Math 350 Exam 2

# Math 350 Exam 2 - AMERICAN RIVER COLLEGE mmmmcs nzunmsm A...

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Unformatted text preview: AMERICAN RIVER COLLEGE mmmmcs nzunmsm- A. swam...” m:- emume gawk Show your work for fun mail. Please use com nomion and pm your answers in due blanks. mm :50: LIFE scxzxc: CALCULUS I Exm 2 Nuvimlzn A, 2010 I Find [be critical numbews and cﬁlica] points for me {allowing fumions. (a) ﬁx) : 25 + 16x — Le Critical numbms): 4 {/{ll} = 04/6 _4¢ Cri1icalpuinl(s): 57 /6—4¢=0 {Mp 25+/¢-4—2-¥L ’ {54¢ =25+¢¥~32 — i _ if = = 5 %¥ * >4 4 7 (h)ﬂx):2f‘~24x‘1+5 cmicalnumbms): o, 4 IL _; 6m=2-;«42«w (0,5), (lb—27) -/a. = “Vii/2% 3,1 In Bil/I'lzfl/z :0 [email protected]) = 20 —¢4.0 +5 » - 1 V; E 31/”(71'0’0 {(4) =2r4z/EQ4‘4 *5 ¢¢o ¢=4 = m—l/ﬂ'b’ ((0)1315: {who =-27 ’I Mmmlﬂeschncuuhllljunl IIIAmImB-Mhsrqezm". 2 FindlhcnuximIMmininufmmﬁlmﬁonﬂx)-x‘¢5xgrlx9 |l fmﬂuimav-lkéJLGiveach mswau-ww(momm,_ Absolmemnx: (4,125) (31—9)(’x+4):o mmum,Ml 1: \$2 7: :’4 Bdrm/enthum (—6)=(—6)3+5(’6)t866)‘” : 73 Abso|memiﬂ=( 3; 21/ 7 5(4) =(—4)3+ 5 6431~8(—<()+ // = 5? J M) = (’éﬁsﬂf— tee/gm i’ = 33:, ' 27 {(4) = 43+5-4‘—8~4*H= as (at) = 5¥’+/0/x—K = o 3 lsvhgylphofdwﬁmcdonjlx)=stxrisingolfnllingwhmx=ﬂlslhempheouuvcupm com-v: down whcnxrﬂhulify you! mum. {a} = ¢qu Checkyomlnswm: {(1) : ¢(oa¢)’+ 0421(4), Wm = ~ 7! Mat +aa¢ I C ‘ g ~ ‘ onnve up 6%) = 46144409 (WM-«J mm mm : —¢M 'mfl “Ml! = _¢w¢-2M¢ o+(—n=—\<0 ’ = _ m1+cn1r = 15(1) 1' —7r(—[) —0 =1Y>O MW ‘5”{10 = —1r cm“ QM“ = Mun :50. um Sci...“ cum.“ 1. Eull 2. 11/4/2011). amen“. mu am a The accompanying ﬁgure shows me graph 13f y = 1m 111: wins comsponding m 111:: value: in the table below an marked M111 don. Fill in the table, using + 7, 0, or DNE 1m posnim negative, m, and "does am «my respectively) forﬂx). ﬁx), and f”(x) :1 mg indicaled values 0fo [For example, an in me am row for 1H). f’H). and ["1—1 1.] mm”._,o..,.m.u.m. x I rm rm max) 4 + 0 — a l + — i 1 J + plus ml: 2 1 + o — 22 l + K o 3.x { o —’ + 45 J e O + 5 Us: a dimmth 10 ﬁnd a good appmximadon for {15.8 . Comm 111: r=sull lo 11.: calcuxawr value. (WriI: boﬂl Mulls In foul decimal places.) {3M1 : Q77? ’ﬁ = [a Diffmliulappmximalinn: 3.?150 6’90 = 2—“? M: ’0‘?“ Calculawranswer: 5‘ 753 {mm a {60+ WW 1/X+d/ﬁ ’3 J? \ «a z W +mw‘” : 4 Jr Jg'C-oa) = 4—0025 = “750 Mm 150. Lik Stich cum“ I. [an 1. mmw. mmellu. hp l on a All foul sidcs ofa squat: are ynwing at a M an cIn/sec. Hnw ras: is m: m oflhe squat: growing VIhEII [he lenth equals 10 cm? (Be sure ID includ: uni's ofmcasulvmenl in your answch A : ,XI 4: [00m Answer: (A M 5%: : 30M/MC It = 904% 46 ._ 2-!0‘3 =éOMn1I/w olﬁ «:0 data 7 Use implicil diﬁ’mlialion with Leibniz "maxim Io ﬁnd Ihe slope Al (71. 