Solution Manual for Calculus for Business, Economics, Life Sciences and Social Sciences, 14th Edition

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SOLUTIONSMANUALCALCULUSFORBUSINESS,ECONOMICS,LIFESCIENCES,ANDSOCIALSCIENCESFOURTEENTHEDITIONANDCALCULUSFORBUSINESS,ECONOMICS,LIFESCIENCES,ANDSOCIALSCIENCESBRIEFVERSIONFOURTEENTHEDITIONRaymond A. BarnettMerritt CollegeMichael R. ZieglerMarquette UniversityKarl E. ByleenMarquette UniversityChristopher J. StockerMarquette University

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1-11FUNCTIONS AND GRAPHSEXERCISE 1-12.4.6.8.10.The table specifies a function, since for each domain value there corresponds one and only one rangevalue.12.The table does not specify a function, since more than one range value corresponds to a given domainvalue.(Range values 1, 2 correspond to domain value 9.)14.This is a function.16.The graph specifies a function; each vertical line in the plane intersects the graph in at most one point.18.The graph does not specify a function. There are vertical lines which intersect the graph in more than onepoint. For example, they-axis intersects the graph in two points.20.The graph does not specify a function.22.14yxxis neither linear nor constant.24.2460xyis linear.26.10xxyis neither linear nor constant.28.321 simplifies to241 constant.2yxxy

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1-2CHAPTER 1: FUNCTIONS AND GRAPHS30.32.34.36.38.f(x) =2232xx. Since the denominator is bigger than 1, we note that the values offare between 0 and 3.Furthermore, the functionfhas the property thatf(–x) =f(x). So, adding pointsx= 3,x= 4,x= 5, we have:x–5–4–3–2 –1012345F(x)2.782.672.45210122.452.672.78The sketch is:40.y=f(4) = 042.y=f(–2) = 344.( )4 at5.fxx46.( )0 at5, 0, 4.fxx 48.Domain:all real numbers.50.Domain: all real numbers exceptx = 2.52.Domain:5 or[ 5,).x 54.Given6721xy. Solving forywe have:67216and3.7yxyxThis equation specifies a function. The domain isR, the set of real numbers.

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EXERCISE 1-11-356.Given()4x xy. Solving forywe have:2244 and.xxyxyxThis equation specifies a function. The domain is all real numbers except 058.Given22.9xySolving forywe have:2229and9.yxyx This equation does not defineyas a function ofx. For example, whenx= 0,y =3.60.Given30.xy. Solving forywe have:31/6and.yxyxThis equation specifies a function. The domain is all nonnegative real numbers, i.e.,0x.62.22( 3 )( 3 )494fxxx64.222(1)(1)421423fxxxxxx66.2633()()44fxxx68.21/ 41/ 24()444fxxxx70.2222(3)( )(3)44541ffhhhh72.222( 3)( 3)496456fhhhhhh 74.2222( 3)( 3)( 3)4( 3)4(964)(94)6fhfhhhhh 76.(A)()3()9339fxhxhxh  (B)()( )339393fxhfxxhxh  (C)()( )33fxhfxhhh 78.(A)22222()3()5()83(2)558363558fxhxhxhxxhhxhxxhhxh(B)2222()( )363558358635fxhfxxxhhxhxxxhhh(C)2()( )635635fxhfxxhhhxhhh80.(A)f(x+h) =x2 + 2xh+h2 + 40x+ 40h(B)f(x+h) –f(x) = 2xh+h2 + 40h(C)()( )fxhfxh= 2x+h+ 40

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1-4CHAPTER 1: FUNCTIONS AND GRAPHS82.GivenA=l w= 81.Thus,w=81l. Now8116222222.PlwllllThe domain isl> 0.84.GivenP= 2+ 2w= 160 or+w= 80 and= 80 –w.NowA=w= (80 –w)wandA= 80w – w2 .The domain is 0 ≤w≤ 80. [Note:w≤ 80 sincew> 80 implies0.]86.(A)(B)p(11) = 1,340 dollars per computerp(18) = 920 dollars per computer88.(A)R(x) =xp(x)=x(2,000 – 60x) thousands of dollarsDomain: 1 ≤x≤ 25(B) Table 11 Revenuex(thousands)R(x)(thousands)1$1,94058,5001014,0001516,5002016,0002512,500(C)90.(A)P(x)=R(x) –C(x)=x(2,000 – 60x) – (4,000 + 500x) thousand dollars= 1,500x– 60x2 – 4,000Domain: 1 ≤x≤ 25(B) Table 13 Profitx(thousands)P(x) (thousands)1–$2,56052,000105,000155,000202,00025–4,000(C)

