MAT265-test-3-solutions - EXAM 3 SOLUTIONS MAT 265 —...

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Unformatted text preview: EXAM 3 .. SOLUTIONS, MAT 265 — Calculus for Engineers—I EXAM 3-A Roman DWILEWICZ PART I - Multiple choice. Choose the response that best completes the statements or answers the question. Please circle your answer choice on the exam and fill in the answer grid on the last page. Show your work on the space provided. 1. Evaluate the integral belgw by interpreting it in terms of areas. I: V 36 — x2 dx. h 1' I 5% Ewe-Kick : gin/7'5”": iéy : 833‘ x o L, E? 6 A) 367: B) 187T ( C) 9n ) D) 18 E) None ofthese choices. 2. Find the absolute minimum ofthe function f(x) = xex on the interval (—00, 0]. Think about the definition of absolute minimum carefully before you pick an answer. N5 £(‘ f): " ”L :T’Q):ex+xex: e"(l+x§:o -> rat-l. e ffiGz): 6‘0”) 4- ex: 9 GEN-x) TOY-F) = e—1 >0 - fig”...(_><e‘) : O J .PfOJnO- K—~>—00 A) does not exist C) 0 D) —e E) None ofthe choices 3. For the function x3 on the interval [1,3], find the value of c that satisfies the conclusion of the Mean Value Theorem. 3 3 fab—{(61): 187¢) we 5 "I 3 3c‘2 ~> 39: 35..) L3: Io—GL 3—! 92 3 C: lag A) 2 B) 2.75 D) W E) Non‘eofthe choices 4. Find the inflection values (x values ofthe inflection points) off(x) = 1n(x2 + 1). ¥I(x):—J~.,2x ‘f/fi): 02(XJ+()- QX‘QX: 02({~X,Z) ’ (xmw (my? l— ><°2=O M» 961/ m “:1 B) ix/E C) ian D) -_I;1.5 E) None ofthese choices. Fall 2014 © School ofMathematical & Statistical Sciences — Arizona State University 2 MAT 265 — Calculus for Engineers-l EXAM 3-A Roman DWILEWICZ 5. Set up but do not evaluate the left sum with n = 4 rectangles for the function f(x) = e" fromx=0tox=2. |___'____.’___’_._—f 6 0,5 l Ls a2 0 05 l (.5 e + e + e + e . _l_ =2 .(e0 +e05 +e1+e15) 21'(e°5 +e1+e15 + e2) C):(e0 +e05 +e1+e15) D) %(e°'5 + e1 + 61'5 + 62) E) None ofthese choices. 6. Iffo5 f(x)dx = 3 and f10f(x)dx— = 4, Whatis‘ the value offs10 f(x)dx? $19604“ 8:}(x3Jx- S4644“ —4 —5 =— ¥ 5 o o A) 7 C) 1 D) -1 E) None ofthe choices 7. You want to build an enclosed rectangular area next to a pre-existing stone wall. You have 1000 meters of fence available for the three sides that need to be built. What is the maximum area (in square meters) you can enclose? a 92K +3: {000 -v> a: (OOO—Jx my Fw'excsénj s-{we Wee £160: (000 — 4x : O K: 9250 M. KOOOO- 02x> ll .F/(XX Cllmmaif'S sch». new x : £50 M149 '— 1131.1. £6250): ,250. 5’00 (WM >‘ = («25 000 A) 62,500 B) 125,000 C) 250,000 D) 31,250 E) None of these choices Fall 2014 © School of Mathematical & Statistical Sciences - Arizona State University 3 MAT 265 — Calculus for Engineers-I EXAM 3-A Roman DWILEWICZ 8. An engineer is given the task of designing a sturdy box-shaped closed container whose base rectangle has one side twice as long as the other side. The base ofthe container must be made from 1" thick titanium plates. For the sides, it suffices to use titanium plating 0.5" thick, and for the top, 0.25” thick. Cost per square meter of titanium plating is proportional to thickness and the 1” thick plates cost $8000 per square meter. What is the minimum cost ifthe volume ofthe container is 2 cubic meters? Round to the nearest cent. .3 \ V: 2xb=jm3-~>b= P 6031‘ 3 ,F[X) : .2 X2~ 3000 + er‘ agOOC) ~—-——-»- I___v—- 130.3? 