Solution Manual For Calculus, 6th Edition

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Section 1.1C01S01.001:Iff(x) = 1x, then:(a)f(−a) =1−a=−1a;(b)f(a−1) =1a−1=a;(c)f(√a) =1√a=1a1/2=a−1/2;(d)f(a2) =1a2=a−2.C01S01.002:Iff(x) =x2+ 5, then:(a)f(−a) = (−a)2+ 5 =a2+ 5;(b)f(a−1) = (a−1)2+ 5 =a−2+ 5 =1a2+ 5 = 1 + 5a2a2;(c)f(√a) = (√a)2+ 5 =a+ 5;(d)f(a2) = (a2)2+ 5 =a4+ 5.C01S01.003:Iff(x) =1x2+ 5 , then:(a)f(−a) =1(−a)2+ 5 =1a2+ 5 ;(b)f(a−1) =1(a−1)2+ 5 =1a−2+ 5 =1·a2a−2·a2+ 5·a2=a21 + 5a2;(c)f(√a) =1(√a)2+ 5 =1a+ 5 ;(d)f(a2) =1(a2)2+ 5 =1a4+ 5 .C01S01.004:Iff(x) =√1 +x2+x4, then:(a)f(−a) =√1 + (−a)2+ (−a)4=√1 +a2+a4;(b)f(a−1) =√1 + (a−1)2+ (a−1)4=√1 +a−2+a−4=√(a4)·(1 +a−2+a−4)a4=√a4+a2+ 1a4=√a4+a2+ 1√a4=√a4+a2+ 1a2;(c)f(√a) =√1 + (√a)2+ (√a)4=√1 +a+a2;(d)f(a2) =√1 + (a2)2+ (a4)2=√1 +a4+a8.C01S01.005:Ifg(x) = 3x+ 4 andg(a) = 5, then 3a+ 4 = 5, so 3a= 1; thereforea=13.C01S01.006:Ifg(x) =12x−1 andg(a) = 5, then:1

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12a−1 = 5;1 = 5·(2a−1);1 = 10a−5;10a= 6;a= 35.C01S01.007:Ifg(x) =√x2+ 16 andg(a) = 5, then:√a2+ 16 = 5;a2+ 16 = 25;a2= 9;a= 3 ora=−3.C01S01.008:Ifg(x) =x3−3 andg(a) = 5, thena3−3 = 5, soa3= 8. Hencea= 2.C01S01.009:Ifg(x) =3√x+ 25 = (x+ 25)1/3andg(a) = 5, then(a+ 25)1/3= 5;a+ 25 = 53= 125;a= 100.C01S01.010:Ifg(x) = 2x2−x+ 4 andg(a) = 5, then:2a2−a+ 4 = 5;2a2−a−1 = 0;(2a+ 1)(a−1) = 0;2a+ 1 = 0 ora−1 = 0;a=−12 ora= 1.C01S01.011:Iff(x) = 3x−2, thenf(a+h)−f(a) = [3(a+h)−2]−[3a−2]= 3a+ 3h−2−3a+ 2 = 3h.C01S01.012:Iff(x) = 1−2x, then2

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f(a+h)−f(a) = [1−2(a+h)]−[1−2a] = 1−2a−2h−1 + 2a=−2h.C01S01.013:Iff(x) =x2, thenf(a+h)−f(a) = (a+h)2−a2=a2+ 2ah+h2−a2= 2ah+h2=h·(2a+h).C01S01.014:Iff(x) =x2+ 2x, thenf(a+h)−f(a) = [(a+h)2+ 2(a+h)]−[a2+ 2a]=a2+ 2ah+h2+ 2a+ 2h−a2−2a= 2ah+h2+ 2h=h·(2a+h+ 2).C01S01.015:Iff(x) = 1x, thenf(a+h)−f(a) =1a+h−1a=aa(a+h)−a+ha(a+h)=a−(a+h)a(a+h)=−ha(a+h).C01S01.016:Iff(x) =2x+ 1 , thenf(a+h)−f(a) =2a+h+ 1−2a+ 1 =2(a+ 1)(a+h+ 1)(a+ 1)−2(a+h+ 1)(a+h+ 1)(a+ 1)=2a+ 2(a+h+ 1)(a+ 1)−2a+ 2h+ 2(a+h+ 1)(a+ 1) = (2a+ 2)−(2a+ 2h+ 2)(a+h+ 1)(a+ 1)= 2a+ 2−2a−2h−2(a+h+ 1)(a+ 1)=−2h(a+h+ 1)(a+ 1).C01S01.017:Ifx >0 thenf(x) =x|x|=xx= 1.Ifx <0 thenf(x) =x|x|=x−x=−1.We are givenf(0) = 0, so the range offis{−1,0,1}. That is, the range offis the set consisting of thethree real numbers−1, 0, and 1.C01S01.018:Givenf(x) = [[3x]], we see that3

