

22  ; an appropriate interval on which to graph one period is [0 ].

¹ 4

¹ 2

3¹ 4

_3

(b) y

4. (a) The sine curve y  a sin k x  b has amplitude a,

period 2k, and horizontal shift b. The sine curve   2 y  4 sin 3 x   6 has amplitude 4  4, period 3 ,

4 2

and horizontal shift  6 ; an appropriate interval on which to   5 .  graph one period is  6 6

5. f x  2  sin x has domain  and range [1 3]. y

0

_2

¹ 6

¹ 3

¹ 2

y 1

x

x _2

7. f x   sin x has domain  and range [1 1]. y

8. f x  2  cos x has domain  and range [1 3]. y

1 ¹

x

1 ¹

9. f x  2  sin x has domain  and range [3 1]. y

¹

x

5¹ 6

6. f x  2  cos x has domain  and range [3 1].

¹

¹

2¹ 3

_4

2

_1

x

¹

x

10. f x  1  cos x has domain  and range [2 0]. y 1

¹

x

x


SECTION 6.3 Trigonometric Graphs

11. g x  3 cos x has domain  and range [3 3]. y

2

12. g x  2 sin x has domain  and range [2 2]. y

2 ¹

x

  13. g x   12 sin x has domain  and range  12  12 . y

1

¹

x

  14. g x   23 cos x has domain  and range  23  23 . y

1 ¹

x

x

15. g x  3  3 cos x has domain  and range [0 6]. y

2

¹

16. g x  4  2 sin x has domain  and range [2 6]. y

2 ¹

x

17. h x  cos x has domain  and range [0 1]. y

1

¹

x

18. h x  sin x has domain  and range [0 1]. y

1 ¹

x

¹

x

13


14

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

19. y  cos 2x has amplitude 1 and period . An appropriate interval on

which to sketch one complete period is [0 ]. To find the key points on

y 1

this interval we divide the interval into four subintervals, each of length 2  1   . Now, we start at the left endpoint of the interval (x  0) 2 4 4

and successively add  4 to obtain the x­coordinates of the key points. We      3  0, and y   1. calculate y 0  1, y 4  0, y  2  1, y 4 20. y   sin 2x has amplitude 1 and period . An appropriate interval on

which to sketch one complete period is [0 ]. To find the key points on

¹ 2

¹ x

¹ 2

¹ x

¹ 4

¹ x 2

_1

y 1

this interval we divide the interval into four subintervals, each of length 2  1   . Now, we start at the left endpoint of the interval (x  0) 2 4 4

and successively add  4 to obtain the x­coordinates of the key points. We      3  1, and y   0. calculate y 0  0, y  4  1, y 2  0, y 4 21. y   cos 4x has amplitude 1 and period  2 . An appropriate interval on   which to sketch one complete period is 0 2 . To find the key points on

_1

y 1

this interval we divide the interval into four subintervals, each of length 2  1   . Now, we start at the left endpoint of the interval (x  0) 4 4 8

and successively add  8 to obtain the x­coordinates of the key points. We      3  0, and  0, y  1, y calculate y 0  1, y  8 4 8  y 2  1. 22. y  sin x has amplitude 1 and period 2. An appropriate interval on

which to sketch one complete period is [0 2]. To find the key points on

_1

y 1

this interval we divide the interval into four subintervals, each of length 2  1  1 . Now, we start at the left endpoint of the interval (x  0) and  4 2 successively add 12 to obtain the x­coordinates of the key points. We

    calculate y 0  0, y 12  1, y 1  0, y 32  1, and y 2  0.

23. y  3 sin 2x has amplitude 3 and period 1. An appropriate interval on which to sketch one complete period is [0 1]. To find the key points on

2 1

x

1 2

1 x

_1

y 3

this interval we divide the interval into four subintervals, each of length 2  1  1 . Now, we start at the left endpoint of the interval (x  0) and 2 4 4 successively add 14 to obtain the x­coordinates of the key points. We

      calculate y 0  0, y 14  3, y 12  0, y 34  3, and y 1  0.

_3


SECTION 6.3 Trigonometric Graphs

24. y  2 cos 8x has amplitude 2 and period  4 . An appropriate interval on   which to sketch one complete period is 0 4 . To find the key points on

y 2

this interval we divide the interval into four subintervals, each of length

¹ 8

  1   . Now, we start at the left endpoint of the interval (x  0) and 4

4

16

 to obtain the x­coordinates of the key points. We successively add 16      3  0, y  calculate y 0  2, y 16 8  2, y 16  0, and   y  4  2.

25. y  10 sin 12 x has amplitude 10 and period 4. An appropriate interval on which to sketch one complete period is [0 4]. To find the key points on

_2

y 10

this interval we divide the interval into four subintervals, each of length

2 1 12  4  . Now, we start at the left endpoint of the interval (x  0) and

successively add  to obtain the x­coordinates of the key points. We calculate y 0  0, y   10, y 2  0, y 3  10, and

¹ x 4

x

8¹ x

6¹ x

¹ 2

¹ x

_10

y 4  0.

26. y  5 cos 14 x has amplitude 5 and period 8. An appropriate interval on

which to sketch one complete period is [0 8]. To find the key points on

y 5

this interval we divide the interval into four subintervals, each of length 2 1 14  4  2. Now, we start at the left endpoint of the interval (x  0)

and successively add 2 to obtain the x­coordinates of the key points. We calculate y 0  5, y 2  0, y 4  5, y 6  0, and

_5

y 8  5.

27. y   13 cos 13 x has amplitude 13 and period 6. An appropriate interval

on which to sketch one complete period is [0 6]. To find the key points

y 1/3

on this interval we divide the interval into four subintervals, each of length 2 1 3 13  4  2 . Now, we start at the left endpoint of the interval (x  0)

and successively add 32 to obtain the x­coordinates of the key points. We     calculate y 0   13 , y 32  0, y 3  13 , y 92  0, and

_1/3

y 6   13 .

28. y  4 sin 2x has amplitude 4 and period . An appropriate interval on which to sketch one complete period is [0 2]. To find the key points on

y 4

this interval we divide the interval into four subintervals, each of length 2  1   . Now, we start at the left endpoint of the interval (x  0) 2 4 4

and successively add  4 to obtain the x­coordinates of the key points. We      3  4, and y   0. calculate y 0  0, y 4  4, y  2  0, y 4

_4

15


16

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

29. y  2 sin 8x has amplitude 2 and period 14 . An appropriate interval on   which to sketch one complete period is 0 14 . To find the key points on

y 2

this interval we divide the interval into four subintervals, each of length

2  1  1 . Now, we start at the left endpoint of the interval (x  0) 8 4 16 1 to obtain the x­coordinates of the key points. We and successively add 16

 

1  2, y 1  0, y 3  2, and calculate y 0  0, y 16 8 16   y 14  0.

30. y  3 sin 4x has amplitude 3 and period 12 . An appropriate interval on   which to sketch one complete period is 0 12 . To find the key points on

1 8

1 x 4

1 4

1 x 2

¹ 3

2¹ 3

_2

y 3

this interval we divide the interval into four subintervals, each of length

2  1  1 . Now, we start at the left endpoint of the interval (x  0) and 4 4 8 successively add 18 to obtain the x­coordinates of the key points. We

 

 

 

 

_3

calculate y 0  0, y 18  3, y 14  0, y 38  3, and y 12  0. 31. y  2 sin 3x has amplitude 2 and period 23 . An appropriate interval on   which to sketch one complete period is 0 23 . To find the key points on

y 2

this interval we divide the interval into four subintervals, each of length

x

2  1   . Now, we start at the left endpoint of the interval (x  0) 3 4 6

and successively add  6 to obtain the x­coordinates of the key points. We     calculate y 0  0, y 6  2, y  3  0, y 2  2, and   y 23  0. 32. y  4 cos 6x has amplitude 4 and period  3 . An appropriate interval on   which to sketch one complete period is 0 3 . To find the key points on

_2

y 4

this interval we divide the interval into four subintervals, each of length

2  1   . Now, we start at the left endpoint of the interval (x  0) 6 4 12

 to obtain the x­coordinates of the key points. We and successively add 12      calculate y 0  4, y 12  0, y  6  4, y 4  0, and y 3  4.

¹ 6

¹ x 3

1

2 x

_4

33. y  1  12 cos x has amplitude 12 and period 2. An appropriate interval

y

this interval we divide the interval into four subintervals, each of length

1

on which to sketch one complete period is [0 2]. To find the key points on

2  1  1 . Now, we start at the left endpoint of the interval (x  0) and  4 2 successively add 12 to obtain the x­coordinates of the key points. We

    calculate y 0  32 , y 12  1, y 1  12 , y 32  1, and y 2  32 .


17

SECTION 6.3 Trigonometric Graphs y

34. y  2  cos 4x has amplitude 1 and period 12 . An appropriate interval   on which to sketch one complete period is 0 12 . To find the key points

0 _1

on this interval we divide the interval into four subintervals, each of length

2  1  1 . Now, we start at the left endpoint of the interval (x  0) and 4 4 8 successively add 18 to obtain the x­coordinates of the key points. We

calculate y 0  1, y   y 12  1.

  1 8

 2, y

  1 4

 3, y

  3 8

1 x 2

1 4

_2 _3

 2, and

   35. y  cos x   2 has amplitude 1, period 2, and horizontal shift 2 . An   5 appropriate interval on which to sketch one complete period is  2  2 . To find

y 1

the key points on this interval we divide the interval into four subintervals, each of

¹

 length 24   2 . Now, we start at the left endpoint of the interval (x  2 ) and

successively add  2 to obtain the x­coordinates of the key points. We calculate       3  1, y 2  0, and y 5  1. y  2  1, y   0, y 2 2

   36. y  2 sin x   3 has amplitude 2, period 2, and horizontal shift 3 . An   7 appropriate interval on which to sketch one complete period is  3  3 . To find

y 2 4¹ 3

the key points on this interval we divide the interval into four subintervals, each of

¹ 3

 length 24   2 . Now, we start at the left endpoint of the interval (x  3 ) and

successively add  3 to obtain the x­coordinates of the key points. We calculate          y 3  0, y 56  2, y 43  0, y 116  2, and y 73  0.

   37. y  2 sin x   6 has amplitude 2, period 2, and horizontal shift 6 . An   13 appropriate interval on which to sketch one complete period is  6  6 . To find

¹ 6

successively add  2 to obtain the x­coordinates of the key points. We calculate          y  4  3, y 4  0, y 34  3, y 54  0, and y 74  3.

7¹ 6

¹

successively add  2 to obtain the x­coordinates of the key points. We calculate          y 6  0, y 23  2, y 76  0, y 53  2, and y 136  0.

 length 24   2 . Now, we start at the left endpoint of the interval (x   4 ) and

x

y

 length 24   2 . Now, we start at the left endpoint of the interval (x  6 ) and

the key points on this interval we divide the interval into four subintervals, each of

¹

7¹ 3

2

the key points on this interval we divide the interval into four subintervals, each of

   38. y  3 cos x   4 has amplitude 3, period 2, and horizontal shift  4 . An   7 . To find appropriate interval on which to sketch one complete period is    4 4

x

13¹ 6

x

y 3 _

¹ 4

3¹ 4

7¹ 4

¹

x


18

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

  39. y  4 sin  x  12 has amplitude 4, period 2, and horizontal shift 12 . An   appropriate interval on which to sketch one complete period is 12  52 . To find the

y 4 2

key points on this interval we divide the interval into four subintervals, each of

length 2  14  12 . Now, we start at the left endpoint of the interval (x  12 ) and

successively add 12 to obtain the x­coordinates of the key points. We calculate       y 12  0, y 1  4, y 32  0, y 2  4, and y 52  0.

0

x

2

1

_2 _4

  40. y  2 cos  x  14 has amplitude 2, period 2, and horizontal shift  14 . An   appropriate interval on which to sketch one complete period is  14  74 . To find

y 1

the key points on this interval we divide the interval into four subintervals, each of

0 _1

length 2  14  12 . Now, we start at the left endpoint of the interval (x   14 ) and successively add 12 to obtain the x­coordinates of the key points. We calculate

x

1

          y  14  2, y 14  0, y 34  2, y 54  0, and y 74  2.

    41. y  2 cos 4 x   4 has amplitude 2, period 2 , and horizontal shift 4 . An   3 appropriate interval on which to sketch one complete period is  4  4 . To find

the key points on this interval we divide the interval into four subintervals, each of  length 24  14   8 . Now, we start at the left endpoint of the interval (x  4 ) and

successively add  8 to obtain the x­coordinates of the key points. We calculate         3  0, y     2, y 5  0, and y 3  2. y  4  2, y 8 2 8 4

  2  42. y  3 cos 3 x   3 has amplitude 3, period 3 , and horizontal shift  3 . An   appropriate interval on which to sketch one complete period is   3  3 . To find

y

2 1 0 _1

¹ 4

y 2 _¹ 3

and successively add  6 to obtain the x­coordinates of the key points. We calculate        y  3  3, y  6  0, y 0  3, y  6  0, and y 3  3.

  43. y  cos 2x    cos 2 x   2 has amplitude 1, period , and horizontal shift     2 . An appropriate interval on which to sketch one complete period is  2  2 .

x

3¹ 4

_2

the key points on this interval we divide the interval into four subintervals, each of  length 23  14   6 . Now, we start at the left endpoint of the interval (x   3 )

¹ 2

0 _2

¹ 3

x

y 1

To find the key points on this interval we divide the interval into four subintervals, each of length 22  14   4 . Now, we start at the left endpoint of the interval

 (x   2 ) and successively add 4 to obtain the x­coordinates of the key points. We        calculate y  2  1, y  4  0, y 0  1, y  4  0, and y 2  1.

¹

_2

0

¹ x 2


19

SECTION 6.3 Trigonometric Graphs

    2 44. y  sin 3x   2  sin 3 x  6 has amplitude 1, period 3 , and horizontal

shift  6 . An appropriate interval on which to sketch one complete period is     5 . To find the key points on this interval we divide the interval into four 6 6

subintervals, each of length 23  14   6 . Now, we start at the left endpoint of the

 interval (x   6 ) and successively add 6 to obtain the x­coordinates of the key       2  1, and points. We calculate y  6  0, y 3  1, y 2  0, y 3   y 56  0.

  2  45. y  2 sin 23 x   6  2 sin 3 x  4 has amplitude 2, period 3, and

horizontal shift  4 . An appropriate interval on which to sketch one complete period   13 is  4  4 . To find the key points on this interval we divide the interval into four

y 1 5¹ 6

0

¹ 6

x

¹ 2

_1

y 2 7¹ 4

¹ 4

2  1  3 . Now, we start at the left endpoint of subintervals, each of length 23 4 4

13¹ 4

x

3 the interval (x   4 ) and successively add 4 to obtain the x­coordinates of the       7  0, y 5  2, and key points. We calculate y  4  0, y   2, y 4 2   y 134  0.

    2 46. y  5 cos 3x   4  5 cos 3 x  12 has amplitude 5, period 3 , and

y 5

 . An appropriate interval on which to sketch one complete horizontal shift 12     3 . To find the key points on this interval we divide the interval period is 12 4

¹ 12

into four subintervals, each of length 23  14   6 . Now, we start at the left

5¹ 12

3¹ 4

x

¹ 3

x

 ) and successively add  to obtain the endpoint of the interval (x  12 6    x­coordinates of the key points. We calculate y 12  5, y  4  0,         5, y 7  0, and y 3  5. y 512 12 4

  2  47. y  2  2 cos 3 x   3 has amplitude 2, period 3 , and horizontal shift  3 . An    appropriate interval on which to sketch one complete period is  3  3 . To find

y

the key points on this interval we divide the interval into four subintervals, each of

2

 length 23  14   6 . Now, we start at the left endpoint of the interval (x   3 )

and successively add  6 to obtain the x­coordinates of the key points. We calculate        y  3  0, y  6  2, y 0  4, y  6  2, and y 3  0.

¹

_3

0


20

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

  48. y  3  sin 2 x  18 has amplitude 1, period 1, and horizontal shift  18 . An   appropriate interval on which to sketch one complete period is  18  78 . To find

y

the key points on this interval we divide the interval into four subintervals, each of 1 1 1 length 22   4  4 . Now, we start at the left endpoint of the interval (x   8 )

and successively add 14 to obtain the x­coordinates of the key points. We calculate           y  18  3, y 18  4, y 38  3, y 58  2, and y 78  3.

   1 1  1 has amplitude 1 , period 1,  49. y  12  12 cos 2x    cos 2 x  3 2 2 6 2

and horizontal shift 16 . An appropriate interval on which to sketch one complete   period is 16  76 . To find the key points on this interval we divide the interval into

1 1 four subintervals, each of length 22   4  4 . Now, we start at the left endpoint of

the interval (x  16 ) and successively add 14 to obtain the x­coordinates of the key         5  1 , y 2  1, y 11  1 , and points. We calculate y 16  0, y 12 2 3 12 2   7 y 6  0.

1 _81

0

7 1 x 8

3 8

y

1

0

1 6

    2 50. y  1  cos 3x   2  1  cos 3 x  6 has amplitude 1, period 3 , and

2 3

1

7 6

x

y 2

horizontal shift   6 . An appropriate interval on which to sketch one complete    period is  6  2 . To find the key points on this interval we divide the interval

1

into four subintervals, each of length 23  14   6 . Now, we start at the left

 endpoint of the interval (x    6 ) and successively add 6 to obtain the     x­coordinates of the key points. We calculate y  6  2, y 0  1, y  6  0,   y 3  1, and y 2  2.

¹

_6

¹ 6

¹ 2

x

51. This function has amplitude a  4 and period 2k  2. As a sine curve its horizontal shift is bs  0, and as a cosine curve   it is bc   2 . Equations are y  a sin k x  bs   4 sin x and y  a cos k x  bc   4 cos x  2 . 52. This curve has amplitude a  2 and period 2k  . As a cosine curve its horizontal shift is bc  0, and as a sine curve it   is bs    4 . Equations are y  a cos k x  bc   2 cos 2x and y  a sin k x  bs   2 sin 2 x  4 .

53. This curve has amplitude a  32 and period 2k  23 . As a cosine curve its horizontal shift is bc  0, and as a sine curve it  3 3  is bs    6 . Equations are y  a cos k x  bc   2 cos 3x and y  a sin k x  bs   2 sin 3 x  6 .

54. This curve has amplitude a  3 and period 2k  4. As a sine curve its horizontal shift is bs  0, and as a cosine curve it is bc  . Equations are y  3 sin 12 x and y  3 cos 12 x  .

pi/6

 , and as a cosine 55. This curve has amplitude a  12 and period 2k  . As a sine curve its horizontal shift is bs   712    - pi/6 1 7 1  curve it is bc    3 . Equations are y   2 sin 2 x  12 and y   2 cos 2 x  3 .

56. This curve has amplitude a  2 and period 2k  . As a sine curve its horizontal shift is bs  0, and as a cosine curve it is   bc    4 . Equations are y  1  2 sin 2x and y  1  2 cos 2 x  4 .

-

Since -7pi/12 is the horizontal shift for the reflected sine curve, lets give these two curves as alternative solutions and instead use -pi/12 for the horizontal shift of the sine curve and pi/6 for the horizontal shift of the cosine curve.

+


SECTION 6.3 Trigonometric Graphs

21

57. This curve has amplitude a  1 and period 2k  . As a sine curve its horizontal shift is bs   2 , and as a cosine curve it     3 is bc  34 . Equations are y  1  sin 2 x   2 and y  1  cos 2 x  4 .

3/4

58. This curve has amplitude a  4 and period 2k  2. As a sine curve its horizontal shift is bs  14 , and as a cosine curve it is     Since -pi/4 is the horizontal - 3/4 bc   14 . Equations are y  4  4 sin  x  14 and y  4  4 cos  x  14 .

+

59. f x  cos 100x, [01 01] by [15 15]

60. f x  3 sin 120x, [01 01] by [4 4]

1

­0.1

0.1

­0.1

0.1

­1

61. f x  sin

x , [250 250] by [15 15] 40

62. f x  cos

1

1

­200

x , [0 500] by [1 1] 80

0

200 ­1

200

­1

63. y  tan 25x, [02 02] by [3 3]

64. y  csc 40x, [01 01] by [10 10] 10

2

­0.2

400

0.2

­0.1

­2

0.1 ­10

65. y  sin2 20x, [05 05] by [02 12]

66. y 

 cos 10x, [04 04] by [01 15]

1.0 1 0.5

­0.5

0.5

­0.4

­0.2

0.0

0.2

0.4

shift for the reflected cossine curve, lets give this as an alternative solution and instead use 3/4 for the horizontal shift of the cosine curve.


22

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

67. f x  x, g x  sin x

68. f x  sin x, g x  sin 2x

6

g ­6

f+g

2 ­4

­2

2

f

4

0 ­2

2

4

6

­6

f

g 1

­4

­2

­4

f+g

0

2

4

6

­1

­6

­2

69. f x  sin 3x, g x  cos 12 x

_5

70. f x  05 sin 5x, g x   cos 2x.

2

f+g

2

1

g

1

0 _1 _2

5

f+g f

_2

2 _1

f

g

_2

71. y  x 2 sin x is a sine curve that lies between the graphs of y  x 2 and y  x 2 .

72. y  x cos x is a cosine curve that lies between the graphs of y  x and y  x.

200

5

­10

­5

10

5 ­5

­200

73. y 

x sin 5x is a sine curve that lies between the   graphs of y  x and y   x. 2

cos 2x is a cosine curve that lies between the 1  x2 1 1 and y   . graphs of y  1  x2 1  x2

74. y 

1 2 ­2

4 ­5

5 ­1


SECTION 6.3 Trigonometric Graphs

23

75. y  cos 3x cos 21x is a cosine curve that lies between

76. y  sin 2x sin 10x is a sine curve that lies between the

1

1

the graphs of y  cos 3x and y   cos 3x.

­0.5

graphs of y  sin 2x and y   sin 2x.

0.5

­0.5

­1

0.5 ­1

77. y  sin x  sin 2x. The period is 2, so we graph the

function over one period,  . Maximum value 176 when x  094  2n, minimum value 176 when x  094  2n, n any integer.

78. y  x  2 sin x, 0  x  2. Maximum value 697 when x  524, minimum value 068 when x  105. 5

2 0 ­2

2

2

4

6

­2

79. y  2 sin x  sin2 x. The period is 2, so we graph the

function over one period,  . Maximum value 300 when x  157  2n, minimum value 100 when x  157  2n, n any integer.

cos x . The period is 2, so we graph the function 2  sin x over one period. Maximum value 058 when

80. y 

x  576  2n (exact value x  116  2n); Minimum

value 058 when x  367  2n (exact value x  76  2n) for any integer n.

2

1 ­2

2 0

­2

2

4

6

­1

81. cos x  04, x  [0 ]. The solution is x  116.

82. tan x  2, x  [0 ]. The solution is x  111.

1 2 0

1

2

3 0

­1

0

1

2

3


24

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

83. csc x  3, x  [0 ]. The solutions are x  034, 280.

84. cos x  x, x  [0 ]. The solution is x  074.

4

2

2 0

85. f x 

0 0

1

2

1

3

2

3

1  cos x x

(a) Since f x  is odd.

1  cos x 1  cos x    f x, the function x x

(c)

1

(b) The function is undefined at x  0, so the x­intercepts occur when

­20

1  cos x  0, x  0  cos x  1, x  0  x  2, 4, 6, 

20 ­1

(d) As x  , f x  0. (e) As x  0, f x  0.

86. f x 

sin 4x 2x

(a) f x 

 sin 4x sin 4x   f x, so the function is even. 2 x 2x

(b) The x­intercepts occur when sin 4x  0, x  0  4x  n x  14 n, n any nonzero integer.

(c) 2

1

(d) As x  , f x  0.

­2

(e) As x  0, f x  2.

2

2  20 seconds. 10 (b) Since h 0  3 and h 10  3, the wave height is 3  3  6 feet.

87. (a) The period of the wave is

88. (a) The period of the vibration is

1 2  second. 880 440

1 of a second, there are 440 vibrations (b) Since each vibration takes 440 per second.

(c)

v 1

0

_1

0.002 0.004 0.006 0.008

t


SECTION 6.3 Trigonometric Graphs

89. (a) The period of p is

1 2  minute. 160 80

(c)

25

p 160

(b) Since each period represents a heart beat, there are 80 heart beats per

140

minute.

120 115

(d) The maximum (or systolic) is 115  25  140 and the minimum (or

100 90 80

diastolic) is 115  25  90. The read would be 14090 mmHg, which is higher than normal.

90. (a) The period of R Leonis is

0

2  312 days. 156

(c)

2 t (s)

1

b 10

(b) The maximum brightness is 79  21  10; the minimum brightness

8 6

is 79  21  58.

4 2 0

x are the points of intersection of the graphs of 91. The solutions to sin x  100

200

400

600

t

1

x . We first find the number of positive solutions. Since y  sin x and y  100

x 1 sin x  1, the y­values of the intersection points must satisfy 100

­20

x  100. We can use a graphing device to see that the graphs intersect at

exactly two points on each of the intervals [0 ], [2 3], and [4 5]. This

20 ­1

pattern holds throughout [0 100], so because 31  100  32, the interval [30 31] is the rightmost interval with two solutions. Thus, there are 32 nonnegative solutions, including x  0, and so by symmetry, there are 31 negative solutions. Therefore, the given equation has 63 solutions on [100 100].     92. (a) y  sin x . This graph looks like a sine function (b) y  sin x 2 . This graph looks like a graph of sin x which which has been stretched horizontally (stretched

more for larger values of x). It is defined only for x  0, so it is neither even nor odd.

has been shrunk for x  1 (shrunk more for larger values of x) and stretched for x  1. It is an even function, whereas

sin x is odd.

1 0 ­1

93. (a) y 

1 200

400

­5

5 ­1

1 is periodic with period 2, and it is neither even nor odd. y 0  12 , and it has graph III. 2  sin x

(b) y  esin 2x is periodic with period , and it is and neither even nor odd. y 0  e0  1, and it has graph I. sin x (c) y  is aperiodic and odd. It has graph IV. 1  x   1 is aperiodic and even. It has graph II. (d) y  cos 1  x2


26

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

6.4

MORE TRIGONOMETRIC GRAPHS

1. The trigonometric function y  tan x has period  and asymptotes x    n, n an integer. 2

2. The trigonometric function y  csc x has period 2 and asymptotes x  n, n an integer. y

y 10

5

¹/2 x

_¹/2

0

_10

¹ x

_5

   3 3. f x  tan x   4 corresponds to Graph II. f is undefined at x  4 and x  4 , and Graph II has the shape of a graph of a tangent function. 3 4. f x  sec 2x corresponds to Graph III. f is undefined at x   4 and x  4 , and Graph III has the shape of a graph of a secant function.

5. f x  cot 4x corresponds to Graph VI.

6. f x   tan x corresponds to Graph I.

7. f x  2 sec x corresponds to Graph IV.

8. f x  1  csc x corresponds to Graph V.

9. y  4 tan x has period .

_2¹

3

10. y  3 tan x has period .

y 10

0

_2¹ ¹

x

_10

0

¹

x

¹

x

_10

11. y   32 tan x has period .

12. y  34 tan x has period . y 5

_2¹

y 10

y 4 2

0

¹

x

_2¹

0

_2 _5

_4


SECTION 6.4 More Trigonometric Graphs

13. y   cot x has period .

27

14. y  2 cot x has period .

y

y 5

4

_2¹

0

x

¹

x

¹

_5

15. y  2 csc x has period 2.

16. y  12 csc x has period 2.

y

y

5

1

¹

17. y  3 sec x has period 2.

x

y

y

5

¹

x

x

19. y  tan 3x has period  3 . An appropriate interval is     6  6 . The graph has x­intercepts where 3x  n  x 3 n, n an integer; and vertical asymptotes where

  3x   2  n  x  6  3 n, n an integer. y

¹

20. y  tan 4x has period  4 . An appropriate interval is     8  8 . The graph has x­intercepts where 4x  n  x 2 n, n an integer; and vertical asymptotes where

  4x   2  n  x  8  4 n, n an integer. y

4

4

2 0

_¹/2

x

18. y  3 sec x has period 2.

5

¹

2

¹ ¹/2

x

0

_¹/2

_2

_2

_4

_4

¹ ¹/2

x


28

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

21. y  5 tan x has period 1. An appropriate interval is    12  12 . The graph has x­intercepts where x  n  x  n, n an integer; and vertical asymptotes where

x    n  x  1  n, n an integer. 2

22. y  3 tan 4x has period 14 . An appropriate interval is    18  18 . The graph has x­intercepts where 4x  n  x  14 n, n an integer; and vertical asymptotes where

1 1 4x   2  n  x  8  4 n, n an integer.

2

y 20

y 10

10 _3

_2

0

_1

5 1

_1

3 x

2

0

_1/2

_10

_5

_20

_10

23. y  2 cot 3x has period 13 . An appropriate interval is   0 13 . The graph has x­intercepts where 3x   2  n  x  16  13 n, n an integer; and vertical asymptotes

where 3x  n  x  13 n, n an integer.

_2/3

_1/3

 x  14  12 n, n an integer; and vertical asymptotes

where 2x  n  x  12 n, n an integer. y 10

5

5 1/3

1 x

2/3

_1

_5

_10

_10

4

is [2 2]. The graph has x­intercepts where  4 x  n  x  4n, n an integer; and vertical asymptotes where  x    n  x  2  2n, n an integer. 2

y

1 x

1/2

     26. y  cot  2 x has period   2. An appropriate 2

interval is [0 2]. The graph has x­intercepts where

 x    n  x  1  2n, n an integer; and vertical 2

2

asymptotes where  2 x  n  x  2n, n an integer. y

2 _2

0

_1/2

_5

   25. y  tan  4 x has period   4. An appropriate interval

4

24. y  3 cot 2x has period 12 . An appropriate interval is   0 12 . The graph has x­intercepts where 2x   2  n

y 10

0

1 x

1/2

2 2

x

_1

1

x


SECTION 6.4 More Trigonometric Graphs

29

   27. y  2 tan 3x has period 13 . An appropriate interval is 28. y  2 tan  2 x has period   2. An appropriate   2  16  16 . The graph has x­intercepts where 3x  n  interval is [1 1]. The graph has x­intercepts where x  13 n, n an integer; and vertical asymptotes where

3x    n  x  1  1 n, n an integer. 2

6

3

y

 x  n  x  2n, n an integer; and vertical 2

 asymptotes where  2 x  2  n  x  1  2n, n an

integer.

y

4 _0.5

4 0.5

x

_1

1

x

2 29. y  csc 4x has period 24   2 . An appropriate interval is 30. y  5 csc 3x has period 3 . An appropriate interval is        4  4 . The graph has vertical asymptotes where  3  3 . The graph has vertical asymptotes where  4x  n  x  4 n, n an integer; maxima where 3x  n  x   3 n, n an integer; maxima where

4x  32  2n  x  38   2 n, n an integer; and

  minima where 4x   2  2n  x  8  2 n, n an

integer.

2 3x  32  2n  x   2  3 n, n an integer; and

 2 minima where 3x   2  2n  x  6  3 n, n an

integer.

y

y

2

10

_¹ 2

¹ 2

x

31. y  sec 2x has period 22  . An appropriate interval is [0 ]. The graph has vertical asymptotes where 2x  n

x   2 n, n an integer; maxima where 2x    2n

x   2  n, n an integer; and minima where 2x  2n  x  n, n an integer.

x

32. y  12 sec 4x has period 12 . An appropriate interval is   0 12 . The graph has vertical asymptotes where 4x  n  x  14 n, n an integer; maxima where 4x    2n  x  14  12 n, n an integer; and

minima where 4x  2n  x  12 n, n an integer.

y

y 3

2 _¹

¹

2 ¹

x

1 _1

_1/2

0 1/8 _1 _2 _3

1/2

1 x


30

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

33. y  5 csc 32 x has period 322  43 . An appropriate   interval is 0 43 . The graph has vertical asymptotes

where 32 x  n  x  23 n, n an integer; maxima where

34. y  5 sec 2x has period 22   1. An appropriate

interval is [0 1]. The graph has vertical asymptotes where 2x  n  x  12 n, n an integer; maxima where

2x    2n  x  12  n, n an integer; and minima

3 x  3  2n  x  1  4 n, n an integer; and 2 2 3 3   minima where 2 x  2  2n  x  13  43 n, n an

where 2x  2n  x  n, n an integer.

y

10

y

integer.

_0.5

10 _1

1

0.5

 x­intercepts where x   4  n  x   4  n, n an

 integer; and vertical asymptotes where x   4  2  n

x   4  n, n an integer. y 4

   36. y  tan x   4 has period  and horizontal shift 4 . An   3 appropriate interval is   4  4 . The graph has  x­intercepts where x   4  n  x   4  n, n an

 integer; and vertical asymptotes where x   4  2  n

x   4  n, n an integer.

y 4

2 _¹

2

0 ¹/4

_2

¹

2¹ x

_2¹

x

_4

  x­intercepts where x   4  2  n  x  4  n, n

an integer; and vertical asymptotes where x   4  n 

   38. y  2 cot x   3 has period  and horizontal shift 3 .   4 An appropriate interval is  3  3 . The graph has

 5 x­intercepts where x   3  2  n  x  6  n, n

an integer; and vertical asymptotes where x   3  n  x 3  n, n an integer.

y 4

4

2 _¹

_2

   37. y  cot x   4 has period  and horizontal shift  4 .   3 An appropriate interval is   4  4 . The graph has

_2¹

¹ 3¹/4

0

_4

x   4  n, n an integer.

x

x

   35. y  tan x   4 has period  and horizontal shift  4 .   5 An appropriate interval is  4  4 . The graph has

_2¹

1

0

¹ 3¹/4

y

2 2¹ x

_2¹

0 ¹/3

_2

_2

_4

_4

¹

x


SECTION 6.4 More Trigonometric Graphs

31

   39. y  csc x   4 has period 2 and horizontal shift 4 .   9 An appropriate interval is  4  4 . The graph has

   40. y  sec x   4 has period 2 and horizontal shift  4 .   7 An appropriate interval is   4  4 . The graph has

3 n an integer; maxima where x   4  2  2n 

n an integer; maxima where x   4    2n 

 3 x 4  2  2n  x  4  2n, n an integer.

 x 4  2n  x   4  2n, n an integer.

 vertical asymptotes where x   4  n  x  4  n,

x  74  2n, n an integer; and minima where

_2¹

 vertical asymptotes where x   4  n  x   4  n,

x  34  2n, n an integer; and minima where

y 4

y

2

2

0 ¹/4

_2

2¹ x

¹

¹

x

_4

   41. y  12 sec x   6 has period 2 and horizontal shift 6 .   13 An appropriate interval is  6  6 . The graph has

  42. y  3 csc x   2 has period 2 and horizontal shift    3  2 . An appropriate interval is  2  2 . The graph has

n an integer; maxima where x   6    2n 

3 n an integer; maxima where x   2  2  2n 

 vertical asymptotes where x   6  n  x  6  n,

x  76  2n, n an integer; and minima where  x 6  2n  x  6  2n, n an integer.

 vertical asymptotes where x   2  n  x   2  n,

x    2n, n an integer; and minima where

 x 2  2  2n  x  2n, n an integer. y

y

6

1 ¹

x

¹

x


Maybe use the same labeling as done in the answer shown here. 32

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

    43. y  tan 2 x   3 has period 2 and horizontal shift 3 .     7 . The graph has An appropriate interval is 12 12     x­intercepts where 2 x  3  n  x   3  2 n, n an integer; and vertical asymptotes where    7 2 x 3  2  n  x  12  n, n an integer. y 4 2 0 _2

5¹/6 ¹/3 ¹/2

     44. y  cot 2x   4  cot 2 x  8 has period 2 and     5 . horizontal shift  . An appropriate interval is 8 8 8     The graph has x­intercepts where 2 x  8  2  n  x  38   2 n, n an integer; and vertical asymptotes     where 2 x   8  n  x  8  2 n, n an integer. y 4 2

x

The label pi/2 looks like it is for the asymptote. Please use different labeling.

_4

     45. y  5 cot 3x   2  5 cot 3 x  6 has period 3 and    horizontal shift   6 . An appropriate interval is  6  6 .    The graph has x­intercepts where 3 x   6  2  n x   3 n, n an integer; and vertical asymptotes where     3 x 6  n  x   6  3 n, n an integer. y 10

_¹/2

_¹/2

¹

x

_4

   46. y  4 tan 4x  2  4 tan 4 x   2 has period 4 and   3 5 horizontal shift  2 . An appropriate interval is 8  8 .   The graph has x­intercepts where 4 x   2  n 

 x  2  4 n, n an integer; and vertical asymptotes where    5  4 x 2  2  n  x  8  4 n, n an integer. y 10 5

0 ¹/6 ¹/2

¹ x

0

_5

x

¹/2

_5

_10

_10

     47. y  cot 2x   2  cot 2 x  4 has period 2 and     3 . horizontal shift  . An appropriate interval is 4 4 4     The graph has x­intercepts where 2 x  4  2  n  x   2  2 n, n an integer; and vertical asymptotes     where 2 x   4  n  x  4  2 n, n an integer. y

48. y  12 tan x    12 tan  x  1 has period   1   and horizontal shift 1. An appropriate interval is 12  32 . The graph has x­intercepts where  x  1  n  x  n, n an integer; and vertical asymptotes where

1  x  1   2  n  x  2  n, n an integer. y 1

2 _¹

¹/2

_2

5 _¹

0 ¹/8

¹ x

0

1/2

1

3/2

x


SECTION 6.4 More Trigonometric Graphs

  1 49. y  2 csc x   3  2 csc  x  3 has period

2  2 and horizontal shift 1 . An appropriate interval is  3

1  7 . The graph has vertical asymptotes where 3 3

   x  13  n  x   3  n, n an integer; maxima

  where  x  13  32  2n  x   16  2n, n an

  integer; and minima where  x  13   2  2n  x  56  2n, n an integer. y

33

1 2 50. y  3 sec 14 x   6  3 sec 4 x  3 has period 8

and horizontal shift 23 . An appropriate interval is   2  26 . The graph has vertical asymptotes where 3 3 1 4

 x  23  n  x  23  4n, n an integer;

  maxima where 14 x  23    2n 

x  143  8n, n an integer; and minima where   1 x  2  2n  x  2  8n, n an integer. 4 3 3 y 10

4

5 1

2

x

_ 4¹ 3

_5

0

8¹ 3

x

20¹ 3

_10

      51. y  sec 2 x   52. y  csc 2 x   4 has period  and horizontal shift 4 . 2 has period  and horizontal shift  2 .     5 An appropriate interval is   An appropriate interval is  2  2 . The graph has 4  4 . The graph has vertical     vertical asymptotes where 2 x      n, n an 2  n  asymptotes where 2 x    n  x  4 4 2    x    2  2 n, n an integer; maxima where integer; maxima where 2 x  4    2n    3  2 x 2  2  2n  x  4  n, n an integer; x  34  n, n an integer; and minima where      and minima where 2 x     n, n an integer. 2  2  2n  2 x  2n  x  4 4 x   4  n, n an integer.

y 5

0

y 5

¹/2

¹

x _¹

0

_5 _5

¹

x


34

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

    2 53. y  5 sec 3x   2  5 sec 3 x  6 has period 3 and     5 . . An appropriate interval is horizontal shift  6 6 6   The graph has vertical asymptotes where 3 x   6  n  x   6  3 n, n an integer; maxima where

   2 3 x 6    2n  x  2  3 n, n an integer; 

2 and minima where 3 x  6  2n  x   6  3 n, n

an integer.

  54. y  12 sec 2x    12 sec 2 x  12 has period

2  1 and horizontal shift 1 . An appropriate interval is 2 2

1  3 . The graph has vertical asymptotes where 2 2

  2 x  12  n  x  12  12 n, n an integer; maxima

  where 2 x  12    2n  x  n, n an integer;

  and minima where 2 x  12  2n  x  12  n, n an integer.

y

y 10 _¹/2

¹/2

1

x

0

  2  55. y  tan 23 x   6  tan 3 x  4 has period

0.5

1

1.5

x

   56. y  tan 12 x   4 has period 12  2 and horizontal

  is   2   . The graph has x­intercepts where

  5  3 . The shift   . An appropriate interval is  4 4 4   graph has x­intercepts where 12 x   4  n 

   vertical asymptotes where 23 x   4  2  n 

   3 where 12 x   4  2  n  x  4  2n, n an

 3  23  2 and horizontal shift 4 . An appropriate interval 2 x     n  x    3 n, n an integer; and 3 4 4 2

x    32 n, n an integer.

x   4  2n, n an integer; and vertical asymptotes

integer.

y

y

2

2 ¹

x

_ 4

¹

3¹ 4

7¹ 4

2¹ x


SECTION 6.4 More Trigonometric Graphs

57. y  3 sec  x  12 has period 2  2 and horizontal   shift  12 . An appropriate interval is  12  32 . The graph   has vertical asymptotes where  x  12  n  x   12  n, n an integer; maxima where

   x  12    2n  x  12  2n, n an integer; and   minima where  x  12  2n  x   12  2n, n an

    2 58. y  sec 3x   2  sec 3 x  6 has period 3 and    horizontal shift  6 . An appropriate interval is  6  2 .   The graph has vertical asymptotes where 3 x   6  n

  x   6  3 n, n an integer; maxima where    2 3 x 6    2n  x  6  3 n, n an integer;    2 and minima where 3 x   6  2n  x   6  3 n,

n an integer.

y

integer.

y

2 6

_¹/2

1

1

¹/2

and horizontal shift  6 . An appropriate interval is     5 . The graph has x­intercepts where  12 12     2 x 6  n  x  6  2 n, n an integer; and    vertical asymptotes where 2 x   6  2  n     n, n an integer. x  512 2

60. y  2 cot 3x  3  2 cot 3 x  1 has period 13 and horizontal shift 1. An appropriate interval is   1  23 . The graph has x­intercepts where

5 1 3 x  1   2  n  x   6  3 n, n an integer;

and vertical asymptotes where 3 x  1  n  x  1  13 n, n an integer.

y

¹/6

y 4 2

4 _¹/3

x

x

     59. y  2 tan 2x   3  2 tan 2 x  6 has period 2

2¹/3 x

0

_1

x

_2 _4

61. y  tan 30x, [03 03] by [3 3]

62. y  csc 50x, [03 03] by [6 6] 5

2

­0.2

0.2 ­2

35

­0.2

0.2 ­5


36

tan 20π x

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

63. y 

 tan 20x, [05 05] by [02 3]

64. y  sec2 10x, [1 1] by [02 5]

0.15

4

Replace "0.5" with "0.15" (4 times)

2

2

­0.5

0.0

­1

0.5

65. (a) d t  3 tan t, so d 015  153,

(b)

d 025  300, and d 045  1894.

d

0

S 8  346, and S 1175  9154.

1

10

(c) d   as t  12 .

   t , so S 2  1039, S 6  0, 66. (a) S t  6 cot 12

0

(b)

0.1

0.2

0.3

0.4

0.5 t

S 5

(c) From the graph, it appears that S t  6 at

approximately t  3 and t  9, corresponding to 9 A . M . and 3 P. M .

0

6

t

(d) As t  12 , the sun approaches the horizon and the person’s shadow grows longer and longer.

67. (a) If f is periodic with period p, then by the definition of a period, f x  p  f x for all x in the domain of f . 1 1 1  for all f x  0. Thus, is also periodic with period p. Therefore, f x  p f x f 1 also has period 2. Similarly, since cos x has (b) Since sin x has period 2, it follows from part (a) that csc x  sin x 1 also has period 2. period 2, we conclude sec x  cos x 68. If f and g are periodic with period p, then f x  p  f x for all x and g x  p  g x for all x. Thus f x  p f x  for all x [unless g x is undefined, in which case both are undefined.] But consider f x  sin x g x  p g x f x sin x (period 2) and g x  cos x (period 2). Their quotient is   tan x, whose period is . g x cos x 69. (a) The graph of y   cot x is the same as the graph of y  tan x shifted  2 units to the right. (b) The graph of y  csc x is the same as the graph of y  sec x shifted  2 units to the right.


SECTION 6.5 Inverse Trigonometric Functions and Their Graphs

6.5

37

INVERSE TRIGONOMETRIC FUNCTIONS AND THEIR GRAPHS

  1. (a) To define the inverse sine function we restrict the domain of sine to the interval   2  2 . On this interval the

sine function is one­to­one and its inverse function sin1 is defined by sin1 x  y  sin y  x. For example,  1 sin1 21   6 because sin 6  2 .

(b) To define the inverse cosine function we restrict the domain of cosine to the interval [0 ]. On this interval the cosine function is one­to­one and its inverse function cos1 is defined by cos1 x  y  cos y  x. For example,  1 cos1 12   3 because cos 3  2 .

    1 sin    2. (a) The cancellation property sin1 sin x  x is valid for x in the interval   2  2 . By this property, sin 4 4    .   and sin1 sin   3 3 (b) If x is not in the interval in part (a), then the cancellation property does not apply. For example,   sin1 sin 56  sin1 12   6.      3. (a) sin1 1   2 because sin 2  1 and 2 lies in  2  2 .      3   (b) sin1 23   3 because sin 3  2 and 3 lies in  2  2 . (c) sin1 2 is undefined because there is no real number x such that sin x  2. 

4. (a) sin1 1    2

(b) sin1 22   4

5. (a) cos1 1     6. (a) cos1 22   4

(b) cos1 21   3

7. (a) tan1 1    4 8. (a) tan1 0  0   9. (a) cos1  12  23

10. (a) cos1 0   2

(b) cos1 1  0  3  3    1  3   (b) tan 3    2  1  2 4 (b) sin (b) tan1

(b) sin1 0  0

11. sin1 23  072973   13. cos1  37  201371

(c) sin1 2 is undefined.    (c) cos1  23  56    (c) cos1  22  34 

(c) tan1 33   6    3 1  3   (c) tan 6

(c) tan1 1   4   (c) sin1  12    6

  12. sin1  89  109491   14. cos1 49  111024

15. cos1 092761  275876

16. sin1 013844  013889

17. tan1 10  147113

18. tan1 26  153235

19. tan1 123456  088998

20. cos1 123456 is undefined because 123456  1.

21. sin1 025713  026005   23. sin sin1 14  14   25. tan tan1 5  5    27. sin sin1 32 is undefined because 32  1.    29. cos cos1  15   15

22. tan1 025713  025168   24. cos cos1 32  23   26. sin sin1 5 is undefined because 5  1.    28. tan tan1 32  32    30. sin sin1  34   34


38

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

     4 31. sin1 sin  4     33. sin1 sin 34 4    5  1 cos 6  56 35. cos     56 37. cos1 cos 76

     4 32. cos1 cos  4     34 34. cos1 cos 34     36. sin1 sin 56 6    7    38. sin1 sin 6 6

        4   39. tan1 tan  40. tan1 tan   4 3 3      2   11   1 1 tan 3   3 sin 4  4 41. tan 42. sin        43. sin cos1 21  23 44. tan sin1 23  3         46. tan cos1  22  tan 34  1 45. cos sin1 1  cos  2 0            23   2 3 1  3 47. sec tan1  33  sec      48. csc tan  csc  6 3 3 3          3 2  1 1 49. csc cot 1  csc 4  2 50. sin sec 2  sin 3  2    51. We want to find sec tan1 x . Let u  tan1 x. Using the Pythagorean identity, we have sec u   1  tan2 u.   Since u lies in the interval   2  2 and sec u is positive on that interval, we can use the cancellation property to write      sec u  1  tan2 u  1  x 2 , and so sec tan1 x  1  x 2 .    52. We want to find cos sin1 x . Let u  sin1 x. Using the Pythagorean identity, we have cos u   1  sin2 u.   Since u lies in the interval   2  2 and cos u is positive on that interval, we can use the cancellation property to write      cos u  1  sin2 u  1  x 2 , and so cos sin1 x  1  x 2 .   53. We want to find tan sin1 x . Let u  sin1 x. Using the Reciprocal and Pythagorean identities, we have     2u  1 1 sin x2 . If x  0, then u  0 and 1 1   tan u   sec2 u  1   2 2 2 cos u 1  x2 1  sin u 1  sin u   x . tan u  0, and similarly if x  0 then tan u  0, so tan sin1 x   1  x2   54. We want to find sin sec1 x . Let u  sec1 x. Using the Pythagorean identity, we have    sec2 u  1 1 2   . Since sin u, u, and x have the same sign, sin u   1  cos u   1  2 sec u sec2 u    x2  1 . sin sec1 x  x 55. If f x  e x and g x  arcsin x, then F x   f  g x  earcsin x .  2 56. If f x  x 2 and g x  tan1 x, then F x   f  g x  tan1 x . 57. If f x  sin1 x and g x  1x, then F x   f  g x  sin1 1x. 1 1 and g x  tan1 x, then F x  58. If f x  . 1x 1  tan1 x

2

59. If f x  e x , g x  arcsin x, and h x  x 2 , then F x   f  g  h x  earcsin x .   60. If f x  tan1 x, g x  x, and h x  x 2  1, then F x   f  g  h x  tan1 x 2  1.   2 61. If f x  tan1 x, g x  e x , and h x  1  x 2 , then then F x   f  g  h x  tan1 e1x .


39

SECTION 6.5 Inverse Trigonometric Functions and Their Graphs

   62. If f x  ln x, g x  arctan x, and h x  x 4 , then F x   f  g  h x  ln arctan x 4 . 63. (a) The arctangent function is defined everywhere, as is x 2 , so the domain of   f x  tan1 x 2 is .

2 1

(b) Because x 2 is even, f x is itself even. (See Exercise 2.7.92.) Also,

x 2  0 for all x and tan1 x  0 for x  0, so f x  0 everywhere.

2 Because tan1 x has a horizontal asymptote at y   2 and x increases

­10

10

without bound, f x has the same asymptote.

64. (a) For f x  sin1 x 2 to be defined we must have 1  x 2  1 

2

1  x  1, so the domain of f x is [1 1].   (b) Because x 2 is even, f x  sin1 x 2 is even. (See Exercise 2.7.92.)

1

Also, x 2  0 for all x and sin1 x  0 for x  0, so f x  0 on its

domain.

­1

1

65. (a) The cosine function has domain  and its range is [1 1], which lies within the domain of arcsin. Thus, f x  sin1 cos x has domain . (b) Because cos x is even, f x  sin1 cos x is even. It is also periodic     x if    x  0 2 2 with period . Note that f x     x if 0  x   2

1

­10

10 ­1

2

  66. (a) The arctangent function is defined everywhere and has range   2  2 , so   tan1 x is never 0 and thus f x  2

1

4

  tan1 x has domain . 2

2

2 is the y­intercept. Because tan1 x approaches  as x (b) f 0   2

1 1    . As  2 2

becomes large, f x has a horizontal asymptote at y  

­4

­2

2

4

x gains negative magnitude, the denominator of f x approaches 0 and so f x grows without bound in the negative x­direction. 67. (a)

From the graph of y  sin1 x  cos1 x, it appears that y  157. We suspect

that the actual value is  2.

2

­1

(b) To show that sin1 x  cos1 x   2 , start with the identity    sin a  2   cos a and take arcsin of both sides to obtain 1

1  cos a. Now let a  cos1 x. Then a 2  sin    1  cos cos1 x cos1 x    sin  sin1 x   sin1 x, so 2

sin1 x  cos1 x   2.


40

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

68. (a)

From the graph of y  tan1 x  tan1 1x , it appears that y  157 for x  0

2

and y  157 for x  0. We suspect that the actual values are   2.

­2

(b) To show that tan1 x  tan1 x1   2 for x  0, we start with the identity   tan a  cot 2  a and take arccot of both sides to obtain

2

cot1 tan a    a. Now use the identity cot1 x  tan1 x1 to write   2 1 1 , we have tan1 tan1 a   x 2  a. Substituting a  tan    1     tan1 1  tan1 1    tan1 1  tan1 1 x x 2 1x 2

­2

tan tan

1x

. For the case x  0, simply note that tan1 x   tan1 x, so for positive x, tan1 x  tan1 x1    2    1   tan1 x  tan1 x1    tan1 x  tan1 x 2.

  69. The domain of f x  sin sin1 x is the same as that of sin1 x, [1 1], and the graph of f is the same as that of y  x on [1 1].

The domain of g x  sin1 sin x is the same as that of sin x,  , because for all x, the value of sin x lies within

 the domain of sin1 x. g x  sin1 sin x  x for   2  x  2 . Because the graph of y  sin x is symmetric about the

 3 line x   2 , we can obtain the part of the graph of g for 2  x  2 by reflecting the graph of y  x about this vertical line. The graph of g is periodic with period 2. y

y

¹/2

1

y=g(x)

y=f(x) _1

1

x

_3¹/2

_¹/2

_1

6.6

¹/2

¹

3¹/2 x

_¹/2

MODELING HARMONIC MOTION

1. (a) Because y  0 at time t  0, y  a sin t is an appropriate model.

(b) Because y  a at time t  0, y  a cos t is an appropriate model.

2. (a) Because y  0 at time t  0, y  aect sin t is an appropriate model.

(b) Because y  a at time t  0, y  aect cos t is an appropriate model.

2 , and the phase 3. (a) For an object in harmonic motion modeled by y  A sin kt  b the amplitude is A, the period is k   b is b. To find the horizontal shift, we factor k to get y  A sin k t  . From this form of the equation we see that the k b horizontal shift is . k (b) For an object in harmonic motion modeled by y  5 sin 4t   the amplitude is 5, the period is  2 , the phase is , and the horizontal shift is  4.

  4. Objects A and B are in harmonic motion modeled by y  3 sin 2t   and y  3 sin 2t   2 . The phase of A is  and  the phase of B is  2 . The phase difference is 2 , so the objects are moving out of phase.


SECTION 6.6 Modeling Harmonic Motion

5. y  2 sin 3t

1 (a) Amplitude 2, period 23 , frequency period  23 .

(b)

6. y  3 cos 12 t 2  4, frequency (a) Amplitude 3, period 12 1

y

1

period  4 .

2

(b) ¹ 3

2¹ 3

y 3

x 4¹ x

7. y   cos 03t

  20 , frequency 3 . (a) Amplitude 1, period 203 3 20

(b)

8. y  24 sin 36t

  5 , frequency 9 . (a) Amplitude 24, period 236 9 5

(b)

y

y 2.4

1

5¹ 18

5¹ 9

x

20¹ x 3

10¹ 3

_1

    3t    025 cos 9. y  025 cos 15t   3  2 3   025 cos 32 t  29

2  4 , frequency 3 . (a) Amplitude 025, period 32 3 4

(b)

10. y   32 sin 02t  14   32 sin 02 t  7   10, frequency 1 . (a) Amplitude 32 , period 202 10

(b)

y

y

1.5

0.25 -7 2¹ 9

8¹ 9

14¹ 9

x

5¹-7

x

10¹-7

_1.5

_0.25

    11. y  5 cos 23 t  34  5 cos 23 t  98

12. y  16 sin t  18

2  3, frequency 1 . (a) Amplitude 5, period 23 3

(b)

y

(a) Amplitude 16, period 2, frequency 21 .

(b)

y 1.6

5

¹

1.8+2 9

_8

3¹ 9 2 _8

3¹ _ 9 8

x

1.8

13. The amplitude is a  10 cm, the period is 2k  3 s, and f 0  0, so f t  10 sin 23 t.

1.8+2 1.8+¹

x

1.8+2¹

41


42

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

14. The amplitude is 24 ft, the period is 2k  2 min, and f 0  0, so f t  24 sin t. 5 Hz, and f 0  0, so f t  6 sin 10t. 15. The amplitude is 6 in., the frequency is 2k  

16. The amplitude is 12 m, the frequency is 2k  05 Hz, and f 0  0, so f t  12 sin t. 17. The amplitude is 60 ft, the period is 2k  05 min, and f 0  60, so f t  60 cos 4t. 18. The amplitude is 35 cm, the period is 2k  8 s, and f 0  35, so f t  35 cos  4 t. 19. The amplitude is 24 m, the frequency is 2k  750 Hz, and f 0  24, so f t  24 cos 1500t. 20. The amplitude is 625 in., the frequency is 2k  60 Hz, and f 0  625, so f t  625 cos 120t. 21. (a) k  2, c  15, and f  3    6, so we have y  2e15t cos 6t.

(b)

have y  100e005t cos  2 t. y 100

5

y  075e3t cos 23 t. y 0.75

x

25. (a) k  7, c  10, and p   6    12, so we have y  7e10t sin 12t.

¹/5

x

26. (a) k  1, c  1, and p  1    2, so we have y  et sin 2t.

(b)

y

x

24. (a) k  075, c  3, and p  3    23 , so we have (b)

4

y 15

x

23. (a) k  100, c  005, and p  4     2 , so we

(b)

have y  15e025t cos 12t.

(b)

y 2

1

(b)

22. (a) k  15, c  025, and f  06    12, so we

y 0.5

1 ¹/30

x

1

x


SECTION 6.6 Modeling Harmonic Motion

27. (a) k  03, c  02, and f  20    40, so we have y  03e02t sin 40t.

28. (a) k  12, c  001, and f  8    16, so we

y

(b)

have y  12e001t sin 16t.

(b)

y 10

0.2 0.2

0.4

0.6

0.8

x

1

x

2

      29. y  5 sin 2t   2  5 sin 2 t  4 has amplitude 5, period , phase 2 , and horizontal shift 4 .     30. y  10 sin t   3 has amplitude 10, period 2, phase 3 , and horizontal shift 3 .

  2  31. y  100 sin 5t    100 sin 5 t   5 has amplitude 100, period 5 , phase , and horizontal shift  5 .     1 2  2 32. y  50 sin 12 t   5  50 sin 2 t  5 has amplitude 50, period 4, phase  5 , and horizontal shift  5 .

      33. y  20 sin 2 t   4  20 sin 2t  2 has amplitude 20, period , phase 2 , and horizontal shift 4 .     8 sin 4t    has amplitude 8, period  , phase   , and horizontal shift   . 34. y  8 sin 4 t  12 3 2 3 12     5 35. y1  10 sin 3t   2 ; y2  10 sin 3t  2 5 (a) y1 has phase  2 and y2 has phase 2 .

y 10

(d)

5 (b) The phase difference is  2  2  2.

yÁ, yª

5

(c) Because the phase difference is a multiple of 2, the curves are in

phase.

0

¹ t

_5 _10

   36. y1  15 sin 2t   3 ; y2  15 sin 2t  6 

 (a) y1 has phase  3 and y2 has phase 6 .

y

(d)

10

  (b) The phase difference is  3  6  6.

(c) Because the phase difference is not a multiple of 2, the curves are out of phase.

0

¹ t

_10

    80 sin 5t   ; y  80 sin 5t    37. y1  80 sin 5 t  10 2 2 3  (a) y1 has phase  2 and y2 has phase 3 .

y 80

(d)

  (b) The phase difference is  2  3  6.

(c) Because the phase difference is not a multiple of 2, the curves are out of phase.

0

_80

¹ t

43


44

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

    3  20 sin 2t  3  20 sin  ; y 38. y1  20 sin 2 t    20 sin 2 t  2t 2 2 2

y 20

(d)

(a) y1 has phase  and y2 has phase 3.

(b) The phase difference is   3  2.

yÁ, yª

(c) Because the phase difference is a multiple of 2, the curves are in

0

phase.

¹ t

_20

39. p t  115  25 sin 160t

2  1  00125, (a) Amplitude 25, period 160  80

frequency

1  80. period

(c) The period decreases and the frequency increases.

P

(b)

140 120 100 80 0

40. y  02 cos 20t  8

 (a) The frequency is 20 2  10 cycles/min.

(c) Since y  02 cos 20t  8  02 1  8  82 and when t  0, y  82, the maximum displacement is 82 m.

(b)

0.01

0.02

t

y 8.4 8.2 8 7.8

0

0.2

0.4

t

41. The graph resembles a sine wave with an amplitude of 5, a period of 25 , and no phase shift. Therefore, a  5,

2 2  5 

  5, and a formula is d t  5 sin 5t. 42. From the graph we see that the amplitude is 6 feet and the period is 12 hours. Also, the sine curve is shifted 6 hours (exactly half its period) to the right, which is equivalent to reflecting the curve about the t­axis. Thus, the equation is    t  6 sin  t. y   6 sin 212 6

43. Since the mass travels from its highest point (compressed spring) to its lowest point in 12 s, it completes half a period in 12 s. 2  12    2. Also, a  5. So y  5 cos 2t. So, 12 one period  12 s  12   2  1    2. So y  2 cos 2t. 44. a  2,  45. Since the Ferris wheel has a radius of 10 m and the bottom of the wheel is 1 m above the ground, the minimum height is 2  , and so y  11  10 sin   t , where t is in 1 m and the maximum height is 21 m. Then a  10 and  20 s    10 10  seconds. 46. Let f t represent the measure of the angle  at time t. The amplitude of this motion is 10. Since the period is 2s, we have 2  2    . Thus f t  10 sin t.   47. a  02, 2  10, b  38     5 . Then y  02 sin 5 t  38.


SECTION 6.6 Modeling Harmonic Motion

48. (a) m  10 g, k  3, a  5 cm. Then f t  5 cos



45

 3 t . 10

 (b) The function f t  a cos kmt describes an object oscillating in simple harmonic motion, so by  comparing it with the general equation y  a cos t we see that   km. This means the frequency is   1  km k    . f  2 2 2 m  k (c) If the mass is increased, then the denominator of increases, so overall the frequency decreases. If the frequency m has decreased, then by definition the oscillations are slower.  k increases. Thus, overall the frequency increases. If (d) If a stiffer spring is used, k is larger and so the numerator of m the frequency has increased, then by definition the oscillations are faster. 49. The amplitude is 12 100  80  10 mmHG, the period is 24 hours, and the phase shift is 8 hours, so   f t  10 sin 12 t  8  90.

50. The signal consists of the seven phases corresponding to the digits 01, 11, 10, 01, 00, 00, and 10. Thus, the encoded string of digits is 01111001000010. 51. (a) The maximum voltage is the amplitude, that is, Vmax  a  45 V.

(b) From the graph we see that 4 cycles are completed every 01 seconds, or equivalently, 40 cycles are completed every second, so f  40.   f  40. (c) The number of revolutions per second of the armature is the frequency, that is, 2  (d) a  45, f   40    80. Then V t  45 cos 80t. 2   1130 52. (a) As the car approaches, the perceived frequency is f  500 1130110  5539 Hz. As it moves away, the perceived   1130 frequency is f  500 1130110  4556 Hz.      1130 1130 (b) The frequency is , so   1000 1130110  11078 (approaching) or 9113 (receding).  500 1130110 2 Thus, models are y  A sin 11078t and A sin 9113t. 53. k  1, c  09, and

  1    . Since f 0  0, f t  e09t sin t. 2 2

  2    4. 2 (a) Since f 0  6, f t  6e28t cos 4t.

54. k  6, c  28, and

1 (b) The amplitude of the vibration is 05 when 6e28t  05  e28t  12 1  ln 12  089 s. t   12 ln 12 28

55.

1  28t  ln 12

kect  4  ectct3  4  e3c  4  3c  ln 4  c  13 ln 4  046. kect3

56. (a)

3 kec0  5  e2c  5  2c  ln 5  c  12 ln 5  080.  06 kec2

(b) f t  3e08t cos t. If the frequency is 165 cycles per second, then f t  3e08t cos 330t.

  165    330. Thus, 2


46

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

57. (a) For fan A, the amplitude is 1, the frequency is 100, and the phase is 0, so y  sin 200t.

  100    200 and an equation is 2

For fan B, the amplitude is 1, the frequency is 100, and the phase is  34 , so again   200 and an equation is   y  sin 200t  34 .

(b) The phase difference is 34 , so the fans are out of phase. If fan A were rotated 34 counterclockwise, the phase difference would become 0 and the fans would be in phase.   58. (a) E I  50 sin 120t, so the voltage phase is 0. E I I  50 sin 120t  54 , so the voltage phase is 54 .

(b) The phase difference is 54 , so the generators are out of phase. If the armature in the second generator were rotated 34 , then the phase difference would become 54  34  2, and the generators would produce voltage in phase.

1  59. From left to right: at t   6 , sin t  2 and the values are increasing; at t  2 , sin t  1 and the values reach a maximum;

at t  56 , sin t  12 and the values are decreasing; at t  , sin t  0 and the values are decreasing; at t  76 , sin t   12 and the values are decreasing; at t  32 , sin t  1 and the values reach a minimum, and at t  116 , sin t   12 and the

values are increasing. 60. From left to right: new moon, waxing crescent moon, waxing gibbous moon, full moon, waning gibbous moon, third quarter moon, waning crescent moon, new moon. Tides and werewolf sightings are in phase with the lunar cycle; paying rent and cellphone bills are out of phase.

CHAPTER 6 REVIEW    2  2   1. (a) Since  23  12  34  14  1, the point P  23  12 lies on the unit circle. 

(b) sin t  12 , cos t   23 , tan t 

1  2   3 . 3  23

   2  2 9  1  1, the point P 3   4 lies on the unit circle. 2. (a) Since 35   45  25 2 5 5 4 (b) sin t   45 , cos t  35 , tan t  35   43 . 5

3. t  23

4. t  53

(a) t    23   3    3 1 (b) P  2  2

   (c) sin t  23 , cos t   12 , tan t   3, csc t  2 3 3 , 

sec t  2, and cot t   33 .

5. t   114

  (a) t  3   114   4     2 2 (b) P  2   2

   (c) sin t   22 , cos t   22 , tan t  1, csc t   2,  sec t   2, and cot t  1.

(a) t  2  53   3    3 1 (b) P 2   2

  (c) sin t   23 , cos t  12 , tan t   3, 

csc t   2 3 3 , sec t  2, and cot t   33 . 6. t   76 (a) t  76     6    3 1 (b) P  2  2

(c) sin t  12 , cos t   23 , tan t   33 , csc t  2,   sec t   2 3 3 , and cot t   3.


CHAPTER 6  2 7. (a) sin 34  sin   4 2

Review

47

 8. (a) tan  3  3    (b) tan   3  3

2 (b) cos 34   cos  4  2

10. (a) cos  5  080902   (b) cos  5  080902

9. (a) sin 11  089121 (b) cos 11  045360 11. (a) cos 92  cos  2 0

12. (a) sin  7  043388

(b) sec 92 is undefined

(b) csc  7  230476

13. (a) tan 52 is undefined

14. (a) sin 2  0

(b) cot 52  cot  2 0

(b) csc 2is undefined.

 15. (a) tan 56   33

1 16. (a) cos  3  2

 (b) cot 56   3

1 (b) sin  6  2

sin t tan t cos t  sin t  sin t  17. cos t cos t cos2 t 1  sin2 t

sin2 t 1  cos2 t 1 1    2 cos t cos t cos t cos3 t sin t sin t sin t    (because t is in quadrant IV, cos t is positive). 19. tan t  cos t  1  sin2 t 1  sin2 t 1 1 1     (because t is in quadrant II, cos t is negative). 20. sec t  2 cos t  1  sin t  1  sin2 t 18. tan2 t  sec t  tan2 t 

5

5 , cos t   12 . Then tan t  13   5 , csc t  13 , sec t   13 , and cot t   12 . 21. sin t  13 13 12 5 12 5  12 13

22. sin t   12 , cos t  0. Since sin t is negative and cos t is positive, t is in quadrant IV. Thus, t determines the terminal point       2 3 3 P 23   12 , and cos t  23 , tan t   , csc t  2, sec t  , cot t   3. 3 3   1 cos t 5 1 23. cot t   2 , csc t  2 . Since csc t  , we know sin t  2  2 5 5 . Now cot t  , so 5 sin t sin t      1 1 1       5   5. cos t  sin t  cot t  2 5 5   12   55 , and tan t     2 while sec t  5 1 5 cos t 2  5

24. cos t   35 , tan t  0. Since cos t and tan t are both negative, t is in quadrant II. Thus, t determines the terminal point   5 P  35 , 45 , and sin t  45 , tan t   43 , csc t  54 , sec t   , cot t   34 . 3  2 1 sin t 1 1   sin t  4 cos t. Thus, cos2 t  sin2 t  1  cos2 t  14 cos t  1  25. tan t  4 , so cot t  4 and cos t 4 

17 17 2 cos2 t  16 17  sec t  16 . Because cosine and secant are negative in Quadrant III, we have sec t   4 , and thus, 

sec t  cot t  4  417 .

  8 , so csc t   17 and cos2 t  1  sin2 t  1   8 2  225  sec2 t  289 . Because cosine and secant 26. sin t   17 8 17 289 225 17 17 119 are positive in Quadrant IV, sec t  17 15 , and so csc t  sec t   8  15   120 .

27. cos t  35 , so because the terminal point is in Quadrant I, sin t  45 . Thus, tan t  43 and sec t  53 , so tan t  sec t  43  53  3.


48

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

28. sin2 t  cos2 t  1 for any value of t.

29. y  10 cos 12 x

2

(a) This function has amplitude 10, period 1  4, 2

and horizontal shift 0. (b)

30. y  4 sin 2x

(a) This function has amplitude 4, period 22   1, and horizontal shift 0.

(b)

y

y 4 10 2¹

0.5

x

31. y   sin 12 x 2  4, and (a) This function has amplitude 1, period 12

(a) This function has amplitude 2, period 2, and horizontal shift  4.

(b)

y

y

2

1

2¹ 2¹

¹/4

x

33. y  3 sin 2x  2  3 sin 2 x  1

(a) This function has amplitude 3, period 22  , and horizontal shift 1.

(b)

x

  32. y  2 sin x   4

horizontal shift 0. (b)

1

x

¹

  34. y  cos 2 x   2

(a) This function has amplitude 1, period 22  , and horizontal shift  2.

y

(b)

y

3 x 1-¹/2

1

1+¹/2

1 ¹/2

¹

x


CHAPTER 6

   1 35. y   cos  2 x  6   cos 2 x  3

(a) This function has amplitude 1, period 2 2  4, and horizontal shift  13 . y

(b)

Review

49

    36. y  10 sin 2x   2  10 sin 2 x  4

(a) This function has amplitude 10, period 22  , and horizontal shift  4.

(b)

y

1

_13/3

11/3 x

_1/3

10 x ¹/2

¹

37. From the graph we see that the amplitude is 5 and the period is  2 . Considered as a sine function, there is no horizontal shift, and considered as a cosine function, the horizontal shift is  8 . Therefore, the function’s equation can be written as    y  5 sin 4x or y  5 cos 4 x  8 . 38. From the graph we see that the amplitude is 2 and the period is 4 (since 14 of the period has been completed at 1 2. Thus, 2  4  k   . Considered as a sine function, there is no horizontal shift, and considered as a cosine function, the k 2

 horizontal shift is 1. Therefore, the function’s equation can be written as y  2 sin  2 x or y  2 cos 2 x  1.

39. From the graph we see that the amplitude is 12 and the period is 1. Considered as a sine function, the horizontal shift is 1 . Therefore, the function’s equation can be written as  13 , and considered as a cosine function, the horizontal shift is  12     1 . y  12 sin 2 x  13 or y  12 cos 2 x  12

40. From the graph we see that the amplitude is 4 and the period is 43 , so 2k  43  k  32 . Considered as a sine function, the horizontal shift is   3 , and considered as a cosine function, the horizontal shift is 0. Therefore, the function’s equation   3 can be written as y  4 sin 32 x   3 or y  4 cos 2 x.

41. y  3 tan x has period .

42. y  tan x has period    1.

y

y

10

¹

2

x _1

1

x


50

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

  1 44. y  sec 12 x   2  sec 2 x   has period

  43. y  2 cot x   2 has period . y

2    4. 1 2

4

y

x

¹

2 _2¹

  45. y  4 csc 2x    4 csc 2 x   2 has period 2  . 2

x

5¹/6

x

  46. y  tan x   6 has period . y

y

2 5

_¹/6 ¹

x

   1  47. y  tan 12 x   8  tan 2 x  4 has period 1  2. 2

1 48. y  4 sec 4x has period 24   2. y

y

10 2 ¹/4

5¹/4

x

_0.5 _0.25

0.25

x

0.5

  49. sin1 1   50. cos1  12  23 2       51. sin1 sin 136   52. tan cos1 12  tan  6 3  3     100 sin 8t    has amplitude 100, period  , phase   , and horizontal shift   . 53. y  100 sin 8 t  16 2 4 2 16     3 has amplitude 80, period 2 , phase 3 , and horizontal shift  . 54. y  80 sin 3 t    80 sin 3t  2 2 3 2 2       3 5 55. y1  25 sin 3 t   2  25 sin 3t  2 ; y2  10 sin 3t  2 (a) y1 has phase 32 and y2 has phase 52 .

y

(d)

20

(b) The phase difference is 32  52  . (c) Because the phase difference is not a multiple of 2, the curves are out of phase.

0 _20

yª ¹ x


CHAPTER 6

      56. y1  50 sin 10t   2 ; y2  50 sin 10 t  20  50 sin 10t  2  (a) y1 has phase  2 and y2 has phase 2 .

y

(d)

yÁ, yª

40

(b) The phase difference is     0. 2

2

(c) Because the phase difference is a multiple of 2, the curves are in

0

_¹/2

phase.

¹/2 x

_40

58. (a) y  sin cos x

57. (a) y  cos x

1

1 ­5 ­5

5 ­1

5

(b) This function has period .

(b) This function has period 2.

(c) This function is even.

(c) This function is even.

  59. (a) y  cos 201x

60. (a) y  1  2cos x 4 1 2

­50

50 ­1

­10

10

(b) This function is not periodic.

(b) This function has period 2.

(c) This function is neither even nor odd.

(c) This function is even.

61. (a) y  x cos 3x

62. (a) y 

x sin 3x (x  0)

5

0 ­5

Review

5

5

­5

(b) This function is not periodic.

(b) This function is not periodic.

(c) This function is even.

(c) This function is neither even nor odd.

10

51


52

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

63. y  x sin x is a sine function whose graph lies between those of y  x and y  x.

64. y  2x cos 4x is a cosine function whose graph lies between the graphs of y  2x and y  2x .

10

5

­10

10

­2

­10

2 ­5

65. y  x  sin 4x is the sum of the two functions y  x and y  sin 4x.

66. y  sin2 x  cos2 x is the sum of the two functions

y  sin2 x and y  cos2 x. Note that sin2 x  cos2 x  1

for all x. 2

1 ­2

2 ­2 ­5

67. We graph y  f x  cos x  sin 2x in the viewing

rectangle [0 2]  [2 2] and see that the function has local maxima of approximately f 063  176 and

5

68. We graph y  f x  cos x  sin2 x in the viewing

rectangle [0 2]  [2 2] and see that the function has local maxima of approximately

f 414  037 and local minima of approximately

f 105  f 524  125 and local minima of

periodic with period 2.

periodic with period 2.

f 251  176 and f 528  037. The function is

f 0  f 2  1 and f   1. The function is 2

2

0

2

4

0

6

69. We want to find solutions to sin x  03 in the interval [0 2], so we plot the functions y  sin x and

y  03 and look for their intersection. We see that the

graphs intersect at x  0305 and at x  2837.

6

70. We want to find solutions to cos 3x  x in the interval [0 ], so we plot the functions y  cos 3x and y  x and

look for their intersection. We see that the graphs intersect at x  0390. 1

1

­1

4

­2

­2

0

2

2

4

6

0

­1

1

2

3


CHAPTER 6

71. y1  cos sin x, y2  sin cos x (a)

1

Review

53

(b) y1 has period , while y2 has period 2.

(c) sin cos x  cos sin x for all x. ­10

0

10

­1

72. The amplitude is a  50 cm. The frequency is 8 Hz, so   8 2  16. Since the mass is at its maximum displacement when t  0, the motion follows a cosine curve. So a function describing the motion of P is f t  50 cos 16t. 73. The amplitude is 12 100  50 cm, the frequency is 4 Hz, so   4 2  8. Since the mass is at its lowest point when t  0, a function describing the distance of the mass from its rest position is f t  50 cos 8t.

74. (a) The initial amplitude is 16 cm and the frequency is 14 Hz, so

(b)

a function describing the motion is y  16e072t cos 28t.

20

(c) When t  10 s, y  16e72 cos 28  00119 cm.

0

2

­20

75. (a) y 

x x  has vertical asymptotes at x  2 and y­intercept 0, so it has graph VII. 2  x 2  x 4  x2

2 (b) y  4 cos  2 x  1 has amplitude 4 and period 2  4, so it has graph I.

(c) y  tan 4x has period 4  4, y­intercept 0, and vertical asymptotes where 4x   2  n  x  4n  2, n an integer, so it has graph V.

(d) y  x 3 is a polynomial of odd degree and increases without bound in the positive direction, so it has graph III.

(e) y  x cos 4x is bounded by y  x since cos 4x  1 for all x. It fluctuates between x and x on every interval of length  2 , so it has graph VI.

(f) y  1  3 sin 2x has amplitude 3, period 2 2  4, and y­intercept 1. Its graph can be obtained by stretching that of y  sin x horizontally by a factor of  2 , then stretching vertically by a factor of 3, then shifting 1 unit upward. This corresponds to graph IV.

x  (g) y  sec 4x has period 2 4  8, y­intercept 1, and vertical asymptotes where 4  2  n  4n  2, so it has

graph II.

4 tan1 x has horizontal asymptotes at y   4    2, so it has graph VIII. (h) y    2


54

CHAPTER 6 Trigonometric Functions: Unit Circle Approach

CHAPTER 6 TEST 

1. Since P x y lies on the unit circle, x 2  y 2  1 y   1  quadrant. Therefore y is negative

y   56 .

  2 11 6

 5   25 36   6 . But P x y lies in the fourth

 2 9 , and so x   3 . From the diagram, 2. Since P is on the unit circle, x 2  y 2  1  x 2  1  y 2 . Thus, x 2  1  45  25 5   3 3 4 x is clearly negative, so x   5 . Therefore, P is the point  5  5 . (a) sin t  45

(b) cos t   35

4

(c) tan t  53   43 5

(d) sec t   53  13   22 4   3 (d) csc 2  1

3. (a) sin 76  05    (c) tan  53  3 4. tan t 

(b) cos

sin t sin t   . But t is in quadrant II cos t  1  sin2 t

Thus, tan t 

sin t  .  1  sin2 t

8 , t in quadrant III 5. cos t   17

quadrant III) 1  6. y  5 cos 4x

cos t is negative, so we choose the negative square root.

tan t  cot t  csc t  1 

1 1 2 .   1  15   15 64  1  289 17

(a) This function has amplitude 5, period 24   2, phase 0, and horizontal shift 0.

(b)

1  (since t is in  1  cos2 t

  1  7. y  2 sin 12 x   6  sin 2 x  3

2 (a) This function has amplitude 2, period 1  4, 2

 phase  6 , and horizontal shift 3 .

y

(b)

5 0

¹/2 x

¹/4

y 2 ¹/3

   9. y  tan 2 x   4 has period 2 .

8. y   csc 2x has period 22  . y

y

2 _¹/2

10. (a) tan1 1   4

13¹/3 x

7¹/3

2 ¹/2

_¹/2 x

   (b) cos1  23  56

¹/2 x

(c) tan1 tan 3  0

     1  (d) cos tan1  3  cos   3  2


CHAPTER 6 2 11. From the graph, we see that the amplitude is 2 and the phase shift is   3 . Also, the period is , so k     k  2  2. Thus, the function is y  2 sin 2 x   3 .       12. y1  30 sin 6t  2 ; y2  30 sin 6t  3

 (a) y1 has phase  2 and y2 has phase 3 .

y 30

(d)

  (b) The phase difference is  2  3  6.

(c) Because the phase difference is not a multiple of 2, the curves are out of phase.

55

0

_¹/2

Test

¹/2 x

_30

13. y  (a)

cos x 1  x2

(b) The function is even. 1.0

(c) The function has a minimum value of approximately 011 when x  254 and a maximum value of 1 when x  0.

0.5

­10

10

14. The amplitude is 12 10  5 cm and the frequency is 2 Hz. Assuming that the mass is at its rest position and moving upward when t  0, a function describing the distance of the mass from its rest position is f t  5 sin 4t.

15. (a) The initial amplitude is 16 in. and the frequency is 12 Hz, so a function describing the motion is y  16e01t cos 24t.

(b)

20

0

­20

0.5

1.0


56

FOCUS ON MODELING

FOCUS ON MODELING Fitting Sinusoidal Curves to Data Replace "x" with "t" in all graphs in #1-5 1. (a) Using the method of Example 1, we find the vertical shift b  12 maximum value  minimum value  12 21  21  0, the

y 2

amplitude

1

a  12 maximum value  minimum value  12 21  21  21, the

0

period

_1

2  2 6  0  12 (so   05236), and the phase shift c  0.  Thus, our model is y  21 cos  6 t.

2

4

6

8

10

12

14 x

2

4

6

8

10

12

14 x

_2

From the graph, we see that the curve fits the data quite well. (b) Using the SinReg command on the TI­84, we find

y  2048714222 sin 05030795477t  1551856108  00089616507.

This model is equivalent to   y  205 cos 050t  155   2  001  205 cos 050t  002  001.

This is the same as the function in part (a), correct to one decimal place.

y 2 1 0 _1 _2

2. (a) Using the method of Example 1, we find the vertical shift b  12 maximum value  minimum value  12 063  010  0265,

y 0.6

the amplitude

0.4

a  12 maximum value  minimum value  12 063  010  0365,

0.2

2  2 55  25  6 (so   1047), and the phase shift the period  c  05. Thus, our model is y  0365 cos 1047 t  05  0265. From the graph, we see that the curve fits the data reasonably well.

(b) Using the SinReg command on the TI­84, we find y  0327038879 sin 1021164911t  2118186963  02896017397.

This model is equivalent to   033 cos 102t  212   2  029  033 cos 102t  055  029.

This is the same as the function in part (a), correct to one decimal place.

0

2

4

6

x

2

4

6

x

_0.2 y 0.6 0.4 0.2 0 _0.2


57

Fitting Sinusoidal Curves to Data

3. (a) Let t be the time since midnight. We find a function of the form y  a sin  t  c  b. a  12 374  366  04. The period is 24 and

  026. b  1 374  366  37. Because the maximum so   224 2

y

37

value occurs at t  16, we get c  16. Thus the function is y  04 cos 026 t  16  37.

36

(b) Using the SinReg command on the TI­84 we obtain the function y  a sin bt  c  d, where a  04, b  026, c  262, and

d  370. Thus we get the model y  04 sin 026t  262  370. 4. (a) Let t be the time in years. We find a function of the form y  a sin  t  c  b, where y is the owl population.

a  12 80  20  30. The period is 2 9  3  12 and so

  052. b  1 80  20  50. Because the values start at the   212 2

middle we have c  0. Thus the function is y  30 sin 052t  50.

(b) Using the SinReg command on the TI­84 we find that for the function y  a sin bt  c  d, where a  258, b  052, c  002, and

d  506. Thus we get the model y  258 sin 052t  002  506. 5. (a) Let t be the time since 1985. We find a function of the form y  a sin  t  c  b. a  12 63  22  205. The period is

0

10

x

20

Time y 80 60 40 20

0

2

4

6

8

10

10

12

12 x

Year y 60

  052. b  1 63  22  425. The 2 15  9  12 and so   220 2

40

y  205 sin 052 t  6  425.

20

average value occurs in the ninth year, so c  6. Thus, our model is

(b) Using the SinReg command on the TI­84, we find that for the function y  a sin bt  c  d, where a  178, b  052, c  311, and

d  424. Thus we get the model y  178 sin 052t  311  424.

0

2

4

6

8

Year since 1985

14

x


CORRECTIONS: p. 8,14,25,40

CHAPTER 7

ANALYTIC TRIGONOMETRY

7.1

Trigonometric Identities 1

7.2 7.3

Addition and Subtraction Formulas 9 Double-Angle, Half-Angle, and Product-Sum Formulas 17

7.4

Basic Trigonometric Equations 27

7.5

More Trigonometric Equations 31 Chapter 7 Review 38 Chapter 7 Test 45

¥

FOCUS ON MODELING: Traveling and Standing Waves 46

1


7

ANALYTIC TRIGONOMETRY

7.1

TRIGONOMETRIC IDENTITIES

1. An equation is called an identity if it is valid for all values of the variable. The equation 2x  x  x is an algebraic identity

and the equation sin2 x  cos2 x  1 is a trigonometric identity. 2. The fact that cos x has the same value as cos x for any x can be expressed as the identity cos x  cos x. sin t  sin t 3. cos t tan t  cos t  cos t 1  cot t 4. cos t csc t  cos t  sin t 1  tan  5. sin  sec   sin   cos  sin  1 1 6. tan  csc      sec  cos  sin  cos  1 sin2 x  1  cos2 x sin2 x 7. tan2 x  sec2 x      1 2 2 2 cos x cos x cos x cos2 x 1 sec x sin x  cos x   tan x 8. 1 csc x cos x sin x   1 2 9. sin 2  y sec y  cos2 y sec y  cos2 y   cos y cos y     sin 2  u cos u   10. tan  2  u sin u  cos   u  sin u  sin u  sin u  cos u 2 sin2 u  cos2 u 1 cos u  cos u    csc u sin u sin u sin u     sin2  12. cos2  1  tan2   cos2  1   cos2   sin2   1 cos2 

11. sin u  cot u cos u  sin u 

1  cos  1  cos2  sin2  sin  sec   cos  cos       tan  13. sin  sin  sin  cos  sin  cos  cos  cos  cot  cos  cos  sin  14.  sec     2 2 1 csc   sin  cos 1  sin   sin  sin  1 sin x  sin x sec x cos x 1  15. cos x tan x sin x 1 cos x  cos x sec x cos x  sin x  16. cos x cot x cos x sin x sin t sin t  tan t tan t   cos t  1 17.  sin t tan t tan t cos t

1


2

CHAPTER 7 Analytic Trigonometry

18.

cos A 1  cot A  1  cot A sin A  sin A  sin A cot A  sin A  sin A  sin A  cos A csc A sin A

    2 3 2 19. sin3  2  x  sin x cos x  cos x  1  cos x cos x  cos x

   20. sin4   cos4   cos2   sin2   cos2  sin2   cos2   cos2   sin2  21.

tan2 x sin2 x sec2 x  1    cos2 x  sin2 x. 2 2 sec x sec x cos2 x 1 sec2 x  1 1  1  cos2 x  sin2 x Another method: 2 sec x sec2 x

cos2 x 1  2 2 sec x  cos x cos x  1  cos x  sin x  sin x 22.  cos x sin x tan x sin x sin x cos x 23.

1  cos y cos y 1  cos y 1  cos y 1  cos y    cos y   cos y  1 1 1  sec y 1 cos y  1 1 cos y cos y

  1  cos y   1  sin y sin y 1  sin y sin y  sin2 y 2 24.    sin y  1 1  csc y sin y  1 1  sin y 1 sin y 25.

cos u 1  sin u2  cos2 u 1  2 sin u  sin2 u  cos2 u 1  2 sin u  1 1  sin u     cos u 1  sin u cos u 1  sin u cos u 1  sin u cos u 1  sin u 2  2 sin u 2 1  sin u 2     2 sec u cos u 1  sin u cos u 1  sin u cos u

26. Note that because sin2 t  1  cos2 t  1  cos t 1  cos t, we can write sin t 1  cos t 1 cos t  csc t     cot t. 1  cos t sin t sin t sin t 27.

cos x cos x   sec x  tan x sec x  tan x

28. Because tan A   tan A, 29.

30.

1  cos t sin t  . Thus, 1  cos t sin t

cos x 1  sin2 x cos2 x 1  sin x 1  sin x    1  sin x  sin x 1 1  sin x 1  sin x 1  sin x  cos x cos x

cot A  1 cot A  1 tan A 1  tan A     cot A 1  tan A 1  tan A tan A tan A 1  tan A

1 1  sin   1  sin  1 2    2 sec2   1  sin  1  sin  cos2  1  sin2  1  1  tan2 x 1 1  tan2 x 1 sec2 x 2  tan2 x 1  1  1  1 2 2 2 2 2 sec x sec x sec x sec x sec x sec2 x 1 1 11  cos2 x  sec2 x sec2 x


SECTION 7.1 Trigonometric Identities

cos x 31. (a)  sec x sin x

sin y tan y 1  cos2 y sin2 y cos y 32. (a)    1 csc y cos y cos y sin y

1  sin2 x cos2 x   sin x sin x

cos x

1  sin x cos x sin2 x 1   csc x  sin x  sin x sin x

(b) We graph each side of the equation and see that the cos x and y  csc x  sin x are graphs of y  sec x sin x identical, confirming that the equation is an identity.

cos2 y 1   sec y  cos y cos y cos y

(b) We graph each side of the equation and see that the tan x graphs of y  and y  sec x  cos x are csc x identical, confirming that the equation is an identity.

1

-5

3

1 5

-5

-1

5 -1

sin  sin  1 tan  cos      sec  34. sin  sin  cos  sin 

cos  cos   33.  cos   cos   cos2  1 sec  cos  1 cos u sec u  cos u cot u  cot u 35. tan u cos u   1 2 37. cos2  2  y csc y  sin y  sin y  sin y

36.

cot x sec x cos x 1    sin x  1 csc x sin x cos x

      sin x      2     cos x     sin x   38. sin x  cos x  2 , so tan x  2 sin x   2 2 cos x  

2

39. sin x  cos x2  sin2 x  2 sin x cos x  cos2 x  1  2 sin x cos x

cos  sin2   cos2  1 sin      sec  csc  cos  sin  cos  sin  cos  sin  41. cos x  sin x  cos x   sin x  cos x  sin x 40. tan   cot  

42. cot  cos   sin   

 cos2   sin2  1 cos  cos   sin      csc  sin  sin  sin 

1  1 cos A sec A  1 1  cos A cos A 43.    1 sec A  1 cos A 1  cos A 1 cos A cos2  1  sin2  1 44.    sin   csc   sin  sin  sin  sin  45. 1  cos  1  cos   1  cos2   sin2  

1 csc2 

sin x cos x   cos2 x  sin2 x  1 sec x csc x 1 1 47.   sec2 y  1  tan2 y 2 cos2 y 1  sin y

46.

48. csc x  sin x 

1  sin2 x cos2 x 1  sin x    cos x cot x sin x sin x sin x

    49. tan x  cot x2  tan2 x  2 tan x cot x  cot2 x  tan2 x  2  cot2 x  tan2 x  1  cot2 x  1  sec2 x  csc2 x


4

CHAPTER 7 Analytic Trigonometry

  50. sin2 y  cos2 y  tan2 y  sin2 y  cos2 y  tan2 y  1  tan2 y  sec2 y 2  2    51. 1  sin2 t  cos2 t  4 sin2 t cos2 t  2 cos2 t  4 sin2 t cos2 t  4 cos2 t cos2 t  sin2 t  4 cos2 t 52.

2 sin x cos x

sin x  cos x2  1

 

2 sin x cos x

sin2 x  cos2 x  2 sin x cos x  1 2 sin x cos x 1 2 sin x cos x  1  1

2 sin x cos x   2 sin x cos x  sin2 x  cos2 x  1

sin2 x 1 cos2 x    csc x sin x sin x sin x     cos2 t 1 2 t  cos2 t 2 t csc2 t  1  cot2 t cos2 t 54. cot2 t  cos2 t   cos  1  cos sin2 t sin2 t 1 1  cos t  cos t cos t 1  cos2 t sec t  cos t cos t cos t    sin2 t   55. 1 1 sec t cos t 1 cos t cos t cos x cos x 1 cos2 x    56. cot x  csc x cos x  1  cot x cos x  cot x  csc x cos x  csc x  sin x sin x sin x sin x cos2 x  1  sin2 x     sin x sin x sin x   57. cos2 x  sin2 x  cos2 x  1  cos2 x  2 cos2 x  1   58. 2 cos2 x  1  2 1  sin2 x  1  2  2 sin2 x  1  1  2 sin2 x  2  2    59. sin4   cos4   sin2   cos2   sin2   cos2  sin2   cos2   sin2   cos2       cos2 x  sin2 x  cos2 x  1 60. 1  cos2 x 1  cot2 x  sin2 x 1  sin2 x 53. csc x cos2 x  sin x 

sin2 t  2 sin t cos t  cos2 t sin2 t  cos2 t 2 sin t cos t 1 sin t  cos t2      2  2  sec t csc t sin t cos t sin t cos t sin t cos t sin t cos t sin t cos t   1 1 sin t cos t 1 1 62. sec t csc t tan t  cot t     2  sec2 t  csc2 t cos t sin t cos t sin t cos2 t sin t

61.

sin2 u sin2 u 1 1 2 2 2 2 1  tan2 u 1 cos u  cos2 u  cos u  cos u  sin u  63.  2 2 2 2 2 2 2 1  tan u cos u  sin u cos u  sin2 u sin u sin u cos u 1 1 2 2 cos u cos u 1  sec2 x 1  sec2 x 1 64.    1  cos2 x  1 1  tan2 x sec2 x sec2 x 1 1 1 1   sec x  csc x cos x sin x cos x sin x  sin x cos x  sin x  cos x  sin x  cos x 65.   cos x cos x sin x cos x sin x sin x tan x  cot x sin2 x  cos2 x   cos x sin x cos x sin x sin x  cos x sin x  cos x sin x  cos x cos x sin x   cos x sin x 66.   sin x  cos x 1 1 sin x  cos x sec x  csc x sin x  cos x  cos x sin x cos x sin x sin x 1  cos x 1  cos x sin x sin x 1  2 cos x  cos2 x  sin2 x 1  cos x       67. sin x 1  cos x sin x 1  cos x 1  cos x sin x sin x 1  cos x 2  2 cos x 2 1  cos x    2 csc x sin x 1  cos x sin x 1  cos x


SECTION 7.1 Trigonometric Identities

68.

csc2 y  cot2 y csc2 y  cot2 y sin2 y 1  cos2 y   sin2 y  2 2 2 sec y sec y tan2 y sin y 1 Another method:

csc2 y  cot2 y sec2 y

sin2 y

cos2 y sin2 y

 cos2 y

cos2 y sin2 y

1 cos2 y

1  cos2 y sin2 y

 cos2 y  cos2 y

 sin2 u sin2 u cos2 u sin2 u  2 u  tan2 u sin2 u   1  cos cos2 u cos2 u cos2 u      4 4 2 2 2 70. sec x  tan x  sec x  tan x sec x  tan2 x  1 sec2 x  tan2 x  sec2 x  tan2 x

69. tan2 u  sin2 u 

sin x 1 1  tan x cos x  cos x  cos x  sin x 71.  sin x cos x 1  tan x cos x  sin x 1 cos x sin2   1 1 sin    cos2  cos  sin   csc  sin   sin     72. cos  cos  sin   cos  cos   cot  cos  sin   1 1  sin  cos   sin  sin  1 sec x  tan x  sec x  tan x 2 sec x 2 sec x 1    2 sec x   73. 2 2 sec x  tan x sec x  tan x 1 sec x  tan x sec x  tan x sec x  tan x 74.

cos2 t  tan2 t  1

1 sin2 t   1  sec2 t  tan2 t cos2 t sin2 t sin2 t sin2 t 1  sin x 1 sin x 4 sin x 4 sin x 1  sin x 1  sin x2  1  sin x2     4 tan x sec x 4   75. 1  sin x 1  sin x cos x cos x 1  sin x 1  sin x cos2 x 1  sin2 x sin x sin x cos y  cos x sin y sin y  tan x  tan y cos x cos y cos x cos y  cos x 76. cos y  cos x sin y  sin x cos y cot x  cot y  sin x sin y sin x sin y    sin x sin y sin x cos y  cos x sin y sin x sin y    tan x tan y cos x cos y cos x sin y  sin x cos y cos x cos y   sin x  cos x sin2 x  sin x cos x  cos2 x sin3 x  cos3 x 77.   sin2  sin x cos x  cos2 x  1  sin x cos x sin x  cos x sin x  cos x   sin y  csc y sin2 y  sin y csc y  csc2 y sin3 y  csc3 y   sin2 y  csc2 y  1 78. sin y  csc y sin y  csc y 79.

1  cos  1  cos  1  cos2  sin2  sin  1  cos       sin  sin  1  cos  sin  1  cos  sin  1  cos  1  cos 

 sin2 t  tan2 t

 1 

sin x  1 sin x  1 sin x  1 sin2 x  1  cos2 x     2 sin x  1 sin x  1 sin x  1 sin x  1 sin x  12 1 sin  sin  cos   tan    81. 1 sin   cos  sin   cos  1  tan  cos  sin A sin A 1  cos A sin A 1  cos A sin A 1  cos A cos A 82.  cot A    cot A    cot A  1  cos A 1  cos A 1  cos A sin A 1  cos2 A sin2 A 1 cos A cos A 1      csc A sin A sin A sin A sin A sec x sec x  tan x sec x sec x  tan x sec x sec x  tan x sec x     sec x sec x  tan x  83. 2 2 sec x  tan x sec x  tan x sec x  tan x 1 sec x  tan x

80.

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6

CHAPTER 7 Analytic Trigonometry

sec2   tan2  1 sec   tan    sec   tan  sec   tan  sec   tan  sin x  cos x sin2 x  cos2 x sin x  cos x sin x  cos x sin x  cos x2 sin x  cos x2  85.    2 2 sin x  cos x sin x  cos x sin x  cos x sin x  cos x sin x  cos x sin x  cos x sin x  cos x2 tan  sin  tan   sin  tan  sin  tan   sin  tan  sin  tan  sin  tan   sin     86.    tan   sin  tan   sin  tan   sin  tan2   sin2  sin2  sec2   1 84. sec   tan   sec   tan  

tan  sin  tan   sin 

tan   sin   tan  sin  tan2  sin2  1  sin x 1  sin x 1  2 sin x  sin2 x 1 2 sin x sin2 x 1  sin x 1  2 sin x  sin2 x       87.  2 2 2 2 1  sin x 1  sin x 1  sin x cos x cos x cos x cos2 x 1  sin x 

 sec2 x  2 sec x tan x  tan2 x  sec x  tan x2

88.

1  sin x 1  sin x 1  sin x 1  sin x2 1  sin x2      1  sin x 1  sin x 1  sin x cos2 x 1  sin2 x

 1  sin x 2  tan x  sec x2 cos x

cos x 1  cos x 1  cos x 1  cos x 1  cos2 x sin2 x 1       sin x sin x sin x sin x 1  cos x sin x 1  cos x sin x 1  cos x 1 1 1    1 1 cos x csc x  cot x  1  cos x sin x sin x sin x 1 sin u  1 sin u  sin u tan u  sin u sec u  1 cos u cos u     90. 1 sin u sec u  1 sin u tan u  sin u 1  sin u cos u cos u sin  sin  x sin    91. x  sin ; then   tan  (since cos   0 for 0      2 ). 2 2 2 cos  cos  1x 1  sin        92. x  tan ; then 1  x 2  1  tan2   1  sec2   1  sec2   sec  (since sec   0 for 0     2 ).      93. x  sec ; then x 2  1  sec2   1  tan2   1  1  tan2   tan  (since tan   0 for 0     2) 89. csc x  cot x 

94. x  2 tan ; then 1 1 1 1          2 2 2 4 tan   2 sec2  x 4x 4 tan2   4 1  tan2  2 tan 2 4  2 tan 2 1 1 cos2  cos      18 cot2  cos  2 8 8 tan   sec  sin2         2 2 2 95. x  3 sin ; then 9  x  9  3 sin   9  9 sin   9 1  sin2   3 cos2   3 cos  cos   0 for 0     ). 2

96. x  5 sec ; then

x 2  25  x

 5 sec 2  25 5 sec 

tan   0 for 0     2 ).

   25 sec2   1 5 sec 

(since

sin   5 tan2  tan  cos   sin  (since    1 5 sec  sec  cos 

97. f x  cos2 x  sin2 x, g x  1  2 sin2 x. From the graph, f x  g x this appears to be an identity. Proof:

1

f x  cos2 x  sin2 x  cos2 x  sin2 x  2 sin2 x  1  2 sin2 x  g x. Since

f x  g x for all x, this is an identity.

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5 -1


SECTION 7.1 Trigonometric Identities

98. f x  tan x 1  sin x, g x 

sin x cos x . From the graph, f x  g x does 1  sin x

not appear to be an identity. In order to show this, let x   4 . Then,      21 . However,   f 4  1  1  1 2

2

7

1

-5

1  1    1 1 2 2 2 . Since f     g   , this is not an     g      4 2 4 4 1 21 2  1  2 2

5 -1

identity.

99. f x  sin x  cos x2 , g x  1. From the graph, f x  g x does not

appear to be an identity. In order to show this, we can set x   4 . Then we have 2  2  2     1  2  2  2  1  g  f 4    1 4 . Since 2 2 2   f 4  g 4 , this is not an identity.

2 1

-5

100. f x  cos4 x  sin4 x, g x  2 cos2 x  1. From the graph, f x  g x appears to be an identity. In order to prove this, simplify the expression f x:    f x  cos4 x  sin4 x  cos2 x  sin2 x cos2 x  sin2 x      cos2 x  sin2 x 1  2 cos2 x  cos2 x  sin2 x    2 cos2 x  cos2 x  sin2 x  2 cos2 x  1  g x

5

1

-5

5 -1

Since f x  g x for all x, this is an identity.

    101. tan x  cot x2  tan2 x  2 tan x cot x  cot2 x  tan2 x  2  cot2 x  tan2 x  1  cot2 x  1  sec2 x  csc2 x 102.

1  cos x  sin x 1  cos2 x  sin2 x  2 cos x  sin x  cos x sin x 1  cos x  sin x 1  cos x  sin x   1  cos x  sin x 1  cos x  sin x 1  cos x  sin x 1  cos2 x  sin2 x  2 cos x 2  2 cos x  2 sin x  2 sin x cos x 1  sin x 1  cos x  sin x 1  cos x    2 cos x 1  cos x cos x 2 cos x  2 cos x

      sin   cos   1 1 103. sin   tan  cos   cot   sin   cos    sin  1   cos  1  cos  sin  cos  sin      1 1 sin  1   cos   1 sin   1  cos  1  cos  sin 

   104. sin6   cos6   sin2   cos2  sin4   sin2  cos2   cos4       sin2  1  cos2   sin2  cos2   cos2  1  sin2   sin2   cos2   3 sin2  cos2   1  3 sin2  cos2 

105.

sin2 y  tan2 y

cos2 y  cot2 y

sin2 y  cos2 y 

  sin2 y sin2 y cos2 y  sin2 y 2 y cos2 y  1 sin 2 2 sin6 y sin2 y cos y cos y       tan6 y 2 2 2 2 2 cos6 y sin y cos y  cos y cos y cos2 y sin2 y  1 cos y sin2 y

sin2 y


8

106.

CHAPTER 7 Analytic Trigonometry

1 sin x  cos x2

1 sin2 x  2 sin x cos x  cos2 x

cos2 x

sec2 x

1

1    cos2 x tan2 x  2 tan x  1 sin x 1 2 cos x cos2 x sin2 x

1  tan x2

   sin x    ln sin x  ln sin x  ln cos x  ln sin x 107. ln tan x sin x  ln tan x  ln sin x  ln  cos x     1    2 ln sin x  ln sec x  2 ln sin x  ln  cos x  108. ln tan x  ln cot x  ln tan x cot x  ln 1  0 2

2

2

2

2

2

2

2

109. esin x etan x  e1cos x esec x1  e1cos xsec x1  esec x e cos x

110. e x2 lnsin x  e x e2 lnsin x  e x elnsin x elnsin x  e x sin2 x

  111. LHS  R cos  sin 2  R sin  sin 2  R cos 2  R sin 2 cos2   sin2   R cos 2  R sin 2  R cos 2  RHS

112. (a) x  y2  x 2  2x y  y 2 is an identity.

(b) x 2  y 2  1 is not an identity. For example, 12  12  1. (c) x y  z  x y  xz is the Distributive Law, an identity. 2

2

(d) This is an identity: LHS  esin xcos x  e1  RHS. 2

(e) xeln x  x  x 2  x 3 is an identity, because it is true for all values of x for which both sides of the equation are defined (x  0).   (f) sin t  cos t  1 is not an identity. For example, sin  4  cos 4  2. (g) x 2  tan2 x  0 is not an identity.

(h) t 2  cos2 t  t  cos t t  cos t (difference of squares).

 113. (a) Choose x   2 . Then sin 2x  sin   0, whereas 2 sin x  2 sin 2  2.  2  2 2 2 2  2  4  1. (b) Choose    4 . Then sec   csc  

replace "θ" with "x" in part (b) (3 times)

    1  1  2 . (c) Choose x   4 and y  4 . Then sin x  y  sin 2  1, whereas sin x  sin y  sin 4  sin 4  2

2

2

Since these are not equal, the equation is not an identity. 1 1 1 (d) Choose x    1 whereas 4 . Then sin x  cos x  sin   cos   1 2  1 4 4 2 2    csc x  sec x  csc  4  sec 4  2  2. Since these are not equal, the equation is not an identity.

114. No. All this proves is that f x  g x for x in the range of the viewing rectangle. It does not prove that these functions

x4 x6 x2   and g x  cos x. In the first viewing rectangle 2 24 720 the graphs of these two functions appear identical. However, when the domain is expanded in the second viewing rectangle, you can see that these two functions are not identical. are equal for all values of x. For example, let f x  1 

1

-2

2 -1

it doesnt really prove anything if the graphs "appear" to be the same

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10 -5 -10

shows


SECTION 7.2 Addition and Subtraction Formulas

9

115. Answers will vary.  116. Label a the side opposite , b the side opposite u, and c the hypotenuse. Since u     2  , we must have u    2    a    2  u. Next we express all six trigonometric function for each angle: cos u  c  sin   cos u  sin 2  u ,     b a sin u  bc  cos   sin u  cos  2  u , tan u  a  cot   tan u  cot 2  u , cot u  b  tan         c  c cot u  tan 2  u , sec u  a  csc   sec u  csc 2  u , and csc u  b  sec   csc u  sec  2 u .

117. From the diagram, we see that sin a  23 , cos b  23 , tan c  23 , and

c

 cot d  23 . So since a  b   2 and c  d  2 ,

a

 sin1 23 cos1 23 tan1 32 cot1 32  abcd   2  2  .

3

3

d

b 2

7.2

2

ADDITION AND SUBTRACTION FORMULAS

1. If we know the values of the sine and cosine of x and y we can find the value of sin x  y using the addition formula for sine, sin x  y  sin x cos y  cos x sin y. 2. If we know the values of the sine and cosine of x and y we can find the value of cos x  y using the subtraction formula for cosine, cos x  y  cos x cos y  sin x sin y.

  6 2 4      2 3 2 6 2 1 4. sin 15  sin 45  30   sin 45 cos 30  cos 45 sin 30  2  2  2  2  4      6 5. cos 105  cos 60  45   cos 60 cos 45  sin 60 sin 45  12  22  23  22  2 4       2 6. cos 195   cos 15   cos 45  30    cos 45 cos 30  sin 45 sin 30    22  23  22  12   6 4   3   tan 30  1  3 3 tan 45 3      7. tan 15  tan 45  30     2 3 3 1  tan 45 tan 30 3  3 11 3   3 1 tan 45  tan 30 33  3       8. tan 165   tan 15   tan 45  30       32 3 1  tan 45 tan 30 3 3 11 3       2 1 2 3 6 2  7      9. sin 19 12   sin 12   sin 4  3   sin 4 cos 3  cos 4 sin 3   2  2  2  2   4        2 3 2 2 6 17  5       1 10. cos 12   cos 12   cos 4  6   cos 4 cos 6  sin 4 sin 6   2  2  2  2  4 

3. sin 75  sin 45  30   sin 45 cos 30  cos 45 sin 30  22  23  22  12 

        tan        tan 3  tan 4  1  3  3  2   tan 12 11. tan  12  3 4 1  tan  1 3 3 tan 4             sin 5   sin      sin  cos   cos  sin    2  3  2  1   6 2 12. sin  512 12 4 6 4 6 4 6 2 2 2 2 4        3 2 2 6 2       1 13. cos 11 12   cos 12   cos 3  4   cos 3 cos 4  sin 3 sin 4   2  2  2  2   4   3      tan 5   tan        tan 4  tan 6   1  3   3  3  2  3 14. tan 712  12 4 6 1  tan  3 3 4 tan 6 11 3 3

15. cos 23 cos 67  sin 23 sin 67  cos 23  67   cos 90  0

16. sin 35 cos 25  cos 35 sin 25  sin 35  25   sin 60  23


10

CHAPTER 7 Analytic Trigonometry

  3 sin   sin 3    sin   1 17. sin 34 cos   cos 4 4 4 4 4 2 18.

 tan 35  tan 15

   3    tan 2   3  tan 5 15 3 

1  tan 35 tan 15

tan 55  tan 10  tan 55  10   tan 45  1 1  tan 55 tan 10     cos   sin 5 sin   cos 5    cos   3 20. cos 518 9 18 9 18 9 6 2       sin 2  u sin 2 cos u  cos 2 sin u 1  cos u  0  sin u cos u   21. tan  2  u  cos   u  cos  cos u  sin  sin u  0  cos u  1  sin u  sin u  cot u 2 2 2      cos 2 cos u  sin 2 sin u  cos 2  u 0  cos u  1  sin u sin u   22. cot  2  u  sin   u  sin  cos u  cos  sin u  1  cos u  0  sin u  cos u  tan u 2 2 2 19.

  23. sec  2 u 

  24. csc  2 u 

1 1 1 1    cos u  sin  sin u  0  cos u  1  sin u  sin u  csc u cos cos   u 2 2 2 1 1 1 1     cos u  cos  sin u  1  cos u  0  sin u  cos u  sec u sin sin   u 2 2 2

    25. sin x   2  sin x cos 2  cos x sin 2  0  sin x  1  cos x   cos x     26. cos x   2  cos x cos 2  sin x sin 2  0  cos x  1  sin x  sin x 27. sin x    sin x cos   cos x sin   1  sin x  0  cos x   sin x

28. cos x    cos x cos   sin x sin   1  cos x  0  sin x   cos x tan x  tan  tan x  0 29. tan x      tan x 1  tan x tan  1  tan x  0    cos x cos    cos x   sin x 2 2  sin x sin 2  30. cot x   2  sin x    sin x cos   cos x sin    cos x   tan x 2 2 2    cos x  cos  sin x  1  cos x  0  sin x  cos x and  x  sin 31. LHS  sin  2 2  2  cos x  cos  sin x  1  cos x  0  sin x  cos x. Therefore, LHS  RHS.  x  sin RHS  sin  2 2 2      1    32. cos x  3  sin x  6  2 cos x  23 sin x  23 sin x  12 cos x  0  tan x  tan    3  tan x 3   33. tan x    3 1  tan x tan  1  3 tan x 3  tan x  tan   4  tan x  1  tan x  1  34. tan x   4 1  tan x tan  1  tan x  1 tan x  1 4

35. sin x  y  sin x  y  sin x cos y  cos x sin y  sin x cos y  cos x sin y  2 cos x sin y

36. cos x  y  cos x  y  cos x cos y  sin x sin y  cos x cos y  sin x sin y  2 cos x cos y 1 1 1 1  tan x tan y cot x cot y  1 1 cot x cot y cot x cot y     37. cot x  y  1 1 tan x  y tan x  tan y cot x cot y cot y  cot x  cot x cot y 1 1 1 1  tan x tan y cot x cot y  1 1 cot x cot y cot x cot y    38. cot x  y   1 1 tan x  y tan x  tan y cot x cot y cot x  cot y  cot x cot y sin y sin x cos y  cos x sin y sin x  y sin x    39. tan x  tan y  cos x cos y cos x cos y cos x cos y sin x sin y cos x cos y  sin x sin y cos x  y 40. 1  tan x tan y  1    cos x cos y cos x cos y cos x cos y


SECTION 7.2 Addition and Subtraction Formulas

41.

sin x  y tan x  tan y sin x cos y  cos x sin y tan x  tan y cos x cos y    1  tan x tan y cos x cos y  sin x sin y cos x  y 1  tan x tan y cos x cos y

42.

sin x  y  sin x  y sin x cos y  cos x sin y  sin x cos y  cos x sin y 2 cos x sin y    tan y cos x  y  cos x  y cos x cos y  sin x sin y  cos x cos y  sin x sin y 2 cos x cos y

11

43. cos x  y cos x  y  cos x cos y  sin x sin y cos x cos y  sin x sin y  cos2 x cos2 y  sin2 x sin2 y      cos2 x 1  sin2 y  1  cos2 x sin2 y  cos2 x  sin2 y cos2 x  sin2 y cos2 x  sin2 y  cos2 x  sin2 y

44. cos x  y cos y  sin x  y sin y  cos x cos y  sin x sin y cos y  sin x cos y  cos x sin y sin y    cos2 y cos x  sin2 y cos x  cos2 y  sin2 y cos x  cos x 45. sin x  y  z  sin x  y  z  sin x  y cos z  cos x  y sin z

 cos z sin x cos y  cos x sin y  sin z cos x cos y  sin x sin y

 sin x cos y cos z  cos x sin y cos z  cos x cos y sin z  sin x sin y sin z 46. The addition formula for the tangent function can be written as tan A  tan B  tan A  B 1  tan A tan B. Also note that tan A   tan A. Using these facts, we get   tan x  y  tan y  z  tan z  x  tan x  y  y  z 1  tan x  y tan y  z  tan z  x    tan x  z 1  tan x  y tan y  z  tan z  x  tan x  z  tan z  x  tan x  y tan y  z tan x  z

 tan x  z  tan x  z  tan x  y tan y  z tan x  z  0  tan x  y tan y  z tan x  z  tan x  y tan y  z tan z  x

47. We want to write cossin1 x  tan1 y in terms of x and y only. We let

1

  sin1 x and   tan1 y and sketch triangles with angles  and  such  that sin   x and tan   y. From the triangles, we have cos   1  x 2 , 1

x

Ï1+y@ ú

¬

Ï1-x@

y

y

1

cos    , and sin    . tan   y sin   x 1  y2 1  y2 From the subtraction formula for cosine we have     1 y 1  x2  x y 1 1 2  x    cos sin x  tan y  cos     cos  cos   sin  sin   1  x   2 2 1y 1y 1  y2

x 48. Let   sin1 x and   cos1 y. From the triangles, tan    and 1  x2  1  y2 , so using the addition formula for tangent, we have tan   y  x 1  y2     y 1  x2  tan sin1 x  cos1 y  tan     1  y2 x 1   y 1  x2   x y  1  x 2 1  y2    y 1  x 2  x 1  y2

1

x

1

Ï1-y@

¬

ú

sin   x

cos   y

Ï1-x@

y


12

CHAPTER 7 Analytic Trigonometry

1 49. Let   tan1 x and   tan1 y. From the triangles, cos    , 1  x2 x 1 y sin    , cos    , and sin    , so using the 1  y2 1  y2 1  x2

Ï1+x@ ¬

subtraction formula for sine, we have   sin tan1 x  tan1 y  sin     sin  cos   cos  sin 

Ï1+y@

x

ú

1

y

1

tan   y

tan   x

x 1 1 y xy       2 2 2 2 1y 1y 1x 1x 1  x 2 1  y2

 50. Let   sin1 x and   cos1 y. From the triangles, cos   1  x 2 and  sin   1  y 2 , so using the addition formula for sine, we have   sin sin1 x  cos1 y  sin     sin  cos   cos  sin       x y  1  x 2  1  y2  x y  1  x 2 1  y2

1

1

x

Ï1-y@

¬

ú

sin   x

cos   y

Ï1-x@

y

1 1   , so the addition formula for sine gives 51. We know that cos1 21   3 and tan 4            sin  cos   cos  sin   3  2  1  2  6  2 . sin cos1 21  tan1 1  sin   3 4 3 4 3 4 2 2 2 2 4   1 3   , so the addition formula for cosine gives 52. We know that sin1 23   3 and cot 6           cos  cos   sin  sin   1  3  3  1  0.  cos sin1 23  cot1 3  cos  3 6 3 6 3 6 2 2 2 2

53. We sketch triangles such that   sin1 43 and   cos1 31 . From the  triangles, we have tan   3 and tan   2 2, so the subtraction formula for 7

4

tangent gives  3  2 2   tan   tan  7  tan sin1 34  cos1 13   1  tan  tan  1  3  2 2 7  3  2 14   76 2

3

3

¬

ú

Ï7

sin   34

2Ï2

1

cos   13

54. We sketch triangles such that   cos1 23 and   tan1 12 . From the 

triangles, we have sin   35 , cos   2 , and sin   1 , so the addition 5 5 formula for sine gives    sin cos1 32  tan1 12  sin  cos   cos  sin   35  2  23  1  2  10  2 5  23   15 3 5

5

5

3

¬

Ï5

Ï5 ú

2

cos   23

2

tan   12

1


13

SECTION 7.2 Addition and Subtraction Formulas

55. As in Example 7, we sketch the angles  and  in standard position

y

(_3, 4)

3

with terminal sides in the appropriate quadrants and find the remaining sides using the Pythagorean Theorem. To find sin   ,

5

4

we use the subtraction formula for sine and the triangles we have

1 (_3, _1)

¬

sketched: sin     sin  cos   cos  sin       3 12  3 4 3 1        5 5 10 10 5 10  3 10  10

sin   45

3

have cos     cos  cos   sin  sin         3 2 4 6  4 5 5      5 3 5 3 15

4

2 Ï5

(_3, _4)

(2, _Ï5)

tan   43 y

cos   23 (12, 5)

13 ¬

x

3

5

(_2Ï5, Ï5)

x

12

y

y

2Ï2

x

(_1, _2Ï2)

3

x

 cos    2 5 5

¬

15

ú

2Ï5

5 sin   13

1

y 5

Ï5

5

have

 2 30  1   15  2 2

ú

x

58. Using the addition formula for tangent and the triangles shown, we    2 2   1 tan   tan  15  tan        1  tan  tan  1  2 2  1

y

y ¬

56. Using the addition formula for cosine and the triangles shown, we

sin     sin  cos   cos  sin        5 2 5 12 2 5 5     13 5 13 5 65

x

Ï10

tan   13

x

3

57. Using the addition formula for sine and the triangles shown, we have

y ú

(_Ï15, 1) 1

4 Ï15

ú x

sin   14

cos    13     2   3  12  4  2. Thus, sin   12 and cos   2 3    56 , so 59. k  A2  B 2      3 sin x  cos x  k sin x    2 sin x  56 .    60. k  A2  B 2  12  12  2 and  satisfies sin   1 , cos   1     4 . Thus, 2 2    sin x  cos x  k sin x    2 sin x   4 .     5   1 and cos    5  1    7 , so 61. k  A2  B 2  52  52  50  5 2. Thus, sin     4 5 2 2 5 2 2       7  7  5 sin 2x  cos 2x  k sin 2x    5 2 sin 2x  4  5 2 sin 2 x  8 .      2   62. k  A2  B 2  32  3 3  36  6 and  satisfies sin   3 6 3  23 , cos   36  12     3 . Thus,      1 3 sin x  3 3 cos x  k sin x    6 sin x   3  6 sin  x  3 .


14

Try to have the graph match the one in the answers CHAPTER 7 Analytic Trigonometry

graph is incorrect for #63

 63. (a) g x  cos 2x  3 sin 2x    2  3  4  2, and  satisfies k  12 

(b) This is a sine curve with amplitude 2, period , and . phase shift  12 y

 sin   12 , cos   23     6 . Thus, we can

2

write

    g x  k sin 2x    2 sin 2x   6  2 sin 2 x  12 .

1 _2¹

0

¹

2¹ x

_1 _2

64. (a) f x  sin x  cos x  k 

  12  12  2, and

 satisfies sin   cos   1     4 . Thus, 2

(b) This is a sine curve with amplitude

 2, period 2,

and phase shift   4. y

we can write

2

   f x  k sin x    2 sin x   4 .

1 _2¹

0

¹

2¹ x

_1 _2

65. f x  cos x. Now

f x  h  f x cos x  h  cos x cos x cos h  sin x sin h  cos x   h h h     1  cos h sin h  cos x 1  cos h  sin h sin x   cos x  sin x  h h h

66. g x  sin x. Now

sin x  h  sin x sin x cos h  cos x sin h  sin x g x  h  g x   h h h     sin h 1  cos h sin h cos x  sin x 1  cos h  cos x  sin x  h h h

    2 67. (a) y  sin2 x   4  sin x  4 . From the graph we see that the value of y seems to always be equal to 1.

1

-5

       2  sin x cos   cos x sin  2 2 (b) y  sin2 x   4  sin x  4  sin x cos 4  cos x sin 4 4 4    2  2  1 sin x  cos x  1 sin x  cos x  12 sin x  cos x2  sin x  cos x2 2 2     1 2  2 sin x  2 sin x cos x  cos2 x  sin2 x  2 sin x cos x  cos2 x  12 [1  2 sin x cos x  1  2 sin x cos x]  12  2  1

5


SECTION 7.2 Addition and Subtraction Formulas

68. (a) y   12 [cos x    cos x  ]. The graph of y appears to be the same as

15

1

that of cos x.

-5

5 -1

(b) y   12 [cos x    cos x  ]

  12 [cos x cos   sin x sin   cos x cos   sin x sin ]

  12 [ cos x  0   cos x  0]   12 2 cos x  cos x  69. If      2 , then     2 . Now if we let y  x  , then     sin x    cos x     2  sin y  cos y  2  sin y   sin y  0.

Therefore, sin x    cos x    0.

70. Let  A and  B be the two angles shown in the diagram. Then 180    A  B, 90    A, and 90    B. Subtracting the second and third equation from the first, we get

6

4

Œ

A

180  90  90    A  B    A    B  4

71. tan1

xy 1  xy

72. cot u   

B

º

4

    . Then

tan   tan    

3

3

8  9  tan   tan  17  6 4 4 3  12 112  2  17 12  6 . 1  tan  tan  1  1

 tan1

6

tan u  tan  1  tan u tan 

4

2

 tan1 tan u    u    tan1 x  tan1 y

x 1x  1 cot  cot u  1   0, so tan1 x tan1 cot   cot u x  1x

   1  u   cot1 cot u    cot1 0  . x 2

y y and tan   . Thus, m  tan . x x tan 2  tan 1 (b) tan   tan 2   1   . From part (a), we have m 1  tan 1 and m 2  tan 2 . Then by 1  tan  2 tan  1 m  m1 . substitution, tan   2 1  m1m2

73. (a) By definition, m 

(c) Let  be the unknown angle as in part (b). Since m 1  13 and m 2  12 , 1

tan  

1

1

2

6

 m2  m1  2 3   67  17    tan1 17  0142 rad  81 . 1  m1m2 1 1 1 3

(d) From part (b), we have cot   have 0 

1  m1m2 . If the two lines are perpendicular then   90 and so cot   0. Thus we m2  m1

1  m1m2  0  1  m 1 m 2  m 1 m 2  1  m 2  1m 1 . Thus m 2 is the negative reciprocal of m 1 . m2  m1

74. Clearly C   4 . Now tan A  B 

1

1

 tan A  tan B     3 1 2 1  1. Thus A  B   4 , so A  B  C  4  4  2 . 1  tan A tan B 1  3

2


16

CHAPTER 7 Analytic Trigonometry

75. (a) y  f 1 t  f 2 t  5 sin t  5 cos t  5 sin t  cos t 10

-5

5

  A2  B 2  52  52  5 2. Therefore, 1 B 5 sin        and 2 2 5 2 2 A B A 1 cos      . Thus,    4. 2 A2  B 2

(b) k 

-10

76. (a) f t  C sin t  C sin t    C sin t  C sin t cos   cos t sin   C 1  cos  sin t  C sin  cos t  A sin t  B cos t

where A  C 1  cos  and B  C sin .     (b) In this case, f t  10 1  cos  3 sin t  10 sin 3 cos t  15 sin t  5 3 cos t. Thus      2    3 and sin   5 3  1 , so    . Therefore, k  152  5 3  10 3 and  has cos   15 2 2 6 10 3 10 3     f t  10 3 sin t  6 .           77. sin s  t  cos  2  s  t  cos 2  s  t  cos 2  s cos t  sin 2  s sin t  sin s cos t  cos s sin t. The last equality comes from again applying the cofunction identities. sin s cos t  cos s sin t sin s  t  78. tan s  t  cos s  t cos s cos t  sin s sin t sin t 1 sin s  sin s cos t  cos s sin t cos s cos t cos t  tan s  tan t   cos s  sin s sin t 1 cos s cos t  sin s sin t 1  tan s tan t  1 cos s cos t cos s cos t 79. Suppose such a triangle exists. We translate it (preserving its supposed equilaterality) so that one of its vertices lies at the origin and its other vertices become P1 m 1  n 1  and P2 m 2  n 2 . n n Then O P1 has slope 1 and O P2 has slope 2 . The angle between these sides is  3 , so from Exercise 73(b), m1 m2 n n2  1    m 1 n2  m 2 n1 m2 m1 3  tan  n 1 n 2  m m  n n —a rational number. But 3 is irrational, so such a triangle cannot exist. 3 1 2 1 2 1 m1 m2    Note: The triangle pictured in the text has side lengths 17, 17, and 3 2 and approximate interior angles 5904 , 5904 , and 6193 . 80. Using the addition formula for tangent twice, we get tan A  tan B  tan C tan A  tan B  1  tan A tan B tan C tan A  B  tan C 1  tan A tan B   . tan A  B  C  tan A  tan B 1  tan A  B tan C 1  tan A tan B  tan  tan B tan C tan C 1 1  tan A tan B But A  B  C  180 and tan 180  0, so the numerator of the last expression must be 0. Thus, tan A  tan B  1  tan A tan B tan C  0  tan A  tan B  tan C  tan A tan B tan C  0  tan A  tan B  tan C  tan A tan B tan C.


