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Unformatted text preview: MAC2311 Test 3' (takehome 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 takehome portion is due by in
Teague’s unit, B230, 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 solicithelp 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 exactform 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 nondifferentiability. 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 X9n(:{)=i‘m —nTx—>— :: £01
x%+o:u x Kg +m Xa+m 32
Lillot, Hal's rule
  ~ . I
ii[SMW Ii] ’ C°‘(‘T"')' '7” z) t .. “rtc043")
X ‘ i 'L
z _, 'tr’cosCE) :: ’21:“) x
x“ 72.;
9 1 ,1 a
X K—atoo b) lim (1+ 20% x—) 0+ 83 Sulosii'iviion ‘iiheorem ibr ilmi‘is QadtcnnZ§ . ii" iim (JnCHXll/X)
imian : Efﬁe iim in<t+><)% )L>0+ ><—M>+ x
X linoF ) : £2 — 9 Lillogf‘ 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 tobe optimized as a function of one variable, state its openinterval 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 27650179 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 dine5 ’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" wPProxiMmkoWfvﬁjw 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.¢‘_kwrﬁw&r a) Use diffei'entiaigto estimate the maximum error, relative error, and percentage error in the AmmHy 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 roundoff )
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 roundoff, 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 signlines 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)=9oo 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 HCix _
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
Kai00 .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 exactform 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 signline work for f ’ and f " . Label the x and ycoordinates 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; mwmsx«urﬁww #10 cm‘ (“Veal #8 ‘ M1 7 ~ b t e. _3
Aegmxt; ll MVélegllsrfa 6.815;) A4
$~kn><xﬁ *3: _5
«5
W Ti” j
W '33 ‘ g; ::...::..r+:ft+]..
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g ((32: w 24:..c2éx if” 6; (0
we“ 4 :1; +mt :52; WWW
u H +r—irxnm, t I }
iwim ? on“; o 3 @211" w
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: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!
; wfvis2ucos(,§!§“) 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'soowi~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
Yelaoowﬁﬂs‘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’Ulp‘
never tﬁ’téfgef‘li xfojft‘xfl 50
V10 ’Pnf Hedge»— ‘ateuca‘tl‘m ‘qs /"' ?ro&abLCQT‘Q ‘ I" H ...
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 Spring '11
 Teague
 Calculus

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