0.1 Concepts Review 1. rational numbers - Calculus 9e Purcell Varberg Rigdon (Solution)

CHAPTER

0.1 Concepts Review

1. rational numbers

3. If not Q then not P.

Problem Set 0.1

Instructor’s Resource Manual

Section 0.1 1

16. ( 5 − 3 ) ( ) ( )( ) ( ) = 5 − 2 5 3 + 3 27.

18. (2 x − 3) = (2 x − 3)(2 x − 3) = xx ( + 2)

4 x 2 − 12 x + 9 28. +

(3 – 9)(2 x x += 1) 6 x 2 + x 3 – 18 – 9 x

19. =

= 6 x 2 – 15 – 9 x

2(3 y − 1) (3 y + 1)(3 y − 1)

2(3 y + 1) 2 = y +

20. (4 x − 11)(3 x − 7) = 12 x 2 − 28 x − 33 x + 77 2(3 y + 1)(3 y − 1) 2(3 y + 1)(3 y − 1)

= 12 x 2 − 61 x + 77 6 y ++ 22 y

2(3 y + 1)(3 y − 1) 2(3 y + 1)(3 y − 1)

21. (3 2 t 2 −+ t 1) = (3 2 −+ 1)(3 t 2 t t −+ t 1)

2(4 y + 1)

= 9 4 3 2 3 2 t 2 − 3 t + 3 t − 3 t +−+ t t 3 t −+ t 1 2(3 y + 1)(3 y − 1) (3 y + 1)(3 y − 1)

= 9 4 − 6 t 3 + 7 t 2 t −+ 2 t 1

29. a. 00 ⋅= 0 b. is undefined.

22. (2 t + 3) = (2 t + 3)(2 t + 3)(2 t + 3)

= (4 2 t + 12 t + 9)(2 t + 3)

c. = 0 d. is undefined.

= 8 t 3 + 12 2 + 2 17 t 0 24 t + 36 t + 18 t + 27 = 8 t 3 + 36 t 2 + 54 t + 27 e. 0 5 = 0 f. 17 0 = 1

23. = =+, x 2 x ≠ 2 0 30. If = a , then 0 =⋅ 0 a , but this is meaningless x –2

x 2 –4 ( – 2)( x x + 2)

x –2

because a could be any real number. No

x 2 , x ≠ 3 single value satisfies =. a

t + 3 t + 3 12 1.000 96

3 38. x = 0.217171717 … 1000 x = 217.171717...

Instructor’s Resource Manual

Section 0.1 3

41. x = 0.199999...

x 1/ 2 = = 54. ( 3.1415 − ) ≈

36 2 57. Let a and b be real numbers with a < b . Let n x =

be a natural number that satisfies

1 / n < b − a . Let S = { k : k n > b } . Since

43. Those rational numbers that can be expressed

a nonempty set of integers that is bounded by a terminating decimal followed by zeros.

below contains a least element, there is a

k 0 ∈ S such that k 0 / n > b but

44. =⎜⎟ p , so we only need to look at . If ( k 0 −/ 1 ) n ≤ b . Then

q ⎝⎠ q

q = 2 ⋅ 5, then

q ⎜⎟⎜⎟ ⎝⎠⎝⎠ 2 5

(0.5) (0.2) . n m The product

0 . If 0 , then choose

of any number of terminating decimals is also a r = n . Otherwise, choose r = n .

terminating decimal, so (0.5) and (0.2) , n m 1 Note that a <−<. b r

and hence their product, , is a terminating

Given a < b , choose r so that a << r 1 b . Then

choose rr 2 , 3 so that a < r 2 <<< r 1 r 3 b , and so decimal. Thus

has a terminating decimal

on.

expansion. Answers will vary. Possible answer: ≈ 120 in 58. 3

45. Answers will vary. Possible answer: 0.000001, 1

12 ≈ 0.0000010819... 59. r = 4000 mi 5280 × = 21,120, 000 ft π

equator = 2 π r = π 2 (21,120, 000)

46. Smallest positive integer: 1; There is no

≈ 132, 700,874 ft

smallest positive rational or irrational number.

60. Answers will vary. Possible answer:

47. Answers will vary. Possible answer:

48. There is no real number between 0.9999…

= 735,840, 000 beats

(repeating 9's) and 1. 0.9999… and 1 represent

61. V =π rh 2 16 =π ⎞ ⎜ ⋅ 12 ⎟ (270 12) ⋅

the same real number.

≈ 93,807, 453.98 in. 3

49. Irrational

volume of one board foot (in inches):

50. Answers will vary. Possible answers:

×× 1 12 12 144 in. = 3

− π and , π − 2 and 2 number of board feet:

51. ( 3 1) + 3 ≈ 20.39230485 ≈ 651, 441 board ft 144

54.3 ft. V 3 = π − π b. Every circle has area less than or equal to

9 π. The original statement is true.

63. a. If I stay home from work today then it

c. Some real number is less than or equal to

rains. If I do not stay home from work,

its square. The negation is true.

then it does not rain.

71. a. True; If x is positive, then 2 x is positive.

b. If the candidate will be hired then she

meets all the qualifications. If the candidate will not be hired then she does

b. False; Take =− 2 . Then x 2 x > but 0

not meet all the qualifications.

x <. 0

64. a. If I pass the course, then I got an A on the

c. False; Take x = . Then x = 1 4 < x

final exam. If I did not pass the course,

thn I did not get an A on the final exam.

If I take off next week, then I finished my

d. True; Let x be any number. Take

y = x 2 + . Then 1 y > x research paper. If I do not take off next 2 . week, then I did not finish my research paper.

e. True; Let y be any positive number. Take

65. a. If a triangle is a right triangle, then x = . Then 0 <<. x y

2 2 a 2 + b = c . If a triangle is not a right

triangle, then a + b ≠ c . ()

2 2 2 72. a. True; x +−<++− x x 1 x :01 <

b. If the measure of angle ABC is greater than

b. False; There are infinitely many prime

0 o and less than 90 o , it is acute. If the

numbers.

measure of angle ABC is less than 0 o or greater than 90 o , then it is not acute.

c. True; Let x

be any number. Take 1 1

y =+ 1 . Then y > .

66. a. If angle ABC is an acute angle, then its

measure is 45 o . If angle ABC is not an acute angle, then its measure is not 45 o .

d. True; 1/ n can be made arbitrarily close to 0.

If 2 < 2 then < . If 2 ≥ 2 b. a b a b a b then

a ≥ b . e. True; 1/ 2 n can be made arbitrarily close

to 0.

67. a. The statement, converse, and

contrapositive are all true.

73. a. If n is odd, then there is an integer k such

that n = 2 k + 1. Then

b. The statement, converse, and

n 2 = (2 k + 1) 2 = 4 k 2 + 4 k + 1

contrapositive are all true.

= 2 2(2 k + 2)1 k +

68. a. The statement and contrapositive are true.

The converse is false.

b. Prove the contrapositive. Suppose n is even. Then there is an integer k such that

b. The statement, converse, and

n = 2. k Then n 2 = (2 ) k 2 = 4 k 2 = 2(2 k 2 ) .

contrapositive are all false.

Thus n 2 is even.

69. a. Some isosceles triangles are not

equilateral. The negation is true.

74. Parts (a) and (b) prove that n is odd if and

only if n 2 is odd.

b. All real numbers are integers. The original

statement is true.

75. a. 243 =⋅⋅⋅⋅ 33333

c. Some natural number is larger than its

square. The original statement is true.

b. 2 =⋅ 4 31 =⋅⋅ 2 2 31 or 2 ⋅ 31

70. a. Some natural number is not rational. The original statement is true.

