{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

Info iconThis preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

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

View Full Document Right Arrow Icon
Background image of page 2
Background image of page 3

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

View Full Document Right Arrow Icon
Background image of page 4
Background image of page 5

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

View Full Document Right Arrow Icon
Background image of page 6
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

{[ snackBarMessage ]}