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EXERCISE 1-21-592.(A)Given 5v– 2s= 1.4. Solving forv, we have:v= 0.4s+ 0.28.Ifs= 0.51, thenv= 0.4(0.51) + 0.28 = 0.484 or 48.4%.(B)Solving the equation fors, we have:s= 2.5v– 0.7.Ifv= 0.51, thens= 2.5(0.51) – 0.7 = 0.575 or 57.5%.EXERCISE 1-22.( )1fxxDomain:[0,); range:[1,).4.2( )10fxxDomain: all real numbers; range:[10,).6.( )53fxxDomain: all real numbers; range: all real numbers.8.( )1520fxxDomain: all real numbers; range:(,15].10.3( )8fxx Domain: all real numbers; range: all real numbers.12.14.16.18.24.20.26.22.

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1-6CHAPTER 1: FUNCTIONS AND GRAPHS28.The graph ofh(x) = –|x– 5| is the graphofy= |x| reflected in thexaxis andshifted 5 units to the right.30.The graph ofm(x) = (x+ 3)2 + 4 is the graph ofy=x2shifted 3 units to the left and 4 units up.32.The graph ofg(x) = –6 +3xis the graph ofy=3xshifted 6 units down.34.The graph ofm(x) = –0.4x2 is the graph ofy=x2 reflected in thexaxis and vertically contracted by a factor of 0.4.36.The graph of the basic functiony= |x| is shifted 3 units to the right and 2 units up.y= |x– 3| + 238.The graph of the basic functiony= |x| is reflected in thexaxis, shifted 2 units to the left and 3 units up.Equation:y= 3 – |x+ 2|40.The graph of the basic function3xis reflected in thexaxis and shifted up 2 units. Equation:y= 2 –3x42.The graph of the basic functiony=x3 is reflected in thexaxis, shifted to the right 3 units and up 1 unit.Equation:y= 1 – (x– 3)3

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EXERCISE 1-21-744.g(x) =33x+ 246.g(x) = –|x– 1|48.g(x) = 4 – (x+ 2)250.1 if1( )22if1xxg xxx  52.102if020( )400.5if20xxh xxx54.420if020( )260if20100360if100xxh xxxxx56.The graph of the basic functiony=xis reflected in thexaxis and vertically expanded by a factor of 2.Equation:y= –2x58.The graph of the basic functiony= |x| is vertically expanded by a factor of 4. Equation:y= 4|x|60.The graph of the basic functiony=x3 is vertically contracted by a factor of 0.25. Equation:y= 0.25x3.62.Vertical shift, reflection inyaxis.Reversing the order does not change the result. Consider a point(a, b) in the plane. A vertical shift ofkunits followed by a reflection inyaxis moves (a, b) to (a,b+k)and then to (–a,b+k). In the reverse order, a reflection inyaxis followed by a vertical shift ofkunitsmoves (a, b) to (–a, b) and then to (–a, b+k). The results are the same.64.Vertical shift, vertical expansion.Reversing the order can change the result. For example, let (a, b) be a point in the plane. A vertical shiftofkunits followed by a vertical expansion ofh(h> 1) moves (a, b) to (a, b+k) and then to (a, bh+kh).In the reverse order, a vertical expansion ofhfollowed by a vertical shift ofkunits moves (a, b) to (a, bh)and then to (a, bh+k); (a, bh+kh) ≠ (a, bh+k).

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1-8CHAPTER 1: FUNCTIONS AND GRAPHS66.Horizontal shift, vertical contraction.Reversing the order does not change the result. Consider a point(a, b) in the plane. A horizontal shift ofkunits followed by a vertical contraction ofh(0 <h< 1) moves(a, b) to (a+k, b) and then to (a+k, bh). In the reverse order, a vertical contraction ofhfollowed by ahorizontal shift ofkunits moves (a, b) to (a, bh) and then to (a+k, bh). The results are the same.68.(A)The graph of the basic functiony=xis vertically expanded bya factor of 4.70.(A)The graph of the basic functiony=x2 is reflected inthexaxis, vertically contracted by a factor of 0.013,and shifted 10 units to the right and 190 units up.(B)(B)72.(A)Letx= number of kwh used in a winter month. For 0 ≤x≤ 700, thecharge is 8.5 + .065x. Atx= 700, the charge is $54. Forx> 700,the charge is54 + .053(x– 700) = 16.9 + 0.053x.Thus,8.5.065if 0700( )16.90.053if700xxW xxx(B)74.(A)Letx= taxable income.If 0 ≤x≤ 12,500, the tax due is $.02x. Atx= 12,500, the taxdue is $250. For 12,500 <x≤ 50,000, the tax due is250 + .04(x– 12,500) = .04x– 250. Forx> 50,000,the tax due is 1,250 + .06(x– 50,000) = .06x– 1,250.Thus,(B)0.02if 012,500( )0.04250if 12,50050, 0000.061, 250if50, 000xxT xxxxx(C)T(32,000) = $1,030T(64,000) = $2,590