Instructor’s Resource Manual

Section 0.1 5

© 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they

=⋅⋅⋅⋅⋅ 2 2 3 5 5 17 or 2 2 ⋅⋅ 3 5 17 2 ⋅

10 x = 24.4444... x = 2.4444...

76. For example, let A =⋅ bc 2 ⋅ d 3 ; then

9 x = 22

A 2 = b 2 ⋅ c 4 ⋅ d 6 , so the square of the number

is the product of primes which occur an even

number of times.

2 d. 1

; 2 q 2 2 2 = p ; Since the prime

e. n = 1: x = 0, n = 2: x = , n = 3: x = –,

factors of p 2 must occur an even number of

times, 2q would not be valid and

= 2 n = 4: x =

must be irrational.

The upper bound is .

78. 3 = ; 3 = 2 ; 3 q 2 = p 2 ; Since the prime

f. 2

q factors of p 2 must occur an even number of

83. a. Answers will vary. Possible answer: An

example

is S = {: xx 2 < x 5, a rational number}. must be irrational.

times, 3q 2 would not be valid and

Here the least upper bound is 5, which is real but irrational.

79. Let a, b, p, and q be natural numbers, so

b b. True

a p aq + bp

and are rational. + =

This

0.2 Concepts Review

sum is the quotient of natural numbers, so it is

also rational.

2. b > 0; b < 0

80. Assume a is irrational,

≠ 0 is rational, and

is rational. Then a =

rational, which is a contradiction.

Problem Set 0.2

81. a. –9 = –3; rational

1. a.

b. 0.375 = ; rational

8 b.

c. (3 2)(5 2) = 15 4 = 30; rational

d. (1 + 3) 2 =+ 1233 +=+ 4 2 3; irrational

3. x −< 7 2 x − 5 −<−< 1 3 x 2

1 < x ; 1, () ∞

2 11. x + 2x – 12 < 0; –2 ± (2) – 4(1)(–12) 2 –2 ± x 52 = =

⎣ x – –1 (

⎦⎣ x – –1 – 13 ⎤ ( )

−<< 2 2 x 1; ( 2, 1) −− 13. 2x + 5x – 3 > 0; (2x – 1)(x + 3) > 0;

Instructor’s Resource Manual

Section 0.2 7

21. ( x + 2)( x − 1)( x −> 3) 0; ( 2,1) − ∪ (3,8)

22. (2 + 3)(3 − 1)( − 2) < 0; ⎛ −∞ − , ⎞⎛ 1 x ⎞ x x ∪ ,2

24. (2 x − 3)( x − 2 1) ( x −> 3) 0;

27. a. False.

b. True.

c. False.

28. a. True.

b. True.

( 1)( 2 33. a. 2 x + x + x 2 – 7) ≥ x – 1

c. False.

3 x 2 + 3 x 2 – 5–7 x ≥ x – 1

2 2 x + 2 x –5–6 x ≥ 0

29. a. ⇒ Let a < b , so ab < b . Also, a < ab .

2 2 2 2 ( x + 3)( x + 1)( – 2) x ≥ Thus, 0

a < ab < b and a < b . ⇐ Let

[ 3, 1] −−∪ [2, ) ∞

a b , so a ≠ Then b

ab − ) = a − 2 + ab b 4 b. x − 2 x 2 ≥ 8

2 2 < 2 b − + ab b = 2 bb ( −

Since b > 0 , we can divide by 2 b to get

b. We can divide or multiply an inequality by

( −∞ − ∪ , 2] [2, ) ∞

any positive number.

a 1 <⇔<⇔<. 1

1 ( a 2 b c. + 1) x 2 − 7( 2 x ++ 1) 10 < 0

[( 2 x 2 +− 1) 5][( x +−< 1) 2] 0

30. (b) and (c) are true. ( 2 − 4)( x 2 x − 1) < 0

(a) is false: Take a =− 1, b = 1 .

( + 2)( + 1)( − 1)( − 2) < 0

(d) is false: if a ≤ b , then a −≥− b .

x > –2 and x < 1; (–2, 1)

2 ( −∞

2, )

2 . 99 x + 5 . 98 < 1 and 1 < 3 . 01 x + 6 . 02

Instructor’s Resource Manual

Section 0.2 9

4 ±− ( 4) x 2 2 − 4(1)(4) −<+< 1 11 46. x − 4 x +≤ 4 0; x =

5 or 5 x > 7 (3x – 1)(x + 6) > 0; (– , – 6) ∞ ∪ ⎜ , ∞

x < 1 or x > ;( −∞ ∪ ,1) ⎜ , ∞

48. 14 x 2 + 11 x − 15 ≤ 0;

42. 2–7 x > 3; −± 11 (11) 2 − 4(14)( 15) − −± x 11 31 = = 2x – 7 < –3 or 2x – 7 > 3

49. x −< 3 0.5 ⇒ 5 x −< 3 5(0.5) ⇒ 5 x − 15 < 2.5

50. x +< 2 0.3 ⇒ 4 x +< 2 4(0.3) ⇒ 4 x + 18 < 1.2

2 ⇒ 2 4 x −< 2 ε (2 x − 1) ≥ (

(5 x + 4)(3 x − 16) < 0;

57. C = π d ⎛

C – 10 ≤ 0.02 ⎜ ⎝ 53 ⎟ ⎠ π d – 10 ≤ 0.02

62. 3 x −< 1 2 x + 6 π ⎛

≤ 0.02 3 x −< 1 2 x + 12

9 2 − 6 +< 1 4 π 2 π x x x + 48 x + 144 We must measure the diameter to an accuracy

5 x 2 − 54 x − 143 < 0

of 0.0064 in.

( 5 x + 11 )( x − 13 ) < 0

58. C − 50 ≤ 1.5, ( F − 32 ) − 50 ≤ 1.5; ⎛ 11 ⎜ ⎞ − ,13

F − 32 ) − 90 ≤ 1.5

9 F − 122 ≤ 2.7

We are allowed an error of 2.7 F.

Instructor’s Resource Manual

Section 0.2 11

63. x < y ⇒ xx ≤ xy and xy < yy Order property: x <⇔ y xz < yz when is positive. z

Transitivity

Conversely,

2 2 ⇒ 2 x – y < 0 Subtract y from each side.

⇒ ( x – y )( x + y ) < 0 Factor the difference of two squares. ⇒ x – y < 0 This is the only factor that can be negative.

Add y t o each side.

64. 0 <<⇒= a b a () a and b =

x + () (–2)

()() a < b , and, by Problem 63,

Since 2 x +≥ 9 9,

b. a – b ≥ a – b ≥ a – b Use Property 4

of absolute values.

c. a ++= b c ( a ++≤++ b ) c a b c x –2

by the Triangular Inequality, and since

1 since x ≤ 1.

+ 3 3 x + 2 2 1 1 1 So 1.9375 1 x 4 + x 3 + x 2 + x + ≤ < 2.

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72. a < b

a +<+ a a b and a +<+ b bb 78. 2 A 2 = 4 π r ; A = π 4 (10) = 400 π

2 a <+< a b 2 b 4 π r 2 − 400 π < 0.01 a + b

4 π r 2 − 100 < 0.01

73. 0 << a b 2 r 0.01 − 100 <

a < ab and ab < b 0.01 0.01

74. ab ≤ ( a + b ) ⇔ ab ≤ ( a + 2 + ab b ) δ ≈ 0.00004 in

⇔≤ 0 a − ab + b = ( a − 2 + ab b

0 ( − ) which is always true. 2 ab 0.3 Concepts Review

75. For a rectangle the area is ab, while for a

2 ⎛ a + square the area is b ⎞

a 2 ⎟ . From 2. (x + 4) + (y – 2) 2 = 25

1 2 a + b ⎛ −+ 2537 +

Problem 74, ab ≤ ( a + b ) ⇔ ab ⎛

so the square has the largest area.

