anddifferentsecondcoordinates.Wehavea function.
32. Sincetheorderedpairsinthegraphof y = x2 3x +7are(x,x2 3x +7),thereare notwoorderedpairswiththesamefirstcoordinateanddifferentsecondcoordinates.We haveafunction.
33. Since y =(x +9)/3,theorderedpairsare (x, (x +9)/3).Thus,therearenotwoordered pairswiththesamefirstcoordinateanddifferentsecondcoordinates.Wehaveafunction.
34. Since y = 3 √x,theorderedpairsare(x, 3 √x). Thus,therearenotwoorderedpairswiththe samefirstcoordinateanddifferentsecondcoordinates.Wehaveafunction.
35. Since y = ±x,theorderedpairsare(x, ±x). Thus,therearetwoorderedpairswiththe samefirstcoordinateanddifferentsecond coordinates.Wedonothaveafunction.
36. Since y = ±√9+ x2,theorderedpairsare (x, ±√9+ x2).Thus,therearetwoordered pairswiththesamefirstcoordinateand differentsecondcoordinates.Wedonot haveafunction.
37. Since y = x2,theorderedpairsare(x,x2). Thus,therearenotwoorderedpairswiththe samefirstcoordinateanddifferentsecond coordinates.Wehaveafunction.
38. Since y = x3,theorderedpairsare(x,x3). Thus,therearenotwoorderedpairswiththe samefirstcoordinateanddifferentsecond coordinates.Wehaveafunction.
39. Since y = |x|− 2,theorderedpairsare (x, |x|− 2).Thus,therearenotwoordered pairswiththesamefirstcoordinateanddifferentsecondcoordinates.Wehaveafunction.
40. Since y =1+ x2,theorderedpairsare (x, 1+ x2).Thus,therearenotwoordered pairswiththesamefirstcoordinateanddifferentsecondcoordinates.Wehaveafunction.
41. Since(2, 1)and(2, 1)aretwoorderedpairs withthesamefirstcoordinateanddifferent secondcoordinates,theequationdoesnot defineafunction.
42. Since(2, 1)and(2, 1)aretwoorderedpairs withthesamefirstcoordinateanddifferent secondcoordinates,theequationdoesnot defineafunction.
43. Domain {−3, 4, 5},range {1, 2, 6}
44. Domain {1, 2, 3, 4},range {2, 4, 8, 16}
45. Domain(−∞, ∞),range {4}
46. Domain {5},range(−∞, ∞)
47. Domain(−∞, ∞); since |x|≥ 0,therangeof y = |x| +5is[5, ∞)
48. Domain(−∞, ∞); since x2 ≥ 0,therangeof y = x2 +8is[8, ∞)
49. Since x = |y|− 3 ≥−3,thedomain of x = |y|− 3is[ 3, ∞);range(−∞, ∞)
50. Since √y 2 ≥−2,thedomainof x = √y 2 is[ 2, ∞);Since √y isarealnumberwhenever y ≥ 0,therangeis[0, ∞).
51. Since √x 4isarealnumberwhenever x ≥ 4, thedomainof y = √x 4is[4, ∞).
Since y = √x 4 ≥ 0for x ≥ 4,therangeis [0, ∞).
52. Since √5 x isarealnumberwhenever x ≤ 5, thedomainof y = √5 x is(−∞, 5].
Since y = √5 x ≥ 0for x ≤ 5,therangeis [0, ∞).
53. Since x = y2 ≤ 0,thedomainof x = y2 is (−∞, 0];rangeis(−∞, ∞);
54. Since x = −|y|≤ 0,thedomainof x = −|y| is(−∞, 0];rangeis(−∞, ∞);
55. 6 56. 5
57. g(2)=3(2)+5=11
58. g(4)=3(4)+5=17
59. Since(3, 8)istheorderedpair,oneobtains f (3)=8.Theansweris x =3.
60. Since(2, 6)istheorderedpair,oneobtains f (2)=6.Theansweris x =2.
61. Solving3x +5=26,wefind x =7.
62. Solving3x +5= 4,wefind x = 3.
63. f (4)+ g(4)=5+17=22
64. f (3) g(3)=8 14= 6
65. 3a2 a 66. 3w2 w
67. 4(a+2) 2=4a+6 68. 4(a 5) 2=4a 22
69. 3(x2 +2x +1) (x +1)=3x2 +5x +2
70. 3(x2 6x +9) (x 3)=3x2 19x +30
71. 4(x + h) 2=4x +4h 2
72. 3(x2 +2xh+h2) x h =3x2 +6xh+3h2 x h
73. 3(x2 +2x +1) (x +1) 3x2 + x =6x +2
74. 4(x +2) 2 4x +2=8
75. 3(x2 +2xh + h2) (x + h) 3x2 + x = 6xh +3h2 h
76. (4x +4h 2) 4x +2=4h
77. Theaveragerateofchangeis
8, 000 20, 000 5 = $2, 400peryear
78. Theaveragerateofchangeasthenumberof cubicyardschangesfrom12to30andfrom30 to60are 528 240 30 12 =$16peryd3 and 948 528 60 30 =$14peryd3,respectively.
79. Theaveragerateofchangeon[0, 2]is
h(2) h(0) 2 0 = 0 64 2 0 = 32ft/sec.
Theaveragerateofchangeon[1, 2]is
h(2) h(1) 2 1 = 0 48 2 1 = 48ft/sec.
Theaveragerateofchangeon[1.9, 2]is
h(2) h(1.9) 2 1 9 = 0 6.24 0 1 = 62 4ft/sec.
Theaveragerateofchangeon[1 99, 2]is
h(2) h(1 99) 2 1 99 = 0 0 6384 0 01 = 63.84ft/sec.
Theaveragerateofchangeon[1.999, 2]is
h(2) h(1 999) 2 1.999 = 0 0 063984 0.001 = 63 984 ft/sec.
80. 6 70 2 0 = 64 2 = 32ft/sec
81. Theaveragerateofchangeis 1768 1970 20 = 10 1millionhectaresperyear.
82. If10 1millionhectaresarelosteachyearand since 1970 10 1 ≈ 195years,thentheforestwill beeliminatedintheyear2183(=1988+195).
83.
86.
f (x + h) f (x) h = 2(x + h)+3+2x 3 h = 2h h = 2
87. Let g(x)= x2 + x.Thenweobtain
g(x + h) g(x) h = (x + h)2 +(x + h) x2 x h = 2xh + h2 + h h = 2x + h +1.
88. Let g(x)= x2 2x.Thenweget
g(x + h) g(x) h =
(x + h)2 2(x + h) x2 +2x h = 2xh + h2 2h h = 2x + h 2.
89. Differencequotientis
= (x + h)2 +(x + h) 2+ x2 x +2 h = 2xh h2 + h h = 2x h +1
90. Differencequotientis
= (x + h)2 (x + h)+3 x2 + x 3 h
= 2xh + h2 h h =2x + h 1
91. Differencequotientis
= 3√x + h 3√x h 3√x + h +3
92. Differencequotientis
= 2√x + h +2√x h 2√x + h 2√x 2√x + h 2√x
= 4(x + h) 4x h( 2√x + h 2√x) = 4h h( 2√x + h 2√x) = 2 √x + h + √x
93. Differencequotientis = √x + h +2 √x +2 h √x + h +2+ √x +2 √x + h +2+ √x +2 = (x + h +2) (x +2) h(√x + h +2+ √x +2)
= h h(√x + h +2+ √x +2) = 1 √x + h +2+ √x +2
94. Differencequotientis = x + h 2 x 2 h x + h 2 + x 2 x + h 2 + x 2 = x + h 2 x 2 h x + h 2 + x 2 = h 2 h x + h 2 + x 2
= 1 2 x + h 2 + x 2 = 1 √2 √x + h + √x
x = 9(x + h) 9x h(3√x + h +3√x) = 9h h(3√x + h +3√x) = 3 √x + h + √x
95. Differencequotientis
= 1 x + h 1 x h · x(x + h) x(x + h) = x (x + h) xh(x + h) = h xh(x + h) = 1 x(x + h)
96. Differencequotientis = 3 x + h 3 x h x(x + h) x(x + h) = 3x 3(x + h) xh(x + h) = 3h xh(x + h) = 3 x(x + h)
97. Differencequotientis = 3 x + h +2 3 x +2 h (x + h +2)(x +2) (x + h +2)(x +2) = 3(x +2) 3(x + h +2) h(x + h +2)(x +2) = 3h h(x + h +2)(x +2)
= 3 (x + h +2)(x +2)
98. Differencequotientis = 2 x + h 1 2 x 1 h · (x + h 1)(x 1) (x + h 1)(x 1) = 2(x 1) 2(x + h 1) h(x + h 1)(x 1) = 2h h(x + h 1)(x 1) = 2 (x + h 1)(x 1)
99.a) A = s2 b) s = √A c) s = d√2 2 d) d = s√2 e) P =4s f) s = P/4 g) A = P 2/16 h) d = √2A
100.a) A = πr2 b) r = A π c) C =2πr
d) d =2r e) d = C π f) A = πd2 4
g) d =2 A π
101. C =50+35n
102.a) When d =100ft,theatmosphericpressure is A(100)= 03(100)+1=4atm.
b) When A =4 9atm,thedepthisfoundby solving4 9=0 03d +1;thedepthis d = 3 9 0 03 =130ft.
103.
(a) Thequantity C(4)=(0 95)(4)+5 8= $9 6billionrepresentstheamountspent oncomputersintheyear2004.
(b) Bysolving0 95n +5 8=15,weobtain n = 9 2 0.95 ≈ 10
Thus,spendingforcomputerswillbe$15 billionin2010.
104.
(a) Thequantity E(4)+ C(4)=[0 5(4)+1]+ 9 6=$12 6billionrepresentsthetotal amountspentonelectronicsandcomputersintheyear2004.
(b) Bysolving
(0.5n +1)+(0.95n +5.8)=20 1.45n =13.2 n ≈ 9
wefindthatthetotalspendingwillreach $20billionintheyear2009(=2000+9).
(c) Theamountspentoncomputersisgrowingfastersincetheslopeof C(n)[whichis 1]isgreaterthantheslopeof E(n)[which is0.95].
105. Let a betheradiusofeachcircle.Note,triangle ABC isanequilateraltrianglewithside 2a andheight √3a
Thus,theheightofthecirclecenteredat C fromthehorizontallineis √3a +2a.Hence, byusingasimilarreasoning,weobtainthat heightofthehighestcirclefromthelineis
2√3a +2a orequivalently(2√3+2)a
106. Inthetrianglebelow, PS bisectsthe90-angle at P and SQ bisectsthe60-angleat Q
107. When x =18and h =0 1,wehave R(18.1) R(18) 0.1 =1, 950
Therevenuefromtheconcertwillincreaseby approximately$1,950ifthepriceofaticketis raisedfrom$18to$19.
If x =22and h =0 1,then R(22.1) R(22) 0 1 = 2, 050.
Therevenuefromtheconcertwilldecreaseby approximately$2,050ifthepriceofaticketis raisedfrom$22to$23.
108. When r =1.4and h =0.1,weobtain A(1 5) A(1 4) 0 1 ≈−16.1
Theamountoftinneededdecreasesbyapproximately16 1in.2 iftheradiusincreasesfrom 1.4in.to2.4in.
If r =2and h =0.1,then A(2 1) A(2) 0.1 ≈ 8 6
Theamountoftinneededincreasesbyabout 8.6in.2 iftheradiusincreasesfrom2in.to3 in.
Inthe45-45-90triangle SPR,wefind
PR = SR = √2d/2
And,inthe30-60-90triangle SQR weget
Since PQ = PR + RQ,weobtain
112. If m isthenumberofmales,then m + 1 2 m =36 3 2 m =36 m =(36) 2 3 x =24males
113. ( 4+6)2 +( 3 3)2 = √4+36= √40=2√10
114. Theslopeis 3 2 5+1 = 1 6 .Thelineisgiven by y = 1 6 x + b forsome b.Substitutethe coordinatesof( 1, 2)asfollows: 2= 1 6 ( 1)+ b 13 6 = b
Thelineisgivenby y = 1 6 x + 13 6 .
115. x 2 x 6=36 x 2 x 42=0 (x 7)(x +6)=0
Thesolutionsetis {−6, 7}
116. Theinequalityisequivalentto 13 < 2x 9 < 13 4 < 2x< 22 2 < 2x< 11
Thesolutionsetis( 2, 11).
(30+25)2 =3025
2.1PopQuiz
1. Yes,since A = πr2 where A istheareaofa circlewithradius r
2. No,sincetheorderedpairs(2, 4)and( 2, 4) havethesamefirstcoordinates.
3. No,sincetheorderedpairs(1, 0)and( 1, 0) havethesamefirstcoordinates.
4. [1, ∞) 5. [2, ∞) 6. 9
7. If2a =1,then a =1/2.
8. 40 20 2008 1998 =$2peryear
9. Thedifferencequotientis
2.1LinkingConcepts
(a) ThefirstgraphshowsU.S.federaldebt D versusyear y
andthesecondgraphshowspopulation P (inmillions)versus y.
(b) Thefirsttableshowstheaverageratesof changefortheU.S.federaldebt
10 yearperiod ave.rateofchange
1940 50 257 51 10 =20.6 1950 60 291 257 10 =3 4
1960 70 381 291 10 =9 0 1970 80 909 381 10 =52.8 1980 90 3207 909 10 =229 8 1990 2000 5666 3207 10 =245.9
Thesecondtableshowstheaverageratesof changefortheU.S.population
10 yearperiod ave.rateofchange
1940 50 150 7 131 7 10 ≈ 1.9 1950 60 179 3 150 7 10 ≈ 2 9 1960 70 203.3 179.3 10 ≈ 2 4 1970 80 226 5 203 3 10 ≈ 2.3 1980 90 248 7 226 5 10 ≈ 2 2 1990 2000 274 8 248 7 10 ≈ 2.6
(c) Thefirsttableshowsthedifferencebetween consecutiveaverageratesofchangeforthe U.S.federaldebt.
