Solution manual for trigonometry 10th edition by larson

a The inequality x < denotes the set of all real 0numbers less than zero.. a The interval [ 4,∞ denotes the set of all real numbers greater than or equal to 4.. a The interval [−5, 2 den

C H A P T E R P Prerequisites

Section P.1 Review of Real Numbers and Their Properties 2

Section P.2 Solving Equations 5

Section P.3 The Cartesian Plane and Graphs of Equations 17

Section P.4 Linear Equations in Two Variables 29

Section P.5 Functions 42

Section P.6 Analyzing Graphs of Functions 51

Section P.7 A Library of Parent Functions 61

Section P.8 Transformations of Functions 66

Section P.9 Combinations of Functions: Composite Functions 76

Section P.10 Inverse Functions 85

Review Exercises 98

Problem Solving 112

Practice Test 118

(e) Irrational numbers: 2

8 5, 7,− −73, 0, 3.14, , 3, 12, 554 −

(a) Natural numbers: 12, 5 (b) Whole numbers: 0, 12, 5 (c) Integers: 7, 0, 3, 12, 5− −(d) Rational numbers: − −7, 73, 0, 3.14, , 3, 12, 554 −

(e) Irrational numbers: 5

9 2.01, 0.6, 13, 0.010110111 , 1, 6− −

(a) Natural numbers: 1 (b) Whole numbers: 1 (c) Integers: 13, 1, 6− − (d) Rational numbers: 2.01, 0.6, 13, 1, 6− − (e) Irrational numbers: 0.010110111

10 25, 17,− −125, 9, 3.12,12π, 7, 11.1, 13−

(a) Natural numbers: 25, 9, 7, 13 (b) Whole numbers: 25, 9, 7, 13 (c) Integers: 25, 17,− 9, 7, 13(d) Rational numbers:

12 5

25, 17,− − , 9, 3.12, 7, 11.1, 13−

(e) Irrational numbers: 12π

11 (a)

(b) (c)

(d)

12 (a)

(b)

(c) (d)

13 − > − 4 8

− 8 − 7 − 6 − 5 − 4

x

16 3

18 (a) The inequality x < denotes the set of all real 0

numbers less than zero

(b) (c) The interval is unbounded

19 (a) The inequality 2− < x < denotes the set of all 2

real numbers greater than −2and less than 2

(b) (c) The interval is bounded

20 (a) The inequality 0 < x ≤ denotes the set of all real 6

numbers greater than zero and less than or equal to 6

(b) (c) The interval is bounded

21 (a) The interval [ )4,∞ denotes the set of all real

numbers greater than or equal to 4

(b) (c) The interval is unbounded

22 (a) (−∞, 2)denotes the set of all real numbers less

than 2

(b) (c) The interval is unbounded

23 (a) The interval [−5, 2) denotes the set of all real

numbers greater than or equal to 5− and less than 2

(b) (c) The interval is bounded

24 (a) The interval (−1, 2]denotes the set of all real

numbers greater than −1 and less than or equal to 2

(b) (c) The interval is bounded

−2 −1

Receipts, R Expenditures, E R− E

49 $2524.0 billion $2982.5 billion 2524.0 −2982.5 = $458.5 billion

50 $2162.7 billion $3457.1 billion 2162.7− 3457.1 = $1294.4 billion

51 $2450.0 billion $3537.0 billion 2450.0− 3537.0 = $1087.0 billion

52 $3021.5 billion $3506.1 billion 3021.5 −3506.1 = $484.6 billion

60 9− 7x

(a) 9− 7 3( )− = 9+ 21 = 30(b) 9− 7 3( ) = 9 −21 = − 12

61 x2 −3x + 2(a) ( )2 ( )

+

−(a) 1 1 2

1 1 0

+ =

−Division by zero is undefined

−+(a) 2 2 0 0

2 2 4

− = =+

(b) 2 2 4

2 2 0

− − = −

− +Division by zero is undefined

65

(h1 6)(h +6) = 1,h ≠ −6+

Multiplicative Inverse Property

66 (x + 3) (− x+ 3) = 0 Additive Inverse Property

67 ( ) ( )