Instructor’s Resource Manual

Section 0.3 13

Problem Set 0.3

d 1 = d 2 so the triangle is isosceles.

2 + 2 a 2 b = c , so the triangle is a right triangle.

x =⇒ 7 () 7, 0

10. midpoint of AB = ⎜

midpoint of CD = ⎜

16. Since the circle is tangent to the x-axis, r = 4. ⎜

center (–1, = 3); radius = 10 ⎛

center (0, = 3); radius = 5 25. =

( x 2 – 12 x + 36) + y 2 = –35 36 + 29. y −=− 2 1( x − 2) ( – 6) x 2 + y 2 = 1 y −=−+ 2 x 2

center = (6, 0); radius = 1 x +−= y 4 0

y −=−+ 4 x 3

( x − 5) 2 x +−= y 7 + 0 ( y + 5) 2 = 50

center 5, 5 ; = ( − ) radius 50 = = 52 31. y = 2 x + 3

2– x y += 3 0

32. y = 0 x + 5 0 x +−= y 5 0

Instructor’s Resource Manual

Section 0.3 15

5–2–4 = 0 m = –; x y 3

35. 3y = –2x + 1; y = – x + ; slope = –; 2

y -intercept =

slope ; =− y -intercept = y = – x –

slope = –5; y-intercept = 4

slope ; =− y -intercept = − 4

2 d. c must be the same as the coefficient of x,

1 3 so c = 3.

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1 –3 x += y 5

perpendicular slope =− ;

41. m = ; –3(–1) += y 5

y += 1 ( x + 2)

2 Point of intersection: (–1, 2)

3 Point of intersection: ( −− 3, 4 ; )

2 3 2 2 − 5 y =−+ 4 x 8

43. y = 3(3) – 1 = 8; (3, 9) is above the line.

Instructor’s Resource Manual

Section 0.3 17

8 x + 12 y = 36 inscribed circle: radius = (4 – 4) 2 + (1 – 3) 2 x

–4 y = –4 circumscribed circle:

radius = (4 – 2) 2 + (1 – 3) 2 = 8

Point of intersection: (3, 1); 3x – 4y = 5;

( – 4) x 2 + ( – 1) y 2 = 8

–4 y = –3 x + 5

50. The radius of each circle is 16 = 4. The centers y

4 4 are ( − 1, 2 and ) ( − 9,10 . ) The length of the belt is

4 m = – the sum of half the circumference of the first 3

circle, half the circumference of the second circle, 4

and twice the distance between their centers. y –1 = – ( – 3) x 3

2 x + 3 y = 6 51. Put the vertex of the right angle at the origin with the other vertices at (a, 0) and (0, b). The

x 15 – 6 y = 15 ⎛ midpoint of the hypotenuse is ab ,. ⎞

distances from the vertices are

Point of intersection: ⎛ , ⎞ ; 2 2 2 ⎜ 2

which are all the same.

m = – 52.

2 From Problem 51, the midpoint of the

hypotenuse, ( 4,3, , ) is equidistant from the

vertices. This is the center of the circle. The y –

radius is

+= 16 9 5. The equation of the

2 54 20 circle is

53. 2 x 2 + y – 4 – 2 – 11 x y = 0 56. The equations of the two circles are 2 2 ( x − R ) 2 + ( y − R ) 2 = R ( 2 x –4 x ++ 4) ( y –2 y +=++ 1) 11 4 1

2 2 2 2 2 ( x − r ) + ( y − r ) = x r ( – 2) + ( – 1) y = 16

Let () aa , denote the point where the two x + y + x 20 – 12 y + 72 = 0 circles touch. This point must satisfy

2 ( 2 x + 20 x + 100) ( + y – 12 y + 36)

x + 10) ( + ( – 6) y = 64 2

center of first circle: (2, 1)

center of second circle: (–10, 6)

a =± ⎜ ⎜ 1 ⎟ R

= (2 10) + 2 d + (1 – 6) 2 = 144 25 +

Since a < R , a =− ⎜ ⎜ 1 ⎟ ⎟ R .

However, the radii only sum to 4 + 8 = 12, so At the same time, the point where the two the circles must not intersect if the distance

circles touch must satisfy

between their centers is 13.

a b =−+ Equating the two expressions for a yields

55. Label the points C, P, Q, and R as shown in the

figure below. Let d = OP , h = OR , and

a = PR . Triangles Δ OPR and Δ CQR are

similar because each contains a right angle and

they share angle ∠ QRC . For an angle of

d 3 a 1 r =− (3 2 2) R ≈ 0.1716 R

and =⇒= h 2 a . Using a

property of similar triangles, QC / RC = 3/2 ,

a =+ 2

By the Pythagorean Theorem, we have

d = h 2 − a 2 = 3 a = 23 +≈ 4 7.464

Instructor’s Resource Manual

Section 0.3 19

60. See the figure below. The angle at T is a right slope m, draw ABC with vertical and

57. Refer to figure 15 in the text. Given ine l 1 with

angle, so the Pythagorean Theorem gives

) 2 + horizontal sides m, 1. 2 PM r PT r

Line l 2 is obtained from l 1 by rotating it

⇔ ( PM ) 2 + 2 rPM + 2 r 2 = (

PT 2 ) + r

around the point A by 90 ° counter-clockwise.

⇔ PM PM ( + 2) r = ( ) 2 PT

Triangle ABC is rotated into triangle AED .

We read off PM + 2 r = PN so this gives ( PM PN )( ) = (

PT 2 )

slope of l 2 =

61. The lengths A, B, and C are the same as the

⎛ corresponding distances between the centers of

the circles:

radius: ⎜ ⎟ C = (8) + (0) = 64 = ⎜ 8

center: , 0 ; ⎛ 1 ⎞

Each circle has radius 2, so the part of the belt around the wheels is

59. Let a, b, and c be the lengths of the sides of the 2(2 π −− a π ) + 2(2 π −− b π ) + 2(2 π −− c π ) right triangle, with c the length of the

= 2[3 - ( π a ++ b c )] = 2(2 ) π = 4 π

hypotenuse. Then the Pythagorean Theorem Since a + b + c = π , the sum of the angles of a says that a 2 + b 2 = c 2 triangle.

+ 8.2 10 8 4 ++π Thus,

2 2 π 2 a π b π c The length of the belt is ≈

2 π⎜ ⎟ is the area of a semicircle with ⎝⎠ 2 diameter x, so the circles on the legs of the

triangle have total area equal to the area of the semicircle on the hypotenuse.

From 2 + 2 a 2 b = c ,

62 As in Problems 50 and 61, the curved portions x is the area of an equilateral triangle 4

of the belt have total length 2 π The lengths r . with sides of length x, so the equilateral

of the straight portions will be the same as the triangles on the legs of the right triangle have

lengths of the sides. The belt will have length total area equal to the area of the equilateral

2 π r + d 1 + d 2 ++ … d n .

triangle on the hypotenuse of the right triangle.

20 Section 0.3 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they 20 Section 0.3 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they

= 3; – ; m = passes through

1 71. Let the origin be at the vertex as shown in the

figure below. The center of the circle is then

1 ( 4 − rr ⎛⎞ , ) , so it has equation 7(0) 5 − ⎜⎟ − 6

( x −− (4 r )) 2 + ( y − r ) 2 = r 2 . Along the side of

length 5, the y-coordinate is always 3 times

−− 23 5 3 the x-coordinate. Thus, we need to find the

69. m =

=− ; m = ; passes through

value of r for which there is exactly one x-

solution to ( x −+ 4 r ) 2 + ⎜ x − r = r 2 ⎟ . ⎜

⎝ 4 ⎠ Solving for x in this equation gives

y −= ⎜ x +

−+ 2 7 r ⎞ 24 r r − 6 . There is

5 5 exactly one solution when −+ 2 r 7 r −= 6 0, that is, when r = 1 or r = 6 . The root r = 6 is

extraneous. Thus, the largest circle that can be inscribed in this triangle has radius r = 1.

