D 4u 2 p. u D 12 C p 144 C 4.72/ 16. D 12 C 12p r 2 D 5 C 3p 3. r D c D

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1 pu D 4u 2 p 4u u C 2 C 36u C 54 SECTION 4.8 Newton s Method 491 Therefore the eqution becomes 8.S.0/=/ 2 D 8.S. 2 /=/ 2 8u 2 C 40u C 48 D 4u 2 4u. p u p u C 2/ C 36u C 54 4u 2 C 4u 6 D 4u. p u p u C 2/ 16u 4 C 32u 3 32u 2 48u C 36 D 16u 2.u/.u C 2/ 16u 4 C 32u 3 32u 2 48u C 36 D 16u 4 C 32u 3 32u 2 48u C 36 D 0 8u 2 C 12u 9 D 0 The lst qudrtic hs positive solution Therefore so u D 12 C p 144 C 4.72/ 16 D 12 C 12p 3 16 r 2 2 D 3 C 3p 3 ; 4 D 3 C 3p 3 4 This gives us tht S. 2 / D S.0/ when r 2 D 5 C 3p 3 4 r D c D s 5 C 3 p 3 4 From prt (e) we know tht for r D 15 < c, S.0/ is the minimum vlue for S, butforr D 2>c, S. 2 / is the minimum vlue. Since r D c is the onl solution of S.0/ D S. 2 / for r> p 2, it follows tht S.0/ provides the minimum vlue for p 2<r<c nd S. 2 / provides the minimum when r>c. 4.8 Newton s Method Preliminr Questions 1. How mn itertions of Newton s Method re required to compute root if f./is liner function? Remember tht Newton s Method uses the liner pproimtion of function to estimte the loction of root. If the originl function is liner, then onl one itertion of Newton s Method will be required to compute the root. 2. Wht hppens in Newton s Method if our initil guess hppens to be zero of f? If 0 hppens to be zero of f,then 1 D 0 f. 0/ f 0. 0 / D 0 0 D 0 I in other words, ever term in the Newton s Method sequence will remin Wht hppens in Newton s Method if our initil guess hppens to be locl min or m of f? Assuming tht the function is differentible, then the derivtive is zero t locl mimum or locl minimum. If Newton s Method is strted with n initil guess such tht f 0. 0 / D 0,thenNewton smethodwillfilinthesensetht 1 will not be defined. Tht is, the tngent line will be prllel to the -is nd will never intersect it. 4. Is the following resonble description of Newton s Method A root of the eqution of the tngent line to f./is used s n pproimtion to root of f./itself? Eplin. Yes, tht is resonble description. The itertion formul for Newton s Method ws derived b solving the eqution of the tngent line to D f./t 0 for its -intercept.

2 492 C HAPTER 4 APPLICATIONS OF THE DERIVATIVE Eercises Inthiseerciseset,llpproimtionsshouldbecrriedoutusingNewton smethod. In Eercises 1 6, ppl Newton s Method to f./nd initil guess 0 to clculte 1 ; 2 ; f./d 2 6, 0 D 2 Let f./d 2 6 nd define nc1 D n f. n/ f 0. n / D n 2 n 6 2 n With 0 D 2, wecompute n f./ D 2 3 C 1, 0 D 3 Let f./d 2 3 C 1 nd define nc1 D n f. n/ f 0. n / D n 2 n 3 n C 1 2 n 3 With 0 D 3, wecompute n f./ D 3 10, 0 D 2 Let f./d 3 10 nd define nc1 D n f. n/ f 0. n / D n 3 n 10 3n 2 With 0 D 2 we compute n f./ D 3 C C 1, 0 D 1 Let f./d 3 C C 1 nd define nc1 D n f. n/ f 0. n / D n 3 n C n C 1 3n 2 C 1 With 0 D 1 we compute n f./ D cos 4, 0 D 1 Let f./d cos 4 nd define nc1 D n f. n/ f 0. n / D n cos n 4 n sin n 4 With 0 D 1 we compute n