2) f0! '1‘: curve deﬁned by ,1 - A gm. :7?— [Jan 33:55: = J28) 100‘ Jr 3’11'34/‘(59 +g“~ﬁ(5¢) +73 = o si+3gf+7y=7 /0—/2 -:i=§ - /»—-'5 5 M-lh :50. Lite Saul“ Clltulus l. 2;... 2. Illa/mm. Barbell». Plge sum. 3 A mungle is 1! unils long by 7 unils wide. A squat: oflcngdl x on esen side is cm from each ofth {our comers ofth lecungle. The sides are folded up w [arm a box. Follow [he given swps :0 ﬁnd Line value ofx mm corresponds m the maximum possible volume of Lh: box. (5) Draw Lb: squares in each corner of 1h: mangle and label the sldes of mu: squat: asxe (b) Find algebralc expressions for {he lenglh, widxh. and helglu a! {he msulling box. Ltnglh= //, 91 Widlh= 7_Q : Heighl= ,lr (c) Find and simplify an =xpr€ssiurl for V0), [he volum: ofvhe box as a ﬁlncﬁon ofx. What an m: crilical numbers fur V(x)7 What an: Lh: dimensmns ohhe box of maximum volume? (Give your answ=rs In two decimal plnces.) V(x)= £3,3zE1—4, Zﬁ: Crznculnumben: /. 39, 46/ Dimcnsionsofmaxlmum-volume box: l.% x 4.2l x 5-2‘ W!) : 6/'77)(7’y¢)(70 L = [/7 1&3?) = (77— ﬂaw/[(46 MWL) K = 5.21 = 77703675‘447? W: 7—205?) / L = 41/ W9!) H276 427” 77:0 H = [‘39 722‘ (my—44277 f 7t : 2/2 V0.59) -— 45,21 1’: L392?) - 73 Mu- m. LII: Sch-c: 0|:qu I. hill 1. Imam». a.me mum. 9 chain. has moddly shlpui plot mm mm is bounded y by we might lines and a 1m. Ifwe m the migm um bedwxmdynxcstheedgeofﬂnlaszoﬂmm uranium)! mm whose graph is given by y =4e’“, (See me ﬁgure.) H: wnms Io pm A rectanguhr puslum an his [and and 1.; wants am hm 1h: pales! possible am Assuming ﬂu! m; units are cxpnssed in miles, whal mthedimuiomand memoflherecunpdupaslule with mum nu? (Work 0mm following slzps m answcring mas quesuan.) (a) mmofm=mmng|dmpmunixgimbyA=xy. Rewrite the masmmian olx don‘emiﬁndm derivative. w; ‘ do 67,-” A=«g=qﬁ_4€—’X/L A'(x)- _ 15 L4. AE'X/L = e—‘F/L / AW) = MOB—"VLYJ’ Ema“) 4/1 = 41K e'ﬂléizﬂ e’m4 = ' 9’3‘8 J‘ 48—M— (b) Find an.- maximum um ohm nu ﬁnmion Aer). wtm me me dimmions 01'th mm pomible mum puma-e7 (1mm: min in yam answels. Whm munding decimal mm. give Iwo decimal Mm.) (.41 Dimensionsofnnxuu: 2 X a e A’m = o , K «a ﬁxewLMér/QO ““'““”'im 464/1002): O ...
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Math 350 Exam 2 - AMERICAN RIVER COLLEGE mmmmcs nzunmsm A...

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