Instructor’s Resource Manual

Section 0.3 21

72. The line tangent to the circle at () ab , will be

The slope of PS is

perpendicular to the line through () ab , and the

center of the circle, which is

. () The slope of

2 4 ( 1 + 2 ) ] 4 through 2 () ab , and () 0, 0 has slope

0, 0 . The line

so ax 2 + by = r has slope − and is

b PS and QR are parallel. The slopes of SR and

perpendicular to the line through () ab , and

PQ are both 3 1 , so PQRS is a

() 0, 0 , so it is tangent to the circle at () ab , . x 3 − x 1

b =± 33 3–33 x y = 36

x –3 y = 12

3 x + 33 y = 36

2 2 x + 3 y = 12 77. x + ( – 6) y = 25; passes through (3, 2) tangent line: 3x – 4y = 1

74. Use the formula given for problems 63-66, for

The dirt hits the wall at y = 8.

()() xy , = 0, 0 .

A = mB , =− 1, C =− Bb ; (0, 0)

m (0) 1(0) − +− Bb Bb −

0.4 Concepts Review

m 2 +− ( 1) 2 m 2 + 1 1. y-axis

2. ( − 4, 2 ) 4. line; parabola

75. The midpoint of the side from (0, 0) to (a, 0) is

Problem Set 0.4

The midpoint of the side from (0, 0) to (b, c) is

⎛ 0 + b 0 + c ⎞⎛ bc ⎞

1. y = –x ,, 2 = + 1; y-intercept = 1; y = (1 + x)(1 – x);

x -intercepts = –1, 1

c –0

c Symmetric with respect to the y-axis

c –0

76. See the figure below. The midpoints of the sides are

, and

5. x 2 + y = 0; y = –x 2 2. x =− y + 1; y -intercepts =− 1,1;

x -intercept = 1 .

x -intercept = 0, y-intercept = 0

Symmetric with respect to the y-axis Symmetric with respect to the x-axis.

2 6. y = x 2 − xy 2 ; -intercept = 0 3. x = –4y – 1; x-intercept = –1 Symmetric with respect to the x-axis

y = x (2 − xx ); -intercepts =

4. y = 4 x 2 − 1; y -intercept =− 1 7. 2 7x 7 + 3y = 0; 3y = –7x 2 ; y = – x 2

3 y = (2 x + 1)(2 x − 1); -intercepts x =− ,

x -intercept = 0, y-intercept = 0

22 Symmetric with respect to the y-axis Symmetric with respect to the y-axis.

8. y = 3 x 2 − 2 x + 2; y -intercept = 2

Instructor’s Resource Manual

Section 0.4 23

2 12. 2 4 x + 3 y = 12; y -intercepts =− 2, 2 x -intercepts = -2, 2; y-intercepts = -2, 2

9. x 2 +y 2 =4

x -intercepts =−

Symmetric with respect to the x-axis, y-axis,

and origin Symmetric with respect to the x-axis, y-axis, and origin

10. 3 x 2 + 4 y 2 = 12; y -intercepts =−

13. x 2 –y 2 =4

x -intercepts =−

2, 2 x- intercept = -2, 2

Symmetric with respect to the x-axis, y-axis, Symmetric with respect to the x-axis, y-axis, and origin

and origin

11. y = –x 2 – 2x + 2: y-intercept = 2

2 14. 2 x + ( y − 1) = 9; -intercepts y =− 2, 4

x -intercepts =

= –1 ± 3 x -intercepts = − 2 2, 2 2

Symmetric with respect to the y-axis

15. 4(x – 1) 2 +y 2 = 36;

18. 4 x + 4 y = 1; -intercepts y =− 1,1 y -intercepts =± 32 =± 42 x -intercepts 1,1 =−

x -intercepts = –2, 4 Symmetric with respect to the x-axis, y-axis, Symmetric with respect to the x-axis

and origin

16. x 2 − 4 x 2 + 3 y =− 2 19. x 4 +y 4 = 16; y-intercepts =− 2, 2 x -intercepts = 2 ± 2 x -intercepts 2, 2 =−

Symmetric with respect to the x-axis Symmetric with respect to the y-axis, x-axis and origin

17. x 2 + 9(y + 2) 2

= 36; y-intercepts = –4, 0

x -intercept = 0

20. y = x 3 – x; y-intercepts = 0;

Symmetric with respect to the y-axis y = x(x 2 – 1) = x(x + 1)(x – 1); x -intercepts = –1, 0, 1 Symmetric with respect to the origin

Instructor’s Resource Manual

Section 0.4 25

21. y =

24. 4 ( x − 5 ) + 9( y + 2) = 36; -intercept x = 5

; y-intercept = 1

x 2 + 1 Symmetric with respect to the y-axis

25. y = (x – 1)(x – 2)(x – 3); y-intercept = –6 x -intercepts = 1, 2, 3

22. y = 2 ; -intercept y = 0

x + 1 x -intercept = 0 Symmetric with respect to the origin

26. y =x 2 (x – 1)(x – 2); y-intercept = 0 x -intercepts = 0, 1, 2

y -intercepts = –2 ±

x -intercept = 1

27. y = x 2 ( x − 2 1) ; y-intercept = 0 x -intercepts = 0, 1

= 4 ( − 4 1) ( + 28. 4 y x x x 1) ; -intercept y = 0 Intersection points: (0, 1) and (–3, 4) x -intercepts 1, 0,1 =− Symmetric with respect to the y-axis

32. 2 2 x +=−− 3 ( x 1)

2 x +=− 3 x 2 + 2 x − 1

29. x + y = 1; y-intercepts = –1, 1;

x 2 += 4 0

x -intercepts = –1, 1

No points of intersection

Symmetric with respect to the x-axis, y-axis and origin

−+=− 2 3 2( − 4) 33. 2 x x

30. x + y = 4; y-intercepts = –4, 4;

−+=− 2 x 3 2 x 2 + 16 x − 32

x -intercepts = –4, 4

2 x 2 − 18 x + 35 = 0

Symmetric with respect to the x-axis, y-axis and origin

Intersection points: ⎜ ⎜

Instructor’s Resource Manual

Section 0.4 27

34. −+= 2 x 3 3 x 2 − 3 x + 12 37. y = 3 x + 1

3 x 2 −+= x 9 0 x 2 + 2 x + (3 x + 1) 2 = 15

No points of intersection

5 Intersection points: ⎛ −− 2 39 −− 1 3 39 ⎞

⎜ and ⎟ 5 5 ⎝ ⎟ ⎠

[ or roughly (–1.65, –3.95) and (0.85, 3.55) ]

35. 2 x 2 + x = 4

x 2 = 2 x =± 2

Intersection points: ( – 2, – 2 , )( 2, 2 )

+ Intersection points:

[ or roughly ( − − 2.88, 8.52 , 1.47,8.88 )( ) ]

Intersection points:

39. a. y = 2 x ; (2) (2 2) 2 2

b. 3 ax 2 + bx + cx + d , with a > 0 : (1) 2 2

c. ax 3 + 2 bx + cx + d , with a < 0 : (3) = 2 13 ≈ 7.21

Four such distances ( d 2 = d 4 and d 1 = d 5 ).

d. y 3 = ax , with a > 0 : (4)

40. x 2 + 2 y = 13; ( 2, 3), ( 2,3), (2, 3), (2, 3) −− − − 0.5 Concepts Review

1 = (2 2) + +−+ ( 3 3) = 4 1. domain; range

Three such distances.