SECTION 7.3 Double-Angle, Half-Angle, and Product-Sum Formulas

7.3

17

DOUBLE-ANGLE, HALF-ANGLE, AND PRODUCT-SUM FORMULAS

1. If we know the values of sin x and cos x, we can find the value of sin 2x using the Double­Angle Formula for Sine: sin 2x  2 sin x cos x.

x x 2. If we know the value of cos x and the quadrant in which lies, we can find the value of sin using the Half­Angle Formula 2 2  x 1  cos x . for Sine: sin   2 2    5 , x in quadrant I  cos x  12 and tan x  5 . Thus, sin 2x  2 sin x cos x  2 5 12  120 , 3. sin x  13 13 12 13 13 169 120  2  2 sin 2x 169 5 119 120 169 120 cos 2x  cos2 x  sin2 x  12  13  14425 13 169  169 , and tan 2x  cos 2x  119  169  119  119 . 169

  4. tan x   43 . Then sin x  45 and cos x   35 (x is in quadrant II). Thus, sin 2x  2 sin x cos x  2  45  35   24 25 , 24  2  2 7 , and tan 2x  sin 2x   25  24  25  24 . cos 2x  cos2 x  sin2 x   35  45  916   7 25 25 25 7 cos 2x 7 25

  5. cos x  45 . Then sin x   35 (csc x  0) and tan x   34 . Thus, sin 2x  2 sin x cos x  2  35  45   24 25 , 24  2  2 7 , and tan 2x  sin 2x   25   24  25   24 .  cos 2x  cos2 x  sin2 x  45   35  169 7 25 25 25 7 7 cos 2x 25

 6. csc x  4. Then sin x  14 , cos x   415 , and tan x   1 (tan x  0). Thus, 15       2  2 7 sin 2x  2 sin x cos x  2  14  415   815 , cos 2x  cos2 x  sin2 x   415  14  151 16  8 , and     15 sin 2x tan 2x   78   815  87   715 . cos 2x 8

   7. sin x   35 . Then, cos x   45 and tan x  34 (x is in quadrant III). Thus, sin 2x  2 sin x cos x  2  35  45  24 25 , 24  2  2 sin 2x 25 7 24 25 24 cos 2x  cos2 x  sin2 x   45   35  169 25  25 , and tan 2x  cos 2x  7  25  7  7 . 25

  8. sec x  2. Then cos x  12 , sin x   23 , and tan x   3 (x is in quadrant IV). Thus,      2     1   3 , cos 2x  cos2 x  sin2 x  1 2   3   12 , and sin 2x  2 sin x cos x  2  23 2 2 2 2    23 sin 2x tan 2x    3. 1 cos 2x 2

9. tan x   13 and cos x  0, so sin x  0. Thus, sin x   1 and cos x  3 . Thus, 10 10        6   3 , cos 2x  cos2 x  sin2 x  3 2   1 2  8  4 , 3   10 sin 2x  2 sin x cos x  2  1 5 10 5 10

3

10

 sin 2x  45   35  54   34 . and tan 2x  cos 2x 5

10

10


18

CHAPTER 7 Analytic Trigonometry

10. cot x  23 .

Then tan x  32 , sin x  3 (sin x  0), and cos x  2 . Thus, 13 13   2  2   5 2 x  sin2 x  2 2 3  12 , cos 2x  cos   49 sin 2x  2 sin x cos x  2 3 13 13   13 , and

13 13 12   sin 2x   12  135  12   13 tan 2x  13 5 5. cos 2x  13  2  1  cos 2x 2  14  12 cos 2x  14 cos2 2x 11. sin4 x  sin2 x  2

13

13

1  cos 4x  14  12 cos 2x  14   14  12 cos 2x  18  18 cos 4x  38  12 cos 2x  18 cos 4x 2   12 34  cos 2x  14 cos 4x  2  1  cos 2x 2 4 2  14  12 cos 2x  14 cos2 2x 12. cos x  cos x  2 1  cos 4x  14  12 cos 2x  14   14  12 cos 2x  18  18 cos 4x  38  12 cos 2x  18 cos 4x 2   12 34  cos 2x  14 cos 4x

13. We use the result of Example 4 to get       1 1  cos 2x  cos 4x  cos 2x cos 4x. cos2 x sin4 x  sin2 x cos2 x sin2 x  18  18 cos 4x  12  12 cos 2x  16

14. Using Example 4, we have   1 1  cos 4x  cos 2x  cos 2x cos 4x. cos4 x sin2 x  cos2 x cos2 x sin2 x  12 1  cos 2x  18 1  cos 4x  16  2 15. Since sin4 x cos4 x  sin2 x cos2 x we can use the result of Example 4 to get sin4 x cos4 x 

 1  1 cos 4x 2  1  1 cos 4x  1 cos2 4x 8 8 64 32 64

1  1 cos 4x  1  1 1  cos 8x  1  1 cos 4x  1  1 cos 8x  64 32 64 2 64 32 128 128   3 1 1 1 3 1  128  32 cos 4x  128 cos 8x  32 4  cos 4x  4 cos 8x

16. Using the result of Exercise 12, we have  2    cos6 x  cos2 x  cos2 x  38  12 cos 2x  18 cos 4x 12  12 cos 2x

3  1 cos 2x  1 cos 4x  3 cos 2x  1 cos2 2x  1 cos 2x cos 4x  16 4 16 16 4 16

Because 14 cos2 2x 

1 cos 4x 1 1  cos 4x    , the last expression is equal to 4 2 8 8

3 1 1 3 1 1 1 16  4 cos 2x  16 cos 4x  16 cos 2x  8  8 cos 4x  16 cos 2x cos 4x 5  7 cos 2x  3 cos 4x  1 cos 2x cos 4x  1 5  7 cos 2x  3 cos 4x  cos 2x cos 4x  16 16 16 16 16            17. sin 15  12 1  cos 30   12 1  23  14 2  3  12 2  3  3

 1 2 1  cos 30 18. tan 15   2 3 1 sin 30 2

  1  22 1  cos 45     21 19. tan 225   2 sin 45 2             20. sin 75  12 1  cos 150   12 1   23  14 2  3  12 2  3


SECTION 7.3 Double-Angle, Half-Angle, and Product-Sum Formulas

19

        1 1    21. cos 165   2 1  cos 330    2 1  cos 30    12 1  23   12 2  3

        22. cos 1125   12 1  cos 225    12 1  cos 45    12 1  22   12 2  2   1  cos 54 1  22 5     21 23. tan 8  5  sin 4  22

24. cos 38  cos 12  34

1  cos 34  2

1  22  2

root because 38 is in quadrant I, so cos 38  0.         1 1  cos    1 1  3  1 2  3 25. cos 12 2 6 2 2 2

   2 2  12 2  2. Note that we have chosen the positive 4

  1  cos 56 1  23 5   2 3 26. tan 12  5  1 sin 6 2

         1 1  cos 9   1 1  2   1 2  2. We have chosen the negative root because 9 is in 2 4 2 2 2 8

27. sin 98  

quadrant III, so sin 98  0.          11  1 11    12 1  23  12 2  3. We have chosen the positive root because 11 28. sin 12  2 1  cos 6 12 is in  quadrant II, so sin 11 12  0.

29. (a) 2 sin 16 cos 16  sin 32 (b) 2 sin 4 cos 4  sin 8 31. (a) cos2 21  sin2 21  cos 42 (b) cos2 9  sin2 9  cos 18 sin 8 8  tan 4  tan  1  cos 8 2 4 1  cos 4  tan  tan 2 (b) sin 4 2

33. (a)

35. sin x  x  sin x cos x  cos x sin x  2 sin x cos x

2 tan 5  tan 10 1  tan2 5 2 tan 5  tan 10 (b) 1  tan2 5   32. (a) cos2  sin2  cos  2 2   (b) 2 sin cos  sin  2 2  1  cos 30 34. (a)  sin 15 2  1  cos 8  sin 4 (b) 2 tan x  tan x 2 tan x 36. tan x  x   1  tan x tan x 1  tan2 x

30. (a)

37. sin x  35 . Since x is in quadrant I, cos x  45 and x2 is also in quadrant I. Thus,          sin x2  12 1  cos x  12 1  45  1  1010 , cos x2  12 1  cos x  12 1  45  3

 3 1010 , and

10 10 x  sin 2  1  310  13 . tan x2  10 cos x2 38. cos x   45 . Since x is in quadrant III, sin x   35 and tan x  34 . Also, since 180  x  270 ,

     90  x2  135 and so x2 is in quadrant II. Thus, sin x2  12 1  cos x  12 1  45  3  3 1010 , 10   x     sin 2 10  3.  3  1 cos x2   12 1  cos x   12 1  45   1   1010 , and tan x2  10 10 cos x2


20

CHAPTER 7 Analytic Trigonometry

2 2 . Since 90  x  180 , we have 39. csc x  3. Then, sin x  13 and since x is in quadrant II, cos x   3          x x x 1   45  2  90 and so 2 is in quadrant I. Thus, sin 2  2 1  cos x  12 1  2 3 2  16 3  2 2 , cos x2 

1 1  cos x  2

     x       1 1  2 2  1 3  2 2 , and tan x  sin 2  322  3  2 2. x 2 3 6 2 32 2 cos 2

  2 2   2 and cos x  2 , since x is in quadrant I. Also, since 0  x  90 ,        0  x2  45 and so x2 is also in quadrant I. Thus, sin x2  12 1  cos x  12 1  22  12 2  2,          1  22 1  cos x 2 x 1 1 1 x   cos 2  2 1  cos x  2 1  2  2 2  2, and tan 2   2  1. 2 sin x 2  41. sec x  32 . Then cos x  23 and since x is in quadrant IV, sin x   35 . Since 270  x  360 , we have      x x x 1   135  2  180 and so 2 is in quadrant II. Thus, sin 2  2 1  cos x  12 1  23  1  66 , 6

40. tan x  1. Then sin x 

        sin x2 30 cos x2   12 1  cos x   12 1  23   5   , and tan x2   1  6   1   55 . x 6 6  5 5 6 cos 2

42. cot x  5.

Then, cos x   5

26

and sin x   1 (csc x  0). 2

Since cot x  0 and

csc x  0, it follows that x is in quadrant III. Thus 180  x  270 and so 90  x2  135 .       26 , x x 1 1 1  5 Thus 2 is in quadrant II. sin 2   12 265 2 1  cos x  2 13 26

     1  5   26 26 , and tan x  1  cos x  cos x2   12 1  cos x   12 1  5   12 265  5  26. 13 2 1 26 sin x  2

7 and find that sin   24 . Thus, using 43. We sketch a triangle with   cos1 25 25

the double-angle formula for sine,   7  sin 2  2 sin  cos   2  24  7  336 . sin 2 cos1 25 25 25 625

12 44. We sketch a triangle with   tan1 12 5 and find that sin   13 . Thus, using a

double-angle formula for cosine,   2  2   1  2 12  cos 2  1  2 sin   119 cos 2 tan1 12 5 13 169 .

25

¬ 7

24

7 cos   25 ¬

13 12

tan   12 5

45. Rewriting the given expression and using a double-angle formula for cosine, we have   1 1 1 8      sec 2 sin1 14   .   2 7 cos 2 sin1 14 1  2 sin2 sin1 41 1  2 14 

46. We sketch a triangle with   cos1 23 and find that sin   35 . Thus, using a half-angle formula for tangent,   5   sin  5  3 2  . tan 12 cos1 23  tan 12   1  cos  5 1 3

3

Ï5

cos   23

¬ 2

5


SECTION 7.3 Double-Angle, Half-Angle, and Product-Sum Formulas

  47. To write sin 2 tan1 x as an algebraic expression in x, we let   tan1 x

Ï1+x@

x and sketch a suitable triangle. We see that sin    and 1  x2 1 cos    , so using the double-angle formula for sine, we have 1  x2   x 1 2x sin 2 tan1 x  sin 2  2 sin  cos   2     . 2 2 1  x2 1x 1x

¬

x

1

  48. To write tan 2 cos1 x as an algebraic expression in x, we let   cos1 x 1  1  x2 and sketch a suitable triangle. We see that tan   , so using the ¬ x x double-angle formula for tangent, we have  1  x2   2   2 tan  2 1  x2 2x 1  x 2 x 1   . tan 2 cos x  tan 2  2    1  tan2  2x 2  1 1  x2 1  x2 x 1 1 x2 x

Ï1-x@

      1  cos cos1 x  1x 1 1 49. Using the half-angle formula for sine, we have sin 2 cos x    . Because cos1 2 2        1 cos1 x is positive. Thus, sin 1 cos1 x  1  x . and so sin has range [0 ], 12 cos1 x lies in 0  2 2 2 2 

    50. Using a double-angle formula for cosine, we have cos 2 sin1 x  1  2 sin2 sin1 x  1  2x 2 .

y

51. To evaluate cos 2, we first sketch the angle  in standard position with 4

terminal side in quadrant III. Using a double-angle formula for cosine, we have  2 7. cos 2  1  2 sin2   1  2  35  25

3

x

5

(_4, _3)

52. To evaluate sin 2, we first sketch the angle  in standard position with

y ¬

terminal side in quadrant IV and find the remaining side using the Pythagorean Theorem. Using the half-angle formula for sine, we have    1  12  1  cos  13   26 . Because 2 lies in  sin   26 2 2 2

¬

12 13

x

5

(12, _5)

quadrant II, where sine is positive, we take the positive value. Thus, sin 2  2626 . 53. To evaluate sin 2, we first sketch the angle  in standard position with terminal side in quadrant II and find the remaining side using the Pythagorean Theorem. Using the double-angle formula for sine, we have      sin 2  2 sin  cos   2 17  4 7 3   8493 .

1

(_4Ï3, 1)

y 7 4Ï3

¬ x

21


22

CHAPTER 7 Analytic Trigonometry

54. To evaluate tan 2, we first sketch the angle  in standard position with terminal

y

(3, 4)

side in quadrant I and find the remaining side using the Pythagorean Theorem. Using the double-angle formula for tangent, we have 2 tan   tan 2  1  tan2 

2  43

24  2   . 7 4

1 3

5 ¬

3

4 x

55. sin 5x cos 4x  12 [sin 5x  4x  sin 5x  4x]  12 sin 9x  sin x

56. sin 2x sin 3x  12 [cos 2x  3x  cos 2x  3x]  12 [cos x  cos 5x]  12 cos x  cos 5x

57. cos x sin 4x  12 [sin x  4x  sin x  4x]  12 [sin 5x  sin 3x]  12 sin 5x  sin 3x 58. cos 5x cos 3x  12 [cos 5x  3x  cos 5x  3x]  12 cos 8x  cos 2x

59. 3 cos 4x cos 7x  3  12 [cos 4x  7x  cos 4x  7x]  32 [cos 11x  cos 3x]  32 cos 11x  cos 3x        3x  sin x sin 60. 11 sin x2 cos x4  11  12 sin x2  x4  sin x2  x4  11 2 4 4     7x  5x 7x  5x cos  2 sin 6x cos x 61. sin 7x  sin 5x  2 sin 2 2     5x  4x 5x  4x 9x x 62. sin 5x  sin 4x  2 cos sin  2 cos sin 2 2 2 2     4x  6x 4x  6x sin  2 sin 5x sin x  2 sin 5x sin x 63. cos 4x  cos 6x  2 sin 2 2     11x 7x 9x  2x 9x  2x 64. cos 9x  cos 2x  2 cos cos  2 cos cos 2 2 2 2       2x  7x 9x 5x 9x 5x 2x  7x sin  2 cos sin   2 cos sin 65. sin 2x  sin 7x  2 cos 2 2 2 2 2 2      x 7x 7x x 3x  4x 3x  4x 66. sin 3x  sin 4x  2 sin cos  2 sin cos   2 sin cos 2 2 2 2 2 2   67. 2 sin 525 sin 975  2  12 cos 525  975   cos 525  975   cos 45   cos 150       cos 45  cos 150  22  23  12 2 3       2 3 68. 3 cos 375 cos 75  32 cos 45  cos 30   32 22  23  34     21 69. cos 375 sin 75  12 sin 45  sin 30   12 22  12  14          75  15 75  15 cos  2 sin 45 cos 30  2  22  23  26 70. sin 75  sin 15  2 sin 2 2           255  195 255  195 sin  2 sin 225 sin 30  2  22 12  22 71. cos 255  cos 195  2 sin 2 2            2 3 6  5  1  5  1  5     2 cos  72. cos 12  cos 12  2 cos 2 12  12 cos 2 12  12 4 cos  6  2 cos 4 cos 6  2  2  2  2 73. cos2 5x  sin2 5x  cos 2  5x  cos 10x 74. sin 8x  sin 2  4x  2 sin 4x cos 4x

75. sin x  cos x2  sin2 x  2 sin x cos x  cos2 x  1  2 sin x cos x  1  sin 2x    76. cos4 x  sin4 x  cos2 x  sin2 x cos2 x  sin2 x  cos2 x  sin2 x  cos 2x 2 tan x 2 tan x sin x  2 cos2 x  2 sin x cos x  sin 2x cos x 1  tan2 x sec2 x   1  1  2 sin2 x 1  cos 2x 2 sin2 x 78.    tan x sin 2x 2 sin x cos x 2 sin x cos x

77.


SECTION 7.3 Double-Angle, Half-Angle, and Product-Sum Formulas

79. tan

x  2 x 

 cos x tan

x  2

1  cos x  cos x sin x

1  cos x sin x

23

1  cos x  cos x  cos2 x sin2 x   sin x sin x sin x

1  cos x 1 2  cos x  csc x    2 sin x sin x sin x sin 4x 2 sin 2x cos 2x 2 2 sin x cos x cos 2x 81.    4 cos x cos 2x sin x sin x sin x 1 1  sin 2x 1  2 sin x cos x  1  1  12 csc x sec x 82. sin 2x 2 sin x cos x 2 sin x cos x 80. tan

83.

84.

cos2 x  sin2 x cos x  sin x 1 cos x cos 2x cos x  sin x cos x  sin x     1  2 sin x cos x cos x  sin x 1 cos x cos x  sin x cos x  sin x cos2 x  sin2 x  2 sin x cos x 1  tan x  1  tan x sin 2x 2 sin x cos x    tan x  1  cos 2x 1  2 cos2 x  1

1 1  tan2 x  2 tan x 2 tan x 1  tan2 x    2   86. sin4 x  cos4 x  sin2 x  cos2 x  2 sin2 x cos2 x  1  12 sin2 2x  1  12 1  cos2 2x  12 1  cos2 2x 85. cot 2x 

1  tan 2x

  2 tan x  tan x 2 tan x  tan x 1  tan2 x 2 tan 2x  tan x  1  tan x 87. tan 3x  tan 2x  x   2 tan x 1  tan 2x tan x 1  tan2 x  2 tan x tan x tan x 1 2 1  tan x 3 tan x  tan3 x  1  3 tan2 x    2x 1 sin 2x 1  2 cos 2 sin 2x  sin 4x 2 sin 2x  2 sin 2x cos 2x sin 2x 1  cos 2x   88.    3x  x 3x  x 2 cos x  2 cos 3x 2 cos 2x cos x 2 cos 2x cos x cos 2 2 cos 2 2

sin 2x cos x sin 2x cos2 x   tan 2x cos x cos 2x cos x cos 2x 2 sin 3x cos 2x sin 3x sin x  sin 5x    tan 3x 89. cos x  cos 5x 2 cos 3x cos 2x cos 3x 

90. 91.

sin 3x  sin 7x 2 sin 5x cos 2x cos 2x    cot 2x cos 3x  cos 7x 2 sin 5x sin 2x sin 2x sin 10x 2 sin 5x cos 5x cos 5x   sin 9x  sin x 2 sin 5x cos 4x cos 4x

sin x  sin 3x  sin 5x sin x  sin 5x  sin 3x 2 sin 3x cos 2x  sin 3x sin 3x 2 cos 2x  1     tan 3x cos x  cos 3x  cos 5x cos x  cos 5x  cos 3x 2 cos 3x cos 2x  cos 3x cos 3x 2 cos 2x  1       xy xy xy   cos 2 sin sin sin x  sin y xy 2 2 2        93.   tan xy xy xy cos x  cos y 2 cos 2 cos cos 2 2 2         xyxy 2x xyxy 2y 2 cos 2 cos sin sin sin y sin x  y  sin x  y 2 2 2 2           tan y 94. xyxy 2x x yx y 2y cos x  y  cos x  y cos y 2 cos 2 cos cos cos 2 2 2 2 92.


24

CHAPTER 7 Analytic Trigonometry

 x  2 4

 . Then 2   x 1  cos x   1  cos 2y  2   1   sin x  1  sin x  tan2    tan2 y  2 4 1  cos 2y 1   sin x 1  sin x 1  cos x   2

95. Let y

 2y

x 

           tan2 x  1  cot2 x  1  2 sin2 2x sec2 x  csc2 x 96. 1  cos 4x 2  tan2 x  cot2 x  1  1  2 sin2 2x       2 4 cos2 x sin2 x sec2 x  csc2 x  8 sin2 x  cos2 x  8 97. sin 130  sin 110  2 cos

  130  110 130  110 sin  2 cos 120 sin 10  2  12 sin 10   sin 10 2 2

100  200 100  200 sin  2 sin 150 sin 50  98. cos 100  cos 200  2 sin 2 2    2 12  sin 50   sin 50 

    45  15 45  15 cos  2 sin 30 cos 15  2  12  cos 15 2 2  cos 15  sin 90  15   sin 75 (applying the cofunction identity)

99. sin 45  sin 15  2 sin

87  33 87  33 cos  2 cos 60 cos 27 2 2  2  12 cos 27  cos 27  sin 90  27   sin 63

100. cos 87  cos 33  2 cos

101.

sin x  sin 2x  sin 3x  sin 4x  sin 5x sin x  sin 5x  sin 2x  sin 4x  sin 3x  cos x  cos 2x  cos 3x  cos 4x  cos 5x cos x  cos 5x  cos 2x  cos 4x  cos 3x sin 3x 2 cos 2x  2 cos x  1 2 sin 3x cos 2x  2 sin 3x cos x  sin 3x   tan 3x  2 cos 3x cos 2x  2 cos 3x cos x  cos 3x cos 3x 2 cos 2x  2 cos x  1

    102. n  1: sin 21 x  2 sin x cos x  21 sin x cos 20 x     n  2: sin 22 x  sin 4x  2 sin 2x cos 2x  2 2 sin x cos x cos 2x  4 sin x cos x cos 2x  22 sin x cos x cos 21 x   n  3: sin 23 x  sin 8x  2 sin 4x cos 4x  2 4 sin x cos x cos 2x cos 4x  8 sin x cos x cos 2x cos 4x    23 sin x cos x cos 2x cos 22 x In general, for n  0 we have         sin 2n x  sin 2 2n1 x  2 sin 2n1 x cos 2n1 x       2 2n1 sin x cos x cos 2x cos 4x cos 8x    cos 2n2 x cos 2n1 x      2n sin x cos x cos 2x cos 4x cos 8x    cos 2n2 x cos 2n1 x

103. With u  sin1 x for 0  x  1, we can write     cos1 1  2x 2  cos1 1  2 sin2 u  cos1 cos 2u  2u  2 sin1 x. 104. With u  tan1

      x2  1 1 1 1 cos 2u  2u  2 tan1 1 , we can write cos  cos x x x2  1


25

SECTION 7.3 Double-Angle, Half-Angle, and Product-Sum Formulas

105. (a) f x 

sin 3x cos 3x  sin x cos x sin 3x  x sin 3x cos x  cos 3x sin x   sin x cos x sin x cos x 2 sin x cos x sin 2x  2  sin x cos x sin x cos x for all x for which the function is defined.

cos 3x sin 3x  sin x cos x

(b) f x 

2 1

-5

5

The function appears to have a constant value of 2 wherever it is defined. 106. (a) f x  cos 2x  2 sin2 x

(b) f x  cos 2x  2 sin2 x    cos2 x  sin2 x  2 sin2 x  cos2 x  sin2 x  1

1

-5

5

The function appears to have a constant value of 1. 107. (a) y  sin 6x  sin 7x

(b) By a sum-to-product formula,

2

-5

5 -2

y  sin 6x  sin 7x     6x  7x 6x  7x  2 sin cos 2 2     13 1  2 sin 2 x cos  2 x

(c) We graph y  sin 6x  sin 7x,   y  2 cos 12 x , and   y  2 cos 12 x . 2

1  2 sin 13 2 x cos 2 x

-5

5 -2

The graph of y  f x lies

between the other two graphs.  1 3 108. From Example 2, we have cos 3x  4 cos3 x  3 cos x. If 3x   3 , then cos 3  2  4 cos x  3 cos x 

1  8 cos3 x  6 cos x  8 cos3 x  6 cos x  1  0. Substituting y  cos x gives 8y 3  6y  1  0.  2 109. (a) cos 4x  cos 2x  2x  2 cos2 2x  1  2 2 cos2 x  1  1  8 cos4 x  8 cos2 x  1. Thus the desired polynomial is P t  8t 4  8t 2  1.

  (b) cos 5x  cos 4x  x  cos 4x cos x  sin 4x sin x  cos x 8 cos4 x  8 cos2 x  1  2 sin 2x cos 2x sin x    8 cos5 x  8 cos3 x  cos x  4 sin x cos x 2 cos2 x  1 sin x [from part (a)]    8 cos5 x  8 cos3 x  cos x  4 cos x 2 cos2 x  1 sin2 x     8 cos5 x  8 cos3 x  cos x  4 cos x 2 cos2 x  1 1  cos2 x  8 cos5 x  8 cos3 x  cos x  8 cos5 x  12 cos3 x  4 cos x  16 cos5 x  20 cos3 x  5 cos x

Thus, the desired polynomial is P t  16t 5  20t 3  5t.

Replace "P" with "Q"


26

CHAPTER 7 Analytic Trigonometry

110. Let c1 and c2 be the lengths of the segments shown in the figure. By the Law of

C b b sin 2x c  or c  . Also by the x sin 2x sin B sin B b a x s c1 s  and Law of Sines applied to BC D and AC D we have sin B sin x B A cÁ cª D s c2 s sin x s sin x  . So c1  and c2  . Since c  c1  c2 , we have sin A sin x sin B  sin A  s sin x s sin x 1 b a b sin 2x 1    s sin x  . By applying the Law of Sines to ABC,  sin B sin B sin A sin B sin A sin B sin A     b b sin 2x 1 b sin x b 1  . Substituting we have  s sin x  s 1   sin A a sin B sin B sin B a sin B sin B a         b b ab 2ab cos x b  2b sin x cos x  sin x 1   2b cos x  s 1  s s  . b sin 2x  s sin x 1  a a a a ab

Sines applied to ABC, we have

111. Using

a

product-to-sum

formula,   RHS  4 sin A sin B sin C  4 sin A 12 [cos B  C  cos B  C]  2 sin A cos B  C  2 sin A cos B  C.

Using another product-to-sum formula, this is equal to     2 12 [sin A  B  C  sin A  B  C]  2 12 [sin A  B  C  sin A  B  C]

 sin A  B  C  sin A  B  C  sin A  B  C  sin A  B  C Now A  B  C  , so A  B  C    2C, A  B  C    2B, and A  B  C  2A  . Thus our expression simplifies to sin A  B  C  sin A  B  C  sin A  B  C  sin A  B  C  sin   2C  sin   2B  0  sin 2A    sin 2C  sin 2B  sin 2A  LHS

112. (a) The length of the base of the inscribed rectangle twice the length of the adjacent side which is 2 5 cos  and the length of the opposite side is 5 sin . Thus the area of the rectangle is modeled by A   2 5 cos  5 sin   25 2 sin  cos   25 sin 2.

  (b) The function y  sin u is maximized when u   2 . So 2  2    4 . Thus the maximum cross-sectional area is   A 4  25 sin 2 4  25 cm2 .       5 2  354 cm. (c) The length of the base is 2 5 cos   5 2  707 cm and the width of the rectangle is 5 sin 4 4 2

113. (a) In both logs the length of the adjacent side is 20 cos  and the length of the opposite side is 20 sin . Thus the cross-sectional area of the beam is modeled by A   20 cos  20 sin   400 sin  cos   200 2 sin  cos   200 sin 2.

  (b) The function y  sin u is maximized when u   2 . So 2  2    4 . Thus the maximum cross-sectional area is   A 4  200 sin 2 4  200.

114. We first label the figure as shown. Because the sheet of paper is folded over,  E AC   C AB  . Thus  BC A   AC E  90  . It follows that  EC D  180   BC A   AC E  180  90    90    2. Also,

from the figure we see that BC  L sin  and C E  L sin , so

A ¬ L

E

DC  EC cos 2  L sin  cos 2. Thus 6  D B  DC 

C B  L sin  cos 2  L sin   L sin  1  cos 2  L sin   2 cos2 . So L

3 6  . 2 sin  cos2  sin  cos2 

D

C

B


27

SECTION 7.4 Basic Trigonometric Equations

115. (a) y  f 1 t  f 2 t  cos 11t  cos 13t

(c) We graph y  cos 11t  cos 13t, y  2 cos t, and y  2 cos t.

2

2 -5

5 -5

-2

5 -2

(b) Using the identity

y cos   cos y  2  cos 2

The graph of f lies between the graphs

  y cos , we have 2     11t  13t 11t  13t f t  cos 11t  cos 13t  2  cos cos 2 2  2  cos 12t  cos t  2 cos 12t cos t 

116. (a) f 1  770 Hz and f 2  1209 Hz, so

of y  2 cos t and y  2 cos t. Thus,

the loudness of the sound varies between y  2 cos t.

(c) 2

y  sin 2  770t  sin 2  1209t  sin 1540t  sin 2418t

0

0.005

(b) Using a sum-to-product formula, we have 1540t  2418t 1540t  2418t cos 2 2  2 sin 1979t cos 439t

-2

y  2 sin

117. We find the area of ABC in two different ways. First, let AB be the base and C D

C

be the height. Since  B OC  2 we see that C D  sin 2. So the area is

1 base height  1  2  sin 2  sin 2. 2 2 On the other hand, in ABC we see that  AC B is a right angle. So BC  2 sin 

and AC  2 cos , and the area is

1

A

¬

1

O

D

B

1 base height  1  2 sin  2 cos   2 sin  cos . 2 2

Equating the two expressions for the area of ABC, we get sin 2  2 sin  cos .

7.4

BASIC TRIGONOMETRIC EQUATIONS

1. Because the trigonometric functions are periodic, if a basic trigonometric equation has one solution, it has infinitely many solutions. 2. The basic equation sin x  2 has no solution (because the sine function has range [1 1]), whereas the basic equation sin x  03 has infinitely many solutions.

3. We can find some of the solutions of sin x  03 graphically by graphing y  sin x and y  03. The solutions shown are x  97, x  60, x  34, x  03, x  28, x  66, and x  91.

4. (a) To find one solution of sin x  03 in the interval [0 2, we take sin1 to get x  sin1 03  030 The other solution in this interval is x    sin1 03  284.

(b) To find all solutions, we add multiples of 2 to the solutions in [0 2. The solutions are x  030  2k and x  284  2k.


28

CHAPTER 7 Analytic Trigonometry

5. Because sine has period 2, we first find the solutions in the interval

y 1

 [0 2. From the unit circle shown, we see that sin   23 in quadrants I 2 and II, so the solutions are    3 and   3 . We get all solutions of the

¬= 3

equation by adding integer multiples of 2 to these solutions: 2   3  2k and   3  2k for any integer k.

y=Ï3/2

_1

0

¹

¬= 3

1

x

6. The sine function is negative in quadrants III and IV, so solutions of sin    22 on the interval [0 2 are   54 and

  74 . Adding integer multiples of 2 to these solutions gives all solutions:   54  2k, 74  2k for any integer k.

7. The cosine function is negative in quadrants II and III, so the solution of cos   1 on the interval [0 2 is   . Adding integer multiples of 2 to this solution gives all solutions:     2k  2k  1  for any integer k. 

8. The cosine function is positive in quadrants I and IV, so the solutions of cos   23 on the interval [0 2 are    6 and 11   116 . Adding integer multiples of 2 to these solutions gives all solutions:    6  2k, 6  2k for any integer k.

9. The cosine function is positive in quadrants I and IV, so the solutions of cos   14 on the interval [0 2 are

  cos1 41  132 and   2  cos1 41  497. Adding integer multiples of 2 to these solutions gives all solutions:

  132  2k, 497  2k for any integer k.

10. The sine function is negative in quadrants III and IV, so the solutions of sin   03 on the interval [0 2 are

    sin1 03  345 and   2  sin1 03  598. Adding integer multiples of 2 to these solutions gives all solutions,   345  2k and   598  2k for any integer k.

11. The sine function is negative in quadrants III and IV, so the solutions of sin   045 on the interval [0 2 are

    sin1 045  361 and   2  sin1 045  582. Adding integer multiples of 2 to these solutions gives all solutions,   361  2k, 582  2k for any integer k.

12. The cosine function is positive in quadrants I and IV, so the solutions of cos   032 on the interval [0 2 are

  cos1 032  125 and   2  cos1 032  504. Adding integer multiples of 2 to these solutions gives all solutions,   125  2k, 504  2k for any integer k.    13. We first find one solution by taking tan1 of each side of the equation:   tan1  3    3 . By definition, this is the    only solution in the interval  2  2 . Since tangent has period , we get all solutions of the equation by adding integer multiples of :     3  k for any integer k.

14. One solution of tan   1 is   tan1 1   4 . Adding integer multiples of  to this solution gives all solutions:   4  k for any integer k.

15. One solution of tan   5 is   tan1 5  137. Adding integer multiples of  to this solution gives all solutions:   137  k for any integer k.   16. One solution of tan    13 is   tan1  13  032. Adding integer multiples of  to this solution gives all solutions:   032  k for any integer k.

    17. One solution of cos    23 is   cos1  23  56 and another is   2  56  76 . All solutions are

  56  2k, 76  2k for any integer k. Specific solutions include   56  2   76 ,   76  2   56 ,   56 ,   76 ,   56  2  176 , and   76  2  196 .

 5  18. One solution of cos   12 is   cos1 12   3 and another is   2  3  3 . All solutions are   3  2k and  5 7 11   53  2k for any integer k. Specific solutions include    53 ,   3 , 3 , 3 , 3 , and 3 .


SECTION 7.4 Basic Trigonometric Equations

29

   3  19. One solution of sin   22 is   sin1 22   4 and another is     4  4 . All solutions are   4  2k and 3 9 11   34  2k for any integer k. Specific solutions include    74 ,  54 ,  4 , 4 , 4 , and 4 .    20. One solution of sin    23 is     sin1 23  43 and another is   2  sin1 23  53 . All solutions are 4 5 10 11   43  2k and   53  2k for any integer k. Specific solutions include    23 ,   3 , 3 , 3 , 3 , and 3 . 21. One solution of cos   028 is   cos1 028  129 and another is   2  cos1 028  500. All solutions are

  129  2k and   500  2k for any integer k. Specific solutions include   500, 129, 129, 500, 757, and 1128.

22. One solution of tan   25 is   tan1 25  119. All solutions are   119  k for any integer k. Specific solutions include   509, 195, 119, 433, 747, and 1061.

23. One solution of tan   10 is   tan1 10  147. All solutions are   147  k for any integer k. Specific solutions include   775, 461, 147, 167, 481, and 795.

24. One solution of sin   09 is     sin1 09  426 and another is   2  sin1 09  516. All solutions are   426  2k and   516  2k for any integer k. Specific solutions include   202, 112, 426, 516, 1054, and 1144. 25. cos   1  0  cos   1. In the interval [0 2 the only solution is   Thus the solutions are   2k  1  for any integer k. 26. sin   1  0  sin   1. In the interval [0 2 the only solution is   32 . Therefore, the solutions are

  32  2k for any integer k.   11 27. 2 cos   3  0  cos   23 . The solutions in the interval [0 2 are    6 , 6 . Thus the solutions are 11   6  2k, 6  2k for any integer k.

28. 2 sin   1  0  sin    12 . The solutions in the interval [0 2 are   76 , 116 . Thus the solutions are   76  2k, 116  2k for any integer k.

29. 3 cos   1  0  cos   13 . The solutions in the interval [0 2 are   cos1 13  123 and   2  cos1 13  505. Thus the solutions are   123  2k, 505  2k for any integer k.

3 . The solutions in the interval [0 2 are     sin1 3  345 and 30. 10 sin   3  0  sin    10 10

3  598. Thus the solutions are   345  2k, 598  2k for any integer k.   2  sin1 10     31. 3 tan2   1  0  tan2   13  tan    33 . The solutions in the interval   2  2 are    6 , so all solutions are

    6  k, 6  k for any integer k.

32. cot   1  0  cot   1. The solution in the interval 0  is   34 . Thus, the solutions are   34  k for any integer k. 3 5 7  33. 2 cos2   1  0  cos2   12  cos    1     4 , 4 , 4 , 4 in [0 2. Thus, the solutions are   4  k, 2

3  k for any integer k. 4

2 4 5  34. 4 sin2   3  0  sin2   34  sin    23     3 , 3 , 3 , 3 in [0 2. Thus, the solutions are   3  k, 2  k for any integer k. 3

    35. 3 sin2   1  0  sin    33     sin1 33  062 in   2  2 . Thus, the solutions are   062  k for any integer k.   36. tan2   9  0  tan   3     tan1 3  125 in   2  2 . Thus, the solutions are   125  k for any integer k.


30

CHAPTER 7 Analytic Trigonometry

 3 5 7 37. sec2   2  0  sec2   2  sec    2. In the interval [0 2 the solutions are    4 , 4 , 4 , 4 . Thus, the solutions are   2k  1  4 for any integer k.

5 7 11 38. csc2   4  0  csc2   4  csc   2. In the interval [0 2 the solutions are    6 , 6 , 6 , 6 . So the

5 solutions are    6  k, 6  k for any integer k.   39. tan2   4 2 cos   1  0  tan2   4 or 2 cos   1.

  If tan2   4  0, then tan2   4  tan   2    tan1 2  111 or   tan1 2  111 in   2 2 .   If 2 cos   1, then cos    12  cos1  12  23 or 43 on [0 2, so   23  2k or 43  2k for any

integer k.

Thus, the original equation has solutions   111  k, 111  k, 23  2k, and 43  2k for any integer k.   1 . The first equation has solution   tan1 2  111 40. tan   2 16 sin2   1  0  tan   2 or sin2   16    on  2  2 . The second equation is equivalent to sin    14 , which has solutions   sin1 41  025,     sin1 41  289,     sin1 41  339, and   2  sin1 41  603 on [0 2.

Thus, the original equation has solutions   111  k, 025  2k, 289  2k, 339  2k, and 603  2k for any integer k. 5 41. 4 cos2   4 cos   1  0  2 cos   12  0  2 cos   1  0  cos   12     3  2k, 3  2k for any

integer k.

42. 2 sin2   sin   1  0  2 sin   1 sin   1  0  2 sin   1  0 or sin   1  0. Since 2 sin   1  0  7 2 sin    12    76 , 116 in [0 2 and sin   1     2 in [0 2. Thus the solutions are   6  2k,

11  2k,   2k for any integer k. 6 2 2 43. tan   tan   6  0  tan   3 tan   2  0  tan   2 or tan   3    tan1 2  111 or

    tan1 3  125 in   2  2 . Thus, the solutions are   111  k, 125  k for any integer k.      44. 3 cos4   5 cos2   2  cos2   1 3 cos2   2  0  cos   1 cos   1 3 cos2   2  0  cos   1 

or cos    36 . The left-hand side of the original equation has period , and the solutions on [0  are   0 or 

cos1 36  062. Thus, the solutions are   k,   062  k for any integer k.

45. 2 cos2   7 cos   3  0  2 cos   1 cos   3  0  cos   12 or cos   3 (which is inadmissible)     3,

5 . Therefore, the solutions are     2k, 5  2k for any integer k. 3 3 3 2 46. sin   sin   2  0  sin   2 sin   1  0  sin   2 (inadmissible) or sin   1. Thus, the solutions are

  32  2k for any integer k.

47. cos2   cos   6  0  cos   2 cos x  3  0  cos x  2 or cos x  3, neither of which has a solution. Thus, the original equation has no solution. 48. 2 sin2   5 sin   12  0  sin x  4 2 sin x  3  0  sin x  4 or sin x  32 , neither of which has a solution. Thus, the original equation has no solution. 49. sin2   2 sin   3  sin2   2 sin   3  0  sin   3 sin   1  0  sin   3  0 or sin   1  0. Since

sin   1 for all , there is no solution for sin   3  0. Hence sin   1  0  sin   1    32  2k for any integer k.

  50. 3 tan3   tan   3 tan3   tan   0  tan  3 tan2   1  0  tan   0 or 3 tan2   1  0. Now tan   0 

5   k and 3 tan2   1  0  tan2   13  tan    1     6  k, 6  k. Thus the solutions are   k, 3

  k, 5  k for any integer k. 6

6


SECTION 7.5 More Trigonometric Equations

31

7 11 51. cos  2 sin   1  0  cos   0 or sin    12     2  k, 6  2k, 6  2k for any integer k.     52. sec  2 cos   2  0  sec   0 or 2 cos   2  0. Since sec   1, sec   0 has no solution. Thus    7  7 2 cos   2  0  2 cos   2  cos   22     4  4 in [0 2. Thus   4  2k, 4  2k for any

integer k.

53. cos  sin   2 cos   0  cos  sin   2  0  cos   0 or sin   2  0. Since sin   1 for all , there is no 3  solution for sin   2  0. Hence, cos   0     2  2k, 2  2k    2  k for any integer k.

54. tan  sin   sin   0  sin  tan   1  0  sin   0 or tan   1  0. Now sin   0 when   k and tan   1  0  tan   1    34  k. Thus, the solutions are   k, 34  k for any integer k.

  55. 3 tan  sin   2 tan   0  tan  3 sin   2  0  tan   0 or sin   23 . tan   0 has solution   0 on   2 2 and sin   23 has solutions   sin1 23  073 and     sin1 32  241 on [0 2, so the original equation has solutions   k,   073  2k, 241  2k for any integer k.

3 56. 4 cos  sin   3 cos   0  cos  4 sin   3  0  cos   0 or sin    34 . cos   0 has solutions    2 , 2 on

[0 2, while sin    34 has solutions     sin1 43  399 and   2  sin1 43  544 on [0 2. Thus, the original equation has solutions    2  k,   399  2k, 544  2k for any integer k.

 sin 70 sin 70  07065   133  sin 2  57. We substitute  1  70 and 1  133 into Snell’s Law to get 2 sin  2 133 2  4495 . sin 1 1  0658, so we substitute 2  90 into Snell’s Law to get  0658 58. The index of refraction from glass to air is 152 sin 90   sin 1  0658  1  411 .

59. (a) F  12 1  cos   0  cos   1    0

(b) F  12 1  cos   025  1  cos   05  cos   05    60 or 360  60  300 (c) F  12 1  cos   05  1  cos   1  cos   0    90 or 270

(d) F  12 1  cos   1  1  cos   2  cos   1    180

60. Statement A is true: every identity is an equation. However, Statement B is false: not every equation is an identity. The difference between an identity and an equation is that an identity is true for all values in the domain, whereas an equation may be true only for certain values in the domain and false for others. For example, x  0 is an equation but not an identity, because it is true for only one value of x.

7.5

MORE TRIGONOMETRIC EQUATIONS

1. Using a Pythagorean identity, we calculate sin x  sin2 x  cos2 x  1  sin x  1  1  sin x  0, whose solutions are x  k for any integer k.

2. Using a double-angle formula, we we see that the equation sin x  sin 2x  0 is equivalent to the equation sin x  2 sin x cos x  0. Factoring the left-hand side as sin x 1  2 cos x, we see that solving this equation is equivalent to solving the two basic equations sin x  0 and 1  2 cos x  0.   3. 2 cos2   sin   1  2 1  sin2   sin   1  0  2 sin2   sin   1  0  2 sin2   sin   1  0. From Exercise 7.4.42, the solutions are   76  2k, 116  2k,  2  2k for any integer k.

4. sin2   4  2 cos2   sin2   cos2   cos2   4  1  cos2   4  cos2   3 Since cos   1 for all , it follows that the original equation has no solution.


32

CHAPTER 7 Analytic Trigonometry

5. tan2   2 sec   2  sec2   1  2 sec   2  sec2   2 sec   3  0  sec   3 sec   1  0  sec   3 or sec   1. If sec   3, then cos   13 , which has solutions   cos1 31  123 and   2  cos1 31  505

on [0 2. If sec   1, then cos   1, which has solution    on [0 2. Thus, solutions are     2k,   123  2k, 505  2k for any integer k.

6. csc2   cot   3  1  cot2   cot   3  cot2   cot   2  0  cot   2 cot   1  0  cot   2 or   cot   1. If cot   2, then tan   12 , which has solution   tan1 12  046 on   2  2 , and if cot   1, then      tan   1, which has solution     4 on  2  2 . Thus, solutions are    4  k,   046  k for any integer k. 7. sin 2  sin   0  2 sin  cos   sin   0  sin  2 cos   1  0  sin   0 or cos   12 . The first equation has 5  solutions   0,  on [0 2, and the second has solutions    3 , 3 on [0 2. Thus, solutions are   k, 3  2k, 5  2k for any integer k. 3

8. 3 sin 2  2 sin   0  3 2 sin  cos   2 sin   0  2 sin  3 cos   1  0  sin   0 or cos   13 . The first equation has solutions   0,  on [0 2, and the second has solutions   cos1 13  123 and

  2  cos1 13  505 on [0 2. Thus, solutions are   k,   123  2k, 505  2k for any integer k.    5 9. 3 cos 2  2 cos2   0  3 2 cos2   1  2 cos2   0  4 cos2   3  cos    23     6 , 6 on [0  5 (the left-hand side of the given equation has period ). Thus, solutions are    6  k, 6  k for any integer k. 

10. cos 2  cos2   12  2 cos2   1  cos2   12  cos2   12  cos    22 . Thus, the solutions are    4  k,

3  k for any integer k. 4   11. 2 sin2   cos   1  2 1  cos2   cos   1  0  2 cos2   cos   1  0  2 cos   1 cos   1  0  5 2 cos   1  0 or cos   1  0  cos   12 or cos   1     3  2k, 3  2k, 2k  1  for any integer k.

12. tan   3 cot   0 

sin  3 cos  sin2   3 cos2  sin2   cos2   4 cos2    0  0  0 cos  sin  cos  sin  cos  sin 

1  4 cos2  2  0  1  4 cos2   0  4 cos2   1  cos    12     3  k, 3  k for any integer k. cos  sin  13. sin   1  cos   sin   cos   1. Squaring both sides, we have sin2   cos2   2 sin  cos   1  sin 2  0,

3  which has solutions   0,  2 , , 2 in [0 2. Checking in the original equation, we see that only   2 and    are

valid. (The extraneous solutions were introduced by squaring both sides.) Thus, the solutions are   2k  1 ,  2  2k for any integer k.

14. Square both sides of cos   sin   1 to get cos2   sin2   2 sin  cos   1  sin 2  0, which has solutions   0,  , , 3 on [0 2. Checking in the original equation, we see that only   0 and   3 are valid. Thus, the solutions 2

2

2

are   2k, 32  2k for any integer k.

1 sin  1   sin cos   1. Squaring both sides, we have sin2 cos2 2 sin  cos   1 cos  cos  3  sin 2  0, which has solutions   0,  2 , , 2 on [0 2. Checking in the original equation, we see that only   0

15. tan 1  sec  

is valid. Thus, the solutions are   2k for any integer k.   16. 2 tan   sec2   4  2 tan   1  tan2   4  tan2   2 tan   3  0  tan   3 tan   1  0    tan   3 or tan   1. The first equation has the solution   tan1 3  125 on   2  2 , and the second has      solution  4 on  2  2 . Thus, the solutions are   125  k,   4  k for any integer k.

5  2 5 2 17. (a) 2 cos 3  1  cos 3  12  3   3 , 3 for 3 in [0 2. Thus, solutions are 9  3 k, 9  3 k for any

integer k.


SECTION 7.5 More Trigonometric Equations

33

5 7 11 13 17 (b) We take k  0, 1, 2 in the expressions in part (a) to obtain the solutions    9 , 9 , 9 , 9 , 9 , 9 in [0 2. 5  5 18. (a) 2 sin 2  1  sin 2  12  2   6 , 6 for 2 in [0 . Thus, solutions are 12  k, 12  k for any integer k.

 , 5 , 13 , 17 in [0 2. (b) Take k  0, 1 in the expressions in part (a) to obtain the solutions   12 12 12 12

2 19. (a) 2 cos 2  1  0  cos 2   12  2  23  2k, 43  2k     3  k, 3  k for any integer k. 2 4 5 (b) The solutions in [0 2 are  3, 3 , 3 , 3 .

  2 k, 11  2 k for any integer k. 20. (a) 2 sin 3  1  0  2 sin 3  1  sin 3   12 , which has solutions   718 3 18 3

 , 11 , 19 , 23 , 31 , 35 . (b) The solutions in [0 2 are   718 18 18 18 18 18  1 5    1 k for any integer k. 21. (a) 3 tan 3  1  0  tan 3     3  6  k    518 3 3 5  11   23 29 35 (b) The solutions in [0 2 are 18 , 18 , 17 18 , 18 , 18 , 18 .

5  1 5 1 22. (a) sec 4  2  0  sec 4  2  4   3  2k, 3  2k    12  2 k, 12  2 k for any integer k.

 , 5 , 7 , 11 , 13 , 17 , 19 , 23 . (b) The solutions in [0 2 are 12 12 12 12 12 12 12 12

23. (a) cos 2  1  0  cos 2  1  2  2k    4k for any integer k.

(b) The only solution in [0 2 is   0.   24. (a) tan 4  3  0  tan 4   3  4  23  k    83  4k for any integer k.

(b) There is no solution in [0 2.    25. (a) 2 sin 3  3  0  2 sin 3   3  sin 3   23  3  43  2k, 53  2k    4  6k, 5  6k for any integer k.

(b) There is no solution in [0 2. 26. (a) sec 2  cos 2  cos2 2  1  cos 2  1  2  k    2k for any integer k. (b) The only solution in [0 2 is   0.

  1 27. (a) sin 2  3 cos 2  tan 2  3    12 tan1 3  062 on   4  4 . Thus, solutions are   062  2 k for any integer k.

(b) The solutions in [0 2 are   062, 219, 376, 533.

 1  5 sin 3  1  5 sin2 3  sin 3   55 , which has solutions 28. (a) csc 3  5 sin 3  sin 3       5 5  089,     1 sin1 5  120, and 1 1 1 1  015,      3 sin 5 3  3 sin 5 3 3 5       23  13 sin1 55  194 on 0 23 . Thus, solutions are   015  13 k, 089  13 k for any integer k.

(b) The solutions in [0 2 are   015, 089, 120, 194, 224, 298, 329, 403, 434, 508, 539, 613.

29. (a) 1  2 sin   cos 2  1  2 sin   1  2 sin2   2 sin2   2 sin   0  2 sin  sin   1  0  sin   0 or  sin   1    0, , or  2 in [0 2. Thus, the solutions are   k,   2  2k for any integer k. (b) The solutions in [0 2 are   0,  2 , .

30. (a) tan 3  1  sec 3  tan 3  12  sec2 3  tan2 3  2 tan 3  1  sec2 3  sec2 3  2 tan 3  sec2 3  2 tan 3  0  3  k for any integer k. Because squaring both sides is an operation that can introduce extraneous solutions, we must check each of the possible solutions in the original equation, and we see that only   23 k are valid solutions.

(b) The solutions in [0 2 are   0, 23 , 43 .


34

CHAPTER 7 Analytic Trigonometry

  31. (a) 3 tan3   3 tan2   tan   1  0  tan   1 3 tan2   1  0  tan   1 or 3 tan2   1  tan   1 or  5 tan    1     6  k, 4  k, 6  k for any integer k. 3

 5 7 5 11 (b) The solutions in [0 2 are    6, 4, 6 , 6 , 4 , 6 .

32. (a) 4 sin  cos   2 sin   2 cos   1  0  2 sin   1 2 cos   1  0  2 sin   1  0 or 2 cos   1  0  5 2 4 sin   12 or cos    12     6  2k, 6  2k, 3  2k, 3  2k for any integer k.

5 2 4 (b) The solutions in [0 2 are    6, 6 , 3 , 3 .

33. (a) 2 sin  tan   tan   1  2 sin   2 sin  tan   tan   2 sin   1  0  2 sin   1 tan   1  0 

5 3 2 sin   1  0 or tan   1  0  sin   12 or tan   1     6  2k, 6  2k, 4  k for any integer k.

3 5 7 (b) The solutions in [0 2 are  6, 4 , 6 , 4 .

cos  sin  1 sin  cos2   cos    sin    sin .  cos  cos  sin  sin  cos2  Multiplying both sides by the common denominator cos2  sin  gives sin2   cos4   sin2  cos2     sin2   cos4   1  cos2  cos2   sin2   cos4   cos2   cos4   sin2   cos2   sin2   1  sin2 

34. (a) sec  tan   cos  cot   sin  

3  2 sin2   1  sin    1     4  k, 4  k for any integer k. Since we multiplied the original 2

equation by cos2  sin  (which could be zero) we must check to see if we have introduced extraneous solutions. However, each of the values of  does indeed satisfy the original equation. 3 5 7 (b) The solutions in [0 2 are    4, 4 , 4 , 4 .

35. (a)

(b) f x  3 cos x  1; g x  cos x  1. f x  g x when

4

3 cos x  1  cos x  1  2 cos x  2  cos x  1  x    2k  2k  1 . The points of intersection are

2

2k  1  2 for any integer k.

-6

-4

-2

2

4

6

-2

The points of intersection are approximately 314 2. 36. (a)

(b) f x  sin 2x  1; gx  2 sin 2x  1. f x  g x when

sin 2x  1  2 sin 2x  1  sin 2x  0  x  12 k. The points of   intersection are 12 k 1 for any integer k.

2

-6

-4

-2

2

4

6

The points of intersection are approximately 628 1, 471 1, 314 1, 157 1, and 0 1.


SECTION 7.5 More Trigonometric Equations

37. (a)

35

  3. f x  g x when tan x  3      x 3  k. The intersection points are 3  k 3 for any integer k.

(b) f x  tan x; g x 

10

-1

1 -10

The point of intersection is approximately 105 173. 38. (a)

(b) f x  sin x  1; g x  cos x. f x  g x when sin x  1  cos x  sin x  12  cos2 x  sin2 x  2 sin x  1  cos2 x 

-6

-4

-2

2

4

6

sin2 x  2 sin x  1  cos2 x  0  2 sin2 x  2 sin x  0 

2 sin x sin x  1  0  sin x  0 or sin x  1  x  k,  2  2k. However, x  k is not a solution when k is even. (The extraneous

-2

solutions were introduced by squaring both sides.) So the solutions are

The points of intersection are approximately 471 0, 314 1, 157 0, and 314 1.

x  2k  1 ,  2  2k, and the intersection points are   2k 1,   2  2k 0 for any integer k.

3 5 7 9 11 13 15 39. cos  cos 3  sin  sin 3  0  cos   3  0  cos 4  0  4   2 , 2 , 2 , 2 , 2 , 2 , 2 , 2 in 3 5 7 9 11 13 15 [0 8     8 , 8 , 8 , 8 , 8 , 8 , 8 , 8 in [0 2.

5 40. cos  cos 2  sin  sin 2  12  cos   2  12  cos   12  cos   12     3 , 3 in [0 2. 

2 41. sin 2 cos   cos 2 sin   23  sin 2    23  sin   23     3 , 3 in [0 2.

42. sin 3 cos   cos 3 sin   0  sin 3    0  sin 2  0  2  0, , 2, 3, 4 in [0 4    0,  2 , , 3 in [0 2. 2

7 43. sin 2  cos   0  2 sin  cos   cos   0  cos  2 sin   1  0  cos   0 or sin    12     2, 6 , 3 , 11 in [0 2. 2 6

 sin   sin   0   sin   0  sin   sin  1  cos   0 (and cos   1    )  2 1  cos  3 sin   cos   0  sin   0 or cos   0    0,  2 , 2 in [0 2 (   is inadmissible).

44. tan

45. cos 2  cos   2  2 cos2   1  cos   2  0  2 cos2   cos   3  0  2 cos   3 cos   1  0 

2 cos   3  0 or cos   1  0  cos    32 (which is impossible) or cos   1    0 in [0 2.   cos  sin  cos  sin    8 sin  cos     sin  cos   8 sin  cos   sin  cos   46. tan   cot   4 sin 2  cos  sin  cos  sin  sin2   cos2   8 sin2  cos2   1  2 2 sin  cos 2  sin 22  12  sin 2   1 . Therefore, 2   4  k

2 3   k  3  k  3 5 7 or 2  4  k    8  2 or   8  2 . Thus on the interval [0 2 the solutions are    8, 8 , 8 , 8 , 9 , 11 , 13 , 15 . 8 8 8 8

47. cos 2  cos2   0  2 cos2   1  cos2   0  cos2   1    k for any integer k. On [0 2, the solutions are   0, .


36

CHAPTER 7 Analytic Trigonometry 

48. 2 sin2   2  cos 2  2 sin2   2  1  2 sin2   4 sin2   3  sin    23    23  k, 43  k for any 2 4 5 integer k. On [0 2, the solutions are  3, 3 , 3 , 3 .

  49. cos 2  cos 4  0  cos 2  2 cos2 2  1  0  cos 2  1 2 cos 2  1  0. The first factor has zeros at

2 4 5  2   0,  and the second has zeros at    3 , 3 , 3 , 3 . Thus, solutions of the original equation are are   0, 3 , 3 ,

, 43 , 53 in [0 2.

50. sin 3  sin 6  0  sin 3  2 sin 3 cos 3  0  sin 3 1  2 cos 3  0. The first factor has zeros at   0,  3,

2 , , 4 , 5 and the second has zeros at  , 5 , 7 , 11 , 13 , 17 , so these are all solutions of the original equation 3 3 3 9 9 9 9 9 9

in [0 2.      1  cos   . Squaring both sides, we have 51. cos   sin   2 sin 2  cos   sin   2  2 cos2   sin2   2 sin  cos   1  cos   1  2 sin  cos   1  cos   either 2 sin   1 or cos   0     6,  , 5 , 3 in [0 2. Of these, only  and 3 satisfy the original equation. 2

6

2

6

2

52. Square both sides of sin   cos   12 to get sin2   cos2   2 sin  cos   14  1  sin 2  14  sin 2  34 .

1 1 3  115,   1 sin1 3  357, and Thus, the possible solutions in [0 2 are   12 sin1 34  042,  2  2 sin 4 2 4 3  1 sin1 3  429. We find that only valid solutions to the original equation on [0 2 are   115, 357. 2 2 4

53. sin   sin 3  0  2 sin 2 cos   0  2 sin 2 cos   0  sin 2  0 or cos   0  2  k or   k  2    12 k for any integer k.

54. cos 5  cos 7  0  2 sin 6 sin   0  sin 6 sin   0  sin 6  0 or sin   0  6  k or   k    16 k for any integer k.

55. cos 4  cos 2  cos   2 cos 3 cos   cos   cos  2 cos 3  1  0  cos   0 or cos 3  12     2 or

5 7 11 13 17   2 5 2 3   3  2k, 3  2k, 3  2k, 3  2k, 3  2k, 3  2k    2  k, 9  3 k, 9  3 k for

any integer k.

56. sin 5  sin 3  cos 4  2 cos 4 sin   cos 4  cos 4 2 sin   1  0  cos 4  0 or sin   12   5  1  5 4   2  k or   6  2k, 6  2k    8  4 k, 6  2k, 6  2k for any integer k.

58. cos x 

57. sin 2x  x

x 3

1

-1

1 1

-1

The three solutions are x  0 and x  095.

-4

-2

2 -1

The three solutions are x  117, 266, and 294.


SECTION 7.5 More Trigonometric Equations

59. 2sin x  x

60. sin x  x 3 1

2 -1 2

1 -1

4

The only solution is x  192. 61.

37

The three solutions are x  0 and x  093.

cos x  x2 1  x2

  62. cos x  12 e x  ex 1

1

-1

-1

1

The two solutions are x  071.

1

The only solution is x  0.

x  2x tan u  tan  63. With u  tan1 x and   tan1 2x, we have u     4  tan u    1  1  tan u tan   1  1  x 2x  1    3  17 3  32  4 2 1 2 2 x  2x  1  2x  2x  3x  1  0  x   . Because tan x is not one-to-one, 2 2 4 we must check both roots, and find that only

 173 is a solution to the original equation. 4

64. With u  sin1 x and   cos1 x, we have 2u      cos 2u    1    cos 2u cos   sin 2u sin   1  1  2 sin2 u cos   2 sin u cos u sin   1. Referring to the diagram, this becomes     1  2x 2 x  2x 1  x 2 1  x 2  1  x  2x 3  2x  2x 3  1  x  1.

22002 sin 2  5000  151250 sin 2  32 sin 2  003308  2  189442 or 2  180  189442  17810558 . If 2  189442 , then   094721 , and if 2  17810558 , then   8905279 .

65. We substitute  0  2200 and R   5000 and solve for . So 5000 

66. Since 4e3t  0, we have 0  4e3t sin 2t  0  sin 2t  2t  0, , 2,     t  0, 12 , 1, 32 ,   .       67. (a) 10  12  283 sin 23 t  80  283 sin 23 t  80  2  sin 23 t  80  070671. Now

sin   070671 and   078484. If 23 t  80  078484  t  80  456  t  344. Now in the interval [0 2, we have     078484  392644 and   2  078484  549834. If 23 t  80  392644 

t  80  2281  t  3081. And if 23 t  80  549834  t  80  3194  t  3994

3994  365  344. So according to this model, there should be 10 hours of sunshine on the 34th day (February 3) and on the 308th day (November 4).   (b) Since L t  12  283 sin 23 t  80  10 for t  [34 308], the number of days with more than 10 hours of daylight is 308  34  1  275 days.


38

CHAPTER 7 Analytic Trigonometry

   . The part of the belt touching the larger 2 2 pulley has length 2    R     R and similarly the part

68. (a) First note that  

touching the smaller belt has length    r. To calculate a and b, we write cot

a b       a  R cot and b  r cot , so 2 R r 2 2

R

a Œ

¬l

Œ

b

R

r r

 the length of the straight parts of the belt is 2a  2b  2 R  r cot . Thus, the total length of the belt is 2     L  and so     2 cot   L     R     r  2 R  r cot  R  r     2 cot 2 2 2 R r L   .   2 cot  2 R r  L 2778 (b) We plot   2 cot and      45113 in the same 2 R r 242  121 10 viewing rectangle. The solution is   1047 rad  60 . 5 0

0

2

69. sin cos x is a function of a function, that is, a composition of trigonometric

y

functions (see Section 2.7). Most of the other equations involve sums, products,

1

differences, or quotients of trigonometric functions. sin cos x  0  cos x  0 or cos x  . However, since cos x  1, the only

solution is cos x  0  x   2  k. The graph of f x  sin cos x is shown.

_2¹

0

_1

CHAPTER 7 REVIEW 

sin  cos   1. sin  cot   tan   sin  sin  cos 

 cos  

sin2  cos2   sin2  1    sec  cos  cos  cos 

2. sec   1 sec   1  sec2   1  tan2    3. cos2 x csc x  csc x  1  sin2 x csc x  csc x  csc x  sin2 x csc x  csc x   sin2 x  4.

1

1  sin2 x

1  sec2 x  1  tan2 x cos2 x

5.

cos2 x  tan2 x

6.

1 1  cos x 1  cos2 x sin2 x 1  sec x   1  1  cos x  1  cos x    sec x sec x 1  cos x 1  cos x 1  cos x

7.

sin2 x

cos2 x  1  sin x

1   sin x sin x

cos2 x sin2 x

tan2 x sin2 x

 cot2 x 

1  cot2 x  sec2 x cos2 x

cos x cos x cos x   sin x 1 1 sec x  tan x  1  sin x cos x cos x cos x

8. 1  tan x 1  cot x  1  cot x  tan x  tan x cot x  2  cot x  tan x  2  2

1 cos2 x  sin2 x 2  2  sec x csc x cos x sin x cos x sin x

sin x cos x  sin x cos x

¹

x


CHAPTER 7

39

sin2 x  cos2 x  sin2 x  1 cos2 x sin2 x  2   2  cos x 2 sin2 x  cos2 x sin x 1 2 10. tan x  cot x      sec x csc x2  csc2 x sec2 x cos x sin x cos x sin x cos x sin x 9. sin2 x cot2 x  cos2 x tan2 x  sin2 x 

cos2 x

Review

 cos2 x 

2 sin x cos x 2 sin x cos x 2 sin x sin 2x   tan x   1  cos 2x 2 cos x 1  2 cos2 x  1 2 cos2 x cos x cos y  sin x sin y cos x cos y sin x sin y cos y sin x cos x  y       cot y  tan x 12. cos x sin y cos x sin y cos x sin y cos x sin y sin y cos x x 1  cos x 13. csc x  tan  csc x   csc x  csc x  cot x  cot x 2 sin x x sin x 1  cos x 1  cos x 1 1 14. 1  tan x tan  1   1 1 1  sec x 2 cos x sin x cos x cos x cos x cos 2x 2 sin x cos x 2 cos2 x  1 1 sin 2x     2 cos x  2 cos x   sec x 15. sin x cos x sin x cos x cos x  tan x  tan   1  tan x 4 16. tan x   4  1  tan x tan   1  tan x 4

11.

1 1 1  1 cos x sec x  1 1  cos x x cos x cos x 17.    tan   1 1 sin x sec x cos x sin x 2 sin x sin x cos x cos x

  18. cos x  cos y2  sin x  sin y2  cos2 x  2 cos x cos y  cos2 y  sin2 x  2 sin x sin y  sin2 y  cos2 x  sin2 x    sin2 y  cos2 y  2 cos x cos y  sin x sin y  2  2 cos x  y   x x x 2 x x x x x x x x 19. cos  sin  1  sin x  cos2  2 sin cos  sin2  sin2  cos2  2 sin cos  1  sin 2  2 2 2 2 2 2 2 2 2 2 2     3x  7x 3x  7x 2 sin sin 2 sin 5x sin 2x sin 2x cos 3x  cos 7x 2 sin 5x sin 2x 2 2          tan 2x 20. 3x  7x 3x  7x sin 3x  sin 7x 2 sin 5x cos 2x 2 sin 5x cos 2x cos 2x cos 2 sin 2 2     x  y  x  y x  y  x  y 2 sin cos 2 sin x cos y sin x sin x  y  sin x  y 2 2        tan x 21. x  y  x  y x  y  x  y cos x  y  cos x  y 2 cos x cos y cos x cos 2 cos 2 2   1 22. sin x  y sin x  y  2 cos x  y  x  y  cos x  y  x  y  12 cos 2y  cos 2x       12 1  2 sin2 y  1  2 sin2 x  12 2 sin2 x  2 sin2 y  sin2 x  sin2 y

2  23. (a) f x  1  cos x2  sin x2 , g x  sin x

5 -1

prove this, expand f x and simplify, using the double-angle formula for sine:

1

-5

(b) The graphs suggest that f x  g x is an identity. To

2  f x  1  cos x2  sin x2    1  cos2 x2  2 cos x2 sin x2  sin2 2x    1  2 cos x2 sin x2  cos2 x2  sin2 2x  1  sin x  1  sin x  g x


40

CHAPTER 7 Analytic Trigonometry

24. (a) f x  sin x  cos x, g x  2

 sin2 x  cos2 x

-5

5

(b) The graphs suggest that f x  g x in general. For example, choose x   6 and evaluate the functions:     1 3 f 6  2  2  12 3 , whereas   1 3  g  6  4  4  1  1, so f x  g x.

-2

25. (a) f x  tan x tan x2 , g x 

1 cos x

(b) The graphs suggest that f x  g x in general. For   example, choose x   3 and evaluate: f 3  tan 3

5

-5

tan  6 

5

   1 3  1  1, whereas g  3  1  2, so 3

2

f x  g x.

-5

26. (a) f x  1  8 sin2 x  8 sin4 x, g x  cos 4x 1

-5

5

(b) The graphs suggest that f x  g x is an identity. To

show this, expand g x by using double-angle identities: g x  cos 4x  cos 2 2x  1  2 sin2 2x    1  2 2 sin x cos x2  1  2 4 sin2 x cos2 x    1  8 sin2 x  1  sin2 x

-1

 1  8 sin2 x  8 sin4 x  f x

27. (a) f x  2 sin2 3x  cos 6x

(b) The graph suggests that f x  1 for all x. To prove this, we use the double angle formula to note that

1

-5

5

cos 6x  cos 2 3x  1  2 sin2 3x, so   f x  2 sin2 3x  1  2 sin2 3x  1.

-1

x 28. (a) f x  sin x cot , g x  cos x 2

(b) Proof: f x  sin x cot x2  sin x 

2

-5

Conjecture: f x  g x  1

5

   2 sin x2 cos x2 

cos x2 sin x2

cos x2 sin x2

Now subtract and add 1, so f x  2 x cos2  1  1  cos x  1  g x  1. 2

Graph for #28 looks kinda jagged

 2 cos2 x2


CHAPTER 7

41

Review

29. 4 sin   3  0  4 sin   3  sin   34    sin1 34  08481 or     sin1 34  22935.     30. 5 cos   3  0  5 cos   3  cos    35    cos1  35  22143 or   2  cos1  35  40689.

31. cos x sin x  sin x  0  sin x cos x  1  0  sin x  0 or cos x  1  x  0,  or x  0. Therefore, the solutions are x  0 and . 5 32. sin x  2 sin2 x  0  sin x 1  2 sin x  0  sin x  0 or sin x  12  x  0,  or x   6 , 6 . Therefore, the 5 solutions in [0 2 are x  0,  6 , 6 , .

33. 2 sin2 x  5 sin x  2  0  2 sin x  1 sin x  2  0  sin x  12 or sin x  2 (which is inadmissible)  x   6, 5 . Thus, the solutions in [0 2 are x   and 5 . 6 6 6

34. sin x  cos x  tan x  1  sin x cos x  cos2 x  sin x   cos x  sin x cos x  sin x  cos2 x  cos x  0  sin x cos x  1  cos x cos x  1  0  sin x  cos x cos x  1  0  sin x  cos x or cos x  1  tan x  1 or 5  5 cos x  1  x   4 , 4 or x  0. Therefore, the solutions in [0 2 are x  0, 4 , 4 .

35. 2 cos2 x  7 cos x  3  0  2 cos x  1 cos x  3  0  cos x  12 or cos x  3 (which is inadmissible)  x   3, 5 . Therefore, the solutions in [0 2 are x   , 5 . 3 3 3

  36. 4 sin2 x  2 cos2 x  3  2 sin2 x  2 sin2 x  cos2 x  3  0  2 sin2 x  2  3  0  2 sin2 x  1  sin x   1 . 2

3 5 7 So the solutions in [0 2 are x   4, 4 , 4 , 4 .

37. Note that x   is not a solution because the denominator is zero. 4 cos x  2  cos x   12  x  23 , 43 in [0 2.

1  cos x  3  1  cos x  3  3 cos x  1  cos x

38. sin x  cos 2x  sin x  1  2 sin2 x  2 sin2 x  sin x  1  0  2 sin x  1 sin x  1  0  sin x  12 or

5 3  3 5 sin x  1  x   6 , 6 or x  2 . Thus, the solutions in [0 2 are x  6 , 2 , 6 .    39. Factor by grouping: tan3 x  tan2 x  3 tan x  3  0  tan x  1 tan2 x  3  0  tan x  1 or tan x   3  2 4 5  2 3 4 5 7 x  34 , 74 or x   3 , 3 , 3 , 3 . Therefore, the solutions in [0 2 are x  3 , 3 , 4 , 3 , 3 , 4 .   40. cos 2x csc2 x  2 cos 2x  cos 2x csc2 x  2 cos 2x  0  cos 2x csc2 x  2  0  cos 2x  0 or csc2 x  2 

cos 2x  0or sin2 x  12  cos 2x  0 or sin x   1 . 2

3 5 7  3 5 7 For cos 2x  0, the solutions in [0 4 are 2x   2 , 2 , 2 , 2  the solutions in [0 2 are x  4 , 4 , 4 , 4 . 3 5 7 For sin x   1 , the solutions in [0 2 are x   4, 4 , 4 , 4 . 2

3 5 7 Thus, the solutions of the equation in [0 2 are x   4, 4 , 4 , 4 .

1  cos x 1 4 sin x cos x   1cos x 4 sin2 x cos x  1  4 sin2 x cos x cos x  0 41. tan 12 x 2 sin 2x  csc x  sin x sin x   3  5 7 11  cos x 4 sin2 x  1  0  cos x  0 or sin x   12  x   2 , 2 or x  6 , 6 , 6 , 6 . Thus, the solutions in  5 7 3 11 [0 2 are x   6, 2, 6 , 6 , 2 , 6 .

42. cos 3xcos 2xcos x  0  cos 2x cos xsin 2x sin xcos 2xcos x  0  cos 2x cos xcos 2xsin 2x sin xcos x  0    cos 2x cos x  1  2 sin2 x cos x  cos x  0  cos 2x cos x  1  cos x 1  2 sin2 x  0  cos 2x cos x  1  cos x cos 2x  0  cos 2x cos x  1  cos x  0  cos 2x 2 cos x  1  0  cos 2x  0 or

3 5 7 2 4  cos x   12  2x   2 , 2 , 2 , 2 (in [0 4) or x  3 , 3 (in [0 2). Thus, the solutions in [0 2 are x  4 , 2 , 3 , 5 , 4 , 7 . 3 4 4 3 4


42

CHAPTER 7 Analytic Trigonometry      1 sin x   3  sin x  1  3 cos x  3 cos x  sin x  1  23 cos x  12 sin x  12 3 cos x cos x  sin x  1  cos x     1  x     , 5  x   , 3 . However, x  3 is  cos  cos x  sin 6 6 2 6 2 6 3 3 6 2 2

43. tan x  sec x 

inadmissible because sec 32 is undefined. Thus, the only solution in [0 2 is x   6.

  sin x  0  2 cos2 x  3 sin x  0 (cos x  0)  2 1  sin2 x  3 sin x  0  cos x 2 sin2 x  3 sin x  2  0  2 sin x  1 sin x  2  0  sin x  12 or sin x  2 (which has no solution)  x   6,

44. 2 cos x  3 tan x  0  2 cos x  3 5 . 6

45. We graph f x  cos x and g x  x 2  1 in the viewing 46. We graph f x  esin x and g x  x in the viewing rectangle [0 65] by [2 2]. The two functions intersect

rectangle [0 65] by [1 3]. The two functions intersect

at only one point, x  118.

at only one point, x  222.

2 2 0

2

4

6

0

-2

2

4

6

4002 sin2   sin2   08  sin   08944    634 64 4002 sin2   2500 sin2   2500. Therefore it is impossible for the projectile to reach a height of 3000 ft. (b) 64 (c) The function M   2500 sin2  is maximized when sin2   1, so   90 . The projectile will travel the highest when it is shot straight up. k 48. Since e02t  0 we have f t  e02t sin 4t  0  sin 4t  0  4t  k  t  where k is any integer. Thus 4 the shock absorber is at equilibrium position every quarter second.      1  cos 30 2 3 2 3   49. Since 15 is in quadrant I, cos 15    . 2 4 2

47. (a) 2000 

 

2 , which is Another method: cos 15  cos 45  30   cos 45 cos 30  sin 45 sin 30  22 23  22 12  6 4   equal to 12 2  3.        3 5 1  1  cos 2 3 2 3 2 6   is in quadrant I, sin 5    . 50. Since 512 12 2 2 4 2       sin       sin  cos   cos  sin   1 3  1 1  31   6 2 , which is Another method: sin 512 4 6 4 6 4 6 2 2 4 2 2 2 2   1 equal to 2 2  3. 1

51. tan  8 

1     1  cos  2 4   1  1 2 21  1 2 sin 4  2

 cos   sin 2     sin   1 52. 2 sin 12 12 12 6 2

53. sin 5 cos 40  cos 5 sin 40  sin 5  40   sin 45  1  22 2    tan 66  tan 6 54.  tan 66  6   tan 60  3 1  tan 66 tan 6


CHAPTER 7

43

Review

    2   cos 2   cos   1  2 55. cos2   sin 8 8 8 4 2 2

    3 sin   sin  cos   cos  sin   sin       sin   1  2 56. 12 cos 12 2 12 6 12 6 12 6 12 4 2 2       57. We use a product-to-sum formula: cos 375 cos 75  12 cos 45  cos 30   12 22  23  14 2 3 .

    1 1  cos 45 675  225 675  225 cos  2 cos 45 cos 225  2    2 2 2 2      1  cos 45  1  1  22 2

58. cos 675  cos 225  2 cos

2

In the solutions to Exercises 59–64, x and y are in quadrant I, so we know that sec x  32  cos x  23 , so sin x  35 and 

1  2. tan x  25 . Also, csc y  3  sin y  13 , and so cos y  2 3 2 , and tan y   4 2 2      59. sin x  y  sin x cos y  cos x sin y  35  2 3 2  23  13  29 1  10 .

      60. cos x  y  cos x cos y  sin x sin y  23  2 3 2  35  13  19 4 2  5 .

        5  2 5  2  2 2 5  2 8  10   tan x  tan y 8  2 5  61. tan x  y     23 2   4    2   4     5 5 2 2 1  tan x tan y 8 8  10 8  10 1 1 2

4

2

4

  62. sin 2x  2 sin x cos x  2  35  23  4 9 5 .

y 63. cos  2

1  cos y  2

     1  2 2  3 2

 2 2

1 3 1  cos y y  64. tan  1 2 sin y 3

 32 2 (since cosine is positive in quadrant I) 6

  32 2 32 2  1 

65. We sketch a triangle such that   cos1 73 . We see that tan   2 310 , and the double-angle formula for tangent gives 2 tan  tan 2   1  tan2 

7

   4 10 2  2 310 12 10 3 .    2  31 1  40 9 1  2 310

¬

2Ï10

3

cos   37

5 . From the 66. We sketch triangles such that   tan1 43 and   cos1 13

¬

triangles, we see that sin   35 , cos   45 , and sin   12 13 , so the addition

5

sin     sin  cos   cos  sin 

3

12

tan   34

5 cos   13

formula for sine gives

5  4  12  63  35  13 5 13 65

  67. The double-angle formula for tangent gives tan 2 tan1 x 

  2 tan tan1 x 2x    . 1  tan2 tan1 x 1  x2

4

13

ú

5


44

CHAPTER 7 Analytic Trigonometry

 68. Let   sin1 x and   cos1 y. From the triangles, cos   1  x 2 and  sin   1  y 2 , so using the addition formula for cosine, we have   cos     cos  cos   sin  sin   1  x 2  y  x 1  y 2    y 1  x 2  x 1  y2   10 10 1    tan 69. (a) tan   x x   10 , for x  0. Since the road sign can first be seen when   2 , (b)   tan1 x   10 10 x  we have 2  tan1  2864 ft. Thus, the sign can first be x tan 2

1

1

x

ú

sin   x

cos   y

Ï1-x@

y

2 1 0

20

seen at a height of 2864 ft.

40

70. (a) Let  be the angle formed by the top of the tower, the car, and the base of the

40

building, and let  be the angle formed by base of the tower, the car, and the base   420 420 of the building, as shown in the diagram. Then tan      tan1 x x   380 380 and tan      tan1 . Thus, x x     420 380       tan1  tan1 . x x (b) We graph  x and find that  is maximized when x  400.

Ï1-y@

¬

380

¬ 

º x

0.055 0.050 0.045 300



71. (a) y  sin 2  x  cos x corresponds to graph VII. (b) y  4 sin x cos x  2 sin 2x corresponds to graph III.   (c) y  1  cos 2x  1  1  2 sin2 x  2 sin2 x corresponds to graph VI.   1  1  2 sin2 x 1  cos 2x    tan2 x corresponds to graph II.  (d) y  1  cos 2x 1  2 cos2 x  1     (e) y  12 sin x  23 cos x  cos  3 sin x  sin 3 cos x  sin x  3 corresponds to graph IV. (f) y  1  tan2 x  sec2 x corresponds to graph VIII.

(g) y  cos2 x  sin2 x  cos 2x corresponds to graph I. sin x  tan 12 x corresponds to graph V. (h) y  1  cos x

400

500


CHAPTER 7

Test

45

CHAPTER 7 TEST sin2  cos2  1 sin  sin   cos      sec  cos  cos  cos  cos  sin x 1  cos x tan x tan x 1  cos x tan x 1  cos x 1  cos x 1 cos x 2.      csc x 1  sec x   2 2 1  cos x 1  cos x 1  cos x sin x cos x 1  cos x sin x 2 tan x 2 tan x 2 sin x 3.  cos2 x  2 sin x cos x  sin 2x   cos x 1  tan2 x sec2 x   x  1  cos x  sin x  1  cos x 4. sin x tan 2 sin x 1. tan  sin   cos  

5. 2 sin2 3x  1  cos 2 3x  1  cos 6x

  6. cos 4x  1  2 sin2 2x  1  2 2 sin x cos x2  1  8 sin2 x 1  sin2 x  1  8 sin2 x  8 sin4 x  2    x   x 2 1  cos x 1  cos x 1  cos x 1  cos x 1  cos2 x 7. sin  cos   2  1  sin x   2 2 2 2 2 2 4  x  x  x   x 2 x  x  x  Another method: sin  cos  2 sin cos  cos2  1  sin 2  1  sin x  sin2 2 2 2 2 2 2 2 sin  x 2 sin  2 sin  2 sin  sin  sin   8.   tan  (because          2 2 2 cos  cos  2 cos 4  x2 4  4 sin 2 1  sin 4  2 sin   cos   0 for   2  2)

9. (a) sin 8 cos 22  cos 8 sin 22  sin 8  22   sin 30  12

      (b) sin 75  sin 45  30   sin 45 cos 30  cos 45 sin 30  22  23  22  12  14 6 2           1   3   1  cos 150 2 3 2 3 2   Another method: Since 75 is in quadrant I, sin 75     , 2 2 4 2    6 2 . which is equal to 14        1  cos  1  23 2 3 6     12 2  3 (c) sin 12  2 2 4

 is in quadrant I, Another method: Since 12        sin       sin  cos   cos  sin   3 2  1 2  1 6  2 , which is equal to sin 12 3 4 3 4 3 4 2 2 2 2 4   1 2  3. 2 

52  102 5 . 10. From the figures, we have cos     cos  cos   sin  sin   2  35  1  23  2  15 5 5 3 5

11. sin 3x cos 5x  12 [sin 3x  5x  sin 3x  5x]  12 sin 8x  sin 2x     7x 3x 2x  5x 2x  5x 12. sin 2x  sin 5x  2 cos sin  2 cos sin 2 2 2 2

  1   35

1 3 53  2.  4 5  4 4 5 5 14. 3 sin   1  0  3 sin   1  sin   13    sin1 13  034 or     sin1 31  280 on [0 2.

13. sin    45 . Since  is in quadrant III, cos    35 . Then tan

1  cos     2 sin 


46

FOCUS ON MODELING

15. 2 cos   1 sin   1  0  cos   12 or sin   1. The first equation has solutions    3  105 and   53  524 on [0 2, while the second has the solution    2  157.

16. 2 cos2   5 cos   2  0  2 cos   1 cos   2  0  cos    12 or cos   2 (which is impossible). So in the interval [0 2, the solutions are   23  209, 43  419.

3 17. sin 2  cos   0  2 sin  cos   cos   0  cos  2 sin   1  0  cos   0 or sin   12     2 , 2 or 5   5 3   6 , 6 . Therefore, the solutions in [0 2 are   6  052, 2  157, 6  262, 2  471.

18. 5 cos 2  2  cos 2  25  2  cos1 04  1159279. The solutions in [0 4 are 2  1159279, 2  1159279,

2  1159279, 4  1159279  2  1159279, 5123906, 7442465, 11407091    057964, 256195, 372123, 570355 in [0 2.   19. 2 cos2 x  cos 2x  0  2 cos2 x  2 cos2 x  1  0  cos x   12 . The solutions in [0 4 are x   3  105, x  23  209, x  43  419, and x  53  524.   x  1 1  cos x 20. 2 tan  csc x  0  2   0  cos x  12 . The solutions in [0 4 are x   3  105 and 2 sin x sin x x  53  524.

9 so tan u  9 . From the triangle, cos u  40 , so using a 21. Let u  tan1 40 40 41  2  1  1519 double-angle formula for cosine, cos 2u  2 cos2 u  1  2 40 41 1681 .

41 u

22. We sketch triangles such that   cos1 x and   tan1 y. From the  y 1 triangles, we have sin   1  x 2 , sin    , and cos    , 1  y2 1  y2

1 ¬

sin     sin  cos   cos  sin   1 y  1  x2   x   2 1y 1  y2

40

Ï1-x@

Ï1+y@ ú

x

so the addition formula for sine gives

9

cos   x

y

1

tan   y

 1  x2  x y  1  y2

FOCUS ON MODELING Traveling and Standing Waves       1. (a) Substituting x  0, we get y 0 t  5 sin 4  0   8 t  5 sin  8 t  5 sin 8 t. (b)

y

6

t=0

3.2

6.4

4 2 _2

x

4¹ 2¹

_4 _6

t=1.6

4.8

     (c) We express the function in the standard form y x t  A sin k x  t: y x t  5 sin 4x   8 t  5 sin 4 x  32 t . . Comparing this to the standard form, we see that this is a traveling wave with velocity   32

2. (a) y  02 sin 1047x  0524t  02 sin 1047x  0524t  2 02 sin 1047x cos 0524t  04 sin 1047x cos 0524t.  m  3m. So the nodes are at 3, 6, 9, 12,   . The nodes occur when 1047x  m  x  1047


Traveling and Standing Waves

(b)

47

y t=0

0.4

t=1

0.2

t=2 1

_0.2

2

3

4

t=3l t=4

5

x

6

t=5

_0.4

t=6

Note that when t  3, cos 0524  3  0001, so y x 3  00004. Thus, the graph of y x 3 cannot be distinguished from the x-axis in the diagram. Yes, this is a standing wave.   068. Since   6, we have 3. From the graph, we see that the amplitude is A  27 and the period is 92, so k  292   6  410, so the equation we seek is y x t  27 sin 068x  410t. k  292

4. (a) We are given A  5, period 23 , and   05. Since the period is 23 , we have k  223  3. Thus, expressing the (b)

function in standard form, we have y x t  5 sin 3 x  05t  5 sin 3x  15t. y

t=0

1

2

4 2

_2

x

¹ ¹/2

3¹/2

_4 t=0.5

1.5

5. From the graphs, we see that the amplitude is A  06. The nodes occur at x  0, 1, 2, 3. Since sin x  0 when x  k (k any integer), we have   . Then since the frequency is 2, we get 20  2    40. Thus, an equation for this model is f x t  06 sin x cos 40t. 6. From the graph, we see that the amplitude is A  7. Now sin x  0 when x  k (k an integer). So for k  1, we must   1 have   2      2. Then since the period is 4, we have 2  4    2 . Thus, an equation for this model is f x t  7 sin 2x cos 12 t.

7. (a) The first standing wave has   1, the second has   2, the third has   3, and the fourth has   4.

(b)  is equal to the number of nodes minus 1. The first string has two nodes and   1; the second string has three nodes and   2, and so forth. Thus, the next two values of  would be   5 and   6, as sketched below.

(c) Since the frequency is 2, we have 440  2    880.

(d) The first standing wave has equation y  sin x cos 880t, the second has equation y  sin 2x cos 880t, the third has equation y  sin 3x cos 880t, and the fourth has equation y  sin 4x cos 880t.

8. (a) The nodes of the tube occur when cos 12 x  0 and 0  x  377. So 12 x  2k  1  2  x  2k  1 . Thus, the nodes are at x  , 3, 5, 7, 9, and 11. We stop there since 13  377. Note that the endpoints of the tube (x  0 and x  377) are not nodes.  (b) In the function y  A cos x cos t, the frequency is 2. In this case, the frequency is 50 2  25 Hz.


CORRECTIONS: p. 27,34,35,39,42

CHAPTER 8

POLAR COORDINATES, PARAMETRIC EQUATIONS, AND VECTORS

8.1 8.2

Polar Coordinates 1 Graphs of Polar Equations 5

8.3

Polar Form of Complex Numbers; De Moivre’s Theorem 12

8.4

Plane Curves and Parametric Equations 24

8.5 8.6

Vectors 35 The Dot Product 41 Chapter 8 Review 45 Chapter 8 Test 53

¥

FOCUS ON MODELING: The Path of a Projectile 56

1


8

POLAR COORDINATES, PARAMETRIC EQUATIONS, AND VECTORS

8.1

POLAR COORDINATES

1. We can describe the location of a point in the plane using different coordinate systems. The point P shown in the figure has   rectangular coordinates 1 1 and polar coordinates 2  4 . 2. (a) If a point P in the plane has polar coordinates r  then it has rectangular coordinates x y where x  r cos  and y  r sin . y (b) If P has rectangular coordinates x y then it has polar coordinates r  where r 2  x 2  y 2 and tan   . x       7 correspond to the point and 2 3 1 in Cartesian coordinates. 3. Yes; both 2  6 6 4. No; adding a multiple of 2 to  gives the same point, as does adding an odd multiple of  and reversing the sign of r .

5.

6.

(2, ¹2 )

O

(1, 0)

7.

(3, _ ¹4 )

¹ 2

O

O

8.

(

5¹ 4, _ 6

)

¹ 4

O

9. 5¹

(_2, 4¹ 3) 4¹ 3

_ 6

10. 7¹ 3

O

O

(_3, 7¹ 3 ) Answers to Exercises 11–16 will vary.       5 3 11. 3  2 has polar coordinates 3 2 or 3 2

(3, ¹2 )

O

(2, 3¹ 4 )1

¹ 2

      5 13. 1 76 has polar coordinates 1  6 or 1  6 . 7¹ 6

1

O

      12. 2 34 has polar coordinates 2 114 or 2 74 .

(_1, 7¹ 6)

3¹ 4

O

      2 5 14. 2   3 has polar coordinates 2 3 or 2 3 .

(_2, _ ¹3 ) O

1

¹ 3

1


2

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

15. 5 0 has polar coordinates 5  or 5 2. (_5, 0)

16. 3 1 has polar coordinates 3 1  2 or 3 1  . (3, 1)

O 1 O

  17. Q has coordinates 4 34 .

    18. R has coordinates 4  34  4 54 .        3    20. P has coordinates 4 134  4 54  4  19. Q has coordinates 4   4  4 4 . 4 .         23  4 4  4 3 . . 22. Q has coordinates 4 21. P has coordinates 4  234  4  4 4 4             4 5  4  .   4 7 . 24. S has coordinates 4 103 23. P has coordinates 4 101 4 4 4 4 4  25. P  3 3 in rectangular coordinates, so r 2  x 2  y 2  32  32  18 and we can take r  3 2.    3 y  1, so since P is in quadrant 2 we take   34 . Thus, polar coordinates for P are 3 2 34 . tan    x 3   26. Q  0 3 in rectangular coordinates, so r  3 and   32 . Polar coordinates for Q are 3 32 .      27. Here r  5 and    23 , so x  r cos   5 cos  23   52 and y  r sin   5 sin  23   5 2 3 . R has    rectangular coordinates  52   5 2 3 .      28. r  2 and   56 , so S has rectangular coordinates r cos  r sin   2 cos 56  2 sin 56   3 1 .     29. r   3  2 . So x  r cos   3 cos 2  3  0  0 and y  r sin   3 sin 2  3  1  3. Thus, the rectangular coordinates are 0 3.     30. r   6 23 . So x  r cos   6 cos 23  6  12  3 and y  r sin   6 sin 23  6  23  3 3. Thus, the    rectangular coordinates are 3 3 3 .        2   2 cos   2  1  1, and 31. r   4 . So x  r cos   4 2      1  y  r sin   2 sin  4  2    1. Thus, the rectangular coordinates are 1 1. 2   32. r   1 52 . So x  r cos   1 cos 52  1  0  0 and y  r sin   1 sin 52  1  1  1. Thus, the rectangular coordinates are 0 1.

33. r   5 5. So x  r cos   5 cos 5  5, and y  r sin   5 sin 5  0. Thus, the rectangular coordinates are 5 0. 34. r   0 13. So x y  0 0 because r  0.        3   3 35. r   3   6 . So x  r cos   3 cos  6  2 and y  r sin   3 sin  6   2 . Thus, the rectangular    coordinates are 32   23 .          36. r   2 2  34 . So x  r cos   2 2 cos  34  2 and y  r sin   2 2 sin  34  2. Thus, the rectangular coordinates are 2 2.

 y 1  1, so, since 37. x y  1 1. Since r 2  x 2  y 2 , we have r 2  12  12  2, so r  2. Now tan    x 1   the point is in the second quadrant,   34 . Thus, polar coordinates are 2 34 .


SECTION 8.1 Polar Coordinates

38. x y 

3

     2 3 3 3 . Since r 2  x 2  y 2 , we have r 2  3 3  32  36, so r  6. Now

  y 1 3     , so, since the point is in the fourth quadrant,   116 . Thus, polar coordinates are 6 116 . 3 3 x 3      2  2 y 39. x y  8 8 . Since r 2  x 2  y 2 , we have r 2  8  8  16, so r  4. Now tan    8  1, so, 8 x   . since the point is in the first quadrant,    . Thus, polar coordinates are 4 4 4     2   2    2 2 2 2 40. x y   6  2 . Since r  x  y , we have r   6   2  8, so r  2 2. Now      2 y tan      1 , so, since the point is in the third quadrant,   76 . Thus, polar coordinates are 2 2 76 . 3 x  6 y 2 41. x y  3 4. Since r  x 2  y 2 , we have r 2  32  42  25, so r  5. Now tan    43 , so, since the point is in x   the first quadrant,   tan1 34 . Thus, polar coordinates are 5 tan1 43 . tan  

 2 y  2, and 5. Now, tan    x 1 since the point is in the fourth quadrant,   2  tan1 2 (since we need 0    2). Thus, polar coordinates are   5 2  tan1 2 .   y 43. x y  6 0. r 2  62  36, so r  6. Now tan    0, so since the point is on the negative x­axis,   . x Thus, polar coordinates are 6 .       44. x y  0  3 . r  3 and since the point is on the negative y­axis,   32 . Thus, polar coordinates are 3 32 . 42. x y  1 2. Since r 2  x 2  y 2 , we have r 2  12  22  2, so r 

45. x  y  r cos   r sin   tan   1, and so    4.

  46. x 2  y 2  9. By substitution, r cos 2  r sin 2  9  r 2 cos2   sin2   9  r 2  9  r  3.

47. x  y 2 . We substitute and then solve for r: r cos   r sin 2  r 2 sin2   cos   r sin2   cos   cot  csc . r sin2  5  5 csc . 48. y  5. By substitution, r sin   5  r  sin  4  4 sec . 49. x  4. We substitute and then solve for r: r cos   4  r  cos    50. x 2  y 2  1. By substitution, r cos 2  r sin 2  1  r 2 cos2   sin2   1  r 2 cos 2  1  r2 

1  sec 2. cos 2

  51. x 2  y 2  y. By substitution, r cos 2  r sin 2  r sin   r 2 cos2   sin2   r sin   r  sin . 32   6x y  r 3  6 r cos  r sin   r  6 sin  cos   3 sin 2. 52. x 2  y 2

53. r  7. But r 2  x 2  y 2 , so x 2  y 2  r 2  49. Hence, the equivalent equation in rectangular coordinates is x 2  y 2  49. 54. r  3  x 2  y 2  r 2  9, so an equivalent equation in rectangular coordinates is x 2  y 2  9.

55.     2  cos   0, so an equivalent equation in rectangular coordinates is x  0. y 56.     tan   0   0  y  0. x 57. r cos   6. But x  r cos , and so x  6 is an equivalent rectangular equation. 2  r sin   2. But r sin   y, so y  2 is an equivalent rectangular equation. 58. r  2 csc   r  sin 


4

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

59. r  4 sin   r 2  4r sin . Thus, x 2  y 2  4y is an equivalent rectangular equation. Completing the square, it can be written as x 2  y  22  4.

  60. r  6 cos   r 2  6r cos . By substitution, x 2  y 2  6x  x 2  6x  9  y 2  9  x  32  y 2  9.

61. r  1  cos . If we multiply both sides of this equation by r we get r 2  r  r cos . Thus r 2  r cos   r, and squaring  2  2 both sides gives r 2  r cos   r 2 , or x 2  y 2  x  x 2  y 2 in rectangular coordinates.

62. r  3 1  sin   3  3 sin   r 2  3r  3r sin   r 2  3r sin   3r. Squaring both sides gives   2  2  r 2  3r sin   9r 2 , or x 2  y 2  3y  9 x 2  y 2 in rectangular coordinates.

63. r  1  2 sin . If we multiply both sides of this equation by r we get r 2  r  2r sin . Thus r 2  2r sin   r, and  2  2 squaring both sides gives r 2  2r sin   r 2 , or x 2  y 2  2y  x 2  y 2 in rectangular coordinates. 64. r  2  cos   r 2  2r  r cos   r 2  r cos   2r  2   x 2  y2  x  4 x 2  y2 1 sin   cos  y  x  1.

65. r 

 2  2  r 2  r cos   2r2  r 2  r cos   4r 2 

r sin   cos   1  r sin   r cos   1, and since r cos   x and r sin   y, we get

1  r 1  sin   1  r  r sin   1. Thus r  1  r sin , and squaring both sides gives 1  sin  r 2  1  r sin 2  x 2  y 2  1  y2  1  2y  y 2  x 2  2y  1  0.

66. r 

4  r 1  2 sin   4  r  2r sin   4. Thus r  4  2r sin . Squaring both sides, we get 1  2 sin  r 2  4  2r sin 2 . Substituting, x 2  y 2  4  2y2  x 2  y 2  16  16y  4y 2  x 2  3y 2  16y  16  0.

67. r 

2  r 1  cos   2  r r cos   2  r  2 r cos . Squaring both sides, we get r 2  2  r cos 2 . 1  cos  Substituting, x 2  y 2  2  x2  4  4x  x 2  y 2  4x  4. y y 69. r 2  tan . Substituting r 2  x 2  y 2 and tan   , we get x 2  y 2  . x x 2  70. r 2  sin 2  2 sin  cos   r 4  2r 2 sin  cos   2 r cos  r sin . By substitution, x 2  y 2  2x y  68. r 

x 4  2x 2 y 2  y 4  2x y  0.

   y 2 2 2 2 71. sec   2  cos   12      3  tan    3  x   3  y   3x  y  3x  y  3x  0. y 72. cos 2  1 means that 2  0    0  tan   0   0  y  0 x 73. (a) In rectangular coordinates, the points r1  1  and r2  2  are x1  y1   r1 cos  1  r1 sin 1  and x2  y2   r2 cos 2  r2 sin 2 . Then, the distance between the points is   D  x1  x2 2  y1  y2 2  r1 cos 1  r2 cos 2 2  r1 sin 1  r2 sin 2 2       r12 cos2  1  sin2 1  r22 cos2 2  sin2 2  2r1r2 cos 1 cos 2  sin 1 sin 2  

 r12  r22  2r1r2 cos 2  1 


SECTION 8.2 Graphs of Polar Equations

5

    (b) The distance between the points 3 34 and 1 76 is D

      340 32  12  2 3 1 cos 76  34  9  1  6 cos 512

          (c) In rectangular coordinates, 3 34 corresponds to 3 cos 34  3 sin 34   3 2 2  3 2 2 and 1 76 corresponds    2    2    3 7  7  1  3 2 2  23  3 2 2  12  340, to  cos 6   sin 6  2  2 , so the distance formula gives D 

as above.

74. (a) Because streets are laid out in a grid, rectangular coordinates are more appropriate when giving driving directions to a friend. (b) Descartes’ famous declaration cogito ergo sum does not apply to pigeons, so polar coordinates are probably more useful in this case.

8.2

GRAPHS OF POLAR EQUATIONS

1. To plot points in polar coordinates we use a grid consisting of circles centered at the pole and rays emanating from the pole. 2. (a) To graph a polar equation r  f  we plot all the points r  that satisfy the equation.

(b) The graph of the polar equation r  3 is a circle with radius 3 centered at the pole. The graph of the polar equation   4 is a line passing through the pole with slope 1.

3

O

3. VI

4. III

5. II

O

6. IV

7. I

8. V

9. Polar axis: 2  sin   2  sin   r, so the graph is not symmetric about the polar axis. Pole: 2  sin     2  sin  cos   cos  sin   2   sin   2  sin   r, so the graph is not symmetric about the pole.  Line    2 : 2  sin     2  sin  cos   cos  sin   2  sin   r, so the graph is symmetric about   2 . 10. Polar axis: 4  8 cos   4  8 cos   r, so the graph is symmetric about the polar axis. Pole: 4  8 cos     4  8 cos  cos   sin  sin   4  8  cos   r, so the graph is not symmetric about the pole. Line    2 : 4  8 cos     4  8 cos  cos   sin  sin   4  8  cos   r, so the graph is not symmetric about    2.

11. Polar axis: 3 sec   3 sec   r, so the graph is symmetric about the polar axis. 3 1 3 Pole: 3 sec        3 sec   r , so the graph is not symmetric cos    cos  cos   sin  sin   cos  about the pole. 1 3 3 Line    2 : 3 sec     cos     cos  cos   sin  sin    cos   3 sec   r, so the graph is not symmetric about    2.


6

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

12. Polar axis: 5 cos  csc   5 cos   csc   r, so the graph is not symmetric about the polar axis. 1 Pole: 5 cos    csc     5 cos    sin    1 1  5 cos  cos   sin  sin   5  cos   5 cos  csc   r , sin  cos   cos  sin   sin  so the graph is symmetric about the pole. 1 Line    2 : 5 cos    csc     5 cos    sin    1 1  5 cos  cos   sin  sin   5  cos   5 cos  csc   r , sin  cos   cos  sin  sin   so the graph is not symmetric about   2 . 4 4   r, so the graph is not symmetric about the polar axis. 3  2 sin  3  2 sin  4 4 4 4 Pole:     r, so the graph is not 3  2 sin    3  2 sin  cos   cos  sin  3  2  sin  3  2 sin  symmetric about the pole. 4 4 4 Line    2 : 3  2 sin     3  2 sin  cos   cos  sin   3  2 sin   r, so the graph is symmetric about   2.

13. Polar axis:

5 5   r , so the graph is symmetric about the polar axis. 1  3 cos  1  3 cos  5 5 5 5 Pole:     r, so the graph is not 1  3 cos    1  3 cos  cos   sin  sin  1  3  cos  1  3 cos  symmetric about the pole. 5 5 5 Line    2 : 1  3 cos     1  3 cos  cos   sin  sin   1  3 cos   r, so the graph is not symmetric about    2.

14. Polar axis:

15. Polar axis: 4 cos 2   4 cos 2  r 2 , so the graph is symmetric about the polar axis. Pole: r2  r 2 , so the graph is symmetric about the pole.

 2 Line    2 : 4 cos 2     4 cos 2  2  4 cos 2  4 cos 2  r , so the graph is symmetric about   2 .

16. Polar axis: 9 sin   9 sin   r 2 , so the graph is not symmetric about the polar axis. Pole: r2  r 2 , so the graph is symmetric about the pole.

 2 Line    2 : 9 sin     9 sin  cos   cos  sin   9 sin   r , so the graph is symmetric about   2 .

17. r  2  r 2  4  x 2  y 2  4 is an equation of a circle with radius 2 centered at the origin.

18. r  1  r 2  1  x 2  y 2  1 is an equation of a circle with radius 1 centered at the origin.

(2, ¹2 ) (2, ¹)

O

(1, ¹2 ) (2, 0)

1

(2, 3¹ 2)

(1, ¹)

(1, 0)

O

1

(1, 3¹ 2 )


SECTION 8.2 Graphs of Polar Equations

19.     2  cos   0  x  0 is an equation of a vertical line.

O

1

O

21. r  6 sin   r 2  6r sin   x 2  y 2  6y 

x 2  y  32  9, a circle of radius 3 centered at 0 3.

(6, ¹2 )

1

22. r  cos   r 2  r cos   x 2  y 2  x  2  x  12  y 2  14 , a circle of radius 12 centered at   10 . 2

(0, ¹2 ) O

(1, 0)

O

1

23. r  2 cos . Circle.

1

24. r  3 sin . Circle.

(3, ¹2 ) (_2, 0)

O

1

O

25. r  2  2 cos . Cardioid.

(

¹ 2, 2

)

26. r  1  sin . Cardioid.

1

(2, ¹2 )

(4, ¹) 1

(2, 3¹ 2 )

7

   y 20.   56  tan    33    33  y   33 x, a x  3 line with slope  3 passing through the origin.

(1, ¹)

(0, 3¹ 2 ) (1, 0) 1


8

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

27. r  3 1  sin . Cardioid.

28. r  cos   1. Cardioid.

(_3, 0)

(_1, 3¹ 2 )

(_3, ¹) 1

(_2, ¹)

(0, 0)

1

(_1, ¹2 ) (_6, ¹2 ) 29. r  sin 2

30. r  2 cos 3

O

O

1

31. r   cos 5

1

32. r  sin 4 1

1

_1

_1

1

O

1

_1

_1

33. r  2 sin 5

O

34. r  3 cos 4

(2, ¹2 )

(_3, 3¹ 2 )

(_3, 0)

O

(_3, ¹)

O

(_3, ¹2 )

35. r 

 3  2 sin 

(Ï3, ¹)

(Ï3, 0)

_1 O

1

_1

(Ï3-2, ¹2 )

36. r  2  sin 

(3, ¹2 )

_2 _3

(2, ¹)

(Ï3+2, 3¹ 2)

O

1

(1, 3¹ 2 )

(2, 0)


SECTION 8.2 Graphs of Polar Equations

37. r 

 3  cos 

2

(Ï3-1, ¹) _2

_1

1

38. r  1  2 cos  (Ï3, ¹2 ) (Ï3+1, 0)

O _1 _2

(1, ¹2 )

1

(3, ¹) (_1, 0)

2

O

(Ï3, 3¹ 2)

 39. r  2  2 2 cos 

40. r  3  6 sin 

1

(1, 3¹ 2 )

(9, ¹2 )

(2, ¹2 ) (2-2Ï2, 0) (2+2Ï2, ¹) O

(_3, 3¹ 2 ) (2, 3¹ 2 )

41. r 2  cos 2

(3, ¹)

(3, 0)

O

42. r 2  4 sin 2

O

O

1

43. r  ,   0

1

44. r  1,   0

( ¹2 , ¹2) (¹, ¹)

O

10 1 , 2¹) ( 2¹

0.4

( ¹1 , ¹)

45. r  2  sec 

2 3¹ , 2) ( 3¹

46. r  sin  tan 

2 O

1

(3, 0)

O

1

9


10

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

47. r  cos

   ,   [0 4] 2

48. r  sin

 8 ,   [0 10] 5

1

1

­1

1

­1

­1

49. r  1  2 sin

1 ­1

   ,   [0 4] 2

50. r 

 1  08 sin2 ,   [0 2] 1

2

­1 ­2

1 ­1

51. r  1  sin n. There are n loops. 2

­2

2

2

­2

­2

n1 52. r  1  c sin 2 2

­2

2 ­2

2

2

­2

2

2

­2

2

2

­2

2

­2

­2

­2

­2

n2

n3

n4

n5

2

2

2

2

­2

2 ­2

­2

2 ­2

­2

2 ­2

­2

2 ­2

c  03 c  06 c1 c  15 c2 3  As c increases, the graph becomes “pinched” along the line   4 , eventually growing more “leaves” with that line as their axis.     1   1 , and so on. is IV, since the graph must contain the points 0 0     53. The graph of r  sin 2 2 2 1 54. The graph of r   is I, since as  increase, r decreases (  0). So this is a spiral into the origin.  5 7  5 7 55. The graph of r   sin  is III, since for    2  2  2     the values of r are also 2  2  2    . Thus the graph must cross the vertical axis at an infinite number of points. 56. The graph of r  1  3 cos 3 is II, since when   0 r  1  3  4 and when    r  1  3  2, so there should be two intercepts on the positive­axis.


SECTION 8.2 Graphs of Polar Equations

57.

x 2  y2

3

 3  4x 2 y 2  r 2  4 r cos 2 r sin 2 

r 6  4r 4 cos2  sin2   r 2  4 cos2  sin2   r  2 cos  sin   sin 2. The equation is r  sin 2, a rose.

58.

x 2  y2

3

O

1

2 2   3   x 2  y 2  r 2  r cos 2  r sin 2 

 2   2 r 6  r 2 cos2   r 2 sin2   r 6  r 2 cos2   sin2  

 2 r 6  r 4 cos2   sin2   r 2  cos 22  r   cos 2. The equations

1

r  cos 2 and r   cos 2 have the same graph, a rose.

59.

 2  x 2  y 2  r 2  r cos 2  r sin 2    r 4  r 2 cos2   r 2 sin2   r 4  r 2 cos2   sin2   

x 2  y2

2

r 2  cos2   sin2   cos 2. The graph is r 2  cos 2, a leminiscate.

O

1

 2  2 60. x 2  y 2  x 2  y 2  x  r 2  r 2  r cos   r 2  [r r  cos ]2  r 2  r 2 r  cos 2  1  r  cos 2  1  r  cos   r  cos   1.

Both r  cos   1 and r  cos   1 give the same graph. To see this, replace 

1

with   : cos     1   cos   1   cos   1. This is a cardioid.

61. (a) r  a cos   b sin   r 2  ar cos   br sin  

(b) r  2 sin   2 cos  has center 1 1   and radius 12 22  22  2.

x 2  y 2  ax  by  x 2  ax  y 2  by  0 

x 2  ax  14 a 2  y 2  by  14 b2  14 a 2  14 b2     2  2 x  12 a  y  12 b  14 a 2  b2 . Thus, in rectangular    coordinates the center is 12 a 12 b and the radius is 12 a 2  b2 .

62. (a)

(b) r  tan  sec   y  x 2.

5

­2

2

(2, ¹2 )

(0, 3¹ 4 )

sin  cos 

1 cos 

yr xr2

O

1

(2, 0)

ry y  2  2 1 x x

11


12

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

64. (a)

63. (a) 5000

­5000

5000

5000

­5000

­5000

5000

­5000

At   0, the satellite is at the “rightmost” point in

The satellite starts out at the same point, 5625 0,

its orbit, 5625 0. As  increases, it travels

but its orbit decays. It narrowly misses the earth at its

counterclockwise. Note that it is moving fastest

first perigee (closest approach to the earth), and

when   .

crashes during its second orbit.

(b) The satellite is closest to earth when   . Its height above the earth’s surface at this point is

(b) The satellite crashes when   837 rad  480 .

22,500 4  cos 3960  45003960  540 mi.     65. The graphs of r  1  sin    6 and r  1  sin   3 have the same shape as r  1  sin , rotated through angles of  6 and

2

 , respectively. Similarly, the graph of r  f    is the 3

1

graph of r  f  rotated by the angle .

­2

2 ­1

66. The circle r  2 (in polar coordinates) has rectangular coordinate equation x 2  y 2  4. The polar coordinate

form is simpler. The graph of the equation r  sin 2 is a four­leafed rose. Multiplying both sides by r 2 , we get 32  r 3  r 2 2 cos  sin   2 r cos  r sin  which is x 2  y 2  2x y in rectangular form. The polar form is definitely simpler.

67. y  2  r sin   2  r  2 csc . The rectangular coordinate system gives the simpler equation here. It is easier to study lines in rectangular coordinates.

8.3

POLAR FORM OF COMPLEX NUMBERS; DE MOIVRE’S THEOREM

1. A complex number z  a  bi has two parts: a is the real part and b is the imaginary part. To graph a  bi we graph the ordered pair a b in the complex plane.  2. (a) The modulus of z is r  a 2  b2 and an argument of z is an angle  satisfying tan   ba.

(b) We can express z in polar form as z  r cos   i sin  where r is the modulus of z and  is the argument of z.    3. (a) The complex number z  1  i in polar form is z  2 cos 34  i sin 34 .    (b) The complex number z  2 cos  6  i sin 6 in rectangular form is z  3  i.    (c) The complex number z can be expressed in rectangular form as 1  i or in polar form as 2 cos  4  i sin 4 .


SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem

4. A nonzero complex number has n different nth roots. The number 16 has four

Im

fourth roots. These roots are 2, 2, 2i, and 2i. In the complex plane, these

2i

roots all lie on a circle of radius 2.

_2

2

0 _2i

5. 4i 

 02  42  4

6. 3i 

Im

 09 3

Im

4i i 0

i 1

0

Re

1

Re

_3i

7. 2 

 402

8. 6  6

Im

Im

i _2

9. 5  2i 

0

i 1

0

Re

  52  22  29

10. 7  3i 

Im

6

1

Re

  49  9  58 Im

5+2i i 0

i 1

Re

0

1

Re 7-3i

     11.  3  i   3  1  2

        12. 1  33 i   1  13  43  2 3 3

Im

Im

0

i

Ï3+i

i 1

0

Re _1- Ï3 i 3

1

Re

Re

13


14

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

   3  4i   9 16  13.  25  25  1 5 

     2  i 2     14.    12  12  1   2

Im

Im

i

3+4i 5

0

1

Im 2z

0

0

Re

15. z  1  i, 2z  2  2i, z  1  i, 12 z  12  12 i

i

i

_Ï2+iÏ2 2

16. z  1 

1

Re

   3i, 2z  2  2 3i, z  1  3i,

 1z  1  3i 2 2 2

Im

z

2z

1z 2

1

Re

z

_z

i

1z 2

0

1

Re

_z

17. z  8  2i, z  8  2i

18. z  5  6i, z  5  6i

Im

z

z

i 0

Im

1

i 0

Re

1

Re

z

z

19. z 1  2  i, z 2  2  i, z 1  z 2  2  i  2  i  4, z 1 z 2  2  i 2  i  4  i 2  5

20. z 1  1  i, z 2  2  3i,

z 1  z 2  1  i  2  3i  1  2i,

z 1 z 2  1  i 2  3i  2  3i  2i  3i 2  1  5i

Im

Im

i 0

zÁzª

zª 1

zÁzª zÁ+zª

Re

zÁ i 0

1

zÁ+zª zª

Re


SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem

21. z  a  bi  a  0 b  0

22. z  a  bi  a  1 b  1

Im

Im

i

i 0

23. z  z  3

1

0

Re

24. z  z  1

Im

25. z  z  2

1

0

Re

26. z  2  z  5

Im

1

Re

Im

i

0

1

0

Re

28. z  a  bi  a  b

Im

i 0

Re

i

i

27. z  a  bi  a  b  2

1

Im

i 0

15

1

Re

1

Re

Im

i 1

0

Re

29. For 1  i, tan   11  1 with  in quadrant I     4 , and r 

     12  12  2. Hence, 1  i  2 cos  4  i sin 4 .

7 30. For 1  i, tan   1 1  1 with  in quadrant IV    4 , and r     1  i  2 cos 74  i sin 74 .

2  1 with  in quadrant II    3 , and r  31. For 2  2i, tan   2 4    2  2i  2 2 cos 34  i sin 34 .

  12  12  2. Hence,

  22  22  2 2. Hence,


16

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

   32. For  2  2i, tan   2  1 with  in quadrant III    54 , and r   2     5   2  2i  2 cos 4  i sin 54 .

    3 with  in quadrant III    7 , and r  33. For  3  i, tan   1 3 6  3    7  7   3  i  2 cos 6  i sin 6 .

  2   2  2   2  2. Hence,   2  3  12  2. Hence,

   34. For 5  5 3i, tan   553   3 with  in quadrant II    23 , and r     5  5 3i  10 cos 23  i sin 23 .

  2 52  5 3  10. Hence,

    2    3 with  in quadrant IV    11 , and r  35. For 2 3  2i, tan   2 2 3  22  4. Hence, 3 6 2 3    11  2 3  2i  4 cos 6  i sin 116 .     2   36. For 3  3 3i, tan   3 3 3  3 with  in quadrant I     , and r  32  3 3  6. Hence, 3    3  3 3i  6 cos  3  i sin 3 .    37. For 2i, tan  is undefined,    2 , and r  2. Hence, 2i  2 cos 2  i sin 2 .   38. For 5i, tan  is undefined,   32 , and r  5. Hence, 5i  5 cos 32  i sin 32 . 39. For 3, tan   0,   , and r  3. Hence, 3  3 cos   i sin .     40. For 2, tan   0,   0, and r  2. Hence, 2  2 cos 0  i sin 0.

    41. For  6  2i, tan   2   33 with  in quadrant II    56 , and r   6       6  2i  2 2 cos 56  i sin 56 .

  2  2   6  2  2 2. Hence,

    42. For  5  15i, tan    15  3 with  in quadrant III    43 , and r   5      Hence,  5  15i  2 5 cos 43  i sin 43 .

  2   2   5   15  2 5.

 43. For 4  3i, tan   34 with  in quadrant I    tan1 34  06435, and r  42  32  5. Hence,      4  3i  5 cos tan1 43  i sin tan1 34 .   44. For 3  2i, tan   23 with  in quadrant I    tan1 23  05880, and r  32  22  13. Hence,       3  2i  13 cos tan1 23  i sin tan1 23 .      2     3 with  in quadrant IV    11 , and r  45. For 4 4 3  42  8. 3  i  4 3  4i, tan   4 3 6 4 3     11  3  i  8 cos 6  i sin 116 . Hence, 4         46. For i 2  6i  6  2i, tan   2  33 with  in quadrant I    tan1 33   6 , and 6        2  2    6  2  2 2. Hence, i 2  6i  2 2 cos  r 6  i sin 6 .   3  1 with  in quadrant II    3 , and r  32  32  3 2. Hence, 47. For 3 1  i  3  3i, tan   3 4    3 1  i  3 2 cos 34  i sin 34 .


SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem

  2  1 with  in quadrant II    3 , and r  22  22  2 2. Hence, 48. For 2i 1  i  2  2i, tan   2 4    2i 1  i  2 2 cos 34  i sin 34 .

17

            6 cos   i sin   49. z 1  3 cos  3  i sin 3 , z 2  2 cos 6  i sin 6 , z 1 z 2  3  2 cos 3  6  i sin 3  6 2 2      i sin       3 cos   i sin   z 1 z 2  32 cos   3 6 3 6 2 6 6    3 cos 54  i sin 54 , z 2  2 cos   i sin ,          z1 z2  3  2 cos 54    i sin 54    2 3 cos  4  i sin 4 ,          z 1 z 2  23 cos 54    i sin 54    23 cos  4  i sin 4

50. z 1 

      2 cos 53  i sin 53 , z 2  2 2 cos 32  i sin 32 ,           4 cos 76  i sin 76 , 2  2 2 cos 53  32  i sin 53  32 z1 z2          z 1 z 2  2 cos 53  32  i sin 53  32  12 cos  6  i sin 6

51. z 1

2 2

     , z z  cos 3    i sin 3    cos 13  i sin 13 ,  i sin 52. z 1  cos 34  i sin 34 , z 2  cos  1 2 3 3 4 3 4 3 12 12     3   3   5  5  z 1 z 2  cos 4  3  i sin 4  3  cos 12  i sin 12 53. z 1  4 cos 120  i sin 120 , z 2  2 cos 30  i sin 30 ,   z 1 z 2  4  2 cos 120  30   i sin 120  30   8 cos 150  i sin 150 ,   z 1 z 2  42 cos 120  30   i sin 120  30   2 cos 90  i sin 90 

  2 cos 75  i sin 75 , z 2  3 2 cos 60  i sin 60 ,     z 1 z 2  2  3 2  cos 75  60   i sin 75  60   6 cos 135  i sin 135 ,    z 1 z 2  2 cos 75  60   i sin 75  60   13 cos 15  i sin 15 

54. z 1 

3 2

55. z 1  4 cos 200  i sin 200 , z 2  25 cos 150  i sin 150 ,   z 1 z 2  4  25 cos 200  150   i sin 200  150   100 cos 350  i sin 350 , 4 cos 200  150   i sin 200  150   4 cos 50  i sin 50  z 1 z 2  25 25

56. z 1  45 cos 25  i sin 25 , z 2  15 cos 155  i sin 155 ,   4 cos 180  i sin 180 , z 1 z 2  45  15  cos 25  155   i sin 25  155   25             z 1 z 2  45 15 cos 25  155   i sin 25  155   4 cos 130   i sin 130   4 cos 130  i sin 130    3  i, so tan  1  1 with 1 in quadrant I  1   6 , and r1  3  1  2. 3    z 2  1  3i, so tan 2  3 with 2 in quadrant I  2   3 , and r1  1  3  2.      Hence, z 1  2 cos  6  i sin 6 and z 2  2 cos 3  i sin 3 .        4 cos   i sin  , Thus, z 1 z 2  2  2 cos  6  3  i sin 6  3 2 2      i sin       cos     i sin   , and 1z  1 cos     i sin   . z 1 z 2  22 cos   1 6 3 6 3 6 6 2 6 6

57. z 1 


18

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

   2  2i, so tan 1  1 with  1 in quadrant IV   1  74 , and r1  2  2  2.   z 2  1  i, so tan 2  1 with 2 in quadrant IV  2  74 , and r2  1  1  2.      Hence, z 1  2 cos 74  i sin 74 and z 2  2 cos 74  i sin 74 .             Thus, z 1 z 2  2  2 cos 74  74  i sin 74  74  2 2 cos 72  i sin 72  2 2 cos 32  i sin 32 ,       z 1 z 2  2 cos 74  74  i sin 74  74  2 cos 0  i sin 0, and 2        1z 1  12 cos  74  i sin  74  12 cos  4  i sin 4 .

