Solution manual for precalculus 7th edition by cohen

natural number, integer, rational number b.. natural number, integer, rational number 9.. We need to find two irrational numbers a and b such that their sum is rational.. We need to find

Chapter 1 Fundamentals

1.1 Sets of Real Numbers

1 a integer, rational number b rational number

2 a rational number b irrational number

3 a natural number, integer, rational number b rational number

4 a rational number b rational number

5 a rational number b rational number

6 a irrational number b irrational number

7 irrational number 8 natural number, integer, rational number

9 irrational number 10 rational number

11 Since 11

4 2.75, sketch the graph: 12 Since 7

8  0.875, sketch the graph:

13 Since 1 2 | 2.4 , sketch the graph: 14 Since 1 2 | 0.4 , sketch the graph:

15 Since 2 1 | 0.4 , sketch the graph: 16 Since  2  1 | 2.4 , sketch the graph:

17 Since 2 3 | 3.1, sketch the graph: 18 Since 2 3 | 0.3 , sketch the graph:

21 Sketching the graph: 22 Sketching the graph:

27 Sketching the graph: 28 Since 32 1.5, sketch the graph:

29 Since 1|S

3, sketch the following graph: 30 Sketching the graph:

33 True (since –2 = –2, it is also true that –”–2) 34 True since 7! 2

35 False 36 True since 0.777  ! 0.7000

37 False (since 2S | 6.2 ) 38 True since S +1

40 True (since S2 | 9.61, using the approximation S | 3.1 )

41 The inequality notation is 2 x  5 : 42 The inequality notation is 2  x  2 :

43 The inequality notation is 1d x d 4 : 44 The inequality notation is 3

2 d x d1

2 :

45 The inequality notation is 0d x  3 : 46 The inequality notation is 4  x d 0 :

47 The inequality notation is 3  x  f : 48 The inequality notation is 2 x  f :

49 The inequality notation is 1 d x  f  50 The inequality notation is 0d x  f :

51 The inequality notation is f  x  1 : 52 The inequality notation is f  x  2 :

54 The inequality notation is f  x S Note that this is the entire number line:

55 a Since 43 4

| 3.16 , it agrees with S to one decimal place

b Since 227 | 3.142, it agrees with S to two decimal places

c Since 355113| 3.1415929, it agrees with S to six decimal places

¹¸| 3.1415926538 , it agrees with S to nine decimal places

56 a We need to find two irrational numbers a and b such that their sum is rational If we choose a 2 and

b  2 , then a  b 2  2 0 , which is rational.

b We need to find two irrational numbers a and b such that their sum is irrational If we choose a 2 and

b 3 , then a  b 2  3 , which is irrational.

57 a We need to find two irrational numbers a and b such that their product is rational If we choose

a 2 and b 8 , then ab 2 • 8 16 4 , which is rational.

b We need to find two irrational numbers a and b such that their product is irrational If we choose

a 2 and b 3 , then ab 2 • 3 6 , which is irrational.

58 a We need to find two irrational numbers a and b such that their quotient is rational If we choose a 12 and

59 a Raising 2 to the 1/2 power results in 21/2 2 , which is irrational

b Raising 2 to the 2 power results in 2 2

2Since the result is rational, we have an example of an irrational number 2 2

raised to an irrational power 2

yielding a rational number

11 Simplify the expression:  8   9 8  9 17 17

12 Simplify the expression:  8   9 8  9  1 1

13 Simplify the expression: 27 5

5 27

22

22  1 1 14 Simplify the expression:

275 527

22

22

22

22 1

15 Simplify the expression: 7(8)  7  8  56  7(8) 56  56 0

16 Simplify the expression: (7)2   72    3 3

1  52

1 52

1  52

1 52

33 Since x < 3, x – 3 < 0 and x – 4 < 0, and thus: x  3  x  4 (x  3)  (x  4)> @  x  3  x  4 2x  7

34 Since x > 4, x – 3 > 0 and x – 4 > 0, and thus: x  3  x  4 (x  3)  (x  4) 2x  7

35 Since 3 < x < 4, x – 3 > 0 and x – 4 < 0, and thus: x  3  x  4 (x  3)  (x  4)> @ x  3  x  4 1

42 The absolute value inequality can be written as x 1 ! 1

2

43 The absolute value inequality can be written as y  (4)  1, or y  4  1

44 The absolute value inequality can be written as x3 (1) d 0.001, or x3 1 d 0.001

45 The absolute value inequality can be written as y  0  3, or y  3

46 The absolute value inequality can be written as y  t  1

47 The absolute value inequality can be written as x2 a2  M

48 The absolute value inequality can be written as a  0  b  0 t a  b  0 , or a  b t a  b

49 Graphing the interval x  4 : 50 Graphing the interval x  2 :

51 Graphing the interval x ! 1 : 52 Graphing the interval x ! 0 , noting that x 

53 Graphing the interval x 5  3 : 54 Graphing the interval x 4  4 :

55 Graphing the interval x 3 d 4 : 56 Graphing the interval x 1 d 1

59 Graphing the interval x 5 t 2 : 60 Graphing the interval x 5 t 2 :

61 a Graphing the interval x 2  1 :

b Graphing the interval 0 x  2  1, noting that x = 2 is excluded:

c The interval in part b does not include 2.

62 Since a = (a – b) + b, a (a  b)  b d a  b  b Subtracting b from each side, a  b d a  b

63 Using the triangle inequality twice: a  b  c a  (b  c) d a  b  c d a  b  c

64 Since the absolute value is always non-negative, x2  4 x t 0 Thus x2  4 x 12 cannot have any real

solutions

65 Consider three cases: a = b, a > b, and a < b

case 1: If a = b, then max (a, b) = a, now verifying: a  b  a  b

Thus the equation is verified.

case 3: If a < b, then max (a, b) = b, and since a – b < 0, we have:

Thus the equation is verified

66 Consider three cases: a = b, a > b, and a < b:

case 1: If a = b, then min(a, b) = a, now verifying: a  b  a  b

Thus the equation is verified

case 2: If a > b, then min(a, b) = b, and since a – b > 0: a  b  a  b

Thus the equation is verified

case 3: If a < b, then min(a, b) = a, and since a – b < 0:

c Since (a)  (b) d a  b , (a  b) d a  b

d Since a  b d a  b and (a  b) d a  b , a  b d a  b , since a  b is either a + b or –(a + b)

1.3 Solving Equations (Review and Preview)

1 Substituting x 2 into the equation: 4 x 5 4(2)  5 8  5 13

So x 2 is a solution to the equation

2 Substituting x 2 into each side of the equation:

1

x

1 2

3

x 1 3

2 1 1 2

So x 2 is a solution to the equation

3 Substituting y 3 into each side of the equation:

7

9 3 127

So y 3 is not a solution to the equation

4 Substituting y 5 into the equation: ( y  1)(y  5) (5  1)(5  5) (4)(10) 40

So y 5 is not a solution to the equation

5 Substituting m 14 into the equation: m2  m  5

16

1 4

2

 1 4

 5 16

1

16 1

4 5 16

5

16 5

16 0

So m 1

4 is a solution to the equation

6 Substitute 1 5 for x in the equation: 1  5 x 3 and then solving for x:

1 55 11

5510 55

x  5  x  5 2x  1

2 x 2x  1

0 1Since this equation is false, there is no solution to the equation

19 Multiplying by x2 – 4 = (x + 2)(x – 2) and then solving for x:

original equation (it produces zeros in two of the denominators) Consequently the original equation has no solutions

20 Multiplying by 2 x2 3x  1 2x  1 x 1 and then solving for x:

2 x 1 

4(2 x  1)(x  1)

x 1

2(2 x  1)(x  1) (2 x  1)(x  1) 3( x  1)  4(2x  1) 2

3x  3  8x  4 2

5x  1 2

5x 3

x 3 5

Check: Replacing x by 3

5 in the original equation yields:

2 16 9

 345

b Multiplying by 3x x  1 and then solving for x:

26 7

32

7 17 3

Check: Replacing x by 7

3 in the original equation yields:

INSTRUCTOR USE ONLY Check: ReplacingCheck: Replacing xx by by 

3

Uin the original equation yields:n the original equation yiel

7

3

7 3

 1

2

7 9

7 1, which is true

23 a Multiplying by 9x(x – 2) and then solving for x:

Since there are no values of x which can make this last statement true, there is no solution.