3. asymptote

41. x 2 + 2x + y 2 – 2y = 20; ( –2, 1 + ) 21 , 4. even; odd; y-axis; origin

( –2, 1 – 21 , 2, 1 )( + )( 13 , 2, 1 – 13 ) Problem Set 0.5

d 1 = (–2 – 2) ++ ⎡ ⎣ 1 21 – 1 ( + 13 ) ⎤ ⎦

1. a. f (1) = 1–1 2 = 0

b. f (–2) = 1 – (–2) 2 = –3

2 c. f (0) = 1–0 = 1

d 2 = (–2 – 2) ++ ⎡ ⎣ 1 21 – 1 – 13 ⎤ ( ) ⎦

d. fk () = 1– k 2

e. f (–5) = 1 – (–5) 2 = –24

d 3 = ( 2 2) −+ ++ 1 21 −− ( 1 21 ) f. f = 1–

2 2 g. f (

2 h. f 1 + h − f 1 =−− 2 h h 2 −=−− 0 2 h h 2

d 4 = ( 2 2) −− 2 +− ⎡ 1 −+ 21 (1 13) ⎤

16 +− 21 − 13 i. f ( 2 + h ) − f () 2 =−+ 1 ( 2 h ) + 3 ( )

=−− 4 h h 2

2. a. F (1) =+⋅= 1 31 4

d 5 = ( 2 2) −− +− 1 21 −− 1 13

c. ⎛⎞⎛⎞ 1 1 ⎛⎞ 1 1 3 F 49 ⎜⎟⎜⎟ = + 3 4 4 ⎜⎟ =

Instructor’s Resource Manual

Section 0.5 29

2 2 2 d. F (

1 + h )( =+ 1 h ) + 31 ( + h )

e. F ( 1 + h 136

2 ) 3 −=+ h + 3 h + h

5. a. f ( 0 . 25 ) =

f. F ( 2 + h ) − F () 2 0 . 25 − 3 − 2 . 75

is not

3 ( defined 2 + h ) + 32 ( + h ) − ⎡ 2 − 32 () ⎤

=+ 8 12 h + 6 h 2 + h 3 ++ 63 h − 14 1 b. fx () = ≈ 2.658 π − 3

b. G (0.999) =

6. a. f(0.79) =

c. G (1.01) =

b. f(12.26) =

y =± 1– x 2 ; not a function

u + 1 u + 1 xy +x=y

8. The graphs on the left are not graphs of

functions, the graphs on the right are graphs of

2 –9 ≥ 0; x functions. 2 x ≥ 9; x ≥ 3

Domain: { x ∈ : x ≥ 3}

fa ( + h )–() fa [2( a + h ) – 1] – (2 2 a 2 – 1)

Domain: { y ∈ : y ≤ 5}

h Domain: { x ∈ : x ≠−

gx 3 ( + h )–() gx

xh + –2 – 11. 3 = x –2

Domain: { y ∈ : y >− 1}

3 x −− 63 x − 3 h + 6 c. φ () u = 2 u + 3

= x 2 − 4 x + hx − 2 h + 4 Domain:

(all real numbers)

2 15. f(x) = –4; f(–x) = –4; even function

a + 8 a + ah + 4 h + 16 16. f(x) = 3x; f(–x) = –3x; odd function

13. a. Fz () = 2 z + 3

2z + 3 ≥ 0; z ≥–

Domain: ⎨ z ∈ : z ≥− ⎬

Domain: ⎨ v ∈ : v ≠ ⎬

Instructor’s Resource Manual

Section 0.5 31

17. F(x) = 2x + 1; F(–x) = –2x + 1; neither

20. gu () =

; (– ) g u = –

; odd function

18. Fx () = x 3 – 2; (– ) F x = –3 – 2; x neither

19. gx () = 3 x 2 + x 2 – 1; (– ) g x = 3 x 2 –2–1 x ; 2 z + 1 –2 z + 1

23. fw () = w – 1; (– ) f w = – – 1; w neither

26. Ft () = – t + 3 ; (– ) F t = –– t + 3; neither

24. 2 2 27. gx () =

;( g −=− x ) ; neither

hx () = x + 4; (– ) h x = x + 4; even

function

28. Gx () = 2 x − 1;( G −=−+ x ) 2 x 1; neither

25. fx () = x 2 ; (– ) f x = –2 x = 2; x even

function

⎧ 1 if t ≤ 0 ⎪

29. gt () = ⎨ t + 1 if 0 << t 2 neither

⎪ 2 ⎩ t – 1 if 2 t ≥

Instructor’s Resource Manual

Section 0.5 33

35. Let y denote the length of the other leg. Then hx () =⎨

2 () 2 = h − x

36. The area is

Ax () = base height × = xh 2 − x 2

37. a. E(x) = 24 + 0.40x

b. 120 = 24 + 0.40x

31. T(x) = 5000 + 805x

0.40x = 96; x = 240 mi

Domain: { x ∈ integers: 0 ≤≤ x 100}

38. The volume of the cylinder is π rh 2 , where h is Tx () 5000 ux () =

+ 805 the height of the cylinder. From the figure, x

Domain: { x ∈ integers: 0 <≤ x 100} r + ⎜ ⎞ ⎟ = (2r) ;

= 3r ;

32. a. Px () = x 6 – (400 5 + xx ( – 4)) h = 12 r = 2 r 3.

() =π = 2 x 6 – 400 – 5 xx ( – 4) Vr r (2 r 3) =π 2 r 3 3

b. P(200) ≈− 190 ; P ( 1000 ) ≈ 610

c. ABC breaks even when P(x) = 0; x 6 – 400 – 5 xx ( – 4) = 0; 390 x ≈

33. Ex () = x – x 2

39. The area of the two semicircular ends is . 4

0.5 1 x

1– π d

The length of each parallel side is

exceeds its square by the maximum amount.

4 Since the track is one mile long, π < 1, so d

d < . Domain: ⎧

⎫ ⎨ d ∈ :0 << d ⎬

34. Each side has length . The height of the

triangle is

40. a. A (1) = 1(1) + (1)(2 1) −=

1 3 42. a. f(x + y) = 2(x + y) = 2x + 2y = f(x) + f(y)

≠ f(x) + f(y)

d. f(x + y) = –3(x + y) = –3x – 3y = f(x) + f(y)

43. For any x, x + 0 = x, so

f (x) = f(x + 0) = f(x) + f(0), hence f(0) = 0. Let m be the value of f(1). For p in N,

p =⋅=+++ p 1 1 1 ... 1, so

b. A (2) = 2(1) + (2)(3 1) −= 4 f (p) = f(1 + 1 + ... + 1) = f(1) + f(1) + ... + f(1)

2 = pf(1) = pm. ⎛⎞ 1 1 1 1

hence f ⎜⎟ = . Any rational number can ⎝⎠ p

be written as

with p, q in N.

f. Domain: { c ∈ : c ≥ 0 so

} f ⎜⎟ f

Range: { y ∈ : y ≥ 0 } ⎛⎞ 1 ⎛⎞ 1 ⎛⎞ 1 = f ⎜⎟ + f ⎜⎟ ++ ... f ⎜⎟

Instructor’s Resource Manual

Section 0.5 35

44. The player has run 10t feet after t seconds. He

46. a. f(1.38) ≈ –76.8204

reaches first base when t = 9, second base when

f (4.12) ≈ 6.7508

t = 18, third base when t = 27, and home plate when t = 36. The player is 10t – 90 feet from

b. x f (x)

first base when 9 ≤ t ≤ 18, hence

–4 –6.1902 player is 10t – 180 feet from second base when

90 2 + (10 t − 90) 2 feet from home plate. The

18 thus he is –3 ≤ t ≤ 27, 0.4118

90 – (10t – 180) = 270 – 10t feet from third base –2 13.7651

and 90 2 + (270 10 ) − t 2 feet from home plate.