10-yearperiods difference
1940-50&1950-60 3 4 20 6= 17 2
1950-60&1960-70 9.0 3.4=5.6
Thesecondtableshowsthedifferencebetween consecutiveaverageratesofchangeforthe U.S.population.
10-yearperiods difference
1940-50&1950-60 2 9 1 9=1 0
1950-60&1960-70 2 4 2 9= 0 5
1960-70&1970-80 2.3 2.4=
(d) ForboththeU.S.federaldebtandpopulation, theaverageratesofchangeareallpositive.
(e) Inpart(c),forthefederaldebtmostofthe differencesarepositiveandforthepopulation mostofthedifferencesarenegative.
(f) TheU.S.federaldebtisgrowingoutofcontrol whencomparedtotheU.S.population.See part(g)foranexplanation.
(g) Sincemostofthedifferencesforthefederal debtinpart(e)arepositive,thefederaldebts areincreasingatanincreasingrate.Whilethe U.S.populationisincreasingatadecreasing ratesincemostofthedifferencesforpopulationinpart(e)arenegative.
ForThought
1. True,sincethegraphisaparabolaopening downwithvertexattheorigin.
2. False,thegraphisdecreasing.
3. True
4. True,since f ( 4.5)=[ 1.5]= 2.
5. False,sincetherangeis {±1}.
6. True 7. True 8. True
9. False,sincetherangeistheinterval[0, 4].
10. True
2.2Exercises
1. parabola
2. piecewise
3. Function y =2x includesthepoints(0, 0), (1, 2), domainandrangeareboth(−∞, ∞) 1 2 x 2 4 y
4. Function x =2y includesthepoints(0, 0), (2, 1), ( 2, 1),domainandrangeareboth (−∞, ∞) 2 -2 x 1 -1 y
5. Function x y =0includesthepoints( 1, 1), (0, 0), (1, 1), domainandrangeareboth (−∞, ∞)
9. Function y =2x2 includesthepoints(0, 0), (±1, 2),domainis(−∞, ∞),rangeis[0, ∞) 1 -1 x 2 8 y
10. Function y = x2 1goesthrough(0, 1), (±1, 0),domainis(−∞, ∞),rangeis[ 1, ∞) 1 -1 x 1 4 y
6. Function x y =2includesthepoints(2, 0), (0, 2), ( 2, 4),domainandrangeareboth (−∞, ∞) 2 -2 x -2 -4 y
7. Function y =5includesthepoints(0, 5), (±2, 5),domainis(−∞, ∞),rangeis {5} 5 -5 x 4 6 y
11. Function y =1 x2 includesthepoints(0, 1), (±1, 0),domainis(−∞, ∞),rangeis(−∞, 1] 1 -1 x 2 -4 y
12. Function y = 1 x2 includesthepoints (0, 1),(±1, 2),domainis(−∞, ∞),range is(−∞, 1] 1 -1 x -4 y
8. x =3isnotafunctionandincludesthepoints (3, 0), (3, 2),domainis {3},rangeis(−∞, ∞)
13. Function y =1+ √x includesthepoints(0, 1), (1, 2), (4, 3),domainis[0, ∞),rangeis[1, ∞) 4 1 x 1 4 y
14. Function y =2 √x includesthepoints(0, 2), (4, 0),domainis[0, ∞),rangeis(−∞, 2]
17. Function x = √y goesthrough (0, 0), (2, 4), (3, 9),domainandrangeis[0, ∞)
15. x = y2 +1isnotafunctionandincludesthe points(1, 0), (2, ±1),domainis[1, ∞),range is(−∞, ∞) 2 1 x 1 -1 y
16. x =1 y2 isnotfunctionandincludesthe points(1, 0), (0, ±1),domainis(−∞, 1],range is(−∞, ∞)
18. Function x 1= √y goesthrough(1, 0), (3, 4), (4, 9),domain[1, ∞),andrange[0, ∞), 1 4 x 2 -2 y
19. Function y = 3 √x +1goesthrough ( 1, 0), (1, 2), (8, 3),domain(−∞, ∞), andrange(−∞, ∞)
20. Function y = 3 √x 2goesthrough ( 1, 3), (1, 1), (8, 0),domain(−∞, ∞), andrange(−∞, ∞) -4 x 1 -2 y
21. Function, x = 3 √y goesthrough (0, 0), (1, 1), (2, 8),domain(−∞, ∞), andrange(−∞, ∞) -8 8 x 2 y
24. Notafunction, x2 + y2 =4goesthrough (2, 0), (0, 2), ( 2, 0),domain[ 2, 2], andrange[ 2, 2] 1 3 x 1 3 y
22. Function, x = 3 √y 1goesthrough (0, 1), (1, 2), ( 1, 0),domain(−∞, ∞), andrange(−∞, ∞) -3 3 x 2 y
25. Function, y = √1 x2 goesthrough (±1, 0), (0, 1),domain[ 1, 1], andrange[0, 1] 2 -2 x -1 y
23. Notafunction, y2 =1 x2 goesthrough (1, 0), (0, 1), ( 1, 0),domain[ 1, 1], andrange[ 1, 1] -2 2 x 2 -2 y
26. Function, y = √25 x2 goesthrough (±5, 0), (0, 5),domain[ 5, 5], andrange[ 5, 0]
4 6 x -4 -6 y
27. Function y = x3 includesthepoints(0, 0), (1, 1), (2, 8),domainandrangeareboth (−∞, ∞) 1 2 x 1 8 y
31. Function y = −|x| includesthepoints(0, 0), (±1, 1),domainis(−∞, ∞),rangeis (−∞, 0]
28. Function y = x3 includesthepoints(0, 0), (1, 1), (2, 8),domainandrangeareboth (−∞, ∞)
32. Function y = −|x +1| includesthepoints ( 1, 0),(0, 1), ( 2, 1),domainis(−∞, ∞), rangeis(−∞, 0] 1 -4 x -4 4 y
29. Function y =2|x| includesthepoints(0, 0), (±1, 2), domainis(−∞, ∞),rangeis[0, ∞) 1 -1 x 2 5 y
33. Notafunction,graphof x = |y| includesthe points(0, 0), (2, 2), (2, 2),domainis[0, ∞), rangeis(−∞, ∞)
30. Function y = |x 1| includesthepoints (0, 1), (1, 0), (2, 1),domainis(−∞, ∞),range is[0, ∞)
34. x = |y| +1isnotafunctionandincludesthe points(1, 0), (2, ±1),domainis[1, ∞),range is(−∞, ∞) 1 2 x
y
35. Domainis(−∞, ∞),rangeis {±2},some pointsare( 3, 2),(1, 2) -1 1 x
39. Domainis[ 2, ∞),rangeis(−∞, 2], somepointsare(2, 2),( 2, 0),(3, 1)
36. Domainis(−∞, ∞),rangeis {1, 3},some pointsare(0, 2),(4, 1)
40. Domainis(−∞, ∞),rangeis( 1, ∞), somepointsare(1, 1),(4, 2),( 1, 1)
-
37. Domainis(−∞, ∞),rangeis (−∞, 2] ∪ (2, ∞),somepointsare(2, 3), (1, 2)
41. Domainis(−∞, ∞),rangeis[0, ∞),some pointsare( 1, 1),( 4, 2),(4, 2)
2 x 4 2 y
42. Domainis(−∞, ∞),rangeis[3, ∞),some pointsare( 1, 3),(0, 3),(1, 4)
38. Domainis(−∞, ∞),rangeis[3, ∞), somepointsare(2, 3),(3, 4) 2 -2 x 1 3 y
43. Domainis(−∞, ∞),rangeis(−∞, ∞), somepointsare( 2, 4),(1, 1) 1 -1 x 1 3 y
44. Domainis[ 2, ∞),rangeis[0, ∞), somepointsare(±2, 0),(3, 1)
47. Domain[0, 4),rangeis {2, 3, 4, 5},somepoints are(0, 2),(1, 3),(1.5, 3)
45. Domainis(−∞, ∞),rangeisthesetofintegers,somepointsare(0, 1),(1, 2),(1.5, 2) -1 1 x 1 2 y
48. Domainis(0, 5],rangeis {−3, 2, 1, 0, 1, 2}, somepointsare(0, 3),(1, 2),(1.5, 2) 1 3 5 x
2 y
46. Domainis(−∞, ∞),rangeisthesetofeven integers,somepointsare(0, 0),(1, 2),(1 5, 2) -1 1 x -1 2 y
49.a. Domainandrangeareboth(−∞, ∞), decreasingon(−∞, ∞)
b. Domainis(−∞, ∞),rangeis(−∞, 4] increasingon(−∞, 0),decreasing on(0, ∞)
50.a. Domainandrangeareboth(−∞, ∞), increasingon(−∞, ∞)
b. Domainis(−∞, ∞),rangeis[ 3, ∞) increasingon(0, ∞),decreasing on(−∞, 0)
51.a. Domainis[ 2, 6],rangeis[3, 7] increasingon( 2, 2),decreasingon(2, 6)
b. Domain(−∞, 2],range(−∞, 3], increasingon(−∞, 2),constant on( 2, 2)
52.a. Domainis[0, 6],rangeis[ 4, 1] increasingon(3, 6),decreasingon(0, 3)
b. Domain(−∞, ∞),range[1, ∞), increasingon(3, ∞),constanton(1, 3), decreasingon(−∞, 1)
53.a. Domainis(−∞, ∞),rangeis[0, ∞) increasingon(0, ∞),decreasing on(−∞, 0)
b. Domainandrangeareboth(−∞, ∞) increasingon( 2, 2/3), decreasingon(−∞, 2)and( 2/3, ∞)
54.a. Domainis[ 4, 4],rangeis[0, 4] increasingon( 4, 0),decreasingon(0, 4)
b. Domainis(−∞, ∞),rangeis[ 2, ∞) increasingon(2, ∞),decreasing on(−∞, 2),constanton( 2, 2)
55.a. Domainandrangeareboth(−∞, ∞), increasingon(−∞, ∞)
b. Domainis[ 2, 5],rangeis[1, 4] increasingon(1, 2),decreasing on( 2, 1),constanton(2, 5)
56.a. Domainis(−∞, ∞),rangeis(−∞, 3] increasingon(−∞, 2),decreasing on(2, ∞)
b. Domainandrangeareboth(−∞, ∞), decreasingon(−∞, ∞)
57. Domainandrangeareboth(−∞, ∞) increasingon(−∞, ∞),somepointsare(0, 1), (1, 3)
y
58. Domainandrangeareboth(−∞, ∞), decreasingon(−∞, ∞),somepointsare(0, 0), (1, 3)
60. Domainis(−∞, ∞),rangeis[1, ∞), increasingon(0, ∞),decreasingon(−∞, 0), somepointsare(0, 1),( 1, 2) 2 -2 x 1 3 y
61. Domainis(−∞, 0) ∪ (0, ∞),rangeis {±1}, constanton(−∞, 0)and(0, ∞), somepointsare(1, 1),( 1, 1)
59. Domainis(−∞, ∞),rangeis[0, ∞), increasingon(1, ∞),decreasingon(−∞, 1), somepointsare(0, 1),(1, 0)
62. Domainis(−∞, 0) ∪ (0, ∞),rangeis {±2}, constanton(−∞, 0)and(0, ∞), somepointsare(1, 2),( 1, 2) 2 -2 x 1 3 y
63. Domainis[ 3, 3],rangeis[0, 3], increasingon( 3, 0),decreasingon(0, 3), somepointsare(±3, 0),(0, 3) 2 4 x
64. Domainis[ 1, 1],rangeis[ 1, 0], increasingon(0, 1),decreasingon( 1, 0), somepointsare(±1, 0),(0, 1)
2 -2 x -2 1 y
65. Domainandrangeareboth(−∞, ∞), increasingon(−∞, 3)and(3, ∞), somepointsare(4, 5),(0, 2) 3 6 x 5 4 y
66. Domainandrangeareboth(−∞, ∞), decreasingon(−∞, ∞), somepointsare( 1, 1),(1, 1)
4 -4 x 2 -2 y
67. Domainis(−∞, ∞),rangeis(−∞, 2], increasingon(−∞, 2)and( 2, 0), decreasingon(0, 2)and(2, ∞),somepoints are( 3, 0),(0, 2),(4, 1)
2 -2 x 1 3 y
68. Domainis(−∞, ∞),rangeis(−∞, 4], increasingon(−∞, 2)and(0, 2),decreasing on( 2, 0)and(2, ∞),somepointsare( 3, 2), (0, 0),(3, 2) 2 -2 x 2 4 y
69. f (x)= 2for x> 1 1for x ≤−1
70. f (x)= 3for x ≤ 1 2for x> 1
71. Thelinejoining( 1, 1)and( 3, 3)is y = x, andthelinejoining( 1, 2)and(3, 2)is y = x 1.Thepiecewisefunctionis
f (x)= x 1for x ≥−1 x for x< 1.
72. Thelinejoining(1, 3)and( 3, 1)is y = x+2,andthelinejoining(1, 1)and(3, 1) is y =2 x.Thepiecewisefunctionis
f (x)= x +2for x ≤ 1 2 x for x> 1
73. Thelinejoining(0, 2)and(2, 2)is y =2x 2, andthelinejoining(0, 2)and( 3, 1)is y = x 2.Thepiecewisefunctionis
f (x)= 2x 2for x ≥ 0 x 2for x< 0
74. Thelinejoining(1, 3)and(3, 1)is y =4 x, andthelinejoining(1, 3)and( 2, 3)is y =2x +1.Thepiecewisefunctionis
f (x)= 2x +1for x ≤ 1 4 x for x> 1.