473/0 L‘ +é2-o9_3<+.2-J‘3>4000 K K . w____________, 5051.63 +4)” 20'0sz + £4.29“? 11> >46): éoocox— 243190 x3“ ’ 3 3: O N 2‘ :2 4 00C) X ‘ ’4 ( £0. > x .quQ > IE, 143 F); ~4,2 69.2 m A) $42,682.72 B) $12,926.60 C) $34,470.96 D) $27,359.61 E) None ofthese choices. 9. Find the absolute maximum and absolute minimum ofthe function x3 — 2x2 — x on the interval [0, 10] and round them to three decimals. f“ 2 Vi" _ 2+ 4 _ '- fiCx): X3“ on‘z-x -> .FG‘): 31(2- 4x —( : O M> Xv 3 sz— ——3—- W GDA’HMJ (057- 0 40 €C2+F$> R5 .4231 ev— "Mn. A abs. max.— - 790, abs. min=0 B) abs. max. = 800, abs min=0 bs. max. = 790, abs. min=-2.631 D) abs. max. = 800, abs min=-2.631 None of these choices. 10. Find the antiderivative F (x) of f (x) - 2x“ that satisfies F (1) - 1. M9 L‘vx‘L'Mcva/g (o, 00> WK): 3—2;} +34%; M> F36<3= fibx — 33—}— + C FOL-éLJ— i+C=l~>C=fL .1 3 5 A)F(x)="::"—1 B)F(x)=xx:x+1 Fa): Egg“):— .L + it C)F(x)=31nx—%+1 3 ' 3x 3. E) None of the choices Fall 2014 © School of Mathematical & Statistical Sciences — Arizona State University 4 MAT 265 — Calculus for Engineers—I EXAM 3-A Roman DWILEWICZ PART II - Free response. Show all your work 1. Suppose that f(x) = 2x6 — 5x4. Use interval notation to indicate whereflx) is concave up and concave down. Justify your answer with the concavity test. [10 pt] 1960: l£x5_20x3 {”(x): 6m”— goxfl 60,568— I). Concave Up: 6- 9°,fl'f), ((J 0") ¥Ilg>>0 4:) an (4WJ*()/ ((1%) ConcaveDown: ("f/O), (Q1!) ,6” (A40 4:“) ”7‘ <' {J O); (O; (J, (A-ccep’fotbfe (Mane/W: [-f, [I #H(K3\<O <=> 671 [-lJ ‘1 (NC-4J0) 2. Consider the function f(x) = 3x5 — 15x4 + 20x3 on [-1,4] . Find the absolute maximum and absolute minimum and the location of each and justify your answers by showing appropriate work. [10 pt] ¥I(x3: 15X 4 _ 60X) 4. QOK'E‘ AbsoluteMaximum: 5192 Occurs at: é‘ ~ '2 2 . . : "'36? : .. ( = ’5 X ( X __ 4 X + 4f > Absolute Minimum Occurs at = «mice-.2)" W: £551.12 04/“ ,f/(KJZU an [—{JA’IJ Aft-Ora? 50 {3(4) : ~39 M W): 5(2. ((2,) = 5192 max 3. Let the function f(x) = ace—2"2 Use interval notation to state where the function is increasing and decreasing. Justify your answer and show all work. [10 pt] ,3_ e / -.2K "2" l 72 6:3: 8 4' X (' 4x) 6 Increasing: (‘2'?) 2) .2. __ = eaix <( _ 4K2) Decreasing: éwl‘éé (3)”). {’64) >0 (=> (—4x°’>o<=> xe< ZIL<:":<X<—' .z_ I ¥G<) <0 <=> (—4x‘z<o (=> 43> ( <:—> xl>¢<m> x <11, ovx>§L Fall 2014 © School of Mathematical & Statistical Sciences — Arizona State University 5 MAT 265 — Calculus for Engineers-I EXAM 3-A Roman DWILEWICZ Name (Last, First): Answer to multi . le choice .art Iuestion number inswers; A, B, C, D, or E Please use ca oital HIHII O>ww> 10 :0 Multiple choice points Free Response pomts Test # 3 total points Fall 2014 © School of Mathematical & Statistical Sciences — Arizona State University 6 ...
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