Precalc Answer

1

1. statement

2. statement

3. not a statement

4. b

5. a.true

b.

c. yes

6. existential

7. universal

8. neither

9. for all, ;; there exists, '

10.holds for all real numbers aand b. Because and

are real, the relationholds when and

. Hence

.

11. Sample:

12. a. squares; x is a rectangle.b. squares; rectanglesc. square; rectangled. x is a square; x is arectangle.

13. a. an even integer; x is prime.b. even integer; primec. even integer; prime

14. ' a real number x such that; real numbers y, .y • x 5 y

x 5 1

60 1 12 5 14775 1(Ï12)2 5Ï12 1

2Ï75 • (Ï75)2 1Ï12)2 5(Ï75 1b 5 Ï12

a 5 Ï75Ï12

Ï75

(a 1 b)2 5 a2 1 2ab 1 b2

(a 1 Ïa 1 1)(Ïa)3 2 1 5 (Ïa 2 1) •

(42 1 4 1 1);p(4) 5 43 2 1 5 (4 2 1) •

15. a. yesb. noc. for 0, 1 and all integersgreater than or equal to 5

16. Sample: All students in mymath class are teenagers.

17. Sample: There exists astudent in my math classwho owns a car.

18. False. Counterexamples:, or , .

19. False, Sample: log

and .

20. x can fool y

21. Sample: Let ; ; realnumbers y, .

22. ; actions x, ' a reaction ysuch that .

23. ; circles x, x is not aparabola.

24. a. , or b.

25. a. 3b.c.

26. a. 1b.c. -1

-1

a2 2 2ab 1 b2 2 a 1 b 2 1c2 2 c 1 1

y2

-2 2 4-2

-4

-6

-8

-10

y = x 2 + 2x – 8

x

x 5 2x 5 -4

x 5 -y

0 Þ 10 • y 5x 5 0

-1 , 0

( 110) 5 -1

b 5 0a 5 0b 5 1a 5 0

Answers for LESSON 1-1 pages 6–13P

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27. a. The sentence is false.b. The sentence is true.c. No, it is neither true nor false.d. Bertrand Russell (1872–1970) was a mathematicianof many talents and wrotebooks on mathematics,philosophy, logic, sociology,and education.

The paradox is: Let B be theset of all sets that are notmembers of themselves. If Xis any set, then

. Now if X is B, then

.Thus, there is acontradiction.

B [ B ↔ B [/ B

X [ B ↔ X [/ X

Answers for LESSON 1-1 pages 6–13 page 2c

Answers for LESSON 1-2 pages 14–20

1. False

2. False

3. False

4. There exists a person whocannot drive a car.

5. There exists a fractionwhich is not a rationalnumber.

6. ; real numbers x, sin x cos x.

7. p

8. c

9. a. people x; a person y; x loves y.b. ' a person x such that ;people y, x does not love y.

10. ;; not p(x, y)

11. ; functions f, ' realnumbers a and b such that

.f(a 1 b) Þ f(a) 1 f(b)

Þ

12. a. ' a real number x suchthat .b. the negation

13. a. There is a man who isnot mortal.b. the given statement

14. d

15. a. ' n in S such that .b. the negation; 11 is in Sand .

16. a. ; even integers m, m isnot in S.b. the negation; S containsno even integers.

17. a. ' a real number x suchthat ; real numbers y, tan .b. the negation. For ,

tan is undefined so ; real

numbers y, tan .π2 Þ y

π2

x 5π2

x Þ y

11 5 11

n $ 11

2x 1 4 # 0

c

3

18. b

19. The flaw occurs after thefourth line. Since ,one cannot divide bothsides of the equation by

.

20. b

21. e

22. a. True. Every student is onat least one sports team.b. True. No one is in theSpanish club.c. True. Every student is inthe math club.d. False. Each academicclub has at least onemember in the sample.e. False. Raul is not in aforeign language club.

x 2 y

x 2 y 5 0

23. Sample:

24. a. ; real numbers a and b,.

b.25. a. 4

b. -3c. 0

26. a. ; postal charges P $ 88¢, ' n and m,nonnegative integers such that .b. ;

; ;.

c. It is true. You can get 92¢with four 23¢ stamps. Byadding 5¢ stamps to 88, 89,90, 91, and 92 cents, youcan get all charges over 88¢.

91 5 2 • 23 1 9 • 590 5 18 • 53 • 23 1 4 • 5

89 588 5 23 1 13 • 55m 5 P23n 1

(2x 1 3)2 5 4x2 1 12x 1 9a2 1 2ab 1 b2(a 1 b)2 5

Ï(-1)2 5 1 Þ -1

Answers for LESSON 1-2 pages 14–20 page 2P

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1. L is greater than 12 or L equals 12.

2. x is greater than 3 and x isless than or equal to 4.

3. False

4. When p is true and q isfalse, or when p is false andq is true, or when both pand q are true.

5. a.

b. or (p and q)

6. False

7. True

8. True

p ; p

9. See below.

10. orand

and

11. andor

or

12. or

13.14.15. exclusive

16. c

17. a.

b. See below.

7 , x # 11

5 , x # 11

x # 7x . 5

x . 4,(x # 4) ; x # 3x # 4) ; ,(3 , x),(3 , x # 4) ; ,(3 , x

L Þ 12 ; L , 12,(L 5 12) ; L # 12L 5 12) ; ,(L . 12),(L $ 12) ; ,(L . 12


p q p xor qT T FT F TF T TF F F

9. p q p or q not (p or q) not p not q (not p) and (not q)T T T F F F FT F T F F T FF T T F T F FF F F T T T T

17. b.(p or q) and

p q p xor q p or q p and q not (p and q) (not (p and q))T T F T T F FT F T T F T TF T T T F T TF F F F F T F

same truth values

same truth values

p q p and q p or (p and q)

T T T TT F F TF T F FF F F F

c

5

18. They are not rectangular, or

are less than inches high

or are less than 5 inches long.

19. a. ; real numbers ,.

b. The statement is true.

20. c

21. sin

22. a. Falseb. True

23. y

x

y = x 21]2

(-2, 2) (2, 2)2

-2 42-4

4

6

-2

Ï32 • Ï2

2 112 • Ï2

2 5Ï6 1 Ï2

4

sin π3 • cos π

4 1 cos π3 • sin π

4 5

(7π12) 5 sin (π

3 1π4) 5

log10 x Þ 0x , 0

312

24. slope , y-intercept ,

x-intercept

25. a. Sample: The waiter givesyou a choice of coffee, tea,or milk. He then comesback to tell you that he hasrun out of all three.Therefore, you can’t havecoffee and you can’t havetea and you can’t havemilk.b. See below.

553

5525 -32


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25. b.p q r q or r p or ,(p or ,p ,q ,r ,q and ,r ,p and

(q or r) (q or r)) (,q and ,r)

T T T T T F F F F F FT T F T T F F F T F FT F T T T F F T F F FT F F F T F F T T T FF T T T T F T F F F FF T F T T F T F T F FF F T T T F T T F F FF F F F F T T T T T T

same truth values

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1. a.

b. See below.c. The output columns foreach network are identical.

2. (p or (not q)) and r

3. a. In 1999, it was 145 yearsago.b. In 1999, it was 62 yearsago.

p q p OR q NOT(p OR q)

1 1 1 01 0 1 00 1 1 00 0 0 1

4. not(not(p and q) or (not r))

5. ((not p) or q) and not((notq) and r)

6. See below.

7. a. 11¢b. 7¢c. the network in Question 6


1. b. outputp q NOT p NOT q ((NOT p) AND (NOT q))1 1 0 0 01 0 0 1 00 1 1 0 00 0 1 1 1

6. The network in Question 6 corresponds to the logicalexpression q or ((not p) and (not r)). Then the input/outputtable for the networks of Questions 5 and 6 is

(NOT (NOT q) NOT output for (NOT p) output forp q r NOT p) NOT AND r ((NOT q) Question NOT AND Question 6

p OR q q AND r) 5 r (NOT r)

1 1 1 0 1 0 0 1 1 0 0 11 1 0 0 1 0 0 1 1 1 0 11 0 1 0 0 1 1 0 0 0 0 01 0 0 0 0 1 0 1 0 1 0 00 1 1 1 1 0 0 1 1 0 0 10 1 0 1 1 0 0 1 1 1 1 10 0 1 1 1 1 1 0 0 0 0 00 0 0 1 1 1 0 1 1 1 1 1

same truth values

c

Thus, they have the same output.

7

8.

9.

10.11. There is a symphony

orchestra with a full-timebanjo player.

12. ' real numbers x and y suchthat .

13. False; counterexample: Let.

14. a.b. Sample: Let . Then z5 z 5 5 Þ -5.

y 5 5

z-5 z 5 -(-5) 5 5

n 5 3

x2 1 y2 # 0

-2 , x # 4

15.

x-intercepts: 2 and 5y-intercept: 10the axis of symmetry:

vertex: (3.5, -2.25)

16. Sample: While working atBell Laboratories (1941–1957), he developed amathematical theory ofcommunication known as“information theory.”

17. Sample: For circuits inseries, consider a string oflights. If one light fails,none of the lights willwork. Each light must workfor the string of lights towork. This is like the ANDgate. For circuits in parallel,consider the lights in ahouse. A light in one roommay work whether or notany other lights in thehouse work. The house iscompletely dark only whenall the lights are off. This islike the OR gate.

x 5 3.5

y

8

12

6

10

2-2 4 6 8

4

2

-2y = x 2 - 7 x + 10

x


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outputinput F input D input M (ON or OFF)

1 0 0 ON1 0 1 OFF1 1 1 ON1 1 0 OFF0 0 0 OFF0 0 1 OFF0 1 1 OFF0 1 0 OFF

p q ,q (p and ,q)T T F FT F T TF T F FF F T F

c

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1. a. antecedent, hypothesis:

conclusion, consequent:

b. True

2. b

3. False

4.

5. True

6. False. Counterexample: Let; cos , which is

not negative.

7. inverse

8. a. If , then the graphof is not anoblique line.b. True

9. a. If a quadrilateral doesnot have two angles ofequal measure, then thequadrilateral does not havetwo sides of equal length.b. False

10. Converse: If it will raintomorrow, then it will raintoday.Inverse: If it does not raintoday, then it will not raintomorrow.

y 5 mx 1 bm 5 0

x 5 1x 5 2π

p and p q p ⇒ q ,(p ⇒ q) ,q (,q)

T T T F F FT F F T T TF T T F F FF F T F T F

2x2 1 3x3 . 1

x . 111. If two supplementary

angles are congruent, thenthey are right angles.If two supplementaryangles are right angles,then they are congruent.

12.13. a. False

b. False

14. a. yesb. noc. yesd. no

15. If one has been convicted ofa felony, then that person isnot allowed to vote.

16. If one can, then one does.

17. a. If Jon wasn’t at the sceneof the crime, then Jondidn’t commit the crime.b. If one has a true alibi,then one is innocent.

18.

19. If a satellite can stay inorbit, then it is at a heightof at least 200 miles abovethe earth.

20. If an integer is of the form2k for some integer k, thenit is even.

not equivalent

p q p ⇒ q q ⇒ pT T T TT F F TF T T FF F T T

log232 5 5


same truth values

c

9

21. If one is elected to thehonor society, then one’sGPA is at least 3.5.

22. a. 1000 3b. -50 THE LOG ISUNDEFINEDc. 0.1 -1

23. See below.

24. a. 1b. 1c. 0

25. a. andb. .c.

26. a. Peggy Sue, Buddy Hollyb. If I Loved You, Carousel(Rodgers and Hammerstein)c. If I Had a Hammer, Peter,Paul and Maryd. Man in the Mirror,Michael Jackson

0-5 2 x

x # -5 or x . 2

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23. p q p AND q NOT q (p AND q) OR (NOT q)1 1 1 0 11 0 0 1 10 1 0 0 00 0 0 1 1

10

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1. a. If an integer is divisibleby 3, then its square isdivisible by 9. 10 is divisibleby 3.b. is divisible by 9.c. ; integers n, if p(n), thenq(n); p(c), for a particular c; ∴ q(c).d. noe. yes

2. True 3. c

4. Invalid: Sample: Anyindividual under 21 years ofage with a driver’s licenseprovides a counterexample.

5. Valid; Law of Transitivity

6. Valid; Law of Detachment

7. Laws of Indirect Reasoningand Transitivity

8. See below.

102

9. a. p ⇒ q,q∴ ,p

b. See below.

10. a. Mary is not at home.b. Let p and q be thestatements:p: Mary is at home.q: Mary answers the phone.Then the argument has theformIf p, then q.not q.∴ not p.c. Law of IndirectReasoning

11. a. If p, then q.If q, then r.∴ If p, then r.

b. Yes, it follows from the Lawof Transitivity.


p q p ⇒ q ,q ((p ⇒ q) and ,q) ,p ((p ⇒ q) and (,q) ⇒ ,pT T T F F F TT F F T F F TF T T F F T TF F T T T T T

9. b.

p q r p ⇒ q q ⇒ r (p ⇒ q) and (q ⇒ r) p ⇒ r ((p ⇒ q) and (q ⇒ r)) ⇒ (p ⇒ r)

T T T T T T T TT T F T F F F TT F T F T F T TT F F F T F F TF T T T T T T TF T F T F F T TF F T T T T T TF F F T T T T T

8.

c

11

12. a. Law of IndirectReasoningb. yes

13. The diagonals of ABCDbisect each other.

14. Sample: If an acrobatic featinvolves a quadruplesomersault, then it is notattempted by the circusacrobats.

15. -3 and -1 are not positivereal numbers, so theuniversal statement doesnot apply.

16. a. yesb. noc. yesd. yese. nof. yes

17. c

18. a

19. yes

20. a. ' a real number y suchthat .b. the statement

21. True

22. a.

b. 2 or -2

23. a. center (3, -5); radius 7;

24. Answers may vary.

y

x2

2

-2-4-6-8

-12

-4 4 6 8 10

x(x 2 2) 1 y(x 1 2)(x 1 2)(x 2 2)

y2 1 3 , 3

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1. True

2. False

3. False

4. a.

b. If an animal is amammal, then it is a whale.Falsec. If an animal is not awhale, then it is not amammal. False

5. a. improper inductionb. invalid

6. a. inverse errorb. invalid

7. a. Law of IndirectReasoningb. valid

8. a. converse errorb. invalid

9. a. improper inductionb. invalid

10. a. q ⇒ pb. ,p ⇒ ,qc. inverse

11. a. p: Peter is not at home.q: The answering machineis on.p ⇒ qq∴ pb. invalid, converse error

Mammals

Whales

12. a. p: q: p ⇒ q,p∴ ,qb. invalid; inverse error

13. a. p(x): x is President of theUnited States.q(x): x is at least 35 yearsold.Let c be Queen Elizabeth.; x, p(x) ⇒ q(x)q(c)

∴ p(c)b. Yes; Noc. invalid; converse error

14. p: the land is covered with ice.q: the land is Antarctica.r: there are researchstations there.s: scientific study is beingconducted.p ⇒ qq ⇒ rr ⇒ ss∴ pInvalid; converse error

x2 5 9x 5 3


c

13

15. p(x): x is a real number.

r(x): x is a pure imaginarynumber.Let ; x, p(x) ⇒ q(x); x, r(x) ⇒ ,q(x)r(c)∴ ,p(c)Valid

16. p: You send a minimumorder of $10 to a mail order house.q: You can return asweepstakes couponenclosed with the catalog. r: Your order is one of thefirst 1000 orders.s: You have a chance ofwinning $100.p ⇒ q(q and r) ⇒ ss∴ pinvalid, converse error

c 5 2i

q(x): x2 $ 017. a. See below.

b. inverse error

18. a. yes; yesb. yes; noc. Arguments I and II bothhave the form belowp ⇒ qp ⇒ r∴ q ⇒ rd. invalid

19. a. Let p: Devin is a boy.Let q: Devin plays baseball.Let r: Devin is a pitcher.p ⇒ qq ⇒ r,r∴ ,pb. The argument correctlyuses the Law of IndirectReasoning and the Law ofTransitivity.

20. p ⇒ q, q ⇒ r, hence: p ⇒ r.r ⇒ q, q ⇒ p, hence: r ⇒ p.∴ p ⇔ r


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p q p ⇒ q ,p (p ⇒ q) and ,p ,q ((p ⇒ q) and (,p) ⇒ ,qT T T F F F TT F F F F T TF T T T T F FF F T T T T T

17. a.

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21. p ⇒ q (4)q ⇒ r (2)r ⇒ s contrapositive of (5)s ⇒ t contrapositive of (3)p ⇒ t Law of Transitivityp (1)∴ t Law of Detachment

22. Sample: Let p: 2 , 1. Let q: 3 , 2Both statements are false.(p and q) is false, but (p ⇒ q) is true.

23. ; integers a and b, if

then .

24. Vanna White is the hostessand the show is not Wheelof Fortune.

25. a. ; real numbers x and a,.

b. i. Let . Bysubstitution,

.c. -400

ii. Let and .By substitution,

.(3y2 2 z)(3y2 1 z)9y 4 2 z2 5

a 5 zx 5 3y2

(x 2 4)(x 1 4)x2 2 16 5

a 5 4x2 2 a2 5 (x 2 a)(x 1 a)

a2

b2 5 2

ab 5 Ï2

26. a. iiib. vc. ivd. iie. i

27. a.i. ii.

iii.

b. Answers will vary.

symmetry

scaleneisoscelestriangle

valid

laws

treaties

all things ratified water-

colors oils

artists

invalidvalid


15

1. A proof is a chain oflogically valid deductionsusing agreed-uponassumptions, definitions, orpreviously provedstatements.

2. ; ; .

3. See below.

4. Let x be the regular hourlywage. From the given,

. By theAddition Property ofEquality (adding -460 toeach side), . Then,using the MultiplicationProperty of Equality(multiplying both sides by

), . So the worker

earns $11.50 per hour.

5.6. or 47

7.8. a. p(x): x is the difference

in the two solutions to thequadratic equation

, .(a Þ 0)bx 1 c 5 0ax2 1

y 5-1 6 Ï69

4

n 5 -46

x 5 6 2Ï30

x 5 11.5112

12x 5 138

8(1.5x) 1 460 5 598

c 518b 5 240a 5 8x

q(x): the difference is

or .

b. By the QuadraticFormula, the solutions tothe quadratic equation

are

and

.

Let and

.

Then

.

If we let

and ,

Then

.

9. ; See below.y , 90

-Ïb2 2 4aca

x 5 x1 2 x2 5

x2 5-b 1 Ïb2 2 4ac

2a

x1 5-b 2 Ïb2 2 4ac

2a

Ïb2 2 4aca

x 5 x1 2 x2 5

x2 5-b 2 Ïb2 2 4ac

2a

x1 5-b 1 Ïb2 2 4ac

2a

-b 2 Ïb2 2 4ac2a

-b 1 Ïb2 2 4ac2a

ax2 1 bx 1 c 5 0

x 5 -Ïb2 2 4aca

x 5Ïb2 2 4ac

a


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3. Conclusions JustificationsGivenAddition Property of Equality (add 80)

Multiplication Property of Equality (mult. by )

Reflexive Property of Equality

9. Conclusions Justifications

Given

Addition Property of Inequality

Multiplication Property of Inequalityy , 90

16 y , 15

12y 2 5 ,

13y 1 10

2m 5 72

11572 5 2m

1080 5 30m1000 5 30m 2 80

c

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10. About 150 feet from thebatter; it is the mean of

and .

11. converse error

12. improper induction

13. invalid; inverse error

14. a. Sample: If you haveoutstanding school grades,then you will receivefinancial aid.b. Law of Transitivity

15. False. x could be -3.

16. Inverse: If the temperatureinside a refrigerator is notabove 40˚F, then thecooling system is notactivated.Converse: If a refrigerator’scooling system is activated,then the temperatureinside is above 40˚F.Contrapositive: If arefrigerator’s coolingsystem is not activated,then the temperatureinside is not above 40˚F.

x2 5 300x1 5 0

17. a. HELLOb. IF (( ) or ( or

)) THEN PRINT“HELLO” ELSE PRINT“GOODBYE”c. ( and )

18. a. 0 b. 0 c. 0

19. See below.

20. ' real numbers x and y suchthat .

21. ; real numbers x and y,.

22. a. ' a real number value ofu such that cos u 0.b. existential

23. a. -84b.c.

24. Student B is correct,because all steps he usedare valid. Both students Aand C are wrong. Student Amultiplied both sides by

, which is invalid if

. Student C made amistake when he equated

to .0 • t3 2 3t

t 5 1

11 2 t

-4t2 2 11t 2 6-4h2 2 19h 2 21

5

xy Þ 0

xy Þ 2

B . 4A 5 7

B , 4B 5 4A Þ 7

Answers for LESSON 1-8 pages 61–66 page 2

Converse Inversep q q p ,p ,q ,p → ,qT T T F F TT F T F T TF T F T F FF F T T T T

⇒19.

same truth values

c

17

1. universal

2. neither

3. existential

4. a. inverseb. contrapositivec. converse

5. ; countries c, c has notlanded people on Mars.

6. ; intelligence memos m, 'a government official g,such that g reads m.

7. ; composite numbers n, ' apositive integer y, such that

, , and y is afactor of n.

8. If you never practice yourpiano lessons, you will notlearn to play piano.

9. b, c

10. False

11. If one passes a state’s barexam, then one can practicelaw in the state. If one canpractice law in a state, thenone has passed the state’sbar exam.

12. a.b. ( or ) and ( or )

13. If , then log .

14. a. Sue is not wearing a bluesweater.b. Sue is wearing a bluesweater and she doesn’thave brown eyes.

x . 0x . 1

x , 3x 5 3x . -3x 5 -3

-3 # x # 3

y Þ 1y Þ n

15. Some British bobby carries a gun.

16. Some President is notguarded by any SecretService agent.

17. A person wants to travelfrom the U.S. to Europe anddoesn’t travel by plane orby ship.

18. and isobtuse.

19. Excessive bail shall berequired, or excessive finesshall be imposed, or crueland unusual punishmentsshall be inflicted.

20. True

21. p and (not q)

22. True

23. False

24. False; Counterexample:rhombus. The diagonals ofa rhombus bisect eachother, but a rhombus is notalways a rectangle.

25. True

26. a. ' a real number x suchthat .b. the statement

27. yes

28. no

29. False

30. When z is negative

31. False

32. True

sin2 x 1 cos2 x Þ 1

/Am/A 5 40°

Answers for Chapter Review pages 71–75P

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33.

34. a.b. 125

35. invalid by IV

36. valid by II

37. valid by III

38. invalid by V

39. a. yesb. no

40. True

41. c

42. All even numbers are realnumbers by Law ofTransitivity.

43., by the Law of

Substitution.

44. Multiplication by is

meaningless when .

45.

46. See below.

4ac4a2 5

ca

b2 2 (b2 2 4ac)4a2 5

-b 2 Ïb2 2 4ac2a 5

-b 1 Ïb2 2 4ac2a •

x 5 0

1x2

π 1 13zπ 1 -13 z # zπz1z-13 z 5

(a 1 b)2 5 a2 1 2ab 1 b2

144x 1 64(3x 1 4)3 5 27x3 1 108x2 1 47. True; All campers

participate in a sportsactivity.

48. False; No camperparticipates in all arts andcrafts activities

49. False; No camperparticipates in all sportsactivities

50. False; Oscar does notparticipate in arts & crafts.

51. False; Oscar participates innature identification, buthe does not participate inarts & crafts.

52. True; Kevin participates inentomology, Jennifer, Rubyparticipate in hiking.

53. False; No camperparticipates in both jewelry design and inswimming

54. invalid by IV

55. valid by I

56. valid by II

57. valid by II and III

Answers for Chapter Review pages 71–75 page 2

46. Conclusions JustificationsGivenAddition Property of InequalityMultiplication Property of Inequalitym . 25

-2m , -5046 1 9m , 11m 2 4

c

c

19

58.

59. a. (p AND q) OR (q AND r)b. 0

60. a. NOT(p AND (NOT q))b. (NOT p) OR qc. See below.d. not(p and (not q))(not p) or (not(not q))(not p) or q

61. p q p or qT T TT F TF T TF F F

;;

outputp q r G not G (not G) and r

1 1 1 0 1 11 1 0 0 1 01 0 1 0 1 11 0 0 0 1 00 1 1 1 0 00 1 0 1 0 00 0 1 1 0 00 0 0 1 0 0

62.

63.

64.

65. See below.

p q r q or r p and (q or r)

T T T T TT T F T FT F T T TT F F F FF T T T FF T F T FF F T T FF F F F F

p q r p ⇒ q (p ⇒ q) ⇒ r

T T T T TT T F T FT F T F TT F F F TF T T T TF T F T FF F T T TF F F T F

p q p ⇒ qT T TT F FF T TF F T

Answers for Chapter Review pages 71–75 page 3P

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60. c. p q not q p and (not q) not(p and (not q)) not p (not p) or q

1 1 0 0 1 0 11 0 1 1 0 0 00 1 0 0 1 1 10 0 1 0 1 1 1

same truth values

65. p q p and q not(p and q) not p not q (not p) or (not q)

T T T F F F FT F F T F T TF T F T T F TF F F T T T T

same truth values

20

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1. The range of a function isthe set of all possible valuesof the dependent variableof the function.Example: The range of thefunction is the setof all real numbers.

2. A real function is a functionwhose domain and rangeare sets of real numbers.Example: is areal function.

3. A real-valued function is afunction whose range is aset of real numbers.Example:

1, if x is a male0,if x is a female

4. a. the correspondencebetween the number ofhours spent driving and thenumber of miles traveledb. the number of hoursspent drivingc. the set of times between0 and 8.5 hoursd. The dependent variableis the number of milestravelede. the set of distancesbetween 0 and 520 miles

5. {(-`, -2), (-2, 0), (0, 2), (2, `)}

6. (-π, π]

f(x) 5 H

f(x) 5 2x 1 1

f(x) 5 x3

7. (-`, 1] or [5, `)

8. a.

b. dependent variable: rindependent variable: Vc. domain: (0, `)range: (0, `)

9.

Domain interval: [-2π, 2π]

10. a. The alphabet can be putinto 1-1 correspondencewith the set of positiveintegers from 1 to 26, which is a subset of the setof integers.b. Any finite set can be putinto 1-1 correspondencewith a subset of the set ofintegers.c. Because a finite set is adiscrete set as explained inpart b.

11. a. 3b. {0, 1, 3, 5, 7, 9, 11, 13, 15}c. Its domain is a finite setwhich is discrete.

12. (-`, `)

13. [-2, 2]

14. (-∞, -3), (3, ∞)

15. (-∞, -2), (-2, 1), (1, ∞)

-2π # x # 2π, x-scale = -2 # y # 2, y-scale = 1

π]2

r 53

Ï3V4π


c

21

16. a. $8.25b. The exact length of timecan not be determined, butis at least 1 hour and lessthan 2 hours.c. $11.25d. $1.50e. No. There are elements cin C for which f(c) has morethan one value.f. Yes. It assigns to eachelement of T exactly oneelement of C.g. i. t

ii. ciii. {t: hours}

17. 8a 1 23

0 , t # 24

18.

19. a.

b. The height will be of

its original value.

20.

21. F9(2, -1), I9(1, 0), R9(1, 1),E9(2, 0)

22. Answers may vary. Sample:; ; y 5 z-10x zy 5 z10x zy 5 x2

=R'

I '= (1, 0)F '= (2, -1)

E '= (2, 0)

F = (-1, 2) E = (0, 2)

I = (0, 1) R = (1, 1)

y

x

23

14

h 5kVr2

(34, -1

8)P

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In-class Activity1.

2. a. Sample: 4.5b. Sample: [4.5, `)

3. Sample: [4.45, `)

4. Sample: [4.4375, `)

5. a.

b.

Lesson

1.

2. ' z;

3. range: [-1.5, 1.5]maximum: 1.5minimum: -1.5

4. a. f(2.05) 82.990f(2.15) 82.998

b. Because the minimumvalue of f is approximately82.9, which exceeds 82.5.

5. a. 58.2 in3

b. 83.1 in2

c. The surface area of a canwith these dimensions isonly slightly greater thanthe minimum.

6. a.

b. r 5.4 cm; h 10.9 cmøø

S(r) 5 2πr2 12000

r

øø

M $ g(z)

(-`, 54G

F7116, )̀(38, 71

16)

-5 # x # 5, x-scale = 1 2 # y # 10, y-scale = 1

7.

Estimated range: [-0.1, 4]

8.

Estimated range: [-1.1, 1.2]

9. a.

b.

c. P(130) 567.7P(140) 565.7P(150) 566.7

d.

(141.4, 565.7)e. 565.7 m

10. radius 2.6 in.height 2.7 in.

11. a. minimum: -3; maximum: 3b. domain: [-5, 6]

range: [-3, 3]c. and d. [-5, -4) and (-1, 4)

x 5 3x 5 0

øø

0 # x # 250, x-scale = 500 # y # 4000, y-scale = 1000

øøø

P 540000

w 1 2w

, 520000

w

-1 # x # 4, x-scale = 1-2 # y # 2, y-scale = 1

0 # x # 2, x-scale = -2 # y # 5, y-scale = 1

1]2


c

23

12. (-`, ]

13. a. Yes, because eachelement in R corresponds toexactly one element in P.b. No, because eachelement in P corresponds to more than one elementin R.c. i. 8

ii. not possibleiii. 15

14. a.b.c. (1, `)

15. a. Switch 1 Switch 2 Light

1 1 11 0 00 1 00 0 1

b. p q p ⇔ qT T TT F FF T FF F T

A “1” in the first or secondcolumn means that theindicated switch is up, anda “0” means that it is down.

If 1 corresponds to T and 0to F, it is apparent that thetwo truth tables areequivalent. Hence, thestairway light situation is aphysical representation of p ⇔ q.

y . 1y $ -1

3Ï2 16. ' y in B such that ; x in A,.

17. Sample:radius: 1.25 inches;height: 5.75 inches;volume: 28.23 cubicinches;surface area:54.98 square inches;optimal height:3.30 inches;optimal radius:1.65 inchesEconomical dimensions arenot used to obtain thisvolume.

ø

ø

ø

ø

f(x) Þ y

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1. a.b.c. Sample: , therelative maximum: 74.28

, the relativemaximum: 78.90d. 70.52

2. a.

b.3. increasing: [0, 0.7];

decreasing: [-1, 0]; [0.7, 1].

4. increasing: [0, 2.0], [5.0, 8.0];decreasing: [2.0, 5.0], [8.0, 10].

5. increasing: [0, 2.0];decreasing: [-3, 0]; [2.0, 6].

6. a. [0, 3] b. (-`, 0] c. [3, `)d. relative minimum values:0, 3; relative maximumvalue: 3

7. a. [-2 , 0], [3, `) b. (-`, -2], [0, 3]c. Noned. relative maximum value:3; relative minimum value: 1

8. a. [-3, -2], [-1, 1]b. [1, 6] c. [-2, -1]d. relative maximumvalues: -1, 2relative minimum value: -1

log x1 , log x2

-2 # x # 12, x-scale = 1-3 # y # 3, y-scale = 1

x2 5 1979

x1 5 19731979 # x # 1983x 5 1975 9. a.

b. positive real numbers;

;

10. a. [1954, 1957], [1958, 1970], [1975, 1980], [1986, 1990],[1993, 1995]b. [1972, 1975], [1980, 1983],[1984, 1986], [1990, 1993]c. 41.65, 62.07, 62.42,64.76, 65.96, 70.75, 71.16d. 35.13, 38.81, 61.29,59.86, 61.28, 64.35, 68.32

11. a. secb. At seconds,the object reachesmaximum height and starts to descend.

12. a.

b.

c. md. ;

13. Its domain is a discrete set, {integers x: 1954 x 1995}.

14. OR

15. 2105 16. a5b5

17. 18.

19. Answers may vary. Sample:increased effort atconservation in bothdecades; increased cost offuel in the middle 1970s

p(1 1 p)1 2 p

65 5 1.2

(Y $ Z )(X $ Y )

##

S.A. < 22.1m22.7 m 3 2.7 m 3 1.4 ms ø 2.7

A(s) 540s 1 s2

h 510s2

t 5 1.2755t $ 1.2775

.x2 • 1x1x2

x1 • 1x1x2

,

1x1

.1x2


25

1. x approaches infinity.

2. x approaches negativeinfinity.

3. The limit of f(x) as xapproaches negativeinfinity is 4.

4. As n approaches infinity,g(n) increases withoutbound.

5. a. As , . So

decreases through

smaller and smaller valuesas the denominator increases. ∴ as ,

.b. `c. Yes; the equation is

.

6. a. As z increases without

bound, so does z2. So

comes closer and closer to 0.b. Sample:

7. The limit does not exist.

8.9. a. Sample:

b. -`

y

x2-2 4

-2

2

-4

4

-4

y 5 0.5

f(z) 5 -12z2

4000z2

y 5 0

3-x → 0x → `

3x

3-x 513x

3x → `x → `

10.

as as

11. a.

b. 5c.d. Sample:

12. a. the even integer powerfunctionsb. the odd integer powerfunctions

13. a.

b. Sample:

-6 # x # 6, x-scale = 1-4 # y # 4, y-scale = 1

limx→`

h(x) 5 -3

x 5 10.5y 5 5

0 # x # 20, x-scale = 2-2 # y # 20, y-scale = 2

x → -`, y → `x → `, y → -`

-20 # x # 20, x-scale = 5-10 # y # 10, y-scale = 2


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14. a. See belowb. or c.

15. a. [-2, 0], [4, 5]b. relative maximum: 3,occurs at ; 2.5, occursat ; 0, occurs at relative minimum: -3, occursat ; -2, occurs in theinterval (2, 3); -3, occurs at

c. [-3, 3]d. -3.1; -1; 1.5; 4.8e. , and

16. (-`, ]

17. domain: (-∞, -3), (-3, 2), (2, ∞); range: (-∞, -1.12], (0, ∞)

-23

32 # x # 4.8-3 , x , -1

x 5 4

x 5 -2

x 5 5x 5 0x 5 -5

0 # x # 100, x-scale = 10-2 # y # 10, y-scale = 2

y 5 ey 5 2.7182818. 2

19.

20.21. ; Law of Indirect

Reasoning

22. If as , or -`,then as , .If as , ,

then as , .

If as , , thenas , or -`depending on the functiong. If ,

, and

, for any real

number L, then

does not exist. Similarrelationships occur when

.x → -`

limx→`

f(x)

limx→`

g(x) Þ L

limx→`

g(x) Þ -`

limx→`

g(x) Þ `

f(x) → `x → `g(x) → 0x → `

f(x) → 1Lx → `

g(x) → L Þ 0x → `f(x) → 0x → `

g(x) → `x → `

n $ 5

81-14

x y

76


14. a. x 100 1000 10,000 100,000 1,000,000

2.70481 2.71692 2.71815 2.71827 2.71828(1 11x)x

27

1. a. 12.08 ftb. very close (differs by 0.06 ft from maximumheight shown in the data)

2. a. 12.11 ftb. very closely (differs by0.03 ft from maximumheight shown in the data)

3. a.

b. (1.50, 7.0)c. (16.50, 4.0)d. T is the time when theball’s motion is altered bystriking the backboard,basket, or floor.

4. a. The horizontal velocity is

b.

c.

d. 1.58 sec

5. No. It takes about 0.995 sec for the ball to travel 12.94 fthorizontally to a point 14 ft from the free throw line. Atthat time it will be 9.5 fthigh and too low to gothrough the hoop.

6. a. b.

-2 # t # 2, t-step = 0.5-10 # x # 5, x-scale = 1-10 # y # 5, y-scale = 1

HI

G

F

t ø

-4.9 msec2

6.5 msec

5.9 msec

13 ftsec

7. a.

b. yesc. Let , then ;so y-intercept is

.Similarly, let , then

or 3; so x-interceptsare , and

.

8. a.

b.

c. They are symmetricabout the line .

9. From , ;

therefore,

. This meansthat the parametricequations are equivalent tothe equation .y 5 2x 1 19

5 5 2x 1 19

y 5 6 • x 1 73 1

t 5x 1 7

3x 5 3t 2 7

y 5 x

-2 # t # 10, t-step = 1-10 # x # 10, x-scale = 1 -2 # y # 10, y-scale = 1

-2 # t # 10, t-step = 1 -2 # x # 10, x-scale = 1-10 # y # 10, y-scale = 1

x(3) 5 3 2 1 5 2x(0) 5 0 2 1 5 -1

t 5 0y(t) 5 0

12 2 3 • 1 5 -2y(1) 5

t 5 1x(t) 5 0

0 # t # 5, t-step = 1-2 # x # 5, x-scale = 1-3 # y # 10, y-scale = 1

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10. a.

b. yesc. no

11. Sample: ,

12. a.

b. They both correspond tothe equation ,but the first graph limitsthe domain of x to [0, `),while the second has thedomain (-`, `).

y 5 3x2 2 2

Hx(u) 5 uy(u) 5 3u2 2 2

-2 # t # 2, t-step = 1-5 # x # 5, x-scale = 1-5 # y # 5, y-scale = 1

Hx(t) 5 Ïty(t) 5 3t 2 2

-2 # t # 2, t-step = 1-5 # x # 5, x-scale = 1-5 # y # 5, y-scale = 1

vy 5 14.7 msec

vx 5 18.5 msec

10 30 5040 6020

10

20

30

0

40

y

xGolf hole

Hx 5 15ty 5 -4.9t2 1 20t 13. Sample:

The graph of the aboveparametric equations is theimage of the graph inExample 2 under the size change with center (0, 0) and magnitude

(a contraction).

14. a. odd; Because n is odd,

b. `

15. a.

b. (0, 2), (2, `) c. (2, `)

16. 17.

18.

19.

20. a.

b.

c.

21. a.

b.

c. Ï22

Ï32

12

54 5 1.25

4Ï41

< .6247

5Ï41

< .7809

4x2 1 1

32n 1 2

x3

y 2x $ -2

-1 # x # 5, x-scale = 1-5 # y # 5, y-scale = 1

f(-x) 5 a(-x)n 5 -f(x)

12

-1 # t # 1

y(t) 512(1 2 t)(1 2 t2)

x(t) 512(1 1 t)(1 2 t2)H


c

29

22. a. Yesb. Consider any point (c, d)that corresponds to ,where is in the interval [-a, a]. Then

, andd 5 (1 2 t0)(1 2 t0

2)c 5 (1 1 t0)(1 2 t0

2)

t0

t 5 t0

The image of (c, d) under areflection over the line

is the point (d, c) andis on the graph, since itcorresponds to ,which is also in the interval[-a, a]. That is,

,and

c 5 [1 2 (-t0)][1 2 (-t0)2]

d 5 [1 1 (-t0)][1 2 (-t0)2]

t 5 -t0

y 5 x

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1. 2. -105°

3. a. b. c.

d. e. f.

4. a. b. c. -1

5. a. nob. Because , but

6.

7. a. b. c. 23

23

23

6Ï527

24 < 60.9565

0.12 1 0.92 Þ 1sin2 u 1 cos2 u 5 1

Ï22-Ï2

2

-Ï3Ï312

Ï32-1

212

5π9 < 1.75 8. with

9. Sample: (0, 1), , ,

(π, -1),

10. domain: (-`, `)range: [-1, 1]

11. a. decreasingb. decreasingc. increasing

12. maximum: 1; minimum: -1

13. cos x oscillates as andas . Therefore,

cos x and cos x do

not exist.

limx→-`

limx→`

x → -`x → `

(3π2 , 0)

(π2, 0)(π

3, 12)0 # t # 2πHx(t) 5 cos t

y(t) 5 sin t

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14. a.

b. Sample: (0, 0), ,

, ,

c. π d.

e. Sample:

15.

Because (cos u, sin u) is theimage of (1, 0) under acounterclockwise rotationof u about the origin and[cos (-u), sin (-u)] can beconsidered as the image of(1, 0) under a clockwiserotation of u about theorigin.cos (-u) cos usin (-u) -sin u

16. a.

b. -125

-1213

55

y

xu-u

(x, y) = (cos u, sin u)

(x, -y) = (cos u, -sin u)

(1, 0)

x 5π2

(-π2, π2)

(-3π4 , 1)(-π

4, -1)(3π4 , -1)

(π4, 1)

-7 # x # 7, x-scale = 1-5 # y # 5, y-scale = 1

17. a. Sample: ;

b.

c. maximum: 23; minimum: -7

18. a.

b. From ,

we have

therefore

. Namely,

c. 2 is half the length ofthe minor axis; 3 is half thelength of the major axis;and (1, -4) is the center ofthe ellipse.

19. d

20. invalid; inverse error

21. 22.23. Cannot be simplified.


qp(x 1 y)2

(x 2 1)2

4 1(y 1 4)2

9 5 1

sin2 t 1 cos2 t 5 1

(y 1 43 )2

5(x 2 12 )21

Hx(t) 2 12 5 sin t

y(t) 1 43 5 cos t

Hx(t) 5 2 sin t 1 1y(t) 5 3 cos t 2 4

0 # t # 7, t-step = 0.1-2 # x # 4, x-scale = 1-7 # y # 1, y-scale = 1

π2

-7 # y # 230 # x # π


31

1. b

2. a. the set of all realnumbersb. the set of positive realnumbersc. ;

3. where x is apositive integer

4. Sample: is given.Thus

by AdditionProperty of Inequality.

by property (3)on page 118.

by basic laws

of exponents.

So by substitution.

Since by property(1) on page 118,multiplication by yields

.Since , .

5. $46,966.66

6. a. Falseb. Sample: It was proved inQuestion 4 that isdecreasing on the set (-`, `), if and

, has this form,

where and .b 534a 5 1

(34)x

a . 0

0 , b , 1

abx

abx1abx2 ,a . 0

bx2 , bx1

bx1

bx1 . 0

bx2

bx1 , 1

bx22x1 5bx2

bx1

bx22x1 , 1

x2 2 x1 . 0

x1 , x2

f(x) 5 3x

f(x) 5 0limx→-`

limx→`

f(x) 5 `

7. a.

b. The graphs are reflectionimages of each other overthe y-axis.

8. Domain: set of realnumbersRange: set of positive realnumbersMaxima and minima: noneIncreasing or decreasing:decreasing over its entiredomainEnd behavior:

Model: decaySpecial properties: Values ofthe function are related bythe laws of exponents.

9. a.b. 1990.6 millionc. i. 23.29 million

ii. 102.58 millioniii. 451.89 millionThe formula is a

reasonable accuratepredictor of the populationfor the period from 1790 to1850.

P(t) 5 (3.93) (e(.029655)t)

limx→-`

g(x) 5 `limx→`

g(x) 5 0

y

x

y = ( )x 1

]{10 y = 10x

-2 -1 1 2-2

2

4

6

8

10

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10. a.b. The values increase lessrapidly then the data up toabout 1860, but grow muchmore rapidly after about1890. The values producedby the continuous modelalways exceed thoseproduced by the discretemodel.

11. < 32% of the originalamount

12. a. increasing (-`, 0],decreasing [0, `)

b. relative maximum:

c.

d.

13. a.

b.

c.

d. -1

14. a. the set of real numbersb.

c.

d. no limit

15. 1

π2, 3π

2

0 # y # 1

Ï22

-Ï32

12

0 , y #1

Ï2π

limx→6`

f(x) 5 0

1Ï2π

Pn 5 3.93(1.029655)n 16.17. 12, 16, 20, 24, 28

18. 4, 3.2, 2.56, 2.048, 1.6384

19. Sample:is not necessarily

exponential (if , where and ,

then , which isnot possible for anexponential function).

is not necessarilyexponential (if

, where and.5, then ,

which is not possible for an exponential function).

is not necessarilyexponential

where .

is always exponential,since there is no function

with such

that .(bc)x

5 1

b Þ cf4(x) 5bx

cx

f4(x)

c 51b → f3(x) 5 1 ;x)

(f3(x) 5 bx • cxf3(x)

f2(-x) 5 -f2(x)c 5b 5 2bx 2 cxf2(x) 5

f2(x)

f1(-x) 5 f(x)c 5 .5b 5 2cx

f1(x) 5 bx 1f1(x)

x # -2.29, x $ 0.29


33

1. the set of all integersgreater than or equal to afixed integer a, if thesequence is infinite, or theset of all integers greaterthan or equal to a but notgreater than another fixedinteger b, where , ifthe sequence is finite

2. , , ,

3. a. neither

b. 2, , , ,

4. a. geometricb. 3, 9, 27, 81, 243

5. a. geometricb. 3, -6, 12, -24, 48

6. a. geometricb. 6, 24, 96, 384, 1536

7. ; integers

8. ; integers

9. a. ; integers,

b.10. a. recursive

b.

c. AA

P4 P2P3 P1B

Hn 5 10(0.8)n21H1 5 10n $ 2

Hn11 5 0.8(Hn)

n $ 1Sn 5 3 2 2(n 2 1)

n $ 1Sn 5 7(12)n21

65

54

43

32

f5 5 18f4 5 11,f3 5 7f2 5 4f1 5 3

b $ a

11. difference equation: ; integers

explicit formula: ; integers

12. a.

b.

13. a. 0, 0, 0, 0, 0; lim 0

b. 1, 0, , 0, ; lim 0

14. ;

15. An ellipse with center (5, -4), minor axis withvertices (3, -4) and (7, -4),and major axis with vertices(5, -1) and (5, -7).

16. a. Trueb. Falsec. Trued. False

17. a. 4b. 2

18. Sample: The resultsdecrease to about 30,000and then level off. Thisincreases faith in themodel, because thepopulation decreases whenthe lake is overstocked.

limx→-`

(23)x

5 `limx→`

(23)x

5 0

515-13

5

Pn2 2 0,03Pn

Pn11 5 1.2Pn 20.2

30000 •

Pn2 2 500

Pn11 5 1.2Pn 20.2

30000 •

n $ 1(n 2 1)dan 5 b 1n $ 1an 1 d

an11 5

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34

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1. 4 2. 3. -1

4. 5 5.

6. 0, 1, < 1.585, 2, < 2.322, < 2.585, < 2.807, 3, < 3.170

7. has domain: range: set of realsmaxima or minima: none

;

model: sound intensity,logarithmic scalesincreasing over entiredomainspecial properties: relationto exponential functions,Change of Base Theorem

8. : Let and

; therefore, r and

9. 10. 5 log 2 log 3

11. log 2 log 3 log 5

12. t < 1.256

13.

14.

15. about 24.2 years

16. a. < 1.845 b. < .542

14logbn 1

34logbw

logb5 1 2 logbn 2 logbw

11

2x 5 25

logb(uv) 5 logbu 2 logbv

logb(uv) 5 r 2 s

uv 5 br2s

logbv 5 slogbu 5v 5 bs

u 5 brbr

bs 5 br2s

limx→0

5 -`limx→`

5 `

x . 0logb x

49, 23, 2

32 c.

Proof: Let

17.18. a. r < 0.00012

b. about 13,000 years

19. recursive: ; integers , explicit: ; integers

20. a. 2, 1.5, 1.417, 1.414, 1.414b. 2, 10.5, 7.060, 6.221, 6.165

c. , which is the limit of part a; 6.164, which is the limit of part b.

