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Sol-161E3-F2007

# Sol-161E3-F2007 - MA 161\‘EXAM 3 Fall 2007 Sowmus Name...

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Unformatted text preview: MA 161 \‘EXAM 3 Fall 2007 Sowmus Name ten—digit Student ID’number RECITATION Division and Section Numbers Recitation Instructor Instructions: 1. Fill in all the information requested above and on the scantron sheet. 2. This booklet contains 12 problems, each worth 8% points. The maximum score is 100 points. 3. For each problem mark your answer on the scantron sheet and also circle it in this booklet. 4. Work only on the pages of this booklet. 5. Books, notes, calculators are not to be used on this test. 6. At the end turn in your exam and scantron sheet to your recitation instructor. MA 161 EXAM 3 Fall 2007 1. The most common linear approximation of 70155 that uses the reciprocal function is M 1%“)? '32); 603W!) ' A. 01001 W «PM ‘7" ”V53“ B. 0.001003 LC%:1[(I000) + ‘5 \HION 306-1069) C. 0000997009 : 0.0m ..... 0,00000\ {XPIOOM D. 0.00099 1,0009) : 0.00|—— 0.0aooo\ [3) @ 0.000997 ’ 0. 0005397 2. The ratio ME— is identical to l—tanhm ML. Wm M WWW-~15WW :a_,___.W:__ﬂ A ,h Pin/«W \: M 050505100?" ' 8m \1\ >’~ “Sky“ X W B. cosh. .... Y\ ——>« .. +Q, e, ’8 2,6 _ m c...- 8, 1 + L _‘ T @62 e)“ are/”’7‘ 69!“ 1%? M2 6% D. (3—258 W “W ””3 2., x 2.99. E. 1 ., 74 _._ ,. é: -— (6 Met) ., e 8 A 3. If f (—5) = —1 and f’ (x) g —3, then the mean value theorem guarantees that 099 Mame [—9 J41 . nil/>3) 998+: ”3 «7" Lanﬁnmg. “W W” WW =ﬁ> Harries :‘r (a) @::2<_10 -Z- (”S”) 22—10 ’ C f(—2)g—8 l .... ﬁg (H‘Ll’w’ﬁpg) 4 _..., 13E”; 3 ,2..(_§) ‘ 3 D. f(—2)2—8 ”J? ”Li 1565‘) < .43 ,5; (Raﬁ _4(.,§\ é “C? E None of the above MA 161 EXAM 3 Fall 2007 4. Given the graph of y : f’(m) below, it follows that l '1 A. f is increasing on (O7 3) B. f is concave down on (0, 6) O. f has a local minimum at cc % 4.4 l \e l f has an inﬂection point at :c = 3 W“) WWW-(WM) K W E. None of the above O 5. 111% 1-230?” s l V \/ A. :0 \ X W (PM) x 1m ' .7: MM :64 X’>C> (l 174) XﬂBO (e ‘ A C. 2—2 h 4“ '52“ in 445;) '5: M : — keg.) ( 6 MFWJWMWWW D — 1 wn‘WWMﬁ—‘vw’ﬂ WW:2MM m #2; 13- does not exist “W “V“ MW” : ”M W 1: i :42» , W0 7 X >60 \ l MMWWWM l w v; ‘ —z 3 MA 161 EXAM 3 Fall 2007 6. Find the points on the ellipse 4:132 + y2 = 4 that are farthest away from the point 2 2., “WW L 2, ‘9: (361) +076} mm (>99) is A. (m atlwy96. ( B. < ’5 l)”: /@~leJc(Ll—W) [email protected]<_%,_ﬂ) and<_%,£_2_> ' ( < 7. A rain gutter is to be constructed from a metal sheet of Width 20 cm by bending up I one fourth of the sheet on each side through an angle 9. The cosine of the angle (9 that will result in the gutter capable of carrying the maximum amount of water is cos 6 = Chis; Sec/5&1); ﬂaw ”b: F f" kg» A_ g 619; A ‘0 8’ waectw m A r 46%; 666566663 +00)(§swi 0- l US! l E + H N (5 Z 4:WM3Q4 s 1- wWQ’Q @ «32% [136(er :: 2; C05 QSMQ «r Wéﬁmﬁ' O .1: €96 6‘77?” E. \/§2+1 l NM} ‘3 LL? (wémzﬂdrwszeb) +9) 6666 : 2,6 ( (I—Qsler) {'COJY’8) + Q) [65 a :zg ("240516 1’ 26463674) , M6320 a L099 <2 Wﬂﬂﬁl w: 6663365; a H ”Wit 9—6 @3636? “"l :36 6 yl+§f§6 62 W I 6496 m MA 161 EXAM 3 Fall 2007 8. Use Newton’s method With initial approximation 51:1 2 l to ﬁnd 332, the second ap— proximation to the root of the equation x4 ~ :1: — 1 z 0. Then, \$2 = Lﬁﬁ (awe? XL(W%W‘ \ A. % 3 W 1A\(ﬂ;tf* '4" @§ Lt Owl—a ( >< ~ ~ XLix‘nﬁix‘ .293 11% (gm) Li%.”\ 2 13.3 4 :3“ ._., I Z: i X\-‘:\ “’5 X1 ‘ 3 3 9. Given f”(:c) : sin6 + c089, f(0) = —1, f’(0) : 4, it follows that 4“??qu '27.; \\ 18 (VJ “S Sin 9' i (2.039 x> PM : ewe J;— some a» c, 490% F; Lt : —LosD +9“? at»; ~22”: 4 ﬂ”; A Q :g f\ “90"» :: «-\$M9 / (Aﬂﬁl Jr‘fg‘a 1. C] ngq : ”SMO-WCQSO +£15ng :— 4+9 ﬁC\:C> Vm’j" VHN? I 9M9» «C93 9’ ”fgﬁ” +<Exﬁrmiivwi+9(iig’%~Eg+%“” 52+? 5 10. The left—endpoint Riemann sum to estimate the area under the graph of f (at) = sing: from :c 2 O to a: = 7r using four approximating rectangles is MA 161 EXAM 3 Fall 2007 l E E t l A. 1+4\/§ B. i 'T fl E g as Tf Sm O 'i‘ 9M ‘f '1’ SIM 2/ + M "i C. 2 ; r’“ r F D’ 12‘: 1-; r“ J)» - ”1: i] : I: D L] , Lonwiwi +| + L Lf +6” @w(1:\/§) 1: 2 1 ‘é 11. The deﬁnite integral / 1 + a: div is the limit of Which Riemann sums? ‘ W ww‘tewwg O n Mamba Z _ 6 3F? 2(‘l “'(Zil' ‘13?) 2’ r ' i=1n+21 46¢} Mr (15%| W¢l¢YJf 8%; \ﬁ‘ gulomhnrav? M B n 2 21 Mama) .1». WW 2 H I ._1 TL +i “€( 7r\ 3 \+‘%—;‘: ’ n+3? n+2. 1; 2- , \A N C' l 2 ganAgﬁm 2,“?(3-‘3‘l‘zﬁl 1217“” d yr) 9° 9 Y‘ n H‘ﬁk ‘ g! 1 i . D- Z _: M 3” & (EA i=1 n+1 ‘ WEB 0° 5“” mail; E None of the above 12. Given g(:1:)=/2w glidu the value Qty/(1) is ‘04») : "(1%)an \ r at \A Wt“ A. 2 % ((2%)ZTL ML} m r ‘ B' 3 i 1 l 2: Lin; 1 4 A _,_ 0‘ § l mm (m W (a >e ® , E g ...
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