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EXERCISE 1-31-976.(A) The graph of the basic functiony=x3 is verticallyexpanded by a factor of 463.78.(A) The graph of the basic functiony=3xis reflectedin thexaxis and shifted up 10 units.(B)(B)EXERCISE 1-32.y=2x+ 1012243xy4.8x– 3y= 24083068xy6.Slope:3m8.Slope:103m y-intercept:b= 2y-intercept:b= 410.Slope:4;m x-intercept: 312.Slope:3;m x-intercept: 214.Slope:9 / 2m ;x-intercept: 4/916.5yx18.6969,;7272mbyx 20.x-intercept: 1;y-intercept: 3;33yx 22.x-intercept: 2,y-intercept:– 1;112yx24.(A)g(B)m(C)n(D)f26.(A)x-intercepts: – 5, – 1;y-intercept: – 5(B) Vertex: ( – 3, 4)(C) Maximum: 4(D) Range:y≤ 4 or ( – ∞, 4]28.(A)x-intercepts: 1, 5;y-intercept: 5(B) Vertex: (3, – 4)(C) Minimum: – 4(D) Range:y≥ – 4 or [ – 4, ∞)

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1-10CHAPTER 1: FUNCTIONS AND GRAPHS30.g(x) = – (x+ 2)2 + 3(A)x-intercepts:– (x+ 2)2 + 3 = 0(x+ 2)2 = 3x+ 2 = ±3x= – 2 –3, – 2 +3y-intercept: – 1(B)Vertex: ( – 2, 3)(C) Maximum: 3(D) Range:y≤ 3 or ( – ∞, 3]32.n(x) = (x– 4)2 – 3(A)x-intercepts: (x– 4)2 – 3 = 0(x– 4)2 = 3x– 4 = ±3x= 4 –3, 4 +3y-intercept: 13(B)Vertex: (4, – 3)(C) Minimum: – 3(D) Range:y≥ – 3 or [ – 3, ∞)34.(A) Slope:523312m.(B) Point-slope form:32(1)2yx.(C) Slope-intercept form:3122yx.(D) Standard form:321xy .36.(A) Slope:734325m .(B) Point-slope form:43(2)5yx .(C) Slope-intercept form:42355yx .(D) Standard form:4523xy.38.(A) Slope:44001m.(B) Point-slope form:40y.(C) Slope-intercept form:4y.(D) Standard form:4y.40.(A) Slope:30mnot defined(B) Point-slope form: none.(C) Slope-intercept form: none.(D) Standard form:2x.42.g(x) =x2 – 6x+ 5 =x2 – 6x+ 9 – 4 = (x– 3)2 – 4(A)x-intercepts: (x– 3)2 – 4 = 0(x– 3)2 = 4x– 3 = ±2x= 1, 5y-intercept: 5(B)Vertex: (3, – 4)(C) Minimum: – 4(D) Range:y≥ – 4 or [ – 4, ∞)44.s(x) = – 4x2 – 8x– 3 = – 42324xx= – 421214xx= – 421(1)4x= – 4(x+ 1)2 + 1

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EXERCISE 1-31-11(A)x-intercepts:– 4(x+ 1)2 + 1 = 04(x+ 1)2 = 1(x+ 1)2 =14x+ 1 = ±12x= –32, –12y-intercept:– 3(B)Vertex: ( – 1, 1)(C) Maximum: 1(D) Range:y≤ 1 or ( – ∞, 1]46.v(x) = 0.5x2 + 4x+ 10 = 0.5[x2 + 8x+ 20] = 0.5[x2 + 8x+ 16 + 4] = 0.5[(x+ 4)2 + 4] = 0.5(x+ 4)2 + 2(A)x-intercepts: noney-intercept: 10(B) Vertex: ( – 4, 2)(C) Minimum: 2(D) Range:y≥ 2 or [2, ∞)48.549540,6;solution set:6,6xx  50.4440,444,11;solution set: 11,xxx 52.(6)(3)0xxTherefore, either(6)0 and (3)0or(6)0 and (3)0.xxxxThe first case isimpossible. The second case implies63.xSolution set:( 6,3).54.2712(3)(4)0xxxxTherefore, either(3)0 and (4)0 or(3)0 and (4)0.xxxxThe first case implies3x and the second case implies4.x Solution set:(,4][ 3,).  56.58.(A)(B) Setf(x) = 0,– 0.8x+ 5.2 = 0,x= 6.5.Setx= 0,y= 5.2.(C)(D)x-intercept:x= 6.5,y-intercept:y= 5.2(E)– 0.8x+ 5.2 < 0x> 6.5