3 SECTION 4.8 Newton s Method f./d 1 sin, 0 D 7 Let f./d 1 sin nd define With 0 D 7 we compute nc1 D n f. n/ f 0. n / D 1 n sin n n n cos n sin n n Use Figure 1 to choose n initil guess 0 to the unique rel root of 3 C 2 C 5 D 0 nd compute the first three Newton itertes FIGURE 1 Grph of D 3 C 2 C 5. Let f./d 3 C 2 C 5 nd define We tke 0 D 14, bsedonthefigure,ndthenclculte nc1 D n f. n/ f 0. n / D n 3 n C 2 n C 5 3n 2 C 2 n Approimte solution of sin D cos 2 in the intervl 0; 2 tothreedecimlplces.thenfindtheectsolutionnd compre with our pproimtion. Let f./d sin cos 2 nd define With 0 D 05 we find nc1 D n f. n/ f 0. n / D n sin n cos 2 n cos n C 2 sin 2 n n 1 2 n The root, to three deciml plces, is Theectrootis 6, which is equl to 0524 to three deciml plces. 9. Approimte both solutions of e D 5 to three deciml plces (Figure 2) = e = FIGURE 2 Grphs of e nd 5.

4 494 C HAPTER 4 APPLICATIONS OF THE DERIVATIVE clculte We need to solve e 5 D 0,soletf./D e 5.Thenf 0./ D e 5. Withninitilguessof 0 D 02, we Newton s Method (First root) 1 D 02 f.02/ f 0.02/ 2 D f / f / 3 D f / f / 0 D 02.guess/ For the second root, we use n initil guess of 0 D 25. Newton s Method (Second root) 1 D 25 f.25/ f 0.25/ 2 D f / f / 3 D f / f / 0 D 25.guess/ Thus the two solutions of e D 5 re pproimtel r nd r The first positive solution of sin D 0 is D. UseNewton smethodtoclculte to four deciml plces. Let f./d sin.tking 0 D 3, wehve n Hence, to four deciml plces. In Eercises 11 14, pproimte to three deciml plces using Newton s Method nd compre with the vlue from clcultor. 11. p 11 Let f./d 2 11, ndlet 0 D 3. Newton smethodields n Aclcultorields =3 Let f./d 3 5, ndlet 0 D 2. Hererepproimtionstotherootoff./,whichis5 1=3. 4 n Aclcultorields =3 Note tht 2 7=3 D 4 2 1=3.Letf./D 3 2, ndlet 0 D 1. Newton smethodields n Thus, 2 7= D Aclcultorields

5 SECTION 4.8 Newton s Method =4 Let f./d 4 3,ndlet 0 D 08. Hererepproimtionstotherootoff./,whichis3 1=4. Aclcultorields n Approimte the lrgest positive root of f./d C C 5 to within n error of t most 10 4.RefertoFigure5. Figure 5 from the tet suggests the lrgest positive root of f./ D C C 5 is ner 2. So let f./ D C C 5 nd tke 0 D 2. 4 n The lrgest positive root of C C 5 is pproimtel In Eercises 16 19, pproimte the root specified to three deciml plces using Newton s Method. Use plot to choose n initil guess. 16. Lrgest positive root of f./d 3 5 C 1. Let f./d 3 5 C 1. Thegrphoff./shown below suggests the lrgest positive root is ner D 22. Tking 0 D 22, Newton smethodgives n The lrgest positive root of 3 5 C 1 is pproimtel Negtive root of f./d 5 20 C 10. Let f./d 5 20 C 10. Thegrphoff./shown below suggests tking 0 D 22. Strtingfrom 0 D 22, the first three itertes of Newton s Method re n Thus, to three deciml plces, the negtive root of f./ D 5 20 C 10 is Positive solution of sin D 08. From the grph below, we see tht the positive solution to the eqution sin D 08 is pproimtel D 11. Now, let f./d sin 08 nd define With 0 D 11 we find nc1 D n f. n/ f 0. n / D n sin n 08 n cos n 08 n

6 496 C HAPTER 4 APPLICATIONS OF THE DERIVATIVE Thus, to three deciml plces, the positive solution to the eqution sin D 08 is Solution of ln. C 4/ D. From the grph below, we see tht the positive solution to the eqution ln. C 4/ D is pproimtel D 2.Now, let f./d ln. C 4/ nd define nc1 D n f. n/ f 0. n / D n ln. n C 4/ n 1 n C4 1 With 0 D 2 we find n Thus, to three deciml plces, the positive solution to the eqution ln. C 4/ D is = ln( + 4) = Let 1 ; 2 be the estimtes to root obtined b ppling Newton s Method with 0 D 1 to the function grphed in Figure 3. Estimte the numericl vlues of 1 nd 2,nddrwthetngentlinesusedtoobtinthem FIGURE 3 The grph with tngent lines drwn on it ppers below. The tngent line to the curve t. 0 ;f. 0 // hs n -intercept t pproimtel 1 D 30. Thetngentlinetothecurvet. 1 ;f. 1 // hs n -intercept t pproimtel 2 D Find the smllest positive vlue of t which D nd D tn intersect. Hint Drw plot. Here is plot of tn nd on the sme es The first intersection with >0lies on the second brnch of D tn, between D 5 4 nd D 3 2. Let f./ D tn. The grph suggests n initil guess 0 D 5 4, from which we get the following tble