75. increasingontheinterval[0.83, ∞), decreasingon(−∞, 0.83]
76. increasingontheinterval(−∞, 0.17], decreasingon[0.17, ∞)
77. increasingon(−∞, 1]and[1, ∞), decreasingon[ 1, 1]
78. increasingon[ 2.35, 0)and[2.35, ∞), decreasingon(−∞, 2.35]and(0, 2.35]
79. increasingon[ 1.73, 0)and[1.73, ∞), decreasingon(−∞, 1 73]and(0, 1 73]
80. increasingon(−∞, 2 59], [ 1 03, 1 03],and[2 59, ∞), decreasingon[ 2 59, 1 03]and[1 03, 2 59]
81. increasingon[30, 50],and[70, ∞), decreasingon(−∞, 30]and[50, 70]
82. increasingon(−∞, 50],and[ 30, ∞), decreasingon[ 50, 30]
83. c,graphwasincreasingatfirst,thensuddenly droppedandbecameconstant,thenincreased slightly
84. a
85. d,graphwasdecreasingatfirst,then fluctuatedbetweenincreasesanddecreases, thenthemarketincreased
86. b
87. Theindependentvariableistime t where t isthenumberofminutesafter7:45andthe dependentvariableisdistance D fromthe holodeck.
D isincreasingontheintervals[0, 3]and [6, 15],decreasingon[3, 6]and[30, 39],and constanton[15, 30].
3 6 t D
88. Independentvariableistime t inseconds,dependentvariableisdistance D fromthepit D isincreasingontheintervals(0, 20), (40, 60),(80, 100),(150, 170),(190, 210),and (230, 250); D isdecreasingontheintervals(20, 40),(60, 80),(100, 120),(170, 190), (210, 230); D isconstanton(120, 150)and (250, 300).
40 80 120 190 250 300 t
89. Independentvariableistime t inyears, dependentvariableissavings s indollars s isincreasingontheinterval[0, 2]; s isconstanton[2, 2 5]; s isdecreasingon[2 5, 4 5].
90. Independentvariableistime t indays,dependentvariableistheamount,a,(indollars)in thecheckingaccount.
a isdecreasingontheinterval[0, 40]; a isconstanton[40, 45]; a isincreasingon[45, 65].
91. In1988,therewere M (18)=565millioncars. In2010,itisprojectedthattherewillbe M (40)=800millioncars.
Theaveragerateofchangefrom1984to1994 is M (24) M (14) 10 =14 5millioncarsper year.
92. IndevelopingcountriesandEasternEurope, theaveragerateofchangeofmotorvehicle ownershipis M (40) M (20) 20 =6 25million vehiclesperyear.
Indevelopedcountries,theaveragerateof changeofmotorvehicleownershipfrom1990 to2010is10millionvehiclesperyear(seepreviousmodel).Thenvehicleownershipisexpectedtogrowfasterindevelopedcountries.
93. Constanton[0, 104],increasingon [104 , ∞)
94. Decreasingsinceacarhasalowermileage athighspeeds.
95. Thecostisover$235for t in[5, ∞).
96. If200+37[w/100] < 862then [w/100] < 662/37 ≈ 17.9.Then w/100 < 18 andthevaluesof w liein(0, 1800).
97. f (x)= 150if0 <x< 3 50x if3 ≤ x ≤ 10
98. f (x)= 4[ x]if0 <x ≤ 3 15if3 <x ≤ 8
99. Amechanic’sfeeis$20foreachhalf-hourof workwithanyfractionofahalf-hourcharged asahalf-hour.If y isthefeeindollarsand x isthenumberofhours,then y = 20[ 2x]. .5 1
100. Forexample,chooseany a,b satisfying a<b andgraphthefunctiondefinedby f (x)=
2x if x ≤ a 5x 3a if a<x<b 5b 3a if x ≥ b
101. Since x 2 ≥ 0,thedomainis[2, ∞). Since √x 2isnonnegative,wefind √x 2+ 3isatleastthree.Thus,therangeis[3, ∞).
102. Sinceanabsolutevalueisnonnegative,the equation |13x 55| = 9hasnosolution.The solutionsetis ∅.
103. Note,wehave13x 55=0. Thesolutionsetis {55/13}
104. Rewritewithouttheabsolutevalues: 13x 55= ±9 13x =55 ± 9=46, 64
Thesolutionsetis 46 13 , 64 13
105. Sincewehaveaperfectsquare (2x 5)2 =0 thesolutionsetis {5/2}.
106. Applyingthequadraticformulato 2w2 5w 9=0,wefind w = 5 ± √25+72 4
Thesolutionsetis 5 ± √97 4 .
ThinkingOutsidetheBoxXXIII
a) Considerthecircle x2 +(y r)2 = r2 where r> 0.Supposethecircleintersectstheparabola y = x2 onlyattheorigin.Substituting y = x2 , weobtain
y +(y r)2 = r 2 y 2 + y(1 2r)=0
Thus,1 2r =0sincethecircleandthe parabolahasexactlyonepointofintersection. Theradiusofthecircleis1/2.
b) Considerthecircle x2 +(y r)2 = r2 where r> 0.Ifthecircleintersectstheparabola y = ax2 onlyattheorigin,thentheequation belowmusthaveexactlyonesolution,namely, y =0. 1 a y +(y r)2 = r 2 y 2 + y 1 a 2r =0
Necessarily,wehave 1 a 2r =0or a = 1 2r .Thus,if r =3then a =1/6.
2.2PopQuiz
1. Domain[0, ∞),range(−∞, 1]
2. Domain[ 3, 3],range[0, 3]
3. Range[2, ∞)
4. Increasingon[0, ∞)
5. Decreasingon(−∞, 3]
2.2LinkingConcepts
(a) Agraphofthebenefitfunctionisgivenbelow.
(b) B(64)=804(64) 41, 064=$10392
(c) If B =$14, 880,then 960a 51, 360=14, 880 960a =66, 240 a =69. Atage69years,theannualbenefitis$14,880.
(d) Sincethefunctionispiecewisedefinedconsistingoflinearfunctions,theaveragerateof changeistheslopeofthelinearfunction.
Thentheaveragerateofchangeforages62-63 is600.
Theaveragerateofchangeforages64-66is 804.
Theaveragerateofchangeforages67-70is 960.
(e) Yes,theanswerstopart(d)aretheslopesof thethreelinesinthepiecewisefunctiondefiningthebenefitformula.
(f) Note, B(62)=$9000.Thenthetotalamount Bobexpectstowithdrawis
(67 0166(1 00308)62 62)$9000 ≈ $171, 800
(g) Note, B(70)=$15, 840.Thenthetotal amountBillexpectstowithdrawis
(67.0166(1.00308)70 70)$15, 840 ≈ $207, 700.
ForThought
1. False,itisareflectioninthey-axis.
2. True 3. False,ratheritisalefttranslation.
4. True 5. True
6. False,thedownshiftshouldcomeafterthe reflection. 7. True
8. False,sincetheirdomainsaredifferent.
9. True 10. True
2.3Exercises
1. rigid
2. nonrigid
3. parabola
4. translation
5. reflection
6. identity
7. linear
8. constant
9. odd
10. even
11. f (x)= |x|,g(x)= |x|− 4
19. f (x)= √x, g(x)= √x 2 4 x
20. f (x)= x,g(x)= x
43. y = 3|x 7| +9
44. y = 2(x +6) 8or y = 2x 20
45. y =(x 1)2 +2;rightby1,upby2, domain(−∞, ∞),range[2, ∞) 1 -2 x 2 11 y
46. y =(x +5)2 4;leftby5,downby4, domain(−∞, ∞),range[ 4, ∞)
49. y =3x 40, domainandrangeareboth(
50. y = 4x +200, domainandrangeareboth(
47. y = |x 1| +3;rightby1,upby3 domain(−∞, ∞),range[3, ∞) 1 3 x 3 5 y
48. y = |x +3|− 4;leftby3,downby4 domain(−∞, ∞),range[ 4, ∞)
51. y = 1 2 x 20, domainandrangeareboth(−∞, ∞)
52. y = 1 2 x +40, domainandrangeareboth(−∞, ∞)
53. y = 1 2 |x| +40,shrinkby1/2, reflectabout x-axis,upby40, domain(−∞, ∞),range(−∞, 40]
y
x
54. y =3|x|− 200,stretchby3,downby200, domain(−∞, ∞),range[ 200, ∞)
y
55. y = 1 2 |x +4|,leftby4, reflectabout x-axis,shrinkby1/2, domain(−∞, ∞),range(−∞, 0] -3 -5 1 x -1 -3 1 y
56. y =3|x 2|,rightby2,stretchby3, domain(−∞, ∞),range[0, ∞) 2 1 x
57. y = √x 3+1,rightby3, reflectabout x-axis,upby1, domain[3, ∞),range(−∞, 1] 3 12 x
58. y = √x +2 4,leftby2, reflectabout x-axis,downby4, domain[ 2, ∞),range(−∞, 4]
x -4 -6 y
59. y = 2√x +3+2,leftby3,stretchby2, reflectabout x-axis,upby2, domain[ 3, ∞),range(−∞, 2] 6
60. y = 1 2 √x +2+4,leftby2,shrinkby1/2, reflectabout x-axis,upby4, domain[ 2, ∞),range(−∞, 4] 2 -2 x 4 2 y
61. Symmetricabouty-axis,evenfunction since f ( x)= f (x)
62. Symmetricabouty-axis,evenfunction since f ( x)= f (x)
63. Nosymmetry,neitherevennorodd since f ( x) = f (x)and f ( x) = f (x)
64. Symmetricabouttheorigin,oddfunction since f ( x)= f (x)
65. Symmetricabout x = 3,neitherevennor oddsince f ( x) = f (x)and f ( x) = f (x)
66. Symmetricabout x =1,neitherevennorodd since f ( x) = f (x)and f ( x) = f (x)
67. Symmetryabout x =2,notanevenorodd functionsince f ( x) = f (x)and f ( x) = f (x)
68. Symmetricaboutthey-axis,evenfunction since f ( x)= f (x)
69. Symmetricabouttheorigin,oddfunction since f ( x)= f (x)
70. Symmetricabouttheorigin,oddfunction since f ( x)= f (x)
71. Nosymmetry,notanevenoroddfunction since f ( x) = f (x)and f ( x) = f (x)
72. Nosymmetry,notanevenoroddfunction since f ( x) = f (x)and f ( x) = f (x)
73. Nosymmetry,notanevenoroddfunction since f ( x) = f (x)and f ( x) = f (x)
74. Symmetricaboutthey-axis,evenfunction since f ( x)= f (x)
75. Symmetricaboutthey-axis,evenfunction since f ( x)= f (x)
76. Symmetricaboutthey-axis,evenfunction since f ( x)= f (x)
77. Nosymmetry,notanevenoroddfunction since f ( x)= f (x)and f ( x) = f (x)
78. Symmetricaboutthey-axis,evenfunction since f ( x)= f (x)
79. Symmetricaboutthey-axis,evenfunction since f ( x)= f (x)
80. Symmetricaboutthey-axis,evenfunction since f ( x)= f (x)
81. e 82. a 83. g 84. h
85. b 86. d 87. c 88. f
89. (−∞, 1] ∪ [1, ∞)
(−∞, 1) ∪ (5, ∞)
93. Usingthegraphof y =(x 1)2 9,wefind thatthesolutionis( 2, 4). -2 4 x -9 4 y
94. Graphof y = x 1 2 2 9 4 showssolution is(−∞, 1] ∪ [2, ∞) -1 2 x -2 4 y
95. Fromthegraphof y =5 √x,wefindthat thesolutionis[0, 25]. 25 x 5 y
96. Graphof y = √x +3 2showssolutionis [1, ∞)
97. Note,thepointsofintersectionof y =3and y =(x 2)2 are(2 ± √3, 3).Thesolutionset of(x 2)2 > 3is
101. Fromthegraphof y =
98. Thepointsofintersectionof y =4and y =(x 1)2 are( 1, 4)and(3, 4).The solutionsetof(x 1)2 < 4istheinterval ( 1, 3).
weobservethatthesolutionsetof √3x2 + πx 9 < 0is( 3.36, 1.55).
102. Fromthegraphof y = x3 5x2 +6x 1, 4 1 x 2 -1 y thesolutionsetof x3 5x2 +6x 1 > 0 is(0.20, 1.55) ∪ (3.25, ∞).
103.a. Stretchthegraphof f byafactorof2.
99. Fromthegraphof y = √25 x2,weconclude thatthesolutionis( 5, 5). -5 5 x -5 4 y
100. Graphof y = √4 x2 showssolutionis [ 2, 2]
b. Reflectthegraphof f aboutthe x-axis.
c. Translatethegraphof f totheleftby1unit.
g. Translatethegraphof f totherightby 1-unitandupby3-units.
d. Translatethegraphof f totherightby 3-units. 2 3 4
e. Stretchthegraphof f byafactorof3and reflectaboutthe x-axis.
h. Translatethegraphof f totherightby 2-units,stretchbyafactorof3,andup by1-unit.
104.a. Reflectthegraphof f aboutthe x-axis.
f. Translatethegraphof f totheleftby 2-unitsanddownby1-unit.
b. Stretchthegraphof f beafactorof2.
c. Stretchthegraphof f byafactorof3and reflectaboutthe x-axis.
g. Translatethegraphof f totheleftby4unitsandreflectaboutthe x-axis.
d. Translatethegraphof f totheleftby 2-units.
e. Translatethegraphof f totherightby 1-unit.
f. Translatethegraphof f totherightby2unitsandupby1-unit.
h. Translatethegraphof f totherightby3units,stretchbyafactorof2,andupby 1-unit.
105. N (x)= x +2000
106. N (x)=1 05x +3000.Yes,ifthemeritincreaseisfollowedbythecostoflivingraise thenthenewsalarybecomeshigherandis N (x)=1.05(x +3000)=1.05x +3150.
107. Ifinflationrateislessthan50%,then 1 √x< 1 2 . Thissimplifiesto 1 2 < √x. After squaringwehave 1 4 <x andso x> 25%
108. Ifproductionisatleast28windows, then1 75√x ≥ 28.Theyneedatleast x = 28 1.75 2 =256hrs.
109.
