Solution Manual For Thomas' Calculus, 12th Edition

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CHAPTER 1 FUNCTIONS1.1 FUNCTIONS AND THEIR GRAPHS1.domain(); range[1)2.domain[0); range(1]œ_ß _œß _œß _œ_ß3.domain2); y in range and y5x10y can be any positive real numberrange).œ Ò ß _œ! ÊÊœ Ò!ß _È4.domain(03,); y in range and yx3xy can be any positive real numberrange).œ_ßÓ Ò_œ! ÊÊœ Ò!ß _È25.domain(33,); y in range and y, now if t33t, or if t3œ_ßÑ Ð_œÊ ! Ê !443t3t3ty can be any nonzero real numberrange0).Ê ! Ê ! ÊÊœ Ð_ßÑ Ð!ß _43t6.domain(4, 44,); y in range and y, now if tt16, or ifœ_ß %Ñ ÐÑ Ð_œ % Ê ! Ê !22t16t16222t416t16, or if tt16y can be any% Ê Ÿ ! Ê Ÿ ! % Ê ! Ê ! Ê2222t16t16#"'22nonzero real numberrange).Êœ Ð_ß Ó Ð!ß _187.(a)Not the graph of a function of x since it fails the vertical line test.(b)Is the graph of a function of x since any vertical line intersects the graph at most once.8.(a)Not the graph of a function of x since it fails the vertical line test.(b)Not the graph of a function of x since it fails the vertical line test.9.basex; (height)xheightx; area is a(x)(base)(height)(x)xx ;œœÊœœœœ#########""ˆ ‰Š‹x3334ÈÈÈperimeter is p(x)xxx3x.œœ10. sside lengthssds; and area is asadœÊœÊœœÊœ#####"#d2È11. Let Ddiagonal length of a face of the cube andthe length of an edge. ThenDdandœj œjœ###D23d. The surface area is 62dand the volume is.######$$Î#œjÊjœÊ j œjœœjœœd6ddd33333ÈÈ##$Š‹12. The coordinates of P arexxso the slope of the line joining P to the origin is m(x0). Thus,ˆ‰ÈßœœÈÈxxx"x,x,.ˆ‰ˆ‰Èœ""mm#13. 2x4y5yx; Lx0y0xxxxxœÊœ œÐÑ ÐÑœ ÐÑœ"""##55525444416222222ÈÉÉxxœœœÉÉ552520x20x254416164220x20x2522È14. yx3y3x; Lx4y0y34yy1yœÊœœÐÑ ÐÑœÐÑœÐÑÈÈÈÈ222222222y2y1yyy1œœÈÈ42242

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2Chapter 1 Functions15. The domain is.16. The domain is.abab_ß __ß _17.The domain is.18. The domain is.ab_ß _Ð_ß !Ó19.The domain is.20. The domain is.abababab_ß !!ß __ß !!ß _21. The domain is5533, 55,22. The range is 2, 3 .abab_ß Ð ß Ó ÒÑ _ÒÑ23. Neither graph passes the vertical line test(a)(b)

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Section 1.1 Functions and Their Graphs324. Neither graph passes the vertical line test(a)(b)xy1xyy1xororxyyxkkÚÞÚÞÛßÛßÜàÜàœÍÍœ "œœ "œ " 25.x01226.x012y010y10027. F x28. G x4x ,x1x2x, x1, x0x, 0xa ba bœœœœŸŸ22x"29. (a)Line throughand: yx; Line throughand: yx2abababab!ß !"ß "œ"ß "#ß !œ f(x)x, 0x1x2, 1x2œŸŸŸœ(b)f(x)2,xx2xxœ! Ÿ "!ß" Ÿ #ß# Ÿ $!ß$ ŸŸ %ÚÝÝÛÝÝÜ30. (a)Line through2and: yx2abab!ß#ß !œ Line through2and: m, so yx2xabababß "&ß !œœœ œ " œ ! """""&& #$$$$$f(x)x, 0xx,xœ #Ÿ ## Ÿ &œ"&$$(b)Line throughand: m, so yxabab"ß !!ß $œœ $œ $ $$ !! Ð"ÑLine throughand: m, so yxabab!ß $#ß "œœœ #œ # $" $%# !#f(x)x,xx,xœ$ $" Ÿ !# $! Ÿ #œ