7 SECTION 4.8 Newton s Method n This is clerl leding nowhere, so we need to tr better initil guess. Note This hppens with Newton s Method it is sometimes difficult to choose n initil guess. Wetrthepointdirectlbetween 5 4 nd 3 2, 0 D n The first point where D nd D tn cross is t pproimtel D , whichispproimtel In 1535, the mthemticin Antonio Fior chllenged his rivl Niccolo Trtgli to solve this problem A tree stnds 12 brcci high; it is broken into two prts t such point tht the height of the prt left stnding is the cube root of the length of the prt cut w. Wht is the height of the prt left stnding? Show tht this is equivlent to solving 3 C D 12 nd find the height to three deciml plces. Trtgli, who hd discovered the secret of the cubic eqution, ws ble to determine the ect nswer D q q p p2919 3p C Suppose tht is the prt of the tree left stnding, so tht 3 is the prt cut w. Since the tree is 12 brcci high, this gives tht C 3 D 12. Letf./D C Werelookingforpointwheref./D 0. Usingtheinitilguess D 2 (it seems tht most of the tree is cut w), we get the following tble 4 n Hence Trtgli sectnsweris , sothe4thnewton smethodpproimtionisccurte to t lest 11 deciml plces. 23. Find (to two deciml plces) the coordintes of the point P in Figure 4 where the tngent line to D cos psses through the origin. 1 P = cos 2π FIGURE 4 Let. r ; cos. r // be the coordintes of the point P.Theslopeofthetngentlineis sin. r /,sowerelookingfor tngentline such tht D 0 when D 0. Thisgivesustheeqution D sin. r /. r / C cos. r / sin. r /. r / C cos. r / D 0 Let f./d cos C sin. Werelookingforthefirstpoint D r where f.r/d 0. Thesketchgivenindictestht 0 D 3=4 would be good initil guess. The following tble gives successive Newton s Method pproimtions 4 n The point P hs pproimte coordintes.27984; /. Newton s Method is often used to determine interest rtes in finncil clcultions. In Eercises 24 26, r denotes erl interest rte epressed s deciml (rther thn s percent). 24. If P dollrs re deposited ever month in n ccount erning interest t the erl rte r, thenthevlues of the ccount fter N ers is! S D P b12n C1 b where b D 1 C r b 1 12 You hve decided to deposit P D 100 dollrs per month.

8 498 C HAPTER 4 APPLICATIONS OF THE DERIVATIVE () Determine S fter 5 ers if r D 007 (tht is, 7%). (b) Show tht to sve $10,000 fter 5 ers, ou must ern interest t rte r determined b the eqution b b C 100 D 0. Use Newton s Method to solve for b. Thenfindr.Notethtb D 1 is root, but ou wnt the root stisfing b>1. () If r D 007, b D 1 C r= , nd S D 100.b61 b/ b 1 D (b) If our gol is to get $10,000 fter five ers, we need S D 10;000 when N D 5.! 10;000 D 100 b61 b ; b 1 Sotht 10;000.b 1/ D 100 b 61 b 100b 100 D b 61 b b b C 100 D 0 b D 1 is root, but, since b 1 ppers in the denomintor of our originl eqution, it does not stisf the originl eqution. Let f.b/d b b C 100. Let susetheinitilguessr D 02, sotht 0 D 1 C r=12 D n The solution is pproimtel b D Theinterestrter required stisfies 1 C r=12 D , sothtr D D Annnulinterestrteof18.828%isrequiredtohve$10,000fterfiveers. 25. If ou borrow L dollrs for N ers t erl interest rte r, ourmonthlpmentofp dollrs is clculted using the eqution! 1 b 12N L D P where b D 1 C r b 1 12 () Find P if L D $5000, N D 3, ndr D 008 (8%). (b) You re offered lon of L D $5000 to be pid bck over 3 ers with monthl pments of P D $200.UseNewton smethod to compute b nd find the implied interest rte r of this lon. Hint Show tht.l=p /b 12N C1.1 C L=P /b 12N C 1 D 0. () b D.1 C 008=12/ D (b) Strting from P D L b b 12N D $15669 L D P! 1 b 12N ; b 1 divide b P,multiplbb 1, multiplbb 12N nd collect like terms to rrive t Since L=P D 5000=200 D 25, wemustsolve Newton s Method gives b nd So the interest rte is round 2545%..L=P /b 12N C1.1 C L=P /b 12N C 1 D 0 25b 37 26b 36 C 1 D 0 r D 12.b 1/ D /

