Solutions to Exam II, Ma123,F04

Calculus With Analytic Geometry Seventh Edition

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Unformatted text preview: [SOLUCDJHJ EY/b/K 11:"? l. (36 pts.) f(x) = x4 --4Jc3 +1 f’(x)=4x3-—12x2=4x2(x—3):0 @ 7C:0, 7c=3 ” _ 2 _ _ f (x)—12x —24x—-12x(x—2)—.o e 530/ 7C: 7— a. Where is f increasing? Where is f decreasing? {3/ ‘0) (“Do/0) ,__ ...._ 4. (o, 3) h—l—Fl’" D 9 7‘ b. Where is f concave up? Where is f concave down? 0. State any inflection points on the graph off. (OI’F{”)) = (0, I) d. Find any relative maximum and relative minimum values off. Where do they occur? 37/ Finsroammma new, 12mng M 3 bF- 43(3): -2; (M; psalms: MA—x Vfiwfi’; (W 4‘) 6. Find th maximal value of f on [-1, 4] Where does it occur“? $ @ 7C‘:—/ 1 ~24 I f. Find the in1ma va ue of f on [-1,4]. 7‘ @ 7c=.3 *26' 4:770: 2. (12 pts) The Regiomontanus Problem. A painting is hung flat against an art museum wall, with its bottom and top edges at distances 6 ft and 10 ft, respectively, from the floor (so the painting is 4 ft tall). The painting is viewed by a tourist whose eye level is 5 ft from the floor. The Viewing angle of the painting by the tourist depends on the distance, x, she is from the wall. In particular, it can be shown1 that the viewing 'angle is biggest (making the painting appear as large as possible) when the quantity 4x x2 +5 f(x)= X20 51522 'K /3 d AW: is maximized. Determine the distance (in feet) that the tourist should stand from the wall to maximize her viewing angle of the painting. (wuquL «9(a)» aw fF{R)-?o 4‘5 K—aoo) («1+ 5)." S‘KK‘I-S)“ 4712-16) (704- 5)‘ 2—D @‘K [5“ l (WWAM‘YH 7c: “If 45 1 (3 GAGA-«42.) (C— i; Mylmn‘x-é 44'- 7(7'5 1 Nahin, Paul (2004), When Least is Best, Princeton University Press, Princeton, NJ, pp. 73-77, 147-148. 3. (21 pts.) Determine the following antiderivatives. a. j3x6—4x4+7x2—9dx 22:7...fifif- E-?’)¢ 4’6: 7 5 3 b. f(x+1)(3x—2)dx —; S (gm-mm xx) 47‘ 1 'Kg—f-ff—‘LF‘FC z = 3‘... + 1. +,_ (5/27) 0*») (’3) 2. 5’1 3/1- ??C +—>< +1.7: +< 4. (10 pts.) Approximate the value of «[57 using differentials (note that «fit-5 = 7). Wm.) c. cox). km.) 1/704; L770: L‘X—I/q‘ 5') 9- ‘l 7.6- ——L j a. ': 7+—% 28 5. (12 pts.) r False (circle one). (PM M m an: WESW ///2.) b. There - e no general formulas for zeroes (in radicals) of polynomials of degree 5 or bi gher False (Circle one). (pa, 555 F- 11!) / c. If f is continuous and f (a) f (b) < 0, then f has a zero in the interval [a, b]@ or False (circle one). I ihere are general formulas for zeroes (in radicals) of polynomials of degree 4 or less. 6 INTEAWI/mz mutt. WW) - u u - I . A, d. All antlderivatlves of a given functlon must differ at a most a constan m r False (circle one). [5% (.242. mums Mam) 6. (14 pts.) A running track consists of a rectangle with a semicircle at each end, as shown below. If the total perimeter is to be exactly 400 meters, find the dimensions gx and r) that maximize the area of the rectangle. (Note: A circle of radius r has area m2 and perimeter 2m.) 7c 7-) 7<= IUD DIWEMSIW 7M7~ M/A’Ximue fLEc’mUGw/m M64} 7C:- /DD m. ...
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This homework help was uploaded on 01/22/2008 for the course MATH 123 taught by Professor Johnson during the Fall '06 term at SDSMT.

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Solutions to Exam - [SOLUCDJHJ EY/b/K 11" l(36 pts f(x = x4-4Jc3 1 f’(x)=4x3-—12x2=4x2(x—3:0 7C:0 7c=3 ” 2 f(x)—12x

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