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4Chapter 1 Functions31. (a)Line throughand: yxabab"ß "!ß !œ Line throughand: yabab!ß ""ß "œ "Line throughand: m, so yxxababab"ß "$ß !œœœ œ " " œ !"""""$$"#####f(x)xxxxxœ" Ÿ !"! Ÿ "" $ÚÛÜ"$##(b)Line through21and00 : yxabab ß ßœ12Line through02and10 : y2x2ababßßœ Line through11and31 : y1ababß ß œ f xx2x02x20x111x3a bÚÛÜœŸŸŸŸ1232. (a)Line throughandT: m, so yx0xˆ‰ˆ‰abTTTTTTT#Î##"!###ß !ß "œœœœ "abf x, 0xx,xTa bœ!ŸŸ "ŸTTT###(b)f xA,xAxTATxAxTa bÚÝÝÝÛÝÝÝÜœ! ŸßŸßŸßŸŸ #TTTT##$#$#33. (a)x0 for x[0 1)(b)x0 for x(1 0]Ú Û œ−ßÜ Ý œ− ß34.xxonly when x is an integer.Ú Û œ Ü Ý35. For any real number x, nxn, where n is an integer. Now: nxnnxn. ByŸŸ "ŸŸ " Ê Ð "Ñ Ÿ Ÿ definition:xn andxnxn. Soxxfor all x.Ü Ý œ Ú Û œÊ Ú Û œ Ü Ý œ Ú Û− d36. To find f(x) you delete the decimal orfractional portion of x, leaving onlythe integer part.

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Section 1.1 Functions and Their Graphs537. Symmetric about the origin38. Symmetric about the y-axisDec:xDec:x_ __ !Inc: nowhereInc:x! _39. Symmetric about the origin40. Symmetric about the y-axisDec: nowhereDec:x! _Inc:xInc:x_ !_ !x! _41. Symmetric about the y-axis42. No symmetryDec:xDec:x_ Ÿ !_ Ÿ !Inc:xInc: nowhere! _

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6Chapter 1 Functions43. Symmetric about the origin44. No symmetryDec: nowhereDec:x! Ÿ _Inc:xInc: nowhere_ _45. No symmetry46. Symmetric about the y-axisDec:xDec:x! Ÿ __ Ÿ !Inc: nowhereInc:x! _47. Since a horizontal line not through the origin is symmetric with respect to the y-axis, but not with respect to the origin, thefunction is even.48. f xxand fxxf x . Thus the function is odd.a bababa bˆ‰œœœœœ œ &"""&xxx&&&ab49. Since f xxxf x . The function is even.a baba bœ " œ " œ ##50. Since f xxxfxxx and f xxxf xxx the function is neither even norÒœÓ Á ÒœÓÒœÓ Á Òœ Óa bababa ba ba b####odd.51. Since g xxx, gxxxxxg x . So the function is odd.a bababa bœœ œ œ $$$52. g xxxxxgxthus the function is even.a babababœ $ " œ $ " œß%#%#53. g xgx . Thus the function is even.a babœœœ"" ""xx##ab54. g x; gxg x . So the function is odd.a baba bœœ œ xxxx## ""55. h t; ht;h t. Since h th tand h tht , the function is neither even nor odd.a baba ba ba ba babœœœÁ Á""" " "" ttt56. Sincet |t|, h thtand the function is even.lœ l œ$$aba bab

