solution(common)

solution(common) - 1[30 pt Let f(m = Edwin/m:3(3 Using an e...

This preview shows pages 1–16. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1. [30 pt] Let f(m) ={ Edwin/m), :3: (3) Using an e — 6 argument, show f is not continuous at .1: = 0 when a = 0 . sol.) I A when a = 0 , ‘ __ sin(1/:1:) ifa:>0, f(m)‘{0 ifng. we’ll show that discontinuity of f at 0, that is, 350 = %,V6 > O, 3.7:; 3.1;. [\$5 —— 0| < 6 and |sin(1/m5) —- 0| 2 60 = (1) In particular, V6, 3n 6 N s.t. < 6.(by Archemedes’ thm. ) %(2n — 1) we can take 1:5 = m, which is depending on any small 6 , and we also 2 get |sin(1/x5) — 0| = |sin(§(2n —— 1)) — 0| = 1 Z :7 = so . Thus (1) holds. So, f is not continuous at a: = O. . wegawwmneeeaHessea2m%wwrsssaa «m see was we. JHWHM4QNﬁﬁ%%iﬁﬂ%ﬁﬁdEe—6%¢Wﬁﬁﬂﬂ% we see-:— Haiti‘s, 9mg see-es we. 5‘3,sin(1/w) 7} 73% a7] Ira-eon see #2121 %§}D}% we :4 exam %°l°ﬂ we a ~%%HNEHH%ﬂﬂﬂﬂ%ﬂﬁﬂﬂﬁfa£izﬁﬁreﬂﬁﬂi Ix E 94:32} asaTs —’r‘— g. (b) Determine all a that make f continuous at a: = 0 . _ m“sin(1/m), an > 0, sol.) Suppose f(ac) — { O, x S 0 At ﬁrst, obviously limznO— f = 0 and f (0) = 0 . Since f is continuous at 0, limz_.0+ x“ sin(1/av) must be zero. To satisfy that , we’ll show that the unknown value a must be lager than zero. Otherwise, the limit limm_,0+ x“ sin(1/x) diverges. We consider that a is divided into three cases. is continuous. Case1:a > 0 ) We have |sin(:-:-)| < 1 for all :1: 7E 0. Since 1 —w“ 3 9:“ sin(;) 5 1:“ for all a: > 0, By the sandwich thm. ,limz_,0+ x“ sin(1 = 0. L Case2 : a = 0 ) Through the previous problem (a), we have already considered. We have 1' a ' 1 = 1' ' 1 '11 t . Elgar s1n( /m) mirng s1n( /:1:)osm a es Case3: a < 0 ) We have Since |sin(%)| < 1 for all a: 0,‘ limzq0+ as“ sin(1/a:) = 00. Thus, f is continuous at 0 only in Casel 2 a > 0 . (EH3 773) 78%" a > 0 E’l-é— \$513 57%. %°lll-7§7lel ERR“; 7E. (c) Determine all a that make f differentiable at as =0. sol.) Suppose that f is differentiable at 0. the“ f( ) 1(0) I __ . a: —‘ . f (0) — along) x _ 0 ex1st . Obviously, m—vO“ (II—0 3-90“ (II-0 . By the way, —— a . —- lim M = lim M9 = lim xa‘l sin(l) must be zero , :c—+0+ ‘ (B — 0 mﬁ0+ (I! — 0 z—>0+ (1: since f’ (0) exist . To satisfy that , we’ll show that the unknown value a > 1. Otherwise, the limit limzq0+ ma‘l sin(1/:r) diverges. We consider that a is divided into three cases. Case1:a >1 ) We have lim :1: =0 |sin(%)l < 1 for all a: 75 0. Since —a:"’1 S 3‘14 Sing) 5 (pa—1 for all a: > 0, limm_,o+ era—1 sin(1/:c) = O by the sandwich thm. Case2: a = 1 ) We have lim 27“"1 sin(1/a:) = lim sin(1/:c) oscillates. x—90+ x—+0+ Case3: a < 1 ) We have . _ . 