24 Factoring then solving for x: 25 Factoring then solving for x:

x2 5x  6 0 ( x  6)(x  1) 0

x 6 or x 1

x2 5x  6 0 ( x  3)(x  2) 0

x 3 or x 2

26 Factoring then solving for z: 27 Factoring then solving for t:

10 z2 13z  3 0 (5 z  1)(2z  3) 0

z 1

5or z 23

3t2 t  4 0 (3t  4)(t  1) 0

x 1 or x 3

x2  3x  40 0 ( x  8)(x  5) 0

x 12or x 6

x(3x 23) 8

3x2  23x  8 0 (3x  1)(x  8) 0

32 Factoring then solving for x: 33 Factoring then solving for x:

x( x 1) 156

x2 x  156 0 ( x  13)(x  12) 0

8 r 2 106

25r 724

23 r 7

48 5

8,1 3

42 Taking square roots yields: 43 Taking square roots yields:

44 Taking square roots yields: 45 Taking square roots yields:

t r 1

2 2 •

22

t r 24

b First factor the equation as x x2 6x  1 0 So x = 0 is one solution Apply the quadratic

formula to find the other two solutions: x (6) r (6)2 4(1)(1)

2(1)

6r 322

x ab

b  a

INSTRUCTOR USE ONLY x 11 b  aa

53 Multiplying by x and solving for x yields: 54 Multiplying by abcx and solving for x yields:

x ac  bc ab

55 Multiplying by abx and solving for x yields:

57 Solving for x by factoring:

x 0 or x p  q

63 Multiplying by x  a x  b and solving for x:

67 Multiplying by 1 rt and solving for r:

x 5

2 or x 4Both of these values check in the original equation

70 Multiplying through by the least common denominator 5(x + 2) gives us:

25 (x  2)(2x  1) 0

2x2  3x  27 0

2 x2  3x  27 0 (2 x  9)(x  3) 0

x 9

2 or x 3Both values check in the original equation

INSTRUCTOR USE ONLY

71 Multiplying through by the least common denominator 6x + 1 yields:

6x2 5x  1 0

6 x2 5x  1 0 (3x  1)(2x  1) 0

x 1

3 or x 1

2Both values check in the original equation

72 Multiplying through by the least common denominator x + 1 gives us:

x2  3x 4

x2  3x  4 0 ( x  4)(x  1) 0

x 4 or x 1

Of these two values, only x = 4 checks in the original equation.

73 Multiplying through by the least common denominator x  2 x 2 yields:

However, neither value satisfies the original equation Thus the original equation has no solution

74 Multiplying through by the least common denominator (x – 1)(x + 1)(x – 3) gives us:

75 a Multiplying each side of the equation by abx:

abx 1x

1

a1

b a

1.4 Rectangular Coordinates Visualizing Data

3 a Draw the right triangle PQR:

b Since the base is b = 5 – 1 = 4 and the height is h = 3 – 0 = 3, the area is given by:

A 12bh 12(4 )(3) 6 square units

4 a Draw the trapezoid ABCD:

b Since a = 7 – 0 = 7, b = 6 – 4 = 2, and h = 4 – 0 = 4, the area is given by:

b Here (x1, y1) = (2, 1) and (x2, y2) = (7, 13), so: d 7 2 2 13  1 2 25  144 169 13

6 a Here (x1, y1) = (–1, –3) and (x2, y2) = (–5, 4), so: d >5  (1)@2 4  (3)> @2 16  49 65

b Here (x1, y1) = (6, –2) and (x2, y2) = (–1, 1), so: d 1  6 2 1  2ª¬ º¼2 49  9 58

7 a Here (x1, y1) = (–5, 0) and (x2, y2) = (5, 0), so: d 5  (5)> @2  0  0 ... by x and solving for x yields: 54 Multiplying by abcx and solving for x yields:

x ac  bc ab

55 Multiplying by abx and solving for. .. data-page="7">

9 Solving for m: 10 Solving for x:

15 x3

15 Multiplying by x and then solving for x:

4...

4 is a solution to the equation

6 Substitute 1 for x in the equation:  5 2

  5          0Now substitute 1 for x in