( )

3 3 Associative Property of Multiplication

3 Commutative Property of Multiplication

x y

= ⋅

=

68. 1( ) ( )1

7 7 12 7 7 12 Associative Property of Multiplication

1 12 Multiplicative Inverse Property

12 Multiplicative Identity Property

73 False Because zero is nonnegative but not positive, not

every nonnegative number is positive

74 False Two numbers with different signs will always

have a product less than zero

75 The product of two negative numbers is positive

76 (a) Because the price can only be a positive rational

number with at most two decimal places, the description matches graph (ii)

(b) Because the distance is a positive real number, the description matches graph (i)

A range of prices can only include zero and positive numbers with at most two decimal places So, a range of prices can be represented by whole numbers and some noninteger positive fractions

A range of lengths can only include positive numbers So, a range of lengths can be represented

by positive real numbers

77 (a)

(b) (i) As n approaches 0, the value of 5 n increases

without bound (approaches infinity)

(ii) As n increases without bound (approaches

infinity), the value of 5 n approaches 0

Section P.2 Solving Equations

=

x x

x x x x x

+ =+ − = −

x x x

of all real numbers

x x

x x

− =+

x x

x x

+ =+

424

1 2

+ = −

+ = −+

= −/

x

x x

Because 1 = −2is a contradiction, the equation has no solution

A check reveals that x = 4 yields a denominator of zero So, x = 4is an extraneous solution, and the original equation has

Multiply each term by (x −1)(x +3 )

A check reveals that x = 1yields a denominator of zero So, x = 1is an extraneous solution, and the original equation has no real solution

Multiply each term by(x +3)(x−3 )

− =

=

=

1 2

1 4

x x

3 4

2 2 2 2

3 4 3 4

35 2 49

7

x x

=

= ±

36 2 43

436.56

=

= ±

≈

x x x

37 2 2

3 8127

3 35.20

38 2 2

9 364

4 2

x x x

+ =+ = ±

= − ±

≈ − −

x x x

− = + − = − −

= The only solution of the equation is x = 2

43

( )

2 2

+ − =+ =

+ =+ = ±

= − ±

= = −

44

( ) ( ) ( )

2 2

+ + =+ = −+ + = − ++ =+ = ±

= − ±

46

( )

2 2

+ + =+ = −+ + = − ++ =+ = ±

= − ±

47.

( )

2 2

2

1221

2112112112212

x x x x

+ − =+ =+ =

   + +   = +  

   

 +  =

 

 + = ±

= − ±

− ±

=

50 2

2 2

10 74 10 (5) (5) 74

4

10 25 494

1( 2 1) 41

2

2

42

1 1 4 2 1

2 2

1 1 84

2

2

42

2

2

42

59 2x2 −7x + =1 0

( ) ( ) ( )( )

( ) ( )

2

2

42

( )( ) ( )

2

2

42

2

2

42

2

2

42

2

2

42

2 2 4 1 2

2 1

2 2 32

( ) ( ) ( )( )

( ) ( )

2

2

42

8 2 62

2 4 62

2

2

42

66 25h2 +80h+ 61 = 0

( )( ) ( )

2

2

42

80 80 4 25 61

2 25

80 6400 610050

12 12 4 1 25

2 1

12 2 11

6 112

( )( ) ( )

2

2

42

14 14 4 1 36

2 1

14 52 7 132

21.7 1.7 4 5.1 3.2

2 5.10.976, 0.643

22.53 2.53 4 2 0.42

2 22.53 9.760941.414, 0.149

2 0.0050.101 0.0063410.012.137, 18.063

2 3.220.08 369.031 2.995, 2.9716.44

75 ( )2

2 64 Extract square roots

2 8

x x

+ =+ = ±

x x x

2

2

11 4 11 4

0 Complete the square

2

3 4

3 2 3 2 3 2

3 0 Complete the square

3

333

x x x

+ − =

+ =+ = ±

2 2

2 2

1 2

1 Extract square roots

11For 1 :

0 1 No solutionFor 1 :

81

( )2 2

1 1

31

− =  =+ =  = −

2 2

=  =+ =  = −

( )( )( )

3 2

3 2

3 2 2

+ − =+ =+ =+ =

=

x x x x x

93.

