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

Page 1 of 16

Solution Manual for Calculus for Business, Economics, Life Sciences and Social Sciences, 14th Edition - Page 1 preview image

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

Page 2 of 16

Solution Manual for Calculus for Business, Economics, Life Sciences and Social Sciences, 14th Edition - Page 2 preview image

Page 3 of 16

Solution Manual for Calculus for Business, Economics, Life Sciences and Social Sciences, 14th Edition - Page 3 preview image

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

Page 4 of 16

Solution Manual for Calculus for Business, Economics, Life Sciences and Social Sciences, 14th Edition - Page 4 preview image

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.

Page 5 of 16

Solution Manual for Calculus for Business, Economics, Life Sciences and Social Sciences, 14th Edition - Page 5 preview image

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

Page 6 of 16

Solution Manual for Calculus for Business, Economics, Life Sciences and Social Sciences, 14th Edition - Page 6 preview image

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)

Page 7 of 16

Solution Manual for Calculus for Business, Economics, Life Sciences and Social Sciences, 14th Edition - Page 7 preview image

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.

Page 8 of 16

Solution Manual for Calculus for Business, Economics, Life Sciences and Social Sciences, 14th Edition - Page 8 preview image

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

Page 9 of 16

Solution Manual for Calculus for Business, Economics, Life Sciences and Social Sciences, 14th Edition - Page 9 preview image

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).

Page 10 of 16

Solution Manual for Calculus for Business, Economics, Life Sciences and Social Sciences, 14th Edition - Page 10 preview image

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

Page 11 of 16

Solution Manual for Calculus for Business, Economics, Life Sciences and Social Sciences, 14th Edition - Page 11 preview image

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, ∞)

Page 12 of 16

Solution Manual for Calculus for Business, Economics, Life Sciences and Social Sciences, 14th Edition - Page 12 preview image

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

Page 13 of 16

Solution Manual for Calculus for Business, Economics, Life Sciences and Social Sciences, 14th Edition - Page 13 preview image

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

Page 14 of 16

Solution Manual for Calculus for Business, Economics, Life Sciences and Social Sciences, 14th Edition - Page 14 preview image

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

Page 15 of 16

Solution Manual for Calculus for Business, Economics, Life Sciences and Social Sciences, 14th Edition - Page 15 preview image

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

Page 16 of 16

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).

Preview Mode

This document has 651 pages. Sign in to access the full document!

Report

Learning Tools

Writing Tools

Browse Resources

Solution Manual for Calculus for Business, Economics, Life Sciences and Social Sciences, 14th Edition - Page 1

Study Now!

Document Details

Related Documents

Test Bank for Calculus for Business, Economics, Life Sciences, and Social Sciences, 14th Edition

Precalculus - Polynomial and Rational Functions

Precalculus - Functions

Precalculus - Exponential and Logarithmic Functions

Math Word Problems - Translating Expressions

Math Word Problems - Equations

Algebra II – Word Problems

Algebra II – Sequences and Series

Algebra II - Segments Lines and Inequalities

Algebra II - Relations and Functions

Company

Explore

Study Tools