2311_T3A_solutions_F11

2311_T3A_solutions_F11 - MAC2311 Test 3' (take-home...

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Unformatted text preview: MAC2311 Test 3' (take-home portion) Name: S0 i (Nil 0‘0; Fall, 2011 - Section (circle yours) MWF 9:00 MWF 10:00 Directions: Show all work and answers on these pages. This take-home portion is due by in Teague’s unit, B-230, by Thursday, 11/10/11, at 3:00PM. Converge is required for #4, and DERIVE will be useful for the other problems. You may use your text, notes, and any other class materials, but you may solicit-help from nobody. 1. Consider f(x) = xv3 x—4 : x(x—4)%. a) Find the absolute extrema on the interval [0, 6]. You can find the derivative by hand or use DERIVE, but either way you must show the work to find critical values, and report final answers in exact form. Fill in blanks below. . I 1434) l: 1 ¥(% 3&(nL0/3 t": 0 ENE x=5 X=Lt , (5‘ Absolute mammum value (0 2- occurs at x = Q Absolute minimum value "‘ 3 occurs at x = 3 b) Sketch the graph of f , labeling local extreme points and inflection points with exact-form x— and y—coordinates. Your f ' work from part a) does not have to be rewritten here, but you should use limits of f ’ to determine the graphic behavior at points of non-differentiability. You may find f” by hand or with DERIVE. However, you must write down f " and show the work to analyze where f ” ‘= 0 or does not exist. Use the back of this sheet as needed. ‘ work space f01'#1b) f'ro X13 $\0Pe. 0‘5 x—p- 1+ 2. Each limit below is either an indeterminate product or indeterminate power. Identify the indeterminate form, and find the limit. Of course, each limit must first be rewritten in a form that allows L’I’lospital’s Rule to be applied. See classroom examples in your notes, MG? #30, and pp.302— 304 in the text. "IT . . 51 ('“ a) lim xsin[£) 2: +00 ° 0 itm X9|n(:{)=i‘m -—n-Tx—>—- :: £01- x%+o:u x Kg +m Xa+m 32 Lil-lot, Hal's rule - - ~ .- I ii-[SMW Ii] ’ C°‘(‘T"')' '7” z) t .. “rt-c043") X ‘ i 'L z _, 'tr’cosCE) :: ’21:“) x x“- 72.; 9| -1 ,1 a X K—a-too b) lim (1+ 20% x—) 0+ 83 Sulosii'ivi-ion ‘iiheorem ibr ilmi‘is QadtcnnZ§ . ii" iim (JnCHXll/X) imian : Effie iim in<t+><)% )L->0+ ><—M>+ x X lino-F ) : £2 — 9- Lil-logf‘ Rule x—m“ X o O _.l__ ilm lkH'x __ 2— K—ao‘l‘ I '— 3. A closed rectangular box with square base must have fixed Volume 9000 cubic centimeters. If material for the top and bottom costs $2 per square centimeter, and material for the sides costs $6 per square centimeter, find the dimensions that minimize the cost of materials. Report answers using appropriate units, and circle your final answers. Note: To earn full credit you must express the function to-be optimized as a function of one variable, state its open-interval domain, find the critical number-(s) in the domain, display enough of the cost graph to'show that the local minimum is actually a global minimum, and state the finai answers in a complete and organized way. You should use the following language to state your answers: “When the length and width are a.) 0 CM and height is CM the cost of materials is minimized at S g 0 O .” _ i arém of ' +0“ area. COS—F '— ahci \ao'l'l‘om + 63 UP 4" sides ‘ C = 2 (2x1) + C, (414(k) Suioiec'l‘ 'l‘o consirdmi volume (Moo;- x 90k C = -Su.do§gc’l' i1) Consivoim+ ClOOO :: xii} 1 ' ' - in: Clooo , and as >(-'> to: cm .—.-. 