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Section 1.1 Functions and Their Graphs757. h t2t, ht2t. So h tht .h t2t, so h th t . The function is neither even nora baba baba ba ba bœ "œ "Áœ "Á odd.58. h t2 tand ht2t2 t. So h thtand the function is even.a baba babœll "œl l " œll "œ59. skt25k 75kst; 60tt180œÊœÐÑ ÊœÊœœÊœ"""33360. Kc v12960c 18c40K40v ; K40 104000 joulesœÊœÊœÊœœœ###abab261. r6k24r; 10sœÊœÊœÊœœÊœkk242412s4ss562. P14.7k14700P; 23.4v628.2 inœÊœÊœÊœœÊœ¸kk147001470024500v1000vv39363. vf(x)x2x222xx72xx;x7œœÐ"% ÑÐÑ œ % $!)! Þ$#64. (a)Let hheight of the triangle. Since the triangle is isosceles, ABAB2AB2 So,œœÊœÞ###Èh2hB is atslope of ABThe equation of AB is### "œÊœ " Ê!ß "Êœ " ÊŠ‹Èabyf(x)x; x.œœ "− Ò!ß "Ó(b)A x2x y2xx2xx; x.Ð Ñ œœÐ "Ñ œ #− Ò!ß "Ó#65. (a)Graph h because it is an even function and rises less rapidly than does Graph g.(b)Graph f because it is an odd function.(c)Graph g because it is an even function and rises more rapidly than does Graph h.66. (a)Graph f because it is linear.(b)Graph g because it contains.ab!ß "(c)Graph h because it is a nonlinear odd function.67. (a)From the graph,1x(2 0)()x4x#Ê− ß %ß _(b)110x4x4xx##Êx0:1000ÊÊx4x2x8x2xx(x4)(x2)###x4 since x is positive;Êx0:1000ÊÊx4x2x82x2xx(x4)(x2)##x2 since x is negative;Ê sign of (x4)(x2)2ïïïïïðïïïïïðïïïïî%Solution interval: (0)()#ß %ß _

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8Chapter 1 Functions68. (a)From the graph,x(5)(1 1)32x1x1Ê−_ß ß(b)x1:2Case Ê32x1x1x13(x1)3x32x2x5.ÊÊ Thus, x(5) solves the inequality.−_ß 1x1:2CaseÊ32x1x1x13(x1)3x32x2x5 which is trueÊÊ if x1. Thus, x(1 1) solves the − ßinequality.1x:3x32x2x5CaseÊÊ 32x1x1which is never true if 1x, so no solution here.In conclusion, x(5)(1 1).−_ß ß69. A curve symmetric about the x-axis will not pass the vertical line test because the pointsx, yandx,ylie on the samababevertical line. The graph of the function yf xis the x-axis, a horizontal line for which there is a single y-value,,œœ !!a bfor any x.70. price405x, quantity30025xR x405x30025xœœÊœa babab71. xxhx; cost5 2x10hC h1010h5h22222h22 h2 h22œÊœœœÊœœÈÈÈaba bŠ‹Š‹È72. (a)Note that 2 mi = 10,560 ft, so there are800xfeet of river cable at $180 per foot and10,560xfeet of landÈab##cable at $100 per foot. The cost is C x180800x100 10,560x .a babÈœ##(b)C$a b!œ"ß #!!ß !!!C$ab&!!¸"ß "(&ß )"#C$ab"!!!¸"ß ")'ß &"#C$ab"&!!¸"ß #"#ß !!!C$ab#!!!¸"ß #%$ß ($#C$ab#&!!¸"ß #()ß %(*C$ab$!!!¸"ß $"%ß )(!Values beyond this are all larger. It would appear that the least expensive location is less than 2000 feet from thepoint P.1.2 COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS1.D :x, D : x1DD : x1. R :y, R : y0, R: y1, R : y0fgfgfgfgfgfg_ _Êœ_ _2.D : x10x1, D : x10x1. Therefore DD : x1.fgfgfgÊÊœRR : y0, R: y2, R : y0fgfgfgœÈ3.D :x, D :x, D:x, D:x, R : y2, R :y1,fgf gg ffg_ __ __ __ _œÎÎR: 0y2, R:yf gÎŸŸ _g fÎ"#4.D :x, D : x0 , D: x0, D: x0; R : y1, R :y1, R: 0y1, R: 1yfgf gg ffgf g_ _œŸŸ _ÎÎÎg fÎ5.(a)2(b)22(c)x2#(d)(x5)3x10x22(e)5(f)2œ##(g)x10(h)(x3)3x6x6œ##%#

