What is not acceptable is an application of L Hospital s Rule,

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1 Final Exam Solutions Math 8, Dcmbr 3, 004 Problm. Calculat th following its. If th it is infinit, indicat whthr it is + or. (a x 4 + 4x x 4 x 6 (b x 4 x (c (d t + 6t t x + 6x x t + 4 t 3 x ( t 0 Solution. t (a As x 4 +, x 4 0 and x 4 rmains positiv; on th othr hand, 4x 6. Thus 4x/(x 4 +, as dos th squar root of (4x/(x 4: Answr (a: + (b As x 4, th dnominator x 0; th numrator also 0. Thus w ar first going to hav to do som algbraic simplifications. A dirct wa is to not that and thus x 6 (x 4(x + 4 ( x ( x + (x + 4, x 6 x x 6 x ( x + (x + 4. As x 4 w hav x (thrfor x + 4 and x +4 8; thus b th it of a product is th product of th its rul, x 4 x 4 x What is not accptabl is an application of L Hospital s ul, x 4 x 4 x x 4 x x 8 /4 3. (If ou don t undrstand this computation, don t worr about it: L Hospital s ul isn t taught in Math 8, which is wh no crdit was givn for this solution. Answr (b: 3 (c Onc again, th numrator approachs 0 and th dnominator approachs 0, so w simplif th fraction b factoring th numrator: t + 6t + 9 (t t (t 3(t + 3 t t. This qualit is valid for all t 3, which is sufficint for computing th its: t + 6t + 9 t + 3 t 3 9 t t 3 3 t Answr (c: 0 (d Looks similar to part (c, but th it is as x. Th trick now is to divid numrator and dnominator b x, so x + 6x x + 6/x + 9/x 9/x. Sinc /x 0 and /x 0 as x, th numrator of this nw xprssion convrgs to and th dnominator to. Thus b th quotint of th its is th it of th quotints rul, Answr (d:

2 ( This on is a littl trick, and gos somwhat against th grain of th wa ou wr taught to simplif things in high-school: w rationaliz th numrator, not th dnominator! ( ( t + 4 t + 4 t t t ( t (t t ( t t t ( t t This trick gts rid of th singularit which division b zro would caus, b canclling a t from numrator and dnominator. At an rat, as t 0 w hav t and thrfor t + 4 4, hnc t 4 Answr (: 4 t + 4 Problm. Diffrntiat f (x, whr: (a f (x x + 3x + ; (b f (x ln(x + ; (c f (x x 4 x ; (d f (x ln(x 3 ; ( f (x x ; x Solution. (a W hav t t 4 t f (x (3x + ( 6x(x + (3x + 6x 6x + (3x +. Answr (a; 6x 6x + (3x + (b mmbr that th drivativ of ln u with rspct to u is /u, and th chain rul: Answr (b: f (x x + (x x x + x x +. (c Thr ruls ar applid: for th it of a product, th fact that du du u, and th chain rul: f (x 4x 3 x + x 4 x x (4x 3 + x 5 x x 3 (x + x. Answr (c: x 3 (x + x Your instructions wr to simplif all xprssions. Points wr takn off for not factoring. (d Thr ar two was to do this problm; it was put on in th hop that good studnts would s that and thrfor ln x 3 3 ln x, f (x d dx (3 ln x 3 x. Altrnativl, it s not much hardr to us th chain rul: f (x x 3 (3x 3 x. Points wr takn off if ou didn t simplif th last xprssion to 3/x. Answr (d: 3 x ( This is th rul for diffrntiating a quotint: f (x xx ( x x (x x + x. Originall, th thought was to ask studnts to show that ( x /x is an incrasing function. (This is a gnral proprt of functions: if g

3 is concav up, thn (g(x g(a/(x a is incrasing. To do that, ou nd to show th drivativ (x x + x is positiv for x > 0. Sinc th dnominator is positiv, that mans w hav to show th numrator, call it h(x, h(x (x x +, is positiv for x > 0. W ll show h(x is incrasing and h(0 0; that would man that h(x > h(0 for x > 0, i.. h(x > 0 for x > 0. But h (x x + (x x x x, which is crtainl positiv for x > 0, and this provs th rsult. Sinc no similar problms wr assignd in homwork (thr arn t an similar problms in th book, w lft th problm at th bar minimum: Answr (: (x x + x W solv this last for d/dx, Substituting x, 0, w obtain d dx 3x 9. m Thrfor th answr is (x x +. Answr: x + Figur shows th solution curv for th quation (not that it s not th graph of a function; b wandring about lik this, thr ar svral placs whr th tangnt lin is vrtical and th graph cuts back undr or ovr itslf. But nar x, 0 thr is indd a sction of th graph which is th graph of a function. W hav plottd th tangnt lin as a dashd lin. Problm 3. Find th quation of th tangnt lin to th curv at x, 0. x x + Solution. You rall should chck that x, 0 is a point on th curv: Yup. (Nvr trust a profssor. OK, th quation of th tangnt lin is m(x, whr th slop m is th drivativ d/dx valuatd at x 0. To valuat this w implicitl diffrntiat th quation with rspct to x: d ( x x + d dx dx i.. 3x 9 d d + dx dx Figur. Implicit plot of x x + on x,, togthr with th tangnt lin at x, 0.