4x 4'— 143((9002?) * ,2— x \n—ao) $0 ciomaCm oi: C00 2 lib—lit?) tor 04><<+°°~ Cm ' as o<>< <+°°- C’zo "llGODO O 2 3X "" _ ‘XL 0: the -~-2\(9000 - 30 x32 27650-179 X: 30X=§>ln= goof =10 and C(3o)=‘|osoo ' ' ' 30 ‘ - 4. Consider Let f (x) = 2):3 e132? + 241w 10. There is no written work to Show for this part a) ' problefii. . '. it’s entirely a problem to be done using CONVERGE. _ ' a) Use Newtoh’s Method to find the “final estimate” for the root of 'f in the interval {2, 3}. Choose your Own starting value. Report all significant digits provided by CONVERGE. X02291). +lf\e"l:ina.l esidofidi‘efi 2.5.0000 ‘ b) Over its natural domain of (—oo,+oo), f has three roots. IfNewton’s Method is applied using start value 'xo : 1 , Will it find the smallest root, the middle root, or the largest Toot? ' (smallest, middle, or largest? ) Kalil dine-5 ’MIS r6 0):. - g Answer: I anllfisl— . ‘ n MAC2311 Test 3 (form A) Name; Lug—m Fall, 2011 Section (circle yours) MWF 9:00 MWF 10:00 Use a calculator as needed. Record answers in blanks where provided, and circle answers elsewhere. Some questions require that work be shown. 5. Let L(x) be the linear approximation function for f x) = «f; at a = 25. Show the work to find the M w ' «l24.967. formula for L , and Esediit to approximate 1 Maw H shown by the default setting of the calculator. 1-” Haw filaflxeo.) Do not round eff. Record all significant digits ' N w 'M was _ \jg NLCK)3{2_§“‘3 ) drama 2‘]? Ungmhiswfi 7- ” b :(2 as” (2 W3 .3, g i- sang“ Z‘ffle :3 Lewis): ,D If, 3.. 34m ,g, 5“ LLN .gL +§Z m _ . By linear approximation, #2496 x ( A5 0, cheek 4:”— ouw" wPProxiMmkoW-fvfijw micwim‘ivr’; £7" > 6. The volume of a hemisphere (half of a sphere) of radius r is given by V 2: gm} . Suppose the radius of a particular hemisphere is measured to be 40 inches with a maximum error of 0.2 inches. _ ,_§WW.¢‘_kwrfi-w&r a) Use diffei'entiaigto estimate the maximum error, relative error, and percentage error in the AmmH-y arml‘r w volume calculation when r = 40 inches. Follow round—off directions under answer bianks. ‘ I EW3£W9sWeL4Qx Clig’érenii‘fl‘fip :gbdciK maximum error in volume z i 2910' M ‘i in’ ( three decimal places) 2. . we He we. ) cl V =31 27V?” air“ relatlve error e I 0/»5 (no round-off ) GiV eflo ram'meii‘he mm: e were 'm w [mm 6 'a 1' Em- AV 3 2W (#9) c: MIDI)“ {'3 go‘o'éfiggcentage error z I I b % 7 d M F" Thrownawx :: iWL a. r” v F‘é‘fla r3 r ‘“ 4mg ” m '0’”? b) Uptional [Extra credit (1 percentage point): '1 he answer tor maxxmum error 1n part a) IS an estimate for AV , the exact change in volume as radius varies from 40 inches to 40.2 . in decimal form without round-off, the calculator gives AV = 10 2,9 . G» 961W m3 3 . av: 40.2) "" $1” (40):? Z “W’M‘g'” 6 '5 A 7. The graphs of f, f ' , arid f " are shown; Fill in the blanks with letters A, B, and C to identify which is which: ' ‘ 7 - _. - _ . . _ f f ' I f " " . v is s c: " B is e; z£evivwi~iv€ «gr A z“ f r “Emcrmsfiag aver Mien/stave imwj S0 t’i'S C E; as. a’eflvcij‘vqfl gr, fiemie iawmgjfiw. mwgf‘ tag, f3fisg§€w QVQW C 35 cg:(i‘§v~e NWVQ % infirm ) flm‘ifirfi . kiwflfififlkwmfl gm“; 93‘ “em. A f) : l E wcg ' ggtm ,_Cx@®Mwi¥DiQV@§_ } a graph“ :rfegffiwgéw 1E1) W Egaa’fi m i. 25 ‘ WWW?» 0% fifaaSEEa 3193 amt i e t m AM as 8. Sketch a possible fgraph satisfying the given conditions. Use dashed lines to indicate asymptotes that occur away from the axes. No work is _ required, but for partial credit you might use the Space below to diSpiay sign-lines for f’ and f ". I. The domain of f is (—oo,2) U(2,+oo) , and the graph is continuous for, all 3: ¢ 2. H: f(1)=-1,f(4)=1,f(5)=3, and f(7)$1- III. fimf(x):%f(x)=0. IV. ‘ 1513f(x)=9-oo and mgfaharoo' V. f’(5) doesnot exist,butf’(x)=+oo, and Higf'(x)=--oo .- ' VI. f'<0foxx<2,2<x<4',andx>s. f’>0for4<x<5. VII. f”>0 for 2<x<5,e.nd x>5. f”<0forx<2. - 9 Identify the indetelminate form, and find the limit. a) :f L: “g— L‘H‘DSFI'LJJ‘S JHOtan x) ' Z ‘ P k H-Cix _ hm I "— m 3 "‘ >L—>o 1’3” K40 " HGX) . 