58. z 1 

     1 with  1 in quadrant IV   1  11 , and r1  12  4  4. 59. z 1  2 3  2i, so tan 1  2 6 2 3 3   z 2  1  i, so tan 2  1 with  2 in quadrant II  2  34 , and r2  1  1  2.      Hence, z 1  4 cos 116  i sin 116 and z 2  2 cos 34  i sin 34 .            i sin 7 , Thus, z 1 z 2  4  2 cos 116  34  i sin 116  34  4 2 cos 712 12       4    13 z 1 z 2   cos 116  34  i sin 116  34  2 2 cos 13 12  i sin 12 , and 2       1   14 cos  1z 1  4 cos  116  i sin  116 6  i sin 6 .   60. z 1   2i, so  1  32 , and r1  2.    z 2  3  3 3i, so tan  2  3 with 2 in quadrant IV  2  43 , and r2  9  27  6.      Hence, z 1  2 cos 32  i sin 32 and z 2  6 cos 43  i sin 43 .          Thus, z 1 z 2   6 2 cos 56  i sin 56 , 2  6 cos 32  43  i sin 32  43          z 1 z 2  62 cos 32  43  i sin 32  43  62 cos  6  i sin 6 , and         1z 1  1 cos  32  i sin  32  22 cos  2  i sin 2 . 2

61. z 1  5  5i, so tan 1  55  1with 1 in quadrant I  1   4 , and r1 

  25  25  5 2.

z 2  4, so  2  0, and r2  4.    Hence, z 1  5 2 cos  4  i sin 4 and z 2  4 cos 0  i sin 0.            i sin  , z z  5 2 cos   i sin  , and  0  i sin  0  20 2 cos Thus, z 1 z 2  5 2  4 cos  1 2 4 4 4 4 4 4 4  1 cos     i sin     2 cos     i sin   . 1z 1   4 4 10 4 4 5 2

      3 with  1 in quadrant III   1  11 , and r1  48  16  8. 62. z 1  4 3  4i, so tan 1  4 3 6 4 3

, and r2  8. z 2  8i, so  2     2   Hence, z 1  8 cos 116  i sin 116 and z 2  8 cos  2  i sin 2 .       11     64 cos  Thus, z 1 z 2  8  8 cos 116   2  i sin 6 2 3  i sin 3 ,       11    z 1 z 2  88 cos 116    cos 43  i sin 43 , and 1z 1  18 cos  2  i sin 6 2 6  i sin 6 .


SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem

19

63. z 1  20, so 1  , and r1  20.   z 2  3  i, so tan  2  1 with 2 in quadrant I  2   6 , and r2  3  1  2. 3   Hence, z 1  20 cos   i sin  and z 2  2 cos  6  i sin 6 .          40 cos 7  i sin 7 , Thus, z 1 z 2  20  2 cos    6  i sin   6 6 6          10 cos 5  i sin 5 , and z 1 z 2  20 2 cos   6  i sin   6 6 6 1 [cos   i sin ]  1 cos   i sin . 1z 1  20 20

 64. z 1  3  4i, so tan 1  43 with 1 in quadrant I  1  tan1 43 , and r1  9  16  5.   z 2  2  2i, so tan 2  1 with 2 in quadrant IV  2  74 , and r2  4  4  2 2.         Hence, z 1  5 cos tan1 43  i sin tan1 43  5 cos 0927  i sin 0927 and z 2  2 2 cos 74  i sin 74 .        Thus, z 1 z 2  5  2 2 cos tan1 43  74  i sin tan1 34  74  10 2 cos 0142  i sin 0142,            5 cos tan1 34  74  i sin tan1 34  74  5 4 2 cos tan1 43  74  i sin tan1 43  74 z 1 z 2   22

 5 4 2 [cos 457  i sin 457], and      1z 1  15 cos  tan1 43  i sin  tan1 43  15 [cos 0927  i sin 0927].

      6  65.  3  i  2 cos 56  i sin 56 , so  3  i  26 cos 306  i sin 306  64 cos 5  i sin 5  64. 66. 1  i 

     10    cos 704  i sin 704  32 cos 32  i sin 32  32i. 2 cos 74  i sin 74 , so 1  i10  2

          5  67.  2  2i  2 cos 54  i sin 54 , so  2  2i  25 cos 254  i sin 254  16 2  16 2i.

  7     7 68. 1  i  2 cos  cos 74  i sin 74  8  8i. 2 4  i sin 4 , so 1  i     1  1  1 and tan   1     . Thus 2  2 i  cos   i sin  . Therefore, 69. r  2 2 4 2 2 4 4  12      2 2   cos 12  2  2 i 4  i sin 12 4  cos 3  i sin 3  1.     1 with  in quadrant IV    11 . Thus 3  i  2 cos 11  i sin 11 , so 70. r  3  1  2 and tan    6 6 6 3       10  10         3   i sin 110  1 cos    i sin    1 1 1 cos 110 3i  12 6 6 1024 3 3 1024 2  2 i  2048 1  3i .      71. r  4  4  4 2 and tan   1 with  in quadrant IV    74 . Thus 2  2i  2 2 cos 74  i sin 74 , so   8 2  2i8  2 2 cos 14  i sin 14  4096 1  0i  4096.    72. r  14  34  1 and tan   3 with  in quadrant III     43 . Thus  12  23 i  cos 43  i sin 43 , so  15   12  23 i  cos 603  i sin 603  cos 20  i sin 20  1.      73. r  1  1  2 and tan   1 with  in quadrant III    54 . Thus 1  i  2 cos 54  i sin 54 , so     7      cos 354  i sin 354  8 2 cos 34  i sin 34  8 2 1  i 1  8 1  i. 2 1  i7  2

2

        74. r  9  3  2 3 and tan   33 with  in quadrant I     6 . Thus 3  3i  2 3 cos 6  i sin 6 , so         4   3  3i  144 cos 23  i sin 23  144  12  23 i  72 1  3i .


20

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

  2  1     . Thus 2 3  2i  4 cos   i sin  , so 75. r  12  4  4 and tan    6 6 6 2 3 3        5  5    1 1 2 3  2i cos 56  i sin 56  1024  23  12 i  2048  3i  14

     1  1  2 and tan   1 with  in quadrant IV    74 . Thus 1  i  2 cos 74  i sin 74 , so   1 cos 56  i sin 56  1 cos 14  i sin 14  1 . 1  i8  16 4 4 16 16

76. r 

 4  1     . Thus 77. r  48  16  8 and tan    6 4 3 3     4 3  4i  8 cos  6  i sin 6 . So,      12    6  2k 6  2k  i sin for k  0, 1.  8 cos 4 3  4i 2 2     i sin   and Thus the two roots are 0  2 2 cos 12 12    13  13  1  2 2 cos 12  i sin 12 .

Im

i 0 wÁ

 4  1     . Thus 78. r  48  16  8 and tan    6 4 3 3     4 3  4i  8 cos  6  i sin 6 . So      13   6  2k 6  2k  i sin for k  0, 1, 2.  2 cos 4 3  4i 3 3    i sin  , Thus the three roots are 0  2 cos 18 18     13  13    i sin 25 . 1  2 cos 18  i sin 8 , and 2  2 cos 25 18 18   79. 81i  81 cos 32  i sin 32 . Thus,      32  2k 32  2k  i sin for k  0, 1, 81i14  8114 cos 4 4   2, 3. The four roots are 0  3 cos 38  i sin 38 ,     1  3 cos 78  i sin 78 , 2  3 cos 118  i sin 118 , and   3  3 cos 158  i sin 158 .   2k 2k cos  i sin for k  0, 1, 2, 3, 5 5   4. Thus the five roots are 0  2 cos 0  i sin 0, 1  2 cos 25  i sin 25 ,     2  2 cos 45  i sin 45 , 3  2 cos 65  i sin 65 , and   4  2 cos 85  i sin 85 .

wü 1

Re

Im wÁ i 0

wü 1

Re

Im

i 0

1

Re w£

80. 32  32 cos 0  i sin 0. Thus, 3215

Im wÁ wª

i 0

wü 1

Re


SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem

2k 2k  i sin , for k  0, 1, 2, 3, 4, 5, 6, 8 8 7. So the eight roots are 0  cos 0  i sin 0  1,

81. 1  cos 0  i sin 0. Thus, 118  cos

  2 2    1  cos  4  i sin 4  2  i 2 , 2  cos 2  i sin 2  i,   3  cos 34  i sin 34   22  i 22 , 4  cos   i sin   1,   5  cos 54  i sin 54   22  i 22 , 6  cos 32  i sin 32  i, and   7  cos 74  i sin 74  22  i 22 .

Im wª

wÁ wü

0

1

Re

   2 cos  4  i sin 4 . So    13   4  2k  4  2k cos  i sin for k  0, 1, 2 1  i13  3 3    i sin  , 2. Thus the three roots are 0  216 cos 12 12       i sin 9 , and   216 cos 17  i sin 17 . 1  216 cos 912 2 12 12 12

i

82. 1  i 

Im wÁ

i wü 0

1

Re

 13  cos 83. i  cos  2  i sin 2 , so i

2  2k 3

 i sin

2  2k 3 

for

Im

3  1 k  0, 1, 2. Thus the three roots are 0  cos  6  i sin 6  2  2 i,

i

1  cos 56  i sin 56   23  12 i, and 2  cos 32  i sin 32  i.

0

1

Re

 15  cos 84. i  cos  2  i sin 2 , so i

2  2k 5

 i sin

2  2k 5

  i sin  , k  0, 1, 2, 3, 4. Thus the five roots are 0  cos 10 10

for

Im i

 9 9 13 13 1  cos  2  i sin 2 , 2  cos 10  i sin 10 , 3  cos 10  i sin 10 , and

wª 0

 17 4  cos 17 10  i sin 10 .

  2k 4

  2k  i sin 4 

2 2  k  0, 1, 2, 3. So the four roots are 0  cos  4  i sin 4  2  i 2 ,

for

    1  cos 34  i sin 34   22  i 22 , 2  cos 54  i sin 54   22  i 22 ,   and 3  cos 74  i sin 74  22  i 22 .

1

Re

1

Re

85. 1  cos   i sin . Then 114  cos

Im wÁ

i

0 wª

21


22

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors    162 1  3  32 and tan   1616 3  3 with  in quadrant III       43 . Thus 16  16 3i  32 cos 43  i sin 43 . So        15 43  2k 43  2k 16  16 3i  i sin for  3215 cos 5 5     i sin 4 , k  0, 1, 2, 3, 4. The five roots are 0  2 cos 415 15       i sin 16 , 1  2 cos 23  i sin 23 , 2  2 cos 16 15 15     22  22  28   3  2 cos 15  i sin 15 , and 4  2 cos 15  i sin 28 15 .

86. r 

Im wÁ

wü i

0

1

Re w¢

87. z 4  1  0  z  114  z  22  22 i, z   22  22 i (from Exercise 85)      , then, z  i 18  cos 2  2k  i sin 2  2k for  i sin 88. z 8  i  0  x  i 18 . Since i  cos  2 2 8 8

  i sin  , cos 5  i sin 5 , cos 9  i sin 9 , k  0, 1, 2, 3, 4, 5, 6, 7. Thus there are eight solutions: z  cos 16 16 16 16 16 16

 13 17 17 21 21 25 25 29 29 cos 13 16  i sin 16 , cos 16  i sin 16 , cos 16  i sin 16 , cos 16  i sin 16 , and cos 16  i sin 16 .   13     89. z 3  4 3  4i  0  z  4 3  4i . Since 4 3  4i  8 cos  6  i sin 6 ,      13   6  2k 6  2k 13 cos  i sin , for k  0, 1, 2. Thus the three roots are  8 4 3  4i 3 3         i sin  , z  2 cos 13  i sin 13 , and z  2 cos 25  i sin 25 . z  2 cos 18 18 18 8 18 18

90. z 6  1  0  z  116 . Since 1  cos 0  i sin 0 z  116  cos 2k6  i sin 2k6 for k  0, 1, 2, 3, 4, 5. Thus there are 

six solutions: z  1, 12  23 i,  12  23 i.

   2 cos 54  i sin 54 , 91. z 3  1  i  z  1  i13 . Since 1  i       54  2k 54  2k  i sin for k  0, 1, 2. Thus the three solutions to this z  1  i13  216 cos 3 3         i sin 5 , 216 cos 13  i sin 13 , and 216 cos 21  i sin 21 . equation are z  216 cos 512 12 12 12 12 12

92. z 3  1  0  z  113 . Since 1  cos 0  i sin 0 z  113  cos

2k 2k  i sin for k  0, 1, 2. Thus the three 3 3 

solutions to this equation are z  cos 0  i sin 0, cos 23  i sin 23 , and cos 43  i sin 43 or z  1,  12  23 i,  12  23 i.     i  i2  4 1 1 i  5 1 5 2   i 93. z  i z  1  0  z  2 1 2 2   i  9 i  i 2  4 1 2   2i or i 94. z 2  i z  2  0  z  2 2    2i  2i2  4 1 2 2i  4 95. z 2  2i z  2  0  z   i 1 2 2 96. z 2  1  i z  i  z  i z  1  0, so z  1 or i. 97. 1

1 cos 0  i sin 0, so by De Moivre’s Theorem, its n roots are      2k 2k 0  2k 0  2k  i sin  cos  i sin for k  0 1 2    n  1. So z 0  1, z k  11n cos n n n n 2 4 2 4  i sin  , z 2  cos  i sin  2 , and so on. z 1  cos n n n n 


23

SECTION 8.3 Polar Form of Complex Numbers; De Moivre’s Theorem

 n 98. From Exercise 97, k  1 for 0  k  n  1. Multiplying both sides by s n and noting that s n  z, we have  n n  s n k  s n  sk  z for 0  k  n  1, showing that 1 s s2      sn1 are nth roots of z. 99. Let   a  bi and z  c  di.

  (a) zz  c  di c  di  c2  d 2 i 2  c2  d 2  z2 .

(b) z  a  bi c  di  ac  bd  ad  bc i, so   z  ac  bd2  ad  bc2  ac2  2abcd  bd2  ad2  2abcd  bc2   ac2  ad2  bc2  bd2 and        z  a 2  b2 c2  d 2  a 2  b2 c2  d 2  ac2  ad2  bc2  bd2  z. (c)

(d)

1 c  di c  di 1 c d    2  2  2 i, so 2 2 z c  di c  di c d c d c  d2       2  2  1  d2 1 1 c2 1 c d          . 2  2 z 2  d2 2 2 z 2 2 2 2 c2  d 2 c2  d 2 c c d c d c d

a  bi c  di ac  bd bc  ad  a  bi c  di     2  2 i, so z c  di c  di c2  d 2 c  d2 c  d2          ac  bd 2 bc  ad 2  ac2  bd2  2abcd  bc2  ad2  2abcd       2 z c2  d 2 c2  d 2 c2  d 2         a 2  b2 c2  d 2 2 2  ac  bd  bc  ad   a 2  b2       2  2 2 2 z 2 2 2 2 c  d c d c d 

100. The cube roots of 1 are 0  1, 1  cos 23  i sin 23   12  23 i, and 2  cos 43  i sin 43  12  23 i, so their       sum is 0  1  2  1   12  23 i  12  23 i  0.

The fourth roots of 1 are 0  1, 1  i, 2  1, and 3  i, so their sum is 0  1  2  3  1  i  1  i  0. The fifth roots of 1 are 0  1, 1  cos 25  i sin 25 , 2  cos 45  i sin 45 , 3  cos 65  i sin 65 , and

4  cos 85  i sin 85 , so their sum is         1  cos 25  i sin 25  cos 45  i sin 45  cos 65  i sin 65  cos 85  i sin 85       5 5 1 1  1  2 cos 25  2 cos 65 (most terms cancel)  1  2 0 4  4   4  4 

3 3  1 2 2 1 2 The sixth roots of 1 are 0  1, 1  cos  3  i sin 3  2  2 i,   cos 3  i sin 3   2  2 i, 

3  1, 4  cos 43  i sin 43   12  23 i, and 5  cos 53  i sin 53  12  23 i, so their sum is    1          1  12  23 i    23 i  1   12  23 i  12  23 i  0. 2 

2 2 2 2  3 3 2 3 The eight roots of 1 are 0  1, 1  cos  4  i sin 4  2  2 i,   i,   cos 4  i sin 4   2  2 i, 

4  1, 5  cos 54  i sin 54   22  22 i, 6  i, and 7  cos 74  i sin 74  22  22 i, so their sum is               1  22  22 i  i   22  22  1   22  22 i  i  22  22 i  0. It seems that the sum of any set of nth roots is 0.    To prove this, factor n  1  1  1 1  1  2  3      n1 . Since this is 0 and 1  0, we must have

1  1  2  3      n1  0.


24

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

101. The cube roots of 1 are 0  1, 1  cos 23  i sin 23 , and 2  cos 43  i sin 43 , so their product is    0  1  2  1 cos 23  i sin 23 cos 43  i sin 43  cos 2  i sin 2  1. The fourth roots of 1 are 0  1, 1  i, 2  1, and 3  i, so their product is 0  1  2  3  1  i  1  i  i 2  1.

The fifth roots of 1 are 0  1, 1  cos 25  i sin 25 , 2  cos 45  i sin 45 , 3  cos 65  i sin 65 , and 4  cos 85  i sin 85 , so their product is      1 cos 25  i sin 25 cos 45  i sin 45 cos 65  i sin 65 cos 85  i sin 85  cos 4  i sin 4  1. 

3  2 2 1 2 3 The sixth roots of 1 are 0  1, 1  cos  3  i sin 3 ,   cos 3  i sin 3   2  2 i,   1, 

4  cos 43  i sin 43   12  23 i, and 5  cos 53  i sin 53  12  23 i, so their product is         cos 2  i sin 2 1 cos 4  i sin 4 5  i sin 5  cos 5  i sin 5  1. 1 cos  cos  i sin 3 3 3 3 3 3 3 3  3 3 2 3 The eight roots of 1 are 0  1, 1  cos  4  i sin 4 ,   i,   cos 4  i sin 4 ,

4  1, 5  cos 54  i sin 54 , 6  i, 7  cos 74  i sin 74 , so their product is         3 3 5 5 7 7  i2  1 cos  4  i sin 4 i cos 4  i sin 4 1 cos 4  i sin 4 i cos 4  i sin 4

cos 2  i sin 2  1. The product of the nth roots of 1 is 1 if n is even and 1 if n is odd. m m  1 . The proof requires the fact that the sum of the first m integers is 2 2 2 2k 2k Let   cos  i sin . Then k  cos  i sin for k  0 1 2     n  1. The argument of the n n n n product of the n roots of unity can be found by adding the arguments of each k . So the argument of the product is 2 n  2  2 n  1  2 2 1  2 2  2 3  [0  1  2  3      n  2  n  1].        0 n n n n n n 2 n  1 n   n  1 . Thus the product of the n roots of Since this is the sum of the first n  1 integers, this sum is n 2 unity is cos n  1   i sin n  1   1 if n is even and 1 if n is odd.

102.

z1 r cos 1  i sin 1  cos 2  i sin 2 r cos 1 cos 2  i 2 sin 1 cos 2  i sin 1 cos 2  i sin 2 cos 1  1  1  z2 r2 cos 2  i sin 2  cos 2  i sin 2 r2 cos2 2  i 2 sin2  2   r  1 cos 1  2   i sin  1  2  r2

103. Let   1  i and z  a  bi. Then z  1  i a  bi  a  b  a  b i, 2  12  12  2, and z2  a 2  b2 .   Then by Exercise 99(b), z2  a  b2  a  b2  2 z2  2 a 2  b2 . Thus, for any given a and b, we have   c  a  b and d  a  b such that c2  d 2  2 a 2  b2 .

8.4

PLANE CURVES AND PARAMETRIC EQUATIONS

1. (a) The parametric equations x  f t and y  g t give the coordinates of a point x y   f t  g t for appropriate values of t. The variable t is called a parameter.     (b) When t  0 the object is at 0 02  0 0 and when t  1 the object is at 1 12  1 1. (c) If we eliminate the parameter in part (b) we get the equation y  x 2 . We see from this equation that the path of the moving object is a parabola.


SECTION 8.4 Plane Curves and Parametric Equations

25

2. (a) It is true that the same curve can be described by parametric equations in many different ways. For instance, the  parabola y  x 2 can be represented by x  t, y  t 2 or by x  t, y  t 2 or by x   t, y  t.

(b) The parametric equations x  2t, y  2t2 model the position of a

y

moving object at time t. When t  0 the object is at 0 0, and when t  1 the object is at 2 4.

t=1 [Ex. 2(b)]

(c) If we eliminate the parameter we get the equation y  x 2 , which is the same equation as in Exercise 1(b). So the objects in Exercises 1(b) and

1

2(b) move along the same path, but traverse the path differently.

t=1 [Ex. 1(b)]

0

1

2

t=0 [1(b) and 2(b)]

3. (a) x  2t, y  t  6

4. (a) x  6t  4, y  3t, t  0

y

y

1 1

1 2

(b) Since x  2t, t 

3 x

x

x

x x and so y   6  2 2

x  2y  12  0.

y (b) Since y  3t, t  and so 3 y  4  2y  4  x  2y  4  0, y  0. x 6 3

 2 6. (a) x  2t  1, y  t  12

5. (a) x  t 2 , y  t  2, 2  t  4 y

y

1

1 4

x

(b) Since y  t  2  t  y  2, we have x  t 2 

x  y  22 , and since 2  t  4, we have

4  x  16.

1

x

2 2   (b) Since y  t  12 , 4y  4 t  12  2t  12 ,

and since x  2t  1, we have x 2  2t  12  4y 

y  14 x 2 .


26

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

7. (a) x 

 t, y  1  t  t  0 y

8. (a) x  t 2 , y  t 4  1 y

1

x

20 10

_10

x

1

(b) Since x 

 t, we have x 2  t, and so y  1  x 2

with x  0.

9. (a) x 

1 , y t 1 t

(b) Since x  t 2 , we have x 2  t 4 and so y  x 2  1, x  0.

10. (a) x  t  1, y  y

t t 1

1

(b) Since x 

1 1 1 we have t  and so y   1. t x x

11. (a) x  4t 2 , y  8t 3

1

x

1

y

x

1

(b) Since x  t  1, we have t  x  1, so y 

x 1 . x

12. (a) x  t, y  1  t

y

y

1 1

x 1 x

1

(b) Since y  8t 3 have y 2  x 3 .

 3 y 2  64t 6  4t 2  x 3 , we

(b) Since x  t, we have y  1  x, where x  0.

14. (a) x  2 cos t, y  3 sin t, 0  t  2

13. (a) x  2 sin t, y  2 cos t, 0  t   y

y

1

1 1

x

(b) x 2  2 sin t2  4 sin2 t and y 2  4 cos2 t. Hence, x 2  y 2  4 sin2 t  4 cos2 t  4  x 2  y 2  4,

where x  0.

1

x

(b) We have cos t  x2 and sin t  y3, so

x22  y32  cos2 t  sin2 t  1  1 x 2  1 y 2  1. 4 9


SECTION 8.4 Plane Curves and Parametric Equations

15. (a) x  sin2 t, y  sin4 t

16. (a) x  sin2 t, y  cos t

y

27

y 1

1

1

x

1

(b) Since x  sin2 t we have x 2  sin4 t and so y  x 2 . But since 0  sin2 t  1 we only get the part of this parabola for which 0  x  1.

17. (a) x  cos t, y  cos 2t

x

(b) Since y  cos t, we have y 2  cos2 t and so

x  y 2  sin2 t  cos2 t  1 or x  1  y 2 (actually  y   1  x), where 0  x  1.

18. (a) x  cos 2t, y  sin 2t

y 1

y

This curve is goimg counterclockwise

1

1

x

(b) Since x  cos t we have x 2  cos2 t, so

2x 2  1  2 cos2 t  1  cos 2t  y. Hence, the

rectangular equation is y  2x 2  1, 1  x  1.

19. (a) x  sec t, y  tan t, 0  t   2  x  1 and y  0.

1

x

(b) x 2  cos2 2t and y 2  sin2 2t. Then

x 2  y 2  cos2 2t  sin2 2t  1  x 2  y 2  1.

20. (a) x  cot t, y  csc t, 0  t   so y  1. y

y

1 1

1 1

x

x

(b) x 2  sec2 t, y 2  tan2 t, and

y 2  1  tan2 t  1  sec2 t  x 2 . Therefore, y 2  1  x 2  x 2  y 2  1, x  1, y  0.

(b) x 2  cot2 t, y 2  csc2 t, and so

x 2  1  cot2 t  1  csc2 t  y 2 . Therefore,

y 2  x 2  1, with y  1. This is the top half of the

hyperbola y 2  x 2  1.


28

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

21. (a) x  tan t, y  cot t, 0  t   2  x  0.

22. (a) x  et , y  et , so y  0.

y

y

1

1 x

1

x

1

(b) x y  tan t  cot t  1, so y  1x for x  0. 23. (a) x  e2t , y  et , so y  0.

(b) x y  et et  1, so y  1x, x  0. 24. (a) x  sec t, y  tan2 t, 0  t   2 , so y  0 and

y

x  1.

2

y

1 _1 0

1

2

3

4

5

7 x

6

2

 2 (b) x  e2t  et  y 2 , y  0.

x

1

(b) x 2  sec2 t  1  tan2 t, so x 2  1  y, x  1, y  0.

25. (a) x  cos2 t, y  sin2 t, so 0  x  1 and 0  y  1. y

26. (a) x  cos3 t, y  sin3 t, 0  t  2 y

1

1

1

x

(b) x  y  cos2 t  sin2 t  1. Hence, the equation is x  y  1 for 0  x  1 and 0  y  1.

1

x

(b) x 23  cos2 t and y 23  sin2 t, and so x 23  y 23  1.

27. x  3 cos t, y  3 sin t. The radius of the circle is 3, the position at time 0 is x 0  y 0  3 cos 0 3 sin 0  3 0 and the orientation is counterclockwise (because x is decreasing and y is increasing initially). x y  3 0 again when t  2, so it takes 2 units of time to complete one revolution.

28. x  2 sin t, y  2 cos t. The radius is 2, the position at time 0 is 0 2, the orientation is clockwise (because x is increasing and y is decreasing initially), and it takes 2 units of time to complete one revolution.

29. x  sin 2t, y  cos 2t. The radius of the circle is 1, the position at time 0 is x 0  y 0  sin 0 cos 0  0 1 and the orientation is clockwise (because x is increasing and y is decreasing initially). x y  0 1 again when t  , so it takes  units of time to complete one revolution. 30. x  4 cos 3t, y  4 sin 3t. The radius is 4, the position at t  0 is 4 0, the orientation is counterclockwise, and it takes 2 units of time to complete one revolution. 3

Answers to Exercises 31–36 will vary.


SECTION 8.4 Plane Curves and Parametric Equations

29

31. The radius is 5 and it takes 4 seconds to complete a clockwise revolution, so if the position at t  0 is 0 5, then parametric equations are x  5 sin 12 t, y  5 cos 12 t.

32. The radius is 1 and it takes 2 seconds to complete a counterclockwise revolution, so if the position at t  0 is 1 0, parametric equations are x  cos t, y  sin t. 33. Since the line passes through the point 4 1 and has slope 12 , parametric equations for the line are x  4  t, y  1  12 t.

34. Since the line passes through the points 6 7 and 7 8, its slope is x  6  t, y  7  t.

87  1. Thus, parametric equations for the line are 76

35. Since cos2 t  sin2 t  1, we have a 2 cos2 t  a 2 sin2 t  a 2 . If we let x  a cos t and y  a sin t, then x 2  y 2  a 2 . Hence, parametric equations for the circle are x  a cos t, y  a sin t.

a 2 cos2 t b2 sin2 t x2 y2   1. If we let x  a cos t and y  b sin t, then 2  2  1. 2 2 a b a b Hence, parametric equations for the ellipse are x  a cos t, y  b sin t. x 37. x   0 cos  t, y   0 sin  t  16t 2 . From the equation for x, t  . Substituting into the equation for y gives  0 cos   2 x x 16x 2 . Thus the equation is of the form y  c1 x  c2 x 2 ,  16 y   0 sin   x tan   2  0 cos   0 cos   0 cos2 

36. Since cos2 t  sin2 t  1, we have

where c1 and c2 are constants, so its graph is a parabola. 38.  0  2048 fts and   30 . (a) The projectile will hit the ground when y  0. Since y   0 sin  t  16t 2 , we have 0  2048  sin 30  t  16t 2  0  1024t  16t 2 64 seconds.

 16t t  64  0  t  0 or t  64. Hence, the projectile will hit the ground after

 (b) After 64 seconds, the projectile will hit the ground at x  2048  cos 30 64  65,536 3  113,5117 ft  215 miles. (c) The maximum height attained by the projectile is the maximum value of y. Since     y  1024t  16t 2  16 t 2  64t  16 t 2  64t  1024  16,384  16 t  322  16,384, y is maximized when t  32, and therefore the maximum height is 16,384 ft  31 miles. 40. x  2 sin t, y  cos 4t

39. x  sin t, y  2 cos 3t

2 1 ­1

1

­2

­1

2

­2

41. x  3 sin 5t, y  5 cos 3t

42. x  sin 4t, y  cos 3t 5

1

­1 ­5

1 ­1


30

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

43. x  sin cos t, y  cos t 32 , 0  t  2

44. x  2 cos t  cos 2t, y  2 sin t  sin 2t 2

1

­1

­2

1

2

­1

­2

45. (a) r  212 , 0    4  x  2t12 cos t, y  2t12 sin t

46. (a) r  sin   2 cos   x  sin t  2 cos t cos t, y  sin t  2 cos t sin t

(b)

(b)

2

2

­2

2 ­2

47. (a) r  (b)

1

4 cos t 4 sin t 4 x  ,y 2  cos  2  cos t 2  cos t

48. (a) r  2sin   x  2sin t cos t, y  2sin t sin t (b)

2

­2

2

2 ­2

2

4 ­2

2

 2   49. x  t 3  2t, y  t 2  t is Graph III, since y  t 2  t  t 2  t  14  14  t  12  14 , and so y   14 on this curve, while x is unbounded.

50. x  sin 3t, y  sin 4t is Graph IV, since when t  0 and t   the curve passes through 0 0. Thus this curve must pass through 0 0 twice.

51. x  t  sin 2t, y  t  sin 3t is Graph II, since the values of x and y oscillate about their values on the line x  t, y  t  y  x. 52. x  sin t  sin t, y  cos t  cos t is Graph I, since this curve does not pass through the point 0 0.


31

SECTION 8.4 Plane Curves and Parametric Equations

53. (a) It is apparent that x  O Q and

(b) The curve is graphed with a  3 and b  2.

y  Q P  ST . From the diagram,

x  O Q  a cos  and y  ST   b sin .

2

Thus, parametric equations are x  a cos  and y  b sin .

­4

­2

y

a

b O

2

4

­2 S ¬ T

(c) To eliminate  we rearrange: sin   yb

P Q

sin2   yb2 and cos   xa

x

cos2   xa2 . Adding the two equations:

sin2   cos2   1  x 2 a 2  y 2 b2 . As

indicated in part (b), the curve is an ellipse. 54. (a) A has coordinates a cos  a sin . Since O A is perpendicular to AB, 2

O AB is a right triangle and B has coordinates a sec  0. It follows that P has coordinates a sec  b sin . Thus, the parametric equations are x  a sec , y  b sin .

­10

10

(b) The curve is graphed at right with a  3, b  2, and     . 55. (a) If we modify Figure 8 so that PC  b, then by the

­2

(b) 5

same reasoning as in Example 6, we see that x  OT   P Q  a  b sin  and y  T C  C Q  a  b cos .

We graph the case where a  3 and b  2.

­20

20

56. (a) From the figure, we see that x  OT   P Q  a  b sin  and y  T C  C Q  a  b cos . When a  1 and b  2, the parametric equations become x    2 sin , y  1  2 cos . The curve is graphed for 0    4. (b)

y

C(a¬,a)

b

P(x, y) O

¬

­10

Q

T

10

x

2 x2 y 2   y . Since  sec . Also, y  b sec   sec   b a2 b2 2 2 2 2 x y y x tan2   sec2   1, we have 2  2  1  2  2  1, which is the equation of a hyperbola. a b b a

57. x  a tan   tan  

x a

tan2  


32

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

58. Substituting the given values for x and y into the equation we derived in Exercise 57, we get     2 2 b t 1 a t   t  1  t  1. Thus the points on this curve satisfy the equation, which is that of a hyperbola. b2 a2 x2 y2 However, this hyperbola is only the part of 2  2  1 for which x  0 and y  0. b a

59. x  t cos t, y  t sin t, t  0 t

x

y

t

0

0

0

5 4 3 2 7 4

  2 8

  2 8

0

4

2 3 4

61. x 

2 3 2 8

 38 2

2

x

y

y

y

1

   58 2  58 2 0  32   7 2  7 2 8 8

2

60. x  sin t, y  sin 2t

x

1

0

0



3t 3t 2 ,y , t  1 3 1t 1  t3 t

09

x

y

t

x

y

t

996 897

2

067 133

11

075 389 292

25 045 113

05

171 086

3

032 096

0

4

018 074

05

133 067

5

012 060

1

15

6

008 050

15

103 154

0

0

x

1

1

15

x

y

t

997 1097

x

y

4

019 076 012 060

125 393

492

45 015 067

15

284

5

171

6

128

7

104

8

2 25 3 35

189 086 051 035 025

087

y 1

1

x

008 050 006 043 005 038

As t  1 we have x   and y  . As t  1 we have x   and y  . As t   we have x  0 and y  0 . As t   we have x  0 and y  0 .

62. x  cot t, y  2 sin2 t, 0  t  

y

2

1

1

x


SECTION 8.4 Plane Curves and Parametric Equations y

63. (a) We first note that the center of circle C (the small circle) has coordinates [a  b] cos  [a  b] sin . Now the arc P Q has the same length as the arc

a ab a , and so         . Thus the b b b x­coordinate of P is the x­coordinate of the center of circle C plus   ab  , and the y­coordinate of P is the b cos     b cos b   ab y­coordinate of the center of circle C minus b  sin     b sin  . b   ab So x  a  b cos   b cos  and b   ab y  a  b sin   b sin  . b P  Q, so b  a   

(b) If a  4b, b 

33

¬ ú

Q

¬

a , and x  34 a cos   14 a cos 3, y  34 a sin   14 a sin 3. 4

(a, 0) P x

y

a

From Example 2 in Section 7.3, cos 3  4 cos3   3 cos . Similarly, one can prove that sin 3  3 sin   4 sin3 . Substituting, we get   x  34 a cos   14 a 4 cos3   3 cos   a cos3    y  34 a sin   14 a 3 sin   4 sin3   a sin3 . Thus,

a x

x 23  y 23  a 23 cos2   a 23 sin2   a 23 , so x 23  y 23  a 23 .

64. The coordinates of C are [a  b] cos  [a  b] sin . Let  be  OC P as shown in the figure. Then the arcs traced a out by rolling the circle along the outside are equal, that is, b  a    . P is displaced from C by amounts b equal to the legs of the right triangle C PT , where C P is the hypotenuse. Since OC Q is a right triangle, it follows that    a    ab  OC Q    . Thus,               . So 2 2 b 2 b 2          ab ab   a  b cos   b sin   x  a  b cos   b sin b 2 2 b   ab   a  b cos   b cos b and          ab ab   a  b sin   b cos   y  a  b sin   b cos b 2 2 b   ab   a  b sin   b sin b Therefore, parametric equations for the epicycloid are     ab ab x  a  b cos   b cos  and y  a  b sin   b sin . b b

y

C a ¬

ú

Œ

T

P Q

x


34

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

65. A polar equation for the circle is r  2a sin . Thus the coordinates of Q are x  r cos   2a sin  cos  and

y  r sin   2a sin2 . The coordinates of R are x  2a cot  and y  2a. Since P is the midpoint of Q R, we use the   midpoint formula to get x  a sin  cos   cot  and y  a 1  sin2  .

66. (a) C  2a cot  2a, so the x­coordinate of P is x  2a cot . Let B  0 2a. Then  O AB is a right angle and  O B A  , so   O A  2a sin  and A  2a sin  cos  2a sin2  . Thus, the y­coordinate of P is y  2a sin2 .

5

Exercise asks for a = 3

(b) The curve is graphed with a  1.

­10

10

ay  cos   67. We use the equation for y from Example 6 and solve for . Thus for 0    , y  a 1  cos   a         ay ay ay   cos1 . Substituting into the equation for x, we get x  a cos1  sin cos1 . a a a             2ay  y 2 2ay  y 2 ay ay ay 2 However, sin cos1  1 . Thus, x  a cos1  ,  a a a a a      2ay  y 2  x 2ay  y 2  x ay y 1  cos  1  cos  and we have a a a a    2ay  y 2  x . y  a 1  cos a 68. (a)

2 ­2

5

2 ­2

2 ­2

­2

2

5

­5

­5

5

5 ­5

­5

R  05 R1 R3 (b) The graph with R  5 seems to most closely resemble the profile of the engine housing.

R5

69. (a) In the figure, since O Q and QT are perpendicular and OT and T D are perpendicular, the angles formed by their intersections are equal, that is,    DT Q. Now the coordinates of T are cos  sin . Since T D is the length of the string that has been unwound from the circle, it must also have arc length , so T D  . Thus the x­displacement from T to D is   sin  while the y­displacement from T to D is   cos . So the coordinates of D are x  cos    sin  and y  sin    cos . (b) y 10

T 1 ¬

D Q

1

x

­20

20 ­10

70. (a) It takes 2 units of time. The parametric equations of the particle that moves twice as fast around the circle are x  sin 2t, y  cos 2t.


35

SECTION 8.5 Vectors

(b) From the first table, we see that the particle travels counterclockwise. If we want the particle to travel in a clockwise direction, then we want the second table to apply. Possible parametric equations for clockwise traversal of the unit circle are x   sin t, y  cos t. t

x  sin t

0

0

1

2

0

3 2

1

2

y

y  cos t

t

x

y

1

0

0

1

0

1

0

t=¹

 

equation.

2

x

0 3¹ t= 2

1

 71. C: x  t, y  t 2 ; D: x  t, y  t, t  0 (a) For C, x  t, y  t 2  y  x 2 .  For D, x  t, y  t  y  x 2 . For F, x  et

t=0

1

1

0

For E, x  sin t

¹

1 t= 2

E: x  sin t, y  1  cos2 t

0

3 2

1

2

0

1

0 1

F: x  et , y  e2t

In #71, replace "e" with "3" (4 times)

x 2  sin2 t  1  cos2 t  y and so y  x 2 .

x 2  e2t  y and so y  x 2 . Therefore, the points on all four curves satisfy the same rectangular

(b) Curve C is the entire parabola y  x 2 . Curve D is the right half of the parabola because t  0 and so x  0. Curve E is the portion of the parabola for 1  x  1. Curve F is the portion of the parabola where x  0, since et  0 for all t. y

y

y

y

1

1

1

1

0

1

0

x

C

8.5

1

D

x

0

E

1

x

0

1

x

F

VECTORS

1. (a) The vector u has initial point A and terminal point B.

(b) The vector u has initial point 2 1 and terminal point 4 3. In component form we write u  2 2

v

u

2u

u+v

and v  3 6. Then 2u  4 4 and u  v  1 8.

u

2. (a) The length of a vector w  a1  a 2  is w     u  22  22  8  2 2.

 a12  a22 , so the length of the vector u in Figure II is

(b) If we know the length w and direction  of a vector w then we can express the vector in component form as w  w cos  w sin .


36

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

3. 2u  2 2 3  4 6

4. v   3 4  3 4

y

y

2u v

u 1

v

u 1 x

1

5. u  v  2 3  3 4  2  3 3  4  1 7 y

1

_v

x

6. u  v  2 3  3 4  2  3 3  4  5 1 y

v u+v _v

u 1 x

1

7. v  2u  3 4  2 2 3  3  2 2  4  2 3

u 1 1

u-v

x

8. 2u  v  2 2 3  3 4  2 2  3 2 3  4  1 10

 7 2 y

y

v 2u+v _2u

v

1

2u

x

1

1

v-2u

1

x

In Solutions 9–18, v represents the vector with initial point P and terminal point Q.

9. P 2 1, Q 5 4. v  5  2 4  1  3 3

10. P 2 1, Q 3 4. v  3  2 4  1  5 3

11. P 1 2, Q 4 1. v  4  1 1  2  3 1

12. P 3 1, Q 1 2. v  1  3 2  1  2 3

13. P 1 3, Q 4 5. v  4  1 5  3  3 2

14. P 2 5, Q 3 1. v  3  2 1  5  1 4

15. P 5 3, Q 1 0. v  1  5 0  3  4 3

16. P 1 3, Q 6 1.

17. P 1 1, Q 1 1. v  1  1  1  1  0 2

18. P 8 6, Q 1 1. v  1  8  1  6  7 5

v  6  1  1  3  5 4


SECTION 8.5 Vectors y

19.

(6, 7)

y

20.

(3, 5)

u

u (4, 3)

(4, 3)

1

1 x

1

The terminal point is 4  2 3  4  6 7. y

21.

The terminal point is 4  1 3  2  3 5. y

22.

(4, 3)

u

(4, 3)

u

1

(_4, 2)

1

(8, 0) x

1

y

x

1

The terminal point is 4  4 3  3  8 0. 23.

x

1

The terminal point is 4  8 3  1  4 2. y

24. (_3, 5)

(2, 3)

u u 1

u

(_3, 5) u

1

x

1

u

x

u

1

25.

u

(2, 3)

y

26.

(_3, 5)

y (2, 3)

(2, 3)

u u

(_3, 5)

u

1 1

1 1

x u

27. u  2 3  2i  3j

28. u  1 0  i

29. u  0 2  2j

30. u  4 5  4i  5j

u

x

37


38

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

31. u  1 4, v  1 2. 2u  2  1 4  2 8; 3v  3  1 2  3 6; u  v  1 4  1 2  0 6; 3u  4v  3 12  4 8  7 4 32. u  2 5  v  2 8. 2u  2  2 5  4 10; 3v  3  2 8  6 24; u  v  2 5  2 8  0 3; 3u  4v  6 15  8 32  14 47 33. u  0 1, v  2 0. 2u  20 1  0 2; 3v  32 0  6 0; uv  0 12 0  2 1; 3u  4v  0 3  8 0  8 3 34. u  i, v  2j. 2u  2i; 3v  3 2j  6j; u  v  i  2j; 3u  4v  3i  8j 35. u  2i  j, v  j. 2u  2 2i  j  4i  2j; 3v  3j; u  v  2i  j  j  2i; 3u  4v  3 2i  j  4j  6i  7j 36. u  i  j, v  i  j. 2u  2i  2j 3v  3i  3j u  v  i  j  i  j  2i; 3u  4v  3 i  j  4 i  j  i  7j     37. u  3i  j v  2i  3j. Then u  32  12  10; v  22  32  13; 2u  6i  2j;      2       1 3 2 2 2u  6  2  2 10; 2 v  i  2 j;  12 v  12  32  12 13; u  v  5i  2j; u  v  52  22  29;     u  v  3i  j  2i  3j  i  4j; u  v  12  42  17; u  v  10  13       38. u  2i  3j v  i  2j. Then u  4  9  13; v  1  4  5; 2u  4i  6j; 2u  16  36  2 13;      1 v  1 i  j;  1 v  1  1  1 5; u  v  i  j; u  v  1  1  2; u  v  3i  5j; 2  2 2 4 2     u  v  9  25  34; u  v  13  5     39. u  10 1, v  2 2. Then u  102  12  101; v  22  22  2 2; 2u  20 2;          2u  202  22  404  2 101; 12 v  1 1;  12 v  12  12  2; u  v  8 3;       u  v  82  32  73; u  v  12 1; u  v  122  12  145; u  v  101  2 2     40. u  6 6, v  2 1. Then u  36  36  6 2; v  4  1  5; 2u  12 12;             2u  144  144  12 2; 12 v  1  12 ;  12 v  1  14  12 5; u  v  8 5; u  v  64  25  89;     u  v  4 7; u  v  16  49  65; u  v  6 2  5 In Solutions 41–46, x represents the horizontal component and y the vertical component.   41. v  10, direction   60 . x  10 cos 60  5 and y  10 sin 60  5 3. Thus, v  xi  yj  5i  5 3j.   42. v  20, direction   150 . x  20 cos 150  10 3 and y  20 sin 150  10. Thus, v  xi  yj  10 3i  10j. 43. v  1, direction   225 . x  cos 225   1 and y  sin 225   1 . Thus,

2 2   2 2 1 1 v  xi  yj    i   j   2 i  2 j. 2 2 44. v  800, direction   125 . x  800 cos 125  45886 and y  800 sin 125  65532. Thus, v  xi  yj  800 cos 125  i  800 sin 125  j  45886i  65532j.

45. v  4, direction   10 . x  4 cos 10  394 and y  4 sin 10  069. Thus, v  xi  yj  4 cos 10  i  4 sin 10  j  394i  069j.      46. v  3, direction   300 . x  3 cos 300  23 and y  3 sin 300   32 . Thus, v  xi  yj  23 i  32 j.    47. v  3 4. The magnitude is v  32  42  5. The direction is , where tan   43    tan1 43  5313 .      48. v   22   22 . The magnitude is v  12  12  1. The direction is , where tan   1 with  in quadrant III    180  tan1 1  225 .

  5 with  in 49. v  12 5. The magnitude is v  122  52  169  13. The direction is , where tan    12   5  15738 . quadrant II      tan1  12


SECTION 8.5 Vectors

50. v  40 9. The magnitude is v  9  1268 .   tan1 40

39

 9 with  in quadrant I  1600  81  41. The direction is , where tan   40

  2   3  2. The direction is , where tan   3 with  in quadrant I  51. v  i  3j. The magnitude is v  12     tan1 3  60 .   52. v  i  j. The magnitude is v  1  1  2. The direction is , where tan   1 with  in quadrant I    tan1 1  45 .

 3 53. v  30, direction   30 . x  30 cos 30  30   2598, y  30 sin 30  15. So the horizontal component of 2  force is 15 3 lb and the vertical component is 15 lb. 54. v  500, direction   70 . x  500 cos 70  17101, y  500 sin 70  46985. So the east component of the velocity is 17101 mi/h and the north component is 46985 mi/h. 55. The flow of the river can be represented by the vector v  3j and the swimmer can be represented by the vector u  2i. Therefore the true velocity is u  v  2i  3j. 56. If the current is 12 mi/h, it can be represented by the vector v  12j. If the swimmer heads in a direction  north of east, his velocity relative to the water is u  2 cos  i  2 sin  j and so his velocity relative to land is u  v  2 cos  i  2 sin   12 j. We want the y­component of his velocity to be 0, so we calculate 2 sin   12  0  sin   35    sin1 53  369 . Therefore, he should swim 369 north of east, or N 531 E.

57. The ocean currents can be represented by the vector v  3i and the salmon can be represented by the vector       u  5 2 2 i  5 2 2 j. Therefore u  v  5 2 2  3 i  5 2 2 j represents the true velocity of the fish.   58. The direction of the plane is N 45 W, so its velocity relative to the air is u  ux  u y , where ux  585 cos 135  4137 and u y  585 sin 135  4137. The wind direction is W, so its velocity is w  wx  0  40 0. Thus, the actual flight path is v  u  w  4137  40 4137  4537 4137, the airplane’s true speed is    v  45372  41372  614 mih, and its true direction is   180  tan1  4137 4537  1376 or approximately N 476 W.

= <0, 40>

59. (a) The velocity of the wind is 40j.

In #71, the exercise asks for component form

(b) The velocity of the jet relative to the air is 425i.

= <425, 0>

(c) The true velocity of the jet is v  425i  40j  425 40.    40  54   is (d) The true speed of the jet is v  4252  402  427 mi/h, and the true direction is   tan1 425 N 846 E.

   55 3 60. (a) The velocity of the wind is w  55 cos 60  55 sin 60   55 2 2 .     (b) The velocity of the jet is w  765 cos 45  765 sin 45   7652 2  7652 2 .      765 2  55 3  765 2  56844 58857. (c) The true velocity of the jet is v  w  55 2  2 2 2  (d) The true speed of the jet is w  v  568442  588572  818 mih. The direction of the vector w  v is       tan1 58857 56844  46 . Thus the true direction of the jet is approximately N 44 E.

61. Let v be the velocity vector of the jet and let  be the direction of this vector. Thus v  765 cos  i  765 sin  j. If w      55 3 j. is the velocity vector of the wind then the true course of the jet is v  w  765 cos   55 i  765 sin   2 2 To achieve a course of true north, the east­west component (the i component) of the jet’s velocity vector must be 0. That   275    cos1  275  921 . Thus the pilot should head his plane in the is 765 cos   55  0  cos    2 765 765

direction N 21 W.


40

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

u

62. The speed of the jet is 300 mi/h, so its velocity relative to the air is v  300 cos  i  300 sin  j. The wind has velocity w  30j, so

w

¬

v

the true course of the jet is given by

u  v  w  300 cos  i  300 sin   30 j. We want the y­component of the jet’s velocity to be 0, so we solve

1    574 . Therefore, the jet should head in the direction 18574 (or S 8426 W). 300 sin   30  0  sin   10

63. (a) The velocity of the river is represented by the vector r  10 0  10i.

(b) Since the boater direction is 60 from the shore at 20 mih, the velocity of the boat is represented by the vector     b  20 cos 60  20 sin 60   10 10 3  10i  10 3j  10 1732.     (c) w  r  b  10  10 0  10 3  20i  10 3j  20 1732  (d) The true speed of the boat is w  202  17322  265 mih, and the true direction is    409  N 491 E.   tan1 1732 20

64. Let w be the velocity vector of the water, let v be the velocity vector of the boat, and let  be the direction of v. Then v  20 cos  i  20 sin  j and w  10i. The true course of the boat is v  w  20 cos   10 i  20 sin  j. To achieve a course of true north, the east­west component of the boat’s velocity vector must be 0. Thus, 20 cos   10  0    10   12    cos1  12  120 . Thus the boater should head the boat in the direction N 30 W. cos   20   65. (a) Let b  bx  b y represent the velocity of the boat relative to the water. Then b  24 cos 18  24 sin 18   24 cos 18 i  24 sin 18 j  228 74.   (b) Let w  x   y represent the velocity of the water. Then w  0  where  is the speed of the water. So the true velocity of the boat is b  w  24 cos 18  24 sin 18    24 cos 18 i  24 sin 18   j. For the direction to be due east, we must have 24 sin 18    0    742 mih. Therefore, the true speed of the water is 74 mi/h. Since b  w  24 cos 18  0  24 cos 18 i, the true speed of the boat is b  w  24 cos 18  228 mih. 66. Let  represent the velocity of the sailor and l the velocity of the ocean liner. Then w  2 0, and l  0 25, and so  r  2  0 0  25. Hence, relative to the water, the sailor’s speed is r  4  625  2508 mih, and their direction 25  9457 or approximately N 457 W. is   tan1 2

67. F1  2 5 and F2  3 8. (a) F1  F2  2  3 5  8  5 3

(b) The additional force required is F3  0 0  5 3  5 3.

68. F1  3 7, F2  4 2, and F3  7 9. (a) F1  F2  F3  3  4  7 7  2  9  0 0 (b) No additional force is required.

69. F1  4i  j, F2  3i  7j, F3  8i  3j, and F4  i  j. (a) F1  F2  F3  F4  4  3  8  1 i  1  7  3  1 j  0i  4j  4j (b) The additional force required is F5  0i  0j  0i  4j  4j.

70. F1  i  j, F2  i  j, and F3  2i  j. (a) F1  F2  F3  1  1  2 i  1  1  1 j  0i  j  j

(b) The additional force required is F4  0i  0j  0i  j  j.       71. F1  10 cos 60  10 sin 60   5 5 3 , F2  8 cos 30  8 sin 30   4 3 4 , and F3  6 cos 20  6 sin 20   5638 2052.    (a) F1  F2  F3  5  4 3  5638 5 3  4  2052  757 1061.

(b) The additional force required is F4  0 0  757 1061  757 1061.


SECTION 8.6 The Dot Product

41

72. F1  3 1, F2  1 2, F3  2 1, and F4  0 4. (a) F1  F2  F3  F4  3  1  2  0 1  2  1  4  2 4 (b) The additional force required is F5  0 0  2 4  2 4.

73. From the figure we see that T1   T1  cos 50 i  T1  sin 50 j and T2  T2  cos 30 i  T2  sin 30 j. Since T1  T2  100j we get  T1  cos 50  T2  cos 30  0 and T1  sin 50  T2  sin 30  100. From the first cos 50 cos 50 sin 30 equation, T2   T1  , and substituting into the second equation gives T1  sin 50  T1   100  cos 30 cos 30  T1  sin 50 cos 30  cos 50 sin 30   100 cos 30  T1  sin 50  30   100 cos 30  cos 30 T1   100  879385. sin 80 cos 30 Similarly, solving for T1  in the first equation gives T1   T2  and substituting gives cos 50   cos 30 sin 50 T2   T2  sin 30  100  T2  cos 30 sin 50  cos 50 sin 30   100 cos 50  cos 50 100 cos 50 T2    652704. Thus, T1  879416 cos 50  i  879416 sin 50  j  565i  674j and sin 80 T2  652704 cos 30  i  652704 sin 30  j  565i  326j. 74. From the figure we see that T1   T1  cos 223 i  T1  sin 223 j and T2  T2  cos 415 i  T2  sin 415 j. Since T1  T2  18 278j, we get  T1  cos 223  T2  cos 415  0 and T1  sin 223  T2  sin 415  18 278. From the cos 223 first equation, T2   T1  , and substituting into the second equation gives cos 415  cos 223 sin 415 T1  sin 223  T1   18 278  cos 415 T1  sin 223 cos 415  cos 223 sin 415   18 278 cos 415  cos 415 T1  sin 223  415   18,278 cos 415  T1   18,278  15 257. sin 638 cos 415 and substituting gives Similarly, solving for T1  in the first equation gives T1   T2  cos 223   cos 415 sin 223 T2   T2  sin 415  18,278  T2  sin 223 cos 415  sin 415 cos 223   18,278 cos 223 cos 223 18,278 cos 223  T2    18,847. sin 638 Thus, T1  15,257 cos 223  i  15,257 sin 223  j  14,116i  5,789j and T2  18,847 cos 415  i  18,847 sin 415  j  14,116i  12,488j. 75. When we add two vectors, the resultant vector can be found by first placing the initial point of the second vector at the terminal point of the first vector. The resultant vector can then found by using the new terminal point of the second vector and the initial point of the first vector. When these n vectors are placed head to tail in the plane so that they form a polygon, the initial point and the terminal point are the same. Thus the sum of the n vectors is the zero vector.

8.6

THE DOT PRODUCT

1. The dot product of u  a1  a2  and v  b1  b2  is defined by u  v  a1 a2  b1 b2 . The dot product of two vectors is a real number, or scalar, not a vector.


42

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

uv . So if u  v  0, the vectors are perpendicular. u v To find the angle  between the vectors u and v in the figure, we first find     4 3  3 2 6 13 12  6 , and so   cos1  6 6513  109 , rounded to the cos       4 3 3 2 65 42  32 32  22

2. The angle  satisfies cos  

nearest degree.

u 3. (a) The component of u along v is the scalar u cos  and can be uv expressed in terms of the dot product as . v projv u ¬   v uv compv u v. (b) The projection of u onto v is the vector projv u  v2 4. The work done by a force F in moving an object along a vector D is W  F  D.         5. (a) u  v  2 0  1 1  2  0  2 6. (a) u  v  1 3   3 1   3  3  0 uv uv 2  1    45   (b) cos    0    90 (b) cos   2 2 2 u v u v    7. (a) u  v  1 0  1 13  1  0  1 replace "13" 8. (a) u  v  6 6  1 1  6  6  12 with "3" uv uv 1    60 (b) cos    12  1    180 (b) cos    12 6 2 2 u v u v

9. (a) u  v  3 2  1 2  3  4  1 uv   1    97 (b) cos   13 5 u v      11. (a) u  v  0 5  1  3  0  5 3  5 3

10. (a) u  v  3 4  4 3  12  12  24 uv 24    163  55 (b) cos   u v

13. (a) u  v  i  3j  4i  j  4  3  1 uv (b) cos     1    86 10 17 u v

14. (a) u  v  3i  4j  2i  j  6  4  10 uv   (b) cos    10 5 5 u v     153   cos1 10

  uv (b) cos    5523  23    30 u v

      15. (a) u  v  1 3  1  3  1  3  2

uv 1   2 22   2    120 u v 17. u  v  12  12  0  vectors are orthogonal (b) cos  

12. (a) u  v  1 1  1 1  1  1  0 uv (b) cos     0  0    90 2 2 u v

5 5

16. (a) u  v  6 8  3 4  18  32  50 uv 50  1    0  105 (b) cos   u v

18. u  v  0 4  4 0  0  0  0  vectors are orthogonal

replace "orthogonal" with "perpendicular"

19. u  v  8  12  4  0  vectors are not orthogonal

20. u  v  2 0  0 7  0  0  0  vectors are orthogonal 21. u  v  24  24  0  vectors are orthogonal

22. u  v  4 1  0 3  4  0  4  vectors are not orthogonal

#17-22 use "perpendicular"

23. u  v  u  w  2 1  1 3  2 1  3 4  2  3  6  4  9

24. u  v  w  2 1  [1 3  3 4]  2 1  4 1  8  1  9

25. u  v  u  v  [2 1  1 3]  [2 1  1 3]  3 2  1 4  3  8  5

26. u  v u  w  2 1  1 3 2 1  3 4  1 10  10 12 u  v 4 6  3 4 12  24  27. x    v 3 4 5 5


SECTION 8.6 The Dot Product

28. x 

1 1  3  5  u  v 3 5  2  2 2 2 2      2  1 1   v 2   1  1  2

2

2

43

2

u  v 7i  24j  j 0  24  24   v j 1 uv 28 56 7i  8i  6j 56  0 30. x      v 8i  6j 10 5 64  36     2 4  1 1 uv 1 1  1 1. v  31. (a) u1  projv u  12  12 v2 (b) u2  u  u1  2 4  1 1  3 3. We resolve the vector u as u1  u2 , where u1  1 1 and u2  3 3.     7 4  2 1 uv 2 1  2 2 1  4 2 32. (a) u1  projv u  v  22  12 v2 (b) u2  u  u1  7 4  4 2  3 6. We resolve the vector u as u1  u2 , where u1  4 2 and u2  3 6.       1 2  1 3 uv 1 3   12 1 3   12  32 v  33. (a) u1  projv u  2 2 v 12  3         1 3 (b) u2  u  u1  1 2   2  2  32  12 . We resolve the vector u as u1  u2 , where u1   12  32 and u2  32  12 .     11 3  3 2 uv 3 2  3 3 2  9 6 34. (a) u1  projv u  v  v2 32  22 (b) u2  u  u1  11 3  9 6  2 3. We resolve the vector u as u1  u2 , where u1  9 6 and u2  2 3.       2 9  3 4 uv 6 3 4   18  24 3 35. (a) u1  projv u  v  4  5 5 5 v2 33  42       24 28 21 18 24 (b) u2  u  u1  2 9   18 5  5  5  5 . We resolve the vector u as u1  u2 , where u1   5  5 and   21 . u2  28  5 5       1 1  2 1 uv 1 2 1  2   1 2 36. (a) u1  projv u  v  1  5 5 5 v2 22  12         (b) u2  u  u1  1 1  25   15  35  65 . We resolve the vector u as u1  u2 , where u1  25   15 and u2  35  65 .

29. x 

37. W  F  d  4 5  3 8  28

38. W  F  d  400 50  201 0  80 400

39. W  F  d  10 3  4 5  25

40. W  F  d  4 20  5 15  280

41. Let u  u1  u2  and v  v1  v2 . Then u  v  u 1  u 2    1   2   u 1  1  u 2  2   1 u 1   2 u 2   1   2   u 1  u 2   v  u 42. Let u  u1  u2  and v  v1  v2 . Then

cu  v  c u 1  u 2    1   2   cu 1  cu 2    1   2   cu 1  1  cu 2  2  c u 1  1  u 2  2   c u  v  u 1 c 1  u 2 c 2  u 1  u 2   c 1  c 2   u  cv

43. Let u  u 1  u 2 , v   1   2 , and w  1  2 . Then u  v  w  u 1  u 2    1   2   1  2   u 1   1  u 2   2   1  2 

 u 1 1   1 1  u 2 2   2 2  u 1 1  u 2 2   1 1   2 2  u 1  u 2   1  2    1   2   1  2   u  w  v  w

44. Let u  u 1  u 2  and v   1   2 . Then u  v  u  v  u 1  u 2    1   2   u 1  u 2    1   2   u 1   1  u 2   2   u 1   1  u 2   2  2  2      u 21  u 22   12   22  u2  v2  u 21   12  u 22   22  u 21  u 22   12   22 


44

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

uv

v. Then v2               uv uv uv uv uv projv u  u  projv u  v  u  v   u  v  v v v2 v2 v2 v2 v2

45. We use the definition that projv u 

u  v2 v2

u  v2 v4

v2 

u  v2 v2

u  v2

Thus u and u  projv u are orthogonal.   uv u  v v  v u  v v2 46. v  projv u  v  v    u  v. v2 v2 v2

v2

0

47. W  F  d  4 7  4 0  16 ft­lb

48. The displacement of the object is D  11 13  2 5  9 8. Hence, the work done is W  F  D  2 8  9 8  18  64  82 ft­lb.

49. The distance vector is D  200 0 and the force vector is F  50 cos 30  50 sin 30 . Hence, the work done is W  F  D  200 0  50 cos 30  50 sin 30   200  50 cos 30  8660 ft­lb.

50. W  F  d, and in this problem F  0 2500, d  500 cos 12  500 sin 12  Thus,

W  0 2500  500 cos 12  500 sin 12   0  2500 1040  260,000 ft­lb. This is the work done by gravity; the

car does (positive) work in overcoming the force of gravity. So the work done by the car is 260,000 ft­lb. 51. (a) Since the force parallel to the driveway is 490  w sin 10  w  is about 2822 lb.

490  28218, and thus the weight of the car sin 10

(b) The force exerted against the driveway is 28218 cos 10  2779 lb.

52. Since the weight of the car is 2755 lb, the force exerted perpendicular to the earth is 2755 lb. Resolving this into a force u perpendicular to the driveway gives u  2766 cos 65  1164 lb. Thus, a force of about 1164 lb is required.

53. Since the force required parallel to the plane is 80 lb and the weight of the package is 200 lb, it follows that 80  200 sin ,   80  2358 , and so the angle of inclination is where  is the angle of inclination of the plane. Then   sin1 200 approximately 236 .

54. Let R represent the force exerted by the rope and d the force causing the cart to roll down the ramp. Gravity acting on the cart exerts a force w of 40 lb directly downward. So the magnitude of the part of that force causing the cart to roll down the ramp is d  40 sin 15 . The angle between R and d is 45 , so the magnitude of the force holding the cart up is R cos 45 . Equating these two, we have R cos 45  40 sin 15 , so R 

40 sin 15  1464 lb. cos 45

55. (a) 2 0  4 2  8, so Q 0 2 lies on L. 2 2  4 1  4  4  8, so R 2 1 lies on L.  (b) u  Q P  0 2  3 4  3 2.  v  Q R  0 2  2 1  2 1.   3 2  2 1 uv 2 1 v w  projv u  2 2 v 2  2  1 8   85 2 1  16 5 5

(c) From the graph, we can see that u  w is

orthogonal to v (and thus to L). Thus, the

distance from P to L is u  w. y

L

P u

Q 0

w v

u-w R x


CHAPTER 8

CHAPTER 8 REVIEW

1. (a)

O

(8, _ 3¹ 4)

¹/6

  3 (b) x  12 cos   12  6 2  6 3,

1 y  12 sin  6  12  2  6. Thus, the rectangular    coordinates of P are 6 3 6 .

3. (a)

2¹ 3

O

O

 

 (b) x  3 cos 74  3 22   3 2 2 ,     y  3 sin 74  3  22  3 2 2 . Thus, the     rectangular coordinates of P are  3 2 2  3 2 2 .

5. (a)

(_Ï3, 2¹ 3 )      (b) x   3 cos 23   3  12  23 ,     y   3 sin 23   3 23   32 . Thus, the   rectangular coordinates of P are 23   32 .

6. (a)

y

y P

P

1

2 2

(b) r 

_ 3¹ 4

      (b) x  8 cos  34  8  22  4 2,       y  8 sin  34  8  22  4 2. Thus, the     rectangular coordinates of P are 4 2 4 2 .

4. (a)

7¹ 4

(_3, 7¹ 4 )

O

2. (a)

(12, ¹6)

Review

x

   82  82  128  8 2 and   tan1 88 .

Since P is in quadrant I,    4 . Polar coordinates    for P are 8 2  4 .    (c) 8 2 54

1 x

(b) r 

  2  2    2  6  8  2 2 and

  tan1

    6  tan1  3    . Since x is 3  2

negative and y is positive,   23 . Polar    coordinates for P are 2 2 23 .    (c) 2 2 53

45


46

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

7. (a)

8. (a)

y

y

2 2

1

x

x

1

P

P

  2   2  6 2  6 2  144  12 and (b) r  

  tan1 62   4 . Since P is in quadrant III, 6 2     54 . Polar coordinates for P are 12 54 .   (c) 12  4

(b) r 

   42  42  32  4 2 and

7   tan1 4 4 . Since P is in quadrant IV,   4 .    Polar coordinates for P are 4 2 74 .    (c) 4 2 34

9. (a) x  y  4  r cos   r sin   4 

10. (a) x y  1  r cos  r sin   1 

r cos   sin   4  r 

r 2 cos  sin   1  r 2 

4 cos   sin 

(b) The rectangular equation is easier to graph. y

2 1  or cos  sin  sin 2

r 2  2 csc 2. (b) The rectangular equation is easier to graph. y

1

1

11. (a) x 2  y 2  4x  4y  r 2  4r cos   4r sin   r 2  r 4 cos   4 sin   r  4 cos   4 sin 

(b) The polar equation is easier to graph.

x

1

x

1

2   2 12. (a) x 2  y 2  2x y  r 2  2 r cos  r sin   r 4  2r 2 cos  sin   r 2  2 cos  sin  

r 2  sin 2

(b) The polar equation is easier to graph.

O

1

O

1


CHAPTER 8

13. (a)

(3, ¹2 )

14. (a)

(3, ¹2 )

Review

(6, 0)

2

(3, 3¹ 2 )

O

(b) r  3  3 cos   r 2  3r  3r cos , which gives  x 2  y 2  3 x 2  y 2  3x   x 2  3x  y 2  3 x 2  y 2 . Squaring both sides 2    gives x 2  3x  y 2  9 x 2  y 2 .

15. (a)

1

(b) r  3 sin   r 2  3r sin , so x 2  y 2  3y   2   x 2  y 2  3y  94  94  x 2  y  32  94 .

16. (a)

O

O

2

(b) r  2 sin 2  r  2  2 sin  cos  

r 3  4r 2 sin  cos    32  4 r sin  r cos  and so, since r2 x  r cos  and y  r sin , we get 3  x 2  y 2  16x 2 y 2 .

2

(b) r  4 cos 3  4 cos 2 cos   sin 2 sin       4 cos  cos2   sin2   2 sin2  cos     r  4 cos3   3 sin2  cos     r 4  4r 3 cos3   3 sin2  cos  , which gives 

x 2  y2

2

 4x 3  12x y 2 .

18. (a)

17. (a) (1, ¹)

O

1

(1, 0)

1 1   2 cos 2 cos   sin2    r 2 cos2   sin2   1 

(b) r 2  sec 2 

r 2 cos2   r 2 sin2   1 

r cos 2  r sin 2  1  x 2  y 2  1.

O

1

(b) r 2  4 sin 2  8 sin  cos   2  r 4  8r 2 sin  cos , so x 2  y 2  8x y.

47


48

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

19. (a)

(b) r  sin   cos   r 2  r sin   r cos , so x 2  y 2  y  x     2  2   x 2  x  14  y 2  y  14  12  x  12  y  12  12 .

(1, ¹2) (1, 0)

O

1

(b) r 

20. (a)

(

¹ 2, 2

(4, ¹)

)

 2 x 2  y 2  4  x. Squaring both sides gives   4 x 2  y 2  4  x2  16  8x  x 2 

1 ( 4, 0)

O

 4  2r  r cos   4, so 2 x 2  y 2  x  4 2  cos 

3

3x 2  8x  4y 2  16.

(2, 3¹ 2 )

21. r  cos 3,   [0 3].

22. r  sin 94,   [0 8]

1

1

­1

1

­1

1

­1

­1

24. r  sin 2  2,   [0 2]. This graph is bounded.

23. r  1  4 cos 3,   [0 6]. 5

2

5

­2

2 ­2

­5


CHAPTER 8

25. (a)

26. (a)

Im

Review

49

Im 2 2

4+4i 1 1

(b) 4  4i has r 

_10i

Re

  16  16  4 2, and

(b) 10i has r  10 and   32 .   (c) 10i  10 cos 32  i sin 32

  tan1 44   4 (in quadrant I).    (c) 4  4i  4 2 cos  4  i sin 4

27. (a)

Re

28. (a)

Im

Im 1+Ï3 i 1

5+3i 1 1

1

Re

  (b) 5  3i. Then r  25  9  34, and

  tan1 53 .       (c) 5  3i  34 cos tan1 53  i sin tan1 53

29. (a)

Im

_1+i

(b) 1  i has r 

30. (a)

Im

5

1

_1

Re

  (b) 1  3i has r  1  3  2, and    tan1 3   3 (since  is in quadrant I).    (c) 1  3i  2 cos  3  i sin 3

_20

5

Re

Re

  1 with 1  1  2 and tan   1

 in quadrant II    34 .    (c) 1  i  2 cos 34  i sin 34

(b) 20 has r  20 and   . (c) 20  20 cos   i sin 

   1  3  2 and tan   1 3   3 with  in    quadrant III    53 . Therefore, 1  3i  2 cos 53  i sin 53 , and so             4 1  3i  24 cos 203  i sin 203  16 cos 23  i sin 23  16  12  i 23  8 1  i 3 .      32. 1  i has r  11  2 and    2 cos  4 . Thus, 1  i  4  i sin 4 , and so  8 2 cos 2  i sin 2  16 1  i  0  16. 1  i8        33. 3  i has r  3  1  2 and tan   1 with  in quadrant I     6 . Therefore, 3  i  2 cos 6  i sin 6 ,

31. 1 

 3i has r

3

and so         4 1 cos 2  i sin 2  1  1  i 3 3i  24 cos 46  i sin 46  16 3 3 16 2 2       1 1  i 3   1 1  i 3  32 32


50

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

    32 1  3 i  cos   i sin  . Thus 34. 12  23 i has r  14  34  1 and   tan1 12  (since  is in quadrant I). So 3 2 2 3 3  20        1  3i  120 cos 203  i sin 203  cos 23  i sin 23   12  23 i  12 1  i 3 2 2   35. 16i has r  16 and   32 . Thus, 16i  16 cos 32  i sin 32 and so      3  4k 3  4k  i sin for k  0, 1. Thus 16i12  1612 cos 4 4       4  1  i 1  2 2 1  i and the roots are 0  4 cos 34  i sin 34 2 2      1  4 cos 74  i sin 74  4 1  i 1  2 2 1  i. 2

2

       36. 4  4 3i has r  16  48  8 and   tan1 4 4 3   3 . Therefore, 4  4 3i  8 cos 3  i sin 3 .         13   6k   6k Thus 4  4 3i  i sin for k  0, 1, 2. Thus the three roots are  3 8 cos 9 9        7 7 13 13 0  2 cos  9  i sin 9 , 1  2 cos 9  i sin 9 , and 2  2 cos 9  i sin 9 .

2k 2k 37. 1  cos 0  i sin 0. Then 116  1 cos  i sin

for k  0, 1, 2, 3, 4, 5. Thus the six roots are      3  1 2 2   1  i 3 , 0  1 cos 0  i sin 0  1, 1  1 cos  3  i sin 3  2  i 2 , 2  1 cos 3  i sin 3 2 2       3  1 cos   i sin   1, 4  1 cos 43  i sin 43   12  i 23 , and 5  1 cos 53  i sin 53  12  i 23 .

 18  cos 38. i  cos  2  i sin 2 . Then i

6

2  2k 8

6

 i sin

2  2k 8

 cos

  4k 16

 i sin

  4k 16

for

  i sin  ,   cos 5  i sin 5 ,   cos 9  i sin 9 , k  0, 1, 2, 3, 4, 5, 6, 7. Thus the eight roots are 0  cos 16 1 2 16 16 16 16 16

 13 17 17 21 21 25 25 3  cos 13 16  i sin 16 , 4  cos 16  i sin 16 , 5  cos 16  i sin 6 , 6  cos 16  i sin 16 , and

 29 7  cos 29 16  i sin 16 .

39. (a)

40. (a)

y

y

1 1 x

1 x

1

(b) x  1  t 2 , y  1  t  t  y  1. Substituting for t gives x  1  y  12  x  2y  y 2 in

rectangular coordinates.

(b) x  t 2  1, y  t 2  1

t 2  y  1.

Substituting for t 2 gives x  y  1  1 y  x  2 where x  1 and y  1.


CHAPTER 8

41. (a)

42. (a)

y

Review

51

y

1 1

x

1

1

(b) x  1  cos t  cos t  x  1, and y  1  sin t  sin t  1  y. Since cos2 t  sin2 t  1, it follows that x  12  1  y2  1 

x  12  y  12  1. Since t is restricted by  0t   2 , 1  cos 0  x  1  cos 2

1  x  2, and similarly, 0  y  1. (This is the lower right quarter of the circle.) 43. x  cos 2t, y  sin 3t

x

1 2 1 (b) x   2   x  2, y  2 . Substituting for t t t 1 gives y  2 x  22 . Since t is restricted by t 1 1 0  t  2, we have  , so x  52 and y  12 . t 2 The rectangular coordinate equation is y  2 x  22 , x  52 . 44. x  sin t  cos 2t, y  cos t  sin 3t

1

­1

1

1 ­1

­1

1 ­1

45. The coordinates of Q are x  cos  and y  sin . The coordinates of R are x  1 and y  tan . Hence, the midpoint P   1  cos  sin   tan  1  cos  sin   tan   , so parametric equations for the curve are x  and y  . is 2 2 2 2 46. Q has coordinates a cos  a sin  and O R  a cos   Q P  P has coordinates 2a cos  a sin . Thus, parametric equations are x  2a cos  y  a sin , 0    2.   47. u  2 3, v  8 1. u  22  32  13, u  v  2  8 3  1  6 4,

u  v  2  8 3  1  10 2, 2u  2 2  2 3  4 6, and 3u  2v  3 2  2 8  3 3  2 1  22 7.   48. u  2i  j, v  i  2j. u  22  12  5, u  v  2  1 i  1  2 j  3i  j, u  v  2  1 i  1  2 j  i  3j, 2u  4i  2j, and 3u  2v  3 2i  j  2 i  2j  4i  7j. 49. The vector with initial point P 0 3 and terminal point Q 3 1 is 3  0 1  3  3 4.

50. If the vector 5i  8j is placed in the plane with its initial point at P 5 6, its terminal point is 5  5 6  8  10 2.      2    2 3 2 51. u  2 2 3 has length 2  2 3  4. Its direction is given by tan     3 with  in quadrant II, 2    so     tan1  3  120 .   52. v  2i  5j has length 22  52  29. Its direction is given by tan    52 with  in quadrant IV, so     2  tan1  52  2918 .    53. u  u cos  u sin   20 cos 60  20 sin 60   10 10 3 .

54. u  u cos  u sin   135 cos 125  135 sin 125   135 cos 125  135 sin 125   77 111


52

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

55. (a) The force exerted by the first tugboat can be expressed in component form as   u  20  104 cos 40  20  104 sin 40  15321 12856, and that of the second tugboat is   34  104 cos 15   34  104 sin 15   32841 8800. Therefore, the resultant force is w  u  v  15321  32841 12856  8800  48162 4056.  (b) The magnitude of the resultant force is 481622  40562  48,332 lb. Its direction is given by 4056  0084, so   tan1 0084  48 or N 852 E. tan   48,162

56. (a) The velocity of the airplane is the sum of its velocity relative to the air, which is       u  600 cos 30  600 sin 30   300 3 300 , and the wind velocity v  50 cos 120  50 sin 120   25 25 3 .     Thus, its velocity is w  300 3  25 300  25 3 .  2    2 300 3  25  300  25 3  602 mi/h, and its direction is given by (b) The true speed of the airplane is w   300  25 3 tan    0694, so   tan1 0694  348 , or N 552 E.  300 3  25  57. u  4 3, v  9 8. u  42  32  5, u  u  42  32  25, and u  v  4 9  3 8  60.  58. u  5 12, v  10 4. u  52  122  13, u  u  52  122  169, and u  v  5 10  12 4  2.   59. u  2i  2j, v  i  j. u  22  22  2 2, u  u  22  22  8, and u  v  2 1  2 1  0.  60. u  10j, v  5i  3j. u  102  10, u  u  102  100, and u  v  10 3  30. 61. u  v  4 2  3 6  4 3  2 6  0, so the vectors are perpendicular.

62. u  v  5 3  2 6  5 2  3 6  8  0, so the vectors are not perpendicular. The angle between them is given by   uv 340 8 340  775 .  cos   , so   cos1 85   u v 85 2 52  32 2  62

63. u  v  2i  j  i  3j  2 1  1 3  5, so the vectors are not perpendicular. The angle between them is given by   uv 2 5  cos   , so   cos1 22  45 .   u v 2 22  12 12  32

64. u  v  i  j  i  j  1 1  1 1  0, so the vectors are perpendicular.

 3 6  1 1 uv 17 37   65. (a) u  3 1, v  6 1. The component of u along v is  . v 37 62  12     uv 17 6 1  102   17 v  (b) projv u  37 37 37 v2       17 and u  u  proj u  3 1  102   17  9  54 .   (c) u1  projv u  102 2 v 37 37 37 37 37 37  1 4  2 9 14 97 uv   .  66. (a) u  i  2j, v  4i  9j. The component of u along v is v 97 42  92   uv 56 126 (b) projv u  v  14 97 4i  9j   97 i  97 j v2   126 56 126 153 68 (c) u1  projv u   56 97 i  97 j and u2  u  projv u  i  2j   97 i  97 j  97 i  97 j. 67. The distance vector is D  1  7 1  1  6 2, so the work done is W  F  D  2i  9j  6 2  12  18  6 ft­lb.


CHAPTER 8

Test

53

68. Here W  F  D  F D cos , where  is the angle between F and D. Thus, we have 3800  250  20 cos   3800  405 .   cos1 5000

CHAPTER 8 TEST         1. (a) x  8 cos 54  8  22  4 2, y  8 sin 54  8  22  4 2. So the point has rectangular coordinates     4 2 4 2 .     (b) P  6 2 3 in rectangular coordinates. So tan   263 and the reference angle is    6 . Since P is in   2  quadrant II, we have   56 . Next, r 2  62  2 3  36  12  48, so r  4 3. Thus, polar coordinates for       the point are 4 3 56 or 4 3 116 . (b) r  8 cos 

2. (a)

x 2  8x  y 2  0

O

x  42  y 2  16

1

r 2  8r cos  

x 2  y 2  8x

x 2  8x  16  y 2  16

The curve is a circle.

(9, ¹2)

3.

The curve is a limaçon.

(_3,3¹ 2) (3, ¹)

4. (a)

O

2 (3, 0)

1+Ï3 i 1

1

    3i has r  1  3  2 and   tan1 3  3 . So, in    trigonometric form, 1  3i  2 cos  3  i sin 3 .     (c) z  1  3i  2 cos  3  i sin 3   z 9  29 cos 93  i sin 93  512 cos 3  i sin 3

(b) 1 

Im

Re

 512 1  i 0  512

      i sin 7 and z  2 cos 5  i sin 5 . 5. z 1  4 cos 712 2 12 12 12      7  5 7  5  i sin  8 cos   i sin   8 and Then z 1 z 2  4  2 cos 12 12         7  5 7  5 3  1 i  3  i.   2  i sin z 1 z 2  42 cos  i sin  2 cos  6 6 2 2 12 12


54

CHAPTER 8 Polar Coordinates, Parametric Equations, and Vectors

   6. 27i has r  27 and    2 , so 27i  27 cos 2  i sin 2 . Thus,        2k  2k 3 2 2 13  i sin  27 cos 27i 3 3        4k   4k  i sin  3 cos 6 6

Im wÁ

wü 1 0

for k  0, 1, 2. Thus, the three roots are      3  1 i  3 3  i ,   3  i sin 0  3 cos  6 6 2 2 2         1  3 cos 56  i sin 56  3  23  12 i  32  3  i , and   2  3 cos 96  i sin 96  3i.

7. (a) x  3 sin t  3, y  2 cos t, 0  t  . From the

work of part (b), we see that this is the half­ellipse y

(b) x  3 sin t  3  x  3  3 sin t 

y  2 cos t 

x 3  sin t. 3

x  32  sin2 t. Similarly, 9

y  cos t, and squaring both sides gives 2

y2  cos2 t. Since sin2 t  cos2 t  1, it follows that 4

1 x

1

Re

Squaring both sides gives

shown.

1

y2 x  32   1. Since 0  t  , sin t  0, so 9 4 3 sin t  0  3 sin t  3  3, and so x  3. Thus the curve consists of only the right half of the ellipse.

8. We start at the point 3 5. Because the line has slope 2, for every 1 unit we move to the right, we must move up 2 units. Therefore, parametric equations are x  3  t, y  5  2t.

9. (a) x  3 sin 2t, y  3 cos 2t. The radius is 3 and the position at time t  0 is 3 sin 2 0  3 cos 2 0  0 3. Initially x is increasing and y is decreasing, so the motion is clockwise. At time t   the object is back at 0 3, so it takes  units of time to complete one revolution. (b) If the speed is doubled, the time to complete one revolution is halved, to  2 . Parametric equations modeling this motion are x  3 sin 4t, y  3 cos 4t.   (c) x 2  y 2  3 sin 2t2  3 cos 2t2  9 sin2 2t  cos2 2t  9, so an equation in rectangular coordinates is x 2  y 2  9.

(d) In polar coordinates, an equation is r  3. 10. (a)

y

(b) u  3  3 i  [9  1] j  6i  10j   (c) u  62  102  2 34

(_3, 9)

u

1 1

(3, _1)

x

11. (a) u  3v  1 3  3 6 2  1  3 6  3  3 2  19 3


CHAPTER 8

Test

55

  (b) u  v  1 3  6 2  5 5  52  52  5 2

(c) u  v  1 3  6 2  1 6  3 2  0

(d) Because u  v  0, u and v are perpendicular. y

12. (a)

(b) The length of u is u 

(_4Ï3, 4)

 4   3 with  in quadrant II, so 3 4 3     180  tan1 33  150 .

tan  

u

1 1

x

  2 4 3  42  8. Its direction is given by

13. (a) The river’s current can be represented by the vector u  8 0 and the motorboat’s velocity relative to the water by the       vector v  12 cos 60  12 sin 60   6 6 3 . Thus, the true velocity is w  u  v  14 6 3 .     2 6 13  0742, so (b) The true speed is w  142  6 3  174 mi/h. The direction is given by tan   14   tan1 0742  366 or N 534 E.    uv 338 3 5  2 1 338  cos1 2  45 .  , so   cos1 26  14. (a) cos    2 u v 26 32  22 52  12  uv 26 13 (b) The component of u along v is .   v 2 26   uv 5 1 v  13 (c) projv u  26 5i  j  2 i  2 j v2   15. The work is W  F  d  3i  5j  7  2 i  13  2 j  3i  5j  5i  15j  3 5  5 15  90 ft­lb.


56

FOCUS ON MODELING

FOCUS ON MODELING The Path of a Projectile x . Substituting this value for t into the equation for y, we get  0 cos    2  x g x x 2 . This shows y   0 sin  t  12 gt 2  y   0 sin   12 g  y  tan  x  2  0 cos   0 cos  2 0 cos2 

1. From x   0 cos  t, we get t 

that y is a quadratic function of x, so its graph is a parabola as long as   90 . When   90 , the path of the projectile is a straight line up (then down). 2. (a) Applying the given values, we get x   0 cos  t  15t and  y  4   0 sin  t  12 gt 2  4  15 3t  16t 2 as the parametric equations for the path of the baseball.

y 10

0

10

x

20

(b) The baseball will hit the ground when y  0  4  2598t  16t 2 . Using the Quadratic Formula,  2598 25982 4164 t   177 seconds (since t must be positive). So the baseball travels 32 x  15  177  265 ft (horizontally) before hitting the ground after 177 s.

3. (a) We find the time at which the projectile hits the ground using the equation t  2 0 gsin  , where  0  1000 and g  16. Since the rocket is fired 5 from the vertical axis and  is measured from the horizontal, we have

sin 85  6226, so the rocket is in the air for 6226 seconds.   90  5  85 . Then t  200032 

(b) Substituting the given values into y   0 sin  t  12 gt 2 , we get

(d)

y  1000 sin 85  t  16t 2  996t  16t 2 . Then

y 15,000

y  f t  at 2  bt  c, where a  16, b  996, and c  0. So y is a

10,000

b   996  31125, and the maximum value is t   2a 216

5000

quadratic function whose maximum value is attained at

f 31125  996 31125  16 311252  15,500 (see Section 3.1 for a

guide to finding the maximum value of a quadratic function). Thus, the

0

2000

4000

6000 x

rocket reaches a maximum height of 15,500 feet. (c) We use the equation x   0 cos  t to find the horizontal distance traveled after t seconds. Since the rocket hits the ground after

6226 seconds, substituting into the equation for horizontal distance gives x  1000 cos 85  6226  5426. Thus, the rocket travels a horizontal

distance of 5426 feet.

2 0 sin  , so the projectile travels g

y (m)

2 2 sin  x   0 cos  0  0 sin 2 meters. Substituting  0  330 ms,

3000

4. (a) The projectile hits the ground when t  g

g

3302 sin 2  sin 2  08999. x  10 km, and g  98 m/s2 , we get 10000 

98 So 2  6415    3208 or 2  11585    5793 .

4000

2000 1000 0

4000

8000

x (m)


The Path of a Projectile

57

(b) The projectile fired at 3208 will hit the target in 357 seconds, while the projectile fired at 5793 reaches the target in 571 seconds. 5. We use the equation of the parabola from Exercise 1 and find its vertex:   2 02 sin  cos x g g 2 2 x y 2 x   y  tan  x  2 g 2 0 cos2  2 0 cos2       2 2 sin  cos x 2 sin  cos  2  02 sin  cos   2 g g 0 0 2     y   2   2 x  g g g 2 0 cos2  2 0 cos2   2    02 sin  cos   02 sin  cos   02 sin2   02 sin2  g y 2 x . Thus the vertex is at  , so the maximum  g 2g g 2g 2 cos2  0

 02 sin2 

. 2g 6. Since the horizontal component of the projectile’s velocity has been reduced by , the parametric equations become height is

x   0 cos    t, y   0 sin  t  12 gt 2 . 7. In Exercise 6 we derived the equations x   0 cos    t,

y

y   0 sin  t  12 gt 2 . We plot the graphs for the given values of

10

 0 , , and  in the figure to the right. The projectile will be blown

8

¬=60¡

backwards if the horizontal component of its velocity is less than the

6

speed of the wind, that is, 32 cos   24  cos   34    414 .

4 ¬=45¡

¬=40¡ ¬=30¡

2

_14 _12 _10 _8 _6 _4 _2 0

The optimal firing angle appears to be between 15 and 30 . We

y

graph the trajectory for   20 ,   23 , and   25 . The solution

3

2

4 x

¬=15¡ ¬=5¡

¬=25¡

appears to be close to 23 .

¬=23¡

2

¬=20¡ 1

0

8. (a) Here  0  200 and   60 , so parametric equations are x   0 cos  t  200 cos 60  t  100t,

 y   0 sin  t  12 gt 2  200 sin 60  t  12 98 t 2  100 3t  49t 2 .    (b) Here vt   0 cos  i   0 sin   gt j  100i  100 3  98t j. At its highest point, the vertical component is 0, so the velocity is 100i. It  lands when y  100 3t  49t 2  0  t  353 s, at which point its    2  v speed is 353  1002  100 3  98  353  200 ms. Notice

that the speed of the rocket when it hits the ground is the same as its initial speed.

(c)

1

2

3

y 1200 800

x

vÁö

1400 1000

4

vÁü v§ v£ü

600 400 200 0

1000

2000

3000

x


CORRECTIONS: p. 8,9,25,44,48,49,56

CHAPTER 9

SYSTEMS OF EQUATIONS AND INEQUALITIES

9.1

Systems of Linear Equations in Two Variables 1

9.2

Systems of Linear Equations in Several Variables 9

9.3 9.4

Partial Fractions 16 Systems of Nonlinear Equations 27

9.5

Systems of Inequalities 35 Chapter 9 Review 47 Chapter 9 Test 55

¥

FOCUS ON MODELING: Linear Programming 58

1


9

SYSTEMS OF EQUATIONS AND INEQUALITIES

9.1

SYSTEMS OF LINEAR EQUATIONS IN TWO VARIABLES

1. The given system is a system of two equations in the two variables x and y. To check if 5 1 is a solution of this system, we check if x  5 and y  1 satisfy each equation in the system. The only solution of the given system is 2 1. 2. A system of equations in two variables can be solved by the substitution method, the elimination method, or the graphical method.  2x  3y  7 Using the substitution method with we solve the second equation for y in terms of x: 5x  y  9  5x  y  9 5x  y  9  y  5x  9. Substituting this into the first equation gives 2x  3 5x  9  7  2x  15x  27  7  17x  34  x  2. Substituting this into the first equation then gives 2 2  3y  7  3y  3  y  1. Using the elimination method, we multiply the second equation by 3 to get 15x  3y  27, then add this to the first equation:

4

2x  3y  15x  3y  7  27  17x  34  x  2. Substituting, we find that y  1, as above.

Using the graphical method, we graph 2x  3y  7  y  73  23 x and

5x  y  9  y  5x  9 on the same screen and see that the lines intersect at

2 -4

-2

-2

2

4

-4

x y  2 1.

3. A system of two linear equations in two variables can have one solution, no solution, or infinitely many solutions. 4. For the given system, the graph of the first equation is the same as the graph of the second equation, so the system has  x t where t is any real number. Some of the infinitely many solutions. We express these solutions by writing y 1t solutions of this system are 1 0, 3 4, and 5 4.

5.

x  y 2

2x  4y  16

Solving the first equation for x, we get x  2  y, and substituting this into the second equation gives

2 2  y  4y  16  4  2y  4y  16  6y  12  y  2. Substituting for y we get x  2  2  4. Thus, the solution is 4 2. 6.

x  y 8

5x  4y  35

Solving the first equation for y, we get y  x  8, and substituting this into the second equation gives

5x  4 x  8  35  5x  4x  32  35  x  3. Substituting for x we get y  3  8  5. Thus, the solution is 3 5. 7.

x  3y  11

3x  5y  17

Solving the first equation for x, we get x  3y  11, and substituting this into the second equation gives

3 3y  11  5y  17  9y  33  5y  17  4y  16  y  4. Substituting for y we get x  3 4  11  1. Thus, the solution is 1 4. 1


2

CHAPTER 9 Systems of Equations and Inequalities

8.

2x  y  7

Solving the first equation for y, we get y  7  2x, and substituting this into the second equation gives

x  2y  2

x  2 7  2x  2  x  14  4x  2  3x  12  x  4. Substituting for x we get y  7  2x  7  2 4  1. Thus, the solution is 4 1. 9.

2x  3y  7

Solving the second equation for x, we get x  5y, and substituting into the first equation gives

x  5y  0

2 5y  3y  7  7y  7  y  1. Thus, x  5 1  5, and the solution is 5 1. 10.

4x  y  5

Solving the first equation for y gives y  5  4x, and substituting into the second equation, we get

5x  2y  4

5x  2 5  4x  4  5x  10  8x  4  3x  6  x  2. Then y  5  4 2  3, and the solution is 2 3. 11.

3x  2y  13

6x  5y 

28

Multiplying the first equation by 2 gives the system

6x  4y  26

6x  5y 

28

Adding, we get y  2,

and substituting into the first equation in the original system gives 3x  2 2  13  3x  9  x  3. The solution is 3 2. 12.

2x  5y  18

Multiplying the first equation by 3 and the second by 2 gives the 3x  4y  19  6x  15y  54 system Subtracting the equations gives 23y  92  y  4. Substituting this value into the first 6x  8y  38

equation in the original system gives 2x  5 4  18  2x  2  x  1. Thus, the solution is 1 4. 13.

2x  y  1

x  2y  8

By inspection of the graph, it appears that 2 3 is the solution to the system. We check this in both

equations to verify that it is a solution. 2 2  3  4  3  1 and 2  2 3  2  6  8. Since both equations are satisfied, the solution is 2 3. 14.

xy2

2x  y  5

By inspection of the graph, it appears that 3 1 is the solution to the system. We check this in both

equations to verify that it is a solution. 3  1  2 and 2 3  1  6  1  5. Since both equations are satisfied, the solution is 3 1. 15.

xy4

16.

2x  y  2

2x  y  4

3x  y  6

The solution is x  2, y  0.

The solution is x  2, y  2. y

y

1 1

x

1 1

x


SECTION 9.1 Systems of Linear Equations in Two Variables

17.

2x  3y  12

18.

x  32 y  4

2x  6y  0

3x  9y  18

The lines are parallel, so there is no intersection and hence

The lines are parallel, so there is no intersection and hence

no solution.

no solution. y

y

1 1

19.

  x  1 y  5 2

 2x 

12x  15y  18

2x  52 y  3

y

1

5x  3y  12

x

There are infinitely many solutions. The lines are the same.

1

5x  3y  18

1

20.

y  10 y

1

x

There are infinitely many solutions.

21.

3

x

1 1

x

Adding the two equations gives 10x  30  x  3. Substituting for x in the first equation gives

5 3  3y  18  3y  3  y  1. Hence, the solution is 3 1.  2x  y  10 22. Solving the second equation for x gives x  3y  16, and substituting into the first equation, we x  3y  16

have 2 3y  16  y  10  6y  32  y  10  7y  42  y  6. Substituting, we have x  3 6  16  2, and so the solution is 2 6.  2x  3y  9 23. Adding the two equations gives 6x  18  x  3. Substituting for x in the second equation gives 4x  3y  9 4 3  3y  9  12  3y  9  3y  3  x  1. Hence, the solution is 3 1.  3x  2y  0 24. Adding the two equations gives 2x  8  x  4. Substituting for x in the second equation gives x  2y  8

3 4  2y  0  12  2y  0  y  6. Hence, the solution is 4 6.  x  3y  5 25. Solving the first equation for x gives x  3y  5. Substituting for x in the second equation gives 2x  y  3

2 3y  5  y  3  6y  10  y  3  7y  7  y  1. Then x  3 1  5  2. Hence, the solution is 2 1.  x  y 7 Adding 3 times the first equation to the second equation gives 5x  20  x  4. So 26. 2x  3y  1 4  y  7  y  3, and the solution is 4 3.


4

CHAPTER 9 Systems of Equations and Inequalities

27. x  y  2  y  x  2. Substituting for y into 4x  3y  3 gives 4x  3 x  2  3  4x  3x  6  3  x  3, and so y  3  2  5. Hence, the solution is 3 5. 28. 9x  y  6  y  9x  6. Substituting for y into 4x  3y  28 gives 4x  3 9x  6  28  23x  46  x  2, and so y  9 2  6  12. Thus, the solution is 2 12. 29. x 2y  7  x  72y. Substituting for x into 5x  y  2 gives 5 7  2y y  2  3510y  y  2  11y  33  y  3, and so x  7  2 3  1. Hence, the solution is 1 3. 30. 4x  12y  0  x  3y. Substituting for x into 12x  4y  160 gives 12 3y  4y  160  40y  160  y  4, and so x  3 4  12. Therefore, the solution is 12 4. 31.  13 x  16 y  1  2x  y  6  y  2x  6. Substituting for y into 23 x  16 y  3 gives 23 x  16 2x  6  3

 4x  2x  6  18  2x  12  x  6, and so y  2 6  6  6. Hence, the solution is 6 6.   32. 34 x  12 y  5  y  10  32 x. Substituting for y into  14 x  32 y  1 gives  14 x  32 10  32 x  1  x  60  9x  4  8x  64  x  8, and so y  10  32 8  2. Hence, the solution is 8 2.

  33. 12 x  13 y  2  x  23 y  4  x  4  23 y. Substituting for x into 15 x  23 y  8 gives 15 4  23 y  23 y  8 

4  2 y  10 y  8  12  2y  10y  120  y  9, and so x  4  2 9  10. Hence, the solution is 10 9. 5 15 15 3

34. 02x 02y  18  x  y9. Substituting for x into 03x 05y  33 gives 03 y  905y  33  02y  06  y  3, and so x  3  9  6. Hence, the solution is 6 3.   3x  2y  8 3x  2y  8 Multiplying the second equation by 3 gives the system Subtracting the second 35. x  2y  0 3x  6y  0

equation from the first gives 8y  8  y  1. Substituting into the first equation we get 3x  2 1  8  3x  6  x  2. Thus, the solution is 2 1.  4x  2y  16 36. Adding the first equation to 4 times the second equation gives 22y  264  y  12, so x  5y  70

4x  2 12  16  x  10, and the solution is 10 12.  x  4y  8 37. Adding 3 times the first equation to the second equation gives 0  22, which is never true. Thus, 3x  12y  2 the system has no solution.  3x  5y  2 Adding 3 times the first equation to the second equation gives 0  12, which is false. Therefore, 38. 9x  15y  6

there is no solution to this system.  2x  6y  10 39. Adding 3 times the first equation to 2 times the second equation gives 0  0. Writing the equation 3x  9y  15 in slope-intercept form, we have 2x  6y  10  6y  2x  10  y  13 x  53 , so the solutions are all pairs of the   form t 13 t  53 where t is a real number.  2x  3y  8 Adding 7 times the first equation to 1 times the second equation gives 0  59, which is false. 40. 14x  21y  3 Therefore, there is no solution to this system.  6x  4y  12 41. Adding 3 times the first equation to 2 times the second equation gives 0  0. Writing the equation in 9x  6y  18

slope-intercept form, we have 6x  4y  12  4y  6x  12  y   32 x  3, so the solutions are all pairs of the form   t  32 t  3 where t is a real number.


SECTION 9.1 Systems of Linear Equations in Two Variables

42.

25x  75y  100

5

1 times the first equation to 1 times the second equation gives 0  0, which is always Adding 25 10

10x  30y  40

true. We now put the equation in slope-intercept form. We have x  3y  4  3y  x  4  y  13 x  43 , so the   solutions are all pairs of the form t 13 t  43 where t is a real number. 43.

8s  3t  3

Adding 2 times the first equation to 3 times the second equation gives s  3, so

5s  2t  1

8 3  3t  3  24  3t  3  t  7. Thus, the solution is 3 7. 44.

u  30  5

3u  80 

Adding 3 times the first equation to the second equation gives 10  10    1, so

5

u  30 1  5  u  25. Thus, the solution is u   25 1.   1x  3y  3 2 5 45.  5 x  2y  10 3

Adding 10 times the first equation to 3 times the second equation gives 0  0. Writing the equation

in slope-intercept form, we have 12 x  35 y  3  35 y   12 x  3  y   56 x  5, so the solutions are all pairs of the   form t  56 t  5 where t is a real number.

46.

  3x  1y 

1 3 2  2x  1 y   1 2 2 2

Adding 6 times the first equation to 4 times the second equation gives x  5  x  5. So

9 5  2y  3  y  21. Thus, the solution is 5 21. 47.

04x  12y  14 12x 

Adding 30 times the first equation to 1 times the second equation gives 41y  410  y  10, so

5y  10

12x  5 10  10  12x  60  x  5. Thus, the solution is 5 10. 48.

26x  10y  4

06x  12y 

Adding 3 times the first equation to 25 times the second equation gives 63x  63  x  1, so

3

26 1  10y  4  10y  30  y  3. Thus, the solution is 1 3.   1x  1y  2 3 4 49.  8x  6y  10

Adding 24 times the first equation to the second equation gives 0  58, which is never true. Thus,

the system has no solution.

  1 x  1y  4 10 2 50.  2x  10y  80

Adding 20 times the first equation to the second equation gives 0  0, which is always true. We

now put the equation in slope-intercept form. We have 2x  10y  80  10y  2x  80  y  15 x  8, so the   solutions are all pairs of the form t 15 t  8 where t is a real number.


6

CHAPTER 9 Systems of Equations and Inequalities

51.

021x  317y  951

52.

235x  117y  589

The solution is approximately 387 274.

1872x  1491y  1233

621x  1292y  1782

The solution is approximately 071 172.

5

-5

5

5

-5

5

-5

53.

-5

2371x  6552y  13,591

54.

9815x  992y  618,555

The solution is approximately 6100 2000.

435x  912y 

0

132x  455y  994

The solution is approximately 285 136. 5

20 -5 0

50

5 -5

1 , a  1. So 55. Subtracting the first equation from the second, we get ay  y  1  y a  1  1  y  a1     1 1 1 1 1 x 0x   . Thus, the solution is   . a1 1a a1 a1 a1   a a , a  b. So x  1 56. Subtracting the first equation from a times the second, we get a  b y  a  y  ab ab   b a b . Hence, the solution is  . x  ba ba ab   ab 1 57. Subtracting b times the first equation from a times the second, we get a 2  b2 y  a  b  y  2 ,  2 ab a b   a 1 1 b 1  1  ax  x  . Thus, the solution is  . a 2  b2  0. So ax  ab ab ab ab ab

1 b 58. Subtracting a times the first equation from the second, we get b b  a y  1  y  . So ax  0 b b  a b b  a     1 1 1 1 1 . Hence, the solution is     2 . x  2 a b  a a b  a b b  a a  ab b  ab  x  y  34 Adding these two equations gives 2x  44  x  22. So 59. Let the two numbers be x and y. Then x  y  10 22  y  34  y  12. Therefore, the two numbers are 22 and 12.   x  3y  0 x  y  2 x  y  60. Let x be the larger number and y be the other number. This gives x  2y  6 x  6  2y two equations gives y  6, so x  6  2 6  18. Therefore, the two numbers are 18 and 6.

Adding these


SECTION 9.1 Systems of Linear Equations in Two Variables

61. Let d be the number of dimes and q be the number of quarters. This gives

d 

q  14

7

Subtracting the

010d  025q  275

first equation from 10 times the second gives 15q  135  q  9. So d  9  14  d  5. Thus, the number of dimes is 5 and the number of quarters is 9. 62. Let c be the number of children and a be the number of adults. This gives

c

a  2200

150c  400a  5050

Subtracting 3 times

the first equation from 2 times the second gives 5a  3500  a  700, so c  700  2200  c  1500. Therefore, the number of children admitted was 1500 and the number of adults was 700.  r  p  185 63. Let r be the amount of regular gas sold and p the amount of premium gas sold. Then 360r  400 p  690

Subtracting the second equation from 4 times the first equation gives 4r  360r  4 185  690  04r  50  r  125. Substituting this value of r into the original first equation gives 125  p  185  p  60. Thus, 125 gallons of regular gas and 60 gallons of premium were sold.

64. Let s be the number of boxes of regular strawberries sold and d the number of deluxe strawberries sold. Then  s  d  135 Subtracting 7 times the first equation from the second equation gives 10d  7d  1110  7 135 7s  10d  1110

 3d  165  d  55. Substituting this value of d into the original first equation gives s  55  135  s  80. Thus, 80 boxes of standard strawberries and 55 boxes of deluxe strawberries were sold.  2x  2y  180 65. Let x be the speed of the plane in still air and y be the speed of the wind. This gives 12x  12y  180 Subtracting 6 times the first equation from 10 times the second gives 24x  2880  x  120, so 2 120  2y  180  2y  60  y  30. Therefore, the speed of the plane is 120 mi/h and the wind speed is 30 mi/h.

66. Let x be speed of the boat in still water and y be speed of the river flow.

Downriver:

x 

y  20

Upriver: 25x  25y  20

Adding 5 times the first equation to 2 times the second gives 10x  140  x  14, so 14  y  20  y  6. Therefore, the boat’s speed relative to the water is 14 mi/h and the current in the river flows at 6 mi/h. 67. Let a and b be the number of grams of food A and food B. Then

012a  020b  100a 

32

50b  22,000

Subtracting 250 times the first equation from the second, we get 70a  14,000  a  200, so 012 200  020b  32  020b  8  b  40. Thus, 200 grams of food A and 40 grams of food B is fed to the rats each day. 68. Let x be the number of pounds of Kenyan coffee and y be the number of pounds of Sri Lankan coffee. This gives  700x  1120y  2310 Adding the first equation and 7 times the second gives 420y  210  y  05, so x  y 3 x  05  3  x  25. Thus, 25 pounds of Kenyan coffee and 05 pounds of Sri Lankan coffee should be mixed.

69. Let x and y be the sulfuric acid concentrations in the first and second containers.  300x  600y  900 015 Subtracting the first equation from 3 times the second gives 900y  90  y  010, so 100x  500y  600 0125 100x  500 010  75  x  025. Thus, the concentrations of sulfuric acid are 25% in the first container and 10% in the second.


8

CHAPTER 9 Systems of Equations and Inequalities

70. Let x and y be the number of milliliters of the two brine solutions.  Quantity: x  y  1000 Subtracting the first equation from 20 times the second gives 3y  1800 Concentrations: 005x  020y  014  y  600, so x  600  1000  x  400. Therefore, 400 milliliters of the 5% solution and 600 milliliters of the 20% solution should be mixed.  Total invested: x  y  20,000 71. Let x be the amount invested at 5% and y the amount invested at 8%. Interest earned: 005x  008y  1180

Subtracting 5 times the first equation from 100 times the second gives 3y  18,000  y  6,000, so x  6,000  20,000  x  14,000. Thus, $14,000 is invested at 5% and $6,000 at 8%. 72. Let x be the amount invested at 6% and y the amount invested at 10%. The ratio of the amounts invested gives x  2y. Then the interest earned is 006x  010y  3520  6x  10y  352,000. Substituting gives 6 2y  10y  352,000  22y  352,000  y  16,000. Then x  2 16,000  32,000. Thus, $32,000 is invested at 6% and $16,000 at 10%. 73. Let x be the length of time the truck travels and y be the length of time the SUV travels. Then (since 15 min  025 hr), y  x  025, so x  y  025. Multiplying by 40, we get 40x  40y  10. Comparing the distances, we get  40x  40y  10 60x  40y  35, or 60x  40y  35. This gives the system Adding, we get 20x  45  x  225, 60x  40y  35

76

so y  225  025  25. Thus, the truck travels for 2 14 hours and the SUV travels for 2 12 hours.

74. Let x be the weight of the gold in the crown and y the weight of the silver. Writing equations representing the weight and   x  y  235 volume of the crown, we have Adding the first equation to 193 times the second, we have x y    14 193 105

88 y  193 105 y  235  14 193   105 y  352  y  42. Thus, x  235  42  193, and so the crown contains 193 g

of gold and 42 g of silver.

75. Let x be the tens digit and y be the ones digit of the number.

xy7

Adding 9 times the first

10y  x  27  10x  y

74

equation to the second gives 18x  36  x  2, so 2  y  7  y  5. Thus, the number is 25. 76. First let us find the intersection point of the two lines. The y-coordinate of the intersection point is the height of the triangle.  y  2x  4 We have Adding 2 times the first equation to the second gives 3y  12, so the triangle has height 4. y  4x  20

78

length 5  2  3. Therefore, the area of the triangle is A  12 bh  12  3  4  6.   77. n  5, so nk1 xk  1  2  3  5  7  18, nk1 yk  3  5  6  6  9  29, n k1 xk yk  1 3  2 5  3 6  5 6  7 9  124, and n 2 2 2 2 2 2 k1 xk  1  2  3  5  7  88. Thus we get the system  18a  5b  29 Subtracting 18 times the first equation from 5 times the 88a  18b  124

Furthermore, y  2x  4 intersects the x-axis at x  2, and y  4x  20 intersects the x-axis at x  5. Thus the base has

second, we get 116a  98  a  0845. Then

b  15 [18 0845  29]  2758. So the regression line is y  0845x  2758.

Please give solution to #77, as done in CA-8 solutions to Exercise 5.1.77

Note that no corrections are required for CA-8 or PMFC-8 for the corresponding exercise solutions.

y 10 8 6 4 2 0

1

2 3 4 5 6 7 x


SECTION 9.2 Systems of Linear Equations in Several Variables

9.2

9

SYSTEMS OF LINEAR EQUATIONS IN SEVERAL VARIABLES

1. If we add 2 times the first equation to the second equation, the second equation becomes x  3z  1. 2. To eliminate x from the third equation, we add 3 times the first equation to the third equation. The third equation becomes 4y  5z  4.  3. The equation 6x  3y  12 z  0 is linear. 4. The equation x 2  y 2  z 2  4 is not linear, since it contains squares of variables.     x y  3y  z  5 is not a linear system, since the first equation contains a product of variables. In fact 5. The system x  y 2  5z  0    move to the left, as in the 2x  yz  3

both the second and the third equation are not linear.     x  2y  3z  10 6. The system 2x  5y is linear.  2    y  2z  4

    x  2y  z  5 7. yz  2    z 4

statement of the exercise

Substituting z  4 into the second equation gives y  4  2  y  2. Substituting z  4 and

y  2 into the first equation gives x  2 2  4  5  x  3. Thus, the solution is 3 2 4.     3x  3y  z  0 Substituting z  3 into the second equation gives y  4 3  10  y  2. Substituting z  3 8. y  4z  10    z 3 and y  2 into the first equation gives 3x  3 2  3  0  x  3. Thus, the solution is 3 2 3.     x  2y  z  7 9. Solving we get 2z  6  z  3. Substituting z  3 into the second equation gives y  3z  9    2z  6

y  3 3  9  y  0. Substituting z  3 and y  0 into the first equation gives x  2 0  3  7  x  4. Thus, the solution is 4 0 3.     x  2y  3z  10 Solving we get 3z  12  z  4. Substituting z  4 into the second equation gives 10. 2y  z  2    3z  12

2y  4  2  y  3. Substituting z  4 and y  3 into the first equation gives x  2 3  3 4  10  x  4. Thus, the solution is 4 3 4.     2x  y  6z  5 Solving we get 2z  1  z   12 . Substituting z   12 into the second equation gives 11. y  4z  0    2z  1     y  4  12  0  y  2. Substituting z   12 and y  2 into the first equation gives 2x  2  6  12  5  x  5.   Thus, the solution is 5 2  12 .


10

12.

CHAPTER 9 Systems of Equations and Inequalities

    4x    

2y 

3z  10

z  6

1z  2

4

Solving we get 12 z  4  z  8 . Substituting z  8 into the second equation gives

2y  8  6  2y  2  y  1. Substituting z  8 into the first equation gives 4x  3 8  10  4x  14    x   72 . Thus, the solution is  72  1 8 .

    3x  y  z  4 Add the third equation to the second equation: y  z  1    x  2y  z  1     3x  y  z  4 Or, add the first equation to three times the second equation: 4y  7z  4    x  2y  z  1

    3x  y  z  4 13. x  y  2z  0    x  2y  z  1

  3   5x  2y  3z  14. 10x  3y  z  20    x  3y  z  8

  3   5x  2y  3z  Add twice the first equation to the second equation: y  5z  14    x  3y  z  8     5x  2y  3z  3 Or, add 10 times the third equation to the second equation: 27y  11z  60    x  3y  z  8

    2x  y  3z  5 15. 2x  3y  z  13    6x  5y  z  7

    2x  y  3z  5 Add 3 times the first equation to the third equation: 2x  3y  z  13    8y  8z  8   5   2x  y  3z  Or, add 3 times the second equation to the third equation: 2x  3y  z  13    14y  4z  32

    x  3y  2z  1 Add 2 times the first equation to 3 times the third equation: y  z  1    2x  z 1     x  3y  2z  1 Or, add 2 times the second equation to the third equation: y  z  1    3z  3

    x  3y  2z  1 16. y  z  1    2y  z  1

    x  2y  z  3     x  2y  z  3 17. y  z  4  y  z  4      x  2y  3z  9  4z  12

Eq. 1  Eq. 3

So z  3 and y  3  4  y  1. Thus, x  2 1  3  3  x  2. So the solution is 2 1 3.           x  5y  3z  4  x  5y  3z  4  x  5y  3z  4  18. 3y  5z  11 3y  5z  11 3y  5z  11           x  2y  z  11 3y  4z  7 z 4 Eq. 2  Eq. 3 1  Eq. 1  Eq. 3 So 3y  5 4  11  y  3. Then x  5 3  3 4  4  x  1. So the solution is 1 3 4.


SECTION 9.2 Systems of Linear Equations in Several Variables

      x  3y  2z  5  x  3y  2z  5 19. 3x  2y  z  8  7y  7z  7      y  3z  1 y  3z  1

   x  3y  2z  5 Eq. 2  3  Eq. 1  7y  7z  7   28z  0

11

Eq. 2  7  Eq. 3

So z  0 and 7y  0  7  y  1. Then x  3 1  2 0  5  x  2. So the solution is 2 1 0.         x  2y  3z  10  x  2y  3z  10  x  2y  3z  10  20.  3y  z  7 3y  z  7 3y  z  7       x  y  z  3y  4z  17 Eq. 2  1  Eq. 3 5z  10 1  Eq. 1Eq. 3 7 So z  2 and 3y  2  7  y  3. Then x  2 3  3 2  10  x  2. So the solution is 2 3 2.          x  y  z  4  x y z 4 x  y  z  4 21. 2y  2z  6 x  3y  3z  10  y  3z  5 1  Eq. 1  Eq. 2  Eq. 3          2x  y  z  3  y  3z  5 2  Eq. 1  1  Eq. 3 2y  2z  6 Eq. 2    x  y  z  4 y  3z  5    4z  4 2  Eq. 2  Eq. 3

So z  1 and y  3 1  8  y  2. Then x  2  1  4  x  1. So the solution is 1 2 1.       x  y  z  0 x  y  z  0      x  y  z  0 22. x  2y  5z  3  y  2z  1 13  Eq. 1  13  Eq. 2  y  2z  1        3x  y   6 5z  10 4  Eq. 2  Eq. 3 4y  3z  6 3  Eq. 1  Eq. 3 So z  2 and y  2 2  1  y  3. Then x  3  2  0  x  1. So the solution is 1 3 2.       4z  1  4z  1  4z  1     x  x x 23. 2x  y  6z  4  y  2z  2 2  Eq. 1  Eq. 2  y  2z  2        2x  3y  2z  8  3y  6z  6 2  Eq. 1  Eq. 3  12z  12 3  Eq. 2  Eq. 3

So z  1 and y  2 1  2  y  0. Then x  4 1  1  x  5. So the solution is 5 0 1.       x  y  2z  2 x  y  2z  2 2       x  y  2z   24.  y  6z  11 3  Eq. 1  Eq. 3 3x  y  5z  8 y  6z  11        4y  z   2x  y  2z  7  23z  46 4  Eq. 2  Eq. 2 2 2  Eq. 1  Eq. 2

So z  2 and y  6 2  11  y  1. Then x  1  2  2  x  1. So the solution is 1 1 2.           2x  4y  z  2  2x  4y  z  2  x  2y  3z  4 25. 5z  10 x  2y  3z  4  Eq. 1  2  Eq. 2  14y  5z  4 Eq. 2  Eq. 3        14y  5z  4 3  Eq. 1  2  Eq. 3  3x  y  z  1  5z  10 Eq. 2  Eq. 3

So z  2 and 14y  5 2  4  y  1. Then x  2 1  3 2  4  x  0. So the solution is 0 1 2.           2x  y  z  8  2x  y  z  8  2x  y  z  8 26. 3y  z  2 Eq. 1  2  Eq. 2  x  y  z  3  3y  z  2         2x   4z  18 8z  32 Eq. 2  3  Eq. 3 y  3z  10 Eq. 1  Eq. 3

So z  4 and 3y  4  2  y  2. Then 2x  2  4  8  x  1. So the solution is 1 2 4.     Eq. 2 2y  4z  1     2x  y  2z  1 27. 2x  y  2z  1  2y  4z  1 Eq. 1      4x  2y   0 4z  2 2Eq. 2  Eq. 3       So z   12 and 2y  4  12  1  y  12 . Then 2x  12  2  12  1  x  14 . So the solution is 14  12   12 .


12

CHAPTER 9 Systems of Equations and Inequalities

   

   Eq. 2 2   z 2  6x  2y  z   6x  2y  28. y  z  1 Eq. 1  6x  2y  z  2  y z  1        x  y  3z  2  21z  14 4  Eq. 2  Eq. 3 4y  17z  10 Eq. 2  6  Eq. 3     So z  23 and y  23  1  y   13 . Then 6x  2  13  23  2  x  13 . So the solution is 13   13  23 .     x  3y  z  2     x  3y  z  2 29. Since 0  7 is false, this system is inconsistent. 3x  4y  2z  1  13y  5z  5 3  Eq. 1  Eq. 2      2x  6y  2z  3  0 7 2  Eq. 1  Eq. 3 y  z  1

       2z  0 Eq. 2  2z  0     x  2y  5z  4  x x 30. x  2z  0  x  2y  5z  4 Eq. 1  2y  3z  4        4x  2y  11z  2  4x  2y  11z  2  2y  3z  2    2z  0  x Since 0  6 is false, this system is inconsistent. 2y  3z  4    0  6 Eq. 2  Eq. 3

Eq. 1  Eq. 2

4  Eq. 1  Eq. 3

       3 Eq. 2  3     x  2y  x  2y  2x  3y  z  1 31. y  z  5 Eq. 2  2  Eq. 1  x  2y  3  2x  3y  z  1 Eq. 1         x  3y  z  4   x  3y  z  4 yz 1 Eq. 3  Eq. 1    3   x  2y Since 0  4 is false, this system is inconsistent. y  z  5    0  4 Eq. 2  Eq. 3     5    x  2y  3z   x  2y  3z  5 32. 2x  y  z  5  5y  5z  5       4x  3y  7z  5 5y  5z  5

  5   x  2y  3z  5y  5z  5 Eq. 2  2  Eq. 1     0  10 2  Eq. 2  Eq. 3

Since 0  10 is false, this system is inconsistent.           x y z0 x  y  z  0 x  y  z  0 33. x  2y  3z  3  y  2z  3 y  2z  3 Eq. 2  Eq. 1         2x  3y  4z  3   00 y  2z  3 2  Eq. 1  Eq. 3

Eq. 3  Eq. 2

Eq. 2  Eq. 3

So z  t and y  2t  3  y  2t  3. Then x  2t  3  t  0  x  t  3. So the solutions are t  3 2t  3 t, where t is any real number.       x  2y  z  3 x  2y  z  3     x  2y  z  3    34. 2x  5y  6z  7  y  4z  1 y  4z  1 Eq. 2  2  Eq. 1         2x  3y  2z  5  0  0 Eq. 2  Eq. 3 y  4z  1 Eq. 3  2  Eq. 1 So z  t and we have y  4t  1  y  4t  1. Substituting into the first equation, we have x  2 4t  1  t  3  x  7t  1. So the solutions are 7t  1 4t  1 t, where t is any real number.           x  3y  2z  0  x  3y  2z  0  x  3y  2z  0 35. 2x  4z  4  6y  8z  4 6y  8z  4 Eq. 2  2  Eq. 1          4x  6y  4 0  0 Eq. 2  Eq. 3 6y  8z  4 Eq. 3  4  Eq. 1   So z  t and 6y  8t  4  6y  8t  4  y  43 t  23 . Then x  3 43 t  23  2t  0  x  2t  2. So the   solutions are 2t  2 43 t  23  t , where t is any real number.


SECTION 9.2 Systems of Linear Equations in Several Variables

       2x  4y  z  3  x  2y  2z  0 36. x  2y  4z  6  2x  4y  z  3      x  2y  2z  0  x  2y  4z  6     x  2y  2z  0 3z  3    0  0 2  Eq. 2  Eq. 3

Eq. 3 Eq. 1 Eq. 2

    x  2y  2z  0  3z  3    6z  6

2  Eq. 1  Eq. 2

13

Eq. 3  Eq. 1

So z  1 and y  t, and substituting into the first equation we have x  2t  2 1  0  x  2t  2. Thus, the solutions are 2t  2 t 1, where t is any real number.     x  z  2  6 x  z  2  6         y  2z y  2z  3  3  37.    x  2y  z  2 2y  2z  2  8 Eq. 3  Eq. 1         2x  y  3z  2  0 y  z  6  12 Eq. 4  2  Eq. 1     x  z  2  6 x  z  2  6        y  2z   3 y  2z  3  1 Eq. 3   2z  2  2 z    1 Eq. 3  2  Eq. 2   2       3z  6  9 Eq. 4  Eq. 2 6  12 3  Eq. 3  2  Eq. 4

So   2 and z  2  1  z  1. Then y  2 1  3  y  1 and x  1  2 2  6  x  1. Thus, the solution is 1 1 1 2.       x y z 0 x y z 0 x y z 0            x  y  2z  2  0  z 0 Eq. 2  Eq. 1 y  z    1 Eq. 4  38.     2x  2y  3z  4  1  z  2  1 Eq. 3  2  Eq. 1 z    0 Eq. 2             2x  3y  4z  5  2 y z 1 Eq. 4  Eq. 3 z  2  1 Eq. 3   x y z 0     yz 1  z 0       1 Eq. 4  Eq. 3 So   1 and z  1  0  z  1. Then y  1  1  1  y  1 and x  1  1  1  0  x  1. So the solution is 1 1 1 1.

39. Let x be the amount invested at 4%, y the amount invested at 5%, and z the amount invested at 6%. We set   x  y  z  100,000   Total money: up a model and get the following equations: Annual income: 004x  005y  006z  0051 100,000    Equal amounts: xy        x  y  z  100,000  x  y  z  100,000  4x  5y  6z  510,000  y  2z  110,000 Eq. 2  4  Eq. 1       x  y  2y  z  100,000  0 Eq. 3  Eq. 1     x  y  z  100,000 y  2z  110,000    3z  120,000 2  Eq. 2  Eq. 3

So z  40,000 and y  2 40,000  110,000  y  30,000. Since x  y, x  30,000. the financial planner should invest $30,000 in short-term bonds, $30,000 in intermediate-term bonds, and $40,000 in long-term bonds.


14

CHAPTER 9 Systems of Equations and Inequalities

40. Let x be the amount invested at 3%, y the amount invested at 5 12 %, and z the amount invested at 9%. We   x  y  z  50,000   Total investment: set up a model and get the following equations: Annual income: 003x  0055y  009z  2540    Twice as much: x  2z        x  y  z  50,000  x  y  z  50,000  6x  11y  18z  508,000 200  Eq. 2  5x  7z  42,000 Eq. 2  11  Eq. 1        x x  2z  0  2z  0     x  y  z  50,000 5x  7z  42,000    3z  42,000 Eq. 2  5  Eq. 3

So z  14,000 and x  2z  28,000. Since x  y  z  50,000, we have y  50,000  14,000  28,000  8000. Thus, $28,000 should be invested in the least risky account, $8,000 in the intermediate account, and 14,000 in the highest-yielding account.