21. a. ; integers

b. .55, .505, .5005 c. 0.5

22. a.b. , but

23. slope: ,

24. a. log 2385 3.377; log 238.5 2.377; log 23.85 1.377; log 2.385 .377; log .2385 -.623

b.

c. log(a • 10n) 5 log a 1 nlog a 2 1log 10 5

log a 2log( 110a) 5

øøøø

ø(b 2 d

a 2 c )-15 ,-1,9 5

a 2 cb 2 d 5

, 5 (b 2 da 2 c )

x Þ 2x2 5 4x 5 -2

n $ 1

hn 5n 1 1

2n

Ï38 øÏ2 ø 1.414

n $ 1an 5 104 2 4n

a1 5 100n $ 1an11 5 an 2 4

x 5 20

logxy 51

logy x

c 51

logy x

c logyx 5 1logyxc 5 logyy

xc 5 ylogxy 5 c

logxy 51

logy x ;x Þ 1


35

1. a. increasing on [1900, 1930] and [1950, 1970]; decreasing on [1930, 1950] and [1970,1990]b. No; 1930 , 1940 but25,678 . 25,111

2. a. max: (1930, 25,678);(1970, 45,619)min: (1950, 25,111)b. Sample: [1930, 1990]

3. a. increasing on , ]

decreasing on

b. relative maximum:

4. a. 0, 1, 0, -1, 0, 1, 0, -1, 0b.c. or d. 1, 5, 9e. 3, 7, 11

5. a. arithmeticb. decreasing

6. a. neitherb. neither

7. a. neitherb. decreasing

8. relative minimum

9.10.11.

12.

13.

14.

15. 2 log N 3 log M log P21

log 12log 7 5 1.277x 5 log712 5

3-2 52z5 ; z 5

518

t 5 log642 5log 42log 6 ø 2.086

b2 5 9; b 5 3

28 5 x; x 5 256

2x 5 8; x 5 3

5 # n # 71 # n # 33 # n # 5

73

F23, `G23G(-`

16.

17. No, each element in Scorresponds to more thanone element in R.

18. Yes, each element in Rcorresponds to exactly oneelement in S. Yes, it isdiscrete.domain: {x: x an integer,

}range: {18, 56.25, 45}minimum: 18maximum: 56.25

19. a. Yes, it is a function. It isnot discrete since L can takeany real values from 0 to320.b. L, length of skid marksc. speed of the car, sd. {L: }e. {s: }

20. the set of real numbers

21. the set of real numbersexcept r 5 and

22. {z: }

23. minimum: -11maximum: 19y: {-11, -3, -1, 9, 19}

24. minimum: 1no maximumrange: the set of positiveodd integers

25. range: (-`, `)no maximum or minimum

26. a. [-17, -4.75]b. (-`, -4.75]

z $ 7

-4

0 # s # 800 # L # 320

1 # x # 25

12 log N 1 log M 2

32 log P

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Answers for Chapter Review pages 142–145

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27. [126, 1001]

28. ,

29. no limit

30. a.b.c.

31. no limit exists 32. 0

33. a. 1 b.c. 1 d.

34. a. ,

b.35. a. ,

b. none

36. -10

37. a. ,

b.c.

38. a. yesb. ; integers

c. 12.8 mmd. 42 folds

39. a. < 10.22%b. < 13.33%

40. a. < 11.6b. moles/liter

41. 20 42. 3 planes

1 3 1027

n $ 0g(n) 5 2n(.1)

limx→`

bx 5 `, limx→-`

bx 5 0

limx→`

bx 5 limx→-`

bx 5 1

limx→-`

bx 5 `limx→`

bx 5 0

limy→-`

g(y) 5 -`

limy→`

g(y) 5 -`

f(y) 5 -4

limy→-`

f(y) 5 -4

limy→`

f(y) 5 -4

y 5 1zx z $ 13

-`n $ 503n $ 53

limx→-`

2ex 5 0limx→`

2ex 5 `

43. a. ; integers , with

b.; integers , with

c.

; integers , with

44. a.

b.

c. about 4.54 in. by 9.06 in.

45. a.b.c. price $7.50

profit $62.50

46. a. increasing over: [-4, -2],[0, 3]decreasing over: (-`, -4], [-2, 0], [3, `)b. relative minima: (-4, -3),(0, -4.5)relative maxima: (-2, 4), (3, 0.5)c. neither

47. a. increasing over: [-4, 0]decreasing over: [0, 4]b. relative maxima at (0, 4)c. even

48. a. increasing over: (-`, -4],[-3, -2], [2, 3], [4, `)decreasing over: [-4, -3], [-1, 1], [3, 4]

55

-10p2 1 150p 2 500p 2 5

A 5 (b 1 1)(25b 1 2)

h 525b

P1 5 200n $ 1

Pn11 5 (1.25)Pn 2.25Pn

2

15000

P1 5 200n $ 1

Pn11 5 (1.25)Pn 1 500P1 5 200n $ 1

Pn11 5 (1.25)Pn

Answers for Chapter Review pages 142–145 page 2c

c

37

b. relative maxima at (-4, 3), (3, -1.5), ,

, and ,

relative minima at (-3, 1.5),(4, -3), , ,and , c. odd

49. a. {-6, -3, -1, 2, 4}b. in the intervals (-6, -3), (-1, 2), (4, `)c. in the intervals (`, -6), (-3, -1), (2, 4)d. , , e. x in the intervals (-`, -7),(-2.5, -1.5)

50. {y: }

51. a.b. {0, 1, 2, 3, 4, 5}c. Falsed. True

52. Sample:

53. a. decreasing over: (-`, 2.1]increasing over: [2.1, `)

b. relative minimum: c. ,

d. { }e. neither

y: y $ -4

limx→`

f(x) 5 `limx→-`

f(x) 5 `y 5 -4

-5 # x # 5, x-scale = 1-10 # y # 10, y-scale = 1

{x: -6 , x , 6}

0 # y # 4

x 5 -7x 5 -2.5x 5 -1.5

x 5

y 5 -21 # x # 2y 5 2-2 # x # -1

y 5 2-2 # x # -1y 5 -2

-1 # x # 254. a. increasing on the reals;

decreasing nowhereb. none

c. ,

d.e. odd

55. a. i. Domain: the set of realnumbersii. range: iii. increasing over: [0, `)

decreasing over: (-`, 0]iv. no maximum value

minimum 0

v.

vi. models: optics, acoustics(subject to restrictions ondomain)vii. properties: evenb. i. Domain: the set of realnumbersii. range: iii. increasing over: (-`, 0]decreasing over: [0, `)iv. maximum 0no minimum value

v.

vi. model: projectile motion(subject to restrictions ondomain)vii. properties: even, areflection image of

over the x-axisza zx2y 5

limx→6`

f(x) 5 -`

5

{y: y # 0}

limx→6`

f(x) 5 `

5

{y: y $ 0}

{y: -1 , y , 1}

limx→-`

f(x) 5 -1limx→`

f(x) 5 1

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56. i. Domain: the set of realnumbersii. range: iii. increasing over [ , ], ; integers ndecreasing over

; integers niv. maximum 5minimum 1v. no limitsvi. models: sound waves,periodic phenomenavii. properties: evenfunction, period 2π

57.

-10 # t # 10, t-step = 1 -5 # x # 5, x-scale = 1 -5 # y # 5, y-scale = 1

5

55

[-π 1 2nπ, 2nπ]

π 1 2nπ2nπ

{y: 1 # y # 5}

58.

59. a.

b.

Answers will vary. It willdepend on the width of theneighboring building.

0 # t # 10, t-step = 10 # x # 120, x-scale = 100 # y # 50, y-scale = 10

25t 1 15y 5 -4.9t2 1x 5 20t,

0 # t # 2π, t-step = -2 # x # 5, x-scale = 1-5 # y # 1, y-scale = 1

π]8


39

1. Thisvalue is the amount thefamily had to take out ofsavings or borrow for theyear 1999.

2. For the given year Y, $65,000, ,

, $54,000.

3. The annual expenses lessthe annual taxes is equal tothe annual after taxexpenses.

4. The annual fraction ofexpenses due to taxes.

5. a. False b. Truec. True d. False

6. The domain of f is the setof real numbers. Let cos x, then, since ,whenever cos ,

, so the rangeof f is the set of realnumbers. The graph of fcrosses the x-axis whenever

or cos .f is an odd function withrelative maxima andminima that get larger inabsolute value as x isfarther from the origin.

7. a. ; {x: }

b. ; {x: }

c. ; {x: }

d. ; {x: , }x Þ -1x Þ 0x3 2 1x3

1 1

x Þ 0x4 21x2

x Þ 0-2x

x Þ 02x2

x 5 0h(x) 5I(x) 5 0

f(x) 5 I(x) 5 xx 5 1h(x) 5

f 5 I • hh(x) 5

A(Y ) 5T(Y ) 5 $11,000E(Y ) 5 $65,000

I(Y ) 5

S(1999) 5 -$10,000; 8. a.

b. A and B are bothpositive over the entiredomain, so mustalways be less than A.c.

9.

10.

11. a.

b. range: {y: };amplitude: 0.5; period: π

12. Counterexample: Consider, . Then

which is increasing onlyfor .

13. .

As , .

14. a. sin (0.5t) t

b. a linear function

21

600s(t) 5

an

bn→ 0n → `

H12, 1, 98, 1, 25

32, 916, 49

128, 14Jx $ 0

2x2f • g 5g 5 2xf 5 x

-0.5 # y # 0.5

-2π # x # 2π, x-scale = -1 # y # 1, y-scale = 0.1

π2

y = sin x y = sin x cos x

y = cos x

I(Y ) 2 D(Y ) 2 E(Y )P(Y ) 5 P(Y 2 1) 1 B(Y ) 1

3 6 9 120

3

6

yf

f + g

f – g

g

x

A • B(x) 5 1

A 2 B

-5 # x # 5, x-scale = 1-5 # y # 5, y-scale = 1

BA • B

A – B

A

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15. a. As , .

b.

c.

16. for any realnumber x; {x: }

17. the line through P tangentto the circle

18. a. ' some satellite s suchthat s is not a military spysatellite.b. ; persons p, p was not aleader of a trade union or phas not received the NobelPeace Prize.

x . 0g(f(x)) 5 x

f(g(x)) 5 x

53 1 01 2 0 5 3

3 11n

1 21n2

limn→`

3 11n

1 21n2

5

3n2

n2 1nn2

n2

n2 21n2

5

1n2

1n2

•(u 1 v)n 53n2 1 nn2

2 1

5 10 15 20 25

1

2

3

4

5y

x

(u 1 v)n → 3n → ` 19. a. 22b. 485

20.

21.

22.23. Sample: For ,

; so . As xincreases in the interval {x: }, I(x) increasesfaster than R(x) decreases.As x decreases in theinterval {x: }, R(x)increases faster than I(x)decreases.

2 4 6 8 10

2

4

6

8

10

y

x

y = 1]x y = x

x # 1

x $ 1

I(x) 1 R(x) 5 2I(x) 5 R(x) 5 1

x 5 1

k 5 20

c 594

-3 -2 -1 1 2 3

1

2

3

4

y

x


41

1. a. b.

c.

d. 7320.5 e. No

2. Sample: ;

;

3. ; domain: all reals

4. , domain:

{x: , 0, 1}

5.

6. ,

,

therefore .

7. a. all realsb.

8. 9.10. 11. x 5 Ï2y ø 8.854

x 5 6Ï7x 5 65,536

1 2 3

1

2

3y

x

f (x)

g(x)

$ 0

f ° g(x) 5 g ° f 5 I

g ° f(x) 5 g(f(x)) 5 (kxk ) 5 x

f(g(x)) 5 k(xk) 5 xf ° g(x) 5

2 4

2

4

6

8

10y

x

f (x)

g(x)

x Þ -1

1

x2 11x2 2 2

6x3 1 1

(cos π2)3 5 0

(cos x)3 5g ° f(x) 5

≈ -.7424cos (x3) 5 cos π3

8

f ° g(x) 5x 5π2

12(-2x 1 3)4

-2398-x4 1 3 12. h is not a 1-1 function.

13.

k is a 1-1 function.

14. :

15. x3 16. 3x

17. a. Yes, 5x is 1-1 for all realnumbers. Therefore itsinverse, log5x is 1-1 over thepositive real numbers.b. 1

18. a.

b. This is the graph of tan x.

c. π19. a. {x: x is real and }

b. x-intercept ,

y-intercept

c. ,

20. a. b. invalidq

p[

p ⇒ q

f(x) 5 2limx→-`

f(x) 5 2limx→`

5 (0, 74)5 (7

2, 0)x Þ 4

-2π -π π 2π

-3

-1

3

12

-2

y

x

-3π2 , -π2 , π2, 3π

2

x 5

x → 32x 2 9f -1

y

xy = k(x)

y = k-1(x)

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21. Let n be odd. Then for someinteger m, .Therefore

.Because the integers areclosed under addition and multiplication,

is aninteger. Therefore,

, where p isan integer. Therefore is even.

n2 1 1n2 1 1 5 2 • p

(2m2 1 2m 1 1)

2 • (2m2 1 2m 1 1) 1)2 1 1 5 4m2 1 4m 1 2 5

n2 1 1 5 (2m 1n 5 2m 1 1

22. a. , , ,

b.

c. k insures ,

h makes the graph quicklyapproach the limit, g makesthe function symmetricabout the y-axis.

limx→` f(x) 5 0

-3 -2 -1 1 2 3

0.2

0.4

0.6

0.8

1.0

y

x

f 5 k °h °g

k 51xh 5 exg 5 x2


43

1. GivenA-2x

A2

Therefore, by the Transitive

Property

2. When the justification is an if-and-only-if (biconditional) statement

3. True 4. yes

5. no 6. yes 7. no

8. a. i. ⇒ ii. ⇔iii. ⇔ iv. ⇔v. ⇔ vi. ⇔

b.9. f: is a 1-1 function

over its entire domain.

10. False. Sample: and

11. 6 12. -2

13. 2 14. ,

15. no real solution

16. When , ,

which is undefined. Thesecond step should bequalified to exclude

.

17. 2640 miles

18. a.b.c. d. 4e. 4 is a solution.

19. a. b. 5116-10x3 2 4

x 5x 536x 5 144

6Ïx 5 12x 2 3 5 9 2 6Ïx 1 x

h 5

4x 2 1 5 0

14x 2 1 5

10x 5

14

t 5 2t 5 0x 5

x 5x 5

(-2)6 5 64(2)6 5 64

x → x3

y 5 3

⇔ x 5 72x 1 5 5 3x 2 2

7 5 x5 5 x 2 22x 1 5 5 3x 2 2 c. d. 170,368

e. No

20. a. -(log10 x)2; all realsb. for all realnumbers. g(x) is undefinedfor . Hence, isundefined.

21. a. b. -2

c. They are reflections ofeach other over the line

.

22. a.

b. x: x is real and

23. a.

b. Sample: and are zeros; zeros of f(x) andg(x) are zeros of h(x)

24. (p AND q) OR (NOT r)

25. a. b.

26. Sample:

2x 2 3 5 Ï4x 1 2

1Ï4x 1 2

51

2x 2 3 ⇔

π6, 5π

6 , 7π6 , 11π

6π6, 5π

6

x 5 4x 5 0

-2 # x # 10, x-scale = 2 0 # y # 35, y-scale = 5

x Þ12JH

f • g(x) 5 -12

-2 2 4 6 8 10-2

2

4

6

8

10y

f (x)

g(x)

x

y 5 x

1100

g ° fx # 0

f(x) # 0. 0

-2(5x 2 4)3

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44

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1.

2. a. School B ($33,000 vs.$31,907) b. 7 years

3. a. Yes b. No

4. a. Yes b. Noc. Yes d. No

5. a. -1 and 0 or 0 and 1b. 2c. -2 and -1 or 2 and 3

6. ; real numbers y0 betweenf(a) and f(b), ' at least onereal number x0 between aand b such that .

7. sin x and e-x are bothcontinuous over the realnumbers. Function additionpreserves continuity so sin is continuous over the real numbers. iscontinuous over the realnumbers and is never equalto zero. Since (nonzero)function division preserves

continuity, is

continuous over the realnumbers.

8. [-1, 0] or [1, 2] or [4, 5]

9. a. [1, 2] b. [1.5, 1.6]

sin x 1 e-x

x21 1

x2 1 1x 1 e-x

f(x0) 5 y0

10. a. 3b. values between [2.347,2.366] are acceptable

11. a. Yesb. No. g is discontinuous at

.

12. a. No

b. is not continuous

on [1,3] so the IntermediateValue Theorem does notapply

13. after minutes

14. a. is tangent to (C at Bso ABC is a right triangle.So

b. 169 miles

15. ; domain: all reals

16. ; domain:

{x: -1, 0, 1}

17. a. Yesb. (middle C)c. 554.4 hertz

18. ,

19. b

20. An infinite number:

21. a. {-0.488, 0.748, 7.33}b. Answers will vary.

x ≈x 5 H1

π, 12π, 1

3π ...J

limx→-`

f(x) 5 `f(x) 5 `limx→`

≈n 5 -5

x Þ

x2

x42 2x2

1 1

e( 1

41x2)≈

d 5 Ï7920h 1 h2

d 5 Ï2rh 1 h2

d2 5 2rh 1 h2r2 1 d2 5 r2 1 2rh 1 h2r2 1 d2 5 (r 1 h)2

AB

ø 1.9

xx 2 2

x 5 0


x f(x) g(x) h(x)

6 34,164 35,000 -836

7 35,530 36,000 -470

8 36,951 37,000 -49

9 38,429 38,000 429

10 39,967 39,000 967

45

1. GivenAdditionProperty ofInequality

MultiplicationProperty ofInequality

2. Let f be a real function thatis decreasing on its entiredomain. Let x and y be twovalues in the domain of f.Because f is decreasing, if

, then for allx and y in the domain of f.Similarly, if , then

for all x and y inthe domain of f. Thus,whenever , .By the Law of theContrapositive,

, that is, f is a 1-1function.

3. Let f be a real function thatis decreasing on its entiredomain. Assume that onsome portion of its domainf -1 is increasing. Then thereexist some values u and vsuch that implies f -1(u) f -1(v). Since f iseverywhere decreasing, it isdecreasing on [f -1(u), f -1(v)].Therefore we must have f(f -1(u)) f(f -1(v)) or,equivalently, . Thiscontradicts our assumptionthat . Therefore f -1

cannot be increasing on anyu , v

u . v.

,u , v

x 5 yf(x) 5 f(y) ⇒

f(x) Þ f(y)x Þ y

f(x) , f(y)x . y

f(x) . f(y)x , y

y . 12

-4y , -4860 2 4y , 12 portion of its domain.

Hence, f -1 must bedecreasing on its entiredomain.

4. b

5. 1.63

6.7. 0.794

8. 6.31

9. 13 pieces

10. 1,000,000, where x is an integer

11. or

12. 10,000 to 13,000 years old

13. -ln and ln . Thus, by the

Intermediate ValueTheorem, g must have azero between 2 and 3.

14. a. [ , ], [-1, 0], [0, 1]b. any interval of length0.25 that includes theinterval (0.5706, 0.5707)Sample:

15. a. No b. Yesc. 3.00002x ø

.5 # x # .75

-2-3

3 . 05 2g(3) 52 , 0g(2) 5

0 , t , 5t .403

0 , x ,

0 # t ,

x .

y # -4

x ,

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16. domain: all real numbers;range: all positive realnumbers ; maximum: 1; minimum: no minimum;increasing on ;decreasing on ;

17. b

limx→6`

g(x) 5 0x $ 0

x # 0

# 1

18. a. If , then .b. They have the sametruth value; they are bothfalse.

19. c

20. This conjecture is false. Themaximum value of occurswhen 4.134. At thispoint, 4127, whichis not equal to 1000x.

x(102x) øx ø

xy

x # 06x # 1



1. a. 8x

b.2. a.

b.

3. ; functions f, g, h, if, then ; x,

if and only if or

4. f:

5. {-2, -1, 0}

6. {-1, 1}

7. {0, }

8.

9. {4, 256}

10.

where n is any integer

11. 99°C

u 5 Hπ6 1 2πn, 5π

6 1 -πnJx 5

H2, (-43)1/3Jx 5

-1 6 Ï3n 5

t 5

x 5

x → 2x 1 1 2 3x

g(x) 5 0f(x) 5 0

h(x) 5 0h 5 f • g

d 5 H2, 52, 3J(2d 2 5)

x 5 -13

12. where n is aninteger

13.

14. 2.161

15.

16. a.b. , c. , d. The distance (in feet)that the runner is ahead ofthe car.e. . The car catchesthe runner 4 seconds afterthe runner started.

17. a. -ln b. (-0.693, 2)

18. a. 2.861 b. x , 2.861x ≈

2 ø -0.693

t 5 4

t $ 2t2 2 5t 1 4 5 0t $ 220t 5 20(t 2 2)2

r(t) 5 20t

n 5 H34, 78J

x ≈

x 5 H61

Ï2J

x 5 {πn}

c

47

19. a. -2 and -1, and -1 and 0b.

20. log , ; x in

the domain of g;

x, ; x in thedomain of f

21. a.

b.

22. a.b. During this interval, theobject is rising and reachesits maximum height at

.t 5 .625

0 # t # 0.625

x Þ12

2x 2 1-4x 1 2 5 -12

10 log x 5g°f(x) 5 g(f(x)) 510log x1525 5

1 5 5 x 2 5 1 5 5 x10(x25)f(g(x)) 5f °g(x) 5

ø -3.423. a.

b. 6c.

d. {-2.06. 2.06}

24. a. Answers will vary.b.

c. Answers will vary. Itshould factor as

.d. The differences arisebecause this equation has 2integer roots, 2 irrationalroots, and 2 complex roots.

(x 1 3i)(x 2 3i)(x2 1 9)

(x 1 Ï7)(x2 1 9)(x 2 2)(x 1 2)(x 2 Ï7) •

x 5

-3 -2 -1 1 2 3

-3-2-1

123456

f(-x) 5 8 2 (e-x 1 ex) 5 f(x)P

reca

lcul

us a

nd D

iscr

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48

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1.2.3.4.

5. a.

b. or

c.

6. a. or b. {-1, 1}c. or

7.8. Yes, this is true for values

on the interval (-4, 0).

9. or

10. or

11. Yes, this is true for valueson the interval (-1, 0).

12.

13. or radians

14. or

15. between 4.21 and14.5 seconds

16. g(x) is discontinuous at, therefore the

Function InequalityTheorem does not apply.

x 5 0

øø

x . 22 , x , -1

0 , x , 1.9x , -1.9

47 , x , 1

b . 40 , b , 3

c . 1c , -4

2 , x , 6

-1 , x , 1x , -4x 5

x . 1-4 , x , -1

23 # x # 5

x . 5x ,23

x 5 H23, 5J

1 , x , 2

ex 1 3 . x

z , 0

x . 0 17.

18.19. a. 3

b. at -2; between -2 and -1;between 0 and 1c. Sample: (0.7, 0.8)

20. a.

b. after 1.875 seconds,18.75 units from thestarting position

21. For all integers x and y,whenever or

( and y is even).

y

x

5

-5

-5 5

x , 0x . 0xy . 0

2468

101214

5 10 15 20

y

x

π3 , u #

π2

x 5 H-1, (45)1/3

ø .928JAnswers for LESSON 3-7 pages 188-193

49

1. a. 7020(1.04)x

b. 7020(1.04)(x 2 1996)

c. $9,992d. $18,304

2. a. They are congruent andrelated by the translation T-77,0.b. -67

3.4. (3x, 4y)

5. a.

b.

6. a. S4,2/3, T-8,5

b.

c.

7.

8. a.b.

c.

9. a. b. c. -12-7818

y 5x

10 1165

y

x

x 2 – y 2 = 1

( )2 – ( )

2 = 1x + 2

]]]2

y – 3]]]1]5

-4 -2 2 4

-4

-2

2

4

(x 1 22 )2

2 (5y 2 15)2 5 1

y 58

3(x 1 5) 2 1

-12 -10 -8 -6 -4 -2

2

4

6

y

x

(x 1 8)2

16 19(y 2 5)2

4 5 1

3y 55x

y 51

x 2 3 1 2

(x, y) →(x 2 50)2 1 (y 2 30)2 5 625

x .

c 5c 5 10. all positive values

11. Apply the scale change

(x, y)

12. a. or b. The solutions are theintervals in which the graphlies below the x-axis.

13. 19,683

14. Let and, where we

assume and . Bydefinition,

. Since (because and ),

has a nonzero quad-ratic term and, hence,cannot be a line.

15. Let and. By definition,

. This is aline for all values of m andn. Therefore, is always aline.

16. 0

17. amplitude 1; period 2π; phase shift 0

18. ' a real number x such thatand and

.

19. a. True b. Falsec. True d. True

20. a. a section of a parabola

b. y 534x2 2 3x 1 1

x $ 2x # 6zx 2 4 z . 3

555

f °g

mnx 1 (mc 1 b)f °g 5 (m(nx 1 c) 1 b) 5g(x) 5 nx 1 c

f(x) 5 mn 1 b

f • gn Þ 0m Þ 0

mn Þ 0mc)x 1 bcmnx2 1 (bn 1b)(nx 1 c) 5

f • g 5 (mx 1n Þ 0m Þ 0

g(x) 5 nx 1 cf(x) 5 mx 1 b

A 5

0 , x , 5x , -5

→ (Ï2x, 1Ï2πy)

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1. The distance from x to 0 ona number line is 7 units.

2. The distance from y to 11on a number line is 2 units.

3. The distance from an to 6on a number line is lessthan 0.01 units.

4. The distance from x to 7 ona number line is greaterthan 3 units.

5. a

6. b

7.8. cm to

reject the part

9. 997

10.

11. a.b. (-3, -4.2)

12. a and b.

13. zx 2 57 z . 2

zp 2 .31 z , 0.03p , 0.34p . 0.28

m 523

t , -107 , t . -87

n .

zL 2 15.7 z . 0.5

zT 2 k z # 3°

14. a. , , 2, ,

, , ,

b. 998, 3

15. a. or b.

16. False; Counterexample: Let, . ; x.

, but ; x,.

17. See below.

18. a. No b. [0, `]

zf(x) z $ zg(x) zf(x) , g(x)

g(x) 5 -1f(x) 5 -2

-15 # x # 5, x-scale = 5 0 # y # 75, y-scale = 10

y = 24

y = |x 2 + 10x|

-12 # x # -6-4 # x # 2

5321, 28

11

4719, 52

198 , 41

17, 229

136 , 29

13, 167 , 73

2311

117 , 74, 17

9-13, 12, 1, 43


17. if and only if and if and only if and )

if and only if ) or )if and only if or .

Considering only the case when , if and only if or x . ax , -azx z . a

zx z . ax $ ax # -azx z $ a

~(x , a~(-a , xzx z $ ax , a~(-a , x~( zx z , a)

x , a-a , xzx z , a

c

1st Theorem ofthe LessonDeMorgan’s Law

51

19. a.b. The solution set of

is . Scaling

this by changes the

solution set to .

Translating this by gives

, which is the

solution set of .

20. a. Yesb. or

21. a.

b. No. There is adiscontinuity at eachintegral value.

22. a.

b. all real values

c.23. Let m and n be even

integers and p be an oddinteger. Then, by definition,

and for someintegers r and s. Likewise,by definition, for some integer q. Then

.1) 5 2(2rs 2 q 2 1) 1 1(2q 1mn 2 p 5 (2r)(2s) 2

p 5 2q 1 1

n 5 2sm 5 2r

x 5 3

x $ -32

(h 2 g)x 5 Ï2x 1 3 2 x

y

x1 2-2 -1-1

-2

2

1

x . 3x , -2

, 4_x 2

52_

12

, x ,92

12

52

-2 , x , 2

12

-4 , x , 4zx z , 4

12, 52 Since the integers are closed

under both multiplicationand subtraction, is an integer, making

an oddinteger. Therefore, is an odd integer.

24. a. Sample:

b. Sample:

c. Case 1: Let and, then and

and

Case 2: Let and, if then

so


so


, if , thenso

Thus, zx z 1 zy z $ zx 1 y z.zx 1 y z

zx z 1 zy z .zy z . yy , 0zx z . x

x , 0y , 0x , 0

zx 1 y zzx z 1 zy z .zx z . x

x , 0y $ 0x , 0

zx 1 y zzx z 1 zy z .zy z . y

y , 0y , 0x $ 0

x 1 y 5 zx 1 y zzx z 1 zy z 5zy z 5 yzx z 5 xy $ 0

x $ 0z0 1 3 z

z0 z 1 z3 z 5z1 1 -1 z

z-1 z 1 z1 z Þ

mn 2 p2(2rs 2 q 2 1) 1 1

2rs 2 q 2 1

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1. :0

2. :0

3. c

4. ,

5. The function ,where n is a positive integeris a 1-1 correspondencebetween the positiveintegers and the positiveodd integers. Therefore thepositive odd integers havethe same cardinality as thepositive integers, which is :0.

6. The union of the set ofpositive odd integers andthe set of positive evenintegers is the set of allpositive integers. All threeof these sets havecardinality :0. Therefore,

.

7. Suppose a hotel has acountably infinite numberof rooms, all full, and 100new guests wish to checkin. If the guests in Room 1move to Room 101, theguests in Room 2 to Room102, and so on, there willbe 100 vacant rooms for thenew guests. All of the newguests have beenaccommodated withoutdisplacing any of the oldguests. Therefore,

.:0 1 100 5 :0

:0 1 :0 5 :0

f(n) 5 2n 2 1

45, 54, 72, 81

17, 35, 53, 71, 18, 27

8. Sample: 0.211111

9. Label the segments and. Extend a line through B

and D and a line through Aand C. Since and have different lengths,these lines intersect at somepoint P. To establish a 1-1correspondence, pair eachpoint E on with the

intersection of and .

10.11. The distinct rays originating

at the center of thesemicircle and passingthrough each distinct pointof the semicircle eachintersect the real numberline at a distinct point.Therefore, the set of pointsof the semicircle are in a 1-1correspondence with thereal numbers, and, hence,have cardinality c.

(x, y) → (3x, 3y)

P

A E B

C F D

CDPE←→

AB

CDAB

CDAB


c

53

12. a. For any two values x1

and x2 in the interval ,

if , then tan tan x2.So tan x is one-to-one

on the interval . For

any real number y, arctan y lies in the

interval and tan x.

So the range of the function tan x with domain

is the entire real line.

Therefore, tan x is a 1-1 correspondence between

and the reals.

b. c

13. Time has cardinality c, and.

14. a.

b. 4c.

15.

16. or

17. a. 2.5b. -7.5c. 316.2

x . 1x , -14

zT 2 M z , 0.001

n . 50

an

n20 4 6 8 103

3.54

4.55

c 1 10,000 5 c

(-π2, π2)

f(x) 5

(-π2, π2)f(x) 5

y 5(-π2, π2)x 5

(-π2, π2)f(x) 5

x1 Þx1 Þ x2

(-π2, π2)18. a.

b.19. ,

20. 10

21. Sample:

22. The Cantor set is the subsetof the real interval [0, 1]consisting of all numbers of

the form , where ei is 0

or 2. Geometrically, it maybe described as follows: oneremoves from [0, 1] its

middle third interval ,

then the middle thirds of

and , and so on.

Among the properties ofthis set is that it is nowheredense in the real line, butdoes have cardinality c.

F23, 1GF0, 13G

(13,23)

ei

3io`

i51

-5-4-3-2-1

-20 -10 10 20

h(x)x

m 5

2 , x , 50 , x , 1

(x, y) → (x 2 1, 10y 1 24).

f (x)

x2 4 6 8 10 12

12141618202224

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1. 5 2. 3

3. 1

4. 3

5. -1, 2

6.

7.

8. 3

9. 10

10. 18, 50

11. -1, 7

12. -2,

13. 1 14. , 1

15. : (-x3)(ln x);domain: {x: }

: ;

domain: {x: }: ln(-x3);

domain: {x: }: -(ln x)3;

domain: {x: }

16. : ;

domain: {x: }

: ;

domain: {x: , }

: ;

domain: {x: }

: ;

domain: {x: }

17. x2 1 6Ïx

x Þ 0

x → 1x 2 6g ° f

x Þ 6

x → 1x 2 6f °g

x Þ 6x Þ 0

x → 1x2 2 6x

fg

x Þ 0

x → x 2 6xf • g

x . 0x →g ° f

x , 0x →f °g

x . 0

-ln xx3x →f

g

x . 0x →f • g

t 5t 557s 5

x 532x 5

y 5y 5

p 5p 5

z 5

w 5

v 529

x 5 -32

x 5x 5

x 5

t 5

y 5x 5 18. 19.

20.

21. {-2, -1, 3}

22. {-5, -4, 0}

23.

24. 26 or

25. 0.896

26. ,

27.

28. or

29.

30. a., b. x: ,

31. 32. b, c

33. a. The second step, whereboth sides of the equationwere divided by x3, isincorrect. This step is validonly if 0, but 0 is asolution to this equation.b. { -1, 0, 2 }

34. a. No. h(2) h(-2) 16,but .b. No, it is not a reversibleoperation because it is nota 1-1 function.

35. a. ⇒b. ⇔c. ⇔d. ⇔e. ⇔f. ⇔

2 Þ -255

x 5

x 5x Þ

t . -1

x . 2Jx , -13H-8 6-6 2 40-4 -2 x

-7 , x , 5

z .13z , -1

-5 # w # 1

x . 12 , x , 0

x ø

x 565x 5

x 5 H0, π2, 3π2 , 2πJ

z 5

y 5

x 5log2

log3

x3 2 x2 2 6Ïx3


c

55

36. True

37. ; . Since

, f and g areinverse functions.

38. log (10x 2 7);

.Since , h and m are inversefunctions.

39. By definition, a decreasingfunction f is such that forall , .Hence, for ,

. Therefore, f is1-1 and has an inverse.

40. a.b.c. No, g is not continuouson the interval [-3, -1]because it is undefined at -2. The Intermediate ValueTheorem does not apply.

41. a. No b. No c. Yes

42. h has a zero between a and b.

43. a.

b. -1 and 0, 0 and 1

-2 -1 1 2

-1

-0.5

0.5

1f (x)

x

g(-1) 5 1g(-3) 5 -1

f(x1) Þ f(x2)x1 Þ x2

f(x1) . (f(x2)x1 , x2

h °m 5 m °h 5 I10log x 5 x10log x1727 5

(m °h)(x) 5x 2 7 1 7 5 x1 7 5(h °m)(x) 5

f °g 5 g ° f 5 I(g ° f )(z) 5 (z3/5)5/3 5 z(f °g)(z) 5 (z5/3)3/5 5 z

44. a. -3b. 2 and 3, 7 and 8, 8 and 9

45.46. 6400 km

47. 1.369 sec

48. about 8 years

49. a. i. C: 10,000 55xii. S: 150 0.06xiii. R:

iv. P:

b.

c.

20000

40000

60000

80000

100000

250 500 x

P

C

R

dollars

20000

40000

60000

80000

100000

250 500

dollars

x

C

R

10,000-0.06x2 1 95x 2

x → R 2 C 5150x 2 0.06x2

S(x) • x 5x →2x →

1x →

t øh 5

v 5cÏ3

2

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50. a. cos

b.

c. The ship’s height will nolonger increase once it iscompletely unloaded.

51.52. a.

b.53. possible

54. Not possible. The functionshown is not 1-1 so it doesnot have an inverse.

55. a.

b.

2 4 6 8 10

-8-6-4-2

2468y

x

g

f

f • g

y

x2 4 6 8 10

2

4

6

8

g

f + g

y

x

0.56 , p , 0.60zp 2 0.58 z , 0.02

0 # t # 3.56

5 10 15 20

-0.4

-0.2

0.2

0.4

0.6s(t )

t

(0.5t) 2 0.4s(t) 5 0.05t 1 0.4 c. Because the range of

sin x is [-1, 1], sin x isbounded by the lines

and .Similarly, x sin x is boundedby the lines , .

56.

57.

58.

59. a.

b.

60.

61. -5, -9

62. d

63. Sample:

64. x ø 60.83

.975 # x # 1.025

x2

0.16 1(y 1 1)2

4 5 1

(x 2 2)2

9 1 9(y 1 5)2 5 1

x2

9 1 9y2 5 1

1 2 3 4 5 6

-6

-4

-2

2

4

6y

x

(g + k)(x)

1 2 3 4 5 6

-10-7.5

-5-2.5

2.55

7.510

y

x

(h – k)(x)

(h • f )(x)

x1 2 3 4 5 6 7

-0.4-0.2

0.20.40.60.8

1

y 5 -xy 5 x

y 5 x 2 1y 5 x 1 1

x 1


57

1. True; there exists aninteger, 12, such that

2. True; there exists aninteger, 1, such that

3. True; there exists aninteger, , such that

4. True; there exists apolynomial, , suchthat

5. (1) If d is a factor of n, thenthere is an integer q suchthat .(2) If there is an integer qsuch that , then dis a factor of n.Statement (1) is used tojustify the substitution of a m for b and b n for c.(2) is used to conclude thata is a factor of c.

6. a. 12 b. 10c. 10 d. 8

7. a. a factor b.c. d. e.f. m is a factor of

8. By substitution, b ca q a r which equalsa(q r) by the DistributiveProperty. Since q and r areintegers, (q r) is aninteger. Then, by thedefinition of factor, a is afactor of (b c).1

1

1•1•

51

n • pq • p(q • p)m • q

n 5 m • q

••

n 5 q • d

n 5 q • d

(n 2 6)(n 2 11).n2 2 17n 1 66 5

n 2 11

4n 5 (nm 1 3n) • 4.2n(2m 1 6) 5 (m 1 3) •

nm 1 3n

17 5 1 • 17.

132 5 12 • 11.

9. Sample: 96

10. a. a(x) is a factor of b(x)because

.a(x) is a factor of c(x)because

.b.

c. The Factor of aPolynomial Sum Theorem

11. a. Sample: , b. ' integers a and b suchthat a is divisible by b and bis divisible by a and .

12. Suppose that a(x), b(x), andc(x) are polynomials suchthat a(x) is a factor of b(x)and b(x) is a factor of c(x).By the definition of factor,there exist polynomials n(x)and m(x) such that

and . By substitution

because polynomialmultiplication is associative.Since polynomials areclosed under multiplication,

is a polynomial;so, by the definition offactor, a(x) is a factor ofc(x).

(n(x) • m(x))

a(x)) 5 (n(x) • m(x)) • a(x)(m(x) • b(x) 5 n(x) • n(x) •

c(x) 5b(x)c(x) 5 n(x) • m(x) • a(x)

b(x) 5

a Þ b

b 5 -2a 5 2

(3x 1 3) • a(x)(3x 1 3)(x 2 6) 5

3x2 2 15x 2 18 5(x 1 3) • a(x)(x 1 3)(x 2 6) 5

x2 2 3x 2 18 5

2x • (x 2 6) 5 2x • a(x)2x2 2 12x 5

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13. a. degree 6; degree 4. b. The degree of theproduct of two polynomialsis equal to the sum of thedegrees of the individual polynomials. The degree ofthe sum of two polynomialsis less than or equal to themaximum of the degrees ofthe individual polynomials.c. The conjecture does notneed to be modified.d. The degree of apolynomial is determined byits term of greatest degree.The product of a polynomialof degree n and apolynomial of degree m willhave a term of greatestdegree of the form

so theproduct has degree .If the individual polynomialsare both of degree n, theirsum will have a term ofgreatest degree of the form making the sum a degree npolynomial—unless in which case the degree willbe less than n. If theindividual polynomials arenot of the same degree, thelarger degree being n, thenthe term of the greatestdegree of the sum will be

making the sum ofdegree n.an • xn

an 5 -bn

(an 1 bn) • xn

n 1 m(an • bm) • xn1m

14. 4

15. Yes, m! m ( ) ( ) K 4 3 2 1.Because the integers areclosed under multiplication, n m ( ) ( ) K 4 2 1 is an integer.Substituting, we have m! n 3. Thus, by thedefinition of factor, m! isdivisible by 3.

16. For any integer n, the sum

. By the distributivelaw, . Since the integers are closed under addition, is aninteger. Thus, by thedefinition of factor, the sumof any three consecutiveintegers is divisible by 3.

17. a.b.

18.19. a. domain: all real

numbers, range: b.c. neitherd.

20. a.

b. 3

20 40 60 80 100 120

0.51

1.52

2.53y

n

T2,3

{x: x $ 2}{y: y $ 3}

(x 2 3)(x 1 3)(x2 1 9)

32x3 1 180x2 1 97x 2 15(8x 2 1)(4x 1 3)(x 1 5)

n 1 1

3n 1 3 5 (n 1 1) • 33n 1 3n 1 (n 1 1) 1 (n 1 2) 5

•5

••••m 2 2•m 2 1•5

•••••m 2 2•m 2 1•5


c

59

21. b is not divisible by a and cis not divisible by a.

22. a. ((NOT p) OR q) AND NOT rb. 1

23. 28 5 1 1 2 1 4 1 7 1 14

24. Sample: If two integershave n and m digits,respectively, and if theyhave the same sign, thenthe number of digits intheir sum is either thelarger of m and n or thelarger of and .If the two integers haveopposite signs, then thenumber of digits in the sumis less than or equal to thelarger of m and n. If bothintegers are nonzero, thenthe number of digits intheir product is either

or .m 1 nm 1 n 2 1

n 1 1m 1 1

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1. a. If Ms. Smith makes 43 copies, she will have 1¢ left.b. , , ,

c.2. The division is in error

because the remainder (12)is greater than the divisor(8). The correct division is

.

3. a. 12.4b. ,

4. a. -7.25b. ,

5. a. 27.9736b. , r 5 37q 5 27

K

r 5 3q 5 -8

r 5 2q 5 12

60 5 7 • 8 1 4

130 5 43 • 3 1 1r 5 1

q 5 43d 5 3n 5 130

6. a. 1186.5454b. ,

7. (integer division,polynomial division),(rational number division,rational expression division)

8. , degree 1

9. a.b.

10. ,

11. a.

b. As , the valuesof h(x) become closer andcloser to those of the linearfunction .L(x) 5 x 2 12

x → 6 `x2 2 9x 1 20

(x 2 12)(x 1 3) 1 56 5

r 5 5q 5 -43

(x 1 2) 2 15(x2 2 3x 1 7)x3 2 x2 1 x 2 1 5r(x) 5 -15

5r(x) 5 x 1 3

r 5 36q 5 1186K

c

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12. a. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9,10, 11, 12b. any value of the form

(for example, 17,30, 43)c. any value of the form

(for example, 13)d. any value of the form

, (forexample, 27)

13. integer

14. real number

15. integer

16. not a division problem

17. a. 8 pointsb. $2150c. ; the quotient is 8, theremainder is 2150.

18. False.

19. False. .

20. Suppose we have integers aand b such that a is divisibleby b. Then, by the definitionof factor, there is an integerq such that .Squaring both sides gives

. Sinceinteger multiplication isclosed, is an integer.Therefore, by the definition of factor, is divisible by b2.a2

q2

q2 • b2a2 5 (q • b)2 5

q • ba 5

x • (x2 1 x 1 1) 1 1x3 1 x2 1 x 1 1 5

93 5 18 • 5 1 3

22150 5 8 • 2500 1 2150

0 # r , 132 • 13 1 r

q • 13

q • 13 1 4

21.

22.

23.

24.25. a. The vessel that spotted

the boat at an angle of 42°b. < 0.4 mi

26. domain: all reals, range: allreals greater than 0,continuously decreasingover its entire domain,

, .

27. If the citrus crop is notruined, the temperature didnot stay below 28°.

28. a.b.c.d. q 5 6, r 5 1

q 5 -6, r 5 1q 5 -5, r 5 2q 5 5, r 5 2

limx→-`

5 `limx→`

5 0

2Ï23 ø 94.3%

3y2 114

z3 1 z2 1 z 1 1z8 1 z7 1 z6 1 z5 1 z4 1

4x 1 83x5 1 4x4 2 8x3 2 3x2 2


61

1. 10

2.3. ,

4.

which can be rewritten in the desired form.

5. By the Remainder Theorem,.

6. ,

7.8.9. a. 2355

b. p(13)

10. p(2) 0

11. ;

12.

13. As , the values off(x) become closer and closerto those of the function

.

14. By the Remainder Theorem,since , is afactor.

15.. Both

and are factors of .x4 2 x2 2 2x 2 1

(x2 1 x 1 1)(x2 2 x 2 1)x4 2 x2 2 2x 2 1(x2 2 x 2 1)(x2 1 x 1 1) 5

(x 2 2)p(2) 5 0

q(x) 5 3x3 2 2x2 1 x 212

x → 6 `

h(x) 5 7 1-18

2x 1 1

x Þ -3f(x) 5 (x 1 7)

5

r(x) 5 p(4) 5 623

r(x) 5 p(-2) 5 -73

r(x) 512 x2 1

34 x 2

74

q(x) 552 x2 1

34

r(x) 5 p(-1) 5 -9

(-3x 1 2)(x2 1 1) 1(3x 2 1)3x3 2 x2 1 1 5

r(x) 5 x2 1 x 2 2q(x) 5 x2 2 3

q(x) 5 3x 1 8, r(x) 5 20

16. a.b. ,

17. True, this is a statement ofthe Quotient-RemainderTheorem for .

18.

19.

20.

21. a.

b.

c. The minimum is at,

d. ;

22. Invalid, this is an exampleof the converse error. As acounterexample, observethat 2 10 is divisible by 4but not by 6.

•

lims→`

A(s) 5 `lims→0

A(s) 5 `

7.5 # x # 8.5, x-scale = 0.1397.8 # y # 399.4, y-scale = 0.2

h ø 8.14s ø 8.14

A(s) 52160

s 1 2s2

h 5540s2

5(1/5) ø 1.38, 4(1/5) ø 1.32

t $ 9, t #-133

(x2 1 13x 1 54)(x 2 5)2(x 1 3)4(x 1 7) •

d 5 5

r(x) 5 0q(x) 5 x2 2 xy 1 y2x3 1 0x2y 1 0xy2 1 y3

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23. Sample: In general, whenand

for somevalue a, and

. In Example 1,,

,

, and

. At ,

, butr(5) 5 8.5q(5) 5 329.5

x 5 5r(x) 532x 1 1

3x3 2 2x2 1 x 212

q(x) 5d(x) 5 2x2 1 xx4 1 x 1 1n(x) 5 6x 5 2

r(a) Þ rq(a) Þ q

q • d(a) 1 rn(a) 5q(x) • d(x) 1 r(x)n(x) 5

and so . InExample 2,

, ,

, and -3. At , , -3 but and 4 so .n(2) 5 6 • d(2) 1 1

d(2) 5n(2) 5 25r(2) 5q(2) 5 7x 5 2

r(x) 510x2 2 12x 1 15q(x) 5 x 4 2 5x3 1x 1 2

d(x) 53x 4 1 8x2 2 9x 1 27n(x) 5 x5 2

n(5) 5 329 • d(5) 1 36d(5) 5 55n(5) 5 18,131



1. Step 1: definition of factorStep 2: The RemainderTheoremStep 3: definition of Zero

2. evaluated atis 0. By the Factor

Theorem, we can concludethat is a factor of

.

3. a. Nob. Yes

4. When n is an odd positiveinteger, has a zeroat . By the FactorTheorem, ( ) is a factorof .

5.

6. a.b.

7. a. , b.c. 6d. 0

(x 2 2), (x 1 3)x 5 -3x 5 2

x 5 -1 1 i, x 5 -1 2 ip(-1) 5 0, p(3) 5 0

y 5-32 , y 5

53

cn 1 dnc 1 d

c 5 -dcn 1 dn

t4 2 5t 2 600t 2 5

t 5 5t 4 2 5t 2 600

8. b and e. These are the onlygraphs where a horizontalline can be drawn thatcrosses the graph morethan 4 times.

9.

10. a. {3, 15, 63, 255, 1023}b. Since is a factor of

, is alwaysa factor of . Hence,no value of canbe prime.