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f(x) = 0if0x <13,f(x) = 1if13x <23,f(2) = 2if23x <1;moreover,f(x) =−3if−1x <−23,f(x) =−2if−23x <−13,f(x) =−1if−13x <0.In general, ifmis any integer, thenf(x) = 3mifmx < m+13,f(x) = 3m+ 1ifm+13x < m+23,f(x) = 3m+ 2ifm+23x < m+ 1.Because every integer is equal to 3mor to 3m+ 1 or to 3m+ 2 for some integerm, we see that the rangeoffincludes the setZof all integers. Becausefcan assume no values other than integers, we can concludethat the range offis exactlyZ.C01S01.019:Givenf(x) = (−1)[[x]], we first note that the values of the exponent [[x]] consist of all theintegers and no other numbers. So all that matters about the exponent is whether it is an even integer oran odd integer, for if even thenf(x) = 1 and if odd thenf(x) =−1. No other values off(x) are possible,so the range offis the set consisting of the two numbers−1 and 1.C01S01.020:If 0< x1, thenf(x) = 34. If 1< x2 thenf(x) = 34 + 21 = 55. If 2< x3 thenf(x) = 34 + 2·21 = 76. We continue in this way and conclude with the observation that if 11< x <12 thenf(x) = 34 + 11·21 = 265. So the range offis the set{34,55,76,97,118,139,160,181,202,223,244,265}.C01S01.021:Givenf(x) = 10−x2, note that for every real numberx,x2is defined, and for every suchreal numberx2, 10−x2is also defined. Therefore the domain offis the setRof all real numbers.C01S01.022:Givenf(x) =x3+ 5, we note that for each real numberx,x3is defined; moreover, for eachsuch real numberx3,x3+ 5 is also defined. Thus the domain offis the setRof all real numbers.C01S01.023:Givenf(t) =√t2, we observe that for every real numbert,t2is defined and nonnegative,and hence that√t2is defined as well. Therefore the domain offis the setRof all real numbers.C01S01.024:Giveng(t) =(√t)2, we observe that√tis defined exactly whent0. In this case,(√t)2is also defined, and hence the domain ofgis the set [0,+∞) of all nonnegative real numbers.C01S01.025:Givenf(x) =√3x−5, we note that 3x−5 is defined for all real numbersx, but that itssquare root will be defined when and only when 3x−5 is nonnegative; that is, when 3x−50, so thatx53. So the domain offconsists of all those real numbersxin the interval[53,+∞).4

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C01S01.026:Giveng(t) =3√t+ 4 = (t+ 4)1/3, we note thatt+ 4 is defined for every real numbertandthe cube root oft+ 4 is defined for every possible resulting value oft+ 4. Therefore the domain ofgis thesetRof all real numbers.C01S01.027:Givenf(t) =√1−2t, we observe that 1−2tis defined for every real numbert, but that itssquare root is defined only when 1−2tis nonnegative. We solve the inequality 1−2t0 to find thatf(t)is defined exactly whent12. Hence the domain offis the interval(−∞,12].C01S01.028:Giveng(x) =1(x+ 2)2,we see that (x+ 2)2is defined for every real numberx, but thatg(x), its reciprocal, will be defined onlywhen (x+ 2)2= 0; that is, whenx+ 2= 0. So the domain ofgconsists of those real numbersx=−2.C01S01.029:Givenf(x) =23−x ,we see that 3−xis defined for all real values ofx, but thatf(x), double its reciprocal, is defined only when3−x= 0. So the domain offconsists of those real numbersx= 3.C01S01.030:Giveng(t) =√23−t ,it is necessary that 3−tbe both nonzero (so that its reciprocal is defined) and nonnegative (so that thesquare root is defined). Thus 3−t >0, and therefore the domain ofgconsists of those real numberst <3.C01S01.031:Givenf(x) =√x2+ 9, observe that for each real numberx,x2+ 9 is defined and, moreover,is positive. So its square root is defined for every real numberx. Hence the domain offis the setRof allreal numbers.C01S01.032:Givenh(z) =1√4−z2,we note that 4−z2is defined for every real numberz, but that its square root will be defined only if4−z20. Moreover, the square root cannot be zero, else its reciprocal will be undefined, so we need tosolve the inequality 4−z2>0; that is,z2<4. The solution is−2< z <2, so the domain ofhis the openinterval (−2,2).C01S01.033:Givenf(x) =√4− √x, note first that we requirex0 in order that√xbe defined. Inaddition, we require 4− √x0 so thatitssquare root will be defined as well. So we solve [simultaneously]x0 and√x4 to find that 0x16. So the domain offis the closed interval [0,16].C01S01.034:Givenf(x) =√x+ 1x−1,5

