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Sol-165E3-F2011

Sol-165E3-F2011 - MA 165 Exam 3 01 Fall 2011 NAME...

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Unformatted text preview: MA 165 Exam 3 01 Fall 2011 NAME earn-tress 10—DIGIT PUID REC. INSTR. _.—__.____. REC. TIME LECTURER INSTRUCTIONS: 1. There are 8 different test pages (including this cover page). Make sure you have a complete test. 2. Fill in the above items in print. Also write your name at the top of pages 2—8. 3. Do any necessary work for each problem on the space provided or on the back of the pages of this test booklet. Circle your answers in this test booklet. No partial credit will be given. 4. No books, notes, calculators or any electronic devices may be used on this exam. 5. Each problem has 8 points assigned. 4 points are given for taking the exam. The maximum possible score is 96+4=100 points. 6. Using a #2 pencil, fill in each of the following items on your scantron sheet: (a) On the top left side, write your name (last name, first name), and fill in the little circles. (b) On the bottom left side, under SECTION NUMBER, put 0 in the first column and then enter the 3—digit section number. For example, for section 016 write 0016. Fill in the little circles. (c) On the bottom, under TEST/ QUIZ NUMBER, write 01 and fill in the little circles. (d) On the bottom, under STUDENT IDENTIFICATION NUMBER, write in your 10—digit PUID, and fill in the little circles. (6) Using a #2 pencil, put your answers to questions 1-12 on your scantron sheet by filling in the circle of the letter of your response. Double check that you have filled in the circles you intended. If more than one circle is filled in for any question, your response will be considered incorrect. Use a #2 pencil. 7. After you have finished the exam, hand in your scantron sheet @131 your test booklet to your recitation instructor. MA 165 Exam3 01 Fa112011 Name:__;____n.___ Page 2/7 (8 pts) 1. Find the absolute maximum and absolute minimum of the function f(x) = 3m4 — 4033 — 12.ch2 +1 on the interval [~2, 3], without specifying the value of :L' which attains the absolute maximum or absolute minimum. / 3 2_ 95‘ A. absolute max 33 absolute min —31 §(X)*12x .. 12X 247‘ : 1 2X ()6: X _ 2) B. absolute max 28 absolute min —4 I 12X OH 1) [702) C. absolute max 28 absolute min —31 / g 09:0 : x:-1) 0/ 2. D. absolute max 33 absolute min 1 {[4} :‘ 3 +4, ,2 +1 ; - 4- E. NO absolute max absolute min —35 O) r 1. g1):3rlé ‘4'8”2 ”42' 4+1‘ ' , .2 4'8" ,_ 32 _1'2‘4. +1 =*-»31:’Ié"a)2§x ml”? {(—2}: 3.15— 498 ~124+L ; 43 1- 32-428 +4 =33 ‘1’“‘3’5’ “my {(3) I '3. 81— 4- -27 «19*? +1 : 24.3 ' 108~M8+ 1 $23 1:2 (8 pts) 2. If f (cc) 2 cc _ 1, then find the values of a: at which f has a local maximum or a local minimumh‘ Damian W we 3‘04 3 XL“ % 7. A. local max at m = 2 local min at a: = O gray; W 7:2,.)‘4: ”333.: B. localmaxataszl localminatmzZ [X’ifL 60") TX") / "76' C. local max at x = 0 local min at a: = 2 SEKW/aw ; ><:0, ><=L I D. local max at a: z 1 local min at a: = 0 $67 +9~+ +0-...w’lwumio +++n a E. localmax atsc=0 NO localmin ‘ WM —W.i;rx.+m»WH-nwwem 0 i if; 2 >6 (8 pts) (8 pts) MA 165 Exam 3 01 Fall 2011 Name: 3. If g(:c) : 4:123 — 33:4, which of the following statements are true? (1) g is decreasing on (1,00). (2) g has a local minimum at a: = 0. (3) The graph of g is concave downward on (—00, 0). 