(a) Bothfunctionsareevenfunctionsand thegraphsareidentical
(b) Onegraphisareflectionoftheother aboutthe y-axis.Bothfunctionsareodd functions.
(c) Thesecondgraphisobtainedbyshifting thefirstonetotheleftby1unit.
If f (x)= x
(d) Thesecondgraphisobtainedbytranslatingthefirstonetotherightby2units and3unitsup.
112. Thegraphisalowersemicircleofradius6 withcenterattheorigin.Thenthedomainis [ 6, 6]andtherangeis[ 6, 0].
113. Note |x|≥ 2/3isequivalentto x ≥ 2/3or x ≤−2/3.Thenthesolutionsetis (−∞, 2/3] ∪ [2/3, ∞).
110. Thegraphof y = x3 +6x2 +12x+8orequivalently y =(x +2)3 canbeobtainedbyshifting thegraphof y = x3 totherightby2units.
ThinkingOutsidetheBoxXXIV
Note,if(x,x + h)issuchanorderedpairthenthe averageis x + h/2.Sincetheaverageisnotawhole number,then h =1.Thus,theorderedpairsare (4, 5),(49, 50),(499, 500),and(4999, 5000).
2.3PopQuiz
1. y = √x +8 2. y =(x 9)2
3. y =( x)3 or y = x3
4. Domain[1, ∞),range(−∞, 5]
5. y = 3(x 6)2 +4
6. Evenfunction
2.3LinkingConcepts
(a) Fromthegraphof y =16 96+9 8(17 67)1/3 20 85 1 25√S
weobtain S mustliein(0, 299 58].
(b) Fromthegraphof
3 y weconclude D mustliein[19.97, ∞).
(c) Givenaportionofthegraphof y =16.96+9.8(17.26)1/3 L 1.25√310.28
y
itfollowsthat L mustliein(0, 20.27].
(d) Thevariableswithnegativecoefficientshave maximumvalues.And,thevariable D (with apositivecoefficient)hasaminimumvalue.
ForThought
1. False,since f + g hasanemptydomain.
2. True 3. True 4. True
5. True,since A = P 2/16. 6. True
7. False,since(f ◦ g)(x)= √x 2 8. True
9. False,since(h ◦ g)(x)= x2 9
10. True,since x belongstothedomainif √x 2 isarealnumber,i.e.,if x ≥ 2.
2.4Exercises
1. sum
2. composition
3. 1+2=1 4. 6+0=6
5. 5 6= 11 6. 42 ( 9)=51
7. ( 4) 2= 8 8. ( 3) 0=0
9. 1/12 10. 12
11. (a 3)+(a2 a)= a2 3
12. (b 3) (b2 b)=2b 3 b2
13. (a 3)(a2 a)= a3 4a2 +3a
14. (b 3)/(b2 b)
15. f + g = {( 3, 1+2), (2, 0+6)} = {( 3, 3), (2, 6)},domain {−3, 2}
16. f + h = {(2, 0+4)} = {(2, 4)}, domain {2}
17. f g = {( 3, 1 2), (2, 0 6)} = {( 3, 1), (2, 6)},domain {−3, 2}
18. f h = {(2, 0 4)} = {(2, 4)}, domain {2}
19. f · g = {( 3, 1 · 2), (2, 0 · 6)= {( 3, 2), (2, 0)}, domain {−3, 2}
20. f h = {(2, 0 4)} = {(2, 0)},domain {2}
21. g/f = {( 3, 2/1)} = {( 3, 2)}, domain {−3}
22. f/g = {( 3, 1/2), (2, 0/6)= {( 3, 1/2), (2, 0)},domain {−3, 2}
23. (f + g)(x)= √x + x 4,domainis[0, ∞)
24. (f + h)(x)= √x + 1 x 2 , domainis[0, 2) ∪ (2, ∞)
25. (f h)(x)= √x 1 x 2 , domainis[0, 2) ∪ (2, ∞)
26. (h g)(x)= 1 x 2 x +4,domain is(−∞, 2) ∪ (2, ∞)
27. (g · h)(x)= x 4 x 2 ,domainis(−∞, 2) ∪ (2, ∞)
28. (f · h)(x)= √x x 2 ,domainis[0, 2) ∪ (2, ∞)
29. g f (x)= x 4 √x ,domainis(0, ∞)
30. f g (x)= √x x 4 ,domainis[0, 4) ∪ (4, ∞)
31. {( 3, 0), (1, 0), (4, 4)} 32. {( 3, 2), (0, 0)}
33. {(1, 4)} 34. {( 3, 0)}
35. {( 3, 4), (1, 4)} 36. {(2, 0)}
37. f (2)=5 38. g( 4)=17
39. f (2)=5 40. h( 22)= 7
41. f (20 2721)=59 8163
42. ≈ g( 2 95667) ≈ 9 742
43. (g ◦ h ◦ f )(2)=(g ◦ h)(5)= g(2)=5
44. (h ◦ f ◦ g)(3)=(h ◦ f )(10)= h(29)=10
45. (f ◦ g ◦ h)(2)=(f ◦ g)(1)= f (2)=5
46. (h ◦ g ◦ f )(0)=(h ◦ g)( 1)= h(2)=1
47. (f ◦ h)(a)= f a +1 3 = 3 a +1 3 1=(a +1) 1= a
48. (h ◦ f )(w)= h (3w 1)= (3w 1)+1 3 = 3w 3 = w
49. (f ◦ g)(t)= f t2 +1 = 3(t2 +1) 1=3t2 +2
50. (g ◦ f )(m)= g (3m 1)= (3m 1)2 +1=9m2 6m +2
51. (f ◦ g)(x)= √x 2,domain[0, ∞)
52. (g ◦ f )(x)= √x 2,domain[2, ∞)
53. (f ◦ h)(x)= 1 x 2,domain(−∞, 0) ∪ (0, ∞)
54. (h ◦ f )(x)= 1 x 2 ,domain(−∞, 2) ∪ (2, ∞)
55. (h ◦ g)(x)= 1 √x ,domain(0, ∞)
56. (g ◦ h)(x)= 1 x = 1 √x ,domain(0, ∞)
57. (f ◦ f )(x)=(x 2) 2= x 4, domain(−∞, ∞)
58. (g ◦ g)(x)= √x = 4 √x,domain[0, ∞)
59. (h ◦ g ◦ f )(x)= h(√x 2)= 1 √x 2 , domain(2, ∞)
60. (f ◦ g ◦ h)(x)= f 1 √x = 1 √x 2, domain(0, ∞)
61. (h ◦ f ◦ g)(x)= h (√x 2)= 1 √x 2 , domain(0, 4) ∪ (4, ∞)
62. (g ◦ h ◦ f )(x)= g 1 x 2 = 1 √x 2 , domain(2, ∞)
63. F = g ◦ h 64. G = g ◦ f
65. H = h ◦ g 66. M = f ◦ g
67. N = h ◦ g ◦ f 68. R = f ◦ g ◦ h
69. P = g ◦ f ◦ g 70. C = h ◦ g ◦ h
71. S = g ◦ g 72. T = h ◦ h
73. If g(x)= x3 and h(x)= x 2,then
(h ◦ g)(x)= g(x) 2= x 3 2= f (x).
74. If g(x)= x 2and h(x)= x3,then
(h ◦ g)(x)= g(x)3 =(x 2)3
75. If g(x)= x +5and h(x)= √x,then
(h ◦ g)(x)= g(x)= √x +5= f (x).
76. If g(x)= √x and h(x)= x +5,then
(h ◦ g)(x)= g(x)+5= √x +5= f (x)
77. If g(x)=3x 1and h(x)= √x,then
(h ◦ g)(x)= g(x)= √3x 1= f (x).
78. If g(x)= √x and h(x)=3x 1,then
(h ◦ g)(x)=3g(x) 1=3√x 1= f (x)
79. If g(x)= |x| and h(x)=4x +5,then
(h ◦ g)(x)=4g(x)+5=4|x| +5= f (x).
80. If g(x)=4x +5and h(x)= |x|,then
(h ◦ g)(x)= |g(x)| = |4x +5| = f (x)
81. y =2(3x +1) 3=6x 1
82. y = 4( 3x 2) 1=12x +7
83. y =(x2 +6x +9) 2= x2 +6x +7
84. y =3(x2 2x +1) 3=3x2 6x
85. y =3 · x +1 3 1= x +1 1= x
86. y =2 1 2 x 5 2 +5= x 5+5= x
87. Since m = n 4and y = m2 , y =(n 4)2 .
88. Since u = t +9and v = u 3 , v = t +9 3
89. Since w = x +16, z = √w,and y = z 8 , weobtain y = √x +16 8
90. Since a = b3 , c = a +25,and d = √c, wehave d = √b3 +25.
91. Aftermultiplying y by x +1 x +1 wehave y = x 1 x +1 +1 x 1 x +1 1 = (x 1)+(x +1) (x 1) (x +1) = x
Thedomainoftheoriginalfunctionis (−∞, 1) ∪ ( 1, ∞)whilethedomainof thesimplifiedfunctionis(−∞, ∞). Thetwofunctionsarenotthesame. -1 2 x -2 1 y
92. Aftermultiplying y by x 1 x 1 wehave y = 3x +1 x 1 +1 3x +1 x 1 3 = (3x +1)+(x 1) (3x +1) 3(x 1) = x
Thedomainoftheoriginalfunctionis (−∞, 1) ∪ (1, ∞)whilethedomainofthe simplifiedfunctionis(−∞, ∞). Thetwofunctionsarenotthesame.
93. Domain[ 1, ∞),range[ 7, ∞)
94. Domain[ 1, ∞),range[ 4, ∞)
97. Domain[0, ∞),range[4, ∞)
95. Domain[1, ∞),range[0, ∞)
98. Domain[ 64, ∞),range[0, ∞)
96. Domain[ 1, ∞),range[1, ∞)
99. P (x)=68x (40x +200)=28x 200. Since200/28 ≈ 7 1,theprofitispositivewhen thenumberoftrimmerssatisfies x ≥ 8.
100. P (x)=(3000x 20x2) (600x +4000)= 20x2 +2400x 4000.
101. A = d2/2 102. P =4√A
103. (f ◦f )(x)=0 899x and(f ◦f ◦f )(x)=0 852x aretheamountsofforestlandatthestartof 2002and2003,respectively.
104. (f ◦f )(x)=2(2x)=4x and(f ◦f ◦f )(x)=8x arethevaluesof x dollarsinvestedinbonds after24and36years,respectively.
105. Totalcostis(T ◦ C)(x)=1 05(1 20x)= 1 26x
106.a) Theslopeofthelinearfunctionis
3200 2200 30 20 =100
Using C(x) 2200=100(x 20),thecost beforetaxesis C(x)=100x +200.
b) T (x)=1 09x c) Totalcostof x palletswithtaxincludedis (T ◦ C)(x)=1.09(100x +200)= 109x +218.
107. Note, D = d/2240 x = d/2240 L3/1003 = 1003d 2240L3 = 1003(26000) 2240L3 = 1004(26) 224L3 = 1004(13) 112L3
Expressing D asafunctionof L,we
write D = (13)1004 112L3 or D = 1.16 × 107 L3
108. Note, S = 6500 (d/64)2/3 = 6500 d2/3/16 = 16(6500) d2/3 .
Expressing S asafunctionof d,we
obtain S = 16(6500) d2/3 or S =104, 000d 2/3
109. Theareaofasemicirclewithradius s/2is (1/2)π(s/2)2 = πs2/8.Theareaofthesquare is s2.Theareaofthewindowis W = s 2 + πs2 8 = (8+ π)s2 8 .
110. Theareaofthesquareis A = s2 andthearea ofthesemicircleis S = 1 2 π s 2 2
Then s2 = 8S π and A = 8S π
111. Formarighttrianglewithtwosidesoflength s andahypotenuseoflength d.Bythe PythagoreanTheorem,weobtain d2 = s 2 + s 2
Solvingfor s,wehave s = d√2 2 .
112. Constructanequilateraltrianglewherethe lengthofonesideis d/2andthealtitudeis p/2.BythePythagoreanTheorem, (p/2)2 +(d/4)2 =(d/2)2 .
Solvingfor p,weget p = √3 2 d.
113. Ifacoatisonsaleat25%offandthereis anadditional10%off,thenthecoatwillcost 0.90(.75x)=0.675x where x istheregular price.Thus,thediscountsaleis32.5%offand not35%off.
114. Additionandmultiplicationoffunctionsare commutative,i.e.,(f + g)(x)=(g + f )(x)and (f g)(x)=(g f )(x).Addition,multiplication, andcompositionoffunctionsareassociative, i.e.,
Q((f + g)+ h)(x)=(f +(g + h))(x), ((f g) h)(x)=(f (g h))(x), and ((f ◦ g) ◦ h)(x)=(f ◦ (g ◦ h))(x)
115. Thedifferencequotientis 1+ 3 x+h 1 3 x h = 3 x+h 3 x h = 3x 3x 3h xh(x + h) = 3 x(x + h) =
116. Fromthegraphbelow,weconcludethedomainis(−∞, ∞),therangeis( 1, ∞),and increasingon[1, ∞).
117. Since x 3 ≥ 0,thedomainis[3, ∞).Since 5√x 3hasrange(−∞, 0],therangeof f (x)= 5√x 3+2is(−∞, 2].
118. Bythesquarerootproperty,wefind (x 3)2 = 3 2 x 3= ± √6 2 x = 6 ± √6
119. Since5x> 1,thesolutionsetis( 1/5, ∞).
120. Since y = 1 3 x 7 9 ,theslopeis 1 3
ThinkingOutsidetheBox
XXV. 1 (0.7)(0.7)2(0.7)3(0.7)4 ≈ 0.97175or 97.2%
XXVI. Since20=3(3)+11(1)and1=3(4)+ 11( 1),allintegersatleast20canbeexpressedintheform3x +11y.
Note,therearenowholenumbers x and y that satisfy19=3x +11y.Thus,19isthelargest wholenumber N thatcannotbeexpressedin theform3x +11y.