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we require thatx= 1 so that the fraction is defined. In addition we require that the fraction be nonnegativeso that its square root will be defined.These conditions imply that both numerator and denominator bepositive or that both be negative; moreover, the numerator may also be zero.But if the denominator ispositive then the [larger] numerator will be positive as well; if the numerator is nonpositive then the [smaller]denominator will be negative. So the domain offconsists of those real numbers for whicheitherx−1>0orx+ 10; that is, eitherx >1 orx−1. So the domain offis the union of the two intervals (−∞,−1]and (1,+∞). Alternatively, it consists of those real numbersxnotin the interval (−1,1].C01S01.035:Given:g(t) =t|t|.This fraction will be defined whenever its denominator is nonzero, thus for all real numberst= 0.Sothe domain ofgconsists of the nonzero real numbers; that is, the union of the two intervals (−∞,0) and(0,+∞).C01S01.036:If a square has edge lengthx, then its areaAis given byA=x2and its perimeterPis givenbyP= 4x. To expressAin terms ofP:x=14P;A=x2=(14P)2=116P2.Thus to expressAas a function ofP, we writeA(P) =116P2,0P <+∞.(It will be convenient later in the course to allow the possibility thatP,x, andAare zero. If this producesan answer that fails to meet real-world criteria for a solution, then that possibility can simply be eliminatedwhen the answer to the problem is stated.)C01S01.037:If a circle has radiusr, then its circumferenceCis given byC= 2πrand its areaAbyA=πr2. To expressCin terms ofA, we first expressrin terms ofA, then substitute in the formula forC:A=πr2;r=√Aπ;C= 2πr= 2π√Aπ= 2√π2Aπ= 2√πA.Therefore to expressCas a function ofA, we writeC(A) = 2√πA,0A <+∞.It is also permissible simply to writeC(A) = 2√πAwithout mentioning the domain, because the “default”domain is correct. In the first displayed equation we do not writer=±√A/πbecause we know thatrisnever negative.C01S01.38:Ifrdenotes the radius of the sphere, then its volume is given byV=43πr3and its surfacearea byS= 4πr2. Hence6

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r= 12√Sπ;V= 43πr3= 43π·18(Sπ)3/2= 16π(Sπ)3/2.Answer:V(S) = 16π(Sπ)3/2,0S <+∞.C01S01.039:To avoid decimals, we note that a change of 5◦C is the same as a change of 9◦F, so whenthe temperature is 10◦C it is 32 + 18 = 50◦F; when the temperature is 20◦C then it is 32 + 2·18 = 68◦F.In general we get the Fahrenheit temperatureFby adding 32 to the product of110Cand 18, whereCis theCelsius temperature. That is,F= 32 + 95C,and thereforeC=59(F−32). Answer:C(F) = 59 (F−32),F >−459.67.C01S01.040:Suppose that a rectangle has base lengthxand perimeter 100. Lethdenote the height ofsuch a rectangle. Then 2x+ 2h= 100, so thath= 50−x. Becausex0 andh0, we see that 0x50.The areaAof the rectangle isxh, so thatA(x) =x(50−x),0x50.C01S01.041:Letydenote the height of such a rectangle. The rectangle is inscribed in a circle of diameter4, so the bottom sidexand the left sideyare the two legs of a right triangle with hypotenuse 4. Consequentlyx2+y2= 16, soy=√16−x2(not−√16−x2becausey0).Becausex0 andy0, we see that0x4. The rectangle has areaA=xy, soA(x) =x√16−x2,0x4.C01S042.042:We take the problem to mean that current production is 200 barrels per day per well, thatif one new well is drilled then the 21 wells will produce 195 barrels per day per well; in general, that ifxnewwells are drilled then the 20 +xwells will produce 200−5xbarrels per day per well. So total productionwould bep= (20 +x)(200−5x) barrels per day. But because 200−5x0, we see thatx40. Becausex0 as well (you don’t “undrill” wells), here’s the answer:p(x) = 4000 + 100x−5x2,0x40,xan integer.C01S01.043:The square base of the box measuresxbyxcentimeters; letydenote its height (in centime-ters). Because the volume of the box is 324 cm3, we see thatx2y= 324. The base of the box costs 2x2cents,each of its four sides costsxycents, and its top costsx2cents. So the total cost of the box isC= 2x2+ 4xy+x2= 3x2+ 4xy.(1)7