4 Problm 4. Th graph of th drivativ f (x of a function f (x is givn blow. f x x Figur. Graph of th drivativ of f, f (x. Indicat b writing th answr in th providd boxs: (a Th x-coordinat(s of all th rlativ maxima of th function f (x. (b Th x-coordinat(s of all rlativ minima of th function f (x. (c Th x-coordinat(s of all inflction points of th function f (x. (d Spcif th intrval(s on which th function f (x is concav down. ( Givn that f (0 0 and that th shadd rgion has ara 08/5, find th valu of f (. Solution. W mphasiz that this is th graph of th drivativ of f, not th graph of f!! Th critical numbrs of f ar whr f (x 0. From th graph w rad off that ths ar at x, x 0, and x 3. W s that f is ngativ on th intrval (, 0, and is positiv (or zro vrwhr ls. Thrfor f is incrasing on (, ] (bcaus f is positiv thr!, dcrasing on [, 0], and incrasing on [0, +. Thus th critical numbr x is a rlativ maximum point (th function is incrasing just to th lft, dcrasing just to th right; th critical numbr x 0 is a rlativ minimum point (th function is dcrasing just to th lft, incrasing just to th right; and th critical numbr x 3 is nithr a rlativ maximum nor a rlativ minimum point. As for inflction points and intrvals of concavit: f is concav up in an intrval whr f > 0, i.. whr f is incrasing, and is concav down whr f < 0, i.. whr f is dcrasing. B balling th graph of f, w s that f is dcrasing from to ; incrasing from to ; dcrasing from to 3; thnc incrasing again from 3 to +. Thrfor f is concav down on (, ] and on [, 3]. Points of inflction ar at x ( f changs from bing concav down to bing concav up and at x (changs from concav up to concav down and at x 3 (changs from concav down to concav up. Cavat: w hav usd and + as though th graph continud to th lft and right. But th instructions ar that this is th graph of f. Th domain of th function appars to run from about.5 to about 4.5. Thus whr w hav usd abov,.5 is mor corrct; and whr w hav usd + th valu of 4.5 is mor corrct. Points wrn t takn off for this subtlt. As for th valu of f ( : w ar told that th shadd rgion has ara 08/5; sinc th intgral intrprts rgions blow th axis as having ngativ ara, this mans w ar told that 0 f (x dx B th Scond Fundamntal Thorm of Calculus 0 and w ar told that f (0 0; thus Thrfor f ( f (x dx f (0 f (, f (. Answr (a: lativ maximum at x Answr (b: lativ minimum at x 0 Inflction points at x, x, x 3 Concav down on [.5, ] and on [, 3].

5 f ( 08 5 An asid: how was th graph in Figur cratd? It is actuall th graphs of two functions joind at x 0. Th lft half is th graph of (9/x(x + This crats an upright parabola crossing th axis at x and x 0 (I ll xplain th factor 9/ in a momnt. Th right half is th graph of x(x 3 (this has roots at x 0 and x 3, and th squar (x 3 guarants that th graph has a root at x 3 but dosn t go ngativ nar x 3. Finall, to look right, ths two graphs should hav th sam drivativ at x 0. Sinc d x(x + x + whn x 0, dx whil d dx x(x + 3 3x x whn x 0, w s that a multipling factor of 4.5 is appropriat for th first function. (Not multipling b 4.5 wouldn t hav changd an of th answrs of th problm, and th mismatch of tangnts at x 0 wouldn t vn b noticabl to th, but I m a prfctionist. Problm 5. A local TV station has mad a surv of th watching habits of childrn on Sundas btwn th hours of 8:00 AM and :00 AM. Th surv indicats that th prcntag of childrn that ar watching th station x hours aftr 8:00 AM is givn b f (x (x3 3x x + 7. (a At what tim btwn 8:00 AM and :00 AM ar th most childrn watching th station? What prcntag of childrn is watching at that tim? (b At what tim btwn 8:00 AM and :00 AM ar th fwst childrn watching th station? What prcntag of childrn is watching at that tim? Solution. This is a max/min problm on a closd intrval [0, 3] (bcaus 3 8. Sinc a continuous function alwas has an absolut maximum and absolut minimum on a closd boundd intrval, both parts hav an answr; furthrmor, th max/min must b takn ithr at an ndpoint or an intrior point, in which cas it will b a critical point. So our first job is to find th critical points: f (x 3x 3x 6 3(x + (x, thrfor th critical points ar x and x. Th onl on of ths in th intrval [0, 3] is x. Thus w considr a minitabl x f (x confidnt that th absolut maximum and absolut minimum must appar in this list. Obviousl th maximum is 36, takn whn x 0, and th minimum is 6, takn whn x. Translating to clock tim, th answrs ar: Maximum numbr at: 8:00 AM Minimum numbr at: 0:00 AM You wrn t askd to plot th graph this illustrats th powr of th analtic mthod, ou don t nd th graph but I hav includd it in Figur x Figur 3. Prcntag of childrn watching x hours aftr 8:00 AM, f (x (x3 3x x + 7. Problm 6. Jack and Jill hav st a wdding dat for thr ars from now. Jill s parnts want to st asid nough mon now so th will hav 0,000 availabl for th wdding. If th invst it in an account arning 6% a ar, how much do th nd to st asid in ordr to hav 0,000 in th account at th nd of thr ars if:

6 (a Th intrst is compoundd monthl; (b Th intrst is compoundd continuousl. Solution. This is a prsnt valu problm. Th answrs ar: Answr (a: $0, Answr (b: $0, (Th monthl intrst rat is 0.06/ xprssd as a dcimal; th 0.8 is 3 ( Th numrical answrs ar $8, and $8, 35.70, but undr th conditions of th xam (no calculators ou would hav bn unabl to comput ths. If ou did, our ntir final would hav bn xamind vr closl for othr signs of chating. Problm 7. Evaluat th intgrals, or indicat if th do not xist: (a (b (c Solution. /x dx. x ln 0 x x dx. x x dx. (a This is a substitution problm. Put u /x; thn du/dx /x, and thus x du x dx. Thus th intgral bcoms /x dx u du Answr (a: /x + C u + C /x + C. (b This is a dfinit intgral. Th bst ida is to comput th indfinit intgral, that is, find an antidrivativ for th intgrand; thn substitut th valus x ln and x 0 and subtract. To comput th indfinit intgral, do an intgration b parts with dv u du; a valu of v which dos this is v u (which officiall ou should gt b a substitution; or rmmbr that d kx /dx k kx. Thus u u du ud ( u u ( u u du u u + u du. Finall, rcall that u du u (rmmbr how w computd v from dv, to gt u u du u u ( 4 u u + u. 4 Howvr, th problm asks for th dfinit intgral from 0 to ln. (This is wh w didn t bothr to add th +C to th indfinit intgral. Th final answr is thrfor which works out to ( u + u ln, Answr (b: 6 8 ln An altrnativ is to us th form of th indfinit intgral from th tabls: u au du a (au au + C. This dosn t xcit much admiration among our loving profssors, but if ou wrot out this gnral form, it would hav bn accptd.

7 (c Substitut u x, so that x + u and du dx, i.. dx du. Thus x x dx + u u du 4u + 4 du 4 ln u + 4 u 4 ln x + (x 4 4 ln x + x, whr th /4 has bn absorbd into th constant: Answr (c: x + 4 ln x + C Problm 8. Local nvironmntalists studing Lak Plntiful stimat that thr ar prsntl 500 trout in th lak. Howvr, bcaus of an insufficint oxgn suppl, th xpct th trout to di at th rat of 5 t/0 trout pr da, whr t dnots th numbr of das that hav passd. Find th population of trout in Lak Plntiful 5 das from now. Solution. If P(t dnots th population of th trout at tim t (in das that hav passd, thn w ar told that Thrfor P (t 5 t/0, P( P(t P(0 t 0 5 x/0 dx 500 x/0 xt x0 500 t/ W hav usd th fact that kx dx k kx, and that whn t 0 is substitutd into 500 t/0 w gt 500. Adding P(0 to both sids, thrfor P(t t/0. W chck that which is as it should b; and that P( , P (t t/0 5 t/0, which is also as it should b. (Big sigh of rlif. Now, what was th qustion? Oh, ah, what s th population aftr 5 das. That s P( / /4, and that s as far as w can tak it. (If ou wrot 389, which is th numrical valu, w d b inclind to think ou usd a calculator... Answr: /4 Problm 9. Lt f (x, x 3 + 3x Find all th critical points of f (x,, and classif thm as to whthr th ild rlativ maxima, rlativ minima, or saddl points. Solution. W comput f x 3x 3 f 3x 4 f xx 6x f x 3 f x 3 f. (Of cours, I know that f x f x, but it s alwas a good ida to comput both anwa, as a partial chck on our work. Th critical points ar thrfor solutions of 3x 3 0 3x 4 0. W can tak th scond of ths, 3x + 4, and substitut it into th first, so as to inat : 3x 3(3x This simplifis to x 3x 4 0,