11. a. p(x) intersects the linewhenever or

. Since p(x) is athird degree polynomial, thepolynomial is also a third degreepolynomial and thus has, atmost, three zeros. Therefore,p(x) intersects at, atmost, three points.b. A polynomial of degree

will intersect the linein, at most, n points.y 5 x

n $ 2

y 5 x

p(x) 2 xp1(x) 5

p(x) 2 x 5 0p(x) 5 xy 5 x

4n 2 1 . 34n 2 1

4 2 1 5 3xn 2 anx 2 a

x 5 -74, x 5 -15, x 5 1

c

63

12. Suppose we have twopolynomials of degree n,

and , thatintersect at more than npoints. Because polynomialaddition is closed,

is also apolynomial. Assume p(x) is not the zeropolynomial. Because thedegree of the sum of twopolynomials is less than orequal to the larger of thedegrees of the twopolynomials, the degree ofp(x) is # n. By the Numberof Zeros Theorem, p(x) hasat most n zeros. Therefore,

and intersect in, at most, n points—contradicting our initialassumption. Therefore, p(x)must equal 0, and hence,

must equal .

13.

14.(x2 2 x 2 1) • (x2 1 1) 1 2xx4 2 x3 1 x 2 1 5

8x 1 0p(x) 5 15x 4 1 47x3 2 50x2 1

p2(x)p1(x)

p2(x)p1(x)

p1(x) 1 -p2(x)p(x) 5

p2(x)p1(x)

15. ,

16.

17. followed by

18. a.

b.

19.20. a. ,

b.

c.d. f is 1-1, g is not

21. a. ; x, b. the statement

22. a. 4b. 4c.

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

p(x) 5 x(x 2 4)(x 2 3) •

log2 x $ 0

{x: x . -2}

g ° f(x) 51

x 1 2

g: {x: x Þ 0}f: {x: x $ -2}

-2 , x , -1

-2 -1 1 2 3 4 5

-10

-5

5

10

15y

x

-12 , x , 4

T-6,-2S1, 14

2xy 1 7y 2)(x 2 y)3(x 1 3y)(3x2 1

r(x) 5 -101q(x) 5 x 3 1 4x2 2 12x 1 34

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In-class Activity1. Sample:

R0 { , -12, -9, -6, -3, 0, 3,6, 9, 12, 15, }R1 { , -11, -8, -5, -2, 1, 4,7, 10, 13, 16, }R2 { , -10, -7, -4, -1, 2, 5,8, 11, 14, 17, }

2. a: the sum is always in theset R0.



5. The values in Rn are givenby , where a is anyinteger and .The sum of any value fromRn plus any value from Rmis

.When this sum is divided by3, it has a fixed remainder:n m (or n m 3 if n m 3). Thus, suchsums are always membersof a single set.

6. a: The product is always inset R2.

$1211

3 • (a 1 b) 1 (n 1 m)(3a 1 n) 1 (3b 1 m) 5

0 # n # 23a 1 n

KK5

KK5

KK5

7. Following the notationused in problem 5, theproduct of any value fromRn and any value from Rmis

.When this product isdivided by 3, it has a fixedremainder: nm (or nm 3if ). Thus, suchproducts are alwaysmembers of a single set.

Lesson 1. R1 2. R0 3. R1

4. a. x is congruent to ymodulo 4b. There are four disjointsets: ,for any integer n},for ,1, 2, 3.

5. Monday

6. a. True b. False c. True

7. 2 8. 16 9. 16

10. 9 11. 4

12. a. 624 b. 357

13. Five hours after 9:00, it is2:00.

14. (mod 2)

15. Sample: x(mod 360)

16. 04

x 1 360 ;x ; 0

k 5 0Rk 5 {x: x 5 4 • n 1 k

nm $ 32

9ab 1 3bn 1 3am 1 nm(3a 1 n) • (3b 1 m) 5


c

65

17. By the Quotient-RemainderTheorem, for any integer nthere exist unique integersq and r such that

where . If, then, by definition, n

is divisible by 3. If ,either so

orso

. So one of thethree consecutive integers,n, , or isdivisible by 3.

18. Suppose (mod m) and(mod m). The

Congruence Theoremrequires that m be a factorof both and .Therefore, by the definitionof factor, there existintegers and such that

and . By the Subtraction

Property of Equality,

. By theAssociative andCommutative Properties ofSubtraction and theDistributive Property

. Because integersare closed undersubtraction, is aninteger. Therefore, bydefinition, m is a factor of

.(a 2 c) 2 (b 2 d)

(k1 2 k2)

(k1 2 k2)m(a 2 c) 2 (b 2 d) 5

k1m 2 k2m(a 2 b) 5 (c 2 d) 5

k2mc 2 d 5a 2 b 5 k1m

k2k1

c 2 da 2 b

c ; da ; b

n 1 2n 1 1

1 5 q • 3 1 3q • 3 1 r 1n 1 1 5r 5 2

q • 3 1 3q • 3 1 r 1 2 5n 1 2 5r 5 1

r Þ 0r 5 0

0 # r , 3q • 3 1 rn 5

Therefore, by theCongruence Theorem,

.

19. a.

b. 3

20. a. {x: , for allintegers n}b. Their difference is aninteger.

21. , ,

22. a. 8 b.23. a.

b.

c.

24. Assume that a, b, and c areany positive integers suchthat a divides b and adivides . Then by thedefinition of factor, thereexist integers and suchthat and

. By theSubtraction Property ofEquality

. Using theDistributive Property, thiscan be written as

. Since isan integer, by the definitionof factor, a divides c.

q2 2 q1(q2 2 q1) • ac 5

q1 • aq2 • a 2(b 1 c) 2 b 5c 5

q2 • a(b 1 c) 5b 5 q1 • a

q2q1

(b 1 c)

a 5 0, a 5 -52

30a(2a 1 5)60a2 1 150a

x8 1 3x5 2 1

v 5 -1 2 iv 5 -1 1 iv 5 -12

x 5 π 1 n

5-5 10-10

3M3(n)

n

(a 2 c) ; (b 2 d) (mod m)

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25. a.

b. ,

26. Valid by the Law ofDetachment

27. a.b.c. 10z2

2y 43x2

x $12x # -3

y

y = 3

y = 2x2 + 5x

x-4 -3 -2 -1 1-4-2

2468 28. For the system of

congruence classes modulo3, both R1 and R2 are theirown reciprocals, since

and R2. R0 does not have areciprocal, since

, for , 1, and2. For the system ofcongruence classes modulo4, both R1 and R3 are theirown reciprocals, since

and R1. R2 does not have areciprocal, since R0, ,

, and R2. Similarly R0 does nothave a reciprocal.

R2 • R3 5R2 • R2 5 R0R2 • R1 5 R2

R2 • R0 5

R3 • R3 5R1 • R1 5 R1

n 5 0R0 Þ R1R0 • Rn 5

R2 • R2 5R1 • R1 5 R1


67

1. 6000200300

2. d

3. 2 is not a digit in base 2.

4. 1111112, 778, 3F16

5. 7

6. 50

7. 375

8. 164

9. a. 7b. 10010002

c.10. 173,2558, AD16

11. a. 4 b. 100

12. 1100012

13. 1001012

14. Carry digit is the output ofthe lower AND gate, whichis 0. Since the output of theOR gate is 0 and is one ofthe inputs to the upperAND gate, the sum digit,which is the output of thatgate, is also 0.

15. e

16. It is even if the last digit is 0and odd if that digit is 1.

17. 43

18. a. 0, 1, 2b. 46c. 12013

26 1 23 5 72

19. Addition (Base 8)

Multiplication (Base 8)

20. R0: -10, -5, 0, 5, 10 R1: -9, -4, 1, 6, 11R2: -8, -3, 2, 7, 12R3: -7, -2, 3, 8, 13R4: -6, -1, 4, 9, 14

21. a.b.

22. , ,

23.

24. ,r(x) 5 -1.78125q(x) 5 2x2 2 .5x 1 3.375

(x2 1 4x 1 9)(5x 1 2)5x3 1 22x2 1 53x 1 18) 5

t 5 -3t 552t 5

19

y 5 4x 5 12

2 0 1 2 3 4 5 6 7

0 0 0 0 0 0 0 0 0

1 0 1 2 3 4 5 6 7

2 0 2 4 6 10 12 14 16

3 0 3 6 11 14 17 22 25

4 0 4 10 14 20 24 30 34

5 0 5 12 17 24 31 36 43

6 0 6 14 22 30 36 44 52

7 0 7 16 25 34 43 52 61

1 0 1 2 3 4 5 6 7

0 0 1 2 3 4 5 6 7

1 1 2 3 4 5 6 7 10

2 2 3 4 5 6 7 10 11

3 3 4 5 6 7 10 11 12

4 4 5 6 7 10 11 12 13

5 5 6 7 10 11 12 13 14

6 6 7 10 11 12 13 14 15

7 7 10 11 12 13 14 15 16

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25. Suppose m is any integer.When m is divided by 4, thepossible remainders are 0,1, 2, 3. By the Quotient-Remainder Theorem, thereexists an integer k such that

, or , or, or .

26. ,

27. a. ab. dc. cd. b

r 5 66q 5 662

m 5 4k 1 3m 5 4k 1 2m 5 4k 1 1m 5 4k

28. a. no (converse error)b. yes (contrapositive)c. no (inverse error)

29. False; .

30. < 23 mph

31. a.

b. 0.001001c. 0.00110011K2

K2

2-2

1 2 2-2 5132-6 1 K 5

2-2 1 2-4 10.0101K2 5

z-1 z Þ -1


69

1. 2, 3, 5, 7, 11, 13, 17, 19, 23,29, 31, 37, 41, 43, 47

2. 420921

3. There are no unicorns.

4. a. There exists a largestpositive integer.b. cannot be aninteger.c. Since n and 1 areintegers and the integersare closed under addition,

must be an integer.This contradicts part b.Therefore, our initialassumption must be false.Therefore, there is nolargest positive integer.

5. a. the product of all theprimesb. and

6. True

7. a. 23 b. No

8.

9.

10.11.12.13. True

14. If and n is even, thenn cannot be a primebecause it has 2 as a factor.

15. Assume there exists anumber n which is thelargest multiple of 5.

n . 2

x(x 1 1)(x2 1 1)

(z 2 Ï17)(z 1 Ï17)

y 98(3y 2 1)(3y 1 1)

(x 211 2 Ï217

3 )3(x 2

11 1 Ï2173 ) •

25 • 3 • 5

p . 1p 5 1

n 1 1

n 1 1

Then, by the definition ofmultiple, , where mis an integer. Because theintegers are closed underaddition, is aninteger. Because theintegers are closed undermultiplication, isan integer which, by thedefinition of multiple, is amultiple of 5. By thedistributive law,

which is an integerlarger than . Thiscontradicts the assumptionthat n is the largestmultiple of 5. Therefore,this assumption is false, sothere is no largest multipleof 5.

16. Assume there exists anumber x which is thesmallest positive realnumber. Since the realnumbers are closed under

multiplication, is a real

number. Furthermore,

because both and x are

positive, is positive.

However, is less than x.

This contradicts theassumption that x is thesmallest positive realnumber. Therefore, thisassumption is false, so thereis no smallest positive realnumber.

12x

12x

12

12x

5m 5 n5m 1 5

5(m 1 1) 5

5(m 1 1)

m 1 1

n 5 5m

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17.18.19. a.

b.c.

20. The prime factorization of1,000,000,000 29 59.Since the FundamentalTheorem of Arithmeticguarantees this factorizationis unique and 11 is prime, 11cannot be a factor.

21. 44

22. c

23. a. (mod 6)0 (mod 6)b. 10 9 90 0 (mod 6);16 15 240 0 (mod 6)

24. Assume for integers a, b,and p that p is a factor ofboth a and b. Then thereexist integers m and n suchthat and , bydefinition of factor. So

, by thedistributive law. Since

is an integer byclosure properties, p is afactor of .a 2 b

m 2 n

(m 2 n)pa 2 b 5 mp 2 np 5

b 5 npa 5 mp

;5•;5•

;xy ; 12

•5

x 5 2, x 5 5, x 5 -7(x 2 2)(x 2 5)(x 1 7)x2 1 2x 2 35

6(y2 1 1)2(y 1 1)2(y 2 1)2

(x 1 2)(x 2 2)(x 1 3)(x 2 3) 25. a. sine function

b. period second,

amplitude 15 amperesc. sin(120πt)d. 0 amperes

26.27. odd

28. a.

b.

c. inches, inches

29. Assume n is an eveninteger. Then, by thedefinition of even,

for some integer m.Therefore

becauseinteger multiplication isassociative. Therefore, bythe definition of an evennumber, if n is an evennumber then is an evennumber. Consequently, by the Law of theContrapositive, if is notan even number then n isnot even.

30. Answers will vary.


n2

n2

2m • 2m 5 2(2m2)n2 5 (2m)2 5

2 • mn 5

h ø 5.14s ø 10.29

A(s) 5 s2 12176

s

h 5544s2

x 5 4, x 5 8

c(t) 5 155

5160


71

1. , 2. ,

3. ,

4. , 5. 66

6. , degree 1

7.8. ,

9. ,

10. ,

11. ,

12. ,

13. ,

14.

,

15. 4 16. 12 17. 10

18. 18 19. R2 20. 2

21.22.23.24.

25.26.27.28.

29. 3, ,

30. , , 31. ,

32. -2, , , 483-15

52-37

13

34-52

114(-3 2 Ï37)1

14(-3 1 Ï37)

66t 1 11)(7t 2 2)(8t 1 1)(-13t2 2

(2x 1 1)(2x2 2 4x 2 9)

(w2 1 z2)(w 1 z)(w 2 z)

2(3v2 1 5)2

(y 1 3)(2y 2 1)(2y 1 1)(y 2 3) •

x(3x 2 5)(x 1 2)

9(t 1 2)(t 1 3)

5(x 2 y)(x 1 y)

r(x) 5 -6732

138 x 2

2916

q(x) 5 x4 112x3 2

54x2 1

r(x) 5 1082q(x) 5 5x2 1 30x 1 181

r(x) 5 24q(x) 5 x3 2 8x2 1 11x 2 15

r(x) 5 x 1 1q(x) 5 2x 2 5

r(x) 5 5q(x) 5 x3 1 3x2

r(x) 5 21x 1 7q(x) 5 7x 2 8

r(x) 5 8q(x) 5 3x2 2 7x 1 2

a 5 -5

5r(x) 5 2x 1 5

r 5 37q 5 -351

r 5 63q 5 83

r 5 2q 5 5r 5 6q 5 5 33. Because and

has no real zeros, 2 is the only realnumber with this property.

34. , , ,

35. True; .

36. False; .

37. True; if a and b are even,then , forsome integers n, m. Then

by the DistributiveProperty. Because theintegers are closed underaddition, is aninteger. Therefore, by thedefinition of factor, is divisible by 4.

38. Let n be an odd integer.Because it is odd,

for someinteger m. Then

by the DistributiveProperty. Because theintegers are closed underboth multiplication andaddition, isan integer. Therefore, by thedefinition of factor, 2 is afactor of . Since anumber is even if it isdivisible by 2, is even.n2 1 n

n2 1 n

2m2 1 3m 1 1

3m 1 1)4m2 1 6m 1 2 5 2(2m2 1(2m 1 1)2 1 (2m 1 1) 5

n2 1 n 5n 5 2m 1 1

2a 1 2b

m 1 n

4(m 1 n)2a 1 2b 5 4m 1 4n 5

b 5 2na 5 2m

156 5 17 • 9 1 3

90 5 5 • 18

Ï1112-12-Ï11

x2 1 x 1 1(x 2 2)(x2 1 x 1 1)

x3 2 x2 2 x 2 2 5

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Answers for CHAPTER REVIEW pages 278–281

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39. b, c, and e

40. Let a, b, c, and d be anyintegers such that a and care divisible by d and

. Then there existsintegers r and s such that

and bydefinition. Then,

.Since is an integer byclosure properties, b isdivisible by d.

41. This is false. For example,is not divisible by 4.

42. Let a be an odd integer.Then for some integer n,

. The next higherodd integer is

. The number onegreater than the product oftwo consecutive oddintegers is then

by the DistributiveProperty. Because theintegers are closed under

2n 1 1)4n2 1 8n 1 4 5 4(n2 5

1 51 5 (2n 1 1)(2n 1 3) 1a(a 1 2) 1

2n 1 3a 1 2 5

a 5 2n 1 1

33 2 1

(r 1 s)d(r 1 s)c 5 d • r 1 d • s 5

b 5 a 1c 5 d • sa 5 d • r

a 5 b 2 c

both multiplication andaddition, is aninteger. Therefore, by thedefinition of factor, thenumber one greater thanproduct of two consecutiveodd integers is divisible by 4.

43. 11 is a factor of

44. (mod 5)

45. If sin(x) , then

(mod 2π) or

(mod 2π)

46. (mod 11)9 (mod 11)

47. (mod 11)

48. (mod 12)

49. Either r(x) is the zeropolynomial or the degreeof r(x) is less than thedegree of d(x).

50.51. See below

52. d

53.54. 7

(x 2 7)

d 5 5

x ; 1

cd ; 2

;c 2 d ; -2

11π6

x ;7π6

x ;5 -12

x ; 0

x 5 y

n2 1 2n 1 1

Answers for CHAPTER REVIEW pages 278–281 page 2c

51.

01 212 7x41 212 7x4

2 6x2x52 6x 1 212x5 2 7x4

2 3x2x6q x6 1 2x5 2 7x4 2 3x2 2 6x 1 21x4 2 3x2 1 2x 2 7

c

73

55. Assume there exists asmallest integer.

56. Assume that there exists aprime p that is a factor of

, but not a factor of n.

57. Assume that n is thesmallest integer. Becausethe integers are closedunder addition, isalso an integer. Since

, n cannot be thesmallest integer. This is acontradiction. Therefore,there is no smallest integer.

58. Assume there are finitelymany prime numbers. Theycan be listed from smallestto largest: , ,

, pm. Multiply these:. . . pm. Consider

the integer . Sinceis larger than the

assumed largest prime pm, itis not prime. By the primeFactor Theorem, musthave a prime factor, p, onthe list of primes. Then p is a factor of both n and . It is also a factor of thedifference .Therefore, p must equal 1.But p is a prime number, sop cannot equal 1. Thiscontradiction proves thatthe original assumption isfalse. Hence, there areinfinitely many primenumbers.

59. 2311

(n 1 1) 2 n 5 1

n 1 1

n 1 1

n 1 1n 1 1

n 5 p1p2p3

Kp2 5 3p1 5 2

n 2 1 , n

n 2 1

n2

60. True

61. six (2, 3, 5, 7, 11, 13)

62. 1000

63. a

64. c

65. Prime

66.67.68. a. 88

b. 47 c.

69. 22 laps, 49.5 m left

70. a. 9 b. 27,500 c.27,500

71. 73 months of 5 days or 5 months of 73 days, 365 months of 1 day, or 1 month of 365 days

72. 5625

73. 143

74. 7

75. 5

76. 10000012

77. 101001012

78. 31

79. 42

80. 1011112, 21 26 47

81. 1111102, 31 31 62

82. d

83. In base 2, a number divisibleby 16 ends in 0000.

51

51

387,500 5 9 • 40,000 1

5789 5 87 • 66 1 47

2 • 3 • 5 • 281

19 • 29

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74

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1. False, counterexample: is

a rational number, but it isnot an integer.

2. True

3. False, counterexample:

4. Suppose r and s are any tworational numbers. Bydefinition, there existintegers a, b, c, and d,where and such

that and . So,

Since by closure propertiesthe product of two integersand the difference betweentwo integers are integers,

and bd are integers.Also, since and

. Therefore, since

is a ratio of integers,

is a rational number bydefinition.r 2 s

ad 2 cbbd

d Þ 0b Þ 0bd Þ 0

ad 2 cb

Addition offractions5

ad 2 cbbd

Multiplicationof fractions5

adbd 2

cbbd

MultiplicationProp. of 15

ab • dd 2

cd • bb

r 2 s 5ab 2

cd

s 5cdr 5

ab

d Þ 0b Þ 0

N 5 0

23 5. The sausage pizza is cut

into N pieces; the area of

each slice is of a pizza.

The pepperoni pizza is cutinto pieces; the area

of each slice is of a

pizza. Since the sausagepizza is cut into morepieces, its slices are smallerthan those of thepepperoni pizza. Thedifference in the areas ofthe sausage and pepperonislices of pizza is:

of a pizza.

6. {N: }

7. {N: }

8. {N: }

9.

{M: and }

10. {N: }

11. {K: }

12. {K: }

13. {a: and }

14. b

15. aa 2 1

a Þ 1a Þ 02a 2 1a2 2 a

K Þ -1K2 1 K 2 1K 1 1

K Þ -1KK 1 1

N Þ -K1N2 1 2NK 1 K2

M Þ 2M Þ 1

M2 2 M 2 1M2 2 3M 1 2

N Þ 0N3 1 KN3

N2

N Þ 0N2 1 KN

N Þ 0N 1 KN

52

N2 2 2N

N 2 2N2 2 2N5

NN2 2 2N 2

(N 2 2N 2 2)1

N1

N 2 2 (NN) 2

1N 2 2 2

1N 5

1N 2 2

N 2 2

1N


c

75

16.

17.

18. Suppose r and s are any tworational numbers. Then,there exist integers a, b, c,and d, where and

such that and

.

Because sums and productsof integers are integers,

and 2bd are

integers. Therefore, is

a rational number bydefinition.

r 1 s2

ad 1 bc

Multiplicationof fractions5

ad 1 bc2bd

Definition ofdivision offractions

5ad 1 cb

bd • 12

DistributiveProperty

ad 1 bcbd2

5

Multiplicationof fractions

adbd 1

bcbd

25

MultiplicationProp. of 1

ab • dd 1

cd • bb

25

ab 1

cd

2

r 1 s2 5

s 5cd

r 5abd Þ 0

b Þ 0

2a

2a3b 19. Since the average of two

rational numbers is rational,between any two numbersthere exist a rationalnumber. If r and s are bothrational numbers where

and with

and , then their

average is . If ,

then .

20. The sausage pizza is cutinto pieces, so each

slice is of a pizza. The

pepperoni pizza is still cut into pieces, so each

slice is of a pizza.

Each student who has oneslice from each pizza would

have

of a whole pizza. The twostudents who have 2 slicesof the popular pizza would

have

of a whole pizza.

21. and

22. , x Þ 3Ï2, x Þ 05x3 2 14 2 2x3

x Þ 0x2

x 1 h2 x Þ -h2

1N 1 2 1

1N 1 2 5

2N 1 2

2NN2

2 41

N 2 2 11

N 1 2 5

1N 2 2

N 2 2

1N 1 2

N 1 2

ab ,

ad 1 bc2bd ,

cd

r , sad 1 bc2bd

d Þ 0

b Þ 0s 5cdr 5

ab

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23.

Yes, both sides are equalfor all values where bothsides are defined.

24. ; Check: Let and

, then

and

25. ; Check: Let ,

then

and

26. a. domain: all real numbersrange: {y: }

b. 4.0c. (-`, 2.5]

0 , y # 4

KK 1 4 5

59

59

89 5 7

8 • 67 • 5

6 • K 1 3K 1 4 5

K 1 2K 1 3 • K 1 1

K 1 2 • KK 1 1 •

K 5 5KK 1 4

727

2(9)(4)42 2 9 5

2A2NN2 2 9 5

727

97 1 9 5

A2

N 2 3 5

A2

N 1 3 1A 5 3

N 5 42A2NN2 2 9

5K

K 1 1

5K 1 1K 1 1 2

1K 1 1

KK 1 1 5 1 2

1K 1 1 27. a. If n is not an even

integer, then n2 is not aneven integer.b. Suppose n is any oddinteger. By definition, thereexists an integer r such that

. Then

1. Since is aninteger by closureproperties, n2 is not an eveninteger.c. Yes

28. ' x such that p(x) and notq(x).

29. a. 1b. A line with slope 1 willbe 45° from the x-axis.

30. 5, 12, 13; 7, 24, 25; and 20,21, 29Patterns will vary.

5

2r2 1 2r2r) 11)2 5 4r2 1 4r 1 1 5 2(2r2 1

n2 5 (2r 1n 5 2r 1 1


77

1.

for

2. No, the expression containsnon-integer powers of thevariable.

3. Yes,

4. Yes, -1, and 3

5.

6. when

-5 and 5

7.

when -3 and -1

8.

when , ,

and

9.

when and x Þ 5x Þ 1

5x 1 5x 2 1

• 1(X 2 5)

5(X 1 5)(X 2 5)

(X 2 1)

x 2 2 25x 2 1x 2 5

p Þ 3

12p Þp Þ 01

2p Þ -

5p 2 1

2p2 2 p

(2p 1 1)p(p 2 3) •

(p 2 3)(p 2 1)(2p 1 1)(2p 2 1)

p2 2 4p 1 34p2 2 1 •

2p 1 1p2 2 3p 5

a Þa Þ

a2 1 4a 1 1a2

1 4a 1 3a

a 1 11

1a 1 3

5

x Þx Þ

-10x2 2 25

1x 1 5 2

1x 2 5 5

5(2x 1 7)(7x 1 2)

(2x 1 7)(x 2 1)(7x 1 2)(x 2 1)

2x2 1 5x 2 77x2 2 5x 2 2 5

x Þx Þ 0x Þ

73x Þ

2e 2 2e2 2 2e

2e 2 2e2 2 2e 5

e 2 2e2 2 2e 1

ee2 2 2e 5

1e 2 2 • ee 5

1e • e 2 2

e 2 2 1

2e 2 2e2 2 2e

1e 1

1e 2 2 5

x 5 e

2x 2 2x2 2 2x

1x 1

1x 2 2 5 10.

when

11.

when , and

12. com denom

13.

if 0, 0, and -h

14. ,

15. , and -1

16. , and

17. a, 1

18. , , and

19. Suppose that r is anyrational number. Thenthere exist integers a and bwith and so

that .

1r 5

ba

1ab

1r 5

r 5ab

abr 5

b Þ 0a Þ 0

x Þ32

x Þ 0x Þ -43x 2 11

x

a Þ

x Þ 0x Þ 6 12x2 1 1x2 2 1

y Þy Þ 03y

2y 1 2

x Þ 03x8

x Þx Þh Þ

5 - 1x2 1 hx

5 - hx2 1 hx • 1h

5 ( xx2 1 hx 2

x 1 hx2 1 hx) 4 h

1x 1 h • xx 2

1x • x 1 h

x 1 hh

51

x 1 h 21x

x 1 h 2 x

1(x 2 5))( 1

(x 1 5) 2

x Þ 6x Þ -3x Þ -4

2y 1 4

y 2 6y 1 3 5

2y 1 6y2 2 2y 2 24 •

x Þ 65

2x 1 13x2

2 252

x 2 51

3x2

2 255

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27.28. a. All graphs open up

and have minimum pointsat (0, 0).b. Odd power functionsinclude negative values andhave no upper or lowerlimits. Even powerfunctions have minimumequal to 0, and include nonegative values.

29. The functions are identical.Relative Maxima (-1.097, -1.486); Relative minimum:none;

. The

functions do not exist when2m, 1, or -1.x 5x 5x 5

g(x) 5 0limx→-`

f(x) 5limx→-`

(7Ï2)2 5 49(2) 5 98


must be a rational number

since it is the ratio of twointegers.

20.

21.

22.

23.

24.

25. a.

b.

26.

ln 3 1.099, ln 2 .693≈x 5

≈x 55 (ex 2 3)(ex 2 2)

e2x 2 5ex 1 6 5 0

400N 2

400N 1 2 5

800N2 1 2N

400N

x2 2 xx2 1 1

361

-(3 1 a)a2

x 2 32x 1 1

H 1 HK2HK2

ba

c

79

1. True

2. d

3.

4. a. The limit of f(x) as xapproaches 0 from the leftis negative infinity.Sample:

b. The limit of f(x) as xapproaches 0 from the leftis positive infinity.Sample:

5. a. T-6, 0

b. -6c. and

6. a. Trueb. Falsec. 0d. i. -.05 .05

ii. -.002 x .002,,, x ,

x 5

h(x) 5 -`limx→-62

h(x) 5 `limx→-61

x 5

-4 -2 2 4-2

2

4

y

x

y

x-4 -2 2 4

-4

-2

2

4

y

x-1 1 2 3 4 5 6

-2-1

1234

7. a. and

b. and

c.

8. a. domain: {x: 0};range: {y: 0}b. decreasing on theintervals (-`, 0) and (0, `)c. and

d. and

e. odd

f.

9. The sun’s brightness seenfrom Earth is 2.25 times thebrightness seen from Mars.

10. a.

-16 # x # 0, x-scale = 2 -6 # y # 0, y-scale = 2

-5 # x # 5, x-scale = 1-5 # y # 5, y-scale = 1

f(x) 5 -`limx→0-

f(x) 5 `limx→01

f(x) 5 0limx→-`

f(x) 5 0limx→`

y Þx Þ

-5 # x # 5, x-scale = 1-1 # y # 7, y-scale = 1

g(t) 5 0limt→-`

g(t) 5 0limx→`

g(t) 5 `limt→02

g(t) 5 `limt→01

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80

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b. The graph of

given in Example 1.

11. a.b. 3

12. a.

b. , -2, and

13. a.

b. 0

14. a.b. 0, 3, -3

15. a.b. 0

16. Suppose a, b, and c are anyintegers such that a is afactor of b and a is a factorof . By definition,there exist integers r and ssuch that and

. Then,

. Since is an integer by closureproperties, a is a factor of cby definition.

s 2 ra • r 5 a(s 2 r)(b 1 c) 2 b 5 a • s 2

c 5b 1 c 5 a • sb 5 a • r

b 1 c

x Þ

(x2 1 22x )

x Þx Þx Þ

13 – x

x Þ

-x1

a Þ -52a Þa Þ13

a2a 1 5

x Þ

4x 2 3

h(x) 51

x 2 417. a.

b. domain: the set of realnumbers

18. a.

b. domain: {t: 1}

19. a. and

b. Sample:y

x-4 -2 2 4-2

2

4

6

f(x) 5 `limx→-1-

f(x) 5 -`limx→-11

t Þ

fg 5 t 1 1

f °g 5 t2 2 2t


x y

-5 2.25-4 2.33-3 2.50-2 3.00-1 ERROR0 1.001 1.502 1.663 1.754 1.805 1.83

c

81

1. $300 per year

2. a.

b.c. Sample: The greater theinitial gas mileage, the lessthe savings for an increaseof 15 mpg.

3. Yes, {x: -1}

4. No

5. Yes, domain: the set of realnumbers.

6. essential

7. a. removableb.

8. False

9. a. y:

b. essential

c.d. f (y)

y-4 -2 2

-2

2

y 5 -53

y Þ -53JH

g(u)

u

(-10, 3)

4

21

-5-15 5 15

x Þ

limx→` S15(x) 5 0

-2.5 # x # 45, x-scale = 10-0.02 # y # 0.07, y-scale = 0.01

10. a. {x: 1 and -1}b. essentialc. 1 and -1d.

11. a. the set of real numbersb. nonec. noned.

12.13. g(x) is a function with a

removable discontinuity at. Since

, defining

makes g(x)continuous for all x.

14. The limit of f(x) as xapproaches 7 from the rightis negative infinity.

g(1) 5 6

(x 1 5) 5 6limx→`

limx→1

g(x) 5x 5 1

y 5x2 2 6x 1 8

2x 2 8

-10 -5 5 10-0.2

0.2

0.4f (k)

k

-4 -2 2 4-2

2

4

f (x)

x

x 5x 5

x Þx Þ

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15. a. domain: {x: 0};range: {y: }b. ;

c. ;

d. ; x,

h(x)e.

16.

17. a. ,

rationalb. irrational

18. 57

19. ;

20. a. or

b. or x $75x #

-115

x 5-115x 5

75

r(x) 5 2x 2 8q(x) 5 x3 1 3x2 1 2x 2 2

5(1001)1000

(1000)1000(1 11

1000)1000

-414(a 2 b)

-5 # x # 5, x-scale = 1-1 # y # 9, y-scale = 3

5x4 5h(-x) 5

5(-x)4 5

h(x) 5 0.limx→-`

h(x) 5 0limx→`

h(x) 5 `limx→0-

h(x) 5 `limx→01

y . 0x Þ 21. a. Let cos 0.2x2.

Since , 0,and d(x) is continuous, theIntermediate Value Theoremensures that there exists azero of d(x) in the intervalfrom 1 to 1.5. Where

, cos ,so cos , and hencethere is a solution to theequation cos in theinterval from 1 to 1.5.b. 1.26

22. a. The duplicating machineworks, and it is not Sunday.b. The car is not Americanmade or it is an electric car.

23. a.

b. Answers will vary.

-2 # x # 2, x-scale = 1-10 # y # 10, y-scale = 5

1.25 # x #

x 5 0.2x2

x 5 0.2x20.2x2 5 0x 2d(x) 5 0

,d(1.5)d(1) . 0x 2d(x) 5


7. 2v2

5v2

134

2214

212 v

2 2212 v

5v2 2 52v

2 21 8v4v3 2 2v2

q 4v3 1 3v2 1 8v 2 2 2v 2 1

12141

52v

83

1.2. a. -4x5

b. f(x) -4x5

3. a.b. The end behavior of r is

like the function ;

,

4. The end behavior of f is like

the function ;

f(k) 0, f(k) 0.

5. The end behavior of g islike the function ;

, .

6. t 5 3w 2112

g(x) 5 `limx→-`

g(x) 5 `limx→`

f(x) 5 x4

5limk→-`

5limk→`

2kg(k) 5

r(x) 513.lim

x→ -`r(x) 5

13lim

x→`

13f(x) 5

y 513

5

4 211x

16x2 2

2x3 9.

10. Sample:

11. a.

b. As the number of totalpoints increases, Viola’sgrade approaches 100%.

12.13. a.

b.

is

always undefined at .

14.

15. , 0, 0, and

-h

16.

17.

18. no real solution.

19. a. b. -

c. -1 d. -Ï32

12

Ï22

3t 2 7y 2 13ty 2 5y2)(-8t2 1(3t 2 7y)3(8t 1 5y)3

(x2 1 14x 1 17)(x 1 11)2(x 2 5)2(x 1 2) •

x Þ

h Þx Þ-4

x(x 1 h)

9x2 1 13x 2 1

z 5 -4

z 1 2z 1 4

; [ t(z)(z 2 3)(z 1 2)(z 2 3)(z 1 4) 5

t(z) 5z2 2 z 2 6z2

1 z 2 125

z 5 3

74

y

x20(0, .15)

40 60 80 100

1

f(x) 53x 2 11x3(x 2 5)

-2500 # x # 12500, x-scale = 2500 -25 # y # 125, y-scale = 25

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8. a. 69.76 kgb. 66.02 kgc. They were 0.00028 timestheir weight on Earth.

c

84

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20. a. The total distancetraveled is 2d. The firstaverage speed is 20 mphand the second is x mph.Since the distance traveledeither way, d, is equal, the 2rates divided into d willgive the total time spent.When this value is dividedinto 2d the average speed isobtained.

b.c. 40; the return

trip approaches 40 mph.

f(x) 5limx→`

40xdxd 1 20d


8.an10m1n 1 K 1 a110m11 1 a010m 1 a-110m21 1 a-210m22 1 K 1 a-m100

10m

c


1. True 2. False 3. True

4. Assume the negation of theoriginal statement.

5. a.b. and a have a factor of 3.c.d. and b have a factor of 3.e. Both a and b have afactor of 3. However, fromthe beginning of the proof,

is assumed in lowest

terms, thus having nocommon factors. There is aninherent contradiction, sothe negated statement isfalse, and the originalstatement, is irrational,must be true.

Ï3

ab

b2b2 5 3k2a2a2 5 3b2

6. a. Yes, .

b. No, 2 is a rationalnumber.

7. a. irrational; The decimalexpansion neitherterminates nor repeats.b. rational; The decimalexpansion repeats.c. rational; The decimalexpansion terminates.

8. See below.

9.

10.

11. True

3,615,8814950

7,304,81010000

Ï4 521

c

85

12. Assume the negation of theoriginal statement is true.Thus, there is a rationalnumber x and an irrationalnumber y whose quotient

is rational. Then since the

quotient of x and y isrational, both x and y mustbe rational, but thiscontradicts the assumptionthat y is irrational. Hence the assumption is false, and,

must be irrational.

13. a.

b.

c. 1.4854315 and

1.4854315

14.

15. False; counterexample: πand 3 π sum to 3, whichis rational.

16. False; counterexample: Let and . Then,

4, but 4 is a rationalnumber.

17. a. when is zero ora perfect squareb. when andnot a perfect square

b2 2 4ac . 0

b2 2 4ac

a • b 5 Ï2 • Ï8 5 Ï16 5b 5 Ï8a 5 Ï2

2

72 1 26Ï541

ø25 2 5Ï311

ø105 1 Ï3

25 2 5Ï311

5 2 Ï3

xy

xy

18. reciprocal of

19.

20. a. The Quotient-RemainderTheorem states that theremainder r for this divisionis an integer in the range

. Therefore, after54 steps at least oneremainder must repeatsince there are only 54unique remainders.b. The number of stepstaken before repeat of theremainder is equal to thenumber of digits in the repeating number sequencein the decimal expansion.Each time the remainderrepeats, the repeating digitsequence is begun again.c. The Quotient-RemainderTheorem requires theremainder to be an integerr in the range forany long division. Hence,after d steps there must bea zero or a repeated remainder. If there is a zeroremainder, then the decimalexpansion terminates. Ifthere is a repeatedremainder, then the decimalexpansion is repeating.

0 # r , d

0 # r , 54

f(z) 56

11z

Ï7 2 Ï61 5 Ï7 2 Ï6

Ï7 2 Ï6Ï7 2 Ï6

5Ï7 2 Ï6

7 2 6 5

1Ï7 1 Ï6

51

Ï7 1 Ï6 •

Ï7 1 Ï6 5

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c

c

86

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d.

21. ; 0, 4

22. ; where

23. ; where n is an odd integer

24. R 5R1R2

R1 1 R2

nπ2

t Þ 614tt2 2 1

x Þx Þ3(x2 1 x 2 4)

x(x 2 4)

2954 5 5.370 25. A number is said to be

algebraic if it is a root ofsome polynomial havingrational coefficients;otherwise it is said to betranscendental.



1. tan ; cot ;

sec ; csc

2. a. tan

b. cot

c. sec

d. csc

3. a. b. 2 c. d.

4. Sample: , ,

,

5. a. , n is an

integerb.

6. a. , n an integerx 5 nπ

-3π # x # 3π, x-scale = π -4 # y # 4, y-scale = 2

x 5π2 1 nπ

(3π2 , -1)(3π

4 , Ï2)(π2, 1)(π

4, Ï2)Ï5Ï5

212

u 5hypotenuse

side opposite u

u 5hypotenuse

side adjacent to u

u 5side adjacent to u

side opposite u

u 5side opposite u

side adjacent to u

x 554x 5

53

x 534x 5

43 b.

7. Sample: (0, 0), ,

,

8. a. sin, csc, and tan are oddfunctions. cos and sec areeven functions. cot is neither.b. nonec. sin and cos

d.

sin

cos

tan 1

cot 1

sec 2

csc 2 2Ï33Ï2

Ï22Ï33

Ï33Ï3

Ï3Ï33

12

Ï22

Ï32

Ï32

Ï22

12

π3

π4

π6

(π3, Ï3)(π

4, 1)(π6, Ï3

3 )


c

87

9. d

10. tan ; cot ;

sec ; csc

11. a. The area of the triangle

ABO is (AB)h. So, the area

of the regular n-gon is thesum of the areas of n

congruent triangles,

n (AB)h. .

The altitude h splits into two smaller angles

measuring . tan .

So, tan .Hence,

the area of the n-gon is

.

b. Let the radius, r, be 6and . Then ,and by thePythagorean Theorem.Using the formula frompart a, the area is

tan 72. The area of the

square is 72 and so the formula checks.

12. a. definition of tangentb. sine is an odd functionand cosine an even functionc. defintion of tangentd. transitive property ofequality and definition ofodd function

(AB)2 5 (6Ï2)2 5

π4 5

4(3Ï2)2 •

h 5 3Ï2AB 5 6Ï2n 5 4

nh2tan πnn12 (2h tan πn)h 5

πnAB 5 2h

12 AB

h(πn) 5

πn

/AOB

m/AOB 52πn

12

12

(-π4) 5 -Ï2(-π

4) 5 Ï2

(-π4) 5 -1(-π

4) 5 -1

13.

14. ;

15. a.

b. ;

c. ;

16. a. . So 17 1(mod 2); 1b. , so1000 10(mod 11);

17. No real solution.

18. a. -1 x 4b. All increasing functionsare 1–1 functions, so theinverse of log5x could beused to simplify the equation without changingthe solutions.

19.

6.32

10 sin 25°sin 123°a 5

10sin 123°5

asin 25°

ø

(10)(.5299).8387ø

10 sin 32°sin 123°b 5

10sin 123°5

bsin 32°

10

25˚

25˚65˚ 58˚

32˚

32˚A

ab

C

B

,,

t 5 10;1000 5 90 • 11 1 10

y 5;17 5 8 • 2 1 1

limx→-`

(5x 1 10) 5 `

limx→`

(5x 1 10) 5 `

limx→2-f(x) 5 -`lim

x→21f(x) 5 `

y

x-4 -2 2 4

-60

-40

-20

20

40

60

x = 2

limy→-`

g(y) 578lim

y→` g(y) 5

78

-2Ï10 1 105

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5.04The hiker is < 6.32 milesfrom the first and < 5.04miles from the second.

ø

(10)(.4226).8387ø 20. a. iv

b. ic. iid. iiie. vif. v



In-Class Activity1. .2 hour = 12 minutes

2. .25 hour = 15 minutes

3. a. 444.44

b. average speed speedin still air

4. ; ;

444.44 ; average speed

speed in still air

Lesson 1. 42

2. 303 mph

3. a. 20 mphb. 60 mphc. not possibled. not possible

4. a.b. 5 and -5

5. a. Yesb. No: the first equation isundefined when 2, but

2 is a solution of thesecond equation.x 5

x 5

x 5x 54(x 1 5)(x 2 5)

kmhr

,mihr

5400d9d 5

2d900d

20,000

d400

d500

,

mihr

6. No; there are no rational numbers x, , and , but there are irrationalsolutions

; and

7. 16

8. 0

9. 5 or 6

10.11. a. 48 mph

b. Yes, the currentdecreases the length oftravel about 0.05 milesevery 15 minutes.

12. a. < 171 mphb. < 342 miles

13. Your speed returning is not 0 mph.

14. If you go some place at 20mph, you cannot average40 mph for the total trip.

15. Your return rate does notdepend on the distancetraveled.

t 5 -4

x 5x 5

x 5

x 5

h-Ï8, -Ï8 1 2, -Ï8 1 4jhÏ8, Ï8 1 2, Ï8 1 4j

x 1 4x 1 2

c

89

16. a. R 2.73 ohmsb. R2 12 ohms

17. ;

18. 20 minutes

19.

20. tan ; cot

21. sec ; csc 2

22. a. The decimal expansionof x is not terminating andthe decimal expansion of xis not repeating.b. If the decimal expansionof x is not terminating andnot repeating, then x is nota rational number.c. i. rational

ii. irrational

x 5-2Ï3

3x 5

247x 5

724x 5

3(r 3 2 16r)3r 2 2 2r 2 8

AB 5 Ï2 1 2BC 5 Ï2 1 1

5ø 23.

24. 2 3

25. a. ;

b. ;

26. a.

b.

27. 7.0%

28. The legislature did not passthis bill.

29. a.

b. , ,

c. a 532

a 5 3a 5 -1a 5 1

2a – 1x 5

x Þ -12

2x 2 154(2x 1 1)(2x 2 1)

4(2x 1 1)

516x2 – 48x 1 4 5

4(4x2 2 1)4(2x 1 1)

limx→-`

f(x) 5 0limx→`

f(x) 5 0

limx→02

f(x) 5 -`

limx→01

f(x) 5 `

x 1

3Ï105

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90

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1. Sample:

2. Sample:

3. , , , ,

, , , ,

, , ,

4.

5. a.

b. 60°, 90°, 108°, 120°,

128 °

c. 180°

6. a. 5b. all 5

7. a. regular tetrahedronb. regular octahedronc. regular icosahedrond. cubee. regular dodecahedron

8. The area of a rectangle is itslength times its width, ,w.Its perimeter is then

. By setting theseequal, the problem is tofind positive integers , andw such that .,w 5 2, 1 2w

2, 1 2w

47

an 5(n – 2)180°

n

t 5 6

(18, 13)(1

7, 13)(16, 13)(1

5, 13)(14, 14)(1

4, 13)(13, 18)(1

3, 17)(13, 16)(1

3, 15)(13, 14)(1

3, 13)

Dividing both sides by 2,w(which is nonzero) yields

, which

simplifies to . This

is equivalent to theequation in Problem 2,which is also equivalent toProblem 1.

9. Let n be an integer greaterthan 2. Suppose ( ) is afactor of 2n. Then

for some

integer k. Then ,

since . By taking the

reciprocal, . So

, or .

This is equivalent to theequation in Problem 2,which is equivalent toProblem 1.

10. of an hour < 33 minutes

11. 2.744 cm x 2.856 cm

12. a. 3 cmb. 3.75 cm from the mirror

13. ; ,

; csc 5π3 5

-2Ï33sec 5π

3 5 2

cot 5π3 5

-Ï33tan 5π

3 5 -Ï3

,,

611

12 5

1k 1

1n

12 2

1n 5

1k

1k5

n – 22n

n Þ 2

2nn 2 2 5 k

2n 5 (n 2 2)k

n 2 2

12 5

1w 1

1,

,w2,w 5

2, 1 2w2,w


c

91

14.

15. a. {x: and }

b. near 5: ;

near -5,

c. ;

d. x-intercept: (-2, 0);

y-intercept:

e.

16. a.

b. , -1, ,

and x Þ -2

25x Þx Þ-17x Þ

x 2 4x 1 2

(7x 1 1)(x 2 4)(5x 2 2)(x 1 2) 5

•(5x 2 2)(x 1 1)(7x 1 1)(x 1 1)

7x2 2 27x 2 45x2 1 8x 2 4 5

•5x2 1 3x 2 27x2 1 8x 1 1

-8 -4

x = 5

x = -5

4 8

-1

1y

x

(0, -225)

limx→`

h(x) 5 0limx→`

h(x) 5 0

limx→-51

h(x) 5 `

limx→ -52

h(x) 5 -`x 5

limx→51

h(x) 5 `

limx→52

h(x) 5 -`x 5

x Þ -5x Þ 5

-10Ï10 1 6013 17.

18. Suppose p and q areintegers such that 4 is afactor of p, and q is even.By definition, there existintegers r and s such that

and . Then. Since

rs is an integer by closureproperties, 8 is a factor ofpq by definition.

19. If two or more differentregular polygons areallowed as faces, there are13 polyhedra for whicheach vertex is surroundedby the same arrangementof polygons. These arecalled semi-regularpolyhedra, and they areconvex.

pq 5 (4r)(2s) 5 8(rs)q 5 2sp 5 4r

22 • 1097P

reca

lcul

us a

nd D

iscr

ete

Mat

hem

atic

s©

Scot

t For

esm

an A

ddis

on W

esle

y


92

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1. a. -1

b.

2. a.

b. , and

3. a.

b. -1, 2, 3, and-4

4. a.

b. , , -1,

, and -6

5. a.

b. 5 and -4

6. a.

b. 1 and -1

7.

8.