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Becausex >0 andy >0 (the box has positive volume), but becauseycan be arbitrarily close to zero (aswell asx), we see also that 0< x <+∞. We use the equationx2y= 324 to eliminateyfrom Eq. (1) andthereby find thatC(x) = 3x2+ 1296x,0< x <+∞.C01S01.044:If the rectangle is rotated around its sideSof lengthxto produce a cylinder, thenxwill alsobe the height of the cylinder. Letydenote the length of the two sides perpendicular toS; thenywill be theradius of the cylinder; moreover, the perimeter of the original rectangle is 2x+ 2y= 36. Hencey= 18−x.Note also thatx0 and thatx18 (becausey0). The volume of the cylinder isV=πy2x, and soV(x) =πx(18−x)2,0x18.C01S01.045:Lethdenote the height of the cylinder. Its radius isr, so its volume isπr2h= 1000. Thetotal surface area of the cylinder isA= 2πr2+ 2πrh(look inside the front cover of the book);h= 1000πr2,soA= 2πr2+ 2πr·1000πr2= 2πr2+ 2000r.Nowrcannot be negative;rcannot be zero, elseπr2h= 1000. Butrcan be arbitrarily small positive aswell as arbitrarily large positive (by makinghsufficiently close to zero). Answer:A(r) = 2πr2+ 2000r,0< r <+∞.C01S01.046:Letydenote the height of the box (in centimeters). Then2x2+ 4xy= 600,so thaty= 600−2x24x.(1)The volume of the box isV=x2y= (600−2x2)·x24x= 14 (600x−2x3) = 12 (300x−x3)by Eq. (1). Alsox >0 by Eq. (1), but the maximum value ofxis attained when Eq. (1) forcesyto be zero,at which pointx=√300 = 10√3. Answer:V(x) = 300x−x32,0< x10√3.C01S01.047:The base of the box will be a square measuring 50−2xin. on each side, so the open-toppedbox will have that square as its base and four rectangular sides each measuring 50−2xbyx(the height ofthe box). Clearly 0xand 2x50. So the volume of the box will beV(x) =x(50−2x)2,0x25.8

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C01S01.048:Recall thatA(x) =x(50−x),0x50. Here is a table of a few values of the functionAat some special numbers in its domain:x05101520253035404550A02254005256006256005254002250It appears that whenx= 25 (so the rectangle is a square), the rectangle has maximum area 625.C01S01.049:Recall that the total daily production of the oil field isp(x) = (20 +x)(200−5x) ifxnewwells are drilled (wherexis an integer satisfying 0x40).Here is a table ofallof the values of theproduction functionp:x01234567p40004095418042554320437544204455x89101112131415p44804495450044954480445544204375x1617181920212223p43204255418040954000389537803655x2425262728293031p35203375322030552880269525002295x3233343536373839p20801855162013751120855580295and, finally,p(40) = 0. Answer: Drill ten new wells.C01S01.050:The surface areaAof the box of Example 8 wasA(x) = 2x2+ 500x ,0< x <∞.The restrictionsx1 andy1 imply that 1x√125.A small number of values ofA, rounded tothree places, are given in the following table.x1234567891011A502258185157150155169191218250287It appears thatAis minimized whenx=y= 5.C01S01.051:Ifxis an integer, thenCeiling(x) =xand−Floor(−x) =−(−x) =x.Ifxis not aninteger, then choose the integernso thatn < x < n+ 1. ThenCeiling(x) =n+ 1,−(n+ 1)<−x <−n,and−Floor(−x) =−[−(n+ 1)] =n+ 1.In both cases we see thatCeiling(x) =−Floor(−x).C01S01.052:The range ofRound(x) is the setZof all integers.Ifkis a nonzero constant, then asxvaries through all real number values, so doeskx. Hence the range ofRound(kx) isZifk= 0. Ifk= 0then the range ofRound(kx) consists of the single number zero.9

