06exam1sol

06exam1sol - [Cg/ll (QOOG ‘ Examj So/u’f'forzs 1. (6...

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Unformatted text preview: [Cg/ll (QOOG ‘ Examj So/u’f'forzs 1. (6 points) Mark each statement below as true or false by circling T or F. No justification is necessary. If f is an even function that has all real numbers in its domain, then f is not one—to—one. 11ach “P60 ‘ ’(Y'X) MEWS fil/W‘f 7452'? an: M Jr'tqoevevn‘ X~V4/I/($ ’Mad‘ ham’h same y-Va/ue a? @629) namel/ x=a and x= ~4. (6wq¢a) Via/ates Aer/M/ 151%. If the graph of y : f has a vertical asymptote at :r = 57 then the graph of y = f (23: + 1) must have a vertical asymptote at :r = 2. (’th‘AL «goal Low Staph +mms€rM$-) If the graph of y = f has a horizontal asymptote at y = —3, then the graph of y = f (—13) — 1 must have a horizontal asymptote at y = 2. W CONECT aSympf’o‘fg in p/acg a“ uyca‘z" SAW“ 61 “/y:—-§¢f/‘. age/h +£8.24 about Mnform'ls graph A0 The function 9(23) : is continuous at as = 0. T @ Since the function f = W is equal to 2 when a: = 0, and is equal to 35 ——4~ when r = 3, the Intermediate Value Theorem can be used to conclude that f (c) = 0 for some 0 between 0 and 3. ’P lies a ae‘SNM’ylinUr'fily off X5!) 50 flifij‘V/‘Q‘Md 56 “lop/160?. The graph of y = $2006 has an inflection point at m = 0. 2) y’aQOngwos. mag y’/:;?ooe~;zoo5x”0" >0 4;, 41/ #0 (Because 2004 (3' $173”) $0 CDWMWI'FL/ ages 227‘ 6401an / a-t X—TO. Comcave opt/wag an MA xii/e; y‘axr's 2. (15 points) Find each of the following limits, with justification. If there is an infinite limit, then explain Whether it is 00 or —00. (a) hm 3:07 — 5m + 1 sis—>0 4x2 +3: — 3 SdM¥Ufim @a‘f/om/{quéns Codhuaus «1/0; (rm 3’7’5’M‘l ‘ 0 “0+ l = ~ 1‘ “'30 ‘fx‘+x~?> - 0mg 3 ’. 00813 1' 14 I» " (b)xir71ri_$_7r glfl XL;_ COSX— “1(ZFC) m? "1“- X—Tf == ’ ' ‘ ‘ Ya” O) Mani 15‘ GO. 43> Jgérmlwe Egg : mem‘or ;s n {at FQV Xflfar’zr .J Olenwmrlcr is m m‘wie ‘f’c; x—afl- (since. x<77 g xqfioj J 50 7U0¥Iénf II: @265 . G) Was In“? is {+00 ((3) lim 3:4 cos (~) 02—>0 44%) £7 fig" SWMWW”) 3/ flay/>40) {This {Ail/0W5 because, KHZ [I’M {205g} :0 m {~90 3. (5 points) Show that lim(4x ~ 10) = ——2 by finding a 6 > 0 such that (3—62 |(4sc —— 10) * (~2)] < 6 Whenever 0 < lac ~ 2| < 6. Shce [LIX‘IO "(‘2)’ 2 (fofi/O‘AR/ : (9x~8’{= Q/X-R/ ) we («ave {qx‘b‘é2)/<E W00 OM51 a? ‘ +44% is) {fart} out] {X—R/(E. ‘1 $03] $32? Works) Eemu-Se whenever O< {X—al < Szziz. 8 . 12 ' t t = . ( pom 8) Le f(:L') 3 + 61 (a) Find the domain of f. QM fix) gg gagged? 0“! H“ 3+e“:o (a, 5m x, )’ 30+ 8)( is G’IWVS ’00sz) $0 3+e¥>0 a/wa s. Don/min: (-ocs 0%) ) (b) Find, with complete mathematical justification, the equations of all vertical asymptotes of f , or explain why none exist. At OK vent}ch QS/VWFA’ILE xza o‘l‘.