3 3

3 33

94

3 3

3 33

x x x x x x

2 2

3 4

− =  =+  = −

( )

3 2

3 23

2 2

1 3 4 3

1 3

1 3

3 5

( ) 2

2 2

4 0

1 172

− +

= Only 3x = and x = − −1 2 17are solutions of the original equation x = − and 2 1 17

− =  =+ =  = −

107 Let y =18

0.514 14.75

18 0.514 14.7532.75 0.514

32.750.51463.7

So, the height of the female is about 63.7 inches

108 Let y = 23

0.532 17.03

23 0.532 17.0340.03 0.532

40.030.53275.2

The height of the missing man is about 75.2 inches

109 False

( ) ( )

2 2

112. (a) The formula for volume of the glass cube is

V = Length × Width × Height

The volume of water in the cube is the length × width × height of the water

So, the volume is x x⋅ ⋅(x −3) = x x2( − 3)(b) Given the equation x x2( −3) = 320 The dimensions of the glass cube can be found by solving for x Then, substitute that value into the

expression x3to find the volume of the cube

15. x > and 0 y < in Quadrant IV 0

16 x < and 0 y < in Quadrant III 0

17 x = − and 4 y > in Quadrant II 0

18 x < 0and 7y = in Quadrant II

19 x + y = 0, 0, 0x ≠ y ≠ means or x = −y y = −x

This occurs in Quadrant II or IV

20 ( )x y xy, , > means x and y have the same signs 0This occurs in Quadrant I or III

(2, 4) (−6, 2)

(−4, 0)

(3, −1) (1.5, −3.5)

3.9 9.5 8.2 2.613.4 10.8

179.56 116.64296.2

Distance = −1 13 = −12 =12(b) 52 +122 = 25+144 =169 =132

28 (a) The distance between (−1, 1) and ( )9, 1 is 10

The distance between ( )9, 1 and ( )9, 4 is 3

The distance between (−1, 1) and ( )9, 4 is ( )

8000 9000 10,000

7500 8500 9500 10,500 11,000 11,500

2 6 8 10 12

40

0

y

2 2+ +

d =  +  +  − 

= + =(c) ( ) ( ) ( )5 2 1 2 4 3 1 7

=

=

≈ The plane flies about 192 kilometers

x

(1, 1)

(9, 7)

12 10 8 6 4 2

1 3 4

2

2 2 3

1 2 5

3 5

,

−

y

38 ( )2 ( )2

2 2

42 18 50 12

24 382020

2 50545

39

1 2 1 2midpoint ,

51 y = 5x − 6

x-intercept:

6 5

0 5 6

6 5

x x x

= −

=

= ( )6

5, 0

y-intercept: y = 5 0( )−6 = − 6 (0, 6− )

52 y = −8 3x

x-intercept:

8 3

0 8 3

3 8

x x

x

= −

=

= ( )8

3, 0

y-intercept: y = −8 3 0( ) = 8 ( )0, 8

= +

= +

− = (−4, 0)

y-intercept: y = 0+ 4 = 2 ( )0, 2

−1

−2

−3

4 7

2 1

−1

2 3 5

x y

1 2

3 4 5

6

4

x y

54 y = 2x− 1

x-intercept:

1 2

2 1 0

x x

x

− =

= ( )1

y-intercept

55 y = 3x −7

x-intercept:

7 3

0 3 7

0 3 70

x x

= −

= −

= ( )7

3, 0

y-intercept: y = 3 0( )− 7 = 7 ( )0, 7

56 y = − x +10

x-intercept: 0 10

10 010

x x

x

= − ++ =

= − (−10, 0)

y-intercept: 0 10

10 10

y = − +

= − = − (0, 10− )

57 y = 2x3 − 4x2

x-intercept:

( )

3 2 2

= ( )0, 0

= −

=

= ± = ± (± 5, 0)

= −

= ( )6, 0

y-intercepts: 2 6 0

6

y y

= −

= ± (0, 6 , 0,) ( − 6)

