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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‘IS)“ 471216)
(704 5)‘ 2—D @‘K [5“ l
(WWAM‘YH 7c: “If
45 1 (3 GAGA«42.) (C— i; Mylmn‘xé 44' 7(7'5 1 Nahin, Paul (2004), When Least is Best, Princeton University Press, Princeton, NJ, pp. 7377, 147148. 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 (gmmm xx) 47‘ 1 'Kg—fff—‘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 «ﬁt5 = 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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 Summer '06
 JOHNSON
 Calculus

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