2.4PopQuiz
1. A = πr2 = π(d/2)2 or A = πd2 4
2. (f + g)(3)=9+1=10
3. (f g)(4)=16 2=32
4. (f ◦ g)(5)= f (3)=9
5. {(4, 8+9)} = {(4, 17)}
6. Since(n ◦ m)(1)= n(3)=5,wefind {(1, 5)}.
7. Since(h + j)(x)= x2 + √x +2,thedomainis [ 2, ∞).
8. Since(h ◦ j)(x)= √x +2 2 ,thedomainis [ 2, ∞).
9. Since(j ◦ h)(x)= √x2 +2,thedomainis (−∞, ∞).
2.4LinkingConcepts
a) Let n bethenumberofbeadsinonefootand suppose n isalsothenumberofspaces.Since thesumofthediametersofthenotchesand thespacesis1foot,wehave1=(2n) d 12
Solvingfor n,weobtain n = 6 d
b) If c istheareainsquareinchesofacross sectionofabead,then c = 1 2 π d 2 2 = πd2 8
Thus, c = πd2 8 sqin.
c) Note,aparallelbeadisahalf-circularcylinder. Sincethelengthofabeadis12inchesandthe areaofacrosssectionofabeadisknown(as inpart(b)),thevolumeofabeadisgivenby v1 = c · 12= πd2 8 (12)= 3πd2 2 cubicinches.
d) Thevolume v2 ofglueon1ft2 offlooris v2 = v1 · n.Thus, v2 = v1 · n = 3πd2 2 · 6 d =9πd cubicinches.
e) Note,1ft3 =1728in3 and1gal= 1728 7.5 cubic inches.Let A bethenumberofsquareinches onegallonofgluewillcover.Then A = 1728 7.5 ÷ v2 = 1728 7.5 ÷ (9πd)
Hence, A = 25.6 πd squarefeet.
f) Forthesquarenotchwhosesideis d inchesin length,wefindthecorrespondingvaluesof n, c, v1, v2,and A asinparts(a)-(e).
i) n = 6 d
ii) c = d2 squareinches
iii) v1 = c · 12=12d2 cubicinches
iv) v2 = v1 · n =12d2 · 6 d =72d cubicinches
v) Asinparte),wehave A = 1728 7.5 ÷ v2 = 1728 7.5 ÷ (72d)
Thus, A = 3.2 d squarefeet.
ForThought
1. False,sincetheinversefunctionis {(3, 2), (5, 5)}.
2. False,sinceitisnotone-to-one.
3. False, g 1(x)doesnotexistsince g isnot one-to-one.
4. True
5. False,afunctionthatfailsthehorizontalline testhasnoinverse.
6. False,sinceitfailsthehorizontallinetest.
7. False,since f 1(x)= x 3 2 +2where x ≥ 0
8. False, f 1(x)doesnotexistsince f isnot one-to-one.
9. False,since y = |x| is V -shapedandthe horizontallinetestfails.
10. True
2.5Exercises
1. one-to-one
2. invertible
3. inverse
4. symmetric
5. Yes,sinceallsecondcoordinatesaredistinct.
6. Yes,sinceallsecondcoordinatesaredistinct.
7. No,sincetherearerepeatedsecondcoordinates suchas( 1, 1)and(1, 1).
8. No,sincetherearerepeatedsecondcoordinates asin(3, 2)and(5, 2).
9. No,sincetherearerepeatedsecondcoordinates suchas(1, 99)and(5, 99).
10. No,sincetherearerepeatedsecondcoordinatesasin( 1, 9)and(1, 9).
11. Notone-to-one 12. Notone-to-one
13. One-to-one 14. One-to-one
15. Notone-to-one 16. One-to-one
17. One-to-one;sincethegraphof y =2x 3shows y =2x 3isanincreasingfunction,the HorizontalLineTestimplies y =2x 3is one-to-one.
18. One-to-one;sincethegraphof y =4x 9shows y =4x 9isanincreasingfunction,the HorizontalLineTestimplies y =4x 9is one-to-one.
19. One-to-one;forif q(x1)= q(x2)then 1 x
x2 +5
4(x1 x2)=0 x1 x2 =0
Thus,if q(x1)= q(x2)then x1 = x2. Hence, q isone-to-one.
20. One-to-one;forif g(x1)= g(x2)then x1 +2 x1 3 = x2 +2 x2 3 (x1 +2)(x2 3)=(x2 +2)(x1 3) x1x2 3x1 +2x2 6=
5(x2 x1)=0 x2 x1 =0
Thus,if g(x1)= g(x2)then x1 = x2 Hence, g isone-to-one.
21. Notone-to-onefor p( 2)= p(0)=1.
22. Notone-to-onefor r(0)= r(2)=2.
23. Notone-to-onefor w(1)= w( 1)=4.
24. Notone-to-onefor v(1)= v( 1)=1.
25. One-to-one;forif k(x1)= k(x2)then 3 √x1 +9= 3 √x2 +9 ( 3 √x1 +9)3 =( 3 √x2 +9)3 x1 +9= x2 +9 x1 = x2.
Thus,if k(x1)= k(x2)then x1 = x2. Hence, k isone-to-one.
26. One-to-one;forif t(x1)= t(x2)then √x1 +3= √x2 +3 (√x1 +3)2 =(√x2 +3)2 x1 +3= x2 +3 x1 = x2.
Thus,if t(x1)= t(x2)then x1 = x2. Hence, t isone-to-one.
27. Invertible, {(3, 9), (2, 2)}
28. Invertible, {(5, 4), (6, 5)}
29. Notinvertible
30. Notinvertible
31. Invertible, {(3, 3), (2, 2), (4, 4), (7, 7)}
32. Invertible, {(1, 1), (4, 2), (16, 4), (49, 7)}
33. Notinvertible
34. Notinvertible
35. Notinvertible,therecanbetwodifferentitems withthesameprice.
36. Notinvertiblesincethenumberofyears(given asawholenumber)cannotdeterminethe numberofdayssinceyourbirth.
37. Invertible,sincetheplayingtimeisafunction ofthelengthoftheVCRtape.
38. Invertible,since1 6km ≈ 1mile
39. Invertible,assumingthatcostissimplyamultipleofthenumberofdays.Ifcostincludes extracharges,thenthefunctionmaynotbe invertible.
40. Invertible,sincetheinterestisuniquelydeterminedbythenumberofdays.
41. f 1 = {(1, 2), (5, 3)}, f 1(5)=3, (f 1 ◦ f )(2)=2
42. f 1 = {(5, 1), (0, 0), (6, 2)}, f 1(5)= 1, (f 1 ◦ f )(2)=2
43. f 1 = {( 3, 3), (5, 0), ( 7, 2)}, f 1(5)=0, (f 1 ◦ f )(2)=2
44. f 1 = {(5, 3.2), (1.99, 2)}, f 1(5)=3.2, (f 1 ◦ f )(2)=2
45. NotinvertiblesinceitfailstheHorizontalLine Test. -.02 -.01 -.03 x .000001 -.000001 y
46. NotinvertiblesinceitfailstheHorizontalLine Test. .4 .2 -.1 x .0005 -.0005 y
47. NotinvertiblesinceitfailstheHorizontalLine Test. 2 5 1 x -2 3 -4 y
48. NotinvertiblesinceitfailstheHorizontalLine Test.
50.a) f (x)istheoperationofdividing x by2. Reversingtheoperation,theinverseis
f 1(x)=2x
b) f (x)istheoperationofadding99to x. Reversingtheoperation,theinverseis
f 1(x)= x 99
c) f (x)isthecompositionofmultiplying x by 5,thenadding1.
Reversingtheoperations,theinverseis
f 1(x)=(x 1)/5
49.a) f (x)isthecompositionofmultiplying x by 5,thensubtracting1.
Reversingtheoperations,theinverseis
f 1(x)= x 1 5
b) f (x)isthecompositionofmultiplying x by 3,thensubtracting88.
Reversingtheoperations,theinverseis
f 1(x)= x +88 3
c) f 1(x)=(x +7)/3
d) f 1(x)= x 4 3
e) f 1(x)=2(x +9)=2x +18
f) f 1(x)= x
g) f (x)isthecompositionoftakingthecube rootof x,thensubtracting9.
Reversingtheoperations,theinverseis
f 1(x)=(x +9)3
h) f (x)isthecompositionofcubing x,multiplyingtheresultby3,thensubtracting 7.
Reversingtheoperations,theinverseis
f 1(x)= 3 x +7 3
i) f (x)isthecompositionofsubtracting1 from x,takingthecuberootoftheresult,thenadding5.
Reversingtheoperations,theinverseis
f 1(x)=(x 5)3 +1
j) f (x)isthecompositionofsubtracting7 from x,takingthecuberootoftheresult,thenmultiplyingby2.
Reversingtheoperations,theinverseis
f 1(x)= x 2 3 +7
d) f (x)isthecompositionofmultiplying x by 2,thenadding5.
Reversingtheoperations,theinverseis
f 1(x)= x 5 2
e) f (x)isthecompositionofdividing x by3, thenadding6.Reversingtheoperations, theinverseis
f 1(x)=3(x 6)=3x 18
f) f (x)istheoperationoftakingthemultiplicativeinverseof x.Sincetakingthe multiplicativeinversetwicereturnstothe originalnumber,theinverseistakingthe multiplicativeinverse,i.e., f 1(x)=1/x
g) f (x)isthecompositionofsubtracting9 from x,thentakingthecuberootofthe result.Reversingtheoperations,theinverseis f 1(x)= x3 +9.
h) f (x)isthecompositionofcubing x,multiplyingtheresultby 1,thenadding4. Reversingtheoperations,theinverseis f 1(x)= 3 (x 4)= 3 √4 x.
i) f (x)isthecompositionofadding4to x, takingthecuberootoftheresult,then multiplyingby3.Reversingtheoperations,theinverseis f 1(x)= x 3 3 4
j) f (x)isthecompositionofadding3to x, takingthecuberootoftheresult,then subtracting9.
Reversingtheoperations,theinverseis f 1(x)=(x +9)3 3.
51. No,sincetheyfailtheHorizontalLineTest.
52. Yes,sincetheyaresymmetricaboutthe line y = x.
53. Yes,sincethegraphsaresymmetricabout theline y = x.
54. Nosince f (x)= x2 failstheHorizontal LineTest.
55. Graphof f 1
Graphof
63. f 1(x)= 3 √x 1 2 x 1 y
64. f 1(x)= 3 √x 2 -2 x -2 2 y
65. f 1(x)=(x +3)2 for x ≥−3
4 -3 x -3 4 y
66. f 1(x)= x2 +3for x ≥ 0
9 3 x 3 9 y
67. Interchange x and y thensolvefor y.
x =3y 7
x +7 3 = y x +7 3 = f 1(x)
68. Interchange x and y thensolvefor y
x = 2y +5
2y =5 x
f 1(x)= 5 x 2
69. Interchange x and y thensolvefor y.
x =2+ y 3for x ≥ 2
(x 2)2 = y 3for x ≥ 2
f 1(x)=(x 2)2 +3for x ≥ 2
70. Interchange x and y thensolvefor y
x = 3y 1for x ≥ 0 x 2 +1=3y for x ≥ 0
f 1(x)= x2 +1 3 for x ≥ 0
71. Interchange x and y thensolvefor y. x = y 9 y = x 9
f 1(x)= x 9
72. Interchange x and y thensolvefor y. x = y +3 y = x +3
f 1(x)= x +3
73. Interchange x and y thensolvefor y. x = y +3 y 5
xy 5x = y +3
xy y =5x +3
y(x 1)=5x +3
f 1(x)= 5x +3 x 1
74. Interchange x and y thensolvefor y
x = 2y 1 y 6
xy 6x =2y 1
xy 2y =6x 1
y(x 2)=6x 1
f 1(x)= 6x 1 x 2
75. Interchange x and y thensolvefor y. x = 1 y xy = 1 f 1(x)= 1 x
76. Clearly f 1(x)= x
77. Interchange x and y thensolvefor y.
x = 3 y 9+5 x 5= 3 y 9 (x 5)3 = y 9
f 1(x)=(x 5)3 +9
78. Interchange x and y thensolvefor y x = 3 y 2 +5 x 5= 3 y 2 (x 5)3 = y 2
f 1(x)=2(x 5)3
79. Interchange x and y thensolvefor y. x =(y 2)2 x ≥ 0 √x = y 2
f 1(x)= √x +2
80. Interchange x and y thensolvefor y x = y 2 x ≥ 0 √x = y sincedomain of f is(−∞, 0] f 1(x)= √xx ≥ 0
81. Note,(g ◦ f )(x)=0.25(4x +4) 1= x and (f ◦ g)(x)=4(0.25x 1)+4= x.
Yes, g and f areinversefunctionsofeach other.
82. Note,(g ◦ f )(x)= 0 2(20 5x)+4= x and (f ◦ g)(x)=20 5( 0.2x +4)= x.
Yes, g and f areinversefunctionsofeach other.
83. Since(f ◦ g)(x)= √x 1 2 +1= x and and(g ◦ f )(x)= √x2 +1 1= √x2 = |x|, g and f arenotinversefunctionsofeachother.
84. Since(f ◦ g)(x)= 4 √x4 = |x| and and(g ◦ f )(x)=( 4 √x)4 = x, g and f arenotinversefunctionsofeachother.
85. Wefind
(f ◦ g)(x)= 1 1/(x 3) +3 = x 3+3
(f ◦ g)(x)= x and (g ◦ f )(x)= 1 1 x +3 3 = 1 1/x
(g ◦ f )(x)= x.
Then g and f areinversefunctionsofeach other.
86. Weobtain
(f ◦ g)(x)=4 1 1/(4 x) =4 (4 x)
(f ◦ g)(x)= x and (g ◦ f )(x)= 1 4 4 1 x = 1 1/x
(g ◦ f )(x)= x
Thus, g and f areinversefunctionsofeach other.