9 SECTION 4.8 Newton s Method If ou deposit P dollrs in retirement fund ever er for N ers with the intention of then withdrwing Q dollrs per er for M ers, ou must ern interest t rte r stisfing P.b N 1/ D Q.1 b M /,whereb D 1 C r. Assumetht$2,000is deposited ech er for 30 ers nd the gol is to withdrw $10,000 per er for 25 ers. Use Newton s Method to compute b nd then find r.notethtb D 1 is root, but ou wnt the root stisfing b>1. Substituting P D 2000, Q D 10;000, N D 30 nd M D 25 into the eqution P.b N 1/ D Q.1 b M / nd then rerrnging terms, we find tht b must stisf the eqution b 55 6b 25 C 5 D 0. Newton smethodwithstrtingvlueof b 0 D 11 ields b Thus,r D 5217%. 27. There is no simple formul for the position t time t of plnet P in its orbit (n ellipse) round the sun. Introduce the uilir circle nd ngle in Figure 5 (note tht P determines becuse it is the centrl ngle of point B on the circle). Let D OA nd e D OS=OA (the eccentricit of the orbit). () Show tht sector BSA hs re. 2 =2/. e sin /. (b) B Kepler s Second Lw, the re of sector BSA is proportionl to the time t elpsed since the plnet pssed point A, nd becuse the circle hs re 2, BSA hs re. 2 /.t=t /,wheret is the period of the orbit. Deduce Kepler s Eqution 2t T D e sin (c) Theeccentricit of Mercur s orbit ispproimtel e D 02.UseNewton s Method to find fter qurter of Mercur s er hs elpsed (t D T=4). Convert to degrees. Hs Mercur covered more thn qurter of its orbit t t D T=4? Auilir circle B P O S Sun A Ellipticl orbit FIGURE 5 () The sector SAB is the slice OAB with the tringle OPS removed. OAB is centrl sector with rc nd rdius OA D, nd therefore hs re 2 2. OPS is tringle with height sin nd bse length OS D e. Hence,thereofthesectoris e2 sin D 2 2. e sin / (b) Since Kepler s second lw indictes tht the re of the sector is proportionl to the time t since the plnet pssed point A, we get 2.t=T / D 2 =2. e sin / 2 t T D e sin (c) If t D T=4,thelstequtionin(b)gives 2 D e sin D 2 sin Let f./ D 2 sin 2. We will use Newton s Method to find the point where f./ D 0. Sincequrteroftheeron Mercur hs pssed, good first estimte 0 would be 2. 4 n From the point of view of the Sun, Mercur hs trversed n ngle ofpproimtel rdins D ı.mercurhs therefore trveled more thn one fourth of the w round (from the point of view of centrl ngle) during this time. 28. The roots of f./ D C 1 to three deciml plces re 3583, 0.251,nd3.332(Figure6).Determinetherootto which Newton s Method converges for the initil choices 0 D 185, 1.7,nd1.55.Thenswershowsthtsmllchngein 0 cn hve significnt effect on the outcome of Newton s Method.