41. Let x, y, and z be the number of acres of land planted with corn, wheat, and soybeans. We set up a model and   x  y  z  1200   Total acres: Substituting 2x for y, we get get the following equations: Market demand: 2x  y    Total cost: 45x  60y  50z  63,750       x  2x  z  1200  z  1200  z  1200      3x  3x  2x  y 2x  y  0 2x  y  0         45x  60 2x  50z  63,750  15x  165x  3750 Eq. 3  50  Eq. 1  50z  63,750

So 15x  3,750  x  250 and y  2 250  500. Substituting into the original equation, we have 250  500  z  1200  z  450. Thus the farmer should plant 250 acres of corn, 500 acres of wheat, and 450 acres of soybeans.

42. Let a, b, and c be the number of gallons of Regular, Performance Plus, and Premium gas sold. The information provided     a  b  c  6500 b c  6500    a  gives the following system: 300a  320b  330c  20,050  02b  03c  550 Eq. 2  3  Eq. 1        a  3c  0 b  4c  6500 Eq. 1  Eq. 3   b c  6500  a  02b  03c  550    5c  7500 10  Eq. 2  2  Eq. 3

Thus, c  1500, so 02b  03 1500  550  b  500 and a  500  1500  6500  a  4500. The gas station sold 4500 gallons of Regular, 500 gallons of Performance Plus, and 1500 gallons of Premium gas.

43. Let a, b, and c be the number of ounces of Type A, Type B, and Type C pellets used. The requirements for the different        2a  3b  c  9  2a  3b  c  9 vitamins gives the following system: 3a  b  3c  14  Equations 2 7b  3c  1 2  Eq. 2  3  Eq. 1       8a  5b  7c  32 7b  3c  4 Eq. 3  4  Eq. 1 and 3 are inconsistent, so there is no solution.


SECTION 9.2 Systems of Linear Equations in Several Variables

15

44. Let a, b, and c represent the number of servings of toast, cottage cheese, and fruit, respectively. Then    2c  6   2a the given dietary requirements give rise to the following system: a  5b  11     100a  120b  60c  460     2a  2c  6  c 6     2a 10b  2c  16 10b  2c  16 2  Eq. 2  Eq. 1        16c  32 Eq. 3  12  Eq. 2 120b  40c  160 Eq. 3  50  Eq. 1

Thus, c  2, so 10b  2 2  16  10b  20  b  2 and 2a  2 2  6  a  1. The patient should eat one serving of toast and two each of cottage cheese and fruit.

45. Let a, b, and c represent the number of Midnight Mango, Tropical Torrent, and Pineapple Power smoothies sold. The        8a  6b  2c  820  8a  6b  2c  820 given information leads to the system 3a  5b  8c  690  22b  58c  3060 8  Eq. 2  3  Eq. 1        3a  3b  4c  450 2b  4c  240 Eq. 2  Eq. 3     8a  6b  2c  820 22b  58c  3060    14c  420 Eq. 2  11  Eq. 3

Thus, c  30, so 22b  58 30  3060  22b  1320  b  60 and 8a  6 60  2 30  820  a  50. Thus, The Juice Company sold 50 Midnight Mango, 60 Tropical Torrent, and 30 Pineapple Power smoothies on that particular day.

46. Let a, b, and c represent the number of days required at each plant. The given information leads        8a  10b  14c  110  8a  10b  14c  110 to the system 16a  12b  10c  150  8b  18c  70 2  Eq. 1  1  Eq. 2        10a  18b  6c  114 84b  2c  162 5  Eq. 2  8  Eq. 3     8a  10b  14c  110 8b  18c  70    382c  1146 21  Eq. 2  2  Eq. 3

Thus, c  3, so 8b  18 3  70  8b  16  b  2 and 8a  10 2  14 3  110  a  6. Thus, Factory A should be scheduled for 6 days, Factory B for 2 days, and Factory C for 3 days.

47. Let a, b, and c be the number of shares of Stock A, Stock B, and Stock C in the investor’s portfolio. Since the total value remains unchanged, we get the following system:     10a  25b  29c  74,000     10a  25b  29c  74,000 12a  20b  32c  74,000  50b  14c  74,000 6  Eq. 1  5  Eq. 2       16a  15b  32c  74,000  125b  72c  222,000 8  Eq. 1  5  Eq. 3     10a  25b  29c  74,000   

50b  14c 

74,000

74c  74,000

5  Eq. 2  2  Eq. 3

So c  1,000. Back-substituting we have 50b  14 1000  74,000  50b  60,000  b  1,200 . And finally 10a  25 1200  29 1000  74,000 10a  30,000  29,000  74,000  10a  15,000  a  1,500. Thus the portfolio consists of 1,500 shares of Stock A, 1,200 shares of Stock B, and 1,000 shares of Stock C.


16

CHAPTER 9 Systems of Equations and Inequalities

    I1  I2  I3  0 48. 16I1  8I2 4  24I2  16I3  4 16  Eq. 1  Eq. 2        8I2  4I3  5 8I2  4I3  5   I2  I3  0   I1  24I2  16I3  4    28I3  19 Eq. 2  3  Eq. 3   19 2 2 19 11 So I3  19 28  068 and 24I2  16 28  4  I2  7  029. Then I1  7  28  0  I1  28  039.    

I 1  I2  I3  0

x  x1 y0  y1 z  z1 49. (a) We begin by substituting 0 , , and 0 into the left-hand side of the first equation: 2 2       2   x0  x1 y0  y1 z  z1  b1  c1 0  12 a1 x0  b1 y0  c1 z 0   a1 x1  b1 y1  c1 z 1  a1 2 2 2    12 d1  d1  d1

Thus the given ordered triple satisfies the first equation. We can show that it satisfies the second and the third in exactly the same way. Thus it is a solution of the system.

(b) We have shown in part (a) that if the system has two different solutions, we can find a third one by averaging the two solutions. But then we can find a fourth and a fifth solution by averaging the new one with each of the previous two. Then we can find four more by repeating this process with these new solutions, and so on. Clearly this process can continue indefinitely, so there are infinitely many solutions.

9.3

PARTIAL FRACTIONS

1. (iii): r x  2. (ii): r x  3. 5. 7. 8.

4 x x  22

B C A   x x  2 x  22

Bx  C A 2x  8     2 x 1 x 4 x  1 x 2  4

B A 1   x 1 x 2 x  1 x  2 x 2  3x  5

x  22 x  4

B C A   x  2 x  22 x 4

Bx  C x2 A    2 x 3 x 4 x  3 x 2  4

x A B x 4. 2    x  1 x  4 x  1 x  4 x  3x  4 1 B 1 C D A 6. 4  3   2  3  x x 1 x  x3 x x  1 x x

B Cx  D 1 1 A 1        2  2 x 1 x 1 x  1 x2  1 x4  1 x 1 x  1 x  1 x 2  1

Ax  B Cx  D x 3  4x 2  2     2 9.  2 x  1 x2  2 x2  1 x 2

10.

B A Cx  D x4  x2  1 Ex  F   2  x  2  2 2 2 2 x x  4 x2  4 x x 4

B C x3  x  1 A D Ex  F Gx  H   2   2  x  2x  5  2 x  2x  5 2x  52 2x  53 x 2  2x  5 x 2x  53 x 2  2x  5        12. Since x 3  1 x 2  1  x  1 x 2  x  1 x  1 x  1  x  12 x  1 x 2  x  1 , we have 11.

B Dx  E C 1 A 1      2  .    x  1 x  12 x 1 x3  1 x2  1 x x 1 x  12 x  1 x 2  x  1


17

SECTION 9.3 Partial Fractions

13.

B 2 A  . Multiplying by x  1 x  1, we get 2  A x  1  B x  1   x 1 x 1 x  1 x  1  AB 0 Adding we get 2A  2  A  1. Now A  B  0  B  A, so 2  Ax  A  Bx  B. Thus AB 2 B  1. Thus, the required partial fraction decomposition is

14.

B A 2x  . Multiplying by x  1 x  1, we get 2x  A x  1  B x  1   x 1 x 1 x  1 x  1  AB 2 2x  Ax  A  Bx  B. Thus Adding we 2A  2  A  1. Since A  B  0  B  A, B  1. The AB 0 required partial fraction decomposition is

15.

2 1 1  .  x 1 x 1 x  1 x  1

1 2x 1  .  x 1 x 1 x  1 x  1

5 B A  . Multiplying by x  1 x  4, we get 5  A x  4  B x  1   x 1 x 4 x  1 x  4  AB 0 5  Ax  4A  Bx  B. Thus Now A  B  0  B  A, so substituting,we get 4A  A  5  4A  B  5 5A  5  A  1and B  1. The required partial fraction decomposition is

16.

A B x 6   . x x  3 x x 3

Multiplying by x x  3we get x  6  A x  3  x B   AB1 x  6  Ax  3A  Bx  A  B x  3A. Thus Now 3A  6  A  2, and 2  B  1  B  1. 3A 6 The required partial fraction decomposition is

17.

1 1 5  .  x 1 x 4 x  1 x  4

x 6 2 1   . x x  3 x x 3

B 12 A 12  . Multiplying by x  3 x  3, we get 12  A x  3  B x  3    x 3 x 3 x  3 x  3 x2  9   A B  0 AB 0 12  Ax  3A  Bx  3B. Thus  Adding, we get 2A  4  A  2. So 2  B  0 3A  3B  12 AB 4 2 2 12   .  B  2. The required partial fraction decomposition is 2 x  3 x  3 x 9

18.

x  12 A B x  12   . Multiplying by x x  4, we get  x x  4 x x 4 x 2  4x x 12  A x  4 Bx  Ax 4A  Bx  A  B x 4A. Thus we must solve the system 2 3 x  12 .   gives 4A  12  A  3, and 3  B  1  B  2. Thus 2 x x 4 x  4x

19.

AB 4A

4 B 4 A  . Multiplying by x 2  4, we get   x 2 x 2 x  2 x  2 x2  4   A B 0 AB 0 4  A x  2  B x  2  A  B x  2A  2B, and so  2A  2B  4 AB 2 1 1 4  .   A  1, and B  1. Therefore, 2 x 2 x 2 x 4

1

This

 12

Adding we get 2A  2


18

20.

CHAPTER 9 Systems of Equations and Inequalities

2x  1

x2  x  2 and so

B 2x  1 A  . Thus, 2x  1  A x  1  B x  2  A  B x  A  2B,  x 2 x 1 x  2 x  1

A B 2

A  2B  1

Adding the two equations, we get 3B  3  B  1. Thus A  1  2  A  1. Therefore,

2x  1 1 1  .  x 2 x 1 x2  x  2 21.

x  14 B x  14 A  . Hence, x  14  A x  2  B x  4  A  B x  2A  4B,   x 4 x 2 x  4 x  2 x 2  2x  8   A B  1 2A  2B  2 and so  Adding, we get 3A  9  A  3. So 3  B  1  B  2. 2A  4B  14 A  2B  7 2 3 x  14  .  Therefore, 2 x 4 x 2 x  2x  8

22.

23.

A B 8x  3 8x  3   . Hence, 8x  3  A 2x  1  Bx  2A  B x  A, giving  x 2x  1 x 2x  1 2x 2  x  2A  B  8 2 3 8x  3 .   So A  3  A  3, and 2 3  B  8  B  2. Therefore, 2 x 2x  1 2x  x A  3 B x A x  . Hence,   4x  3 2x  1 4x  3 2x  1 8x 2  10x  3 x  A 2x  1  B 4x  3  2A  4B x  A  3B, and so Adding, we get 2B  1  B   12 , and A  32 . Therefore,

24.

2A  4B  1

A  3B  0

3 2

2A  4B  1

2A  6B  0

1 x 2  .  4x  3 2x  1 8x 2  10x  3

A B C 7x  3 7x  3    . Hence,  x x  3 x  1 x x 3 x 1 x 3  2x 2  3x       7x  3  A x  3 x  1  Bx x  1  C x x  3  A x 2  2x  3  B x 2  x  C x 2  3x Thus 

   

 A  B  C x 2  2A  B  3C x  3A 0

Coefficients of x 2

2A  B  3C  7    3A  3

Coefficients of x

AB C

1 B C 0

2  B  3C  7

Constant terms

So 3A  3  A  1. Substituting, the system reduces to

Adding these two equations, we get 3  4C  7  C  1. Thus 1  B  1  0  B  2.

7x  3 2 1 1 Therefore, 3  .   x x 3 x 1 x  2x 2  3x 25.

B C A 9x 2  9x  6 9x 2  9x  6   . Thus,   x  2 x  2 2x  1 x  2 x  2 2x  1 2x 3  x 2  8x  4

9x 2  9x  6  A x  2 2x  1  B x  2 2x  1  C x  2 x  2        A 2x 2  3x  2  B 2x 2  5x  2  C x 2  4  2A  2B  C x 2  3A  5B x  2A  2B  4C


SECTION 9.3 Partial Fractions

19

    Coefficients of x 2    2A  2B  C  9  2A  2B  C  9 This leads to the system 16B  3C  45  3A  5B  9 Coefficients of x        2A  2B  4C  6 4B  3C  15 Constant terms   9   2A  2B  C  Hence, 15C  15  C  1; 16B  3  45  B  3; and 2A  6  1  9  A  2. 16B  3C  45    15C  15 Therefore,

26.

9x 2  9x  6

2x 3  x 2  8x  4

3 1 2   . x  2 x  2 2x  1

B C 3x 2  3x  27 A 3x 2  3x  27      . Thus,  2 x  2 2x  3 x  3 x  2 2x  3 x  3 x  2 2x  3x  9

3x 2  3x  27  A 2x  3 x  3  B x  2 x  3  C x  2 2x  3        A 2x 2  3x  9  B x 2  5x  6  C 2x 2  x  6

    2A  B  2C  3 So 3A  5B  C  3    9A  6B  6C  27

 2A  B  2C x 2  3A  5B  C x  9A  6B  6C     Coefficients of x 2    2A  B  2C  3  2A  B  2C  3 7B  4C  3  7B  6C  3 Coefficients of x        21B  6C  27 24C  24 Constant terms

Hence, 24C  24  C  1; then 7B  4  3  B  1; and 2A  1  2  3  A  3. Therefore,

27.

1 1 3 3x 2  3x  27     . 2 x  2 2x  3 x  3 x  2 2x  3x  9

x2  1 B x2  1 C A . Hence,    2  x x 1 x3  x2 x 2 x  1 x

x 2  1  Ax x  1  B x  1  C x 2  A  C x 2  A  B x  B, and so B  1; A  1  0  A  1; and

1 x2  1 1 2  2  . 1  C  1  C  2. Therefore, 3  2 x x  1 x x x 28.

3x 2  5x  13 B C A 3x 2  5x  13       . Thus, 2 2 3x  2 x  2 3x  2 x  4x  4 3x  2 x  2 x  22

3x 2  5x  13  A x  22  B 3x  2 x  2  C 3x  2      A x 2  4x  4  B 3x 2  4x  4  C 3x  2

 A  3B x 2  4A  4B  3C x  4A  4B  2C     A  3B  3 Coefficients of x 2  3     A  3B This leads to the following system: 4A  4B  3C  5 Coefficients of x  8B  3C  17      4A  4B  2C  13  Constant terms 8B  5C  8    3   A  3B   109 Hence, 8C  9  C  98 ; 8B  3 98  8B  27  8B  3C  17 8  17  B  64 ; and    8C  9 135 109 9   2 64 64  8 327  3  A   135 . Therefore, 3x  5x  13  A  3 109  A   . 64 64 64 3x  2 x  2 x  22 3x  2 x  22


20

29.

CHAPTER 9 Systems of Equations and Inequalities

B A  . Hence, 2x  A 2x  3  B  2Ax  3A  B. So 2A  2  2x  3 2x  32 2x 3 1 A  1; and 3 1  B  0  B  3. Therefore, 2 .   2x  3 2x  32 4x  12x  9 2x

4x 2  12x  9

2x

2x  32

B A   . Hence, x  4  A 2x  5  B  2Ax  5A  B, and so 30. 2 2x  5 2x  5 2x  52 x 4

  A  12 and 5 12  B  4  B   32 . Therefore,

31.

x 4

1

3

2A

1

5A  B  4

2 2   . 2x  5 2x  52 2x  52

B 4x 2  x  2 C D A 4x 2  x  2 . Hence,  3   2  3  4 3 x x 2 x  2x x x  2 x x 4x 2  x  2  Ax 2 x  2  Bx x  2  C x  2  Dx 3  A  D x 3  2A  B x 2  2B  C x  2C So 2C  2  C  1; 2B  1  1  B  0; 2A  0  4  A  2; and 2  D  0  D  2. Therefore, 1 2 2 4x 2  x  2 .   3  4 3 x x  2 x  2x x

32.

x 3  2x 2  4x  3 B A C D   2  3  4 . Hence, x 3  2x 2  4x  3  Ax 3  Bx 2  C x  D. Thus A  1; B  2; 4 x x x x x C  4; and D  3. Therefore,

33.

10x 2  27x  14 x  13 x  2

2 1 4 3 x 3  2x 2  4x  3   2  3  4. x x4 x x x

B C D A    . Thus, x  2 x  1 x  12 x  13

10x 2  27x  14  A x  13  B x  2 x  12  C x  2 x  1  D x  2

       A x 3  3x 2  3x  1  B x  2 x 2  2x  1  C x 2  x  2  D x  2        A x 3  3x 2  3x  1  B x 3  3x  2  C x 2  x  2  D x  2

 A  B x 3  3A  C x 2  3A  3B  C  D x  A  2B  2C  2D


SECTION 9.3 Partial Fractions

21

which leads to the system   A B  0     3A  C  10  3A  3B  C  D  27     A  2B  2C  2D  14

  A B  0     2 3B  C  10 Coefficients of x    3B  2C  D  17 Coefficients of x     3B  5C  7D  15 Constant terms Coefficients of x 3

    A B  0 A B  0         3B  C  10 3B  C  10    3C  D  7 3C  D  7         3C  8D  2 9D  9

Hence, 9D  9  D  1, 3C  1  7  C  2, 3B  2  10  B  4, and A  4  0  A  4. Therefore, 10x 2  27x  14 x  13 x  2

34.

4 1 4 2  .   x  2 x  1 x  12 x  13

2x 2  5x  1 B C 2x 2  5x  1 2x 2  5x  1 D A        . Thus,  x  1 x  1 x  12 x 4  2x 3  2x  1 x  1 x 3  x 2  x  1 x  13 x  1 x  13 2x 2  5x  1  A x  13  B x  1 x  12  C x  1 x  1  D x  1        A x 3  3x 2  3x  1  B x  1 x 2  2x  1  C x 2  1  D x  1        A x 3  3x 2  3x  1  B x 3  x 2  x  1  C x 2  1  D x  1

 A  B x 3  3A  B  C x 2  3A  B  D x  A  B  C  D  AB      3A  B  C

0

  A B  0     2B  C  2 Coefficients of x 2    2B  C  D  3 Coefficients of x     2B  3C  4D  2 Constant terms Coefficients of x 3

 2  3A  B  D 5     A  B  C  D  1     A B  0 A B  0         2B  C  2 2B  C  2    2C  D  1 2C  D  1         2C  5D  5 6D  6 which leads to the system

Hence, 6D  6  D  1, 2C  1  1  C  0,

2x 2  5x  1 1 1 1 2B  0  2  B  1, and A  1  0  A  1. Therefore, 4   .  x  1 x  1 x  13 x  2x 3  2x  1


22

35.

CHAPTER 9 Systems of Equations and Inequalities

3x 3  22x 2  53x  41 x  22 x  32

B D A C    . Thus, x  2 x  22 x  3 x  32

3x 3  22x 2  53x  41  A x  2 x  32  B x  32  C x  22 x  3  D x  22      A x 3  8x 2  21x  18  B x 2  6x  9

    C x 3  7x 2  16x  12  D x 2  4x  4

 A  C x 3  8A  B  7C  D x 2

 21A  6B  16C  4D x  18A  9B  12C  4D     A  C  3 A  C  3 Coefficients of x 3         8A  B  7C  D  22 2 B  C  D  2 Coefficients of x  so we must solve the system   6B  5C  4D  10 21A  6B  16C  4D  53 Coefficients of x         9B  6C  4D  13 18A  9B  12C  4D  41 Constant terms     A  C  3 A C  3         B  C  D  2 B  C  D  2  Hence, D  1, C  2  2  C  0, B  0  1  2    C  2D  2 C  2D  2         3C  5D  5 D  1  B  1, and A  0  3  A  3. Therefore,

36.

3x 3  22x 2  53x  41 x  22 x  32

1 3 1   . x  2 x  22 x  32

B D 3x 2  12x  20 A C 3x 2  12x  20 3x 2  12x  20      . Thus,    2 2 x  22 2 x  2 x  2 2 x 4  8x 2  16  2  2  22 x x x x 4

3x 2  12x  20  A x  2 x  22  B x  22  C x  22 x  2  D x  22          A x 3  2x 2  4x  8  B x 2  4x  4  C x 3  2x 2  4x  8  D x 2  4x  4

 A  C x 3  2A  B  2C  D x 2  4A  4B  4C  4D x  8A  4B  8C  4D     A  C  0 Coefficients of x 3 A  C  0         2A  B  2C  D  2 B  4C  D  3 3 Coefficients of x  which leads to the system   4A  4B  4C  4D  12 6B  8C  2D  6 Coefficients of x         4B  16C  12D  4 8A  4B  8C  4D  20 Constant terms     A  C  0 A  C  0         B  4C  D  3 B  4C  D  3  Hence, 48D  48  D  1,    16C  8D  24 16C  8D  24         32C  32D  0 48D  48 16C  8  24  C  1; B  4  1  3  B  2, and A  1  0  A  1. Therefore, 3x 2  12x  20 2 1 1 1    .  x  2 x  22 x  2 x  22 x 4  8x 2  16

37.

  Bx  C x 3 A x 3    2   2 . Hence, x  3  A x 2  3  Bx 2  C x  A  B x 2  C x  3A. So 3 x x  3x x x 3 x 3 x 3 x 1 1 3A  3  A  1; C  1; and 1  B  0  B  1. Therefore, 3   2 . x x  3x x 3


SECTION 9.3 Partial Fractions

38.

23

3x 2  2x  8 3x 2  2x  8 Ax  B C   . Thus,   2  x 1 x 2  2 x  1 x 3  x 2  2x  2 x 2   3x 2  2x  8  Ax  B x  1  C x 2  2  A  C x 2  A  B x  B  2C, which leads to the system   2   C3 Coefficients of x 2 A  C  3 Coefficients of x   A   A  B  2 Coefficients of x  B C1 Coefficients of x       B  2C  8 Constant terms 2C  6 Constant terms Hence, 2C  6  C  3, B  3  1  B  2; and A  3  3  A  0. Therefore, 3x 2  2x  8 2 3 .  2  x 3  x 2  2x  2 x 2 x 1

Ax  B Cx  D 2x 3  7x  5    2 . Thus, 39.  2 2 2 x x 2 x 1 x x 2 x 1     2x 3  7x  5  Ax  B x 2  1  C x  D x 2  x  2

 Ax 3  Ax  Bx 2  B  C x 3  C x 2  2C x  Dx 2  Dx  2D

 A  C x 3  B  C  D x 2  A  2C  D x  B  2D

We must solve the system  A C 2      B C  D0  A  2C  D  7     B  2D  5

Coefficients of x 3 Coefficients of x 2 Coefficients of x Constant terms

  A C  2     BC  D 0 0   C  D 5 C D 5     2D  10 C  D  5

  A C      BC D     

2

Hence, 2D  10  D  5, C  5  5  C  0, B  0  5  0  B  5, and A  0  2  A  2. Therefore, 

40.

2x  5 5 2x 3  7x  5    . x2  x  2 x2  1 x2  x  2 x2  1

x2  x  1 Ax  B Cx  D x2  x  1      2 . Thus, 2x 4  3x 2  1 2x 2  1 x 1 2x 2  1 x 2  1     x 2  x  1  Ax  B x 2  1  C x  D 2x 2  1  Ax 3  Ax  Bx 2  B  2C x 3  2Dx 2  C x  D  A  2C x 3  B  2D x 2  A  C x  B  D

which leads to the system   A  2C 0     B  2D  1   C 1 A    B  D1

  A  2C  0     2 B  2D  1 Coefficients of x   C  1 Coefficients of x     D 0 Constant terms Coefficients of x 3

Hence, D  0, C  1, B  0  1  B  1, and A  2  0  A  2. Therefore,

x2  x  1 2x  1 x  2  2 . 4 2 2x  3x  1 2x  1 x  1


24

41.

CHAPTER 9 Systems of Equations and Inequalities

Bx  C x4  x3  x2  x  1 A Dx  E   2   2 2 . Hence, x 2 x  1 x2  1 x x 1

 2   x 4  x 3  x 2  x  1  A x 2  1  Bx  C x x 2  1  x Dx  E       A x 4  2x 2  1  Bx 2  C x x 2  1  Dx 2  E x    A x 4  2x 2  1  Bx 4  Bx 2  C x 3  C x  Dx 2  E x  A  B x 4  C x 3  2A  B  D x 2  C  E x  A

So A  1, 1  B  1  B  0; C  1; 2  0  D  1  D  1; and 1  E  1  E  2. Therefore, x4  x3  x2  x  1 1 1 x 2   2   2 2 . x x 1 x2  1 x x2  1

42.

Cx  D Ax  B 2x 2  x  8   2  2 2 . Thus, 2 x  4 x 4 x2  4   2x 2  x  8  Ax  B x 2  4  C x  D  Ax 3  4Ax  Bx 2  4B  C x  D  Ax 3  Bx 2  4A  C x  4B  D

2x 2  x  8 2 1 and so A  0, B  2, 0  C  1  C  1, and 8  D  8  D  0. Therefore,   2  2 2 . x 4 x2  4 x2  4

43. We must first get a proper rational function. Using long division, we find that 2 2x 2  x  5 x 5  2x 4  x 3  x  5 2 2  2x  x  5   x 2  A  Bx  C . Hence,  x  x x 2 x 3  2x 2  x  2 x 3  2x 2  x  2 x2  1 x  2 x 2  1   2x 2  x  5  A x 2  1  Bx  C x  2  Ax 2  A  Bx 2  C x  2Bx  2C

 A  B x 2  C  2B x  A  2C

Equating coefficients, we get the system       2 Coefficients of x 2  2  2    A  B A  B A  B 2B  C  1 Coefficients of x  2B  C  1  2B  C  1       A    2C  5 Constant terms B  2C  3 5C  5

Therefore, 5C  5  C  1, 2B  1  1  B  1, and A  1  2  A  3, so x 1 3 x 5  2x 4  x 3  x  5  2  x2  . 3 2 x 2 x 1 x  2x  x  2


This is the solution to the wrong problem (it was changed in CA-8 and AT-5). The correct solution is given in CA-8 Exercise #5.3.44. [Note that in PMFC-8 the corresponding problem was not changed and the solution is correct.] 44.

SECTION 9.3 Partial Fractions

25

x 5  3x 4  3x 3  4x 2  4x  12 x 5  3x 4  3x 3  4x 2  4x  12 . We use long division to get a proper rational    x 4  4x 3  6x 2  8x  8 x  22 x 2  2

function:

x 4  4x 3  6x 2  8x  8

x 

1

x 5  3x 4  3x 3  4x 2  4x  12

x 5  4x 4  6x 3  8x 2  8x

x 4  3x 3  4x 2  4x  12

x 4  4x 3  6x 2  8x  8 x 3  2x 2  4x  4

B x 5  3x 4  3x 3  4x 2  4x  12 Cx  D x 3  2x 2  4x  4 A  , so  2  x  1       x 1 2 2 2 2 2 x  2 x 2 x  2 x  2 x  2 x  2 x  2     x 3  2x 2  4x  4  A x  2 x 2  2  B x 2  2  C x  D x  22

Thus,

       A x 3  2x 2  2x  4  B x 2  2  C x  D x 2  4x  4

 Ax 3  2Ax 2  2Ax  4A  Bx 2  2B  C x 3  4C x 2  4C x  Dx 2  4Dx  4D

 A  C x 3  2A  B  4C  D x 2  2A  4C  4D x  4A  2B  4D   A  C  1 Coefficients of x 3     2A  B  4C  D  2 Coefficients of x 2  which leads to the system  2A  4C  4D  4 Coefficients of x     4A  2B  4D  4 Constant terms     A  C  1 A  C 1         B  2C  D  0 2  Eq. 1  Eq. 2 B  2C  D  0     2C  4D  2 B  3D  2 Eq. 2  Eq. 3 Eq. 2  Eq. 3         2B  8C  4D  12 2  Eq. 3  Eq. 4 8C  2D  8 2  Eq. 3  Eq. 4   A  C 1     B  2C  D  0 Then D  0; 2C  2  C  1; B  2  0  B  2; and A  1  1   2C  4D  2     18D  0 4  Eq. 3  Eq. 4 A  0. Therefore,

45.

x x 5  3x 4  3x 3  4x 2  4x  12 2  2 .  x 1   2 2 2 x 2 x  2 x  2 x  2

A B ax  b   . Hence, ax  b  A x  1  B x  1  A  B x  A  B. x 1 x 1 x2  1  AB a ab . So Adding, we get 2A  a  b  A  2 AB b Substituting, we get B  a  A 

ab ab ab ab 2a   . Therefore, A  and B  . 2 2 2 2 2


26

CHAPTER 9 Systems of Equations and Inequalities

ax 3  bx 2 Ax  B Cx  D 46.   2  2 2 . Hence, 2 x  1 x 1 x2  1   ax 3  bx 2  Ax  B x 2  1  C x  D  Ax 3  Ax  Bx 2  B  C x  D  Ax 3  Bx 2  A  C x  B  D

and so A  a, B  b, a  C  0  C  a, and b  D  0  D  b. Therefore, A  a, B  b, C  a, and D  b.

1 x is already a partial fraction decomposition. The denominator in the first term is a  47. (a) The expression 2 x 1 x 1 quadratic which cannot be factored and the degree of the numerator is less than 2. The denominator of the second term is linear and the numerator is a constant. x (b) The term can be decomposed further, since the numerator and denominator both have linear factors. x  12 B A x   . Hence, x  A x  1  B  Ax  A  B. So A  1, B  1, and 2 x 1 x  1 x  12 1 1 x   . x  1 x  12 x  12 (c) The expression

1 2  is already a partial fraction decomposition, since each numerator is constant. x  1 x  12

x 2 2 is already a partial fraction decomposition, since the denominator is the square of a quadratic x2  1 which cannot be factored, and the degree of the numerator is less than 2.

(d) The expression 

48. Combining the terms, we have     1 x 2  2x  1 2 x2  1 2 1 x  1 1 1      x  1 x  12 x 1 x  12 x  1 x  12 x  1 x  12 x  1 

2x 2  2  x  1  x 2  2x  1 x  12 x  1

3x 2  x

x  12 x  1

3x 2  x

A B C   , and so  x  1 x  12 x 1 x  12 x  1     3x 2  x  A x  1 x  1  B x  1  C x  12  A x 2  1  B x  1  C x 2  2x  1

Now to find the partial fraction decomposition, we have

 Ax 2  A  Bx  B  C x 2  2C x  C  A  C x 2  B  2C x  A  B  C   2    C 3 A  C  3 Coefficients of x   A  which result in the system  B  2C  1 B  2C  1 Coefficients of x        A  B  C  0 B  2C  3 Eq. 1  Eq. 3 Constant terms

   A   

 C

3

4C 

4

B  2C  1

so 4C  4  C  1, B  2 1  1  B  1, and A  1  3  A  2. 1  Eq. 2  Eq. 3

Therefore, we get back the same expression:

3x 2  x

x  12 x  1

1 1 2  .  x  1 x  12 x 1


SECTION 9.4 Systems of Nonlinear Equations

9.4

27

SYSTEMS OF NONLINEAR EQUATIONS

1. The solutions of the system are the points of intersection of the two graphs, namely 2 2 and 4 8. 2. For 2 2: 2y  x 2  2 2  22  0 and y  x  2  2  4, so 2 2 is a solution.

For 4 8: 2y  x 2  2 8  42  0 and y  x  8  4  4, so 4 8 is a solution.  y  x2 Substituting y  x 2 into the second equation gives x 2  x  12  3. y  x  12

0  x 2  x  12  x  4 x  3  x  4 or x  3. So since y  x 2 , the solutions are 3 9 and 4 16.   x 2  y 2  25 Substituting for y in the first equation gives x 2  2x2  25  5x 2  25  x 2  5  x   5 4. y  2x           5 2 5 and When x  5 then y  2 5, and when x   5 then y  2  5  2 5. Thus the solutions are      5 2 5 .

5.

x 2  y2  8 x  y0

Solving the second equation for y gives y  x, and substituting this into the first equation gives

x 2  x2  8  2x 2  8  x  2. So since y  x, the solutions are 2 2 and 2 2.  x2  y  9 Solving the first equation for y, we get y  9  x 2 . Substituting this into the second equation gives 6. x  y 3  0   x  9  x 2  3  0  x 2  x  6  0  x  3 x  2  0  x  3 or x  2. If x  3, then y  9  32  0,

and if x  2, then y  9  22  5. Thus the solutions are 3 0 and 2 5.  x  y2  0 7. Solving the first equation for x gives x  y 2 , and substituting this into the second equation gives 2x  5y 2  75   2 y 2  5y 2  75  3y 2  75  y 2  25  y  5. So since x  y 2 , the solutions are 25 5 and 25 5. 8.

x2  y  1

Solving the first equation for y, we get y  x 2  1. Substituting this into the second equation gives 2x 2  3y  17   2x 2  3 x 2  1  17  2x 2  3x 2  3  17  5x 2  20  x 2  4  x  2. If x  2, then y  22  1  3,

and if x  2, then y  22  1  3. Thus the solutions are 2 3 and 2 3.  x 2  2y  1 9. Subtracting the first equation from the second equation gives 7y  28  y  4. Substituting y  4 into x 2  5y  29

the first equation of the original system gives x 2  2 4  1  x 2  9  x  3. The solutions are 3 4 and 3 4.  3x 2  4y  17 10. Multiplying the first equation by 2 and the second by 3 gives the system 2x 2  5y  2  6x 2  8y  34 Adding we get 7y  28  y  4. Substituting this value into the second equation gives 6x 2  15y  6       2x 2  5 4  2  2x 2  22  x 2  11  x   11. Thus the solutions are 11 4 and  11 4 .


28

11.

CHAPTER 9 Systems of Equations and Inequalities

3x 2  y 2  11

Multiplying the first equation by 4 gives the system

x 2  4y 2  8

12x 2  4y 2  44 x 2  4y 2  8

Adding the equations

gives 13x 2  52  x  2. Substituting into the first equation we get 3 4  y 2  11  y  1. Thus, the solutions are 2 1, 2 1, 2 1, and 2 1.

12.

2x 2  4y  13

Multiplying the second equation by 2 gives the system

x 2  y 2  72

2x 2  4y  13

2x 2  2y 2  7

Subtracting the

equations gives 4y  2y 2  6  y 2  2y  3  0  y  3 y  1  0  y  3, y  1. If y  3, then   5 2 2x 2  4 3  13  x 2  25 . If y  1, then 2x 2  4 1  13  x 2  92  x   3 2 2 . Hence, the  x   2 2       5 2 3 2 solutions are  2  3 and  2  1 . 13.

x  y2  3  0

14.

x 2  y2  1

Adding the two equations gives 2x 2  x  1  0. Using the Quadratic Formula we have 2x 2  y 2  4  0   1  1  4 2 1 1  9 1  3 1  3 1  3 x   . So x   1 or x   12 . Substituting x  1 2 2 4 4 4 4  into the first equation gives 1  y 2  3  0  y 2  2  y   2. Substituting x  12 into the first equation gives        1  y 2  3  0  y 2  7  y   7 . Thus the solutions are 1  2 and 1   7 . 2 2 2 2 2

2x 2  y 2  x  3

Subtracting the first equation from the second equation gives x 2  x  2 

x 2  x  2  0  x  2 x  1  0  x  2, x  1. Solving the first equation for y 2 we have y 2  x 2  1. When  x  1, y 2  12  1  0 so y  0 and when x  2, y 2  22  1  3 so y   3. Thus, the solutions are 1 0,       2  3 , and 2 3 . 15.

x2  y 

8

x  2y  6

By inspection of the graph, it appears that 2 4 is a solution, but is difficult to get accurate values

for the other point. Multiplying the first equation by 2 gives the system

2x 2  2y  16 x  2y  6

Adding the equations gives

2x 2  x  10  2x 2  x  10  0  2x  5 x  2  0. So x   52 or x  2. If x   52 , then  52  2y  6    2y   72  y  74 , and if x  2, then 2  2y  6  2y  8  y  4. Hence, the solutions are  52  74 and

2 4.

16.

x  y 2  4

x  y

2

By inspection of the graph, it appears that 0 2 and 5 3 are solutions to the system. We check

each point in both equations to verify that it is a solution. For 0 2: 0  22  4 and 0  2  2. For 5 3: 5  32  5  9  4 and 5  3  2. Thus, the solutions are 0 2 and 5 3.


SECTION 9.4 Systems of Nonlinear Equations

17.

x2 

y0

x 3  2x  y  0

29

By inspection of the graph, it appears that 2 4, 0 0, and 1 1 are solutions to the system.

We check each point in both equations to verify that it is a solution. For 2 4: 22  4  4  4  0 and 23  2 2  4  8  4  4  0.

For 0 0: 02  0  0 and 03  2 0  0  0.

For 1 1: 12  1  1  1  0 and 13  2 1  1  1  2  1  0. Thus, the solutions are 2 4, 0 0, and 1 1.  x 2  y 2  4x 18. By inspection of the graph, it appears that 0 0 is a solution, but is difficult to get accurate values for x  y2

the other points. Substituting for y 2 we have x 2  x  4x  x 2  3x  0  x x  3  0. So x  0 or x  3. If x  0,        then y 2  0 so y  0. And is x  3 then y 2  3 so y   3. Hence, the solutions are 0 0, 3  3 , and 3 3 .  y  x 2  4x Subtracting the second equation from the first equation gives x 2  4x  4x  16  19. y  4x  16

x 2  8x  16  0  x  42  0  x  4. Substituting this value for x into either of the original equations gives y  0. Therefore, the solution is 4 0.   x  y2  0 x  y2 20. Substituting for y in the Solving the first equation for x and the second equation for y gives 2 yx  0 y  x2   first equation gives x  x 4  x x 3  1  0  x  0, x  1. Thus, the solutions are 0 0 and 1 1.  x  2y  2 21. Now x  2y  2  x  2y  2. Substituting for x gives y 2  x 2  2x  4  y 2  x 2  2x  4

y 2  2y  22  2 2y  2  4  y 2  4y 2  8y  4  4y  4  4  y 2  4y  4  0  y  22  0  y  2. Since x  2y  2, we have x  2 2  2  2. Thus, the solution is 2 2.  y  4  x2 22. Setting the two equations equal, we get 4  x 2  x 2  4  2x 2  8  x  2. Therefore, the solutions y  x2  4 are 2 0 and 2 0.  xy 4 23. Now x  y  4  x  4  y. Substituting for x gives x y  12  4  y y  12  y 2  4y  12  0 x y  12  y  6 y  2  0  y  6, y  2. Since x  4  y, the solutions are 2 6 and 6 2.  x y  24 24 Since x  0 is not a solution, from the first equation we get y  24. . Substituting into the 2 2 x 2x  y  4  0  2    24  2x 4  4x 2  576  x 4  2x 2  288  0  x 2  18 x 2  16  0. second equation, we get 2x 2  4  x

Since x 2  18 cannot be 0 if x is real, we have x 2  16  0  x  4. When x  4, we have y  24 4  6 and when 24  6. Thus the solutions are 4 6 and 4 6. x  4, we have y  4  x 2 y  16 16 16 . Substituting for x 2 gives  4y  16  0  4y 2  16y  16  0 25. Now x 2 y  16  x 2  y y x 2  4y  16  0  y 2  4y  4  0  y  22  0  y  2. Therefore, x 2  has no solution.

16  8, which has no real solution, and so the system 2


30

26.

27.

CHAPTER 9 Systems of Equations and Inequalities

x

x 2  y2  9

y 0  Solving the first equation for x, we get x   y. Substituting for x gives y 2  4x 2  12   2  y 2  4  y  12  y 2  4y  12  0  y  6 y  2  0  y  6, y  2. Since x   2 is not a real    solution, the only solution is  6 6 . x 2  y2  1

  Adding the equations gives 2x 2  10  x 2  5  x   5. Now x   5  y 2  9  5  4 

y  2, and so the solutions are

28.

x 2  2y 2  2

          5 2 , 5 2 ,  5 2 , and  5 2 .

Multiplying the first equation by 2 gives the system

2x 2  3y  15

equations gives 4y 2  3y  11  4y 2  3y  11  0  y 

3 

2x 2  4y 2  4

2x 2  3y  15

Subtracting the two

 9  4 4 11 which is not a real number. 2 4

Therefore, there are no real solutions.

29.

2x 2  8y 3  19

4x 2  16y 3  34

Multiplying the first equation by 2 gives the system

4x 2  16y 3  38 4x 2  16y 3  34

Adding the two

equations gives 8x 2  72  x  3, and then substituting into the first equation we have 2 9  8y 3  19      y 3   18  y   12 . Therefore, the solutions are 3  12 and 3  12 . 30.

x 4  y 3  17

3x 4  5y 3  53

Multiplying the first equation by 3 gives the system

3x 4  3y 3  51 3x 4  5y 3  53

Subtracting the equations

gives 2y 3  2  y 3  1  y  1, and then x 4  1  15  x  2. Therefore, the solutions are 2 1 and 2 1.   

2 3  1 x y 31. 4 7     1 x y

1 1 If we let u  and   ,the system is equivalent to x y

equation by 4 gives the system

4u  6  2

4u  7  1

2u  3  1

4u  7  1

Multiplying the first

Adding the equations gives   3, and then substituting into the first

  equation gives 2u  9  1  u  5. Thus, the solution is 15  13 .

 4 6 7    2  4  2 x y 32. 2 1    2  4 0 x y

  4u  6  7 1 1 2 , and multiplying the If we let u  2 and   4 , the system is equivalent to  u  2  0 x y

second equation by 3, gives

  4u  6  7 2

 3u  6  0

Adding the equations gives 7u  72  u  12 , and   14 . Therefore,

            1 1 2 2 , 2  2 ,  2 2 , x 2   2  x   2, and y 4   4  y   2. Thus, the solutions are   u   and  2  2 .


SECTION 9.4 Systems of Nonlinear Equations

33.

y  x 2  8x

34.

y  2x  16

The solutions are 8 0 and 2 20.

y  x 2  4x

2x  y  2

y  x 2  4x

y  2x  2

The solutions are approximately 035 130 and 565 930.

20

10

-10

10 -20

35.

x 2  y 2  25

x  3y  2

5

 y   25  x 2

36.

y   13 x  23

The solutions are 451 217 and 491 097.

x 2  y 2  17

x 2  2x  y 2  13

 y   17  x 2  y   13  2x  x 2

The solutions are approximately 2 361.

5

5

-5

5

-5

-5

5 -5

   2  y2 x y   18  2x 2  1 37.  9 18  y  x 2  6x  2  y  x 2  6x  2

38.

x 2  y2  3

y  x 2  2x  8

 y   x2  3

y  x 2  2x  8

The solutions are approximately 222 140,

The solutions are 123 387 and 035 421.

188 072, 345 299, and 465 431.

5

5

-5

5 -5

5

-5 -5

39.

  

 4 32  x 4 y  2  x 2  2x  y  0  y  x 2  2x x 4  16y 4  32

The solutions are 230 070 and 048 119.

40.

y  e x  ex y  5  x2

119 359 and 119 359. 5

2

-2

The solution are approximately

2 -2

-2

2

31


32

CHAPTER 9 Systems of Equations and Inequalities

  log x  log y  3 2 41.  2 log x  log y  0

Adding the two equations gives 3 log x  32  log x  12  x 

 10. Substituting into the

second equation we get 2 log 1012  log y  0  log 10  log y  0  log y  1  y  10. Thus, the solution is   10 10 .

42.

2x  2 y  10

4x  4 y  68

2x  2 y  10

22x  22y  68

If we let u  2x and   2 y , the system becomes

u    10

u 2   2  68

Solving the first equation for u, and substituting this into the second equation gives u    10  u  10  , so 10  2   2  68  100  20   2   2  68   2  10  16  0    8   2  0    2 or   8. If   2, then u  8, and so y  1 and x  3. If   8, then u  2, and so y  3 and x  1. Thus, the solutions are 1 3 and 3 1.  xy 3 43. Solving the first equation for x gives x  3  y and using the hint, x 3  y 3  387  3 x  y 3  387     x  y x 2  x y  y 2  387. Next, substituting for x, we get 3 3  y2  y 3  y  y 2  387  9  6y  y 2  3y  y 2  y 2  129  3y 2  9y  9  129  y  8 y  5  0  y  8 or y  5. If y  8, then

x  3  8  5, and if y  5, then x  3  5  8. Thus the solutions are 5 8 and 8 5.  x2  xy  1 44. Adding the equations gives x 2 x yx y y 2  4  x 2 2x y y 2  4  x  y2  4  x  y  2. x y  y2  3 If x  y  2, then from the first equation we get x x  y  1  x  2  1  x  12 , and so y  2  12  32 . If   x  y  2, then from the first equation we get x x  y  1  x  2  1  x   12 , and so y  2   12   32 .     Thus the solutions are 12  32 and  12   32 . 45. Let  and l be the lengths of the sides, in cm. Then we have the system

l  180

2l  2  54

We solve the second equation

for  giving,   27  l, and substitute into the first equation to get l 27  l  180  l 2  27l  180  0 

l  15 l  12  0  l  15 or l  12. If l  15, then   27  15  12, and if l  12, then   27  12  15. Therefore, the dimensions of the rectangle are 12 cm by 15 cm.

46. Let b be the length of the base of the triangle, in feet, and h be the height of the triangle,  1 bh  84  168 2 . By substitution, The first equation gives b  in feet. Then  b2  h 2  252  625 h 

    168 2  h 2  625  h 4  625h 2  1682  0  h 2  49 h 2  576  0  h  7 or h  24. Thus, the lengths of h

the other two sides are 7 ft and 24 ft.

47. Let l and  be the length and width, respectively, of the rectangle. Then, the system of equations is  2l  2  70  Solving the first equation for l, we have l  35  , and substituting into the second gives l 2  2  25  l 2  2  25  l 2  2  625  35  2  2  625  1225  70  2  2  625  22  70  600  0    15   20  0    15 or   20. So the dimensions of the rectangle are 15 and 20.


SECTION 9.4 Systems of Nonlinear Equations

33

48. Let  be the width and l be the length of the rectangle, in inches. From the figure, the diagonals of the rectangle are  l  160 1602 160 simply diameters of the circle. Then, . By substitution, 2  l 2  400    2 2 2 l l   l  20  400        l 4  400l 2  1602  0  l 2  80 l 2  320  0  l  80  4 5 or l  320  8 5. Therefore, the dimensions   of the rectangle are 4 5 in. and 8 5 in..

49. At the points where the rocket path and the hillside meet, we have

  y  1x 2

 y  x 2  401x

Substituting for y in the second

  801  0  x  0, x  801 . When x  0, the rocket has x  0  x x  equation gives 12 x  x 2  401x  x 2  801 2 2 2     801 1 801 801 801 not left the pad. When x  2 , then y  2 2  4 . So the rocket lands at the point 801 2  4 . The distance from    

the base of the hill is

801 2  2

801 2  44777 meters. 4

50. Let x be the circumference and y be length of the stove pipe. Using the circumference we can determine  x 2 1 2 x . Thus the volume is  x y. So the system is given by y  the radius, 2r  x  r  2 2 4   x y  1200 1 1 1 2 x y x x y  x 1200  600  Substituting for x y in the second equation gives 1  4 4 4 x 2 y  600 4 1200 600 600 1200   . Thus the dimensions of the sheet metal are 2  63 in and  1910 in. x  2. So y  x 2  


34

CHAPTER 9 Systems of Equations and Inequalities

51. The point P is at an intersection of the circle of radius 26 centered at A 22 32

y

40 and the circle of radius 20 centered at B 28 20. We have the system A  B x  222  y  322  262 20  2 2 2 x  28  y  20  20  0 _20 20 40 x x 2  44x  484  y 2  64y  1024  676  2 2 x  56x  784  y  40y  400  400 _20  x 2  44x  y 2  64y  832 Subtracting the two equations, we get 12x  24y  48  x  2y  4, x 2  56x  y 2  40y  784

which is the equation of a line. Solving for x, we have x  2y  4. Substituting into the first equation gives

2y  42 44 2y  4 y 2 64y  832  4y 2 16y1688y176 y 2 64y  832  5y 2 168y192  832   2 451024  5y 2  168y  1024  0. Using the Quadratic Formula, we have y  168 16825  16810 7744  16888 10

 y  8 or y  2560. Since the y-coordinate of the point P must be less than that of point A, we have y  8. Then x  2 8  4  12. So the coordinates of P are 12 8.

To solve graphically, we must solve each equation for y. This gives x  222  y  322  262    y  322  262  x  222  y  32   676  x  222  y  32  676  x  222 . We use the function  y  32  676  x  222 because the intersection we at interested in is below the point A. Likewise, solving the second  equation for y, we would get the function y  20  400  x  282 . In a three-dimensional situation, you would need a minimum of three satellites, since a point on the earth can be uniquely specified as the intersection of three spheres centered at the satellites. 52. The graphs of y  x 2 and y  x  k for various values of k are shown. If we solve  y  x2 we get x 2  x  k  0. Using the Quadratic Formula, the system y  x k   1  1  4k . So there is no solution if 1  4k is undefined, that we have x  2 is, if 1  4k  0  k   14 . There is exactly one solution if 1  4k  0  k   14 , and there are two solutions if 1  4k  0  k   14 .

y k=2 1

k=_ 4 1

1

x

k=_3


SECTION 9.5 Systems of Inequalities

9.5

35

SYSTEMS OF INEQUALITIES

1. If the point 2 3 is a solution of an inequality in x and y, then the inequality is satisfied when we replace x by 2 and y by 3. Because 4 2  2 3  8  6  2  1, the point 2 3 is a solution of the inequality 4x  2y  1.

2. To graph an inequality we first graph the corresponding equation. So to

y

graph y  x  1, we first graph the equation y  x  1. To decide which

side is the graph of the inequality we use test points. Test Point

Inequality y  x  1 ?

0 0

Part of graph

201

Not part of graph

x

1

y=x+1

001X ?

0 2

1

Conclusion

3. If the point 2 3 is a solution of a system of inequalities in x and y, then each inequality is satisfied when we replace x by 2 and y by 3. Because 2 2  4 3  16  17 and 6 2  5 3  27  29, the point 2 3 is a solution of the given system.

4. (a)

xy0 xy2

(b)

xy0 xy2

y

(c)

y

1 x

(d)

xy0 xy2 y

x-y=0

x+y=2 1

xy0 xy2 y

x-y=0

1

x-y=0

1

x+y=2 1

5.

x

x-y=0

1

x+y=2 1

x

x+y=2 1

6. Test Point Inequality x  5y  3 ?

Conclusion

1 2 1  5 2  3 X

Solution

1 2

1  5 2  3 X

Solution

1 2

1  5 2  3 

Not a solution

8  5 1  3 

Not a solution

8 1

?

? ?

Test Point Inequality 3x  2y  2 2 1 1 3

?

3 2  2 1  2 X ?

3 1  2 3  2  ?

Conclusion Solution Not a solution

1 3

3 1  2 3  2 X

Solution

0 1

3 0  2 1  2 X

Solution

?

x


36

CHAPTER 9 Systems of Equations and Inequalities

7. Test Point System

8. 3x  2y  5 2x  y  3

 ?  3 0  2 0  5X

0 0

?  2 0  0  3  ?  3 1  2 2  5X

1 2

?  2 1  2  3X  ?  3 1  2 1  5 X

1 1

?  2 1  1  3X  ?  3 3  2 1  5 

3 1

?

2 3  1  3 X

9. y  2x. The test point 1 0 satisfies the

Conclusion

Test Point System

Not a solution

0 0

Solution

1 3

Solution

3 0

Not a solution

1 2

10. y  3x. The test point

 

x  2y  4

4x  3y  11 ?

0  2 0  4 

?  4 0  3 0  11   ?  1  2 3  4 X ?  4 1  3 3  11 X  ?  3  2 0  4  ?  4 3  3 0  11   ?  1  2 2  4 X ?  4 1  3 2  11 

11. y  2. The test point

1 0 satisfies the

Conclusion Not a solution

Solution

Not a solution

Not a solution

12. x  1. The test point

0 3 satisfies the

2 0 satisfies the

inequality.

inequality.

inequality.

inequality.

y

y

y

y

y=3x 1

1

x

1

x

1

1

x=_1

y=2 1

1

x

1

x

y=_2x

13. x  2. The test point

14. y  1. The test point

0 0 satisfies the

0 2 satisfies the

inequality.

inequality.

inequality.

inequality.

y

y

y

y

x=2

1 1

x

1

15. y  x  3. The test

point 0 0 satisfies the

point 0 0 satisfies the

y=x-3

y=1 1

16. y  1  x. The test

1 x

1

x

1 1

x y=1-x


SECTION 9.5 Systems of Inequalities

17. 2x  y  4. The test

point 0 0 satisfies the

point 0 0 satisfies the

inequality.

1

inequality. y

y

2 2

x

10

x

_x@+y=5

3x-y-9=0

21. x 2  y 2  9. The test point 0 4 satisfies the inequality.

y=x@+1

1

x

1

y

point 0 2 satisfies the

inequality. y

1

20. y  x 2  1. The test

point 0 6 satisfies the

inequality.

y

2x-y=_4

19. x 2  y  5. The test

18. 3x  y  9  0. The test

37

1

x

22. x 2  y  22  4. The test point 0 1 satisfies the inequality.

y

x@+y@=9

x@+(y-2)@=4

1 x

1

1 x

1

23. 3x  2y  18

24. 4x  3y  9

4 -2 0

2

4

6

8

8 4

10 12

-4

-4

-8

-2 0

-12

25. 5x  2y  8

2

4

6

8

-4

26. 5x  3y  15

8

4

4 0 -2

0 -4

2

4

-8

6

5

-4 -8

27. The boundary is a solid curve, so we have the inequality y  12 x  1. We take the test point 0 2 and verify that it satisfies the inequality: 2  12 0  1.

28. The boundary is a solid curve, so we have the inequality y  x 2  2. We take the test point 0 0 and verify that it satisfies the inequality: 0  02  2.

29. The boundary is a broken curve, so we have the inequality x 2  y 2  4. We take the test point 0 4 and verify that it satisfies the inequality: 02  42  4.

30. The boundary is a solid curve, so we have the inequality y  x 3  4x. We take the test point 1 1 and verify that it satisfies the inequality: 1  13  4 1.


38

31.

CHAPTER 9 Systems of Equations and Inequalities

xy4 yx

The vertices occur where

xy4 yx

y

Substituting, we have

y=x 1

2x  4  x  2. Since y  x, the vertex is 2 2, and the solution set is not

32.

2x  3y  12

2x  3y  12

3x  y  21

The vertices occur where

x

1

bounded. The test point 0 1 satisfies each inequality.

x+y=4

2x  3y  12 3x  y  21

y

2x+3y=12 1

and adding the two equations gives 11x  75  x  75 11 .

1

9x  3y  63   132150   18  y   6 . Therefore, the Then 2 75 11  3y  12  3y  11 11 11   6 vertex is 75 11   11 , and the solution set is not bounded. The test point 0 5

x

3x-y=21

satisfies each inequality.

33.

  y  1x  2 4

 y  2x  5

The vertex occurs where

  y  1x  2

y

4

Substituting for y

 y  2x  5

y=41 x+2

gives 14 x  2  2x  5  74 x  7  x  4, so y  3. Hence, the vertex is 4 3,

1

34.

xy0

4  y  2x

The vertices occur where

y x

4  y  2x

y

set is not bounded. The test point 3 0 satisfies each inequality.

One vertex occurs where

y  2x  8

y   12 x  5

4+y=2x 2 x-y=0

x

2

y

Substituting for

y gives 2x  8   12 x  5   32 x  3  x  2, so y  2 2  8  4.  y  2x  8 Hence, this vertex is 2 4. Another vertex occurs where  y0 2x  8  0  x  4; this vertex is 4 0. Another occurs where  y   12 x  5  y  5; this gives the vertex 0 5. The origin is another x 0 vertex, and the solution set is bounded. The test point 1 1 satisfies each inequality.

y=2x-5

Substituting for y

gives 4  x  2x  x  4, so y  4. Hence, the vertex is 4 4, and the solution

    y  2x  8 35. y   12 x  5    x  0, y  0

x

1

and the solution is not bounded. The test point 0 1 satisfies each inequality.

y=_2x+8 1

y=_ 2 x+5 x

1 1


SECTION 9.5 Systems of Inequalities

    4x  3y  18 36. 2x  y  8    x  0, y  0

One vertex occurs where

4x  3y  18 2x  y  8

y

Subtracting twice

2x+y=8

the second equation from the first gives y  2, so 2x  2  8  x  3 and the vertex is 3 2. Other vertices occur at the x-intercept of 2x  y  8 and the

4x+3y=18

1

x

1

y-intercept of 4x  3y  18; these are 4 0 and 0 6, respectively. The origin is

another vertex, and the solution set is bounded. The test point 1 1 satisfies each inequality.

37.

     

x 0

y 0  3x  5y  15     3x  2y  9

From the graph, the points 3 0, 0 3 and 0 0 are vertices,

y 3x+2y=9

and the fourth vertex occurs where the lines 3x  5y  15 and 3x  2y  9

intersect. Subtracting these two equations gives 3y  6  y  2, and so x  53 .   Thus, the fourth vertex is 53  2 , and the solution set is bounded. The test point

3x+5y=15

1

x

1

1 1 satisfies each inequality.

38.

   

y

x 2 y  12

   2x  4y  8

From the graph, the vertices occur at 2 1 and 28 12.

39.

y  9  x2

x  0, y  0

y=12

6

2x-4y=8 x

6

The solution set is not bounded. The test point 6 0 satisfies each inequality.

x=2

y

From the graph, the vertices occur at 0 0, 3 0, and 0 9.

y=9-x@

The solution set is bounded. The test point 1 1 satisfies each inequality. y=0

1 x

1 x=0

 2   y  x 40. y  4   x  0

From the graph, the vertices occur at 0 0, 0 2, and 2 4. The

y x=0 y=4

solution set is bounded. The test point 1 3 satisfies each inequality. y=x@

1 1

x

39


40

41.

CHAPTER 9 Systems of Equations and Inequalities

y  9  x2 y  x 3

The vertices occur where

y  9  x2

y  x 3

Substituting for y

y=9-x@

y y=x+3

gives 9  x 2  x  3  x 2  x  6  0  x  2 x  3  0  x  3, x  2. Therefore, the vertices are 3 0 and 2 5, and the solution set is

1

42.

y

y  x2

xy6

The vertices occur where

y  x2

xy6

x

1

bounded. The test point 0 4 satisfies each inequality.

Substituting for y y=x@

gives x 2  x  6  x 2  x  6  0  x  3 x  2  0  x  3, x  2. Since y  x 2 , the vertices are 3 9 and 2 4, and the solution set is not

43.

x 2  y2  4 xy0

The vertices occur where

x 2  y2  4 xy0

x+y=6

2

bounded. The test point 0 7 satisfies each inequality.

1 y

x-y=0

Since x  y  0 1

  x  y, substituting for x gives y 2  y 2  4  y 2  2  y   2, and         x   2. Therefore, the vertices are  2  2 and 2 2 , and the

44.

x@+y@=4

y

x 0 y 0

 x  y  10     2 x  y2  9

x

1

solution set is bounded. The test point 1 0 satisfies each inequality.      

x

From the graph, the vertices are 0 3, 0 10, 3 0, and

10 0. The solution set is bounded. The test point 4 1 satisfies each inequality.

x+y=10

x=0

1

x

y=0

1

x@+y@=9

45.

 

x2  y  0

2x 2  y  12 2x 2  2y  0

2x 2  y  12

The vertices occur where

x2  y  0

2x 2  y  12

y

x@-y=0

Subtracting the equations gives 3y  12  y  4, and

x  2. Thus, the vertices are 2 4 and 2 4, and the solution set is bounded.

46.

2x 2  y  4 x2  y  8

The vertices occur where

x

1

2x 2  y  4 x2  y  8

y

Adding the

equations gives 2x 2  x 2  12  x 2  4  x  2. Therefore, the vertices are 2 4 and 2 4. The solution set is not bounded. The test point 0 6 satisfies each inequality.

2x@+y=12

1

The test point 0 1 satisfies each inequality.

2 x@-y=8

2x@+y=4

(_2, _4)

1

x (2, _4)


41

SECTION 9.5 Systems of Inequalities

47.

x 2  y2  9

The vertices occur where

2x  y 2  1

x 2  y2  9

Subtracting the

2x  y 2  1

equations gives x 2  2x  8  x 2  2x  8  x  2 x  4  0. Therefore,

y

2x+y@=1

1

      the vertices are 2  5 and 2 5 . The solution set is bounded. The test

point 0 0 satisfies each inequality.

48.

x 2  y2  4

x 2  y2  4

Subtracting the

x 2  2y  1

equations gives y 2  2y  3  y 2  2y  3  y  3 y  1  0. However,      y  3 is extraneous, and the vertices are  3 1 and 3 1 . The solution

y x@-2y=1

(_Ï3, 1)

(Ï3, 1)

1

We find the vertices of the region by solving pairs of the

corresponding equations: 

3x  y  0 xy2

x  2y  14

x y 2

3x  y 

0

2x  y  2

x  2y  14

3y  12

x

1

set is bounded. The test point 0 2 satisfies each inequality.     x  2y  14 49. 3x  y  0    x y 2

x

1

(_2, _Ï5)

The vertices occur where

x 2  2y  1

x@+y@=9

(_2, Ï5)

x@+y@=4

x+2y=14

y

3x-y=0

 y  4 and x  6.

 x  1 and y  3. Therefore, the vertices

1 x

1 x-y=2

are 6 4 and 1 3, and the solution set is not bounded. The test point 0 4 satisfies each inequality.

50.

   

y  x 6

3x  2y  12

   x  2y  2

To find where the line y  x  6 intersects the lines

y 3x+2y=12

3x  2y  12 and x  2y  2, we substitute for y: 3x  2 x  6  12  x  0 and y  6; x  2 x  6  2  x  14 and y  8. Next, adding the equations 3x  2y  12 and x  2y  2 gives 4x  14  x  72 , so these lines   intersect at the point 72  34 . Since the vertex 14 8 is not part of the solution   set, the vertices are 0 6 and 72  34 , and the solution set is not bounded. The test

y=x+6

2 2

x

x-2y=2

point 4 2 satisfies each inequality.

51.

x  0, y  0

x  5, x  y  7

y

The points of intersection are 0 7, 0 0, 7 0, 5 2,

and 5 0. However, the point 7 0 is not in the solution set. Therefore, the

x=5 x=0

vertices are 0 7, 0 0, 5 0, and 5 2, and the solution set is bounded. The test point 1 1 satisfies each inequality.

x+y=7

1 y=0

1

x


42

52.

CHAPTER 9 Systems of Equations and Inequalities

x  0, y  0

y  4, 2x  y  8

y

2x+y=8

The points of intersection are 0 8, 0 4, 4 0, 2 4,

and 3 2. However, the point 0 8 is not in the solution set. Therefore, the

y=4 x=0 1

vertices are 0 4, 2 4, 4 0, and 0 0. The solution set is bounded. The test point 1 1 satisfies each inequality.

53.

   

y  x 1

1

We find the vertices of the region by solving pairs of the

x  2y  12    x 1 0

corresponding equations. Using x  1 and substituting for x in the line y  x  1 gives the point 1 0. Substituting for x in the line x  2y  12 gives    y  x 1 13 the point 1 2 .  x  y  1 and y  1  2y  12  x  2y  12   10 13 3y  13  y  13 3 and x  3 . So the vertices are 1 0, 1 2 , and   10  13 , and none of these vertices is in the solution set. The solution set is 3 3

y

x

y=0

x+2y=12

x+1=0 y=x+1

1

x

1

bounded. The test point 0 2 satisfies each inequality.

    x  y  12 54. y  12 x  6    3x  y  6

y

Graphing these inequalities, we see that there are no points 3x+y=6

that satisfy all three, and hence the solution set is empty.

x+y=12 2

y=21 x-6

55.

x 2  y2  8

x  2, y  0

   The intersection points are 2 2, 2 0, and 2 2 0 .

x@+y@=8

x

2 1 2

y

x=2

1

However, since 2 2 is not part of the solution set, the vertices are 2 2, 2 0,    and 2 2 0 . The solution set is bounded. The test point 21 1 satisfies each

y=0

1

x

inequality.

 2   x  y  0 56. xy6    xy6

Adding the equations x  y  6 and x  y  6 yields 2x  12

 x  6. So these curves intersect at the point 6 0. To find where x 2  y  0

and x  y  6 intersect, we solve the first for y, giving y  x 2 , and then substitute

x+y=6

y

x@-y=0

2 1

x-y=6

x

into the second equation to get x  x 2  6  x 2  x  6  0  x  3 x  2  0  x  3 or x  2. When x  3, we have

y  9, and when x  2, we have y  4, so the points of intersection are 3 9 and 2 4. Substituting y  x 2 into the

equation x  y  6 gives x  x 2  6, which has no solution. Thus the vertices (which are not in the solution set) are 6 0, 3 9, and 2 4. The solution set is not bounded. The test point 1 0 satisfies each inequality.


SECTION 9.5 Systems of Inequalities

57.

x 2  y2  9

x  y  0, x  0

Substituting x  0 into the equations x 2  y 2  9 and

43

x+y=0 y

1

x  y  0 gives the vertices 0 3 and 0 0. To find the points of intersection

x=0

for the equations x 2  y 2  9 and x  y  0, we solve for x  y and substitute

x 1  3 2 2 2 into the first equation. This gives y  y  9  y   2 . The points     x@+y@=9 0 3and 3 2 2   3 2 2 lie away from the solution set, so the vertices are 0 0,     0 3, and  3 2 2  3 2 2 . Note that the vertices are not solutions in this case. The solution set is bounded. The test point

0 1 satisfies each inequality.

58.

   

y  x3

y  2x  4   x  y  0

y

The curves y  x 3 and x  y  0 intersect when

y=2x+4

x 3  x  0  x  0  y  0. The lines x  y  0 and y  2x  4 intersect

y=x#

1

when x  2x  4  3x  4  x   43  y  43 . To find where y  x 3 and y  2x  4 intersect, we substitute for y and get x 3  2x  4 

x

1 x+y=0

  x 3  2x  4  0  x  2 x 2  2x  2  0  x  2 (the other factor has no

real solution). When x  2 we have y  8Thus, the vertices of the region are   0 0,  43  43 , and 2 8. The solution set is bounded. The test point 0 2 satisfies each inequality.

    x  2y  14 59. 3x  y  0    x y2

The lines x  2y  14 and 3x  y  0 intersect when

14  x  6x  x  2 and y  6. The lines 3x  y  0 and x  y  2 intersect where 3x  2  x  x  1 and y  3. The lines x  2y  14 and x  y  2 intersect where 14  2y  2  y  y  4 and x  6. Thus, the vertices of the region are 1 3, 2 6, and 6 4. The solution set is bounded. The test point 1 0 satisfies each inequality.

    x  2y  14 60. 3x  y  0    x y2

y (2, 6) 2 (_1, _3)

2

(6, 4)

x

y

The vertices are 1 3, 2 6, and 6 4. In this case the

(2, 6)

solution set is not bounded. The test point 3 0 satisfies each inequality. 2 (_1, _3)

2

(6, 4)

x


44

CHAPTER 9 Systems of Equations and Inequalities

    x  y  12 61. y  12 x  6    y  2x  6

The lines x  y  12 and y  12 x  6 intersect where

y (2, 10)

12  x  12 x  6  x  12 and y  0. The lines y  12 x  6 and y  2x  6 intersect where 12 x  6  2x  6  x  8 and y  10. The lines x  y  12

and y  2x  6 intersect where 12  x  2x  6  x  2 and y  10. The solution set is not bounded. The test point 0 8 satisfies each inequality.

62.

   

y  x 1

x  2y  12

   x 1 0

The lines y  x  1 and x  2y  12 intersect where

2 2

(12, 0)

(_8, _10)

(_1, 132 )

y

( 103 , 133 )

10 y  1  12  2y  y  13 3 and x  3 . The lines y  x  1 and x  1  0

intersect at 1 0, and the lines x  2y  12 and x  1  0 intersect where

12  2y  1  y  13 2 and x  1. Thus, the vertices of the region are     10  13 . The solution set is bounded. The test point , and 1 0, 1 13 2 3 3

(_1, 0)

    3x  y  5 30x  10y  50        x  2y  5  10x  20y  50  63.   x  6y  9 10x  60y  90         x  0, y  0 x  0 y  0

1

y (0, 5)

(1, 2)

The vertices are 1 5,

point 4 3 satisfies each inequality.    y  x 3 65. y  2x  6 Using a graphing device, we find the region shown. The   y 8 vertices are 3 0, 1 8, and 11 8.

    x  y  12 66. 2x  y  24    x  y  6

1 y

17  7 , and 8 1. The solution set is bounded. The test 4 4

Using the graphing device, we find the region shown. The

(3, 1)

1

The vertices are 0 5, 1 2, 3 1, and 9 0. The solution set is not bounded. The test point 1 3 satisfies each inequality.

Add dots for the graphs of #62-64. [see #61 graph] x

1

0 2 satisfies each inequality.

  x y6     4x  7y  39 64.  x  5y  13     x  0 y  0

x

1

(9, 0)

x

(1, 5)

( 174 , 74 )

(8, 1) x

1

10 8 6 4 2 0 -2 -4

10

10

vertices are 3 9, 6 12, and 12 0. 0

10


SECTION 9.5 Systems of Inequalities

67.

y  6x  x 2

xy4

10 8 6 4 2

Using a graphing device, we find the region shown. The

vertices are 06 34 and 64 24.

-4 -2 -2 0 2 -4

68.

      

y  x3

2x  y  0

45

4

6

8 10

2

4

12 10 8 6 4 2

Using a graphing device, we find the region shown. The

y  2x  6

vertices are 0 0, 22 103 and 15 3. -4

-2

0 -2

y

69. (a) Let x and y be the numbers of acres of potatoes and corn, respectively.   x  y  500     90x  50y  40,000 A system describing the possibilities is  30x  80y  30,000     x  0 y  0

(0, 375) (200, 300) (375, 125)

100

x ( 4000 9 , 0)

100

(b) Because the point 300 180 lies in the feasible region, the farmer can plant 300 acres of potatoes and 180 acres of corn. (c) Because the point 150 325 lies outside the feasible region, the farmer cannot plant this combination of crops.

70. (a) Let x and y be the numbers of acres of cauliflower and cabbage, respectively.   x  y  300     70x  35y  17,500 A system describing the possibilities is  25x  55y  12,000     x  0 y  0

y

(0, 2400 11 ) (150, 150)

(200, 100)

50 50

(250, 0)

(b) Because the point 155 115 lies in the feasible region, the farmer can plant 155 acres of cauliflower and 115 acres of cabbage. (c) Because the point 115 175 lies outside the feasible region, the farmer cannot plant this combination of crops. 71. Let x be the number of fiction books published in a year and y the number of nonfiction books. Then the following system of inequalities holds:     x  0, y  0 From the graph, we see that the vertices are 50 50, 80 20 x  y  100    y  20, x  y and 20 20.

y

120 100 80 60 40 20

x+y=100 x=y y=20 20 40 60 80 100

x

x


46

CHAPTER 9 Systems of Equations and Inequalities

72. Let x be the number of chairs made and y the number of tables. Then the following     2x  3y  12 system of inequalities holds: 2x  y  8 The intersection points are 0 4,    x  0, y  0

y 2x+y=8

2x+3y=12

1

0 8, 6 0, 4 0, 0 0, and 3 2. Since the points 0 8 and 6 0 are not in

73. Let x be the number of Standard Blend packages and y be the number of Deluxe

y

200

Blend packages. Since there are 16 ounces per pound, we get the following system

3 x+38 y=90 4

of inequalities:      

x

1

the solution set, the vertices are 0 4, 0 0, 4 0, and 3 2.

1 x+58 y=80 4

100

x 0

y 0 1 x  5 y  80   4 8   3 3 4 x  8 y  90

200 x

100

From the graph, we see that the vertices are 0 0, 120 0, 70 100 and 0 128. 74. Let x be the amount of fish and y the amount of beef (in ounces) in each can. Then

y

the following system of inequalities holds:     12x  6y  60   

12x+6y=60

3x  9y  45 x  0, y  0

3x+9y=45

2

x

2

From the graph, the vertices are 15 0, 3 4, and 0 10.

75. x  2y  4, x  y  1, x  3y  9, x  3. Method 1: We shade the solution to each inequality with lines perpendicular to the boundary. As you can see, as the number of inequalities in the system increases, it gets harder to locate the region where all of the shaded parts overlap. Method 2: Here, if a region is shaded then it fails to satisfy at least one inequality. As a result, the region that is left unshaded satisfies each inequality, and is the solution to the system of inequalities. In this case, this method makes it easier to identify the solution set.   To finish, we find the vertices of the solution set. The line x  3 intersects the line x  2y  4 at 3 12 and the line

x  3y  9 at 3 2. To find where the lines x  y  1 and x  2y  4 intersect, we add the two equations, which gives

3y  5  y  53 , and x  23 . To find where the lines x  y  1 and x  3y  9 intersect, we add the two equations,       5 3 1 2 5 3 5 which gives 4y  10  y  10 4  2 , and x  2 . The vertices are 3 2 , 3 2, 3  3 , and 2  2 , and the solution set is bounded.

_x+y=1

y

1

x+3y=9 x

1 x=3

Method 1

Method 2

Solution Set

x+2y=4


CHAPTER 9

Review

CHAPTER 9 REVIEW 1.

  3x  y  5  2x  y  5

y

Adding, we get 5x  10  x  2. So 2 2  y  5  y  1. 1

Thus, the solution is 2 1.

2.

  y  2x  6

 y  x  3

3.

  2x  7y  28

 2x  7y  28

  6x  8y  15 6x  8y  15 4.   3 x  2y  4 6x  8y  16

y

Since these equations represent the

2 2

x

1

x

1

x

y

Adding gives 0  1 which is

false. Hence, there is no solution. The lines are parallel.

    2x  y  1 5. x  3y  10    3x  4y  15

x

1

same line, any point on this line will satisfy the system. Thus the solution are   t 27 t  4 , where t is any real number.

2

1

y

y  27 x  4

x

Subtracting the second equation from the first, we get

0  3x  3  x  1. So y   1  3  4. Thus, the solution is 1 4.

  2x  7y  28

1

1

y

Solving the first equation for y, we get y  2x  1.

Substituting into the second equation gives x  3 2x  1  10  5x  7      7 12 x   75 . So y    75  1  12 5 Checking the point  5  5 in the third     ? 21 48 equation we have 3  75  4 12 5  15 but  5  5  15. Thus, there is no

solution, and the lines do not intersect at one point.

1

47


48

CHAPTER 9 Systems of Equations and Inequalities

    2x  5y  9 6. x  3y  1    7x  2y  14

y

Adding the first equation to twice the second equation gives

11y  11  y  1. Substituting back into the second equation, we get

1

x  3 1  1  x  4. Checking point 4 1 in the third equation gives

7 4  2 1  26  14. Thus there is no solution, and the lines do not intersect at

7.

  y  x 2  2x y 6x

x

1

one point.

Substituting for y gives 6  x  x 2  2x  x 2  x  6  0. Factoring, we have x  2 x  3  0.

Thus x  2 or 3. If x  2, then y  8, and if x  3, then y  3. Thus the solutions are 3 3 and 2 8.

8.

  x 2  y2  8 

y  x 2

Substituting for y in the first equation gives x 2  x  22  8 

  2x 2  4x  4  0  2 x 2  2x  2  0. Using the Quadratic Formula, we have          2  2 3 2  4  8   1  3. If x  1  3. then y  1  3  2  1  3, and if x  1  3, x 2 2             then y  1  3  2  1  3. Thus, the solutions are 1  3 1  3 and 1  3 1  3 .  4   3x   6 y 9. 8   x 4 y

Adding twice the first equation to the second gives 7x  16  x  16 7 . So

16 8   4 7 y

  16 14 16y  56  28y  12y  56  y   14 3 . Thus, the solution is 7   3 .

10.

 

x 2  y 2  10

 x 2  2y 2  7y  0

Subtracting the first equation from the second gives y 2  7y  10 

 y 2  7y  10  0  y  2 y  5  0  y  2, y  5. If y  2, then x 2  4  10  x 2  6  x   6, and if      y  5, then x 2  25  10  x 2  15, which leads to no real solution. Thus the solutions are 6 2 and  6 2 .

 32x  y  0 43 11.   7x  12y  341   y  7x  341 12 The solution is approximately 2141 1593.   032x  043y 

12.

  12x  32y 

660

 7137x  3931y  20,000

   y  6 x  1102 3

 y   7137 x  20,000 3931

3931

The solution is approximately 6104 10573.

Graphs missing in #11,12

Note that in CA-8 and PMFC-8 the corresponding graphs are fine.


CHAPTER 9

13.

  x  y 2  10 

1 y  12 x  22

  y  x  10

 y  22 x  12

The solutions are 1194 139 and 1207 144.

14.

  y  5x  x

 y  x5  5

Review

49

The solutions are approximately

145 135, 1 6, and 151 1293.

graphs missing in #13,14 15.

   

x  2y  z 

8

   x 

4x z 9       2x  y  z  8 

  2y  z  8 8   x  2y  z   8y  3z  23 8y  3z  23    3y  z  8 z  5

Therefore, z  5, 8y 3 5  23

 y  1, and x  2 1  5  8  x  1. Hence, the solution is 1 1 5.       x  y  3z  4 x  y  3z  4       x  y  3z  4 16. 4x  2y  z  11  6y  13z  27  6y  13z  27        5x  y  z  16   3z  9 6y  16z  36

Therefore, z  3,

6y  13 3  27  y  2, and x  2  3 3  4  x  3. Hence, the solution is 3 2 3.           x  y  2z  6  x  y  2z  6  x  y  2z  6 17. 2x Therefore, 3z  6  z  2, 2y  2  0   5z  12  2y  z  0  2y  z  0        x  2y  3z  9   4y  z  6 3z  6 y  1, and x  1  2 2  6  x  1. Hence, the solution is 1 1 2.       x  2y  3z  1 x  2y  3z  1     x  2y  3z  1     18. y  4z  1 y  4z  1 x  3y  z  0          2x 28z  0 6y  4z  6  6z  6

 x  3, and so the solution is 3 1 0.           x  2y  3z  1  x  2y  3z  1  x  2y  3z  1 19. 2x  y  z  3  3y  5z  1  3y  5z  1        2x  7y  11z  2   6y  10z  1 0  1

Thus, z  0, y  1, and x  2 1  3 0  1

which is impossible. Therefore, the

system has no solution.       x  y z 2 x  y z  2 x  y z  2              2x    3z 5 2y  5z  2  1 2y  5z  2  1 20.        4  9 3y  z  3  7 13z  12  11  x  2y             x  y  2z  3  5   z  2  3 z  2  3   x  y z  2      2y  5z  2  1 Therefore, 14  28    2, 13z12 2  11  z  1; 2y5 12 2  1  13z  12  11      14  28  2y  1  1  y  0; and x  1  2  2  x  1. So the solution is 1 0 1 2.


50

21.

CHAPTER 9 Systems of Equations and Inequalities

  x  3y 

z4

 4x  y  15z  5

  x  3y  z  

4

y  z  1

Thus, the system has infinitely many solutions given by z  t,

y  t  1  y  1  t, and x  3 1  t  t  4  x  1  4t. Therefore, the solutions are 1  4t 1  t t, where t is any real number.       2x  3y  4z  3 2x  3y  4z  3 3       2x  3y  4z  22. 4x  5y  9z  13  y z 7  y z 7        2x    7z  0 3y  3z  3 0  24

Since this last equation is impossible, the system is inconsistent and has no solution.       x  z  2 x  z    2 x  z    2               2x  y  y  2z  4  8 y  2z  4  8  2  12   23.     3y  z    4 5z  13  20 3y  z    4               x  yz  z  4  20 y  8  10   x  z  2      y  2z  4    8 Therefore, 33  120     40  20  z   60 , 5z  13  40 11 11 11 ,  5z  13  20      33  120         40  8  y  48 ; and x   60   40  2  x  2 . Hence, the solution is y  2  60  4  11 11 11 11 11 11   2  48   60   40 . 11 11 11 11       x  4y  z  8 x  4y  z  8       x  4y  z  8 24. 2x  6y  z  9  2y  3z  7  2y  3z  7        x  6y  4z  15   6y  9z  21 00

Thus, the system has infinitely many solutions. Letting z  t, we find 2y  3t  7  y  72  32 t, and     x  4 72  32 t  t  8  x  6  5t. Therefore, the solutions are 6  5t 72  32 t t , where t is any real number.

25. Let the age of the younger child be x, so the older child’s age is x  4. Then because the sum of their ages is 22, we have x  x  4  22  2x  4  22  2x  18  x  9. Thus, the younger child is 9 years old and the older child is 9  4  13 years old. 26. Let x be the amount in the account yielding 6% and y the amount in the one yielding 7%. The system is   y  2x Substituting gives 006x  007 2x  600  02x  600  x  3000, so  006x  007y  600 y  2 3000  6000. Hence, $3000 is invested at 6% and $6000 is invested at 7%.

27. Let the time spent driving be x and the time spent flying be y. The total travel time is 6 hours, so x  y  6. The distance driven is 70x and the distance flown is 550y. The total distance traveled is 2700 miles, so 70x  550y  2700.   x y 6 We have the system Subtracting 70 times the first equation from the second, we have  70x  550y  2700 550y  70y  2700  70 6  480y  2280  y  19 4 . The traveler drove for 4 hours 15 minutes and flew for 1 hour 45 minutes.


CHAPTER 9

Review

51

28. Let the volume of the 15% solution be x mL and the volume of the 25% solution be y mL. Then x  y  500, and the volume of acid in the mixture is 015x  025y  022 500  110.   x y  500 Thus, we solve the system Subtracting 015 times the first equation from the second, we have  015x  025y  110 025y  015y  110  015 500  01y  35  y  350. Thus, x  500  350  150, so the chemist should mix 150 mL of the 15% solution and 500  150  350 mL of the 25% solution.

29. Let n, d, and q be the numbers of nickels, dimes, and quarters in the piggy bank. We get the following system:   50   n d  q Since 10d  25n, we have d  52 n, so substituting into the first equation we get 5n  10d  25z  560    10d  5 5n

n  52 n  q  50  72 n  q  50  q  50  72 n. Now substituting this into the second equation we have     115 115 5n  10 52 n  25 50  72 n  560  5n  25n  1250  175 2 n  560  1250  2 n  560  2 n  650 

n  12. Then d  52 12  30 and q  50  n  d  50  12  30  8. Thus the piggy bank contains 12 nickels, 30 dimes, and 8 quarters.

30. Let c, s, and p be the number of each kind of salmon caught. The given information leads to the system              c  s  p  25  c  s  p  25  c  s  p  25  c  s  p  25 c  s p3  c  s  p  3  2s  2 p  22  2s  2 p  22  p  4, so             c  2s  0 4 p  16 3s  p  25 c  2s 2s  2 4  22  s  7 and c  7  4  25  c  14. The fisherman caught 14 coho, 7 sockeye, and 4 pink salmon.

31.

3x  1 B A 3x  1  .Thus, 3x  1  A x  3  B x  5  x A  B  3A  5B,   x 5 x 3 x  5 x  3 x 2  2x  15    AB 3  3A  3B  9 and so  Adding, we have 8B  8  B  1, and A  2. Hence,  3A  5B  1  3A  5B  1 3x  1 1 2  .  x 5 x 3 x 2  2x  15

32.

8

x 3  4x

8

 

x x2  4

  A B C 8    . Then 8  A x 2  4  Bx x  2  x x  2 x  2 x x 2 x 2

C x x  2  x 2 A  B  C  x 2B  2C  4A. Thus, 4A  8  A  2, 2B  2C  0  B  C, and 1 1 8 2  . 2  2B  0  B  1, so C  1. Therefore, 3   x x 2 x 2 x  4x 33.

2x  4

x x  12

B C A   . Then 2x  4  A x  12  Bx x  1  C x  Ax 2  2Ax  A  Bx 2  x x 1 x  12

Bx  C x  x 2 A  B  x 2A  B  C  A. So A  4, 4  B  0  B  4, and 8  4  C  2  C  2. 4 2 4 2x  4    . Therefore, x x  1 x  12 x x  12 34.

x 6 Ax  B C x 6  . Thus,  2  2  x 2 x  4 x  2 x 3  2x 2  4x  8 x 4   x  6  Ax  B x  2  C x 2  4  Ax 2  2Ax  Bx  2B  C x 2  4C  x 2 A  C  x 2A  B  2B  4C


52

CHAPTER 9 Systems of Equations and Inequalities

and we get the system

      

A 2A  B

 C0

1 

2B  4C  6

   A   

 C0

B  2C  1 

2B  4C  6

   A   

 C0

B  2C  1

8C  8

x 6 x 1 1 Thus, 8C  8  C  1, B  2  1  B  1, and A  1  0  A  1. So 3 .  2  x  2x 2  4x  8 x 4 x 2

35.

  2x  1 2x  1 A Bx  C 2  1 Bx  C x  Ax 2  A Bx 2 C x  A  B x 2      . Then 2x 1  A x x x3  x x x2  1 x2  1 x 2 1 2x  1   2 . C x  A. So A  1, C  2, and A  B  0 gives us B  1. Thus 3 x x x x 1

     B Cx  D A 5x 2  3x  10   2  . 36. Since x 4 x 2 2  x 2  1 x 2  2  x  1 x  1 x 2  2 , we have 4 2 x  1 x  1 x x 2 x 2 Thus       5x 2  3x  10  x  1 x 2  2 A  x  1 x 2  2 B  x 2  1 C x  D  Ax 3  Ax 2  2Ax  2A  Bx 3  Bx 2  2Bx  2B  C x 3  Dx 2  C x  D

 A  B  C x 3  A  B  D x 2  2A  2B  C x  2A  2B  D     A BC  0 Coefficients of x 3 A B C  0          A B  2 D 5 Coefficients of x 2B  C  D  5 This leads to the system     2A  2B  C  3 Coefficients of x 3C  3          2A  2B   D  10 Constant terms 4B  C  D  13   A B C  0      2B  C  D  5 Thus C  1, 3 1  3D  3  D  0, 2B  1  0  5  B  3, and  C  1      3C  3D  3 A  3  1  0  A  2. Thus,

37.

3 x 2 5x 2  3x  10   2  . x 1 x 1 x4  x2  2 x 2

Cx  D Ax  B 3x 2  x  6   2 2 . Thus x 2  22 x 2 x2  2 3x 2  x  6 

 x 2  2 Ax  B  C x  D  Ax 3  Bx 2  2Ax  2B  C x  D

 x 3 A  x 2 B  x 2A  C  2B  D     A  0 A  0         B  3 B  3  This leads to the system   C  1 2A C  1         D 0 2B D6

Thus

3x 2  x  6 x 3   2 2 . 2 2 2 x  2 x 2 x 2


CHAPTER 9

Review

53

Bx  C x2  x  1 A Dx  E   2  2 . Thus x xx 2  12 x 1 x2  1  2   x 2  x  1  x 2  1 A  x x 2  1 Bx  C  x Dx  E  Ax 4  2Ax 2  A  Bx 4  C x 3  Bx 2  C x  Dx 2  E x 38.

 x 4 A  B  x 3 C  x 2 2A  B  D  x C  E  A   AB  0      C  0   Thus A  1, B  1, C  0, D  0, and E  1, so This leads to the system 2A  B D  1     C E 1     A 1 x2  x  1 x 1 1    2 2 . x 2 xx 2  12 x 1 x 1

39.

  2x  3y  7  x  2y  0

By inspection of the graph, it appears that 2 1 is the solution to the system. We check this in both

equations to verify that it is the solution. 2 2  3 1  4  3  7 and 2  2 1  2  2  0. Since both equations are satisfied, the solution is indeed 2 1.

40.

  3x  y  8 

y  x 2  5x

By inspection of the graph, it appears that 2 14 and 4 4 are solutions to the system.

We check each possible solution in both equations to verify that it is a solution: 3 2  14  6  14  8 and

22  5 2  4  10  14; also 3 4  4  12  4  8 and 42  5 4  16  20  4. Since both points satisfy both equations, the solutions are 2 14 and 4 4. 41.

 

x2  y  2

 x 2  3x  y  0

By inspection of the graph, it appears that 2 2 is a solution to the system, but is difficult

to get accurate values for the other point. Adding the equations, we get 2x 2  3x  2  2x 2  3x  2  0   2 2x  1 x  2  0. So 2x  1  0  x   12 or x  2. If x   12 , then  12  y  2  y  74 . If x  2, then   22  y  2  y  2. Thus, the solutions are  12  74 and 2 2. 42.

 

x  y  2  x 2  y 2  4y  4

By inspection of the graph, it appears that 2 0 and 2 4 are solutions to the system. We

check each possible solution in both equations to verify that it is a solution: 2  0  2 and 22  02  4 0  4;

also 2  4  2 and 22  42  4 4  4  16  16  4. Since both points satisfy both equations, the solutions are 2 0 and 2 4.

43. The boundary is a solid curve, so we have the inequality x  y 2  4. We take the test point 0 0 and verify that it satisfies the inequality: 0  02  4.

44. The boundary is a solid curve, so we have the inequality x 2  y 2  8. We take the test point 0 3 and verify that it satisfies the inequality: 02  32  8.


54

CHAPTER 9 Systems of Equations and Inequalities

46. y  x 2  3

45. 3x  y  6

y

47. x 2  y 2  9

48. x  y 2  4

y

y

y

1

1 1 1

49.

x

50.

 y  1x  1 3

0

   x  y  2 51. y  x  2    x 3

 y  x 1

 x 2  y2  1 y

y

x

0

1

x

53.

xy0

2

y 4

1 1

  x 2  y2  9

    y  2x 52. y  2x    y  1x  2

y

1

1

4

x

y

The vertices occur where y  x. By substitution, 1

x 2  x 2  9  x   3 , and so y   3 . Therefore, the vertices are 2 2     3   3 and  3  3 and the solution set is bounded. The test point 2

2

x

1

x

x

1

  y  x 2  3x

1

1

1

2

x

1

2

1 1 satisfies each inequality.

  y  x2  4 54.  y  20

y

The vertices occur where y  x 2  4 and y  2. By

substitution, x 2  4  20  x 2  16  x  4 and y  20. Thus, the vertices

(_4, 20)

(4, 20)

are 4 20 and the solution set is bounded. The test point 0 10 satisfies each inequality.

2 2

    x  0, y  0 55. x  2y  12    y  x 4

The intersection points are 4 0, 0 4,

y

4  16 , 0 6, 3 3

0 0, and 12 0. Since the points 4 0 and 0 6 are not in the solution set,   the vertices are 0 4, 43  16 3 , 12 0, and 0 0. The solution set is bounded. The test point 1 1 satisfies each inequality.

x

y=x+4

x+2y=12 1 1

x

x


CHAPTER 9

56.

      

x4

x  y  24

x  2y  12

y

point 20 4. The lines x  4 and x  2y  12 intersect at the point 4 4, but intersect at the point 4 20. Hence, the vertices are 4 20 and 20 4. The solution set is not bounded. The test point 12 12 satisfies each inequality.

55

x=4

The lines x  y  24 and x  2y  12 intersect at the

this vertex does not satisfy the other inequality. The lines x  y  24 and x  4

Test

x+y=24 1 1

x=2y+12 x

       x  y  z  a  x  y  z  a ab ac ab ac ,z , and x   a 57. x yz b  2z  a  b Thus, y    2 2 2 2    x yz c  2y  ac   bc bc ac ab . The solution is   . x 2 2 2 2        ax  by  cz  a  b  c  ax  by  cz  a  b  c 58. bx  by  cz  a  b  c  0  bx  by  cz  Subtracting the second equation from c c      cx  cy  cz   c x yz  1 the first gives a  b x  a  b  x  1Subtracting the third equation from the second, b  c x  b  c y  0  y  x  1. So 1  1  z  1  z  1, and the solution is 1 1 1.

59. Solving the second equation for y, we have y  kx. Substituting for yin the first equation gives us 12 x  kx  12  1  k x  12  x  . Substituting for y in the third equation gives us kx  x  2k  k1 2k 12 2k . These points of intersection are the same when the x-values are equal. Thus,  k  1 x  2k  x  k 1 k1 k 1    12 k  1  2k k  1  12k  12  2k 2  2k  0  2k 2  10k  12  2 k 2  5k  6  2 k  3 k  2. Hence, k  2 or k  3.

60. The system will have infinitely many solutions when the system has solutions other than 0 0 0. So we solve the system with x  0, y  0, and z  0:     kx  y  z0    kx  y  z  0 kx  y  z  0        2y  k  3 z  0 x  2y  kz  0  2y  k  3 z  0          x    3z  0 y  3k  1 z  0 [k  3  2 3k  1] z  0 Since z  0, we must have k  3  2 3k  1  5k  1  0  k  15 .

CHAPTER 9 TEST 1. (a) The system is linear.   x  3y  7 (b) Multiplying the first equation by 5 and then adding gives 13y  39  y  3. So  5x  2y  4 x  3 3  7  x  2. Thus, the solution is 2 3.

2. (a) The system is nonlinear.


56

CHAPTER 9 Systems of Equations and Inequalities

(b)

  6x  y 2  10  3x  y  5

  6x  y 2  10  y 2  2y  0

Thus y 2  2y  y y  2  0, so either y  0 or y  2. If y  0, then

  3x  5  x  53 and if y  2, then 3x  2  5  3x  3  x  1. Thus the solutions are 53  0 and 1 2.

3. (a) The system is nonlinear.   x 2  y 2  100 Substituting y  3x into the first equation gives x 2  3x2  100  x 2  10  x   10. (b) y  3x         If x   10, then y  3 10, and if x  10, then y  3 10. We can verify that  10 3 10 and    10 3 10 are valid solutions to the first equation in the given system.

graph missing in #4

4.

  x  2y  1 

Note that in CA-8 and PMFC-8 the corresponding graph is not missing

y  x 3  2x 2

The solutions are approximately 055 078, 043 029, and 212 056.

    x  2y  z  3     x  2y  z  3 5. (a) x  3y  2z  3  y z 0      2x  3y  z  8  y  3z  2

    x  2y  z  3 y z 0 Eq. 1  Eq. 2     2z  2 2  Eq. 1  Eq. 3

 z  1, Eq. 2  Eq. 3

so y  1  0  y  1 and z  2 1  1  3  x  2. Thus, the solution is 2 1 1.

(b) The system is neither inconsistent nor dependent.

       x  y  9z  3  x  y  9z  3 6. (a) x  4z  7  y  13z  10      3x  y  z  5  2y  26z  14

    x  y  9z  3 y  13z  10 Eq. 1  Eq. 2     0 6 3  Eq. 1  Eq. 3

2  Eq. 2  Eq. 3

The last equation cannot be satisfied, so the system has no solution.

(b) The system is inconsistent.

    2x  y  z  0 7y  9z  2 3  Eq. 1  2  Eq. 2     0  0 Eq. 2  Eq. 3 Eq. 1  2  Eq. 3   Letting z  t, we have 7y  9t  2  y  27  97 t, so 2x  27  97 t  t  0  x  17  17 t. The solutions are   1  1 t 2  9 t t where t is any real number. 7 7 7 7

    2x  y  z  0     2x  y  z  0 7. (a) 3x  2y  3z  1  7y  9z  2      x  4y  5z  1  7y  9z  2

(b) The system is dependent.

       x  y  2z  8  x  y  2z  8 8. (a) 2x  y  20  3y  4z  4      2x  2y  5z  15  z  1

2  Eq. 1  Eq. 2 2  Eq. 1  Eq. 3

x  0  2 1  8  x  10. Thus, the solution is 10 0 1.

(b) The system is neither inconsistent nor dependent.

 z  1, so 3y  4 1  4  y  0 and


CHAPTER 9

Test

57

9. Let  be the speed of the wind and a the speed of the airplane in still air, in kilometers per hour. Then the speed of the of the plane flying against the wind is a   and the speed of the plane flying with the wind is a  . Using distance  rate  time,    600  25 a    240  a   Adding the two equations, we get 600  2a  a  300. we get the system  50  300   360  a   a   60

So 360  300      60Thus the speed of the airplane in still air is 300 km/h and the speed of the wind is 60 km/h.

10. Let x, y, and z represent the price in dollars for coffee, juice, and donuts respectively. Then the system of equations is       2x  y  2z  625 Friend A 2x  y  2z  625       2x  y  2z  625 x  3z  375 Friend B  y  4z  125  y  4z  125        3x  y  4z  925   Friend C y  5z  200 z  075

Thus, z  075, y  4 075  125  y  175, and 2x  175  2 075  625  x  15. Thus coffee costs $150, juice costs $175, and donuts cost $075.

11. 3x  4y  6 x@+y²5

y

12. x 2  y  3 y=2x+5

1

1 1

    2x  y  8 13. x  y  2    x  2y  4

x

y

these two equations gives 3x  6  x  2, and so y  8  2 2  4 . Thus, the

third vertex is 2 4.   x2  y  5 

x

1

From the graph, the points 4 0 and 0 2 are vertices. The

third vertex occurs where the lines 2x  y  8 and x  y  2 intersect. Adding

14.

y

y  2x  5

1 x

1

Substituting y  2x  5 into the first equation gives

x@+5²5

y

y=2x+5

x 2  2x  5  5  x 2  2x  0  x x  2  0  x  0 or x  2. If

x  0, then y  5  2 0  5, and if x  2, then y  5  2 2  1. Thus, the

vertices are 0 5 and 2 1.

1 1

15.

4x  1

x  12 x  2

B C A  . Thus,  x  1 x  12 x 2

x

    4x  1  A x  1 x  2  B x  2  C x  12  A x 2  x  2  B x  2  C x 2  2x  1  A  C x 2  A  B  2C x  2A  2B  C


58

FOCUS ON MODELING

which leads to the system of equations       A  C  0 A  C  0  C 0      A A  B  2C  4  B  3C  4  B  3C  4        2A  2B  C  1   2B  3C  1 9C  9

Therefore, 9C  9  C  1, B  3 1  4  B  1, and A  1  0  A  1. Therefore, 4x  1

x  12 x  2

16.

1 1 1  .  x  1 x  12 x 2

2x  3 Bx  C 2x  3 A      2 . Then x x 3  3x x x 3 x2  3   2x  3  A x 2  3  Bx  C x  Ax 2  3A  Bx 2  C x  A  B x 2  C x  3A.

x 2 1 2x  3   2 . So 3A  3  A  1, C  2 and A  B  0 gives us B  1. Thus 3 x x x x 3

FOCUS ON MODELING Linear Programming

2.

1. Vertex 0 2 0 5 4 0

M  200  x  y

200  0  2  198 200  0  5  195 200  4  0  196

Thus, the maximum value is 198 and the minimum value is 195.

    x  0, y  0 3. 2x  y  10    2x  4y  28

N  12 x  14 y  40

Vertex 1 0   1 1 2 2 2 2 4 0

1 2

1 1  1 0  40  405 4  2  1  1 1  40  40375 2 4 2 1 2  1 2  40  415 2 4 1 4  1 0  40  42 2 4

Thus, the maximum value is 42 and the minimum value is 40375.

The objective function is P  140  x  3y. From the graph,

y

the vertices are 0 0, 5 0, 2 6, and 0 7. Vertex 0 0 5 0 2 6 0 7

2x+y=10

P  140  x  3y

140  0  3 0  140 140  5  3 0  135 140  2  3 6  156 140  0  3 7  161

Thus the maximum value is 161, and the minimum value is 135.

2x+4y=28 1 1

x


Linear Programming

  x  0, y  0      x  10, y  20 4.  xy5      x  2y  18

The objective function is Q  70x  82y. From the graph,

y

x+2y=18

the vertices are at 0 9, 0 5, 5 0, 10 0, and 10 4. Note that the restriction y  20 is irrelevant, superseded by x  2y  18 and x  0. Vertex 0 9 0 5 5 0 10 0 10 4

59

Q  70x  82y

x+y=5

1

70 0  82 9  738

x

1

70 0  82 5  410

x=10

70 5  82 0  350

70 10  82 0  700

70 10  82 4  1028

Thus, the maximum value of Q is 1028 and the minimum value is 350.

5. Let t be the number of tables made daily and c be the number of chairs made daily. Then the data given can be summarized by the following table: Tables t

Chairs c

Available time

2h

3h

108 h

Finishing

1h

1 h 2

20 h

Profit

$35

$20

Carpentry

    2t  3c  108 Thus we wish to maximize the total profit P  35t  20c subject to the constraints t  12 c  20    t  0, c  0 From the graph, the vertices occur at 0 0, 20 0, 0 36, and 3 34.

y

Vertex 0 0 20 0 0 36 3 34

P  35t  20c

35 0  20 0 

0

1

t+2 c=20

35 20  20 0  700

35 0  20 36  720 35 3  20 34  785

2t+3c=108 10 10

Hence, 3 tables and 34 chairs should be produced daily for a maximum profit of $785.

x


60

FOCUS ON MODELING

6. Let c be the number of colonial homes built and r the number of ranch homes built. Since there are 100 lots available, c  r  100. From the capital restriction, we get 30,000c  40,000r  3,600,000, or 3c  4r  360. Thus, we wish to     c  0, r  0 maximize the profit P  4000c  8000r subject to the constraints c  r  100    3c  4r  360 From the graph, the vertices occur at 0 0, 100 0, 40 60, and 0 90.

y

Vertex 0 0 100 0 40 60 0 90

P  4000c  8000r

4000 0  8000 0 

c+r=100

0

4000 100  8000 0  400,000

4000 40  8000 60  640,000

3c+4r=360

20

4000 0  8000 90  720,000

x

20

Therefore, the contractor should build 90 ranch style houses for a maximum profit of $720,000. Note that ten of the lots will be left vacant.

7. Let x be the number of crates of oranges and y the number of crates of grapefruit. Then the data given can be summarized by the following table: Oranges

Grapefruit

Available

Volume

4 ft3

6 ft3

300 ft3

Weight

80 lb

100 lb

5600 lb

Profit

$250

$400

In addition, x  y. Thus we wish to maximize the total profit P  25x  4y subject to the constraints     x  0, y  0, x  y 4x  6y  300    80x  100y  5600 From the graph, the vertices occur at 0 0, 30 30, 45 20, and 70 0.

y

Vertex 0 0 30 30 45 20 70 0

P  25x  4y

25 0  4 0 

80x+100y=5600

0

25 30  4 30  195

25 45  4 20  1925 25 70  4 0  175

x=y

4x+6y=300 10 10

x

Thus, the truck should carry 30 crates of oranges and 30 crates of grapefruit for a maximum profit of $195.


Linear Programming

8. Let x be the daily production of standard calculators and y the daily production of     x  100, y  80 scientific calculators. Then the inequalities x  200, y  170 describe the    x  y  200

y

180 120 90

100 100 100 170 200 170 200 80

C  5x  7y

x+y=200

30

30 60 90 120 150 180 x

(b) Maximize the objective function P  2x  5y: Vertex

5 100  7 100  1200

100 100

5 200  7 170  2190

200 170

5 120  7 80  1160

120 80

5 100  7 170  1690

100 170

5 200  7 80  1560

200 80

120 80

y=80

60

200 170, 200 80, and 120 80.

Vertex

x=200

150

constraints. From the graph, the vertices occur at 100 100, 100 170,

(a) Minimize the objective function C  5x  7y:

x=100 y=170

61

P  2x  5y

2 100  5 100  300 2 100  5 170  650 2 200  5 170  450 2 200  5 80 

0

2 120  5 80  160

So to minimize cost, they should produce

So to maximize profit, they should produce

120 standard and 80 scientific calculators.

100 standard and 170 scientific calculators.

9. Let x be the number of television sets shipped from Long Beach to Santa Monica and y the number of television sets shipped from Long Beach to El Toro. Thus, 15  x sets must be shipped to Santa Monica from Pasadena and 19  y sets

to El Toro from Pasadena. Thus, x  0, y  0, 15  x  0, 19  y  0, x  y  24, and 15  x  19  y  18.   x  0, y  0      x  15, y  19 Simplifying, we get the constraints  x  y  24      x  y  16 The objective function is the cost C  5x  6y  4 15  x  55 19  y  x  05y  1645, which we wish to

minimize. From the graph, the vertices occur at 0 16, 0 19, 5 19, 15 9, and 15 1. Vertex 0 16 0 19 5 19 15 9 15 1

C  x  05y  1645

y

x=15

0  05 16  1645  1725 0  05 19  1645  174 5  05 19  1645  179 15  05 9  1645  184 15  05 1  1645  180

10 x+y=16 10

x+y=24 x

The minimum cost is $17250 and occurs when x  0 and y  16. Hence, 16 TVs should be shipped from Long Beach to

El Toro, 15 from Pasadena to Santa Monica, and 3 from Pasadena to El Toro.


62

FOCUS ON MODELING

10. Let x be the number of sheets shipped from the east side warehouse to customer A and y be the number of sheets shipped from the east side warehouse to customer B. Then 50  x sheets must be shipped to customer A from the west side

warehouse and 70  y sheets must be shipped to customer B from the west side warehouse. Thus, we obtain the constraints     x  0, y  0 x  0, y  0          x  50, y  70  x  50, y  70     x  y  80  x  y  80        x  y  75  50  x  70  y  45

The objective function is the cost C  05x  06y  04 50  x  055 70  y  01x  005y  585, which we wish to minimize. From the graph, the vertices occur at 5 70, 10 70, 50 30, and 50 25. y

Vertex 5 70 10 70 50 30 50 25

x=70

C  01x  005y  585

01 5  005 70  585  625

y=70

01 10  005 70  585  63

x+y=75

01 50  005 30  585  65

01 50  005 25  585  6475

x+y=80

10

x

10

Therefore, the minimum cost is $6250 and occurs when x  5 and y  70. So 5 sheets should be shipped from the east side warehouse to customer A, 70 sheets from the east side warehouse to customer B, and 45 sheets from the west side warehouse to customer A.

11. Let x be the number of bags of standard mixtures and y be the number of bags of deluxe mixtures. Then the data can be summarized by the following table: Standard

Deluxe

Available

Cashews

100 g

150 g

15 kg

Peanuts

200 g

50 g

20 kg

Selling price

$195

$220

Thus the total revenue, which we want to maximize, is given by R  195x  225yWe have the constraints        x  0, y  0, x  y  x  0, y  0, x  y  01x  015y  15 10x  15y  1500      02x  005y  20  20x  5y  2000

From the graph, the vertices occur at 0 0, 60 60, 90 40, and 100 0.

y

Vertex 0 0 60 60 90 40 100 0

R  195x  225y

196 0  225 0  0

10x+15y=1500

195 60  225 60  252

195 90  225 40  2655 195 100  225 0  195

x=y 20x+5y=2000

10 10

x

Hence, the confectioner should pack 90 bags of standard and 40 bags of deluxe mixture for a maximum revenue of $26550.


Linear Programming

63

12. Let x be the quantity of type I food and y the quantity of type II food, in ounces. Then the data can be summarized by the following table: Type I

Type II

Required

Fat

8g

12 g

24 g

Carbohydrate

12 g

12 g

36 g

Protein

2g

1g

4g

Cost

$020

$030

    x  0, y  0, Also, the total amount of food must be no more than 5 oz. Thus, the constraints are x  y  5, 8x  12y  24    12x  12y  36, 2x  y  4

The objective function is the cost C  02x  03y, which we wish to minimize. From the graph, the vertices occur at 1 2, 0 4, 0 5, 5 0, and 3 0. Vertex 1 2 0 4 0 5 5 0 3 0

y

C  02x  03y

x+y=5

02 1  03 2  08 02 0  03 4  12 02 0  03 5  15

2x+y=4 12x+12y=36

02 5  03 0  10 02 3  03 0  06

8x+12y=24 1 x

1

Hence, the rabbits should be fed 3 oz of type I food and no type II food, for a minimum cost of $060.

13. Let x be the amount in municipal bonds and y the amount in bank certificates, both in dollars. Then 12000  x  y is the amount in high-risk bonds. So our constraints can be stated as     x  0, y  0, x  3y     x  0, y  0, x  3y  x  y  12,000 12,000  x  y  0      12,000  x  y  2000  x  y  10,000

From the graph, the vertices occur at 7500 2500, 10000 0, 12000 0, and 9000 3000. The objective function is P  007x  008y  012 12000  x  y  1440  005x  004y, which we wish to maximize. y

Vertex 7500 2500 10000 0 12000 0 9000 3000

P  1440  005x  004y

x+y=12,000

1440  005 7500  004 2500  965

1440  005 10,000  004 0  940 1440  005 12,000  004 0  840

1440  005 9000  004 3000  870

x+y=10,000

x=3y

2000 2000

x

Hence, the advisor should invest $7500 in municipal bonds, $2500 in bank certificates, and the remaining $2000 in high-risk bonds for a maximum yield of $965.


64

FOCUS ON MODELING

14. The only change that needs to be made to the constraints in Exercise 13 is that the 2000 in the last inequality becomes 3000. Then we have     x  0, y  0, x  3y     x  0, y  0, x  3y  x  y  12,000 12,000  x  y  0      12,000  x  y  3000  x  y  9000

From the graph, the vertices occur at 6750 2250, 9000 0, 12000 0, and 9000 3000. The objective function is Y  007x  008y  012 12000  x  y  1440  005x  004y, which we wish to maximize. y

Vertex 6750 2250

Y  1440  005x  004y

1440  005 9000  004 0  990

9000 0 12000 0 9000 3000

x+y=12,000

1440  005 6750  004 2250  10125 1440  005 12,000  004 0  840

1440  005 9000  004 3000  870

x=3y

x+y=9000 2000

x

2000

Hence, she should invest $6750 in municipal bonds, $2250 in bank certificates, and $3000 in high-risk bonds for a maximum yield of $101250, which is an increase of $4750, over her yield in Exercise 13.

15. Let g be the number of games published and e be the number of educational programs published. Then the number of utility programs published is 36  g  e. Hence we wish to maximize profit, P  5000g  8000e  6000 36  g  e  216,000  1000g  2000e, subject to the constraints        g  4, e  0  g  4, e  0 36  g  e  0  g  e  36      36  g  e  2e  g  3e  36.

  From the graph, the vertices are at 4 32 3 , 4 32, and 36 0. The objective function is P  216,000  1000g  2000e. e

Vertex   4 32 3

P  216,000  1000g  2000e   216,000  1000 4  2000 32 3  233,33333

36 0

216,000  1000 36  2000 0  180,000

4 32

216,000  1000 4  2000 32  276,000

g=4 g+e=36 g+3e=36

10 10

g

So, they should publish 4 games, 32 educational programs, and no utility program for a maximum profit of $276,000 annually.


Linear Programming

    4y  3x  5 4x  y  15    x  3y  7

16.

The minimum value of P for which the line P  x  y intersects the feasible region is P  3, and the maximum value is P  10. These are the extreme values of P on the feasible region because the range of P on that region is

[3 10]. The minimum value occurs at 1 2 and the maximum value occurs at 5 5.

P=5 y 5 P=4 4 P=3 3 P=2 2 P=1 1 0

P=11 P=10 P=9 P=8 P=7 P=6 1

2

3

4

5

x

65


CORRECTIONS p. 19

CHAPTER 10

MATRICES AND DETERMINANTS

10.1

Matrices and Systems of Linear Equations 1

10.2

The Algebra of Matrices 12

10.3

Inverses of Matrices and Matrix Equations 20

10.4

Determinants and Cramer’s Rule 29 Chapter 10 Review 43 Chapter 10 Test 53

¥

FOCUS ON MODELING: Computer Graphics 56

1


10 MATRICES AND DETERMINANTS 10.1 MATRICES AND SYSTEMS OF LINEAR EQUATIONS 1. A system of linear equations with infinitely many solutions is called dependent. A system of linear equations with no solution is called inconsistent.     1 1 1 1  x  y  z  1    2. x  2z  3    1 0 2 3     2y  z  3 0 2 1 3

3. (a) The leading variables are x and y. (b) The system is dependent.

(c) The solution of the system is x  3  t y  5  2t z  t.    x  2    4. (a)   y1 0 1 0 1   z3 0 0 1 3 

1 0 0 2

1 0 1 2

The solution is 2 1 3.

   x z2   The solution is z  t, y  1  t, x  2  t. (b)  0 1 1 1   yz1 0 0 0 0     1 0 0 2 2  x    This system is inconsistent and has no solution. (c)   y 1 0 1 0 1    03 0 0 0 3

5. 3  2 

3

6. 2  4

1 1 2

 11.   2 1 1

7. 2  1

 0 1 

0 1 3

13. (a) Yes, this matrix is in row­echelon form. (b) Yes, this matrix is in reduced row­echelon form.   x  3 (c) y  5

8. 3  1 

 12.  

1 0

9. 1  3

0

1 1

3 2

1 1

3

10. 2  2

 7  3

14. (a) Yes, this matrix is in row­echelon form. (b) No, this matrix not in reduced row­echelon form. The entry above the leading 1 in the second row is not 0.   x  3y  3 (c)  y 5

1


2

CHAPTER 10 Matrices and Determinants

16. (a) Yes, the matrix is in row­echelon form.

15. (a) Yes, this matrix is in row­echelon form.

(b) Yes, the matrix is in reduced row­echelon form.    7z  0  x (c) y  3z  0    01

(b) No, this matrix is not in reduced row­echelon form, since the leading 1 in the second row does not have a zero above it.     x  2y  8z  0 (c) y  3z  2    00 17. (a) No, this matrix is not in row­echelon form, since the row of zeros is not at the bottom.

18. (a) Yes, the matrix is in row­echelon form. (b) Yes, the matrix is in reduced row­echelon form.   1  x (c) y 2    z3

(b) No, this matrix is not in reduced row­echelon form.   0  x (c) 00    y  5z  1 19. (a) Yes, this matrix is in row­echelon form.

20. (a) No, this matrix is not in row­echelon form, since the fourth column has the leading 1 of two rows.

(b) Yes, this matrix is in reduced row­echelon form.   x  3y    0      z  2  0 (c)  01      00

(b) No, this matrix is not in reduced row­echelon form.   x  3y   0      y  4 0 (c)  u2       0

Notice that this system has no solution.

 21.   

1 3

1

2

1

1

 4 

5

2 3

3

1

3

8

2

1 3

5

6 5 1

7

 23.  2 

3R1  R2  R2 

1 2 1 1

 22.   10 3 

0

3

1 3

 24.  0 0

1

 1 20  

1

 1 13  

2 1

 1 1  

2 1

1

   

2R2  R3  R3 

0

1

2

4

7

0

 4 

1 2 1 1

   

2R1  R2  R2 

3R1  R3  R3 

1

5 2 3

3

1 3

8

 0 1 5 14  

2

1 3

1

 2 3  0 8

5

 1 13   8 8

2 1

 0  0

1 3

1

 1 1  

0 3

3


3

SECTION 10.1 Matrices and Systems of Linear Equations

    x  2y  4z  3 25. (a) y  2z  7    z2

    x  y  3z  8 26. (a) y  3z  5    z  1

(b) y  2 2  7  y  3, so x  2 3  4 2  3 

(b) y  3 1  5  y  2, so x  2  3 1  8 

x  1. The solution is 1 3 2.

x  3. The solution is 3 2 1.

  x  2y  3z    7      y  2z 5 27. (a)  z  2  5      3

  x  2z  2  5      y  3z  1 28. (a)  z  0        1

(b) z  2 3  5  z  1, so y  2 1  5 

(b) z  1  0  z  1, so y  3 1  1 

y  3 and x  2 3  3 1  3  7  x  7. The

y  2 and x  2 1  2 1  5  x  5. The

solution is 5 2 1 1.

solution is 7 3 1 3. 

1 2 1 1

 29.  0

 1 2 5 

1

R3  R1  R3



1 3 8

1 2 1 1

 0  0

 1 2 5 

1 1

6 3

 0  0

R3  3R2  R3



3 2 7

1 2

1

1

1

 5 . Thus, 4z  8 

2

0 4 8

z  2; y  2 2  5  y  1; and x  2 1  2  1  x  1. Therefore, the solution is 1 1 2.

1 1 6 3

   30.  1 1 3 3

R2  R1  R2



1 2 4 7

 13 R2

   0 0 3 0    1 2 4 7

 R3  R1  R3

1 1

6 3

  0 0 1 0   0 1 2 4

1 1

6 3

   0 1 2 4 .   0 0 1 0

R3  R2



Thus, z  0; y  2z  y  0  y  4; and x  y  6z  3  x  4  0  3  x  1. Therefore, the solution is 1 4 0. 

 31.  

1 2 1

3

1 2

1

 4 

2 3 1 5

1 2

1

1

   0 7 1 1    0 7 3 3

R2  3R1  R2  R3  2R1  R3

R3  R2  R3



1 2

 0  0

1

1

 1 .

7 1 0

2 2

Thus, 2z  2  z  1; 7y  1  1  y  0; and x  2 0  1  1  x  2. Therefore, the solution is 2 0 1. 

1

0

1

7

5

3 1

2

1 1 2

0

   32.   2 1 2 5 

1

0

1

7

   0 1 4 19    0 3 6 33

R2  2R1  R2  R3  5R1  R3

R2  R2  R3  3R2  R3

1 0

1

7

  0 1 4 19   . 0 0 18 90

Thus, 18z  90  z  5; y  4 5  19  y  1; and x  5  7  x  2. The solution is 2 1 5. 

 33.  2 3

3

1

 5  

1 8 16 

R1  15 R2  R1 6R2  R3  R2

 1 30 R3

R2  2R1  R2  R3  3R1  R3

1 0 1 1    0 30 0 90    0 0 1 2

1 1 2

 0  0

5

5

0

 5  

4 2 16 

R1  R3  R1

 1 30 R2

4R2  5R3  R3 

1 1 2

 0  0

5

0

 5 5  

0 30 60

1 0 0 1    0 1 0 3 . Therefore, the solution is 1 3 2.   0 0 1 2


4

CHAPTER 10 Matrices and Determinants

1

 34.  1

0 2

1 3

 1 3 

R2  R1  R2  2R1  R3  R3

1 0 1 3

  0 2 0 0   0 3 4 4

1 2 R2

 2R3  3R2  R3

2 3 2 2   1 0 0 2    0 1 0 0 . The solution is 2 0 1.   0 0 1 1     1 2 1 9 1 2 1 9     R2  2R1  R2  0 4 1 20    35.     2 0 1 2   R3  3R1  R3 0 1 5 5 3 5 2 22

1 0 1 3

  0 1 0 0   0 0 8 8

4R3  R2  R3 

1

R1  18 R3  R1



2 1

1R 8 3

9

   0 4 1 20 .   0 0 19 0

Thus, 19x3  0  x3  0; 4x2  20  x2  5; and x1  2 5  9  x1  1. Therefore, the solution is 1 5 0.       2 1 0 7 2 1 0 7 2 1 0 7       R2  R1  R2 R3  7R2  R3   0 2 1 1  2  0 2 1 1 . Then, 9x3  9   36.    2 1 1 6  2     R3  3R1  R3 3 2 4 11 0 7 8 1 0 0 9 9 x3  1; 2x2  1  1  x2  1; and 2x1  1  7  x1  3. Therefore, the solution is x1  x2  x3   3 1 1.       2 3 1 13 2 3 1 13 2 3 1 13       2R2  R1  R2 3  13R2  R3   0 1 11 25  R  0 1 11  37.  25    1 2 5 6  2R3    . 5R1  R3 5 1 1 49 0 13 3 33 0 0 146 292

Thus, 146z  292  z  2; y  11 2  25  y  3; and 2x  3  3  2  13  x  10. Therefore, the solution is 10 3 2.       10 10 20 60 10 10 20 60 10 10 20 60   2R2  3R1  R2     3  7R2  R3    0 10 120 230  R   38.   15 20 30 25  2R3     0 10 120 230 . R1  R3 5 30 10 45 0 70 40 150 0 0 880 1760

Thus, 880z  1760  z  2; 10y  120 2  230  y  1; and 10x  10  40  60  x  1. Therefore, the solution is 1 1 2.       1 2 1 3 1 2 1 3 1 0 11 11   R2  3R1  R2     1  2R2  R1   0 1 5 4  R 0 1   39.  5 4  3 7 2 5    R3  . The third row of the 2R1  R3  R3 R2  R3 2 3 7 4 0 1 5 2 0 0 0 6

matrix states 0  6, which is impossible. Hence, the system is inconsistent, and there is no solution.       1 2 3 4 1 2 3 4 3 0 1 3   R2  3R1  R2     R1  R2  R1   0 6 10 15  3  0 6 10 15 .   40.   3 0 1 3  R3     R1  R3 R3  R2  R3 1 4 7 2 0 6 10 2 0 0 0 13

The system is inconsistent, and there is no solution.       2 3 9 5 1 0 3 2 1 0 3 2   R R   R2  2R1  R2    1 R2 1 2    3   0 3 15 9  41.   1 0 3 2    2 3 9 5      R3  3R1  R3 3 1 4 3 3 1 4 3 0 1 5 3     1 0 3 2 1 0 3 2     3  R2  R3  0 1 5 3 R      0 1 5 3 . Therefore, this system has infinitely many solutions, given by x  3t  2 0 1 5 3 0 0 0 0  x  2  3t, and y  5t  3  y  3  5t. Hence, the solutions are 2  3t 3  5t t, where t is any real number.


5

SECTION 10.1 Matrices and Systems of Linear Equations

1  42.  2 

5

2

3

 1 

6 11

3 16

R2  2R1  R2  R3  3R1  R3

20 26

1 2 5 3   0 2 1 7   0 10 5 35

R3  5R2  R3  

1 2

5 3  2 1 7  .

  0

0

0

0 0

The system is dependent; there are infinitely many solutions, given by 2y  t  7  2y  t  7  y  12 t  72 ; and     x  2 12 t  72  5t  3  x  t  7  5t  3  x  4t  10. The solutions are 4t  10 12 t  72  t , where t is any real number. 

1 1

3

3

   43.   4 8 32 24  2 3 11

3

3

 14 R2

   0 4 20 12    0 1 5 2

R2  4R1  R2  R3  2R1  R3

4

1 1

1 1

 0  0

 R3  R2  R3

3

3

 1 5 3  . The third row of 0

0 5

the matrix states 0  5, which is impossible. Hence, the system is inconsistent, and there is no solution.

 44.  

6 2 12

2

1 3 3

1

  1 R1   2 10    

2

2

6

1 3 1

6

1 3 1

 R2  R1  R2   1 3 2 10     0

1

3 2

6

R3  R1  R3

0

0 1

6

1 3 1 6

 R  3R  R  3 2 3  4    0

0 3 12

0

 0 1 4 . 0 0 0

The system is dependent; there are infinitely many solutions, given by z  4, and x  3y  z  6, so x  3t  4  6  x  3t  2. The solutions are 3t  2 t 4, where t is any real number. 