87. Weobtain
(f ◦ g)(x)= 3 5x3 +2 2 5
= 3 5x3 5 = 3 √x3
(f ◦ g)(x)= x and (g ◦ f )(x)=5 3 x 2 5 3 +2 =5 x 2 5 +2 =(x 2)+2
(g ◦ f )(x)= x.
Thus, g and f areinversefunctionsofeach other.
88. Wenote (f ◦ g)(x)= 3 √x +3 3 27 = x and (g ◦ f )(x)= 3 x3 27+3 = x.
Then g and f arenotinversefunctionsofeach other.
89. y1 and y2 areinversefunctionsofeachother and y3 = y2 ◦ y1. -4 4 x -4 -2 y
90. y1 and y2 areinversefunctionsofeachother and y3 = y1 ◦ y2 -4 4 x -4 -2 y
91. C =1 08P expressesthetotalcostasa functionofthepurchaseprice;and P = C/1.08isthepurchasepriceasafunction ofthetotalcost.
92. V (x)= x3,S(x)= 3 √x
93. Thegraphof t asafunctionof r satisfiesthe HorizontalLineTestandisinvertible.Solving for r wefind, t 7 89= 0 39r r = t 7 89 0 39 andtheinversefunctionis r = 7.89 t 0 39 . If t =5 55min.,then r =
=6 rowers.
94. Solvingfor F ,weobtain F 32= 9C 5 F = 9C 5 +32
andtheinversefunctionis F = 9C 5 +32;a formulathatcanconvertCelsiustemperature toFahrenheittemperature.
95. Solvingfor w,weobtain 1.496w = V 2 w
andtheinversefunctionis w =
.If V =115ft./sec.,then w =
96. r = 5.625 × 10 5 V 500 where 0 ≤ V ≤ 0.028125
97.a) Let V =$28, 000.Thedepreciationrateis r =1 28, 000 50, 000 1/5 ≈ 0 109 or r ≈ 10.9%.
b) Writing V asafunctionof r wefind 1 r = V 50, 000 1/5 (1 r)5 = V 50, 000 and V =50, 000(1 r)5
98. Let P =80, 558.Then r = 22, 402 10, 000 1/10 1 ≈ 0 0839
Theaverageannualgrowthrateis r ≈ 8 4%.
Solvingfor P ,weobtain
1+ r = P 10, 000 1/10
(1+ r)10 = P 10, 000 and P =10, 000(1+ r)10
99. Since g 1(x)= x +5 3 and f 1(x)= x 1 2 , wehave g 1 ◦ f 1(x)= x 1 2 +5 3 = x +9 6
Likewise,since(f ◦ g)(x)=6x 9,weget (f ◦ g) 1(x)= x +9 6
Hence,(f ◦ g) 1 = g 1 ◦ f 1
100. Since f ◦ g ◦ g 1 ◦ f 1 (x)= f ◦ (g ◦ g 1) (f 1(x))= f (f 1(x))= x and therangeofonefunctionisthedomainof theotherfunction,wehave (f ◦ g) 1 = g 1 ◦ f 1 .
101. Onecaneasilyseethattheslopeoftheline joining(a,b)to(b,a)is 1,andthattheirmidpointis a + b 2 , a + b 2 Thismidpointlieson theline y = x whoseslopeis1. Then y = x is theperpendicularbisectorofthelinesegment joiningthepoints(a,b)and(b,a)
102. Itisdifficulttofindtheinversementallysince thetwosteps x 3and x+2aredoneseparately andsimultaneously.
103. Dividingweget x 3 x +2 =1 5 x +2
104. f 1(x)= x 1 5 1 2or f 1(x)= 5 1 x 2
105. (f ◦ g)(2)= f (g(2))= f (1)= 2+3 5 =1 (f g)(2)= f (2)g(2))= 7 5 1= 7 5
106. y = 2√x 5
107. Observe,thegraphof f (x)= √9 x2 isa lowersemicircleofradius3centeredat(0, 0). Thendomainis[ 3, 3],therangeis[ 3, 0], andincreasingon[0, 3]
108. No,twoorderedpairshavethesamefirstcoordinates.
109. 0 75+0 80=0 225x 0 125x 1 55=0 1x 15.5= x
Thesolutionsetis {15.5}.
110. Theslopeoftheperpendicularlineis 1/2. Using y = mx + b andthepoint(2, 4),wefind 4= 1 2 (2)+ b 4= 1+ b 5= b
Theperpendicularlineis y = 1 2 x +5.
Since640, 000=210 54,wehaveeither x =210 and y =54,or x =54 and y =210.Ineithercase, |x y| =399.
2.5PopQuiz
1. No,sincethesecondcoordinateisrepeatedin (1, 3)and(2, 3).
2. 2,sincetheorderpair(5, 2)belongto f 1 .
3. Since f 1(x)= x/2, f 1(8)=8/2=4.
4. No,since f (1)=1= f ( 1)orthesecondcoordinateisrepeatedin(1, 1)and(1, 1).
5. f 1(x)= x +1 2
6. Interchange x and y thensolvefor y x = 3 y +1 4 x +4= 3 y +1 (x +4)3 = y +1 (x +4)3 1= y Theinverseis g 1(x)=(x +4)3 1.
7. (h ◦ j)(x)= h( 3 √x +5) = 3 √x +5 3 5 =(x +5) 5 (h ◦ j)(x)= x
2.5LinkingConcepts
a) r =365 A P 1/n 1
b) Forthefirstloan, r =365 229 200 1/30 1 ≈ 1 651
or r ≈ 165 1%annually.
Forthesecondloan,
r =365 339 300 1/30 1 ≈ 1.49 or r ≈ 149%annually.
Forthethirdloan,
r =365 449 400 1/30 1 ≈ 1 409
or r ≈ 140.9%annually.
c) Ifoneborrowed$200atanannualrateof r =165.1%compoundeddaily,thenafterone yearonewillhavetopayback 200 1+ 1.651 365 365 =$1, 038 56
e) Itchargeshighratesbecauseofhighrisks.
ForThought
1. False
2. False,sincecostvariesdirectlywiththenumber ofpoundspurchased.
3. True 4. True
5. True,sincetheareaofacirclevariesdirectly withthesquareofitsradius.
6. False,since y = k/x isundefinedwhen x =0
7. True 8. True 9. True
10. False,thesurfaceareaisnotequalto (SurfaceArea)= k length width height forsomeconstant k
2.6Exercises
1. variesdirectly
2. variation
3. variesinversely
4. variesjointlly
5. G = kn 6. T = kP 7. V = k/P 8. m1 = k/m2 9. C = khr 10. V = khr2
11. Y = kx √z 12. W = krt/v
13. Avariesdirectlyasthesquareofr
14. CvariesdirectlyasD
15. y variesinverselyas x 16. m1 variesinverselyas m2
17. Notavariationexpression
18. Notavariationexpression
19. a variesjointlyas z and w
20. V variesjointlyas L,W ,and H
21. H variesdirectlyasthesquarerootof t and inverselyas s
22. B variesdirectlyasthesquareof y and inverselyasthesquarerootof x
23. D variesjointlyas L and J andinverselyas W
24. E variesjointlyas m andthesquareof c.
25. Since y = kx and5= k 9, k =5/9. Then y =5x/9.
26. Since h = kz and210= k 200, k =21/20. So h =21z/20.
27. Since T = k/y and 30= k/5, k = 150 Thus, T = 150/y
28. Since H = k/n and9= k/( 6), k = 54 So H = 54/n
29. Since m = kt2 and54= k · 18, k =3 Thus, m =3t2
30. Since p = k 3 √w and 3 √2 2 = k 3 √4,we get k = 3 √2
31. Since y = kx/√z and2.192= k(2.4)/√2.25, weobtain k =1.37.Hence, y =1.37x/√z.
32. Since n = kx√b and 18.954= k( 1 35)√15 21,wefind k =3 6. So n =3 6x√b
33. Since y = kx and9= k(2),weobtain y = 9 2 · ( 3)= 27/2.
34. Since y = kz and6= k√12,weobtain y = 3 √3 √75=15
35. Since P = k/w and2/3= k 1/4 ,wefind k = 2 3 ·· 1 4 = 1 6 .Thus, P = 1/6 1/6 =1.
36. Since H = k/q and0 03= k 0.01 ,weget P = 0.0003 0.05 =0 006
37. Since A = kLW and30= k(3)(5√2), weobtain A = √2(2√3) 1 2 = √6.
38. Since J = kGV and √3= k√2√8,we
find J = √3 4 √6 · 8=2√18=6√2.
39. Since y = ku/v2 and7= k 9/36, wefind y =28 · 4/64=7/4.
40. Since q = k√h/j3 and18= k√9/8, weget q =48 4 1/8 =1536
41. Let Li and Lf bethelengthininchesandfeet, respectively.Then Li =12Lf isadirectvariation.
42. Let Ts and Tm bethetimeinsecondsandminutes,respectively.Then Ts =60Tm isadirect variation.
43. Let P and n bethecostperpersonandthe numberofpersons,respectively.Then P =20/n isaninversevariation.
44. Let n and w bethenumberofrodsandthe weightofarod,respectively.
Then n = 40, 000 w isaninversevariation.
45. Let Sm and Sk bethespeedsofthecarinmph andkph,respectively.Then Sm ≈ Sk/1.6 ≈ 0 6Sk isadirectvariation.
46. Let Wp and Wk betheweightinpoundsand kilograms,respectively.Then Wp =2.2Wk is adirectvariation.
47. Notavariation 48. Notavariation
49. Let A and W betheareaandwidth,respectively.Then A =30W isadirectvariation.
50. Let A and b betheareaandbase,respectively. Then A = 1 2 10b =5b isadirectvariation.
51. Let n and p bethenumberofgallonsandprice pergallon,respectively.Since np =5,we obtainthat n = 5 p isaninversevariation.
52. Let L and W bethelengthandwidth,respectively.Then L =40/W isaninversevariation.
53. If p isthepressureatdepth d,then p = kd. Since4.34= k(10), k =0.434.At d =6000ft, thepressureis p =0.434(6000)=2604lbper squareinch.
54. Solvingforthedepth d in2170=0.434d, wefind d = 2170 0.434 =5000feet.
55. If h isthenumberofhours, p isthenumberof pounds,and w isthenumberofworkersthen h = kp/w.Since8= k(3000)/6, k =0 016. Fiveworkerscanprocess4000poundsin h =(0 016)(4000)/5=12 8hours.
56. Since V = k√A and154= k√16000,wefind k ≈ 1 21747.Theviewfrom36000feet is V =(1 21747)√36000 ≈ 231miles.
57. Since I = kPt and20 80= k(4000)(16),we find k =0 000325.Theinterestfromadeposit of$6500for24daysis
I =(0 000325)(6500)(24) ≈ $50 70
58. Since d = kt2 and16= k(12), k =16.Thus, after2secondsthecabfalls d =(16)4=64 feet.
59. Since C = kDL and18 60= k(6)(20),we obtain k =0 155.Thecostofa16ftpipe withadiameterof8inchesis
C =0 155(8)(16)=$19 84
60. Since C = kDL and36.60= k(0.5)(20),we find k =3.66.Thecostofa100ftcopper tubingwitha 3 4 -inchdiameteris
C =3.66(0.75)(100)=$274.50.
61. Since w = khd2 and14 5= k(4)(62),wefind k = 14 5 144 .Thena5-inchhighcanwitha diameterof6incheshasweight w = 14 5 144 (5)(62)=18.125oz.
62. Since V = kt/w and10= k(80)/600,weget k =75.At90oFand800poundsthevolume is V =75(90)/(800) ≈ 8.4375cubicinches.
63. Since V = kh/l and10= k(50)/(200),weget k =40.Thevelocity,iftheheadis60ftand thelengthis300ft,is V =(40)(60)/(300)=8 ft/year.
64. Since V = kiA and3= k(0.3)(10), k =1. Ifthehydraulicgradientis0.4andthe dischargeis5gallonsperminutethen 5=(1)(0.4)A.Thecross-sectionalareais A =12.5ft2 .
65. No,itisnotdirectlyproportionalotherwise thefollowingratios 42, 506 1.34 ≈ 31, 720, 59, 085 0.295 ≈ 200, 288,and 738, 781 0 958 ≈ 771, 170wouldbeall thesamebuttheyarenot.
66. Let r = knw c .Tofind k,wesolve
54= k(50)(27) 25
Weobtain k =1andconsequently r = nw c If n =40, c =13,and w =26,thenthe gearratiois r = 40(26) 13 =80. If n =45, w =27,and r =67 5,thenby solving67 5= 45(27) c oneobtains c =18.
67. Since g = ks/p and76= k(12)/(10), k = 190 3 . IfCalvinstudiesfor9hoursandplaysfor 15hours,thenhisscoreis
g = 190 3 · s p = 190 3 · 9 15 =38
68. Since c = klw and263 40= k(12)(9),weget k = 263.4 108 .Acarpetwhichis12ftwideand thatcosts$482.90(i.e.482.90= 263.4 108 12 l) haslength l =16 5ft.
69. Since h = kv2 and16= k(32)2,weget k = 1 64 . Toreachaheightof20 2 5 ,thevelocity v mustsatisfy
20+ 2 5 12 = 1 64 v 2 .
Solvingfor v,wefind v ≈ 35.96ft/sec.
73. Interchange x and y thensolvefor y. x = 3 y 9+1 (x 1)3 = y 9 (x 1)3 +9= y f 1(x)=(x 1)3 +9
74. If s isthesideofthesquare,thenthediagonal is d = √2s bythePythagoreanTheorem.If A istheareaofthesquare,then A = s2.Since s = √A,weobtain d = √2s = √2√A or d = √2A
75. Let x betheaveragespeedintherain.Then 3x +5(x +5)=425
Solvingfor x,wefind x =50mph.
76. Since f ( x)= f (x),thegraphissymmetric abouttheorigin.
77. Theslopeofthelineis1/2.Using y = mx + b andthepoint( 4, 2),wefind
2= 1 2 ( 4)+ b 2= 2+ b 4= b
Thelineisgivenby y = 1 2 x +4,or2y = x +8. Astandardformis x 2y = 8.