The player is 10t – 270 feet from third base –1 9.9579 when 27 ≤≤ t 36, thus he is

0 0 plate.

90 – (10t – 270) = 360 – 10t feet from home

a. s =⎨ ⎪ 90 2 + (270 10 ) − t 2 3 if 18 –0.4521 <≤ t 27 ⎪ ⎪⎩ 360 – 10 t

if 27 <≤ t 36 4 4.4378 ⎧ 180 − 180 10 − t

90 ⎪ 2 + (270 10 ) − t 2 if 18 <≤ t 27

45. a. f(1.38) ≈ 0.2994

a. Range: {y ∈ R: –22 ≤ y ≤ 13}

b. f(x) = 0 when x ≈ –1.1, 1.7, 4.3

f (x) ≥ 0 on [–1.1, 1.7] ∪ [4.3, 5] –1 –1.8

a. f(x) = g(x) at x ≈ –0.6, 3.0, 4.6

36 Section 0.5 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they 36 Section 0.5 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they

()–() gx g () −= 6 ≈ 4.3333 to ∞ . On (

decreased from ∞ to

x 3 –7 x 2 + 9 x + 9 (

− 3, 2 ) the maximum occurs around Largest value f (–2) – (–2) g = 45 x = 0.1451 with value 0.6748 . Thus, the

range is ( −∞ , 0.6748 49. ⎤⎡ ⎦⎣ ∪ 2.8889, ∞ ) .

c. x 2 + x –6 = (x + 3)(x – 2) = 0 0; Vertical asymptotes at x = –3, x = 2

d. Horizontal asymptote at y = 3

0.6 Concepts Review

1. ( x 2 + 1) 3

a. x-intercept: 3x – 4 = 0; x =

3 2. f(g(x))

y -intercept: 2 = 3. 2; left

4. a quotient of two polynomial functions

c. x 2 + x –6 = 0; (x + 3)(x – 2) = 0

Problem Set 0.6

Vertical asymptotes at x = –3, x = 2

1. a. ( f + g )(2) = (2 3) 2 ++ 2 = 9

d. Horizontal asymptote at y = 0

b. ( fg ⋅ )(0) = (0 3)(0 ) + 2 = 0

c. ( gf )(3) =

a. ( – )(2) f g = (2 2 +

a. x-intercepts:

y -intercept:

Instructor’s Resource Manual

Section 0.6 37

5. ( f g ) () x = f ( 1 + x ) = 1 + x − 4

2 2 2 ( g f ) () x = g ⎛ 2 ⎜ x − 4 ⎞ ⎟ =+ 1 x 2 − 4

5 6. g 3 (x) = (x 2 3 4 2 ⎝ 2 ⎠ 3 3 3 +1) = (x + 2x + 1)(x + 1) =x 6 4 + 3x 2 + 3x +1

3. a. ( Φ+Ψ )( ) t t 3 1 ( g g g )( x ) = ( g g )( x 2 + 1 =++ )

2 2 4 = g[(x 2 + 1) + 1] = g( x + 2x + 2)

r 3 8. + g(2.03) ≈ 0.000205 1

3 1 ≈ 4.789 e. (Φ – Ψ )(5t) = [(5t)

3 1 10. [ g 3 ( ) – ( )] π g π 1/ 3 = [(6 – 11) – (6 – 11)] π 3 π 1/ 3 = 125t + 1–

Domain: (– ∞, – 1]∪ [1, ∞) b. fx () = , g (x) = x 3 + 3x x

13. p = f g h b. () () gx () = x ,

⎟ if f(x) =1/ x , ⎜⎟

hx 2 () = x + 1 = ( 2 x 16 – 1) 2 + p = f g h if fx ( ) 1/ = x , g(x) = x + 1,

x 4 hx = Domain: 2 (– ∞, 0 ) ∪ (0, ∞) () x

17. Translate the graph of y =x to the right 2 units and down 3 units.

15. Translate the graph of 2

g ( x ) = x to the right 2

units and down 4 units.

18. Translate the graph of y =x 3 to the left 1 unit

and down 3 units.

16. Translate the graph of hx () = x to the left 3

units and down 4 units.

Instructor’s Resource Manual

Section 0.6 39

24. a. F(x) – F(–x) is odd because

21. Ft () = t

F (–x) – F(x) = –[F(x) – F(–x)]

b. F(x) + F(–x) is even because

F (–x) + F(–(–x)) = F(–x) + F(x) = F(x) + F(–x)

25. Not every polynomial of even degree is an even function. For example f ( x ) = x 2 + x is

22. Gt () =− t t neither even nor odd. Not every polynomial of

odd degree is an odd function. For example

g ( x ) = x 3 + x 2 is neither even nor odd.

26. a. Neither

23. a. Even; f. Neither

(f + g)(–x) = f(–x) + g(–x) = f(x) + g(x) = (f + g)(x) if f and g are both even

29 – 3(2 + t ) (2 ++ t ) functions. 2

27. a. P =

= t + t + 27

b. Odd; (f + g)(–x) = f(–x) + g(–x) = –f(x) – g(x) = –(f + g)(x) if f and g are both odd

+ 15 27 ≈ 6.773 functions.

b. When t = 15, P = 15 +

28. 2 (t) = (120 + 2t + 3t )(6000 + 700t )

c. Even; = 2100 t 3 + 19, 400 t 2 + 96, 000 t + 720, 000

( fg ⋅ )( −= x )[( f − x )][ ( g − x )]

= [ ( )][ ( )] ( fx gx = fgx ⋅ )( )

if 0 << t 1 if f and g are both even functions.

(400 ) t 2 + [300( t − 1)] 2

if 1 t ≥

d. Even;

if 0 << t 1 =− [ fx ( )][ − gx ( )] [ ( )][ ( )] = fx gx

( f ⋅ g )( − x ) = [ f ( − x )][ g ( − x )] ⎧⎪ 400 t

2 250, 000 t − 180, 000 t + 90, 000 if 1 t ≥ = ( fgx ⋅ )( )

Dt () =⎨

if f and g are both odd functions.

30. D(2.5) ≈ 1097 mi

e. Odd;

( f ⋅ g )( − x ) = [ f ( − x )][ g ( − x )]

= [ ( )][ fx − gx ( )] =− [ ( )][ ( )] fx gx =−⋅ ( fgx )( )

if f is an even function and g is an odd function.

40 Section 0.6 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they 40 Section 0.6 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they

ax + ⎛ b

f ( f ( x )) = f ⎜

⎝ cx – a ax ⎠ b c () + cx – a – a f 1 1 ( f 2 ( )) x = ;

+ bc = 0 , f(f(x)) is undefined, while if

f 1 ( f 4 ( )) x =

x =a c , f(x) is undefined.

f 1 ( f 5 ( )) x =

⎛ x ⎛ –3 ⎛ x –3 ⎞ ⎞ –3 ⎞

32. fffx ( ( ( ))) = f ⎜ f ⎜

⎟ ⎟ = f ⎜ x + 1 ⎟ f 1 ( f 6 ( )) x =

⎜ x –3 + 1 ⎟

⎛ x –3–3–3 x ⎞

If x = –1, f(x) is undefined, while if x = 1,

f 2 ( f 3 ( )) x =

f (f(x)) is undefined.