1

4 2 3

 45.   2 1 8

 5 12   5 11 30

R2  2R1  R2  R3  8R1  R3

1

4 2 3

   0 9 9 18    0 27 27 54

R3  3R2  R3



1

4 2 3

  0 9  0 0

 9 18  . 0 0

Therefore, this system has infinitely many solutions, given by 9y  9t  18  y  2  t, and x  4 2  t  2t  3  x  5  2t. Hence, the solutions are 5  2t 2  t t, where t is any real number. 

3

2 3 10

   46.   1 1 1 5  1

4 1 20

R1  R2



1 1 1 5

 3  1

 2 3 10   4 1 20

R2  3R1  R2  R3  R1  R3

1 1 1 5

 0  0

5 5

 0 25   0 25

R3  R2  R3



1 1 1 5    0 5 0 25 . The system is dependent; there are infinitely many solutions, given by 5s  25  s  5, and   0 0 0 0

r  5  t  5  r  t. Hence, the solutions are t 5 t, where t is any real number. 

2

1 2 12

3

3 3 2

 1 47.   1  2

 1 6   18

R1  R2  R1

1 12 1

6

   2 1 2 12    3 32 3 18

R2  2R1  R2  R3  3R1  R3

1 12 1 6

 0 0  0 0

 0 0 . 0 0

Therefore, this system has infinitely many solutions, given by x  12 s  t  6  x  6  12 s  t. Hence, the solutions are   6  12 s  t s t , where s and t are any real numbers. 

 0 1 5 7   R R 2  1 48.   3 2 0 12   3 0 10 80

 3 2 0 12   3  R1  R3  0 1 5 7  R    3 0 10 80

Therefore, the system is inconsistent and has no solution.

 3 2 0 12   3  2R2  R3  0 1 5 7  R    0 2 10 68

 3 2 0 12    0 1 5 7 .   0 0 0 82


6

CHAPTER 10 Matrices and Determinants

4 3

 49.   2

 1 3 4  

1 1

1 8

2

1 1

2

3

1 1

2

3

    R 2  2R1  R2 2  2 1 3 4  R  0 1 1  2       R3  4R1  R3 4 3 1 8 0 1 7 20   1 1 2 3   R3  R2  R3  . Therefore, 6z  18  z  3; y  3  2  0 1 1 2    0 0 6 18 R1  R3



3

 1 1 2 3    0 1 1 2    0 1 7 20

y  1; and x  1  2 3  3  x  2. Hence, the solution is 2 1 3.

2 3

5

14

1

1 1

3

1

1 1 3

 R2  4R1  R2    R1  R3      50.   4 1 2 17    4 1 2 17    0 5 1 1

1

3 1

1 3

R1

3

2 3

5

14

R3  2R1  R3

1

1 1 3

0

0

 R R R  3 2 3  2 5     0 5

0 5

7 20

 2 5  . 5 25

Thus 5z  25  z  5; 5y  2 5  5  5y  15  y  3; and x  3  5  3  x  1. Hence, the solution is 1 3 5. 

 51.  1

 2 4  

0

2

 0  0

3R2  R1  R2  R3  2R2  R3

1 11 1

3 1

1

3

 9 

1 7

 0  0

R2  R3  R3



1 7 7

Therefore, the system is inconsistent and there is no solution. 

 52.  

1 3

2

5

 2 3 2 2   0

1

4

7

1 3

2

5

   0 3 6 12    0 3 6 12

R2  2R1  R2  R3  R1  R3

3 1

1 3 R2

 R2  R3  R3

1

3

 9 .

1 7 0

0 16

1 3

2

5

0

0

0

 0  0

 1 2 4  . Therefore,

the system is dependent. Let z  t. Then y  2t  4  y  4  2t and x  3 4  2t  2t  5  x  3 4  2t  2t  5  4t  7. The solutions are 4t  7 4  2t t, where t is any real number. 

1

2 3

 53.   2 4 6 3

5

 10  

7 2 13

1 2

3 5

  0 0 12  0 1 7

R2  2R1  R2  R3  3R1  R3

 0 

2

1 2

3 5

 0 1 7  0 0 12

R2  R3



 2 . Therefore, 0

12z  0  z  0; y  7 0  2  y  2; and x  2 2  3 0  5  x  9. Hence, the solution is 9 2 0. 

3 1 0 2

   54.   4 3 1 4 

3R2  4R1  R2  2R3  R2  R3

2 5 1 0

3

1 0

2

   0 13 3 20    0 13 3 4

R2  R3  R3



Therefore, the system is inconsistent and there is no solution. 

1 1

 55.  1 1

0

6 1

8

 5 

3 14 4

R2  R1  R2  R3  R1  R3

1 1

 0  0

1

6 5

8

 3  

4 20 12

3

1 0

2

1 1

6

8

0

0

0

   0 13 3 20 .   0 0 0 16

R3  4R2  R3



 0  0

 1 5 3  . Therefore,

the system is dependent. Let z  t. Then y 5t  3  y  35t and x  y 6t  8  x  83  5t6t  5t. The solutions are 5  t 3  5t t, where t is any real number.


7

SECTION 10.1 Matrices and Systems of Linear Equations

 56.   

3 1

4 2 1

2 1  1 7  

3 2 1

 1 3 2 1    0 2 1 1    0 10 7 11

1 3 2 1 1 3 2 1    R2  1 R3 2  4R1  R2   4 2 1 7  R 4   0 10 7 11    R3  3R1  R3    3 1 2 1 0 8 4 4   1 3 2 1   R3  5R2  R3  0 2 1 1    . Thus 2z  6  z  3; 2y  3  1  2y  2  0 0 2 6 R1  R3

 R1

y  1; and x  3 1  2 3  1  x  2. Hence, the solution is 2 1 3.       1 2 1 3 3 1 2 1 3 3 1 2 1 3 3       R2  3R1  R2  3 4 1 1 9   0 2 4 8 18   3 4 1 1 9     R1    R3  R1  R3 57.         0 3 0 4 3   1 1 1 1 0    1 1 1 1 0  R4  2R1  R4       2 1 4 2 3 0 5 6 8 9 2 1 4 2 3     1 2 1 3 3 1 2 1 3 3      0 1 2 4 9  R  3R  R  0 1 2 4 1 9   323   3R4  2R3  R4 2 R2       0 3 0 4 3  R4  5R2  R4  0 0 6 8 24       0 5 6 8 9 0 0 4 12 36   1 2 1 3 3    0 1 2 4 9    . Therefore, 20  60    3; 6z  24  24  z  0. Then y  12  9  y  3 and  0 0 6 8 24    0 0 0 20 60

x  6  9  3  x  0. Hence, the solution is 0 3 0 3.     1 1 1 1 6 1 1 1 1 6     R2  R4  R2 R2  2R1  R2  2 0 1 3  0 2 3 1 4  8 R3  R4  R3     R3 R1  R3 58.       1  1 1 0 4 10   0 2 1 5 16  R4  3R1  R4 2 R4     3 5 1 1 20 0 2 2 2 2     1 1 1 1 6 1 1 1 1 6     0 1 1 1 0 1 1 1 1 1 R4  R2     5 R  3 R  R 4 3 4     . R3  R4  0 0 5 1 2    0 0 5 1 2      0 0 3 7 14 0 0 0 32 64

1 1 1 1

 0 0   0 0  0 1

5 3

1

 2    7 14   1 1 1

Thus 32  64    2; 5z  2  2  5z  0  z  0; y  0  2  1  y  3; and x  3  0  2  6  x  1. Hence the solution is 1 3 0 2.       1 1 2 1 2 1 1 2 1 2 1 1 2 1 2        0 3 1 2 2 0 3 1 2 2 0 3 1 2 2 R3  R1  R3       R  R2  R4 4           59.         1 1 0 3 2  R4  3R1  R4  0 0 2 4 4   0 0 2 4 4        0 3 7 1 1 0 0 6 3 3 3 0 1 2 5   1 1 2 1 2   0 3 1 2 2   R4  3R3  R4 . Therefore, 9  9    1; 2z  4 1  4  z  0. Then    0 0 2 4 4    0 0 0 9 9 3y  0  2 1  2  y  0and x  0  2 0  1  2  x  1. Hence, the solution is 1 0 0 1.

6


8

CHAPTER 10 Matrices and Determinants

1 3 2

1

2

   1 2 0 2 10    60.    0 0 1 5 15    3

0 2

1

R4  14R3  R4



1 3

 0   0 

R2  R1  R2  R4  3R1  R4

2

1 2

 1 2 3 8    0 1 5 15   9 4 2 3

R4  9R2  R4



1 3

 0   0 

2

1 2

 1 2 3 8    0 1 5 15   0 14 25 75

3 0 0   1 3 2 1 2    0 1 2 3 8    . Thus, 45  135    3; z  5 3  15  z  0;   0 0 1 5 15   0 0 0 45 135

y  2 0  3 3  8  y  1; and x  3 1  2 0  3  2  x  2. Hence, the solution is 2 1 0 3.

1 1

 61.  3

0 1 0

 0 1 2 0  

1 4

R2  3R1  R2  R3  R1  R3

1 2 0

1 1

0

1 0

   0 3 1 1 0    0 3 1 1 0

R3  R2  R3



1 1

0

1 0

0

0

0 0

 0  0

 3 1 1 0  .

Therefore, the system has infinitely many solutions, given by 3y  s  t  0  y  13 s  t and x  13 s  t  t  0    x  13 s  2t. So the solutions are 13 s  2t  13 s  t  s t , where s and t are any real numbers. 

     2 1 2 1 5 1 1 4 1 3 1 1 4 1 3   R1  R2     2  2R1  R2   2 1 2 1 5  R  0 1 10 1 11    62.   1 1 4 1 3        R1 R3  3R1  R3 3 2 1 0 0 3 2 1 0 0 0 1 11 3 9   1 1 4 1 3   R3  R2  R3  0 1 10 1 11 . Thus, the system has infinitely many solutions, given by z  2t  2     0 0 1 2 2

z  2  2t; y  10 2  2t  t  11  y  31  19t; and x  31  19t  4 2  2t  t  3  x  20  12t. Hence, the solutions are 20  12t 31  19t 2  2t t, where t is any real number.

1

0

1 1

4

   0 1 1 0 4    63.    1 2 3 1 12    2

R3  R1  R3



R4  2R1  R4

1

0

1 1

4

   0 1 1 0 4       0 2 2 0 8   

R3  2R2  R3



1 0

1 1

4

   0 1 1 0 4      0 0 0 0 0  

0 2 5 1 0 0 4 3 9 0 0 4 3 9   1 0 1 1 4    0 1 1 0 4    R3  R4  . Therefore, 4z  3t  9  4z  9  3t  z  94  34 t. Then we have   0 0 4 3 9    0 0 0 0 0     3 9 3 7 7 y  94  34 t  4  y  7 4  4 t and x  4  4 t  t  4  x  4  4 t. Hence, the solutions are   7  7 t  7  3 t 9  3 t t , where t is any real number. 4 4 4 4 4 4


SECTION 10.1 Matrices and Systems of Linear Equations

9

0 0 1 1 2 0 1 0 0 2 6      R2  R3  R2  1 2 0 3 12   0 2 0 5 18  0 R  R R4  R3  R4 1 3             1  R2  R1  R2  1 0 0 2  0 1 1 2 2 0 0 4 12  6 0 R 3 2      2 0 2 5 6 0 0 2 9 18 0 0 2 9 18     1 0 0 2 6 1 0 0 2 6      0 1 1 2  0 1 1 2 0 0 R2  2R3  R2     R  R  R 4 3 4   .   R2  R3  0 0 2 9 18    0 0 2 9 18      0 0 2 9 18 0 0 0 0 0   Thus, the system has infinitely many solutions given by 2z  9t  18  z  92 t  9; y  92 t  9  2t  0    y  52 t  9; and x  2t  6  x  2t  6. So the solutions are 2t  6 52 t  9 92 t  9 t , where t is any real number.

   64.   

0 1 1 2

3 2

0 1

075 375 295 40875

 65. Using a graphing device to find the reduced row­echelon form of   095 875 0 is x  125, y  025, z  075.

 3375  , we find that the solution 125 015 275 36625

131 272 371 139534

 66. Using a graphing device to find the reduced row­echelon form of   021 0 0

solution is x  371, y  172, z  381.

0

373

 134322  , we find that the

234 456 213984

42 31

  6  67. Using a graphing device to find the reduced row­echelon form of   35  solution is x  12, y  34, z  52,   13.

0 42

04

 45   , we find that the 0 67 32 3488   31 48 52 766 0

49 27 52

0

9

0 145

   0 27 0 43 1187    68. Using a graphing device to find the reduced row­echelon form of  , we find that the solution  0 31 42 0 721    73 54 0 0 1327 is x  13, y  07, z  12,   32.

69. Let x, y, z represent the number of VitaMax, Vitron, and VitaPlus pills taken daily. The matrix representation for the system of equations is   1       5 10 15 50 1 2 3 10 1 2 3 10 1 2 3 10 5 R1   1 2   2  3R1  R2     3  R2  R3   15 20 0 50  5R  3 4 0 10  R    0 2 9 20  R   1       0 2 9 20 . R3  2R1  R3  5 R3 10 10 10 50 2 2 2 10 0 2 4 10 0 0 5 10 Thus, 5z  10  z  2; 2y  18  20  y  1; and x  2  6  10  x  2. Hence, the patient should take 2 VitaMax, 1 Vitron, and 2 VitaPlus pills daily.


10

CHAPTER 10 Matrices and Determinants

70. Let x be the quantity, in mL, of 10% acid, y the quantity of 20% acid, and z the quantity of 40% acid. Then     1 1 1 100  10R2  01x  02y  04z  18   R2 R1  R2   Writing this equation in matrix form, we get  x  y  z  100 01 02 04 18       R   3 R1  R3   x  4z  0 1 0 4 0 

1

1

1

100

  0 1 3 80    0 1 5 100

1 1 1 100

 0 1 3  0 0 2

R3  R2  R3  R3

 80  . Thus 2z  20  z  10; y  30  80  y  50; and 20

x  50  10  100  x  40. So the chemist should mix together 40 mL of 10% acid, 50 mL of 20% acid, and 10 mL of 40% acid.

71. Let x, y, and z represent the distance, in miles, of the run, swim, and cycle parts of the race respectively. Then, since distance time  , we get the following equations from the three contestants’ race times: speed        y x z       10    4    20   25  2x  5y  z  50 x  y  z 3  4x  5y  2z  90 which has the following matrix representation: 75 6 15       x    y    z   175  8x  40y  3z  210 

15

2

3

5 1

50

   4 5 2 90    8 40 3 210

40

R2  2R1  R2  R3  4R1  R3

2

5

1

50

   0 5 0 10    0 20 1 10

2

5

1

50

   0 5 0 10 .   0 0 1 30

R3  4R2  R3



Thus, z  30  z  30; 5y  10  y  2; and 2x  10  30  50  x  5. So the race consists of a 5­mile run, a 2­mile swim, and a 30­mile cycle.

72. Let a, b, and c be the number of students in classrooms A, B, and C, respectively, where a, b, c  0. Then, the        a  b  c  100  a  b  c  100 1a  1b  By substitution, a  52 a  32 a  100  a  20; system of equations is b  52 a 2 5      c  3b  3a  1b  1c 5 3 5 2 5 3 b  2 20  50; and c  2 20  30. So there are 20 students in classroom A, 50 students in classroom B, and

30 students in classroom C.

73. Let t be the number of tables produced, c the number of chairs, and a the number of armoires. Then, the system of equations   1      t  2c  2a  600  2 t  c  a  300 1 3 and a matrix representation is is 2 t  2 c  a  400  t  3c  2a  800      2t  3c  4a  1180  t  3 c  2a  590 

2

1 2 2

600

   1 3 2 800    2 3 4 1180

R2  R1  R2  R3  2R1  R3

1

2 2 600

   0 1 0 200    0 1 0 20

R3  R2  R3



1 2 2 600

   0 1 0 200 .   0 0 0 180

The third row states 0  180, which is impossible, and so the system is inconsistent. Therefore, it is impossible to use all of the available labor­hours.


11

SECTION 10.1 Matrices and Systems of Linear Equations

74. The number of cars entering each intersection must equal the number of cars leaving that intersection. This leads   200  180  x  z      x  70  20   to the following equations: Simplifying and writing this in matrix form, we get     200  y  30     y  z  400  200       1 0 1 0 380 1 0 1 0 380 1 0 1 0 380        1 0 0 1 50  R  R  R  0 0 1 1 430   0 1 0 1 170  2 1 2        R  R 2 3    .     0 1 0 1 170  R4  R3  R4  0 1 0 1 170    0 0 1 1 430        0 1 1 0 600 0 0 1 1 430 0 0 1 1 430

Therefore, z  t  430  z  430  t; y  t  170  y  170  t; and x  430  t  380  x  t  50. Since x, y, z,   0, it follows that 50  t  430, and so the solutions are t  50 170  t 430  t t, where 50  t  430.

75. Line containing the points 0 0 and 1 12: Using the general form of a line, y  ax  b, we substitute for x and y and solve for a and b. The point 0 0 gives 0  a 0  b  b  0; the point 1 12 gives 12  a 1  b  a  12. Since a  12 and b  0, the equation of the line is y  12x. Quadratic containing the points 0 0, 1 12, and 3 6: Using the general form of a quadratic, y  ax 2  bx  c, we substitute for x and y and solve for a, b, and c. The point 0 0 gives 0  a 02  b 0  c  c  0; the point 1 12

gives 12  a 12  b 1  c  a  b  12; the point 3 6 gives 6  a 32  b 3  c  9a  3b  6. Subtracting the third equation from 3 times the third gives 6a  30  a  5. So a  b  12  b  12  a  b  17. Since a  5, b  17, and c  0, the equation of the quadratic is y  5x 2  17x.

Cubic containing the points 0 0, 1 12, 2 40, and 3 6: Using the general form of a cubic, y  ax 3  bx 2  cx  d,

we substitute for x and y and solve for a, b, c, and d. The point 0 0 gives 0  a 03  b 02  c 0  d  d  0; the point the point 1 12 gives 12  a 13  b 12  c 1  d  a  b  c  d  12; the point 2 40 gives

40  a 23  b 22  c 2  d  8a  4b  2c  d  40; the point 3 6 gives 6  a 33  b 32  c 3  d      a  b  c  12 27a  9b  3c  d  6. Since d  0, the system reduces to 8a  4b  2c  40 which has representation    27a  9b  3c  6 

1 1 1 12

1

1

1

12

 12 R2

1 1 1 12

1 1

1 12

   R  3R  R     2  8R1  R2  3 2 3   8 4 2 40  R   0 4 6 56    0 2 3 28  2 3 28    0 2     .   R3 1  27R1  R3  6 R3 0 18 24 318 0 3 4 53 0 0 1 22 27 9 3 6

So c  22 and back­substituting we have 2b  3 22  28  b  47 and a  47  22  0  a  13. So the

cubic is y  13x 3  47x 2  22x. Fourth­degree polynomial containing the points 0 0, 1 12, 2 40, 3 6, and 1 14: Using the general form of a

fourth­degree polynomial, y  ax 4  bx 3  cx 2  dx  e, we substitute for x and y and solve for a, b, c, d, and e. The point

0 0 gives 0  a 04 b 03 c 02 d 0e  e  0; the point 1 12 gives 12  a 14 b 13 c 12 d 1e;

the point 2 40 gives 40  a 24 b 23 c 22 d 2e; the point 3 6 gives 6  a 34 b 33 c 32 d 3e; the point 1 14 gives 14  a 14  b 13  c 12  d 1  e.


12

CHAPTER 10 Matrices and Determinants

Because the first equation is e  0, we eliminate e from the other equations to get       a  b  c  d  12 1 1 1 1 12 1 1 1 1 12        1   16a  8b  4c  2d  40  16 8 4 2 40  R2  16R1  R2  0 8 12 14 152   2 R4  R2   R3 81R1  R3   R2  R3            R  R  R 6 6 4  81a  27b  9c  3d  1 4   81 27 9 3  0 54 72 78 966  R3  R4    a  b  c  d  14 1 1 1 1 14 0 2 0 2 26       1 1 1 1 12 1 1 1 1 12 1 1 1 1 12        0  0 1 1 0 1 13  0 1 13  0 1 13 3  8R2  R3  0 1  R     R  6 R  R     434  .   0 8 12 14 152  R4  54R2  R4  0 0 12 6 48   0 0 12 6 48        0 54 72 78 966 0 0 72 24 264 0 0 0 12 24

So d  2. Then 12c  6 2  48  c  3and b  2  13  b  11.

Finally, a  11  3  2  12  a  4. So the fourth­degree polynomial

40

containing these points is y  4x 4  11x 3  3x 2  2x.

20

­2

2

4

­20

10.2 THE ALGEBRA OF MATRICES 1. We can add (or subtract) two matrices only if they have the same dimension. 2. (a) We can multiply two matrices only if the number of columns in the first matrix is the same as the number of rows in the second matrix. (b) If A is a 3  3 matrix and B is a 4  3 matrix, then (ii) B A and (iii) A A are defined, but (i) AB and (iv) B B are not.

3. (i) A  A and (ii) 2A exist for all matrices A, but (iii) A  A is not defined when A is not square.

4. The entry in the first row, second column is a12  3 3  1 2  2 1  9; the entry in the second row, third column is a23  1 2  2 1  0 0  0; the entry in the third row, first column is a31  1 1  3 3  2 2  4, and      3 1 2 1 3 2 4 9 7          so on:   1 2 0   3 2 1    7 7 0  1 3 2 2 1 0 4 5 5 5. The matrices have different dimensions, so they cannot be equal.  6. Because 14  025, ln 1  0, 2  4, and 3  62 , the corresponding entries are equal, so the matrices are equal.

7. All corresponding entries must be equal, so a  5 and b  3.

8. All corresponding entries must be equal, so a  5 and b  7.             2 6 1 3 1 3 0 1 1 2 1 1 2 0 2       9.  10.  5 3 6 2 1 5 1 1 0 1 3 2 0 2 2         1 1 0 1 1 1 2 3 6               11. 3  12. 2   1 0 1    2 1  is undefined because these  4 1    12 3  0 1 1 3 1 1 0 3 0 matrices have incompatible dimensions.


SECTION 10.2 The Algebra of Matrices



2 6 1 2      3 6  is undefined because these 13.  1 3    2 4 2 0 matrices have incompatible dimensions.



1 2

3



5

2

1

2 1 2

  14.  6 3 4  

2 3

   1 2   6  7 6 0

1 2

3 2 

13

7

   5  1 1    1 4 2 2 1 7 10 7 1 7 1 2         1  2 5   4 6  1  2 1   1  12  1 17. 2X  A  B  X  2 B  A  .    2 2 1 2 3 7 2 4 1 3

15. 

1 2



 16.  0

18. 3X  B  C  3X  C  B. but C is a 3  2 matrix and B is a 2  2 matrix, so C  B is impossible. Thus, there is no solution. 19. 2 B  X  D. Since B is a 2  2 matrix, B  X is defined only when X is a 2  2 matrix, so 2 B  X is a 2  2 matrix. But D is a 3  2 matrix. Thus, there is no solution. 20. 5 X  C  D  X  C  15 D            10 20 2 3 2 4 2 3 4 7          1         X  15 D  C   30 20     1 0    6 4    1 0    7 4 . 5 10 0 0 2 2 0 0 2 2 2

21. 15 X  D  C  X  D  5C            2 3 10 20 10 15 10 20 0 5                    X  5C  D  5   1 0    30 20    5 0    30 20    25 20 . 0 2 10 0 0 10 10 0 10 10 22. 2A  B  3X  2A  B  3X 

X   13 2A  B              1  8 12   2 5  1  6 7   2  73  1   4 6   2 5  2       1 1 3 3 3 1 1 1 3 3 7 2 6 3 7

In Solutions 23–36, the matrices A, B, C, D, E, F, G, and H are defined as follows:       0 2 5 2  52 3 12 5    A C  B 1 1 3 0 2 3 0 7       1 1 0 0 5 3 10          E  F  G 2 0 1 0  6 1 0 0 0 0 1 5 2 2       0 3 12 5 2  52 5 2 5    23. (a) B  C   1 1 3 0 2 3 1 1 0

(b) B  F is undefined because B 2  3 and F 3  3 don’t have the same dimensions.       0 3 12 5 1 3 5 2  52    24. (a) C  B   0 2 3 1 1 3 1 3 6

D

3

3

7 3

3

1

H 

2 1

 


14

CHAPTER 10 Matrices and Determinants

(b) 2C  6B  2  

0

2 5

25. (a) 5A  5 

0

10 25

2  52

7

0

2 3



  6

0

35

1 5 14 8 30 2  

3

1 1 3

6 10 24

(b) C  5A is undefined because C 2  3 and A 2  2 don’t have the same dimensions.       2  52 0 13  72 15 3 12 5  2   26. (a) 3B  2C  3  1 1 3 0 2 3 3 1 3

(b) 2H  D is undefined because 2H 2  2 and D 1  2 don’t have the same dimensions.

27. (a) AD is undefined because A 2  2 and D 1  2 have incompatible dimensions.     2 5     14 14 (b) D A  7 3  0 7 28. (a) D H 

7 3 

3

1

2 1



27 4

(b) H D is undefined because H and D have incompatible dimensions.      4 7 2 5 3 1    29. (a) AH   14 7 0 7 2 1      3 1 2 5 6 8    (b) H A   2 1 0 7 4 17

30. (a) BC is undefined because B 2  3 and C 2  3 have incompatible dimensions.    1 0 0    1  3 12 5 3 2 5   0 1 0     (b) B F    1 1 3  1 1 3 0 0 1 

 31. (a) G F   

5 3 10 1

1 0 0

      0  0 1 0   

5 3 10

6

1

0 0 1 5    5 3 10 1 1           (b) G E     6 1 0  2   8  5 2 2 0 1

2

6



2

5

2



32. (a) B 2 is undefined because B 2  3 is not square.      1 0 0 1 0 0 1 0 0          (b) F 2    0 1 0  0 1 0    0 1 0  0 0 1 0 0 1 0 0 1 

33. (a) A2  

2 5

0

7

 

2 5

0

7



4 45

0

49

 

 0  2


SECTION 10.2 The Algebra of Matrices

(b) A3  

2 5



0

34. (a) D A B 

7



2 5

0 

7 3 

7

 

2 5

2 5

0  

7

3

 

4 45

0

49

 

2 5 0

7

1 5   3 2   14 14 



8 335

0 

343

15

 

1 5   2   28 21 28

1 1 3    1 6 5   2 5   3 12 5  7 3    28 21 28  (b) D AB  7 3  1 1 3 7 7 21 0 7      1   1     1 1 6 5   2 5 3 2 5   13  2     2      35. (a) AB E   1 1 3   7 7 21   0 7 7 0 0 0

7



1 1 3

(b) AH E is undefined because the dimensions of AH 2  2 and E 3  1 are incompatible.         5 2 5     0 2  52 3 12 5    7 3    38 11 35 36. (a) D BDC  D B  C  7 3  1 1 3 0 2 3 1 1 0 (b) B F  F E is undefined because the dimensions 2  3  3  3  2  3 and 3  3  3  1  3  1 are incompatible.

In Solutions 37–42, the matrices A, B, and C are defined as follows:       12 01 03 11 24     02 02 01    C  B A  0 05   09 01 04  11 21 21 05 21 07 03 05   156 562    37. AB   128 088  38. B has 2 columns and A has 3 rows, so B A is undefined.  109 097     035 003 033   019 029   39. BC   40. C B    055 105 105  027 325 241 431 446   06 094 004   41. B and C have different dimensions, so B  C is undefined. 42. A2   112 192   01  041 095 131     x 2        2y  2 x 2y 2 2  . Thus we must solve the system So x  2 and 2y  2  y  1. Since 43.   4  2x 4 6 2x 6y      6  6y these values for x and y also satisfy the last two equations, the solution is x  2, y  1.

  3x  6        x y 6 9 x y 3x 3y 3y  9  . Since 3   , we must solve the system 44. 3   y x 9 6 y x 3y 3x 3x  6      3y  9 

x  2 and 3y  9  y  3. Thus, the solution is x  2, y  3.

So 3x  6 


16

CHAPTER 10 Matrices and Determinants

45. 2 

x

y

2 4

x

y

2x

2y

 . Since 2   , Thus we must solve the xy xy 2 6 xy xy 2 x  y 2 x  y   2x  2      2y  4 So x  1 and y  2. Since these values for x and y also satisfy the last two equations, the system  2 x  y  2      2 x  y  6

solution is x  1, y  2.           x y y x 4 4 x y yx 4 4     . Thus we must solve the system 46.  y x x y 6 6 x  y x  y 6 6   xy 4      y  x  4 Adding the first equation to the last equation gives 2x  10  x  5 and 5  y  6  y  1. So the   x  y  6     xy 6 solution is x  5, y  1.        2x  5y  7 2 5 x 7      . 47. written as a matrix equation is   3x  2y  4 3 2 y 4     6x  y  z  12 48. 2x  z 7    y  2z  4

 written as a matrix equation is  2

3

  x y z2      4x  2y  z  2 50.   x  y  5z  2     x  y  z  2 

4

1



x

12

        1   y    7 .

0

0

    3x1  2x2  x3  x4  0 49. x1  x3 5    3x  x x  4 2

6 1

1 2

4

z

x1

2

    0    x2      written as a matrix equation is     5 .  1 0 1 0    x3      4 0 3 1 1 x4 

   written as a matrix equation is   

1 1

3 2 1

1

1

 x      4 2 1   2      y   . 2 1 1 5  z   1 1 1 2 1

   0    . ABC is undefined because the dimensions of A (2  4) 1 7 9 2 , and C    1  2 12 4 0   2      3  3 21 27 6  1 7 9 2   . B AC is undefined because and B (1  4) are not compatible. AC B   2 2 14 18 4

51. A  

1 0 6 1

, B 

the dimensions of B (1  4) and A (2  4) are not compatible. BC A is undefined because the dimensions of C (4  1) and A (2  4) are not compatible. C AB is undefined because the dimensions of C (4  1) and A (2  4) are not compatible. C B A is undefined because the dimensions of B (1  4) and A (2  4) are not compatible.


SECTION 10.2 The Algebra of Matrices

52. (a) Let A  

a b c d 

 and B  

e f g h



. Then A  B  



a b

a b

A2  

AB  

c d

c d

cg d h

cg d h

a 2  bc ab  bd

e2  f g e f  f h

ae  bg a f  bh

ae  c f be  d f

a 2  bc  ae  bg  ae  c f  e2  f g ab  bd  e f  f h  a f  bh  be  d f

c  g a  e  d  h c  g



a b



e f



g h

A2  AB  B A  B 2  





c d

, and

a  e b  f   b  f  d  h

a  e2  b  f  c  g



cg d h

ae b f ae b f   A  B2   

ae b f



ac  cd bc  d 2

ce  dg c f  dh

c  g b  f   d  h2

; B 2  

; B A  

e f g h e f g h

 

 

e f g h a b c d

;

 

eg  gh f g  h 2

ag  ch bg  dh

ac  cd  eg  gh  ce  dg  ag  ch bc  d 2  f g  h 2  c f  dh  bg  dh a 2  2ae  e2  b c  g  f c  g

;

. Then 

a b  f   e b  f   b d  h  f d  h

c b  f   g b  f   d 2  2dh  h 2   a  e b  f   b  f  d  h a  e2  b  f  c  g   A  B2  c  g a  e  d  h c  g c  g b  f   d  h2 c a  e  g a  e  d c  g  h c  g

17

(b) No. From part (a), A  B2  A  B A  B  A2  AB  B A  B 2  A2  2AB  B 2 unless AB  B A which is not true in general, as we saw in Example 5.      5 075 010 0 4          53. (a) AB    025 070 070   20    22  7 0 020 030 10

(b) Five members of the group have no postsecondary education, 22 have 1 to 4 years, and seven have more than 4 years.      075 020 005 80 96          54. (a) AB    060 030 010   170    103  040 030 030 40 95 (b) 96 students slept less than 4 hours, 103 slept 4 to 7 hours, and 95 slept more than 7 hours.      50 20 15 350 35375          55. (a) AB    40 75 20   575    65625  35 60 100 425 89250 (b) The total revenue for Monday is the 1 1th entry of the product matrix, $35375.

(c) The total revenue is the sum of the three entries in the product matrix, $190250.      4000 1000 3500      $4690 $1690 $13,210 56. (a) B A  $090 $080 $110  400 300 200   700 500 9000

 


18

CHAPTER 10 Matrices and Determinants

(b) The entries in the product matrix represent the total food sales in Santa Monica, Long Beach, and Anaheim, respectively.      12 10 0 $1000 $500 $32,000 $18,000          57. (a) AB    4 4 20   $2000 $1200    $42,000 $26,800  8 9 12 $1500 $1000 $44,000 $26,800 (b) The daily profit in January from the Biloxi plant is the 2 1 matrix entry, namely $42,000.

(c) The total daily profit from all three plants in February was $18,000  $26,800  $26,800  $71,600.   2000 2500      3000 1500      58. (a) AB  6 10 14 28    105,000 58,000  2500 1000    1000 500

(b) That day they canned 105,000 ounces of tomato sauce and 58,000 ounces of tomato paste.      120 50 60 010 9700          59. (a) AC    40 25 30   050    4650  Ashton’s stand sold $97 worth of produce on Saturday, Bryn’s 60 30 20 100 4100

stand sold $4650 worth, and Cimeron’s stand sold $41 worth.      7000 100 60 30 010          (b) BC    35 20 20   050    3350  Ashton’s stand sold $70 worth of produce on Sunday, Bryn’s stand 4850 60 25 30 100 sold $3350 worth, and Cimeron’s stand sold $4850 worth.       120 50 60 100 60 30 220 110 90            (c) A  B    40 25 30    35 20 20    75 45 50  60 30 20 60 25 30 120 55 50

This represents the melons, squash, and

tomatoes they sold during the weekend.            16700 120 50 60 100 60 30 010 220 110 90 010                       (d) A  B C    40 25 30    35 20 20   050    75 45 50   050    8000  100 8950 60 30 20 60 25 30 100 120 55 50 During the weekend, Ashton’s stand sold $167 worth, Bryn’s stand sold $80 worth, and Cimeron’s stand sold       9700 7000 16700            $8950 worth of produce. Notice that A  B C  AC  BC    4650    3350    8000 . 4100 4850 8950

60. Suppose A is n  m and B is i  j. If the product AB is defined, then m  i. If the product B A is defined, then j  n. Thus if both products are defined and if A is n  m, then B must be m  n.         1 1 1 2 1 3 1 n 2 3 n  A   A      A    61. (a) A   0 1 0 1 0 1 0 1           n1 2n1 2 2 4 4 8 8 2 1 1    A3     A4         An     A2   (b) A   2n1 2n1 2 2 4 4 8 8 1 1


SECTION 10.2 The Algebra of Matrices

62. Let A

 

a b c d

.

For the first matrix, we have A2

 

a b c d

 

19

insert exponent (2 times)

  a 2  bc  4              b a  d  0 4 0 a 2  bc b a  d a b a 2  bc ab  bd   . So A2     2 2  c a  d  0 ac  cd bc  d c a  d bc  d 0 9 c d      bc  d 2  9

If a  d  0, then a  d, so 4  a 2  bc  d2  bc  d 2  bc  9, which is a contradiction. Thus a  d  0. Since b a  d  0 and c a  d  0, we must have b  0 and c  0. So the first equation becomes a 2  4  a  2, and the

fourth equation becomes d 2  9  d  3.           4 0 2 0 2 0 2 0 2 0  are A1   , A2   , A3   , and A4   . Thus the square roots of  0 9 0 3 0 3 0 3 0 3   a 2  bc  1        b a  d  5 1 5  For the second matrix, we have A2   Since a  d  0 and c a  d  0, we must have  c a  d  0 0 9      bc  d 2  9     a2  1 a  1     c  0. The equations then simplify into the system b a  d  5  b a  d  5       d2  9 d  3 We consider the four possible values of a and d. If a  1 and d  3, then b a  d  5  b 4  5  b  54 . If a  1

and d  3, then b a  d  5  b 2  5  b   52 . If a  1 and d  3, then b a  d  5  b 2  5    1 5  are b  52 . If a  1 and d  3, then b a  d  5  b 4  5  b   54 . Thus, the square roots of  0 9         1 54 1  52 1 52 1  54 , A2   , A3   , and A4   . A1   0 3 0 3 0 3 0 3


20

CHAPTER 10 Matrices and Determinants

10.3 INVERSES OF MATRICES AND MATRIX EQUATIONS 

1. (a) The matrix I  

1 0 0 1

 is called an identity matrix.

(b) If A is a 2  2 matrix, then AI  A and I A  A.

(c) If A and B are 2  2 matrices with AB  I then B is the inverse of A. 

2. (a) 

5 3 3 2



x



y



4 3

 

      d b 2 3 2 3 1 1     . ad  bc c a 5  2  3  3 3 5 3 5      2 3 4 1     . (c) The solution of the matrix equation is X  A1 B   3 5 3 3

(b) The inverse of A is A1 

(d) The solution of the system is x  1, y  3.     4 1 2 1 ; B   . 3. A   7 2 7 4           1 0 4 1 2 1 1 0 2 1 4 1 .    and B A     AB   0 1 7 2 7 4 0 1 7 4 7 2 

4. A  

7 3 2 . ; B   2

2 3

2 1      7 3 2 3 1 0 2 2   . BA   2 1 4 7 0 1 

 5. A   

4 7

1 1

3 1 4

8 3

   0 ; B   2

AB  

4

2 3



1

4 7

   1 1  . AB  

3 2

4

9 10 8

     6. A    1 1 6 ; B   12



8 3

4

1

1

1

2 1

3 1 4

1 3 1 0 1 1 3 2      8 3 4 1 3 1 1 0 0           B A   2 1 1   1 4 0    0 1 0   1 0 1 1 3 2 0 0 1 

7 3 2    1 0  and  2

3 2

4

  0   2

2



0 1

1 0 0

     1 1     0 1 0  and

0

9 10 8

     14 11  . AB   1 1 6   12 1 1 2 1 12  12 2 1 12  12 2 2      9 10 8 3 2 4 1 0 0           B A   12 14 11   1 1 6    0 1 0  . 1 1 1 2 2 1 12 0 0 1 2 2

0 0 1

1 0 0

     14 11     0 1 0  and 1 1 0 0 1 2 2


21

SECTION 10.3 Inverses of Matrices and Matrix Equations

7 4

5 7 3 4

1

2

5



2 4 7 4 1 2 1 0 1 2 1 . Then, A A1        3 3 7 7 14  12 3 7 3 2 3 2 2 2 0 1 2 2      7 4 1 2 1 0  . and A1 A   3 7   3 2 2 2 0 1   1 3 2    8. B    0 2 2  We begin with a 3  6 matrix whose left half is B and whose right half is I3 . 2 1 0       1 3 2 1 0 0 1 3 2 1 0 0 1 3 2 1 0 0   2R  5R  R     3  2R1  R3 3 2 3   0 2 2 0 1 0  0 2 2 0 1 0 R       0 2 2 0 1 0   0 5 4 2 0 1 0 0 2 4 5 2 2 1 0 0 0 1     0 1 3 2 1 0 0 1 0 1 1  32 1  R1  3R2  R1  R R R   2 R2 1 3 1 0 1 1 0 1  0 1 0 2 2   1 0      2 1 R2  R3  R2  2 R3 5 5 0 0 1 2 2 1 0 0 1 2 2 1          1 1 1 1 1 1 1 3 2 1 0 0 1 0 0 1 1 1           0 1 0 2 2 1 . Then B 1   2 2 1 ; B 1 B   2 2 1   0 2 2    0 1 0 ;          2 52 1 2 52 1 2 1 0 0 0 1 0 0 1 2 52 1      1 3 2 1 1 1 1 0 0          and B B 1    0 2 2   2 2 1    0 1 0 . 2 1 0 2 52 1 0 0 1   1 1 2  and verify that A1 A  A A1  I . 9. Using a graphing device, we find A1   3 2 2 2   1 1 0   1 1  10. Using a graphing device, we find B 1    33 31 3  and verify that B B  B B  I3 . 13 12 1  1     3 2 9 2 9 2 1      11.  27  26 13 3 13 9 13 3 7. A  

12.  13. 

14. 

15. 

5 13 7

4

8 5 6 3

8

4

  A1 

      4 7 4 7 4 7 1        20  21 3 5 3 5 3 5

1 

1 

1 

    13 5 13 5 1     26  25 5 2 5 2

    5 4  53  43 1     35  32 8 7  83  73

  4 3 1  , which is not defined, and so there is no inverse.  24  24 8 6


22

CHAPTER 10 Matrices and Determinants

1 1 3  

1 16.  2

17. 

5 4

04 12

03

06

1 2  53

1 

4 2 3 1 0 0

 

4  13 5

1 2



12 1 15

3 2

 

    1 2 06 12 1     024  036 03 04 1 2 2

   18.  3 3 2 0 1 0

4

2

3

3

1 0 0

4 0

4

0 0

4

1 4 R1

   R1  R3  R1   0 6 1 3 4 0    0 6 1 3 4 0    3R3  R2  R3  1 0 1 0 0 1 0 2 1 1 0 4 0 0 2 6 4 12       1 0 1 0 0 1 1 0 0 3 2 5 1 0 0 3 2 5 1      1  R3  R1   0 6 1 3 4 0  R   6 R2   0 6 0 6 6 6       0 1 0 1 1 1 . R2  R3  R2  0 0 1 3 2 6 0 0 1 3 2 6 0 0 1 3 2 6   3 2 5    Therefore, the inverse matrix is  1 1 1  . 3 2 6 

2 4

4R2  3R1  R2  4R3  R1  R3

1 1 0 0

2 4

1

1 0 0

6 0

5

0

1 1 0

 1 2 R3

1 4 0

    3R3  2R2  R3  0 6 1 1 2 0   0 6 1 1 2 0       3R1  2R2  R1 0 4 1 1 0 2 0 0 1 5 4 6 1 4 0 0 0 1     6 0 0 24 24 30 1 0 0 4 4 5 1     R1  5R3  R1 6 R1 0 6 0  0 1 0 1 1 1 .   6 6 6      1 R2  R3  R2 6 R2 , R3 0 0 1 5 4 6 0 0 1 5 4 6   4 4 5    Therefore, the inverse matrix is   1 1 1 . 5 4 6

   19.   1 1 1 0 1 0 

5

7 4 1 0 0

2R2  R1  R2  2R3  R1  R3

1

0 1 1 0 1

1

   2  3R1  R2  3 1 3 0 1 0  R  0 1 0      R3  5R1  R3 5 7 4 1 0 0 0 7 1 6 7 5 0 0 1     1 0 1 1 0 1 1 0 0 26 7 25     1  R3  R1 R3  7R2  R3  0 1 0 3 1 3  R  0 1 0 3 1  3    R  . 2 , R3 0 0 1 27 7 26 0 0 1 27 7 26   26 7 25    Therefore, the inverse matrix is  3 1 3 . 27 7 26

   20.   3 1 3 0 1 0 

 1 2 3 1 0 0    21.   4 5 1 0 1 0  1 1 10 0 0 1

R3  R1  R3  R1  R3

R2  4R1  R2  R3  R1  R3

 1 2 3 1 0 0    0 3 13 4 1 0    0 3 13 1 0 1

R3  R2  R3



Since the left half of the last row consists entirely of zeros, there is no inverse matrix.

1

 3 1 3  

6 0 5

 1 2 3 1 0 0    0 3 13 4 1 0 .   0 0 0 3 1 1


SECTION 10.3 Inverses of Matrices and Matrix Equations

0 2 2 1 0 0

3 2 0 1 0 0

2 1 0 1 0 0    22.  1 1 4 0 1 0

23

1 1 4 0 1 0 1 1 4 0 1 0   R2  2R1  R2   R1  R2  R1                 2 1 0 1 0 0    R3  2R1  R3  0 1 8 1 2 0   R3  R2  R3 2 1 2 0 0 1 2 1 2 0 0 1 0 1 6 0 2 1       1 0 4 1 1 0 1 0 0 1 1 2 1 0 0 1 1 2   R1  2R3  R1   R2    0 1 8 1 2 0    0 1 0 3 2 4    0 1 0 3 2 4  1   R2  4R3  R2    . 2 R3 1 1 0 2 0 0 2 1 0 1 0 0 2 1 0 1 0 0 1 2   1 1 2    Therefore, the inverse matrix is  3 2 4  .  12 0 12 R1  R2

1 2 3 0 0 1

1

1 2

3 0 0

1

1

0 0

1 0 1

    2  3R1  R2 3 1 3 0 1 0 R       0 7 6 0 1 3  0 2 2 1 0 0 0 2 2 1 0 0 1 2 3 0 0 1     1 0 1 1 0 1 1 0 1 1 0 1     1 R1  R3  R1 3  2R2  R3  0 1 0 3 1 3  R   2 R3      0 1 0 3 1 3   R2  3R3  R2 0 2 2 1 0 0 0 0 2 7 2 6     1 0 1 1 0 1 1 0 0  92 1 4     1  R3  R1  0 1 0 3 1 3  R  3 1 3     0 1 0 . Therefore, the inverse matrix is  7 7 0 0 1 1 3 0 0 1 2 1 3 2   9  1 4  2   3 1 3 .   7 1 3 2

 23.  3

 24.  5

 1 3 0 1 0 

R1  R3



 R R R  1 3 1  1 1 0 1 0    5

0 0 1 0 1 1 1 0 1

 R2  5R1  R2   0    0 R3  2R1  R3

1 1 5 1

 R  2R  R 3 2 3 5  

2 2 0 0 0 1 0 2 0 2 0 3     1 0 0 1 0 1 1 0 0 1 0 1 1 0 1 1       2 R3  0 1 1 5 1 5   0 1 0 1 0  3 . Therefore, the inverse matrix is  1 0  3 .   R2 R     2 2  R 3 2 13 0 0 1 6 1 13 6 1 0 0 2 12 2 13 2 2 

2 2 0 0 0 1


24

CHAPTER 10 Matrices and Determinants

1 2 0 3 1 0 0 0

  0 1 1 1 0 1 0 0   25.   0 1 0 1 0 0 1 0  

R3  R2  R3  R4  R1  R4

1 2 0 2 0 0 0 1

1 2 0 3 1 0

0

0

 0   0  0 1

1 2

0

 0 1 1    0 0 1  0 0 

3

1

1

0

0 0 0

 1 0 0   0 0 1 1 0   0 1 1 0 0 1

1 0 2 1 1 2

0

R3  R4

0

 0   1 0 0 1 1 0   0 0 0 1 1 0 0 0 0 1 1 0 0 1     1 0 0 1 1 0 2 0 1 0 0 0 0 0 2 1      0 1 0 0 1 0 1 1   0 1 0 0 1 0 1 1    R1  R1  R4      .  0 0 1 0 0 1 1 0    0 0 1 0 0 1 1 0      0 0 0 1 1 0 0 1 0 0 0 1 1 0 0 1   0 0 2 1    1 0 1 1    Therefore, the inverse matrix is  .  0 1 1 0    1 0 0 1  0 1 1 1 0 1 0    0 0 1 0 0 1 1 

R1  2R2  R1  R2  R3  R2

 0 1   0 0 

0 1 0

0

1

R1  2R3  R1  R2  R4  R2

     1 0 1 0 1 0 0 0 1 0 1 0 1 0 0 0 1 0 1 0 1 0 0 0        0 1 0 1 0 1 0 0  R R R  0 1 0 1 0 1 0 0  R R R  0 1 0 1 0 1 0 0    414   424   26.      .  1 1 1 0 0 0 1 0  R3  R1  R3  0 1 0 0 1 0 1 0  R3  R2  R3  0 0 0 1 1 1 1 0        1 1 1 1 0 0 0 1

Therefore, there is no inverse matrix. 

 27.  

   29.    

3

2

0 1

1

0 2

 3  0

2  3

1

1

 0   4 1 2   2 2 0 1

1 0 0 0

0 0 0 7

2 1 3

3 1 2

2

   0 1 0 1    2    2 2 1 2    1 1 1 0

1 1  72 6   1 1   0  2 6  1 0 0 3

1

  0 2 0 0   33.   0 0 4 0  

4 3 3 

   1 1 3 1 3

0 1

1 7 3    31.  0 2 1   0 0 3 

1

0 2

1 4 1

3

0 1 0 1 1 0 0 1

1 0 0 0

 28.   

   30.   

0 0 0

5 5

2

1

 0 

1

0 1 2 3

0 1

1

  0 1 0 0    2   0 0 1 0    4 0 0 0 17

3

2 3 1 0 0 0

1

  2 5 0 0   32.   4 2 3 0   

 34.  

1 0 0 2

1

2  25

1

  

 1 1    0 1  1 0

3 1

5 1 2 1

1

0

0 0 3

2 5  15

1 3 25  25  2 1  5 5  1  5  35

 1 16  16    1 1 0 0    4  2    1 1 3 3   3 5 6

13 6

2 56

1

0

0 0

1 0

0

1 6

  1  2 0 0   5 5   16  2 1 0   15 3  15  1 2 1  37 15 15 3

1

 0 

0 1 1 0 1

   0 0

 0   0  13 1 2


SECTION 10.3 Inverses of Matrices and Matrix Equations

In Solutions 35–38, the matrices A and B are defined as follows:   1 0 2    A  0 2 1  4 2 1 

 14

3 3 4 4   7 23 3  35. A1 B     16  16  16  7 1 5 8 8 8

7 3 4

 22 2 37. B AB 1    7 7 50 7

39.

  3x  2y  1

40.

  5x  7y  9

26 7

2 1 2

 B  

0

3

1

0

 1  2

0

0

1

25 7

13 7

22 7

  1 5 2  36. AB 1    7 7 7  

 

16  7  37 7

4 8 13 7 7   7 1 8 17 6  38. B AB    7  7 7  31 9  10 7 7 7

3 2



x

1

5 7



x

9

2

5



x

2

7

4



x

0

1



x

is equivalent to the matrix equation 

25

     . 13 9 y 3        x 1 9 2 3     . Therefore, x  3 and y  4. Using the inverse from Exercise 11,     y 13 3 3 4

 13x  9y  3

is equivalent to the matrix equation 

    . 3 4 y 6        x 4 7 9 6   . Therefore, x  6 and y  3. Using the inverse from Exercise 12,     y 3 5 6 3  3x  4y  6

  2x  5y  2 41.  5x  13y  20

.     5 13 y 20        x 13 5 2 126   . Therefore, x  126 and y  50. Using the inverse from Exercise 13,     y 5 2 20 50

42.

  7x  4y 

0

 8x  5y  100

is equivalent to the matrix equation 

is equivalent to the matrix equation  

Using the inverse from Exercise 14, 

    2x  4y  z  7 43. x  y  z  0    x  4y  2

x

y



 53

 83

    . 8 5 y 100      43 0  400 3 . Therefore, x   400 and y   700 .   3 3  73  700 100 3 

2 4

       is equivalent to the matrix equation   1 1 1   y   

1 4 0 z    x 4 4 5 7 38             Using the inverse from Exercise 19,  9  y    1 1 1   0    . z 5 4 6 2 47 

Therefore, x  38, y  9, and z  47.



7

 0 .

2


26

CHAPTER 10 Matrices and Determinants

    5x  7y  4z  1 44. 3x  y  3z  1    6x  7y  5z  1

5

7 4



x

1

         is equivalent to the matrix equation   3 1 3   y    1 . 6

7 5 z 1     x 26 7 25 1 8           3 1   1    1 . Using the inverse from Exercise 20,  y 3        z 27 7 26 1 8 

Therefore, x  8, y  1, and z  8.        2y  2z  12 0 2 2 x 12            45. 3x  y  3z  2 is equivalent to the matrix equation   3 1 3   y    2 .    x  2y  3z  8 1 2 3 z 8 

x

     Using the inverse from Exercise 23,   y 

 92 1

z

3

7 2

4



12

      1 3    2    1 3

8

20

 10  . 16

Therefore, x  20, y  10, and z  16.        x  2y  3  0 1 2 0 3 x 0            0 1 1 1  y   1  yz 1      is equivalent to the matrix equation  46.      .  0 1 0 1  z   2   y    2           x  2y  2  3 1 2 0 2  3        1 0 0 2 1 0 x         y   1 0 1 1   1   5         Using the inverse from Exercise 25,         .  z   0 1 1 0   2   1         3 1 0 0 1 3  Therefore, x  1, y  5, z  1, and   3.

47. Using a calculator, we get the result 3 2 1.

48. Using a calculator, we get the result 1 2 3.

49. Using a calculator, we get the result 3 2 2.

50. Using a calculator, we get the result 6 12 24.

51. Using a calculator, we get the result 8 1 0 3.

52. Using a calculator, we get the result 8 4 2 1.   53. This has the form M X  C, so M 1 M X  M 1 C and M 1 M X  M 1 M X  X.  1     3 2 3 2 3 2 1      . Since X  M 1 C, we get Now M 1    98 4 3 4 3 4 3        x y z 3 2 1 0 1 7 2 3     . u   4 3 2 1 3 10 3 5      39  92 1 4 3 6  39 2          54. Using the inverse matrix from Exercise 23, we see that   3 1 3   6 12    15 30 . 7 33 1 3 0 0 33 2 2     39 39 x u      2    Hence,  y     15 30  . 33 33 z  2


SECTION 10.3 Inverses of Matrices and Matrix Equations

1

      a a 1 1 a a 1 1 1          55.  2a 1 1 a 2  a 2 2a 2 a a a a a a     1 a 0 0 0 1 0 0 0 1 0 0 0 1a 0 0 0 R 1 a     1 0 b 0 0 0 1 0 0  0 1 0 0 0 1b 0 0  b R2        56.    . 1 0 0 c 0 0 0 1 0  0 0 1 0 0 R3 0 1c 0 c     1 R d 4 0 0 0 d 0 0 0 1 0 0 0 1 0 0 0 1d     1a 0 0 0 a 0 0 0      0 1b 0 0 b 0 0 0      Thus the matrix  .  has inverse   0  0 0 c 0 0 1c 0     0 0 0 1d 0 0 0 d     1    2 x 1 1x x 2 x x 2 x 1 1       . 57.  2x 2  x 2 x 2 x 2 x 2 x x2 1x 2x 2 

a a

The inverse does not exist when x  0. 1      x e2x e3x e2x e e x e2x 1     1 . The inverse exists for all x. 58.  2 e4x  e4x e2x e x e2x e3x e2x e3x     0 1 0 0 1 ex 1 ex 0 1 0 0  R  ex R  R   12 e2x R2   2 1 2 x e2x 0 0 1 0   0 2e2x 0 e x 1 0    59.  e                 1 2 R3 0 0 2 0 0 1 0 0 2 0 0 1     1 1 ex 1 0 0 0 1 ex 0 1 0 0 2 2     x 1 x  1 e2x 0 . 1  e R2  R1  0 1 0 1 ex  1 e2x 0  R      0 1 0 2 e  2 2 2 1 1 0 0 1 0 0 0 0 1 0 0 2 2   1 1 ex 0 2  2  1 ex  1 e2x 0 . The inverse exists for all x. Therefore, the inverse matrix is   2  2 1 0 0 2    1 1    x 1  1 1 x 1  1 1 1  x 1  x 1   x2   x 1  x2  60.  .   x 1 x   1   x 2  1  x x x x x x x x 1 x 1 x x The inverse exists for x  0, 1.       3 1 3 1 0 0 4 2 4 0 1 0 1 1 1 1 1 0   R3  R1  R3     1  R2  R1  3 1 3 1 0 0 R    61. (a)   4 2 4 0 1 0       3 1 3 1 0 0  R1  R2 3 2 4 0 0 1 0 1 1 1 0 1 0 1 1 1 0 1     R1  12 R2  R1 1 1 1 1 1 0 1 0 1 1  12 0     R R R  12 R2 R2  3R1  R2 3 0 1 3 1  0 2 0 4 3 0    0 1 0 2        2 1 R3  2 R2  R3 0 0 1 1  32 1 0 1 1 1 0 1     0 1 1 1 0 0 0 1 1     3 3   0 1 0 2 0 0 . Therefore, the inverse of the matrix is  2 .  2 2 3 3 0 0 1 1 2 1 2 1 1

27


28

CHAPTER 10 Matrices and Determinants

A

0

     (b)   B    2 C

1 1



10

1

        0   14    1 .

3 2 1  32

1

13

2

Therefore, the rats should be fed 1 oz of food A, 1 oz of food B, and 2 oz of food C.        A 0 1 1 9 2        3       (c)    0   12   0   B   2 . 2 3 C 1 2 1 10 1 Therefore, the rats should be fed 2 oz of food A, no food B, and 1 oz of food C.        2 7 A 0 1 1        3       (d)  0   4    2 .  B    2 2 1  32 1 11 7 C Since A  0, there is no such combination of foods.

     3 1 4 1 0 0 1 1 2 1 1 0 3 1 4 1 0 0   R R     R2  R1  R2  1 1 2 1 1 0  12 3 1 4 1 0 0   62.      4 2 6 0 1 0  R3  R1  R3 0 1 1 1 0 1 0 1 1 1 0 1 3 2 5 0 0 1     1 1 2 1 1 0 1 1 2 1 1 0     1   0 2 2 4 3 0  R3  2 R2  R3      0 2 2 4 3 0 . 0 1 1 1 0 1 0 0 0 1  32 1 

R2  3R1  R2



Since the inverse matrix does not exist, it would not be possible to use matrix inversion in the solutions of parts (b), (c), and (d).

    9x  11y  8z  740 63. (a) 13x  15y  16z  1204    8x  7y  14z  828 

9 11 8 1 0 0   (c)  13 15 16 0 1 0   

9 11



x

740

         (b)   13 15 16   y    1204  8

8

7 14

z

828

9 11 8 1 0 0   R3  31R2  R3  0 8 40 13 9 0  20    8 7 14 0 0 1 0 25 62 8 0 9     9 11 8 1 0 0 9 11 8 1 0 0 1   R R    252 R2 2 3 0    8 40 13 9 0     0 252 0 243 279 180  63 R3  2R2  R3 0 252 0 243 279 180 0 8 40 13 9 0     9 11 8 1 0 0 9 11 8 1 0 0 1    R  11R  8R  R  R3 27 5 27 5 31 31 1 2 3 1 0 1    2520 7  0  28    0 1 0  28 28  7   28 1 25 0 0 2520 1305 1125 360 0 0 1  29 56 56  7     63  63 7 7 9 1 9 0 0 1 0 0 4 4 4 4 1     31  5   0 1 0  27  0 1 0  27 31  5 . 9 R1    7  7  28 28 28 28 25  1 25  1 0 0 1  29 0 0 1  29 7 7 56 56 56 56 9R2  13R1  R2  9R3  8R1  R3


SECTION 10.4 Determinants and Cramer’s Rule

29

7 7 1 4 4   27 31  5 . (You could also use a calculator to find the inverse.) Therefore, Thus, the inverse of the matrix is    28 28 7  25  1  29 7 56 56        7 7 1 x 740 16 4 4         y     27 31  5   1204    28 . The salesperson earns $16 on a standard phone, $28 on a deluxe    28 28    7  25  1  29 z 828 36 7 56 56

phone, and $36 on a super deluxe phone.

64. No. Consider the following counterexample: A  

0 1 0 0

 and B  

0 2 0 0

. Then, AB  O, but neither A  O nor 

0 1 . Then, A2  O, but A  O. B  O. There are infinitely many matrices for which A2  O. One example is A   0 0

65. To show that B 1 A1 is the inverse of AB, we need to show that the inverse property holds. Using the associative     property of matrices, we calculate AB B 1 A1  A B 1 B A1  AIn A1  A A1  In and     B 1 A1 AB  B 1 A1 A B  B 1 In B  B 1 B  In . This shows that the inverse property of matrices holds for AB and B 1 A1 , and so the two matrices are inverses of one another.

10.4 DETERMINANTS AND CRAMER’S RULE 1. True. det A is defined only for a square matrix A. 2. True. det A is a number not a matrix. 3. True. If det A  0 then A is not invertible.      2 1   2 4  1 3  11 4. (a)    3 4    1 0 2     (b)  3 2 1   1 [2 4  1 3]  0 [3 4  1 0]  2 [3 3  2 0]  8  3  2 9  7    0 3 4    2 0  has determinant D  2 3  0 0  6. 5. The matrix  0 3   0 1  has determinant D  0 0  1 2  2. 6. The matrix  2 0   3   1 2  has determinant D  3  2  1 1  0. 7. The matrix  2 3 2 1  3   02 04  has determinant D  02 08  04 04  0. 8. The matrix  04 08   4 5  has determinant D  4 1  5 0  4. 9. The matrix  0 1


30

CHAPTER 10 Matrices and Determinants

10. The matrix 

1

2

3 2

 has determinant D  2 2  1 3  1.

 2 5 does not have a determinant because it is not square.   3 12. The matrix   does not have a determinant because it is not square. 0 11. The matrix

1 13. The matrix  2

14. The matrix 

1 8  has determinant D  1  1  1  1  1  1  1 . 2 2 8 4 8 8 1 12

22 14

05

10 

 has determinant D  22 10  05 14  22  07  29. 1 0 12

   In Solutions 15–20, A    3 5 2 . 0 0 4

15. M11  5  4  0  2  20, A11  12 M11  20

16. M33  1  5  3  0  5, A33  16 M33  5

17. M12  3  4  0  2  12, A12  13 M12  12

18. M13  3  0  0  5  0, A13  14 M13  0

19. M23  1  0  0  0  0, A23  15 M23  0 20. M32  1  2  3  12  72 , A32  15 M32   72     2 1 0      2 4    2 6  4  4. Since M  0, the . Therefore, expanding by the first column, M  2  21. M   0 2 4      1 3  0 1 3

matrix has an inverse.   1 2 5    22. M    2 3 2 . Therefore, 3 5 3              2 2   3 2   2 3    2   9  10  2 6  6  5 10  9  19  24  5  0, and so   5 M  1        3 3  5 3  3 5

the matrix does not have an inverse.   30 0 20    23. M    0 10 20 . Therefore, expanding by the first row, 40 0 10          0 10   10 20    20    30 100  0  20 0  400  3000  8000  5000, and so M 1 exists. M  30     0 0 10   40  

 24. M   

2

1 2

         2  32   2 4  1      1 4  2  8  3  1  5  4, and the  M  1 . Therefore,  4 0   2 2  1 2  2 4   2 2 1

2  32 12

matrix has an inverse.


SECTION 10.4 Determinants and Cramer’s Rule

31

    1 3 7     1 3   3 7   2 6  14  16  0. Since 8 . Therefore, expanding by the second row, M  2  25. M   2 0 8       0 2 2 2 0 2 2 M  0, the matrix does not have an inverse.     0 1 0     2 4   6  4  2. Since M  0, the matrix . Therefore, expanding by the first row, M   1  26. M   2 6 4     1 3 1 0 3

has an inverse.   1 3 3 0    0 2 0 1 . Therefore, expanding by the third row, 27. M      1 0 0 2  1 6 4 1           1 3 3 3 3 0           1 3 3 3 3 3       6  6  4  4, and so M 1 exists.   4  1 M  1  2 0 1   2  0 2 0   1       1 4 2 0 6 4     1 6 4 6 4 1 

        1 2 2  2 0 2           1 2  2 2      3 4 0 4    6  16  2  2  92, 2 . Therefore, M  1  4 0 4   2  3 4 4   6  28. M             3 4  4 4      0 1 6 0 0 1 0  1 6 0 1 0 2 0 1

2 0 2

and so M 1 exists.      10 20 31   1 2 1          30.  10 11 45   1080. The matrix has an inverse. 29.  2 2 1   6. The matrix has an inverse.      20 40 50  1 2 2         7  1 3 2 5   1 10 2      3 9 11 5   2 18 18 13      31.    0. The matrix has no inverse.   12. The matrix has an inverse. 32.    2 6  3 30 4 24  0 31          5 15 10 39   1 10 2 10          3 5 10   2  4 3 2 10       2 2 26  8 6 24 1  3     34.  33.    8. The matrix has an inverse.   0. The matrix has no inverse.  6  20 15 3 27  9 16 45          8 12 20 36   12 9 6 1      0 0 4 6 0 0 4 6         2 1 1 3 2 1 1 3    , by replacing R3 with R3 R2 . Then, expanding by the third row, 35. M      2 1 2 3 0 0 1 0     3 0 1 7 3 0 1 7     0 0 6     2 1     6 2  0  3  1  18. M  1  2 1 3   6   3 0   3 0 7


32

CHAPTER 10 Matrices and Determinants

  36. M    

2

3 1 7

 6 2 3  .  7 7 0 5 3 12 4 0 4

     2 0 1 7   2 3 1 7          4 0 2 3    4 6 2 3 ,    M Then       7 0 5  7 7 0 5  7      3 0 4 0  3 12 4 0 

by replacing C2 with C2  3C3 . So expanding about the second column,       2 1 7           2 1    4 2     7 7  22  3  5  1183.   3 M  7  4 2 3   7 7      3 4  3 4   3 4 0     1 2 3 4 5       1 2 3 4   1 2 3 0 2 4 6 8         0 2 4 6     1 2       60  2  120. 37. M   0 0 3 6 9 , so M  5    5  4  0 2 4   20  3     0 0 3 6 0 2   0 0 0 4 8   0 0 3     0 0 0 4 0 0 0 0 5           2 1 6 4  2 1 6 4   0 1 6 4         7 2 2 5   11 2 2 5   7 2 2 5        38. M   . Then, M   , by replacing C1 with C1  2C2 . So   4 2 10 8   0 2 10 8   4 2 10 8            6 1 1 4  8 1 1 4 6 1 1 4             1 6 0   1 6 0   1 6 4    1 6 4               1 6           1 6    104   M  11  2 10 8   8  2 2 5   11  2 10 0   8  2 2 13   88      2 10           2 10   2 10 0   1 1 8  2 10 8   1 1 4  88  2  104  2  32   4 1 0    39. B    2 1 1  4  0 3           4 1 4 0 1 0   6  12  4  2   1   1 (a) B  2       4 0 4 3 0 3          4 1 4 1   4  6  2  3 (b) B  1      2 1  4 0 (c) Yes, as expected, the results agree.

40. If we expand along the first row of each submatrix, we see that the determinant is 210  1024.               2x  y  9    2 9   9 1   2 1       25.     41. Then D     10, and D y     5, Dx      x  2y  8  8 2 1 2 1 8    Dy  Dx  10 25 Hence, x     2, y   5, and so the solution is 2 5. D D 5 5               6x  12y  33    6 33   33 12   6 12    9, and  D y      6, Dx      12. 42. Then D         4x  7y  20  20 7  4 7  4 20       Dy  Dx  3 9 12 Hence, x   and y   2, and so the solution is 32  2 .   D D 6 2 6


SECTION 10.4 Determinants and Cramer’s Rule

33

              1 3  3 6   1 6     8.       43. Then, D      12, and D y     20, Dx     3x  2y  1 3 1 1 2 3 2    Dy  Dx  8 12 Hence, x   04,and so the solution is 06 04.  20  06, y   D D 20         1   1  1  1   1x  1y  1       1 1 1 1 2 3 2 3 3 2         1.  44. Then, D   1    6 , Dx    3   3 , and D y   1  1 1 3  1x  1 y  3          4 6 2 4 6 2 6 4 2    1  Dy  Dx  1 Hence, x   31  2, y   1  6, and so the solution is 2 6. D D     x  6y  3

6

6

               04 04   04 12   04 12    08.   32, and  D y      08, Dx    45. Then, D         12x  16y  32  12 32   32 16   12 16     Dy  Dx  32 08 Hence, x   4, y   1,and so the solution is 4 1.   D D 08 08               10x  17y  21    10 21   21 17   10 17    30.   12, and  D y      30, Dx    46.  Then, D         20x  31y  39  20 39   39 31   20 31        Dx  2 , y  D y  30  1, and so the solution is 2  1 . Hence, x    12 30 5 5 D D 30     x  y  2z  0 47. Then expanding by the second row, 3x  z  11    x  2y  0   04x  12y  04

           0 1 2   1 1 2             1 1   1 2   1 2        12  1  11, Dx    11 0 1   11    44,   1 D   3 0 1   3          1 2   2 0  2 0      0 2 0  1 2 0           1 1 0   1 0 2             1 1   1 2     D y    3 11 1   11    11.   22, and Dz    3 0 11   11           1 2   1 0       1 2 0   1 0 0 

22 11 Therefore, x  44 11  4, y  11  2, z  11  1, and so the solution is 4 2 1.

       5 3 1         5 3  0 4     28  426  398,   6 Then D   0 4 6   1      7 10   7 10     7 10 0             5 6 3 1  6 1               22 4   0 22     6 3    272  126  398,  D y    0 22 6   1  6  Dx    22 4 6   1          13 10   13 10   7 13       13 10 0   7 13 0           5 3 6          5 3  5 5   6 6   428  1562  1990.        6    154  642  796, and Dz    0 4 22   4    22   7 10   7 13   7 13     7 10 13 

  6   5x  3y  z  48. 4y  6z  22    7x  10y  13

796 1990 Therefore, x  398 398  1, y  398  2, and z  398  5, and so the solution is 1 2 5.


34

CHAPTER 10 Matrices and Determinants

    2x1  3x2  5x3  1 49. x 1  x 2  x3  2    2x  x  8 2

3

Then, expanding by the third row,        2 3 5         2 5   2 3      6  1  7,  D   1 1 1   2      1 1   1 1    0 2 1

        1 3 5            2 1  2 1    1 1      3  30  20  7,  Dx    2 1 1      5   3 1         8 2 8 1  2 1  8 2 1

       2 1 5          2 5   2 1      Dx    1 2 1   8    24  3  21, and  2        1 1   1 2    0 8 1

      2 3 1        2 3 2 1     Dx    1 1 2   2    6  8  14.   8 3       1 1 1 2   0 2 8

21 14 Thus, x1  7 7  1, x2  7  3, x 3  7  2, and so the solution is 1 3 2.

   c 2   2a 50. a  2b  c  9    3a  5b  2c  22

       2 0 1        1 2  2 1      18  1  19,   1 Then D   1 2 1   2     3 5 5 2    3 5 2        2 0 1        9 2  2 1      18  1  19,   1 Da    9 2 1   2      22 5  5 2    22 5 2 

         2 2 1          1 9  1 1   9 1      80  10  5  95, and   1   2 Db    1 9 1   2        3 22  3 2  22 2     3 22 2 

       2 0 2        1 2 2 9     2  2  0.  2 Dc    1 2 9   2     3 5  5 22     3 5 22 

Hence, a  1, b  5, and c  0, and so the solution is 1 5 0.


SECTION 10.4 Determinants and Cramer’s Rule

   

 1x  1y  1z  7   3 5 2 10  10x  6y  15z  21 2 2 3 11 51.  3 x  5 y  2 z  10  20x  12y  45z  33 Then       4 9 5x  4y  5z  9 x  5y  z  5          10 6 15           20 12   20 45   12 45      15   6   2400  1950  300  750, D   20 12 45   10       5 4  5 5    4 5     5 4 5           21 6 15           33 12   33 45   12 45      5040  1440  3600  0,   15    6 Dx    33 12 45   21        9 4   9 5  4 5     9 4 5           10 21 15              20 33   20 45   33 45    D y    20 33 45   10    2400  6825  5175  750, and   15    21          5 9 5 5    9 5    5 9 5          10 6 21           20 12   20 33   12 33      2400  2070  420  750.   21    6 Dz    20 12 33   10       5 4  5 9    4 9     5 4 9 

Therefore, x  0, y  1, z  1, and so the solution   is 0 1 1.         5       2 1 0   2x  y 5 3 0 3     24  35  11,  1 52. 5x Then D   5 0 3   2   3z  19      0 7 4 7    0 4 7 4y  7z  17        5 1 0         19 3  0 3     60  82  22,   1 Dx    19 0 3   5      17 7  4 7    17 4 7        2 5 0          5 3  19 3    D y    5 19 3   2    164  175  11, and   5       0 7  17 7     0 17 7         2 1 5         2 1  2 5     52  85  33. Thus, x  2, y  1, and z  3, and so the   17  Dz    5 0 19   4     5 0  5 19     0 4 17 

solution is 2 1 3.        0 3 5  3y  5z  4         2 0  2 1      12  70  58,   5 53. 2x Then D   2 0 1   3   z  10      4 7 4 0    4x  7y 4 7 0  0         0 4 5  4 3 5            4 5  4 5       216, and   378,  D y    2 10 1   4  Dx    10 0 1   7        10 1   10 1      4 0 0  0 7 0       0 3 4       0 4 3 4     120  56  176.   7 Dz    2 0 10   4      2 10   0 10    4 7 0   108 88 189 108 88 Thus, x  189 29 , y   29 , and z  29 , and so the solution is 29   29  29 .

35


36

CHAPTER 10 Matrices and Determinants

  4   2x  5y 54. x  y z8    3x  5z  0

       2 5 0         1 1   1 1      10  40  50,   5 Then D   1 1 1   2     3 5 0 5   3 0 5

          2 4 0  4 5 0             1 1      8 1   4 5       80  32  48, and   4   220,  D y    1 8 1   2  Dx    8 1 1   5         3 5 0 5 8 1     3 0 5 0 0 5

     2 5 4         5 4      132. Thus, x  22 , y  24 , and z   66 , and so the solution is 22  24   66 .    Dz    1 1 8   3   5 25 25 5 25 25  1 8   3 0 0

55.

  x y z0      2z    0      x

y z

0

Then

 2z 1         1 1 1 1 1 1 1   2 0 1                       2 0 0 1             1 1 2 1 0 1 1 0      2   1    1 D     1  0 1 0   1  2 0 1    2          0 1 1 0  2 1 1 0 1 0 2 0               1 2 0 1 2 0   1 0 2 0  2 0  1 1  1 1  2 1  4,       0 1 1 1   1 1 1       0 0 0 1 1 1       2,    Dx      1  0 0 1   1 1    0 1 1 0   1 1        1 1 0   1 0 2 0

      1 0 1 1     1 1 1             2 0 0 1  1 1  0 1     1  2 1  1,  Dy           1 2 0 1   1    2  0 0 1 0   1 0   1 0        0 1 0   1 1 2 0

      1 1 0 1     1 1 1         2 0 0 1 1 1 0 1       1 1  2 1  1, and      Dz     2  1  1      2 0 1 0 1 0 0 1 0 1 0       0 1 0   1 0 1 0

      1 1 1 0   1 1 1       2 0 0 0 1 1       2 2  4. Hence, we have x  Dx   2  1 ,    D     2  1    2 0 0  0 1 1 0  D 4 2 1 1         0 1 1   1 0 2 1 y

   Dy  D

  Dz  D  1 1 1 1 4  ,z  , and    1, and the solution is 12  14  14  1 .   D D 4 4 4 4 4


SECTION 10.4 Determinants and Cramer’s Rule

          xy1   1 1 0 0    0 1 0 1 1 0                 yz 2  0 1 1 0         1 1 1 0     1    1  1  2, 56. Then D     1  0 1 1   1  0 1 1   1        0 1 1 1 0 0 1 1 z    3              1 0 1  0 0 1       x  4  1 0 0 1          1 1 0 0        2 1 0 1 1 0             2 1 1 0 3 1 1 1 1 1         1  3  2,           Dx      1 0 1 1   1 3 1 1       1   2 3 0 1 1 4 1 0 1 0 1                 4 0 1 0 0 1   4 0 0 1          1 1 0 0      0 1 0 2 1 0                 0 2 1 0 1 0 3 1  1 1                   3  1  4,  Dy        1  2  1  1 1        0 1 1 3 1 1   0 3 1 1 1 1 4 1  0 1          1 0 1 4 0 1    1 4 0 1           1 1 1 0     1 1 0 1 2 0            0 1 2 0         1 1 3 1     1  1  0,   1 Dz      1  0 3 1   1  1 2 0   1      0 0 3 1 1 2 4 1       0 3 1 0 4 1    1 0 4 1           1 1 0 1       1 0 1  1 1 2              0 1 1 2           0 1 1 2 1 3     4  2  6.   1   1  D      1  0 1 3   1  1 1 2          0 0 1 3 1 3 1 3 0 4             0 1 3 0 0 4    1 0 0 4     Dy  Dx  Dz  D  2 4 0 6 Hence, the solution is x   1, y     2, z    0, and     3. D D D D 2 2 2 2

   0 0 1       6 2 1 1      1 48  6  1 42  21  57. Area    6 2 1     2 2 2 2  3 8   3 8 1

y

  12 [5  2  6  10]   12 [3  16]  12 19  19 2

(6, 2)

1 (0, 0)

     1 0 1            3 5 5 1 1 1        58. Area    3 5 1    1    1 2 2   2 2  2 1  2 2 1 

(3, 8)

x

1

y (_2, 2)

(3, 5)

1 (1, 0)

x

37


38

CHAPTER 10 Matrices and Determinants

       1 3 1             2 9  2 1  9 1 1 1     1   3 59. Area    2 9 1    1       2 2  5 6   5 1  6 1   5 6 1 

y (_1, 3)

  12 [1 9  6  3 2  5  1 12  45]

2 1

  12 [15  3 3  57]   12 63  63 2

       2 5 1             7 2  7 1  2 1 1 1     1   5 60. Area    7 2 1    2       2 2  3 4   3 1  4 1   3 4 1 

(2, 9)

x (5, _6)

(_2, 5)

  12 [2 2  4  5 7  3  28  6]   12 [2 6  5 4  34]   12 12  20  34  12 66  33

y (7, 2)

1

x

1 (3, _4)

    a 0 0 0 0       b 0 0 0     c 0 0   0 b 0 0 0     0 c 0 0   d 0         abcde 61.  0 0 c 0 0   a    ab  0 d 0   abc   0 0 d 0   0 e     0 0 0 d 0 0 0 e     0 0 0 e   0 0 0 0 e   a a a a a         a a a a    a a a    0 a a a a     0 a a a       a a     2 3    a5   62.  0 0 a a a   a  a  0 a a a   0 0 a a   0 a       0 0 0 a a   0 0 a     0 0 0 a   0 0 0 0 a

     x 12 13     x   12    0  x  2  x x  1  0  x  0, 1, or 2 63.  0 x  1 23   0  x  2    0 x 1    0 0 x 2 

     x 1 1              1 x  1 x  1 1      x 0  x  x 2  1  x  x 2  2x  1  0  x  12  0  x  1  64.  1 1 x   x       1 x  x x  x 1   x 1 x 

      1 0 x    2      x 1 1 0   x   0  1  x 2  0  x 2  1  x  1 65.  x 2 1 0   0  1      x 0 0 1    x 0 1

      a b x  a        a x a   a   b    1 [ax  x x  a]  a x  b  bx  66.  x x  b x   1     x   x x b  x   0 1 1 

 ax  x 2  ax  ax  ab  bx  x 2  ax  bx  ab  x  a x  b  0  x  a or x  b


SECTION 10.4 Determinants and Cramer’s Rule

39

         1 x x2               y y2   x x2   x x2    2   1   1   yz 2  y 2 z  xz 2  x 2 z  x y 2  x y 2 67.  1 y y   1       2 2 2  z z   z z  y y     1 z z2 

 yz 2  y 2 z  x z 2  x 2 z  x y 2  x y 2  x yz  x yz  x yz  xz 2  y 2 z  yz 2  x 2 y  x 2 z  zy 2  x yz        z x y  xz  y 2  yz  x x y  xz  y 2  yz  z  x x y  x z  y 2  yz    z  x x y  z  y y  z  z  x x  y y  z

68.

   

x  2y  6z  5

3x  6y  5z  8    2x  6y  9z  7 (a) If x  1, y  0, and z  1, then x 2y 6z  12 06 1  5, 3x 6y 5z  3 16 05 1  8, and 2x  6y  9z  2 1  6 0  9 1  7. Therefore, x  1, y  0, z  1 is a solution of the system.        1 0 6  1 2 6 1 2 6           . Then, M   3 6 5    3 0 5  (replacing C2 with C2  2C1 ), so (b) M   3 6 5            2 2 9  2 6 9 2 6 9      1 6   2 5  18  46.  M  2    3 5       1 2 6 x 5           (c) We can write the system as a matrix equation:    3 6 5   y   8  or M X  B. Since M  0, M has 2 6 9 z 7 an inverse. If we multiply both sides of the matrix equation by M 1 , then we get a unique solution for X, given by X  M 1 B. Thus, the equation has no other solution.

(d) Yes, since M  0.

  a b 1  1 1   1 69. (a) If three points lie on a line then the area of the “triangle” they determine is 0, that is,  12 Q    a2 b2 1   0  2   a3 b3 1  Q  0. If the points are not collinear, then the point form a triangle, and the area of the triangle determined by these

points is nonzero. If Q  0, then  12 Q   12 0  0, so the “triangle” has no area, and the points are collinear.         6 4 1           y   2 10   6 4   6 4      (6, 13) (b) (i)  2 10 1           6 13   6 13   2 10    6 13 1  (2, 10)  26  60  78  24  60  8  34  104  68  0

Thus, these points are collinear.

(_6, 4) 1 1

x


40

CHAPTER 10 Matrices and Determinants

        5 10 1             2 6   5 10   5 10      (ii)  2 6 1           15 2   15 2   2 6    15 2 1 

y (_5, 10) (2, 6)

 4  90  10  150  30  20  94  140  50  4

These points are not collinear. Note that this is difficult to determine from the diagram.

1 2

(15, _2)

x

   x y 1      70. (a) Let M   x1 y1 1 . Then, expanding by the third column,    x2 y2 1               x1 y1   x y   x y      x1 y2  x2 y1   x y2  x2 y  x y1  x1 y M         x2 y2   x2 y2   x1 y1   x1 y2  x2 y1  x y2  x2 y  x y1  x1 y  x2 y  x1 y  x y2  x y1  x1 y2  x2 y1

 x2  x1  y  y2  y1  x  x1 y2  x2 y1

So M  0  x2  x1  y  y2  y1  x  x1 y2  x2 y1  0  x2  x1  y  y2  y1  x  x1 y2  x2 y1  y  y1 x y  y1  y x  x1  x 1 2  1 2  x2  x1  y  y2  y1  x  x1 y2  x1 y1  x1 y1  x2 y1  y  2 x2  x1 x2  x1 x2  x1 y  y1 y  y1 y 2 x  x1   y1  y  y1  2 x  x1 , which is the “two­point” form of the equation for the line x2  x1 x2  x1 passing through the points x1  y1  and x2  y2 . (b) Using the result of part (a), the line has equation          x y 1          20 50     x y  x y 0    20 50 1   0            10 25   10 25   20 50     10 25 1 

500  500  25x  10y  50x  20y  0  25x  30y  1000  0  5x  6y  200  0.

71. (a) Let x, y, and z be the weights (in pounds) of apples, peaches, and pears, respectively.   x  y z  18   We get the model 075x  090y  060z  1380    075x  090y  060z  180


SECTION 10.4 Determinants and Cramer’s Rule

41

          1 1 1          075 090   075 060   090 060      0  090  135  045, 1 1 (b) D   075 090 060   1         075 090   075 060   090 060     075 090 060            18 1 1          1380 090   1380 060   090 060      0  72  108  36, 1 1 Dx    1380 090 060   18         180 090   180 060   090 060     180 090 060            1 18 1          075 1380   075 060   1380 060       1   18    D y    075 1380 060   1            075 180   075 060   180 060     075 180 060   72  162  117  27, and           1 1 18          075 090   075 1380   090 1380       18   1 Dz    075 090 1380   1         075 090   075 180   090 180     075 090 180 

 108  117  243  18.   Dy  Dx  Dz  36 27 18 So x   8; y   6; and z   4.    D D D 045 045 045 Thus, the customer bought 8 pounds of apples, 6 pounds of peaches, and 4 pounds of pears.

72. (a) Using the points 10 25, 15 3375, and 40 40,we substitute for x and y and get the system     100a  10b  c  25 225a  15b  c  3375    1600a  40b  c  40         100 10 1            225 15   100 10     100 10  1  1 (b) D   225 15 1   1         1600 40   225 15     1600 40   1600 40 1 

 9000  24,000  4000  16,000  1500  2250  15,000  12,000  750  3750,          25 10 1          25  25 10   3375 15    10   1 1 Da    3375 15 1   1        40   3375 15   40 40   40    40 40 1 

 1350  600  1000  400  375  3375  750  600  375  1875,          100 25 1           100 25  100 25   225 3375     1 1 Db    225 3375 1   1         225 3375   1600 40    1600 40    1600 40 1

 9000  54,000  4000  40,000  3375  5625  45,000  36,000  2250  11,250, and


42

CHAPTER 10 Matrices and Determinants

          100 10 25          100 10   100 10   225 15       40     3375   Dc    225 15 3375   25         225 15   1600 40   1600 40      1600 40 40  25  9,000  24,000  3375  4,000  16,000  40  1,500  2,250

 25  15,000  3375  12,000  40  750  375,000  405,000  30,000  0.

Thus, a 

Db  Dc  Da  1875 11,250 0     005, b   3, and c   0. The model is D D D 3750 3,750 3,750

y  005x 2  3x.

73. Using the determinant formula for the area of a triangle, we have        1000 2000 1              1000 2000    1000 2000   5000 4000  1 1   1 1 Area    5000 4000 1    1        2 2  5000 4000    2000 6000   2000 6000   2000 6000 1    12  22,000,000  2,000,000  6,000,000   12  14,000,000  7,000,000

Thus, the area is 7,000,000 ft2 .

74. (a) The coordinates of the vertices of the surrounding rectangle are a1  b1 , a2  b1 , a2  b3 , and a1  b3 . The area of the surrounding rectangle is given by a2  a1   b3  b1   a2 b3  a1 b1  a2 b1  a1 b3  a1 b1  a2 b3  a1 b3  a2 b1 . (b) The area of the three blue triangles are as follows:

Area of  a1  b1   a2  b1   a2  b2 : 12 a2  a1   b2  b1   12 a2 b2  a1 b1  a2 b1  a1 b2  Area of  a2  b2   a2  b3   a3  b3 : 12 a2  a3   b3  b2   12 a2 b3  a3 b2  a2 b2  a3 b3  Area of  a1  b  a1  b3   a3  b3 : 12 a3  a1   b3  b1   12 a3 b3  a1 b1  a3 b1  a1 b3 . Thus the sum of the areas of the blue triangles, B, is

B  12 a2 b2  a1 b1  a2 b1  a1 b2   12 a2 b3  a3 b2  a2 b2  a3 b3   12 a3 b3  a1 b1  a3 b1  a1 b3   12 a1 b1  a1 b1  a2 b2  a2 b3  a3 b2  a3 b3   12 a1 b2  a1 b3  a2 b1  a2 b2  a3 b1  a3 b3   a1 b1  12 a2 b3  a3 b2   12 a1 b2  a1 b3  a2 b1  a3 b1 

So the area of the red triangle A is the area of the rectangle minus the sum of the areas of the blue triangles, that is,   A  a1 b1  a2 b3  a1 b3  a2 b1   a1 b1  12 a2 b3  a3 b2   12 a1 b2  a1 b3  a2 b1  a3 b1   a1 b1  a2 b3  a1 b3  a2 b1  a1 b1  12 a2 b3  a3 b2   12 a1 b2  a1 b3  a2 b1  a3 b1 

 12 a1 b2  a2 b3  a3 b1   12 a1 b3  a2 b1  a3 b2    a b 1  1 1    (c) We first find Q   a2 b2 1  by expanding about the third column.    a3 b3 1               a1 b1   a1 b1   a2 b2    1   1   a2 b3  a3 b2  a1 b3  a3 b1   a1 b2  a2 b1 Q  1        a3 b3   a2 b2   a3 b3   a1 b2  a2 b3  a3 b1  a1 b3  a2 b1  a3 b2

So 12 Q  12 a1 b2  a2 b3  a3 b1   12 a1 b3  a2 b1  a3 b2 , the area of the red triangle. Since 12 Q is not always positive, the area is  12 Q.

75. (a) If A is a matrix with a row or column consisting entirely of zeros, then if we expand the determinant by this row or       column, we get A  0   A1 j   0   A2 j       0   Anj   0.


CHAPTER 10

Review

43

(b) Use the principle that if matrix B is a square matrix obtained from A by adding a multiple of one row to another, or a multiple of one column to another, then A  B. If we let B be the matrix obtained by subtracting the two rows (or columns) that are the same, then matrix B will have a row or column that consists entirely of zeros. So B  0  A  0.

(c) Again use the principle that if matrix B is a square matrix obtained from A by adding a multiple of one row to another, or a multiple of one column to another, then A  B. If we let B be the matrix obtained by subtracting the proper multiple of the row (or column) from the other similar row (or column), then matrix B will have a row or column that consists entirely of zeros. So B  0  A  0.

76. Gaussian elimination is superior, since it takes much longer to evaluate six 5  5 determinants than it does to perform one five­equation Gaussian elimination.   77. Let B  A1 . Then, using the given formula, det A A1  det A det A1  det I  det A det A1  1  det A det A1  det A1 

1 . det A

CHAPTER 10 REVIEW 1. (a) 2  3

2. (a) 2  3

(b) Yes, this matrix is in row­echelon form.

(b) Yes, this matrix is in row­echelon form.

(c) No, this matrix is not in reduced row­echelon form, since the leading 1 in the second row does not have a

(c) Yes, this matrix is in reduced row­echelon form.  x  6 (d) y 0

0 above it.   x  2y  5 (d)  y 3 3. (a) 3  4 (b) Yes, this matrix is in row­echelon form. (c) Yes, this matrix is in reduced row­echelon form.    8z  0  x (d) y  5z  1    0 0

4. (a) 3  4 (b) No, this matrix is not in row­echelon form, since the leading 1 in the second row is not to the left of the one above it. (c) Since this matrix is not in row­echelon form, it is not in reduced row­echelon form.     x  3y  6z  2 (d) 2x  y 5    z0


44

CHAPTER 10 Matrices and Determinants

6. (a) 4  4

5. (a) 3  4 (b) No, this matrix is not in row­echelon form. The leading 1 in the second row is not to the left of the

above it.

one above it. (c) No, this matrix is not in reduced row­echelon form.   y  3z  4   (d) x  y 7    x  2y  z  2

1

2 2

6

(b) No, this matrix is not in row­echelon form. The leading 1 in the fourth row is not to the left of the one

1 1 0 1

  R R  1   2 7.   1 1 0 1    1 2 1 3 7 2

2 2 1 3

(c) No, this matrix is not in reduced row­echelon form.   x  8y  6z  4      y  3z  5 (d)  2z  7     x  y  z  0 

1 1 0 1

 R2  R1  R2    0 6  R3  2R1  R3  7 0

3 2 3 3

1 1 0 1

 R R R  3 2 3  7    0 9 0

3 2 0 1

 7 . 2

Thus, z  2, 3y  2 2  7  3y  3  y  1, and x  1  1  x  0, and so the solution is 0 1 2.       1 1 1 2 1 1 1 2 1 1 1 2   R R R   R R R   2 1 2 3 2 3      8.   1 1 3 6    0 2 2 4    0 2 2 4 . Thus, z  1; 2y  2 1  4 0 2 3 5 0 2 3 5 0 0 1 1

 y  1; and x  1  1  2  x  2, and so the solution is 2 1 1.     1 2 3 2 1 2 3 2     R2  2R1  R2 3  R2  R3  0 3 5 6  R   9.      2 1 1 2   R3  2R1  R3 0 3 5 5 2 7 11 9

1 2

 0  0

3 2

3 5 0

0

 6 . 1

The last row corresponds to the equation 0  1, which is always false. Thus, there is no solution.         1 1 1 2 1 1 1 2 1 1 1 2 1 1 1 2   1       R2  R1  R2 3  R2  R3  0 2 2 4 R 0 2 2 4 0 1 1 2 2 R2  10.       1 1 3 6  R3        3R1  R3 3 1 5 10 0 2 2 4 0 0 0 0 0 0 0 0   1 0 2 4   R1  R2  R1  0 1 1 2 . Let z  t. Then y  t  2  y  2  t and x  2t  4  x  4  2t, and so the    0 0 0 0

solutions are 4  2t 2  t t, where t is any real number.       1 1 1 1 0 1 1 1 1 0 1 1 1 1 0       R2  R1  R2  1 1 4 1 1   0 2 5 2 1   0 1 4 5 6        R3  R1  R3 R3  R2  R3 11.          1 2 0 4 7   0 3 1 3 7   0 3 1 3 7  R4  2R1  R4       2 2 3 4 3 0 0 1 2 3 0 0 1 2 3       1 1 1 1 0 1 1 1 1 0 1 1 1 1 0           6 6  R3  R4  0 1 4 5 6  R4  13R3  R4  0 1 4 5  R3  3R2  R3  0 1 4 5      .    0 0 13 12 11    0 0   0 0 1 2 3  1 2 3       0 0 1 2 3 0 0 13 12 11 0 0 0 14 28 Therefore, 14  28    2, z  2 2  3  z  1, y  4 1  5 2  6  y  0, and x  0  1  2  0  x  1. So the solution is 1 0 1 2.


CHAPTER 10

1 0 3 0 1    0 1 0 4 5    12.   0 2 1 1 0  

1 0  0 1   0 2 

R4  2R1  R4



3

0 1

 5   0  6

0 4 1

1

1 0  0 1   0 0 

R3  2R2  R3  R4  R2  R4

3

0

45

 5   9 10   0 1

0 4 1

1

Review

2 1 5 4 4 0 1 1 4 0 0 1     1 0 3 0 1 1 0 3 0 1        0 1 0 4  R3 5  R4  3R3  R4   0 1 0 4 5   R3  R4    1       0 0 1 0  1 9 R4   0 0 1 0 1  0 0 1 9 10 0 0 0 9 9     1 0 3 0 1 1 0 0 0 2      0 1 0 4 5  0 1 0 0 1 R1  3R3  R1         . Therefore, the solution is 2 1 1 1. R2  4R4  R2  0 0 1 0 1   0 0 1 0 1      0 0 0 1 1 0 0 0 1 1 

1 1 3 2

 13.  2

 1 1 2 

1 1

0

1 1

3

2

1 1

3

2

1     R3  R2  R3  0 3 5 2   0 3 5 2  3 R2       3 0 4 4 0 3 5 2 0 0 0 0     4 4 1 1 3 2 1 0 3 3     5 2 . The system is dependent, so let z  t: y  5 t   2  1  R2  R1  0 1 5 2  R  0 1             3 3   3 3  3 3  0 0 0 0 0 0 0 0   y  53 t  23 and x  43 t  43  x   43 t  43 . So the solution is  43 t  43  53 t  23  t , where t is any real number.

 14.  1

1

2

R2  2R1  R2  R3  3R1  R3

1

 3 

1 1

0

1

1 1 0 1

1     3  R2  R3 0 2 2 2 R   2 R4     0 2 2 2   0 2 2 2 0 0 0 0   1 0 1 2    0 1 1 1 . Since the system is dependent, let z  t. Then y  t  1    0 0 0 0

R2  R1  R2  R3  R1  R3

1 3 2 1  1 1 0 1   1  R2  R1 0 1 1 1 R    0 0 0 0 

y  t  1 and x  t  2  x  t  2. So, the solution is t  2 t  1 t where t is any real number.

1 1

1 1 0

1 1

1 1 0

1 2 R4

1 1

1 1 0

2  3R1  R2 1  R2  R1      R  R   0 2 4 2 2 0 1 2 1 1 3 1 1 1 2   1 0 1 0 1  . Since the system is dependent, Let z  s and   t. Then y  2s  t  1  y  2s  t  1 and 0 1 2 1 1

15. 

x  s  1  x  s  1. So the solution is s  1 2s  t  1 s t, where s and t are any real numbers.

 1 1 3    16.  2 1 6 1 2 9

R2  2R1  R2  R3  R1  R3

 1 1 3   0 3 0   0 1 6

R2  3R3  R2



 1 1 3    0 0 18 . Since the second row corresponds   0 1 6

to the equation 0  18, which is always false, this system has no solution.


46

CHAPTER 10 Matrices and Determinants

1 1

 17.  3

1

1 0

 2 1 6  

R2  3R1  R2  R3  R1  R3

4 3 3

1 1

 0  0

1 0

 5 4 6   5 4 3

1 1

1 0

0

0 3

 0  0

R3  R2  R3



 5 4 6  . The last row of this

matrix corresponds to the equation 0  3, which is always false. Hence there is no solution.       1 2 3 2 1 2 3 2 1 2 3 2       R2  2R1  R2 3  R2  R3  0 5 11 3  R     18.      0 5 11 3 . Since the third  2 1 5 1   R3  4R1  R3 0 5 11 2 0 0 0 1 4 3 1 6

row corresponds to the equation 0  1, which is always false, this system has no solution.       1 0 0 1 1 1 0 0 1 1 1 1 1 1 2       R2  R1  R2  1 1 1 1 0   0 1 1 2 1   1 1 1 1 0       R1  12 R3  R3  R1  R3 19.         0 1 1 2 1   2 0 0 2 2    1 1 1 1 2  R4  2R1  R4       2 4 4 2 6 0 4 4 4 4 2 4 4 2 6     1 0 0 1 1 1 0 0 1 1     R1  R3  R1  0 1 1 2 1   0 1 1 2 1   14 R3 R3  R2  R3     R2  2R3  R2          R4  4R2  R4  0 0 0 4 0   121 R4 0 0 0 0 0 R4  R3  R4     0 0 0 12 0 0 0 0 1 0   1 0 0 0 1    0 1 1 0 1     . This system is dependent. Let z  t, so y  t  1  y  t  1; x  1  x  1. So the 0 0 0 1 0   0 0 0 0 0 solution is 1 t  1 t 0, where t is any real number.     1 1 2 3 0 1 1 2 3 0   R  3R  R   3 1 3    20.   0 1 1 1 1    0 1 1 1 1  3 2 7 10 2 0 1 1 1 2

R3  R2  R3



 1 1 2 3 0    0 1 1 1 1 . Since the   0 0 0 0 1

third row corresponds to the equation 0  1, which is always false, this system has no solution.

21. A 3  3 and B 2  3 have different dimensions, so they are not equal.       25 1 5 1 5 e0     B, so A and B are equal. 22. A   0 21 log 1 12 0 12 In Solutions 23–34, the matrices A, B, C, D, E, F, and G are defined as follows: A 

1

2 0 1 4

   D  0 1  2

0

B 

E 

2 1

 12

1

 

1 2 4 2 1 0

 

4 0 2

   F   1 1 0 

 C  

7 5 0

23. A  B is not defined because the matrix dimensions 1  3 and 2  3 are not compatible.

3

G

1 2

 2 32   2 1 5


CHAPTER 10

 24. C  D   

1 2

3

1

1

4

   2 32     0 1   

2 1 

 25. 2C  3D  2  

1 2

3

2

0

4

 12 1 2

4

5  2 

1

Review

1 6

3 12

4 18

               2 32    3  0 1    4 3    0 3    4 2 6 0 4 2 2 0 2 1

 0  2

26. 5B  2C is not defined because the matrix dimensions 2  3 and 3  2 are not compatible.      27. G A  5 2 0 1  10 0 5

28. AG is undefined because the matrix dimensions 1  3 and 1  1 are not compatible.           1 3  1 3  4 2  11 2 2 2 7      1 2 4   1 2 4 10     1 11  2 3    2  29. BC   30. C B   2 32  8      2  2 9 2 1 0 2 1 0 1 2 4 3 8 2 1 2 1            1 3 4 0 2 4 0 2 2 14 2       1 2 4  30 22 2  2 3    3 3    1 1 0     31. B F   32. FC   1 1 0        2  2 2  2 1 0 9 1 4 27 57 7 5 0 7 5 0 2 1 2 2             1 3 3 7  1 11 1 4  2  2  2 2      2 1 2 1 15 3 1 3              33. C  D E   2 2    0 1   1  2 2  1  4 2  1 1 2 2 2 1 2 0 0 1  12 1            4 0 2 1 6 1 4 4 0 2 0 2 12 12                     34. F 2C  D   4 2  1 1 0   4 3    0 1    1 1 0   4 4     7 5 0 4 2 2 0 7 5 0 6 2 20 34     27 0 21 6 42 24       35. AB 2   36. A2 B    20 5 13   3 37 22  5 22 7 3 42 27     19 14 26 8  32 4 4     7 1 11  7  38. B AB 1   37. A1 B A    3  3   2 1  4  3  35 13 18 80 7 16 3 3 2 39. AB  12 42.

1 1  A 3 

45. AB   

40. B A  12

2 5 2

6

 

2 1 3

3 52 1 1



 32

   46. AB    2 2 1   1 0

1 1

 2

1

    43. A1 B A  4

1 0 0 1 

5 2 

 and B A   

1 0 0

3 52

1 1

 

2 5

6

2

 32

      2    0 1 0  and B A   1

1 1 1

0 0 1

  1   41. A1   3     44. A1  B A  4

2

5 2   2 1 3 

 1

1 0 0 1



.

1 0 0

      2   2 2 1    0 1 0 .

1 1 1

0

1 1

0 0 1

47


48

CHAPTER 10 Matrices and Determinants

In Solutions 47–52, A  

2 1 3 2

, B  

1 2 2

4

, and C  

0 1 3

.

2 4 0 

47. A  3X  B  3X  B  A  X  13 B  A. Thus, X  13 

1 2 4

2



2 1 3 2

48. 12 X  2B  A  X  2B  2A  X  2A  2B  2 A  B.         2 1 1 2 3 1 6 2    2   . X  2  3 2 2 4 1 6 2 12



1 3 .   1  3 5

2

Thus,

49. 2 X  A  3B  X  A  32 B  X  A  32 B. Thus,           3 3 7 2 1 2 2 1 2 1  2 .   3   2 X  2 3 6 0 8 2 4 3 2 3 2 50. 2X  C  5A  2X  5A  C, but the difference 5A  C is not defined because the dimensions of 5A and C are not the same. 1 AX  X  A1 C. Now 51. AX  C  A   

A1 

     2 1 0 1 3 2 2 6 1  2 1   2 1  1   .  . Thus, X  A C   4  3 3 2 3 2 2 4 0 4 5 9 3 2

52. AX  B  A1 AX  X  A1 B. From Exercise 65,        2 1 2 1 1 2 4 8 . Thus, X  A1 B     . A1   3 2 3 2 2 4 7 14 

1 4 2 9

2

2

53. D   54. D   

55. D   

. Then D  1 9  2 4  1, and so D 1  

1 3

2

1

.

. Then D  2 3  1 2  8, and so D 1   1  8

4 12 2

9 4

6

1

3 1 4 .  8 1 1 2 8 4

3 2

. Then D  4 6  2 12  0, and so D has no inverse. 

         1 2    1 2   2 2  6  4 2  0, and so D has no inverse.   4 . Then D  2  56. D   1 1 2        0 2 3 2 0 3 2 2 4 0


CHAPTER 10

Review

49

    3 0 1     3 0    2 3    4  12  9  1. So D 1 exists.   1 . Then, D  1  57. D   2 3 0        2 3   4 2  4 2 1       1 3 1 1 1 0 1 3 1 1 1 0 3 0 1 1 0 0   R2  2R1  R2     R2 1  R2  R1     0 9 2 2 3 0  3  2 3 0 0 1 0  R        2 3 0 0 1 0   2R3 R3  4R1  R3  4 2 1 0 0 1 0 14 3 4 4 1 4 2 1 0 0 1       1 3 1 1 1 0 1 3 1 1 1 0 1 3 1 1 1 0     R3  R2   R3  9R2  R3 3  R2  R3   0 27 6 6 9 0  R          0 27 6 6 9 0  1 R3  0 1 0 2 1 2   R1  3R2  R1 3 0 28 6 8 8 2 0 1 0 2 1 2 0 9 2 2 3 0       3 2 3 1 0 1 5 4 6 1 0 0 3 2 3 1       2 R3 0 1 0 . Thus, D 1     2 1 2  2 1 2  0 1 0 2 1 2   .     R1  R3  R1 8 6 9 0 0 2 16 12 18 0 0 1 8 6 9           1 2 3 1 0 0 1 2 3       2 4 2 5     4 5 1          58. D    2 4 5 . Then, D  1  5 6   2  2 6   3  2 5   1  4  6  1. So D exists.  2 4 5 0 1 0        2 5 6 0 0 1 2 5 6     1 2 3 1 0 0 1 2 3 1 0 0   R  R   R  3R  R R2  2R1  R2 2 3 1 3 1       0 0 1 2 1 0    0 1 0 2 0 1   R3  2R1  R3 0 1 0 2 0 1 0 0 1 2 1 0       1 3 2 1 2 0 5 3 0 1 0 0 1 3 2       1 1  2R2  R1    0 1 0 2 0 1  R       0 1 0 2 0 1 . Thus, D   2 0 1 . 2 1 0 0 0 1 2 1 0 0 0 1 2 1 0       1 0 0 1 1 0 0 1 1 0 0 0 1         2 R2   2 0 2 0 2 0 2 0 2 0 2 0 1 0 0 1 3 3   R3     1     59. D   . Thus, D   0 3 3   2   3   24 and D exists.  0 0 3 3   1 0 4     0 0 3 3 0 0 1 0 4 R4 0 0 4 0 0 0 4 0 0 0 4 0 0 0 1       1 0 0  14 1 0 0 1 1 0 0 0 1 0 0 0 1 0 0  14       R1  R4  R1  0 1 0 1  0 1 0 1 0 1 0 0  0 1 0 0 0 1 0 1        R2 R4  R2 2 2 4 2 4 1     . Therefore, D   .  0 0 1 1  0 0 1 1 0 0 1 0  0 0 1 0 0 0 1 1  R3  R4  R3 3 3 4  3 4      1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 14 4 4   1 0 1 0   0 1 0 1   60. D   . Thus, 1 1 1 2   1 2 1 2           1 0 1   0 1 1                  1 2 1 1  1 2 1 1           0  1  0  1  0.  D   1 1 2    1 1 2           1 2 1 2  1 2 2 1    2 1 2 1 2 2 Hence, D 1 does not exist.


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CHAPTER 10 Matrices and Determinants

    2 5 2 5 1       , and so . If we let A   , then A1  61.  24  25 5 12 5 2 y 5 12 17 5 2        x 2 5 10 65     . Therefore, the solution is 65 154. y 5 12 17 154        6 5 x 1 6 5     . , then 62.  If we let A   8 7 y 1 8 7              7 5 7 5 1 7 5 7 5 x 6 1 2 , and so     2 2     .  2    1  A1  2 42  40 8 6 4 3 4 3 1 8 6 y 7 

12 5



x

10

12 5

Therefore, the solution is 6 7.            1 2 1 5 2 1 5 1 0 0 1 2 2 0 1 0 2 1 5 x 3     R R        1 2          1  63.   1 2 2   y    4 . Let A   1 2 2 . Then  1 2 2 0 1 0    2 1 5 1 0 0  1 1 0 3 1 0 3 0 0 1 1 0 3 0 0 1 1 0 3 z 6     1 2 2 0 1 0 1 2 2 0 1 0   R  2R  R   R1  2R2  R1 R2  2R1  R2 2 3 2        0 3 1 1 2 0    0 1 1 1 0 2   R3  R1  R3 R3  R3  2R2 0 2 1 0 1 1 0 2 1 0 1 1       1 0 4 2 1 4 1 0 4 2 1 4 1 0 0 6 3 8     R1  4R3  R1   R3  0 1 1 1 0 2      0 1 0 1 1 1 .      0 1 1 1 0 2  R2   R3  R2 0 0 1 2 1 3 0 0 1 2 1 3 0 0 1 2 1 3          1 1 6 3 8 x 6 3 8  12 3                1   1  Hence, A1    1 1 1  and  y    1 1 1   4    12 , and so the solution is 1 1 2 1 3 z 2 1 3 6 12   1 1  1 .  12 12 12          2 0 3 x 5 2 0 3 2 0 3 1 0 0          R R 1 2         64.   1 1 6   y    0 . Let A   1 1 6 . Then  1 1 6 0 1 0   3 1 1 z 5 3 1 1 3 1 1 0 0 1       1 1 6 0 1 0 1 1 6 0 1 0 1 1 6 0 1 0       2  2R1  R2 2  2R3  R2 2 0 3 1 0 0 R  0 2 9 1 2 0  R           0 2 9 1 2 0  R3  3R1  R3 3 1 1 0 0 1 0 4 17 0 3 1 0 0 1 2 1 1     1 1 0 12 5 6 1 1 0 12 5 6  1    R1  9R3  R1 1  R2  R1  0 2 0 17 7 9   2 R2  0 1 0 17  7  9  R           2 2 2 R2  6R3  2R2 0 0 1 2 1 1 0 0 1 2 1 1            7 3 3 7 3 3 x 10 1 0 0 72  32  32 2 2 2  2 2 2  5           0 1 0 17  7  9 . Hence, A1   17  7  9  and  y    17  7  9   0    20 , and    2    2  2 2 2  2 2  2 2   z 5 0 0 1 2 1 1 2 1 1 2 1 1 5 so the solution is 10 20 5.

65. (a) The i jth entry of A represents how many pounds of vegetable j were sold on day i, and the ith entry of B represents the price of vegetable i.


CHAPTER 10

Review

51

  150   685   100    . The jth entry of AB represents the total revenue on day j. (b) AB    14 12 16  410 050 25 16 30

66. (a) We are given that x  y  18, and since the total is $600, 20x  50y  600.      20 50 x 600      (b)  1 1 y 18     1 5 1 50  1 3 , so     30 (c) A1  1 2 20 1  50 1 1 20 30 3          1 1 5 18 5   20  30 10 600  600 3  3   . The customer    30 X  A1 B   30 1 2 1 600  2 18 20  12 8 18 30 3 30 3 received 10 $20 bills and 8 $50 bills.

               2 13   13 7  2 7   60  78  18.   208  210  2, and  D y      32  42  10, Dx    67. D         6 30   30 16   6 16    2  1 and y  18  9 , and so the solution is 1  9 . Therefore, x  10 5 10 5 5 5          140 11   12 11    1260  220  1480, and   108  77  185, Dx    68. D      9 9  20  7        12 140   Dy      240  980  740. Therefore, x  1480  8 and y  740  4, and so the solution is 8 4.   185 185 7 20          2 1 5         2 1   1 7      195  39  156,   3 69. D   1 7 0   5      1 7   5 4    5 4 3        0 1 5         0 1   9 7     495  27  522,   3 Dx    9 7 0   5     9 7  9 4     9 4 3         2 0 5           2 0  1 9    D y    1 9 0   5    180  54  126, and   3        1 9   5 9     5 9 3         2 1 0         2 1   2 1      117  117  234.  9 Dz    1 7 9   9      1 7  5 4    5 4 9 

  522   87 , y  126  21 , and z  234  3 , and so the solution is  87  21  3 . Therefore, x  156 26 156 26 156 2 26 26 2


52

CHAPTER 10 Matrices and Determinants

       3 4 1         3 1   1 4      52  11  41,   1 70. D   1 0 4   4      1 4  2 5   2 1 5        10 4 1         10 1   20 4      880  20  860,   1 Dx    20 0 4   4      20 4   30 5     30 1 5           3 10 1           10 1      10 1   20 4     660  80  40  540, and  D y    1 20 4   3    2   1          20 4   30 5   30 5     2 30 5         3 4 10         3 10   1 20      40  50  10.   1 Dz    1 0 20   4      1 20   2 30     2 1 30    860 540 10 860 540 10 Therefore, x  860 41  41 , y   41 , and z  41 , and so the solution is 41   41  41 .       1 3 1               3 1   1 3   1 3  1 1     1 4  8  10  11.   71. The area is   3 1 1     2 2 2  2 2   2 2   3 1    2 2 1        5 2 1               1 5   5 2   5 2  1 1     1 21  3  27  51 .   72. The area is   1 5 1     2 2 2 2  4 1   4 1   1 5    4 1 1 

73. Let x be the amount invested in Bank A, y the amount invested in Bank B, and z the amount invested in Bank C.     x  y z  60,000 y z  60,000     x  We get the following system: 002x  0025y 003z  1575  2x  25y 3z  157,500 which      2x   2x  2z  y y 2z  0 

1

1 1

60,000

   has matrix representation   2 25 3 157,500  

1 1

1 60,000

   0 1 0 40,000    0 05 1 37,500

2 1 2

R1 R2  R1  R3 05R2  R3

0

   0 1 0 40,000    0 0 1 17,500

1

1

1

60,000

  R2   1 R3  0 05 1 3 37,500     0 3 0 120,000   1 0 1 2500   R1  R3  R1  0 1 0 40,000 . Thus, $2500 is    0 0 1 17,500

R2  2R1  R2  R3  2R1  R3

1 0 1 20,000

invested in Bank A, $40,000 in Bank B, and $17,500 in Bank C. 74. Let x, y, and z be the weights (in pounds) of haddock, sea bass, and red snapper. Our system has the matrix     1 1 1 560 1 1 1 560 1     R2 R2  R3  R2  225     representation   0 225 0 765    375 225 600 1725   R3  375R1  R3 375R2  225R3  R3 0 375 225 1140 375 0 600 960       1 1 1 560 1 0 1 220 1 0 0 160   R1  R2  R1     1  R3  R1 0 1 0    0 1 0 340  R   340   0 1 0 340 . Thus, he caught 160 lb 1     R 50625 3 0 0 50625 30375 0 0 1 60 0 0 1 60 of haddock, 340 lb of sea bass, and 60 lb of red snapper.


CHAPTER 10

Test

53

CHAPTER 10 TEST 

1 8 0

0

0 0 0

0

   1.   0 1 7 10  is in row­echelon form, but not reduced row­echelon form because the 1 in the second row does not have a 0 above it.   0 0 0 4   0 0 2 5   2.   is in neither row­echelon nor reduced row­echelon form.  0 1 2 7    1 0 3 0   1 0 0  is in reduced row­echelon form. 3.  0 0 1 

1 0 0

3

0 0 1

3 2

   4.   0 1 0 2  is in reduced row­echelon form.

    x  y  2z  0 5. 2x  4y  5z  5    2y  3z  5

1 1

 has the matrix representation   2 4

0

 5 5  

R2  R1  R2

1 1

2

0

   0 2 1 5    0 2 3 5



2 3 5  1 1 2 0 1 1 2 0 1     R3  R2  R3 5  0 2 1 5   2 R3        0 2 1 5 . Thus z  0, 2y  0  5  y  2 , and 0 0 2 0 0 0 1 0   5 5 5 x  2  2 0  0  x  2 . Thus, the solution is 2  52  0 . 

    2x  3y  z  3 6. x  2y  2z  1    4x  y  5z  4 R2  2R1  R2  R3  4R1  R3

0

2

2 3 1

 has the matrix representation  1 

1 2 2 1    0 7 3 5    0 7 3 8

R3  R2  R3



4 

3

 2 2 1   1 5

R1  R2



4

1 2 2 1    0 7 3 5 .   0 0 0 3

Since the last row corresponds to the equation 0  3, this system has no solution.     x  2y  3 1 2 0 3     R R R 3 1 3 7. has the matrix representation  3y  z  2 0 3 1 2       x  2y  z  2 1 2 1 2 3R3  4R2  R3 

 1 2 0 3   0 3 1 2   0 0 7 7

1 7 R3



 1 2 0 3    0 3 1 2 .   0 0 1 1

1

2 2 1

  2 3 1  4 1 5

1

2 0

 3  4

3

  0 3 1 2   0 4 1 5

Thus z  1, 3y  1  2  y  1, and x  2 1  3  x  1. Hence, the solution is 1 1 1.


54

CHAPTER 10 Matrices and Determinants

    x  3y  z  0 8. 3x  4y  2z  1    x  2y  1

 has the matrix representation  

1 3 1 0    0 5 1 1    0 0 0 0

R3  R2  R3



1 3 1

0

 3 4 2 1  

 15 R2 

1

3 1

0

   0 5 1 1    0 5 1 1   1 0  25  35   1 .  0 1 1  5 5  0 0 0 0

R2  3R1  R2  R3  R1  R3

1 2 0 1  1 3 1 0   1  3R2  R1  0 1 1 1  R   5 5  0 0 0 0 

Since this system is dependent, let z  t. Then y  15 t  15  y  15 t  15 and x  25 t   35  x  25 t  35 . Thus, the   solution is 25 t  35  15 t  15  t . 

In Solutions 9–16, A  

2 3 2 4

2 4

1 0 4

    , B   1 1 , and C   1 1 2 .     3 0

0 1 3

9. A  B is undefined because A is 2  2 and B is 3  2, so they have incompatible dimensions. 10. AB is undefined because A is 2  2 and B is 3  2, so they have incompatible dimensions.            12 22 6 12 6 10 2 4  2 4          2 3     3  1 1    0 1    3 3    3 2  11. B A  3B           1 1   2 4 3 0 6 9 9 0 3 9 3 0 

1 0 4

2 3



2 4

14

4

36 58

   2 3   2 3           12. C B A    1 1 2   1 1  2 4   3 3  2 4   0 3  0 1 3 3 0 8 1 18 28 

13. A  

2 4

  A1 

    1  4 3   2  32   8  6 2 2 1 1

14. B 1 does not exist because B is not a square matrix.

15. det B is not defined because B is not a square matrix.        1 0 4        1 1  1 2     1  4  3  4 16. det C   1 1 2   1      0 1 1 3    0 1 3 17. (a) The system

  4x  3y  10

 3x  2y  30

is equivalent to the matrix equation 

4 3

3 2

 

x y



10 30

.

              4 3  2 3 x 2 3 10 70   4 2  3 3  1. So D 1    and       . (b) We have D    3 4 y 3 4 30 90  3 2  Therefore, x  70 and y  90.


CHAPTER 10

Test

55

       1 4 0   1 4 1          1 0   1 1     2. Since A  0, A does not have an inverse, and   0, B   0 2 0   2  18. A   0 2 0   2        3 1    1 1    3 0 1  1 0 1 

 since B  0, B does have an inverse.   

1 0 0 1 2 0

  0 2 0 0 1 0   0 0 1 3 6 1

   z  14   2x 19. 3x  y  5z  0    4x  2y  3z  2

1 2 R2



1 4 0 1 0 0

 0 2 0 0 1 0 



1

4 0 1 0 0

  0 2 0 0 1 0   0 12 1 3 0 1

R3  3R1  R3

3 0 1 0 0 1

1 0 0 1 2 0

1 2 0

R1  2R2  R1  R3  6R2  R3

     0 1 0 0 1 0 . Therefore, B 1   0 1 0 .     2 2 0 0 1 3 6 1 3 6 1

       2 0 1         3 1   1 5      26  10  36,   1 Then D   3 1 5   2     4 2  2 3   4 2 3

       14 0 1         0 1   1 5      182  2  180,   1 Dx    0 1 5   14     4 2  2 3    2 2 3 

       2 14 1            2 14   14 1    D y    3 0 5   3    120  300  180, and  5        4 2   2 3     4 2 3         2 0 14         3 1   1 0      4  140  144.      Dz    3 1 0   2     14  2 2  2 2     4 2 2 

180 144 Therefore, x  180 36  5, y  36  5, z  36  4, and so the solution is 5 5 4.

20. Let x and y represent the number of pounds of almonds and walnuts respectively. Then the problem is modeled by the system

of equations

x 

y 3

475x  345y  1191

     1 1   345  475  13, , so det D   Then D     475 345  475 345  

1

1

          3   1 3 1   1191  1425  234. Then    det Dx       1035  1191  156, and det D y    475 1191   1191 345  x

   D y  234 Dx  156    12 and y   18, so the customer bought 12 pounds of almonds and 18 pounds D D 13 13

of walnuts.


56

FOCUS ON MODELING

FOCUS ON MODELING Computer Graphics 

1. The data matrix D  

0 1 1 0

 represents the gray

0 0 1 1

square.

Reflection using T   

TD 

y

2

1

0

0 1



1

0

0 1

:

0 1 1 0



0 1



0 0 1 1

y

1

1

0

0 0 1 1

 

2

0

1

1

x

2

_1

1

0

2

x

1 1

2

x

_1

Expansion with c  2 using T  

2 0

: 0 1      2 0 0 1 1 0 0 2 2 0    TD  0 1 0 0 1 1 0 0 1 1 y

Shearing with c  1 using T  

: 0 1      1 1 0 1 1 0 0 1 2 1    TD  0 1 0 0 1 1 0 0 1 1 y

2

2

1

1

0

1

0

x

2

_1

_1

2. The data matrix D  

0 1 1 0 0 0 1 1

square.

 represents the

Reflection in y­axis using T1   

T1 D  

y

2

1

0

0 1

 

1

0

1

0 1 1 0 0 0 1 1 y

1 0

0 1   

3 1

2

2

x

_1

1

_1

0

1

x

:

0 1 1 0

0

0

1 1

 


57

Computer Graphics

Expansion in y­direction with c  3 using T2   

T2 D  

1 0 0 3

 

0 1 1 0 0 0 1 1 y

3. (a) T  

1 15



0 1 1 0 0 0 3 3

0 3 

:

Shear in y­direction with c  1 using T3  

y

3

2

2

1

1

x

1

1 0

: 1 1      1 0 0 1 1 0 0 1 1 0    T3 D   1 1 0 0 1 1 0 1 2 1

3

0

_1

1 0

_1

0

x

1

 is a rightward shear in the x­direction. 0 1     1  1 15   1 15  1  (b) T  1 0 1 0 1

(c) T 1 is a leftward shear in the x­direction.

  (d) The result is the original matrix. Algebraically, T 1 T D  T 1 T D  I D  D where I is the 2  2 identity           1 15 1 15 0 1 1 0 1 15 0 1 25 15 0 1 1 0          matrix:  0 1 0 1 0 0 1 1 0 1 0 0 1 1 0 0 1 1 

4. (a) T  

3 0 0 1

x­direction.

(b) S  

 is an expansion by a factor of 3 in the

1 0 0 2

y­direction.

 is an expansion by a factor of 2 in the

y

y

2

2

1

1

0

1

2

3

x

0

_1

3 0

0 3 3 0

(c) T S D  



0 1

 

1 0

0 2  

 

0 1 1 0 0 0 1 1



  

3 0 0 1

 

0 1 1 0 0 0 2 2

 

This corresponds to expansion by a factor of 3 in the x­direction and a factor of 2 in the y­direction.

x

3

_1

 

0 0 2 2

2

1

y

2 1

0 _1

1

2

3

x


58

FOCUS ON MODELING

(d) W  T S  

3 0

 

1 0



3 0

 0 2     3 0 0 1 1 0 0 3 3 0   . It is the same as T S D. (e) W D   0 2 0 0 1 1 0 0 2 2   0 1 1 4 4 1 1 6 6 0 0  5. (a) D   0 0 4 4 5 5 7 7 8 8 0   075 0 , (b) T   0 1      075 0 0 1 1 4 4 1 1 6 6 0 0 0 075 075 3 3 075 075 45 45 0 0    TD  0 0 4 4 5 5 7 7 8 8 0 0 1 0 0 4 4 5 5 7 7 8 8 0   1 025 , (c) S   0 1      1 025 0 1 1 4 4 1 1 6 6 0 0 0 1 2 5 525 225 275 775 8 2 0    SD   0 1 0 0 4 4 5 5 7 7 8 8 0 0 0 4 4 5 5 7 7 8 8 0   0 1 2 1 0 0 y  represents the 6. (a) The data matrix D   4 0 0 2 4 4 0 0 1 

0 2

figure at right.