9. , with and

10. , 6, -3,

-2, 5, and -5

11. a.

b. -1, 5, 2, and-3z Þ

z Þz Þz Þ

1z 1 3

z 2 5(z 2 2)(z 1 3) 5

(z 1 1)(z 2 2)(z 1 1)(z 2 5) •

z 2 5z2 1 z 2 6 5 • z2 2 z 2 2

z2 2 4z 2 5

x Þx Þx Þ

x Þx Þx2 1 6x 1 8x2 2 x 2 30

a Þ 0a Þ -3a 2 3

z 1 1z(z 1 3)

3(x 2 5)(x 2 6)

r Þr Þ

-2r2 2 1

t Þt Þ

-2t2 1 20t 1 13(t 2 5)(t 1 4)

p Þp Þ -5

p Þp Þ -23p Þ12

p 1 63p 1 2

z Þz Þz Þz Þ

3(z 1 1)(z 1 4)

y Þ -3y Þ -4, y Þ -5

13y 1 61(y 1 4)(y 1 5)(y 1 3)

x Þ23

12. a.

b. -3, -1, and 3

13.

; ,

,

14. irrational, since isirrational for any nonzeronumber x

15. rational, because -7 is aninteger

16. irrational, the decimalexpansion neitherterminates nor repeats

17. rational, since the decimal

expression of repeats,

.

18. rational, since 0 is aninteger

19. rational, equals

20. rational, equals

21. irrational, is not aninteger

22.

23. 24 2 6Ï65

15 1 5Ï54

Ï3

263

3733396

3.142857

227

ex

x Þ 2x Þ -2

x Þ 01x 2 2

x 1 2(x 1 2)(x 2 2) 5

5x 1 2x2 2 4 5

1x 1

2x2

1 24x2

y Þy Þy Þ

12y 1 26(y 1 3)(y 2 3)(y 1 1)

55y 1 5 1 7y 1 21

(y 1 3)(y 2 3)(y 1 1)

5(y 1 3)(y 2 3) 1

7(y 1 1)(y 2 3) 5

5y2 2 9 1

7y2 2 2y 2 3 5


c

93

24.25.

26.

27.

28. Irrational; there are nopolynomials whosequotient equals this.

29. rational

30. Irrational; cos u and sin uare not polynomialfunctions.

31. rational

32. -2 and 3

33. 0 and 1

34. 1

35. no solutions

36. or

37. 2

38. or

39.

40.

41.

42. -1k2 2 k

35839000

24599

8931000

14y 5

12y 5

v 5

t 5 134t 5 -

x 5

x Þx Þ

y Þy Þ

1Ïx 1 h 1 Ïx

715 1 5Ï2

-3Ï10 2 6

12 2 4Ï6 43. Assume the negation is true, that is rational.By definition, there existintegers a and b, with

such that , where

is in lowest terms. Then,

. Thus,

has a factor of 13,because if a is an integerand is divisible by aprime, then a is divisible bythat prime. Therefore, let

for some integer k.Then,

. So b2 and b have afactor of 13 by similarargument. Thus, a and bhave a common factor of13. This is a contradiction

since is in lowest terms.

Hence, the assumptionmust be false, and so isirrational.

44. True, every integer a can be

expressed as .

45. True, if and are two

rational numbers, where

and , then

. ac and bd are

both integers and since and .

Hence, is rational.acbd

d Þ 0b Þ 0bd Þ 0

acbd5

cd•a

b

d Þ 0b Þ 0

cd

ab

a1

Ï13

ab

13k213b2 5 (13k)2 ⇒ b2 5

a 5 13k

a2

a2

13 5a2

b2 ⇒ a2 5 13b2

ab

5abÏ13

b Þ 0

Ï13

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c

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46. False; counterexample:

47. Assume the negation istrue. Thus, the difference ofa rational number p and anirrational number q is arational number r. Then

. So .However, by the closureproperty of rationalnumbers the differencebetween two rationalnumbers is another rationalnumber. Hence, there is acontradiction, and so theassumption is false, whichproves the originalstatement.

48. d

49. The limit of f(x) as xdecreases without bound is1.3.

50. a.b. i. ` ii. -` iii. 0 iv. 0

51. a.b.c. ;

52. a. and

b. ; ,

;

c. ; limt→ -`

f(t) 5 3limt→`

f(t) 5 3

limt→ -21

f(t) 5 -`limx→-22

f(t) 5 `

limx→21

f(t) 5 `limx→22

f(t) 5 -`

t 5 -2t 5 2

limx→-`

h(x) 5 2limx→`

h(x) 5 2

limx→62

h(x) 5 -`

limx→61

h(x) 5 `

T-2, 0

p 2 r 5 qp 2 q 5 r

Ï2 • Ï2 5 2

d.

53.

54. 9t 24 55. h(y) y2

56. p(t) 57. q(z)

58. True

59. a. 5b. -5c. f(-5) -0.1

60. (2n 1) , n an integer

61. Sample:

62. fλ 63. $179

64. 61.6 mph

65. a. b.

c. d. 2.4 hr

66. a. 3b. 3: essentialc. noned. x-intercept: ; y-intercept: -2e. ;

f.

-8 8x = 3

y

x-100

100

limx→-`

f(x) 5 `limx→0

f(x) 5 `

3Ï6 ø -1.8

x 5x 5

c 1 hch

1h

1c

øv 5

x2 2 16x 2 4

π21x 5

5

14z5

1195

7352y 5

34x 1

18y 5

-4 4

-4

4y

x

x = -2 x = 2


c

95

67. a. 1 and -1b. 1 and -1: essentialc. 2d. x-intercept: none; y-intercept: -3e. ;

f.

68. a. 0 and -4b. 0: essential; -4:removablec. 0d. x-intercept: none; y-interecept: nonee. ;

f.

69. a. ;

b. ;

c. ;

d. 4, -3, and 3y 5x 5x 5

limx→-`

f(x) 5 3limx→`

f(x) 5 3

limx→-32

f(x) 5 `

limx→-31

f(x) 5 -`

limx→42

f(x) 5 -`limx→41

f(x) 5 `

-4 -2 2 4

-4

-2

2

4y

x

limx→-`

g(x) 5 0limx→`

g(x) 5 0

y 5

x 5x 5x 5x 5

y

x-8x = -1 x = 1

y = 2

-4 4 8

-10

5

10

limx→-`

h(x) 5 2

limx→`

h(x) 5 2

y 5x 5x 5x 5x 5 70. a.

b.71. a.

b. where n is an integer

72.

0; -1

73. 74.

75.

76. a. b. c. d.77. csc 3

78. tan = 1

79. cot

80. sec π 5 -1

π6 5 Ï3

π4

x 5

-53

54

-43

-34

13Ï169 2 x2

135

125

x 5y 5

-3 -1 3

-8

-4

yx

x 5π2 1 nπ,

-3

3

6

π]2

3π]]2- π

]2- 3π]]2

y

x

x 5 5, y 5 2

2 4 6 8

-4

-2

2

4

y

x

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96

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1. An identity is an equationwhich is true for all of thevalues of the variable forwhich both sides aredefined.

2. {x: ; integers n}

3. all real numbers

4. {t: }

5. 2 sin x sin y cos ( )cos ( )

6. 11,398,150,000,000

7. a.

b. identity; domain:

{x: ; integers n}

8. a.

b. This is not an identity.Sample counterexample:

For , sin ,

but sin .

9.10. csc x tan x sin x tan x

11. tan x csc x sec x

12. cos sin x(3π2 1 x) 5

5

5

sin2 x 1 cos2 x 5 1

(0 2π2) 5 -1

(π2 2 0) 5 1x 5 0

-5 # x # 3, x-scale = 1-3 # y # 3, y-scale = 1

x Þ(2n 1 1)π

2

-5 # x # 5, x-scale = 1-3 # y # 3, y-scale = 1

øx 1 y

2x 2 y5

t . 0

x Þ nπ

13. a. This is not an identity.

b. Sample counterexample:For , sin (0.2π)0.588, but

.

14. a. sin and

2 sin cos 2

; sin and

2 sin cos

; sin and

2 sin π cos π (2)(0)(-1) 0b. Yes, they seem to haveidentical graphs.

15. a.

b.


π]2


π]2


π]2

55

(2 • π) 5 0Ï32

π3 5 2 • Ï3

2 • 12 5π3

(2 • π3) 5Ï32

Ï22 5 1

π4 5 2 • Ï2

2 • π4

(2 • π4) 5 1

(1 2 0.2) 5 0.644 • 0.2 •

øx 5 0.2


c

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c.

d. Yes

16. 17. 18.19. {x: (mod π)}

20. , 21. c

22. ,;

, ;

, { };

,

{ and }x Þ -1x: x Þ 0

( fg)(x) 5

2x 2 1x2 2 x 1 1

x: x Þ 0x 1 2 11x 2

1x2

(f • g)(x) 5 2x3 1 x2 2

{x: x Þ 0}2x 1 1 2 x2 22x

(f 2 g)(x) 5{x: x Þ 0)(f 1 g)(x) 5 x2 1 2x 1 1

t 5 62t 512

x ò 0

-Ï3-Ï32-Ï3

2


π]2

x sin x

-3.0 -0.141 -0.525 0.384-2.5 -0.598 -0.710 0.111-2.0 -0.909 -0.933 0.024-1.5 -0.997 -1.001 0.003-1.0 -0.841 -0.842 0.0002-0.5 -0.479 -0.479 0.00000 0 0 0.00000.5 0.479 0.479 0.00001.0 0.841 0.842 0.00021.5 0.997 1.001 0.0032.0 0.909 0.933 0.0242.5 0.598 0.710 0.1113.0 0.141 0.525 0.384

zsin x 2 (x 2x3

6 1x5

120)zx 2x3

6 1x5

12026. a.

c

23. , ; t; , ; t;

; t; ,

; t

24. and ,since these segments areradii of the same respectivecircles. So by SAS,

. Therefore, ,since corresponding parts incongruent figures arecongruent.

25. angles in Quandrant 2:

; angles in

Quadrant 3: ;

angles in Quadrant 4:

26. a. See below.

b. -2 # x # 2

3π2 , u , 2π

π , u ,3π2

π2 , u , π

AB 5 CDnAOBnCOD ù

OD 5 OBOC 5 OA

( fg)(t) 5 e2t(f • g)(t) 5 1

(f 2 g)(t) 5 et 2 e-t(f 1 g)(t) 5 et 1 e-t

1. a. (left side)

definition of cot

addition of fractions

Pythagorean Identity

definition of csc(right side)

for all real numbers x for which both sides are defined.b. { , ; integers n}

2.

definition of cot definition of csc

addition of fractions


3. Technique 3:


⇔ ; provided sin

⇔ definitions of cot and csc

4. a. , ; integers n}b. (left side)

definition of cot

simplification(right side)

for all real numbers x for which bothsides are defined.

5. cos x tan x sin x

definition of tan

simplification sin x

domain: { , ; intenters n}x: x Þ(2n 1 1)

2 π[ cos x tan x 5 sin x

5

5 cos x • sin xcos x

[ sin x cot x 5 cos x

5 cos x

sin x cot x 5 sin x • cos xsin x

{x: x Þ nπ[ cot2 x 1 1 5 csc2 x

5 csc2 xcot2 x 1 1

x Þ 0M 1sin2 x

51

sin2 xcos2 xsin2 x 1

sin2 xsin2 x

5 1sin2 x 1 cos2 x

[ cot2 x 1 1 5 csc2 x

51

sin2 x

5cos2 x 1 sin2 x

sin2 x

51

sin2 x5cos2 xsin2 x 1

sin2 xsin2 x

csc2 xcot2 x 1 1

x: x Þ nπ

[ cot2 x 1 1 5 csc2 x

5 csc2 x

51

sin2 x

5cos2 x 1 sin2 x

sin2 x

cos2 xsin2 x 1

sin2 xsin2 xcot2 x 1 1 5

98

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6.

definitions of tan Pythagorean and cot Identitysimplification

domain:

7.

definition of csc

simplification simplification

domain , ; integers n}

8.definitions oftan and cotaddition of multiplication of fractions fractionsPythagoreanIdentity

domain:

9. an identity: (left side)

definition of cot

Distributive Property

simplificationPythgorean Identity

(right side)5 15 cos2 x 1 sin2 x

5sin2 x cos2 x

sin2 x 1 sin2 x

sin2 x (cot2 x 1 1) 5 sin2 x (cos2 xsin2 x 1 1)

{x: x Þnπ2 , ; integers n}

[ tan x 1 cot x 5 sec x • csc x

51

sin x cos x

51

sin x cos x5sin2 x 1 cos2 x

sin x cos x

definitions of secand csc5

1cos x • 1

sin x5sin xcos x 1

cos xsin x

sec x • csc xtan x 1 cot x

{x: x Þnπ2

[ csc2x sin x 5sec2x 2 tan2x

sin x

simplification51

sin x

PythagoreanIdentity5

cos2 xsin x cos2 x

51 2 sin2 xsin x cos2 x5

1sin x

definitions ofsec and tan

1cos2 x 2 sin2 x

cos2 xsin x55

1sin2 x • sin x

sec2 x 2 tan2 xsin xcsc2 x sin x

{x: x Þnπ2 , ; integers n}

[ tan x • cot x 5 cos2 x 1 sin2 x5 1

5 1sin xcos x • cos x

sin x

cos2 x 1 sin2 xtan x • cot x


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10. a.

b.

c.

11.

12.13. a. sin x, tan x, csc x, cot x

b. cos x, sec x

14.

15. a. {n: for anyinteger q}b. The set in part a is theset of all integers withremainder of 3 whendivided by 7.

n 5 3 1 7q

9(x 2 3)x(1 2 3x), x Þ 0, x Þ

13

x 5 5 or x 5 -4

2468

10

3π]]2

-π]]2

-3π]]2

π]2

y

x

-3Ï7373

Ï738

-83 16. a.

b.c.

17.

18. or

19.20. invalid; inverse error

21. b

22. a. hyperbolic cosine:

;

hyperbolic sine:

b. Samples:;

;

1 2 tanh2 x 5 sech2 x

sinh xcosh x 5 tanh x

cosh2 x 2 sinh2 x 5 1

sinh x 512(ex 2 e-x)

cosh x 512(ex 1 e-x)

ø 429 m

Ï2 2 2 cos u

d 5 Ï(a 2 1)2 1 b2

y 5 sin (x 1π2)

2x13 5 2x • 23 5 8(2x)y 5 2x13y 5 8(2x)


101

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1.

2.

3.

4. a.

b.

c. amplitude: 2; vertical

shift: -5; period: ;

phase shift:

5. Sample:

cos

6. All of them

7. a. 335 rabbits; 20 foxesb. : (320, 22) and

: (335, 23)

8. a. The average of themaximum and minimumrabbit populations:

.

b. Half the difference ofthe maximum and minimumrabbit populations:

.

9. a. The average of themaximum and minimumfox populations:

.20.5 530 1 11

2

95 5335 2 145

2

240 5335 1 145

2

t 5 5t 5 1

2 2(x 12π3

3 )y 5 9

π2

2π3

-π]]3

-2π]]3

2π]]3

π]3

-π π

-6

-4

-2

2

y

x

y 5 2 sin (3x 23π2 ) 2 5

y 5 sin (2x 2 3π) 2 6

y 5 4 sin (x3)

y 5 sin (x 2 π) b. Half the difference ofthe maximum andminimum fox populations:

.

c. To give function F theappropriate phase shift,since the fox populationhas maxima at years 2 and6, but the function

has maxima at

0 and 4.

10. a. 2 b. 2πc. Sample: πd. Sample:

11. a. amplitude: 5; period: ;

b. phase shift:

12.

13.

14. a. amplitude: 2; period:

b.

c. 7 in. d. 3 in.

2π]]3

π]3

5π]]3

4π]]3π0

2

4

6

8

y

t

π3

Hx 5 3 cos t 1 4y 5 4 sin t 1 1

Hx 5 5 sin t 1 2y 5

13 cos t 2 8

y

x-π π

-6

-3

3

6

- 2π]]3 - π

]32π]]3

π]3

-π2

2π3

y 5 2 sin (x 1 π)

y 5 cos (π2 t)

9.5 530 2 11

2


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15. See below. 16. See below.

17. a.

b. No. Counterexample: For ,

, but

18. a.

b. ; limx→-`

h(x) 5 4limx→`

h(x) 5 4

4 111

x2 1 3

1 1 2 5 3

x2 1 2 521 1 2-1 552

2x 1 2-x 5x 5 1

-1 # x # 1, x-scale = 0.2-1 # y # 4, y-scale = 0.5

y = x2 + 2

y = 2x + 2-x

c. h has no discontinuitiesbecause the denominator isnever equal to 0.

d.

Yes. The graphasymptotically approaches

as and .19.20. (cos u, sin u)

21. 1

T(x, y) 5 (x 2 6, y 2 41)

x → -`x → `y 5 4

-20 # x # 20, x-scale = 2 -2 # y # 10, y-scale = 1


15. It is an identity. Proof:

definition of cot definition of csc

simplification simplificationPythagorean Identity

; ; real numbers x such that ; integers n

x Þ nπ[ cos x cot x 5 csc x 2 sin x

cos2 xsin x

1 2 sin2 xsin x

cos2 xsin x

1sin x 2 sin xcos x • cos x

sin x

csc x 2 sin xcos x cot x

16. a. addition of fractions

simplification


definition of sec

b. ; ; integers n}{x: x Þ(2n 1 1)π

2

[ 11 1 sin x 1

11 2 sin x 5 2 sec2 x

5 2 sec2 x

52

cos2 x

52

1 2 sin2 x

51 2 sin x 1 1 1 sin x

1 2 sin2 x1

1 1 sin x 11

1 2 sin x

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103

22. Sample: (Source:http://water.dnr.state.sc.us/climate/sercc/products/normals/317069_30yr_norm.html)

amplitude: 20°; period; 1 year

2 4 6Month (Jan = 1)

Average MonthlyHigh Temperature

Raleigh, NC

Tem

pera

ture

(F°)

8 10 120

20406080

100


c

c


1. OP, OQ, OR, and OS are allradii of circle O, so they areall equal.

. So, (Side-Angle-

Side Congruence). Hence.

2. cos (x y) cos x cos ysin x sin y; cos (x y)cos x cos y sin x sin y

3.

4. 75°

5. cos cos x cos

sin x sin sin x

sin x.

6. The graph of isidentical to the graph of

phase shifted

by .π2

y 5 cos x

y 5 sin x

5π2 5 0 1

π2 1(x 2

π2) 5

Ï6 1 Ï24

152

251

RP 5 QS

nROP ù nQOSa 1 b 5 m/QOS

m/POR 5

7. cos cos ;

cos cos sin sin

8. cos cos cos x

sin sin sin x

sin x

9. cos ( ) cos ( )cos x cos y sin x sin y(cos x cos y sin x sin y)2 sin x sin y

10. The formula becomes cos 0 cos2 .

11. sin cos x(π2 2 x) 5

a 1 sin2 a 5 15

5221

5x 1 y2x 2 y

5x 5 0 2 (-1)3π2

23π2(3π

2 1 x) 5

12 2

12 5 0

Ï22 • Ï2

2 2Ï22 • Ï2

2 5

π4 5

π4

π4 2

π4

π2 5 0(π

4 1π4) 5

104

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12. a.

b.

c. cos x •

sin x sin

cos sin x

13.

14. a. amplitude: ; period:

; phase shift: 0

b. for n an integer

such that

15. a.

b.(x 2 5)2

9 1(y 1 2)2

16 5 1

y

x-4 -2 2 4 6 8

-8

-6

-4

-2

2

4

0 # n # 10

u 5nπ5

2π5

12

Ï2 2 Ï64

x 112

Ï32

π6 5cos π6 1

cos (x 2π6) 5

f(x) 5 cos (x 2π6)

3π # x # 3π, x-scale = -1.5 # y # 1.5, y-scale = 1

π]2

16.

17.

domain: ,

; integers n}

18. a. -1b. 0c. 0d. undefinede. -1f. undefined

19. a.

b.

c. (i)

and cos

(ii)

but cos

d. Ï1 1 cos 2x2 5 z cos x z

3π4 5 -Ï2

2

5Ï22Ï1 1 cos (2 • 3π

4 )2

π4 5

Ï22

5Ï22Ï1 1 cos (2 • π4)

2

-π2 # x #π2

-π # x # π, x-scale = -2 # y # 2, y-scale = 1

y = cos x

y =

π]2

1 + cos 2x]]]]]]]]2

{x: x Þnπ2

5cos xsin x 5 cot x

1sin x

1cos x

csc xsec x 5

-Ï343


105

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1. sin sin x cos cos x sin y; sinsin x cos cos x sin y

2. sin(α ( β)) sin α cos( β)cos α sin( β)

sin α cos β cos α sin β

3. , ;

,

4. , ; , ;

5. sin sin cos

cos sin cos

sin cos x

6. Step 1: Definition oftangent; Step 2: Identitiesfor sin(α β) and cos(αβ); Step 4: Simplify andDefinition of tangent

11

x 5

x 1 0 • x 5 1 • π2

x 1π2(π

2 1 x) 5

E 5 (Ï6 2 Ï24 , Ï6 1 Ï2

4 )Ï32 )D 5 (12Ï2

2 )C 5 (Ï22

12)B 5 (Ï3

2

Ï6 2 Ï24 )A 5 (Ï2 1 Ï6

4

25-1

-5-1

y 2(x 2 y) 5

y 1(x 1 y) 5 7.

8.9. a. tan

tan x

b. The period is no largerthan π.

10. tan(α β) tan(α (-β))

5

11. The identity becomes tan (β β)

12. 268 mm

tan β 2 tan β1 1 tan β tan β 5 0

52

tan α 2 tan β1 1 tan α tan β

tan α 1 tan(-β)1 2 tan α tan(-β) 5

152

tan x 1 01 2 (tan x) • 0 5

5tan x 1 tan π

1 2 tan x tan π

(x 1 π) 5

Ï6 1 Ï24

1 1 Ï31 2 Ï3

5 -2 2 Ï3


13. ;

14. cos cos x cos sin x sin (cos x)

(sin x) 1 sin x

15. a.MultiplicationFactoring

Distributive propertyFactoring5 c2 1 d2 1 (a 1 k)2 1 (b 1 k)2

5 c2 1 d2 1 a2 1 2ak 1 k2 1 b2 1 2kb 1 b2

a 1 b 5 c 1 d5 c2 1 d2 1 a2 1 b2 1 2k(a 1 b) 1 2k2

5 c2 1 d2 1 a2 1 b2 1 2k(c 1 d) 1 2k2

5 a2 1 b2 1 c2 1 2ck 1 k2 1 d2 1 2dk 1 k2

a2 1 b2 1 (c 1 k)2 1 (d 1 k)2

-5 •

• 0 2π2 5

π2 2(x 1

π2) 5

4-cos θ sin φ 2 sin θ cos φ-sin θ sin φ 1 cos θ cos φ5 3 cos θ cos φ 2 sin θ sin φ

sin θ cos φ 1 cos θ sin φ

-sin φcos φ43cos φ

sin φ-sin θcos θ43cos θ

sin θRθ?Rφ 5

3cos(θ 1 φ)sin(θ 1 φ)

-sin(θ 1 φ)cos(θ 1 φ)4Rθ1φ 5

c

b. Sample: For , sumsof squares are equal tothe sums of differentsquares. If ,

16. 44.3 hoursø22 1 32 1 32 1 62 5 58

512 1 42 1 42 1 52k 5 2

k . 0

106

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17. a. rationalb. irrationalc. rational

18.

19. 18 ft

20. See below.

(40x2 2 41x 2 20)(8x 1 3)(5x 2 7)2 •

c

20. a. sin(α β γ) sin α cos β cos γ cos α sin β cos γ 1cos α cos β sin γ sin α sin β sin γb. Sample: cos(α β γ) cos α cos β cos γsin α sin β cos γ sin α cos β sin γ cos α sin β sin γc. Proof: (Chunk (α β) and apply cosine and sine of sumidentities.) cos[(α β) γ]

cos(α β) cos γ sin(α β) sin γ(cos α cos β sin α sin β) cos γ (sin α cos β cos α sin β) sin γcos α cos β cos γ sin α sin β cos γ sin α cos β sin γ cos αsin β sin γ

2225

12251215

11

122

25112

1511

107

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1. cos

2. sin sin x cos x cos x sin x2 sin x cos x

3. a. cos

sin2 ;

sin 2 sin cos 5

b. cos cos ;

sin sin

4. a.

b.

5. a.

b.6. a. 253 ft

b. 4 secøø- Ï15

8

-78

120169

119169

π3 5

Ï32(2 • π6) 5

π3 5

12(2 • π6) 5

Ï32 5

Ï322 • 12 •

(π6)(π

6)(2?π6) 5

34 2

14 5

12(π

6) 5

2cos2(π6)(2 • π6) 5

512x 5 sin(x 1 x) 5

1 2 2sin2 x(1 2 sin2 x) 2 sin2 x 5

2x 5 cos2 x 2 sin2 x 5 7.

8.

9. Both f and g are thefunction x → cos 2x

10. a.

b. 303 ftc. 75.7 ft

11. See Below

12. a.

b.

c. Since the answers toparts a and b both

represent cos , they are

equal.

π12

Ï2 1 Ï32

Ï6 1 Ï24

π4

-2π # x # 2π, x-scale = -2 # y # 2, y-scale = 0.5

π]2

Ï2 1 Ï22

-Ï2626


11. Left sideSine of a sum identityDouble angle identitiesMultiplicationAdditionPythagorean IdentityMultiplicationAddition

Right side55 3 sin x 2 4 sin3 x5 3 sin x 2 3 sin3 x 2 sin3 x5 3 sin x(1 2 sin2 x) 2 sin3 x5 3 sin x cos2 x 2 sin3 x5 2 sin x cos2 x 1 cos2 x sin x 2 sin3 x5 (2 sin x cos x)cos x 1 (cos2 x 2 sin2 x)sin x5 sin 2x cos x 1 cos 2x sin x

5 sin 3x 5 sin (2x 1 x)

c

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13. a.

b.

c.

d.

e. Does sin2

cos2 ? Does

5 1?

Does

? Does

? Yes

14. See Below

15. 0

16. t 5 63, 62

225225 5 1

88 2 48Ï2225 5 1

1137 1 48Ï2

225

1 (4 2 6Ï215 )2(3 1 8Ï2

15 )2

(x 1 y) 5 1(x 1 y) 1

4 2 6Ï215

3 1 8Ï215

2Ï23

-35 17.

18. a.

b.

c.

d. 2.214

19. invalid, converse error

20. a. ' a real number x suchthat is not real.b. the negation

21. a. cos 3x 4 cos3 x3 cos x; cos 4 x 8 cos4 x8 cos2 x 1b. Sample: For cos (nx), theleading terms are always2n–1 cosn x.

125

25

Ïx 2 2

ø

-43

45

-35

f -1(x) 5x 2 2

3


14. tan2 θ(cot2 θ cot4 θ) csc2 θDef. of tan and cot Def. of csc

Simplifying

Adding fractions


∴ tan2 θ(cot2 θ cot4 θ) csc2 θdomain: , ∀ integers nx Þ

nπ2

51

51

sin2 θ

5sin2 θ 1 cos2 θ

sin2 θ

5 1 1cos2 θsin2 θ

51

sin2 θ5sin2 θcos2 θ (cos2 θ

sin2 θ 1cos4 θsin4 θ)

1

c

109

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In-Class Activity1. a. 0 θ π

b., c.

d. (1, 0), , ,

(-1, π)e. Answers may vary, butgraphs should be similar.

2. a.b., c.

d. (0, 0), , ,

e. Answers may vary, butgraphs should be similar.

(-1, -π2)(1, π2)(Ï22 , π4)

y = xy = sin x

y = sin-1 x

-π

π

- π]2

1 π]2-1-1

- π]2

1

π]2

x

y

-π2 # θ #π2

(Ï22 , π4)(0, π2)

1-1 π-1

1

πy = cos-1 x

y = cos x

y = x

y

x

##

3. a.b., c.

d. (0, 0), , ,

e. Answers may vary, butgraphs should be similar.

Lesson

1. a. sin y and

b. tan y and

2. a. θ b. 53°

3. a. See Belowb. Domain: ;

Range:

c. increasing

4. a. all real numbers

b.

c. increasing d. π2; -π2

-π2 , y ,π2

-π2 # y #π2

-1 # x # 1

ø5 tan-1(h6)

π2-π2 # y #x 5

π2-π2 # y #x 5

(Ï3, π3)(-1, -π4)(-1, -π4)

yy = tan x

y = tan-1 xx

y = x

π

-π

ππ]2

π]2

1

1 1-1-π- π]2

- π]2

-π2 # θ #π2


points on y 5 sin x

correspondingpoints y 5 sin-1x

3. a.

(0, 0)

(0, 0) (1, π2)(Ï32 , π3)(Ï2

2 , π4)(12, π6)(-1

2, -π6)(-Ï2

2 , -π4)(-Ï3

2 , -π3)(-1, -π

2)

(π2, 1)(π

3, Ï32 )(π

4, Ï22 )(π

6, 12)(-π6, -1

2)(-π4, -Ï2

2 )(-π3, -Ï3

3 )(-π2, -1)

c

110

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5. ; -45° 6. ; 90°

7. ; -60° 8. 0.955

9. -1.120 10. 1.107

11. a. the sine of the number

whose cosine is

b.

12. .707

13. sin

14. sin

15.

16. 1.2

17. .8

18. 74°

19. θ , where h

altitude (in kilometers)

20.

21. sin , cos

22. 2 sin αsin 2αcos α 5

2 sin α cos αcos α 5

2x 5192x 5 -4Ï5

9

Ï2 1 Ï22

55 tan-1( h16)

ø

-Ï22

3

x

9 – x 2

(cos-1 x3) 5

Ï9 2 x2

3

a

ba2 + b2

(tan-1 ba) 5

bÏa2 1 b2

øÏ22

45

35

-π3

π2-π4 23. a. Sample: Let . Then

sin 1 sin 5

sin 0 cos cos 0 sin 1

sin 0 cos 2 cos 0 sin

5 0

5 sin 0

b. sin sin 5

sin x cos sin cos x 1

sin x cos 2 sin cos x 5

2 sin x sin x

24. The graphs of

sin sin and

y 5 sin x are identical.

Sample: Let x 5 ,

sin 1 sin 5

sin cos cos sin

sin cos 2 cos sin 5

sin

25. a. Sample: 15, 26, 37, 48,59, 70, 81, 92, 103, 114

b. Sample:

26. a.b. sin (cos-1 x)cos (sin-1 x)

c. Sample: tan (cot-1 x) 5cot (tan-1 x) for x . 0

5 Ï1 2 x2 5-1 # x # 1

21π2

17π2 ,13π

2 ,5π2 , 9π

2 ,

π25

12 1

12 5 1 5

1 1(12) 2 0(Ï3

2 )1(12 1 0(Ï3

2 )π3

π2

π3

π2

π3 1

π2

π3 1

π2

(π2 2

π3)(π

2 1π3)

π2

(x 2π3)1(x 1

π3)

y 5

12 5••

π3

π3

π3

π3 1

(x 2π3)(x 1

π3) 1

0 1Ï22 1 0 2

Ï22

π4 5

π4

π4

π4 1

(0 2π4)(0 1

π4)

x 5 0


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1. , n an integer,

or equivalently

(mod 2π)

2. , n an integer or

, n an integer or

equivalently (mod π)

or (mod π)

3. , n an integer or

(mod π)

4. a. 0.841 and 5.442b. ±0.841 1 2nπ, n aninteger

5. a.

b. or

1 2nπ, where n is anyinteger, or equivalently

(mod 2π) or

mod 2π.

6. , n an integerπ2 1 2πnx 5

u ; 11π6

u ; 7π6

u 511π6u 5

7π6 1 2nπ

7π6 , 11π

6

z ; π4

z 5π4 1 πn

y ; 5π3

y ; π3

y 55π3 12πn

y 5π3 1 2πn

x ; π2

x 5π2 1 2πn 7. .848 or

2.29 2π8. or

9.

10. 6.78° θ 83.22°

11. or

2πn, n an integer

12. a. No

b. or

13. a. 42.22°b. 45.85°

14. a. θ 63.43°

b. cos θ

15. a. b. c.

16. See below.

17. a.

b.

c. 5 -2 2 Ï3Ï6 1 Ï2Ï2 2 Ï6

Ï6 1 Ï24

Ï2 2 Ï64

π3

π3-π4

51

Ï5

øøø

7π4 # x # 2π0 # x #

3π4

x 55π3 1x 5

π3 1 2πn

,,

π3 # x #

5π3

4π3 # x ,

3π2

π3 # x ,

π2

# x #0 # x #


16. Right side

Distributive properties


Left side

∴ 2 sin3 x 1 sin 2x cos x4 sin2 x 1 2 cos 2x 5 sin x

5

52 sin x

2 5 sin x

52 sin x(sin2 x 1 cos2 x)4 sin2 x 1 2 2 4 sin2 x

52 sin3 x 1 2 sin x cos x cos x

4 sin2 x 1 2(1 2 2 sin2 x)

52 sin3 x 1 sin 2x cos x

4 sin2 x 1 2 cos 2x Formulas for sin 2x and cos 2x

c

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18.

a. 2πb. Values of the functionrepeat every interval oflength 2π; that is, f(x 2π) f(x), ∀ realnumbers x.

19. a. True

b.

c.d. 6

e.

f(x ) = 6

x = 4-10 10 20

-5

5

10

15

y

x

x 5 4

x 5 -16

51


π]2

20. 29

21. or

22. a. If a parallelogram hasone right angle, then it is arectangle.

b. True

23. a. i. ,

ii.

iii.

b. The solutions of 2 sin2 [f(x)] 2 5 sin [f(x)] 2 3 5 0 are all numbers xsuch that f(x) is equal to the solutions of 2 sin2 x 25 sin x 2 3 5 0.

( 111π6 1 2πn)( 1

7π6 1 2πn)

5π6 1 2πnπ

6 1 2πn,

2πn3

1118 1

7π18 1

2πn3

t . 4t , 1


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1.

2.

3.

4.

5.

6.7. b

8.

9.

10.

11.

12. a. , ,

sin

b.

c. Does

? Does 5

? Does 5

? Yes

13.

14. 0

15. Ï22

-π4

2 1 Ï34

Ï3 1 24

8 1 4Ï316

Ï3 1 24(Ï6 1 Ï2

4 )2

(ÏÏ3 1 22 )2

5

x 5ÏÏ3 1 2

2

Ï6 1 Ï24

5π12 5

y 5π4x 5

23π

Ï63

-4Ï29

2Ï42 1 215

-4Ï2 1 Ï2115

2 2 Ï3

Ï2 1 Ï22

Ï2 1 Ï22

-83

3Ï5555

-Ï558 16.

17.

18.

19.

20.

21. n an

integer

22. a. , π,

b. , (2n + 1)π,

or , n an integer

23. 0.644 or 5.640 x2π, approximately

24. ,

25.26. 1

27. sin (y 2 x)

28. 2 sin x cos x

29. -sin2 x

30. sin x cos y 1 cos x sin y

π6 , x ,

5π6

3π2 , x # 2π0 # x ,

π2

##0 # x #

5π3 1 2πn

x 5π3 1 2πn

5π3x 5

π3

x 5 -π2 1 2πn, π4 1 πn,

x 54π3 , 5π

3

x 53π4 , 7π

4

x 5π3, 5π

3

2

5 21

Ï215

-12


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31. Left side

Identity for the cosine of a sum

Evaluating trigonometric functionssin x MultiplicaitonRight side

∴ cos ; domain: all real numbers

32. Left side 5 sec x cot x

Def. of secant and cotangent

Multiplication of fractions

5 csc x Def. of cosecant5 Right side

∴ sec x cot x 5 csc x; domain: , n an integer

33. Left side

1

sin x5 cos x5 Right side

∴ sin ; domain: all real numbers

34. Left side

Adding fractions


csc2 α Def. of csc5 Right side

∴ 5 2 csc2 α; domain: , n an integerα Þ nπ11 1 cos α 1

11 2 cos α

5 2

52

sin2 α

52

1 2 cos2 α

51

1 1 cos α 11

1 2 cos α

(π2 1 x) 5 cos x

5 1 • cos x 1 0 •

cos π2 • sin x5 sin π2 • cos x

5 sin (π2 1 x)

x Þnπ2

51

sin x

51

cos x • cos xsin x

(3π2 1 x) 5 sin x

555 0 2 (-sin x)

sin 3π2 sin x5 cos 3π

2 cos x 2

5 cos (3π2 1 x)

Formula for sine of a sum

Evaluating trigonometric functionsMultiplication

c

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35. Left side 5 cos(α 2 β) 2 cos(α 1 β)5 cos α cos β 1 sin α sin β 2 Sum and difference

cos α cos β 1 sin α sin β identities5 2 sin α sin β Addition5 Right side∴ cos(α 2 β) 2 cos(α 1 β) 5 2 sin α sin β; domain: all real numbers

36. sec x 1 cot x csc x sec x csc2 x

5

Addition of fractions 5

Pythagorean Identity 5

∴ sec x 1 cot x csc x 5 sec x csc2 x; domain: , n an integer

37. Left side 5 cos 4x5 cos2 2x 2 sin2 2x Identity for cos 2x

5 (cos2 x 2 sin2 x)2 2 (2 sin x cos x)2 Identities for cos 2x and sin 2x

5 cos4 x 2 2 sin2 x cos2 x 1 sin4 x 24 sin2 x cos2 x Multiplication

5 cos4 x 2 6 sin2 x cos2 x 1 sin4 x Addition5 Right side

∴ cos 4x 5 cos4 x 2 6 sin2 x cos2 x 1 sin4 xdomain: all real numbers

38. Right side

Identities for cos 2x

Simplification

5 tan2 x Simplification and def. of tan5 Left side

∴ 5 tan2 x; domain: , n an integer

39. θ 5 sin , where h 5 height of the kite

above the ground in feet.

(h 2 3200 )-1

x Þπ2 1 nπ1 2 cos 2x

1 1 cos 2x

52 sin2 x2 cos2 x

51 2 (1 2 2 sin2 x)1 1 (2 cos2 x 2 1)

51 2 cos 2x1 1 cos 2x

x Þnπ2

1cos x sin2 x

sin2 x 1 cos2 xcos x sin2 x

51

cos x sin2 x1

cos x 1cos xsin2 x


Trigonometricdefinitions

Trigonometricdefinitions

c

c

40. θ 5 tan , where x isthe distance traveled east in km

41. cos

42. 34.8° or 55.2°

43. 22.3° θ 67.7°,approximately

44. a. θ 38.0°b. θ 49.1° or θ 49.1°,approximately

45. sin (2x)

46. cos

47. a. sin

b.

-6 # x # 6, x-scale = 1-3 # y # 3, y-scale = 1

(πx 2π4) 2 1y 5 2

(2x 2π2)y 5 -12

y 5 -12

., -5 6

##

ø(d7)-1t 5

15π

( x100)-1 c. amplitude 5 2, period

5 2, phase shift ,

vertical shift

48. a. g(x) 5 3 sin (8x)b.

49. x 5 9 sin t 2 7, y 5 6 cos t 1 4

50. x 5 5 sin t 1 4, y 5 2 cos t 2 1

51. See below.

52. not an identity;Counterexample: Let .Then cos but 2 cos .(0) 5 2

(2 • 0) 5 1,x 5 0

- # x # , x-scale = -3 # y # 3, y-scale = 1

π]4

π]4

3π]4

5 -1

5π4

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51. an identity;1 1 cot2 x csc2 x

Def. of cot 5 Def. of csc

Addition 5

Pythagorean Identity 5

∴ 1 1 cot2 x 5 csc2 x, , n an integerx Þ nπ

1sin2 x

sin2 x 1 cos2 xsin2 x

51

sin2 x1 1cos2 xsin2 x

c

c

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53. See below.

54. approximately

55. You can check variousdifferent cases by holding αconstant and graphing theresulting functions in β. For

example, when α , you

could graph y 5 sin

and y 5 sin cos β 2

cos sin β.π3

π3

(π3 1 β)

5π3

-2 # x # 2

56. a.

b. y 5 tan xc. sin x sec x 5 tan x

57. a. x 1.1 or x 3.6b. 1.1, 3.6 2π

58. a. , , ,

b.59. 0 0.675# x #

π4 , x ,

3π4

67π46

5π46

3π4x 5 6

π4

, x #0 # x ,øø


π]2


53. an identity;Left side 5 tan (π 1 γ)

Def. of tan

5 tan γ Def. of tan5 Right side

∴ tan γ 5 tan (π 1 γ)

5-sin γ-cos γ

5sin π cos γ 1 sin γ cos πcos π cos γ 2 sin π sin γ

5sin (π 1 γ)cos (π 1 γ)

Identities for sine and cosineof a sumEvaluating specific values oftrignometric functions

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1. , , integers

2. , , integers

3. Sample: S 5, 7, 10, 14, 20,28, K

4. There is more than onesequence satisfying therecurrence relation.

5. a. 1, 2, 5, 10, 17, 26b.c. , sothe initial condition is met.

, sothe recursive relationship issatisfied. Therefore, theexplicit formula is correct.

6. a. 3, 7, 11, 15,19, 23b.c. , so theinitial condition is met.

, so therecursive relationship issatisfied. Therefore, theexplicit formula is correct.

7. a. Sample:10 FOR N 1 TO 5020 TERM SIN(3.1415*N/2)30 PRINT TERM40 NEXT N50 ENDb. 1, 0, , 0, 1, 0, , 0, 1, 0-1-1

55

tn 1 44n 1 4 2 1 54(n 1 1) 2 1 5tn11 5

4(1) 2 1 1 3t1 5tn 5 4n 2 1

tn 1 2n 2 12n 2 1 5(n2 2 2n 1 2) 12 1 2 5

2 5 n2 1 2n 1 1 2 2n 2tn11 5 (n 1 1)2 2 2(n 1 1) 1

t1 5 12 2 2(1) 1 2 5 1tn 5 n2 2 2n 1 2

5

k $ 1;tk11 5 tk 1 2k 1 1t1 5 0

k $ 1;sk11 5 sk 1 4s1 5 11 8. a. 1, 1, 2, 3, 5, 8, 13, 21, 34,

55b. 1, 3, 4, 7, 11, 18, 29, 47,76, 123c. 1, 3, 8, 21, 55, 144, 377,987, 2584, 6765. Thesequence is the even termsof the Fibonacci sequence.

9. 1, , , , , ; ,

integers

10. a. 1, 1, 1, 1, 1, 1; , integers

b. 1, 2, 1, 2, 1, 2;

c. 1, , 1, , 1, ;

; even integers

11. a. ; , integers

b. , integers, conjectured from 2,

8, 26, 80, . . . .c. , so theinitial condition is met.

, so the recursiverelationship is met.Therefore, the explicitformula is correct.

12. a. , , , , ,

b. ;

; integers n $ 1

sn 52n 2 1

2n

6364

3132

1516

78

34

12

3an 1 23(3n 2 1) 1 2 53 2 1 5

3n 2 3 13 • 3n 2 1 53 • 3n11 2 1 5an11 5

a1 5 31 2 1 5 2

n $ 1;an 5 3n 2 1

k $ 1;ak11 5 3ak 1 2a1 5 2

n . 1

Han 5 1 ; odd integers n . 0 an 5 c 2 1

c 2 1c 2 1c 2 1

Han 5 1 ; odd integers n . 0an 5 2 ; even integers n . 1

n $ 1;an 5 1

n $ 1;

an 51n

16

15

14

13

12


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13. a. ;

b. There are essentialdiscontinuities at and

.c. , ,

14.

15. , for ,

16. a. Sample:

b. 2

-3 -2 -1 1 2 3

-3-2-1

12y

x

n Þ -2n Þ -1n 1 1n 1 2

(k 1 1)2 (k 1 2)4

y 5 2x 5 -3x 5 3x 5 -3

x 5 3

5 2limx→-`

f(x)

5 2limx→`

f(x) 17. a. If there exists an x suchthat p(x) is true and q(x) isfalse.b. False

18. a. i. , , , ;

);

; integers

ii. , , , ;

),

; integers

b. );

Yes, the formula works forall noninteger numbers

or 1.c Þ 0

an 51

c 2 1 • (1 2 (1c)n

n $ 1

an 513(1 2 (1

4)n

85256

2164

516

14

n $ 1

an 512(1 2 (1

3)n

4081

1327

49

13



1. a.b. 24

2. a.

b.

3. True, the only difference isthat different letters areused for the indices.

4.

5. a.

b.Sn11 5 S1 1

1n 1 1

S1 512H

ok

j51

1j 1 1

o43

i51i2

1516

2-4 1 2-3 1 2-2 1 2-1

-2 1 2 1 8 1 16 6. a. b.

7. a.

b.

8. a. , ,

,

b.

(oki51

i2) 1 (k 1 1)2

K 1 k2) 1 (k 1 1)2 5(12 1 22 1k2 1 (k 1 1)2 5

ok11

i51i2 5 12 1 22 1 K 1

o4

i51ai 5 30o

3

i51ai 5 14

o2

i51ai 5 5o

1

i51ai 5 1

n 1 12n 2 1 1

n 1 12n 1

n 1 12n 1 1

3715 ø 2.47

on21

j50(12) j

o4

k50(12)k

c

9.

10.

11. It does.

12.

13. Sample:Let . Then

, and

.

14.

15. explicit; -1, 1, -1, 1, -1, 1

16. a. x2 is not defined since

.

b. The definition definesthe first term and defineseach successive term interms of the preceedingterm.

17. ; ,

; integers n $ 1

an 51

n 1 112, 13, 14, 15, 16, 17

x2 510

11000 o

5

i5-5i2

(1 1 1 1 1 1 1)2 5 16

( o4

n51an)2

512 5 4

12 1 12 1 12 1o4

n51(an

2) 5

an 5 1 ; n

0 1 0 216 5 1 2

16

1 1 0 1 0 1(-15 115) 2

16 5

(-14 114) 1(-13 1

13) 1

(-12 112) 1(1

5 216) 5 1 1

(14 2

15) 1(1

3 214) 1(1

2 213) 1

(1 212) 1o

5

k51(1k 2

1k 1 1) 5

o4

j51( j2 1 j 2 4)

(oki51

i(i 2 1)) 1 (k 1 1) k 18. a.3.75, ; integers b. 279° F

19. Left side

Right side; real numbers x and y,

20.

21.

22. Sample: The Arrow Paradoxstates that an arrow nevermoves, because at eachinstant the arrow is in afixed position. Another ofZeno’s paradoxes, known asthe Paradox of Achilles andthe Tortoise, is frequentlysummarized as follows.Achilles, who could run 10yards per second, competedagainst a tortoise which ran1 yard per second. In orderto make the race more fair,the tortoise was given aheadstart of 10 yards.Zeno’s argument, thatAchilles could never passthe tortoise, was based onthe “fact” that wheneverAchilles reached a certainpoint where the tortoisehad been, the tortoisewould have moved aheadof that point.

(g o f )(k) 5(k 1 1)(k 1 2)

2

2k2 2 k 1 1(2k 1 1)(2k 2 1)

cos2 y 2 cos2 x.sin2 x 2 sin2 y 5[55 cos2 y 2 cos2 x5 1 2 cos2 x 2 1 1 cos2 y5 (1 2 cos2 x) 2 (1 2 cos2 y)

5 sin2 x 2 sin2 y

k $ 1T1 5 325, Tk11 5 0.95Tk 1


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121

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1. a.b. Assume that S(k) is truefor a particular butarbitrarily chosen integer

where

.

2. a. and

, so S(1) is true.

b. ;

:

c.

, so istrue.d. S(n) is true ; integers

.

3. a. S(1): 3 is a factor of 3;S(3): 3 is a factor of 33; S(5): 3 is a factor of 135.b. All are true.c. : 3 is a factor of

.