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C01S01.053:By the result of Problem 52, the range ofRound(10x) is the set of all integers, so the rangeofg(x) =110Round(10x) is the set of all integral multiple of110.C01S01.054:What works forπwill work for every real number; letRound2(x) =1100Round(100x). Tobe certain that this is correct (we will verify it only for positive numbers), write the [positive] real numberxin the formx=k+t10 +h100 +m1000 +r,wherekis a nonnnegative integer,t(the “tenths” digit) is a nonnegative integer between 0 and 9,h(the“hundredths” digit) is a nonnegative integer between 0 and 9, as ism, and 0r <0.001. ThenRound2(x) =1100Floor(100x+ 0.5)=1100Floor(100k+ 10t+h+110(m+ 5) + 100r).If 0m4, the last expression becomes1100 (100k+ 10t+h) =k+t10 +h100,which is the correct two-digit rounding ofx. If 5m9, it becomes1100 (100k+ 10t+h+ 1) =k+t10 +h+ 1100,also the correct two-digit rounding ofxin this case.C01S01.055:LetRound4(x) =110000Round(10000x). To verify thatRound4 has the desired propertyfor [say] positive values ofx, write such a numberxin the formx=k+d110 +d2100 +d31000 +d410000 +d5100,000 +r,wherekis a nonnegative integer, eachdiis an integer between 0 and 9, and 0r <0.00001. ApplicationofRound4 toxthen produces110000Floor(10000k+ 1000d1+ 100d2+ 10d3+d4+110(d5+ 5) + 10000r).Then consideration of the two cases 0d54 and 5d59 will show thatRound4 produces the correctfour-place rounding ofxin both cases.C01S01.056:LetChop4(x) =110000Floor(10000x). Suppose thatx >0. Writexin the formx=k+d110 +d2100 +d31000 +d410000 +r,wherekis a nonnegative integer, each of thediis an integer between 0 and 9, and 0r <0.0001. ThenChop4(x) produces110000Floor(10000k+ 1000d1+ 100d2+ 10d3+d4+ 10000r)=110000(10000k+ 1000d1+ 100d2+ 10d3+d4)10

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because 10000r <1. It follows thatChop4 has the desired effect.C01S01.057:x0.00.20.40.60.81.0y1.00.44−0.04−0.44−0.76−1.0The sign change occurs betweenx= 0.2 andx= 0.4.x0.200.250.300.350.40y0.440.31250.190.0725−0.04The sign change occurs betweenx= 0.35 andx= 0.40.x0.350.360.370.380.390.40y0.07250.04960.02690.0044−0.0179−0.04From this point on, the data forywill be rounded.x0.3800.3820.3840.3860.3880.390y0.0044−0.0001−0.0045−0.0090−0.0135−0.0179Answer (rounded to two places): 0.38. The quadratic formula yields the two roots12(3± √5); the smallerof these is approximately0.381966011250105151795.Problems 58 through 66 are worked in the same way as Problem 57.C01S01.058:The sign change intervals are [2,3], [2.6,2.8], [2.60,2.64], and [2.616,2.624].Answer:12(3 +√5)≈2.62.C01S01.059:The sign change intervals are [1,2], [1.2,1.4], [1.20,1.24], [1.232,1.240], and [1.2352,1.2368].Answer:−1 +√5≈1.24.C01S01.060:The sign change intervals are [−4,−3], [−3.4,−3.2], [−3.24,−3.20], [−3.240,−3.232], and[−3.2368,−3.2352]. Answer:−1− √5≈ −3.24.C01S01.061:The sign change intervals are [0,1], [0.6,0.8], [0.68,0.72], [0.712,0.720], and [0.7184,0.7200].Answer:14(7− √17)≈0.72.C01S01.062:The sign change intervals are [2,3], [2.6,2.8], [2.76,2.80], [2.776,2.784], and [2.7792,2.7808].Answer:14(7 +√17)≈2.78.C01S01.063:The sign change intervals are [3,4], [3.2,3.4], [3.20,3.24], [3.208,3.216], and [3.2080,3.2096].Answer:12(11− √21)≈3.21.C01S01.064:The sign change intervals are [7,8], [7.6,7.8], [7.76,7.80], [7.784,7.792], and [7.7904,7.7920].Answer:12(11 +√21)≈7.79.C01S01.065:The sign change intervals are [1,2], [1.6,1.8], [1.60,1.64], [1.608,1.616], [1.6144,1.6160],and [1.61568,1.61600]. Answer:16(−23 +√1069)≈1.62.C01S01.066:The sign change intervals are [−10,−9], [−9.4,−9.2], [−9.32,−9.28], [−9.288,−9.280], and[−9.2832,−9.2816]. Answer:16(−23− √1069)≈ −9.28.11