‘ ‘8 we?! Lave, a {I’m‘l‘ M53 Such qs (i‘M :2 ll: 0% 63" 62+ /ca5+ 14W X—FQJ‘ or x405). Xaa gal“ +/4a‘5 Mob? 2»: par/Jeazar' I‘M/54' float has a ad'stfimajx at xcq, NOW) “1053 5% ggo’lleml OJ) +WO C°H7£5I700U5 “QM low; 50 QdUQ/é/ 2 Cowl‘MUOUS 8W1" Mien on #13 Jammy) “Mick ['5 7% eu'fim X_QX5'S ‘/ WUS/ [MS no o/lSCoH’finUl¥i@5) WM! [:1 ParficU/ar/J Gamma” Mas’l‘aéef f1“: “01" SU‘QFlciQVt—l' 4% Bay {5247‘ 10 his M0 vert. “SyM’o’lLo‘feS SFMP’)’ M523 ’l‘lm‘lL H” is never UMd’Q’ngeoé. W‘ancf/om 5 whose graph [oaks 5E1 gzj‘i/‘fl is 95611947 EWW‘WVQ/ 50+ [MS a w‘fi’ml asywprl’o‘l'e a” M same ‘ asprt'fieg ape, [1341305 (nemrily) ’l’o ‘HQ presech o'lf {mama/e limits) ETC/‘13 #16 pregame 0‘0 UM of 0541‘s (of even Zero dawn/tinder: : rememéaer ‘Hm‘t numerator: 41/50 ,0/4/ a fat/ef) . {If Wr‘fiml “Symfia‘faf, (c) Find7 With complete mathematical justification, the equations of all horizontal asymp— totes of f, or explain Why none exist. We eomFufig 56% III/mils a‘ I mif/z “M “€60 = g X c O [55“? cél/IOMWIKOF 923+!) 00 WLJE Mumua+r 37L“ Saki/Ewe) {—9 an X-) W 3*5 3 e 7 J 3 <8 I, 2 I’M A -"-" I“ 5 ,2: s. [I'M 7‘ : awe XT‘OQ’QI‘) >66 "00 3+9} 3+0 5 ( mm 709-0139, 0)) So 4'th me it? horizth asyupfofest [E and! :3; , (d) It is a fact that f is a one—to—one function. Find an expression for the function f “1(w), the inverse of f. , 2/ ‘ y.— 3+€X Sill/rial? +5184 Solve, “Eh/I 3 X5“ <1? y” 8 3+9} 3+6 " 3? ‘3 l" .3“... < 6 _’ X 2 5. (8 points) Let f(:E) = x :3. tive. Show the steps of your computation. Find a formula for f’ using the limit definition of the deriva- $1) 2 [1M ’F(x+i\) 4kg [Mo L , 26%) ’ RX :: I’M s<+h+3 x+3 Leo L, g.‘ [I'M 33+“),21 film's ><+3 4 m ,L, {12(7‘+5’1)(X*3)'2x(>(+14+3) MO L‘ {x+h+3)(x+3) ’ [I‘M a(KZ+KL|+gX+3A’KZ’XL1-’5XE [4‘70 Mx—rh +3)(x+3) z: (Em L A60 MHQSfifiB} M “M Q Q .—- n (mo (x+h+3)(x+3) (3%?" 6. (6 points) The graph of the function f is given below, as well as the graphs of the function’s first and second derivatives, f’ and f”, respectively. Indicate Which graph belongs to which function, and give your reasoning. Wf’depffl’ficqqtfomj ONE, 5150qu alcove ‘ /E> €XP/Q"n/ M‘s din ways 4‘0 Emmi. Here; one matted? : W defleoe curve u—“‘ [la/03$ no'fc Adi/e. i743 Olefin/476% 3MPA806/ Since we’d new? 'fo see a curve 'fliq'l' crosses “th woods of P and 0 (Mai 23 nefer bail/wen 'fl‘éSQ +1“. Pain—f5.) 09 an 7%) W t, w AS 61 Comsefaeme War-Haj WA 7%“; 4.256) we note Ma‘f twan only be "flu; dam/{five 6 U“ " dude “0+ 0.63 ‘q'flwwfl' W11: )5 chquse “1&1 J line is decrease} an Same [fiat/ails {w ,‘5 «934.5%; amofi I'Acmséufl 0w '31 5am edema/s 7%?" —(‘” ,3 “SHWE‘ We +403 (MW 7% flu salvo? line, "403+ log 70’, gy pruoess O‘F dim/Wm) 'qfihm” "403+ {Lgs OTC o’mer COMfleChoMS make ‘flris Cféar: absent/3 4554+ fig ,Ql-fgmak Meme, "—- —-'-=~ ” is Mamas/v5 We +111 36mm as ‘fliose wheml‘-——” .3 PosfiLA/‘e Moi So "193wa 7. (12 points) Suppose the function g is defined by 1 ifx=2 kx+(1—2k) ifx<2 VIE—1 if$>2, Where k is a constant. (a) Find lim (Your answer may depend on ac—>2‘ a: — 2 = “M (kx-JPQk -—- f Xag‘ X “ R c: “M kx—Qk = (I: k(xr69_ . K631— K-ZL gag“ Xmg — I‘M— k: k (SQch ar‘ no‘f—so’secrafiy/ We’re COMfzflLeoK flit damn a0 3 “am m ea” 421‘ x== a . ) — 2 (b) Find 11m (Again, your answer may depend on x——>2+ a: —— 2 z [W WIx—f ~( 2 I,‘M(\1x—t—U(J§THD x-aéfl x—gg >902“ (w; M +0 J GYM“, +h;5 "5 lfl'e o’erwa‘fim [Clem fie (/ a1; 7C: 2 (c) Find k so that 9(112) is differentiable at a: = 2. (Hint: how do parts (a) and (b) help?) 3 is oQfiéPerenfinLJk aha-“5L <29 ("M W exists. X—ea ><~ a 77%, ’a‘H‘er {r‘mi‘t WI” 6031‘ if {1‘s Ht- am! m‘jfifiéanoe limh‘s