4. a. S(1): ; S(3): ;S(5): b. S(1) is false. S(3) and S(5)are true.c.4 . (k 1 2)2

(k 1 1)2 1S(k 1 1):

29 , 3613 , 165 , 4

(k 1 1)3 1 2(k 1 1)S(k 1 1)

n $ 1

S(k 1 1)(k 1 1)(k 1 2)k(k 1 1) 1 2(k 1 1) 5

ok11

i512i 5 o

k

i512i 1 2(k 1 1) 5

(k 1 2)(k 1 1)

ok11

i512i 5S(k 1 1)

S(k): ok

i512i 5 k(k 1 1)

1(1 1 1) 5 2

o1

i512i 5 2(1) 5 2

S(k): ok

i51(2i 2 1) 5 k2

k $ 1

1 5 12 5. a. ;

b. All are true.

c.

6. a.

b.

c.

d.

e. use inductive assumption

f.

g.

h. The Principle ofMathematical Induction

7. a. , so S(1) is true.b. Assume that S(k) is truefor a particular butarbitarily chosen integer

. S(k):

. Show

is (k 1 1)(k 1 2)(2(k 1 1) 1 1)

6

k2 1 (k 1 1)2 512 1 22 1 K 1S(k 1 1):

k2 5k(k 1 1)(2k 1 1)

6

12 1 22 1 K 1k $ 1

S(1): 12 51(2)(3)

6

(k 1 1)(k 1 2)2

k(k 1 1) 1 2(k 1 1)2

k 1 11 1 2 1 3 1 K 1 k 1

k 1 1 5(k 1 1)(k 1 2)

2

1 1 2 1 3 1 K 1 k 1

k 5k(k 1 1)

2

1 1 2 1 3 1 K 1

1(1 1 1)2 5 1

(k 1 1)(3k 1 2)2

ok11

i51(3i 2 2) 5S(k 1 1):

13 55(14)

2

1 1 4 1 7 1 10 1S(5):

1 1 4 1 7 53(8)

2 ;S(3):

1 51(2)

2S(1):


c

true.

c. Since S(1) is true and , then by

mathematical induction,S(n) is true ; integers .

8. a. 1, 4, 13, 40, 121b. 1, 4, 13, 40, 121

c. Let S(n):

; integers .

(1) . This

agrees with the recursiveformula, hence S(1) is true.

(2) Assume S(k): is

true for some integer .Show :

is true.

. 3k11 2 12

3k11 2 3 1 22 5

(3k 212 ) 1 1 53ak 1 1 5 3

ak11 53k11 2 1

2

ak11 5S(k 1 1)k $ 1

ak 53k 21

2

a1 531 2 1

2 5 1

n $ 1

an 5(3n 2 1)

2

n $ 1

S(k) ⇒ S(k 1 1)

(k 1 1)(k 1 2)(2(k 1 1) 1 1)6

(k 1 1)(k 1 2)(2k 1 3)6 5

(k 1 1)(2k2 1 7k 1 6)6 5

(k 1 1)(k(2k 1 1) 1 6(k 1 1))6 5

k(k 1 1)(2k 1 1) 1 6(k 1 1)2

6 5

(k 1 1)2 5

(k 1 1)2 5k(k 1 1)(2k 1 1)

6 1

12 1 22 1 K 1 k2 1 Therefore, for all integers, if S(k) is true, then

is true. Thus, from(1) and (2) above, using thePrinciple of MathematicalInduction, S(n) is true for allintegers . Hence, theexplicit formula describesthe same sequence as therecursive formula.

9. S(n): ;integers . (1) from the recursive definition;

fromthe explicit definition.Hence, S(1) is true. (2) Assume S(k):

for some integer. Show :

is true.

. Therefore,is true if S(k) is

true, and (1) and (2) proveby the Principle ofMathematical Inductionthat the explicit formuladoes describe the sequence.

S(k 1 1)2(k 1 1)2 2 12(k2 1 2k 1 1) 2 1 52k2 2 1 1 4k 1 2 5

ak 1 4k 1 2 5ak11 5ak11 5 2(k 1 1)2 2 1

S(k 1 1)k $ 12k2 2 1

ak 5

12 2 1 5 1a1 5 2 •

a1 5 1n $ 1an 5 2n2 2 1

n $ 1

S(k 1 1)k $ 1


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c

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10. (1) .

. Hence, S(1) is

true. (2) Assume S(k):

is true

for some integer .Show :

is true.

Now

, so theinductive step is true. Thus,(1) and (2) prove by thePrinciple of MathematicalInduction that S(n) is truefor all integers .

11.

12. a. (-3)2 -3 (-2)2 -2(-1)2 -1 n2 nb. 16

13. a. 1, 2, 3, 4, 5, 6b. Sample: , for allintegers .n $ 1

an 5 n

11K111111

on

i51

in

n $ 1

(k 1 1)(3(k 1 1) 2 1)2

(k 1 1)(3k 1 2)2 5

3k2 1 5k 1 22 5

3k2 2 k 1 6k 1 22 52 5

ok

i51(3i 2 2) 1 3(k 1 1) 2

ok11

i51(3i 2 2) 5

(k 1 1)(3(k 1 1) 2 1)2

ok11

i51(3i 2 2) 5

S(k 1 1)k $ 1

ok

i51(3i 2 2) 5

k(3k 2 1)2

1 • (3 2 1)2 5 1

o1

i51(3i 2 2) 5 1 14. a.–c.

d. i. ii. iii.

15. a. Since is a factor of, and is a

factor of , then by the

Transitive Property ofPolynomial Factors, isa factor of .b.c. Since is a factor of

by part a and is a factor of bypart b, then is a factorof bythe Factor of a PolynomialSum Theorem.

16. True

17. The program contains aninitial condition in line 10and a recurrence relation inline 30. Given an infiniteamount of time andcomputer memory, it wouldprint all the integersgreater than N 1. 2

(x5 2 xy 4) 1 (xy 4 2 y5)x 2 yxy 4 2 y5

x 2 yx5 2 xy 4x 2 y

xy 4 2 y5 5 y 4(x 2 y)x5 2 xy4

x 2 y

x(x4 2 y 4)x5 2 xy4 5

x4 2 y4x4 2 y4x 2 y

5π6

π3

π4

-1 1 3

-1

1

3y

x

y = cos-1 x

y = cos x

y = x


1., so 16 is a factor.

2.

. It is clear that 2 isa factor of . Since 2is also a factor of ,then 2 will be a factor oftheir sum,

, by the Factor ofa Polynomial Sum Theorem.

3. a. S(1): 2 is a factor of 2;since , S(1) is true.S(13): 2 is a factor of 158;since , S(13) istrue. S(20): 2 is a factor of382; since ,S(20) is true.b. S(k): 2 is a factor of

. : 2 is a factor of

.c.d.

; 2 is afactor of and 2k, so 2 is a factor of

by the Factor of an IntegerSum Theorem.e. 2 is a factor of ; integers .

4. a. Since 5 is a factor of, S(1) is true.

b. S(k): 5 is a factor of.

c. : 5 is a factor of.6k11 2 1

S(k 1 1)6k 2 1

61 2 1 5 5

n $ 1n2 2 n 1 2

(k 1 1) 1 2(k 1 1)2 2

k2 2 k 1 22) 1 2k(k2 2 k 1

k2 1 k 1 2 5k2 1 k 1 2

(k 1 1) 1 2(k 1 1)2 2

S(k 1 1)k2 2 k 1 2

2 • 191 5 382

2 • 79 5 158

2 • 1 5 2

(2(n 1 1))(n2 1 n) 1

n2 1 n2(n 1 1)

2(n 1 1)(n2 1 n) 1n2 1 3n 1 2 5

(n 1 1)2 1 (n 1 1) 5

16 • 753 2 4 • 3 2 1 5 112 5 d.

; 5 is a factorof and of 5, so 5 is afactor of their sum. Thus,

is true. Hence, S(n):5 is a factor of ;

is true by thePrinciple of MathematicalInduction.

5. Since ,substituting and

into the result ofExample 3 yields is a factor of ;

.

6. Since , substituting and into the result of Example 3yields is a factorof ; .

7. S(1) is true since . Assume

S(k): 3 is a factor of . Show :

3 is a factor of is true.

. Since 3 is a factorof and

), istrue. Hence, 3 is a factor of

; by thePrinciple of MathematicalInduction.

n $ 1n3 1 14n 1 3

S(k 1 1)3(k2 1 k 1 5k3 1 14k 1 3

3k 1 15(k3 1 14k 1 3) 1 3k2 1k3 1 3k2 1 17k 1 18 5(k 1 1)3 1 14(k 1 1) 1 3 514(k 1 1) 1 3

(k 1 1)3 1S(k 1 1)k3 1 14k 1 3

1 1 3 5 18 5 3 • 613 1 14 •

n $ 113n 2 1n 5 13n 2 1

13 2 1 5 12

y 5 1x 5 1313 2 1 5 12

n $ 19n 2 3n

9 2 3 5 6y 5 3

x 5 99 2 3 5 6

n $ 16n 2 1

S(k 1 1)

6k 2 16(6k 2 1) 1 56 • 6k 2 6 1 5 5

6 • 6k 2 1 56k11 2 1 5


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8. (1) 6 is factor of , hence S(1) is

true. (2) Assume S(k): 6 is afactor of is truefor some positive integer k.Show : 6 is a factorof istrue. Now

.Because 6 is a factor of

by the inductiveassumption, a factor of

by the theorem,and a factor of 12, itfollows that 6 is a factor of

. SinceS(1) is true and S(k)

, by the Principle ofMathematical Induction, 6 is a factor of , ;

.

9. Sample: By Example 3 withand ,

is a factor of.

10. a. S(1) is false. S(2) and S(3)are true.b. Sample: All we canconclude is that S(2) andS(3) are true.c. . 8 is a factor of if .Hence, S(n) holds if .n $ 2

n $ 24n12n 2 8n 5 4n(3n 2 2n)

xn 2 yn 5 22n 2 1

x 2 y 5 3y 5 1x 5 22

n $ 1n3 1 11n

S(k 1 1)⇒

(k 1 1)3 1 11(k 1 1)

3k(k 1 1)

k3 1 11k

3k(k 1 1) 1 12(k3 1 11k) 111k 1 11 53k 1 1 1

k3 1 3k2 111(k 1 1) 5(k 1 1)3 1

(k 1 1)3 1 11(k 1 1)S(k 1 1)

k3 1 11k

12 5 2 • 613 1 11 5 11. S(1) is true because

and

. Assume

S(k): is true

for some integer . Show that :

is

true. Now

. Since

S(1) is true and S(k), by the Principle of

Mathematical Induction,S(n) is true for all integers

.n $ 1

S(k 1 1)⇒

[(k 1 1)(k 1 2)]2

4

(k 1 1)2[k2 1 4(k 1 1)]4 5

4(k 1 1)3

4 5

(k 1 1)3 5k2(k 1 1)2

4 1

(k 1 1)3 5 Fk(k 1 1)2 G2

1

ok

i51i3 1o

k11

i51i3 5

ok11

i51i3 5 F(k 1 1)(k 1 2)

2 G2

S(k 1 1)k $ 1

ok

i51i3 5 Fk(k 1 1)

2 G2

F1(1 1 1)2 G2

5 12 5 1

o1

i51i3 5 13 5 1


c

12. Let S(n): .which

agrees with the recursivedefinition, hence S(1) istrue. Assume S(k):

is true for someinteger k. Show :

is true.

.Since S(1) is true andS(k) , by thePrinciple of MathematicalInduction, S(n) is true for all

. Hence, the explicitformula correctly definesthe sequence.

13. True

n $ 1

S(k 1 1)⇒

2k11 1 32k11 1 6 2 3 52(2k 1 3) 2 3 5Tk11 5 2Tk 2 3 5Tk11 5 2k11 1 3

S(k 1 1)Tk 5 2k 1 3

T1 5 21 1 3 5 5Tn 5 2n 1 3 14. See below.

15. See below.

16. a.

b. 0c. 13

d.17.18. Samples: , ,

n3 1 3n2 1 2nn(n 1 1)(n 1 2) 5

n3 1 14nn4 1 2n2

f(t) 5 30 • 3t

103

-2 # x # 20, x-scale = 5-.5 # y # 2.5, y-scale = 1


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14. Right side cos (2x)cos2 x sin2 x Formula for cos(2x)(cos2 x sin2 x) 1 Multiplication by 1(cos2 x sin2 x)(cos2 x sin2 x) Pythagorean Identitycos4 x sin4 x MultiplicationLeft side

Therefore, cos4 x sin4 x cos (2x) by the Transitive Property

15. Left side

Factoring

Factoring

DivisionMultiplication

Therefore, when by theTransitive Property.

a Þ ba4 2 b4

a 2 b 5 a3 1 a2b 1 ab2 1 b3

5 a3 1 a2b 1 ab2 1 b35 (a 1 b)(a2 1 b2)

5(a 2 b)(a 1 b)(a2 1 b2)

a 2 b

5(a2 2 b2)(a2 1 b2)

a 2 b

a4 2 b4

a 2 b5

525

25125

• 2525

5

c

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1. If and , then.

2. a. basis step: S(1) is thestatement: If ,then .b. S(3): If , then

.c. inductive step: AssumeS(k). So if , bymultiplication, which implies .But so by theTransitive Property .Thus .

3. The sense of inequality isnot changed because .

4. Let S(n) be the statement :if , then . Letcase 1 be that and case 2 be that .For case 1 and 2 we have,

, since ,S(1) is true. Assume S(k).So if , then xk . 0x . 0

0 , xS(1): xn 5 x1 5 x

x $ 10 , x , 1

xn . 0x . 0

x . 0

S(k) ⇒ S(k 1 1)xk11 , 1

x # 1xk 1 1 , xx • xk , x • 1

0 , x , 1

x3 , 10 , x , 1

x1 , 10 , x , 1

a , cb , ca , b Mx

Product ofPowersProperty

Thus, . So, by the Principle ofMathematical Induction,S(n) is true for all integers

.

5. a. basis step: S(1):

b. S(2):

c. inductive step: S(k):;

by multiplication.But

.Thus

6. Let . Then and, so gives

us a contradiction. So wecan conclude that thestatement is false.

7. See below.

3 . 31 1 2(1) 5 331 5 3n 5 1

S(k) ⇒ S(k 1 1)3(1 1 2k) $ 1 1 2(k 1 1)

3 • 3k 5 3k11 $3(1 1 2k)

3 • 3k $3k $ 1 1 2k

9 $ 532 $ 1 1 2(2) ⇒

31 $ 1 1 2(1) ⇒ 3 $ 3

n $ 1

S(k) ⇒ S(k 1 1)

xk11 . 0x • xk . 0 • x


c

7. Let S(m) be the statement: , so S(1) is true.

Assume S(k). Then Definition of S(k)M2

Product of Powers andDistributive PropertyDistributive PropertyTransitive Property, since

Thus, . So by the Principle of MathematicalInduction, S(m) is true for all integers .m $ 1

S(k) ⇒ S(k 1 1)k $ 12k11 $ 1 1 (k 1 1)

2k11 $ 1 1 (k 1 1) 1 k

2k11 $ 2 1 2k2 • 2k $ 2(1 1 k)

2k $ k 1 1S(1): 21 5 2 $ 2 5 1 1 1

2m $ 1 1 m

8. Let S(n) be the statement , so S(9) is true.

Assume S(k). Then Definition of S(k)M2

Product of PowersDistributive PropertyTransitive Property since

Thus, . So by the Principle of MathematicalInduction, S(n) is true for all integers .

9. Let S(t) be the statement: S(4): , so and S(3) is true.Assume S(k). Then

Addition Property of InequalityFactorTransitive Property since

Thus . So by the Principle of MathematicalInduction, S(t) is true for all integers .

10. Let S(n) be the statement: 3 is a factor of .S(1): ; since 3 is a factor of 15, S(1) is true.Assume S(k). Then 3 is a factor of . Show : 3 is a factor of is true.

, By the Factor of an IntegerSum Theorem .

Thus, by the Principle of Mathematical Induction S(n) is true forall integers greater than zero.

S(k) ⇒ S(k 1 1)5 (k3 1 14k) 1 3(k2 1 k 1 5)5 k3 1 3k2 1 17k 1 15(k 1 1)3 1 14(k 1 1)

(k 1 1)3 1 14(k 1 1)S(k 1 1)k3 1 14k

13 1 14(1) 5 15n3 1 14n

t $ 3S(k) ⇒ S(k 1 1)

6k 2 6 . 03(k 1 1)2 $ 9(k 1 1)3(k 1 1)2 $ 9(k 1 1) 1 6k 2 63k2 1 6k 1 3 $ 9k 1 6k 1 3

3k2 $ 9k27 $ 273(32) 5 27, and 9 • 3 5 27

3t2 $ 9t

n $ 9S(k) ⇒ S(k 1 1)

(k 2 1) . 02k11 . 40(k 1 1)2k11 . 40(k 1 1) 1 40(k 2 1)2k11 . 40(2k)2 • 2k . 2(40k)

2k . 40k

S(9): 29 5 512 . 360 5 40(9)2n . 40n


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11. a.

; integers b. $6639.79

12.

The graphs are related

because

so cot x represents the

transformation applied

to tan x.

T-1 π2

-tan (x 2π2)cot x 5

-2π -π π 2π- 2π3

2π3- π

3

y

xy = cot x

y = tan x

π3

n . 1

HA1 5 8000An11 5 (1.008)an 2 400 13. a. Sample: and

, and120(mod 7) 1(mod 7)b. Yes. xy(mod7)1(mod 7)

14. The inequality in Example 1does not hold for allnegative integers n, since if

and then, which is greater

than 1.The inequality in Example 2is true for all negativeintegers n, since 3n is alwayspositive for and

is always negativefor . So, for all integers .n , 0

3n $ 1 1 2nn , 01 1 2n

n , 0

xn 5 2n 5 -1x 5 .5

;;

xy 5 120y 5 10x 5 12


1. < 1016.6

2.

3. , ,

4. a.b. < 0.00098c.

5. a. 5 2b. 6

6. 500

7. No, it diverges since the

ratio and

.

8. a. Sample:

b. Sample:

9. a.

b. ,

Left side,

Right side

; , :

Left side

,

Right side

c. right side d. left side

10.p2(3n12 2 1)p(3n12 2 1

3 2 1 ) 5

2 2 2(12)n((12)n

2 1

-( 12) ) 51 •

5

2 2 2(12)n(1 2 (12)n

12

) 51 •

5

r 512a 5 12n 2 1

1 • (2n 2 11 ) 55

2n 2 11 • (1 2 2n

-1 ) 55

r 5 2:a 5 1

a(rn 2 1r 2 1 )1 2 rn

1 2 r 5

a • -1-1 • a(1 2 rn

1 2 r ) 5

o`

n50 (3

2)n

o`

n50 15n

limk→`

(109 )k

5 ` Þ 0

r 5109 . 1

n 5 688

n 5 10

S5 513760S4 5

2512S3 5

116

6 2 6(12)n

11. a. , , for all

integers .

b.

c.

d. < 5.99999982e. 6f. The result confirms partsd and e.

12. a.

b.

13. Let S(n): if , then; integers .

S(1): , is true sincex is greater than 1.Assume S(k). If , then

.Mx

Product ofPowersTransitiveProperty ofInequality,since

Thus . So by the Principle ofMathematical Induction,S(n) is true for all positiveintegers.

S(k) ⇒ S(k 1 1)x . 1

xk11 . 1

xk11 . xx • xk . x • 1

xk . 1x . 1

x1 5 x . 1n $ 1xn . 1

x . 1

620099990

K 171

10413(n21)

6.2 171104 1

71107 1

711010 1

o25

j513(1

2) j 21

an 5 3(12)n21

2 # k # 25

ak 512ak21a1 5 3


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14. a. S(3): .

Does ? Does ?

Does ? Yes. Hence,S(3) is true.

b. S(1): ,which is true.

Assume S(k):

is true for

some integer . Show

that :

is true.

. Hence,

S(n) holds ; n $ 1

(k 1 1)(k 1 2)(k 1 3)3

(k 1 1)(k 1 2)(k3 1 1) 5

(k 1 1)(k 1 2) 5

k(k 1 1)(k 1 2)3 1(k 1 1)(k 1 2) 5

ok11

i51i(i 1 1) 1o

k11

i51i(i 1 1) 5

(k 1 1)(k 1 2)(k 1 3)3

ok11

i51i(i 1 1) 5S(k 1 1)

k $ 1

k(k 1 1)(k 1 2)3

ok

i51i(i 1 1) 5

1 • 2 51 • 2 • 3

3

20 5 202 1 6 1 12 5 204 • 5

1 • 2 1 2 • 3 1 3 • 4 5

3 • 4 • 53o

3

i51i(i 1 1) 5 15. a.

The graph has shifted

units to the right.

b. or

16.

17. a. about 9.867b. 3.141c. π

18. a. 4,

b. 10 SUM 020 FOR TERM 1 TO

10030 LET A (-1)^

(TERM 1)*4/(2*TERM 1)

40 SUM SUM A50 PRINT SUM60 NEXT TERM70 END

c. 3.141594d. π

152

15

55

-43, 45, -4

7, 49, - 411

503

0 , u #3π4-π , u # -π

4

π2

y

x-π π

-10

-5

5

10

- π]2

π]2


1. In the Strong Form ofMathematical Induction,the assumption that each ofS(1), S(2), , S(k) are true isused to show that is true. In the original form,only the assumption thatS(k) is true is used to prove

is true.

2. a. Sample: Combine 1 and2, then join 3, then join 4;combine 1 and 2, then 3and 4, then the two blocks.b. 3

3. The 7-piece and 13-pieceblocks needed 6 and 12steps respectively, by theinductive assumption. Thatis 18 steps; joining them isthe 19th step.

4. a. 2, 2, 4, 8, 14b. i. S(n): an is an eveninteger.ii. areeven integers.iii. a1, a2, , ak are all evenintegers. Prove that isan even integer.iv. .ak, , and are alleven integers by theinductive assumption. Sothere exist integers p, q,and r such that

. By theFactor of an Integer SumTheorem, isthen also an even integer.So is an even integer.ak11

ak 1 ak21 1 ak22

ak22 5 2p 1 2q 1 2rak 1 ak21 1

ak22ak21

ak 2 2ak11 5 ak 1 ak21 1

ak11

K

a1 5 2, a2 5 2, a3 5 4

S(k 1 1)

S(k 1 1)K

v. The sum of any evenintegers is even.

5. a. 5, 15, 20, 35b. and aremultiples of 5. Assume a1,a2, , ak are all multiplesof 5. Show that is amultiple of 5. Now

. 5 is a factor of ak

and , so by the Factor ofan Integer Sum Theorem itis factor of their sum, .Hence, by the Strong Formof Mathematical Induction, an is a multiple of 5 ; integers

.

6. and are oddintegers. Assume a1, a2, ,ak are all odd integers.

, whereand

for some integers qand r. Then

sois odd. Hence, by the

Strong Form ofMathematical Induction,every term in the sequenceis an odd integer.

7. a. S(2): ., so S(2) is true.

S(3): . ,so S(3) is true.b. ; integers j such that

, c. : d.

(Fk 1 Fk22)(Fk11 1 Fk21) 1Lk11 5 Lk 1 Lk21 5

Lk11 5 Fk12 1 FkS(k 1 1)Lj 5 Fj11 1 Fj212 # j # k

4 5 3 1 1L3 5 F4 1 F2

3 5 2 1 1L2 5 F3 1 F1

ak11

3 5 2(q 1 2r 1 1) 1 14r 12q 11 1 2(2r 1 1) 5

ak11 5 2q 12r 1 1

ak 5ak21 5 2q 1 1ak21 1 2akak11 5

Ka2 5 5a1 5 3

n $ 1

ak11

ak21

ak 1 ak21

ak11 5ak11

K

a2 5 15a1 5 5


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e.

8. a. 9 b. 9

9. Let S(n): isdivisible by 3. Then S(1):

is divisible by 3. , which is divisible by 3, soS(1) is true. Assume S(k):

is divisible by3 is true for some integer

. Show :

is divisible by 3 istrue.

3k1

. isdivisible by 3 by theinductive assumption, and

is divisible by3; thus their sum is divisible3(k2 1 3k 1 2)

k3 1 3k2 1 2k3k 1 2)3(k2 12k) 13k2 1(k3 1

6 511k 16k2 12 5 k3 16k 1 3 1 2k 13k2 11

1k3 1 3k2 12(k 1 1) 53(k 1 1)2 1(k 1 1)3 1

2(k 1 1)3(k 1 1)2 1(k 1 1)3 1

S(k 1 1)k $ 1

k3 1 3k2 1 2k

1 1 3 1 2 5 613 1 3 • 12 1 2 • 1

n3 1 3n2 1 2n

Fk12 1 Fk(Fk11 1 Fk) 5Lk11 5 (Fk21 1 Fk22) 1 by 3. Therefore, by

mathematical induction,is divisible by

3 ; integers .

10. S(n): . S(1): a1

4(3)0. a1 4, so S(1) is true.Assume S(k): ak 4(3)k 2 1 istrue. Show S(k 1): ak 1 1

4(3)k is true. ak 1 1 3ak

3(4 3k 2 1) 4 3k. Bymathematical induction,S(n) is true for all .Hence, is anexplicit formula for thesequence.

11. a.

b. tan x

c. See below.


an 5 4(3)n21n $ 1

•5•55

515

55an 5 4(3)n21

n $ 1n3 1 3n2 1 2n


11. c. Left sideDefinition of cosecant,secant, and cotangent

Multiplication

Subtraction of Fractions


Simplifying fractions

Definition of tangentRight side

Therefore, csc x sec x cot x tan x ; x for which both sidesare defined.

5255 tan x

5sin xcos x

5sin2 x

sin x cos x

51 2 cos2 xsin x cos x

51

sin x cos x 2cos2 x

sin x cos x

51

sin x • 1cos x 2

cos xsin x

5 csc x sec x 2 cot x

c

c

12. a.b. 100 3 100 3 100c. Yesd. Yes

13. , 2

14. a. a1 and a2 must bemultiples of 7.b. a1 and a2 must bemultiples of m.c. a1 and a2 are multiples ofm. Assume a1, a2, K , ak aremultiples of m.

. Because and are multiples of m, theirsum, , is also a multipleof m by the Factor of aPolynomial Sum Theorem.By the Strong Form ofMathematical Induction,every term in the sequenceis a multiple of m.

15. a. Suppose that arecurrence relation defines

in terms of , , ,, and n for each integer

. Then there is exactlyone sequence defined bythis recurrence relation andthe initial condition .x1 5 a

n $ 1x1

Kxn21xnxn11

ak11

akak21ak21

ak 1ak11 5

x 5 -5

P(x) 5 0.15x3 2 1.5x2 b. Suppose there is asecond sequence y forwhich and isdefined by the samerecurrence relation as .Let S(n) be the statement

. (1) Because and , . So S(1)is true. (2) Assume S(1), S(2),

, S(k) are true. Then, , , .

The sequences have thesame recurrence relation;

is defined in terms of , , and k; and is

defined in terms of , ,and k; so .

Thus, S( ) is true. By (1),(2), and the Strong Form ofMathematical Induction,S(n) is true for all .Therefore, the adaptedRecursion Principle isproved.

n $ 1

k 1 1xk11 5 yk11xk

Kx1

xk11ykKy1yk11

xk 5 ykKx2 5 y2x1 5 y1

K

x1 5 y1y1 5 ax1 5 axn 5 yn

xn11

yn11y1 5 a


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1. a. 2 and 5b. 2 and 5 are notexchanged.

2. a. 2 and 5b. 5 is placed in Lr

3. 1, 6, 4, 9

4. , ,

5. 5, -7, 1.5, -1, 13, 6 Initialorder-7, 1.5, -1, 5, 6, 13 After firstpass-7, -1, 1.5, 5, 6, 13 Aftersecond pass

6. See below. , -1, 1.5, 5,6, 13

7. If no interchanges arenecessary, adjacentnumbers are in order.Hence, by the TransitiveProperty, the entire list is inorder.

8. a. 7, 9, 4, 6, -4, 5, 0 initialorder4, 7, 9, 6, -4, 5, 0-4, 4, 7, 9, 6, 5, 0

L 5 -7

Lr 5 [L, 5 {1, 6, 4}f 5 9

b. Apply the Filterdownalgorithm to the list. Thesmallest number is now infront. Apply Filterdown tothe sublist which includesall but the first number.Apply Filterdown to thesublists wich aresuccessively one elementsmaller. Continue until thelist is exausted.

9. Sample: 6, 5, 4, 3, 2, 1

10. Quicksort

11. Quicksort

12. Let S(n) be the statement: If, then .

S(1): , which istrue.Assume S(k).

Inductiveassumption

Mx

Product of Powersand SimplificationTransitiveProperty since

for Therefore, S(k) S( ).Hence, by the Principle ofMathematical Induction,S(n) is true for all integers

.n $ 1

k 1 1⇒x $ 1x2 . x

xk11 $ x

xk11 $ x2x • xx • xk $

xk $ x

x1 5 x $ xxn $ xx $ 1


6. L = {5, -7, 1.5, -1, 13, 6}

f = 5 Lr = {13, 6} L, = {-7, 1.5, -1}

(L,)r = {1.5, -1}(L,), = Ø f = -7 f = 13 (Lr)r = Ø(Lr), = {6}

((L,)r)r = Ø((L,)r), = {-1} f = 1.5

c

13. ,

14. and are even. Assume, , are all even.

Show that is even.

forsome integers p and r. Since

is an integer byclosure properties, iseven. Therefore, is evenfor all integers by the Strong Form ofMathematical Induction.

15. a.

b. , for all integers

c.

d.

16. Prove S(n):

for all integers . S(1):

is true.

Assume S(k):

is true. Show that kk 1 1

ok

i51

1i(i 1 1) 5

o1

i51

1i(i 1 1) 5

11 1 1

n $ 1

on

i51

1i(i 1 1) 5

nn 1 1

43

(43)(1 2 (1

4)n)k $ 1

Ak 51

4k21

A2 514, A3 5

116

n $ 1an

ak11

cp 1 r

c(2p) 1 2r 5 2(cp 1 r)ak11 5 c • ak 1 ak21 5

ak11

akKa2a1

a2a1

r 5 1q 5 14 : is

true.

. So by mathematical

induction, S(n) is true ;integers .

17. a.

b. ohms

18. The argument is valid. The three premises are: (1) , (2) , (3) .By the Transitive Propertyand statements (1) and (2),

. Then by modustollens and (3), . This isthe conclusion, and so theargument is valid.

19. Sample: e, d, c, b, a; thereare 24 such orderings. Theonly restriction is that amust be the last letter inthe list.

,pp ⇒ r

,rq ⇒ rp ⇒ q

7017 ø 4.12

R 5R1R2

R1 1 R2

n $ 1

k 1 1k 1 2

(k 1 1)(k 1 1)(k 1 2) 5

1(k 1 1)(k 1 2) 5

k(k 1 2)(k 1 1)(k 1 2) 1

1(k 1 1)(k 1 2) 5

kk 1 1 1

1(k 1 1)(k 1 2) 5

ok

i51

1i(i 1 1) 1o

k11

i51

1i(i 1 1 5

ok11

i51

1i(i 1 1) 5

k 1 1k 1 2S(k 1 1)


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1.2.3. a. repeated multiplication:

9; sum of powers of 2: 6b. repeated multiplication:9999; sum of powers of 2:26

4. a. pb.

5. E(10) 499,500

6. a. 1 secondb. 1,000 seconds or16.67 minutesc. 1012 seconds or31,710 years

7.

8. S(n): is divisibleby 16. S(1): isdivisible by 16. S(1) is truesince 0 is divisible by 16.Assume S(k): isdivisible by 16 for somepositive integer k. Showthat 16 is a factor of

.

. 16 isa factor of (bythe inductive assumption)and a factor of 16k. Hence,

5k 2 4k 2 15(5k 2 4k 2 1) 1 16k1) 1 20k 1 5 2 4(k 1 1) 2 1 5

5(5k 2 4k 24(k 1 1) 2 1 54k 1 1) 25(5k 2 4k 2 1 1

5k11 2 4(k 1 1) 2 1 55k11 2 4(k 1 1) 2 1

5k 2 4k 2 1

51 2 4(1) 2 15n 2 4n 2 1

L = {7, -3, 2, -6, 10, 5}

f = 7 Lr = {10} L, = {-3, 2, -6, 5}

(L,)r = {2, 5}(L,), = {-6} f = -3

((L,)r)r = {5}((L,)r), = Ø f = 2

ø

ø

5

3log2n

E(n) 5 2n 2 1

E(n ) 5 n by the Factor of an IntegerSum Theorem, 16 is also afactor of 16k. Therefore, bymathematical induction,

is divisible by16 for all positive integersn.

9. Let S(n): .For , the formulayields

. This agrees with theinitial condition, so S(1) istrue. Assume S(k):

is true forsome positive integer k.Show that :

istrue.

.This agrees with therecursive formula, so

is true. Hence, bymathematical induction,S(n) is true for all integers

, and so the explicitformula for the sequence is

.

10. Left side

Right side 2(0 1 49 16) 3(0 1 23 4) (1 1 1 11) 2(30) 3(10) 5 85

11. a.

b. o27

n51(34 1 5(n 2 1))

an 5 34 1 5(n 2 1)

5215111121

111111115

26 1 43 5 855 -1 1 4 1 13 1

3n2 1 5n 2 3

n $ 1

S(k 1 1)

ak 1 6k 1 83) 1 6k 1 8 5(3k2 1 5k 25(k 1 1) 2 3 5

ak11 5 3(k 1 1)2 13(k 1 1)2 1 5(k 1 1) 2 3

ak11 5S(k 1 1)

3k2 1 5k 2 3ak 5

3 5 55 • 1 2a1 5 3 • 12 1

n 5 1an 5 3n2 1 5n 2 3

5n 2 4n 2 1

5(5k 2 4k 2 1) 1


c

12. a.

b.c.d.

13. a. NOT(p AND (q OR r))b. Sample: , ,

14. a.b.c. -26 1 26i

11 2 5i1 2 3i

r 5 0q 5 1p 5 0

-5 5 10-10

10

y

x

x 5 4y 5 5

h(x) 5 5 123

x 2 4 15. a. 2b. 3c. Each of the n digits ofthe second number aremultiplied by up to n digitsof the first number. Hence,there are at most n2

multiplications.d. 2n column additionse. ; For theproblem, , so theefficiency should be 8. Allfour multiplication stepsshown, must be carried out,as well as the four columnadditions. , so thealgorithm checks.

16. See below.

4 1 4 5 8

n 5 2E(n) 5 n2 1 2n


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16. For Merge sort, . For Selection sort, .Approximate :E(n)

E(n) 5 2nE(n) 5 nlog2n

Merge sort 33 664 133,000

Selection sort 20 200 20,000

Bubblesort 45 4950

Quicksort 13 464 113,000 1.8 3 107

5 3 10115 3 107

2 3 106

2 3 107

n 5 1,000,000n 5 10,000n 5 100n 5 10

139

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1. 7, 13, 31, 85, 247

2. -2, , 110

3. 0, 2, 2, 4, 4

4. 2,

5. 3, 5, -2, -24, -40

6. 51

7. -3c

8. a. 3, 5, 7, 9, 11b. for allintegers

9. a. 1,

b. for all integers

10. a. 3, -1, 3, -1, 3

b.

11.12. a. 1, 2, 3, 4, 5

b. ; integers c. 2, 4, 6, 8, 10; ; integers

13. a.

b. 9

14. False 15.

16. 17.

18. a. S(n):

n(n 1 1)(n 1 2)3

on

i51i(i 1 1) 5

o-1

i5-n

1io

6

i512i

17

K 1 (n 1 n)(n 2 3) 1 (n 2 2) 1

n $ 1an 5 2n

n $ 1an 5 n

In 5 n(n 1 1)

an 5 H 3 when n is odd-1 when n is even

n $ 1

an 51n!

12, 16, 1

24, 1120

n $ 1an 5 2n 1 1

23, 13, 15, 2

15

52, 56

3 , 1052

b. S(4): ;

1 • 2 2 • 3 3 • 4

4 • 5 40 and ,

so S(4) is true.

19. a. S(k):

b. 202c.and

20. a. 3 9 27 81 120

and

, so the

formula works for .

b.

c.

, which agrees

with part b.

21.

22. < 2.9653

23. a.

b. < 2.6667

24.

25. a.

b. < 7.9029c. 10

Sn 5 10(1 2 (45)n)

154

4t3 (1 2 (1

4)n11)

(k)(2k 1 3)

Fokj51

( j 2 1)(2j 1 1)G 1

ok11

j51( j 2 1)(2j 1 1) 5

32 • 242 5 363

32(243 2 1) 5

32(35 2 1) 5

120 1 243 5 363

n 5 4

32(81 2 1) 5 120

32(34 2 1) 5

5111

101(102) 5 1030210100 1 202 5 10302

ok

i512i 5 k(k 1 1)

4(5)(6)3 5 405

111

o4

i51i(i 1 1) 5

4(5)(6)3


c

26. Let S(n): . (1) S(1): .

, so S(1) is true. (2) Assume S(k):

is true for somearbitrary integer .Show that :

is true. Fromthe recursive definition,

Hence, by mathematicalinduction, S(n) is true for allintegers , and theexplicit formula is correct.

27. a. 0, 2, 6, 12, 20b. Let S(n): .(1) S(1): .

, so S(1) is true. (2) Assume S(k):

is true for somearbitrary integer .Show that :

is true.From the recursivedefinition,

.Hence, by mathematicalinduction, S(n) is true ;integers , and theexplicit formula is correct.

n $ 1

5 (k 1 1)((k 1 1) 2 1)5 (k 1 1)k5 k2 1 k5 k(k 2 1) 1 2k

bk11 5 bk 1 2k

(k 1 1)((k 1 1) 2 1)bk11 5S(k 1 1)

k $ 1k(k 2 1)

bk 5b1 5 0

b1 5 1(1 2 1)bn 5 n(n 2 1)

n $ 1

5 2 • 3k11 2 2.5 2 • 3k11 2 6 1 45 3(2 • 3k 2 2) 1 4ak11 5 3ak 1 4

2 • 3k11 2 2ak11 5S(k 1 1)

k $ 12 • 3k 2 2

ak 5a1 5 4

a1 5 2 • 31 2 2an 5 2 • 3n 2 2 28. Let S(n): 3 7 11 K

. (1) S(1): .S(1) is true. (2) Assume S(k):3 7 11 K (4k 1)k(2k 1) is true for somearbitrary integer .Show that S(k 1): 37 11 K (4(k 1)1) (k 1)(2(k 1) 1) istrue. From the inductiveassumption, 3 7 11K (4k 1) (4(k1) 1)

Hence, by mathemeticalinduction, S(n) is true for allintegers .

29. (1) S(1):

. 3 • 1(1 2) 9 and

, so S(1) is true.

(2) Assume that S(k):

is true for some arbitraryinteger .

Show S(k 1):

5is true.

(k 1 1)(k 1 2)(2(k 1 1) 1 7)2

ok11

i513i(i 1 2)1

k $ 1

k(k 1 1)(2k 1 7)2o

k

i513i(i 1 2) 5

1 • 2 • 92 5 9

511 • 2 • 9

2

o1

i513i(i 1 2) 5

n $ 1

5 (k 1 1)(2(k 1 1) 1 1)5 (k 1 1)(2k 1 3)5 2k2 1 k 1 4k 1 35 k(2k 1 1) 1 (4(k 1 1) 2 1)

21121

111

111521111

11k $ 1

1521111

3 5 1(2 • 1 1 1)n(2n 1 1)(4n 2 1) 5

1111


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(2k2 7k 6k 18)

5 (k 2)(2k 9)

5

Hence, by the Principle ofMathematical Induction,S(n) is true ; integers .

30. (1) S(1): 3 is a factor of. ,

so S(1) is true. (2) AssumeS(k): 3 is a factor of

is true for someinteger . Show that

: 3 is a factor ofis true.

Expanding,

(k3 3k2 3k 1)(14k 14)

(k3 14k) (3k2

3k 15)

(k3 14k) 3(k2 k 5).11115

11115

111115

14(k 1 1)(k 1 1)3 1

(k 1 1)3 1 14(k 1 1)S(k 1 1)

k $ 1k3 1 14k

13 1 14(1) 5 1513 1 14(1)

n $ 1

(k 1 1)(k 1 2)(2(k 1 2) 1 7)2

11(k 1 12 )

111(k 1 12 )

2(k 1 1)(3k 1 9)2 5

k(k 1 1)(2k 1 7)2 1

(k 1 1)(3k 1 9) 5

k(k 1 1)(2k 1 7)2 1

3(k 1 1)(k 1 3) 5

ok11

i513i(i 1 2) 5 ( o

k

i513i(i 1 2)) 1 By the inductive

assumption, 3 is a factor ofk3 14k, and 3 is a factor of3(k2 k 5). Therefore, 3 isa factor of their sum by theFactor of an Integer SumTheorem. Hence, bymathematical induction,S(n) is true ; integers .

31. (1) S(2): 3 is a factor of . ,

so S(2) is true. (2) Assumethat S(k): is a factor of

is true for someinteger . Prove

: 3 is a factor ofis true.

2(k 1)3 5(k 1)2k3 6k2 6k 25k 5 (2k3 5k)3(2k2 2k 1)Since 3 is factor of 2k3 5kand 3(2k2 2k 1). 3 is afactor of their sum by theFactor of an Integer SumTheorem. Hence, bymathematical induction,S(n) is true ; integers .n $ 2

212

211252

21115121

2(k 1 1)3 2 5(k 1 1)S(k 1 1)

k $ 22k3 2 5k

16 2 10 5 62 • 23 2 5 • 2

n $ 1

111


c

32. See below.

33. See below.

34. a. S(1): ,

so S(1) is true.

b. Assume S(k):

for some integer .

c. :

1k 1 2

(121

k 1 2) 5(121

k 1 1)(12

13) K (12

12)S(k 1 1)

k $ 11k 1 1

(1 213) K (1 2

1k 1 1) 5

(1 212)

11 1 1

12 5(1 2

12) 5


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c

d.

Hence, .Therefore, by the Principleof Mathematical Induction,S(n) is true for all integers

.n $ 1

S(k) ⇒ S(k 1 1)

51

k 1 2( 1k 1 1)(k 1 1

k 1 2)5

5 ( 1k 1 1)((k 1 2

k 1 2) 2 ( 1k 1 2))

5 ( 1k 1 1)(1 2

1k 1 2)

5 S(k)(1 21

k 1 2)(12

1k 1 2)(12

1k 1 1)

(1213) K (12

12)

32. Let S(n) be the statement: If , then .(1) S(1): , so S(1) is true, since by the givenstatement. (2) Assume S(k) for some arbitrary integer. Show S(k 1): If , then .

by the inductive assumptionMx

Product of powersGiven SimplificationGiven

Hence, . Therefore, by mathematical inductionS(n) is true for integers .

33. (1) S(2): . So S(2) is true.(2) Assume S(k). Show .

by the inductive assumptionM4

SimplificationDistributive PropertyGiven

So S(k) S(k 1). Therefore by the Principle of MathematicalInduction, S(n) is true for all integers .n $ 2

1⇒4(k 1 1) , 4(k 1 1) 1 204k11 . 4(k 1 1)

k $ 24k11 . 4(k 1 1) 1 12(2) 2 44k11 . 4(k 1 1) 1 12k 2 44k11 . 16k4 • 4k . 4 • 4k4k . 4k

S(k 1 1): 4k11 . 4(k 1 1)42 5 16 . 8 5 4(2)

n # 1S(k) ⇒ S(k 1 1)

x # 10 # xk11 # x0 # xk11 # 1

x # 10 # xk11 # 1 • 10 # xk11 # x2x • 0 # x • xk # x • x0 # xk # x

0 # xk11 # x0 # x # 11

x $ 0x1 5 x0 # xn # x0 # x # 1

c

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35. a. 0, 4, 4, 16, 28b. Let S(n): 4 is a factor of

.(1) S(1): 4 is a factor of .

, so S(1) is true. S(2): 4 is factor of .

, so S(2) is true.(2) Assume S(1), S(2), K ,and S(k) are true for someinteger . So 4 is afactor of b1, b2, K , bk.Show S(k 1): 4 is a factorof is true. Since and have 4 as a factor,there exist integers p and qsuch that and

. Substituting intothe recurrence relation,

4q 3(4p) 4q12p 4(q 3p). q 3p isan integer by closureproperties, so 4 is a factorof . Hence, by theStrong Form ofMathematical Induction,S(n) is true for all integers

.

36. a.b.c. 28

37. a.

for all integers b. $79,489.90

k . 1

HA1 5 80,000Ak11 5 1.01Ak 2 900

Ck11 5 k 1 Ck

C2 5 1, C3 5 3, C4 5 6

n $ 1

bk11

115151bk11 5

bk 5 4qbk21 5 4p

bk

bk21bk11

1

k $ 1

b2 5 4b2

b1 5 0b1

bn

38. a. , ; integers

b. 256

39. a. initial order: 1, 3, 5, 2, 4First pass: 1, 3, 2, 4, 5Second pass: 1, 2, 3, 4, 5b. 5, 4, 3, 2, 1

40.

41. a. , for all

integers

b.

; integers

c.

d.

42. a. 10, 13, 16, 19, 22b. explicit; the nth term ofthe sequence is given as afunction of n in line 20.c. an 5 3n 1 7

S20 5 2(1 2 (12)20)

o20

i51(12)i21

n $ 1

an 5 (12)n21k $ 1

ak11 512aka1 5 1

L = {21, 1, 8, 13, 1, 5}

f = 21 Lr = Ø L, = {1, 8, 13, 1, 5}

(L,)r = {8, 13, 5}(L,), = {1} f = 1

((L,)r)r = {13}((L,)r), = {5} f = 8

k $ 1

Hb1 5 2bk11 5 2bk


1. a. Leonhard Euler; 18thb. Cardano; 16thc. Wessel; 18th

2. a. 8i b.3. a. 21 b. 4

4. real 5 8, imaginary 5 7

5. 10 1 15i

6. a. True b. No

7. 3 2 4i

8.9. a. 8 1 i b. 8 2 i

c. 2 1 i d. 3 2 2ie. 8 2 i; they are equal.

10.11. a-d.

12. a. 4 1 2i ohms

b.13. a. 5 2 6i

b.

c. parallelogram; Sample:slopes of opposite sides areequal.

imaginary

real0

z + w

z

w

4

-4

-8

4 8-4

65 2

35 i

-12 -9 -6 -3 3 6 9 12

-9-6-3

369

imaginary

real

d (0, -7)

c (-3, 0)

b (-4, 7) a (12, 8)

-35 1

45 i

2 167 i

-

--

Ï20 i 5 2Ï5 i

14. x 5 7, y 5 9

15. a. ohms

b. 5i ohms

16. For two imaginary numbersmi and ni, mi 1 ni 5(m 1 n)i. Since m and n arereal, m 1 n is real, and (m 1 n)i is an imaginarynumber.

17.

5

So 1 1 i is a solution of.

18. 63i, 62i

19.

20. Let . Then 5a 2 bi. By definition ofequality for complexnumbers, ifand only if and b 5

then 2 and .So , andz is a real number.