Solution Manual For Calculus, 6th Edition - Page 13 preview image

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Section 1.2C01S02.001:The slope ofLism= (3−0)/(2−0) =32, soLhas equationy−0 = 32 (x−0);that is,2y= 3x.C01S02.002:BecauseLis vertical and (7,0) lies onL, every point onLhas Cartesian coordinates (7, y)for some numbery(and every such point lies onL). Hence an equation ofLisx= 7.C01S02.003:BecauseLis horizontal, it has slope zero. Hence an equation ofLisy−(−5) = 0·(x−3);that is,y=−5.C01S02.004:Because (2,0) and (0,−3) lie onL, it has slope (0 + 3)/(2−0) =32. Hence an equation ofLisy−0 =32(x−2);that is,y=32x−3.C01S02.005:The slope ofLis (3−(−3))/(5−2) = 2, so an equation ofLisy−3 = 2(x−5);that is,y= 2x−7.C01S02.006:An equation ofLisy−(−4) =12(x−(−1));that is, 2y+ 7 =x.C01S02.007:The slope ofLis tan(135◦) =−1, soLhas equationy−2 =−1·(x−4);that is,x+y= 6.C01S02.008:Equation:y−7 = 6(x−0);that is,y= 6x+ 7.C01S02.009:The second line’s equation can be written in the formy=−2x+ 10 to show that it has slope−2. BecauseLis parallel to the second line,Lalso has slope−2 and thus equationy−5 =−2(x−1).C01S02.010:The equation of the second line can be rewritten asy=−12x+172to show that it has slope−12. BecauseLis perpendicular to the second line,Lhas slope 2 and thus equationy−4 = 2(x+ 2).C01S02.011:x2−4x+ 4 +y2= 4: (x−2)2+ (y−0)2= 22. Center (2,0), radius 2.C01S02.012:x2+y2+ 6y+ 9 = 9: (x−0)2+ (y+ 3)2= 32. Center (0,−3), radius 3.C01S02.013:x2+ 2x+ 1 +y2+ 2y+ 1 = 4:(x+ 1)2+ (y+ 1)2= 22. Center (−1,−1), radius 2.C01S02.014:x2+ 10x+ 25 +y2−20y+ 100 = 25:(x+ 5)2+ (y−10)2= 52.Center (−5,10), radius 5.C01S02.015:x2+y2+x−y=12:x2+x+14+y2−y+14= 1; (x+12)2+ (y−12)2= 1. Center: (−12,12),radius 1.C01S02.016:x2+y2−23x−43y=119:x2−23x+19+y2−43y+49=169; (x−13)2+ (y−23)2= (43)2. Center(13,23), radius43.1

