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exam1sol - SOLUTIONS Math 41 Autumn 2009 First Exam —...

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Unformatted text preview: [ SOLUTIONS ' Math 41, Autumn 2009 First Exam — October 13, 2009 Page 1 of 11 1. (13 points) Find each of the following limits, with justification. If the limit does not exist, explain why. If there is an infinite limit, then explain whether it is 00 or —00. <1 a im ——————x2_4 :2 ‘71-.” O i [Lle‘] ( ) 3L2 x2+a3—2 aiz+a_;\ 2 7; 2 O {sulnsifb’lhm M/e) - . 5 + 10 , _ -— [Lilia (b) 33 2132+ m <6— sulxfiflfim yielné % I wind. is q Q. mfiémgdg: lam 5”” slim SEE-2.. x—a—gf x3+Lix2+le X94 x[x+z)7‘ = lim 5 Xa—rfi x{x+2.) ' W *Numem‘l'or is mnzere Mile Anomimr/Lor approaches ZEN/*0) 50 1%}; //yy,»,+ will Milo/Vt Irfhmbz Siflflrl‘lca‘l‘ian '5 ‘ll‘Sljmépl Wart]: 42p x‘a-{f} we 1mm X<O 129+ x+Z >0) 5:» denomMa’vl'or i3 Mégzfim win/p. HUMfronLor i; fesi’IL/wej' 5‘th ’Haus) fl/Jhen‘t is ne affine (a: a) . Math 41 Autumn 2009 First Exam — October 13, 2009 Page 2 of 11 [5P1‘5]( ()c $13; (f o g)( 92,) Where f (ac) = sin(7r:v) and 9 satisfies lim g(a: )2 1 and lim g(:r)— — 5 :c->3+ z-—>3— We m L"; ($03M) == fig 10/360): i9 ”54 3w) = 10(1): 5/1177"st X-r 3* SMCQ I'M" 360 9(1ny (W0! “mafia/Va fig (”H—5,46! WIS/a4 ”(L/£6 /mn‘ m/e Ag? 29ng (pm/gosfi/OHSJ am! also #4 *Qn‘ +547! ‘P is mrflLI'nULx/S) ‘ Sinai/ar/y) 11,3 (3%)", ”’4 £660) 10 ['M 36%: 4Y5)" - 57/2571: 0. x93 x»? 171%) Jig/3; (1003) (X) 1‘ O Amuse ‘Ha +wo anew/ea? bur/5 exisllqu gm 670M. [IL/01% 77:: fi/O’SiC/QJ vars/om 010 +42 aim/009%” ru/Q flamélifgfixn= 101mg“) Jcesn'é af’P/Y Idem; ECCQUS-e—‘I'Aa I’m? 1”” (X) 49mm?" exiwf Bd‘f as seem. [Jove %+ )(‘93 deem/é" wean hm £660) ages/é flis‘f ,] >69 3 Math 41, Autumn 2009 First Exam —— October 13, 2009 Page 3 of 11 2. (20 points) Consider the function —s1n:c 1fac27r a: sinus ifzzz<7r f(a:)= 7r, . For each of the statements below, circle Whether it is TRUE or FALSE. To receive credit, please also explain your reasoning completely. [Qf£5(a flat) is a one—to—one function. TRUE 3,453 43(4) 2 cfin/‘71) =0 amt)? 4Y0) fiSInOf-‘OJ We 03:4 42,40! ’I‘WO olidima‘ 7<~Vn/(Ies Mam/)7 ,7: "rr 477:1 xzo) flan‘ )n‘eH 7% Same “gm-Pm Value I 50 ‘P is £61: oneiomng [2M5] (b) The graph of y = f (:13) has a vertical asymptote. TRUE @ A ””0041 MWF+°+€ X“? "f 1‘; W’SfS COO/0’ 01/ 1% pmsern‘ #122190 firaone—xugol vars/'07,.) “Va/V83 l ”“7- We /, ru/e omi' ”MP/9%? ”’7 ”SSE/”y Mai 7%: limo/Deng" sz'y/ II? ‘4‘ 1" 540» ‘flmfib 19%) 7/775 exists: “it M 4"“ Law ’3" 3w Sin/a) 57mg 7: MW 41”: Viewer ’"VGIV‘ES Wank/>1 -_ x]! ‘n . in , thsw x7321?” M </ of? n>77 +119” 6m +“{y)=xi.1:\ 1115’“ :Tfik) SW9 7%): ,3 “I'm/"5 Mhaii ""‘1 "‘ POWlicular MMWUS on 141a Men/«l {17} Ha) , and It); Value is ”(ways icnn‘e‘Qrt/me a. "WIhvo/ws f:l:n‘ny c 1? 