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Section 1.2 Combining Functions; Shifting and Scaling Graphs96.(a)(b)2(c)1œ""3x1x1x(d)(e)0(f)"x43(g)x2(h)œœ"" " #" #x1x1x1xx7.f g hxf g h xf g 4xf 3 4xf 123x123x1133xaba bababababababababa b‰ ‰œœœœœœ8.f g hxf g h xf g xf 2 x1f 2x13 2x146x1aba bababababababababa b‰ ‰œœœœœœ222229.f g hxf g h xf gffaba babÉaba bˆ‰ˆ‰ˆ ‰Š‹É‰ ‰œœœœœ " œ11xx5xx14x14x14x1x % "10.f g hxf g h xf g2xffaba bababa bŠ‹Š‹Èˆ‰‰ ‰œœœœœœŠ‹ÈŠ‹È2x2x12x83xx72x23$ 222x2xxx$ $ 11. (a)f gx(b)j gx(c)g gxaba baba baba b‰‰‰(d)j jx(e)g h fx(f)h j fxaba baba baba b‰‰ ‰‰ ‰12. (a)f jx(b)g hx(c)h hxaba baba baba b‰‰‰(d)f fx(e)j g fx(f)g f hxaba baba baba b‰‰ ‰‰ ‰13.g(x)f(x)(fg)(x)‰(a)x7xx7ÈÈ(b)x23x3(x2)3x6œ(c)xx5x5##ÈÈ(d)xxxxx1x11x(x1)xx1xx1œœ(e)1x""x1x(f)x""xx14. (a)f gxg x.aba ba b‰œ ll œ"l "lx(b)f gxso g xx.aba ba b‰œœÊ " œÊ " œÊœßœ "g xg xxg xxxg xxg xxxxa ba ba ba ba b" " " " """""(c)Sincef gxg xx , g xx .aba ba ba bÈ‰œœ l lœ#(d)Sincef gxfxx , f xx . (Note that the domain of the composite is.)aba ba bˆ‰È‰œœ llœÒ!ß _Ñ#The completed table is shown. Note that the absolute value sign in part (d) is optional.g xf xf gxxxxxxxxxa ba baba bÈÈ‰l l "l lll"" "l "l " "##xxxxxx15. (a)f g1f 11(b)g f 0g22(c)f f1f 02aba babababa baba babœœœœœœ (d)g g 2g 00(e)g f2g 11(f)f g 1f10aba baba bababa baba bœœœœ œœ16. (a)f g 0f1213, where g 0011abababa ba bœœœœœ (b)g f 3g111, where f 3231abababa ba bœœ œœœ (c)g g1g 1110, where g111aba babababœœœœ œ

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10Chapter 1 Functions(d)f f 2f 0202, where f 2220aba ba ba bœœœœœ(e)g f 0g 2211, where f 0202aba ba ba bœœœœœ(f)f gf2, where g1ˆ‰ˆ‰ˆ‰ˆ ‰ˆ ‰""""""#######œœœœœ 517. (a)f gxf g x1aba baba bÉÉ‰œœœ11xxxg fxg f xaba baba b‰œœ1x1È(b)Domainf g :,10,, domaing f :1,abab‰Ð_ Ó Ð_Ñ‰Ð_Ñ(c)Rangef g :1,, rangeg f :0,abab‰Ð_Ñ‰Ð_Ñ18. (a)f gxf g x12xxaba baba bÈ‰œœg fxg f x1xaba babk ka b‰œœ(b)Domainf g :0,, domaing f :,abab‰Ò_Ñ‰Ð_ _Ñ(c)Rangef g :0,, rangeg f :, 1abab‰Ð_Ñ‰Ð_Ó19.f gxxf g xxxg xg x2 xxg x2xaba baba baba ba ba b‰œÊœÊœÊœœ†g xg x2a ba bg xxg x2xg xÊ†œ Êœ œa ba ba b2x2x1xx120.f gxx2f g xx22 g x4x2g xg xaba babababa ba ba ba bÉ‰œÊœÊœÊœÊœ33x6x622321. (a)y(x7)(b)y(x4)œ œ ##22. (a)yx3(b)yx5œœ##23. (a)Position 4(b)Position 1(c)Position 2(d)Position 324. (a)y(x1)4(b)y(x2)3(c)y(x4)1(d)y(x2)œ œ œ œ ####25.26.27.28.