4 :4xfi-a 3X (30’) "a A- (1), (2); and (3) ( ‘2“ V 1 I? " We)” i” - fl" (1 7° B. (3) only () (If? 1...: + + + 1 1' giiiiiagw:w;:~;;w“§ C. (1) only 0 4— (1) Q) T 2. D. (1) and (2) only ) 0,7 mot iau. a ('2- ? $613. (1) and (3) only ’3 0!): May—aw =.4<2><(2—3>(), // ...- gm; ~ ~ -_.._..9::i::$.;:..;; 1;. o a V (53;) in {rug 3 4. Suppose that the second derivative of the function f (as) is given by ‘5‘” 3, ‘, r-—~--#«~A f”(w)=(w+p3):*(wi1)3(w 7 1)6Q:§)7(x —. 5). How many inflection points does the graph of y = f(a:) have ? A. None // L, ‘; +""a+’ gm 41,111-11... gee—ea B_ 1 -1 3 C. 2 961). 3 E4 Page 3/7 MA 165 Exam 3 01 Fall 2011 Name: ________ Page 4/7 (8 pts) 5 If f’(1:)= g’(m) for all a: on the interval (0, 8) and f(1)— 9(1) 2 2, then determine the value for f(5)~— 9(5 ). A. 5 ){32 C —5 W F(>€?~7C€'>1/~3W D 1 Wilma/n {w == 5" mi :3 (a; weE. can not be determined from the above information only £09 Lgm): covwf 0m [0,?) {;(1)'%(1) : 2 30(52'56) : ‘2 (8 pts) 6. Compute the following limit _ 62“ — 1 —— cc 11m ilk—+0 :62 Warning: In the numerator, the power of e is 233 and not a3. e A. DNE, gm 69 X~l x 11+. ’11 96 1 XHD Z I: m M W B- “ 0 X X—veo 2X 2 __¢ C. 1 o /,1 62"! X L’H :2 D 2 'M .. (L : m 11.35:}... 1; E x’rOr >< x~+m 2p 0° 0 \m - l D ‘23; ) ’ gxwwl/P , e «by if , 26 a NH» 771 w w...“ maWW-asv— , X—fiO' >1 XMG" (2'3) 0 \écr (8 pts) (8 pts) MA 165 Exam 3 01 Fall 2011 7. Compute the following limit Name: lim (1 (IF->00 2' X gmfllvfiy _ (its: : € X—wm me LH/ifg): Lm X—v-ao )(ww (3:17;; C! 8. Compute the following limit . . ., g L??? (I+~§'/)>{ :gfiiafivx {W 2>$ +_ SE €[email protected](it"§) A. DNE B. e (ifwmlljg— 6% C. 0 _, X—l bay ‘, . 1 x 1 oo — e0 X92 OM) X 1 , .12. 96B — LH , o 2 2 {WM X' 17 MWX~1 1 sz (X4)! “ C' ‘ I, +f—MX 3 L" m “an y L/IL/ Ti— D. l X‘Ii /—_L + W: 3% .L 4 X ’ J. 0 _, 1m X 4. , M :— X-WL, l+ X 2 Page 5/7 (8 ptS) MA 165 Exam 3 01 Fall 2011 Name: 9. Which of the following is the graph of the function fix) 2 A. {132 Page 6/8 ? $2+9' $615. ‘7‘ SW": 03ml ”FM“ 45930 (8 pts) MA 165 Exam 3 01 Fall 2011 Name: 10. Find the point on the line y = 2m + 3 that is Closest to the origin. 3" ' >6 m . ‘2. am 9?: \{x“++\x"t+.n><+<? 1% B W 3;} -: Ex‘z‘wzx 5%” C (42‘: ”ifiimw (10X +4“: d” ‘2 5W+22~r+$ D iii-“£1 5 tax +1210 11 xx”? d‘x d? Q‘fi1’wfi’ «v «3‘: “A E. due -5; ”>2. ’ € 10? ‘ G n. (1me 6 f: I” m~w+g‘ Xi'? Lt) 1??! 5, +3 g Page 7/8 332 yZ . Z‘i‘g—l A:{2><)§?~*9) Aré’x‘g A. 20 '3 L 1, B. 16 :9 g... 2; :2 4w><> 8’ f ‘1’) ”1 9&0. 12 (9:..A 4px?“ 2. D. 8 A14y.3 4-?(1 Msz E 4 «5%, + + + + 0 “ :‘M‘ “with...m.,.,,u..,.,.,_.,-§ T {:2 when hwflt‘m ”Wiggy W’Mli)! lg :36 .E E ;. I2 MA 165 Exam 3 01 Fall 2011 Name: ___—.________ Page 8/8 (8 pts) 12. A cylindrical can (With both top and bottom lids) is to hold ZOOOCm3 of oil. Find the radius of the can that Will minimize the cost of the metal to manufacture the can. v V0 lame V2: tweak Trvrtkzwoo $5133. ”3% 7r (:3 {3‘ AWOL A: ZWVm-t ‘2th 10 14 '4’» , w" 25?qu B. — 3:33 “ ALZ‘W‘V +277 "Tm 7r 20 3 M30 0 11“ :ng 0%" V ‘7 D. __ 7r id; «:3: Igpcg m it”, {0 d?“ '37 REF E 3&9 7r fill w=~«_+_j~f+++ CDC m fi/‘C '3 flat m M ...
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