01:1") ‘Men parv'fc) lie/ow shows xgfitfiFO, 40W not Mflnfiée‘ [HPEXW )is continuous at a:- — 7r. FALSE We @331 7% 30 Each ”fin 79% Jag/004 010 Man‘ ,‘7‘ Wms- 7L0 5Q Wm are ‘HWQ #11ij 1'3: C/Prkf a We have {(77) 2* isim'f: C so In ”Wm/gr 707%) ,5 49/121800 / Cm'f/‘Woug off a, [obi/7f, I (1/9 llama [W [60- AM Smx 3 3/7177 :0 mm! xarr" x977 Hm I, f x—mv“ t6!) Xsnij «Si/0h farm: 0 5051;71:96710J “PM, in fiffi'CU/qr- ”/AI‘J‘ [M1776 80.5“ ___,2- / ’Eflaliy) 10 (71’) ’ hm swig") beau/sq 50744 We Z,9m_ / 17MB; ‘p )5 Confinuops‘ Kl X=7T, Math 41, Autumn 2009 First Exam —— October 13, 2009 Page 4 of 11 For quick reference, here again is the definition of f: f($) = 71' . . —sma; 1fm27r a: [Ad-5:“ :((d) f’( ((x) i. e., the derivative of f) has a horizontal asymptote. FALSE fimf‘ note 4‘7"): 37?:- SW +¥cesx Memever >071". 14/87/56” #WL ximoopé‘) :0, ”MA % wi/i m/[zw US$23 emit/ole ‘flm‘i‘ 4V [my #11 AOI‘IEM’ILQI aaymflafg £9 {sinx ifac<7r 4:25 see‘htis/ LE3?” WI”! +171 $645 ‘Hrq‘l’ ’Is’smxsl, m! /Smsxsl £ra/l x>77‘. [170%.],er ’pm‘ sysfeml: / l: ) and flze seem)” £7 [ 601% owl‘w/ucln me ,Oos‘ff‘five 7wm‘ifig 14,} >07) WE OL‘hpun i."- 1T 1T STZS’XQSMX 4,- GA ’XES'TIm3X\( g i?» a“ X>T. AGUHU %S{ SIS-lends.) M ’6 a! - < 333m)”. geosx 5 17+ 1 (4341.4)441'3254‘01- a” X>7T) , _ [30" 71:91:) (‘14-?) C‘- 0+OZO) amo( Xi”; (1+1):O+O=O flu: Af'fla S7UeEZQ’TLeaNn/1) W9— Ctm COMO/vole #347‘ ”M 1T " 4 x—aoo (KNWr x “’3" fl 0, ”AC4 ’5 “’49; we mm 412me [Ergo )f”( )> 0 Whenever 7r < ac < 2w TRUE FALSE 1T fird: “0‘69, 4AA“ will” x>fi) since ‘P/X): ~"“SMX+3("COS)( 1W [mtg ‘wa‘f' -6 , 1T, .. 3217' 7T 77 "’ ’F/rx) :— 21rx,"sim( + gm” + garcosf‘ firm/1y ,. 7F 5,”.— i‘zcosy ”ysinx ‘gr X>(/., d!- 174B seems I'm"; ’{0 anal/23 yyfi’m‘lim/lk Alf We (DOM Wain/7 558 WAN! Afl’f‘aflj' a‘ll fir (War) ’flQ Mme mp ‘i/ze infirva/ m 7063mm. ‘ // EMS‘WJ ‘0 [3"): (9)33Mb") (5:) “171C043” ‘ $940») = (an): {—3 = ZIKO. C ( cs ”M I, , fitb} 5/1482 ‘P60 is @n‘h'nuaoj 42w a” x>7T We I’M/Q flm‘l' x193" 4" ({x‘) 5 10/91r)<0) 0W1," if r/ . _ FN‘fiCE/{er x42)”- $AJ<O flmL/s‘) “gr all x SVkan‘f/l: (/9)? +0, ‘2‘” é‘f/ kssfivm «277) 10%)‘(0 1+ gnaw; 4114+ #1”! WE SURE/7 x in #22 i am] 17"“er Mam {\I/flf) Eflf‘ Pfigfvt‘ Math 41, Autumn 2009 First Exam —— October 13, 2009 Page 5 of 11 3. (8 points) Let f (cc) 2 V x2 + a: + 1. Find a formula for f’(x) using the limit definition of the deriva- tive. Show the steps of your computation. i. [W (x+A)Z+ M +I ~ X2+x+/ [1—90 [1 x 3m x/(mfi-(wuw !