Instructor’s Resource Manual

Section 0.6 41

f 5 ( ( )) fx 1 =

= (((( f 3 f 3 ) f 3 ) f 3 ) f 3 )

= ((((( f f ) f ) f ) f ) f )

1 = (((( f 2 f 3 ) f 4 ) x f 5 ) f 6 ) 1 –

F f = f , then F = f

d. If G f 3 f 6 = f 1 , then G f 4 = f 1 so

e. If f 2 f 5 H = f 5 , then f 6 H = f 5 so

Problem Set 0.7

0.7 Concepts Review

3. odd; even

f. 11

4. r = (–4) + 3 2 = 5; cos θ == – ⎝ 180 ⎠

Instructor’s Resource Manual

Section 0.7 43

. The results

f. 36. 0

π were derived in the text. ⎠

sin = cos =

If the angle is / 3 π then the triangle in the 56. 4 tan34. 2°

5. a. ≈ 68.37 π 1 sin 34.1°

figure below is equilateral. Thus cos = 3 2

b. π ≈ 0.8845 3 sin 3.1°+ cot 23.5°

5.34 tan 21.3°

and by the Pythagorean Identity, sin = . 3 2

c. tan (0.452) ≈ 0.4855

d. sin (–0.361) ≈ –0.3532 234.1sin(1.56)

6. a. ≈ 248.3

cos(0.34 )

b. sin (2.51) 2 + cos(0.51) ≈ 1.2828 56. 3 tan34. 2°

7. a. ≈ 46.097

sin 56.1° π

3 Referring to Figure 2, it is clear that sin = 1 ⎛

and cos = 0 . The rest of the values are

8. Referring to Figure 2, it is clear that obtained using the same kind of reasoning in

sin = 0 0 and cos = 0 1 . If the angle is π / 6 ,

the second quadrant.

then the triangle in the figure below is

equilateral. Thus, PQ = OP = . This

implies that sin = . By the Pythagorean

Identity, cos

=− 1 sin

6 6 ⎝⎠ 2 4 1 b. sec( ) π= = –1

cos( ) π

c. sec ⎜ ⎟ ⎞=

⎝ 4 ⎠ cos 3 () π 4

d. csc ⎜⎟ ⎛⎞=

⎝⎠ 2 sin π () 2

44 Section 0.7 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they 44 Section 0.7 Instructor’s Resource Manual © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they

b. cos 3 t = cos(2 t += t ) cos 2 cos – sin 2 sin t t t t

e. cot ⎜⎟ ⎛⎞=

= (2 cos ⎝⎠ 2 4 sin t – 1) cos – 2sin t 2 t cos t ()

= 2 cos 3 t – cos – 2(1 – cos t 2 t ) cos t

sin – ⎛ π π ()

cos – 3 π ()

4 = 4 cos t – 3cos t

c. sin 4 x = sin[2(2 )] 2sin 2 cos 2 x = x x

sin π

()

10. a. tan ⎛⎞=

⎝⎠ 3 cos () 3

⎜⎟ 3 = 3

= 2(2sin cos )(2 cos x x 2 x –1)

= 2(4sin cos x 3 x – 2sin cos ) x x

b. sec ⎜⎟ ⎛⎞=

1 = 8sin cos x 3 x – 4sin cos x x

⎝⎠ 3 cos π () 3

d. (1 cos )(1 cos ) 1 cos + θ − θ =− 2 θ = sin 2 θ

π π cos ()

⎝⎠ 3 sin () 3 3 csc u sec u

= sin 2 u + cos 2 u = 1

d. csc ⎜⎟ ⎛⎞=

b. (1 cos − 2 x )(1 cot + 2 x ) = (sin π 2 1 x )(csc 2 x )

⎝⎠ 4 sin π ()

⎛ π π sin – ()

⎜ 1 ⎞= 6 ⎛

e. tan –

c. sin (csc – sin ) t t t = sin t ⎜ – sin t ⎟ ⎝

6 ⎠ cos – () 3

b. (sec –1)(sec t t += 1) sec 2 t –1 tan = 2 t 14. a. y = sin 2x

Instructor’s Resource Manual

Section 0.7 45

16. y = 3cos

2 Period = 4 π , amplitude = 3

17. y = 2 sin 2x

21. y = 21 + 7 sin( 2 x + 3 )

Period = π , amplitude = 2 Period = π , amplitude = 7, shift: 21 units up,

3 units left 2

= tan x 22. y = 3cos ⎜ x – ⎟ –1

18. y

Period = π π Period = 2 π , amplitude = 3, shifts:

units 2

right and 1 unit down.

19. y = 2 + cot( 2 x )

Period =

, shift: 2 units up

23. y = tan 2x – ⎜ ⎟ ⎝

Period = , shift:

units right

20. y =+ 3 sec( x − π )

Period = 2 π , shift: 3 units up, π units right

Instructor’s Resource Manual

Section 0.7 47

1 cos 2 π () + 1 cos π

24. a. and g.: y = sin

⎜ x + ⎟ = cos x = – cos( – ) π x 2 π

b. and e.: y = cos ⎜ x + ⎟ = sin( x +π ) 2 + 3

=− sin( π− x )

1 – cos 2 ()

=− sin( x +π )

d. and h.: y = sin ⎜ x − ⎟ = cos( x +π ) 4

= cos( x −π )

32. a. sin(x – y) = sin x cos(–y) + cos x sin(–y) = sin x cos y – cos x sin y

25. a. –t sin (–t) = t sin t; even

b. cos(x – y) = cos x cos(–y) – sin x sin (–y) = cos x cos y + sin x sin y

b. sin (– ) 2 t = sin ; 2 t even

tan x + tan(– ) y

1 c. tan( – ) x y =

c. csc(– t ) =

= –csc t; odd

1 – tan tan(– ) x y

sin(– t ) tan – tan x = y

d. sin( ) −= t – sin t = sin ; t even

+ 1 tan tan x y

e. sin(cos(–t)) = sin(cos t); even

tan t + tan π

tan t + 0

33. tan( t +π= )

1 – tan tan t π

1 – (tan )(0) t

f. –x + sin(–x) = –x – sin x = –(x + sin x); odd

= tan t

26. a. cot(–t) + sin(–t) = –cot t – sin t = –(cot t + sin t); odd

34. cos( x − π ) = cos cos( x −− π ) sin sin( x − π ) = –cos x – 0 · sin x = –cos x

b. sin (– ) 3 t = – sin ; 3 t odd

35. s = rt = (2.5 ft)( 2 π rad) = 5 π ft, so the tire

1 goes 5 π feet per revolution, or 1

c. sec(– t) =

= sect; even

5 π revolutions

cos(–t )

per foot.

t ; even

≈ 336 rev/min

e. cos(sin(–t)) = cos(–sin t) = cos(sin t); even

36. s = rt = (2 ft)(150 rev)( 2 π rad/rev) ≈ 1885 ft

f. (– x ) 2 + sin(–x ) = x 2 – sin x; neither

37. rt 11 = rt 22 ; 6(2 ) π t 1 = 8(2 )(21) π

π ⎞ ⎛⎞ 1 1 t 1 = 28 rev/sec

39. a. tan α = 3 44. Divide the polygon into n isosceles triangles by

drawing lines from the center of the circle to

the corners of the polygon. If the base of each triangle is on the perimeter of the polygon, then

b. 3 x + 3 y = 6 2 the angle opposite each base has measure π . n

3 y = –3 x + 6 Bisect this angle to divide the triangle into two

x + 2; m = –

3 3 right triangles (See figure).

tan α = –

40. m 1 = tan θ 1 and m 2 = tan θ 2 tan θ + tan( − θ )

tan θ = tan( θ 2 − θ )

− 1 tan θ 2 tan( − θ 1 )

1 + 3(2 ) 7 h = r cos . θ ≈ 0.1419 n

45. The base of the triangle is the side opposite the

angle t. Then the base has length r 2 sin

2 y = x –2 (similar to Problem 44). The radius of the

m 1 = , m 2 = –2 semicircle is r sin and the height of the

tan θ =

3 triangle is r cos . =

42. Recall that the area of the circle is π . The r

sin 2 Observe that the ratios of the vertex angles

measure of the vertex angle of the circle is 2 π.