21. Suppose there is a smallestinteger, s. Then ,and is an integer. Thiscontradicts the assumptionthat s is the smallestinteger. So the assumptionis false, and there is nosmallest integer.

s 2 1s 2 1 , s

z 5 a 1 bi 5 a 1 0ib 5 0b 5 0-b,

a 5 aa 1 bi 5 a 2 bi

zz 5 a 1 bi

429 i2 5

2529 1

429 5 110

29 i 2

( 529 1

229 i) 5

2529 1

1029 i 2

z • 1z 5 (5 2 2i) •529 1

229 i;

z2 2 2z 1 2 5 0

2 5 01 1 i 1 i 2 1 2 2 2 2i 1

1 1 i 1 i 1 i2 2 2 2 2i 1 2(1 1 i)2 2 2(1 1 i) 1 2 5

52 1 3i


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22. a. . 5b. , 3

23. a. 40°b. 51.25°F 28.75°F

24. a. domain: {x: 2 5};range: {y: 1 2}b.

c. domain: {x: 1 2};range: {y: 2 5}d. They are inverses.

# y ## x #

y = f (x)

reflection imageof graph of

y = f (x)

1

1

2

3

4

5

6

2 3 4 5

y

6 x

# y ## x #

# t # ---

25. a.

b. They are complexconjugates.

c.

d. For , the

roots are

and .

So

1

5

. 5ca5

4ac4a2

b2 2 (b2 2 4ac)4a2

-b 2 Ïb2 2 4ac2a

• -b 1 Ïb2 2 4ac2az1 • z2 5

-ba .5-2b2a 5

-b 2 Ïb2 2 4ac2a

-b 1 Ïb2 2 4ac2az1 1 z2 5

-b 2 Ïb2 2 4ac2az2 5

-b 1 Ïb2 2 4ac2az1 5

ax2 1 bx 1 c 5 0

z1 • z2 5 2z1 1 z2 5 -32,

z2 5 -34 2Ï23

4 i

z1 5 -34 1

Ï234 i,


1. a-d.

2. a. Sample:

b. Sample:

c. Sample:

3. a. By case b of the polarrepresentation theorem in

this lesson,

1

Then by case b again,

for any

integer k is a coordinate

representation of .

b. By case c of the polarrepresentation theorem in

this lesson,

. So by

case b, for

any integer k is acoordinate representation

of .F4, π3G

F-4, 4π3 1 2kπG

πG 5 F-4, 4π3 Gπ

3 1F-4,

F4, π3G 5

F4, π3G

F4, -5π3 1 2kπG

2(-1)πG 5 F4, -5π3 G.F4, π3

F4, π3G 5

F2, -5π6 G

F-2, π6GF2, 19π

6 G

2

ad

c

b

4

4. a., b.

5. Given any point [r, θ].First plot P and Q 5 [1, θ] 5(cos θ, sin θ).

Because [r, θ] is r times asfar from the origin as Q, itsrectangular coordinates are(r cos θ, r sin θ). Thus, therectangular coordinates ofP are given by cos θand sin θ.

6. (0, 4)

7. (1.7, 1.5)

8.9. [ , 56.3°]

10. a. The ship should sail 26.3°East of South.b. .38 hours or

23 minutes

11. a. Sample:

b. Sample:

c. Sample: F3, 5π6 G

F3, 5π6 G

F3, 11π6 G

øø

-Ï13ø[Ï29, 21.8°]ø

-ø-

P = [r, u]

u

Q

y

x

y 5 rx 5 r

P 5

a

b

2 4

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12.13. a. Sample: [1, 0°]

b. Sample: [1, 90°] c. Sample: [11, 90°]

14.15. P1 5 [3, 0°], P2 5 [3, 30°],

P3 5 [3, 90°], P4 5 [3, 120°],P5 5 [3, 195°], P6 5 [3, 240°],P7 5 [3, 285°]; r 5 3

16. Q1 5 4, , Q2 5 3, ,

Q3 5 2, , Q4 5 1, ,

Q5 5 0, , Q6 5 1, ,

Q7 5 2, , Q8 5 3, ,

Q9 5 4, ; θ 5

17.

18. amps

19. a., so

. and . So

5 .b.5 . So

.

c. , so

0 1 5 .z w( z

w)[32i.

56 1 3i2 2 4i 5

z w5 0 1

32i.( z

w)zw 5

(6 2 3i)(2 1 4i) 5 0 2

32 i

w • z[ z • w 524 2 18i.(2 2 4i) 5(6 1 3i)

5w • z24 2 18i.z • w 524 1 18i

(6 2 3i)(2 1 4i)5w • zw1z5[ z 1 w8 2 i.

1 (2 2 4i)(6 1 3i)5w1z5 2 2 4iw

5 6 1 3iz5 8 2 iz 1 w5 8 1 i(2 1 4i)

z 1 w 5 (6 2 3i) 1

-52

-1.5 1 (-1.5Ï3)i

π6

π6GF

π6GFπ

6GFπ6GFπ

6GFπ6GF-π

6GF-π6GF-π

6GF-

F10Ï33 , π6G; (5, 5Ï3

3 )

-3 1 4i 20. a. cos (θ 1 φ) 5 cos θ cos φ2 sin θ sin φb. sin (θ 1 φ) 5 sin θ cos φ 1cos θ sin φ

21. a.

b. 6.5 mpg

22. Let the vertices in clockwiseorder be A( 1, 0), B( 5, 5),C( 11, 5), and D( 7, 0). Theslope of 5 0; the slope of 5 0. The slope of

5 ; the slope of

5 . Since

ABCD is composed of twopairs of parallel lines, it is aparallelogram.

23. The north magnetic polelies just north of NorthAmerica and west ofGreenland and is aboutlatitude 76° N andlongitude 101° W onBathurst Island in Canada.The south magnetic pole isin Antarctica and is aboutlatitude 66° S and longitude140° E, just off the coast ofAntarctica due south ofAustralia. The location ofboth magnetic poles varyover time.

-5 2 0-11 2 (-7) 5

54DC

-5 2 0-5 2 (-1) 5

54

ABBCAD

------

400h26 1 h


1. ;

5

The slope of

the slope of 5

and so .

The slope of 5

the slope of 5

and so .

Therefore, the figure is aparallelogram.

2. ;;;

. The

slope of ;

the slope of 5

5 ; and so .

The slope of

; the slope of

; and so

. Therefore, thefigure is a parallelogram.

3. a.

b. 3Ï2(cos3π4 1 i sin 3π

4 )F3Ï2, 3π

4 GZP z zOW

d 2 0c 2 0 5

dcOW 5

dc

b 1 d 2 ba 1 c 2 a 5

5ZP

WPiOZba

b 1 d 2 da 1 c 2 cWP

5b 2 0a 2 0 5

baOZ

5 (a 1 c, b 1 d) 5 Pz 1 w 5 (a 1 c) 1 (b 1 d)iw 5 c 1 di 5 (c, d) 5 W0 5 0 1 0i 5 (0, 0) 5 Oz 5 a 1 bi 5 (a, b) 5 Z

OWiZP5 -94;

-9 2 04 2 0OW-94;

5-6 2 311 2 7ZP

WPiOZ-6 2 (-9)11 2 4 5

37;

WP

53 2 07 2 0 5

37;OZ

P.z 1 w 5 11 2 6i 5 (11, -6)w 5 4 2 9i 5 (4, -9) 5 W;

5 O,0 5 0 1 0i 5 (0, 0)z 5 7 1 3i 5 (7, 3) 5 Z 4. a.

b.

5. a. [5, 127°]b. (cos 127° 1 i sin 127°)

6.

7.

8.

9. [6, 210°]

10. 50 cos sin

11.12.

13. modulus: 3, argument: 270°

imaginary

real

z

zw

3 65˚

15

40˚

-2 2

10

-11 2 10i

7π6 )7π

6 1 i(

4 62

6 93 12

2 4

øø

Ï33 (cos 30° 1 i sin 30°)

FÏ33 , 30°G

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14. The midpoint of 5

.

The midpoint of 5

. Since

the diagonals have thesame midpoint, ZOWP is aparallelogram.

15. a.b.

16. a.

b. , and, while , and

Then

, ; and

5 . So the

triangles are similar by theSSS Similarity Theorem.c.

17. a.

b. , and, while , and

. So EFG bythe SSS CongruenceTheorem.

nE′F′G′ùn2Ï2F′G′ 5E′G′ 5 Ï29E′F′ 5 5,FG 5 2Ï2

EF 5 5, EG 5 Ï29G′ 5 3 1 6i

E′ 5 5 1 i, F′ 5 1 1 4i,

Ï5

Ï5B′C′BC 5

Ï202

A′B′AB 5

5Ï5

5 Ï5Ï5

A′C′AC 5

Ï85Ï17

5Ï85.

A′C′ 5B′C′ 5 Ï20A′B′ 5 5,AC 5 Ï17

AB 5 Ï5, BC 5 2C′ 5 8 1 i

B′ 5 4 1 3i,A′ 5 1 1 7i,

-π # x # 2π, x-scale =

-2.5 # y # 2.5, y-scale = 1

π]2

zw z 5 Ï5, -θ ø -63.4°

(7 1 02 , 5 1 0

2 ) 5 (72, 52)PO

(4 1 32 , 6 1 (-1)

2 ) 5 (72, 52)

ZW c.18. a.

5

5 5

and 5

5 .b. ; real numbers c andcomplex numbers

, Proof:

5 5

19. a.

b.20.21. Let , where a and

b are real numbers. Then

1 . Since a isreal, 2a is real. So for allcomplex numbers, the sumof the number and itscomplex conjugate is a realnumber.

22.

23. a. Yesb. domain: the set of realnumbers; range: the set ofintegers

3π8

(a 2 bi) 5 2a(a 1 bi)z 1 z 5z 5 a 2 bi.

z 5 a 1 bi

14 1 8i

x2 1 y2 5 16

4 8

zc z zz z .zc z za 1 bi zÏa2 1 b2zc zÏc2a2 1 c2b2 5

zcz z 5 zca 1 cbi z 5zcz z 5 zc z zz z .a 1 biz 5

3Ï413Ï16 1 253Ï42 1 (-5)2

3 z4 2 5i z 53Ï41,Ï369Ï144 1 225

Ï122 1 (-15)2

z12 2 15i z 5

T1,1


c

24. center , radius 5 3

25. Geometric SubtractionTheorem: Let and be twocomplex numbers that arenot collinear with theorigin. Then the pointrepresenting is thefourth vertex of aparallelogram withconsecutive vertices

, 0, and .w 5 c 1 dia 1 biz 5

z 2 w

w 5 c 1 diz 5 a 1 bi

5 (-1, 2) 26. Geometric DivisionTheorem: Let z and w becomplex numbers. If

, and ,

then .

That is, dividing a complexnumber z by w applies to za size change of magnitude

and a rotation of about

the origin.

-φ1s

zw 5 Frs, θ 2 φG (s Þ 0)

w 5 [s, φ][r, θ]z 5

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1.

θ 0 π 2π

r 0 3 0 -3 0

2.θ 0 π 2π

r 6 3 0 -6 0 6

2 4 6

-3Ï23Ï23Ï3

3π2

3π4

π2

π3

π4

π6

1 2 3

3Ï32

3Ï22

32

3π2

π2

π3

π4

π6

c

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c

c

3. sin θ⇒ Conversion ⇒ formula⇒

Conversion formula

⇒This verifies that the curve in Question 1 is a circle and

has center and radius .

4.

5.

0 # x # 2π, x-scale = 1-1 # y # 6, y-scale = 1

1 2 3 4 5

0 # x # 2π, x-scale = 10 # y # 5, y-scale = 1

1 2

32(0, 32)

(x)2 1 (y 232)2

5 (32)2

x2 1 y2 5 3yr2 5 3y

r 53yr

r 5 3 6. limaçon

7. a.b.

8. not periodic

9. periodic, Sample: 0 θ 2π

10.

Cardio- is a prefix meaning“heart.” Cardioid curvesresemble hearts.

11. a. The polar graph of sin θ is the graph of

sin θ taken through a scale change of k.b. It is a circle with radius

and center .(0, k2)k2

r 5r 5 k

1 2

##

1 2

y 1 x 5 1

12.

13. a.

b. c. 5

14. a. [2, 90°], [1, 195°], [4, 330°]

b.

2 4

A'AB

B'

C'

C

C′ 5B′ 5A′ 5

T-7/2, -8

Z′ 5 (-32, -9), W′ 5 (-9

2, -5)

imaginary

real-10

-4

4

40°

zw

z 150°

15. a.

b. They are complexconjugates.

c.

their arguments areopposites.

16. a.b.c.d. The sum, , is equal to the answer to part a.

17.

18. No, the graph of 9 is a circle and it fails thevertical line test forfunctions.

x2 1 y2 5

o10

k51[k(2k 1 1)]

-.88 1 .16i-.48 1 .36i-.4 2 .2i-.88 1 .16i

z 5 F1, π3G, w 5 F1, -π3G;

12 2

Ï32 i

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19. a.

b.

c. i. Around θ , r has negative values, causing a loop.ii. If , there are no negative values for θ.iii. for all θr . 0

a 5 b5 π

a + b

a

r = a + b cos u, where a . b

a + b

a

r = a + b cos u, where a = b

a + b

a

r = a = b cos u, where a , b

r = a + b cos u, where a . br

a + b

a – b

π 2π u

r = a + b cos u, where a = br

a + b

π 2π u

r = a + b cos u, where a , br

a + b

a – b π 2π u

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In-class Activity1. Sample: Put calculator in

Polar mode and enterequation using the key.

2. a.

The graph is rotated clockwise.b. There is a horizontal

translation to the right.

c. It would have a rotation

of counterclockwise.

d. The graph will be thesame as the graph inExample 3 but rotatedclockwise if the angle is (θ 1 x) and counterclockwise if (θ 2 x).

3.

0 # u # 2π, u-step = -2 # x # 2, x-scale = 1-2 # y # 2, y-scale = 1

π]12

r = cos u

π3

π4

π4


π]12

y 5

sin θ 5 cos (90 2 θ)

4. a.

b. , c. tan θ gives therectangular equation

which

simplifies to

The rectangular equation isundefined when or

. So these areequations of the asymptotes.

5. a.

b. sec θ 5 , so ;

, which is the

equation for a vertical linethrough (1, 0).

5 1x 5rr

r 5rx

rxr 5


π]6

x 5 -1x 5 1

-x4

x2 2 1 .y2 5

5yxÏx2 1 y2

r 5x 5 -1x 5 1


π]12


π]12

r = sin u


c

5. c. The graph of csc θwill be a horizontal linethrough the point (0, 1)because by definition,

csc θ 5 so,

which is the equation of ahorizontal line through thepoint (0, 1).

Lesson 8–51.

2.

3.

4. It will be the same graphsince cos (θ 2 π) cos θ.

5. Sample: [1, 0], ,

, , [π+1, π]Fπ 1 22 , π2GFπ 1 4

4 , π4GFπ 1 6

6 , π6G5 -

1

1 2 3

2 4

y 5r1

y 5rr

r 5ry

ry

r 5 6. Sample: [1, 0], ,

, , [2π, π]

7. a. cos 5θ or sin 5θ

b. Sample: r 5 2 cos 5θ

8.

There are horizontalasymptotes because cot θ isundefined when sin θ .

9. a. cos θ sin θcos

- # x # 2π, x-scale = 3 -2 # y # 2, y-scale = 1

π]4

(θ 2π4)r 5 Ï2

1r 5

5 0

1 2

1 2

r 5 2r 5 2

F2π/2, π2GF2π/4, π4GF2π/6, π6G

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11. a.

b. [0, 0],

12. a.

b.

c.

d.

13. a. voltsb. volts

14. about 26.6° and 63.4°

15. The chambered nautilus isalso known as the pearlynautilus. It has a smooth,coiled shell 15–25 cm indiameter, consisting of 30to 36 chambers; it lives inthe outermost chamber.

-9 2 2i-20 2 10i

-4 -2 2 4 6 8

-4

-2

2

4z

u

w

u'

v'

v

w'

z '

imaginary

real

z9 512 1 Ï3 1 (Ï3

2 2 1)i(-1 2

3Ï32 )i,

w9 5 -32 1 Ï3 1

v9 5 -32 23Ï3

2 i,

u9 512 1

Ï32 i,

-14 2Ï34 i

512 (cos 4π

3 1 i sin 4π3 )F12, 4π

3 GF2Ï2, π4G

2 4

r = 4 sin u

r = 4 cos u


c

9. b. cos

cos θ sin θc. cos θ sin θ

⇒

⇒ ⇒ ⇒ ⇒ ⇒

⇒

Hence, the graph is a circle.

10.

2 4 6

1 2

5 ( 1Ï2)2

(x 212)2

1 (y 212)2

y 114 5

12

x2 2 x 114 1 y2 2

x2 2 x 1 y2 2 y 5 0x2 1 y2 5 x 1 y

r2 5 x 1 yx 1 y

r

r 5xr 1

yr

1r 5

15

sin θ • Ï22 )

5 Ï2(cos u • Ï22 1

sin π4)sin θ

15 Ï2(cos θ cos π4

(θ 2π4)r 5 Ï2

12. 625

13. a.

b.

c. 2; 128; 8192; 524,288;33,554,432

14. Sample: Using DeMoivre’sTheorem, for [r, θ] with

, , and 5 r. So and

. Sample: Usingmathematical induction, letS(n): . S(1): , so S(1) is true. Assume S(k): . Then

5So S(k) ⇒ S( ), and S(n)is true for all integers n.

15. a.

b. rose curve

2 3

k 1 1zz zk11.zzk z zz z 5 zz zk zz z1

zzk11 z 5 zzk • z z 5zzk z 5 zz zk

zz1 z 5 zz z 5 zz z 1zzn z 5 zz zn

zz zn 5 rnzzn z 5 zrn z 5 rn,

zz zzn 5 [rn, nθ]r $ 0z 5

50π3 5

2π3 1 2(8)π

38π3 5

2π3 1 2(6)π;

26π3 5

2π3 1 2(4)π;

14π3 5

2π3 1 2(2)π;

z25 5 F225, 50π3 G

z19 5 F219, 38π3 G;

z13 5 F213, 26π3 G;

z7 5 F27, 14π3 G;

-

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1.

2. 216

3.4. closer

5. farther

6.

7.

8.9.

10.

11. a.

b.

-128 2 128Ï3 i

F4, π3G45 F256, π3G 5

-128 2 128Ï3 i

real

imaginary

z 3z 4

z 5

z 6

zz2

z 10z 7

z 8z 9

1

w9 ø F.39, π4Gw6 ø F.53, 3π

2 G;w3 5 F.729, 3π

4 G;r 5 (Ï2)4θ/π

real

imaginary

w3

w 4

w5

ww 2

2 4-4

-4

-2

2

4

-2

real

imaginary

z 3

z4

z 5z 6

zz2

z 10z 9

z 8

z71

512 2 512Ï3 i(cos12π

7 1 i sin12π7 )

F81, 4π5 G


c

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16. a.

b. limaçon

17. Let . Then and 2

5 which is animaginary number.

18. amplitude 5 3, period 5

phase shift

-3.5 # x # 3.5, x-scale = 1 -3 # y # 3, y-scale = 1

5 -π3

2π3 ,

0 1 2bia 1 bi(a 2 bi) 5 a 1 bi 2a 1 biz 2 z 5a 2 bi

z 5z 5 a 1 bi

1 2

19. b 20. c

21. a.b.c.d. no real solution

22. a.

b. . . . , will repeatthe values of , . . . , in such a way as

. . . , .c. See graph. ,

, d. The exponents of termsthat are equal arecongruent to each othermod 12.

z -3 5 z9z -2 5 z10z -1 5 z11

z12 5 z24z2 5 z14,z 5 z13,

z12z, z2z24z13 z14

real

imaginary

z 3z 4

z 5

z 6

zz 2

z 10 = z -2z 11 = z -1

z12

z 7

z 8

z 9 = z -3

4Ï134Ï13, -

3Ï13-

3Ï13


4. [2, 228°]5 5 [25, 5 228°] 5[32, 1140°] 5 [32, 60°] 532(cos 60° 1 i sin 60°) 516 1

5. a. squareb.

6. 1 real solution and 4 nonreal solutions.

7. 1 real solution and 8 nonreal solutions.

8. 2 real solutions and 162 nonreal solutions.

9. 729; 9, 9(cos 300° 1i sin 300°)

--

real1-1-1

1

imaginary

[2, 15˚]

[2, 285˚]

[2, 195˚]

[2, 105˚]

16Ï3 i

•

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1. For , 5 , which is z1.

2. ; ;

3. 3[cos (63°) 1 i sin (63°)]; 3[cos (135°) 1 i sin (135°)]; 3[cos (207°) 1 i sin (207°)]; 3[cos (279°) 1 i sin (279°)]; 3[cos (351°) 1 i sin (351°)]

real13

3

-1-1

1

imaginary

z0

z4

z1

z2

z3

z4 5z3 5z2 5z1 5z0 5

z0

z1

z5

z2

z3

z4

real1-1-1

1

imaginary

z5 5 Ï2 2 iÏ2z4 ø -.52 2 1.93i;z3 ø -1.93 2 .52i;z2 5 -Ï2 1 iÏ2;z1 ø .52 1 1.93iz0 ø 1.93 1 .52i

5 F3, 5π6 GF3, 17π

6 GF3, π6 12πk3 Gk 5 4

10. a. 16b. ,

11. a.

b.

c. They are the vertices of aregular n-gon, with centerat the origin, and a vertexat (1, 0).

imaginary

real

1

1

[1, 72°]

imaginary

real

1

1

Ï2-Ï2i, 1 2 i,-1 2 i,-Ï2Ï2 i, -1 1 i,

c

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[(.7)5, 400°], [(.7)6, 480°],[(.7)7, 560°], [(.7)8, 640°]b.

c. 0

17.

The graph of 3θ is aspiral of Archimedes, andthat of 3θ is alogarithmic spiral.

18. ; thus and

Therefore,

.824 2 .294i;

;

.

19. a.b. S(n): . S(1): , so S(1) is true. Assume S(k): . Then

5 2k11 1 15 2k11 1 2 2 15 2(2k 1 1) 2 1

ak11 5 2ak 2 12k 1 1ak 5

a1 5 21 1 1 5 3an 5 2n 1 1

an 5 2n 1 1

-.824 2 .294i 5zwø( z

w)-.824 1 .294iø3 2 2i

-4 1 izw 5

-øzw 5

3 1 2i-4 2 i

w 5 -4 2 i.w 5 -4 1 i,z 5 3 1 2i,z 5 3 2 2i

r 5

r 5

60

r = 3u

r = 3u

real

imaginary

a6a2

a3

a5a4

a1

1


12. a. 2, ,

b. ,

13. a.

b.

14. 1

15. The 3 cube roots of 8 are 2,their

sum is 1

16. a. [.7, 80°], [(.7)2, 160°],[(.7)3, 240°], [(.7)4, 320°],

(-1 2 Ï3 i) 5 02 1 (-1 1 Ï3 i)

-1 2 Ï3 i;-1 1 Ï3 i,

-

imaginary

real

3

3

-3Ï22 6

3Ï22 i3Ï2

2 63Ï2

2 i,

imaginary

real

3

3

z 5 63, 63i

real1-1-1

1

imaginary

[2, 180˚]

[2, 300˚]

[2, 60˚]

1 2 Ï3 i-2, 1 1 Ï3 i

real1-1-1

1

imaginary

[2, 240˚]

[2, 120˚]

[2, 0˚]

-1 2 Ï3 i-1 1 Ï3 i

c

c

So S(k) ⇒ S( ).Therefore by mathematicalinduction, S(n) is true for allpositive integers n.

20. a. ,

b. ;

; f(x) 5 -`limx→2/31

f(x) 5 `limx→`

f(x) 5 -`limx→-`

y 5 2x 1 3x 523

k 1 1 21. a.b. 4c. 1d. 1 and -4, by theTransitive Property ofPolynomial Factors

-x 2 1

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22. Sample:5 REM ENTER A COMPLEX NUMBER, A + BI, AND DESIRED

ROOT, N

10 PRINT “WHAT IS THE REAL COMPONENT, A, OF THECOMPLEX NUMBER”;

20 INPUT A30 PRINT “WHAT IS THE IMAGINARY COMPONENT, B”;40 INPUT B50 PRINT “WHICH ROOT DO YOU WANT”;60 INPUT N70 LET PI 5 3.1415926535980 IF A 5 0 AND B 5 0 THEN PRINT “0 IS THE ONLY

ROOT.”:GOTO 190

85 REM CALCULATE THE ARGUMENT, D, IN RADIANS OF A 1 BI

90 IF A 5 0 AND B 0 THEN LET D PI/2100 IF A 5 0 AND B 0 THEN LET D 5 PI/2110 IF A 0 THEN LET D 5 ATN(B/A)120 IF A 0 THEN LET D 5 ATN(B/A) 1 PI125 REM CALCULATE THE ABSOLUTE VALUE OF A + BI130 LET L = SQR(A * A + B * B)135 REM OUTPUT THE N NTH ROOTS OF A + BI140 PRINT “THE ABSOLUTE VALUE OF EACH ROOT IS”;L^(1/N)150 PRINT “THE ARGUMENTS ARE”160 FOR I 5 0 TO (N 2 1)170 PRINT (D/N) 1 I*(2 * PI/N)180 NEXT I190 END

,.

.5 -,

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1. a. , and

b. Since for a smallenough real number x,

for a large enoughreal number x, and p iscontinuous, theIntermediate ValueTheorem ensures thereexists a real number c such that .

2. 11

3. zeros: 0 (with multiplicity 4), 1, 5

4. zeros: 0, , all withmultiplicity 1

5. zeros: i, , both with

multiplicity 1

6. zeros: , both withmultiplicity 2

7. a. Sample:

b. Sample:

8. Sample:

9. a. 2b. 3c. 5d. 7

10. zeros: 2 (with multiplicity 2),

11. It has three more zeros.There can be three moresimple zeros, or one zero

1 6 Ï5 i

(x 2 (1 2 i))(x 2 (1 1 i)) •(x 1 1)

p(x) 5 (x 2 3)2 •

(x 2 4i)p(x) 5 8x(x 2 6)

(x 2 4i)xp(x) 5 (x 2 6)

6 i

-12

6Ï6

-

p(c) 5 0

p(x) . 0

p(x) , 0

p(x) 5 `limx→`

p(x) 5 -`limx→-`

with multiplicity 3, or onesimple zero and one zerowith multiplicity 2.

12. a. 7 b. 12 c. # 5

13. a. 256b.

14. a. [rn, nθ]b. [r, -θ]c. [rn, -nθ]d. [rn, -nθ]e. the conjugate of the nthpower of the complexnumber

15.5 0;

16. ∞17. a. 1, i, -1, -i, 1, i, -1, -i, 1

b. , and

c. i. -i ii. -1 iii. 1

18. a.

b.

c.

19. a. i. 14 ii. 1b. n DIV 7c. c MOD 12

p(x) 5 `limx→`

p(x) 5 `,limx→-`

p(x) 5 3x4(1 223x 1

13x3)

f(x) 5 3x4

i4k13 5 -ii4k12 5 -1,i4k11 5 i,i4k 5 1

4 2 4i 2 1 2 8 1 4i 1 5 5 0(2 2 i)2 2 4(2 2 i) 1 5 54 1 4i 2 1 2 8 2 4i 1 5(2 1 i)2 2 4(2 1 i) 1 5 5

[2, 45˚]

[2, 0˚]

[2, 315˚][2, 270˚]

[2, 225˚]

[2, 180˚]

z = [2, 135˚] [2, 90˚]


c

20. any one of: ;; or

21. -0.4301, -0.7849 1.307i

22. Sample: They all have 6complex zeros. They each will have 1 withmultiplicity m and 2x 5

x 5

6ø3 , x , 4-2 , x , -1

-5 , x , -4 with multiplicity n. The endbehavior of each function is the same: ,

. The graphs

differ in their “spread”depending on the choicesfor m and n.

p(x) 5 `limx→`

p(x) 5 `limx→-`

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c

c


1. a.b. 0

2. a.b.

3. a.5 5 0;

5

b. No, the coefficients of pare not all real numbers.

4. False

5.

6. (with multiplicity 2)

7. Sample: 1 15

8. Sample:

9. Sample: y

x

y

x

p(x) 5 x3 2 x2 2 7x

2 2 i, -2

1 2 iÏ32 , 4 2 5i

-25 1 20 1 5 5 020i2 1 525i2 2

p(-5i) 5-1 2 4 15 5p(i) 5 i2 1 4i2 1

-23 2 77i-23 1 77i

4 1 7i 10. Sample:

a. There are four nonrealzeros: either two conjugatepairs (each zero ofmultiplicity 1), or one conjugate pair (each zeroof multiplicity 2).b. There is one pair ofnonreal conjugate zeros,each of multiplicity 1.c. There are no nonrealzeros.d. There is one pair ofnonreal conjugate zeros,each of multiplicity 1.

11. 1, -1, i, -i

y

d.

c.

b.

a.

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12. a. Sample: 1

b. If p(x) does not have realcoefficients, then isnot necessarily a zero ofp(x), and p(x) may havedegree 2.

13. Sample: 1

14. a. i. 1, -1ii. 0iii. i, -i

b. They get closer togetheruntil they coincide at 0,then they split apart inopposite directions alongthe imaginary axis.

15. zeros: (each withmultiplicity 2), 1

16. a. 3b. -2 (with multiplicity 3), 3i (with multiplicity 2)

17.

18. a., b.

c. is the reflectionimage of over thereal axis, so 5 .z 1 wz 1 w

z 1 wz 1 w

y

x-2

-4

-2

2

4

6

z + ww

z

z + ww

z

≈ [1.59, 1.65]

≈ [1.59, 3.22]

≈ [1.59, .08]

≈ [1.59, 4.79]

6Ï3 i

c 5

3x 2 28x212x3 2p(x) 5

1 1 3i

14x 2 20x3 2 4x2p(x) 5 19. a. 3

b.20. not ((p and q) or r)

21. a. For , the zeros are, ; for 0, the

zeros are 0, ; for 3,the zeros are , . Asc slides from -5 to 0, its tworeal zeros move closer tothe origin, and its twoimaginary zeros convergeat the origin. As c thenslides from 0 to 3, thepolynomial has 4 real zeros.b. For 4, the zeros are

; for , the zeros

3 are . As c slides from

3 to 4, its positive zeros converge to , and itsnegative zeros converge to - . As c slides from 4 to

, its four complex zerosconverge to two complexzeros.

254

Ï2

Ï2

3 6 i2

c 52546Ï2

c 5

616Ï3c 562

c 56i6Ï5c 5 -5

2t2 2 4t 1 2


c

22.

For , the zeros areapproximately 2.50, 2.34,and .08 .93i. As cincreases, the real zerosmove slowly toward theorigin and the nonrealzeros move toward eachother until they merge into

6-

c 5 -5

-1 1 real

imaginary

-1

1

p-5p0

p-1p5

p5 p1p-5

p0p5

p5 p-1

p-5

p-5

p-1

approximately .11 (a zeroof multiplicity 2) when c-.05; at that time, the otherzeros are -2.33 and 2.13. Asc increases, the zero at .11splits into two real zeroswhich move apart alongthe real axis, while theother real zeros continuemoving toward the origin.When c 4.69, the twolargest zeros merge into1.53, then split intocomplex conjugates.

ø

ø

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c approximate zeros

5 2.50, .08 2 .93i, .08 1 .93i, 2.341 2.36, .10 2 .39i, .10 1 .39i, 2.17

.05 2.33, .11, .11, 2.130 2.32, 0.00, .20, 2.12

4.5 2.11, .93, 1.37, 1.674.69 2.10, .96, 1.53, 1.53

5 2.08, 1.00, 1.54 2 .19i, 1.54 1 .19i------

-------

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1. To 4 digits, the sequence is.5000, .7071, .8409, .9170, … . The sequenceapproaches 1.

2.

3. 7; 49; 2401; 5,764,801

4. [1, 10°], [1, 20°], [1, 40°], [1, 80°], [1, 160°], [1, 320°],[1, 280°], [1, 200°], [1, 40°],[1, 80°], …

5. [1, 1], [1, 2], [1, 4], [1, 8], [1, 16], [1, 32]

6. a. Let be the kth numberobtained (by pressing thecosine key k times). Then isapproaching x for larger andlarger k. But cos ,and cos approaches cos xsince the cosine function iscontinuous. Therefore,

cos x.b. 0.739

7. 1, 0

x 5

ak21

ak21ak 5

ak

ak

real

imaginary

a2

a3

a5

a4

a1 a0

1

real

imaginary

a6

a7

a2

a3

a5

a4a1a01

-12, 23, 3, -12, 23, 3

8. For all , let .

Then . And if

,

and

So

, hence the period is 3.

9. a. 0 b. No

10.

11. Sample: p(x) 510x

12. zeros: (with multiplicity 3), 2,

13.

;

;

2-2 1-1 real

imaginary

1

-2

2

-1

z3 5 2 (cos17π10 1 i sin17π

10 )z2 5 2 (cos6π

5 1 i sin6π5 );

z1 5 2 (cos7π10 1 i sin7π

10)z0 5 2 (cosπ

5 1 i sinπ5);

6Ï2i-2

x4 5 4x3 1 9x2 2

-4 -2 2 4

-4

-2

2

4

imaginary

real

a0

a3 5x

x 2 (x 2 1) 5x1 5 x.

511 2

x 2 1x

a3 5

x 2 1x ,1 2 x

1 2 x 2 1 51 2 x

-x 5

511 2

11 2 x

a2 5x Þ 0

a1 51

1 2 x

a0 5 xx Þ 1


c

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14. a. i. Sample: [6, 110°]ii. Sample: [243, 200°]iii. Sample: [32, 350°]

b.[7776, 550°].

[243, 200°][32, 350°] [(243)(32), 200°1 350°] [7776, 550°]. So .

15. a.

b. θ 5 0°

16.

17. 116.7 feetø1152π2

Ht1 5 1tn11 5 tn 2 4 for n $ 1

2

(zw)5 5 z5 • w55

5 • z5 • w5 5

[65, 5 • 110°] 5(zw)5 5 [6, 110°]5 5

18. If m is odd, then for some integer k.

Since is aninteger, is anodd integer by definition.∴ If m is any odd integer,then is an oddinteger.

19. f(z) is constructed bysquaring the absolute valueof the complex number zand doubling its argumentto obtain . The point isthen translated by adding cto give f(z).

z2

m2 1 m 2 3

m2 1 m 2 3(2k2 1 3k 2 1)

5 2(2k2 1 3k 2 1) 1 15 (4k2 1 6k 2 2) 1 11 (2k 1 1) 2 35 (4k2 1 4k 1 1)(2k 1 1)2 1 (2k 1 1) 2 3m2 1 m 2 3 5

m 5 2k 1 1

c

c

Answers for CHAPTER REVIEW pages 547–549

1.

2.

3. 2.5 cos 35° 1 2.5i sin 35°,(2.5 cos 35°, 2.5 sin 35°),2.5(cos 35° 1 i sin 35°)

4. ( 4, 0), [4, π], 4(cos π 1 i sin π)

5. ( 7, 5), , (cos 144° 1 i sin 144°)Ï74ø

[Ï74, 144°]ø-

-

F8, 7π4 G

(4Ï2, -4Ï2),4Ï2 2 4Ï2 i,

12(cos11π6 1 i sin11π

6 )(6Ï3, -6), F12, 11π

6 G, 6.

7. 25, θ 163.7°

8. b, θ

9. a. Sample:

b. Sample:

c. , n an

integer.

10. 11.

12. 13. 15 2 3i

14. 15.

16. 92 2 16i 17. 125

526 2

713 i-65 2 72i

64Ï5 i

23 2

Ï33 i4Ï3 i

F4, π2 1 2nπGF4, -3π

2 GF4, π2G

5π2zr z 5

øzz z 5

i sin3π2 )1

2 (cos3π2 1-12i, (0, -12),

167

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18. 16 1 36i 19. i

20. [20, 190°]

21. (cos 62° 1 i sin 62°)

22. [8, 220°]

23. a.

b. True

24. Sample: [ , 130.6°]

25.

26.27. 24.1 1 0i

28. 28 1 96i

29.

30.

i sin for 0, 1,

2, …, 5

31. 2 cos for

0, 1, 2, …, 9

32. a.

b.

33. a. [rn, nθ]b. (r(cos θ 1 i sin q))n 5rn(cos nθ 1 i sin nθ)

34. 243(cos 150° 1 i sin 150°) 5[243, 150°]

35. zeros: 0, 3, (both withmultiplicity 2)

36. zeros: 3, 2i (all withmultiplicity 1)

37. 1 1 i, 3-

66

F5, 11π12 G, F5, 19π

12 G-125Ï2

2 1125Ï2

2 i

n 5

πn5 1 2i sin πn

5 ,

n 5( π36 1

πn3 ))

3(cos( π36 1

πn3 ) 1

F4, π6G, F4, 5π6 G, F4, 3π

2 Gør 5 4, x 5 -2Ï3

F2, 11π6 G

Ï85ø

Q 5 (12, -Ï3

2 )P 5 (12, Ï3

2 ),

Ï21

38. a. with multiplicity 2

b. No, the Conjugate ZerosTheorem does not apply ifthe coefficients of thepolynomial are not all real numbers.

39. , so 5

So 5 .

40.(3 1 2i) , and zv 1

5 , so.

41. If and di, then 55 , which is a

real number.

42. cos θ 1 (r sin θ)i andcos( θ) 1 (r sin( θ))i 5

r cos θ 2 (r sin θ)i. So z andw are complex conjugates.

43. a. ; ;

b.5

44. 7

45. True

46.47. p(x) has real coefficients, so

the conjugate of 2i would

5 2 2i

(n 1 m)θ] 5 zn1m[rn1m,[rn • rm, nθ 1 mθ] 5

mθ][rm,zn • zm 5 [rn, nθ] •

(n 1 m)θ][rn1m,zn1m 5[rm, mθ]zm 5

zn 5 [rn, nθ]

--w 5 rz 5 r

-bd 1 0i-bdzw 5 bdi20 1

w 5z 5 0 1 bi

zv 1 zwz(v 1 w) 55 1 12i(14 1 5i)

zw 5 (-9 1 7i) 15 5 1 12i

z(v 1 w) 5 (3 1 2i) •

z 2 wz 2 w(4 1 i) 5 -1 2 3i.

5 (3 2 2i) 2z 2 w-1 2 3i.z 2 wz 2 w 5 -1 1 3i

i2


c

also be a zero. But p(x) hasdegree 3 and so cannothave 4 zeros.

48. a. 2 real, 2 nonrealb. i. cannot bedetermined

ii. 0iii. 1 2 2i

49. Its multiplicity is at least 2.

50. 1 2 3i volts

51. amps

52. volts

53. a.-d.

54. a.-e.

55. True

56. a. a parallelogramb.

slope of is

slope of

slope of ;

slope of

57.AB 5

-6 2 02 2 0 5 -3

515DB 5

-5 2 (-6)7 2 2

5 -3;-5 2 17 2 5CD 5

1 2 05 2 0 5

15;CA

D 5 B 1 C 5 7 2 5i

2 4

ad

c

e

b

2 8

2

4

-4

-2

4

imaginary

a

c

d

b

real6

-15 1 40i

15 2

185 i

58. a.,b.

c.

So 5

and

The ratio of similitude is . In polar

form,

and so all distances are multiplied by .d. The argument of is

348.7°, which is 225°greater than the argumentof F, which is 123.7°.

123.7°] 225°]5 348.7°] or applying a size change of and arotation of 225° to F toobtain .F9

Ï2[Ï26,

• [Ï2,[Ï13,(-2 1 3i) • (-1 2 i) 5

ø

øF9

Ï2

FÏ2, 5π4 G,z 5

Ï25 Ï2.

F9L9FL 5

Ï40Ï20

Ï2,Ï505 5

L9Y9LY 5

Ï74Ï37

5 Ï2,F9Y9FY

F9L9 5 Ï40.Ï50,L9Y9 5

F9Y9 5 Ï74,Ï20,FL 5

FY 5 Ï37, LY 5 5,

imaginary

real-2 4

-4

4F

F '

Y '

L

YL'

imaginary

zw

z

real4

20˚

70˚4

8

12

16

20

8

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Answers for CHAPTER REVIEW pages 547–549 page 3

c

c

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59. ; vertices: A 5 (0, 0), B 5 (2, 5), C 5 (3, 1), D 5 (1, 4);

slope of is slope of

slope of slope of 5

Since the slopes of opposite sides are equal, ABCD is aparallelogram.

60.θ 0° 30° 45° 60° 90° 120° 135° 180° 240° 270°

r 6 5.2 4.2 3 0 -3 4.2 -6 -3 0

61.θ 2 π 2π

r 1 3.8 1.3 .64 .42 .32

62.θ 0° 30° 60° 120° 180° 330°

r 5 5.8 10 -10 -5 5.8

1 3 5 7

øø

øøøøø 2 41 3

3π2

π2

π6

2 4 6 8

-øøø

-5 2 (-1)2 2 3 5 4BC4 2 (-1)

1 2 3 5 -52,DC 5

4 2 01 2 0 5 4,AD 55 -52,-5 2 0

2 2 0AB

--z 1 w 5 3 2 i


c

c

63.

8-leafed rose curve

64.

limaçon

0 # x # 2π, x-scale = -1 # y # 5, y-scale = 1

π]2

5

5

0 # x # 2π, x-scale = π-5 # y # 5, y-scale = 1

65.

is a logarithmicspiral, and θ is aspiral of Archimedes.

66. a.

b. farther

67. a.

b. closer

68. a. a pentagonb.

69. imaginary

real1

[1, ]2π]]3

[1, ]5π]]3[1, ]4π

]]3

[1, ]π]3

[1, π]

real1-1-1

1

imaginary

[2, ]37π}]20

[2, ]13π}]20

[2, ]21π}]20

[2, ]29π}]20

[2, ]π]4

1]2

1]4

w 4

w 3w 2 w 1

-2 2

-2

2

4

imaginary

real

z 5

z4

z 3z 2

z 1

r 5 4 1r 5 4θ

304

r = 4u

r = 4 + u

170

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171

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1. < -3.35 minutes/day

2. 3.5; longer

3.

4. , ,

5. ,

6. C to D or D to E

7. < -2.5

8. 5

9. A and F

10. a. 464 ft/secb. 475.2 ft/secc. 480.16 ft/sec

11. a.

b. 224 ft/secc. ft/sec

12. 4

13. 1

14. a.

b. The slope of the linethrough any two points onthe graph of f is always

.12

12

352 2 16∆t

-15 # x # 40, x-scale = 5 0 # y # 5000, y-scale = 1000

∆y 5 5∆x 5 2

f(x1 1 ∆x) 2 f(x1)∆x

f(x2) 2 f(x1)x2 2 x1

y2 2 y1

x2 2 x1

f(x1 1 ∆x); x1; x1 1 ∆x

15. a.

b. $25/computerc. $12.50/computer

16. a. 1:30 P.M.b. i, ii, iii could be true orfalse; iv is true, v is false

17. a. 639.984 ft/secb. 640 ft/secc. ft/sec

18.

19.

20. c

21. a. or b.c. -3, 3d. {y: }y $ -9

-3 , x , 3x , -3x . 3

9x 2 x2

3 2 27x2

π6 , x #

π2

800 2 32t

n s0 3001 3002 3003 3004 3005 3006 3257 3508 3759 400

10 425

1 2 3 4 5 6 7 8 9 10

100

200

300

400

500y

x

Sala

ry (d

olla

rs)

Number of Computers


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22. a. f(1) < 488.5; that is veryclose to the actual value of 490b. f(305) < 566; that is 17 minutes less than theactual value, or within 3%of the actual time.

c. For the shortest day, Dec. 21, , and forthe longest day, June 21,

. f(355) <483 minutes and f(172) <983 minutes.d. Answers will varydepending on students’latitude.

x 5 172

x 5 355



1. 480 ft/sec

2. a. Samples: 17.5 ft/sec, 22.5 ft/sec, 30 ft/secb.

slope ≈ 40 ft/secc. Sample: 40 ft/sec

3. a.

b.

-3 3

-3

3y

x

y 5 -52x 2 2

2Time (seconds)

Dis

tanc

e (fe

et)

4 6

30

60

90

120

150

d

t

Q1Q 2Q 3

P

c. Sample:

4. The derivative of a realfunction f at a point x is

, provided

this limit exists and is finite.

5. True

6. a. , ,

b. , ;

, ; ,

7. For a discrete function, youcannot find values of x sothat .∆x → 0

y 512x 2 3

at x 5 4y 5 2at x 5 0

y 5 -12x 2 3at x 5 -4

f9(4) ø 12

f9(0) 5 0f9(-4) ø -12

lim∆ x→0

f(x) 1 ∆x) 2 f(x)∆x

-3 3

-3

3y

x

c

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8. a.,b.

c. 2, 1.5, 1.25d. 1e.

9.

10. h9(30) 1 ft/sec; h9(60)

-.7 ft/sec; h9(100) ft/sec;

The fastest change is at.t 5 30

12ø

øø

5 16π

43π(12 1 6∆x 1 ∆x2)5 lim

∆x→0

43π(12∆x 1 6∆x2 1 ∆x3)

∆x5 lim∆x→0

43π(8112∆x16∆x21 ∆x)32

43π(8)

∆x

5 lim∆x→0

43π(2 1 ∆x)3 2

43π(2)3

∆x5 lim∆x→0

V(2 1 ∆x) 2 V(2)∆x5 lim

∆x→0

V9(2)

-1 # x # 4, x-scale = 1-2 # y # 10, y-scale = 2

-1 # x # 4, x-scale = 1-2 # y # 10, y-scale = 2

11. a. 1b. -1c. A9(0) is ;

this limit does not existbecause it has differentvalues when approachingzero from the right andfrom the left.

12. a. i. during the attack andslope times

ii. during the decay andrelease time

iii. Sample: at the breakpoint and during thesustain time

b. i. sound getting louderii. sound getting softeriii. constant volume

13. a. f9(7:05) ft/min;

f9(7:15) 1 ft/minb. At 7:05, the water level is falling at about 1.3 ft/min; at 7:15, the water level isrising at about 1 ft/min.

14. a. 3b. The average rate ofchange is the slope of thesecant line through and , but f is aline. Hence, the slope of thesecant line is the slope of f.

15. a.

b. ,

c. 34

f(24) 5 165f(20) 5 162

f(n) 5180(n 2 2)

n

x 5 2 1 ∆xx 5 2

ø

ø -97 ø -1.3

A(01 ∆x)2A(0)∆xlim

∆x→0


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16. Let S be the sequencedefined by , ; positive integers n. ShowS satisfies the recursivedefinition: ;

, ; positiveintegers .

; so S has the sameinitial condition as T. For alln,

. So, S satisfiesthe same recurrencerelation as T. Since in thisrelation is defined interms of by the RecursionPrinciple, they are the samesequence.

17. -0.5

18. 1

19.

20. -4

21. π

π3

Sn

Sn11

Sn 1 4n 1 2(2n2 1 1) 1 4n 1 2 52n2 1 4n 1 2 1 1 52(n2 1 2n 1 1) 1 1 5

Sn11 5 2(n 1 1)2 1 1 5

1 5 3S1 5 2 • 12 1k $ 1

Tk 1 4k 1 2Tk11 5T1 5 3

Sn 5 2n2 1 122. a. degrees Fahrenheit/min

b. i. The oven is heatingup slowly.

ii. The oven is heatingup quickly.

iii. The oven is coolingoff.

iv. The oven ismaintaining aconstanttemperature.

23. a. inches/secondb. i. when the bar is being

slowly liftedii. during rapid lifting

of the bariii. when the bar is being

lowerediv. when the bar is being

held steady, or whileit is on the floor

24. a. feet/secondb. i. when the jogger is

moving slowlyii. when the jogger is

running at top speediii. neveriv. when the jogger

stops or runs in place


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In-class Activity 1. 0

2. a. negativeb.

c. < .416

3. Sample: a. .456b. .496c. .500d. .500

e.4. See below.