‘\lx‘+x+l)(W+W) he 0 I" . (dmfiwfil + \lxzv‘w I) 2 I,” ((X+A)1+{x+h)+! ran“ 0 IMO I”(1’(X+A)L+6¢+A)+/ + V¥2+X+/) 2 [I-M xZ+ka+Hz+x+ln+lfxz-x—i [”0 1“ (1/(x+h)1-+6(+A)+I " +W) :: hm szxmhk W WWW) I; M 5mm “90 W+W :7 3x+( ,— ’0 4/ xz+y+l + JXQWM Math 41, Autumn 2009 First Exam — October 13, 2009 Page 6 of 11 4. (8 points) 1 + 2 continuous on its domain? Explain your answer. m [Luff] (a) Is the function h<aj) : léS — [460 is a mv‘iml ”film/ion) MIA/c4 we [We Seem l5 Ci’Wfinwvs on if: 0km)“. (Wham/y) [HQ ’5 Ol Zia/anew} mp rel/mama, éncéiomSj Sinc’e po/ynamia/s We. arrénuous 07%" 6W7 fem—é a“! 7V07£ien7£5 0"” @"fifibous ‘Chcéohs are Codhz/ous Where 0’9“;an MO /5 cow/inuous an 17‘: ”MW/2.) [315%] (b) Suppose the function 9 satisfies |g’ (x)| g 4 for every value of ac. Must g be continuous at every value of :10? If so, explain Why this is the case; if not, give a counterexample showing why not. Yes "‘ SNCQ {q/fidfxdl 14,— W X) #IJ if! ffllflgcular— IMP/KT find" 3’60 }5 JJAJ fire-m7 7:) my “”7471“ 350 if Jfimflfihge «+ Every Fob/ff. BvlL +5}: Mfliésfiadj in prucoiar) 360 is (km/Mews n+ 8W7 Pei/Til. Math 41, Autumn 2009 First Exam — October 13, 2009 Page 7 of 11 5. (6 points) Using the graph below of y = f (:2), list the following quantities in increasing order (from smallest to largest). No justification is necessary. 0 NZ) f’(4) f”(2) f(3)—f(2) gum—fen M m < o «W < awaken) < 1%) W3 ‘ ‘pfl/Q)< O [momma 4/74 3me 09 NO {5 clearly comma Jam! 00‘ X=Q _ L ’flnoflwrfiur Var/Hes [m a” Pow/I‘m) 57 Wfie 010 1542 “lact- ‘fiflr/ 101’s ’7/4 ”(Maj Imam (my? 155m 7% 1N>o a mix 4; WW“) >0 Wl'enevw" l9“). We can Wang “/Agge fizr MM; fly W‘évowfij P” 0:: 5/9025 SE)? elm ? . ‘p(&) is ’Mi 5/an (5p #{ 4172244 ,r/ ”M: 4‘)1 XCQ If?“ 3% SMffl O'P‘flw, ’fznjm‘f: [1m (if x:‘{ 4T3" ‘95”): 51$?” Is ‘Ma Slope a“ #1 seem} /ML We” X"6{ am“! X‘? ”“400 l . iffiW—‘Mzfl L WIS “fill 3/709 (f {A (emf; /n< ée‘ll’wm X=6Z anal x24 121 (WNW at #< graph pmnwml’) or: [454] armies 7445f 59‘”: .f P:)~”g‘)< t2}:{2)< «Pl/Q) (W 75% H4}: #Z')& (a) Math 41, Autumn 2009 First Exam — October 13, 2009 Page 8 of 11 6. (10 points) The period P of a pendulum (time for one full swing and back), measured in seconds, is observed to be a function of the length ac of the pendulum, measured in inches, according to the following chart: m 4 6|8 10|12114l P(m) 0.64 0.78|0.90 1.01]1.11 1.20 [391‘s] (a) Estimate the value of P’ (10). What are its units? PM a "M LN“). X4) ‘0 x.