60 y2 = x + 1

x-intercept: 0 1

1

x x

= +

= − (−1, 0)

y-intercepts: 2 0 1

1

y y

= +

= ± ( ) (0, 1 , 0, 1− )

61 x2 − y = 0

( ) ( ) ( ) ( )

62 x− y2 = 0

( ) ( ) ( ) ( )

x x

1 1 -axis symmetry

11

1

=+

x x

x x

67

( ) ( ) ( )( )

−2

3 4

x y

75 y = x2 − 2x

x-intercepts: ( ) ( )0, 0 , 2, 0

y-intercept: ( )0, 0

No symmetry

x y

x y

x y

4 5

−1

2 3 4

x

(0, −1) (−1, 0) (0, 1)

y

81 y = x−6

x-intercept: ( )6, 0

y-intercept: ( )0, 6

No symmetry

1 2 3 4 x

(0, 3) −3, 0

x

(1, 0) (0, 1)

y

2

−2 2

−2

4 6 8 10 12

4 6 8 10 12 x

(6, 0) (0, 6)

y

x

2 3 1

(−1, 0) (1, 0)

(0, 1)

y

144 1002

2 2

1

3 11 15 52

1 8 202

1

64 4002

2 3 4 6 x y

(0, 0)

90 2 ( )2

1 1

x + y− = Center: ( )0, 1 , Radius: 1

x

y

( )1

2,1 2

2000 4000 6000 10,000

Diameter of wire (in mils)

y

20 40 60 80 100 50

100 150 200 250 300 350 400 450

96 (a)

The model fits the data well

(b) Graphically: The point (50, 74.7 represents a life )expectancy of 74.7 years in 1990

Algebraically: ( )

( )

63.6 0.97 50

1 0.01 50112.11.574.7

+

=+

=

=

y

So, the life expectancy in 1990 was about 74.7 years

(c) Graphically: The point (24.2, 70.1 represents a life )expectancy of 70.1 years during the year 1964

Algebraically:

63.6 0.97

1 0.0163.6 0.9770.1

1 0.0170.1 1 0.01 63.6 0.9770.1 0.701 63.6 0.97

6.5 0.26924.2

+

=++

=+

t t t

t t

When 70.1,y = t = 24.2which represents the year

+

=+

y

The y-intercept is (0, 63.6 In 1940, the life )expectancy of a child (at birth) was 63.6 years

(e) Answers will vary

97 False, you would have to use the Midpoint Formula

100. True Depending upon the center and radius, the graph of

a circle could intersect one, both, or neither axis

101. Answers will vary Sample answer: When the x-values are much larger or smaller than the y-values, different

scales for the coordinate axes should be used

102 The y-coordinate of a point on the x-axis is 0 The

x-coordinates of a point on the y-axis is 0

103 Use the Midpoint Formula to prove the diagonals of the

parallelogram bisect each other

104 (a) Because (x y lies in Quadrant II, 0, 0) (x0,−y0)must

lie in Quadrant III Matches (ii)

(b) Because (x y lies in Quadrant II, 0, 0) (−2 ,x y0 0)must lie in Quadrant I Matches (iii)

(c) Because (x y lies in Quadrant II, 0, 0) ( 0 1 0)

2

,

x y must

lie in Quadrant II Matches (iv)

(d) Because (x y lies in Quadrant II, 0, 0) (−x0,−y0)must lie in Quadrant IV Matches (i)

100

Section P.4 Linear Equations in Two Variables

y-intercept: ( )0, 3

16

( )

Slope: 1-intercept: 0, 10

m y

= −

−

17

( )

3 4 3 4

1Slope:

undefined

m = y

x

1 2

18

( )

2 3 2 3

2Slope:

( )

Slope: 0-intercept: 0, 5

m y

=

Slope: undefined-intercept: none

y

22

5 3

3 5 0

y y y

+ =

= −

= − Slope: 0m =

y-intercept: ( )5

3

0,−

23

7 6

7 6

Slope:

-intercept: 0, 5

m y

=

−

−2

−1

1 2

x y

(0, −5)