3

2 1

(b) T  

1

1

, 0 1    1 1 0 1 2 1 0 0   TD  0 1 0 0 2 4 4 0   0 1 4 5 4 0   0 0 2 4 4 0

The transformation is a reflection about the x­axis and a shear in the x­direction.      1 1 1 0 1 1    (c) T   0 1 0 1 0 1

0

1

2

3

4

5

x

1

2

3

4

5

x

y 0 _1 _2 _3 _4


CORRECTIONS: p. 3,25,31,39,42,59,50,53,67

CHAPTER 11

CONIC SECTIONS

11.1 11.2

Parabolas 1 Ellipses 5

11.3

Hyperbolas 14

11.4 11.5 11.6

Shifted Conics 22 Rotation of Axes 34 Polar Equations of Conics 46 Chapter 11 Review 54 Chapter 11 Test 69

¥

FOCUS ON MODELING: Conics in Architecture 72

1


11 CONIC SECTIONS 11.1 PARABOLAS 1. A parabola is the set of all points in the plane equidistant from a fixed point called the focus and a fixed line called the directrix of the parabola. 2. The graph of the equation x 2  4 py is a parabola with focus F 0 p, directrix y   p, and vertical axis. So the graph of x 2  12y is a parabola with focus F 0 3 and directrix y  3.

3. The graph of the equation y 2  4 px is a parabola with focus F  p 0, directrix x   p, and horizontal axis. So the graph of y 2  12x is a parabola with focus F 3 0 and directrix x  3.

4. (a) From top to bottom: focus 0 3, vertex 0 0, directrix y  3. (b) From left to right: directrix x  3, vertex 0 0, focus 3 0.

5. y 2  2x is Graph III, which opens to the right and is not as wide as the graph for Exercise 9.

6. y 2   14 x is Graph V, the only graph that opens to the left.

7. x 2  6y is Graph II, which opens downward and is narrower than the graph for Exercise 10. 8. 2x 2  y  x 2  12 y is Graph I, the only graph that opens upward.

9. y 2  8x  0  y 2  8x is Graph VI, which opens to the right and is wider than the graph for Exercise 5.

10. 12y  x 2  0  x 2  12y is Graph IV, which opens downward and is wider than the graph for Exercise 7. 11. (a) x 2  16y, so 4 p  16  p  4. The focus is 0 4, the directrix is y  4, and the focal diameter is 16.

12. (a) x 2  8y, so 4 p  8  p  2. The focus is

0 2, the directrix is y  2, and the focal diameter is 8.

y

(b)

(b)

1

2 x

2

13. (a) y 2  4x, so 4 p  4  p  1. The focus is

1 0, the directrix is x  1, and the focal diameter is 4.

(b)

y

the directrix is x  6, and the focal diameter is 24. y

2 1

1

x

14. (a) y 2  24x, so 4 p  24  p  6. The focus is 6 0, (b)

y

1

1

x

x

1


2

CHAPTER 11 Conic Sections

1 y 2  y 2  16x, so 4 p  16  p  4. The 15. (a) x  16

focus is 4 0, the directrix is x  4, and the focal diameter is 16.

16. (a) y  12 x 2  x 2  2y, so 4 p  2  p  12 . The   focus is 0 12 , the directrix is y   12 , and the focal diameter is 2.

y

(b)

y

(b) 2

1

x

1

x

1

17. (a) y  2x 2  x 2   12 y, so 4 p   12  p   18 .   The focus is 0  18 , the directrix is y  18 , and the focal diameter is 12 .

(b)

1 y 2  y 2  12x, so 4 p  12  18. (a) x   12

p  3. The focus is 3 0, the directrix is x  3,

and the focal diameter is 12. y

(b)

y 1

x

2

  19. (a) 5y  x 2 , so 4 p  5  p  54 . The focus is 0 54 , the directrix is y   54 , and the focal diameter is 5.

(b)

x

1

_1

  20. (a) 9x  y 2 , so 4 p  9  p  94 . The focus is 94  0 , the directrix is x   94 , and the focal diameter is 9.

(b)

y

y

2 1 1

21. (a) x 2  12y  0  x 2  12y, so 4 p  12 

p  3. The focus is 0 3, the directrix is y  3,

and the focal diameter is 12. (b)

x

1

x

22. (a) x  15 y 2  0  y 2  5x, so 4 p  5    p   54 . The focus is  54  0 , the directrix is x  54 , and the focal diameter is 5.

y

(b) 1 2

y

x 1 1

x


SECTION 11.1 Parabolas

23. (a) 5x  3y 2  0  y 2   53 x. Then 4 p   53    5 . The focus is  5  0 , the directrix is p   12 12 5 , and the focal diameter is 5 . x  12 3 y

(b)

2

x= 5 12 1

24. (a) 8x 2  12y  0  x 2   32 y. Then 4 p   32    p   38 . The focus is 0  38 , the directrix is y  38 , and the focal diameter is 32 . y

(b)

1

25. x 2  20y

x

1

x

26. x 2  8y

5

-5

5 -2

-10

10

27. y 2   13 x

28. 8y 2  x 2

1

5

-2

10

-2

-1

29. 4x  y 2  0

30. x  2y 2  0 2

In #29, ndicate y-axis scale. -2

-1

1

2

4

-2

31. Since the focus is 0 3, p  3  4 p  12. Hence, the standard equation of the parabola is x 2  12y.   32. Since the focus is 0  18 , p   18  4 p   12 . Hence, the standard equation of the parabola is x 2   12 y. 33. Since the focus is 8 0, p  8  4 p  32. Hence, the standard equation of the parabola is y 2  32x.

34. Since the focus is 5 0, p  5  4 p  20. Hence, the standard equation of the parabola is y 2  20x.   35. Since the focus is 0  34 , p   34  4 p  3. Hence, the standard equation of the parabola is x 2  3y.   1  0 , p   1  4 p   1 . Hence, the standard equation of the parabola is y 2   1 x. 36. Since the focus is  12 12 3 3 37. Since the directrix is x  2, p  2  4 p  8. Hence, the standard equation of the parabola is y 2  8x.

3


4

CHAPTER 11 Conic Sections

38. Since the directrix is y  14 , p   14  4 p  1. Hence, the standard equation of the parabola is x 2  y.

1 , p   1  4 p   2 . Hence, the standard equation of the parabola is x 2   2 y. 39. Since the directrix is y  10 10 5 5

40. Since the directrix is x   18 , p  18  4 p  12 . Hence, the standard equation of the parabola is y 2  12 x.

1 , p   1  4 p   1 . Hence, the standard equation of the parabola is y 2   1 x. 41. Since the directrix is x  20 20 5 5

42. Since the directrix is y  5, p  5  4 p  20. Hence, the standard equation of the parabola is x 2  20y.

43. The focus is on the positive x-axis, so the parabola opens horizontally with 2 p  2  4 p  4. Hence, the standard equation of the parabola is y 2  4x.

44. The focus is on the negative y-axis, so the parabola opens vertically with 2 p  6  4 p  12. Thus, the standard equation of the parabola is x 2  12y.

45. The parabola opens downward with focus 10 units from 0 0, so p  10  4 p  40 and the standard equation of the parabola is x 2  40y.

46. Since the parabola opens upward with focus 5 units from the vertex, the focus is 5 0. So p  5  4 p  20. Thus, the standard equation of the parabola is x 2  20y.

47. The directrix has y-intercept 6, and so p  6  4 p  24. Therefore, the standard equation of the parabola is x 2  24y.

48. Since the focal diameter is 8 and the focus is on the negative y-axis, 4 p  8. So the standard equation is x 2  8y.

49. p  6  4 p  24. Since the parabola opens upward, its standard equation is x 2  24y.

50. The directrix is x  2, and so p  2  4 p  8. Since the parabola opens to the left, its standard equation is y 2  8x.

51. p  4  4 p  16. Since the parabola opens to the left, its standard equation is y 2  16x.

52. p  3  4 p  12. Since the parabola opens downward, its standard equation is x 2  12y.

53. The focal diameter is 4 p  32  32  3. Since the parabola opens to the left, its standard equation is y 2  3x.

54. The focal diameter is 4 p  2 5  10. Since the parabola opens upward, its standard equation is x 2  10y.

55. The equation of the parabola has the form y 2  4 px. Since the parabola passes through the point 4 2, 22  4 p 4  4 p  1, and so its standard equation is y 2  x.

56. Since the directrix is x   p, we have p2  16, so p  4, and the standard equation is y 2  4 px or y 2  16x.  57. The area of the shaded region is width  height  4 p  p  8, and so p2  2  p   2 (because the parabola opens   downward). Therefore, its standard equation is x 2  4 py  4 2y  x 2  4 2y. 58. The focus is 0 p. Since the line has slope 12 , an equation of the line is y  12 x  p. Therefore, the point where the

line intersects the parabola has y-coordinate 12 2  p  p  1. The parabola’s equation is of the form x 2  4 py, so  1  5 2 2 (since p  0). Hence, the standard equation of the parabola is 2  4 p  p  1  p  p  1  0  p  2   x2  2 5  1 y.

59. (a) A parabola with directrix y   p has equation x 2  4 py. If the directrix   is y  12 , then p   12 , so an equation is x 2  4  12 y  x 2  2y.

If the directrix is y  1, then p  1, so an equation is x 2  4 1 y  x 2  4y. If the directrix is y  4, then p  4, so an equation is

x 2  4 4 y  x 2  16y. If the directrix is y  8, then p  8, so

the standard equation is x 2  4 8 y  x 2  32y.

(b)

-4

-2

1

0 -1 -2

2

4 x@=_32y x@=_16y

-3 -4 -5

x@=_4y x@=_2y

As the directrix moves further from the vertex, the parabolas get flatter.


SECTION 11.2 Ellipses

60. (a) If the focal diameter of a parabola is 4 p, it has standard equation

(b)

10

x 2  4 py. If the focal diameter is 4 p  1, the standard equation is

2y=x@

8 4y=x@

6

x 2  y. If the focal diameter is 4 p  2, the standard equation is x 2  2y.

4

If the focal diameter is 4 p  4, the standard equation is x 2  4y. If the

focal diameter is 4 p  8, the standard equation is x 2  8y.

y=x@

5

8y=x@

2 -4

-2

0

2

4

As the focal diameter increases, the parabolas get flatter.

61. (a) Since the focal diameter is 12 cm, 4 p  12. Hence, the parabola has standard equation y 2  12x.

(b) At a point 20 cm horizontally from the vertex, the parabola passes through the point 20 y, and hence from part (a),   y 2  12 20  y 2  240  y  4 15. Thus, C D  8 15  31 cm.

62. The equation of the parabola has the form x 2  4 py. From the diagram, the parabola passes through the point 10 1, and so 102  4 p 1  4 p  100 and p  25. Therefore, the receiver is 25 ft from the vertex.

63. With the vertex at the origin, the top of one tower will be at the point 300 150. Inserting this point into the equation x 2  4 py gives 3002  4 p 150  90000  600 p  p  150. So the standard equation of the parabolic part of the

cables is x 2  4 150 y  x 2  600y.

64. The equation of the parabola has the form x 2  4 py. From the diagram, the parabola passes through the point 100 379, and so 1002  4 p 379  1516 p  10000 and p  65963. Therefore, the receiver is about 65963 inches  55 feet from the vertex.

65. Many answers are possible: satellite dish TV antennas, sound surveillance equipment, solar collectors for hot water heating or electricity generation, bridge pillars, etc. 66. Yes. If a cone intersects a plane that is parallel to a line on the cone, the resulting curve is a parabola, as shown in the text.

11.2 ELLIPSES 1. An ellipse is the set of all points in the plane for which the sum of the distances from two fixed points F1 and F2 is constant. The points F1 and F2 are called the foci of the ellipse. x2 y2 2. The graph of the equation 2  2  1 with a  b  0 is an ellipse with horizontal major axis, vertices a 0 and a 0 a b  x2 y2 and foci c 0, where c  a 2  b2 . So the graph of 2  2  1 is an ellipse with vertices 5 0 and 5 0 and foci 5 4 3 0 and 3 0. x2 y2 3. The graph of the equation 2  2  1 with a  b  0 is an ellipse with vertical major axis, vertices 0 a and 0 a b a  x2 y2 and foci 0 c, where c  a 2  b2 . So the graph of 2  2  1 is an ellipse with vertices 0 5 and 0 5 and foci 4 5 0 3 and 0 3. 4. (a) From left to right: vertex 5 0, focus 3 0, focus 3 0, vertex 5 0.

(b) From top to bottom: vertex 0 5, focus 0 3, focus 0 3, vertex 0 5.

y2 x2   1 is Graph II. The major axis is horizontal and the vertices are 4 0. 16 4 y2 6. x 2   1 is Graph IV. The major axis is vertical and the vertices are 0 3. 9 7. 4x 2  y 2  4  x 2  14 y 2  1 is Graph I. The major axis is vertical and the vertices are 0 2. 5.


6

CHAPTER 11 Conic Sections 1 x 2  1 y 2  1 is Graph III. The major axis is horizontal and the vertices are 5 0. 8. 16x 2  25y 2  400  25 16

9.

y2 x2   1. 25 9

(c)

(a) This ellipse has a  5, b  3, and so c2  a 2  b2  16  c  4. The vertices c are 5 0, the foci are 4 0, and the eccentricity is e   45  08. a

y

1 1

x

1

x

2

x

(b) The length of the major axis is 2a  10, and the length of the minor axis is 2b  6.

10.

y2 x2  1 16 25

(c)

(a) This ellipse has a  5, b  4, and so c2  25  16  9  c  3. The vertices

y

1

are 0 5, the foci are 0 3, and the eccentricity is e  ac  35  06.

(b) The length of the major axis is 2a  10, and the length of the minor axis is 2b  8.

11.

y2 x2  1 36 81

(c)

 (a) This ellipse has a  9, b  6, and so c2  81  36  45  c  3 5. The    vertices are 0 9, the foci are 0 3 5 , and the eccentricity is

y

2

 e  ac  35 .

(b) The length of the major axis is 2a  18 and the length of the minor axis is 2b  12.

12.

x2  y2  1 4

(c)

 (a) This ellipse has a  2, b  1, and so c2  4  1  3  c  3. The vertices     are 2 0, the foci are  3 0 , and the eccentricity is e  ac  23 .

y

1 1

x

(b) The length of the major axis is 2a  4 and the length of the minor axis is 2b  2.

13.

y2 x2  1 49 25  (a) This ellipse has a  7, b  5, and so c2  49  25  24  c  2 6. The    vertices are 7 0, the foci are 2 6 0 , and the eccentricity is  e  ac  2 7 6 .

(b) The length of the major axis is 2a  14 and the length of the minor axis is 2b  10.

(c)

y

2 2

x


7

SECTION 11.2 Ellipses

14.

y2 x2  1 9 64

(c)

 (a) This ellipse has a  8, b  3, and so c2  64  9  55  c  55. The    vertices are 0 8, the foci are 0  55 , and the eccentricity is

y

2 x

2

 e  ac  855 .

(b) The length of the major axis is 2a  16 and the length of the minor axis is 2b  6.

15. 9x 2  4y 2  36 

y2 x2  1 4 9

(c)

 (a) This ellipse has a  3, b  2, and so c2  9  4  5  c  5. The vertices     are 0 3, the foci are 0  5 , and the eccentricity is e  ac  35 .

y

1 1

x

(b) The length of the major axis is 2a  6, and the length of the minor axis is 2b  4.

16. 4x 2  25y 2  100 

y2 x2  1 25 4

(c)

 (a) This ellipse has a  5, b  2, and so c2  25  4  21  c  21. The    vertices are 5 0, the foci are  21 0 , and the eccentricity is

y

1 1

x

1

x

1

x

 e  ac  521 .

(b) The length of the major axis is 2a  10, and the length of the minor axis is 2b  4.

17. x 2  4y 2  16 

y2 x2  1 16 4

(c)

 (a) This ellipse has a  4, b  2, and so c2  16  4  12  c  2 3.The    vertices are 4 0, the foci are 2 3 0 , and the eccentricity is

y

1

  e  ac  2 4 3  23 .

(b) The length of the major axis is 2a  8, and the length of the minor axis is 2b  4.

18. 4x 2  y 2  16 

y2 x2  1 4 16

 (a) This ellipse has a  4, b  2, and so c2  16  4  12  c  2 3. The    vertices are 0 4, the foci are 0 2 3 , and the eccentricity is   e  ac  2 4 3  23 .

(b) The length of the major axis is 2a  8, and the length of the minor axis is 2b  4.

(c)

y

1


8

CHAPTER 11 Conic Sections

19. 16x 2  25y 2  1600 

y2 x2  1 100 64

(c)

(a) This ellipse has a  10, b  8, and so c2  100  64  36  c  6. The

y

2

vertices are 10 0, the foci are 6 0, and the eccentricity is e  ac  35 .

x

2

(b) The length of the major axis is 2a  20 and the length of the minor axis is 2b  16.

y2 x2  1 49 2   (a) This ellipse has a  7, b  2, and so c2  49  2  47  c  47. The    vertices are 7 0, the foci are  47 0 , and the eccentricity is

20. 2x 2  49y 2  98 

(c)

y

1 x

1

 e  ac  747 .

(b) The length of the major axis is 2a  14 and the length of the minor axis is  2b  2 2. y2 x2  1 3 9   (a) This ellipse has a  3, b  3, and so c2  9  3  6  c  6. The    vertices are 0 3, the foci are 0  6 , and the eccentricity is

21. 3x 2  y 2  9 

(c)

y

1 1

x

1

x

1

x

 e  ac  36 .

(b) The length of the major axis is 2a  6 and the length of the minor axis is  2b  2 3. y2 x2  1 9 3   (a) This ellipse has a  3, b  3, and so c2  9  3  6  c  6. The    vertices are 3 0, the foci are  6 0 , and the eccentricity is

22. x 2  3y 2  9 

(c)

y

1

 e  ac  36 .

(b) The length of the major axis is 2a  6 and the length of the minor axis is  2b  2 3. y2 x2  1 2 4   (a) This ellipse has a  2, b  2, and so c2  4  2  2  c  2. The    vertices are 0 2, the foci are 0  2 , and the eccentricity is

23. 2x 2  y 2  4 

 e  ac  22 .

(b) The length of the major axis is 2a  4 and the length of the minor axis is  2b  2 2.

(c)

y

1


9

SECTION 11.2 Ellipses

y2 x2  1 4 3  (a) This ellipse has a  2, b  3, and so c2  4  3  c  1. The vertices are

24. 3x 2  4y 2  12 

(c)

y

1

2 0, the foci are 1 0, and the eccentricity is e  ac  12 .

x

1

(b) The length of the major axis is 2a  4 and the length of the minor axis is  2b  2 3. 25. x 2  4y 2  1 

y2 x2  1 1 1

(c)

y 1

4

 (a) This ellipse has a  1, b  12 , and so c2  1  14  34  c  23 . The vertices    are 1 0, the foci are  23  0 , and the eccentricity is   3 e  ac  32 1  2 .

1

x

(b) The length of the major axis is 2a  2, and the length of the minor axis is 2b  1.

26. 9x 2  4y 2  1 

y2 x2  1 19 14

(c)

 5  c  5 . The vertices are (a) This ellipse has a  12 , and so c2  14  19  36 6        56 5 1 0  2 , the foci are 0  6 , and the eccentricity is e  ac  12  35 .

y 0.5

0.5 x

(b) The length of the major axis is 2a  1, and the length of the minor axis is 2b  23 .

27. x 2  4  2y 2  x 2  2y 2  4 

y2 x2  1 4 2

(c)

  2, and so c2  4  2  2  c  2. The    vertices are 2 0, the foci are  2 0 , and the eccentricity is

(a) This ellipse has a  2, b 

y

1 x

1

 e  ac  22 .

(b) The length of the major axis is 2a  4, and the length of the minor axis is  2b  2 2. x2 y2 1 28. y 2  1  2x 2  2x 2  y 2  1  1  1

2   2 (a) This ellipse has a  1, b  2 , and so c2  1  12  12  c  22 . The    vertices are 0 1, the foci are 0  22 , and the eccentricity is   e  ac  11 2  22 .

(b) The length of the major axis is 2a  2, and the length of the minor axis is  2b  2.

(c)

y 1

1

x


10

CHAPTER 11 Conic Sections

29. This ellipse has a horizontal major axis with a  5 and b  4, so its standard equation is

x2

y2 x 2 y2   1.   1  25 16 52 42

y2 y2 x2 x2   1. 30. This ellipse has a vertical major axis with a  5 and b  2. Thus, its standard equation is 2  2  1  4 25 2 5  31. This ellipse has a vertical major axis with c  2 and b  2. So a 2  c2  b2  22  22  8  a  2 2. So its standard equation is

x2

y2 y2 x2   1.   1     2 4 8 22 2 2

32. This ellipse has a vertical major axis with a  4 and c  3. So c2  a 2  b2  9  16  b2  b2  7. Thus, its standard equation is

x2 y2 y2 x2  2 1   1. 7 7 16 4

x2 y2   1. Substituting the 162 b2 64  36  1  36  1  1  36  3  b2  4 36  48. Thus, the standard point 8 6 into the equation, we get 256 4 4 3 b2 b2 b2 2 2 y x   1. equation of the ellipse is 256 48

33. This ellipse has a horizontal major axis with a  16, so its standard equation has the form

x2 y2 34. This ellipse has a vertical major axis with b  2, so its standard equation has the form 2  2  1. Substituting the 2 a 4 16 4 4 1 3 4 4 1  . Thus, the standard point 1 2 into the equation, we get 4  2  1  2  1   2   a 2  4 4 3 3 a a a y2 x2 3y 2 x2  1   1. equation of the ellipse is 4 163 4 16 35.

y2 y2 x2 4x 2 x2  1 1  y 2  20   25 20 20 25 5  4x 2 . y   20  5

y2 y2 1  1  x 2  y 2  12  12x 2  12 12  y   12  12x 2 .

36. x 2 

4 2

5 -2 -5

-1

5

-2

1

2

-4

-5

 37. 6x 2  y 2  36  y 2  36  6x 2  y   36  6x 2 . 5

-10

x2  38. x 2  2y 2  8  2y 2  8  x 2  y 2  4  2  x2 y  4 . 2

10

2

-5 -2

2 -2


SECTION 11.2 Ellipses

11

39. The foci are 4 0, and the vertices are 5 0. Thus, c  4 and a  5, and so b2  25  16  9. Therefore, the standard equation of the ellipse is

y2 x2   1. 25 9

40. The foci are 0 3 and the vertices are 0 5. Thus, c  3 and a  5, and so c2  a 2  b2  9  25  b2  b2  25  9  16. Therefore, the standard equation of the ellipse is

y2 x2   1. 16 25

41. The foci are 1 0 and the vertices are 2 0. Thus, c  1 and a  2, so c2  a 2  b2  1  4  b2  b2  4  1  3. Therefore, the standard equation of the ellipse is

y2 x2   1. 4 3

42. The foci are 0 2 and the vertices are 0 3. Thus, c  2 and a  3, so c2  a 2  b2  4  9  b2  b2  9  4  5. Therefore, the standard equation of the ellipse is

y2 x2   1. 5 9

    43. The foci are 0  10 and the vertices are 0 7. Thus, c  10 and a  7, so c2  a 2  b2  10  49  b2  b2  49  10  39. Therefore, the standard equation of the ellipse is

y2 x2   1. 39 49

    44. The foci are  15 0 and the vertices are 6 0. Thus, c  15 and a  6, so c2  a 2  b2  15  36  b2  b2  36  15  21. Therefore, the standard equation of the ellipse is

y2 x2   1. 36 21

45. The length of the major axis is 2a  4  a  2, the length of the minor axis is 2b  2  b  1, and the foci are on the y-axis. Therefore, the standard equation of the ellipse is x 2 

y2  1. 4

46. The length of the major axis is 2a  6  a  3, the length of the minor axis is 2b  4  b  2, and the foci are on the x-axis. Therefore, the standard equation of the ellipse is

y2 x2   1. 9 4

47. The foci are 0 2, and the length of the minor axis is 2b  6  b  3. Thus, a 2  4  9  13. Since the foci are on the y-axis, the standard equation is

x2 y2   1. 9 13

48. The foci are 5 0, and the length of the major axis is 2a  12  a  6. Thus, c2  a 2  b2  25  36  b2  b2  36  25  11. Since the foci are on the x-axis, the standard equation is

y2 x2   1. 36 11

49. The endpoints of the major axis are 10 0  a  10, and the distance between the foci is 2c  6  c  3. Therefore, b2  100  9  91, and so the standard equation of the ellipse is

x2 y2   1. 100 91

50. Since the endpoints of the minor axis are 0 3, we have b  3. The distance between the foci is 2c  8, so c  4. Thus, a 2  b2  c2  9  16  25, and the standard equation of the ellipse is

x2 y2   1. 25 9

51. The length of the major axis is 10, so 2a  10  a  5, and the foci are on the x-axis, so the form of the equation is  2   5 x2 y2 4 5 4 4 22  2  1. Since the ellipse passes through  2 1  2 1 2   5 2 , we have 25 25 25 5 b b b b y2 x2 2   1. b  5, and so the standard equation is 25 5


12

CHAPTER 11 Conic Sections

52. The length of the minor axis is 10, so 2b  10  b  5, and the foci are on the y-axis, so the form of the equation is  2  2    5 40 40 5 40 4 x 2 y2  2  1. Since the ellipse passes through   2 1 2  5 40 , we have 1 2 25 a 25 25 a 5 a a 2 2 y x   1.  a 2  50, and so the standard equation is 25 50 2  6 and 53. The eccentricity is 13 , so e  13 , and the foci are 0 2, so c  2. Thus, e  ac  a  ce  13

b2  a 2  c2  36  4  32. The major axis lies on the y-axis, so the standard equation is 54. The eccentricity is e  075  34 and the foci are 15 0 

y2 x2   1. 32 36

3  0 , so c  3 . Thus, a  c  32  2 and e 2 2 34

b2  a 2  c2  4  94  74 . The major axis lies on the x-axis, so the standard equation is

y2 x2  7  1. 4

4   55. Since the length of the major axis is 2a  4, we have a  2. The eccentricity is 23  ac  2c , so c  3. Then y2 b2  a 2  c2  4  3  1, and since the foci are on the y-axis, the standard equation of the ellipse is x 2   1.

4

  56. The eccentricity is e  35 and the major axis has length 2a  12, so a  6. Thus, c  ae  2 5 and y2 x2   1. b2  a 2  c2  36  20  16. The foci are on the x-axis, so the standard equation of the ellipse is

36

  4x 2  y 2  4 57.  4x 2  9y 2  36

16

y

Subtracting the first equation from the second gives

(0, 2) 1

8y 2  32  y 2  4  y  2. Substituting y  2 in the first equation gives

(0, _2)

 2   y2 x    1   9x 2  16y 2  144  144x 2  256y 2  2304 16 9 58.     16x 2  9y 2  144  144x 2  81y 2  1296 y2  x2   1 9 16

Adding gives 175y 2  1008  y   12 5 . Substituting for y gives  2 1296 12  144  9x 2  144  2304 9x 2  16 12 5 25  25  x   5 , and so the   12 four points of intersection are  12 5  5 .

   100x 2  25y 2  100 59. y2    1 x 2 9

Dividing the first equation by 100 gives x 2 

y

(_ 125 , 125)

( 125 , 125)

1

x

1

(

12 12 _5 ,_5

)

(

12 12 5,_5

y

y2  1. 4

y2 y2  0 Subtracting this equation from the second equation gives 9 4   1  1 y 2  0  y  0. Substituting y  0 in the second equation gives 9 4 x 2  02  1  x  1, and so the points of intersection are 1 0.

x

1

4x 2  22  4  x  0, and so the points of intersection are 0 2.

(_1, 0)

1

(1, 0) 1

x

)


SECTION 11.2 Ellipses

60.

  25x 2  144y 2  3600

 144x 2  25y 2  3600

  3600x 2  20,736y 2  518,400  3600x 2 

625y 2  90,000

61. (a) The ellipse x 2  4y 2  16 

y

Subtracting

3600  the second equation from the first, we have 20,111y 2  428,400, so y 2  169  2 60 2 y   60  3600  x   60 13 . Substituting for y gives 25x  144 13 13 , and so   60 the four points of intersection are  60 13   13 .

13

(_ 6013 , 6013 )

10

( 6013 , 6013 ) 10 x

(

_ 60 , _ 60 13 13

)

(

60 , _ 60 13 13

)

y2 x2   1 has a  4 and b  2. Thus, an equation of the ancillary circle is 16 4

x 2  y 2  4.

(b) If s t is a point on the ancillary circle, then s 2  t 2  4  4s 2  4t 2  16  2s2  4 t2  16, which implies that 2s t is a point on the ellipse. 1 k=4 100  x 2 . For the top 62. (a) x 2  ky 2  100  ky 2  100  x 2  y   k 2 1 half, we graph y  100  x 2 for k  4, 10, 25, and 50. k=10 k 1 (b) This family of ellipses have common major axes and vertices, and the eccentricity increases as k increases. 63.

k=50

-10-8 -6 -4 -2 0

k=25

2 4 6 8 10

y2 x2   1 is an ellipse for k  0. Then a 2  4  k, b2  k, and so c2  4  k  k  4  c  2. Therefore, all of k 4k the ellipses’ foci are 0 2 regardless of the value of k.

64. The foci are c 0, where c2  a 2  b2 . The endpoints of one latus rectum are the points c k, and the length is 2k.   b2 a 2  c2 k2 k2 c2 a 2  c2 c2  k2  . Since Substituting this point into the equation, we get 2  2  1  2  1  2  a b b a a2 a2   2b2 b4 b2 b2 b2  a 2  c2 , the last equation becomes k 2  2  k  . Thus, the length of the latus rectum is 2k  2  . a a a a 65. Using the perihelion, a  c  147,000,000, while using the aphelion, a  c  153,000,000. Adding, we have 2  2  2a  300,000,000  a  150,000,000. So b2  a 2 c2  150  106  3  106  22,4911012  224911016 .

y2 x2   1. 16 22500  10 22491  1016 c 66. Using the eccentricity, e  025   c  025a. Using the length of the minor axis, 2b  10,000,000,000  a 18 b  5  109 . Since a 2  c2  b2 , a 2  025a2  25  1018  15 a 2  25  1018  a 2  80 3  10    16     5 5 9 9 9  5  109 . Since the Sun is at one focus of the ellipse, a  80 3  10  4 3  10 . Then c  025 4 3  10 3    the distance from Pluto to the Sun at perihelion is a  c  4 53  109  53  109  3 53  109  387  109 km; the    distance from Pluto to the Sun at aphelion is a  c  4 53  109  53  109  5 53  109  645  109 km. Thus, an equation of the orbit is

67. Using the perilune, a  c  1075  68  1143, and using the apolune, a  c  1075  195  1270. Adding, we get

2a  2413  a  12065. So c  1270  12065  c  635. Therefore, b2  120652  6352  1,451,610. Since

a 2  1,455,642, an equation of Apollo 11’s orbit is

y2 x2   1. 1,455,642 1,451,610


14

CHAPTER 11 Conic Sections

68. Placing the origin at the center of the sheet of plywood and letting the x-axis be the long central axis, we have 2a  8, so  that a  4, and 2b  4, so that b  2. So c2  a 2  b2  42  22  12  c  2 3  346. So the tacks should be located 2 346  692 feet apart and the string should be 2a  8 feet long. 69. From the diagram, a  40 and b  20, and so an equation of the ellipse whose top half is the window is

x2 y2   1. 1600 400

252 h2 Since the ellipse passes through the point 25 h, by substituting, we have   1  625  4y 2  1600  1600 400   5 39 975   1561 in. Therefore, the window is approximately 156 inches high at the specified point. y 2 2 70. Have each friend hold one end of the string on the blackboard. These fixed points will be the foci. Then, keeping the string taut with the chalk, draw the ellipse. 71. We start with the flashlight perpendicular to the wall; this shape is a circle. As the angle of elevation increases, the shape of the light changes to an ellipse. When the flashlight is angled so that the outer edge of the light cone is parallel to the wall, the shape of the light is a parabola. Finally, as the angle of elevation increases further, the shape of the light is hyperbolic. 72. The shape drawn on the paper is almost, but not quite, an ellipse. For example, when the bottle has radius 1 unit and the compass legs are set 1 unit apart, then it can be shown that an equation of the resulting curve is 1  y 2  2 cos x. The graph of this curve differs very slightly from the ellipse with the same major and minor axis. This example shows that in mathematics, things are not always as they appear to be.

11.3 HYPERBOLAS 1. A hyperbola is the set of all points in the plane for which the difference of the distances from two fixed point F1 and F2 is constant. The points F1 and F2 are called the foci of the hyperbola. x2 y2 2. The graph of the equation 2  2  1 with a  0, b  0 is a hyperbola with horizontal transverse axis, vertices a 0 a b  x2 y2 and a 0 and foci c 0, where c  a 2  b2 . So the graph of 2  2  1 is a hyperbola with vertices 4 0 and 4 3 4 0 and foci 5 0 and 5 0. y2 x2 3. The graph of the equation 2  2  1 with a  0, b  0 is a hyperbola with vertical transverse axis, vertices 0 a a b  y2 x2 and 0 a and foci 0 c, where c  a 2  b2 . So the graph of 2  2  1 is a hyperbola with vertices 0 4 and 4 3 0 4 and foci 0 5 and 0 5. 4. (a) From left to right: focus 5 0, vertex 4 0, asymptote y   34 x, asymptote y  34 x, vertex 4 0, focus 5 0.

(b) From top to bottom: focus 0 5, vertex 0 4, asymptote y   43 x, asymptote y  43 x, vertex 0 4, focus 0 5.

5.

x2  y 2  1 is Graph III, which opens horizontally and has vertices at 2 0. 4

6. y 2 

x2  1 is Graph IV, which opens vertically and has vertices at 0 1. 9

1 x 2  1 is Graph II, which opens vertically and has vertices at 0 3. 7. 16y 2  x 2  144  19 y 2  144 1 x 2  1 y 2  1 is Graph I, which opens horizontally and has vertices at 5 0. 8. 9x 2  25y 2  225  25 9


15

SECTION 11.3 Hyperbolas

9.

y2 x2  1 4 16

(c)

 (a) The hyperbola has a  2, b  4, and c2  16  4  c  2 5. The vertices    are 2 0, the foci are 2 5 0 , and the asymptotes are y   42 x 

y

1 1

x

1

x

1

x

2

x

1

x

y  2x.

(b) The transverse axis has length 2a  4. 10.

x2 y2  1 9 16

(c)

(a) The hyperbola has a  3, b  4, and c2  9  16  25  c  5. The vertices

y

1

are 0 3, the foci are 0 5, and the asymptotes are y   34 x.

(b) The transverse axis has length 2a  6.

11.

x2 y2  1 36 4

(c)

 (a) The hyperbola has a  6, b  2, and c2  36  4  40  c  2 10. The    vertices are 0 6, the foci are 0 2 10 , and the asymptotes are y  3x.

y

2

(b) The transverse axis has length 2a  12.

12.

y2 x2  1 9 64

(c)

 (a) The hyperbola has a  3, b  8, and c2  9  64  73  c  73. The    vertices are 3 0, the foci are  73 0 , and the asymptotes are y   83 x.

y

5

(b) The transverse axis has length 2a  6.

13.

x2 y2  1 1 25

(c)

y 1

 (a) The hyperbola has a  1, b  5, and c2  1  25  26  c  26. The    vertices are 0 1, the foci are 0  26 , and the asymptotes are y   15 x.

(b) The transverse axis has length 2a  2.

14.

y2 x2  1 2 1   (a) The hyperbola has a  2, b  1, and c2  2  1  3  c  3. The       vertices are  2 0 , the foci are  3 0 , and the asymptotes are  y   22 x.

 (b) The transverse axis has length 2a  2 2.

(c)

y

1 1

x


16

CHAPTER 11 Conic Sections

15. x 2  y 2  1

(c)

 (a) The hyperbola has a  1, b  1, and c2  1  1  2  c  2. The vertices    are 1 0, the foci are  2 0 , and the asymptotes are y  x.

y

1 1

x

(b) The transverse axis has length 2a  2.

16.

y2 x2  1 16 12

(c)

 (a) The hyperbola has a  4, b  2 3, and c2  16  12  28       c  28  2 7. The vertices are 4 0, the foci are 2 7 0 , and the

y

2 2

x

1

x

2

x

2

x

1

x

 asymptotes are y   23 x.

(b) The transverse axis has length 2a  8. 17. 9x 2  4y 2  36 

y2 x2  1 4 9

(c)

 (a) The hyperbola has a  2, b  3, and c2  4  9  13  c  13. The    vertices are 2 0, the foci are  13 0 , and the asymptotes are y   32 x.

y

1

(b) The transverse axis has length 2a  4. 18. 25y 2  9x 2  225 

x2 y2  1 9 25

(c)

 (a) The hyperbola has a  3, b  5, and c2  25  9  34  c  34. The    vertices are 0 3, the foci are 0  34 , and the asymptotes are y   35 x.

y

2

(b) The transverse axis has length 2a  6. 19. 4y 2  9x 2  144 

x2 y2  1 36 16

(c)

 (a) The hyperbola has a  6, b  4, and c2  a 2  b2  52  c  2 13. The    vertices are 0 6, the foci are 0 2 13 , and the asymptotes are

y

2

y   32 x.

(b) The transverse axis has length 2a  12. 20. y 2  25x 2  100 

x2 y2  1 100 4

 (a) The hyperbola has a  10, b  2, and c2  100  4  104  c  2 26. The    vertices are 0 10, the foci are 0 2 26 , and the asymptotes are y  5x.

(b) The transverse axis has length 2a  20.

(c)

y

5


17

SECTION 11.3 Hyperbolas

y2 x2  1 8 2    (a) The hyperbola has a  2 2, b  2, and c2  8  2  10  c  10. The       vertices are 2 2 0 , the foci are  10 0 , and the asymptotes are

21. x 2  4y 2  8  0 

(c)

y

1 1

x

1

x

1

x

1

x

 y   2 x   12 x. 8

 (b) The transverse axis has length 2a  4 2. x2 y2  1 3 9   (a) The hyperbola has a  3, b  3, and c2  3  9  12  c  2 3. The       vertices are 0  3 , the foci are 0 2 3 , and the asymptotes are

22. 3y 2  x 2  9  0 

(c)

y

1

 y   33 x.

 (b) The transverse axis has length 2a  2 3.

23. x 2  y 2  4  0  y 2  x 2  4 

x2 y2  1 4 4

(c)

 (a) The hyperbola has a  2, b  2, and c2  4  4  8  2 2. The vertices are    0 2, the foci are 0 2 2 , and the asymptotes are y  x.

y

1

(b) The transverse axis has length 2a  4.

x2 y2  1 4 12  (a) The hyperbola has a  2, b  2 3, and c2  4  12  16  c  4. The

24. x 2  3y 2  12  0 

(c)

y

1

vertices are 0 2, the foci are 0 4, and the asymptotes are 

2 x   3 x. y  3 2 3

(b) The transverse axis has length 2a  4. y2 25. 4y 2  x 2  1  1  x 2  1 4

 (a) The hyperbola has a  12 , b  1, and c2  14  1  54  c  25 . The      vertices are 0  12 , the foci are 0  25 , and the asymptotes are 1 y   12 1 x   2 x.

(b) The transverse axis has length 2a  1.

(c)

y 1

1

x


18

CHAPTER 11 Conic Sections

26. 9x 2  16y 2  1 

y2 x2  1 19 116

y

(c)

1

1  25  c  5 . The (a) The hyperbola has a  13 , b  14 , and c2  19  16 144 12     1 5 vertices are  3  0 , the foci are  12  0 , and the asymptotes are

1

x

3 y   14 13 x   4 x.

(b) The transverse axis has length 2a  23 . 27. From the graph, the foci are 4 0, and the vertices are 2 0, so c  4 and a  2. Thus, b2  16  4  12, and since the vertices are on the x-axis, the standard equation of the hyperbola is

x2 y2   1. 4 12

28. From the graph, the foci are 0 13 and the vertices are 0 12, so c  13 and a  12. Then

b2  c2  a 2  169  144  25, and since the vertices are on the y-axis, the standard equation of the hyperbola is x2 y2   1. 144 25

29. From the graph, the vertices are 0 4, the foci are on the y-axis, and the hyperbola passes through the point 3 5. So

y2 x2 9 25 32 52  2  1. Substituting the point 3 5, we have  2 1 1  2  16 16 16 b b b 9 y2 9 x2 2  2  b  16. Thus, the standard equation of the hyperbola is   1. 16 16 16 b

the equation is of the form

    x2 y2 30. The vertices are 2 3 0 , so a  2 3, so the standard equation of the hyperbola is of the form   2  2  1. b 2 3 Substituting the point 4 4 into the equation, we get

standard equation of the hyperbola is

16 16 16 16 16 4  1 2   b2  48. Thus, the 1 2  12 b2 12 12 b b

y2 x2   1. 12 48

a 3 31. From the graph, the vertices are 0 3, so a  3. Since the asymptotes are y  3x   x, we have  3  b  1. b b y2 x2 y2 Since the vertices are on the x-axis, the standard equation is 2  2  1   x 2  1. 9 3 1 b 1 3 b 32. The vertices are 3 0, so a  3. Since the asymptotes are y   12 x   x, we have   b  . Since the a 3 2 2 4y 2 y2 x2 x2   1. 1 vertices are on the x-axis, the standard equation is 2  9 9 3 322 33. x 2  2y 2  8  2y 2  x 2  8  y 2  12 x 2  4   y   12 x 2  4

34. 3y 2  4x 2  24  3y 2  4x 2  24  y 2  43 x 2  8   y   43 x 2  8 10

5

-5

5 -5

-10

10 -10


SECTION 11.3 Hyperbolas

35.

x2 y2 x2 y2 x2  1   1  y2  2 2 6 2 6 3  x2 2 y 3 5

-5

36.

19

y2 x2   1  16x 2  25y 2  1600  100 64 2 25y 2  16x 2  1600  y 2  16 25 x  64   2 y   16 25 x  64

20

5

-20

20

-5 -20

37. The foci are 5 0 and the vertices are 3 0, so c  5 and a  3. Then b2  25  9  16, and since the vertices are on the x-axis, the standard equation of the hyperbola is

y2 x2   1. 9 16

38. The foci are 0 10 and the vertices are 0 8, so c  10 and a  8. Then b2  c2  a 2  100  64  36, and since the vertices are on the y-axis, the standard equation of the hyperbola is

x2 y2   1. 64 36

39. The foci are 0 2 and the vertices are 0 1, so c  2 and a  1. Then b2  4  1  3, and since the vertices are on the y-axis, the standard equation is y 2 

x2  1. 3

40. The foci are 6 0 and the vertices are 2 0, so c  6 and a  2. Then b2  c2  a 2  36  4  32, and since the vertices are on the x-axis, the standard equation is

y2 x2   1. 4 32

b b 41. The vertices are 1 0 and the asymptotes are y  5x, so a  1. The asymptotes are y   x, so  5  b  5. a 1 2 y  1. Therefore, the standard equation of the hyperbola is x 2  25 6 a 42. The vertices are 0 6, so a  6. The asymptotes are y   13 x   x   13  b  18. Since the vertices are on b b y2 x2 the y-axis, the standard equation of the hyperbola is   1. 36 324 43. The vertices are 0 6, so a  6. Since the vertices are on the y-axis, the standard equation of the hyperbola has the form y2 81 25 45 x2 25  2  1. Since the hyperbola passes through the point 5 9, we have  2 1 2   b2  20. 36 36 36 b b b x2 y2   1. Thus, the standard equation is 36 20 44. The vertices are 2 0, so a  2. Since the vertices are on the x-axis, the hyperbola has a standard equation of the form    y2 30 30 9 5 x2  2  1. Since the hyperbola passes through the point 3 30 , we have  2  1   2  b2  24. 4 4 4 b b b x2 y2 Thus, the standard equation is   1. 4 24 45. The asymptotes of the hyperbola are y  x, so b  a. Since the hyperbola passes through the point 5 3, its foci are x2 y2 25 9 on the x-axis, and its standard equation has the form, 2  2  1, so it follows that 2  2  1  a 2  16  b2 . a a a a y2 x2   1. Therefore, the standard equation of the hyperbola is 16 16


20

CHAPTER 11 Conic Sections

46. The asymptotes of the hyperbola are y  x, so b  a. Since the hyperbola passes through the point 1 2, its foci are on

y2 x2 4 1 the y-axis, and its standard equation has the form 2  2  1, so it follows that 2  2  1  a 2  3. Therefore, the a a a a x2 y2   1. standard equation of the hyperbola is 3 3

x2 42 12 y2 47. The foci are 0 3, so c  3 and an equation is 2  2  1. The hyperbola passes through 1 4, so 2  2  1  a b a b        16b2  a 2  a 2 b2  16b2  9  b2  9  b2 b2  b4  8b2  9  0  b2  1 b2  9  0, Thus, b2  1, a 2  8, and the standard equation is

y2  x 2  1. 8

       x2 y2  1. The hyperbola passes through 4 18 , 48. The foci are  10 0 , so c  10 and an equation is 2  b a     18 16 2 2 2 2 2 so 2  2  1  16b  18a  a b  16b  18 10  b2  10  b2 b2  b4  24b2  180  0  a b    y2 x2 b2  30 b2  6  0. Thus, b2  6, a 2  4, and the standard equation is   1. 4 6 49. The foci are 5 0, and the length of the transverse axis is 6, so c  5 and 2a  6  a  3. Thus, b2  25  9  16, and the standard equation is

x2 y2   1. 9 16

50. The foci are 0 1, and the length of the transverse axis is 1, so c  1 and 2a  1  a  12 . Then b2  c2  a 2  1  14  34 , and since the foci are on the y-axis, the standard equation is

x2 y2 4x 2   1  4y 2   1. 14 34 3

  y2 x2   1 has a  5 and b  5. Thus, the asymptotes are y  x, and their 5 5 slopes are m 1  1 and m 2  1. Since m 1  m 2  1, the asymptotes are perpendicular.

51. (a) The hyperbola x 2  y 2  5 

(b) Since the asymptotes are perpendicular, they must have slopes 1, so a  b. Therefore, c2  2a 2  a 2 

y2 c2 x2 since the vertices are on the x-axis, the hyperbola’s standard equation is 1  1  1  x 2  y 2  . 2 2 2 2c 2c

y2 x2 y2 x2 52. The hyperbolas 2  2  1 and 2  2  1 are conjugate to each other. a b a b (a) x 2  4y 2  16  0 

y2 x2   1 and 4y 2  x 2  16  0  16 4

y2 x2   1. So the hyperbolas are conjugate to each other. 16 4 (b) They both have the same asymptotes, y   12 x. (c) The two general conjugate hyperbolas both have asymptotes b y   x. a

y

1 1

x

c2 , and 2


SECTION 11.3 Hyperbolas

21

  53. x  c2  y 2  x  c2  y 2  2a. Let us consider the positive case only. Then   x  c2  y 2  2a  x  c2  y 2 , and squaring both sides gives x 2  2cx  c2  y 2  4a 2    4a x  c2  y 2  x 2  2cx  c2  y 2  4a x  c2  y 2  4cx  4a 2 . Dividing by 4 and squaring both sides   gives a 2 x 2  2cx  c2  y 2  c2 x 2  2a 2 cx  a 4  a 2 x 2  2a 2 cx  a 2 c2  a 2 y 2  c2 x 2  2a 2 cx  a 4  a 2 x 2  a 2 c2  a 2 y 2  c2 x 2  a 4 . Rearranging the order, we have c2 x 2  a 2 x 2  a 2 y 2  a 2 c2  a 4      c2  a 2 x 2  a 2 y 2  a 2 c2  a 2 . The negative case gives the same result.

54. (a) The hyperbola F2 5 0.

x2 y2   1 has a  3, b  4, so c2  9  16  25, and c  5. Therefore, the foci are F1 5 0 and 9 16

  25 2569 25 16 (b) Substituting P 5 16 3 into an equation of the hyperbola, we get 9  16  9  9  1, so P lies on the hyperbola.    (c) d P F1   5  52  163  02  16 , and d F  P  5  52  163  02  13 900  256  34 2 3 3 .

16 (d) d P F2   d P F1   34 3  3  6  2 3  2a.

55. (a) From the equation, we have a 2  k and b2  16  k. Thus, c2  a 2  b2  k  16  k  16  c  4. Thus the foci of the family of hyperbolas are 0 4.    x2 kx 2 y2 x2 2 (b)  1 y k 1 y k . For k=12 k 16  k 16  k 16  k 8 k=8  6 2 kx , k  1, 4, 8, 12. As k the top branch, we graph y  k  k=4 4 16  k increases, the asymptotes get steeper and the vertices move further apart.

-10

2

k=1

0

10

56. d A B  500  2c  c  250. (a) Since t  2640 s and   980 fts, we have d  d P A  d P B  t  980 fts  2640 s  2,587,200 ft  490 mi.

(b) c  250, 2a  490  a  245 and the foci are on the y-axis. Then b2  2502  2452  2475. Hence, the standard equation of the hyperbola is

x2 y2   1. 60,025 2475

2502 x2 (c) Since P is due east of A, c  250 is the y-coordinate of P. Therefore, P is at x 250, and so 1  2475 2452   2502 x 2  2475  1  10205. Then x  101, and so P is approximately 101 miles from A. 2452 57. Since the asymptotes are perpendicular, a  b. Also, since the sun is a focus and the closest distance is 2  109 , it follows   2  109 and that c  a  2  109 . Now c2  a 2  b2  2a 2 , and so c  2a. Thus, 2a  a  2  109  a   21 4  1018 x2 y2 a 2  b2   1   23  1019 . Therefore, the standard equation of the hyperbola is 23  1019 23  1019 32 2 x 2  y 2  23  1019 .


22

CHAPTER 11 Conic Sections

58. (a) These equally spaced concentric circles can be used as a kind of measure where we count the number of rings. In the case of the red dots, the sum of the number of wave crests from each center is a constant, in this case 17. As you move out one wave crest from the left stone you move in one wave crest from the right stone. Therefore this satisfies the geometric definition of an ellipse. (b) Similarly, in the case of the blue dots, the difference of the number of wave crests from each center is a constant. As you move out one wave crest from the left center you also move out one wave crest from the right stone. Therefore this satisfies the geometric definition of a hyperbola. 59. Some possible answers are: as cross-sections of nuclear power plant cooling towers, or as reflectors for camouflaging the location of secret installations. 60. The wall is parallel to the axis of the cone of light coming from the top of the shade, so the intersection of the wall and the cone of light is a hyperbola. In the case of the flashlight, hold it parallel to the ground to form a hyperbola.

11.4 SHIFTED CONICS 1. (a) If we replace x by x  3 the graph of the equation is shifted to the right by 3 units. If we replace x by x  3 the graph is shifted to the left by 3 units. (b) If we replace y by y  1 the graph of the equation is shifted upward by 1 unit. If we replace y by y  1 the graph is shifted downward by 1 unit. 2. x 2  12y, from top to bottom: focus 0 3, vertex 0 0, directrix y  3. x  32  12 y  1, from top to bottom: focus 3 4, vertex 3 1, directrix y  2. 3.

4.

5.

x2 y2 x  32 y  12  2  1, from left to right: vertex 5 0, focus 3 0, focus 3 0, vertex 5 0.   1, from 2 2 5 4 5 42 left to right: vertex 2 1, focus 0 1, focus 6 1, vertex 8 1. y2 x2   1, from left to right: focus 5 0, vertex 4 0, asymptote y   34 x, asymptote y  34 x, vertex 4 0, 42 32 x  32 y  12 focus 5 0.   1, from left to right: focus 2 1, vertex 1 1, asymptote y   34 x  13 4 , 2 4 32 asymptote y  34 x  54 , vertex 7 1, focus 8 1. x  22 y  12  1 9 4 x2 y2   1 by shifting it 2 units to 9 4   the right and 1 unit upward. So a  3, b  2, and c  9  4  5. The

(a) This ellipse is obtained from the ellipse

center is 2 1, the vertices are 2  3 1  1 1 and 5 1, and the foci are    2  5 1 .

(b) The length of the major axis is 2a  6 and the length of the minor axis is 2b  4.

(c)

y

1 1

x


23

SECTION 11.4 Shifted Conics

6.

x  32  y  32  1 16

(c) x2

1 x

1

 y 2  1 by shifting to the right   3 units and downward 3 units. So a  4, b  1, and c  16  1  15. The center is 3 3, the vertices are 3  4 3  1 3 and 7 3, and the    foci are 3  15 3 .

(a) This ellipse is obtained from the ellipse

y

16

(b) The length of the major axis is 2a  8 and the length of the minor axis is 2b  2.

7.

x2 y  52  1 9 25

(c)

y

1 x

1

x2 y2   1 by shifting it 5 units (a) This ellipse is obtained from the ellipse 9 25  downward. So a  5, b  3, and c  25  9  4. The center is 0 5, the vertices are 0 5  5  0 10 and 0 0, and the foci are 0 5  4  0 9 and 0 1.

(b) The length of the major axis is 2a  10 and the length of the minor axis is 2b  6.

8. x 2 

y  22 1 4

(c)

y2 (a) This ellipse is obtained from the ellipse x 2   1 by shifting it 2 units 4   downward. So a  2, b  1, and c  4  1  3. The center is 0 2, the

y

1 x

1

vertices are 0 2  2  0 4 and 0 0, and the foci are          0 2  3  0 2  3 and 0 2  3 .

(b) The length of the major axis is 2a  4 and the length of the minor axis is 2b  2.

9.

y  12 x  52  1 16 4 y2 x2   1 by shifting it 5 units to 16 4   the left and 1 units upward. So a  4, b  2, and c  16  4  2 3. The

(c)

y

(a) This ellipse is obtained from the ellipse

center is 5 1, the vertices are 5  4 1  9 1 and 1 1, and the          foci are 5  2 3 1  5  2 3 1 and 5  2 3 1 .

(b) The length of the major axis is 2a  8 and the length of the minor axis is 2b  4.

1 1

x


24

CHAPTER 11 Conic Sections

10.

y  1 x  12  1 36 64

y

(c)

y2 x2   1 by shifting it 1 unit to the 36 64   left and 1 unit downward. So a  8, b  6, and c  64  36  2 7. The center is 1 1, the vertices are 1 1  8  1 9 and 1 7, and          the foci are 1 1  2 7  1 1  2 7 and 1 2 7  1 .

(a) This ellipse is obtained from the ellipse

2 2

x

(b) The length of the major axis is 2a  16 and the length of the minor axis is 2b  12.

11. 4x 2  25y 2  50y  75  4x 2  25 y  12  25  75 

x2 y  12  1 25 4

y

(c)

y2 x2   1 by shifting it 1 unit 25 4   upward. So a  5, b  2, and c  25  4  21. The center is 0 1, the

(a) This ellipse is obtained from the ellipse

1 1

vertices are 5 1  5 1 and 5 1, and the foci are          21 1   21 1 and 21 1 .

x

(b) The length of the major axis is 2a  10 and the length of the minor axis is 2b  4.

12. 9x 2  54x  y 2  2y  46  0  9 x  32  81  y  12  1  46  0 

(c)

x  32 y  12  1 4 36

y

1 2

y2 x2   1 by shifting it 3 units to 4 36   the right and 1 unit downward. So a  6, b  2, and c  36  4  4 2. The

(a) This ellipse is obtained from the ellipse

x

center is 3 1, the vertices are 3 1  6  3 7 and 3 5, and the foci          are 3 1  4 2  3 1  4 2 and 3 4 2  1 .

(b) The length of the major axis is 2a  12 and the length of the minor axis is 2b  4.

13. x  32  8 y  1

(a) This parabola is obtained from the parabola x 2  8y by shifting it 3 units to the right and 1 unit down. So 4 p  8  p  2. The vertex is 3 1, the focus is 3 1  2  3 1, and the directrix is y  1  2  3.

(b)

y

1 1

x


25

SECTION 11.4 Shifted Conics

14. y  12  16 x  3

y

(b)

(a) This parabola is obtained from the parabola y 2  16x by shifting it 3 units to

2

the right and 1 unit down. So 4 p  16  p  4. The vertex is 3 1, the

15. y  52  6x  12  6 x  2

x

1

focus is 3  4 1  7 1, and the directrix is x  3  4  1.

y

(b)

1 1

(a) This parabola is obtained from the parabola y 2  6x by shifting to the right

x

2 units and down 5 units. So 4 p  6  p   32 . The vertex is 2 5, the     focus is 2  32  5  12  5 , and the directrix is x  2  32  72 .

  16. y 2  16x  8  16 x  12

"1" seems to be in the wrong location.

y

(b)

(a) This parabola is obtained from the parabola y 2  16x by shifting to the right 12   unit. So 4 p  16  p  4. The vertex is 12  0 , the focus is     1  4 0  9  0 , and the directrix is x  1  4   7 . 2 2 2 2

2 x

1

The directrix is at x = -7/2

17. 2 x  12  y  x  12  12 y

(b)

y

(a) This parabola is obtained from the parabola x 2  12 y by shifting it 1 unit to the   right. So 4 p  12  p  18 . The vertex is 1 0, the focus is 1 18 , and the directrix is y   18 .

1 x

1

2 2   18. 4 x  12  y  x  12   14 y

y

(b)

1

(a) This parabola is obtained from the parabola x 2   14 y by shifting it 12 unit to   1 . The vertex is  1  0 , the focus is the left, so 4 p   14  p   16 2     1 1 1 1 1  1 .  2  0  16   2   16 , and the directrix is y  0  16 16

19. y 2  6y  12x  33  0  y  32  9  12x  33  0  y  32  12 x  2

1

(b)

x

y

(a) This parabola is obtained from the parabola y 2  12x by shifting it 2 units to the right and 3 units upward. So 4 p  12  p  3. The vertex is 2 3, the focus is 2  3 3  5 3, and the directrix is x  2  3  1.

2 1

x

In #16, something went wrong with the tick spacing.


26

CHAPTER 11 Conic Sections

20. x 2  2x  20y  41  0  x  12  1  20y  41  0  x  12  20 y  2

y

(b)

(a) This parabola is obtained from the parabola x 2  20y by shifting it 1 unit to the left and 2 units upward, so 4 p  20  p  5. The vertex is 1 2, the focus

1

is 1 2  5  1 7, and the directrix is y  2  5  3.

21.

y  32 x  12  1 9 16

1

x

1

x

10

20

1

x

y

(b)

x2 y2   1 by shifting it 9 16  1 unit to the left and 3 units up. So a  3, b  4, and c  9  16  5. The

(a) This hyperbola is obtained from the hyperbola

1

center is 1 3, the vertices are 1  3 3  4 3 and 2 3, the foci are 1  5 3  6 3 and 4 3, and the asymptotes are

y  3   43 x  1  y   43 x  1  3  y  43 x  13 3 and y   43 x  53 .

22. x  82  y  62  1

(b)

y

(a) This hyperbola is obtained from the hyperbola x 2  y 2  1 by shifting to the right 8 units and downward 6 units. So a  1, b  1, and c 

  1  1  2.

The center is 8 6, the vertices are 8  1 6  7 6 and 9 6, the    foci are 8  2 6 , and the asymptotes are y  6   x  8 

_10

_20

y  x  14 and y  x  2.

23. y 2 

x  12 1 4

(b)

x2 (a) This hyperbola is obtained from the hyperbola y 2   1 by shifting it 1 unit 4   to the left. So a  1, b  2, and c  1  4  5. The center is 1 0, the

y

1

vertices are 1 1  1 1 and 1 1, the foci are          1  5  1  5 and 1 5 , and the asymptotes are y   12 x  1  y  12 x  12 and y   12 x  12 .

24.

y  12  x  32  1 25 y2  x 2  1 by shifting to the 25   left 3 units and upward 1 unit. So a  5, b  1, and c  25  1  26. The

(a) This hyperbola is obtained from the hyperbola

center is 3 1, the vertices are 3 1  5  3 4 and 3 6, the foci    are 3 1  26 , and the asymptotes are y  1  5 x  3  y  5x  16 and y  5x  14.

(b)

y

5 1

x

x


27

SECTION 11.4 Shifted Conics

25.

y  12 x  12  1 9 4

y

(b)

x 2 y2   1 by shifting it 1 unit 9 4   to the left and 1 unit downward, so a  3, b  2, and c  9  4  13. The center is 1 1, the vertices are 1  3 1  4 1 and 2 1, the         foci are 1  13 1  1  13 1 and 13  1 1 , and the

1

(a) This hyperbola is obtained from the hyperbola

1

x

5

x

asymptotes are y  1   23 x  1  y  23 x  13 and y   23 x  53 .

26.

x2 y  22  1 36 64

(b)

x2 y2   1 by shifting it (a) This hyperbola is obtained from the hyperbola 36 64  2 unit downward. So a  6, b  8, and c  36  64  10. The center is 0 2, the vertices are 0 2  6  0 8 and 0 4, the foci are

y 5

0 2  10  0 12 and 0 8, and the asymptotes are y  2   34 x  y  34 x  2 and y   34 x  2.

27. 36x 2  72x  4y 2  32y  116  0  36 x  12  36  4 y  42  64  116  0 

y

(b) x  12 y  42  1 36 4

y2 x 2   1 by shifting it 1 unit 36 4   to the left and 4 units upward. So a  6, b  2, and c  36  4  2 10.

2

(a) This hyperbola is obtained from the hyperbola

1

x

The center is 1 4, the vertices are 1 4  6  1 2 and 1 10,          the foci are 1 4  2 10  1 2 10  4 and 1 4  2 10 , and

the asymptotes are y  4  3 x  1  y  3x  7 and y  3x  1. 28. 25x 2  9y 2  54y  306  25x 2  9 y  32  81  306 

x 2 y  32  1 9 25

x2 y2   1 by shifting it 9 25   3 units downward. So a  3, b  5, and c  9  25  34. The center is 0 3, the vertices are 3 3  3 3 and 3 3, the foci are         34 3 , and the asymptotes are  34 3   34 3 and

(a) This hyperbola is obtained from the hyperbola

(b)

y

2 1

x

y  3   53 x  y  53 x  3 and y   53 x  3.

29. This is a parabola that opens down with its vertex at 0 4, so its equation is of the form x 2  a y  4. Since 1 0 is a point on this parabola, we have 12  a 0  4  1  4a  a   14 . Thus, an equation is x 2   14 y  4.

30. This is a parabola that opens to the right with its vertex at 6 0, so its equation is of the form y 2  4 p x  6, with

p  0. Since the distance from the vertex to the directrix is p  6  12  6, an equation is y 2  4 6 x  6 

y 2  24 x  6


28

CHAPTER 11 Conic Sections

31. This is an ellipse with the major axis parallel to the x-axis, with one vertex at 0 0, the other vertex at 10 0, and one   focus at 8 0. The center is at 010 2  0  5 0, a  5, and c  3 (the distance from one focus to the center). So b2  a 2  c2  25  9  16. Thus, an equation is

y2 x  52   1. 25 16

32. This is an ellipse with the major axis parallel to the y-axis. From the graph the center is 2 3, with a  3 and b  2. Thus, an equation is

x  22 y  32   1. 4 9

33. This is a hyperbola with center 0 1 and vertices 0 0 and 0 2. Since a is the distance form the center to a vertex, a we have a  1. The slope of the given asymptote is 1, so  1  b  1. Thus, an equation of the hyperbola is b 2 2 y  1  x  1. 34. From the graph, the vertices are 2 0 and 6 0. The center is the midpoint between the vertices, so the center is   26 00   4 0. Since a is the distance from the center to a vertex, a  2. Since the vertices are on the x-axis, the 2 2 y2 x  42 42 0  42  2  1. Since the point 0 4 lies on the hyperbola, we have  2 1 4 4 b b 2 2 2 16 16  4 y 3y 2 16 16  4 x x  2  1  3  2  b2  . Thus, an equation of the hyperbola is  16  1    1. 4 3 4 4 16 b b

equation is of the form

3

35. The ellipse with center C 2 3, vertices V1 8 3 and V2 12 3, and foci F1 4 3 and F2 8 3 has

x  22 y  32   1. The distance between the vertices a2 b2 is 2a  12  8  20, so a  10. Also, the distance from the center to each focus is c  2  4  6, so a horizontal major axis, so its equation has the form

b2  a 2  c2  100  36  64. Thus, an equation is

y  32 x  22   1. 100 64

36. The ellipse with vertices V1 1 4 and V2 1 6 and foci F1 1 3 and F2 1 5 is centered midway between the   vertices; that is, it has C 1 12 4  6  1 1. The distance between the vertices is 2a  6  4  10, so a  5.

Also, the distance from the center to each focus is c  1  3  4, so b2  a 2  c2  25  16  9. Thus, an equation is x  12 y  12   1. 9 25

37. The hyperbola with center C 1 4, vertices V1 1 3 and V2 1 11, and foci F1 1 5 and F2 1 13 has x  12 y  42   1. The distance between the vertices 2 a b2 is 2a  11  3  14, so a  7. Also, the distance from the center to each focus is c  4  5  9, so a vertical transverse axis, so its equation has the form

b2  c2  a 2  81  49  32. Thus, an equation is

x  12 y  42   1. 49 32

38. The hyperbola with vertices V1 1 1 and V2 5 1, and foci F1 4 1 and F2 8 1 is centered midway between   the vertices; that is, it has C 12 1  5  1  2 1. It has a horizontal transverse axis, so its equation has the form

x  22 y  12   1. The distance between the vertices is 2a  5  1  6, so a  3. Also, the distance from the 2 a b2 x  22 y  12   1. center to each focus is c  2  4  6, so b2  c2  a 2  36  9  27. Thus, an equation is 9 27

39. The parabola with vertex V 3 5 and directrix y  2 has an equation of the form x  32  4 p y  5. The distance from the vertex to the directrix is p  5  2  3, so an equation is x  32  12 y  5.


SECTION 11.4 Shifted Conics

29

40. The parabola with focus F 1 3 and directrix x  3 has vertex midway between the focus and the directrix; that is, it has   V 12 3  1  3  2 3. The distance from the vertex to the directrix is p  2  3  1. (Since the directrix is to the right of the focus, p is negative.) Thus, an equation is y  32  4 1 x  2  4 x  2.

41. The hyperbola with foci F1 1 5 and F2 1 5 that passes through the point 1 4 is centered midway between the foci;   y2 x  12 that is, it has C 1 12 5  5  1 0. It has a vertical transverse axis, so its equation has the form 2   1. a b2 The point 1 4 lies on the transverse axis, so it is a vertex and we have a  4  0  4. Also, the distance from the center to each focus is c  5  0  5, so b2  c2  a 2  25  16  9. Thus, an equation is

y2 x  12   1. 16 9

42. The hyperbola with foci F1 2 2 and F2 4 2 that passes through the point 3 2 is centered midway between   the foci; that is, it has C 12 2  4  2  1 2. It has a horizontal transverse axis, so its equation has the form

y  22 x  12   1. The point 3 2 lies on the transverse axis, so it is a vertex and we have a  3  1  2. a2 b2 Also, the distance from the center to each focus is c  4  1  3, so b2  c2  a 2  9  4  5. Thus, an equation is x  12 y  22   1. 4 5

43. The ellipse with foci F1 1 4 and F2 5 4 that passes through the point 3 1 is centered midway between the foci;   y  42 x  32 that is, it has C 12 1  5  4  3 4, and so its equation has the form   1. The distance from 2 a b2 the center to each focus is c  3  1  2, so c2  a 2  b2  a 2  b2  4. Substituting the point x y  3 1 into the 3  32 1  42   1  b2  25. From above, we have a 2  b2  4  29. Thus, an 2 a b2 y  42 x  32   1. equation of the ellipse is 29 25 equation of the ellipse, we have

44. The ellipse with foci F1 3 4 and F2 3 4 and x-intercepts 0 and 6 is centered midway between the foci; that is, it has   x  32 y 2 C 3 12 4  4  3 0, and so its equation has the form  2  1. Because 0 is an x-intercept, we substitute b2 a 02 32  2  1  b  3. We also have c  4  0  4, so a 2  b2  c2  25 and an 0 0 into this equation, obtaining b2 a y2 x  32   1. equation is 9 25 45. The parabola that passes through the point 6 1, with vertex V 1 2 and horizontal axis of symmetry has an equation of 1 . the form y  22  4 p x  1. Substituting the point 6 1 into this equation, we have 1  22  4 p 6  1  p  28

Thus, an equation is y  22  17 x  1.

46. Because 6 2 is lower than V 4 1, the parabola opens downward, so an equation is x  42  4 p y  1.

Substituting the point 6 2, we have 6  42  4 p 2  1  p  1, so an equation is x  42  4 y  1.

47. y 2  4 x  2y  y 2  8y  4x  y 2  8y  16  4x  16 

y

y  42  4 x  4. This is a parabola with 4 p  4  p  1. The vertex is

4 4, the focus is 4  1 4  3 4, and the directrix is x  4  1  5. 1 1

x


30

CHAPTER 11 Conic Sections

  48. 9x 2  36x  4y 2  0  9 x 2  4x  4  36  4y 2  0 

y

y2 x  22   1. This is an ellipse with a  3, 4 9      b  2, and c  9  4  5. The center is 2 0, the foci are 2  5 , the

9 x  22  4y 2  36 

1 x

1

vertices are 2 3, the length of the major axis is 2a  6, and the length of the minor axis is 2b  4.

    49. x 2  5y 2  2x  20y  44  x 2  2x  1  5 y 2  4y  4  44  1  20

y

x  12 y  22  x  12  5 y  22  25    1. This is a hyperbola 25 5    with a  5, b  5, and c  25  5  30. The center is 1 2, the foci are    1  30 2 , the vertices are 1  5 2  4 2 and 6 2, and the asymptotes 

1 x

2

are y  2   55 x  1  y   55 x  1  2  y   55 x  2  55 and 

y  55 x  2  55 . y

50. x 2  6x  12y  9  0  x 2  6x  9  12y  x  32  12y. This is a

parabola with 4 p  12  p  3. The vertex is 3 0, the focus is 3 3,

_1

and the directrix is y  3.

51. 4x 2  25y 2  24x  250y  561  0      4 x 2  6x  9  25 y 2  10y  25  561  36  625  4 x  32  25 y  52  100 

x  32

1

x

y 1 x

1

y  52

  1. This is an ellipse 25 4   with a  5, b  2, and c  25  4  21. The center is 3 5, the foci are    3  21 5 , the vertices are 3  5 5  2 5 and 8 5, the length of the major axis is 2a  10, and the length of the minor axis is 2b  4.

  52. 2x 2  y 2  2y  1  2x 2  y 2  2y  1  1  1  2x 2  y  12  2 

  y  12  1. This is an ellipse with a  2, b  1, and c  2  1  1. 2 The center is 0 1, the foci are 0 1  1  0 0 and 0 2, the vertices are     0 1  2 , the length of the major axis is 2a  2 2, and the length of the minor x2 

axis is 2b  2.

y

1 1

x


SECTION 11.4 Shifted Conics

  53. 16x 2  9y 2  96x  288  0  16 x 2  6x  9y 2  288  0    16 x 2  6x  9  9y 2  144  288  16 x  32  9y 2  144 

31

y

2

y2 x  32   1. This is a hyperbola with a  4, b  3, and 16 9  c  16  9  5. The center is 3 0, the foci are 3 5, the vertices are

x

1

3 4, and the asymptotes are y   43 x  3  y  43 x  4 and y  4  43 x.

  54. 4x 2  4x  8y  9  0  4 x 2  x  8y  9  0 

y

   2 4 x 2  x  14  8y  9  1  4 x  12  8 y  1  

2 x  12  2 y  1. This is a parabola with 4 p  2  p  12 . The vertex is     1  1 , the focus is 1  3 , and the directrix is y   1  1  1 . 2 2 2 2 2

1 x

1

  55. x 2  16  4 y 2  2x  x 2  8x  4y 2  16  0    x 2  8x  16  4y 2  16  16  4y 2  x  42  y   12 x  4.

In #54, vertex should be at (1/2, 1), not at (1/2, 1/2).

y

1

Thus, the conic is degenerate, and its graph is the pair of lines y  12 x  4 and

x

1

y   12 x  4.

56. x 2  y 2  10 x  y  1  x 2  10x  y 2  10y  1      x 2  10x  25  y 2  10y  25  1  25  25  x  52  y  52  1.   This is a hyperbola with a  1, b  1, and c  1  1  2. The center is 5 5,    the foci are 5  2 5 , the vertices are 5  1 5  4 5 and 6 5, and the asymptotes are y  5   x  5  y  x and y  x  10.

y

1 x

1

    57. 3x 2  4y 2  6x  24y  39  0  3 x 2  2x  4 y 2  6y  39      3 x 2  2x  1  4 y 2  6y  9  39  3  36 

3 x  12  4 y  32  0  x  1 and y  3. This is a degenerate conic whose

y

(1, 3)

1 1

x

graph is the point 1 3.

      58. x 2  4y 2  20x  40y  300  0  x 2  20x  4 y 2  10y  300  x 2  20x  100    4 y 2  10y  25  300  100  100  x  102  4 y  52  100. Since u 2   2  0 for all u,   , there is no x y such that x  102  4 y  52  100. So there is no solution, and the graph is empty.


32

CHAPTER 11 Conic Sections

59. 2x 2  4x  y  5  0  y  2x 2  4x  5.

-2

2

4

-5 -10

    60. 4x 2  9y 2  36y  0  4x 2  9 y 2  4y  0  4x 2  9 y 2  4y  4  36    9 y 2  4y  4  36  4x 2  y 2  4y  4  4  49 x 2    y  2   4  49 x 2  y  2  4  49 x 2

5

-4

61. 9x 2  36  y 2  36x  6y  x 2  36x  36  y 2  6y    9x 2  36x  45  y 2  6y  9  9 x 2  4x  5  y  32      y  3   9 x 2  4x  5  y  3  3 x 2  4x  5

-2

2

2

4

4

-10

  62. x 2  4y 2  4x  8y  0  x 2  4x  4 y 2  2y  0      x 2  4x  4  4 y 2  2y  1  0  4 y  12  x  22 

2

2 y  1   x  2  y  1   12 x  2  y  1  12 x  2.So

y  1  12 x  2  12 x  2 and y  1  12 x  2   12 x. This is a degenerate

-5

conic.

    63. 4x 2  y 2  4 x  2y  F  0  4 x 2  x  y 2  8y  F    2    4 x 2  x  14  y 2  8y  16  16  1  F  4 x  12  y  12  17  F (a) For an ellipse, 17  F  0  F  17.

(b) For a single point, 17  F  0  F  17. (c) For the empty set, 17  F  0  F  17.

64. The parabola x 2  y  100  x 2   y  100 has 4 p  1  p   14 . The vertex is 0 100 and so the focus is       399 0 100  14  0 399 4 . Thus, one vertex of the ellipse is 0 100, and one focus is 0 4 . Since the second focus of   401 the ellipse is 0 0, the second vertex is 0  14 . So 2a  100  14  401 4  a  8 . Since 2c is the distance between 2

2

399 401 399 2 2 2 the foci of the ellipse, 2c  399 4 , c  8 , and then b  a  c  64  64  25. The center of the ellipse is  2   y  399 2 x x2 8y  3992 8 0 399 and so its equation is     1. 2  1, which simplifies to 8 25 25 160,801 401 8


SECTION 11.4 Shifted Conics

65. (a) x 2  4 p y  p, for

p  2,  32 , 1,  12 , 12 , 1, 32 , 2. 3

1 2

p= 2

p= 1

2

8

-4

0 -4

4

8

3

1

_2

x 2  4 py vertically  p units so that the vertex is at 0  p. The

focus of x 2  4 py is at 0 p, so this point is also shifted  p units

vertically to the point 0 p  p  0 0. Thus, the focus is located

(c) The parabolas become narrower as the vertex moves toward the origin.

-8 p= _ 2

(b) The graph of x 2  4 p y  p is obtained by shifting the graph of

at the origin.

4 -8

33

p= _1 _2

66. Since 0 0 and 1600 0 are both points on the parabola, the x-coordinate of the vertex is 800. And since the highest point it reaches is 3200, the y-coordinate of the vertex is 3200. Thus the vertex is 800 3200, and the equation is of the form x  8002  4 p y  3200. Substituting the point 0 0, we get 0  8002  4 p 0  3200  640,000  12,800 p

 p  50. So an equation is x  8002  4 50 y  3200  x  8002  200 y  3200. 67. Since the height of the satellite above the earth varies between 140 and 440, the

y

length of the major axis is 2a  140  2 3960  440  8500  a  4250. Since the center of the earth is at one focus, we have a  c  earth radius  140  3960  140  4100 

c  a  4100  4250  4100  150. Thus, the center of the ellipse is 150 0. So b2  a 2  c2  42502  1502  18,062,500  22500  18,040,000. Hence,

an equation is

410

Path of satellite Earth

140

3960

x

Center of ellipse is (_150, 0)

y2 x  1502   1. 18,062,500 18,040,000

68. (a) We assume that 0 1 is the focus closer to the vertex 0 0, as shown in the figure in the text. Then the center of the ellipse is 0 a and 1  a  c. So c  a  1 and a  12  a 2  b2  a 2  2a  1  a 2  b2  b2  2a  1.

x2 y  a2 y  22 x2    1. If we choose  1. If we choose a  2, then we get 2a  1 3 4 a2 x2 y  52   1. (Answers will vary, depending on your choices of a  1.) a  5, then we get 9 25 (b) Since a vertex is at 0 0 and a focus is at 0 1, we must have c  a  1 (a  0), and the center of the hyperbola Thus, the equation is

is 0 a. So c  a  1 and a  12  a 2  b2  a 2  2a  1  a 2  b2  b2  2a  1. Thus the equation x2 x2 y  22 y  a2  1. If we let a  2, then we get   1. If we let a  5, then we get  2 2a  1 4 5 a x2 y  52   1. (Answers will vary, depending on your choices of a.) 25 11

is

(c) Since the vertex is at 0 0 and the focus is at 0 1, we must have

y

p  1. So x  02  4 1 y  0  x 2  4y, and there is no other

possible parabola.

(d) Graphs will vary, depending on the choices of a in parts (a) and (b). (e) The ellipses are inside the parabola and the hyperbolas are outside the parabola. All touch at the origin.

1 1

x


34

CHAPTER 11 Conic Sections

  69. (a) Factored form: y  2x 2  8x  6  2 x 2  4x  3  2 x  3 x  1     Vertex form: y  2 x 2  4x  6  2 x 2  4x  4  6  8  2 x  22  2

   Standard form: y  2 x  22  2  12 y  x  22  1  x  22  12 y  2  x  22  4 18 y  2

(b) To sketch the graph, we use the vertex form or the standard form. The maximum or minimum value is immediately evident from the vertex form. The standard form gives the focus. The factored form gives the real and complex zeroes. 70. Consider the blue circle, centered at P x y with radius r, tangent to the two red circles. Because the blue circle is tangent to the inner red circle, a line from 1 0 to P passes through the point of tangency, and so its length is the sum of the radii of the two circles, that is, 3  r. Similarly, the length of the line segment from 1 0 to P is the radius of the outer red circle minus the radius of the blue circle, that is, 5  r . The sum of these distances—that is, the distance from 1 0 to P to 1 0—is thus 3  r  5  r  8. But by definition, an ellipse is a set of points whose distances from two fixed points (the foci of the ellipse) sum to a constant. To find the equation of this ellipse, noting that it is centered at the origin, we substitute the points 4 0 and 3 0 into the x2 y2 42 32 standard equation 2  2  1, obtaining 2  1  a  4 and  1  b  3. Thus, an equation of the ellipse is a b a b2 y2 x2   1. 16 9

11.5 ROTATION OF AXES 1. If the x- and y-axes are rotated through an acute angle  to produce the new X- and Y -axes, then the x y-coordinates x y and the XY -coordinates X Y  of a point P in the plane are related by the formulas x  X cos   Y sin , y  X sin   Y cos , X  x cos   y sin , and Y  x sin   y cos . 2. (a) In general, the graph of the equation Ax 2  Bx y  C y 2  Dx  E y  F  0 is a conic section.

AC . B (c) The discriminant of this equation is B 2  4AC. If the discriminant is 0 the graph is a parabola, if it is positive the graph is an ellipse, and if it is negative the graph is a hyperbola.  2 and 3. x y  1 1,   45 . Then X  x cos   y sin   1  1  1  1  2 2   Y  x sin   y cos   1  1  1  1  0. Therefore, the XY -coordinates of the given point are X Y   2 0 . (b) To eliminate the x y-term from this equation we rotate the axes through an angle  that satisfies cot 2 

2

2

    4. x y  2 1,   30 . Then X  x cos   y sin   2  23  1  12  12 1  2 3 and     Y  x sin   y cos   2  12  1  23  12 2  3 . Therefore, the XY -coordinates of the given point are        X Y   12 1  2 3  12 2  3 .      3  3 ,   60 . Then X  x cos   y sin   3  12  3  23  0 and    Y  x sin   y cos   3  23  3  12  2 3. Therefore, the XY -coordinates of the given point are    X Y   0 2 3 .

5. x y 

6. x y  2 0,   15 . Then X  x cos   y sin   2 cos 15  0 sin 15  19319 and Y  x sin   y cos   2 sin 15  0 cos 15  05176. Therefore, the XY -coordinates of the given point are approximately X Y   19319 05176.


35

SECTION 11.5 Rotation of Axes

7. x y  0 2,   55 . Then X  x cos   y sin   0 cos 55  2 sin 55  16383 and Y  x sin   y cos   0 sin 55  2 cos 55  11472. Therefore, the XY -coordinates of the given point are approximately X Y   16383 11472.      2 4 2 ,   45 . Then X  x cos   y sin   2  1  4 2  1  5 and 8. x y  2 2   Y  x sin   y cos    2  1  4 2  1  3. Therefore, the XY -coordinates of the given point are 2

2

X Y   5 3.

9. x 2  3y 2  4,   60 . Then x  X cos 60  Y sin 60  12 X  23 Y and y  X sin 60  Y cos 60  23 X  12 Y .  2  2   3 23 X  12 Y  4  Substituting these values into the equation, we get 12 X  23 Y       3Y 2 Y2 X2 9 2 3Y 2 3Y 2 3 3XY X2 3XY 3XY 3XY 3X 2   3   4  X     4 4 2 4 4 2 4 4 4 4 4 2 2   2X 2  2 3XY  4  X 2  3XY  2. 

10. y  x  12 ,   45 . Then x  X cos 45  Y sin 45  22 X  22 Y and y  X sin 45  Y cos 45  22 X  22 Y . Substituting these values into the equation, we get       2    2 X2 2 X  2Y  2 X  2Y  1 2X 2Y  1  2Y  1  2  2 2 2 2 2 2 2 2    Y2  X2  2 22 X 22 Y  1   2Y  1   2 2    Y2      X2  22 X  2X 22 Y  1   2Y  1  22 Y  0  X 2  Y 2  2XY  3 2X  2Y  2  0. 2 2   11. x 2  y 2  2y,   cos1 35 . So cos   35 and sin   45 . Then  2  2   X cos   Y sin 2  X sin   Y cos 2  2 X sin   Y cos   35 X  45 Y  45 X  35 Y  2 45 X  35 Y

24XY 16Y 2 16X 2 24XY 9Y 2 8X 6Y 7X 2 48XY 7Y 2 8X 6Y 9X 2             0 25 25 25 25 25 25 5 5 25 25 25 5 5 7Y 2  48XY  7X 2  40X  30Y  0.   12. x 2  2y 2  16,   sin1 35 . So sin   35 and cos   45 . Then x  45 X  35 Y and y  35 X  45 Y . Substituting these 

values into the equation, we get       4 X  3 Y 2  2 3 X  4 Y 2  16  16 X 2  24 XY  9 Y 2  2 9 X 2  24 XY  16 Y 2  16  5 5 5 5 25 25 25 25 25 25

16 X 2  18 X 2  9 Y 2  32 Y 2  24 XY  48 XY  16  34 X 2  41 Y 2  24 XY  16  34X 2  41Y 2  24XY  400. 25 25 25 25 25 25 25 25 25

    3X  Y 13. x 2  2 3x y  y 2  4,   30 . Then x  X cos 30  Y sin 30  23 X  12 Y  12     and y  X sin 30  Y cos 30  12 X  23 Y  12 X  3Y . Substituting these values into the 2            1   2 1 X  3Y 3X  Y  2 3 12 3X  Y  2 X  3Y  4 equation, we get 12 2 2         2 3X  Y  2 3 3X  Y X  3Y  X  3Y  16           3X 2  2 3XY  Y 2  6X 2  4 3XY  6Y 2  X 2  2 3XY  3Y 2  16  8X 2 8Y 2  16 

X2 Y2   1. This is a hyperbola. 2 2


36

CHAPTER 11 Conic Sections

  1 X  1 Y  1 X  Y  14. x y  x  y,    4 . Therefore, x  X cos 4  Y sin 4  2

2

2

 1 X  1 Y  1 X  Y . Substituting into the equation gives and y  X sin  4  Y cos 4 



1 X  Y  2   2

X

2

2

1 X  Y  2

2

2

2

 1 X  Y   1 X  Y   12 2 2

Y2  1. This is a hyperbola. 2

  X 2  Y 2  2 X  X 2  Y 2  2 2X 

15. (a) x y  8  0x 2  x y  0y 2  8. So A  0, B  1, and C  0, and so the

2

(c)

discriminant is B 2  4AC  12  4 0 0  1. Since the discriminant is

positive, the equation represents a hyperbola. (b) cot 2  

y

1

x

1

AC  0  2  90    45 . Therefore, B 

x  22 X  22 Y and y  22 X  22 Y . After substitution, the original      2 2 equation becomes 22 X  22 Y 2 X  2 Y 8

 X2 Y2 X  Y  X  Y  8   1. This is a hyperbola with a  4, b  4, and c  4 2. Hence, the vertices 2 16 16    are V 4 0 and the foci are F 4 2 0 .

16. (a) x y  4  0  0x 2  x y  0y 2  4  0. So A  0, B  1, and C  0,

y

(c)

and so the discriminant is B 2  4AC  12  4 0 0  1. Since the discriminant is positive, the equation represents a hyperbola.

1

x

1

AC  0  2  90    45 . Therefore, (b) cot 2  B 

x  22 X  22 Y and y  22 X  22 Y . After substitution, the original      2 2 equation becomes 22 X  22 Y 2 X  2 Y 40

  Y2 X2 X  Y  X  Y   4    1. This is a hyperbola with a  2 2, b  2 2, and c  4. Hence, the 2 8 8    vertices are V 0 2 2 and the foci are F 0 4.

  17. (a) x 2  2 3x y  y 2  2  0. So A  1, B  2 3, and C  1, and so the   2 discriminant is B 2  4AC  2 3  4 1 1  0. Since the discriminant is positive, the equation represents a hyperbola.

(c)

y

1 1

11 AC 1      2  60    30 . Therefore, (b) cot 2  B 2 3 3 

x  23 X  12 Y and y  12 X  23 Y . After substitution, the original equation becomes     2  2    3 1 1 X  3Y  1 X  3Y  2 3 23 X  12 Y 20 2 X  2Y 2 2 2 2       3 X 2  3 XY  1 Y 2  3 3X 2  2XY  3Y 2  14 X 2  23 XY  34 Y 2  2  0  4 2 4 2          X 2 34  32  14  XY  23  3  23  Y 2 14  32  34  2  2X 2  2Y 2  2  Y 2  X 2  1.

x


37

SECTION 11.5 Rotation of Axes

  18. (a) 13x 2  6 3x y  7y 2  16. Then A  13, B  6 3, and C  7, and so   2 the discriminant is B 2  4AC  6 3  4 13 7  256. Since the

(c)

y

1

discriminant is negative, the equation represents an ellipse.

1

13  7 AC 1      2  60    30 . Therefore, (b) cot 2  B 6 3 2 

x

x  23 X  12 Y and y  12 X  23 Y . After substitution, the original equation becomes    2  2     1 X  3Y  7 1 X  3Y 13 23 X  12 Y  6 3 23 X  12 Y  16  2 2 2 2           13 3X 2  2 3XY  Y 2  3 3 3X 2  2XY  3Y 2  74 X 2  2 3XY  3Y 2  16  4 2          9  7  XY  13 3  6 3  7 3  Y 2 13  9  21  16  16X 2  4Y 2  16  X 2 39  4 2 4 2 2 2 4 2 4

  Y2  1. This is an ellipse with a  2, b  1, and c  4  1  3. Thus, the vertices are V 0 2 and the 4    foci are F 0  3 .̀ X2 

19. (a) 11x 2  24x y  4y 2  20  0. So A  11, B  24, and C  4, and so the discriminant is B 2  4AC  242  4 11 4  0. Since the

discriminant is positive, the equation represents a hyperbola.

AC 11  4 7 7 . Therefore,     cos 2   25 B 24 24   cos   1725  35 and sin   1725  45 . Hence, 2 2

(b) cot 2 

(c)

y

1 1

x

7 , we have 3X Since cos 2   25  45 Y and y  45 X  35 Y . After substitution, the original 5 2  10626 , so   53 . equation becomes 2       4 X  3 Y  4 4 X  3 Y 2  20  0  11 35 X  45 Y  24 35 X  45 Y 5 5 5 5       11 9X 2  24XY  16Y 2  24 12X 2  7XY  12Y 2  4 16X 2  24XY  9Y 2  20  0  25 25 25

x

X 2 99  288  64  XY 264  168  96  Y 2 176  288  36  500  125X 2  500Y 2  500  1 X 2  Y 2  1. 4


38

CHAPTER 11 Conic Sections

  20. (a) 21x 2  10 3x y  31y 2  144. Then A  21, B  10 3, and C  31,   2 and so the discriminant is B 2  4AC  10 3  4 21 31  2304.

(c)

y

1

Since the discriminant is negative, the equation represents an ellipse.

1

21  31 AC 1  (b) cot 2       2  120    60 . B 10 3 3 

x

Therefore, x  12 X  23 Y and y  23 X  12 Y . After substitution, the original equation becomes  2       2   3 3 1 1 21 12 X  23 Y  10 3 12 X  23 Y  144  2 X  2 Y  31 2 X  2 Y           21 X 2  2 3XY  3Y 2  5 3 2 2  144  3X 2  2XY  3Y 2  31 4 2 4 3X  2 3XY  Y          15  93  XY  21 3  10 3  31 3  Y 2 63  15  31  144  36X 2  16Y 2  144  X 2 21  4 2 4 2 2 2 4 2 4  1 X 2  1 Y 2  1. This is an ellipse with a  3, b  2, and c  9  4  5. 4 9 21. (a)

  2 3x  3x y  3. So A  3, B  3, and C  0, and so the discriminant   3 0  9. Since the discriminant is positive, is B 2  4AC  32  4

(c)

y

1

the equation represents a hyperbola.

x

1

1 AC    2  60    30 . Therefore, (b) cot 2  B 3 

x  23 X  12 Y and y  12 X  23 Y . After substitution, the equation   2     1 X  3Y  3 becomes 3 23 X  12 Y  3 23 X  12 Y 2 2        3 3  4 3X 2  2 3XY  Y 2  4 3X 2  2XY  3Y 2  3              3  3 3  3  3 3 X2  3 Y 2  3  3 X2   6  Y2 1 Y 2  1. This  X 2 3 4 3  3 4 3  XY 6 4 4 4 4 2 2 2 2 3    is a hyperbola with a  2 and b  2 3. 3

22. (a) 153x 2  192x y  97y 2  225. Then A  153, B  192, and C  97, and so the discriminant is B 2  4AC  1922  4 153 97  22500.

Since the discriminant is negative, the equation represents an ellipse.

AC 153  97 56 56 (b) cot 2     cos 2  . Therefore, B 192 192 200   156200 4 3  16  12 cos   156200 2 20  5 and sin   2 20  5 

(c)

y

1

  cos1 54  369 . Substituting gives  2      3 X  4 Y  97 3 X  4 Y 2  225  153 45 X  35 Y  192 45 X  35 Y 5 5 5 5       153 16X 2  24XY  9Y 2  192 12X 2  7XY  12Y 2  97 9X 2  24XY  16Y 2  225  25 25 25

1

X 2 2448  2304  873  XY 3672  1344  2328  Y 2 1377  2304  1552  5625    5625X 2  625Y 2  5625  X 2  19 Y 2  1. This is an ellipse with a  3, b  1, and c  9  1  2 2.

x


39

SECTION 11.5 Rotation of Axes

23. (a) x 2  2x y  y 2  x  y  0. So A  1, B  2, and C  1, and so the

y

(c)

discriminant is B 2  4AC  22  4 1 1  0. Since the discriminant is

zero, the equation represents a parabola.

1 1

AC (b) cot 2   0  2  90    45 . Therefore, B 

x

x  22 X  22 Y and y  22 X  22 Y . After substitution, the original equation becomes  2       2 X  2Y 2 X  2Y 2 X  2Y  2 2 2 2 2 2 2

 2       2 2  22 X  22 Y  22 X  22 Y  0  2 X 2 Y    1 X 2  XY  1 Y 2  X 2  Y 2  1 X 2  XY  Y 2  2Y  0  2X 2  2Y  0  X 2  2 Y . This is a 2 2 2 2



  1 . parabola with 4 p  1 and hence the focus is F 0  2

4 2

In #24(c), the parabola should appear to be symmetric with respect to the X-axis. 24. (a) 25x 2  120x y  144y 2  156x  65y  0. Then A  25, B  120, and

(c)

y

C  144, and so the discriminant is

B 2  4AC  1202  4 25 144  0. Since the discriminant is zero,

the equation represents a parabola.

AC 25  144 119    cos 2  119 169 . Therefore, B 120 120   1119169 5 . Hence, cos   1119169  12  13 2 13 and sin   2

(b) cot 2 

2 2

x

Since cos 2  119 169 , we have

12X 5Y 5X 12Y 2  452 , so   23 . x  and y   . Substituting gives 13 13 13 13          5Y 2 5Y 12Y 12Y 2 5Y 5X 5X 12X 12X 12X     144     120  156 25 13 13 13 13 13 13 13 13 13 13     Y2  12Y X2  5X 65  0 25  122  120  12  5  144  52  25  52  120 5 12  144  122  13 13 169 169 Y X 156  12  65  5  156  5  65  12  0  169Y 2  169X  0  X  Y 2 . This is a parabola with 13 13 4 p  1.


40

CHAPTER 11 Conic Sections

   25. (a) 2 3x 2  6x y  3x  3y  0. So A  2 3, B  6, and C  0, and    so the discriminant is B 2  4AC  62  4 2 3 0  36. Since the

y

(c)

2

discriminant is positive, the equation represents a hyperbola.  AC 2 3 1 (b) cot 2       2  120    60 . Therefore, B 6 3 

2

x

x  12 X  23 Y and y  23 X  12 Y , and substituting gives  2             3 3 3 1 1 1 2 3 12 X  23 Y  6 12 X  23 Y 2 X  2Y  3 2 X  2 Y  3 2 X  2Y  0            3 X 2  23XY  3Y 2  3 3X 2  2XY  3Y 2  3 X  3Y  3 3X  Y  0  2 2 2 2             X 2 23  3 2 3  X 23  3 2 3  XY 3  3  Y 2 3 2 3  3 2 3  Y  32  32  0        3X 2  2 3X  3 3Y 2  0  X 2  2X  3Y 2  0  3Y 2  X 2  2X  1  1    X  12  3Y 2  1. This is a hyperbola with a  1, b  33 , c  1  13  2 , and C 1 0. 3

26. (a) 9x 2  24x y  16y 2  100 x  y  1. Then A  9, B  24, and

C  16, and so the discriminant is B 2  4AC  242  4 9 16  0. Since the discriminant is zero, the equation represents a parabola.

(c)

y 1 1

x

AC 9  16 7 7    369 . Now    cos 2  25 B 24 24   4 and sin   1725  3 , and so cos   1725  2 5 2 5

(b) cot 2 

x  45 X  35 Y , y  35 X  45 Y . By substitution,  2        3 X  4 Y  16 3 X  4 Y 2  100 4 X  3 Y  3 X  4 Y  1  9 45 X  35 Y  24 45 X  35 Y 5 5 5 5 5 5 5 5         9 2 2  24 12X 2  7XY  12Y 2  16 9X 2  24XY  16Y 2  100 1 X  7 Y  1 25 16X  24XY  9Y 25 25 5 5  625Y 2  500X  3500Y  2500  5Y 2  28Y  4X  20          196 196 96 24 14 2  4 X  24 . This is a 5 Y 2  28 5 Y  25  4X  20  5  4X  5  4 X  5  Y  5 5 5   14 . parabola with 4 p  45 and V  24   5 5

27. (a) 52x 2  72x y  73y 2  40x  30y  75. So A  52, B  72, and C  73, and so the discriminant is

B 2  4AC  722  4 52 73  10,000. Since the discriminant is decidedly negative, the equation represents an ellipse.


SECTION 11.5 Rotation of Axes

41

52  73 7 AC    . Therefore, as in Exercise 19(b), we get cos   35 , sin   45 , and B 72 24 x  35 X  45 Y , y  45 X  35 Y . By substitution,  2      4 X  3 Y  73 4 X  3 Y 2 52 35 X  45 Y  72 35 X  45 Y 5 5 5 5      40 35 X  45 Y  30 45 X  35 Y  75        52 9X 2  24XY  16Y 2  72 12X 2  7XY  12Y 2  73 16X 2  24XY  9Y 2 25 25 25

(b) cot 2 

 24X  32Y  24X  18Y  75  468X 2  832Y 2  864X 2  864Y 2  1168X 2  657Y 2  1250Y  1875  2500X 2  625Y 2  1250Y  1875    100X 2  25Y 2  50Y  75  X 2  14 Y  12  1. This is an ellipse with a  2, b  1, c  4  1  3, and center C 0 1.

(c)

y

1 x

1

  7 , we have 2  cos1  7  10626 and so   53 . Since cos 2   25 25

28. (a) 7x  24y2  49x 2  336x y  576y 2  600x  175y  25. So A  49, B  336, and C  576, and so the

discriminant is B 2  4AC  3362  4 49 576  0. Since the discriminant is zero, the equation represents a parabola.  AC 49  576 527 1527625 7 and (b) cot 2  . Therefore, cos    25    cos 2   527 625 2 B 336 336  7 24 24 7  24 sin   1527625 2 25 . Substituting x  25 X  25 Y and y  25 X  25 Y gives        7 X  24 Y 2  336 7 X  24 Y 24 X  7 Y  576 24 X  7 Y 2 49 25 25 25 25 25 25 25 25     7 X  24 Y  175 24 X  7 Y  25   600 25 25 25 25       49 2 2  336 168X 2  527XY  168Y 2  576 576X 2  336XY  49Y 2 625 49X  336XY  576Y 625 625

 168X  576Y  168X  49Y  25  X 2 2401  56,448  331,776  XY 16,464  177,072  193,536  Y 2 28,224  56,448  28,224

(c)

 6252 Y  15,625    1   Y  1 . This is a parabola with 390,625X 2  390,625Y  15,625  25X 2  25Y  1  X 2  Y  25 25   1 4 p  1 and vertex 0 25 . y

1 1

x

  1  527  14748 and so   74 . Since cos 2   527 625 , we have 2  cos 625


42

CHAPTER 11 Conic Sections

29. (a) The discriminant is B 2  4AC  42  4 2 2  0. Since the discriminant is 0, the equation represents a parabola. (b) 2x 2  4x y  2y 2  5x  5  0  2y 2  4x y  2x 2  5x  5    2 y 2  2x y  2x 2  5x  5    2 y 2  2x y  x 2  2x 2  5x  5  2x 2  2 y  x2  5x  5    y  x2  52 x  52  y  x   52 x  52  y  x  52 x  52

5

5

30. (a) The discriminant is B 2  4AC  22  4 1 3  8  0. Since the discriminant is negative, the equation represents an ellipse.   (b) x 2  2x y  3y 2  8  3y 2  2x y  8  x 2  3 y 2  23 x y  8  x 2 

 3 y 2  23 x y  19 x 2  8  x 2  13 x 2  3 y  13 x   83  29 x 2  y  13 x   83  29 x 2   y  13 x  83  29 x 2 

y  13 x

2

2

5

 8  23 x 2 

-5

5 -5

31. (a) The discriminant is B 2  4AC  102  4 6 3  28  0. Since the discriminant is positive, the equation represents a hyperbola. (b) 6x 2  10x y  3y 2  6y  36  3y 2  10x y  6y  36  6x 2    3y 2  2 5x  3 y  36  6x 2  y 2  2 53 x  1 y  12  2x 2     2  2 y 2  2 53 x  1 y  53 x  1  53 x  1  12  2x 2 

10

-10

10

 2 10 2 2 y  53 x  1  25 9 x  3 x  1  12  2x 

-10

 2 y  53 x  1  79 x 2  10 3 x  13     y  53 x  1   79 x 2  10 3 x  13   y   53 x  1  79 x 2  10 3 x  13

(or a degenerate conic)

32. (a) The discriminant is B 2  4AC  62  4 9 1  0. Since the discriminant is 0, the equation represents a parabola. (b) 9x 2  6x y  y 2  6x  2y  0  y 2  6x y  2y  9x 2  6x 

5

y 2  2 3x  1 y  9x 2  6x 

y 2  2 3x  1 y  3x  12  3x  12  9x 2  6x   2 y  3x  1  1  y  3x  1  1  y  3x  1  1. So y  3x or y  3x  2. This is a degenerate conic.

Note that the conic can't be both a parabola and a degenerate.

-2

2 -5


SECTION 11.5 Rotation of Axes

43

33. (a) 7x 2  48x y  7y 2  200x  150y  600  0. Then A  7, B  48, and C  7, and so the discriminant is

B 2  4AC  482  4 7 7  0. Since the discriminant is positive, the equation represents a hyperbola. We 7 AC   cos   45 and sin   35 . now find the equation in terms of XY -coordinates. We have cot 2  B 24 Therefore, x  45 X  35 Y and y  35 X  45 Y , and substitution gives

2           3 X  4 Y  7 3 X  4 Y 2  200 4 X  3 Y  150 3 X  4 Y  600  0 7 45 X  35 Y  48 45 X  35 Y 5 5 5 5 5 5 5 5       7 16X 2  24XY  9Y 2  48 12X 2  7XY  12Y 2  7 9X 2  24XY  16Y 2  25 25 25

 160X  120Y  90X  120Y  600  0  112X 2  168XY  63Y 2  576X 2  336XY  576Y 2  63X 2  168XY  112Y 2  6250X  15,000  0    25X 2  25Y 2  250X  600  0  25 X 2  10X  25  25Y 2  600  625  X  52  Y 2  1. This is a   hyperbola with a  1, b  1, c  1  1  2, and center C 5 0.

(b) In the XY -plane, the center is C 5 0, the vertices are V 5  1 0  V1 4 0 and V2 6 0, and the foci      are F 5  2 0 . In the x y-plane, the center is C 45  5  35  0 35  5  45  0  C 4 3, the vertices are         12 4 3 3 4 24 18 V1 45  4  35  0 35  4  45  0  V1 16 5  5 and V2 5  6  5  0 5  6  5  0  V2 5  5 , and the foci are         F1 4  45 2 3  35 2 and F2 4  45 2 3  35 2 . (c) In the XY -plane, the equations of the asymptotes are Y  X  5 and Y  X  5. In the x y-plane, these equations

become x  35  y  45  x  45  y  35  5  7x  y  25  0. Similarly, x  35  y  45  x  45  y  35  5  x  7y  25  0.

       34. (a) 2 2 x  y2  7x  9y  2 2x 2  4 2x y  2 2y 2  7x  9y. Therefore, A  2 2, B  4 2, and C  2 2,  X Y X Y AC  x  y  2X. Thus the  0  2  90    45 . Thus x   , y   and so cot 2  B 2 2     2 16X  2Y  8X 2  16X  2Y  4X 2  8X  Y  4 X 2  2X  1  Y  4 equation becomes 2 2 2X   2  X  12  14 Y  4. This is a parabola.

  1 . Thus, in XY -coordinates the vertex is V 1 4 and the focus is F 1  63 . In x y-coordinates, (b) 4 p  14  p  16 16         5 2 3 2 79 2 47 2 V and F 32   32 . 2  2 (c) The directrix is Y   65 16 . Thus

 65 x  y   or y  x  6516 2 .  16 2

35. We use the hint and eliminate Y by adding: x  X cos   Y sin   x cos   X cos2   Y sin  cos  and

y  X sin   Y cos   y sin   X sin2   Y sin  cos , and adding these two equations gives x cos   y   sin   X cos2   sin2   x cos   y sin   X. In a similar manner, we eliminate X by subtracting:

x  X cos   Y sin   x sin   X cos  sin   Y sin2  and y  X sin   Y cos     y cos   X sin  cos   Y cos2 , so x sin   y cos   Y cos2   sin2   x sin   y cos   Y . Thus, X  x cos   y sin  and Y  x sin   y cos .


44

36.

CHAPTER 11 Conic Sections

x

  y  1. Squaring both sides gives x  2 x y  y  1  2 x y  1  x  y, and squaring both sides again

gives 4x y  1  x  y2  1  x  y  x  x 2  x y  y  x y  y 2  4x y  x 2  y 2  2x y  2x  2y  1  AC x 2  y 2  2x y  2x  2y  1  0. Then A  1, B  2, and C  1, and so cot 2   0  2  90  B 

  45 . Therefore, x  22 X  22 Y and y  22 X  22 Y , and substituting gives  2       2 X  2Y 2 X  2Y 2 X  2Y  2 2 2 2 2 2 2  2         22 X  22 Y  2 22 X  22 Y  2 22 X  22 Y  1  0           1 X 2  2XY  Y 2  X 2  Y 2  1 X 2  2XY  Y 2  2 2X  1  0  X 2 1  1  1  XY 1  1  2 2 2 2         1 . This is a parabola with Y 2 12  1  12  2 2X  1  0  2Y 2  2 2X  1  Y 2  2X  12  2 X   2 2    1 4 p  2 and vertex V   0 . However, in the original equation we must have x  0 and y  0, so we get only the 2 2 part of the parabola that lies in the first quadrant. 

37. (a) Z  

x y

X cos   sin  , Z    , and R   . Y

cos      x cos   sin  X X cos   Y sin Y      . Equating the entries in this Thus Z  R Z      y sin  cos  Y X sin   Y cos  

sin 

matrix equation gives the first pair of rotation of axes formulas. Now     cos  sin  cos  sin  1 1      and so Z   R 1 Z  R  cos2   sin2   sin  cos   sin  cos         X cos  sin  x x cos   y sin        . Equating the entries in this matrix equation gives Y  sin  cos  y x sin   y cos 

the second pair of rotation of axes formulas.    cos 1  sin 1 cos 2  sin 2   (b) R1 R2   sin 1 cos 1 sin 2 cos 2        cos 1 cos 2  sin 1 sin 2  cos 1 sin 2  sin 1 cos 2 cos 1  2  sin 1  2       sin 1  2 cos 1  2 sin 1 cos 2  cos 1 sin 2  sin 1 sin 2  cos 1 cos 2   38. (a) Using A  A cos2   B sin  cos   C sin2 , B   2 C  a sin  cos   B cos2   sin2  , and C   A sin2   B sin  cos   C cos2 , we first expand the terms.   2   2 B  2 C  a sin  cos   B cos2   sin2 

   2  4 C  A2 sin2  cos2   4B C  A sin  cos  cos2   sin2   B 2 cos2   sin2   4 C  A2 sin2  cos2   4B C  A sin  cos3   4B C  A sin3  cos 

 B 2 cos4   2B 2 sin2  cos2   B 2 sin4 


SECTION 11.5 Rotation of Axes

  4A C   4 A cos2   B sin  cos   C sin2  A sin2   B sin  cos   C cos2 

 4A2 sin2  cos2   4AB sin  cos3   4AC cos4   4AB sin3  cos   4B 2 sin2  cos2 

 4BC sin  cos3   4AC sin4   4BC sin3  cos   4C 2 sin2  cos2 

Since there are many terms in this expansion, we find the coefficients of like trignometric terms.   2 B Term 4A C  Sum cos4 

B2

sin  cos3 

4B C  A

sin2  cos2  sin3  cos  sin4  So 2

B   4A C  

4 C  A2  2B 2 4B C  A B2

4AC

4AB  4BC

B 2  4AC

4A2  4B 2  4C 2

4BC  4AB  4AB  4BC  0   8AC  2B 2  2 B 2  4AC

4AC

B 2  4AC

4AB  4BC

4BC  AB  4AB  4BC  0

     B 2  4AC cos4   2 B 2  4AC sin2  cos2   B 2  4AC sin4 

    2 cos4   2 sin2  cos2   sin4   B 2  4AC cos2   sin2      B 2  4AC 12  B 2  4AC 

B 2  4AC

So B 2  4AC  B 2  4A C  .

(b) Using A  A cos2   B sin  cos   C sin2  and C   A sin2   B sin  cos   C cos2 , we have A  C   A cos2   B sin  cos   C sin2   A sin2   B sin  cos   C cos2       A sin2   cos2   C sin2   cos2   AC

(c) Since F   F, F is also invariant under rotation. 39. Let P be the point x1  y1  and Q be the point x2  y2  and let P  X 1  Y1  and Q  X 2  Y2  be the images of P and Q under the rotation of . So X 1  x1 cos   y1 sin  Y1  x1 sin   y1 cos , X 2  x2 cos   y2 sin and    Y2  x2 sin   y2 cos . Thus d P   Q   X 2  X 1 2  Y2  Y1 2 , where X 2  X 1 2 

 2  2 x2 cos   y2 sin   x1 cos   y1 sin   x2  x1  cos   y2  y1  sin 

 x2  x1 2 cos2   x2  x1  y2  y1  sin  cos   y2  y1 2 sin2 

and Y2  Y1 2 

 2  2 x2 sin   y2 cos   x1 sin   y1 cos    x2  x1  sin   y2  y1  cos 

 x2  x1 2 sin2   x2  x1  y2  y1  sin  cos   y2  y1 2 cos2 

So X 2  X 1 2  Y2  Y1 2  x2  x1 2 cos2   x2  x1  y2  y1  sin  cos   y2  y1 2 sin2 

 x2  x1 2 sin2   x2  x1  y2  y1  sin  cos   y2  y1 2 cos2 

 x2  x1 2 cos2   y2  y1 2 sin2   x2  x1 2 sin2   y2  y1 2 cos2       x2  x1 2 cos2   sin2   y2  y1 2 sin2   cos2   x2  x1 2  y2  y1 2

45


46

CHAPTER 11 Conic Sections

    Putting these equations together gives d P   Q   X 2  X 1 2  Y2  Y1 2  x2  x1 2  y2  y1 2  d P Q.

11.6 POLAR EQUATIONS OF CONICS 1. All conics can be described geometrically using a fixed point F called the focus and a fixed line  called the directrix. For a distance from P to F fixed positive number e the set of all points P satisfying  e is a conic section. If e  1 the conic is a distance from P to  parabola, if e  1 the conic is an ellipse, and if e  1 the conic is a hyperbola. The number e is called the eccentricity of the conic. ed ed 2. The polar equation of a conic with eccentricity e has one of the following forms: r  or r  . 1  e cos  1  e sin  3. Substituting e  23 and d  3 into the general equation of a conic with vertical directrix, we get r  r

6 . 3  2 cos 

4. Substituting e  43 and d  3 into the general equation of a conic with vertical directrix, we get r  r

12 . 3  4 cos 

2 3 3  1  23 cos 

4 3 3  1  43 cos 

5. Substituting e  1 and d  2 into the general equation of a conic with horizontal directrix, we get r  r

2 . 1  sin 

6. Substituting e  12 and d  4 into the general equation of a conic with horizontal directrix, we get r  r

4 . 2  sin 

12  1  sin  1 4 2  1  12 sin 

20 45 r  . 1  4 cos  1  4 cos  12 06  2 r  . 8. r  2 csc   r sin   2  y  2. So d  2 and e  06 gives r  1  06 sin  1  06 sin  9. Since this is a parabola whose focus is at the origin and vertex at 5 2, the directrix must be y  10. So d  10 and 10 1  10  . e  1 gives r  1  sin  1  sin  d P F 2 2 10. Since the vertex is at 2 0 we have d P F  2. Now, since  e we get  04  d P    5. d P  d P  04 04  7 28 The directrix is x  7, so substituting e  04 and d  7 we get r  r  . 1  04 cos  1  04 cos  6 is Graph II. The eccentricity is 1, so this is a parabola. When   0, we have r  3 and when    11. r  2 , we 1  cos  have r  6. 2 1 12. r  is Graph III. r  , so e  12 and this is an ellipse. When   0, r  2, and when   , r  23 . 2  cos  1  1 cos  7. r  5 sec   r cos   5  x  5. So d  5 and e  4 gives r 

2

3 13. r  is Graph VI. e  2, so this is a hyperbola. When   0, r  3, and when   , r  3. 1  2 sin  5

14. r 

5 3 is Graph I. r  , so e  1 and this is a parabola. When   0, r  53 and when   , r  53 . 3  3 sin  1  sin 


SECTION 11.6 Polar Equations of Conics

47

15. r 

4 12 , so e  23 and this is an ellipse. When   0, r  4, and when   , r  4. is Graph IV. r  2 3  2 sin  1  sin 

16. r 

6 12 is Graph V. r  , so e  32 and this is a hyperbola. When   0, r  12 5 , and when   , 2  3 cos  1  3 cos 

3

2

we have r  12.

4 has e  1 and d  4, so 1  sin  it represents a parabola.

17. (a) The equation r 

3

18. (a) The equation r 

3 2  has e  1 2  2 sin  1  sin 

and d  32 , so it represents a parabola. V(3/4, ¹/2) _4

_8

_4 y=_4

O

4 V(2, 3¹/2)

ed , 1  e sin  the directrix is parallel to the polar axis and has   equation y  4. The vertex is 2 32 .

e  1 and d  53 , so it represents a parabola. x=5/3

O

O

4

ed , 1  e sin  the directrix is parallel to the polar axis and has   equation y  d  32 . The vertex is 34   2 .

(b) Because the equation is of the form r 

5

5 3  has 3  3 cos  1  cos 

1

y=3/2 2

8

(b) Because the equation is of the form r 

19. (a) The equation r 

_2

2

V(5/6, 0)

3

ed , 1  e cos  the directrix is parallel to the polar axis and has   equation x  d  53 . The vertex is 56  0 .

(b) Because the equation is of the form r 

2

20. (a) The equation r 

2 5  has 5  5 cos  1  cos 

e  1 and d  25 , so it represents a parabola. x=_2/5

1 V(1/5, ¹) O

1

ed , 1  e cos  the directrix is parallel to the polar axis and has   equation x  d   25 . The vertex is 15   .

(b) Because the equation is of the form r 


48

CHAPTER 11 Conic Sections

21. (a) The equation r  ellipse.

2 4 has e  12  1, so it represents an  1 2  cos  1  cos  2

ed (b) Because the equation is of the form r  with d  4, the directrix is 1  e cos    vertical and has equation x  4. Thus, the vertices are V1 4 0 and V2 43   .

x=_4 1 Vª(4/3, ¹)

1

O

VÁ(4, 0)

(c) The length of the major axis is 2a  V1 V2   4  43  16 3 and the center is at the   midpoint of V1 V2 , 43  0 . The minor axis has length 2b where 2  2  b2  a 2  c2  a 2  ae2  83  83  12  16 3 , so   8 3 2b  2  16 3  3  462.

6 2 has e  23  1, so it represents an  2 3  2 sin  1  sin 

VÁ(6, ¹/2)

ed with d  3, the directrix is 1  e sin    horizontal and has equation y  3. Thus, the vertices are V1 6  2 and   V2 65  32 .

O

22. (a) The equation r  ellipse.

3

(b) Because the equation is of the form r 

1 Vª(6/5, 3¹/2)

1 y=_3

(c) The length of the major axis is 2a  V1 V2   6  65  36 5 and the center is at the    midpoint of V1 V2 , 12 5  2 . The minor axis has length 2b where   2  18  2 2  36 , so b2  a 2  c2  a 2  ae2  18  5 5 3 5   12 5 2b  2  36 5  5  537.

23. (a) The equation r  ellipse.

3 12 has e  34  1, so it represents an  4  3 sin  1  3 sin  4

ed (b) Because the equation is of the form r  with d  4, the directrix is 1  e sin     horizontal and has equation y  4. Thus, the vertices are V1 12 7  2 and   V2 12 32 .

96 (c) The length of the major axis is 2a  V1 V2   12 7  12  7 and the center is at   3 the midpoint of V1 V2 , 36 7  2 . The minor axis has length 2b where

  2  48  3 2  144 , so b2  a 2  c2  a 2  ae2  48  7 7 4 7   24 7 2b  2  144 7  7  907.

VÁ(12/7, ¹/2) O 1 4 8 Vª(12, 3¹/2)

y=4


SECTION 11.6 Polar Equations of Conics

49

9

24. (a) The equation r  ellipse.

18 2 has e  34  1, so it represents an  4  3 cos  1  3 cos 

x=6

4

VÁ(18/7, 0)

ed (b) Because the equation is of the form r  with d  6, the directrix is 1  e cos     0 and V2 18 . vertical and has equation x  6. Thus, the vertices are V1 18 7

Vª(18, ¹)

2 O

5

144 (c) The length of the major axis is 2a  V1 V2   18 7  18  7 and the center is at   the midpoint of V1 V2 , 54 7   . The minor axis has length 2b where

  2  72  3 2  324 , so b2  a 2  c2  a 2  ae2  72  7 7 4 7   36 7 2b  2  324 7  7  1361.

In #25, indicate vertical axis scale 8 has e  2  1, so it represents a hyperbola. 1  2 cos  ed with d  4, the transverse axis is (b) Because the equation has the form r  1  cos    horizontal and the directrix has equation x  4. The vertices are V1 83  0 and

x=4

25. (a) The equation r 

VÁ(8/3, 0)

Vª(_8, ¹)

O

V2 8   8 0.

  (c) The center is the midpoint of V1 V2 , 16 3  0 . To sketch the central box and the

asymptotes, we find a and b. The length of the transverse axis is 2a  16 3 , and so  2  2 a  83 , and b2  c2  a 2  ae2  a 2  83  2  83  64 3 , so   8 3 b  64 3  3  462.

10 has e  4  1, so it represents a hyperbola. 1  4 sin  ed with d  52 , the transverse axis (b) Because the equation has the form r  1  cos 

In #26, indicate scale in both axes; I believe that ticks are every 2 units for the x-axis, but every 1 unit for the y-axis

26. (a) The equation r 

VÁ(2, 3¹/2)

is vertical and the directrix has equation y   52 . The vertices are

       10 3 3 V1  10 3  2  3  2 and V2 2 2 .   (c) The center is the midpoint of V1 V2 , 83  32 . To sketch the central box and the

asymptotes, we find a and b. The length of the transverse axis is 2a  43 , and so  2  2 a  23 , and b2  c2  a 2  ae2  a 2  23  4  23  20 3 , so  b  20 3  258.

y=_5/2 Vª(_10/3, ¹/2)


50

CHAPTER 11 Conic Sections

27. (a) The equation r  hyperbola.

10 20 has e  32  1, so it represents a  2  3 sin  1  3 sin 

VÁ(4, 3¹/2)

2

(b) Because the equation has the form r 

O y=_20/3

ed with d  20 3 , the transverse axis 1  cos 

is vertical and the directrix has equation y   20 3 . The vertices are       3 3 V1 20  2  20 2 and V2 4 2 .   (c) The center is the midpoint of V1 V2 , 12 32 . To sketch the central box and the

Vª(_20, ¹/2)

In #27, indicate scale in both axes.

asymptotes, we find a and b. The length of the transverse axis is 2a  16, and so 2  a  8, and b2  c2  a 2  ae2  a 2  8  32  82  80, so   b  80  4 5  894.

28. (a) The equation r  hyperbola.

3 6  has e  72  1, so it represents a 7 2  7 cos  1  cos 

x=6/7

2

1

ed with d  67 , the transverse axis (b) Because the equation has the form r  1  cos    is horizontal and the directrix has equation x  67 . The vertices are V1 23  0 and     V2  65    65  0 .   (c) The center is the midpoint of V1 V2 , 14 15  0 . To sketch the central box and the

O VÁ(2/3, 0)

Vª(_6/5, ¹)

8 , and so asymptotes, we find a and b. The length of the transverse axis is 2a  15     4 , and b2  c2  a 2  ae2  a 2  4  7 2  4 2  4 , so a  15 15 2 15 5   b  45  2 5 5  089.

29. (a) r 

4  e  3, so the conic is a hyperbola. 1  3 cos 

(b) The vertices occur where   0 and   . Now   0  r 

4  1, 1  3 cos 0

4 4 and     r    2. Thus the vertices are 1 0 and 1  3 cos  2 2 .

In #29 indicate scale in both axes.

(_2, ¹)

(1, 0)

8

8 3 r   e  1, so the conic is a parabola. 3  3 cos  1  cos    8 8 4 (b) Substituting   0, we have r    . Thus the vertex is 43  0 . 3  3 cos 0 6 3

30. (a) r 

1

( 43 , 0)


SECTION 11.6 Polar Equations of Conics

2  e  1, so the conic is a parabola. 1  cos  2 (b) Substituting   , we have r   22  1. Thus the vertex is 1 . 1  cos 

31. (a) r 

(1, ¹)

1

10

32. (a) r 

10 3 r   e  23 , so the conic is an ellipse. 3  2 sin  1  2 sin 

(10, ¹2 )

3

3  (b) The vertices occur where    2 and   2 . Now   2 

10 10 10 10  10 and   32  r   2. Thus,   3  3  2 sin  1 5 3  2 sin 2 2     3 . the vertices are 10  and 2 2 2

r

33. (a) r 

1

(2, 3¹ 2 )

1 6 6 2 r   e  12 , so the conic is an ellipse. 2  sin  1  1 sin 

(2, ¹2)

2

3  (b) The vertices occur where    2 and   2 . Now   2 

1

6 6 6 6   2 and   32  r    6. Thus, the  3  2  sin 2 3 1 2  sin 2     vertices are 2 2 and 6 32 .

r

(6, 3¹ 2 )

5

34. (a) r 

5 2  e  32 , so the conic is a hyperbola. r  2  3 sin  1  3 sin 

1

2

3  (b) The vertices occur where    2 and   2 . Now   2 

(

5 5 5  5 and   32  r    55  1. Thus, 3 2  3 sin  1 2  3 sin 2 2      3  the vertices are 5 2 and 1 2 .

r

¹ _5, 2

)

(1, 3¹ 2)

7

35. (a) r 

7 2 r   e  52 , so the conic is a hyperbola. 2  5 sin  1  5 sin  2

7 7   7 and  3 3 2  5 sin  2    77  1. Thus, the vertices are  73   2 and 3

3 (b) The vertices occur where    2 and   2 . r 

  32  r    1 32 . 36. (a) r 

7

(_ 73 , ¹2 )

1

(1, 3¹ 2)

2  5 sin 2

8

8 3 r   e  13 , so the conic is an ellipse. 3  cos  1  1 cos  3

(b) The vertices occur where   0 and   . Now   0  8 8 r  84  2 and     r   82  4. Thus, the vertices 3  cos 0 3  cos  are 2 0 and 4 .

(4, ¹)

(2, 0) 1

51


52

CHAPTER 11 Conic Sections 1

37. (a) r 

1 4  e  34 , so the  4  3 cos  1  3 cos  4

conic is an ellipse. The vertices occur where   0 and   . Now   0  r 

1  1 and 4  3 cos 0

(b) If the ellipse is rotated through  3 , the equation of the resulting conic is r  1.0

1  17 . Thus, the vertices  r  4  3 cos    are 1 0 and 17   . We have d  13 , so the

1 .  4  3 cos    3

0.5

directrix is x   13 .