— '0 Fa;— XSlZ) Afflgleme (fucked i5 P02)‘f’(laj; LII-LO] 0/0 ’1‘“) a : E : 005 SEC/inch ' {5r X: 3) 0%de i: ire-PM 0,904.0, - ,, 8‘10 — Aa - ”a“ 1‘ 0.055 SEC/"welt flag) file 505:6 answr We am We is Mt 473.2% awMjeS 79152 m affmximq’l‘ionf 50 53710) ‘A 0.05625 Sec/inch . [4N5] (b) What is the practical meaning of the quantity P’ (10)? Give a brief but specific one— or two— sentence explanation that is understandable to someone who is not familiar with calculus. We» 4’“ FEM/[UM ’5 ,0 inc/’95 In bat/7) any Mcmse in H: legfl w/// 65:7 aéaut 6% [ACWRSQ i‘m like. PeNOCl) 0+ 4’15 M‘ffl ONO 0590+ 0.05625 Secom/sfer "ACI‘J- Sim;/£?r/J a M456 in [eryfi‘ (9w: '0 inches) («IOU/J Amy aéOUlL q OkaPOISfe ('14 fer/002) 0+ ‘Me 36.1446. mfg . [Effs] (c) The ideal clock pendulum requires a period of exactly one second. According to the available ' information, is an ideal clock pendulum shorter or longer than 9.9 inches? Explain your answer. Am, .3 a at we P’fioM 0.05523 EL m we warm 7% mat/m 41,744 PM Incl. ) [OM +0 qql") Le. £7 0" Mali) We O‘Pfid 0-" approximate MR in Per/000 o‘p (0.055135rzc/inclx)(O-anl‘) = 0‘005a5wc; tux, PM?) s? (LO/4.005%) Seminar. 45:3 13 Sh” aflv&r ‘flmm I fecomr‘fl) 50 WW. QXch‘i 7"le We’a/ ”emf +0 Wk ‘fl’c pehdi/II/m ever? flan 99 [ma/(5 ff, mild/E 72¢ I'fi/fm/ 0/0ch [9744}. Math 41, Autumn 2009 First Exam — October 13, 2009 Page 9 of 11 7. (16 points) Find the derivative, using any method you like. You do not need to simplify your answers. v5 1+7r P%0: 7&€+—(§-~ C) [##JQ) ))p(:c 2m 7+$\/_— Sin t 1 + cos t 1:ng1“) ))=T(t 1 + cost _ sin 75 rl/{Jz W [F S)"‘/%‘ir:f)~(®y9(H.C-DS+ ] ([+afiflz ’mzf flfilw —(1+t+2fixmm¢—cmw [1(6) 2 (i ’1‘ qéflfiifie (03“) + (”‘9‘ 2f z)(fi(‘tsiw{*@fl‘)) ‘ f I + ‘Hflw «054) + [/+{+2#)( Sm {cont 7.5m!) Math 41, Autumn 2009 First Exam — October 13, 2009 Page 10 of 11 8. (6 points) Find (with justification) values of constants a, b, and c so that the functions f(a:) = cosac and g(ac) = a + bx: + 6:132 simultaneously satisfy all of the following conditions: f(0) = 9(0), f’(0) = g'(0), f”(0) = 9"(0)- We, have ‘97:) = 003x an»? 36:): art‘bw- (xZ J so ‘PIM': —$'M)( m1 37*): [0+ 90 ) Z59 Ari/(x): “€051 anal 3"(Y)= 62g ' flag 4}”): @502 l 3(0)a arL-Omo‘zc a ‘ anal / I J «Pl/6))2—Si/10‘O) 810): [”3005 h) tame «(\"flikmoc—l ) 3’10): 2c , We Nyw‘m i=‘l‘WejW‘RJ so 3;! J owl O‘i‘70)53’(a):(9) 5:2 5:0 and ”l rip/[(0); 3//(0):(2C) go C: ”1/? .7 ,- 2 (17mm) mm 359212;) Math 41, Autumn 2009 First Exam — October 13, 2009 Page 11 of 11 9. (13 points) Sketch the graph of a function g with all of the following properties. Be sure to label the scales on your axes. o The domain ofg is (—00, oo) o g is continuous everywhere except at x = 3 (where g is not continuous) - 9(—3) = 0 0 lim g(:1:) = ~—oo $—*—OO 0 lim g(m) = 1 $400 0 Both lim g(ac) and lim g'(x) exist 93—>3 12—»3 og’($)<01fx>3or2<[a§|<3 og’(m)>01f:c<-—30r|ml<2 - $13513 Ig’(x)l=oo og”(:1:)<0if0<:r<3 og”(:c)>0if—3<:r<00r|a:|>3 . / flaw {’de d‘ 3 bf disawihwzg Concav'rf/ mail-cite; ~ I hole I'MSMPL ...
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