Solution Manual For Calculus, 6th Edition - Page 14 preview image

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C01S02.017:y= (x−3)2: Opens upward, vertex at (3,0).C01S02.018:y−16 =−x2: Opens downward, vertex at (0,16).C01S02.019:y−3 = (x+ 1)2: Opens upward, vertex at (−1,3).C01S02.020:2y=x2−4x+ 4 + 4:y−2 =12(x−2)2. Opens upward, vertex at (2,2).C01S02.021:y= 5(x2+ 4x+ 4) + 3 = 5(x+ 2)2+ 3: Opens upward, vertex at (−2,3).C01S02.022:y=−(x2−x) =−(x2−x+14) +14:y−14=−(x−12)2. Opens downward, vertex at (12,14).C01S02.023:x2−6x+ 9 +y2+ 8y+ 16 = 25: (x−3)2+ (y+ 4)2= 55. Circle, center (3,−4), radius 5.C01S02.024:(x−1)2+ (y+ 1)2= 0: The graph consists of the single point (1,−1).C01S02.025:(x+ 1)2+ (y+ 3)2=−10: There are no points on the graph.C01S02.026:x2+y2−x+ 3y+ 2.5 = 0:x2−x+ 0.25 +y2+ 3y+ 2.25 = 0: (x−0.5)2+ (y+ 1.5)2= 0.The graph consists of the single point (0.5,−1.5).C01S02.027:The graph is the straight line segment connecting the two points (−1,7) and (1,−3)(including those two points).C01S02.028:The graph is the straight line segment connecting the two points (0,2) and (2,−8), includingthe first of these two points but not the second.C01S02.029:The graph is the parabola that opens downward, symmetric around they-axis, with vertexat (0,10) andx-intercepts±√10.C01S02.030:The graph ofy= 1 + 2x2is a parabola that opens upwards, is symmetric around they-axis,and has vertex at (0,1).C01S02.031:The graph ofy=x3can be visualized by modifying the familiar graph of the parabola withequationy=x2: The former may be obtained by multiplying they-coordinate of the latter’s point (x, x2)byx. Thus both have flat spots at the origin. For 0< x <1, the graph ofy=x3is below that ofy=x2.They cross at (1,1), and forx >1 the graph ofy=x2is below that ofy=x3, with the difference becomingarbitrarily large asxincreases without bound. If the graph ofy=x3forx0 is rotated 180◦around thepoint (0,0), the graph ofy=x3forx <0 is the result.C01S02.032:The graph off(x) =x4can be visualized by first visualizing the graph ofy=x2.If they-coordinate of each point on this graph is replaced with its square (x4), the result is the graph off. Theeffect on the graph ofy=x2is to multiply they-coordinate byx2, which is between 0 and 1 for 0<|x|<1and which is larger than 1 for|x|>1. Thus the graph offsuperficially resembles that ofy=x2, but ismuch closer to thex-axis for|x|<1 and much farther away for|x|>1.The two graphs cross at (0,0)(where each has a flat spot) and at (±1,1), but the graph offis much steeper at the latter two points.C01S02.033:To graphy=f(x) =√4−x2, note thaty0 and thaty2= 4−x2;that is,x2+y2= 4.Hence the graph offis theupper halfof the circle with center (0,0) and radius 2.C01S02.034:To graphy=f(x) =−√9−x2, note thaty0 and thaty2= 9−x2;that is, thatx2+y2= 9. Hence the graph offis thelower halfof the circle with center (0,0) and radius 3.2

Solution Manual For Calculus, 6th Edition - Page 15 preview image

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-4-2246-4-224C01S02.035:To graphf(x) =√x2−9, note that there is no graph for−3< x <3, thatf(±3) = 0, andthatf(x)>0 forx <−3 and forx >3. Ifxis large positive, then√x2−9≈ √x2=x, so the graph offhasx-intercept (3,0) and rises asxincreases, nearly coinciding with the graph ofy=xforxlarge positive.The casex <−3 is trickier. In this case, ifxis a large negative number, thenf(x) =√x2−9≈ √x2=−x(Note the minus sign!).So forx−3, the graph offhasx-intercept (−3,0) and, forxlarge negative,almost coincides with the graph ofy=−x. Later we will see that the graph offbecomes arbitrarily steepasxgets closer and closer to±3.C01S02.036:Asxincreases without bound—either positively or negatively—f(x) gets arbitrarily closeto zero. Moreover, ifxis large positive thenf(x) is negative and close to zero, so the graph offlies justbelow thex-axis for suchx.Similarly, the graph offlies just above thex-axis forxlarge negative.Ifxis slightly less than 1 but very close to 1, thenf(x) is the reciprocal of a tiny positive number, hence is alarge positive number. So the graph offjust to the left of the vertical linex= 1 almost coincides with thetop half of that line. Similarly, just to the right of the linex= 1, then graph offalmost coincides with thebottomhalf of that line. There is no graph wherex= 1, so the graph resembles the one in the next figure.The only intercept is they-intercept (0,1). The graph correctly shows that the graph offis increasing forx <1 and forx >1.C01S02.037:Note thatf(x) is positive and close to zero forxlarge positive, so that the graph offis justabove thex-axis—and nearly coincides with it—for suchx. Similarly, the graph offis just below thex-axisand nearly coincides with it forxlarge negative. There is no graph wherex=−2, but ifxis slightly greaterthan−2 thenf(x) is the reciprocal of a very small positive number, sof(x) is large and nearly coincideswith the upper half of the vertical linex=−2.Similarly, ifxis slightly less than−2, then the graph off(x) is large negative and nearly coincides with the the lower half of the linex=−3.The graph offisdecreasing forx <−2 and forx >−2 and its only intercept is they-intercept(0,12).C01S02.038:Note thatf(x) is very small but positive ifxis either large positive or large negative. Thereis no graph forx= 0, but ifxis very close to zero, thenf(x) is the reciprocal of a very small positivenumber, and hence is large positive. So the graph offis just above thex-axis and almost coincides with itif|x|is large, whereas the graph offalmost coincides with the positivey-axis forxnear zero. There are nointercepts; the graph offis increasing forx <0 and is decreasing forx >0.3