10 500 C HAPTER 4 APPLICATIONS OF THE DERIVATIVE FIGURE 6 Grph of f./d C 1. Let f./d C 1, nddefine Tking 0 D 185, wehve nc1 D n f. 1 n/ f 0. n / D 3 n 3 n 4 n C 1 n n Tking 0 D 17, wehve n Tking 0 D 155, wehve n n n Wht hppens when ou ppl Newton s Method to find zero of f./d 1=3?Notetht D 0 is the onl zero. Let f./d 1=3.Define nc1 D n f. n/ f 0. n / D n 1=3 n D n 3 n D 2 n 1 3 2=3 n Tke 0 D 05.Thenthesequenceofitertesis 1; 2; 4; 8; 16; 32; 64; Tht is, for n nonzero strting vlue, the sequence of itertes diverges spectculrl, since n D. 2/ n 0.Thuslim n!1 j n j D lim n!1 2 n j 0 j D Wht hppens when ou ppl Newton s Method to the eqution 3 20 D 0 with the unluck initil guess 0 D 2? Let f./d Define nc1 D n f. n/ f 0. n / D n 3 n 20 n 3 2 n 20 Tke 0 D 2.Thenthesequenceofitertesis 2; 2; 2; 2;,whichdivergesboscilltion. Further Insights nd Chllenges 31. Newton s Method cn be used to compute reciprocls without performing division. Let c>0nd set f./d 1 c. () Show tht.f./=f 0.// D 2 c 2. (b) Clculte the first three itertes of Newton s Method with c D 103 nd the two initil guesses 0 D 01 nd 0 D 05. (c) Eplin grphicll wh 0 D 05 does not ield sequence converging to 1=103. () Let f./d 1 c. Then f./ f 0./ D 1 c 2 D 2 c2

11 SECTION 4.8 Newton s Method 501 (b) For c D 103, wehvef./d nd thus nc1 D 2 n n. Tke 0 D 01. Tke 0 D 05. n n (c) The grph is disconnected. If 0 D 5,. 1 ;f. 1 // is on the other portion of the grph, which will never converge to n point under Newton s Method. In Eercises 32 nd 33, consider metl rod of length L fstened t both ends. If ou cut the rod nd weld on n dditionl segment of length m,levingtheendsfied,therodwillbowupintocirculrrcofrdiusr (unknown), s indicted in Figure 7. L h R 32. Let h be the mimum verticl displcement of the rod. () Show tht L D 2R sin nd conclude tht FIGURE 7 The bold circulr rc hs length L C m. (b) Show tht L C m D 2R nd then prove h D sin L.1 cos / 2 sin D L L C m 2 () From the figure, we hve sin D L=2 R, so tht L D 2R sin.hence h D R R cos D R.1 cos / D 1 2 L L.1 cos /.1 cos / D sin 2 sin (b) The rc length L C m is lso given b rdius ngle D R 2.Thus,L C m D 2R.DividingL D 2R sin b L C m D 2R ields L 2R sin D D sin L C m 2R 33. Let L D 3 nd m D 1. ApplNewton smethodtoeq.(2)toestimte, ndusethistoestimteh. We let L D 3 nd m D 1.Wewntthesolutionof Let f./d sin 3 4. sin sin D L L C m D 0 sin 3 4 D 0 L L C m

12 502 C HAPTER 4 APPLICATIONS OF THE DERIVATIVE The figure bove suggests tht 0 D 15 would be good initil guess. The Newton s Method pproimtions for the solution follow The ngle where sin D 4 n LCm L ispproimtel Hence 1 cos h D L 2 sin Qudrtic Convergence to Squre Roots Let f./d 2 c nd let e n D n p c be theerrorin n. () Show tht nc1 D 1 2. n C c= n / nd e nc1 D e 2 n =2 n. (b) Show tht if 0 > p c, then n > p c for ll n.eplingrphicll. (c) Show tht if 0 > p c, then e nc1 e 2 n =.2p c/. () Let f./d 2 c. Then s longs n 0. Now n C 1 D n f. n/ f 0. n / D n 2 n c D 2 n C c D 1 n C c ; 2 n 2 n 2 n en 2 n p c 2 D 2 n 2 n D 1 2 n C c n D 2 n 2 p n c C c D 1 2 n 2 n p c C c 2 n p c D nc1 p c D e nc1 (b) Since 0 > p c 0, we hve e 0 D 0 p c>0. Now ssume tht e k >0for k D n. Then0<e k D e n D n p c, whence n > p c 0; i.e., n >0nd e n >0.Bprt(),wehvefork D n C 1 tht e k D e nc1 D e2 n 2 n > 0 since n >0. Thus e nc1 >0.Thereforebinductione n >0for ll n 0. Hencee n D n p c>0for ll n 0. Therefore n > p c for ll n 0. The figure below shows the grph of f./ D 2 c. The-intercept of the grph is, of course, D p c. We see tht for n n > p c, the tngent line to the grph of f intersects the -is t vlue nc1 > p c. (c) B prt(b),if 0 > p c, then n > p c for ll n 0. Accordingl,forlln 0 we hve e nc1 D e2 n < e2 n 2 n 2 p. In other c words, e nc1 < e2 n 2 p for ll n 0. c In Eercises 35 37, fleible chin of length L is suspended between two poles of equl height seprted b distnce 2M (Figure 8). B Newton s lws, the chin describes ctenr D cosh, where is the number such tht L D 2 sinh M. The sg s is the verticl distnce from the highest to the lowest point on the chin. = cosh(/) s 2 M FIGURE 8 Chin hnging between two poles.