8 which has roots x 4, x. W comput th corrsponding valus of from th quation w usd to inat, 3x + 4, to find that x 4, 6 is on critical point and x, is th othr. To classif ths w must comput th discriminant, D f xx f f x 6x ( 3 6x 9, at th two critical points. At x 4, 6 w gt D 5, which is positiv; chcking f xx or f at ths valus, th r both positiv; thus (4, 6 is a local minimum point of f. At th scond critical point (, w comput D 5, which is ngativ. Th scond drivativ tst immdiatl tlls us that (, is a saddl point. Thr can b no local maximum point, bcaus it would hav to b a critical point, and w v alrad usd up all th critical points. Thus th answrs to th problm ar: Critical points: x 4, 4 and x, lativ maxima: Non lativ minima: x 4, 4 Saddl points: x, Problm 0. Evaluat th doubl intgrals blow ovr th spcifid rctangular rgion. Choos th ordr of intgration carfull, whr appropriat. (a x + d A, : 0 x, 0. (b Solution. ln(x d A, : x,. ( (a W comput th doubl intgral as an itratd intgral: x + d A x x0 ( x + d 0 and bgin working on th innrmost intgral first: 0 ( x + d x + 0 ( ( x + x x +. Substituting into quation (??, w gt And that s th answr: x x + d A x0 x + dx ( x 6 x3 + x Answr (a: 7 6 (b W will giv two solutions. Th first utilizs th fact that ln(x ln x + ln, a fact which a good studnt might utiliz. This lads to ln(x d A x0 ln x d A + ln d A. Now, ach of th intgrands in ( is what is calld sparabl : it is th product of a function f (x and a function g(. It s as to find th doubl intgral of a sparabl function f (xg( on a rctangl

9 ( (3 : a x b, c d, sinc b ( d f (xg( d A a b a c ( d f (x f (xg( d dx ( d ( b g( d c c g( d dx a f (x dx. (Th f (x factors out of th innr intgral bcaus it s a constant as far as th variabl of intgration is concrnd; in th nxt lin, th d c g( d factors out of th intgral with rspct to x, bcaus it dosn t dpnd on x. In othr words, th doubl intgral of f (xg( is simpl th product of th two singl intgrals f (x dx and g( d, valuatd on th corrsponding intrvals. With rfrnc to (, w s that th first intgral is sparabl, ( ln x ( d A ln x dx d ( ln (ln (( ln ( ln (ln ln sinc ln and ln 0. W hav don th dfault intgration b parts, ln d ln d(ln ln d ln d ln. Th scond intgral in ( is also sparabl, ( ln ( d A ln d dx ( (ln ( whr w hav usd th substitution v ln to comput ln d v dv v (ln. Substituting ( and (3 into (, w find ln(x d A + +. Answr (b: + Th mor dirct wa of doing this problm dosn t utiliz sparabilit, but rquirs computing a slightl mor complicatd intgral. (On th othr hand, it s onl on intgral, not two. W bgin b writing th doubl intgral as an itratd intgral: ln(x d A x x ( ln(x d dx. W work on th innr intgral first. In indfinit form it is ln(x d, which suggsts a chang of variabl u ln(x (whr x is fixd and is variabl. W hav du d (compar with Problm (d. W obtain ln(x d u du u (ln(x. Intgrating btwn and thrfor rsults in ln(x d (ln(x (ln x (( + ln x (ln x ( + ln x + ln x.

10 (It s almost miraculous that th (ln x trm disappars at last vr fortunat, sinc it would b nast to intgrat. Substituting into th original intgral, w obtain ln(x d A + ln x dx ( + (x ln x x ( + +. What would hav happnd if w had intgratd first with rspct to x, thn with rspct to? Th rsult is slightl mor complicatd. It turns out that ln(x dx x ln(x x. Whichvr wa w do th ordr of intgration, ( ( ln(x ln(x d dx dx d x ln(x x. You v sn both solutions; which do ou think is asir?

6.1 Integration by Parts and Present Value. Copyright Cengage Learning. All rights reserved.

6.1 Integration by Parts and Present Value. Copyright Cengage Learning. All rights reserved. 6.1 Intgration by Parts and Prsnt Valu Copyright Cngag Larning. All rights rsrvd. Warm-Up: Find f () 1. F() = ln(+1). F() = 3 3. F() =. F() = ln ( 1) 5. F() = 6. F() = - Objctivs, Day #1 Studnts will b