each 645% am? one M, We 165%)? ‘lizese “I‘wo (\lfialT-{fimLfSfl M (a) mat (a) so we km W new? ksi- _ [Nate ’fl’iS QWDMQ’fi'C‘I/ly 30mm: ’flm‘fj I'S Com‘fmoous at?" Ya?) So H‘é "0+ nacesmr/ 1‘0 make +1113 check) (d) Find an equation of the tangent line to the graph of g at 1' = 2, for the k that you found in f l _. “M 3% S __,L 1? hi) then 3639*“; Kn; kuaj 50 +112 +them+ (Me 405 6704/1011 y»3(a) :- i x—~5Z) <79 y“ ( c 1L(/><*QV) [Mo Mezai‘tlo loaf“ i“ may O'H’e" {QM-I] 8. (6 points) Below is shown the graph of y = f By interpreting some of the following six expressions as slopes of secant and tangent lines, list these quantities in increasing order (from smallest number to largest). No explanation is necessary. f(0-5) - MD M) — W) o 0.5 f”(0-5) flamed "filer 55‘ “W < 1°70 “Ftp—Ff!) < N’IQF O < €705). Ragweij “PW/0.5) I’s £€C¢WS€ ‘9 l5 (“Cal/Q 0,0 11+ X‘O.§_ We Value: anal am nfijgire éecause. 1f is JequsiLrJ sf X4 X 7&2. 1% Values ‘F/Q—WQ ,Ml «WW—1%] 0.3 G‘VQ We a’fi‘m flISO E€Cau5€ ~F rs Jewish: mar.) m3 rev-rm) are.” TO amid/mm ‘7‘le “gar avath numbers) ‘flriné 019 4/] 070 75% as Slopes) drawn atom. (vileCUPl/af/‘ure @010 allows WS/opes ’f‘o [1e ranked . who! NEIL slopes of‘lfi’fi‘zfj [Mes "if? [K a? a X: X: ‘ may: . 1% -r = 1%)- ‘(Q . The fieepefifwegafimy ) a“ l ’5 file Slam JP 1418 Sea-“ML [line Slopes? line canes mi; . M +6 m sml/eyé P0 m Pow: (51%)) m! @411”)- flumerical Value} I —F(0.S) #9?» f_ +703 ~Fo . "33"— " 0.5{00 lS 5(on line ‘flim [git/(fl m4(0.;¥{a$). 9. (17 points) Differentiate, using any method you like. You do not need to simplify your answers. (a) f(a:) = 4232f -— 893 + \3/5 = foa'; " 8K 'f‘ ’sfg‘l J, so) «NW +695)th - 2’ + o = WK 8 = 10. (5 points) Let 9(1)) be the fuel efficiency, in miles per gallon, of a car going 21 miles per hour. (a) What are the units of g’(90)? Miles V) milesperflwr’ (b) What is the practical meaning of the statement 9' (55) = —0.54? Give a brief one— or two—sentence explanation that is understandable to someone Who is not familiar With calculus. Matty answers are acceptnéla. t‘ére is (134E, PCS-535% Pl‘MSIL’S 5 When file Cari: 509601 ‘5 55 miles per hour; fig rm? éfcmfige Q70 £6] QTQlflcley‘zC/ W5,th P85339011 +0 SW08 l5 “06,7, Maj/Mp5 1‘3) Speed will result {w a Jewse g€ 79d Q‘Fficieyzcy (anal vice llama/3) a‘f an appmfima‘fe m'l'g OTC 0.5771 mp3 Per MFA. any small immense m (Coulal say “513860” or “Klocl’fy 11. (8 points) A function f has all of the following characteristics simultaneously. o The domain is all real numbers except at = 2, and the range is (—2, +00) 0 lint; f = +00 0 lim f(:c) 2—2 and lim f(m) =~2 x—>+oo :I:—>——oo o f is continuous on (—002) and (2, +00) 0 f is increasing on (—00, 2) and decreasing on (2, +00) (a) Sketch a possible graph of f below. Be sure to label the scales on your axes. One possible answer: (b) Give a possible formula for f. (Hint: think about transforming the graph of a familiar shape.) fir “Hoe shape drawn «haveJ we Cooloi say : l y: (Xi-(2)2, ~12 looks lféfl Sin?!“ 0‘? )1: ...
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06exam1sol - [Cg/ll (QOOG ‘ Examj So/u’f'forzs 1. (6...

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