0.5

x=-1/3

O

1

2

38. (a) r 

2 5   e  35 , so the 5  3 sin  1  3 sin  5

conic is an ellipse. The vertices occur where    2 and   32 . Now    2 r  and   32  r 

2

(b) If the ellipse is rotated through   23 , the equation of the resulting conic is r 

2 1 5  3 sin  2

2 .  5  3 sin   23 0.5

 14 . Thus, the 5  3 sin 32

    1 3 2 vertices are 1  2 and 4  2 . We have d  3 , so the directrix is y   23 .

1

O y=_2/3

-1.0

-0.5 -0.5


SECTION 11.6 Polar Equations of Conics

2  e  1, so the conic is a parabola. 1  sin  2 Substituting    2 , we have r  t 1  sin   1, 2   so the vertex is 1  . Because d  2, the directrix 2

39. (a) r 

In #39 indicate scale in both axes.The ticks are every 2 units for both axes.

is y  2.

(b) If the ellipse is rotated through     4 , the equation of the resulting conic is

r

2 .  1  sin    4 2

y=2

-8

O

-6

-4

-2

2 -2 -4 -6 -8

9 9 2   e  1, so the conic 2  2 cos  1  cos  is a parabola. Substituting   0, we have   9 9  , so the vertex is 94  0 . r 2  2 cos 0 4

40. (a) r 

Because d  92 , the directrix is x  92 .

(b) If the ellipse is rotated through    56 , the equation of the resulting conic is

r

9 .  2  2 cos   56

x=9/2

O

10

5 1

5

In #40 indicate scale in y-axis .

-5

41. The ellipse is nearly circular when e is close to 0 and becomes more elongated as

4

e  1 . At e  1, the curve becomes a parabola.

e=0.4 e=0.6

2

2 -2

4 e=0.8

e=1

-4

d where d  12 , d  2, and 1  sin  d  10. As d increases, the parabolas get flatter while the vertex moves further

42. (a) Shown are the graphs of the conics r 

10 d=10

from the focus at the origin.

-10

10 d=2 -10

1

d=2

53


54

CHAPTER 11 Conic Sections

e where e  05, e  1, and 1  e sin  e  10. As e increases, the conic changes from an ellipse to a parabola, and

(b) Shown are the graphs of the conics r 

finally to a hyperbola. The vertex gets closer to the directrix (shown as a

2 e=10 -2

2 e=0.5 e=1 -2

dashed line in the figure).

  ed we need to show that ed  a 1  e2 . 1  e cos  e2 d 2 From the proof of the Equivalent Description of Conics we have a 2   2 . Since the conic is an ellipse, e  1 1  e2   and so the quantities a, d, and 1  e2 are all positive. Thus we can take the square roots of both sides and maintain     2d 2 a 1  e2 ed e ed equality. Thus a 2   r  .  ed  a 1  e2 . As a result, r  2  a  1  e cos  1  e cos  1  e2 1  e2

43. (a) Since the polar form of an ellipse with directrix x  d is r 

(b) Since 2a  299  108 we have a  1495  108 , so a polar equation for the earth’s orbit (using e  0017) is   1495  108 1  00172 149  108  . r 1  0017 cos  1  0017 cos 

44. (a) Using the form of the equation from Exercise 27, the perihelion distance occurs when    and the aphelion distance occurs when   0. Since the focus is at the origin, the perihelion distance is     2 a 1  e a 1  e2 a 1  e 1  e a 1  e 1  e   a 1  e and the aphelion distance is   a 1  e. 1e 1e 1e 1e (b) Given e  0017 and 2a  299  108 we have a  1495  108 . Thus the perihelion distance is

1495  108 [1  0017]  1468  108 km and the aphelion distance is 1495  108 1  0017  1520  108 km.

45. From Exercise 44, we know that at perihelion r  443  109  a 1  e and at aphelion r  737  109  a 1  e. 737  109 1e a 1  e  1664   1664 1  e  1  e   9 a 1  e 1e 443  10 0664 1664  1  e  1664  0664  2664e  e   025. 2664 Dividing these equations gives

46. Since the focus is at the origin, the distance from the focus to any point on the conic is the absolute value of the r-coordinate of that point, which we can obtain from the polar equation. 47. The r-coordinate of the satellite will be its distance from the focus (the center of the earth). From the r-coordinate we can easily calculate the height of the satellite.

CHAPTER 11 REVIEW 1. (a) y 2  4x. This is a parabola with 4 p  4  p  1. The vertex is 0 0, the

(b)

y

focus is 1 0, and the directrix is x  1.

1 1

x


CHAPTER 11 1 y 2  y 2  12x. This is a parabola with 4 p  12  p  3. The vertex 2. (a) x  12

Review

55

1

x

y

(b)

is 0 0, the focus is 3 0, and the directrix is x  3.

2

3. (a) 18 x 2  y  x 2  8y. This is a parabola with 4 p  8  p  2. The vertex is

y

(b)

0 0, the focus is 0 2, and the directrix is y  2.

1

4. (a) x 2  8y. This is a parabola with 4 p  8  p  2. The vertex is 0 0,

1

x

1

x

1

y

(b)

vertex is 0 0, the focus is 0 2, and the directrix is y  2.

6. (a) 2x  y 2  0  y 2  2x. This is a parabola with 4 p  2  p  12 . The vertex   is 0 0, the focus is 12  0 , and the directrix is x   12 .

x

y

(b)

the focus is 0 2, and the directrix is y  2.

5. (a) x 2  8y  0  x 2  8y. This is a parabola with 4 p  8  p  2. The

1

1

(b)

y

1 1

x


56

CHAPTER 11 Conic Sections

7. (a) y  22  4 x  2. This is a parabola with 4 p  4  p  1. The vertex is

y

(b)

2 2, the focus is 1 2, and the directrix is x  3.

1 x

1

8. (a) x  32  20 y  2. This is a parabola with 4 p  20  p  5. The

y

(b)

vertex is 3 2, the focus is 3 7, and the directrix is y  3.

1 x

5

9. (a) 12 y  32  x  0  y  32  2x. This is a parabola with 4 p  2    p   12 . The vertex is 0 3, the focus is  12  3 , and the directrix is x  12 .

y

(b)

1

x

1

10. (a) 2 x  12  y  x  12  12 y. This is a parabola with 4 p  12  p  18 .   The vertex is 1 0, the focus is 1 18 , and the directrix is y   18 .

y

(b)

1 1 x

11. (a) 12 x 2  2x  2y  4  x 2  4x  4y  8  x 2  4x  4  4y  8 

y

(b)

x  22  4 y  3. This is a parabola with 4 p  4  p  1. The vertex is 2 3, the focus is 2 3  1  2 2, and the directrix is y  3  1  4.

12. (a) x 2  3 x  y  x 2  3x  3y  x 2  3x  94  3y  94   2   x  32  3 y  34 . This is a parabola with 4 p  3  p  34 . The vertex       is 32   34 , the focus is 32   34  34  32  0 , and the directrix is y   34  34   32 .

1 1

(b)

x

y

2

1

x


CHAPTER 11

13. (a)

 y2 x2   1. This is an ellipse with a  5, b  3, and c  25  9  4. The 9 25 center is 0 0, the vertices are 0 5, and the foci are 0 4.

y

(c)

1

(b) The length of the major axis is 2a  10 and the length of the minor axis is

  y2 x2   1. This is an ellipse with a  7, b  3, and c  49  9  2 10. 49 9    The center is 0 0, the vertices are 7 0, and the foci are 2 10 0 .

x

1

2b  6.

14. (a)

57

Review

y

(c)

1

(b) The length of the major axis is 2a  14 and the length of the minor axis is

x

1

2b  6.

15. (a)

  y2 x2   1. This is an ellipse with a  7, b  2, and c  49  4  3 5. 49 4    The center is 0 0, the vertices are 7 0, and the foci are 3 5 0 .

(c)

y

1

(b) The length of the major axis is 2a  14 and the length of the minor axis is

x

1

2b  4.

16. (a)

  y2 x2   1. This is an ellipse with a  6, b  2, and c  36  4  4 2. 4 36    The center is 0 0, the vertices are 0 6, and the foci are 0 4 2 .

(c)

y

1

(b) The length of the major axis is 2a  12 and the length of the minor axis is

x

1

2b  4.

y2 x2   1. This is an ellipse with a  4, b  2, and 17. (a) x 2  4y 2  16  16 4   c  16  4  2 3. The center is 0 0, the vertices are 4 0, and the foci    are 2 3 0 .

(c)

y

1 x

1

(b) The length of the major axis is 2a  8 and the length of the minor axis is 2b  4.

y2 x2   1. This is an ellipse with a  12 , b  13 , and 19 14     c  14  19  16 5. The center is 0 0, the vertices are 0  12 , and the    foci are 0  16 5 .

18. (a) 9x 2  4y 2  1 

(b) The length of the major axis is 2a  1 and the length of the minor axis is 2b  23 .

(c)

y 0.5

0.5 x


58

CHAPTER 11 Conic Sections

19. (a)

y2 x  32   1This is an ellipse with a  4, b  3, and 9 16   c  16  9  7. The center is 3 0, the vertices are 3 4, and the foci    are 3  7 .

(c)

y

1 1

x

1

x

(b) The length of the major axis is 2a  8 and the length of the minor axis is 2b  6.

20. (a)

y  32 x  22   1. This is an ellipse with a  5, b  4, and 25 16  c  25  16  3. The center is 2 3, the vertices are

(c)

y 1

2  5 3  3 3 and 7 3, and the foci are 2  3 3  1 3 and 5 3.

(b) The length of the major axis is 2a  10 and the length of the minor axis is 2b  8.

21. (a)

y  32 x  22   1. This is an ellipse with a  6, b  3, and 9 36   c  36  9  3 3. The center is 2 3, the vertices are    2 3  6  2 9 and 2 3, and the foci are 2 3  3 3 .

(c)

y 1 x

1

(b) The length of the major axis is 2a  12 and the length of the minor axis is 2b  6.

22. (a)

 x2 y  52   1. This is an ellipse with a  5, b  3, and 3 25   c  25  3  22. The center is 0 5, the vertices are    0 5  5  0 10 and 0 0, and the foci are 0 5  22 .

(c)

y 1

1

x

(b) The length of the major axis is 2a  10 and the length of the minor axis is  2b  2 3.   23. (a) 4x 2  9y 2  36y  4x 2  9 y 2  4y  4  36  4x 2  9 y  22  36 x2 y  22   1. This is an ellipse with a  3, b  2, and 9 4   c  9  4  5. The center is 0 2, the vertices are 3 2, and the foci    are  5 2 .

(c)

y

(b) The length of the major axis is 2a  6 and the length of the minor axis is 2b  4.

1 1

x


CHAPTER 11

24. (a) 2x 2  y 2  2  4 x  y  2x 2  4x  y 2  4y  2      2 x 2  2x  1  y 2  4y  4  2  2  4  2 x  12  y  22  8 x  12

(c)

y

1 x

1

y  22

59

Review

    1. This is an ellipse with a  2 2, b  2, and 4 8     c  8  4  2. The center is 1 2, the vertices are 1 2  2 2 , and

the foci are 1 2  2  1 0 and 1 4.  (b) The length of the major axis is 2a  4 2 and the length of the minor axis is 2b  4.

y2 y2 x2 x2  1   0. This is a hyperbola with a  4, b  3, and 9 16 16 9   c  16  9  25  5. The center is 0 0, the vertices are 0 4, the foci

25. (a) 

(b)

1

are 0 5, and the asymptotes are y   43 x.

26. (a)

 y2 x2   1. This is a hyperbola with a  7, b  4 2, and 49 32  c  49  32  9. The center is 0 0, the vertices are 7 0, the foci are

(b)

28. (a)

x2 y2   1. This is a hyperbola with a  2, b  7, and 4 49   c  4  49  53. The center is 0 0, the vertices are 2 0, the foci are     53 0 , and the asymptotes are y   72 x.

x2 y2   1. This is a hyperbola with a  5, b  2, and 25 4   c  25  4  29. The center is 0 0, the vertices are 0 5, the foci are    0  29 , and the asymptotes are y   52 x.

1

x

2

x

1

x

1

x

y

2

9 0, and the asymptotes are y   4 7 2 x.

27. (a)

y

(b)

y

2

(b)

y

2


60

CHAPTER 11 Conic Sections

 y2 x2   1. This is a hyperbola with a  4, b  2 2, 16 8    and c  16  8  24  2 6. The center is 0 0, the vertices are 4 0,     the foci are 2 6 0 , and the asymptotes are y   2 4 2 x  y   1 x.

29. (a) x 2  2y 2  16 

y

(b)

1

2

x2 y2   1. This is a hyperbola 30. (a) x 2  4y 2  16  0  4y 2  x 2  16  4 16   with a  2, b  4, and c  4  16  2 5. The center is 0 0, the vertices    are 0 2, the foci are 0 2 5 , and the asymptotes are y   24 x 

1

x

1

x

y

(b)

1

y   12 x.

31. (a)

y2 x  42   1. This is a hyperbola with a  4, b  4 and 16 16   c  16  16  4 2. The center is 4 0, the vertices are 4  4 0    which are 8 0 and 0 0, the foci are 4  4 2 0 , and the asymptotes

y

(b)

1

1 x

are y   x  4.

32. (a)

  y  22 x  22   1. This is a hyperbola with a  2 2, b  2 2 and 8 8     c  8  8  4. The center is 2 2, the vertices are 2  2 2 2 , the

(b)

y 1

x

1

foci are 2  4 2  2 2 and 6 2, and the asymptotes are y  2   x  2  y  x and y  x  4.

33. (a)

x  12 y  32   1. This is a hyperbola with a  2, b  6, and 4 36   c  4  36  2 10. The center is 1 3, the vertices are    1 3  2  1 1 and 1 5, the foci are 1 3  2 10 , and the

(b)

1

1 8 asymptotes are y  3   13 x  1  y  13 x  10 3 and y   3 x  3 .

34. (a)

 x2 y  32   1. This is a hyperbola with a  3, b  4, and 3 16      c  3  16  19. The center is 0 3, the vertices are 0 3  3 , the     foci are 0 3  19 , and the asymptotes are y  3   43 x    y  43 x  3 and y   43 x  3.

y

(b)

2

x

2

x

y

1


CHAPTER 11

   35. (a) 9y 2  18y  x 2  6x  18  9 y 2  2y  1  x 2  6x  9  9  9  18

61

Review y

(b)

x  32 y  12  9 y  12  x  32  18    1. This is a 2 18     hyperbola with a  2, b  3 2, and c  2  18  2 5.The center is       3 1, the vertices are 3 1  2 , the foci are 3 1  2 5 , and

1 1

x

the asymptotes are y  1   13 x  3  y  13 x and y   13 x  2.   36. (a) y 2  x 2  6y  y 2  6y  9  x 2  9  y  32  x 2  9 

y

(b)

x2 y  32   1. This is a hyperbola with a  3, b  3, and 9 9   c  9  9  3 2. The center is 0 3, the vertices are 0 3  3  0 6    and 0 0, the foci are 0 3  3 2 , and the asymptotes are y  3  x 

1 x

1

y  x  3 and y  x  3.

37. This is a parabola that opens to the right with its vertex at 0 0 and the focus at 2 0. So p  2, and the equation is y 2  4 2 x  y 2  8x.

38. This is an ellipse with the center at 0 0, a  12, and b  5. The equation is then

y2 y2 x2 x2   1.  2 1 2 144 25 12 5

39. From the graph, the center is 0 0, and the vertices are 0 4 and 0 4. Since a is the distance from the center to a

vertex, we have a  4. Because one focus is 0 5, we have c  5, and since c2  a 2  b2 , we have 25  16  b2 

b2  9. Thus an equation of the hyperbola is

x2 y2   1. 16 9

40. This is a parabola that opens to the left with its vertex at 4 4, so its equation is of the form y  42  4 p x  4 with

p  0. Since 0 0 is a point on this hyperbola, we must have 0  42  4 p 0  4  16  16 p  p  1. Thus the

equation is y  42  4 x  4.

41. From the graph, the center of the ellipse is 4 2, and so a  4 and b  2. The equation is

x  42 y  22  1 2 4 22

x  42 y  22   1. 16 4 42. From the graph, the center is at 1 0, and the vertices are 0 0 and 2 0. Since a is the distance form the center to a b vertex, a  1. From the graph, the slope of one of the asymptotes is  1  b  1. Thus an equation of the hyperbola is a x  12  y 2  1. 43.

x2 x2 y 1   y  1  x 2  12 y  1. This is a parabola with 12 12 4 p  12  p  3. The vertex is 0 1, the focus is 0 1  3  0 2, and the directrix is y  1  3  4.

y 1

1

x


62

44.

CHAPTER 11 Conic Sections

  y2 y x2    12x 2  y 2  12y  0  12x 2  y 2  12y  36  36  12 144 12  x2 y  62   1. This is an ellipse with a  6, b  3, and 3 36      c  36  3  33. The center is 0 6, the foci are 0 6  33 , and the

y

x2 y2   1. This is a hyperbola with a  12, b  12, 144 144      and c  144  144  12 2. The center is 0 0, the foci are 0 12 2 , the

y

1

vertices are 0 6  6  0 0 and 0 12.

x

1

45. x 2  y 2  144  0 

4

vertices are 0 12, and the asymptotes are y  x.

x

4

  x  32  y 2  1. This is a 46. x 2  6x  9y 2  x 2  6x  9  9y 2  9  9   hyperbola with a  3, b  1, and c  9  1  10. The center is 3 0, the    foci are 3  10 0 , the vertices are 3  3 0  6 0 and 0 0, and the

y

1 1

x

asymptotes are y   13 x  3  y  13 x  1 and y   13 x  1.

    47. 4x 2  y 2  8 x  y  4 x 2  2x  y 2  8y  0      4 x 2  2x  1  y 2  8y  16  4  16  4 x  12  y  42  20    x  12 y  42   1. This is an ellipse with a  2 5, b  5, and 5 20      c  20  5  15. The center is 1 4, the foci are 1 4  15 , and the    vertices are 1 4  2 5 .

  48. 3x 2  6 x  y  10  3x 2  6x  6y  10  3 x 2  2x  1  6y  10  3     13 2  3 x  12  6y  13  3 x  12  6 y  13 6  x  1  2 y  6 .   This is a parabola with 4 p  2  p  12 . The vertex is 1  13 6 the focus is       1 10 5 13 1 8 1  13 6  2  1  6  1  3 , and the directrix is y   6  2   3 . 49. x  y 2  16y  x  64  y 2  16y  64  y  82  x  64. This is a parabola with 4 p  1  p  14 . The vertex is 64 8, the focus is     64  14  8   255  8 , and the directrix is x  64  14   257 4 4 .

y

3 3

x

y

1

x

1

y

2 10

x


CHAPTER 11

  50. 2x 2  4  4x  y 2  y 2  2x 2  4x  4  y 2  2 x 2  2x  1  4  2 

 y2  x  12  1. This is a hyperbola with a  2, 2      b  1, and c  2  1  3. The center is 1 0, the foci are 1  3 , the     vertices are 1  2 , and the asymptotes are y   2 x  1      y  2 x  2 and y   2 x  2.     51. 2x 2  12x  y 2  6y  26  0  2 x 2  6x  y 2  6y  26      2 x 2  6x  9  y 2  6y  9  26  18  9  2 x  32  y  32  1 x  32 1 2

63

y

1

y 2  2 x  12  2 

Review

x

1

y x

1

_1

 y  32  1. This is an ellipse with a  1, b  22 , and

     c  1  12  22 . The center is 3 3, the foci are 3 3  22 , and the

vertices are 3 3  1  3 4 and 3 2.     52. 36x 2  4y 2  36x  8y  31  36 x 2  x  4 y 2  2y  31      36 x 2  x  14  4 y 2  2y  1  31  9  4 

y

1

2 2 y  12   1. This is a 36 x  12  4 y  12  36  x  12  9 

1

x

    hyperbola with a  1, b  3, and c  1  9  10. The center is 12  1 , the          foci are 12  10 1 , the vertices are 12  1 1   12  1 and 32  1 ,   and the asymptotes are y  1  3 x  12  y  3x  52 and y  3x  12 .

   2 2    2  y  1  27 25 2  9 x  5 8 y  1 8 y 53. 9x 2 8y 2 15x8y27  0  9 x 2  53 x  25   75 36 4 4 6 2 4 . However, since the left-hand side of the equation is greater than or equal to 0, there is no point that satisfies this equation. The graph is empty.   y 54. x 2  4y 2  4x  8  x 2  4x  4  4y 2  8  4  x  22  4y 2  12    y2 x  22   1. This is an ellipse with a  2 3, b  3, and 12 3  c  12  3  3. The center is 2 0, the foci are 2  3 0  1 0 and    5 0, and the vertices are 2  2 3 0 .

1 1

x

55. The parabola has focus 0 1 and directrix y  1. Therefore, p  1 and so 4 p  4. Since the focus is on the y-axis and the vertex is 0 0, an equation of the parabola is x 2  4y.

56. The parabola with vertex at the origin and focus F 5 0 has p  5, so 4 p  20. The focus is on the x-axis, so an equation is y 2  20x.

57. The ellipse with center at the origin and with x-intercepts 2 and y-intercepts 5 has a vertical major axis, a  5, and b  2, so an equation is

x2 y2   1. 4 25


64

CHAPTER 11 Conic Sections

58. The hyperbola has vertices 0 2 and asymptotes y   12 x. Therefore, a  2, and the foci are on the y-axis. Since the slopes of the asymptotes are  12  

a y2 x2  b  2a  4, an equation of the hyperbola is   1. b 4 16

59. The ellipse has center C 0 4, foci F1 0 0 and F2 0 8, and major axis of length 10. Then 2c  8  0  c  4. Also,

since the length of the major axis is 10, 2a  10  a  5. Therefore, b2  a 2  c2  25  16  9. Since the foci are on

the y-axis, the vertices are on the y-axis, and an equation of the ellipse is

x2 y  42   1. 9 25

60. The hyperbola has center C 2 4, foci F1 2 7 and F2 2 1, and vertices V1 2 6 and V2 2 2. Thus, 2a  6  2  4  a  2. Also, 2c  7  1  6  c  3. So b2  9  4  5. Since the hyperbola has center C 2 4, its equation is y  42 x  22   1. 4 5

61. The ellipse has foci F1 1 1 and F2 1 3, and one vertex is on the x-axis. Thus, 2c  3  1  2  c  1, and so the

center of the ellipse is C 1 2. Also, since one vertex is on the x-axis, a  2  0  2, and thus b2  4  1  3. So an

equation of the ellipse is

x  12 y  22   1. 3 4

62. The parabola has vertex V 5 5 and directrix the y-axis. Therefore,  p  0  5  p  5  4 p  20. Since the parabola opens to the right, its equation is y  52  20 x  5.

63. The ellipse has vertices V1 7 12 and V2 7 8 and passes through the point P 1 8. Thus, 2a  12  8  20    8  12 x  72 y  22 a  10, and the center is 7  7 2. Thus an equation of the ellipse has the form  1.  2 2 100 b

1  72 8  22  1  3600  36b2  100b2  64b2  3600  b2  225  4 . 100 b2 4 x  72 x  72 y  22 y  22 Therefore, an equation of the ellipse is  1   1. 2254 100 225 100 Since the point P 1 8 is on the ellipse,

64. The parabola has vertex V 1 0, horizontal axis of symmetry, and crosses the y-axis where y  2. Since the parabola has a horizontal axis of symmetry and V 1 0, its equation is of the form y 2  4 p x  1. Also, since the parabola crosses

the y-axis where y  2, it passes through the point 0 2. Substituting this point gives 22  4 p 0  1  4 p  4.

Therefore, an equation of the parabola is y 2  4 x  1.

65. The length of the major axis is 2a  186,000,000  a  93,000,000. The eccentricity is e  ca  0017, and so c  0017 93,000,000  1,581,000. (a) The earth is closest to the sun when the distance is a  c  93,000,000  1,581,000  91,419,000.

(b) The earth is furthest from the sun when the distance is a  c  93,000,000  1,581,000  94,581,000.

66. We sketch the LORAN station on the y-axis and place the x-axis halfway between them as x2 y2 suggested in the exercise. This gives us the general form 2  2  1. Since the ship is 80 miles a b closer to A than to B we have 2a  80  a  40Since the foci are 0 150, we have c  150. Thus b2  c2  a 2  1502  402  20900. So this places the ship on the hyperbola given by the x2 y2 1600 y2 225 y2   1. When x  40, we get  1   1600 20,900 1600 20,900 1600 209 y  415. (Note that y  0, since A is on the positive y-axis.) Thus, the ship’s position is

equation

approximately 40 415.

y A 150

40 x

150 B


CHAPTER 11

67. (a) The graphs of

x2 y2   1 for k  1, 2, 4, and 8 are shown in 16  k 2 k2

the figure.   (b) c2  16  k 2  k 2  16 

Review

65

y k=8

c  4. Since the center is 0 0,

2

the focus of each of the ellipses is 4 0.

k=4 2 x k=2 k=1

68. (a) The graphs of y  kx 2 for k  12 , 1, 2, and 4 are shown in the figure.     1 1 1 (b) y  kx 2  x 2  y  4 y. Thus the foci are 0 . k 4k 4k

10

k=4

k=2

8

k=1

6 4

(c) As k increases, the focus gets closer to the vertex.

1

2 -2

0

k=2 2

69. (a) x 2  4x y  y 2  1. Then A  1, B  4, and C  1, so the discriminant is 42  4 1 1  12. Since the discriminant is positive, the equation represents a hyperbola.     11 AC   0  2  90    45 . Therefore, x  22 X  22 Y and y  22 X  22 Y . (b) cot 2  B 4 Substituting into the original equation gives  2       2   2 2 2 2 2 2  4 22 X  22 Y 1 2 X 2 Y 2 X 2 Y  2 X 2 Y       y 1 X 2  2XY  Y 2  2 X 2  XY  XY  Y 2  1 X 2  2XY  Y 2  1  (c) 2 2 3X 2  Y 2  1  3X 2  Y 2  1. This is a hyperbola with a  1 , b  1, and 3    c  13  1  2 . Therefore, the hyperbola has vertices V  1  0 and foci 3 3   F  2  0 , in XY -coordinates. 3

1

1

x


66

CHAPTER 11 Conic Sections

  70. (a) 5x 2  6x y  5y 2  8 2x  8 2y  4  0. Then A  5, B  6, and C  5, so the discriminant is

62  4 5 5  64. Since the discriminant is negative, the equation represents an ellipse.     AC (b) cot 2   0  2  90    45 . Therefore, x  22 X  22 Y and y  22 X  22 Y . Substituting B into the original equation gives  2        2 2 5 22 X  22 Y  6 22 X  22 Y 2 X 2 Y  2           5 22 X  22 Y  8 2 22 X  22 Y  8 2 22 X  22 Y  4  0        5 X 2  2XY  Y 2  3 X 2  Y 2  5 X 2  2XY  Y 2  8X  8Y  8X  8Y  4  0  2 2   y 2 2 2 2 2X  8Y  16Y  4  0  X  4 Y  2Y  2  X 2  4 Y  12  6  (c)     X2 Y  12   1. This ellipse has a  6, b  26 , and c  6  32  3 2 2 . 6 32       Therefore, the vertices are V  6 1 and the foci are F  3 2 2  1 .

1 1

   71. (a) 7x 2  6 3x y  13y 2  4 3x  4y  0. Then A  7, B  6 3, and C  13, so the discriminant is   2 6 3  4 7 13  256. Since the discriminant is negative, the equation represents an ellipse.

x

  AC 7  13 1      2  60    30 . Therefore, x  23 X  12 Y and y  12 X  23 Y . B 6 3 3 Substituting into the original equation gives   2     1 X  3Y 7 23 X  12 Y  6 3 23 X  12 Y 2 2  2         13 12 X  23 Y  4 3 23 X  12 Y  4 12 X  23 Y  0         7 3X 2  2 3XY  Y 2  3 3 3X 2  3XY  XY  3Y 2 4 2      2 2  6X  2 3Y  2X  2 3Y  0   13 4 X  2 3XY  3Y     y 9  13  8X  Y 2 7  9  39  0  4X 2  8X  16Y 2  0  (c) X 2 21  4 2 4 4 2 4 1   4 X 2  2X  1  16Y 2  4  X  12  4Y 2  1. This ellipse has a  1,   x 1 b  12 , and c  1  14  12 3. Therefore, the vertices are    V 1  1 0  V1 0 0 and V2 2 0 and the foci are F 1  12 3 0 .

(b) cot 2 


Note that the conic can't be both a parabola and a degenerate.

CHAPTER 11

67

Review

72. (a) 9x 2  24x y  16y 2  25. Then A  9, B  24, and C  16, so the discriminant is 242  4 9 16  0. Since the discriminant is zero, the equation represents a parabola.

(b) cot 2 

9  16 7 7 AC     cos 2   , so cos   B 24 24 25

(or a degenerate conic).

1725  35 , sin   2

   53 , and thus x  35 X  45 Y and y  45 X  35 Y . Substituting,

1725  45 2 y

(c)

9x 2  24x y  16y 2  25   2      4 X  3 Y  16 4 X  3 Y 2  25  9 35 X  45 Y  24 35 X  45 Y 5 5 5 5

1 x

1

25X 2  25  X 2  1  X  1. Thus the graph is a degenerate conic that consists of two lines. Converting back to x y-coordinates, we see that X  35 x  45 y, so 35 x  45 y  1  3x  4y  5.

73. 5x 2  3y 2  60  3y 2  60  5x 2  y 2  20  53 x 2 . This conic is an ellipse.

5

-5

5 -5

 74. 9x 2  12y 2  36  0  12y 2  9x 2  36  y 2  34 x 2  3  y   34 x 2  3.

5

This conic is a hyperbola.

-5

5 -5

75. 6x  y 2  12y  30  y 2  12y  30  6x  y 2  12y  36  66  6x    y  62  66  6x  y  6   66  6x  y  6  66  6x. This conic is

20

a parabola.

10

-10

76. 52x 2  72x y  73y 2  100  73y 2  72x y  52x 2  100  0. Using the quadratic formula,

   72x 72x2 473 52x 2 100 y  146

1

-2

2 ,000x 2 29,200  72x 10146  72x201467325x  10 2  36 73 x  73 73  25x

This conic is an ellipse.

10

2 -1


68

CHAPTER 11 Conic Sections

1  e  1. Therefore, this is a 1  cos  parabola.

2  e  23 . Therefore, this is an 3  2 sin  ellipse.

77. (a) r 

78. (a) r 

(b)

(b)

( 25 , ¹2) 1

( 21 , ¹)

_1

1

(2, 3¹ 2 ) 4  e  2. Therefore, this is a 1  2 sin  hyperbola.

12  e  4. Therefore, this is a 1  4 cos  hyperbola.

79. (a) r 

80. (a) r 

(b)

(b)

(_4, 3¹ 2 ) ( 43 , ¹2)

(_4, 0)

( 125 , ¹)

1

1

  x2 y2 2 0   1 is the standard equation of an ellipse centered at the origin passing through 2 4 and 0 2, corresponding to graph IV.

81. (a) 2x 2  y 2  4 

(b) 3x 2  4y  12  y   34 x 2  3 is an equation of a parabola opening downward with vertex 0 3, corresponding to graph III. (c) x 2  y 2  4 is the standard equation of a circle centered at the origin with radius 2, corresponding to graph II.

(d) x 2  2x y  y 2  x  y  0 has discriminant B 2  4AC  22  4 1 1  0, so it is an equation of a rotated parabola, corresponding to graph VIII. (e) x y  1  y  1x is an equation of a hyperbola centered at the origin, corresponding to graph I.   x2  1 is an equation of a (f) 4y 2  x 2  8y  0  4 y 2  2y  x 2  0  4 y  12  x 2  4  y  12  4 hyperbola centered at 0 1, corresponding to graph VII.   (g) 2y 2  5x  4y  8  5x  2 y 2  2y  8  2 y  12  8  2  x  2  25 y  12 . This is an equation of a parabola opening rightward with vertex 2 1, corresponding to graph VI.

(h) 153x 2  192x y  97y 2  225 has discriminant B 2  4AC  1922  4 153 97  22,500, so it is an equation of a rotated ellipse, corresponding to graph V.


CHAPTER 11

Test

1

x

1

x

1

x

69

CHAPTER 11 TEST 1. x 2  12y. This is a parabola with 4 p  12  p  3. The focus is 0 3 and the directrix is y  3.

2.

_1

  y2 x2   1. This is an ellipse with a  4, b  2, and c  16  4  2 3. The 16 4    vertices are 4 0, the foci are 2 3 0 , the length of the major axis is 2a  8, and the length of the minor axis is 2b  4.

3.

y

 x2 y2   1. This is a hyperbola with a  3, b  4, and c  9  16  5. The 9 16 vertices are 0 3, the foci are 0 5, and the asymptotes are y   34 x.

y

1

y

1

4. The parabola with vertex 0 0 and focus 4 0 has p  4, so 4 p  16. The focus lies to the right of the vertex, so an equation is y 2  16x.

5. The ellipse with foci 3 0 and vertices 4 0 has a  4 and c  3, so c2  a 2  b2  9  16  b2  b2  16  9  7. Thus, an equation is

x2 y2   1. 16 7

6. The hyperbola has foci 0 5 and asymptotes y   34 x. Since the foci are 0 5, c  5, the foci are on a 3 a  a  34 b. Then the y-axis, and the center is 0 0. Also, since y   34 x   x, it follows that  b b 4  2 3 2 2 c2  52  25  a 2  b2  34 b  b2  25 16 b  b  16, and by substitution, a  4 4  3. Therefore, an equation of the hyperbola is

x2 y2   1. 9 16

7. This is a parabola that opens to the left with its vertex at 0 0. So its equation is of the form y 2  4 px with p  0.   Substituting the point 4 2, we have 22  4 p 4  4  16 p  p   14 . So an equation is y 2  4  14 x  y 2  x.

8. This is an ellipse tangent to the x-axis at 0 0 and with one vertex at the point 4 3. The center is 0 3, and a  4 and b  3. Thus the equation is

x2 y  32   1. 16 9


70

CHAPTER 11 Conic Sections

9. This a hyperbola with a horizontal transverse axis, vertices at 1 0 and 3 0, and foci at 0 0 and 4 0. Thus the center is 2 0, and a  3  2  1 and c  4  2  2. Thus b2  22  12  3. So an equation is x  22 

x  22 y2 1  2 3 1

y2  1. 3

    10. 16x 2  36y 2  96x  36y  9  0  16 x 2  6x  36 y 2  y  9      16 x 2  6x  9  36 y 2  y  14  9  144  9  2  x  32 16 x  32  36 y  12  144  

y  12

2

y 1 x

1

 1. This is an     ellipse with a  3, b  2, and c  9  4  5. The center is 3  12 , the       vertices are 3  3  12  0  12 and 6  12 , and the foci are          h  c k  3  5  12  3  5  12 and 3  5  12 . 9

4

    11. 9x 2  8y 2  36x  64y  164  9 x 2  4x  8 y 2  8y  164      9 x 2  4x  4  8 y 2  8y  16  164  36  128 

y

x  22 y  42 9 x  22  8 y  42  72    1. This is a hyperbola 8 9  with a  2 2, b  3, and c2  a 2  b2  17. The center is 2 4, the foci are       2  17 4 , the vertices are 2  2 2 4 , and the asymptotes are 

1 1

x

y  4   3 4 2 x  2  y  3 4 2 x  4  3 2 2 and y   3 4 2 x  4  3 2 2 .

12. 2x  y 2  8y  8  0  y 2  8y  16  2x  8  16  y  42  2 x  4. This is a parabola with 4 p  2  p   12 . The vertex is 4 4, the focus is     4  12  4  72  4 , and the directrix is x  4  12  92 .

y 1 1

x

13. The ellipse with center 2 0, foci 2 3 and major axis of length 8 has a horizontal major axis with 2a  8  a  4. Also, c  3, so b2  a 2  c2  16  9  7. Thus, an equation is

y2 x  22   1. 7 16

14. The parabola has focus 2 4 and directrix the x-axis (y  0). Therefore, 2 p  4  0  4  p  2  4 p  8, and the vertex is 2 4  p  2 2. Hence, an equation of the parabola is x  22  8 y  2  x 2  4x  4  8y  16  x 2  4x  8y  20  0.

15. We place the vertex of the parabola at the origin, so the parabola contains the points 3 3, and the equation is of the form   y 2  4 px. Substituting the point 3 3, we get 32  4 p 3  9  12 p  p  34 . So the focus is 34  0 , and we should place the light bulb 34 inch from the vertex.


CHAPTER 11

Test

16. (a) 5x 2  4x y  2y 2  18. Then A  5, B  4, and C  2, so the discriminant is 42  4 5 2  24. Since the discriminant is negative, the equation represents an ellipse.   AC 52 3 (b) cot 2    . Thus, cos 2  35 and so cos   135  255, 2 B 4 4       135 5 sin    5 . It follows that x  2 5 5 X  55 Y and y  55 X  2 5 5 Y . By 2    2  2        5 X  2 5Y  2 5 X  2 5Y substitution, 5 2 5 5 X  55 Y  4 2 5 5 X  55 Y  18 5 5 5 5      4X 2  4XY  Y 2  45 2X 2  4XY  XY  2Y 2  25 X 2  4XY  4Y 2  18 

      2 2 8 2 1  8  4  18  6X 2  Y 2  18  X  Y  1. This is an X 2 4  85  25  XY 4  12 5  5 Y 5 5 3 18   ellipse with a  3 2 and b  3.    y (c) (d) In XY -coordinates, the vertices are V 0 3 2 . Therefore, in 

1

x

1

x y-coordinates, the vertices are x   3 2 and y  6 2  5 5       3 2 6 2 3 2 6 2 V1     , and x   and y     5 5 5 5      2 . V2 3 2  6 5

Since cos 2  35 we have

5

2  cos1 35  5313 , so   27 . 17. (a) Since the focus of this conic is the origin and the directrix is x  2, the equation has the form

ed . Subsituting e  12 and d  2 we 1  e cos  1 2 get r  r  . 1 2  cos  1  cos 

(b) r 

3 3 2 r  . So e  12 and the 2  sin  1  1 sin  2

conic is an ellipse.

r

(3, ¹2 )

2

( 32 , ¹) O

1

O

1

(

3¹ 1, 2

( 32 , 0) )

71


72

FOCUS ON MODELING

FOCUS ON MODELING Conics in Architecture 1. Answers will vary. 2. (a) The difference P F2   P A  c is a constant (the length of the string). Also, P F1   P A  d is constant. Subtracting these two equations, we find that P F2   P F1   c  d is a constant. This is the definition of a hyperbola. (b) Reflect the entire apparatus through a line perpendicular to F1 F2 .

4. As the lids are twisted more, the vertices of the hyperbolic cross sections get closer together.   5. (a) The tangent line passes though the point a a 2 , so an equation is y  a 2  m x  a.

  (b) Because the tangent line intersects the parabola at only the one point a a 2 , the system

y  a 2  m x  a y  x2

has

only one solution, namely x  a, y  a 2 .   y  a 2  m x  a y  a 2  m x  a (c)   a 2  m x  a  x 2  x 2  mx  am  a 2  0. This quadratic 2 yx y  x2   has discriminant m2  4 1 am  a 2  m 2  4am  4a 2  m  2a2 . Setting this equal to 0, we find m  2a.

(d) An equation of the tangent line is y  a 2  2a x  a  y  a 2  2ax  2a 2  y  2ax  a 2 .

6. (a) F1 and F2 are points of tangency of the spheres to the plane, and the vertical line through Q 1 and Q 2 is also tangent to each sphere. Since P lies outside of both spheres, we see that P F1  P Q 1 and P F2  P Q 2 . (b) Each sphere is tangent to the enclosing cylinder along a circle in a vertical plane. The distance between these two planes is constant, so P Q 1  P Q 2  Q 1 Q 2 is constant for any choice of P.

(c) Combining the results of parts (a) and (b), we see that P F1  P F2  P Q 1  P Q 2 is constant.

(d) By definition, an ellipse is a curve consisting of all points whose distances to two fixed points F1 and F2 add up to a constant. That is the case here.


CORRECTIONS: 2,4,11

CHAPTER 12

SEQUENCES AND SERIES

12.1

Sequences and Summation Notation 1

12.2

Arithmetic Sequences 6

12.3

Geometric Sequences 11

12.4 12.5

Mathematical Induction 20 The Binomial Theorem 29 Chapter 12 Review 33 Chapter 12 Test 39

¥

FOCUS ON MODELING: Modeling with Recursive Sequences 41

1


12 SEQUENCES AND SERIES 12.1 SEQUENCES AND SUMMATION NOTATION 1. A sequence is a function whose domain is the natural numbers. 2. The nth partial sum of a sequence is the sum of the first n terms of the sequence. So for the sequence an  n 2 the fourth partial sum is S4  12  22  32  42  30.

3. an  n  3. Then a1  1  3  2, a2  2  3  1, a3  3  3  0, a4  4  3  1, and a100  100  3  97.

4. an  2n  1. Then a1  2 1  1  1, a2  2 2  1  3, a3  2 3  1  5, a4  2 4  1  7, and a100  2 100  1  199. 1 1 1 1 1 1 1 1 . Then a1   1, a2   , a3   , a4   , and 5. an  3n  4 3 1  4 3 2  4 2 3 3  4 5 3 4  4 8 1 1 a100   . 3 100  4 296

6. an  n 3  2. Then a1  13  2  3, a2  23  2  10, a3  33  2  29, a4  43  2  66, and a100  1003  2  1,000,002.

7. an  3n . Then a1  31  3, a2  32  9, a3  33  27, a4  34  81, and a100  3100  5154  1047 .  n1  11  21  31   1 , a   1 41   1 , and . Then a1   15  1, a2   15   15 , a3   15  25 8. an   15 4 5 125  1001  6338  1070 . a100   15

1 1 1 1n 11 12 13 14 . Then a1   1, a2   , a3    , a4   , and 2 2 2 2 4 9 16 n 1 2 3 42 1 1100 .  a100  2 10,000 100 1 1 1 1 1 1 1 1 1 1 , and a100  .  1, a2   , a3   , a4    10. an  2 . Then a1  2 2 2 2 2 4 9 16 10,000 n 1 2 3 4 100 9. an 

11. an  1  1n . Then a1  1  11  0, a2  1  12  2, a3  1  13  0, a4  1  14  2, and a100  1  1100  2.

1 2 3 4 1n1 n 12  1 13  2 14  3 15  4 . Then a1   , a2    , a3   , a4    , and n1 11 2 21 3 31 4 41 5 100 1101  100  . a100  101 101 13. an  n n . Then a1  11  1, a2  22  4, a3  33  27, a4  44  256, and a100  100100  10200 . 12. an 

14. an  3. Then a1  3, a2  3, a3  3, a4  3, and a100  3.   15. an  2 an1  3 and a1  4. Then a2  2 4  3  14, a3  2 14  3  34, a4  2 34  3  74, and a5  2 74  3  154.

1 2 2 1 1 a 24 4  4, a3    , a4  3   , and a5  9   . 16. an  n1 and a1  24. Then a2  6 6 6 3 6 9 6 54 17. an  2an1  1 and a1  1. Then a2  2 1  1  3, a3  2 3  1  7, a4  2 7  1  15, and a5  2 15  1  31. 1 1 1 1 2 1 3 1 5  , a3  and a1  1. Then a2   , a4   , and a5   . 18. an  1  an1 11 2 3 5 8 1 1 1 2 1 3 2

3

5

1


2

CHAPTER 12 Sequences and Series

19. an  an1  an2 , a1  1, and a2  2. Then a3  2  1  3, a4  3  2  5, and a5  5  3  8.

20. an  an1  an2  an3 and a1  1, a2  1, and a3  1. Then a4  1  1  1  3 and a5  3  1  1  5. 21. (a) a1  7, a2  11, a3  15, a4  19, a5  23,

a6  27, a7  31, a8  35, a9  39, a10  43

Delete tick marks at 2.5 and 7.5, since the domain is only integers

22. (a) a1  2, a2  6, a3  12, a4  20, a5  30,

a6  42, a7  56, a8  72, a9  90, a10  110

(b)

(b) 40

100

20

50

0

0

5

0

10

12 12 23. (a) a1  12 1  12, a2  2  6, a3  3  4,

12 12 12 a4  12 4  3, a5  5 , a6  6  2, a7  7 ,

0

5

10

24. (a) a1  6, a2  2, a3  6, a4  2, a5  6, a6  2, a7  6, a8  2, a9  6, a10  2

(b)

3 12 4 12 6 a8  12 8  2 , a9  9  3 , a10  10  5

(b)

5 10

0

5 0

0

5

a6  05, a7  2, a8  05, a9  2, a10  05

a6  2, a7  1, a8  3, a9  2, a10  1

(b) 2 1 5

10

26. (a) a1  1, a2  3, a3  2, a4  1, a5  3,

(b)

0

5

10

25. (a) a1  2, a2  05, a3  2, a4  05, a5  2,

0

0

10

4 2 0 -2 -4

5

10

27. 2, 4, 6, 8,   . All are multiples of 2, so a1  2, a2  2  2, a3  3  2, a4  4  2,   . Thus an  2n.

28. 1, 3, 5, 7,   . All are odd numbers, so a1  2  1, a2  2  2  1, a3  3  2  1, a4  4  2  1,   . Thus an  2n  1. 29. 3, 9, 27, 81,   . All are powers of 3, so a1  3, a2  32 , a3  33 , a4  34 ,   . Thus an  3n .

1 , 1 ,   . The denominators are all powers of 3, and the terms alternate in sign. Thus a  30.  13 , 19 ,  27 1 81

11 12 , a  , 2 31 32

13 14 1n , a4  ,   . So an  . 3 4 3n 3 3 31. 4, 9, 14, 19,   . The difference between any two consecutive terms is 5, so a1  5 1  1, a2  5 2  1, a3  5 3  1, a4  5 4  1,   . Thus, an  5n  1. a3 

32. 10, 3, 4, 11,   . The difference between any two consecutive terms is 7, so a1  7 1  17, a2  7 2  17, a3  7 3  17, a4  7 4  17,   . Thus, an  7n  17.

33. 5, 25, 125, 625,   . These terms are powers of 5, and the terms alternate in sign. So a1  12  51 , a2  13  52 , a3  14  53 , a4  15  54 ,   . Thus an  1n1  5n .

34. 3, 03, 003, 0003,   . The ratio of any two consecutive terms is 10, so a1  3  100 , a2  3  101 , a3  3  102 ,  n1 1 a4  3  103 ,   . Thus an  3  10 .


SECTION 12.1 Sequences and Summation Notation

3

7 , 9 ,   . We consider the numerator separately from the denominator. The numerators of the terms differ 35. 1, 34 , 59 , 16 25

by 2, and the denominators are perfect squares. So a1  a5 

2 1  1 2 2  1 2 3  1 2 4  1 , a2  , a3  , a4  , 2 2 2 1 2 3 42

2 5  1 2n  1 ,   . Thus an  . 2 5 n2

36. 34 , 45 , 56 , 67 ,   . Both the numerator and the denominator increase by 1, so a1  a4 

42 n2 ,   . Thus an  . 43 n3

12 22 32 ,a  ,a  , 13 2 23 3 33

37. 0, 2, 0, 2, 0, 2,   . These terms alternate between 0 and 2. So a1  1  1, a2  1  1, a3  1  1, a4  1  1, a5  1  1, a6  1  1,    Thus an  1  1n . n1

38. 1, 12 , 3, 14 , 5, 16 ,   . So a1  1, a2  21 , a3  31 , a4  41 ,   . Thus an  n 1

.

39. a1  2, a2  4, a3  6, a4  8,   . Therefore, an  2n. So S1  2, S2  2  4  6, S3  2  4  6  12, S4  2  4  6  8  20, S5  2  4  6  8  10  30, and S6  2  4  6  8  10  12  42. 40. a1  12 , a2  22 , a3  32 , a4  42 ,   . Therefore, an  n 2 . So S1  12  1, S2  122  5, S3  532  59  14, S4  14  42  14  16  30, S5  30  52  30  25  55, and S6  55  62  55  36  91.

1 1 1 1 1 4 1 1 41. a1  13 , a2  2 , a3  3 , a4  4 ,   . Therefore, an  n . So S1  13 , S2  13  2  , S3  13  2  3  13 27 , 3 9 3 3 3 3 3 3 1 1 1 1 1 1 1 1 1 1 1 1 1 121 1 364 S4  13  2  3  4  40 81 , and S5  3  32  33  34  35  243 , S6  3  32  33  34  35  36  729 . 3 3 3 42. a1  4, a2  4, a3  4, a4  4,   . Therefore, an  4 1n1 . So S1  4, S2  4  4  0, S3  4  4  4  4, S4  0, S5  4, and S6  0. 2 2 2 2 2 2 2 2 , and S4  23  2  3  4  80 43. an  n . So S1  23 , S2   2  89 , S3  23  2  3  26 27 81 . Therefore, 3 3 3 3 3 3 3 3 n 3 1 . Sn  3n       1 1 44. an   . So S1  12  13 , S2  12  13  13  14  12   13  13  14  12  14 , n1 n2           1 1 S3  2  3  13  14  14  15  12   13  13   14  14  15  12  15 , and               S4  12  13  13  14  14  15  15  16  12   13  13   14  14   15  15  16  12  16 . Therefore,           1  1 1  1  1       1  1 1 1 1 Sn  12  13  13  14      n1  n2 2 3 3 n1 n1  n2  2  n2 .                  n  n  1. So S1  1  2  1  2, S2  1 2  2  3  1   2  2  3  1  3,                    1 2  2 3  3  4  1   2  2   3  3  4  1  4, S3              S4  1 2  2 3  3 4  4 5               1  2 2   3 3   4 4  51 5

45. an 

Therefore,          1 2  2  3   n  n  1 Sn                 1   2  2   3  3     n  n  n  1  1  n  1


4

CHAPTER 12 Sequences and Series

In #46, replace "k" with "n" (4 times)

 k  log k  log k  1. So S1  log 1  log 2   log 2, S2   log 2  log 2  log 3   log 3, k 1 S3   log 2  log 2  log 3  log 3  log 4   log 2  log 2   log 3  log 3  log 4   log 4, and

46. an  log

S4   log 2  log 2  log 3  log 3  log 4  log 4  log 5

  log 2  log 2   log 3  log 3   log 4  log 4  log 5   log 5

Therefore, Sn   log n  1. 4 47. k1 k  1  2  3  4  10 4 48. k1 k 2  1  22  32  42  1  4  9  16  30

49. 50. 51.

52.

1 1 1 1 1 25 k1 3k  3  6  9  12  36

4

51

j 1 2 3 4 50 51 j1 1  1  1  1  1      1  1  1  1  1  1  1      1  1  1

8

i i 1 1  1

12

02020202 8

i 4 10  10  10  10  10  10  10  10  10  10  90 53. k1 2k1  20  21  22  23  24  1  2  4  8  16  31  54. i31 i2i  1  21  2  22  3  23  2  8  24  34

5

for #58, I get 0.153446

55. 385 56. 15,550 57. 46,438 58. 0153146 59. 22 60. 0688172 4 3 3 3 3 3 61. k1 k  1  2  3  4  1  8  27  64         4 j 1 0 1 2 3 3 2 15     0   62. j1 j 1 2 3 4 5 3 2 5          63. 6k0 k  4  4  5  6  7  8  9  10  64. 9k6 k k  3  6  9  7  10  8  11  9  12  54  70  88  108  k 3 4 5 100 65. 100 k3 x  x  x  x      x n j1 x j  12 x  13 x 2  14 x 3      1n1 x n  x  x 2  x 3      1n1 x n 66. j1 1  67. 4  8  12  16      48  12 k1 4k 10 68. 2  5  8      29  k1 3k  1  2 69. 12  22  32      102  10 k1 k

 1 1 1 1 1 1k       100 k2 k ln k 2 ln 2 3 ln 3 4 ln 4 5 ln 5 100 ln 100 999 1 1 1 1 1       k1 71. 12 23 34 999  1000 k k  1      1 2 3 n  k 72. 2  2  2      2  nk1 2 1 2 3 n k  k 73. 1  x  x 2  x 3      x 100  100 k0 x  k1  k  x k1 74. 1  2x  3x 2  4x 3  5x 4      100x 99  100 k1 1           75. 2, 2 2, 2 2 2, 2 2 2 2,   . We simplify each term in an attempt to determine a formula for an . So a1  212 ,       a2  2  212  232  234 , a3  2  234  274  278 , a4  2  278  2158  21516 ,   . Thus 70.

n

n

an  22 12 .

76. F1  1, F2  1, F3  2, F4  3, F5  5, F6  8, F7  13, F8  21, F9  34, F10  55

77. (a) A1  $2004, A2  $200801, A3  $201202, A4  $201605, A5  $202008, A6  $202412


SECTION 12.1 Sequences and Summation Notation

5

(b) Since 3 years is 36 months, we get A36  $214916. 78. (a) I1  0, I2  $050, I3  $150, I4  $301, I5  $503, I6  $755 (b) Since 5 years is 60 months, we have I60  $97700.

79. (a) P1  35,700, P2  36,414, P3  37,142, P4  37,885, P5  38,643 (b) Since 2014 is 10 years after 2004, P10  42,665.

80. (a) The amount owed at the end of the month, An , is the amount owed at the beginning of the month, An1 , plus the interest, 0005An1 , minus the $200 monthly repayment. Thus An  An1  0005An1  200  An  1005An1  200.

(b) A1  9850, A2  969925, A3  954774, A4  939548, A5  924246, A6  908867. Thus the amount owing after six months is $908867.

81. (a) The number of catfish at the end of the month, Pn , is the population at the start of the month, Pn1 , plus the increase in population, 008Pn1 , minus the 300 catfish harvested. Thus Pn  Pn1  008Pn1  300  Pn  108Pn1  300. (b) P1  5100, P2  5208, P3  5325, P4  5451, P5  5587, P6  5734, P7  5892, P8  6064, P9  6249, P10  6449, P11  6665, P12  6898. Thus there should be 6898 catfish in the pond at the end of 12 months.

82. (a) Let n be the number of years since 2022, so P0  $240,000. Each month, the median price of a house increases to 106 times its price the previous month. Thus, Pn  106n P0 . (b) Since 2030  2022  8, we calculate P8  382,52353. In 2030, the median price of a house is estimated to be $382,524.

83. (a) Let An be the salary in the nth year. Then A1  $45,000. Since the salary increases by $2000 each year, An  An1  2000. Thus, A1  $45,000 and An  An1  2000.

(b) A5  A4  2000  A3  2000  2000  A2  2000  4000  A1  2000  6000  $53,000. 84. (a) Let n be the days after the experiment starts, so C0  4. Each day the concentration increases by 10%, so after n days the concentration of the brine solution is Cn  110Cn1 with C0  4.

(b) C8  110C7  110 110C6       1108  C0  1108  4  86. Thus, the brine solution concentration is 86 g/L of salt.

85. Let Fn be the number of pairs of rabbits in the nth month. Clearly F1  F2  1. In the nth month each pair that is two or more months old (that is, Fn2 pairs) will add a pair of offspring to the Fn1 pairs already present. Thus Fn  Fn1  Fn2 . So Fn is the Fibonacci sequence. 86. (a) an  n 2 . Then a1  12  1, a2  22  4, a3  32  9, a4  42  16.

(b) an  n 2  n  1 n  2 n  3 n  4, a1  12  1  1 1  2 1  3 1  4  1  0 1 2 3  1, a2  22  2  1 2  2 2  3 2  4  4  1  0 1 2  4, a3  32  3  1 3  2 3  3 3  4  9  2  1  0 1  9,

a4  42  4  1 4  2 4  3 4  4  16  3  2  1  0  16.

Hence, the sequences agree in the first four terms. However, for the second sequence, a5  52 

5  1 5  2 5  3 5  4  25  4  3  2  1  49, and for the first sequence, a5  52  25, and thus the sequences disagree from the fifth term on. (c) an  n 2  n  1 n  2 n  3 n  4 n  5 n  6 agrees with an  n 2 in the first six terms only.

(d) an  2n and bn  2n  n  1 n  2 n  3 n  4.


6

CHAPTER 12 Sequences and Series

a  n if an is even 2 87. an1  3an  1if an is odd

With a1  11, we have a2  34, a3  17, a4  52, a5  26, a6  13, a7  40,

a8  20, a9  10, a10  5, a11  16, a12  8, a13  4, a14  2, a15  1, a16  4, a17  2, a18  1,    (with 4, 2, 1 repeating). So a3n1  4, a3n2  2, and a3n  1, for n  5. With a1  25, we have a2  76, a3  38, a4  19, a5  58, a6  29, a7  88, a8  44, a9  22, a10  11, a11  34, a12  17, a13  52, a14  26, a15  13, a16  40, a17  20, a18  10, a19  5, a20  16, a21  8, a22  4, a23  2, a24  1, a25  4, a26  2, a27  1,    (with 4, 2, 1 repeating). So a3n1  4, a3n2  2, and a3n3  1 for n  7. We conjecture that the sequence will always return to the numbers 4, 2, 1 repeating.

88. an  anan1  anan2 , a1  1, and a2  1. So a3  a31  a31  a2  a2  1  1  2, a4  a42  a41  a2  a3  1  2  3, a5  a53  a52  a2  a3  1  2  3, a6  a63  a63  a3  a3  2  2  4, a7  a74  a73  a3  a4  2  3  5, a8  a85  a84  a3  a4  2  3  5, a9  a95  a95  a4  a4  3  3  6, and a10  a106  a105  a4  a5  3  3  6. The definition of an depends on the values of certain preceding terms. So an is the sum of two preceding terms whose choice depends on the values of an1 and an2 (not on n  1 and n  2).

12.2 ARITHMETIC SEQUENCES 1. An arithmetic sequence is sequence where the difference between successive terms is constant. 2. The sequence an  a  n  1 d is an arithmetic sequence where a is the first term and d is the common difference. So, for the arithmetic sequence an  2  5 n  1 the first term is 2 and the common difference is 5. 3. True. The nth partial sum of an arithmetic sequence is the average of the first and last terms times n. 4. True. If we know the first and second terms of an arithmetic sequence then we can find any other term.

5. (a) a1  7  3 1  1  7, a2  7  3 2  1  10,

a3  7  3 3  1  13, a4  7  3 4  1  16,

6. (a) a1  10  20 1  1  10, a2  10  20 2  1  10, a3  10  20 3  1  30,

a5  7  3 5  1  19

a4  10  20 4  1  50,

(b) The common difference is 3. (c)

a5  10  20 5  1  70

an 20

(b) The common difference is 20.

15

(c)

10

60

5 0

an

40 1

2

3

4

5 n

20 0

1

2

3

4

5 n


SECTION 12.2 Arithmetic Sequences

7. (a) a1  3  5 1  1  3,

8. (a) a1  7  3 1  1  7, a2  7  3 2  1  4,

a2  3  5 2  1  8,

a3  7  3 3  1  1, a4  7  3 4  1  2,

a3  3  5 3  1  13,

a5  7  3 5  1  5

a4  3  5 4  1  18,

(b) The common difference is 3.

a5  3  5 5  1  23

(c)

(b) The common difference is 5. an 5

(c)

1

2

an 10 5

3

4

4

0 _5

5 n

1

2

3

5 n

_10 _20

10. (a) a1  12 1  1  0, a2  12 2  1  12 ,

9. (a) a1  15  05 1  1  15, a2  15  05 2  1  2,

a3  12 3  1  1, a4  12 4  1  32 ,

a3  15  05 3  1  25,

a5  12 5  1  2

a4  15  05 4  1  3,

(b) The common difference is 12 .

a5  15  05 5  1  35

(c)

(b) The common difference is 05. (c)

2

an 3

1

2 1 0

an

0

1

2

3

4

1

2

3

4

5 n

5 n

11. a  10, d  6, an  a  d n  1  10  6 n  1. So a10  10  6 10  1  44. 12. a  5, d  2, an  a  d n  1  5  2 n  1. So a10  5  2 10  1  13.

13. a  06, d  1, an  a  d n  1  06  n  1. So a10  06  10  1  84.

14. a  18, d  02, an  a  d n  1  18  02 n  1. So a10  18  02 10  1  0.

15. a  52 , d   12 , an  a  d n  1  52  12 n  1. So a10  52  12 10  1  2.        16. a  3, d  3, an  a  d n  1  3  3 n  1. So a10  3  3 10  1  10 3.

17. a4  a3  a3  a2  a2  a1  6. The sequence is arithmetic with common difference 6.

18. a4  a3  a3  a2  a2  a1  12. The sequence is arithmetic with common difference 12.

19. Since a3  a2  7 and a4  a3  6, the terms of the sequence do not have a common difference. This sequence is not arithmetic. 20. a4  a3  a3  a2  a2  a1  32. The sequence is arithmetic with common difference 32.

21. Since a2  a1  4  2  2 and a4  a3  16  8  8, the terms of the sequence do not have a common difference. This sequence is not arithmetic.

22. a4  a3  a3  a2  a2  a1  2. This sequence is arithmetic with common difference 2. 23. a4  a3   32  0   32 , a3  a2  0  difference  32 .

3   32 , a2  a1  32  3   32 . This sequence is arithmetic with common 2

7


8

CHAPTER 12 Sequences and Series

8 4 24. a4  a3  ln 16  ln 8  ln 16 8  ln 2, a3  a2  ln 8  ln 4  ln 4  ln 2, a2  a1  ln 4  ln 2  ln 2  ln 2. This

sequence is arithmetic with common difference ln 2.

25. a4  a3  77  60  17, a3  a2  60  43  17, 4  a1  43  26  17. This sequence is arithmetic with common difference 17. 1 and a  a  1  1   1 , the terms of the sequence do not have a common difference. 26. Since a4  a3  15  14   20 3 2 4 3 12

This sequence is not arithmetic.

27. a1  4  7 1  11, a2  4  7 2  18, a3  4  7 3  25, a4  4  7 4  32, a5  4  7 5  39. This sequence is arithmetic, the common difference is d  7 and an  4  7n  4  7n  7  7  11  7 n  1. 28. a1  4  21  6, a2  4  22  8, a3  4  23  12, a4  4  24  20, a5  4  25  36. Since a4  a3  8 and a3  a2  4, the terms of the sequence do not have a common difference. This sequence is not arithmetic. 1 1 1 1 1 1 1 1 1 1  , a2   , a3   , a4   , a5   . Since 29. a1  1  2 1 3 1  2 2 5 1  2 3 7 1  2 4 9 1  2 5 11

2 and a  a  1  1   2 , the terms of the sequence do not have a common difference. This a4  a3  19  17   63 3 2 7 3 21

sequence is not arithmetic.

30. a1  1  12  32 , a2  1  22  2, a3  1  32  52 , a4  1  42  3, a5  1  52  72 . This sequence is arithmetic, the common difference is d  12 and an  1  n2  1  12 n  12  12  32  12 n  1.

31. a1  6 1  10  4, a2  6 2  10  2, a3  6 3  10  8, a4  6 4  10  14, a5  6 5  10  20. This sequence is arithmetic, the common difference is d  6 and an  6n  10  6n  6  6  10  4  6 n  1. 32. a1  3  11 1  2, a2  3  12 2  5, a3  3  13 3  0, a4  3  14 4  7,

a5  3  15 5  2. Since a4  a3  7 and a3  a2  5, the terms of the sequence do not have a common difference. This sequence is not arithmetic.

33. 6, 8, 10, 12,   . Then d  a2  a1  8  6  2, a5  a4  2  12  2  14, an  6  2 n  1, and a100  6  2 99  204.

34. 5, 0, 5, 10,   . Then d  a2  a1  0  5  5, a5  a4  5  10  5  15, an  5  5 n  1, and a100  5  5 99  490.

35. 29, 11, 7, 25,   . Then d  a2  a1  11  29  18, a5  a4  18  25  18  43, an  29  18 n  1, and a100  29  18 99  1753.

36. 64, 49, 34, 19,   . Then d  a2  a1  49  64  15, a5  a4  15  19  15  4, an  64  15 n  1, and a100  64  15 99  1421. 37. 4, 9, 14, 19,   . Then d  a2  a1  9  4  5, a5  a4  5  19  5  24, an  4  5 n  1, and a100  4  5 99  499.

38. 11, 8, 5, 2,   . Then d  a2  a1  8  11  3, a5  a4  3  2  3  1, an  11  3 n  1, and a100  11  3 99  286.

39. 12, 8, 4, 0,   . Then d  a2  a1  8  12  4, a5  a4  4  0  4  4, an  12  4 n  1, and a100  12  4 99  384.

40. 14 , 34 , 54 , 74 ,   . Then d  a2  a1  34  14  12 , a5  a4  12  74  12  94 , an  14  12 n  1, and a100  14  12 99  199 4 .

41. 25, 265, 28, 295,   . Then d  a2  a1  265  25  15, a5  a4  15  295  15  31, an  25  15 n  1, and a100  25  15 99  1735.

42. 15, 123, 96, 69,   . Then d  a2  a1  123  15  27, a5  a4  27  69  27  42, an  15  27 n  1, and a100  15  27 99  2523.

43. 2, 2  s, 2  2s, 2  3s,   . Then d  a2  a1  2  s  2  s, a5  a4  s  2  3s  s  2  4s, an  2  n  1 s, and a100  2  99s.


SECTION 12.2 Arithmetic Sequences

9

44. t, t  3, t  6, t  9,   . Then d  a2  a1  t  3  t  3, a5  a4  3  t  9  3  t  12, an  t  3 n  1, and a100  t  3 99  t  297. 45. a50  1000 and d  6. Thus, a50  a1  d 50  1  1000  a1  6 50  1  a1  1000  294  706 and a2  706  6  712.

46. a100  750 and d  20. Thus, 750  a1  20 99  a1  1230 and a5  1230  20 4  1150.

  1 . Thus, a  1  8 1   5 and 47. a14  23 and a9  14 , so 23  a1  13d and 14  a1  8d  23  14  5d  d  12 1 4 12 12 5  1 n  1. an   12 12

48. a12  118 and a8  146, so 118  a1  11d and 146  a1  7d  118  146  4d  d  7. Thus, a1  146  7 7  195 and an  195  7 n  1.

49. a1  25 and d  18, so an  601  25  18 n  1  601  33. Thus, 601 is the 33rd term of the sequence.

50. a1  3500 and d  15, so an  2795  2795  3500  15 n  1  n  48. Thus, 2795 is the 48th term of the sequence. 51. a  3, d  5, n  20. Then S20  20 2 2  3  19  5  1010.

52. a  10, d  8, n  30. Then S30  30 2 [2  10  29 8]  3180.

53. a  40, d  14, n  15. Then S15  15 2 [2 40  14  14]  870. 54. a  2, d  23, n  25. Then S25  25 2 [2 2  24  23]  6850.

55. a  a1  4 and a3  2  4  3  1 d  d  3, so S15  15 2 [2 4  14 3]  255.

56. a3  45  a  3  1 d  45  a  2d  45 and a7  55  a  7  1 d  55  a  6d  55. Solving the   system, we have a  2d  a  6d  45  55  4d  10  d  52 and a  45  2 52  40. Therefore,   5  4900. 2  40  48  S49  49 2 2

57. 1  5  9      401 is a partial sum of an arithmetic series with a  1 and d  5  1  4. The last term is

401  an  1  4 n  1, so n  1  100  n  101. So the partial sum is S101  101 2 1  401  101  201  20,301.

58. 5  25  0  25      60 is a partial sum of an arithmetic sequence with a  5 and d  25  5  25. The last

65  n  1  n  27. So the partial sum is S  27 5  60  1485  7425. term is 60  an  5  25 n  1, so 25 27 2 2

59. 250  233  216      97 is a partial sum of an arithmetic sequence with a  250 and d  233  250  17. The last

10 term is 97  an  250  17 n  1, so n  1  97250 17  9  n  10. So the partial sum is S10  2 250  97  1735.

60. 89  85  81      13 is a partial sum of an arithmetic sequence with a  89 and d  85  89  4. The last term is 20 13  an  89  4 n  1, so n  1  1389 4  19  n  20. So the partial sum is S20  2 89  13  1020.

61. 07  27  47      567 is a partial sum of an arithmetic sequence with a  07 and d  27  07  2. The last term is 567  an  07  2 n  1  28  n  1  n  29. So the partial sum is S29  29 2 07  567  8323.

62. 10  99  98      01 is a partial sum of an arithmetic sequence with a  10 and d  01. The last term

is 01  an  10  01 n  1, so 99  n  1  n  100. So the partial sum is S100  100 2 10  01  505. 10 63. k0 3  025k is a partial sum of an arithmetic sequence with a  3  025  0  3 and d  025. The last term is a11  3  025  10  55. So the partial sum is S11  11 2 3  55  4675. 20 64. n0 1  2n is a partial sum of an arithmetic sequence where a  1  2  0  1, d  2, and the last term is a21  1  2  20  39. So the partial sum is S21  21 2 1  39  399.

65. We have an arithmetic sequence with a  5 and d  2. We seek n such that 2700  Sn 

n [2a  n  1 d]. Solving for 2

n [10  2 n  1]  5400  10n  2n 2  2n  n 2  4n  2700  0  n  50 n  54  0  2 n  50 or n  54. Since n is a positive integer, 50 terms of the sequence must be added to get 2700.

n, we have 2700 


10

CHAPTER 12 Sequences and Series

n [2  12  n  1  8]  4n 2  8n  2700  0  4 n  27 n  25  0. 2 Thus, 2700 is the sum of the first 25 terms of the sequence.

66. a  12 and d  8, so Sn  2700  2700 

67. Let x denote the length of the side between the length of the other two sides. Then the lengths of the three sides of the triangle are x  a, x, and x  a, for some a  0. Since x  a is the longest side, it is the hypotenuse, and by the Pythagorean Theorem, we know that x  a2  x 2  x  a2  x 2  2ax  a 2  x 2  x 2  2ax  a 2  x 2  4ax  0  x x  4a  0  x  4a (x  0 is not a possible solution). Thus, the lengths of the three sides are x  a  4a  a  3a, x  4a, and x  a  4a  a  5a. The lengths 3a, 4a, 5a are proportional to 3, 4, 5, and so the triangle is similar to a 3-4-5 triangle.

68. P  10110  10210  10310      101910  101231910 . Now, 1  2  3      19 is an arithmetic series with 1  19 a  1, d  1, and n  19. Thus, 1  2  3      19  S19  19  190, and so P  1019010  1019 . 2 69. The sequence 1, 35 , 37 , 13 ,    is harmonic if 1, 53 , 73 , 3,    forms an arithmetic sequence. Since 53  1  73  53  3  73  23 , the sequence of reciprocals is arithmetic and thus the original sequence is harmonic.

70. The two original numbers are 3 and 5. Thus, the reciprocals are 13 and 15 , and their average is     1 1  1  1 5  3  4 . Therefore, the harmonic mean is 15 . 2 3 5 2 15 15 15 4

71. The diminishing values of the computer form an arithmetic sequence with a1  12,500 and common difference d  1875. Thus the value of the computer after 6 years is a7  12,500  7  1 1875  $1250. 72. The number of poles in a layer can be viewed as an arithmetic sequence, where a1  25 and the common difference is 1. The number of poles in the first 12 layers is S12  12 2 [2 25  11 1]  6  39  234.

73. The increasing salary values form an arithmetic sequence with a1  45,000 and common difference d  2000. Then

his total earnings for a ten-year period are S10  10 2 [2 45,000  9 2000]  540,000. Thus, the total earnings for the

10-year period are $540,000.

74. The number of cars that can park in a row can be viewed as an arithmetic sequence, where a1  20 and the common difference is 2. Thus the number of cars that can park in the 21 rows is S21  21 2 [2 20  20 2]  105  80  840.

75. The number of seats in the nth row is given by the nth term of an arithmetic sequence with a1  15 and common n difference d  3. We need to find n such that Sn  870. So we solve 870  Sn  [2 15  n  1 3] for n. We have 2 n 870  27  3n  1740  3n 2  27n  3n 2  27n  1740  0  n 2  9n  580  0  x  20 x  29  0  2 n  20 or n  29. Since the number of rows is positive, the theater must have 20 rows. 76. The sequence is 16, 48, 80,   . This is an arithmetic sequence with a  16 and d  48  16  32. (a) The total distance after 6 seconds is S6  62 32  5  32  3  192  576 ft. (b) The total distance after n seconds is Sn  n2 [32  32 n  1]  16n 2 ft.

77. The number of gifts on the 12th day is 1  2  3  4      12. Since a2  a1  a3  a2  a4  a3      1, the number of gifts on the 12th day is the partial sum of an arithmetic sequence with a  1 and d  1. So the sum is   1  12  6  13  78. S12  12 2

78. (a) We want an arithmetic sequence with 4 terms, so let a1  10 and a4  18. Since the sequence is arithmetic,   8  46 are the a4  a1  3d  18  10  8  3d  8  d  83 . Therefore, a2  10  83  38 and a  10  2 3 3 3 3 two arithmetic means between 10 and 18.

(b) We want an arithmetic sequence with 5 terms, so let a1  10 and a5  18. Since the sequence is arithmetic, a5  a1  4d  18  10  8  4d  8  d  2. Therefore, a2  10  2  12, a3  10  2 2  14, and a4  10  3 2  16 are the three arithmetic means between 10 and 18.


SECTION 12.3 Geometric Sequences

11

(c) We want an arithmetic sequence with 6 terms, with the starting dosage a1  100 and the final dosage a6  300. Since the sequence is arithmetic, a6  a1  5d  300  100  200  5d  200  d  40. Therefore, a2  140, a3  180, a4  220, a5  260, and a6  300. The patient should take 140 mg, then 180 mg, then 220 mg, then 260 mg, and finally arrive at 300 mg.

12.3 GEOMETRIC SEQUENCES 1. A geometric sequence is a sequence where the ratio between successive terms is constant. 2. The sequence an  ar n1 is a geometric sequence where a is the first term and r is the common ratio. So, for the geometric sequence an  2 5n1 the first term is 2 and the common ratio is 5. 3. True. If we know the first and second terms of a geometric sequence then we can find all other terms. 4. (a) The nth partial sum of a geometric sequence an  ar n1 is given by Sn  a

1  rn . 1r

 k1  a  ar  ar 2  ar 3     is is an infinite geometric series. If r  1, then this series (b) The series  k1 ar converges and its sum is S  a 1  r. If r  1 the series diverges.

5. (a) a1  7 30  7, a2  7 31  21,

6. (a) a1  6 050  6, a2  6 051  3,

a3  7 32  63, a4  7 33  189,

a3  6 052  15, a4  6 053  075,

a5  7 34  567

a5  6 054  0375

(b) The common ratio is 3.

(b) The common ratio is 05.

(c)

(c)

an

an

600

6 4

400

2

200

0 _2

500 300

1

2

3

 2  3 a3  8  14  12 , a4  8  14   18 ,  4 1 a5  8  14  32

5 n

8. (a) a1   19 30   19 , a2   19 31   13 , a3   19 32  1, a4   19 33  3,

a5   19 34  9

(b) The common ratio is 3.

(b) The common ratio is  14 . an 5

4

3

5 n

4

 0  1 7. (a) a1  8  14  8, a2  8  14  2,

(c)

2

_4

100

0

1

(c)

an 0

1

_2

2

3

4

5 n

_10 _20

9. a  7, r  4. So an  ar n1  7 4n1 and a4  7  43  448.

10. a   32 , r  3. So an  ar n1   32 3n1 and a4   32 33   81 2.

1

2

3

4

5 n

_4 _6 _8 _10

This is the wrong art for #7


12

CHAPTER 12 Sequences and Series

11. a  5, r  3. So an  ar n1  5 3n1 and a4  5 33  135.     n1  n   3 12. a  3, r  3. So an  ar n1  3 3  3 and a4  3  3  9.

a a 6 12 24 a2  2, and 4   2. Since these ratios are the same, the sequence is geometric with common   2, 3  a1 3 a2 6 a3 12 ratio 2. a 31 a 48 93  16 and 3   . Since these ratios are not the same, this is not a geometric sequence. 14. 2  a1 3 a2 48 16 3 a 4 12 23 a  and 3   . Since these ratios are not the same, this is not a geometric sequence. 15. 2  a1 13 2 a2 12 3 1 a 1 a 1 144 48 16 a  , 3    , and 4    . Since these ratios are the same, the sequence is 16. 2  a1 432 3 a2 144 3 a3 48 3 13.

geometric with common ratio  13 .

17.

a2 32 34 38 1 a 1 a 1   , 3   , and 4   . Since these ratios are the same, the sequence is geometric with a1 3 2 a2 32 2 a3 34 2

common ratio 12 . 18.

a2 103 109 1027 1 a 1 a 1   , 3   , and 4   . Since these ratios are the same, the sequence is geometric with a1 10 3 a2 103 3 a3 109 3 common ratio 13 .

19. 20.

a2 a 14 1 12  1 and 3    . Since these ratios are not the same, this is not a geometric sequence.  a1 12 a2 12 2

a2 e4 a e6 a e8  2  e2 , 3  4  e2 , and 4  6  e2 . Since these ratios are the same, the sequence is geometric with a1 a2 a3 e e e

common ratio e2 . a a 11 121 1331 a  11, 3   11, and 4   11. Since these ratios are the same, the sequence is geometric 21. 2  a1 10 a2 11 a3 121 with common ratio 11. 22.

1

1

2

6

a 1 3 a2  41  and 4  81  . Since these ratios are not the same, this is not a geometric sequence. a1 2 a3 4

23. a1  2 31  6, a2  2 32  18, a3  2 33  54, a4  2 34  162, and a5  2 35  486. This sequence is geometric, the common ratio is r  3, and an  a1 r n1  6 3n1 .

24. a1  4  31  7, a2  4  32  13, a3  4  33  31, a4  4  34  85, and a5  4  35  247. Since a3 a2 31  13 7 and a  13 are different ratios, this is not a geometric sequence. a1 2 1 1 1 1 1 1 1 1 1 , a3  3  , a4  4  , and a5  5  . This sequence is geometric, the 25. a1  , a2  2  4 16 64 256 1024 4 4 4 4  n1 . common ratio is r  14 and an  a1r n1  14 14 26. a1  11 21  2, a2  12 22  4, a3  13 23  8, a4  14 24  16, and

a5  15 25  32. This sequence is geometric, the common ratio is r  2, and an  a1r n1  2 2n1 .         27. Since ln a b  b ln a, we have a1  ln 50  ln 1  0, a2  ln 51  ln 5, a3  ln 52  2 ln 5, a4  ln 53  3 ln 5,   a5  ln 54  4 ln 5. Since a1  0 and a2  0, this sequence is not geometric. a a 4 27 are 28. a1  11  1, a2  22  4, a3  33  27, a4  44  256, and a5  55  3125. Since 2   4 and 3  a1 1 a2 4 different, this is not a geometric sequence.


SECTION 12.3 Geometric Sequences

13

a 29. 2, 6, 18, 54,   . Then r  2  62  3, a5  a4  3  54 3  162, and an  2  3n1 . a1 14  n1 a2 3 28 56 2 2 56 2 112 2 30. 7, 14 . 3 , 9 , 27 ,   . Then r  a  7  3 , a5  a4  3  27  3  81 , and an  7 3 1 009 a  03, a5  a4  03  00081 03  000243, and 31. 03, 009, 0027, 00081,   . Then r  2  a1 03 an  03 03n1 .

  n1      a 2   2, a5  a4  2  2 2  2  4, and an  2, 2, 2 2,   . Then r  2  2 . a1 1       1 ,   . Then r  a2  12   1 , a  a   1   1  1  1 , a  144  1 n1 . 33. 144, 12, 1,  12 n 5 4 12 12 12 12 144 12 a1 144   a 2 1 2 1 , and a  8 1 n1 . 34. 8, 2,  12 ,  18 ,   . Then r   14 , a5  a4    18  14   32  n 4 a1 8 4   a 353 35. 3, 353 , 373 , 27,   . Then r  2   323 , a5  a4  323  27  323  3113 , and a1 3 n1   3  32n23  32n13 . an  3 323

32. 1,

t2  n1 a2 t t4 t t5 t t t2 t3 t4   , and an  t  2  , a5  a4   . 36. t, , , ,   . Then r  2 4 8 a1 t 2 2 8 2 16 2

n1  a s 27 37. 1, s 27 , s 47 , s 67 ,   . Then r  2   s 2n27 .  s 27 , a5  a4 s 27  s 67 s 27  s 87 , and an  s 27 a1 1 38. 5, 5c1 , 52c1 , 53c1 ,   . Then r   n1  5  5cnc  5cnc1 . an  5 5c

a2 5c1  5c , a5  a4  5c  53c1  5c  54c1 , and  a1 5

 3 2 a 4 39. a1  14, a2  4. Thus r  2   and a4  a1r 41  14 27  16 49 . a1 14 7  4 a 3 40. a1  8, a2  6. Thus r  2  , so a5  a1r 51  8 34  81 32 . a1 4 41. a3  

1 a a r5 9 and a6  9. Thus, 6  1 2  r 3  r 3   27, so r  3. Therefore, a1 32   13  3 a3 13 a1r

1 and a   1 3  1 . a1   27 2 27 9

a7 8 2 81 32 329 a 12 . Thus,  , so r  . Therefore, a4  r 3 a1  a1  34    r3  r3  9 a4 12 27 3 827 2 r  n1 2 81 . and the nth term is an  a1r n  2 3

42. a4  12 and a7 

9 a 9216 a 18  512  r  8. Therefore, a1  23   and 43. a3  18 and a6  9216. Thus, r 3  6  a3 18 64 32 r 9 an   8n1 . 32 27 3 a6 729256 a 54 3 44. a3  54 and a6  729   r   . Thus, a1  23   384 and 256 . Thus, r  a  54 512 8 r 382 3   a2  384  38  144. a 729  1728, a2  1728 075  1296, and a3  1296 075  972. 45. r  075 and a4  729, so a1  34  r 0753


14

CHAPTER 12 Sequences and Series

 6 a 18 1 . 46. r  16 and a3  18, so a1  23   648 and a7  648 16  2 72 r 16  n1 47. a  1536 and r  12 , so an  ar n1  6  1536 12  1536 21n  log2 6  log2 1536  1  n  n  1  log2 1536 6  1  log2 256  9. Thus, 6 is the ninth term.

n5  468,750  48. a2  30 and a5  3750, so r 3  3750 30  125 and r  5. Thus, an  468,750  3750  5

,750 n  5  log5 468 3750  log5 125  n  8. Therefore, the eighth term of the sequence is 468,750.

1  26  5 63  315. 12     1 1 4   80 2 3 2 81  80 . 50. a  23 , r  13 , n  4. Then S4   3 81 2 3 1 1

49. a  5, r  2, n  6. Then S6  5

3

3

a ar 5 a 2 51. a3  28, a6  224, n  6. So 6  2  r 3 . So we have r 3  6  224 28  8, and hence r  2. Since a3  a  r , we a3 a3 ar a 28 1  26  7 63  441. get a  23  2  7. So S6  7 12 r 2 012 a 000096 a 52. a2  012, a5  000096, n  4. So r 3  5   0008  r  02, and thus a1  2   06. a2 012 r 02 Therefore, S4  06

1  024  07488. 1  02

a 53. 1  3  9      2187 is a partial sum of a geometric sequence, where a  1 and r  2  31  3. Then the last term is a1 2187  an  1  3n1  n  1  log3 2187  7  n  8. So the partial sum is S8  1

1  38  3280. 13 1

1 is a partial sum of a geometric sequence for which a  1 and r  a2   2   1 . The last 54. 1  12  14  18      512 2 a1 1  10  n1 1   12 1 1    341 term is an , where an   1 2 , so n  10. So the partial sum is S10  1 512 . 512 1  1 2

30 a  2. The 55. 15  30  60      960 is a partial sum of a geometric sequence for which a  15 and r  2  a1 15 last term is an  960  15 2n1  n  7, so the partial sum is S7  15

1  27  645. 1  2

a 1 2560 56. 5120  2560  1280      20 is a partial sum of a geometric sequence for which a  5120 and r  2   . a1 5120 2  9  n1 1  12  n  9, so the partial sum is S9  5120  10,220. The last term is an  20  5120 12 1  12 a 57. 125  125  125      12,500,000 is a partial sum of a geometric sequence for which a  125 and r  2  10. The a1 1  108  13,888,88875. 1  10 a 1 58. 10,800  1080  108      0000108 is a partial sum of a geometric sequence for which a  10,800 and r  2  . a1 10  9 1  n1 1  10 1 The last term is an  0000108  10,800 10  n  9, so the partial sum is S9  10,800 11,999999988. 1 1  10 last term is an  12,500,000  125  10n1  n  8, so the partial sum is S8  125


SECTION 12.3 Geometric Sequences

59.

5 

k1

 k1

3 12

3

 5 1  12 1  12

 5  k1 1   32 55    60. 8  32 8 3 2 k1 1  2 5 

93  16

1  26  105 1  2 k1  5  k1 1  23 5  211 63. 3 23 3  2 27 1 k1

61.

6 

15

1  56  39,060 15 k1  6  k1 1  32 6     1330 64. 64 32  64 k1 1  32

5 2k1  5

62.

3

6 

10  5k1  10

1     is an infinite geometric series with a  1 and r  1 . Therefore, it is convergent with sum 65. 1  13  19  27 3

S

3 a 1    .  1 1r 2 1 3

66. 1  12  14  18     is an infinite geometric series with a  1 and r   12 . Therefore, it is convergent with sum S

2 1 1     . 3 1 3 1  2 2

1     is an infinite geometric series with a  1 and r   1 . Therefore, it is convergent with sum 67. 1  13  19  27 3

S

3 a 1   .  1 1r 4 1  3

4  8     is an infinite geometric series with a  2 and r  2 . Therefore, it is convergent with sum 68. 25  25 125 5 5 2 2 2 S  5 2  53  . 3 1 5 5  2  3 69. 1  32  32  32     is an infinite geometric series with a  1 and r  32  1. Therefore, the series diverges.

70.

1 1 1 1 1 1 1  8  10  12     is an infinite geometric series with a  6 and r  2  . Therefore, it is convergent with 6 9 3 3 3 3 3 3 1

sum S 

1 9 1 a 6 3    .   6 1 1r 648 3 8 1 9

71. 3  32  34  38     is an infinite geometric series with a  3 and r   12 . Therefore, it is convergent with sum S

3    2. 1   12

72. 1  1  1  1     is an infinite geometric series with a  1 and r  1. Because r  1  1  1, the series diverges.

73. 3  3 11  3 112  3 113     is an infinite geometric series with a  3 and r  11  1. Therefore, the series diverges. 10 3 74.  100 9  3  1  10     is an infinite geometric series with a  

 100  100 a 9 10 1000 9   with sum S     100 9  13   117 . 13 1r 1  3 10

10

100 3 3 and r  100 . Therefore, it is convergent  9 10  9

10

1  1     is an infinite geometric series with a  1 and r  1 . Therefore, the sum of the series is 75. 1  12   4 2 2 2 2 2 1   1 2  2  1.  S 1 21 1  2


16

CHAPTER 12 Sequences and Series

    76. 1  2  2  2 2  4     is an infinite geometric series with a  1 and r   2. Because r  2  1, the series diverges. 9  9  9     is an infinite geometric series with a  9 and r  1 . Thus 77. 0999     10 100 1000 10 10 9

0999    

a  10 1  1. 1r 1 10

53  53 53 53 78. 02535353     02  1000 100,000  10,000,000     is an infinite geometric series (after the first term) with a  1000

53 1000  53  100  53 , and so 02535353     2  53  2  99  53  251 . 1000 99 990 1 10 990 990 990 1  100 3  3 3 3 1 79. 0030303     100 10,000  1,000,000     is an infinite geometric series with a  100 and r  100 . Thus 3 1 3 a  1001   . 0030303     1r 99 33 1  100 25 25 80. 211252525     211  1025 ,000  1,000,000  100,000,000     is an infinite geometric series (after 25 25 10,000 1 , and so  the first term) with a  1025 ,000 and r  100 . Thus 000252525     1 9900 1  100 1 . Thus 005353     and r  100

211252525    

25 211  99  25 20,914 10,457 211     . 100 9900 9900 9900 4950

112  112 112 112 81. 0112  0112112112     1000 1,000,000  1,000,000,000     is an infinite geometric series with a  1000 and 1 . Thus 0112112112     r  1000

112

a  10001 1r 1

1000

112 . 999

123  123 123 123 1 82. 0123123123     1000 1,000,000  1,000,000,000     is an infinite geometric series with a  1000 and r  1000 .

Thus 0123123123    

123 123 41 1000   . 1 999 333 1  1000

a 83. Since we have 5 terms, let us denote a1  5 and a5  80. Also, 5  r 4 because the sequence is geometric, and so a1 r 4  80 5  16  r  2 . If r  2, the three geometric means are a2  10, a3  20, and a4  40. (If r  2, the three geometric means are a2  10, a3  20, and a4  40, but these are not between 5 and 80.)

84. The sum is given by         a  b  a 2  2b  a 3  3b      a 10  10b  a  a 2  a 3      a 10  b  2b  3b      10b  a

1  a 10 10 1  a 10   b  10b  a   55b 1a 2 1a

a a3 85. (a) 5 3 5 3    is neither arithmetic nor geometric because a2  a1  a3  a2 and 2   . a1 a2 (b) 13  1 53  73     is arithmetic with d  23 , so the next term is a5  73  23  3.     (c) 3 3 3 3 9    is geometric with r  3, so the next term is a5  9 3.

(d) 3  32  0 32     is arithmetic with d  32 , so the next term is a5  32  32  3.

86. (a) 1 1 1 1    is geometric with r  1, so the next term is a5  1.    1 (b) 5 3 5 6 5 1    is geometric with r  516 , so a5  1  516   . 6 5

a a3 (c) 2 1 12  2 is neither arithmetic nor geometric because a2  a1  a3  a2 and 2   . a1 a2


SECTION 12.3 Geometric Sequences

17

(d) x  1 x x  1 x  2 is arithmetic with d  1, so a5  x  2  1  x  3.

87. (a) The value at the end of the year is equal to the value at beginning less the depreciation, so Vn  Vn1  02Vn1  08Vn1 with V1  160,000. Thus Vn  160,000  08n1 .

(b) Vn  100,000  08n1  160,000  100,000  08n1  0625  n  1 log 08  log 0625  log 0625  211. Thus it will depreciate to below $100,000 during the fourth year. n1 log 08 88. Let an denote the number of ancestors a person has n generations back. Then a1  2, a2  4, a3  8,   . Since 4  8      2, this is a geometric sequence with r  2. Therefore, a  2  214  215  32,768. 15 2 4

89. Since the ball is dropped from a height of 80 feet, a  80 . Also since the ball rebounds three-fourths of the distance  n fallen, r  34 . So on the nth bounce, the ball attains a height of an  80 34 . Hence, on the fifth bounce, the ball goes  5 a5  80 34  80243 1024  19 ft high.

90. a  5000, r  108. After 1 hour, there are 5000  108  5400, after 2 hours, 5400  108  5832, after 3 hours, 5832  108  629856, after 4 hours, 629856  108  68024448, and after 5 hours, 68024448  108  7347 bacteria. After n hours, the number of bacteria is an  5000  108n .

91. Let an be the amount of water remaining at the nth stage. We start with 5 gallons, so a  5. When 1 gallon (that is, 15 of the mixture) is removed, 45 of the mixture (and hence 45 of the water in the mixture) remains. Thus, a1  5  45  a2  5  45  45    ,  n  3 and in general, an  5 45 . The amount of water remaining after 3 repetitions is a3  5 45  64 25 , and after  5 5 repetitions it is a5  5 45  1024 625 .

92. Let aC denote the term of the geometric series that is the frequency of middle C. Then aC  256 and aC1  512. Since aC 256 this is a geometric sequence, r  512 256  2, and so aC2  r 2  22  64. 93. Let an be the height the ball reaches on the nth bounce. From the given information, an is the geometric sequence  n an  9  13 . (Notice that the ball hits the ground for the fifth time after the fourth bounce.)  2  3  4 (a) a0  9, a1  9  13  3, a2  9  13  1, a3  9  13  13 , and a4  9  13  19 . The total distance traveled is 8 a0  2a1  2a2  2a3  2a4  9  2  3  2  1  2  13  2  19  161 9  17 9 ft.

(b) The total distance traveled at the instant the ball hits the ground for the nth time is  2  3  4  n1 Dn  9  2  9  13  2  9  13  2  9  13  2  9  13      2  9  13   2  3  4  n1  9  2 9  9  13  9  13  9  13  9  13      9  13  n   n3   n  1  13    2 9   9  18  13   9  27 1  13 1 1 3 

94. The annual salaries form a geometric sequence with a  45,000 and r  105. Then Sn  45,000

1  105n . At the end of 1  105

1  10510  566,00516, or $566,00516. 1  105 95. Let a1  1 be the man with 7 wives. Also, let a2  7 (the wives), a3  7a2  72 (the sacks), a4  7a3  73 (the cats), 10 years, the total earnings are S10  45,000

and a5  7a4  74 (the kits). The total is a1  a2  a3  a4  a5  1  7  72  73  74 , which is a partial sum of a geometric sequence with a  1 and r  7. Thus, the number in the party is S5  1 

1  75  2801. 17


18

CHAPTER 12 Sequences and Series

96. (a) (b)

10

 k1

1 k1 50 2

 50 

 k1 1  k1 50 2



 10 1  12

50 1  12

1  12

 100 1  00009765  999023 mg

 100 mg

97. Let an be the height the ball reaches on the nth bounce. We have a0  1 and an  12 an1 . Since the total distance d traveled includes the bounce up as well and the distance down, we have  2    3  4 d  a0  2  a1  2  a2      1  2 12  2 12  2 12  2 12       i  2  3  1  1  1  12  12  12      1  1 2 i0

1

1  12

3

Thus the total distance traveled is about 3 m.

 2 98. The time required for the ball to stop bouncing is t  1  1  1     which is an infinite geometric series with 2 2      2 2 1 2 2 21     2  2. Thus the     a  1 and r  1 . The sum of this series is t  1 2 21 21 21 21 1  2  time required for the ball to stop is 2  2 341 s. 99. (a) If a square has side x, then by the Pythagorean Theorem the length of the side of the square formed by    x2 x x2 x 2  x 2    . In our case, x  1 and the side of the   joining the midpoints is, 2 2 4 4 2  2 first inscribed square is 1 , the side of the second inscribed square is 1  1  1 , the side of the 2 2 2 2  3 third inscribed square is 1 , and so on. Since this pattern continues, the total area of all the squares is 2  2  4  6  2  3 1 1 1 2    1      1  12  12  12       2. A1   2 2 2 1  12  2  3 (b) As in part (a), the sides of the squares are 1, 1 , 1 , 1 ,   . Thus the sum of the perimeters is 2 2 2  2  3 1 1 1 4      , which is an infinite geometric series with a  4 and r  1 . Thus S  414  4  2 2 2 2      4 2 424 2 4 2 21 4     8  4 2.  the sum of the perimeters is S  21 21 21 21 1  1 2

 2 2 100. Let An be the area of the disks of paper placed at the nth stage. Then A1   R 2 , A2  2   12 R   2R ,  2 1 1 2 2 2 2 A3  4   14 R   4 R ,   . We see from this pattern that the total area is A   R  2  R  4  R     . Thus, the total area, A, is an infinite geometric series with a1   R 2 and r  12 . So A 

 R2

1  12

 2 R 2 .


SECTION 12.3 Geometric Sequences

19

101. Let an denote the area colored blue at nth stage. Since only the middle squares are colored blue, an  19  area remaining yellow at the n  1 th stage. Also, the area remaining yellow at the nth stage is 89 of the area    2  3 remaining yellow at the preceding stage. So a1  19 , a2  19 89 , a3  19 89 , a4  19 89 ,   . Thus the total area    2  3 colored blue A  19  19 89  19 89  19 89     is an infinite geometric series with a  19 and r  89 . So the total area is A 

1 9

1  89

 1.

102. Let a1 , a2 , a3 ,    be a geometric sequence with common ratio r. Thus a2  a1 r, a3  a1  r 2 ,   , an  a1  r n1 . Hence,    2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 n1 1 1    ,      ,        , and so a2 a1  r a1 r a3 a1 r 2 a1 r an a1 r n1 a1 r a1  r 2 a1  r n1 1 1 1 1 , , ,    is a geometric sequence with common ratio . a1 a2 a3 r 103. a1  a2  a3     is a geometric sequence with common ratio r . Thus a2  a1 r, a3  a1  r 2 ,   , an  a1  r n1 .     Hence log a2  log a1r  log a1  log r, log a3  log a1  r 2  log a1  log r 2  log a1  2 log r,   ,     log an  log a1  r n1  log a1  log r n1  log a1  n  1 log r , and so log a1  log a2  log a3     is an arithmetic sequence with common difference log r.

104. Since a1  a2  a3     is an arithmetic sequence with common difference d, the terms can be expressed as a2  a1  d, 2  a3  a1  2d,   , an  a1  n  1 d. So 10a2  10a1 d  10a1  10d , 10a3  10a1 2d  10a1  10d ,   , n1  , and so 10a1  10a2  10a3     is a geometric sequence with common ratio r  10d . 10an  10a1 n1d  10a1  10d 105. By the partial sum formula with n  2k , we write the left-hand side of the equation to be proved as k

k

1  r  r 2      r 2 1  S2k  By the same formula with n  2k1 , the right-hand side is

1  r2 1r

   k1  k1    1  r 2k1  k1 1 k1 2 2 2 2 1 r r  r r 1  r2  1  S2k1 r 1  1r  k1 2 1  r2  1r k

 These are equal, and so we are done.

1  r2 1r


20

CHAPTER 12 Sequences and Series

12.4 MATHEMATICAL INDUCTION 1. Mathematical induction is a method of proving that a statement P n is true for all natural numbers n. In Step 1 we prove that P 1 is true.

2. (b) (ii) We prove “If P kis true then P k  1 is true.” 3. Let P n denote the statement 2  4  6      2n  n n  1. Step 1: P 1 is the statement that 2  1 1  1, which is true. Step 2: Assume that P k is true; that is, 2  4  6      2k  k k  1. We want to use this to show that P k  1 is true. Now 2  4  6      2k  2 k  1  k k  1  2 k  1

induction hypothesis

 k  1 k  2  k  1 [k  1  1]

Thus, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

4. Let P n denote the statement 1  4  7  10      3n  2 

n 3n  1 . 2

1 [3 1  1] 12  , which is true. 2 2 k 3k  1 Step 2: Assume that P k is true; that is. 1  4  7      3k  2  . We want to use this to show that 2 P k  1 is true. Now k 3k  1  3k  1 induction hypothesis 1  4  7  10      3k  2  [3 k  1  2]  2 3k 2  k  6k  2 3k 2  5k  2 k 3k  1 6k  2 k  1 3k  2 k  1 [3 k  1  1]       2 2 2 2 2 2 Thus, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. Step 1: P 1 is the statement that 1 

n 3n  7 . 2 1  3  1  7 Step 1: We need to show that P 1 is true. But P 1 says that 5  , which is true. 2 k 3k  7 Step 2: Assume that P k is true; that is, 5  8  11      3k  2  . We want to use this to show that 2 P k  1 is true. Now

5. Let P n denote the statement 5  8  11      3n  2 

5  8  11      3k  2  [3 k  1  2] 

k 3k  7  3k  5 2

6k  10 3k 2  13k  10 3k 2  7k   2 2 2 k  1 [3 k  1  7] 3k  10 k  1   2 2 

Thus, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

induction hypothesis


SECTION 12.4 Mathematical Induction

6. Let P n denote the statement 12  22  32      n 2  Step 1: P 1 is the statement that 12 

21

n n  1 2n  1 . 6

123 , which is true. 6

Step 2: Assume that P k is true; that is, 12  22  32      k 2 

k k  1 2k  1 . 6

We want to use this to show that P k  1 is true. Now,

k k  1 2k  1  k  12 induction hypothesis 6     2k 2  k  6k  6 k 2k  1  6 k  1  k  1  k  1 6 6   2k 2  7k  6 k  1 k  2 2k  3   k  1 6 6

12  22  32      k 2  k  12 

k  1 [k  1  1] [2 k  1  1]  6

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 7. Let P n denote the statement 1  2  2  3  3  4      n n  1  Step 1: P 1 is the statement that 1  2 

n n  1 n  2 . 3

1  1  1  1  2 , which is true. 3

Step 2: Assume that P k is true; that is, 1  2  2  3  3  4      k k  1 

k k  1 k  2 . We want to use this to 3

show that P k  1 is true. Now 1  2  2  3  3  4      k k  1  k  1 [k  1  1]

k k  1 k  2  k  1 k  2 3 k k  1 k  2 3 k  1 k  2 k  1 k  2 k  3    3 3 3 

induction hypothesis

Thus, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 8. Let P n denote the statement 1  3  2  4  3  5      n n  2  Step 1: P 1 is the statement that 1  3 

n n  1 2n  7 . 6

129 , which is true. 6

Step 2: Assume that P k is true; that is, 1  3  2  4  3  5      k k  2 

k k  1 2k  7 . We want to use this to 6

show that P k  1 is true. Now 1  3  2  4  3  5      k k  2  k  1 [k  1  2]

k k  1 2k  7  k  1 k  3 6     2k 2  7k  6k  18 k 2k  7 6 k  3   k  1  k  1 6 6 6 

k  1 [k  1  1] [2 k  1  7] 6

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

induction hypothesis


22

CHAPTER 12 Sequences and Series

9. Let P n denote the statement 13  23  33      n 3 

n 2 n  12 . 4

12  1  12 , which is clearly true. 4 k 2 k  12 Step 2: Assume that P k is true; that is, 13  23  33      k 3  . We want to use this to show that 4 P k  1 is true. Now

Step 1: P 1 is the statement that 13 

13  23  33      k 3  k  13   

k 2 k  12  k  13 4   k  12 k 2  4 k  1 4

k  12 k  22

induction hypothesis

  k  12 k 2  4k  4 4

k  12 [k  1  1]2

 4 4 Thus, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

  10. Let P n denote the statement 13  33  53      2n  13  n 2 2n 2  1 .   Step 1: P 1 is the statement that 13  12 2  12  1 , which is clearly true.   Step 2: Assume that P k is true; that is, 13  33  53      2k  13  k 2 2k 2  1 . We want to use this to show that P k  1 is true. Now

  13  33  53      2k  13  2k  13  k 2 2k 2  1  2k  13

induction hypothesis

 2k 4  k 2  8k 3  12k 2  6k  1  2k 4  8k 3  11k 2  6k  1       k 2  2k  1 2k 2  4k  1  k  12 2 k  12  1

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

11. Let P n denote the statement 23  43  63      2n3  2n 2 n  12 . Step 1: P 1 is true since 23  2 12 1  12  2  4  8. Step 2: Assume that P k is true; that is, 23  43  63      2k3  2k 2 k  12 . We want to use this to show that P k  1 is true. Now 23  43  63      2k3  [2 k  1]3  2k 2 k  12  [2 k  1]3

induction hypothesis

   2k 2 k  12  8 k  1 k  12  k  12 2k 2  8k  8  2 k  12 k  22  2 k  12 [k  1  1]2

Thus, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.


SECTION 12.4 Mathematical Induction

23

1 1 1 n 1       . 12 23 34 n n  1 n1 1  12 , which is clearly true. Step 1: P 1 is the statement that 12 1 1 1 1 k Step 2: Assume that P k is true; that is     . We want to use this to show that 12 23 34 k k  1 k 1 P k  1 is true. Now

12. Let P n denote the statement

1 1 1 1 k 1        12 23 k k  1 k  1 k  2 k  1 k  1 k  2  

induction hypothesis

k 2  2k  1 k k  2  1 k  12   k  1 k  2 k  1 k  2 k  1 k  2 k 1 k 1  k 2 k  1  1

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.   13. Let P n denote the statement 1  2  2  22  3  23  4  24      n  2n  2 1  n  1 2n . Step 1: P 1 is the statement that 1  2  2 [1  0], which is clearly true.

  Step 2: Assume that P k is true; that is, 1  2  2  22  3  23  4  24      k  2k  2 1  k  1 2k . We want to use

this to show that P k  1 is true. Now

1  2  2  22  3  23  4  24      k  2k  k  1  2k1    2 1  k  1 2k  k  1  2k1

induction hypothesis

     2 1  k  1  2k  k  1  2k  2 1  2k  2k      2 1  k  2k1  2 1  [k  1  1] 2k1

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 14. Let P n denote the statement 1  2  22      2n1  2n  1.

Step 1: P 1 is the statement that 1  21  1, which is clearly true.

Step 2: Assume that P k is true; that is, 1  2  22      2k1  2k  1. We want to use this to show that P k  1 is true. Now

1  2  22      2k1  2k  2k  1  2k

induction hypothesis

 2  2k  1  2k1  1

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 15. Let P n denote the statement n 2  n is divisible by 2.

Step 1: P 1 is the statement that 12  1  2 is divisible by 2, which is clearly true.

Step 2: Assume that P k is true; that is, k 2  k is divisible by 2. Now     k  12  k  1  k 2  2k  1  k  1  k 2  k  2k  2  k 2  k  2 k  1. By the induction hypothesis, k 2  k is divisible by 2, and clearly 2 k  1 is divisible by 2. Thus, the sum is divisible by 2, so P k  1 is true. Therefore, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.


24

CHAPTER 12 Sequences and Series

16. Let P n denote the statement that 5n  1 is divisible by 4.

Step 1: P 1 is the statement that 51  1  4 is divisible by 4, which is clearly true.

Step 2: Assume that P k is true; that is, 5k  1 is divisible by 4. We want to use this to show that P k  1 is true. Now,     5k1  1  5  5k  1  5  5k  5  4  5 5k  1  4 which is divisible by 4 since 5 5k  1 is divisible by 4 by the induction hypothesis. Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

17. Let P n denote the statement that n 2  n  41 is odd.

Step 1: P 1 is the statement that 12  1  41  41 is odd, which is clearly true.

Step 2: Assume that P k is true; that is, k 2  k  41 is odd. We want to use this to show that P k  1 is true. Now,   k  12  k  1  41  k 2  2k  1  k  1  41  k 2  k  41  2k, which is also odd because k 2  k  41 is odd by the induction hypothesis, 2k is always even, and an odd number plus an even number is always odd. Therefore, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n. 18. Let P n denote the statement that n 3  n  3 is divisible by 3.

Step 1: P 1 is the statement that 13  1  3  3 is divisible by 3, which is true.

Step 2: Assume that P k is true; that is, k 3  k  3 is divisible by 3. We want to use this to show that P k  1 is true.     Now, k  13  k  1  3  k 3  3k 2  3k  1  k  1  3  k 3  k  3  3k 2  3k  k 3  k  3  3 k 2  k ,   which is divisible by 3, since k 3  k  3 is divisible by 3 by the induction hypothesis, and 3 k 2  k is divisible by 3. Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

19. Let P n denote the statement that 8n  3n is divisible by 5.

Step 1: P 1 is the statement that 81  31  5 is divisible by 5, which is clearly true.

Step 2: Assume that P k is true; that is, 8k  3k is divisible by 5. We want to use this to show that P k  1 is true. Now,   8k1  3k1  8  8k  3  3k  8  8k  8  5  3k  8  8k  3k  5  3k , which is divisible by 5 because 8k  3k

is divisible by 5 by our induction hypothesis, and 5  3k is divisible by 5. Thus P k  1 follows from P k. So by the

Principle of Mathematical Induction, P n is true for all n.

20. Let P n denote the statement that 32n  1 is divisible by 8.

Step 1: P 1 is the statement that 32  1  8 is divisible by 8, which is clearly true.

Step 2: Assume that P k is true; that is, 32k  1 is divisible by 8. We want to use this to show that P k  1 is true. Now,   32k1  1  9  32k  1  9  32k  9  8  9 32k  1  8, which is divisible by 8, since 32k  1 is divisible by 8 by the induction hypothesis. Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true

for all n.

21. Let P n denote the statement n  2n . Step 1: P 1 is the statement that 1  21  2, which is clearly true.

Step 2: Assume that P k is true; that is, k  2k . We want to use this to show that P k  1 is true. Adding 1 to both sides

of P k we have k  1  2k  1. Since 1  2k for k  1, we have 2k  1  2k  2k  2  2k  2k1 . Thus k  1  2k1 ,

which is exactly P k  1. Therefore, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.


SECTION 12.4 Mathematical Induction

25

22. Let P n denote the statement n  12  2n 2 , for all n  3.

Step 1: P 3 is the statement that 3  12  2  32 or 16  18, which is true.

Step 2: Assume that P k is true; that is, k  12  2k 2 , k  3. We want to use this to show that P k  1 is true. Now   k  22  k 2  4k  4  k 2  2k  1  2k  3  k  12  2k  3  2k 2  2k  3

induction hypothesis

 2k 2  2k  3  2k  1

because 2k  1  0 for k  3

 2k 2  4k  2  2 k  12 Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n  3. 23. Let P n denote the statement 1  xn  1  nx, if x  1.

Step 1: P 1 is the statement that 1  x1  1  1x, which is clearly true.

Step 2: Assume that P k is true; that is, 1  xk  1  kx. Now, 1  xk1  1  x 1  xk  1  x 1  kx,

by the induction hypothesis. Since 1  x 1  kx  1  k  1 x  kx 2  1  k  1 x (since kx 2  0), we have

1  xk1  1  k  1 x, which is P k  1. Thus P k  1 follows from P k. So the Principle of Mathematical Induction, P n is true for all n. 24. Let P n denote the statement 100n  n 2 , for all n  100.

Step 1: P 100 is the statement that 100 100  1002 , which is true.

Step 2: Assume that P k is true; that is, 100k  k 2 . We want to use this to show that P k  1 is true. Now 100 k  1  100k  100  k 2  100  k 2  2k  1  k  12

induction hypothesis because 2k  1  100 for k  100

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n  100. 25. Let P n be the statement that an  5  3n1 .

Step 1: P 1 is the statement that a1  5  30  5, which is true.

Step 2: Assume that P k is true; that is, ak  5  3k1 . We want to use this to show that P k  1 is true. Now,     ak1  3ak  3  5  3k1 , by the induction hypothesis. Therefore, ak1  3  5  3k1  5  3k , which is exactly P k  1. Thus, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

26. an1  3an  8 and a1  4. Then a2  3  4  8  4, a3  3  4  8  4, a4  3  4  8  4,   , and the conjecture is that an  4. Let P n denote the statement that an  4. Step 1: P 1 is the statement that a1  4, which is true. Step 2: Assume that P k is true; that is, ak  4. We want to use this to show that P k  1 is true. Now, ak1  3  ak  8  3  4  8  4, by the induction hypothesis. This is exactly P k  1, so by the Principle of Mathematical Induction, P n is true for all n. 27. Let P n be the statement that x  y is a factor of x n  y n for all natural numbers n. Step 1: P 1 is the statement that x  y is a factor of x 1  y 1 , which is clearly true.

Step 2: Assume that P k is true; that is, x  y is a factor of x k  y k . We want to use this to show that P k  1 is true.   Now, x k1  y k1  x k1  x k y  x k y  y k1  x k x  y  x k  y k y, for which x  y is a factor because x  y   is a factor of x k x  y, and x  y is a factor of x k  y k y, by the induction hypothesis. Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.


26

CHAPTER 12 Sequences and Series

28. Let P n be the statement that x  y is a factor of x 2n1  y 2n1 .

Step 1: P 1 is the statement that x  y is a factor of x 1  y 1 , which is clearly true.

Step 2: Assume that P k is true; that is, x  y is a factor of x 2k1  y 2k1 . We want to use this to show that P k  1 is true. Now x 2k11  y 2k11  x 2k1  y 2k1  x 2k1  x 2k1 y 2  x 2k1 y 2  y 2k1      x 2k1 x 2  y 2  x 2k1  y 2k1 y 2

for which x  y is a factor. This is because x  y is a factor of x 2  y 2  x  y x  y and x  y is a factor of

x 2k1  y 2k1 by our induction hypothesis. Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n .

29. Let P n denote the statement that F3n is even for all natural numbers n. Step 1: P 1 is the statement that F3 is even. Since F3  F2  F1  1  1  2, this statement is true. Step 2: Assume that P k is true; that is, F3k is even. We want to use this to show that P k  1 is true. Now, F3k1  F3k3  F3k2  F3k1  F3k1  F3k  F3k1  F3k  2  F3k1 , which is even because F3k is even by the induction hypothesis, and 2  F3k1 is even. Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

30. Let P n denote the statement that F1  F2  F3      Fn  Fn2  1. Step 1: P 1 is the statement that F1  F3  1. But F1  1  2  1  F3  1, which is true. Step 2: Assume that P k is true; that is, F1  F2  F3      Fk  Fk2  1. We want to use this to show that P k  1   is true. Now Fk12  1  Fk3  1  Fk2  Fk1  1  Fk2  1  Fk1  F1  F2  F3      Fk  Fk1 by the induction hypothesis. Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

31. Let P n denote the statement that F12  F22  F32      Fn2  Fn  Fn1 . Step 1: P 1 is the statement that F12  F1  F2 or 12  1  1, which is true. Step 2: Assume that P k is true, that is, F12  F22  F32      Fk2  Fk  Fk1 . We want to use this to show that P k  1 is true. Now 2 2  Fk  Fk1  Fk1 F12  F22  F32      Fk2  Fk1

   Fk1 Fk  Fk1  Fk1  Fk2

induction hypothesis

by definition of the Fibonacci sequence

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.


SECTION 12.4 Mathematical Induction

27

32. Let P n denote the statement that F1  F3      F2n1  F2n . Step 1: P 1 is the statement that F1  F2 , which is true since F1  1 and F2  1. Step 2: Assume that P k is true; that is, F1  F3      F2k1  F2k , for some k  1. We want to use this to show that P k  1 is true; that is, F1  F3      F2k11  F2k1 . Now F1  F3      F2k1  F2k1  F2k  F2k1

induction hypothesis

 F2k2

definition of F2k2

 F2k1 Therefore, P k  1 is true. So by the Principle of Mathematical Induction, P n is true for all n.   n  Fn1 Fn 1 1 .   33. Let P n denote the statement  Fn Fn1 1 0 

Step 1: Since 

1 1 1 0

2

 

1 1 1 0

 

2 1

1 1 1 0

k

1 1

k 

Fk1  Fk Fk1

1 1 1 0 

Step 2: Assume that P k is true; that is,   

1 1 1 0

k1 

   



1 1

 

1 0

 

F3 F2



F2 F1

Fk1

Fk

1 1 1 0

Fk  Fk1

Fk Fk1 

 Fk

, it follows that P 2 is true.

. We show that P k  1 follows from this. Now,

Fk1

Fk

Fk Fk1 



 

Fk2 Fk1 Fk1

Fk

1 1 1 0

 

 

induction hypothesis

by definition of the Fibonacci sequence

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n  2. 34. Let a1  1 and an1 

1 Fn , for n  1 . Let P n be the statement that an  , for all n  1. 1  an Fn1

Step 1: P 1 is the statement that a1 

F1 F , which is true since a1  1 and 1  11  1. F2 F2

Step 2: Assume that P k is true; that is, ak  ak1 

1 (definition of ak1 )  1  ak

Fk . We want to use this to show that P k  1 is true. Now Fk1

1 Fk 1 Fk1

(induction hypothesis) 

Fk1 F  k1 (definition of Fk2 ). Fk  Fk1 Fk2

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n  100.

35. Since F1  1, F2  1, F3  2, F4  3, F5  5, F6  8, F7  13,    our conjecture is that Fn  n, for all n  5. Let P n denote the statement that Fn  n.

Step 1: P 5 is the statement that F5  5  5, which is clearly true.

Step 2: Assume that P k is true; that is, Fk  k, for some k  5. We want to use this to show that P k  1 is true. Now, Fk1  Fk  Fk1  k  Fk1 (by the induction hypothesis) k  1 (because Fk1  1). Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n  5.


28

CHAPTER 12 Sequences and Series

36. Since 100  10  103 , 100  11  113 , 100  12  123 ,   , our conjecture is that 100n  n 3 , for all natural numbers n  10. Let P n denote the statement that 100n  n 3 , for n  10.

Step 1: P 10 is the statement that 100  10  1,000  103  1,000, which is true.

Step 2: Assume that P k is true; that is, 100k  k 3 . We want to use this to show that P k  1 is true. Now 100 k  1  100k  100  k 3  100

induction hypothesis

 k3  k2

because k  10

 k 3  3k 2  3k  1  k  13 Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n  10. 37. (a) P n  n 2  n  11 is prime for all n. This is false as the case for n  11 demonstrates: P 11  112  11  11  121, which is not prime since 112  121.

(b) n 2  n, for all n  2. This is true. Let P n denote the statement that n 2  n. Step 1: P 2 is the statement that 22  4  2, which is clearly true.

Step 2: Assume that P k is true; that is, k 2  k. We want to use this to show that P k  1

is true. Now k  12  k 2  2k  1. Using the induction hypothesis (to replace k 2 ), we have

k 2  2k  1  k  2k  1  3k  1  k  1, since k  2. Therefore, k  12  k  1, which is exactly P k  1. Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

(c) 22n1  1 is divisible by 3, for all n  1 . This is true. Let P n denote the statement that 22n1  1 is divisible by 3. Step 1: P 1 is the statement that 23  1  9 is divisible by 3, which is clearly true.

Step 2: Assume that P k is true; that is, 22k1  1 is divisible by 3. We want to use this to show that P k  1 is   true. Now, 22k11  1  22k3  1  4  22k1  1  3  1 22k1  1  3  22k1  22k1  1 , which is

divisible by 3 since 22k1  1 is divisible by 3 by the induction hypothesis, and 3  22k1 is clearly divisible by 3. Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

(d) The statement n 3  n  12 for all n  2 is false. The statement fails when n  2: 23  8  2  12  9.

(e) n 3  n is divisible by 3, for all n  2. This is true. Let P n denote the statement that n 3  n is divisible by 3. Step 1: P 2 is the statement that 23  2  6 is divisible by 3, which is clearly true.

Step 2: Assume that P k is true; that is, k 3 k is divisible by 3. We want to use this to show that P k  1 is true. Now     k  13  k  1  k 3  3k 2  3k  1  k  1  k 3  3k 2  2k  k 3  k  3k 2  2k  k  k 3  k  3 k 2  k .   The term k 3  k is divisible by 3 by our induction hypothesis, and the term 3 k 2  k is clearly divisible by 3. Thus k  13  k  1 is divisible by 3, which is exactly P k  1. So by the Principle of Mathematical Induction, P n is true for all n.

(f) n 3  6n 2  11n is divisible by 6, for all n  1. This is true. Let P n denote the statement that n 3  6n 2  11n is divisible by 6. Step 1: P 1 is the statement that 13  6 12  11 1  6 is divisible by 6, which is clearly true.

Step 2: Assume that P k is true; that is, k 3  6k 2  11k is divisible by 6. We show that P k  1 is then also true. Now

k  13  6 k  12  11 k  1  k 3  3k 2  3k  1  6k 2  12k  6  11k  11    k 3  3k 2  2k  6  k 3  6k 2  11k  3k 2  9k  6        k 3  6k 2  11k  3 k 2  3k  2  k 3  6k 2  11k  3 k  1 k  2


SECTION 12.5 The Binomial Theorem

29

In this last expression, the first term is divisible by 6 by our induction hypothesis. The second term is also divisible by 6. To see this, notice that k  1 and k  2 are consecutive natural numbers, and so one of them must be even (divisible by 2). Since 3 also appears in this second term, it follows that this term is divisible by 2 and 3 and so is divisible by 6. Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

38. The induction step fails when k  2, that is, P 2 does not follow from P 1. If there are only two cats, Tadpole and

Sparky, and we remove Sparky, then only Tadpole remains. So at this point, we still know only that Tadpole is black. Now removing Tadpole and putting Sparky back leaves Sparky (who is not necessarily black) alone. So the induction hypothesis does not allow us to conclude that Sparky is black.

12.5 THE BINOMIAL THEOREM 1. An algebraic expression of the form a  b, which consists of a sum of two terms, is called a binomial.