= r sin cos +

2 2 2 2 must equal the ratios of the areas. Thus,

2 43. 2 A (2)(5) = 25cm

Instructor’s Resource Manual

Section 0.7 49

49. As t increases, the point on the rim of the

46. cos cos cos cos

2 4 8 16 wheel will move around the circle of radius 2. 1 ⎡

t = ; that 1 ⎡

c. The point is at (2, 0) when

16 16 16 16 is , when 5 t = .

50. Both functions have frequency

. When

47. The temperature function is

⎛− ⎜ t ⎟

you add functions that have the same

frequency, the sum has the same frequency.

The normal high temperature for November

a. yt () = 3sin( π t / 5) 5cos( − π t / 5)

15 th is then T ( 10 . 5 ) = 67 . 5 °F.

+ 2sin(( π t / 5) 3) −

48. The water level function is

F ( t ) = 8 . 5 + 3 . 5 sin ⎜

The water level at 5:30 P.M. is then F (17.5) ≈

5.12 ft .

b. yt () = 3cos( π t / 5 2) cos( −+ π t / 5)

+ cos(( π t / 5) 3) −

51. a. C sin( ωφ t + ) = ( cos )sin C φ ω t + ( sin ) cos C φ ω t . Thus A =⋅ C cos φ and B =⋅ C sin φ .

b. A 2 + B 2 = ( cos ) C φ 2 + ( sin ) C φ 2 = C 2 (cos 2 φ ) + C 2 (sin 2 φ ) = C 2

c. A 1 sin( ω t + φ 1 ) + A 2 sin( ω t + φ 2 ) + A 3 (sin ω t + φ 3 ) = A 1 (sin ω t cos φ 1 + cos ω t sin φ 1 ) + A 2 (sin ω t cos φ 2 + cos ω t sin φ 2 )

A 3 (sin ω t cos φ 3 + cos ω t sin φ 3 )

= ( A 1 cos φ 1 + A 2 cos φ 2 + A 3 cos φ 3 ) sin ω t + ( A 1 sin φ 1 + A 2 sin φ 2 + A 3 sin φ 3 ) cos ω t

= C sin ( ωφ t + ) where C and φ can be computed from

A = A 1 cos φ 1 + A 2 cos φ 2 + A 3 cos φ 3

B = A 1 sin φ 1 + A 2 sin φ 2 + A 3 sin φ 3 as in part (b).

d. Written response. Answers will vary.

52. ( a.), (b.), and (c.) all look similar to this:

The plot in (a) shows the long term behavior of The windows in (a)-(c) are not helpful because

the function, but not the short term behavior, the function oscillates too much over the

whereas the plot in (c) shows the short term domain plotted. Plots in (d) or (e) show the

behavior, but not the long term behavior. The behavior of the function.

plot in (b) shows a little of each. Instructor’s Resource Manual

Section 0.7 51

54. a. hx () = ( f g ) () x

56. fx ()( = ⎪ x − 2 n ) , x ∈ ⎢ 2 n − ,2 n + ⎥

cos(100 ) 2 x +

⎩ 2 0.0625, otherwise

⎛ 1 ⎞ ⎜ 2 ⎟ cos (100 ) 1 x +

where n is an integer.

0.8 Chapter Review Concepts Test

1. False:

p and q must be integers.

pqp 1 ,, 1 2 , and q 2 are integers, so are ⎧ pq 12 − pq 21 and qq 12 .

⎪ 4 ( x − x ) + x ∈ 1 : ⎢ nn , + ⎟

55. fx () =⎨

3. False:

If the numbers are opposites

⎪− (– π and π ) then the sum is 0, ⎪

( x − 3 x ) + : x ∈ ⎢ n + , n +⎟ 1 which is rational.

where n is an integer.

4. True:

Between any two distinct real

numbers there are both a rational and an irrational number.

5. False:

0.999... is equal to 1.

()() a = a = a

1 6. True:

(*)* ab c = a bc ; *( * ) a bc = a b −1 c 1 x

7. False:

8. True:

Since x ≤≤ y z and x ≥ zx , == y z x

9. True:

If x was not 0, then ε=

would

be a positive number less than x .

10. True:

y −=−− x ( x y ) so

20. True:

If r > 1, then 1 −< r 0. Thus,

( x − yy )( − x ) = ( x − y )( 1)( − x − y )

1 1 since 1 + r ≥− 1 r , ≤ .

1 − r 1 − r 1 + [] r []

12. True:

ab , and bc , share point b in

common.

21. True:

If x and y are the same sign, then

13. True:

If (a, b) and (c, d) share a point then

x – y = x – y . x – y ≤+ x y

c < b so they share the infinitely

when x and y are the same sign, so

many points between b and c.

x – y ≤+ x y . If x and y have opposite signs then either

For example, if x = − , then 3 (x > 0, y < 0) or

−=−− x () 3 = 3 = which does 3 x – y = –– x y =+ x y

not equal x.

(x < 0, y > 0). In either case x – y =+ x y .

16. False:

For example, take x = and 1

y =−. 2

If either x = 0 or y = 0, the inequality is easily seen to be true.

x < y ⇔ x 4 < y 17. True: 4 22. True:

If y is positive, then x = y satisfies

x 4 = x 4 and 4 y = y 4 , so x 4 < y 4 x 2 2 = () y = y .

18. True:

x +=−+ y ( x y )

23. True:

For every real number y, whether it

=−+−= x ( y ) x + y

is positive, zero, or negative, the cube root x = 3 y satisfies

19. True:

0.1 Concepts Review 1. rational numbers - Calculus 9e Purcell Varberg Rigdon (Solution)

CHAPTER

k 0 ∈ S such that k 0 / n > b but

Intersection points: ( – 2, – 2 , )( 2, 2 )

( –2, 1 – 21 , 2, 1 )( + )( 13 , 2, 1 – 13 ) Problem Set 0.5

= C sin ( ωφ t + ) where C and φ can be computed from

Parts

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0.1 Concepts Review 1. rational numbers - Calculus 9e Purcell Varberg Rigdon (Solution)

CHAPTER

k 0 ∈ S such that k 0 / n > b but

Intersection points: ( – 2, – 2 , )( 2, 2 )

( –2, 1 – 21 , 2, 1 )( + )( 13 , 2, 1 – 13 ) Problem Set 0.5

= C sin ( ωφ t + ) where C and φ can be computed from

Parts

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PEMBELAJARAN BILANGAN BULAT KELAS VII PADA TINGKAT SMP OLEH : SITTI HAFIANAH AZIS,S.Pd ABSTRAK - PEMBELAJARAN BILANGAN BULAT KELAS VII SMP (hafiana)

MENINGKATKAN HASIL BELAJAR SISWA KELAS VIII.A MTs DDI PADANGLAMPE DALAM MELAKUKAN OPERASI PERKALIAN BENTUK ALJABAR DENGAN MENGGUNAKAN TABEL Oleh: SITTI HAFIANAH AZIS,S.Pd ABSTRAK - PENGGUNAAN TABEL PD OPERASI PERKALIAN BENTUK ALJABAR (hafiana)

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