13 SECTION 4.8 Newton s Method Suppose tht L D 120 nd M D 50. () Use Newton s Method to find vlue of (to two deciml plces) stisfing L D 2 sinh.m=/. (b) Compute the sg s. () Let f./d 2 sinh The grph of f shown below suggests 47 is root of f.strtingwith 0 D 47, wefindthefollowingpproimtionsusing Newton s method Thus, to two deciml plces, D D nd 2 D (b) The sg is given b s D.M/.0/ D cosh M C C cosh 0 C C D cosh M Using M D 50 nd D 4695, wefinds D Assume tht M is fied. () Clculte ds d. Note tht s D cosh M. (b) Clculte d dl bimplicitdifferentitionusingthereltionl D 2 sinh M. (c) Use () nd (b) nd the Chin Rule to show tht ds dl D ds d d cosh.m=/.m=/ sinh.m=/ 1 D dl 2 sinh.m=/.2m=/ cosh.m=/ 3 () The sg in the curve is M s D.M/.0/ D cosh ds M d D cosh M M sinh 1 (b) If we differentite the reltion L D 2 sinh C C. cosh 0 C C/ D cosh M with respect to, wefind 0 D 2 d M dl sinh 2M d dl cosh M M Solving for d=dl ields (c) B the Chin Rule, d M dl D 2 sinh 2M cosh M 1 ds dl D ds d d dl The formul for ds=dl follows upon substituting the results from prts () nd (b).

14 504 C HAPTER 4 APPLICATIONS OF THE DERIVATIVE 37. Suppose tht L D 160 nd M D 50. () Use Newton s Method to find vlue of (to two deciml plces) stisfing L D 2 sinh.m=/. (b) Use Eq. (3) nd the Liner Approimtion to estimte the increse in sg s for chnges in length L D 1 nd L D 5. (c) Compute s.161/ s.160/ nd s.165/ s.160/ directl nd compre with our estimtes in (b). () Let f./d 2 sinh.50=/ 160. Usingthegrphbelow,weselectninitilguessof 0 D 30.Newton smethodthenields Thus, to two deciml plces, n = 2 sinh(50/) (b) With M D 50 nd 2846, wefindusingeq.(3)tht B the Liner Approimtion, ds dl D 061 s ds dl L If L increses from 160 to 161, then L D 1 nd s 061; ifl increses from 160 to 165, then L D 5 nd s 305. (c) When L D 160, 2846 nd 50 s.160/ D 2846 cosh I 2846 wheres, when L D 161, 2825 nd s.161/ D 2825 cosh Therefore, s.161/ s.160/ D 062, verclosetothepproimtionobtinedfromthelinerapproimtion.moreover,when L D 165, 2749 nd 50 s.165/ D 2749 cosh I 2749 thus, s.165/ s.160/ D 302, ginverclosetothepproimtionobtinedfromthelinerapproimtion. 4.9 Antiderivtives Preliminr Questions 1. Find n ntiderivtive of the function f./d 0. Since the derivtive of n constnt is zero, n constnt function is n ntiderivtive for the function f./d Is there difference between finding the generl ntiderivtive of function f./nd evluting R f./d? No difference. The indefinite integrl is the smbol for denoting the generl ntiderivtive. 3. Jcques ws told tht f./ nd g./ hve the sme derivtive, nd he wonders whether f./ D g./. DoesJcqueshve sufficient informtion to nswer his question? No. Knowing tht the two functions hve the sme derivtive is onl good enough to tell Jcques tht the functions m differ b t most n dditive constnt. To determine whether the functions re equl for ll,jcquesneedstoknowthevlue of ech function for single vlue of. Ifthetwofunctionsproducethesmeoutputvluefor singleinputvlue,themusttke the sme vlue for ll input vlues.