Lesson

1. a. ,

b. ,

2. a. negativeb. zeroc. positived. zero

3π2 , x # 2π0 # x ,

π2

3π2 , x # 2π0 # x ,

π2

12

x 5π3

∆x 5 .0001, -.416∆x 5 .001, -.417∆x 5 .01, -.421∆x 5 .1, -.461

3. a.b. 408 ft/sec

4. a. < -.990b. cos 3 < -.98999

5. , ,

,

6. a.

b. , so.

7. a.

b. , so.g9(x) 5 10x 1 2

g(x) 5 5x2 1 2x 1 0

(10x 1 ∆x 1 2) 5 10x 1 2lim∆ x→0

10x∆x 1 ∆x2 1 2∆x∆x 5lim

∆ x→0

5(x 1 ∆x)2 1 2(x 1 ∆x) 2 (5x2 1 2x)∆x 5

lim∆ x→0

g9(x) 5

f9(x) 5 2 • 0x 1 3 5 3f(x) 5 0x2 1 3x 1 1

lim∆x→0

3∆x∆x 5 3

f9(x) 5 lim∆x→0

3(x 1 ∆x) 2 3x∆x 5

y

y = k'(x)

y = k (x)x1

2

-2-1-3-5 3 5

k9(2) ø 1k9(-1) ø 110

k9(-3) ø -1k9(-6) 5 0

v(t) 5 h9t 5 600 2 32t


Part x sin x cos x tan x f9(x)

1 1 0 undef. 0

2 2 < .909 < -.416 < -2.19 < -.416

3

4 < .707 < 1 Ï22

Ï22

Ï22

π4

12Ï31

2Ï32

π3

π2

4.

The values in the columns for cos x and f9(x) are either identicalor approximately the same.

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8.9.

10. a.b. 40c.

11. a. Samples:x f9(x)-1 40 21 02 -2

b.c. , ,

,

12. See below.

13. f and g are almost identical

-2π # x # 2π, x-scale = π-1.5 # y # 1.5, y-scale = 0.5

f9(2) 5 -2f9(1) 5 0f9(0) 5 2f9(-1) 5 4

f9(x) 5 -2x 1 2

-2 # x # 10, x-scale = 2-75 # y # 375, y-scale = 75

(5, 97)

g9(x) 5 8x

f9(t) 5 0

f9(x) 5 -6 14. a.b. Shows tangents at

, 0, and 1.

c. for all x. Thederivative of a constant iszero, so if ,then forany real number c.

15. a. ib. Velocity as a function oftime

16. a. square inchesper inchb. If the value is 2 inches ofwidth, then the area of theborder is increasing at therate of 52 square inches ofborder for an inch of width.

17.18. True

19. -2

x , -3

A9(2) 5 52

h9(x) 5 2x 1 0 5 2xh(x) 5 x2 1 c

f9(x) 5 g9(x)

-2 # x # 2, x-scale = 0.5-3 # y # 5, y-scale = 1

x 5 -1

f9(x) 5 2x, g9(x) 5 2x


12.

π5 4πr2

(3r2 1 3r∆r 1 ∆r2)43lim

∆r→05

43π(r3 1 3r2∆r 1 3r∆r2 1 ∆r3) 2

43πr3

∆rlim∆r→0

5

43π(r 1 ∆r)3 2

43πr3

∆rlim∆r→0

5

5(r 1 ∆r)2 1 2(r 1 ∆r) 2 (5r2 1 2r)∆rlim

∆r→0V9(r) 5

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20. a. an odd function; that is,

b. i.

ii. -1

iii. < 32

<

< 32

f(-x) 5 -f(x)21. < 21°

22. As , ; as , .

23. a.-c. If , then.f9(x) 5 3x2

f(x) 5 x3

g(x) → `x → `g(x) → -`x → -`



1. 61,800,000 is the averagerate of change in worldpopulation for the years1960 to 1965. It is found bytaking one-fifth of thedifference between the1965 and 1960 populations.

2. per year

3. 5,363,600,000 people

4. 10 mph/sec

5. True

6. False

7. a. increasingb. decreasingc. Neither; it is always -32 ft/sec2

8. a.15 9.8t m/secb.-9.8 m/sec2

9. per minute

or degrees Fahrenheit/min2

degrees Fahrenheitmin

a(t) 5 v9(t) 5 s0(t) 52v(t) 5 s9(t) 5

-440,000people

year

10.

a. < 8.12 degrees/minb. < -14.63 degrees/min2

c. at the beginning; d. at the end;

11. 7,742,000,000

t 5 10t 5 0

graph of f '

graph of f "

graph of f

-2 # x # 10, x-scale = 2-50 # y # 10, y-scale = 20

-2 # x # 10, x-scale = 2-180 # y # 180, y-scale = 80

-2 # x # 10, x-scale = 2-5 # y # 25, y-scale = 5

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12. , ,

13.14. S(1) 13 2(1) 3 has

3 as a factor so S(1) is true.Assume S(k) k3 2k has a factor of 3. Then S(k 1)k3 3k2 3k 1 2k2 (k3 2k) (3k2

3k 3). k3 2k has afactor of 3 and 3k2 3k3 3(k2 k 1) so it has a factor of 3. Thus, S(k) S(k 1) and byMathematical Induction n3 2n has a factor of 3 forall integers n.

15. See below.

16. a. all real numbers except or

b. removable discontinuityat , essentialdiscontinuity at x 5 -2

x 5 1

x 5 1x 5 -2

1

1⇒

11511

111115

1111151

15

515

12

x

y

-2 2 4 6-1

1

f9(6) < 12f9(4) < 1f9(1) < -2

c.17. a. III

b. IVc. IId. I

18. Sample:a. There were more housesbegun (to be built) in thecurrent period than in theprevious period.b. The unemployment levelstarted dropping moreslowly.c. The principle that anincrease in the demand fora finished product willcreate a greater demandfor capital goods.d. Two bodies attract eachother, and that force causesthem to come together atan ever-increasing velocity.

19. a.–b. Answers will vary.

x 5 -2

-5 # x # 3, x-scale = 1-15 # y # 5, y-scale = 5


15. cos 2x cos2x sin2xcos 2x (1 sin2x) sin2x Pythagorean Identitycos 2x 1 2 sin2x Addition2 sin2x 1 cos 2x

sin2x 1 2 cos 2x25

2525

22525

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1. a. or b.

2. increasing: ;

decreasing:

3. a.

b. all values; f(x) iseverywhere decreasingc. 0d. No; as the graph shows,f(x) merely “flattens out” at

. Since it is everywheredecreasing, f(x) has norelative maxima or minima.

4. a. for integers nb. for all integers n

5. decreasing

6. < 24 ft

7. , 5 w 5P4

h9(nπ) 5 0x 5 nπ

x 5 1

-5 # x # 5, x-scale = 1-25 # y # 25, y-scale = 5

f

f '

x .34

x ,34

-2 , x , 1x . 1x , -2 8. a.

b. f is increasing for or .c. f is decreasing for

.

d. -1, , (3, -4)

e.

9. No; for example, considerthe function graphedbelow.

As x goes from -4 to -3 to -2to -1 to 0, goes from -8to -6 to -4 to -2 to 0. Thoseslopes are increasing, butthe function is decreasing.

10. a

11. b

12. a

f9(x)

-4 -2 2 4

10

20

y

x

y 5 x2

-4 # x # 6, x-scale = 1-10 # y # 10, y-scale = 5

(-1, )20]]3

(3, -4)

)203(

-1 , x , 3

x . 3x , -1

f0(x) 5 2x 2 2


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13. a. The curve is increasingfrom April to July and fromOctober to December.

b. i. Trueii. Falseiii. Trueiv. False

14. a. v(3) 15 ft/secb. a(3) 4 ft/sec2

15.16. -6 1 3∆x; -5.7

f9(1) 5 -2e-1 < -.74

55

24

20

Jan Apr July Oct

17. a. < 4.4°b.

18. a.b.c.

19. .75

20. a. 6b. 7!c. !

21. a. Sample: isincreasing on the set ofreals but at , .b. Sample: isdecreasing on the set ofreals, but at , .f9(x) 5 0x 5 0

f(x) 5 -x3f9(x) 5 0x 5 0

f(x) 5 x3

(n 2 1)

h + k(x) 5 x, x $ 3k(x) 5 ln(x 2 3) 1 2h(x) 5 ex22 1 3

-12 -6 6

-24

-18(-6, -19)

-12

-6

6y

xy = 4x + 5

y = 3x – 1


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1. a. f(4) < 12.6; there areabout 13 bacteria at 4 hoursb. f(6) f(3) 20 1010; f(3) f(0) 10 5 5;so f(6) f(3) is doublef(3) f(0).

2. a. Sample: (0, 1), (1, e), (2, e2), (3, e3)b. 1, e, e2, e3

3. a. < 7.77, < 7.43, < 7.39b. e2 < 7.389

4. Given , let and

. Then, .

Hence, .

5.

6.

7. sect ,34

f9(t) 5 k(f(t) 2 a0)

g9: x → (ln 3)3x

g0.3

-2 # x # 2, x-scale = 0.5-4 # y # 8, y-scale = 2

g0.5

g0.1

f(x)ex ⇒ f9(x) 5f(x) 5

(ln e)f(x)1 • ex ⇒ f9(x) 5f(x) 5b 5 ea 5 1(ln b)f(x)abx ⇒ f9(x) 5f(x) 5

22

52525252

8. a. and b.c.

9. a. -7.5 mph/sec; < -4.2 mph/secb. The car is deceleratingmore rapidly during thefirst four seconds ofbreaking than during thelast six seconds.

10. a. , 1, and 2 secondsb. vertical; over the zeromark

c. and seconds

d. at the extremese. , 1, and 2 secondsf. vertical; over the zeromark

11.12. (-1.5, -8.5)

13. a. at about 9.53 secondsb. about 305 ft/secc. about 208 mph

14. a.b. -13

15. ; z 5 32in 5 5

m 5 -13 2 3∆x

t 5

f9(1) 5 -1

t 5 0

32t 5

12

t 5 0

(3, -3)

3-2

-4

(1, 1 )2]3

(-1, 7 )2]3

4

y

x

-1 , x , 3x . 3x , -1


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16. Sample:

642-4-6 -2-2

8

2

64

y

x

17. 1000, 100000, 11000

18.19. Sample: Kirchoff’s law for

circuits: where E is a voltage source,L is an inductance, R is aresistance, and I is current.

E(t) 5 LI9 1 RI

b 5 e2

Answers for LESSON 9-6 pages 587–591c


1. 9

2. a.b. i. 3.31

ii. 3.0301

3. a.b. -4

4.5.6.7.8.9. False

10. increasing

11. . Since for all real x,

for all real numbers.Since the derivative ispositive, the slopes of thetangents to the curve areall positive, and thefunction is increasing for allreal numbers.

12. decreasing

3 $ 0f9(x) 5 3x2 1

x2 $ 0f9(x) 5 3x2 1 3

k9(x) 5 -6x 1 2

g9(x) 5 -6x

f9(x) 5 2

f9(x) 5 -2x 1 1

f9(1) 5 4, f9(-1) 5 -4

-4t 1 5 2 2∆t

3 1 3∆x 1 (∆x)2

13. a. i.ii.iii.

b.

14. a. 1 min/oz; an extraminute of cooking time isrequired for everyadditional ounce ofpotatoes.

b. min/oz; as weight

increases, the rate ofchange of baking timeneeded decreases for everyadditional ounce ofpotatoes.c. Potatoes weighingbetween 10 and 16 ouncesneed less cooking time perounce than potatoesweighing between 4 and 6 ounces.

d. minutes1623

23

-5 # x # 5, x-scale = 1-50 # y # 50, y-scale = 10

x 5 -2 or x 5 3-2 , x , 3x , -2 or x . 3

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15. a.

b. increasingc. 1983–1984

16. a. : grams/day; : grams/day2

b. < -25.5 grams/day; at 7 days, the amount ofradon present is decreasingby about 25.5 grams/dayc. 5 days

17. a. -64 ft/secb. -32 ft/secc. 2.5 secd. -80 ft/sec

18. a. i. 1 mi/minii. 0 mi/miniii. 1 mi/miniv. 0 mi/min

b. It is stationary.c.

d. positive: 0 x 2, 5 x 8; negative: 2 x 5, 8 x 12

19. a. to the rightb. slowing down

,,,,,,

,,

2 4 6

1

8 10 12Time

Velo

city

A0(t)A9(t)

1980 1985 1990 1995

306090

YearN

umbe

r of S

tock

sTr

aded

(bill

ions

) 20. a. < 119 ft

21. a. (4 cos u)

(4 sin u) 8 sin u cos u4(2 sin u cos u) 4 sin 2u

b.

22. a.

b.

c. to d. Sample: to

23. -3

24. a. 3b. 0c. -1

25. a. , ,, ,

b.

26. a

27. Sample:

2

2 4

4y

x

-4 4

3

-3

g9(5) < -3g9(3) < 0g9(0) < 2

g9(-2) < 0g9(-4) < -1

x 5 1x 5 -4x 5 5x 5 1

32

-13

u 5π4

555

• A(u) 512xy 5

12


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28. a.b.c.d. , ,

e. negativef9(6) < -2

f9(3) < 0f9(0) < 1.5x 5 -1, x 5 3x , -1, x . 3-1 , x , 3 29. a. increasing: ,

; decreasing:

b.c. positive: ;negative: -5 , x , -1

-1 , x , 5x 5 -3, x 5 1

-3 , x , 11 , x , 5

-5 , x , -3


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1. determining whether ornot the order of symbolscounts and whether or notrepetition of symbols isallowed

2. A string is an ordered list ofsymbols.

3. Ordered symbols; repetitionis allowed.

4. Ordered symbols; repetitionnot allowed.


6. Unordered symbols;repetition not allowed.



9. Unordered symbols;repetition is allowed.


11. Offices are ordered, andrepetition is not allowed.

12. a. Samples:

Route A

Home

Route B

School

b. for the Samples in part a,SSSESESSEE, EESESSSESSc. Ordered symbols;repetition is allowed.

13. Ordered symbols; repetition not allowed.

14. a. Because the cosinefunction has period 2π, anyinterval of length 2πcontains all possible valuesof the function.

b.

c. ,

d. Transform the polarequation into a rectangularequation:

which isnot standard form for anellipse.

x 5 6(x2 1 y2)3/2

x2 1 y2 5 xÏx2 1 y2

Ïx2 1 y2 5 6Ï xÏx2 1 y2

r 5 6Ïcos u

0 # u # 2π, u-step =-1.5 # x # 1.5, x-scale = 0.5-1.5 # y # 1.5, y-scale = 0.5

π]6

r 5 -Ïcos ur 5 Ïcos u

π2 , u ,

3π2


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15.

It is an identity.Left side (tan u)(sin ucot u cos u)

(sin u

Right side55 sec u

51

cos u

5sin ucos u ( 1

sin u)(sin2u 1 cos2u

sin u )5sin ucos u

cos u cos usin u15

sin ucos u

15

-2π # x # 2π, x-scale = 1 -10 # y # 10, y-scale = 2

16. c

17. a. 142

b.

18.19. ' a positive integer n that is

not prime.

20. ; positive integers n and m,.

21.22.23. (3) 10,000; (4) 648; (5) 720;

(6) 286; (7) 5,527,200; (8) 360; (9) 625

8x3 2 12x2y 1 6xy2 2 y3

x2 1 2xy 1 y2

nm Þ 11

x , -1, x . 0n

d


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1. 4

2. 9 possible code words

3. 12 different shirts

4. 6 different seatings

5. 5 ways to complete theseriesgame

A

T

A

5

T

A

6

T

A

7

T

4

Chair 1

Luis

Van

Luis

Charisse

Van

Charisse

Charisse

LuisVan

Chair 2 Chair 3

Luis

Van

Van

Charisse

Luis

Charisse

Select colorwhite

blue

red SMLXL

SMLXL

SMLXL

Select size

Selectfirst

letterA

ABC

ABC

C

ABC

B

Selectsecondletter

CACBCC

AAABACBABBBC

CodeWord

6. a. 360b. 1296

7. 8n

8. It would be impractical torepresent all 3,276,000possible outcomes

9. 3168

10. a. 10,000b. dw

11. 1716

12. 84

13. a. 555b. < 0.0054

14. a. There are two choicesfor each element: include in the subset, or don’tinclude it. Since there are n elements, by theMultiplication CountingPrinciple there are n factorsof 2, or 2n subsets.b. 15c. a spoonful of cereal (andmilk) only; no fruit


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15. S(1) is true, because thenumber of ways the firststep can be done is n1.Assume S(k), the number ofways to do the first k steps,is . Let m bethe number of ways to dothe first k steps and let nrepresent the number ofways to do the ststep. Then by the inductivehypothesis, m n1 • n2 •K • nk and so mn n1 • n2 • K • nk • .So S(k 1), the number ofways to do the (k 1) stepsis n1 • n2 • … • nk • .Thus S(n) is true for all n.

16. Ordered symbols; repetitionis allowed.

17. Ordered symbols; repetitionis not allowed.

nk11

11

nk115n 5 nk11

5

(k 1 1)

n1 • n2 • K • nk

18.

19. a. 2485b. 15,050c. 49,495,500

d.

20.

21.

22.


6253

-67

m!n!

n2(n 1 1) 2

m2 (m 1 1)

5 m

5m∆x∆x

5mx 1 m∆x 1 b 2 mx 2 b

∆x

5m(x 1 ∆x) 1 b 2 (mx 1 b)

∆x

∆y∆x 5

f(x 1 ∆x) 2 f(x)∆x


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1. a.

b. Yes

2. 720

3. a. 362,880b. 17,280c. 8640

4. 60,480

5. 210

6. 362,880

7. 60

8.

9. a.

P(10, 10)b. ; integers , P(n, n 1) P(n, n)

10. a. 46,656 b. 42,840

11. a. 120 b. 625

12. 10,080

13. 28,800

14. For , the number ofpossible permutations is1 1! n! Assume thetheorem is true for n k.That is, there are k!permutations of k differentelements. Consider nk 1. There are k!1

5

555

n 5 1

52n $ 2

10!1! 5

10!1 5

10!0! 5

10!(10 2 0)! 5

10!(10 2 9)! 5P(10, 9) 5

n 5 3

amth

amht

h

t

tm

h

atmh

athm

h

m

mta h

ahmt

ahtm

t

m

mh

t

permutations of the first kelements by the inductivehypothesis. Then there arek 1 places to insert the (k 1)st element into anyof these permutations. Soby the MultiplicationCounting Theorem, thereare k!(k 1) (k 1)!permutations. Hence, bymathematical induction,the Permutation Theorem istrue.

15. a. (n 2)! (n 2)(n 3)(n 4)(n 5) K(2)(1) (n 2)[(n 3)(n 4)(n 5) K (2)(1)](n 2)(n 3)!b. n 7

16. 24

17. Sample: n 11, r 9

18. 480

19. 6 trips

20.

21. cm

22. f(x) 2x3 14x2 12x144

23. (x, y) → (x 1, 2(y 7))


12

2215

403

-2429 2

2729 i

Denver

SanFrancisco

M–PP–M

MinneapolisD–PP–D

PhoenixD–MM–D

55

522

522225

222252

151

11


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1. A combination of elementsof a set S is an unorderedsubset without repetitionallowed. A permutation is asubset of S which isordered.

2. a. C(n, r), , nCr

b.

3. a. abcd, abce, abde, acde,bcdeb. 24c. 120

4. 35

5. 792

6. 161,700

7. a. 201,376 b. < .00000497

8. 7

9. a.

b. If there are n objects,there is only one way tochoose all n of them.

10. 241,500

11. 66

12. a. 16b. 32c. 64

13. 301,644,000

n!n! • 1 5 1n!

n!0! 5

n!n!(n 2 n)! 5C(n, n) 5

nCr 5n!

r!(n 2 r)!

(nr ) 5

n!r!(n 2 r)

C(n, r) 5n!

r!(n 2 r)!

(nr )

14. a. i. 10 • 9 • 8 • 75040 (24)(210)(4!)(210)

ii. 33 • 32 • 31 • 30982,080(24)(40,920)(4!)(40,920)

iii. 97 • 96 • 95 • 9483,156,160(24)(3,464,840)(4!)(3,464,840)

b.

, and

C(n, 4) is an integer. So, the product of any 4 consecutive integers n, , ,and is divisible by 4!.

15. a. 161,700b. 152,096

16. the number of different 7-card hands

17. a. 7!b. 2520c. 16,807

18. 117,600

19. Sample:y

x

4

2

2-2 4 6

-2

n 2 3n 2 2n 2 1

n(n 2 1)(n 2 2)(n 2 3)4!

n!4!(n 2 4)! 5C(n, 4) 5

55

5

55

5

555


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20. a0x4 a1x3y a2x2y2

a3xy3 a4y4

21. a. sin (a b) sin (a – b)

(sin a cos b sin b cos a)(sin a cos b – sin b cos a)

sin2 a cos2 b sin a sin bcos b cos a sin a sin bcos a cos b sin2b cos2a

sin2 a cos2 b sin2 b cos2 asin2a cos2 b cos2 a sin2 b

b. sin2 a cos2 b cos2 a sin2 bsin2 a cos2 b sin2 a

sin2 b sin2 a sin2 bcos2 a sin2 b

sin2 a (cos2 b sin2 b)sin2 b (sin2 a cos2 a)

sin2 a sin2 b251

215

22152

2525

21

25

15

1

1111 22. If the pants are not blue,

then the coat is not green.

23. x3 3x2y 3xy2 y3

24. Sample: Suppose there are31 flavors.

a.

b. 302 (1 1 30) 5 465

312 (1 1 31) 5 496

111


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1. a. 1, 7, 21, 35, 35, 21, 7, 1b. a7 7a6b 21a5b2

35a4b3 35a3b4 21a2b5

7ab6 b7

c. a7 7a6b 21a5b2

35a4b3 35a3b4 21a2b5

7ab6 b7

2.

3.

4. a. (2a b)6 64a6

192a5b 240a4b2

160a3b3 60a2b4

12ab5 b6

b. 64 192 240 16060 12 1 729 36

5. 56875s12r3

6. -225, 173, 520 x4y9

7. a. -4b. 16

8. < 2.13 1010

9.10. The outside right diagonal

(the last term in each row)is 1.

(x 1 2)7

3

55111111

11111

151

(99) y9(9

8) xy8 1(97) x2y7 1

(96) x3y6 1(9

5) x 4y5 1

(94) x5y 4 1(9

3) x6y3 1

(92) x7y2 1(9

0) x9 1 (91) x8y 1

x92kyk 5on

k50(9k)(x 1 y)9 5

(3010)

2121

2121

111111

11. a. nCr Calculation Theoremb. forming a least commondenominatorc. addition of fractions andDistributive Propertyd.e. nCr Calculation Theorem

12. is the coefficient of

in the expansion of . But

, so is also the

coefficient of , which is

. So .

13. 126

14. 462

15. 70,073,640

16. a. 10b. 32

17. a. Ordered symbols;repetition is not allowed.b. < 1.0897 1010

18. a.b. removablec. noned.

-4 -2 2 4

-4

-2

2

4

y

x

x 5 3

3

( nn 2 r)(n

r ) 5( nn 2 r)

xn2(n2r)yn2ryn2rx r 5

(nr )(y 1 x)n

(x 1 y)n 5(x 1 y)nxn2ry r

(nr )

(n 1 1)!


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c

19. c

20.row sum of squares

of elements

0 11 22 63 20

The sum of the squares forrow n seems to be themiddle element of row 2n.So the sum of squares ofthe elements of the 12throw would be the middleelement of row 24.



1. a. 16b. {1, 2, 3, 4}, {1, 2, 3}, {1, 2},{1}, { }, {1, 2, 4}, {1, 3}, {2}, {1, 3, 4}, {1, 4}, {3}, {2, 3, 4},{2, 3}, {4}, {2, 4}, {3, 4}

2. a. 56 b. 93

3. < .00597

4. a. 1b. 5c. 10d. 10e. 5f. 1

5. 32

6. If , then 20 1 and

, so

.

7. a. < .0060b. < .0403c. < .1209d. < .2150

8. 64

o0

k50(nk) 5 2n

(00) 5

0!0!0! 5 1o

0

k50 (0

k) 5

5n 5 0

9. < 0.088

10. < 0.790

11. a. The number of 5-element subsets of S1 plus the number of 4-element subsets of S1 isthe number of 5-elementsubsets of S.b. Let S be a set of elements and let S1 be a setof n of these elements. Thenevery r-element subset of Sis either an r-element subsetof S1 or an (r – 1)-element ofS1 along with the left-overelement not in S1. So

.

12. a. < .5177b. < .0278c. < .4914d. No

(n 1 1r ) 5 (n

r ) 1 ( nr 2 1)

n 1 1

c

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13. The expression

represents the coefficientsof the expansion of . Letting , we obtain

, sothe sum of the coefficientsmust be zero.

14. 31

15. 128a7 448a6b672a5b2 560a4b3

280a3b4 84a2b5

14ab6 b7

16. 1.2944 109

17. Each style is available in3000 fabrics.

18. f is increasing when or when .x . 2

x , 0

3

212

1212

(x 2 y)n 5 (1 2 1)n 5 0x 5 y 5 1

(x 2 y)n

(n2) 2 (n

3) 1 K 6 (nn)

(n0) 2 (n

1) 1 19. a.

b. eight-leafed rose curvec. Symmetry with respect

to the origin, , ,

20.


limt→∞

g(t ) = ∞g(t )

t

limt→-∞

g(t ) = -∞

u 5π4

u 53π2u 5

π2

1 2

Answers for LESSON 10-6 pages 633–638 page 2cc

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In Class Activity 1. a. (x y z)2 x2

2xy 2xz y2 2yz z2

If x 2, y 3, and z 5:(2 3 5)2 (10)2 100;(2)2 2(2)(3) 2(2)(5)(3)2 2(3)(5) (5)2 100

b. 6

2. a. x2, xy, xz, y2, yz, z2

b. x2 2xy 2xz y2

2yz z2

c. Ball1 in box x and ball2 inbox z, or ball1 in box z andball2 in box xd. They are the variableparts of the expansion.

3. a. x3 3x2y 3x2z3xy2 6xyz 3xz2 y3

3y2z 3yz2 z3

b. 10c. Same as part a

4. a. x4 4x3y 4x3z6x2y2 12x2yz 6x2z2

4xy3 12xy2z 12xyz2

4xz3 y 4 4y3z 6y2z2

4yz3 z4

b. 15c. 6

5. a.

b. Sample: (x y z)6 has

terms

c. 105

(86) 5 28

11

(n 1 2n )

11111

111111

111

111111

111

11111

511111

55115551111

1511


c

Lesson 1. 105

2. 105

3. 36

4. ❍❍_❍_❍❍❍_❍_❍

5. ❍_ _❍❍❍❍❍_ _❍❍

6. 1, 1, 1, 2, 3

7. 0, 3, 2, 3, 0

8. 1, 0, 3, 3, 1

9. 495

10. 55

11. 1365

12. 84

13. 1820

14. 4, 5

15. 20

16. 126

17. 1771

18. < .394

19. 448

20. 1,792,505

21. a. 120b. 48

22. 19

23. 10

24.

25. (( ) and p) q⇒p ⇒ q

u 5 tan-1 30d

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26. a. 56b. 56c. 56d. 56e. The problem of findingthe number of terms in a–dis equivalent to theproblem of finding thenumber of sequences of 4nonnegative integers whichadd to 5.

f. To find that number,

evaluate for

and .

.8 • 7 • 63 • 2 • 1 5 56

(85) 5(5 1 (4 2 1)

5 ) 5

n 5 4r 5 5

(r 1 (n 2 1)r )



1. a. 21b. 10

2. a. m3 3m2n 3m2p3mn2 6mnp 3mp2

n3 3n2p 3np2 p3

b. 8x3 12x2y 12x2

6xy2 12xy 6x y3

3y2 3y 1

3. a. 7560b. 1030

4. 13,860

5. The coefficient of in the expansion of (x1

x2 x3)n is equal to thenumber of choices in thefollowing countingproblem. A set has nelements. You wish tochoose a1 of them. Then,from the remaining n a1,2

11

x1a1x2

a2x3a3

222112

121111

111111

you choose a2. Then, fromthe n a1 a2 thatremains, you choose a3. Thenumber of ways to makethe 3 selections is

.

,

since

6. For , in the expansionof , the coefficient

of is ,

which is a restatement ofthe Binomial Theorem.

7. a. 900b. 0.0009

(nr )n!

(n 2 r)!r! 5xn2ry r

(x 1 y)nk 5 2

0! 5 1.n 2 a1 2 a2 2 a3)! 5

n!a1!a2!a3!

(n 2 a1 2 a2)!a3!(n 2 a1 2 a2 2 a3)! 5

(n 2 a1)!a2!(n 2 a1 2 a2

n!a1!(n 2 a1)! •

( na1

)(n 2 a1a2

)(n 2 a1 2 a2a3

) 5

22

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8. a. Think of the exponent 6as 6 identical balls. Think ofthe variables x, y, z, and was boxes. Each distributionof all six balls into the fourboxes can be thought of asa term in the expansion of(x y z w)6.b. Suppose that S is the set{x, y, z, w}. The number of6-element collectionsselected from these 4elements of S gives thenumber of terms in theexpansion of (x y zw)6.c. 84

9. 495

10. < 0.19

11. 4,838,400

12. 4,096,000,000

13. ; u < 5.44y < -4.47

111

111

14. S(1): a1 c, so a1 is divisibleby c.S(2): a2 c, so a2 is divisibleby c.Assume S(k): ak is divisibleby c, for all .Then

for someintegers p and q

, where is an integer.So is divisible by c.So S(k) S( ), and,therefore, every term isdivisible by 3.

15. a. The degree of d(x) is oneless than the degree of p(x).b. p(x) (3x 2) d(x)x 2, so some samples are:d(x) p(x)

x 3x2 3x 2x2 3x3 2x2 x 2

x 5 3x2 18x 12

16. Samples:

deed:

noon:

tomtom:

deeded: 6!3!3! 5 20

6!2!2!2! 5 90

4!2!2! 5 6

4!2!2! 5 6

111111

11

1115

k 1 1⇒ak11

q 2 4p5 c(q 2 4p)

5 cq 2 4cpak11 5 ak 2 4ak21

k # n

5

5


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1. unordered symbols;repetition allowed

2. unordered symbols;repetition not allowed

3. ordered symbols; repetition not allowed

4. unordered symbols;repetition not allowed

5. ordered symbols; repetition allowed

6. 4080

7. 2002

8. 18,564

9. 60,480

10. 36

11. x8 8x7y 28x6y2

56x5y3 70x4y4 56x3y5

28x2y6 8xy7 y8

12. 16a4 96a3b 216a2b2

216ab3 81b4

13. -10240

14. 78750x5y 4

15.

16.

17. For each of the C(n, r)combinations of n objectstaken r at a time, there arer! arrangements. So

P(n, r), or P(n, r) r! C(n, r).5C(n, r) • r! 5

n!0! 5 n!P(n, n) 5

n!(n 2 n)! 5

C(n, n 2 r)

n!(n 2 (n 2 r))!(n 2 r)! 5

C(n, r) 5n!

r!(n 2 r)! 5

1212

11111

111

18. Add the coefficients;

.

19.

20. a. 308,915,780b. 255,024

21. 5040

22. 1000

23. < 1.2165 1017

24. < 5.0215 1014

25. < 1.40 1010

26. 87,500

27. 120

28. 36

29. a. 1140b. 8000

30. 47,775

31. 210

32. 50C23 < 1.0804 1014

33. 26

34. a. C(2000, 50)b. C(620, 12) • C(580, 12) •C(450, 13) • C(350, 13)

35. 66

36. < 0.0879

37. a. < 0.0163b. < 0.181c. < 0.6325

3

3

3

3

(204 )

o10

k50(10

k ) 5 210

Answers for Chapter Review pages 653 –655

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38. 14 strings

39. 8 numbers

33

214

44

12

3

11

23

4

22

143

defg

d

e

f

g

defg

deg

def

40. 14 outcomes

AA

A A

A AA

B B BB B B

A A AA A A

B B BB

B BB

Answers for Chapter Review pages 653 –655 page 2c

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1. a.

b. Sample: A, a, B, b, A, e,D, g, C, c, A, d, C; no

2. Sample:

3. edge, vertex

4. Sample:

5. Sample:

start

finis

h

start

finis

h

start

end

C

B

D

g

A e

ba

dc

6. a.

b. does not affect it

7. 320 students

8. .42

9. < .43

10. a.

b. 19%c. 5.8%

11. Euler’s

12. a. 39b. 38

13. a. ; ; ;;

b. .24c. .25d. .10

14. Yes

z 5 .45y 5 .01x 5 .50w 5 .30v 5 .90

Contacts.048

Contacts.010

NoContacts.752

NoContacts.190

AllCustomers

Male

Female

.8.94

.06

.05

.95

.2

3 5 12 3 7

95

15

7

1

4

3

A B

D

H

L

I

K

G J

C E F


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15. Sample:

16. a. Sample:

b. There is a relativeminimum at .x 5 -2

-4 -2 2

2

6

10y

x

polygons

quadrilaterals

parallelograms

squares

kites

rhombuses

trapezoids

isoscelestrapezoids

rectangles

17. a.b. 0c. 4

18. a. -tan xb. tan xc. -tan x

19. 7

20.1. nonreversible2. reversible3. reversible

21. a. Samples: , b. One dimension must beeven for an array to betraversed.

5 3 53 3 3

y 5 53y 5 153y 1 1 5 16

Ï3y 1 1 5 4

3 3 4


1. a. Yes b. 4 edges, 4 vertices

2.

3. a. Yesb. Noc.d. and , and , and e. ,

4. False

5.

6.

7.

8. 11 vertices, 11 edges

110

102

020Gv1

v2

v3

Fv3v2v1

v2

v3

v4

v1

v1

v2

v3

v4

F0001

1001

0110

0110G

v4v3v2v1

e4e3

e6

e5e2e1e4e3

v4

e3

e5e1

e2

e6

e4

v2v1

v5

v3

v4

9.

10. a. Nob. There are two edgesbetween v1 and v3. Thoseedges are parallel, so thegraph is not simple.

11. edge endpointe1 {v1, v2}e2 {v2, v4}e3 {v2, v4}e4 {v3}

12. Yes, they have the samelists of vertices and edgesand the same edge-endpoint function.

13. a. ' a graph G such that Gdoes not have any loopsand G is not simple.b. True: G could haveparallel edges and no loops.

a b

dc

a b

dc

a b

dc

a b

dc

a b

dc

a b

dca b

dc

a b

dca b

dc

a b

dc


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14.

15. The tester prefers A to C,and prefers C to P. But in adirect comparison of A andP, the tester prefers P.

16. a.

b. < .396c. < .474

17. a. 18 b. 17 c.18. Sample:

7x 1 15p(x) 5 x3 2 x2 2

V 5 E 1 1

ChocolateChip

PeanutButter

ChocolateChip

Vanillawafer

redjar

greenjar

5]]12 5

]]24

7]]24

3]]16

5]]16

7]]12

3]8

5]8

1]2

1]2

v4e2

e1

e6

e5

v1 v2

e4

e3

v3

19. quotient: ; remainder:

14,043

20. a.

b.n

21. a. i. 1ii. 4iii. 6iv. 4v. 1

b. i. 1ii. 3iii. 3iv. 1

c. Pascal’s triangle;binomial coefficientsd. 1, 2, 1 is row 3 of Pascal’striangle.

Fcos (nu)sin (nu)

-sin (nu)cos (nu)G

5Fcos usin u

-sin ucos uG

Fcos 2usin 2u

-sin 2ucos 2u

G287x 2 2006

6x3 2 42x2 1


1. a. ; ;;

b. 8

2. The statement does notinclude the case whenedges are loops which arecounted twice.

3. a. 2 b. v1 c. 1d. v2 e. 1 f. v1

g. contributes 1 to thedegree of v3 and 1 to thedegree of v1.h. contributes 2 to thedegree of v3.

4. 253

5.

6. a. 28b.

c. It is equivalent to 8 people shaking handswith each other.

7. 42

8. 8

deg(v4) 5 3deg(v3) 5 0deg(v2) 5 3deg(v1) 5 2 9. Assume that each of the

nine people could shakehands with exactly threeothers. Represent eachperson as the vertex of agraph, and draw an edgejoining each pair of peoplewho shake hands. To saythat a person shakes handswith three other people isequivalent to saying thatthe degree of the vertexrepresenting that person is 3. The graph would thenhave an odd number ofvertices of odd degree. Thiscontradicts Corollary 2 ofthe Total Degree of a GraphTheorem. Thus, the givensituation is impossible.

10. The total degree of anygraph equals twice thenumber of edges in thegraph.

11. counterexample:

12. Impossible; it can’t have anodd number of oddvertices.

13. Sample: v1v3

v2 v4

v1 v2

e2

e1


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14.

15. Impossible; one of thedegree 3 vertices goes toeach of the other vertices.But the other degree 3vertex cannot connect toitself (the graph is simple),to the first degree 3 vertex(no parallel edges), or tothe other two vertices (theyalready have one edge).

16. a.b.

c.

17.

18. a.

b. It is the same problembut with one more person.

19. “Twice the number ofedges” must be an evennumber.

(n 1 1)n2

n(n 2 3)2

n(n 2 1)2

n(n 2 1)n 2 1

v1 v3

v4v2

20. The set of even verticescontributes an evennumber to the totaldegrees of the graph. Sincethat total degree is even,the set of odd vertices mustalso contribute an evennumber to the total degree.An odd number of oddvertices would contributean odd number, so thenumber of odd verticesmust be even.

21. See below.

22. 10 hoursA

B

C

D

E

F

G

H L

I

J

K

2

3

36

3

28

9

4

5

3

4 110

3

1

2


21. F MiamiMilwaukee

Minn./St. Paul

Miami012

Milwaukee107

Minn./St. Paul290G

c

c

23. , , ,

24. a. , b. , c.

25.

26. a. If G is a graph with medges and n vertices and

, then G is not

a simple graph.

m .n(n 2 1)

2

F 718

1231G

-1 , x , 2x . 3-3 , x , 1

1 , x , 3x , -3

-iÏ32

iÏ32-Ï5

5x 5Ï55 b. Proof of statement:

Given a simple graph Gwith n vertices and medges, let maximumnumber of edges possible(that is, ). We knowfrom the lesson that

, so .m #n(n 2 1)

2k 5n(n 2 1)

2

m # k

k 5


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In-Class Activity1. No, edge 3 is repeated

2. Sample: e4, e1; e3, e2; e4, e2

3. a. Sample: e1, e2; e1, e4, e3, e2

b. Sample: e1, e4, e5, e3, e2

4. See below.

Lesson 1. a. No

b. Noc. No

2. a. Yesb. Yesc. No

3. a. If at least one vertex of agraph has an odd degree,then the graph does nothave an Euler circuit.

b. The presence of an oddvertex is sufficient to showthat a graph cannot havean Euler circuit.

4. a. Yesb. Noc. No

5. e1 e2 e9 e8 e7 e6 e5

6. Yes, Sample from vertex A: a b f g h i c e d j

7. a. e1, e2, e3, e4, e5, e6

b. 2

8. a. e1e4, e2e4, e3e4, e1e2e3e4,e1e3e2e4, e2e1e3e4, e2e3e1e4,e3e1e2e4, e3e2e1e4

b. Samples: e1e1e1e4,e1e2e2e4, e1e3e3e4, e1e2e1e4,e1e3e1e4

4. sometimes sometimes sometimesnever sometimes sometimesnever always sometimesnever always always

c

c

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8. c. 3d. No, you can use pairs ofe1, e2, and e3 as many timesas you wish.

9. Yes, if the graph is notconnected, there is no waya circuit could contain everyvertex.

10. Not necessarily; there is aconnected graph (whichmust have an Euler circuit)and a nonconnected graph(which cannot).

11. Yes; think of replacing eachbridge by two bridges. Sucha walk exists by thesufficient condition for anEuler Circuit Theorem, sinceevery vertex will have aneven degree.

12. No

13. Yes, if the walk repeats anedge, then there is a circuit.Remove edges from thegraph until there is nocircuit. Then connect v to w.

v1 v2

v5

v3

v1 v2

v4

v3

v4 v5

14. Sample:

15.

16. a.

b. < 2.48%c. < 19.76%d. < 80.24%

17. Leonhard Euler, eighteenth

18. 120

19. < -.204

20. (x, y) → ( , )

21. n must be an odd number.For the complete graph ofan n-gon, the degree ofeach vertex is . Thatdegree is even if n is odd.

n 2 1

3y 1 4x 1 5

0.005

0.0049

0.02

0.98

0.98

0.02

Hascancer

Test Positive

Test Negative

Test Positive

Test Negative

Doesn't have

cancer

0.995

0.0001

0.0199

0.9751

F0111

1011

1101

1110G

v1 v2

v3v5

v4


1. False

2. 2, 0

3. a. 0b. 0c. 1

4. False

5. False

6. e2 e3; e3 e2; e3 e3; e2 e2; e5 e5

7. e4e1e4, e4e2e5, e4e3e5, e5e2e4,e5e3e4

8. e2e3e3, e3e2e2, e3e3e2

9. 5 walks: e1e2, e1e5, e4e2, e4e5,e6e3

10. ;

11. If the main diagonal is allzeros, there are no loops,and if all other entries arezero or one, there are noparallel edges, so the graphis simple.

12. 105

13. 4

F 39

11

94442

114219

GA3 5

F231

3136

16

10GA2 5

14. a. If the adjacency matrixfor a graph is symmetric,then its graph is notdirected.b. Sample:

15. No, add edge i.

16.

Add d13 between room 5and room 6.

17. Yes

18.

19. imaginary

2

1

-1 1 2-2-1

-2

real

-2 + i

-1 – 2i2 – i

1 + 2i

v1 v2

v4 v3

0 14

5

7

8

6

3

2

d1 d2d3

d6

d4

d12

d7

d5

d8d9 d11

d10

v1 g c

e d

a

f

b

i

h

v2

v6

v5

v3

v4

v1 v2


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20. for , -1

21. Left network , (p and (qor , r)), p or , (qor , r), p or (, qand r)Rightnetwork

Therefore, the left networkis equivalent to the rightnetwork.

22. a. ,

b. There are an infinitenumber of solutions. For all

real x, .y 549x 1

23

y 51711x 5

1411

;

;

;

;z Þ 1z2 1 1

z 2 123. a. ,

,

,

, , .The pattern is .b. The paths betweenvertices are circular.c. v1

v3

v2

An 5 An(mod3)A6 5 A3A5 5 A2A4 5 A1

F001

100

010GA4 5

F100

010

001GA3 5

F010

001

100GA2 5

F001

100

010GA 5


1. a. 15%b. 75%

2. a. 36%b. < 22.9%

3. Yes

4. a. the probability that itwill be cloudy 10 days aftera cloudy dayb. No matter what theweather is today, theprobabilities for theweather in 10 days areabout the same.

5. a. 0.7b. 0.6

6. ;

7.

8.

In 20 generations, of the

seeds will produce pale

flowers, and brilliant

flowers, no matter whatseeds you start with.

9. a.

b. MBC SBS.9.2

.1

.8GMBCSBS F

.9 .8

.1

.2MBC SBS

47

37

F3737

4747GT 20 5

47 5 b

4070 5

2870 5

1270 1

710 (4

7) 5

410 (3

7) 1.4(37) 1 .7(4

7) 5

a 537b 5

47

c. MBC: 67%; SBS: 33%

10. a. 15%b. 27.4%c. 32.1% tall, 32.7%medium, 35.3% short

11. Let , and

. If A and B are

stochastic, then each rowsums to 1 and each entry isnonnegative.

.

Row 1 sums to a1b1

a2b2 a1b2 a2b4 a1b1

a1b2 a2b3 a2b4

a1(b1 b2) a2(b3 b4)a1 a2 1, since b1 b2

b3 b4 a1 a2 1. Row 2 sums to a3b1

a4b3 a3b2 a4b4 a3b1

a3b2 a4b3 a4b4

a3(b1 b2) a4(b3 b4)a3 a4 1, since b1 b2

b3 b4 a3 a4 1. Eachentry of AB is nonnegativesince it is the sum of twoterms, each of which is the product of twononnegative numbers.Hence, the product of two2 2 stochastic matrices isstochastic.

3

51515151

5111511

15111

51515151

5111511

15111

Fa1b1 1 a2b3

a3b1 1 a4b3

a1b2 1 a2b4

a3b2 1 a4b4G

AB 5

Fb1

b3

b2

b4GB 5

Fa1

a3

a2

a4GA 5


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12. a. Yes

b.

c. ,

d. Over the long term, theproportion of occurrencesstabilize to a and b.

13. See below.

14. 24

15. 10

16. a. Yes, because all verticesare even, and it isconnected.b. 4

17. No, a graph cannot have anodd number of oddvertices.

18. a. v(1) 18 ft/sec

b. 2.125 sec

c. 1.5625 sect 55032 5

t 56832 5

5

b 559a 5

49

F.4444.4444

.5556

.5556GT 16 ø

F.4444.4444

.5556

.5556GT 8 ø

F.4445.4444

.5555

.5556GT4 5

F.45.44

.55

.56GT 2 5

d. The time in part c ismidway between the timesin parts a and b.e. Never, it is always -32 ft/sec2.

19. no solution

20. ;

21. Let n, n 1, n 1, and n 3 represent the fourconsecutive integers. By theQuotient-RemainderTheorem, n 4q r whereq is an integer and r 0, 1,2, or 3. If r 0, then n isdivisible by 4. If r 1, n 3 (4q 1) 34(q 1) is divisible by 4. Ifr 2, n 2 (4q 2)2 4(q 1) is divisible by4. If r 3, n 1 (4q3) 1 4(q 1) is divisibleby 4. So, exactly one ofevery four consecutiveintegers is divisible by 4.

22. A good source is MarkovChains: Theory andApplications by DeanIsaacson and RichardMadsen.

1511515

1511515

151151

55

515

111

limn→`

f(n) 513lim

n→-`f(n) 5

13


13. T stabilizes to since if [a b] [a b]

.6a .3b a implies -.4a .3b 0 implies -.4a .3b 0

.4a 7b b a b 1 .4a .4b .4.7b .4

and a 537b 5

47

5515151515151

5F.6.3 .4.7GF3

737

4747G

1.2. a.

b.

c.3. a.

b.

c.4. a.

V 2 E 1 F 5 6 2 9 1 5 5 2

V 2 E 1 F 5 6 212 1 8 5 2

V 2 E 1 F 5 9 2 16 1 9 5 2 b.

c.5.6.7. a. The vertex of degree 1

and its adjacent edge wereremoved.b. Both V and E werereduced by 1, so stayed the same. Since F didnot change, didnot change.

8. a. An edge was removed.b. Both E and F werereduced by 1, so wasnot changed. Since V didnot change, didnot change.

9. True, by the contrapositiveof the second theorem ofthis lesson: Let G be a graphwith at least one edge. If Ghas no vertex of degree 1,then G has a cycle.

10.

5 25 25 1 2 1 1 25 2 2 1 1 1V 2 E 1 FV 2 E 1 F

V 2 E 1 F

-E 1 F

V 2 E 1 F

V 2 E

V 2 E 1 F 5 5 2 8 1 5 5 2

V 2 E 1 F 5 5 2 8 1 5 5 2

V 2 E 1 F 5 7 212 1 7 5 2


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11. a. Sample:

b. Since the graph in part ahas no crossings and 6 edges, holds true. Remove a vertexof degree 1 and its adjacentedge does not change thevalue of . This wasdone in part a, and F didnot change, so holds for the originalgraph.