78. Since 6 < 3x 9 < 6,weobtain
3 < 3x< 15.
Thesolutionsetis(1, 5).
ThinkingOutsidetheBoxXXVIII
Let d bethedistanceSharonwalks.Since2minutesisthedifferenceinthetimesofarrivalat4mph and5mph,weobtain d 4 d 5 = 2 60
Thesolutionoftheaboveequationis d =2/3mile. Since d/4=(2/3)/4=10/60,Sharonmustwalk toschoolin9minutestoarriveontime.Thus, Sharon’sspeedinordertoarriveontimeis r = d t = 2/3 9/60 = 40 9 mph.
2.6PopQuiz
1. Since4= k 20,theconstantofvariationis k =4/20=1/5.
2. a = k/b =10/2=5
3. Since C = kr2 and108= k 32,wefind k =12. Then C =12 42 =$192.
4. Since C = kLW and180= k(6)(2),weobtain k =15.Then C =15(5)(4)=$300.
2.6LinkingConcepts
a) m = kd3 p2 where m isthemassoftheplanet,
d isthemeandistancebetweenthesatellite andtheplanet, p isperiodofrevolutionofthe satellite,and k istheproportionconstant.
b) Solving5.976 × 1024 = k(384 4 × 103)3 27 3222 for k,
wefind k ≈ 7 8539 × 1010 ≈ 7 85 × 1010
c) ThemassofMarsis
7.8539 × 1010(9330)3 (7 65/24)2 ≈ 6 278 × 1023 kg
d) AnapproximateratioofEarth’smassto Mars’massis 5.976 × 1024 6 278 × 1023 ≈ 10 ortheratioisapproximately10to1.
e) Solvingfor d,weobtain (7 8539 × 1010)d3 (42.47/24)2 =1 921 ×
7 8539 × 10
d ≈ 424, 678km d ≈ 4.247 × 105 km.
ReviewExercises
1. Function,domainandrangeareboth {−2, 0, 1}
x -2 1 y
2. Notafunction,domainis {0, 1, 2},rangeis {±1, ±3}
3. y =3 x isafunction,domainandrange areboth(
,
4. 2x + y =5isafunction,domainandrange areboth(−∞, ∞)
5. Notafunction,domainis {2},rangeis(−∞, ∞) 3 1 x 3 -3 y
6. Function,domainis(−∞, ∞),rangeis {3},
7. x2 + y2 =0.01isnotafunction,domain andrangeareboth[ 0.1, 0.1] -0.2 0.2 x 0.2 -0.2 y
8. (x 1)2 +(y +2)2 =5isnotafunction,domain is[1 √5, 1+ √5],rangeis[ 2 √5, 2+ √5]
12. x = √y isafunction,domainandrange areboth[0, ∞)
9. x = y2 +1isnotafunction,domainis[1, ∞), rangeis(−∞, ∞) 1 5 x 2 -2 y
10. y = |x 2| isafunction,domainis(−∞, ∞), rangeis[0, ∞) 2 6 x 2 4 y
11. y = √x 3isafunction,domainis[0, ∞), rangeis[ 3, ∞) 1 9 x -3 2 y
13. 9+3=12 14. 6 7= 1 15. 24 7=17 16. 1+3=4 17. If x2 +3=19,then x2 =16or x = ±4.
18. If2x 7=9,then2x =16or x =8.
19. g(12)=17 20. f ( 1)=4
21. 7+( 3)=4 22. 7 ( 11)=18
23. (4)( 9)= 36 24. 19
25. f ( 3)=12 26. g(7)=7
27. f (g(x))= f (2x 7) =(2x 7)2 +3 =4x 2 28x +52
28. g(f (x))= g(x 2 +3) =2(x 2 +3) 7 =2x 2 1
29. (x2 +3)2 +3= x4 +6x2 +12
30. 2(2x 7) 7=4x 21
31. (a +1)2 +3= a2 +2a +4
32. 2(a +2) 7=2a 3
33. f (3+ h) f (3) h = (9+6h + h2)+3 12 h = 6h + h2 h =6+ h
34. g(5+ h) g(5) h = 2(5+ h) 7 3 h = 2h h =2
35. f (x + h) f (x) h = (x2 +2xh + h2)+3 x2 3 h = 2xh + h2 h = 2x + h =
36. g(x + h) g(x) h = 2(x + h) 7 2x +7 h = 2h h =2
37. g x +7 2 =(x +7) 7= x
38. f √x 3 =(x 3)+3= x
39. g 1(x)= x +7 2
40. Fromnumber39, g 1( 3)= 3+7 2 =2
41. f (x)= √x,g(x)=2√x +3;leftby3, stretchby2 -3 6 x 1 6 y
42. f (x)= √x,g(x)= 2√x +3;stretchby2, reflectabout x-axis,upby3 1 8 x 1 3 y
43. f (x)= |x|,g(x)= 2|x +2| +4;leftby2, stretchby2,reflectabout x-axis,upby4 -2 4 x -2 4 y
44. f (x)= |x|,g(x)= 1 2 |x 1|− 3;rightby1, stretchby 1 2 ,downby3 1 2 x -3 2 y
45. f (x)= x2,g(x)= 1 2 (x 2)2 +1;rightby2, stretchby 1 2 ,upby1 1 2 x 1 3 y
46. f (x)= x2,g(x)= 2x2 +4;stretchby2, reflectabout x-axis,upby4 1 2 x -4 4 y
47. (f ◦ g)(x)= √x 4,domain[0, ∞)
48. (f ◦ g)(x)= √x 4,domain[4, ∞)
49. (f ◦ h)(x)= x2 4,domain(−∞, ∞)
50. (h ◦ f )(x)=(x 4)2 = x2 8x +16,domain (−∞, ∞)
51. (g ◦ f ◦ h)(x)= g(x2 4)= √x2 4. Tofindthedomain,solve x2 4 ≥ 0.Then thedomainis[2, ∞) ∪ (−∞, 2]
52. (h ◦ f ◦ g)(x)= h(√x 4)=(√x 4)2 . Then(h◦f ◦g)(x)= x 8√x+16.Thedomain is[0, ∞).
53. Translatethegraphof f totherightby2units,stretchbyafactorof2,shiftupby1unit.
55. Translatethegraphof f totheleftby1-unit, reflectaboutthe x-axis,shiftdownby3-units.
56. Translatethegraphof f totherightby1-unit, reflectaboutthe x-axis,shiftupby2-units.
57. Translatethegraphof f totheleftby2-units, stretchbyafactorof2,reflectaboutthe xaxis.
54. Translatethegraphof f totheleftby3-units, stretchbyafactorof2,shiftdownby1-unit.
58. Stretchthegraphof f byafactorof3,reflect aboutthe x-axis,shiftupby1-unit.
59. Stretchthegraphof f byafactorof2,reflect aboutthe x-axis,shiftupby3-units.
1 2 1 2 3 3 x 4 5 3 y
60. Translatethegraphof f totherightby1-unit, stretchbyafactorof4,shiftupby3-units.
1 2 3 x 3 5 7 y
61. F = f ◦ g 62. G = g ◦ f
63. H = f ◦ h ◦ g ◦ j 64. M = j ◦ h ◦ g
65. N = h ◦ f ◦ j or N = h ◦ j ◦ f
66. P = g ◦ g ◦ g ◦ g ◦ j
67. R = g ◦ h ◦ j 68. Q = j ◦ g
69. f (x + h) f (x) h = 5(x + h)+9+5x 9 h = 5h h = 5
70. f (x + h) f (x) h = = √x + h 7 √x 7 h √x + h 7+ √x 7 √x + h 7+ √x 7 = (x + h 7) (x 7) h √x + h 7+ √x 7 = 1 √x + h 7+ √x 7
71. f (x + h) f (x) h = = 1 2x +2h 1 2x h · (2x +2h)(2x) (2x +2h)(2x) = (2x) (2x +2h) h(2x +2h)(2x) = 2 (2x +2h)(2x) = 1 (x + h)(2x)
72. f (x + h) f (x) h = = 5(x2 +2xh + h2)+(x + h)+5x2 x h = 10xh 5h2 + h h = 10x 5h +1
73. Domainis[ 10, 10],rangeis[0, 10], increasingon[ 10, 0],decreasingon[0, 10] -10 10 x -10 12 y
74. Domainis[ √7, √7],rangeis[ √7, 0], increasingon[0, √7],decreasingon[ √7, 0] -5 5 x -3 -2 y
75. Domainandrangeareboth(−∞, ∞), increasingon(−∞, ∞) -1 1 x 1 -1 y
76. Domain(−∞, 4],range[0, ∞), increasingon[0, 4],decreasingon(−∞, 0] -2 3 x 3 4 y
77. Domainis(−∞, ∞),rangeis[ 2, ∞), increasingon[ 2, 0]and[2, ∞), decreasingon(−∞, 2]and[0, 2] -2 2 x -2 4 y
78. Domainis(−∞, ∞),rangeis[ 1, ∞), increasingon[1, ∞),decreasingon(−∞, 1], constanton[ 1, 1]
-2 2 x -2 4 y
79. y = |x|− 3,domainis(−∞, ∞), rangeis[ 3, ∞)
80. y = |x 3|− 1,domainis(−∞, ∞), rangeis[ 1, ∞)
81. y = 2|x| +4,domainis(−∞, ∞), rangeis(−∞, 4]
82. y = 1 2 |x| +2,domainis(−∞, ∞), rangeis(−∞, 2]
83. y = |x +2| +1,domainis(−∞, ∞), rangeis[1, ∞)
84. y = −|x +1|− 2,domain(−∞, ∞), range(−∞, 2]
85. Symmetry: y-axis 86. Symmetry: y-axis
87. Symmetricabouttheorigin
88. Symmetricabouttheorigin
89. Neithersymmetry 90. Neithersymmetry
91. Symmetricaboutthe y-axis
92. Symmetricaboutthe y-axis
93. Fromthegraphof y = |x 3|− 1,thesolution setis(−∞, 2] ∪ [4, ∞) 3 6 x 2 -1 y
94. Nosolutionsinceanabsolutevalueis nonnegative
95. Fromthegraphof y = 2x2 +4,thesolution setis( √2, √2)
4 y
x
96. Fromthegraphof y = 1 2 x 2 +2,thesolution setis(−∞, 2] ∪ [2, ∞)
100. Inversefunctions, f (x)=(x 2)3,g(x)= 3 √x +2
101. Inversefunctions, f (x)=2x 4,g(x)= 1 2 x +2
97. Nosolutionsince √x +1 2 ≤−2
3 x -2 1 y
98. Fromthegraphof y = √x 3,thesolutionset is[0, 9)
102. Inversefunctions, f (x)= 1 2 x +4,g(x)= 2x +8
99. Inversefunctions,
103. Notinvertible,sincetherearetwosecond componentsthatarethesame.
104. f 1 = {(1/3, 2), (1/4, 3), (1/5, 4)} with domain {1/3, 1/4, 1/5} and range {−2, 3, 4}
105. Inverseis f 1(x)= x +21 3 with domainandrangeboth(−∞, ∞)
106. Notinvertible 107. Notinvertible
108. Inverseis f 1(x)=7 x with domainandrangeboth(−∞, ∞)
109. Inverseis f 1(x)= x2 +9for x ≥ 0 withdomain[0, ∞)andrange[9, ∞)
110. Inverseis f 1(x)=(x +9)2 for x ≥−9 withdomain[ 9, ∞)andrange[0, ∞)
111. Inverseis f 1(x)= 5x +7 1 x with domain(−∞, 1) ∪ (1, ∞),and range(−∞, 5) ∪ ( 5, ∞)
112. Inverseis f 1(x)= 5x +3 x +2 domain(−∞, 2) ∪ ( 2, ∞),and range(−∞, 5) ∪ (5, ∞)
113. Inverseis f 1(x)= √x 1with domain[1, ∞)andrange(−∞, 0]
114. Inverseis f 1(x)= 4 √x 3for x ≥ 81 withdomain[81, ∞),range[0, ∞)
115. Let x bethenumberofroses.Thecost functionis C(x)=1.20x +40,therevenue functionis R(x)=2x,andtheprofitfunction is P (x)= R(x) C(x)or P (x)=0.80x 40.
Since P (50)=0,tomakeaprofitshemust sellatleast51roses.
116.a) Since V = πr2h and V =1,weobtain h = 1 πr2
b) Since V = πr2h and V =1,weget 1 πh = r 2 or r = 1 √πh .
c) Fromparta),wehave h = 1 πr2 .
Since S =2πr2 +2πrh,weobtain S =2πr 2 +2πr 1 πr2
S =2πr 2 + 2 r
117. Since h(0)=64and h(2)=0,therangeof h = 16t2 +64istheinterval[0, 64]. Thenthedomainoftheinversefunctionis [0, 64].Solvingfor t,weobtain
16t2 =64 h
t2 = 64 h 16 t = √64 h 4
Theinversefunctionis t = √64 h 4
118. Solvingfor S in T =1.05S,we obtain S = T 1 05 .
119. Since A = π d 2 2 , d =2 A π
120. If A istheareaofthesquare,thenthelength ofonesideis √A. Sinceonesideofthesquare istwicetheradius r then √A =2r Then A =4r2
121. Theaveragerateofchangeis 8 6 4 =0.5inch/lb.
122. Theaveragerateofchangeis 130 40 3 =30mph/sec.
123. Since D = kW and9= k · 25,weobtain D = 9 25 100=36.
124. Since t = ku/v and6= k 8/2,we find k =1.5.Then t =(1.5)(19)/3=19/2.
125. Since V = k√h and45= k√1.5,thevelocity ofaTriceratopsis V = 45 √1.5 √2.8 ≈ 61kph.
126. Since R = k/p and21= k/240,wefind k =5040.If p =224,thenheneeds R =5040/224=22.5rows.
127. Since C = kd2 and4.32= k 36,a16-inch diameterglobecosts C = 4 32 36 162 =$30.72.
128. F = k m1m2 d2 where m1,m2 arethemasses and d isthedistancebetweenthecentersof theobjects.
130. No,itisnotafunctionsincetherearetwo orderedpairsinaverticallinewiththesame firstcoordinateanddifferentsecond coordinates.