1 11m x“ 1 = hm 1 -oo. z—>0+ :1:—->0+ (12”a+ Since Isin(%)[ < 1 for all a: aé 0, lim\$_,o+ 11:“‘1 sin(1/:z:) 7.: 00. Thus, f is differentiable at 0 only in Casel : a > 1 . (\$1431 7%) 7d? a > 1 E1173: “£319; 575}. %°l-T'Jr7§7/lxl Elil‘i'l; 77g- ((1) Determine all a that make f’ (as) continuous at :1: = 0 . sol.) Suppose that ama'l sin(1/m) + a3“ cos(1/m)(—1/\$2), f'a) ={ ={ am“*‘sin<1/x>— wa-Zcosa/z); x > o, 0, 0, is continuous at an = 0. At ﬁrst, obviously limmﬁo— f’ = O and f’ (0) = 0 . Since f ’ is continuous at 0, limx_.0+ ama‘l sin(1/x) — #4 cos(1 /x) must be zero. To satisfy that , we’ll show that the unknown value a‘ must be lager than two.(i.e. a > 2 ) Otherwise, the limit limx_,0+ {acca'1 sin(1/av)—:1:“‘2 cos(1/m)} diverges. We consider that a is divided into cases. Case: a > 2 ) We have 11m x“_1 = O art—90+ lim (ca—2 = 0 m-»0+ _ 1 |sm(;)| < 1 for all :1: 74 0. 1 lcos(;)| < 1 for all m 95 0. By the sandwich thm. , limw_,0+ ama'l sin(1/:1:) - wa'2 cos(1/:1:) = 0 The other cases are similar with case2, 3 of the previous problems (b), (c), so that lim\$_,0+ area-1 sin(1/:1:) — #4 cos(1/:c) diverges in that cases. Thus, f’ is continuous at 0 only for a > 2 . p (as 7%?!) 721%}; > 2 "El-3- B’lﬁltﬁl 5s. %°l3l'7§7l7‘l ERIE 7%. \$50 2. [20 Point] (a) [10 Point] -solution 1- 1 + tan29 = sec29 => 1 + (t2)2 = sec20 0 (1‘0) ‘x =9 1+t4=sec20 HE 320“, \$3 @QBI klﬁam 9% to" 3% EDI 2M1 EIII 91-:- -solution 2- tan(7r-9) = ~1— 2 -ta.n0=tT sec20 d0 =— 2tdt 0 (1,0) x —2t _ ~2t \$6030 — 1 + t4 ==> 1+ (— t2)2 = sec20 1 + tan29 = sec20 => 1 +t‘1 = sec20 0 0g 3E0", 5?: E'QBI MESH! 93% I toil 33E l.‘g‘OI gAI EIII Eur-3- (b) [10 Point] -solution 1- 2t _ 2(1-3t‘) ‘ (1+t4)2 position (0,?) ? (0, 7/1?) maximum speed 2! 0|»?— 5183’ -solution 2- solution 101W L?’—§Oi| ﬁ—‘F—ﬁiOi EDH. 51-i- §¢E solution 1 Bi DéOI 73%. I critical point i ﬁiLiEi i%-‘u’- — '3 ——AJ£ '3 tt((°9+. -SIH£)4 I 511%“: et(mg_£ +S7Ht) t J31 ; d‘ﬂd’a : guest—mm) __ (0ch Hm A? C STL +§3K—- o+‘ :— a '5: J. c. z _- M A% £‘E1: (03% _\$.(n:r;_: o _\ ‘ p ) _ A» _ l W“) “M11” ~ Whit) (10—) 1(1)) ‘&(£) -' 93"”; e')(.,( -0) : PDQ + 93"” (§ ‘23 ) &(Jb A1? A ‘clbr _ (ii (1 cl?) _ '3; E; "J?= 75:11?) ﬁ'dt—{Z or~ ———-——'&x V :13; (§@) 4 Aver _ A cochJr Shot (D “(—5)‘ Tea COS£~SWK __ L ~ (ms-E— 9m? (9 = 468°ch ~ sme) 1 2. B CD owA ®,- “‘3 ; (cest~gm—E)”‘ 7 A 76' 8+1 LOSE— (ME) :2, : pr 3 9*(cmrs'wb) LEW E L ~ {’1‘— i=3: eNlLC°S_1é__S,]M .1; 3 : ~— 3.9pr (5‘23) 41 2% ° ° «Mr- c*(cost+s‘mt) ii) I %A%%>= 3% .— 5 (X- W“ 1’? ° 75?“W “13* 673) 4% (1+ V (w?) ‘ Hwy-LE ’ Biff/l U’S) = {TL SQJ‘CLGS‘bgl/Wb)? Jr {EEC bye-51%)}: 4'5 7L = j 93"). it 0 1-—*=Fz%, wag 1% 'Wﬁr/ 7w #4:} ; +1 0% Data 2&5: 73% : —-\ v WM vii} w Gyms? : - 3 A (eves—E) = EJ‘ "SW/Vt L W T : COSt+ 3 WP:— “WW—é; WM Wow; ,<L)01|~1 %% . C073) “)0 T5 bv%% 1W3?) (0721) 'WWW vii}? °&%%t>aﬁ 0'33, F—y» . 5‘4 *2 — COJCm‘mb, ﬁt}. a: fanatic. ( 0.70) bmuwev l). Uéfwé émrh. (040\$ ow; mram Chm U¢EH amt) ® ®[email protected] 2.5, 131:. e #733. 