111x , \im \mg: :29. b) xllfim " x->+oo fo‘fi 4'” . lut- '-‘— [{m -x‘ >érv+vv 2x4" “L. ‘: hm x Kai-00 .2; __ l‘m FL. fx—q —— \ X Z 7%?” i : \\m m xwpca ' and fn(x) = I ('3 e 10. Consider f (x) = if; , with f ’(x) = e The natural domain of f is (Hoo,+oo). Find the limits below and sketch f . If the graph is asymptotic, label the asymptotes with their equations and be sure your artwork shows this asymptotic behavior. Show your sign—line work for f ’ and f ” . Label the x~ and y—coordinates of local extreme points and inflection points using exact-form x- and y~coordinates. (fix ) V‘éfiJf ’7‘ w e 9 l” K 5*" “’3 at} m»: We )6 “F3” 2, h M3919” ‘on f": 0 7C D Polw'l” W WWhm ll Lt (kg) ¥ ex! six is?“ e"! 1: C3 e” Mt"; X- Vigil” [wiv(xs_2_>:f 0 CD (3 XM'Z'? 0 in 2: W7 x2: 2. l 5% Z. a» a , €42: 1 DZ 1 1. Sketch the graph of f (x) = x + 2 cosx over [0,271'] . Show your sign-line work for f ’ and f " . Label the x- and y-coordinates of local extreme points and inflection points using exact—form x- and y—coordinates. I fix)»; HZGSM“) 42x) :1 [*ZSt‘nX I 7:33 $9M; mwms-x-«urfiww #10 cm‘ (“Veal #8 ‘ M1 7 ~ b t e. _3 Aegmxt; ll MVéleglls-rfa 6.815;) A4 $~kn><xfi *3: _5 «5 W Ti” j W '33 ‘ g; ::...::..r+:ft+].. I 0 Ti“ fijf 21? n ..,_ g ((32: w 24:..c2éx if” 6; (0 we“ 4 :1; +mt :52; WWW u H +r—irxn-m, t I } iwim ? on“; o 3 @211" w “A. “:3sz W ‘ ‘ :9” on Q ’Fnhflvef “gill/5 110 mg? 2:“ 3 3E IL 92‘ X {:(x') j: 33: - W M “T? - a: n“ e N (a 2’24; tar“ 1:“ r” 1:“ 2w 5: m 3: ;+2cgs®w 2f (9} “a: P; A5"? é: r ZM§(5_E‘) 33¢? + m .37 6;; (a; e a “if ‘3)11‘_ 3* fiélf+2(o:§11"n¥ ,7! ; wfvi-s2ucos(,§!§“) 2_ ) 1 LI 211 211'4»2c05[21r):: 2113920) =2rr+l % 8,25? 12. A cattle rancher has 6000 teet offence to create a large rectangular enclosure subdivided into three adjacent pens, as shown. Find the dimensions x and y that maximize area of the entire enclosure. Area of a rectangle is the product of its length and width. All segments shown are fenced. 'l‘o earn full credit you must express the optimization function as a function of one variable and state its domain. If the domain is a ciosed interval, use the “closed—interval strategy” to justify that your critical value maximizes area. If the domain is an open interval, sketch the graph of the area function to show that HS local maxnnum 13 I116 only extreme value 11'! tile oomaln. Note: The domain can be expressed either as a closed or open interval, but either way your domain must address the constraint on the amount of fence to he used. Areas—— lengtfitwiei'iit’ and Sum of all sr'x Ema, fajmewt : 69000 x A “3 31y Suhdeczi ‘io €o~6+maiv~+ DOC) :thrlvi): (“1y : QDOC"Z.X A(x):X(l'soow-i~x) yzfigoomwlix 69F A00 :5 Isoox m 3le fir 0.4 X 5 “5906.) (when )4 reaches 3am: y20> Aim): 15130 M X r A’ao AIDME a: we: 0 o o Xetgoo £000 Lagoon. NLmX31500 3000 o Yelaoo-wfifls‘ao) y: 75% Moiet Ii: ‘Hne. 40mm 'ts Viewed (1.); 4%: or“. \n‘i‘eran ‘94 K4 3°00, ’ dine O'F you“ Show 'H/‘g {OCOQ mmx'amum .FO‘. are“, (it vi: \‘5'00 is the Ohij bead actreme Value over 0 4 K 43°00), cmcL hence K: “300 Pro‘uwcf's V‘Hae “\OS'OLGi‘Q MMMWM area. . 1 Answer: II x "—“ / 50 O Ieet and y : 7 5b teet, the total area ot the enclosure Will be I! [1516700 square feet. 13. Let 2 2x3 m13x2 +24xHIO. 3) Starting at x0 = 2 , calcUIate the first Newton iteration algebraically by hand to find x1. Then sketch the tangent line that produces 161. jC(x ):’ix3~13><1+ ZHX'WO 76 ’m = ex”:sz w» x: 2~fl§l= 2e——2-‘—-:?_+l ‘ 43b.) “’1‘1L A x1: g/l or 2‘s; _ b) Show why Newton’s method breaks down if x0 = 3. Grafkicdl’y/ Jrkerefs 0k onfiLormjtu/Q afi‘ X033f'1‘tVYS "Falwsel’U-lp‘ never tfi’téfgef‘li xfojft‘xfl 50 V10 ’P-nf Hedge»— ‘ateuca‘tl‘m ‘qs /"' ?ro&abLCQT‘Q ‘ I" H ...
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2311_T3A_solutions_F11 - MAC2311 Test 3' (take-home...

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