ThinkingOutsidetheBox
XXIX. Thequadrilateralhasvertices A(0, 0), B(0, 9/2), C(82/17, 15/17),and D(7/2, 0). Theareaoftriangle∆ABC is
1 2 9 2 82 17 = 369 34 andtheareaoftriangle∆ACD is
1 2 7 2 15 17 = 105 68
Thesumofareasofthetwotrianglesisthe areaofthequadrilateral,i.e.,
Area= 369 34 + 105 68 = 843 68 squareunits
XXX. Note,thenumberofhandshakesinagroup with n peopleis
1+2+3+ +(n 1)= (n 1)n 2
Inthefirstdelegation,thenumberofhandshakesis (n 1)n 2 =190whichimpliesthat n =20,thenumberofmembersinthefirst delegation.
Sincetherewere480handshakesbetweenthe firstdelegationandseconddelegation,thesize oftheseconddelegationis
480 n = 480 20 =24delegates.
Thus,thetotalnumberofdelegatesis20+24, or44delegates.
Chapter2Test
1. No,since(0, 5)and(0, 5)aretwoordered pairswiththesamefirstcoordinateand differentsecondcoordinates.
2. Yes,sincetoeach x-coordinatethereisexactly one y-coordinate,namely, y = 3x 20 5
3. No,since(1, 1)and(1, 3)aretwoordered pairswiththesamefirstcoordinateand differentsecondcoordinates.
4. Yes,sincetoeach x-coordinatethereisexactly one y-coordinate,namely, y = x3 3x2 +2x 1.
5. Domainis {2, 5},rangeis {−3, 4, 7}
6. Domainis[9, ∞),rangeis[0, ∞)
7. Domainis[0, ∞),rangeis(−∞, ∞)
8. Graphof3x 4y =12includesthe points(4, 0), (0, 3) 4 -4 x -3 -6 y
9. Graphof y =2x 3includesthe points(3/2, 0), (0, 3) 4 2 x -3 1 y
10. y = √25 x2 isasemicirclewithradius5
11. y = (x 2)2 +5isaparabolawithvertex (2, 5)
12. y =2|x|− 4includesthepoints(0, 4), (±3, 2)
2 x 2 -4 y
13. y = √x +3 5includesthepoints (1, 3), (6, 2)
14. Graphincludesthepoints ( 2, 2), (0, 2), (3, 2)
15. √9=3 16. f (5)= √7
17. f (3x 1)= (3x 1)+2= √3x +1.
18. g 1(x)= x +1 3 19. √16+41=45
20. g(x + h) g(x) h = 3(x + h) 1 3x +1 h = 3h h =3
21. Increasingon(3, ∞),decreasingon(−∞, 3)
22. Symmetricaboutthe y-axis
23. Add1to 3 <x 1 < 3toobtain 2 <x< 4. Thus,thesolutionset( 2, 4).
24. f (x)isthecompositionofsubtracting2from x,takingthecuberootoftheresult,then adding3.Reversingtheoperations,theinverseis f 1(x)=(x 3)3 +2.
25. Therangeof f (x)= √x 5is[0, ∞).Then thedomainof f 1(x)is x ≥ 0Note, f (x)is thecompositionofsubtracting5from x,then takingthesquarerootoftheresult.Reversing theoperations,theinverseis f 1(x)= x2 +5 for x ≥ 0.
26. 60 35 200 =$0.125perenvelope
27. Since I = k/d2 and300= k/4,weget k =1200 If d =10,then I =1200/100=12 candlepower.
28. Let s bethelengthofonesideofthecube. BythePythagoreanTheoremwehave s2 + s2 = d2.Then s = d √2 andthevolume is V = d √2 3 = √2d3 4 .
TyingItAllTogether
1. Add3tobothsidesof2x 3=0toobtain 2x =3.Thus,thesolutionsetis 3 2 .
2. Add2x tobothsidesof 2x +6=0toget 6=2x.Thenthesolutionsetis {3}
3. Note, |x| =100isequivalentto x = ±100. Thesolutionsetis {±100}
4. Theequationisequivalentto 1 2 = |x +90| Then x +90= ± 1 2 and x = ± 1 2 90. Thesolutionsetis {−90.5, 89.5}.
5. Note,3=2√x +30.Ifwedivideby2and squarebothsides,weget x +30= 9 4 .
Since x = 9 4 30= 27 75,thesolutionset is {−27 75}
6. Note, √x 3isnotarealnumberif x< 3. Since √x 3isnonnegativefor x ≥ 3,it followsthat √x 3+15isatleast15. Inparticular, √x 3+15=0hasno realsolution.
7. Rewriting,weobtain(x 2)2 = 1 2 .Bythe
squarerootproperty, x 2= ± √2 2
Thus, x =2 ± √2 2 .Thesolutionset is 4 ± √2 2
8. Rewriting,weget(x +2)2 = 1 4 .Bythesquare
rootproperty, x +2= ± 1 2 .Thus, x = 2 ± 1 2
Thesolutionsetis 3 2 , 5 2
9. Squaringbothsidesof √9 x2 =2,weget 9 x2 =4.Then5= x2 Thesolutionsetis ±√5
10. Note, √49 x2 isnotarealnumberif 49 x2 < 0.Since √49 x2 isnonnegative for49 x2 ≥ 0,wegetthat √49 x2 +3is atleast3.Inparticular, √49 x2 +3=0has norealsolution.
11. Domainis(−∞, ∞),rangeis(−∞, ∞), x-intercept(3/2, 0) 1 2 x 3 1 y
12. Domainis(−∞, ∞),rangeis(−∞, ∞), x-intercept(3, 0) 1 3 x 3 6 y
13. Domainis(−∞, ∞),rangeis[ 100, ∞), x-intercepts(±100, 0)
14. Domainis(−∞, ∞),rangeis(−∞, 1], x-intercepts( 90 5, 0)and( 89 5, 0).
15. Domainis[ 30, ∞)sinceweneedtorequire x +30 ≥ 0,rangeis(−∞, 3], x-interceptis ( 27 75, 0)sincethesolutionto 3 2√x +30=0is x = 27.75
16. Domainis[3, ∞)sinceweneedtorequire x 3 ≥ 0,rangeis[15, ∞),no x-intercept 3 7 x
17. Domainis(−∞, ∞),rangeis(−∞, 1]since thevertexis(2, 1),andthe x-interceptsare 4 ± √2 2 , 0 sincethesolutionsof 0= 2(x 2)2 +1are x = 4
20. Domainis[ 7, 7],rangeis[3, 10]sincethe graphisasemi-circlewithcenter(0, 3)and radius7,thereareno x-interceptsasseenfrom thegraph
18. Domainis(−∞, ∞),rangeis[ 1, ∞)sincethe vertexis( 2, 1),andthe x-interceptsare ( 5/2, 0)and( 3/2, 0)forthesolutionsto 0=4(x +2)2 1are x = 5/2, 3/2.
21. Since2x> 3or x> 3/2,thesolutionset is(3/2, ∞).
22. Since 2x ≤−6or x ≥ 3,thesolutionset is[3, ∞).
23. Basedontheportionofthegraphof y = |x|− 100abovethe x-axis,andits x-intercepts,thesolutionsetof |x|− 100 ≥ 0 is(−∞, 100] ∪ [100, ∞).
24. Basedonthepartofthegraphof y =1 2|x +90| abovethe x-axis,andits x-intercepts,thesolutionsetof1 2|x+90| > 0 is( 90 5, 89 5).
25. Basedonthepartofthegraphof y =3 2√x +30belowthe x-axis,andits x-intercepts,thesolutionsetof 3 2√x +30 ≤ 0is[ 27 75, ∞).
19. Domainis[ 3, 3],rangeis[ 2, 1]sincethe graphisasemi-circlewithcenter(0, 2)and radius3.The x-interceptsare(±√5, 0)forthe solutionsto0= √9 x2 2are x = ±√5. Thegraphisshowninthenextcolumn.
26. Solutionsetis(−∞, ∞)sincethegraph isentirelyabovethe x-axis
27. Basedontheportionofthegraphof y = 2(x 2)2 +1belowthe x-axis,and
its x-intercepts,thesolutionsetof 2(x 2)2 +1 < 0is −∞, 4 √2 2 ∪ 4+ √2 2 , ∞
28. Basedonthepartofthegraphof y =4(x +2)2 1abovethe x-axis,and its x-intercepts,thesolutionsetof 4(x +2)2 1 < 0is(−∞, 5/2] ∪ [ 3/2, ∞).
29. Basedonthepartofthegraphof y = √9 x2 2abovethe x-axis,andits x-intercepts,thesolutionsetof √9 x2 2 ≥ 0is[ √5, √5].
30. Sincethegraphof y = √49 x2 +3isentirely abovetheline y =3,thesolutionto √49 x2 +3 ≤ 0istheemptyset ∅
31. f (2)=3(2+1)3 24=3(27) 24=57
32. Thegraphof f (x)=3(x +1)3 24isgiven.
36. Solvingfor x,weobtain y +24=3(x +1)3 y +24 3 =(x +1)3 3 y +24 3 = x +1 3 y +24 3 1= x.
37. Basedontheanswerfromnumber36,the inverseis f 1(x)= 3 x +24 3 1
38. Thegraphof f 1 isgivenbelow.
33. f = K ◦ F ◦ H ◦ G
34. Solvingfor x,weget 3(x +1)3 =24 (x +1)3 =8 x +1=2 x =1 Thesolutionsetis {1}
35. Basedonthepartofthegraphof y =3(x +1)3 24abovethe x-axis,and its x-intercept(1, 0),thesolutionsetof 3(x +1)3 24 ≥ 0is[1, ∞).
39. Basedonthepartofthegraphof f 1 above the x-axis,andits x-intercept( 21, 0),thesolutionsetof f 1(x) > 0is( 21, ∞).
40. Since f = K ◦ F ◦ H ◦ G,weget f 1 = G 1 ◦ H 1 ◦ F 1 ◦ K 1
ConceptsofCalculus
1.a) Let f (x)= x2 f (2+ h) f (2) h = (2+ h)2 22 h = 4+4h + h2 4 h = 4h + h2 h = h(4+ h) h f (2+ h) f (2) h =4+ h
b) Note,2+ h getscloserandcloserto2 as h approaches0.Usingtheresultsfrom parta),weconclude lim h→0 f (2+ h) f (2) h =lim h→0(4+ h) =4.
2.a) Let f (x)= x2 2x. f (x + h) f (x) h = (
Thus,weobtain f (x + h) f (x) h = 1 √x + h + √x
b) Note,as h approaches0,weseethat
getscloserandcloserto
Thatis,bytakingthelimit,wehave
Thus,weobtain f (x + h) f (x) h =2x + h 2
b) Note,2x + h 2getscloserandcloserto 2x 2as h approaches0.Thenweobtain lim h→0 f (x + h) f (x) h =lim h→0(2x + h 2) =2x 2.
Thus,lim h→0 f (x + h) f (x) h =2x 2.
c) Basedonpartb),theinstantaneousrate ofchangeof f (x)= x2 2x is2x 2. And,when x =5theinstantaneousrate ofchangeis2(5) 2or8.
3.a) Let f (x)= √x.Thenwecalculatetheinstantaneousrateofchangeof f . f (x + h) f (x) h = √
c) Thentheinstantaneousrateofchangeof f (x)= √x is 1 2√x .And,when x =9the instantaneousrateofchange 1 2√9 or 1 6
4.a) Let f (t)= 16t2 +128t.Then
f (0)= 16(0)2 +128(0)=0 and f (3)= 16(9)2 +128(3)=240
b) Thedistancetraveledinthefirst threesecondsis
f (3) f (0)=240 0=240ft.
c) Theaveragerateofchangeoftheheightis
f (3) f (0) 3 0 = 240ft 3sec =80ft/sec.
d) Wecalculatethedifferencequotient.
f (t + h) f (t) h =
5.a) Let f (x)=1/x. f (x) f (2)
16(t + h)2 +128(t + h) 16t2 +128t h = 16(t2 +2ht + h2 )+128t +128h 16t2 +128t h =
Thus,wefind
f (t + h) f (t) h = 32t 16h 128.
e) When h becomescloserandcloserto0,we obtainthat
32t 16h +128 getscloserandcloserto 32t +128
Thus, lim h→0
f (t + h) f (t) h = 32t +128
f) Note,basedontheanswerfromparte),the instantaneousvelocityoftheballattime t is
32t +128ft/sec
When t =0,theinstantaneousvelocityis
32(0)+128=128ft/sec
When t =2,theinstantaneousvelocityis
32(2)+128=64ft/sec
When t =4,theinstantaneousvelocityis
32(4)+128=0ft/sec.
When t =6,theinstantaneousvelocityis
32(6)+128= 64ft/sec
When t =8,theinstantaneousvelocityis
32(8)+128= 128ft/sec.
b) If x iscloseto2,then
.
c) Let h = x c.Note, h iscloseto0if x is near c,andconversely.Then lim x→c f (x) f (c) x c =lim h→0 f (c + h) f (c) h
Thus,theinstantaneousrateofchange when x =2is lim x→2
asshowninpartb).
6.a) Let f (x)= x3
b) If x isclosetoc,then lim x→c f (x) f (c) x c =lim x→c (x 2 + cx + c 2) = c 2 + c(c)+ c 2
Thus,lim x→c f (x) f (c) x c =3c 2
c) Itisshownin5c)thattheinstantaneous rateofchangeisalimitofanaveragerate ofchangeas h approaches0.Thus,the instantaneousrateofchangewhen x = c is lim x→c f (x) f (c) x c =3c 2 asshowninpartb).