595%)“ am; => 213 (W33 aloe? 0 UP a” H%&W&» W4L%%mw;thﬁv% Bay (mm (Mg) 4+9??? ahjnqzy 0‘“: 17% : “3&6: (1‘3) x0: logxxogam m oars)- mm M «a. W m. We 4 [03a (logae) ‘ e é “9903". or; (eé Ma 91”“. (“a 1%): wasi— -—~*~—~—\$l<&<68 l warm). (9 "<” H)- _L [email protected] Foam 06;, Wm ‘31) as 6.8 “t +91% ('2) Newer 11) “Dania, 45m 3% 73%. (0M xvr rung aex Emma—u) Wd »; W’X om % alum 9.13 002x 3; H311. 74 \V‘O‘: ‘V‘ 0‘ (mo 21 tat-am“ EM) M0; \nﬁ’Xv‘) ;\ ,én a..94 wuxrmum-‘g Qérui. % QVL 5%. '(170 04mm; T304“ .)£H7=o‘>o. S. 502) \$7 M‘vx, 366(0,0.\) 4qu Jam .. Lo ’- M: Eab 0.1/0 I - - _‘ - 5mg Fm, ﬁg) u, o. cleareasmg "Functmn m (9,o.\)l Vac) 4 Ho) ’4 :3) -\:Lo.\) 4 2.\ ¥Lo.\)‘¥1-H If 456 dam ’WM: ‘F(0.1)¥1.|| 75 szonlee’ 6’?) m 17W '0’? (Lac) we“) 47w. ‘>\e emar (‘4 3*) 4;“). x) a; 14 etc. wrong- 60‘\w\o\'\’jov\ @ ex) ‘FLQQ A _,.L- fl. Bic. b7 M.V-T Q r ,_ JEEP/{’80) = l l f (a) , o Are SM‘IS‘H "g ‘Rﬁbo ‘0\ “we coefﬁcients 4 CMCM 7(4):). +4; ,9 (Xq: 1" T t *- mm 1- iii , w of z . , QB wrong? answer ov- wrongr WWW/mom wrong 7Q ,....- GD l 7. [20pt] Let f(x)=x7: for x>0. [15pt](a) Find all asymptotes for the graph y = f (x) for x > O . [5pt](b) Find the maximum of f (x) . [Solution] 1 i1“ 1 (a) f(x)=x7:=eﬁ "” 1 ' .... .. x. f'(x)=e\$ (—lnx) = * 573, “15' §¥3F5 27% E5 373(ul~‘€:% (b)°lV‘i @1015 %%%i%) To find extreme values, let f '(x)=0 when x > 0. 1 Since x\$¢0 if x>0, 2-lnx=0. .'.x=e In addition, 1 l limx"; =1imeJ? and lim—=-oo. x90 x—)0 x—)0 x 1lnx slimeW =0 ---- --(2) x—)0 * 4x3, 7411:} we] gamut m eagle 27s. 1 1 lnx 11m x5 = lim e7: x—ND x->oo and ﬁrm—13 =lim =lim L0 x—WDJ; x—No 1 ix-)=n,\/;— L. 2J§ l x —l—lnx 1ime~5 =e° =1 ---- ~(3) X —}ao * 4%], 7%} 3l7§°l Ealﬂbl 3l7§-% ‘3}101‘3 2’3. Hence, we can see that y =1 is the only asymptote from the table ---- * 2’9, €347} 5391011. x=0. y=1) 07E}. 1 2 .1. (b) f(ez) =(e2)‘le—2 = (62 )2 = e2 is the global maximum value of f (x) ---- * 573, 73347} %ﬂ% 73-?- ”eljéql “til 07%;] EE 2%] \$1;— 3421. © TN ATstW "gram HM KN Ufa-ﬂ +0 the, cmté, A: \0—~U- 4’ —\\ 3% \I l’wkﬁ“ 77:9 Volume :59 TN. Sohé. ~ :ZIPEO' ~%=&UC’§3 6—3231. 1:434 ~ e—--72a,‘ 723% WM ‘3; WM '- %°—% Wt 25% 7*qu Hm a m% e“ we? ‘ 773‘ 1 f‘ [7Q 77"” W7; 69 A tome ’ " /————% 27% \w ~ 33 ' m Comm-(A 0"? a low Q) I U31: . '3‘; a Totalx [DIN yum 3»: . 437\$. ﬁc't‘" TIL-r1. V I Tim 4:9"me “PROM 1‘44, W “:11 *0 Thee/“Tm? -JE. ,‘\ hm?“ \$1: ‘0 “IT-V» :5 » ’7 27%: S\"—\rL‘-'\" ‘tt—t'z - TM QLr‘QMe’ area 09th, 91,“; :z‘EAr; E...“ .m*23=2ﬁmm+\) a 33* 7n\,<g;u\§, >oN-d6 *WL- wwv Rm) "(4091/ wAtVbTA, m3; SM ‘5‘“ WL. IO' (0‘) 8A(l*7() 9%, eXP(L*7()§ (H (D ChppbmeX) Ert’a DHDHZ), (777g) @ EI'T—lw3 (\$73) (3) a‘qowozqﬁqg (WEI) '(mam‘mde c 14-77(74CL7H2), 0Q, 71:21am 0¢%t+(mv~cle%meA) %%w 7&3)?!“ ) (C) (AhaPP‘Y ( 1‘ 70‘7‘7/0H3Mx) ‘ 7C) (675) Maplcdﬂ (/féﬂ éﬁ‘H éHg C121 0 H? 73% JAM. i% WEI-ML. 95-3 (M7114 2.5% 7,5; 30 2'5 ‘ MIST ,A4 FILE NOTE ...
View Full Document

{[ snackBarMessage ]}

Page1 / 16

solution(common) - 1[30 pt Let f(m = Edwin/m:3(3 Using an e...

This preview shows document pages 1 - 16. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online