12. a.b. It is not connected andcontains crossings.

13. a. 9 b. 5

14. a. 6 vertices, 9 edgesb. impossible c. 5d. In this graph, a facecannot have 1 edge, sincethis would mean a lineconnects a house or utilityto itself. A face cannot have2 edges, since this wouldmean a house and a utilityhave two lines connectingthem. Finally, a face cannothave 3 edges, since therestrictions prevent twohouses or two utilities frombeing connected to eachother. So a face must haveat least 4 edges.

V 2 E 1 F 5 6 2 6 1 8 5 8

V 2 E 1 F 5 2

V 2 E

V 2 E 1 F 5 2

e. Since each edge bordersexactly two faces, if we sumfor every face the numberof edges bordering it, weget 2E. Since there are atleast 4 edges borderingeach face, there must be at

most faces. Hence, .

Multiplying, , or .

f. By part a, . If thereare no crossings, by part c,

. This contradicts parte, since 2(5) 9 does nothold true. Therefore, it isimpossible to connect threehouses and three utilitieswithout lines crossing.

15. a.

b. R U

c. Urban 71%; Rural 29%

d. .95a .02b a-5a 2b 05a 5b 5

7b 5

;

16. 5

a 527 ø .286b 5

57 ø .714

5515151

øø

F.95.02

.05

.98GRU

.95 .98

.02

.05Rural Urban

#F 5 5

E 5 92F # E

4F # 2E

F #2E4

2E4


c

17. a. No, there are twovertices with odd degrees.b. Sample:

18. No; for example, let v1, v2,v3 have degree 3, and v4

have degree 1. Then {v1, v2},{v1, v3}, and {v1, v4} are thethree edges from v1. {v2, v1}and {v2, v3} are 2 edges fromv2. Now there must beanother edge from v2. Butthat edge cannot connect v2

to v4, since v4 must remainwith degree 1. It cannotconnect to v1 or v3, since asimple graph cannotcontain parallel edges. Andit cannot connect to itself,since a simple graph doesnot have any loops. So,there is no such graph.

19. < 20%

School

20. To prove S(n):

.

S(1): ,

and , so S(1) is

true. Assume S(k):

.

Then

.

So, S(k) ⇒ S(k 1), and bymathematical induction S(n)is true for all integers n.

1

(k 1 1)((k 1 1) 1 1)(k 1 1 1 5)3

(k 1 1)(k 1 2)(k 1 6)3 5

(k 1 1)(k2 1 5k 1 3k 1 12)3 5

(k 1 1)[k(k 1 5) 1 3(k 1 4)3 5

3(k 1 1)(k 1 4)3 5

k(k 11)(k 1 5)3 1

(k 1 1)(k 1 4) 5k(k1 1)(k1 5)

3 1

(k 1 1)((k 1 1) 1 3) 5

ok

i51i(i 1 3) 1o

k11

i51i(i 1 3) 5

k(k 1 1)(k 1 5)3o

k

i51i(i 1 3) 5

1(2)(6)3 5 4

1(4) 5 4o1

i51i(i 1 3) 5

n(n 1 1)(n 1 5)3o

n

i51i(i 1 3) 5


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21. a. The faces are n-gons, soeach face has n edges. Forall F faces, each edgeappears in 2 faces, so thetotal number of edges is

, or .

b. At each vertex, m edgesmeet. So for all V vertices,the total degree would bemV. But each edge meets attwo vertices, so the number

of edges is , so .

c. . From

parts a and b, and

. So by substitution,

. Divide

both sides by 2E to obtain

.

Thus, .1n 1

1m 5

12 1

1E

1m 2

12 1

1n 5

1E

2Em 2 E 1

2En 5 2

V 52Em

F 52En

V 2 E 1 F 5 2MV 5 2E

E 5mV2

nF 5 2EE 5nF2

d. A regular n-gon musthave at least 3 sides, hence

. In a polyhedron, atleast 3 edges must meet ata vertex, hence . m and n cannot both begreater than 3, because

.

Thus, either or .Neither m nor n can begreater than 5, because

.

Computing values for Ewhen m and n range from 3to 5, five solutions arefound.

n m E polyhedron3 3 6 tetrahedron3 4 12 octahedron3 5 30 icosahedron4 3 12 cube5 3 30 dodecahedron

e. These solutions relate tothe five polyhedra given inthe column at the rightabove.

12 ,

12 1

1E

1m 1

1n #

16 1

13 5

n 5 3m 5 3

12 ,

12 1

1E

1m 1

1n #

14 1

14 5

m $ 3

n $ 3


1. a. Sample:

b. Yes, Sample:

2. Sample:

3.

4. Sample:

5. v1

v2

v3

v4

v5e 1 e 2

e 3

e4

e5

6. a. Falseb. False

7. Yes

8. a. i. Yes ii. Yes iii. Nob. i. No ii. No iii. Noc.

length path(s)1 e2

2 e1e3

3 none4 e1e4e6e5

5 e1e4e7e6e5

e2e3e4e6e5

e2e5e6e4e3

6 e2e3e4e7e6ee2e5e6e7e4e3

d. Samples: e1e2e5e6e4;e3e2e1; e4e7e6e5e2e1

9. a. v2, v3, v5

b. e2, e4, e6, e7

c. noned. e1 and e8

e. e6

f. Falseg. deg(v1) 4; deg(v2) 3;deg(v3) 2; deg(v4) 4;deg(v5) 3.h. 16

10. a. Sample: e1e5e6e3e4e2

b. Sample: e1e6e2; e1e5e2

c. Sample: e1e1; e1e5e4e4

d. 2e. 3; Sample: e1, e3, e6

555

55


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11.edge endpoints

e1 {v1, v5}e2 {v1, v2}e3 {v1, v3}e4 {v2, v3}e5 {v2, v4}e6 {v4}e7 {v4, v5}e8 {v1, v5}

12. Impossible; a graph cannothave an odd number of oddvertices.

13. Sample:

14. Sample:

15. Impossible; a graph cannothave an odd number of oddvertices.

16. No, the sum of the entriesmust be the total degree ofthe graph, which must beeven.

17. The graph has an Eulercircuit by the sufficientcondition for an EulerCircuit Theorem, since it isconnected and every vertexis of even degree.

18. No, the graph is notconnected.

19. No, because v2 and v3 haveodd degree.

20. It cannot be determinedsince the graph may not beconnected.

21. a.

b. < 61.4%

22. a.

b. ≈ 0.48%

23. a.

b. 33 hours

A6

B39

D5

16C11 E

4G1228

F5

H5

I5

33

Broken into

Not broken into

.96

.04

.02

.98

.0001

.9999

Alarm.000096

No alarm.000004

Alarm.019998

No alarm.979902

Pac

Lux

Rudder .035

Rudder .022

.03

.97

.04

.96

.05

.95

.07

.69

.31

.93

Gauge .02

None .64

Gauge .01

None .28


c

24. No; this situation may berepresented as a graph with25 vertices, each with 5 edges. This is not possiblesince a graph cannot havean odd number of oddvertices.

25. No, a graph cannot have anodd number of oddvertices.

26. Yes; below is a samplegraph:

27. a. Vertices F and G haveodd degrees, so there is notan Euler circuit.b. the edge between F andG

28. a. Yes, sample:

b. No, two of the verticeshave odd degree, so noEuler circuit is possible.

a3

a2

a1

a6

a5

a4

29. a.

b. B NB

c. T8 <

They bowl on about 56% ofthe Tuesdays.d. ≈ 56%

30. 38% Democrat, 36%Republican, 26%Independent

31. v1 v2 v3 v4

32. The diagonal has zeros, andall other entries are eitherzeros or ones.

33. a.

b. No, the matrix is notsymmetric.

34.

35. a. 0 b. 0

v3

v1

v2

v1 v2

v4 v3

F0002

1000

2120

0110Gv1

v2

v3

v4

F.5557.5554

.4443

.4446GF .4.75

.6.25GB

NB

.4.25A

B .75

.6

NB


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36. 2

37. a. v1 v2 v3

b. 39

38. a. v1 v2 v3

b. 9

F110

101

100Gv1

v2

v3

F121

201

110Gv1

v2

v3

39. a. There are no walks oflength 4 or more.b. v1 v2

v4 v3


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1.

2.

3.

4. a. [55, -22.5°]

b. (55 cos(-22.5°), 55 sin(-22.5°)) < (50.8, -21 .0)

5. (cos 218°, sin 218°) <(-0.788, -0.616)

6. a. (5, -6)b.

4-2

-2

-4

-6

2 6

(5, -6)

y

x

4-2

-2

-4

-6

2

(3, -6)

y

x

150

N

E

S

W

180

10˚

N

E

S

W

7. a. length , direction< 308.7°b.

8. units

9.

10. [ , tan-1 ] <[24.8, 130°]

11. The standard position arrowfor the vector from (-1, 2) to (4, -1) has endpoint (4 (-1), -1 2) (5, -3);the standard position arrowfor the vector from (3, -2) to (8, -5) has endpoint (8 3,-5 (-2)) (5, -3). So thevectors are the same.

12. (cos 82°, sin 82°) <(0.139, 0.990)

13. tan-1 < 18.4° or 198.4°

14. or

15. [5, 45°] (5Ï22 , 5Ï2

2 )5w→

5

k 5 14k 5 2

13

522

522

(-1916)Ï617

810˚

y

x

Ï157 ø 12.5

4-2

-2

-4

-6

2 6

(4, -5)

y

x

5 Ï41


c

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16. a. Sample:

b. relative maximum at; relative minimum

at

17. a. ; ;

for .b. < 7449.44

k . 1Ak 5 1.008Ak21 2 200

A2 5 7864A1 5 8000

x 5 -2x 5 2

2

4

-4

-2 2-4 4-6 6

y

x

18. a. iib. i and iii

19. This is a graph of half of aparabola whose intercept is(0, 1).

20. This is a graph of an ellipsewith vertices (1, 0), (0, 2), (-1, 0), and (0, -2).

21. (-1,13), (-5, -1), (5, 1)

22. Any vector of the form (r, 0)with r positive has polarform [r, 0].


1. (12, -14)

2. < [9.73, 72.9°]

3. a. magnitude: < 43.3 lb;direction: < 6.6° N of Eb. The resultant force is43.3 lb in the direction 6.6°N of E.

4. 71.96 lb in the direction of< 7.18° N of E

5. speed: 23.6 mph;

direction tan-1 <47° N of E

6. < 2157 lb in the directionof < 52.5° N of E

7. < 55.2 nautical miles eastand < 151.6 nautical milesnorth of its starting point

8. a. [-6, 20°] or [6, 200°]b. (6 cos 200°, 6 sin 200°) <(-5.64, -2.05)

9. a.

b.

c.v

v

v + v

v

uv + u

v

uu + v

(10Ï316 )u 5

Ï556 ø

10. a.

b.

c.

11. Sample counterexample:

If [1, 45°] and

[1, 45°], then [2, 45°] [2, 90°].

12. Let (v1, v2) be a vectorwith polar representation

[r, u]. Hence, (r cos u, r sin u), and v1 r(cos u), v2 = r(sin u).

[r, u 180°](r[cos(u 180°)], r[sin(u 180°)])5 (r[cos u cos 180° sin usin 180°], r[sin ucos 180°cos u sin 180°])5 (r[(cos u)(-1) (sin u)(0)], r[(sin u)(-1) (cos u)(0)])5 (-r(cos u), -r(sin u))5 (-v1, -v2)

12

12

11

51-v→

5

5v→

5

v→

5

Þu→

1 v→

5

v→

5u→

5

v

-wv – w

-w

v

wv + w


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13. a. If (v1, v2), then

then (v1, v2) (-v1, -v2)(v1, -v1, v2 -v2) (0, 0),which is the zero vector.b. Sample: The arrow for

is a point.

14. a. (3, -4) b. (-9, 6)

15. a. The current is about 6.7knots.b. The boat travels about18.4 knots.

16. The length of (kv1, kv2) is

5

5

5 zk z zv→

zzk z Ï(v1

2 1 v22)

Ïk2(v12 1 v2

2)

Ïk2v12 1 k2v2

2

Ï(kv1)2 1 (kv2)2 5

v→

2 v→

51151

v→

2 v→

5 v→

1 -v→

v→

5 17. 144

18. a. < 11.8 9.84i ampsb. a sine curve withamplitude < 15.4 andphase shift < -39.8°

19.

20. 0.4

21. Samples: a. gravityb. gravity, engine thrust,lift due to wing designc. gravity, torque, initialvelocity of balld. gravity, friction, initialpush

x 1 10(x 1 4)(x 2 2)(x 1 1)

1


1. a.

b.

2. (-12, 18)

3. Samples: (-12, 28), (6, -14);in general, k(-6, 14) where k is any real number.

4. (-48, 18) -6(8, -3), so thevectors are parallel.

5. a. ;

b.

6. a.

b. Sample: ;

7. Sample: ,

8. Sample: t(7, 2)

(x 1 8, y 2 5) 5

y 5 5 2 5tx 5 -1 2 5t

y 5 -1 1 9tx 5 -3 2 4t

y

-4 -2-2

2

4

6

8

P(-3, -1)

-w

(-4, 9)

x

x

y

t = 0

t = -1

t = 2

t = 1

6

12

18

-2-4

t = - 3]4

y 5 8 1 5tx 5 -3 1 2t

5

(54 cos 52°, 54 sin 52°)

F54, 52°G 9. Sample: ;

10. -1 -1(v1, v2)

(-v1, -v2)

11. a. (-15, 20)b. (-30, 40) c. (-30, 40)

d. a(bv1, bv2)(abv1, abv2) ab(v1, v2)

12. a.

b. Sample: The midpoint ofthe points determined by

and is the point

determined by .

13. about 360.6 lb in thedirection about 26.3°

14. about 236.3 mph in thedirection about 138.4°

15. (-2, 14), (2, -14)

16. 4

t 5a 1 b

2

t 5 bt 5 a

y0 1v2

2 )(x0 1v1

2 ,

(2x0 1 v1, 2y0 1 v2)2 5

(x0 1 y0) 1 (x0 1 v1 1 y0 1 v2)2 5

(ab)v→

555a(bv

→) 5

-v→

5

5v→

5

y 5 8 1 tx 5 2 1 3t


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t 0 1 2 -1

x -3 -1 1 -5 -4.5y 8 13 18 3 4.25

-34

t 0 1

x x0

y y0 y0 1 v2y0 1v2

2

x0 1 v1x0 1v1

2

12

c

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17. a. -0.267 words per sec; thesubjects forget 8 words per30-second period when thewait time changes from 0sec to 30 sec.b. -.13 words per sec; thesubjects forget 6 words per30-second period when thewait time changes from 30sec to 60 sec.

18. The initial conditions donot hold, because S(1), S(3),. . . are not true.

19. b

20. and

21. Since and, when ,is determined.

When , ,is determined.

Create a number line with Pat 0 and Q at 1. Then eachvalue of t will determinethe corresponding point onthe number line.

y0 1 v2)Q 5 (x0 1 v1t 5 1

P 5 (x0, y0)t 5 0y 5 y0 1 v2t

x 5 x0 1 v1t

AD←→

AC←→


In-Class Activity

1. a. ;

b.2. a. < .361 b. 68.8°

3. cos u 0; u 90°

4. Sample: The product of theslopes of perpendicularlines is -1, and the ratio ofsecond to first componentsis the slope of the linecontaining the vector. Since

, the lines

containing the vectors areperpendicular.

Lesson 1. 70 2. (-350, -420)

3. 73 4. 150°

5.

6. The meaning of the dotproduct of two vectorsbeing negative is the sameas the meaning of cos ubeing negative, where u isthe angle between the twovectors and 0 u π. Forthese values of u, cos u , 0

only when .

Therefore the dot product

π2 , u # π

##

u 5 cos-1 38 ø 68°

63.5 • 7

-12 5 -1

55

Ï74

zs→

z 5 Ï85r→ 5 Ï13

of two vectors is negativeonly if the angle betweenthem is obtuse or if theyhave opposite directions.

7.

8.

;

9. (-8, 6) or (8, -6)

10. Sample:

11. a.b.c. 0°, 180°d. If ,

; if ,

Hence, zk z cos u 5 k-k • -1 5 kzk z • cos 180° 5

k , 0k • 1 5 kzk z • cos 0° 5k . 0

zk z[(v1)2 1 (v2)2]k(v1)2 1 k(v2)2

k(4, 3)(x 2 6, y 2 2) 5

8Ï17

1-8

Ï175 0( -2

Ï17)(4) 1

s→

• w→

5 ( 8Ï17)(1) 5

-8Ï17

18

Ï175 0( 2

Ï17)(4) 1

( -8Ï17)(1) 5r

→ • w

→5

Ï4 5 2Ï6817 5Ï64

17 1417 5

Ï( 8Ï17)2

1 ( -2Ï17)2

5zs→

z 5

Ï4 5 2Ï6817 5Ï64

17 1417 5

Ï( -8Ï17)2

1 ( 2Ï17)2

5zr→

z 5

y 5 -65


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11. e. cos u Angle Between Vectors Theorem

from part d, cos u

from parts a and b

Distributive Law

simplify

Therefore, the Angle Between Vectors Theorem holds true for parallel vectors.

kzk z

kzk z 5

k[(v1)2 1 (v2)

2]zk z [(v1)2 1 (v2)2]

kzk z 5

k(v1)2 1 k(v2)

2

zk z [(v1)2 1 (v2)2]k

zk z 5

5k

zk zu→ • v→

zu→

z zv→

zk

zk z 5

u→ • v→

zu→

z zv→

z5

11. e. See below.

12. a.

b. ;

13. 8.13°

14. a.

b.

c.5

15. a. (250 cos 55°, 250 sin 55°)< (143, 205)b. Sample: ;y 5 30 1 175t

x 5 -50 1 193t

k(u→

• v→

)k(u1v1 1 u2v2) 5

ku2v2 5ku1v1 1

ku2) • v→

(ku→

) • v→

5 (ku1,

u→

• v→

1 u→

• w→

u1v1 1 u2v2 1 u1w1 1 u2w2 5u1v1 1 u1w1 1 u2v2 1 u2w2 5u1(v1 1 w1) 1 u2(v2 1 w2) 5

(v1 1 w1, v2 1 w2) 5u→

•

u→

• (v→

1 w→

) 5

v→

• u→

v1u1 1 v2u2 5

u→

• v→

5 u1v1 1 u2v2 5

y 5 6 11013t

x 5 -8 12413t

(2413, 10

13)c. (46.5, 117.5); it is 46.5miles east and 117.5 milesnorth of Indianapolis.

16. air speed < 198.7 mph,compass heading < 8.18° Nof E

17. a. is 3; is 4

b. Sample: and

;

;

18. a. <194 b. No, .

19. The units digit of the fourthpower of a number is thefourth power of the unitsdigit, and 24, 44, 64, and 84

each have units digit 6.

20. < 2.162

21. a. -3 b. 4 c. (-3, 4)

d. (v→

• i→

)i→

1 (v→

• j→

) j→

5 v→

r 565 . 1

2(194) 5 388169) 52(25 12( zu

→z2 1 zv

→z2) 5

306 1 82 5 388zu→

2 v→

z2 5

zu→

1 v→

z2 1v→

5 (12, 5)

u→

5 (3, 4)

u→

2 v→

u→

1 v→


1. (3, 6, 2)

2.

3.

4.

5. d

6.

7. and

8. 9. 133Ï10

y 5 0x 5 0

z

2

42

4y = 3

y

x

-4 42-2

-22

2

4

6

(1, 5, -2)

y

x

z

-4 42-2

-22

2

4

6

(3, -2, 4)

y

x

z

-4 42-2

-22

4

2

4

6(4, 2, 8)

y

x

z

10.25

11. center: (1, -3 -4); radius: 6

12. center (0, 7.5, 0); radius 7.5

13. < 79 cm

14.

,

15. y -2 or y 6

16. < 4.1 km

17. False

18. a. ;

b.

c. ;

19. a.

b. limaçon

21 3

y 5 1 1 2tx 5 2 1 t

(3Ï55 , 6Ï5

5 )y 5 1 1 2tx 5 2 2 4t

55

Ï(y2 2 y1)2 1 (x2 2 x1)2 1 (z2 2 z1)2

PQ 5 ÏPR2 1 QR2 5

PR 5 Ï(y2 2 y1)2 1 (x2 2 x1)2

|y2 – y1|

|z2 – z1|

|x2 – x1|P

R

Q

yx

z

5

(z 2 8)2 5(y 2 2)2 1(x 1 1)2 1


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20. For , 3 is a factor of. Assume 3 is a

factor of for .For ,

.As both terms are divisibleby 3, so is the sum, .Hence, by mathematicalinduction, 3 is a factor of

, for all positiveintegers .n $ 1n3 1 2n

n3 1 2n

(k3 1 2k) 1 3(k2 1 k 1 1)k3 1 3k2 1 5k 1 3 5(k 1 1)3 1 2(k 1 1) 5

n3 1 2n 5n 5 k 1 1n 5 kn3 1 2n

n3 1 2n 5 3n 5 1 21. left side

right side

22. an ellipsoid that intersectsthe x-axis at (6a, 0, 0), they-axis at (0, 6b, 0), and thez-axis at (0, 0, 6c)

5tan2x sin2x5

sin2x sin2xcos2x5

sin2x(1 2 cos2x)cos2x5

sin2x 2 sin2x cos2xcos2x5

sin2xcos2x 2 sin2x5

tan2x 2 sin2x5



1. (-927, 98, -414)

2. (3, -1, 6) 3. (-7, 1, 8)

4. 5. -17

6. (4, -2, 19) 7. (0, 0, 0)

8. square root

9. < 81°

10. a. < 83 cm b. < 84.7°

11.12. (4, -10, -11)

(4, -5, 6) 4(4) (-10)(-5)(-11)(6) 0

13. (11, -37, 23);

3 11 4

-37 5 23 0, so is

orthogonal to .u→

3 v→u→

5 • 1

• 1 • u→

• (u→

3 v→

) 5

u→

3 v→

5

5115 • (u

→3 v

→) • v

→5

r→

• s→

5 0

3Ï3

14. a.b. Sample:

15. a. u1v1 u2v2

u3v3 v1u1 v2u2 v3u3

b. Commutative Propertyof the Dot Product

16. (u2v3 u3v2,u3v1 u1v3, u1v2 u2v1)

u1u2v3 u1u3v2

u2u3v1 u1u2v3 u1u3v2

u2u3v1 0; u2v3 u3v2, u3v1 u1v3,

u1v2 u2v1) v1u2v3

v1u3v2 v2u3v1 v2u1v3

v3u1v2 v3u2v1 0521212 • v

→52

22(u

→3 v

→) • v

→55

21212u

→5

• 222(u

→3 v

→) • u

→5

v→

• u→

511511u

→ • v

→5

(-12, 7, 17)

z 512

c

c

17. (au1, au2, au3)(bv1, bv2, bv3) abu1v1

abu2v2 abu3v3 ab(u1v1

u2v2 u3v3) ab

18. a. The angle between

and is

, which is the

angle between and .b. Scalar multiplicationdoes not change themeasure of the anglebetween two vectors.

19. a. (-1, 0, 1), (0, -1, 3)b. Sample: (1, 3, 1)c. (-2, 4, 6)d. < 0.645 e. 164

5v→

5u→

v→

u→

cos-1( u→

• v→

zu→ z zv→ z )5cos-1(ab(u

→ • v

→)

a zu→ zb zv→ z )5cos-1( au

→ • bv

→

zau→ z zbv

→ z )bv→

au→

(u→

• v→

)51

15115

• (au→

) (bv→

) 5 20. center: (2, , 0); radius:

21. ( , ) and ( , )

22. a. Sample: ( , )

b. Sample: ;

23. 255

24. a. {0, 1, 2} b. No

25. (10, -8, 0) whose

length is . The

parallelogram with and

as sides has base

and height 1, so its area is .2Ï41

5zu→

z 5 2Ï41

v→

u→

2Ï41

u→

3 v→

5

y 5 -2 1 7tx 5 1 2 4t

t(-4, 7)5y 1 2x 2 1

-2Ï5-Ï52Ï5Ï5

12Ï17

-32


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1. a. ( , , )

b. ; ;

c. Sample: (-2, 5, 2)2. a. No, when , ,

but then .b. (3, -2, -3)c. Sample: ;

; 3.4.

5. (-4, 2, -6) -2(2, -1, 3);

is parallel to (2, -1, 3),since it is a scalar multipleof it. (2, -1, 3) isperpendicular to the

plane N, and hence isperpendicular to N.

6. Sample: ;;

7.

8. Since and lie in M,

3 gives the vectorperpendicular to M. Thus, Q

is on M (so that is on M)

if and only if is

perpendicular to 3

.P1P→

3

P1P→

2

P1Q→

P1Q→

P1P→

3P1P→

2

P1P→

3P1P→

2

2x 2 y 1 3z 5 -2

z 5 -1 1 3ty 5 3 2 tx 5 2 1 2t

w→

w→

5w→

5

22

2

4

(0, -4, 0)

(2, 0, 0)4

y

x

z

(0, 0, )4]3

2x 1 y 2 2z 5 -1z 5 3 2 3ty 5 2 2 2tx 5 1 1 3t

z Þ -8t 5 4x 5 10

z 5 -2 1 4ty 5 5x 5 1 2 3t

t(-3, 0, 4)5z 1 2y 2 5x 2 1 9. a.

b.10. M1, is perpendicular to

(a1, b1, c1), and M2 isperpendicular to (a2, b2, c2),so < is perpendicular toboth (a1, b1, c1) and

(a2, b2, c2). Since is alsoperpendicular to both (a1, b1, c1) and (a2, b2, c2),

is parallel to <.

11. a. (3, -2, 1) and (1, 2, -1),the vectors perpendicularto the planes M1, and M2

are not parallel.

b. Sample:

c. Sample: ; ;

12.13.14.

(u1, u2, u3) (v1 w1,v2 w2, v3 w3)

(u1v1 u1w1) (u2v2

u2w2) (u3v3 u3w3)(u1v1 u2v2 u3v3)

(u1w1 u2w2 u3w3)

15. e

16. a. W I

b. < 7.4%

F.96.5

.04.5GW

I

u→

• v→

1 u→

• w→

5111115

111115

111 • 5

u→

• (v→

1 w→

)

p 5 6Ï5

5n 1 10d 1 25q 5 1000

z 534 1 8t

y 5 1 1 4tx 574

(74, 1, 34)

w→

w→

7x 2 2y 1 z 5 6

7x 2 2y 1 z 5 6


c

17. a.

b.

18.

19. a. y

x2-1

2

-14

x 567

(n 1 1)(2n 1 1)6n2

on

j51(1n)2

• (1n) b. a cone

c. cubic units

20. a. a plane parallel to and 1unit above the xy-planeb. As a increases, vectorsperpendicular to the planerotate from the z-axistoward the x-axis, so theplane tilts more steeply.

16π3


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1. a. ,

b. When , the point

is generated; when

, the point is

derived. These points arethe same as those on theintersection line shown onpage 768 of the StudentEdition. Since two pointsdetermine a line, this is thesame line.

2. a.

b. Sample:

3. a. Sample: b. See below.

3x 2 2y 1 4z 5 2

Hx 5 t

y 512 2

12t

z 512 2

12t

z

11

1

x

y

y – z = 0

x + y + z = 1

(79, 2, 50

9 )z 5509

(319 , 0, 89)

z 589

y 59z 2 8

21x 5-12z 1 83

21 4. the line

5. the plane

6. z

z

z

x

y

M1 = M2 = M3

x

yM1 = M2

M3

x

yM1 M2

M3

-x 1 3y 2 2z 5 -6

Hx 549 1 11t

21y 5 t

z 57 1 8t

21


3. b. subtracting plane Nsubtracting plane M

Therefore, the system has no solution.0 5 1

3x 2 2y 1 4z 5 13x 2 2y 1 4z 5 2

c

7. a. the first and secondequations, because thevector (1, 1, 3) isperpendicular to bothb. (2, -1, 1) is perpendicularto the third plane, so thatplane is not parallel to thefirst two, and must intersectboth of them.

8. a. Subtracting the secondequation from the first, weget . Subtractingthe third equation fromtwice the second equation,we get , or

. Substituting forx, , whichsimplifies to . Hence,there is no solution to thissystem.b. The first two equationsintersect at the line

, , .

9. the point

10. the line

11. the line

Hx 5 ty 5 7z 5 5 1 t

Hu 5 ty 5 9 2 tw 5 -27 1 8t

(12, 12, 0)

z 5 ty 55 2 t

5x 510 2 3t

5

-8 5 1-(3y 1 8) 1 3y 5 1

x 5 3y 1 8x 2 3y 5 8

-x 1 3y 5 1

12. ,

13. mi, mi, mi

14. a plane parallel to thex-axis containing (0, 0, 10)and (0, 10, 0)

15.

16. a.b. The planes are parallel.

17. is perpendicular to

both and .

if and only

if is also perpendicular to

. Thus , , and ,being all perpendicular tothe same vector at the samepoint, are coplanar.

18. 4

19. Yes, it is an example ofmodus tollens.

20. a.b.

or (u2v3

u3v2)x (u3v1 u1v3)y(u1v2 u2v1)z (u2v3

u3v2)x0 (u3v1 u1v3)y0

(u1v2 u2v1)z02121

252121

2z 2 z0) 5 0(u

→3 v

→) • (x 2 x0, y 2 y0,

13x 1 18y 1 2z 5 55

u3→

u2→

u1→

u2→

3 u3→

u1→

u1→

• (u2→

3 u3→

) 5 0

u3→

u2→

u2→

3 u3→

3x 2 4y 2 z 5 -29

Hx 5 -7 1 3ty 5 2 2 4tz 5 -t

h 5 30g 5 2d 5 10

c 5 0a 5 b 512


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1. magnitude: ;direction: < -15.3°

2. < 18.4° and 198.4°

3. 9 and -3

4. < [ , 292°]

5. (-2, -2) 6. (-9, 6)

7. -10 8. (1, -8)

9. (2, 28)

10. Sample: [4.14, 78.0°]

11. Sample: [3, 285°]

12. Sample: [3, 35°]

13. Sample: [20, 35°]

14. (-5, 13)

15. Sample:

16. [8, 305°]

17. (-7, -1, 5)

18. (-3, 5, -3)

19. (11, -12, 5)

20. 21. 8

22. (11, 18, 19)

23. (-2, -3, 4)(11, 18, 19)

24. Sample: (-2, -13, -6)

25. < 130.4° 26. < 146.8°

27. 180°; they have oppositedirections.

28. (u1, u2)(av1, av2) u1av1 u2av2

a(u1v1 u2v2) a(u→

• v→

)51

515 • u

→ • (a v

→) 5

76 5 0-22 2 54 15

• v→

• (u→

3 v→

) 5

Ï30

(-32, 1)

Ï29

Ï130 29. The vector from (x, y) to , is . Therefore, by

definition, the vectors areparallel.

30. (v1, v2,v3) (kv1 mv1, kv2 mv2,kv3 mv3) (kv1, kv2, kv3)1 (mv1, mv2, mv3)

31. No

32. (ku1, ku2, ku3)(v1, v2, v3) ku1v1

ku2v2 ku3v3 k(u1v1

u2v2 u3v3)

Therefore, is orthogonal

to if is orthogonal to

.

33. neither

34. parallel

35. perpendicular

36. 2

37. -10

38. and

39. a.

b. (14.5, 6.76)

N

S

W E25˚16

(-1513Ï13, -10

13Ï13)(1513Ï13, 10

13Ï13)

v→

u→

v→

ku→

k(u→

• v→

)51

15115

• (ku→

) • v→

5

k v→

1 m v→

551

115(k 1 m)v

→5 (k 1 m)

a(1, m)(a, ma) 5y 1 ma)(x 1 a


c

c. The ship is going 14.5mph towards the east and6.76 mph towards thenorth.

40. a. [50, 52°] b. (30.8, 39.4)c. The kite is 39.4 m abovea spot on the ground,which is 30.8 m away fromthe owner.

41. a. 32.1 lb of force withdirection 80.6°counterclockwise from thepositive x-axisb. Sarah

42. a., b.

c. [360, 141.8°]; Relative tothe ground, the plane’sspeed is 360 km/hr, and itsheading is 38.2° North ofWest.

43. No

44. < 290 mph at 31.5° Southof East

45. (-2.5, 4.3)y

x

[5, 120˚]

EW

N[400, 135˚]

[60, 270˚]

S

ab

46.

47. a. (6, 3)b. length < 6.7; direction < 26.6°

48. a. < (-4.6, 3.4)b.

49. a.-d.

50. a.-c.

51. z

x

y2

4

24

-2-4

4-4u

v

u

v

y

x

u-v

u – v

u + v

[5, 30˚]

[3, -60˚]

u

2 v

u

u + v

u – v

v -u

-v

(-4.6, 3.4)

2

y

x-2

1

3

-4

N

420˚W

S

E


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52. (-8, -5, -2); This vector canbe pictured by an arrowstarting at the endpoint of

and ending at the

endpoint of , providing

and have the sameinitial points; or putting thevector in standard position,it is the diagonal of thefigure having vertices (-3, -4, 1), (-5, - 1, -3), (0, 0, 0), and (-8, -5, -2).

53. a., c.

b. (0, 0, 4)

54. a. i. (1, -8) ii. (4, -7) iii. (-8, -11)

b. Sample: (3, 1)c.

-2 2

(1, -8)

(7, -6)

y

x4 6 8

-8

-6

-4

-2

x

2

4

22-2 y

z

v

u

v •u

v→

u→

v→

u→

55.56. Sample:

57.

58.

59.

60. center: (0, 2, -4); radius:

61. and

62. (6, 0, 0), (0, -4, 0), (0, 0, 12)

63. and

64.

65.66. Sample: (-3, 1, 5) and

(3, -1, -5)

67. ( , , )t(2, -6, 1)

5z 2 2y 1 1x 2 5

2x 2 4y 5 6

Hx 5 3 1 2ty 5 -4tz 5 -1

y 5 0x 5 0

4

4

12

y

z

-8

(6, 0, 0)

(0, -4, 0)

(0, 0, 12)

4

8x

y 5 -3y 5 3

Ï5

Hx 5 5 1 3ty 5 -4t

Hx 5 1 2 5ty 5 -2 2 3t

5x 1 3y 1 1 5 0

t(1, 1)(x 2 1, y 2 2) 5

(x 2 1, y 2 2) 5 t(-5, 5)


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1. 3

2. a. < 58 mi

b.

c. D r1t1 r2t2 r3t3

3. Sample:

i

1 0.5 60 302 1 51 25.53 1.5 57 28.54 2 54 275 2.5 2 16 3 52 267 3.5 58 298 4 57 28.59 4.5 45 22.5

10 5 44 2211 5.5 41 20.5

4. a. 4.4 ft b. 110 ft c. 440 ft

5. 112 m

6. a. this car b. 629.2 ftc.

2 4 6 8 10

44

88

g(x)ft/sec

x (sec)

o11

i51.5f ( i

2) ø 260.5 units2

.5(f ( i2))f ( i

2)i2

o3

i51riti

5115

26

km

2 ] 3

11 2 /

3 km

20 k

m

10

20

40

60

80

30 604020 50time (minutes)

rate

(kph

)13

d. better e. 608.3 ft

7. 105 mi

8. a. ≈ 260 mi b. ≈ 250 mic. the estimate from part a;because it has morerectangles, with a total areathat is closer to that of theactual graph

9. < .9 10. 5525

11. or oror

12. a. ;

b. oblique asymptote:; vertical

asymptote:

13. a., b. Sample:

c. d.

14.

15. a. 3.0, 3.8, 4.6, 5.4, 6.2, 7.0

b. , where 0, 1,

K, n

c.

16. a. 580.8 ft and 589.6 ftb. Answers may vary.

b 2 an

k 53 14kn

88 ftsec

13600 hr

sec 5

5280 ftmi • 60 mi

hr • 60 mihr 5

y 5 bky 5 8

b

a

8642

y

x

x 5 2y 5 2x 1 7

limx→-`

f(x) 5 -`limx→`

f(x) 5 `

x ø 3.61x ø 0.46x ø -2.68x ø -5.82


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1. 39.3525 ft

2. Sample: using g(3): 22.44 ft;using g(3.5): 25.41 ft

3. 668.8 ft

4.

5. 4

6. 588.8631

7. a. -908.0501 (using rightendpoints)b. The region between thegraph of g and the x-axishas more area below thex-axis than above.

8. < 0.8658

9. < 0.8655

10. N Sum10 0.8914150 0.87123

100 0.86864500 0.86655

11. N Sum 10 541.220 564.350 577.808

100 582.252500 585.786

1000 586.227

b 2 an

12. a. ≈ 38b. ≈ 36c. ≈ 37.4

13. a.

b. negative

14. a.

b. negative

15. a.

b. positive

16. < 10,500 ft2

17. distance

18. 22.5 miles ahead of whereit started

y

x1 2 3

-4

-2

2

4

1 2 3

-4

-2

2

4

x

y

y

x2 4 6 8

-4

-2

2

4

c

19. a. Sample:

b. Answers will vary.c. any single edge of thecircuit

20. has multiplicity 1,and has multiplicity 3.

21. 105,625

x 5 2x 5 -1

22. See below.

23. 4 units2

24. 8πunits2

25. a. 0b. The region between thegraph of g and the x-axishas just as much area abovethe x-axis as below it.


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22.

It appears to be an identity.(tan u) (sin u cot u cos u) sec u

definitions of tan, cot, and sec

common denominators


simplifying

(tan u)(sin u cot u cos u) sec u51[

1cos u

( sin ucos u) ( 1

sin u)( sin ucos u) (sin2 u

sin u 1cos2 usin u )

51

cos u( sin ucos u) (sin u 1

cos usin u cos u)

1


π]2

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1. upper sum: ;

lower sum:

2. a. lower sum: 0.328350;upper sum: 0.338350b. lower sum: 0.332829;upper sum: 0.333828

3. a. upper; lower b. the definite integral of f from a to b

4.

5.

6.

7. 28.5

8. 600

9. -4.5π10. 14.67

11. a.

b. positive

12. a.

b. negative

2 4

-8

-4

2

x

y

y

x-4 -2

1

(12t 1 2) dtE2

-4

x2 dxE3

-3

E10

1 (log x) dx

732

1532 13. a.

b. negative

14.

15.16. a. < 156 miles

b. the distance from thestarting to ending point forthe 5-hour period

17. a. 15b. 126

18. a. $9.45/yr b. $18.89/yr c. < .09445P/yr

19.

The graph is a line.

20. 2,358,350

21. 1.4925 < 1.5075

22. Sample: k f(x) dx

f(x) dx: Under a vertical

scale change of magnitudek, the area is multiplied by k.

E0

ak

5Ea

0

##

1 2 3

c(b 2 a)

ma2

2

y

x2 4 6 8

-8

-4

4

8


1.

2. a. 11b. 4 c. 9

3. t hours, f(t) gal/hr, ∆t hours,f(t) ∆t gallons

4. a. i. 11,000 gal ii. 7400 gal iii. 18,400 gal

b. the amount of waterthat flowed through thepipes during the 24-hourperiod for each pipe andfor both pipes togetherc. They are nearly equal.

5.

For each xi, g(xi) 5f(xi), so

g(x) dx, the area between

g(x) and the x-axis, will be

5f(x) dx 5 f(x) dx.

6. x2 dx

7. 2 sin x dxEb

a

E14

6

Eπ

05Eπ

0

Eπ

0

5

π

2

4

y

x

f(x) = sin x

g(x) = 5 sin x

(8 2 2x) dxE4

-236 5

dx 5 35 1 1 5(8 2 2x) E4

3

(8 2 2x) dx 1E3

-28. 3 log x dx

9. a. 7a and

b.c. answer from b:

;

area of trapezoid:

10. a. [f(x) g(x)] dx

b. The area between thegraphs of f and g (from ato b) is the area between f and the x-axis minus thearea between g and the x-axis.

11. Sample: Let f(x) x, g(x) x, and a 1.

Then x dx • x dx

, but

x2 dx .

12. a. (f(zi) g(zi)) ∆x

b. (f(zi) g(zi)) ∆x

c. (f(x) g(x)) dx

13. a. 39.27b. 2.04

14. 175 ft

2E2

-3

2on

i51

2

513

3 513E1

0

12

2 • 12

2 514

5E1

0E1

0

555

2Eb

a

4 • 7 1 192 5 52

3 • 42

2 1 7(4) 5 52

3 • a2

2 1 7a

a2

2

E4

3


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15. Sample:a. (24, 30, 26) b.

c.

16.

< 52%

go to grad school

go tograd school

not go tograd school

not go tograd school

humanities

science

.45 .60

.40

.30

.70

.55

Hx 5 1 1 12ty 5 2 1 15tz 5 -4 1 13t

12x 1 15y 1 13z 5 -10

17. 50π < 157.08 cm3

18. 33.51 ft3

19. ma2

2 1 ca

32π3 ø



1. a. , with i 1, 2, 3, K, 25

b. .3536 units2

2. 9

3.

4.

5. a. x2 dx

b.

6. 45

7. 600

8. 492

2083

E6

2

40003 units2

-10 10 x

y100

24223

5i

25 9. 125

10.

11. a.

b. < 0.162 c. < 0.624 units2

12. 8 ft3

13. 12

14. a. t2 dt

b.

15. 116

16.

(b 1 a)(b 2 a)2

a b

y

x

y = x

73

3 5 114.3

E7

0

π4

a3

c

17.

18. 125,970

19. a. f9(x) 4xb.c.

20. a.b.

21. a. (i2 5i 2)

(-2) (-4) (-4) (-2)2 8 16 14;

i2 140, 5i 140,

2 14, so i2

5i 2 145o7

i511o

7

i51

2o7

i515o

7

i51

5o7

i515o

7

i51

5111111

512o7

i51

imaginary

real

w = 5 + 2i

w = 5 + 2i

-2 2-2

-4

2

4

-4 4

_

5 1 2i

-2 -1 1 2

1

2

3

y

x

x 5 05

v3

v1v2

v4

b. Sample:

(ai2 bi c) ai2

bi c

22. x3

23. 1.5

24.

25. An Archimedean screwconsists of a spiral passagewithin an inclined cylinder.It is used for raising waterto a certain height. This isachieved by rotating thespiral in the cylinder.

26. Using wood with areasonably consistentdensity, Archimedes couldhave weighed a rectangularpiece of wood andmeasured its area. Then hecould have cut out aparabolic region andweighed it. The weights ofthe parabolic region andthe rectangular piecewould be proportional totheir areas.

27. a. Sample:

b. Answers may vary.

a4

4

Ï22

on

i511o

n

i51

1on

i51511o

n

i51


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In-class Activity1. a. A truncated cone with

infinite lateral height.b. A truncated cone withlateral height PQ.c. v πr2h where r distance between x and

and h length of PQ

2. a. sphereb. torus (doughnut)c. circle Cd. a torus with no hole andwidth 2C

3. Sample: Earth rotatesaround its polar axis.

Lesson 1. 41.25π < 129.59 in3

2. a. cylinderb. (15.625π)1.3 < 63.8 in3

3. < 632.5 units3

4. The radius of thecross-section is the heightof the graph where

since , so

.

5. kf(x) dx k f(x) dx for

k a constant. But r is aconstant, and so r2 is a

constant. So r2 dx

r2 1 dx or r2 dx.Er

-rEr

-r

5Er

-r

Ea

b5Ea

b

f(zi ) 5 Ïr2 2 zi2

f(x) 5 Ïr2 2 x2

x 5 zi

604π3

5PQ5

5

6. (-r, 0) and (r, 0) are the leftand right endpoints of thesemicircle.

7. a.

b. cylinderc. 4 units

d. πri2 dx 16π dx

48πunits3

e. πr2h π(42)(3)48πunits3

8. a.

b. 8πunits3

9. a.

b.

10. a. (2x2 16x 33) dx

b. 9 units2

12E5

2

196π3 ø 205.25 units2

y

x2 4 6

2

4

6

2 4

2

y

x

55

5E3

05E3

0

2 4 6

2

4

6y

x = 3

y = 4

x


c

11.

12. 11

13. π

14. log x7 dx or 7 log x dx

15. 21

16. 2ax b a∆x

17. a. -`b. -`c. 4

11

E11

3E11

3

2452 18. See below.

19. a is a factor of for some integer m. So

, so a isa factor of bc. (a is a factorof b is not needed.)

20. 8πunits3

bc 5 b(am) 5 a(bm)

c ⇔ c 5 am


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c

18. left side

2cos2 x 1 2sin2 x

2cos2 x 1 2(1 cos2 x)2cos2 x 1 2 2cos2 x4cos2 x 3 right side5

251225

2225

225

2cos3 x 2 cos x 2 2sin2 x cos xcos x5

(2cos2 x 2 1) cos x 2 (2sin x cos x)cos x5

cos 2x cos x 2 sin 2x sin xcos x5

cos (2x 1 x)cos x5

cos 3xcosx5

formulas for cos (a b)and sin (a b)

; real numbers for which cos x 0since sin2 x 1 cos2 x25

Þ

11

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1. Isaac Newton and GottfriedLeibniz

2. Many different functionshave the same derivative.

3. 1 4. antiderivative

5. ln x 6. c

7. ln 20 ln 5 < 1.386

8. a. 0.5b. 0.5043; the result agreesvery well; the relative erroris only about 0.8%.

9. Fundamental Theorem ofArithmetic: Suppose that nis an integer and that . Then either n is a primenumber or n has a primefactorization which isunique except for the orderof the factors. FundamentalTheorem of Algebra: Everypolynomial of degree with real or complexcoefficients has at least onecomplex zero.

n $ 1

n . 1

2

10. a.

b.

11. 112 ft3

12. -47.5

13. 13

14. (1 sin 2x) dx

15. Sample: e1e2e3e4e5e7e6

16. No Euler circuit exists.


1E0

5π4

E4

0

y

x

(4, 4)

2 6

2

4

6

4


1. 3600

2. 2025

3. a.

b. 0.500

4. 32

5. 4

6.

7. a.

b. < 5.59

8. 72

9. 221.

10. 21

11. 240

12. 2x dx

13. 1 dx

14. False

15. y

x

f (x )

I II

a c b

E3

1

E10

4

6

1 3 5

1

3

5y

x

y = 25 – x2

π4

π600 sin (π

3 1 i π600)o

100

i51

Area I is f(x) dx, Area II is

f (x) dx, and the union of

the regions represented by

Areas I and II is f(x) dx.

Since Area I Area IIequals the area of theunion of the regionsrepresented by Areas I and

II, then f(x) dx

f(x) dx f(x) dx.

16. a. 29.4 m b. 158.4 m c. 240 m

17. 316.8 ft

18. 1295 ft

19. 96 m3

20. a. (f(t) g(t)) dt

b. 69.5 gal

21. 465.75π < 1463 in3

22. a. 60 ft/sec b. 120 ft c. 90 ft d. 210 ft

23. 4635 ft

1E12

0

Eb

a5Eb

c

1Ec

a

1

Eb

a

Eb

c

Ec

a


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24.

25. (-(x 3)2 2) dx

26. a. x3 dx

b. negative

27. a. < 120 units2

b. 93. units2

28. 37. units2

29. a.

b. (3x 2) dx 3251E4

0

2 4

2

6

10

14 (4, 14)

y

x

3

3

E1

-2

12E4

1

E3

-4 zx z dx 30. a.

b. ≈ 130.9 units3

31. a.

b. 2πunits3

y

x1 2

1

2

xy =

125π3

y

x

(4, 5)

2 6

1

3

5

4


Precalc Answer

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