Solution Manual For Calculus, 6th Edition - Page 16 preview image

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C01S02.039:Note thatf(x)>0 for allxother thanx= 1, wherefis not defined. If|x|is large, thenf(x) is near zero, so the graph offalmost coincides with thex-axis for suchx. Ifxis very close to 1, thenf(x) is the reciprocal of a very small positive number, hencef(x) is large positive. So for suchx, the graphoff(x) almost coincides with the upper half of the vertical linex= 1. The only intercept is (0,1).C01S02.040:Note first thatf(x) is undefined atx= 0. To handle the absolute value symbol, we look attwo cases: Ifx >0, thenf(x) = 1; ifx <0, thenf(x) =−1. So the graph offconsists of the part of thehorizontal liney= 1 for whichx >0, together with the part of the horizontal liney=−1 for whichx <0.C01S02.041:Note thatf(x) is undefined when 2x+ 3 = 0;that is, whenx=−32. Ifxis large positive,thenf(x) is positive and close to zero, so the graph offis slightly above thex-axis and almost coincideswith thex-axis. Ifxis large negative, thenf(x) is negative and close to zero, so the graph offis slightlybelow thex-axis and almost coincides with thex-axis.Ifxis slightly greater than−32thenf(x) is verylarge positive, so the graph offalmost coincides with the upper half of the vertical linex=−32.Ifxisslightly less than−32thenf(x) is very large negative, so the graph offalmost coincides with the lower halfof that vertical line. The graph offis decreasing forx <−32and also decreasing forx >−32. The onlyintercept is at(0,13).C01S02.042:Note thatf(x) is undefined when 2x+ 3 = 0;that is, whenx=−32. Ifxis large positiveor large negative, thenf(x) is positive and close to zero, so the graph offis slightly above thex-axis andalmost coincides with thex-axis for|x|large.Ifxis close to−32thenf(x) is very large positive, so thegraph offalmost coincides with the upper half of the vertical linex=−32. The graph offis increasing forx <−32and decreasing forx >−32. The only intercept is at(0,19).C01S02.043:Giveny=f(x) =√1−x, note thaty0 and thaty2= 1−x; that is,x= 1−y2. So thegraph is the part of the parabolax= 1−y2for whichy0. This parabola has horizontal axis of symmetrythey-axis, opens to the left (because the coefficient ofy2is negative), and has vertex (1,0). Therefore thegraph offis the upper half of this parabola.C01S02.044:Note that the intervalx <1 is the domain off, so there is no graph forx1. Ifxis a largenegative number, then the denominator is large positive, so that its reciprocalf(x) is very small positive.Asxgets closer and closer to 1 (whilex <1), the denominator approaches zero, so its reciprocalf(x) takeson arbitrarily large positive values. So the graph offis slightly above thex-axis and almost coincides withthat axis forxlarge negative; the graph offalmost coincides with the upper half of the vertical linex= 1forxnear (and less than) 1. The graph offis increasing for allx <1 and (0,1) is the only intercept.C01S02.045:Note thatf(x) is defined only if 2x+ 3>0;that is, ifx >−32. Note also thatf(x)>0for all suchx. Ifxis large positive, thenf(x) is positive but near zero, so the graph offis just above thex-axis and almost coincides with it. Ifxis very close to−32(but larger), then the denominator inf(x) isvery tiny positive, so the graph offalmost coincides with the upper half of the vertical linex=−32forsuchx. The graph offis decreasing for allx >−32.C01S02.046:Given:f(x) =|2x−2|. Case 1:x1. Then 2x−20, so thatf(x) = 2x−2. Becausef(1) = 0, the graph offforx1 consists of the part of the straight line through (1,0) with slope 2. Case2:x <1. Then 2x−2<0, so thatf(x) =−2x+ 2. The liney=−2x+ 2 passes through (1,0), so thegraph offforx <1 consists of the part of the straight line through (1,0) with slope−2.C01S02.047:Given:f(x) =|x|+x. Ifx0 thenf(x) =x+x= 2x, so ifx0 then the graph offisthe part of the straight line through (0,0) with slope 2 for whichx0. Ifx <0 thenf(x) =−x+x= 0,so the rest of the graph offcoincides with the negativex-axis.4

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