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Section 1.2 Combining Functions; Shifting and Scaling Graphs1129.30.31.32.33.34.35.36.37.38.39.40.

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12Chapter 1 Functions41.42.43.44.45.46.47.48.49.50.51.52.

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Section 1.2 Combining Functions; Shifting and Scaling Graphs1353.54.55. (a)domain: [0 2]; range: [](b)domain: [0 2]; range: [1 0]ß#ß $ß ß(c)domain: [0 2]; range: [0 2](d)domain: [0 2]; range: [1 0]ßßß ß(e)domain: [2 0]; range: [1](f)domain: [1 3]; range: [] ß!ßß!ß "(g)domain: [2 0]; range: [](h)domain: [1 1]; range: [] ß!ß " ß!ß "

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14Chapter 1 Functions56. (a)domain: [0 4]; range: [3 0](b)domain: [4 0]; range: []ß ß ß!ß $(c)domain: [4 0]; range: [](d)domain: [4 0]; range: [] ß!ß $ ß"ß %(e)domain: [4]; range: [3 0](f)domain: [2 2]; range: [3 0]#ß ß ß ß(g)domain: [5]; range: [3 0](h)domain: [0 4]; range: [0 3]"ß ßßß57. y3x358. y2x1x1œœœ %###ab59. y60. y11œ" œœœ"""""*###Î$ˆ‰xxxx####ab61. yx162. y3x1œ%œÈÈ63. y16x64. yxœ% œœ% Éˆ ‰ÈÈx##$#""##65. y3x27x66. yœ " œ " œ " œ " abˆ ‰$$#)$xx$

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Section 1.2 Combining Functions; Shifting and Scaling Graphs1567. Let yxf xand let g xx,œ # " œœÈa ba b"Î#h xx, i xx, anda ba bˆ‰ˆ‰Èœœ#""##"Î#"Î#j xxf. The graph ofa ba b’“È ˆ‰œ #œB"#"Î#h xis the graph of g xshifted leftunit; thea ba b"#graph of i xis the graph of h xstretcheda ba bvertically by a factor of; and the graph ofÈ#j xf xis the graph of i xreflected acrossa ba ba bœthe x-axis.68. Let yf xLet g xx,œ" œÞœÈa ba babx#"Î#h xx, and i xxa baba babœ #œ #"Î#"Î#"#Èf xThe graph of g xis theœ" œÞÈa ba bx#graph of yx reflected across the x-axis.œÈThe graph of h xis the graph of g xshifteda ba bright two units. And the graph of i xis thea bgraph of h xcompressed vertically by a factora bof.È#69. yf xx . Shift f xone unit right followed by aœœa ba b$shift two units up to get g xx.a babœ " #370. yxf x .œ" B # œ Ò "# Ó œabababa b$$Let g xx , h xx, i xx,a ba baba bababœœ "œ "#$$$and j xx. The graph of h xis thea bababa bœ Ò "# Ó$graph of g xshifted right one unit; the graph of i xisa ba bthe graph of h xshifted down two units; and the grapha bof f xis the graph of i xreflected across the x-axis.a ba b71. Compress the graph of f xhorizontally by a factora bœ"xof 2 to get g x. Then shift g xvertically down 1a ba bœ"#xunit to get h x.a bœ ""#x

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