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1-12CHAPTER 1: FUNCTIONS AND GRAPHS60.g(x) = – 0.6x2 + 3x+ 4(A)g(x) = – 2: – 0.6x2 + 3x+ 4 = – 20.6x2 – 3x– 6 = 0x= – 1.53, 6.53(B)g(x) = 5: – 0.6x2 + 3x+ 4 = 5– 0.6x2 + 3x– 1 = 00.6x2 – 3x+ 1 = 0x= 0.36, 4.64(C)g(x) = 8: – 0.6x2 + 3x+ 4 = 8– 0.6x2 + 3x– 4 = 00.6x2 – 3x+ 4 = 0No solution62.Using a graphing utility withy= 100x– 7x2 – 10 and the calculus option with maximum command, weobtain 347.1429 as the maximum value.64.The slope of the line through1122(,)and(,)xyxyis2121yyxx;the slope of the line through1133(, y )and(,)xxyis3131yyxx. By Problem 57,21312131.yyAyyxxBxx66.We have two representations of (d, P): (0, 14.7) and (34, 29.4).(A)A line relatingPtodpasses through the above two points.Its equation is:P– 14.7 =(29.414.7)(340)(d– 0)or0.43214.7Pd(B)The rate of change of pressure with respect to depth is approximately0.432lbs/in2 per foot.(C)Ford= 50,P=0.432(50) + 14.7 ≈ 36.3 lbs/in2

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EXERCISE 1-31-13(D)ForP= 4 atmospheres, we haveP= 4(14.7) = 58.8 lbs/in2and hence58.8 =0.432d+ 14.7ord=58.814.70.432102ft.68.We have two representations of (t, a): (0, 2,880) and (180, 0).(A)The linear model relating altitudeato the time in airthasthe following equation:a– 2,880 =(02,880)(1800)(t– 0)ora= – 16t+ 2,880(B)The rate of descent for an ATPS system parachute is 16 ft/sec.(C)It is 16 ft/sec.70.Letybe daily cost of producingxtennis rackets. Then we have two points for (x, y):(50, 3,855) and (60, 4,245).(A) Sincexandyare linearly related, then the two points(50, 3,855) and (60, 4,245) will lie on the line expressingthe linear relationship betweenxandy. Thereforey– 3,855 =(4, 2453,855)(6050)(x– 50)ory= 39x+ 1,905(B)(C)Theyintercept, $1,905, is the fixed cost and theslope, $39, is the cost per racket.72.We observe that for (t, V) two points are given: (0, 224,000) and (16, 115,200)(A) A linear model will be a line passing through the two points (0, 224,000) and (16, 115,200). Theequation of this line is:V– 115,200 =(224, 000115, 200)(016)(t– 16) orV= – 6,800t+ 224,000(B) Fort= 10V= – 6,800(10) + 224,000 = $156,000

1-14CHAPTER 1: FUNCTIONS AND GRAPHS(C)ForV= $100,000100,000 = – 6,800t+ 224,000ort=(224, 000100, 000)6,80018.24So, during the 19thyear, the depreciated valuefalls below $100,000.(D)74.LetTbe the true airspeed at the altitudeA(thousands of feet). SinceTincreases 1.6% ,(1000)200(1.016)203.2.T(A)A linear relationship betweenAandThas slope(203.2200)3.21m.Therefore,3.2200.TA(B)ForA= 6.5 (6,500 feet),T= 3.2(6.5) + 200 = 20.8+200 = 220.8 mph76.(A) For (x, p) we have two representations: (9,800, 1.94) and (9,400, 1.82).The slope ism=(1.941.82)(9,8009, 400)= 0.0003Using one of the points, say (9,800, 1.94), we findb:1.94 = (0.0003)(9,800) +borb= – 1So, the desired equation is:p= 0.0003x– 1.(B) Here the two representations of (x, p) are: (9,300, 1.94)and (9,500, 1.82). The slope is(1.941.82)0.0006(9,3009,500)m Using one of the points, say (9,300, 1.94) we findb:1.94 = – 0.0006(9,300) +borb= 7.52So, the desired equation is:p= – 0.0006x+ 7.52.(C) To find the equilibrium point, we need to solve0.0003x– 1 = – 0.0006x+ 7.52 forx. Observe that0.0009x= 8.52 orx=8.520.0009= 9,467Substitutingx= 9,467 in either of equations in (A) or (B)we obtain.0003(9, 467)11.84pSo, the desired point is (9,467, 1.84).

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