2. We can find the coefficients in the expansion of a  bn from the nth row of Pascal’s Triangle. So a  b4  1a 4  4a 3 b  6a 2 b2  4ab3  1b4 .

    n n! 4 4! 3. The binomial coefficients can be calculated directly using the formula . So  4.   k! n  k! 3!1! k 3 4. To expand a  bn we can use the Binomial Theorem. Using this theorem we find           4 4 4 3 4 2 2 4 4 4 a  a b a b  ab3  b . a  b4  0 1 2 3 4

5. x  y6  x 6  6x 5 y  15x 4 y 2  20x 3 y 3  15x 2 y 4  6x y 5  y 6

6. 2x  14  2x4  4 2x3  6 2x2  4  2x  1  16x 4  32x 3  24x 2  8x  1  2   3  4  1 1 1 1 4 1 1 4  x 4  4x 3   6x 2  4x   x 4  4x 2  6  2  4 7. x  x x x x x x x 8. x  y5  x 5  5x 4 y  10x 3 y 2  10x 2 y 3  5x y 4  y 5

9. x  15  x 5  5x 4  10x 3  10x 2  5x  1    6      a  b  a 3  6a 2 a b  15a 2 b  20a ab b  15ab2  6 ab2 b  b3 10.     a 3  6a 2 ab  15a 2 b  20ab ab  15ab2  6b2 ab  b3

or a 3  6a 52 b12  15a 2 b  20a 32 b32  15ab2  6a 12 b52  b3 .  5  5  4  3  2 11. x 2 y  1  x 2 y 5 x 2 y 10 x 2 y  10 x 2 y 5x 2 y 1  x 10 y 5 5x 8 y 4  10x 6 y 3 10x 4 y 2 5x 2 y 1   6    12. 1  2  16  6  15  2  15  14  2  20  13  2 2  15  12  4  6  1  4 2  23      1  6 2  30  40 2  60  24 2  8  99  70 2 13. 2x  3y3  2x3  3 2x2 3y  3  2x 3y2  3y3  8x 3  36x 2 y  54x y 2  27y 3 3   2  3 14. 1  x 3  13  3  12  x 3  3  1 x 3  x 3  1  3x 3  3x 6  x 9    5  4  3  2     1  5 1 1  1 1 1  x x2  x2 x 15.  5 x  10 x  10 x x 5 x x x x x x 5 10 10 1  5  72  2  12  5x  x 52 x x x x  x 2  x 3   x 4  x 5 x x 5 1 x5 16. 2   25 5 24 10 23 10 22 5 2   3240x 20x 2 5x 3  58 x 4  32 2 2 2 2 2 2   6 6  5  4! 6! 17.   15  4! 2! 2  1  4! 4


30

CHAPTER 12 Sequences and Series

  8  7  6  5! 8! 8   8  7  56 18.  3! 5! 3  2  1  5! 3   100  99  98! 100 100! 19.   4950  98! 2! 98!  2  1 98   10  9  8  7  6  5! 10 10!   3  2  7  6  252 20.  5! 5! 5  4  3  2  1  5! 5    3  2!  4  3  2! 3 4 3! 4!   18 21.  1! 2! 2! 2! 1  2!  2  1  2! 1 2    5! 5  4  3! 5  4  3! 5 5 5! 22.     10  10  100  2! 3! 3! 2! 2  1  3! 3!  2  1 2 3 5 5 5 5 5 5 23. 0  1  2  3  4  5  1  15  25  32             5 5! 5! 5! 5 5 5 5 5 5! 24.     1  0. Notice that the first and sixth       1 1! 4! 2! 3! 3! 2! 4! 1! 0 1 2 3 4 5 terms cancel, as do the second and fifth terms and the third and fourth terms.           25. x  2y4  40 x 4  41 x 3  2y  42 x 2  4y 2  43 x  8y 3  44 16y 4  x 4  8x 3 y  24x 2 y 2  32x y 3  16y 4             26. 1  x5  50 15  51 14 x  52 13 x 2  53 12 x 3  54 1 x 4  55 x 5  1  5x  10x 2  10x 3  5x 4  x 5            2    3    4    5    6 1 6 6 4 1 6 6 6 5 1 6 3 1 6 2 1 6 6 1 1  27. 1       1 1  1 1 1 1 x x x x x x x 2 0 1 3 4 5 6 15 6 20 15 6 1 1  2  3  4  5  6 x x x x x x 4        2    3    4    28. 2A  B 2  40 2A4  41 2A3 B 2  42 2A2 B 2  43 2A B 2  44 B 2  16A4  32A3 B 2  24A2 B 4  8AB 6  B 8   20   19 29. The first three terms in the expansion of x  2y20 are 20  x 20 , 20  2y  40x 19 y, and 0 x 1 x 20 18 2 18 2 2 x  2y  760x y .  30    12 30    12 29 x x 30. The first four terms in the expansion of x 12  1 are 30  x 15 , 30 1  30x 292 , 0 1 27   30  12 28 2 14 , and 30 x 12 x  435x 1 13  4060x 272 . 2 3 25    23  13 24   253 are 25  a  25a 263 , and 25  a 253 . 31. The last two terms in the expansion of a 23  a 13 24 a 25 a       40   1 40 40 , 40 x 39 1 32. The first three terms in the expansion of x   40x 38 , and are 40  x x 0 1 x x   40 38 1 2 x  780x 36 . 2 x  18 33. The middle term in the expansion of x 2  1 occurs when both terms are raised to the 9th power. So this term is 18  2 9 9 x 1  48,620x 18 . 9   16 4 16 16 34. The fifth term in the expansion of ab  120 is 20 4 ab 1  4845a b .   2 23 2 23 35. The 24th term in the expansion of a  b25 is 25 23 a b  300a b .   3 27 3 27 36. The 28th term in the expansion of A  B30 is 30 27 A B  4060A B .   1 99 99 37. The 100th term in the expansion of 1  y100 is 100 99 1  y  100y .     1 1 25 25  2 24  is 1 x  25x 47 . 38. The second term in the expansion of x 2  x x


SECTION 12.5 The Binomial Theorem

31

39. The term that contains x 4 in the expansion of x  2y10 has exponent r  4. So this term is 10 4 104  13,440x 4 y 6 . 4 x  2y  12    r 40. The rth term in the expansion of 2y is 12 2 y 12r . The term that contains y 3 occurs when 12  r  3  r     9 3 2 y  3520 2y 3 . r  9 . Therefore, the term is 12 9

  r  2 12r 12 r 242r 41. The rth term is 12  r a b . Thus the term that contains b8 occurs where 24  2r  8  r  8. So r a b   8 8 8 8 the term is 12 8 a b  495a b .  8r   8r   8r   xr 1  r8 8r  8r  r8 8r  x 2r8 . So the term that does not contain x occurs 42. The rth term is r8 8xr 2x 2 x 2  4 8 1 when 2r  8  0  r  4. Thus, the term is 4 8x4  17,920. 2x 43. x 4  4x 3 y  6x 2 y 2  4x y 3  y 4  x  y4

44. x  15  5 x  14  10 x  13  10 x  12  5 x  1  1  [x  1  1]5  x 5         45. 8a 3  12a 2 b  6ab2  b3  30 2a3  31 2a2 b  32 2ab2  33 b3  2a  b3  4    4    3    2     46. x 8  4x 6 y  6x 4 y 2  4x 2 y 3  y 4  40 x 2  41 x 2 y  42 x 2 y 2  43 x 2 y 3  44 y 4  x 2  y x  h3  x 3

x 3  3x 2 h  3xh 2  h 3  x 3

3x 2 h  3xh 2  h 3

  h 3x 2  3xh  h 2

  3x 2  3xh  h 2 h h 4 4 4 3       x  1 x h  42 x 2 h 2  43 xh 3  44 h 4  x 4 x 4  4x 3 h  6x 2 h 2  4xh 3  h 4  x 4 x  h4  x 4 0   48. h h h   3 2 2 3 h 4x  6x h  4xh  h 4x 3 h  6x 2 h 2  4xh 3  h 4    4x 3  6x 2 h  4xh 2  h 3 h h   100 49. 101100  1  001100 . Now the first term in the expansion is 100  1, the second term is 0 1 100 98 100 99 2 1 1 001  1, and the third term is 2 1 001  0495. Now each term is nonnegative, so 47.

h

h

101100  1  001100  1  1  0495  2. Thus 101100  2. n  n  n  n  n! n! n! n!   1.    1. Therefore,  . 50.  0!n! 1  n! n n!0! n!  1 0 n 0   n  n! n n  1! n n n! n n  1! 51.     n.    n. Therefore, 1 1! n  1! 1 n  1! 1 n1 n  1! 1! n  1! 1   n  n   n. 1 n1   n  n! n n!   for 0  r  n. 52.  r! n  r ! n r r n  r! r!     n! n n n! .  53. (a)   r 1 r r  1! [n  r  1]! r! n  r! n! r  n! n! n  r  1  n! (b)    r  r  1! n  r  1! r! n  r  1 n  r ! r  1! [n  r  1]! r! n  r ! r  n! n  r  1  n!   r! n  r  1! r! n  r  1! Thus a common denominator is r! n  r  1!.


32

CHAPTER 12 Sequences and Series

(c) Therefore, using the results of parts (a) and (b),     r  n! n! n n n! n  r  1  n!      r! n  r  1! r! n  r  1! r 1 r r  1! [n  r  1]! r! n  r!

  n! r  n  r  1 n! n  1 r  n!  n  r  1  n! n1 n  1!     r! n  r  1! r! n  r  1! r! n  1  r ! r ! n  1  r! r n  54. Let P n be the proposition that is an integer for the number n, 0  r  n. r n  0 Step 1: Suppose n  0. If 0  r  n, then r  0, and so   1, which is obviously an integer. Therefore, P 0 r 0 is true.   k1 Step 2: Suppose that P k is true. We want to use this to show that P k  1 must also be true; that is, is an r       k k k 1  by the key property of binomial coefficients integer for 0  r  k  1. But we know that  r 1 r r     k k (see Exercise 49). Furthermore, and are both integers by the induction hypothesis. Since the sum of two r 1 r   k 1 integers is always an integer, must be an integer. Thus, P k  1 is true if P k is true. So by the Principal of r n Mathematical induction, is an integer for all n  0, 0  r  n. r 

55. By the Binomial Theorem, the volume of a cube of side x  2 inches is         3 3 3 2 3 3 3 3 2 x  x 2  2  x 3  3  2x 2  3  4x  8  x 3  6x 2  12x  8. The volume x 2  x  2  0 1 3 2

of a cube of side x inches is x 3 , so the difference in volumes is x 3  6x 2  12x  8  x 3  6x 2  12x  8 cubic inches.   n pr q nr . In this case, 56. By the Binomial Theorem, the coefficient of pr in the expansion of  p  qn is n r p  09, q  01, n  5, and r  3, so the probability that the archer hits the target exactly 3 times in 5 attempts is     5 5 P 093 012  00729. 093 0153  2 53 57. Notice that 100!101  100!100  100! and 101!100  101  100!100  101100  100!100 . Now

100!  1  2  3  4      99  100 and 101100  101  101  101      101. Thus each of these last two expressions consists of 100 factors multiplied together, and since each factor in the product for 101100 is larger than each factor in the product for

100!, it follows that 100!  101100 . Thus 100!100  100!  100!100  101100 . So 100!101  101!100 . 58.

11 2

1214

13318

1  4  6  4  1  16

1  5  10  10  5  1  32

Conjecture: The sum is 2n . Proof: 2n  1  1n 

n 0

10  1n 

n

11  1n1 

1 n  n  n  n       0 1 2 n

n  2

12  1n2     

n  n

1n  10


CHAPTER 12

n

n 

n 

Review

n 

59. 0  0n  1  1n  10 1n  11 1n1  12 1n2      1n 10 n  n 1 n  2 n  n  n  0      1n         1k k n 0 1 2

CHAPTER 12 REVIEW 1. an 

1 4 9 16 100 n2 12 22 32 42 102 . Then a1   , a2   , a3   , a4   , and a10   . n1 11 2 21 3 31 4 41 5 10  1 11

2n 21 22 23 24 8 . Then a1  11  2, a2  12  2, a3  13   , a4  14  4, and n 1 2 3 3 4 1024 512 210   . a10  110 10 10 5

2. an  1n

1 2 1n  1 11  1 12  1 13  1  , a3  . Then a1   0, a2    0, 3 3 3 8 4 n 1 2 33 2 1 1 14  1 110  1 a4      , and a . 10 64 32 500 43 103

3. an 

n n  1 1 1  1 2 2  1 3 3  1 4 4  1 . Then a1   1, a2   3, a3   6, a4   10, and 2 2 2 2 2 10 10  1 a10   55. 2

4. an 

654 2n! 2  1! 2  2! 2  3! . Then a1  1  15,  1, a2  2  3, a3  3  n 2 n! 8 2  1! 2  2! 2  3! 2  4! 2  10! 8765  105, and a10  10   654,729,075. a4  4 16 2  4! 2  10!         n1 11 21 3! 31 4!  3, a3   6, 6. an  . Then a1   1, a2    2! 1! 2! 2! 2 2 2 2     41 5! 10  1 11!  10, and a10   55. a4    2! 3! 2! 9! 2 2

5. an 

7. an  an1  2n  1 and a1  1. Then a2  a1  4  1  4, a3  a2  6  1  9, a4  a3  8  1  16, a5  a4  10  1  25, a6  a5  12  1  36, and a7  a6  14  1  49. 1 1 1 1 1 a a a a a a , a5  4  , a6  5  , and 8. an  n1 and a1  1. Then a2  1  , a3  2  , a4  3  n 2 2 3 6 4 24 5 120 6 720 1 a . a7  6  7 5040 9. an  an1  2an2 , a1  1 and a2  3. Then a3  a2  2a1  5, a4  a3  2a2  11, a5  a4  2a3  21, a6  a5  2a4  43, and a7  a6  2a5  85. 12       3an1 and a1  3  312 . Then a2  3a1  3 3  3  312  334 ,       a3  3a2  3  334  374  378 , a4  3a3  3  378  3158  31516 ,       a5  3a4  3  31516  33116  33132 , a6  3a5  3  33132  36332  36364 ,    a7  3a6  3  36364  312764  3127128 .

10. an 

33


34

CHAPTER 12 Sequences and Series

5 5 5 5 5 5 12. (a) a1  1  , a2  2  , a3  3  , 2 4 8 2 2 2 5 5 5 5 , a5  5  a4  4  16 32 2 2

11. (a) a1  2 1  5  7, a2  2 2  5  9,

a3  2 3  5  11, a4  2 4  5  13,

a5  2 5  5  15 (b)

(b)

an

an

14 12 10 8 6 4 2 0

2 1

1

2

3

4

0

5 n

1

2

3

4

5 n

(c) S5  7  9  11  13  15  55

5 5 5 5 5 155 (c) S5       2 4 8 16 32 32

(d) This sequence is arithmetic with common difference

(d) This sequence is geometric with common ratio 12 .

2. 31 3 32 9 33 27 , 13. (a) a1  2  , a2  3  , a3  4  4 8 16 2 2 2

7 1 2  , a2  4   3, 2 2 2 5 3 3 4 5 a3  4   , a4  4   2, a5  4   2 2 2 2 2

14. (a) a1  4 

34 81 35 243 , a5  6  a4  5  32 64 2 2 (b)

(b)

an

3

4

2

2

0

(c) S5 

an

1

1

2

3

4

0

5 n

1

2

3

4

5 n

5 3 25 7 (c) S5   3   2   2 2 2 2

3 9 27 81 243 633      4 8 16 32 64 64

(d) This sequence is geometric with common ratio 32 .

(d) This sequence is arithmetic with common difference  12 .

15. 5 55 6 65   . Since 55  5  6  55  65  6  05, this is an arithmetic sequence with a1  5 and d  05. Then a5  a4  05  7.            16. 2 2 2 3 2 4 2   . Since 2 2  2  3 2  2 2  4 2  3 2  2, this is an arithmetic sequence with       a1  2 and d  2. Then a5  a4  2  4 2  2  5 2.

17. t  3 t  2 t  1 t   . Since t  2  t  3  t  1  t  2  t  t  1  1, this is an arithmetic sequence with a1  t  3 and d  1. Then a5  a4  1  t  1.       4  2, this is a geometric sequence with a  2 and r  2. Then 18. 2 2 2 2 4   . Since 2  2 2 2   1 2 2 2   a5  a4  r  4  2  4 2.

1 t2 t 1 1 1 19. t 3  t 2  t 1   . Since 3  2  , this is a geometric sequence with a1  t 3 and r  . Then a5  a4  r  1   . t t t t t t 20. 1  32  2  52    . Since  32  1  2  32 , and 1 21. 34  12  13  29    . Since 23  4 4 . a5  a4  r  29  23  27

 32 2  3 , the series is neither arithmetic nor geometric. 1 

2 2 1 3  9  2 , this is a geometric sequence with a  3 and r  2 . Then 1 3 4 3 1 1 2 3


CHAPTER 12

Review

35

1 1 2 1 1 1 1 1  a and a  , this is a geometric sequence with a1  a and r  . Then 22. a 1  2    . Since 1 a a a 1 a a a 1 1 1 a5  a4  r  2   3 . a a a 12 2 2i 24i 6i  2i,   2  2i,  2i, this is a geometric sequence with common 23. 3 6i 12 24i   . Since 3 6i i 12 i ratio r  2i. a 2  2i 24. The sequence 2, 2  2i, 4i, 4  4i, 8,    is geometric (where i 2  1), since 2   1  i, a1 2 a3 4i 2  2i 8  8i a 1 1 i i 4i 4  4i     1  i, 4     1     1  2  1  1  i,  a2 2  2i 2  2i 2  2i 8 a3 4i i i i i a5 8 4  4i 32  32i 8     1  i. Thus the common ratio is i  1 and the first term is 2. So the  a4 4  4i 4  4i 4  4i 32 nth term is an  a1r n1  2 1  in1 .

25. a6  17  a  5d and a4  11  a  3d. Then, a6  a4  17  11  a  5d  a  3d  6  6  2d  d  3. Substituting into 11  a  3d gives 11  a  3  3, and so a  2. Thus a2  a  2  1 d  2  3  5. 26. a20  96 and d  5. Then 96  a20  a  19  5  a  95  a  96  95  1. Therefore, an  1  5 n  1.  2 27. a3  9 and r  32 . Then a5  a3  r 2  9  32  81 4.

3 28. a2  10 and a5  1250 27 . Then r 

 n1 an  ar n1  6 53 .

a5  a2

1250 27  125  r  5 and a  a  a2  10  6. Therefore, 1 27 3 5 10 r 3

29. (a) An  52,000  104n1

(b) A1  $52,000, A2  52,0001041  $54,080, A3  52,0001042  $56,24320, A4  52,0001043  $58,49293, A5  52,000  1044  $60,83265, A6  52,000  1045  $63,26595

30. (a) An  55,000  1,600 n  1

(b) A1  $55,000, A2  55,000  1600  56,600, A3  55,000  1600 2  58,200, A4  55,000  1600 3  59,800, A5  55,000  1600 4  61,400, A6  55,000  1600 5  63,000. The salary for Position I is higher in the sixth year.

31. Let an be the number of bacteria in the dish at the end of 5n seconds. So a0  3, a1  3  2, a2  3  22 , a3  3  23 ,   . Then, clearly, an is a geometric sequence with r  2 and a  3. Thus at the end of 60  5 12 seconds, the number of bacteria is a12  3  212  12,288.

32. Let d be the common difference in the arithmetic sequence a1 , a2 , a3 ,   , so that an  a1  n  1 d, n  1, 2, 3,   , and let e be the common difference for b1 , b2 , b3 ,   , so that bn  b1  n  1 e. Then     an  bn  a1  n  1 a  b1  n  1 e  a1  b1   n  1 d  e, n  1, 2, 3,   . Thus a1  b1  a2  b2     is an arithmetic sequence with first term a1  b1 and common difference d  e. 33. Suppose that the common ratio in the sequence a1  a2  a3     is r . Also, suppose that the common ratio in the sequence

b1  b2  b3     is s. Then an  a1r n1 and bn  b1 s n1 , n  1 2 3   . Thus an bn  a1r n1 b1 s n1  a1 b1  rsn1 . So the sequence a1 b1  a2 b2  a3 b3     is geometric with first term a1 b1 and common ratio rs. 34. (a) Yes. If the common difference is d, then an  a1  n  1 d. So an  2  a1  2  n  1 d, and thus the sequence a1  2 a2  2 a3  2    is an arithmetic sequence with the common difference d, but with the first term a1  2. (b) Yes. If the common ratio is r, then an  a1  r n1 . So 5an  5a1   r n1 , and the sequence 5a1  5a2  5a3     is also geometric, with common ratio r, but with the first term 5a1 .


36

CHAPTER 12 Sequences and Series

35. (a) 6 x 12    is arithmetic if x  6  12  x  2x  18  x  9.  12 x  x 2  72  x  6 2. (b) 6 x 12    is geometric if  6 x 36. (a) 2 x y 17    is arithmetic. Therefore, 15  17  2  a4  a1  a  3d  a  3d. So d  5, and hence, x  a  d  2  5  7 and y  a  2d  2  2  5  12.  13 a4 ar 3 17 3 . So (b) 2 x y 17    is geometric. Therefore, 17 2  a  a  r  r  2 1   2  13  23 17 13 23 1713 and y  a  ar 2  2  2  2 17  213 1723 . x  a2  ar  2 17 3 2 2 2 37.

38.

6

2 2 2 2 2 k3 k  1  3  1  4  1  5  1  6  1  16  25  36  49  126

2i

4

21

22

23

24

4

6

8

i 1 2i  1  2  1  1  2  2  1  2  3  1  2  4  1  2  3  5  7 

6

210  140  126  120 596  105 105

k1  2  20  3  21  4  22  5  23  6  24  7  25  2  6  16  40  96  224  384 k1 k  1 2  40. 5m1 3m2  31  30  31  32  33  13  1  3  9  27  121 3

39.

41. 42. 43. 44.

10

2 2 2 2 2 2 2 2 2 2 2 k1 k  1  0  1  2  3  4  5  6  7  8  9

1 1 1 1 1 1 1 1 j2 j  1  1  2  3  4  5      98  99

100 50

k1

10

3k

3 32 33 34 349 350           2k1 22 23 24 25 250 251

2 n 2 1 2 2 2 3 2 9 2 10 n1 n 2  1  2  2  2  3  2      9  2  10  2

45. 3  6  9  12      99  3 1  3 2  3 3      3 33   2 46. 12  22  33      1002  100 k1 k

33

k1 3k

47. 1  23  2  24  3  25  4  26      100  2102

48.

 1 212  2 222  3 232  4 242      100 21002  k2  100 k1 k  2

 1 1 1 1 1       999 k1 k k  1 12 23 34 999  1000

49. 1  09  092      095 is a geometric series with a  1 and r 

09  09. Thus, the sum of the series is 1

1  0531441 1  096   468559. 1  09 01 50. 3  37  44      10 is an arithmetic series with a  3 and d  07. Then 10  an  3  07 n  1  07 n  1  7 S6 

143  n  11. So the sum of the series is S11  11 2 3  10  2  715.          51. 5  2 5  3 5      100 5 is an arithmetic series with a  5 and d  5. Then 100 5  an  5  5 n  1         n  100. So the sum is S100  100 5  100 5  50 101 5  5050 5. 2

52. 13  23  1  43      33 is an arithmetic series with a  13 and d  13 . Then an  33  13  13 n  1  n  99. So the   2  99  1  99  100  1650. sum is S99  99 2 3 2 3 3 6 53. n0 3  4n is a geometric series with a  3, r  4, and n  7. Therefore, the sum of the series is S7  3 

  1  47  35 1  47  9831. 1  4


CHAPTER 12

54.

Review

37

8

k2 is a geometric series with a  7, r  512 , and n  9. Thus, the sum of the series is k0 7  5   1  592 1  592 1  5 1  592  5  55 7 S9  7   7

15 1  5  1 5 1  5    74 1  625 5  5  3125  5467  1092 5  79088

4  8     is a geometric series with a  1 and r   2 . Therefore, it is convergent with sum 55. 1  25  25 125 5

S

5 1 a   .  2 1r 7 1  5

56. 01  001  0001  00001     is an infinite geometric series with a  01 and r  01. Therefore, it is convergent with 1 01  . sum S  1  01 9

57. 5  5 101  5 1012  5 1013     is an infinite geometric series with a  5 and r  101. Because r  101  1, the series diverges. 1 1 1 58. 1  12  13  32     is an infinite geometric series with a  1 and r   . Thus, it is convergent with sum 3 3 3     1 3  S  12 3  3 . 31 1  1 3

 2

59. 1  98  98 diverges.

 3  98     is an infinite geometric series with a  1 and r   98 . Because r  98  1, the series

    60. a  ab2  ab4  ab6     is an infinite geometric series with first term a and common ratio b2 . Because b  1, b2   1, a . 1  b2 61. We have an arithmetic sequence with a  7 and d  3. Then n n n Sn  325  [2a  n  1 d]  [14  3 n  1]  11  3n  650  3n 2  11n  3n  50 n  13  0  2 2 2 is inadmissible). Thus, 13 terms must be added. n  13 (because n   50 3 and so the series is convergent with sum S 

62. We have a geometric series with S3  52 and r  3. Then 52  S3  a  3a  9a  13a  a  4, and so the first term is 4.   1  215  2 215  1  65,534, and so the total 63. This is a geometric sequence with a  2 and r  2. Then S15  2  12 number of ancestors is 65,534. n 3n  1 64. Let P n denote the statement that 1  4  7      3n  2  . 2 1 [3 1  1] 12 Step 1: P 1 is the statement that 1   , which is true. 2 2 k 3k  1 Step 2: Assume that P k is true; that is, 1  4  7      3k  2  . We want to use this to show that 2 P k  1 is true. Now 1  4  7  10      3k  2  [3 k  1  2]  

k 3k  1  3k  1 2

3k 2  k  6k  2 k 3k  1 6k  2   2 2 2

3k 2  5k  2 k  1 3k  2  2 2 k  1 [3 k  1  1]  2

induction hypothesis


38

CHAPTER 12 Sequences and Series

Thus, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

1 1 1 1 n      .  13 35 57 2n  1 2n  1 2n  1 1 1 Step 1: P 1 is the statement that  , which is true. 13 211 1 1 1 1 k Step 2: Assume that P k is true; that is,    . We want to use this to  13 35 57 2k  1 2k  1 2k  1 show that P k  1 is true. Now 1 1 1 1 1       13 35 57 2k  1 2k  1 2k  1 2k  3

65. Let P n denote the statement that

  

k 1  2k  1 2k  1 2k  3

induction hypothesis

k 2k  3  1 2k 2  3k  1  2k  1 2k  3 2k  1 2k  3

k 1 k 1 k  1 2k  1   2k  3 2 k  1  1 2k  1 2k  3

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.       1 1 1 1 1 1    1   n  1. 66. Let P n denote the statement that 1  1 2 3 n

Step 1: P 1 is the statement that 1  11  1  1, which is clearly true.       1 1 1 1 1 1    1   k  1. We want to use this to Step 2: Assume that P k is true; that is, 1  1 2 3 k show that P k  1 is true. Now        1 1 1 1 1 1 1    1  1 1 1 2 3 k k 1         1 1 1 1 1 1 1    1  1  1 1 2 3 k k 1   1 induction hypothesis  k  1 1  k1  k  1  1

Thus, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

67. Let P n denote the statement that 7n  1 is divisible by 6.

Step 1: P 1 is the statement that 71  1  6 is divisible by 6, which is clearly true.

Step 2: Assume that P k is true; that is, 7k  1 is divisible by 6. We want to use this to show that P k  1 is true. Now   7k1  1  7  7k  1  7  7k  7  6  7 7k  1  6, which is divisible by 6. This is because 7k  1 is divisible by

6 by the induction hypothesis, and clearly 6 is divisible by 6. Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.

68. Let P n denote the statement that F4n is divisible by 3. Step 1: Show that P 1 is true, but P 1 is true since F4  3 is divisible by 3. Step 2: Assume that P k is true; that is, F4k is divisible by 3. We want to use this to show that P k  1 is true.     Now, F4k1  F4k4  F4k2  F4k3  F4k  F4k1  F4k1  F4k2  F4k  F4k1  F4k1    F4k  F4k1  2  F4k  3  F4k1 , which is divisible by 3 because F4k is divisible by 3 by our induction hypothesis, and 3  F4k1 is clearly divisible by 3. Thus, P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.


CHAPTER 12

Test

39

69. an1  3an  4 and a1  4. Let P n denote the statement that an  2  3n  2. Step 1: P 1 is the statement that a1  2  31  2  4, which is clearly true.

Step 2: Assume that P k is true; that is, ak  2  3k  2. We want to use this to show that P k  1 is true. Now ak1  3ak  4    3 2  3k  2  4

definition of ak1 induction hypothesis

 2  3k1  6  4  2  3k1  2

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.    5! 54 54 5! 5 5     10  10  100 70.  2! 3! 3! 2! 2 2 2 3     10 10 10! 10! 10  9 10  9  8  7 71.       45  210  255 2 6 2! 8! 6! 4! 2 432                  72. 5k0 5k  50  51  52  53  54  55  2 0!5!5!  1!5!4!  2!5!3!  2 1  5  10  32   8                  73. 8k0 8k 8k  2 80 88  2 81 87  2 82 86  2 83 85  84 84  2  2  82  2  282  2  562  702  12,870         74. A  B3  30 A3  31 A2 B  32 AB 2  33 B 3  A3  3A2 B  3AB 2  B 3                    75. x  25  50 x 5  51 x 4 2  52 x 3 22  53 x 2 23  54 x 24  55 25

 x 5  10x 4  40x 3  80x 2  80x  32 6                76. 1  x 2  60 16  61 15 x 2  62 14 x 4  63 13 x 6  64 12 x 8  65 x 10  66 x 12

 1  6x 2  15x 4  20x 6  15x 8  6x 10  x 12           77. 2x  y4  40 2x4  41 2x3 y  42 2x2 y 2  43  2x y 3  44 y 4  16x 4  32x 3 y  24x 2 y 2  8x y 3  y 4   3 19 3 19 78. The 20th term is 22 19 a b  1540a b . 20     23 20    23 19 79. The first three terms in the expansion of b23  b13 b b are 20  b403 , 20 0 1         23 18 13 2 b13  20b373 , and 20 b b  190b343 . 2   r 10r . The term that contains A6 occurs when r  6. Thus, the 80. The rth term in the expansion of A  3B10 is 10 r A 3B 10 6 term is 6 A 3B4  210A6 81B 4  17,010A6 B 4 .

CHAPTER 12 TEST 1. an  2n 2  n  a1  1, a2  6, a3  15, a4  28, a5  45, a6  66, and S6  1  6  15  28  45  66  161. 2. an1  3an  n, a1  2  a2  3 2  1  5, a3  3 5  2  13, a4  3 13  3  36, a5  3 36  4  104, and a6  3 104  5  307. 3. (a) The common difference is d  5  2  3. (b) an  2  n  1 3

(c) a35  2  3 35  1  104

3  1. 4. (a) The common ratio is r  12 4  n1 (b) an  a1r n1  12 14  101 (c) a10  12 14  38  65,3536 4


40

CHAPTER 12 Sequences and Series 1

1 1.  r  15 , so a5  ra4  25 5. (a) a1  25, a4  15 . Then r 3  5  25 125  8 1  15 97,656 58  1   (b) S8  25 1 12,500 3125 1 5

6. (a) a1  10 and a10  2, so 9d  8  d   89 and a100  a1  99d  10  88  78.    10 8  60 (b) S10  10 2 [2a  10  1 d]  2 2  10  9  9

7. Let the common ratio for the geometric series a1  a2  a3     be r, so that an  a1r n1 , n  1, 2, 3,   . Then 2    n1  an2  a1r n1  a12 r 2 . Therefore, the sequence a12 , a22 , a32 ,    is geometric with common ratio r 2 .            5  2  1  12  1  22  1  32  1  42  1  52  0  3  8  15  24  50 8. (a) n1 1  n  (b) 6n3 1n 2n2  13 232  14 242  15 252  16 262  2  4  8  16  10 2

3

9

9. (a) The geometric sum 13  22  23  24      210 has a  13 , r  23 , and n  10. So 3 3 3 3   12310 58,025 1 1 1024 S10  3  123  3  3 1  59,049  59,049 .

1  1  1     has a  1 and r  212  1 . Thus, (b) The infinite geometric series 1  12 2 2 232 2     2 2 21 1    S    2  2. 11 2

21

21

21

n n  1 2n  1 . 6 123 , which is true. Step 1: Show that P 1 is true. But P 1 says that 12  6 k k  1 2k  1 . We want to use this to show that Step 2: Assume that P k is true; that is, 12  22  32      k 2  6 P k  1 is true. Now

10. Let P n denote the statement that 12  22  32      n 2 

12  22  32      k 2  k  12 

k k  1 2k  1  k  12 6

induction hypothesis

  k  1 2k 2  k  6k  6 k  1

k k  1 2k  1  6 k  12  6 6     2 2 k  1 2k  7k  6 k  1 2k  k  6k  6   6 6 k  1 [k  1  1] [2 k  1  1] k  1 k  2 2k  3   6 6 

Thus P k  1 follows from P k. So by the Principle of Mathematical Induction, P n is true for all n.  5    2    3    4    5     11. 2x  y 2  50 2x5  51 2x4 y 2  52 2x3 y 2  53 2x2 y 2  54 2x y 2  55 y 2 12.

10

 32x 5  80x 4 y 2  80x 3 y 4  40x 2 y 6  10x y 8  y 10

3 7 3 3 3 3x 2  120  27x 128  414,720x

13. (a) Each week he gains 24% in weight, that is, 024an . Thus, an1  an  024an  124an for n  1.

a0 is given to be 085 lb. Then a0  085, a1  124 085, a2  124 124 085  1242 085,   a3  124 1242 085  1243 085, and so on. So we can see that an  085 124n .

(b) a6  124a5  124 124a4       1246 a0  1246 085  31 lb (c) The sequence a1  a2  a3     is geometric with common ratio 124.


Modeling with Recursive Sequences

41

FOCUS ON MODELING Modeling with Recursive Sequences 00365  00001. Thus the amount in the account at 365 the end of the nth day is An  10001An1 with A0  $275,000.

1. (a) Since there are 365 days in a year, the interest earned per day is

(b) A0  $275,000, A1  10001A0  10001  275,000  $275,02750, A2  10001A1  10001 10001A0   100012 A0  $275,05500,

A3  10001A2  100013 A0  $275,08251, A4  100014 A0  $275,11002, A5  100015 A0  $275,13753,

A6  100016 A0  $275,16504, A7  100017 A0  $275,19256

(c) An  10001n  275,000

2. (a) Tn  Tn1  15 with T1  5.

(b) T1  5, T2  T1  15  5  15  65, T3  T2  15  5  15  15  5  2  15  80, T4  T3  15  5  2  15  15  5  3  15  95, T5  T4  15  5  3  15  15  5  4  15  110, T6  T5  15  5  4  15  15  5  5  15  125 (c) This is an arithmetic sequence with Tn  5  15 n  1.

(d) Tn  65  5  15n  15  615  15n  n  41. So the student swims 65 minutes on the 41st day.

(e) Using the partial sum of an arithmetic sequence, the student swims for 30 2 [2 5  30  1 15]  15  535  8025 minutes  13 hours 225 minutes.

003  00025. Thus the amount in the account at the 12 end of the nth month is An  10025An1  100 with A0  $100.

3. (a) Since there are 12 months in a year, the interest earned per day is

(b) A0  $100, A1  10025A0  100  10025  100  100  $20025,

A2  10025A1  100  10025 10025  100  100  100  100252  100  10025  100  100  $30075,   A3  10025A2  100  10025 100252  100  10025  100  100  100  100253  100  100252  100  10025  100  100  $40150,   A4  10025A3  100  10025 100253  100  100252  100  10025  100  100  100  100254  100  100253  100  100252  100  10025  100  100  $50251

(c) An  10025n  100      100252  100  10025  100  100, the partial sum of a geometric series, so An  100 

10025n1  1 1  10025n1  100  . 1  10025 00025

(d) Since 5 years is 60 months, we have A60  100 

1002561  1  $658083. 00025

4. (a) The amount An of pollutants in the lake in the nth year is 30% of the amount from the preceding year (030An1 ) plus the amount discharged that year (2400 tons). Thus An  030An1  2400. (b) A0  2400, A1  030 2400  2400  3120,

A2  030 [030 2400  2400]  2400  0302 2400  2400 2400  2400  3336,   A3  030 0302 2400  2400 2400  2400  2400  0033 2400  0302 2400  2400 2400  2400  34008,   A4  030 0033 2400  0302 2400  2400 2400  2400  2400

 0034 2400  0033 2400  0302 2400  2400 2400  2400  34202


42

FOCUS ON MODELING

(c) An is the partial sum of a geometric series, so An  2400 

1  030n1

 2400   1  030  n1  34286 1  030

(e)

4000

1  030n1 070

2000

1  0307 0  34278 tons. The sum of a geometric 0 10 20 070 1 series, is A  2400   34286 tons. 070 5. (a) For Plan I, the interest earned in the first year is 005U0 , and so the additional amount invested is 01 105U0 . Thus, U1  105U0  01 105U0 . The interest earned in the second year is 005U1 and the additional amount invested is   01 105U1 , so U2  105U1  01 105U1 . We see that Un  105Un1  01 105Un1 . For Plan 2, V0 earns 5% interest in the first year, and the additional amount invested is 500n, so V1  105V0  500 1. After the second year, V2  105V1  500 2. Thus, Vn  105Vn1  500n. (d) A6  2400 

(b)

n

Un

Vn

n

Un

Vn

0

5000

5000

5

10,22732

14,40054

1

5775

5750

10

21,12467

40,21234

2

667013

703750

20

89,25030

160,45901

3

770399

888938

30

377,07656

419,21761

4

889811

11,33384

40

1,593,12324

903,59757

From the tables, we see that Un overtakes Vn somewhere between n  30 and n  40. We find that the smallest value of n for which Un  Vn (other than n  1) is n  32 (U32  503,02955 and V32  494,46242).


CORRECTIONS: 2,4,7,18,22,24,32

CHAPTER 13

PROBABILITY AND STATISTICS

13.1

Counting 1

13.2

Probability 9

13.3

Binomial Probability 17

13.4

Expected Value 22 Chapter 13 Review 25 Chapter 13 Test 30

¥

FOCUS ON MODELING: The Monte Carlo Method 32

1


13 COUNTING AND PROBABILITY 13.1 COUNTING 1. The Fundamental Counting Principle says that if one event can occur in m ways and a second event can occurs in n ways, then the two events can occur in order in m  n ways. So if you have two choices for shoes and three choices for hats, then the number of different shoe­hat combinations you can wear is 2  3  6.

2. The number of ways of arranging r objects from n objects in order is called the number of permutations of n objects taken n! r at a time, and is given by the formula P n r  . So, if 3 students are chosen from a class of 10 to serve as n  r! president, vice president, and treasurer, then the order in which these students are assigned these positions matters. Thus, 10!  720. the number of ways of choosing the 3 class officers is P 10 3  10  3! 3. The number of ways of choosing r objects from n objects is called the number of combinations of n objects taken r at n! . So if 3 students are chosen from a class of 10 to serve as a a time, and is given by the formula Cn r  r! n  r! committee, then the order in which these students are chosen doesn’t matter. Thus, the number of ways of choosing the 10!  120. 3 class officers is C 10 3  3! 10  3!

4. (a) False. When counting combinations, order does not matter. (b) True. When counting permutations, order matters.

(c) False. For a set of n distinct objects, the number of different combinations of these objects is less than the number of different permutations. (d) True. If we have a set with five distinct objects then the number of different ways of choosing two members of this set is the same as the number of ways of choosing three members. 5. (a) P 8 3 

8! 8!  8  7  6  336  5! 8  3!

6. (a) P 11 4  7. (a) C 8 3 

8! 8! 876    56 3! 8  3! 3! 5! 321

8. (a) C 11 4  9. (a) P 12 4  10. (a) P 9 6  11. (a) P 8 2 

11! 11!  11  10  9  8  7920  7! 11  4!

11  10  9  8 11!   330 4! 7! 4321

(b) P 9 2 

9! 9!  9  8  72  7! 9  2!

(b) P 10 5  (b) C 9 2 

10! 10!  30,240  5! 10  5!

9! 9! 98    36 2! 9  2! 2! 7! 21

(b) C 10 5 

10! 10!   252 5! 10  5! 5! 5!

12! 12! 12! 12!  12  11  10  9  11,880 (b) C 12 4    495  8! 4! 12  4! 4! 8! 12  4!

9! 9! 9! 9!  9  8  7  6  5  4  60,480 (b) C 9 6    84  3! 6! 9  6! 6! 3! 9  6! 8! 8!  8  7  56  6! 8  2!

12. (a) P 10 3 

10!  720 10  3!

(b) P 8 6 

8! 8!  8  7  6  5  4  3  20,160  2! 8  6!

(b) P 10 7 

10!  604,800 10  7!

1


2

CHAPTER 13 Counting and Probability

In Solutions 13–16, we use the fact that C n p  C n n  p. See Exercise 4(d). 8! 8!   28 (b) C 8 6  C 8 2  28 13. (a) C 8 2  2! 8  2! 2! 6! 14. (a) C 10 3  15. (a) C 50 2 

10!  120 3! 10  3!

50!  1225 2! 50  2!

(b) C 10 7  C 10 3  120 (b) C 50 48  C 50 2  1225

100!  4950 (b) C 100 98  C 100 2  4950 2! 100  2! 17. By the Fundamental Counting Principle, the number of possible single­scoop ice cream cones is     number of ways to number of ways to   4  3  12. choose the flavor choose the type of cone 16. (a) C 100 2 

18. By the Fundamental Counting Principle, the possible number of three­letter words is       number of ways to number of ways to number of ways to   . choose the first letter choose the second letter choose the third letter

(a) Since repetitions are allowed, we have 26 choices for each letter. Thus, there are 26  26  26  17,576 words.

(b) Since repetitions are not allowed, we have 26 choices for the first letter, 25 choices for the second letter, and 24 choices for the third letter. Thus there are 26  25  24  15,600 words.

19. (a) By the Fundamental Counting Principle, the possible number of ways 8 horses can complete a race, assuming no ties in any position, is       number of ways to number of ways to number of ways to     87654321 choose the first finisher choose the second finisher choose the eighth finisher  8!  40,320 (b) By the Fundamental Counting Principle, the possible number of ways the first, second, and third place can be decided, assuming no ties, is       number of ways to number of ways to number of ways to    8  7  6  336. choose the first finisher choose the second finisher choose the third finisher

20. Since there are four choices for each of the five questions, by the Fundamental Counting Principle there are 4  4  4  4  4  1024 different ways the test can be completed.

21. The number of possible seven­digit phone numbers is       number of ways to number of ways to number of ways to     . choose the first digit choose the second digit choose the seventh digit Since the first digit cannot be a 0 or a 1, there are only 8 digits to choose from, while there are 10 digits to choose from for the other 6 digits in the phone number. Thus the number of possible seven­digit phone numbers is 8  10  10  10  10  10  10  8,000,000.

22. Since a runner can only finish once, there are no repetitions. And since we are assuming that there is no tie, the number of different finishes is       number of ways to number of ways to number of ways to      5  4  3  2  1  120. choose the first runner choose the second runner choose the fifth runner

23. Since there are 4 main courses, there are 6 ways to choose a main course. Likewise, there are 5 drinks and 3 desserts so there are 5 ways to choose a drink and 3 ways to choose a dessert. So the number of different meals consisting of a main course, a       number of ways to number of ways to number of ways to drink, and a dessert is    4 5 3  60. choose the main course choose a drink choose a dessert 24. By the Fundamental Counting Principle, the number of different routes from town A to town D via towns B and C is       number of routes number of routes number of routes    4 5 6  120. from A to B from B to C from C to D

4


SECTION 13.1 Counting

3

25. The number of possible sequences of heads and tails when a coin is flipped 5 times is       number of possible number of possible number of possible      2 2 2 2 2 outcomes on the first flip outcomes on the second flip outcomes on the fifth flip  25  32 Here there are only two choices, heads or tails, for each flip.

26. Since each die has six different faces, the number of different outcomes when rolling a red and a white die is 6  6  36.

27. Since there are six different faces on each die, the number of possible outcomes when a red die and a blue die and a white die  are rolled is      number of possible number of possible number of possible    6 6 6  63  216. outcomes on the red die outcomes on the blue die outcomes on the white die

28. The number of possible pants­shirt­shoes outfits is       number of ways number of ways number of ways    5 8 12  480. to choose pants to choose a shirt to choose shoes

29. The number of different California license plates possible is          number of ways to number of ways number of ways    9 263 103  158,184,000. choose a nonzero digit to choose 3 letters to choose 3 digits

30. The number of possible ID numbers consisting of one letter followed by three digits is 26 10 10 10  26,000.

31. Since successive numbers cannot be the same, the number of possible choices for the second number in the combination is only 59. The third number in the combination cannot be the same as the second in the combination, but it can be the same as the first number, so the number of possible choices for the third number in the combination is also 59. So the number of possible combinations consisting of a number in the clockwise direction, a number in the counterclockwise direction, and then a number in the clockwise direction is 60 59 59  208,860. 32. The number of possible license plates of two letters followed by three digits is        number of ways to number of ways   262 103  676,000. Since 676,000  8,000,000, there will not be choose 2 letters to choose 3 digits enough different license plates for the state’s 8 million registered cars.

33. Since a student can hold only one office, the number of ways that a president, a vice president, and a secretary can be chosen from  a class of 30 students  is    number of ways number of ways to number of ways    30 29 28  24,360. to choose a president choose a vice president to choose a secretary 34. The number of ways to choose a graduate president, an undergraduate treasurer, and a vice president is       number of ways

    to choose a president    from the 10 graduates

number of ways

to choose a treasurer

from the 7 undergraduates

number of ways

      to choose a vice president   10 7 15  1050. from the remaining 15 students

35. We have 7 choices for the first digit and 10 choices for each of the other 8 digits. Thus, the number of Social Security numbers is 7  108  700,000,000.

36. The number of possible ways to arrange three girls and four boys is     number of ways to number of ways   3!  4!  144. arrange 3 girls to arrange 4 boys 37. (a) The number of ways to select 5 of the 8 objects is C 8 5 

8!  56. 5! 3!

(b) A set with 8 elements has 28  256 subsets.

38. We may choose any subset of the 8 available brochures. There are 28  256 ways to do this.

39. Each subset of toppings constitutes a different way a hamburger can be ordered. Since a set with 10 elements has 210  1024 subsets, there are 1024 different ways to order a hamburger.


4

CHAPTER 13 Counting and Probability

40. We consider a set of 20 objects (the shoppers in the mall) and a subset that corresponds to those shoppers that enter the store. Since a set of 20 objects has 220  1,048,576 subsets, there are 1,048,576 outcomes to their decisions.

41. (a) The number of ways to seat 10 people in a row of 10 chairs is 10!  3,628,800.

10! 10! 6!   151,200. 6! 4! 4! 42. The number of ways of selecting 3 objects in order (a 3­letter word) from 6 distinct objects (the 6 letters) assuming that the letters cannot be repeated is P 6 3  6  5  4  120. (b) The number of ways to choose 6 out of 10 people and seat them in 6 chairs is C 10 66! 

43. In selecting these officers, order is important and repetition is not allowed, so the number of ways of choosing 3 officers from 15 students is P 15 3  2730.

44. The number of ways of selecting 3 objects in order (a 3­digit number) from 4 distinct objects (the 4 digits) with no repetition of the digits is P 4 3  4  3  2  24. 45. Since the order of finish is important, we want the number of permutations of 8 objects (the contestants) taken 3 at a time, 8! 8! which is P 8 3   8  7  6  336.  5! 8  3!

46. The number of ways of ordering 8 pieces in order (without repeats) is P 8 8  8!  40,320.

47. The number of ways of ordering 9 distinct objects (the contestants) is P 9 9  9!  362,880. Here a runner cannot finish more than once, so no repetitions are allowed, and order is important. 48. The number of ways of ordering three of the five distinct flags is P 5 3  60.

49. The number of ways of ordering 1000 distinct objects (the contestants) taking 3 at a time is P 1000 3  1000  999  998  997,002,000. We are assuming that a person cannot win more than once, that is, there are no repetitions. 50. In selecting these officers, order is important and repetition is not allowed, so the number of ways of choosing 4 officers from 30 students is P 30 4  657,720.

51. We first place Kit in the first seat, and then seat the remaining 4 students. Thus the number of these arrangements is     number of ways to number of ways to seat   P 1 1  P 4 4  1! 4!  24. seat Kit in the first seat the remaining 4 students

52. We start by placing Kit in the middle seat, and then we place the remaining 4 students in the remaining four seats. Thus the     number of ways number of ways to seat number of possible arrangements is   1  P 4 4  1! 4!  24. to seat Kit the remaining 4 students

53. Here we have 6 objects, of which 2 are blue marbles and 4 are red marbles. Thus the number of distinguishable permutations 6  5  4! 6!   15. is 2! 4! 2  4! 54. Here we have 14 objects (the 14 balls) of which 5 are red balls, 2 are white balls, and 7 are blue balls. So the number of 14!  72,072. distinguishable permutations is 5! 2! 7! 55. The number of distinguishable permutations of 12 objects (the 12 coins), from like groups of size 4 (the pennies), size 3 (the 12!  277,200. nickels), size 2 (the dimes) and size 3 (the quarters) is 4! 3! 2! 3! 56. The word ELEEMOSYNARY has 12 letters of which 3 are E, 2 are Y, and the rest are distinct. So we wish to find the number of distinguishable permutations of 12 objects (the 12 letters) from like groups of sizes 3 and 2 and 7 like groups of size 1. 12! replace "12" with "14" (2 times)  39,916,800. We get 3! 2! 1! 1! 1! 1! 1! 1! 57. The number of distinguishable permutations of 12 objects (the 12 ice cream cones) from like groups of size 3 (the vanilla cones), size 2 (the chocolate cones), size 4 (the strawberry cones), and size 5 (the butterscotch cones) is 14!  2,522,520. 3! 2! 4! 5!


SECTION 13.1 Counting

5

58. This is the number of distinguishable permutations of 7 objects (the students) from like groups of size 3 (the ones who stay in the three­person room), size 2 (the ones who stay in the two­person room), size 1 (the one who stays in the one­person 7!  420. room), and size 1 (the one who sleeps in the car). This number is 3! 2! 1! 1! 59. The number of distinguishable permutations of 8 objects (the 8 cleaning tasks) from like groups of sizes 5, 2, and 1 workers 8! is  168. 5! 2! 1! 60. The number of distinguishable permutations of 30 objects (the students) from like groups of sizes 8, 11, and 11 is 30!  4,128 ,840,588,600. 8! 11! 11! 61. Here we are interested in the number of ways of choosing three objects (the three members of the committee) from a set of 25!  2300. 25 objects (the 25 members). The number of combinations of 25 objects taken three at a time is C 25 3  3! 22! 6!  20. 62. We want the number of ways of choosing a group of three from a group of six. This number is C 6 3  3! 3! 12!  220. 63. We want the number of ways of choosing a group of three from a group of 12. This number is C 12 3  3! 9! 64. We want the number of ways of choosing a group of 6 people from a group of 10 people. The number of combinations of 10!  210. 10 objects (people) taken 6 at a time is C 10 6  6! 4! 65. We want the number of ways of choosing a group (the 5­card hand) where order of selection is not important. The number 52! of combinations of 52 objects (the 52 cards) taken 5 at a time is C 52 5   2,598,960. 5! 47! 66. Since order is not important in a 7­card hand, the number of combinations of 52 objects (the 52 cards) taken 7 at a time is 52!  133,784,560. C 52 7  7! 45! 67. The order of selection is not important, hence we must calculate the number of combinations of 10 objects (the 10 questions) 10! taken 7 at a time. This gives C 10 7   120. 7! 3! 68. In this exercise, we assume that the pizza toppings cannot be repeated, so we are interested in the number of ways to select a 16! subset of 3 toppings from a set of 16 toppings. The number of ways this can occur is C 16 3   560. 3! 13! 69. We assume that the order in which the violinist plays the pieces in the recital is not important, so the number of combinations 12! of 12 objects (the 12 pieces) taken 8 at a time is C 12 8   495. 8! 4! 70. The order in which the shirts are selected is not important and no shirt is repeated. So the number of combinations of eight 8! shirts taken five at a time is C 8 5   56. 5! 3! 71. The order in which the pants are selected is not important and no pair is repeated, so the number of combinations of ten 10!  120. pairs of pants taken three at a time is C 10 3  3! 7! 72. (a) Since Kai must go on the field trip, we first pick Kai to go on the field trip, and then select the six other students from 29! the remaining 29 students. Since C 29 6   475,020, there are 475,020 ways to select the students to go on 6! 23! the field trip with Kai. (b) We first take Kai out of the class of 30 students and select the seven students from the remaining 29 students. Thus 29! there are C 29 7   1,560,780 ways to pick the 7 students for the field trip. 7! 22! (c) We are interested only in the group of 7 students taken from the class of 30 students, not the order in which they are 30!  2,035,800. picked. Thus the number is C 30 7  7! 23!


6

CHAPTER 13 Counting and Probability

73. Since the order in which the numbers are selected is not important, the number of combinations of 49 numbers taken 6 at a 49!  13,983,816. time is C 49 6  6! 43! 74. The number of distinguishable permutations of 13 objects (the total number of blocks the jogger must travel) which can be 13!  1287. partitioned into like groups of size 8 (the east blocks) and of size 5 (the north blocks) is 8! 5! 75. (a) The number of ways of choosing 5 students from the 20 students is C 20 5 

20!  15,504. 5! 15!

12!  792. 5! 7! (c) We use the Fundamental Counting Principle to count the number of possible committees with 3 math majors and 2 physics majors. Thus, we get     number of ways to choose number of ways to choose   C 12 3  C 8 2  220 28  6160. 3 of 12 math majors 2 of the 8 physics majors

(b) The number of ways of choosing 5 students for the committee from the 12 math majors is C 12 5 

76. The number of ways 2 of 10 Americans and 2 of 10 Canadians can be chosen is     number of ways to pick number of ways to pick   C 10 2  C 10 2  45 45  2025. 2 of 10 Americans 2 of 10 Canadians 77. The number of ways the committee can be chosen is       number of ways to number of ways to number of ways to    C 20 1  C 19 1  C 18 4 choose a chair choose a secretary choose four other members  20  19  3060  1,162,800 78. (a) We choose 2 of the 9 children and 4 of the 16 adults:     number of ways to choose number of ways to choose   C 9 2  C 16 4  36 1820  65,520. 2 of 9 children 4 of 16 adults (b) Method 1: We consider the number of ways of selecting the group of 6 from the 25 campers and subtract the groups that contain no children and those that contain one child. The number of groups that contain at least two children is C 25 6  C 9 0  C 16 6  C 9 1  C 16 5  177,100  1 8008  9 4368  129,780. Method 2: In the method we construct all the groups that are possible:         groups with 2 children and 4 adults

groups with 3 children and 3 adults

groups with 4 children

and 2 adults

groups with 5 children and 1 adult

 groups with 6 children

 C 9 2  C 16 4  C 9 3  C 16 3  C 9 4  C 16 2  C 9 5  C 16 1  C 9 6  C 16 0  36 1820  84 560  126 120  126 16  84 1

 65,520  47,040  15,120  2,016  84  129,780.

79. The number of ways the committee can be chosen is         number of ways to choose number of ways to choose number of ways to choose number of ways to choose    2 of 6 freshmen 3 of 8 sophomores 4 of 12 juniors 5 of 10 seniors  C 6 2  C 8 3  C 12 4  C 10 5  15  56  495  252  104,781,600 80. The leading and supporting roles are different (order counts), while the extra roles are not (order doesn’t count). Also, the instrumental roles must be filled by the instrumentalists and the singing roles by the singers. Thus, number of ways the ensemble can be chosen is         

number of ways to choose



the leading and supporting instrumentalists

number of ways to choose



the leading and supporting singers

number of ways to choose 5 extras

number of ways to choose 3 extras



from the 8 remaining instrumentalists

from the 10 remaining singers

 P 10 2  P 12 2  C 8 5  C 10 3  90 132 56 120  79,833,600.


SECTION 13.1 Counting

7

81. We choose 3 forwards from the forwards, 2 defensemen from the defensemen, and a goalie from the 2 goalies. Thus the number of ways to pick the 6 starting players is       number of ways to number of ways to number of ways to    C 12 3  C 6 2  C 2 1 pick 3 of 12 forwards pick 2 of 6 defensemen pick 1 of 2 goalies  220 15 2  6600 82. To order a pizza, we must make several choices. First the size (4 choices), the type of crust (2 choices), and then the toppings. Since there are 14 toppings, the number of possible choices is the number of subsets of the 14 toppings, that is 214 choices. So by the Fundamental Counting Principle, the number of possible pizzas is 4 2  214  131,072. 83. We count the total number of committees and subtract the number that contain both Skyler and Riley. The total number of committees possible is C 10 4 and the number that contain both Skyler and Riley is C 8 2, so the number of possible committees is C 10 4  C 8 2  210  28  182. 84. Method 1: We consider the number of five­member committees that can be formed from the 26 interested people and subtract the numbers of those that contain no teacher and those that contain no students. Thus the number of committees is C 26 5  C 12 5  C 14 5  65,780  792  2002  62,986. Method 2: Here we construct all the committees that are possible. Thus the number of committees is         committees with

1 student and 4 teachers

committees with

2 students and 3 teachers

committees with

3 students and 2 teachers

committees with

4 students and 1 teacher

 C 14 1  C 12 4  C 14 2  C 12 3  C 14 3  C 12 2  C 14 4  C 12 1  14 495  91 220  364 66  1001 12  6,930  20,020  24,024  12,012  62,986

85. Since the two algebra books must be next to each other, we first consider them as one object. So we now have four objects to arrange and there are 4! ways to arrange these four objects. Now there are two ways to arrange the two algebra books. Thus the number of ways that five mathematics books may be placed on a shelf if the two algebra books are to be next to each other is 2  4!  48. 86. We treat Fin and Sydney as one object and Riley and Kelly as one object, so we need the number of ways of permuting eight objects. We then multiply this by the number of ways of arranging Fin and Sydney within their group and arranging Riley and Kelly within their group. Thus, the number of possible arrangements is P 8 8  P 2 2  P 2 2  8!  2!  2!  161,280. 87. (a) To find the number of ways the students and teachers can be seated, we first select and place a student in the first seat     select 1 of arrange the and then arrange the other 7 people:   C 4 1  P 7 7  4  7!  20,160. the 4 students remaining 7 people (b) To find the number of ways the students and teachers can be seated, we first select and place teachers in the first and     arrange 2 of arrange the last seats and then arrange the other 6 people:   P 4 2  P 6 6  12  the 4 teachers remaining 6 people grey 6!  8,640. 88. (a) We treat the people wearing white jackets as one object and find the number of ways of permuting five objects, then permute the four wearing gray jackets. Thus the number of arrangements is P 5 5  P 4 4  5!  4!  2880.

(b) Method 1: The number of ways the wearers of white and gray jackets can be seated is       number of ways the number of ways the number of ways to select   white jackets can be arranged gray jackets can be arranged the jacket color of the first seat  P 4 4  P 4 4  C 2 1  4!  4!  2  1152 Method 2: There are eight choices for the first seat, four for the second, three each for the third and fourth, two each for the fifth and sixth, and one each for the seventh and eighth. Thus the number of possibilities is 8  4  3  3  2  2  1  1  1152.


8

CHAPTER 13 Counting and Probability

89. The number of ways the winner can be chosen is       number of ways number of ways number of ways       to choose 6 semifinalists   to choose 2 finalists   to choose the winner from the 30 contestants from the 6 semifinalists from the 2 finalists

 C 30 6  C 6 2  C 2 1  593,775 15 2  17,813,250

90. There are many different possibilities here, so we consider the complement where no professor is chosen for the delegates and subtract this number from the way to select three people from the group of eight people, which is C 8 3. If the professor cannot to be selected, then we must select three people from a group of five, and this can be done in C 5 3 ways. Thus the number of delegations that contain a professor is C 8 3  C 5 3  56  10  46.

91. Since there are 26 letters, the possible number of combinations of the first and the last initials is 26 26  676. Since 677  676, there must be at least two people that have the same first and last initials in any group of 677 people.

92. When 2 objects are chosen from 10 objects, it determines a unique set of 8 objects, those not chosen. So choosing 2 of 10 objects is the same a choosing 8 of the 10. In general, every subset of r objects chosen from a set of n objects determines a corresponding set of n  r objects, namely, those not chosen. Therefore, the total number of combinations for each type are equal. 93. We are only interested in selecting a set of 3 marbles to give to Alex and a set of 2 marbles to give to Sasha, not the order in which we hand out the marbles. Since both C 10 3  C 7 2 and C 10 2  C 8 3 count the number of ways this can be done, these numbers must be equal. (Calculating these values shows that they are indeed equal.) In general, if we wish to find two distinct sets of k and r objects selected from n objects (k  r  n), then we can either first select the k objects from the n objects and then select the r objects from the n  k remaining objects, or we can first select the r objects from         n n r n nk    . the n objects and then the k objects from the n  r remaining objects. Thus, r k k r 94. (a) x  y5  x  y x  y x  y x  y x  y

 x  y x  y x  y x x  x y  yx  yy

 x  y x  y x x x  x x y  x yx  x yy  yx x  yx y  yyx  yyy

 x  y x x x x  x x x y  x x yx  x x yy  x yx x  x yx y  x yyx  x yyy

 yx x x  yx x y  yx yx  yx yy  yyx x  yyx y  yyyx  yyyy

 x x x x x  x x x x y  x x x yx  x x x yy  x x yx x  x x yx y  x x yyx  x x yyy

x yx x x  x yx x y  x yx yx  x yx yy  x yyx x  x yyx y  x yyyx  x yyyy

yx x x x  yx x x y  yx x yx  yx x yy  yx yx x  yx yx y  yx yyx  yx yyy

yyx x x  yyx x y  yyx yx  yyx yy  yyyx x  yyyx y  yyyyx  yyyyy

(b) There are ten terms that contain two x’s and three y’s. Their sum is

x x yyy  x yx yy  x yyx y  x yyyx  yx x yy  yx yx y  yx yyx  yyx x y  yyx yx  yyyx x (c) To count the number of terms with two x’s, we must count the number of ways to pick two of the five positions to contain an x. This number is C 5 2.   n (d) In the Binomial Theorem, the coefficient is the number of ways of picking r positions in a term with n factors to r contain an x. By definition, this is C n r .


SECTION 13.2 Probability

9

13.2 PROBABILITY 1. The set of all possible outcomes of an experiment is called the sample space. A subset of the sample space is called an event. The sample space for the experiment of tossing two coins is S  H H H T T H T T , and the event “getting at least 3 n E  . one head” is E  H H H T T H. The probability of getting at least one head is P E  n S 4 2. (a) The probability of E or F occurring is P E  F  P E  P F  P E  F.

(b) If the events E and F have no outcomes in common (that is, the intersection of E and F is empty), then the events are called mutually exclusive. So in drawing a card from a deck, the event E, “getting a heart,” and the event F, “getting a spade,” are mutually exclusive.

(c) If E and F are mutually exclusive, then the probability of E or F is PE  F  P E  P F. n E  F . So in tossing a die, the conditional 3. The conditional probability of E given that F occurs is P E  F  n F probability of the event E, “getting a six,” given that that the event F, “getting an even number,” has occurred is P E  F  13 .

4. (a) The probability of E and F occurring is PE  F  P E P F  E. So if two marbles are drawn consecutively, without replacement, from a jar that contains six blue and four red marbles, then the probability that the first marble 6  4  4. drawn is blue (E) and the second is red (F) is P E  F  P E  P F  E  10 9 15

(b) If the occurrence of E does not affect the probability of the occurrence F, then the events are called independent. If E and F are independent events, then the probability of E and F is PE  F  P E P F. So if two marbles are drawn consecutively, with replacement, from a jar that contains six blue and four red marbles, then the probability that 6  4  6 . the first marble drawn is blue (E) and the second is red (F) is P E  F  P E  P F  10 10 25

5. (a) S  1 2 3 4 5 6 (b) E  2 4 6 (c) E  5 6

6. (a) There are two possible outcomes of the coin toss and 52 possible outcomes of drawing a card, so n S  2  52  104. (b) H A H A H A H A

(c) T J  T Q T K  T J  T Q T K  T J  T Q T K  T J  T Q T K 

(d) H 2 H 3 H4 H5 H6 H7 H8 H9 H10 H J  H Q H K  H A

7. Let H stand for head and T for tails. (a) The sample space is S  H H H T T H T T .

1 (b) Let E be the event of getting exactly two heads, so E  H H. Then P E  nE nS  4 .

3 (c) Let F be the event of getting at least one head. Then F  H H H T T H, and P F  nF nS  4 .

2 1 (d) Let G be the event of getting exactly one head, that is, G  H T T H . Then P G  nG nS  4  2 .

8. Let H stand for heads and T for tails; the numbers 1, 2,   , 6 are the faces of the die. (a) S  H 1 H2 H3 H 4 H5 H6 T 1 T 2 T 3 T 4 T 5 T 6

(b) Let E be the event of getting heads and rolling an even number. Then E  H2 H4 H6, and 3 1 P E  nE nS  12  4 .

(c) Let F be the event of getting heads and rolling a number greater than 4. Then F  H5 H6, and 2 1 P F  nF nS  12  6 .

3 1 (d) Let G be the event of getting tails and rolling an odd number. Then G  T 1 T 3 T 5, and P G  nG nS  12  4 .


10

CHAPTER 13 Counting and Probability

1 9. (a) Let E be the event of rolling a six. Then P E  nE nS  6 .

3 1 (b) Let F be the event of rolling an even number. Then F  2 4 6. So P F  nF nS  6  2 .

1 (c) Let G be the event of rolling a number greater than 5. Since 6 is the only face greater than 5, P G  nG nS  6 . 2 1 10. (a) Let E be the event of rolling a two or a three. Then P E  nE nS  6  3 .

3 1 (b) Let F be the event of rolling an odd number. So F  1 3 5, and P F  nF nS  6  2 .

1 (c) Let G be the event of rolling a number divisible by 3. Then G  3 6 and P G  nG nS  3 .

4 1 11. (a) Let E be the event of choosing a king. Since a deck has four kings, P E  nE nS  52  13 .

(b) Let F be the event of choosing a face card. Since there are three face cards per suit and four suits, 12 3 P F  nF nS  52  13 .

  3  10 . (c) Let F be the event of choosing a face card. Then P F   1  P F  1  13 13

13 1 12. (a) Let E be the event of choosing a heart. Since there are 13 hearts, P E  nE nS  52  4 .

26 1 (b) Let F be the event of choosing a heart or a spade. Since there are 13 hearts and 13 spades, P F  nE nS  52  2 .

(c) Let G be the event of choosing a heart, a diamond or a spade. Since there are 13 cards in each suit, 39 3 P G  nG nS  52  4 .

5 13. (a) Let E be the event of selecting a red ball. Since the jar contains five red balls, P E  nE nS  8 .

(b) Let F be the event of selecting a yellow ball. Since there is only one yellow ball,   1 7 P F   1  P F  1  nF nS  1  8  8 .

0 (c) Let G be the event of selecting a black ball. Since there are no black balls in the jar, P G  nG nS  8  0.

14. (a) Let E be the event of selecting a white ball or a yellow ball. Since there are two white balls and one yellow ball,   3 5 P E   1  P E  1  nE nS  1  8  8 . (b) Let F be the event of selecting a red ball, a white ball, or a yellow ball. Since all the types of balls are in the jar, P E  1. (c) Let G be the event of selecting a white ball. Since there are two white balls,   2 6 3 P E   1  P E  1  nE nS  1  8  8  4 .

C 13 5 1287   0000495. C 52 5 2,598,960 (b) Let E be the event of choosing five cards of the same suit. Since there are four suits and 13 cards in each suit, 5,148 4  C 13 5   000198. n E  4  C 13 5. Also, n S  C 52 5. Therefore, P E  C 52 5 2,598,960 (c) Let E be the event of dealing five face cards. Since there are 3 face cards in each suit and 4 suits, 792 C 12 5   0000305. P E  C 52 5 2,598,960 (d) Let E be the event of dealing a royal flush (ace, king, queen, jack, and 10 of the same suit). Since there is only one such 4 4   153908  106 . sequence for each suit, there are only 4 royal flushes, so P E  C 52 5 2,598,960

15. (a) Let E be the event of dealing five hearts. Since there are 13 hearts, P E 

16. (a) Let E be the event of choosing three defective charging devices. Since there are four defective charging devices, C 4 3 4 1 n E     0018. n E  C 4 3. Also, n S  C 12 3. Therefore, P E  n S C 12 3 220 55


SECTION 13.2 Probability

11

(b) Let E be the event of choosing three functioning charging devices. Since there are eight functioning charging devices, C 8 3 56 14 n E     0255. n E  C 8 3, so P E  n S C 12 3 220 55 17. (a) Let E be the event of choosing two red balls. Since there are three red balls, n E  C 3 2. Also, n S  C 8 2. C 3 2 3 n E    0107. Therefore, P E  n S C 8 2 28 (b) Let E be the event of choosing two white balls. Since there are five white balls, n E  C 5 2, so C 5 2 10 5 n E     0357. P E  n S C 8 2 28 14 3. 18. (a) Let E be the event of choosing a T. Since 3 of the 16 letters are T’s, P E  16 5. (b) Let F be the event of choosing a vowel. Since there are 5 vowels, P F  16   5  11 . (c) Let F be the event of choosing a vowel. Then P F  1  16 16

19. (a) Let E be the probability that at least one card is a spade. The number of hands that do not contain a spade is the number 7411 C 39 5   0778. of possible five­card hands using the other three suits, that is, C 39 5. Thus, P E  1  C 52 5 9520 (b) Let E be the probability that at least one card is a face card. The number of hands that do not contain a face card is the number of possible five­card hands using the cards of the deck that are not face cards, that is, C 40 5. Thus, 6221 C 40 5   0747. P E  1  C 52 5 8330 20. (a) S  1 1  1 2  1 3  1 4  1 5  1 6  2 1  2 2  2 3  2 4  2 5  2 6  3 1  3 2  3 3  3 4  3 5  3 6  4 1  4 2  4 3  4 4  4 5  4 6  5 1  5 2  5 3  5 4  5 5  5 6  6 1  6 2  6 3  6 4  6 5  6 6 6  1. (b) Let E be the event of getting a sum of 7. Then E  1 6  2 5  3 4  4 3  5 2  6 1, and P E  36 6

4  1. (c) Let F be the event of getting a sum of 9. Then F  3 6  4 5  5 4  6 3, and P F  36 9 6  1. (d) Let E be the event that the two dice show the same number. Then P E  36 6

(e) Let E be the event that the two dice show different numbers. Then E  is the event that the two dice show the same   number. Thus, P E  1  P E   1  16  56 . 5 (f) Let E be the event of getting a sum of 9 or higher. Then P E  10 36  18 .

3 21. (a) Let E be the event that the spinner stops on red. Since 12 of the regions are red, P E  12 16  4 .

(b) Let F be the event that the spinner stops on an even number. Since 8 of the regions are even­numbered, 8  1. P F  16 2

4  1. (c) Since 4 of the even­numbered regions are red, P E  F  P E  P F  P E  F  34  12  16

4  1. 22. (a) Let E be the event that the spinner stops on blue. Since only 4 of the regions are blue, P E  16 4

8  1. (b) Let F be the event that the spinner stops on an odd number. Since 8 of the regions are odd­numbered, P F  16 2

(c) Since none of the odd­numbered regions are blue, P E  F  P E  P F  14  12  34 .

23. (a) Yes, the events are mutually exclusive since the number cannot be both even and odd. So P E  F  P E  P F  36  36  1.

(b) No, the events are not mutually exclusive since 6 is both even and greater than 4. So P E  F  P E  P F  P E  F  36  26  16  23 .

24. (a) No, the events are not mutually exclusive since 4 is greater than 3 and also less than 5. So P E  F  P E  P F  P E  F  36  46  16  1.


12

CHAPTER 13 Counting and Probability

(b) Yes, the events are mutually exclusive since there are only 2 numbers less than 3, namely 1 and 2, but they are not divisible by 3. So P E  F  P E  P F  26  26  23 . 25. (a) No, the events E and F are not mutually exclusive since the jack, queen, and king of spades are both face cards and 12 3 11 spades. So P E  F  P E  P F  P E  F  13 52  52  52  26 .

(b) Yes, the events E and F are mutually exclusive since the card cannot be both a heart and a spade. So 13 1 P E  F  P E  P F  13 52  52  2 .

26. (a) No, events E and F are not mutually exclusive since a king can be a club. So 4 1 4 P E  F  P E  P F  P E  F  13 52  52  52  13 .

(b) No, events E and F are not mutually exclusive since an ace can be a spade. So 4  13  1  4 . P E  F  P E  P F  P E  F  52 52 52 13

27. (a) Let E denote a roll of five and F a roll greater than three. Using the formula for conditional probability, we have 1 n E  F  . P E  F  n F 3 (b) Let E denote a roll of three and F an odd roll. Using the formula for conditional probability, we have 1 n E  F  . P E  F  n F 3 n E  F 4 1   . n F 12 3 n E  F 1 (b) Let E denote drawing a king and F drawing a spade. Then P E  F   . n F 13 n E  F 1 (c) Let E denote drawing a spade and F drawing a king. Then P E  F   . n F 4

28. (a) Let E denote drawing a queen and F drawing a face card. Then P E  F 

29. Let E denote the spinner stopping on an even number and F the spinner stopping on red. Then n E  F 4 1 P E  F    . n F 12 3 30. Let E denote the spinner stopping on a number divisible by 3 and F the spinner stopping on blue. Then 1 n E  F  . P E  F  n F 4 31. (a) There is only one red ball numbered 3 and one green ball numbered 3. If the ball drawn is numbered 3, then the probability it is red is 12 . (b) There is only one ball numbered 7 and it is green, so if the ball drawn is numbered 7, then the probability it is green is 1. (c) There are two even­numbered red balls and three even­numbered green balls, so if the ball is even­numbered, then the probability it is a red ball is 25 . (d) There are five red balls, and two are even­numbered, so if the ball drawn is red, then the probability it is even­numbered is 25 . 32. (a) If the first ball drawn is red, there are four red balls and seven green balls remaining, so the probability that the second 4 . ball is red is 11 5. (b) If the first ball drawn is green, there are five red balls and six green balls remaining, so the required probability is 11

(c) If the first ball drawn is odd­numbered, then there are six odd­numbered and five even­numbered balls remaining, so the 5 . required probability is 11

(d) If the first ball drawn is even­numbered, then there are seven odd­numbered and four even­numbered balls remaining, 4 . so the required probability is 11


SECTION 13.2 Probability

13

33. (a) Let E be the event of drawing a black ball first. Because the jar contains seven black balls and three white balls, the 7 . The probability of the second ball being white is 3  1 , so the probability of the first ball being black is P E  10 9 3

7  1  7. probability of the intersection is 10 3 30

7  2  7 . (b) Here the probability of the second ball being black is 69  23 , so the probability of the intersection is 10 3 15

34. (a) Let E be the event of drawing a red sock. Since three pairs are red, the drawer contains six red socks, and so 6 1 P E  nE nS  18  3 .

5 (b) Let F be the event of drawing another red sock. Since there are 17 socks left of which 5 are red, P F  nF nS  17 .

5  5 . (c) In this case, the probability is P E  F  P E  P F  13  17 51

4 and the probability of the second being a king is 4 , so the 35. (a) The probability of the first card being an ace is 52 51 4  4  4 . probability of the intersection is 52 51 663

4 and the probability of the second being an ace is 3 , so the (b) The probability of the first card being an ace is 52 51 4  3  1 . probability of the intersection is 52 51 221

36. (a) Yes, the first roll does not influence the outcome of the second roll.

   1  1 . (b) The probability of getting a six on both rolls is P E  F  P E  P F  16 6 36

37. Let E be the event of a “one” facing up on the first roll, and let F be the event of an even number facing up on the second 1. roll. Since these events are independent, P E  F  P E  P F  16  36  12

38. (a) Yes, they are independent. The toss of the coin does not influence the roll of the die. (b) The probability of getting a tail is 12 and the probability of getting an even number is 36  12 , so the probability of the intersection is 12  12  14 .

39. (a) Yes. What happens on spinner A does not influence what happens on spinner B.

   2  1. (b) The probability that A stops on red and B stops on yellow is P E  F  P E  P F  24 8 8

40. (a) Let E A and E B be the event that the respective spinners stop on a purple region. Since these events are independent,     1 . P E A  E B  P E A  P E B   14  28  16 (b) Let FA and FB be the event that the respective spinners stop on a blue region. Since these events are independent,     1. P FA  FB  P FA  P FB   14  18  32

41. (a) Let M and F stand for “male”and “female”. Then S  M M M M F M M M M F M M M M F M M M M F F F M M F M F M F M M F M F F M M F M F M M F F M F F F F M F F F F M F F F F M F F F F 1. (b) Let E be the event that the child gets only male goldfish. Then E  M M M M and P E  16

(c) Let F be the event that the child gets two male and two female goldfish. Then

6  3. F  F F M M F M F M F M M F M F F M M F M F M M F F, so P F  16 8

(d) Let F be the event that the child gets four goldfish of the same gender. Then F  M M M M F F F F, and 2  1. P F  16 8

(e) Let H be the event that the child gets at least two female goldfish. Then H  is the event that the child gets fewer   than two female goldfish. Thus, H   M M M M F M M M M F M M M M F M M M M F, so n H   5, and   5  11 . P H  1  P H   1  16 16

42. Let E be the event that a 13­card bridge hand consists of all cards from the same suit. Since there are exactly 4 such hands 4  63  1012 . (one for each suit), P E  C 52 13


14

CHAPTER 13 Counting and Probability

43. Let E be the event that the ball lands in an odd­numbered slot. Since there are 18 odd numbers between 1 and 36, 9 P E  18 38  19 .

44. (a) Let E be the event that the toddler arranges the word FRENCH. Since the letters are distinct, there are P 6 6 ways of 1 1 arranging the blocks of which only one spells the word FRENCH. Thus P E    00014. P 6 6 720 (b) Let E be the event that the toddler arranges the letters in alphabetical order. Since there are P 6 6 ways of arranging 1 1 the blocks of which only one is in alphabetical order, P E    00014. P 6 6 720 45. Let E be the event of selecting the 6 winning numbers. Since there is only one way to select them, 1 1   715  108 . P E  C 49 6 13,983,816 46. Let E be the event that no marketing employee is chosen. The number of ways that no marketing employee is chosen is the C 11 6  000078. same as the number of ways that only product managers are chosen, which is C 11 6. Thus P E  C 30 6 47. The sample space consist of all possible true­false combinations, so n S  210 . Let E be the event that the student answers 1 1 . all 10 questions correctly. Since there is only one way to answer all 10 questions correctly, P E  10  1024 2 48. Let E be the event that the batch will be discarded. Thus, E is the event that at least one defective bulb is found. It is easier to find E  , the event that no defective bulbs are found. Since there are 10 bulbs in the batch of which 8 are non­defective,     C 8 3 C 8 3 P E  . Thus P E  1  P E   1   1  04667  05333. C 10 3 C 10 3

49. (a) Let E be the event that the monkey types HAMLET as the first word. Since HAMLET contains six letters and there are 1 48 typewriter keys, P E  6  818  1011 . 48 (b) Let F be the event that the monkey types TO BE OR NOT TO BE as the first words. Since this phrase has 18 characters 1 (including spaces), P F  18  547  1031 . 48 1 1  00014.  720 6! (b) The probability that the monkey arranges the six blocks to spell HAMLET three consecutive times is the probability of   1 3  268  109 . three independent events E, and hence is equal to [P E]3  720

50. (a) Let E be the event that the monkey arranges the six blocks to spell HAMLET. Then P E 

51. Let E be the event that the toddler will arrange the eight blocks to spell TRIANGLE or INTEGRAL. The number of ways of arranging these blocks is the number of distinguishable permutations of eight blocks. Since no two blocks are the same, the 2  496  105 , number of distinguishable permutations is 8!. Two of these arrangements result in event E, so P E  8! or 00000496. 52. Let E be the event that you predict the correct order for the horses to finish the race. Since there are eight horses, there are P 8 8  8! ways that the horses could finish, of which you predict only one. Thus, 1 1 P E    248  105 . P 8 8 40,320 53. (a) Let E be the event that the pea is tall. Since tall is dominant, E  T T T t t T . So P E  34 .   (b) E  is the event that the pea is short. So P E   1  P E  1  34  14 .


SECTION 13.2 Probability

54.

(a) Let E be the event that the offspring will be tall. Since only offspring

Parent 2

Parent 1

15

t

t

T

Tt

Tt

t

tt

tt

with genotype Tt will be tall, P E  24  12 . (b) E  is the event that the offspring will not be tall (thus, the offspring is   short). So P E   1  P E  1  12  12 .

55. Let E be the event that the player wins on spin 1, and let F be the event that the player wins on spin 2. What happens on the first spin does not influence what happens on the second spin, so the events are independent. Thus, 1  1  1 . P E  F  P E  P F  38 38 1444

56. Let E be the event that the committee consists entirely of juniors and F the event that it consists entirely of seniors. The sample space is the set of all ways that five people can be chosen from the group of 14. These events are mutually exclusive, C 8 5 6  56 31 C 6 5    . so P E  F  P E  P F  C 14 5 C 14 5 2002 1001 57. Let E, F and G denote the events of rolling two ones on the first, second, and third rolls, respectively, of a pair of dice. The 1  1  1  1  214  105 . events are independent, so P E  F  G  P E  P F  P G  36 3 36 36 36

58. Let E be the event that a player has exactly five winning numbers and F be the event that a player has all six winning numbers. These events are mutually exclusive. For a players to have exactly five winning numbers means that the player has five of the six winning numbers and one number that was not selected in the lottery. So n E  C 6 5  C 43 1. C 6 5  C 43 1 1 Thus, P at least five winning numbers  P E  F  P E  P F    00000185. C 49 6 C 49 6 59. Let E be the event that the marble is red and F be the event that the number is odd­numbered. Then E  is the event that the marble is blue, and F  is the event that the marble is even­numbered. 6  3 (a) P E  16 8 8  1 (b) P F  16 2

6  8  3  11 (c) P E  F  P E  P F  P E  F  16 16 16 16          8  5  13 . (d) P E  F   P E   P F   P E   F   10  16 16 16 16

60. The number of ways a set of six numbers can be selected from a set of 49 is C 49 6. Since the games are independent, the  2 1 probability of winning the lottery two times in a row is  511  1015 . C 49 6    4  10 . The probability of selecting a red ball from each jar 61. The probability of selecting two red balls from jar B is 57 6 21       8 7 4 is 37 57  15 49 . The probability of selecting two red balls after putting all balls in one jar is 14 13  13 . Hence, picking both balls from jar B gives the greatest probability.

1 , and is the same for the second and third wheels. The events are 62. (a) The probability that the first wheel has a bar is 11 1  1  1  1 . independent, and so the probability of getting 3 bars is 11 11 11 1331

(b) The probability of getting a number on the first wheel is 10 11 , the probability of getting the same number on the second 1 , and the probability of getting the same number on the third wheel is 1 . Thus, the probability of getting wheel is 11 11 1 1 10 the same number on each wheel is 10 11  11  11  1331 .

(c) We use the complement, no bar, to determine the probability of at least one bar. The probability that the first wheel does not have a bar is 10 11 , and is the same for the second and third wheels. Since the events are independent, the probability 3

10 10 10 1000 1000 331 of getting no bar is 10 11  11  11  3  1331 , and so P at least one bar  1  1331  1331 . 11


16

CHAPTER 13 Counting and Probability

63. Let E be the event that the student opens the lock within an hour. The number of combinations that can be tried in one hour is 10  60  600. The number of possible combinations is P 40 3, assuming that no number is repeated. Thus, 600 5 600    0010. P E  P 40 3 59,280 494 64. Let E be the event that of the three bulbs selected, two are defective and one is not. Then number of ways to pick two defective and one functional bulb C 6 2  C 18 1 15  18 135 P E     . number of ways to select three bulbs C 24 3 2024 1012 65. (a) Let E be the event that the curriculum committee consists of two part­time faculty and four full­time faculty. So N committees with two part­timers and four full­timers C 8 2  C 10 4 28  210 490 P E      0317. N ways to select a six­member committee C 18 6 18564 1547 (b) Let F be the event that curriculum committee consists of two or fewer part­time faculty. Then P F  

1  210  8  252  28  210 C 8 0  C 10 6 C 8 1  C 10 5 C 8 2  C 10 4    C 18 6 C 18 6 C 18 6 18,564 193 8106  18,564 442

  249 (c) F  is the event that the curriculum committee has more that two part­time faculty: P F   1 P F  1 193 442  442 .

138,415 C 95 3   0856. C 100 3 161,700 (b) The probability that at least one of the three fittings inspected is defective is 1  P E  0144.

66. (a) Let E be the event that the plumber inspects three non­defective fittings. Then P E 

67. Let E be the event that Alex stands next to Sydney. To find n E we treat Alex and Sydney as one object and find the number of ways to arrange the 19 objects and then multiply the result by the number of ways to arrange Alex and Sydney. 2 19!  2!   01. So n E  19!  2!. The sample space is all the ways that 20 people can be arranged. Thus, P E  20! 20 68. Let E be the event that the monkey arranges the six blocks to spell BUBBLE. The number of ways of arranging these blocks is the number of distinguishable permutations of six blocks. Since there are three blocks labeled B, 6! the number of distinguishable permutations is . Only one of these arrangements spells the word BUBBLE. Thus, 3! 1 1 3!  . P E   6! 6! 120 3! 69. Let E be the event that the monkey arranges the 11 blocks to spell PROBABILITY. The number of ways of arranging these blocks is the number of distinguishable permutations of 11 blocks. Since there are two blocks labeled B and two 11! . Only one of these arrangements spells the word blocks labeled I, the number of distinguishable permutations is 2! 2! 1 1 2! 2!  . PROBABILITY. Thus P E   11! 11! 9,979,200 2! 2! 70. (a) Because the events are independent, the probability that the family has two boys given that the oldest child is a boy is 12 . (b) There are four equally likely possibilities: boy­boy, boy­girl, girl­boy, and girl­girl. In three cases, at least one of the children is a boy, and in one of those cases both children are boys. Thus, the probability that the family has two boys given that one of the children is a boy is 13 . 71. (a) The sample space S consists of all possible choices of 2 marbles, so n S  C 10 2  45. The event E that both marbles chosen are green has n E  C 4 2  6. Thus, the probability that 2 green marbles are chosen is n E 6 2   . n S 45 15


SECTION 13.3 Binomial Probability

17

4  2 and (b) Let E be the event that the first marble is green and F that the second marble is green. Then P E  10 5 2 . P F  E  39  13 , so the probability that both marbles chosen are green is P E P F  E  25  13  15

The answers are the same, but the thought process involved is different. If 3 marbles are chosen, then using the method of part (a), we find that n S  C 10 3  120 and n E  C 6 1  C 4 2  6  6  36 (where E is the event that 1 red and 2 green marbles are chosen), so 36 3 n E   . P E  n S 120 10 Using the method of part (b), with E the event that the marbles are chosen green­green­red, F the event that they are 4  3  6  1, chosen green­red­green, and G the event that they are chosen red­green­green, we have P E  10 9 8 10

4  6  3  1 , and P G  6  4  3  1 . Thus, the probability that 1 red and 2 green marbles are chosen is P F  10 9 8 10 10 9 8 10 3 , as before. P E  P F  P G  10

13.3 BINOMIAL PROBABILITY 1. A binomial experiment is one in which there are exactly two outcomes. One outcome is called success and the other is called failure. 2. If a binomial experiment has probability p of success then the probability of failure is 1  p. The probability of getting

exactly r successes in n trials of this experiment is C n r pr 1  pnr .    3. P 2 successes in 5  C 5 2  072 033  013230    4. P 3 successes in 5  C 5 3  073 032  030870    5. P 0 success in 5  C 5 0  070 035  000243    6. P 5 successes in 5  C 5 5  075 030  016807    7. P 1 success in 5  C 5 1  071 034  002835    8. P 1 failure in 5  C 5 1  031 074  036015. Note that exactly one failure is the same as exactly 4 successes,    and P 4 successes in 5  C 5 4  074 031  036015. 9. P at least 4 successes  P 4 successes  P 5 successes  C 5 4 074 031  C 5 5 075 030  036015  016807  052822 10. P at least 3 successes  P 3 successes  P 4 successes  P 5 successes

 C 5 3 073 032  C 5 4 074 031  C 5 5 075 030  030870  036015  016807  083692

11. P at most 1 failure  P 0 failure  P 1 failure  P 5 successes  P 4 successes  C 5 5 075 030  C 5 4 074 031

 016807  036015  052822

12. P at most 2 failures  P 0 failure  P 1 failure  P 2 failure

 P 5 successes  P 4 successes  P 3 successes

 C 5 5 075 030  C 5 4 074 031  C 5 3 073 032

 016807  036015  030870  083692


18

CHAPTER 13 Counting and Probability

13. P at least 2 successes  P 2 successes  P 3 successes  P 4 successes  P 5 successes  013230  030870  036015  016807  096922

14. P at most 3 failures  P 0 failure  P 1 failure  P 2 failures  P 3 failures

 P 5 successes  P 4 successes  P 3 successes  P 2 successes  016807  036015  030870  013230  096922 (b)

15. (a) Outcome

Probability

1

02

2

02

3

02

4

02

5

02

0.2

0

1

2

3

4

5

In #16, include tick for 5 16. (a)

(b) Outcome

Probability

1

05

2

03

3

01

4

01

5

0

17. (a)

0.1

0

1

2

3

4

0

1

2

3

4

(b) r

Probability

0

4

1 16 1 4 3 8 1 4 1 16

r

Probability

0

00776

1

02592

2

03456

3

02304

4

00768

5

00102

1 2 3

18. (a)

1 16

(b)

0.5

0

1

2

3

4

5


SECTION 13.3 Binomial Probability

19. (a)

(b) r

Probability

0

02097

1

03670

2

02753

3

01147

4

00287

5

00043

6

000036

7

0000013

r

Probability

0

0000001

1

0000054

2

0001215

3

0014580

4

0098415

5

0354294

6

0531441

20. (a)

0.3

(b)

0

1

2

0

1

2

3

4

5

6

7

5

6

0.5

3

4

 2  4 5 21. Here “success” is “face is 4” and P face is 4  16 . Then P 2 successes in 6  C 6 2  16  020094. 6 22. Here P success  08 and P failure   02.   (a) P 0 success in 7  C 7 0  080 027  00000128    (b) P 7 successes in 7  C 7 7  087 020  0209715

(c) P the archer hits the target more than once  1  [P 0 success in 7  P 1 success in 7]         1  C 7 0  080 027  C 7 1  081 026  099963

(d) P at least 5 successes  P 5 successes  P 6 successes  P 7 successes           C 7 5  085 022  C 7 6  086 021  C 7 7  087 020  085197

   23. P 4 successes in 10  C 10 4  044 066  025082

24. The complement is that none of the raccoons had rabies.

   P at least one had rabies  1  P none had rabies  1  C 4 0  010 094  1  06561  03439

   25. (a) P 5 in 10  C 10 5  0455 0555  023403

(b) P at least 3  1  P at most 2  P 0 in 10  P 1 in 10  P 2 in 10            1  C 10 0  0450 05510  C 10 1  0451 0559  C 10 2  0452 0558  090044

   26. (a) P 12 in 15  C 15 12  0112 093  331695  1010

19


20

CHAPTER 13 Counting and Probability

(b) P at least 12  P 12 in 15  P 13 in 15  P 14 in 15  P 15 in 15        C 15 12  0112 093  C 15 13  0113 092       C 15 14  0114 091  C 15 15  0115 090  340336  1010

27. (a) The complement of at least one seed germinating is no seed germinating, so    P at least 1 germinates  1  P 0 germinates  1  C 4 0  0750 0254  099609.

(b) P at least 2 germinate  P 2 germinates  P 3 germinates  P 4 germinates           C 4 2  0752 0252  C 4 3  0753 0251  C 4 4  0754 0250

 094922    (c) P 4 germinates  C 4 4  0754 0250  031641

28. (a) P at least 3 boys  P 3 boys  P 4 boys  P 5 boys           C 5 3  053 052  C 5 4  054 051  C 5 5  055 050  05

(b) P at least 4 girls  P 4 girls  P 5 girls  P 6 girls  P 7 girls              C 7 4  054 053  C 7 5  055 052  C 7 6  056 051  C 7 7  057 050  05

   29. (a) P all 10 are boys  C 10 10  05210 0480  00014456.    (b) P all 10 are girls  C 10 0  0520 04810  000064925.    (c) P 5 in 10 are boys  C 10 5  0525 0485  024413.    30. (a) P 2 in 12  C 12 2  022 0810  028347. (b) The complement of “at least 3” is “at most 2”, so

P at least 3 in 12  1  P at most 2 in 12  1  [P 0 in 12  P 1 in 12  P 2 in 12]            1  C 12 0  020 0812  C 12 1  021 0811  C 12 2  022 0810

 044165    31. (a) P 3 in 3  C 3 3  00053 09950  0000000125.

(b) The complement of “one or more bulbs is defective” is “none of the bulbs is defective.” So    P at least 1 is defective  1  P none is defective  1  C 3 0  00050 09953  0014925.

   32. P at least 1 in 10  1  P 0 in 10  1  C 10 0  0050 09510  040126

33. The complement of “2 or more workers call in sick” is “fewer than 2 workers calls in sick.” So P 2 or more  1  [P 0 in 8  P 1 in 8]         1  C 8 0  0040 0968  C 8 1  0041 0967  0038147

   34. P 3 in 5 favor  C 5 3  063 042  03456

   35. (a) P 6 in 6  C 6 6  0756 0250  017798    (b) P 0 in 6  C 6 0  0750 0256  000024414    (c) P 3 in 6  C 6 3  0753 0253  013184


SECTION 13.3 Binomial Probability

21

(d) P at least 2 seasick  1  P at most 1 seasick  1  [P 6 in 6 OK  P 5 in 6 OK]         1  C 6 6  0756 0250  C 6 5  0755 0251  046606 36. (a) P machine breaks  P at least 1 component fails  1  P no component fails     1  C 4 0  0010 0994  0039404    (b) P no component fails  C 4 0  0010 0994  0960596    (c) P 3 components fail  C 4 3  0013 0991  000000396 37. (a) The complement of “at least one inherits the disease” is “none inherits the disease.” Then    P at least 1 inherits the disease  1  P none inherits the disease  1  C 4 0  0250 0754  068359. (b) P at least 3 inherit the disease  P 3 inherit the disease  P 4 inherit the disease        C 4 3  0253 0751  C 4 4  0254 0750  005078

38. There are 52 cards in the deck, of which   to each  suit, so P heart  P spade  P diamond  P club  025.  13 belong (a) P 3 in 3 are hearts  C 3 3  0253 0750  0015625    (b) P 2 in 3 are spades  C 3 2  0252 0751  0140625    (c) P 0 in 3 are diamonds  C 3 0  0250 0753  0421875    (d) P at least 1 is a club  1  P none is a club  1  C 3 0  0250 0753  0578125

39. Alex (a nonsmoker) is already in the room, concerns the four other participants assigned to the room.  exercise   so this (a) P 1 in 4 is a smoker  C 4 1  031

073  04116

   (b) P at least one smoker  1  P no smoker  1  C 4 0  030 074  07599

40. (a) P 2 or more  1  [P 0 in 100  P 1 in 100]         1  C 100 0  0020 098100  C 100 1  0021 09899  059673

(b) Since P at least 1 interested  1  P 0 interested and 09835  0507, the telephone consultant needs to make at least 35 calls to ensure at least a 05 probability of reaching one or more interested parties.

41. (a) P 8 or more recover  P 0 dies  P 1 dies  P 2 die  C 10 0 060 0410  C 10 1 061 049  C 10 2 062 048  00123

(b) Yes, the drug appears to be effective.

42. (a) P 5 or more hits  P 5 hits  P 6 hits  P 7 hits  P 8 hits  C 8 5 065 043  C 8 6 066 042  C 8 7 067 041  C 8 8 068 040  0594

(b) No, the coaching does not appear to have made any difference.


22

0.273438

CHAPTER 13 Counting and Probability

0.3

43. (a) Number of heads

Probability

0

0003906

1

003125

2

0109375

3

021875

4

0273475

0.1

5

021875

0.05

6

0109375

7

003125

8

0003906

0.25 0.2 Probability

0.15

0

2

4

6

8

Number of heads

If n  8, then 4 heads has the greatest probability of occurring. If the coin is flipped 100 times, then 50 heads has the greatest probability of occurring.

Height of bars is incorrect. Please fix

0.3

(b) Number of heads

Probability

0

0001953

1

0017578

2

0070313

3

0164063

4

0246094

5

0246094

6

0164063

7

0070313

8

0017578

9

0001953

0.25 0.2 Probability

0.15 0.1 0.05 0

2

4

6

8

Number of heads

If n  9, then 4 and 5 heads are the most likely outcomes. If the coin is flipped 101 times, then 50 and 51 heads are the most likely outcomes.

13.4 EXPECTED VALUE 1. If a game gives payoffs of $10 and $100 with probabilities 09 and 01, respectively, then the expected value of this game is E  10  09  100  01  $19. 2. If you played the game in Exercise 1 many times then you would expect your average payoff per game to be about $19.     3. You win $2 with probability 12 and $1 with probability 12 . Thus, E  2 12  1 12  15, and so your expected winnings are $150 per game.

    4. The probability that you win $10 is 16 , and the probability that you lose $1 is 56 . Thus E  10 16  1 56  0833, and so your expectation is $0833.

1 , the expected value of this game is 5. Since the probability of drawing the ace of spades is 52     1  1 51  49  094. So your expected winnings are $094 per game. E  100 52 52 52     6. The expected value of this game is E  3 12  2 12  52  25. So your expected winnings are $250 per game.


SECTION 13.4 Expected Value

 

23

 

7. Since the probability that you roll a six is 16 , the expected value of this game is E  3 16  050 56  55 6  09167. So you expect to win $092 per game.  2  2 8. The probability that you get two tails is 12  14 , the probability that you get one tail and one head is C 2 1 12  12 ,  2 and the probability that you get two heads is 12  14 . If you get two heads, you will receive $4, if you get one head and one tail, you will get $2 $1  $1, and if you get two tails, you will lose $2. Thus the expected value of this game is       1 1 E  4 4  1  2 14  1. So your expected winnings are $1 per game. 2

9. Since the probability that the die shows an even number equals the probability that that die shows an odd number, the     expected value of this game is E  2 12  2 12  0. So you should expect to break even after playing this game many times.

10. Since there are 4 aces, 12 face cards, and only one 8 of clubs, the expected value of this game is       4  26 12  13 1  $1425. E  104 52 52 52

11. Since it costs $050 to play, if you get a silver dollar, you win only 1  050  $050. Thus the expected value of this game     2  050 8  030. So your expected winnings are $030 per game. In other words, you is E  050 10 10 should expect to lose $030 per game.

8  7  56 , and the probability of not choosing 12. The probability of choosing two white balls (that is, no black ball) is 10 9 90   56 34 two white balls (that is, at least one black ball) is 1  90  90 . Therefore, the expected value of this game is     34 E  5 56 90  0 90  3111. Thus, your expected winnings are $311 per game.     1  1 37   2  00526. 13. You can either win $35 or lose $1, so the expected value of this game is E  35 38 38 38

Thus, the expected value is $00526 per game.

1 . After the first prize winner is selected, then 2  106 1 1 P winning the second prize  . Similarly, P winning the third prize  . So the expected 6 2  10  1 2  106  2             1 1 1  105  104  $0555. value of this game is E  106 6 6 2  10 2  10  1 2  106  2 (b) Since we expect to win $0555 on the average per game, if we pay $100, then our net outcome is a loss of $0445 per game. Hence, it is not worth playing, because on average you will lose $0445 per game.

14. (a) We have P winning the first prize 

15. By the rules of the game, a player can win $10 or $5, break even, or lose $100. Thus the expected value of this game is         10  5 10  100 2 78 E  10 100 100 100  0 100  050. So the expected winnings per game are $050.

16. Since the safe has a six digit combination, there are 106 possible combinations to the safe, of which only one is correct. The     1  106  1 6  0. expected value of this game is E  10  1  1 106 106 17. If the stock goes up to $20, the investors make $20  $5  $15; if the stock falls to $1, they lose $5  $1  $4. So the expected value of the profit is E  15 01  4 09  21. Thus, the investors’ expected profit per share is $210, that is, they should expect to lose $210 per share. They did not make a wise investment.  3 1 18. Since the wheels of the slot machine are independent, the probability that you get three watermelons is 11 . So the     expected value of this game is E  475 13  025 1  13  $0246. 11

11


24

CHAPTER 13 Counting and Probability

19. There are C 49 6 ways to select a set of six numbers from the group of 49 numbers, of which only one is a winning set.      1 1 Thus the expected value of this game is E  106  1  1 1   $093. C 49 6 C 49 6 20. (a) Since the life insurance policy costs $25 per year, we have the expected value E   7500  25 00003  25 09997  2275. (b) The expected yearly income is 450,000 2275  $10,237,500.

21. The expected number is E  2 015  3 045  4 030  5 010  335 hours of TV. 22. The expected number is 005 3  015 2  045 1  035 0  09 foreign languages.

23. The expected number is 3 030  2 045  1 015  0 010  195 times in any given week. 24. (a) Number of girls

Probability

0

1 8 3 8 3 8 1 8

1 2 3

(b) The expected number of girls is         0 18  1 38  2 38  3 18  15.

25. (a) A standard deck contains 1 ace of spades and 48 cards that are not aces. Thus, the expected value is   1 12  48 1   3  $023, and so the game is not fair. 52 52 2 13   1 x  48 1  0  x  $2400. (b) The game would be fair with payout x, where 52 52 2

26. (a) The expected value is 13 20  23 10  0, and so the game is fair.

3070 10 27. (a) The expected value is 16  16 30  35 36 2  36   9 , and so the game is not fair. 1 x  35 2  0  x  $70. (b) The game would be fair with payout x, where 36 36

1 28. (a) The expected value is 16  12 10  11 12 1   12 , and so the game is not fair.

(b) The game would be fair with payout x, where 16  12 x  11 12 1  0  x  $11.   1  1  1 600  1  1  1  1 1   23 , and so the game is not fair. 29. (a) The expected value is 52 6 2 52 6 2 624   1 1 1 1  1  1  0  x  $623. (b) The game would be fair with payout x, where 52  6  2 x  1  52 6 2   30. (a) The expected value is 28 1  68 12   18 , and so the game is not fair.   (b) The game would be fair with payout x, where 28 x  68 12  0  28 x  38  x  $150.

31. If you win, you win $1 million minus the price of the stamp. If you lose, you lose only the price of the stamp (currently

63

44 cents). So the expected value of this game is 999,99956  expect to lose 39 cents on each entry, and so it’s not worth it.

20  106  1 1    039. Thus, you 044 20  106 20  106

-0.63 -0.58 Currently the price of a first class stamp is $0.63.


CHAPTER 13

Review

25

CHAPTER 13 REVIEW 1. The number of possible outcomes is       number of outcomes number of outcomes number of ways    2 6 52  624. when a coin is tossed a die is rolled to draw a card 2. (a) The number of 3­digit numbers that can be formed using the digits 1–6 if a digit can be used any number of times is 6 6 6  216. (b) The number of 3­digit numbers that can be formed using the digits 1–6 if a digit can be used only once is 6 5 4  120.

3. (a) Order is not important, and there are no repetitions, so the number of different two­element subsets is 54 5!   10. C 5 2  2! 3! 2

5!  20. 3! 4. Since the order in which the people are chosen is not important and a person cannot be bumped more than once (no (b) Order is important, and there are no repetitions, so the number of different two­letter words is P 5 2 

repetitions), the number of ways that 7 passengers can be bumped is C 120 7  59488  1010 . 5. You earn a score of 70% by answering exactly 7 of the 10 questions correctly. The number of different ways to answer the 10!  120. questions correctly is C 10 7  7! 3! 6. There 2 ways to answer each of the 10 true­false questions and 4 ways to answer each of the 5 multiple­choice questions.    So the number of ways that this test can be completed is 210 45  1,048,576.

7. You must choose 2 of the 10 questions to omit, and the number of ways of choosing these 2 questions is 10!  45. C 10 2  2! 8! 8. Since the order of the scoops of ice cream is not important and the scoops cannot be repeated, the number of ways to have a banana split is C 15 4  1365. 9. The maximum number of employees using this security system is       number of choices number of choices number of choices    26 26 26  17,576. for the first letter for the second letter for the third letter 10. Since there are n! ways to arrange a group of size n and 5!  120, there are 5 students in this class. 11. We could count the number of ways of choosing 7 of the flips to be heads; equivalently we could count the number of ways of 10!  120. choosing 3 of the flips to be tails. Thus, the number of different ways this can occur is C 10 7  C 10 3  3! 7! 12. The number ways to form a license plate consisting of 2 letters followed by 3 numbers is 26 26 10 10 10  676,000. Since there are fewer possible license plates than 700,000, there must be fewer than 700,000 licensed cars in the Yukon. x! x! 13. Let x be the number of people in the group. Then C x 2  10   10   20  x x  1  20 2! x  2! x  2!  x 2  x  20  0  x  5 x  4  0  x  5 or x  4. So there are 5 people in this group.

14. Each topping corresponds to a subset of a set with n elements. Since a set with n elements has 2n subsets and 211  2048, the pizza parlor offers 11 toppings. 15. A letter can be represented by a sequence of length 1, a sequence of length 2, or a sequence of length 3. Since each symbol is either a dot or a dash, the possible number of letters is       number of letters number of letters number of letters    23  22  2  14. using 3 symbols using 2 symbols using 1 symbol 16. Since the nucleotides can be repeated, the number of possible words of length n is 4n . Since 42  16  20 and 43  64, the minimum length of word needed is 3.


26

CHAPTER 13 Counting and Probability

17. (a) Since we cannot choose a major and a minor in the same subject, the number of ways a student can select a major and a minor is P 16 2  16  15  240. (b) Again, since we cannot have repetitions and the order of selection is important, the number of ways to select a major, a first minor, and a second minor is P 16 3  16  15  14  3360.

(c) When we select a major and 2 minors, the order in which we choose the minors is not important. Thus the number of     number of ways number of ways to ways to select a major and 2 minors is   16  C 15 2  16  105  1680. to select a major select two minors

18. (a) Solution 1: Since the leftmost digit of a three­digit number cannot be zero, there are 9 choices for this first digit and 10 choices for each of the other two digits. Thus, the number of three­digit numbers is 9 10 10  900. Solution 2: Since there are 999 numbers between 1 and 999, of which the numbers between 1 and 99 do not have three digits, there are 999  99  900 three­digit numbers. (b) There are 1001 numbers from 0–1000. From part (a), there are 900 three­digit numbers. Therefore the probability that 900  0899. the number chosen is a three­digit number is P E  1001

19. Because the letters are distinct, the number of anagrams of the word RANDOM is 6!  720. 20. Because two letters are the same, the number of anagrams of the word BLOB is

4!  12. 2!

21. Because three letters are the same, the number of anagrams of the word BUBBLE is

6!  120. 3!

22. Because there are two sets of four indistinguishable letters (I, S) and one set of two indistinguishable letters (P), the number 11! of anagrams of the word MISSISSIPPI is  34,650. 4! 4! 2! 23. (a) The possible number of committees is C 18 7  31,824.

(b) Since we must select the 4 graduates from the group of 10 graduates and the 3 undergraduates from the group of 8 undergraduates, the possible number of committees is     number of ways to number of ways to   C 10 4  C 8 3  210  56  11,760. choose 4 of 10 graduates choose 3 of 8 undergraduates (c) We remove Alex (an undergraduate) from the group of 18, so the possible number of committees is C 17 7  19,448. (d) The possible number of committees is       possible number of possible number of possible number of   committees with 5 undergraduates committees with 6 undergraduates committees with 7 undergraduates  C 8 5  C 10 2  C 8 6  C 10 1  C 8 7  C 10 0  56  45  28  10  8  1  2808

(e) Since the committee is to have 7 members, “at most 2 graduates” is the same as “at least 5 undergraduates,” which we found in part (d). So there are 2808 possibilities. (f) We select the specific offices first, then complete the committee from the remaining members of the group. So the number of possible committees is     number of ways to choose number of     ways to choose  a chairman, a vice­chairman,      P 18 3  C 15 4  4896  1365  6,683,040. and a secretary four other members

24. Method 1: We choose the 5 states first and then one of the two senators from each state. Thus the number of committees is C 50 5  25  67,800,320. Method 2: We choose one of 100 senators, then choose one of the remaining 98 senators (deleting the chosen senator and the other senator form that state), then choose one of the remaining 96 senators, continuing this way until the 5 senators are chosen. Finally, we need to divide by the number of ways to arrange the 5 senators. Thus the number of committees is 10098969492  67,800,320. 5!


CHAPTER 13

Review

27

2 25. (a) The probability that the marble is red is 10 15  3 .

8 . (b) The probability that the marble is even numbered is 15 2. (c) The probability that the marble is white and an odd number is 15

7 5 12 4 (d) The probability that the marble is red or odd numbered is P red P odd P red  odd  10 15  15  15  15  5 .

26. Let Rn denote the event that the nth marble is red and let Wn denote the event that the nth marble is white. 9 3 (a) P both marbles are red  P R1  R2   P R1   P R2  R1   10 15  14  7 . (b) Solution 1: The probability that one is white and that the other is red is C 10 1  C 5 1 10 number of ways to select one white and one red   . number of ways to select two marbles C 15 2 21 Solution 2:

P one white and one red  P W1  R2   P R1  W2   P W1   P R2  W1   P R1   P W2  R1  5  10  10  5  10  15 14 15 14 21

(c) Solution 1: Let E be the event “at least one is red”. Then E  is the event “both are white”.   5  4  2 . Thus P E  1  2  19 . P E   P W1  W2   P W1   P W2  W1   15 14 21 21 21

9 19 Solution 2: P at least one is red  P one red and one white  P both red  10 21  21  21 (from (a) and (b)).

(d) Since 5 of the 15 marbles are both red and even­numbered, the probability that both marbles are red and even­numbered 5  4  2 . is 15 14 21

2  1  1 . (e) Since 2 of the 15 marbles are both white and odd­numbered, the probability that both are white is 15 14 105

27. (a) S  H H H H H T H T H H T T T H H T H T T T H T T T . (b) P H H H  18

(c) P 2 or more heads  P exactly 2 heads  P 3 heads  38  18  48  12 (d) P tails on the first toss  48  12

28. The probability that you select a mathematics book is

4 2 number of ways to select a mathematics book    04. number of ways to select a book 10 5

29. Since rolling a die and selecting a card are independent, 4  1. P both show a six  P die shows a six  P card is a six  16  52 78 4  1 30. (a) P ace  52 13

(b) Let E be the event the card chosen is an ace, and let F be the event the card chosen is a jack. Then 4  4  2. P E  F  P E  P F  52 52 13

(c) Let E be the event the card chosen is an ace, and let F be the event the card chosen is a spade. Then 4  13  1  4 . P E  F  P E  P F  P E  F  52 52 52 13

(d) Let E be the event the card chosen is an ace, and let F be the event the card chosen is a red card. Then n E  F 2  1. P E  F   52 26 n S 1  1  1  1 . 31. (a) Since these events are independent, the probability of getting the ace of spades, a six, and a head is 52 6 2 624 1 1 1 (b) The probability of getting a spade, a six, and a head is 13 52  6  2  48 .

3 1 3 (c) The probability of getting a face card, a number greater than 3, and a head is 12 52  6  2  52 .

32. (a) The probability the first die shows some number is 1, and the probability the second die shows the same number is 16 . So the probability each die shows the same number is 1  16  16 .


28

CHAPTER 13 Counting and Probability

(b) By part (a), the event of showing the same number has probability of 16 , and the complement of this event is that the dice show different numbers. Thus the probability that the dice show different numbers is 1  16  56 . 33. (a) Since there are four kings in a standard deck, P 4 kings 

1 1 C 4 4  52515049   369  106 . C 52 4 270,725

4321 13121110 C 13 4 4321 11  000264. (b) Since there are 13 spades in a standard deck, P 4 spades   52515049  4165 C 52 4 4321 2  26252423 2  C 26 4 4321 92  011044. (c) Since there are 26 red cards and 26 black cards, P all same color   833  52515049 C 52 4 4321

34. In the numbers game lottery, there are 1000 possible “winning” numbers. 1 . (a) The probability that the player wins $500 is 1000 (b) There are P 3 3  6 ways to arrange the digits 1, 5, 9. However, if the player wins only $50, it means that their 5  1 . number 159 was not the winning number. Thus the probability is 1000 200

35. The contestant knows the first digit and must arrange the other four digits. Since only one of the P 4 4  24 arrangements 1. is correct, the probability that the contestant guesses correctly is 24

36. The number of different pizzas is the number of subsets of the set of 12 toppings, that is, 212  4096. The number of pizzas with anchovies is the number of ways of choosing anchovies and then choosing a subset of the 11 remaining toppings, that 1 is, 1  211  2048. Thus, P getting anchovies  2048 4096  2 .

Note that this makes intuitive sense: for each pizza combination without anchovies there is a corresponding one with anchovies, so half will have anchovies and half will not. 37. (a) Since there are only two sock colors, any three socks must contain a matching pair. (b) Method 1: If the two socks drawn form a matching pair then they are either both red or both blue. So     C 20 2 C 30 2 choosing a both red or   051. P P  P both red  P both blue  C 50 2 C 50 2 matching pair both blue Method 2: The complement of choosing a matching pair is choosing one sock of each color. So C 20 1  C 30 1  1  049  051. P choosing a matching pair  1  P different colors  1  C 50 2 38. (a) number of codes  choices for 1st digit  choices for 2nd digit      choices for 5th digit  10  10  10  10  10  105  100,000

(b) Since there are five numbers (0, 1, 6, 8 and 9) that can be read upside down, we have number of codes  choices for 1st digit  choices for 2nd digit      choices for 5th digit  55  3125.

55 n E 1  5  . n S 32 10 (d) Suppose a zip code is turned upside down. Then the middle digit remains the middle digit, so it must be a digit that reads the same when turned upside down, that is, a 0, 1 or 8. Also, the last digit becomes the first digit, and the next to last digit becomes the second digit. Thus, once the first two digits are chosen, the last two are determined. Therefore, the number of zip codes that read the same upside down as right side up is number of codes  choices for 1st digit  choices for 2nd digit      choices for 5th digit  5  5  3  1  1  75. (c) Let E be the event that a zip code can be read upside down. Then by parts (a) and (b), P E 

39. (a) Order is important, and repeats are possible. Thus there are 10 choices for each digit. So the number of different Zip+4 codes is 10  10      10  109 .

(b) If a Zip+4 code is to be a palindrome, the first 5 digits can be chosen arbitrarily. But once chosen, the last 4 digits are determined. Since there are 10 ways to choose each of the first 5 digits, there are 105 palindromes. 5

(c) By parts (a) and (b), the probability that a randomly chosen Zip+4 code is a palindrome is 109  104 . 10


CHAPTER 13

Review

29

40. (a) Using the rule for the number of distinguishable combinations, the number of divisors of N is 7  1 2  1 5  1  144.

(b) An even divisor of N must contain 2 as a factor. Thus we place a 2 as one of the factors and count the number of distinguishable combinations of M  26 32 55 . So using the rule for the number of distinguishable combinations, the number of even divisors of N is 6  1 2  1 5  1  126.

(c) A divisor is a multiple of 6 if 2 is a factor and 3 is a factor. Thus we place a 2 as one of the factors and a 3 as one of the factors. Then we count the number of distinguishable combinations of K  26 31 55 . So using the rule for the number of distinguishable combinations, the number of even divisors of N is 6  1 1  1 5  1  84.

126 7 (d) Let E be the event that the divisor is even. Then using parts (a) and (b), P E  nE nS  144  8 . 4  1 41. (a) P king  52 13

8  2 (b) P king or ace  52 13

4 1 number of kings   . number of face cards 12 3 number of kings 4 1 (d) The probability that the card is a king given that it is not an ace is   . number of non­aces 48 12  3  0 1 1 12 .  42. (a) Because each card is replaced, the probability that all three cards are kings is C 3 3 13 13 2197 4 4 4 1 1 Another method:We can calculate the probability as    3  . 52 52 52 2197 13  2  1 1 36 12 (b) The probability that exactly two cards are jacks is C 3 2 .  13 13 2197  0  3 1000 3 10 .  (c) The probability that none of the cards is a face card is C 3 0 13 13 2197 (d) “At least one of the cards is a face card” is the complement of the event in part (c), so its probability is 1000 1197 1  . 2197 2197  4  4 5 43. (a) P 4 sixes in 8 rolls  C 8 4  16  0026048. 6 (c) The probability that the card is a king given that it is a face card is

(b) There are three even numbers on a die and three odds numbers, so P even  P odd  05. Thus

P 2 or more evens in 8 rolls  1  P fewer than 2 evens  1  [P 0 evens  P 1 even]         1  C 8 0  050 058  C 8 1  051 057  096484

   44. (a) P 5 in 5 are white flesh  C 5 5  035 070  000243    (b) P 0 in 5 are white flesh  C 5 0  030 075  016807    (c) P 2 in 5 are white flesh  C 5 2  032 073  03087

(d) P 3 or more are red flesh  P 3 in 5 are red flesh  P 4 in 5 are red flesh  P 5 in 5 are red flesh           C 5 3  032 073  C 5 4  031 074  C 5 5  030 075  083692

45. (a) The probability that 9 or more patients would have recovered without the drug is C 12 9 0659 0353  C 12 10 06510 0352  C 12 11 06511 0351  C 12 12 06512 0350  0347.

(b) No, the drug does not appear to be effective.


30

CHAPTER 13 Counting and Probability

46. The probabilities are as follows: 0 head: C 4 0 070 034  00081 1 head: C 4 1 071 033  00756

2 heads: C 4 2 072 032  02646 3 heads: C 4 3 073 031  04116 4 heads: C 4 4 074 030  02401

Outcome (heads)

Probability

0

00081

1

00756

2

02646

3

04116

4

02401

47. There are 36 possible outcomes in rolling two dice and 6 ways in which both dice show the same number, namely, 1 1,     6  1 30  0. 2 2, 3 3, 4 4, 5 5, and 6 6. So the expected value of this game is E  5 36 36 1 , 48. Using the same logic as in Exercise 32(a), the probability that all three dice show the same number is 1  16  16  36

1  35 . Thus, the expected value of this game is while the probability they are not all the same is 1  36 36     1 35 30 E  5 36  1 36   36  083. So your expected winnings per game are $083; that is, you expect to lose

$083 per game.

49. Since the contestant makes a guess as to the order of ratification of the 13 original states, the number of such guesses is 1 . Thus, the expected value is P 13 13  13!, while the probability that the contestant guesses the correct order is 13!     1 13!  1 E  1,000,000  0  000016. So the contestant’s expected winnings are $000016. 13! 13!

50. The expected number of times the athlete goes jogging in any given week is 04 3  01 2  02 1  03 0  16.

CHAPTER 13 TEST 1. The order is fixed, but for each grandchild they have three choices of pictures. Thus, the number of possibilities is 3  3  3  3  81.

2. There are 4 main courses, 3 desserts, and 6 drinks to choose from, so the total number of possibilities is 4  3  6  72. 3. (a) If repetition is allowed, then each letter can be chosen in 26 ways and each digit in 10 ways, so the number of possible passwords is 264  103  456,976,000.

(b) If repetition is not allowed, then the first letter can be chosen in 26 ways, the second in 25 ways, the third in 24 ways, and the fourth in 23 ways. The first digit can be chosen in 10 ways, the second in 9 ways, and the third in 8 ways. Thus, in this case the total number of possible passwords is 26  25  24  23  10  9  8  258,336,000. 4. (a) Order is important in the arrangement, therefore the number of ways to arrange P 30 4  657,720.

(b) Here we are interested in the group of books to be taken on vacation, so order is not important. Therefore, the number of ways to choose these books is C 30 4  27,405.

5. There are two choices to be made: Choose a road to travel from Ajax to Barrie, and then choose a different road from Barrie to Ajax. Since there are 4 roads joining the two cities, we need the number of permutations of 4 objects (the roads) taken 2 at a time (the road there and the road back). This number is P 4 2  4  3  12. 6. A customer must choose a size of pizza and must make a choice of toppings. There are 4 sizes of pizza, and each choice

of toppings from the 14 available corresponds to a subset of the 14 objects. Since a set with 14 objects has 214 subsets, the number of different pizzas this parlor offers is 4  214  65,536. 7. (a) We want the number of ways of arranging 4 distinct objects (the letters L, O, V, E). This is the number of permutations of 4 objects taken 4 at a time. Therefore, the number of anagrams of the word LOVE is P 4 4  4!  24.


CHAPTER 13

Test

31

(b) We want the number of distinguishable permutations of 6 objects (the letters K, I, S, S, E, S) consisting of three like groups of size 1 and a like group of size 3 (the S’s). Therefore, the number of different anagrams of the word KISSES is 6! 6!   120. 1! 1! 1! 3! 3! 8. We choose the officers first. Here order is important, because the officers are different. Thus there are P 30 3 ways to do this. Next we choose the other 5 members from the remaining 27 members. Here order is not important, so there are C 27 5 ways to do this. Therefore the number of ways that the board of directors can be chosen is 27!  1,966,582,800. P 30 3  C 27 5  30  29  28  5! 22! 9. One card is drawn from a standard 52­card deck. 1 (a) Since there are 26 red cards, the probability that the card is red is 26 52  2 . 4  1 . (b) Since there are 4 kings, the probability that the card is a king is 52 13

2  1. (c) Since there are 2 red kings, the probability that the card is a red king is 52 26

10. Let R be the event that the ball chosen is red. Let E be the event that the ball chosen is even­numbered. 5  03846. (a) Since 5 of the 13 balls are red, P R  13 6  04615. (b) Since 6 of the 13 balls are even­numbered, P E  13

5  6  2  9  06923. (c) P R or E  P R  P E  P R  E  13 13 13 13

11. Let E be the event of choosing 3 males. Then number of ways to choose 3 males C 5 3 n E    0022. P E  n S number of ways to choose 3 goldfish C 15 3

n E 6 1   . n S 36 6 13. There are 4 students and 12 astrological signs. Let E be the event that at least 2 have the same astrological sign. Then E  is the event that no 2 have the same astrological sign. It is easier to find E  . So   number of ways to assign 4 different astrological signs P 12 4 55 12  11  10  9   .  P E  number of ways to assign 4 astrological signs 12  12  12  12 96 124   41 Therefore, P E  1  P E  1  55 96  96  0427.    14. (a) P 6 heads in 10 tosses  C 10 6  0556 0454  023837. 12. Two dice are rolled. Let E be the event of getting doubles. Since a double may occur in 6 ways, P E 

(b) “Fewer than 3 heads” is the same as “0, 1, or 2 heads.” So

P fewer than 3 heads  P 0 head in 10  P 1 head in 10  P 2 heads in 10           C 10 0  0550 04510  C 10 1  0551 0459  C 10 2  0552 0458  002739

4  1 , the probability 15. A deck of cards contains 4 aces, 12 face cards, and 36 other cards. So the probability of an ace is 52 13

3 , and the probability of a non­ace, non­face card is 36  9 . Thus the expected value of this game of a face card is 12  13 52 13  52     1 3 9 85 is E  10 13  1 13  5 13  13  0654, that is, about $065.


32

FOCUS ON MODELING

FOCUS ON MODELING The Monte Carlo Method 1. (a) You should find that with the switching strategy, you win about 90% of the time. The more games you play, the closer to 90% your winning ratio will be. 1 , since there are 10 doors and (b) The probability that the contestant has selected the winning door to begin with is 10 9 . If the contestant switches, only 1 is a winner. So the probability that the contestant has selected a losing door is 10 1 , and the they exchange a losing door for a winning door (and vice versa), so the probability that they lose is now 10 9 . probability that they win is now 10

2. (a) You should find that you get a combination consisting of one head and one tail about 50% of the time. (b) The possible gender combinations are B B BG G B GG. Thus, the probability of having one child of each sex is 2  1. 4 2

3. (a) You should find that player A wins about 78 of the time. That is, if you play this game 80 times, player A should win approximately 70 times. (b) The game will end when either player A gets one more head or player B gets three more tails. Each toss is independent, and both heads and tails have probability 12 , so we obtain the following probabilities. Outcome

Probability

H

1 2 1  1  1 2 2 4 1  1  1  1 2 2 2 8 1  1  1  1 2 2 2 8

TH TTH TTT

Player A

Since Player A wins for any outcome that ends in heads, the probability that he wins is 12  14  18  78 . 4. (a) If you simulate 80 World Series with coin tosses, you should expect the series to end in 4 games about 10 times, in 5 games about 20 times, in 6 games about 25 times, and in 7 games about 25 times. (b) We first calculate the number of ways that the series can end with team A winning. (Note that a team must win the final game plus three of the preceding games to win the series.) To win in 4 games, team A must win 4 games right off the bat, and there is only 1 way this can happen. To win in 5 games, team A must win the final game plus 3 of the first 4 games, so this can happen in C 4 3  4 ways. To win in 6 games, team A must win the final game plus 3 of the first 54 5 games, so this can happen in C 5 3   10 ways. To win in 7 games, team A must win the final game plus 3 21 654 of the first 6 games, so this can happen in C 6 3   20 ways. By symmetry, it is also true for team B that 321 they can win in 4 games 1 way, in 5 games 4 ways, in 6 games 10 ways, and in 7 games 20 ways. The probability that any particular team wins a given game is 12 ; this fact, together with the assumption that the games are independent allows us to calculate the probabilities in the following table. Series

Number of possibilities

4 games

2

5 games

8

6 games

20

7 games

40

Probability   2  12  12  12  12  18   8  12  12  12  12  12  14   5 20  12  12  12  12  12  12  16   5 40  12  12  12  12  12  12  12  16


The Monte Carlo Method

33

5  6  5  7  5 13  58. Thus, on average, we expect a World Series to end in (c) The expected value is 18  4  14  5  16 16 16

about 58 games.

5. With 1000 trials, you are likely to obtain an estimate for  that is between 31 and 32. 6. Modify the TI­84 program in Problem 5 to the following: PROGRAM:PROB6 :0P :For(N,1,1000) :randX:randY :P+(X2 Y)P :End :Disp “PROBABILITY IS APPROX”,P/1000 You should find that the probability is very close to 13 . 7. (a) We can use the following TI­84 program to model this experiment. It is a minor modification of the one given in Problem 5. PROGRAM:PROB7 :0P :For(N,1,1000) :randX:randY :P+((X+Y)1)P :End :Disp “PROBABILITY IS APPROX”,P/1000 You should find that the probability is very close to 12 . (b) Following the hint, the points in the square for which x  y  1 are the ones that lie below the line x  y  1. This triangle has area 12 (it takes up half the square), so the probability that x  y  1 is 12 .


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