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241ex2dsolution

# 241ex2dsolution - Calculus I(h-‘Iath 241 — Test 2(No...

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Unformatted text preview: Calculus I (h-‘Iath 241) — Test. 2 (No Work — No Credit) Problem 1. [id Points] Show that jet-r] = '3 — I? -+ 3.1: — E: has :1 acre in the interval [13.2] [a complete argunierit is asked tor]. L'se Newton’s nzethod to ﬁnd approximately where ptr] vanishes. Its ﬁrst guess. use on = ‘1. Improve the gusss once. [-5 points extra credit for showing that p{:r] has only one real roots Problem 2. :9 Points: Let .,"'::n::l he deﬁned on an open interned E. State in precise mathematical terms 1:vhat it means that ﬂat} is concave up at e t: I. Problem 3. [213 Points] lConsider a round hex 1:vith top and lie-item. its interior is subdivided into S wedge Like seetors. The total urea of material used is .9. How should its radius r and height. fr. he chosen so that the volume of the he}; is maximized? Find hf}: Problem =1. {'33: Points] Consider the function ﬁx] = mat}? -l— Uta: — El :s': — 3'33 — 23:2. Discuss the graph of f. .lustii'v your answers. 1. Find the intereepts of f with the coordinate axes and the intervals on which the function is pt'isitjve. respeetively negative. 2. Find the critical points of the function and the intervals on which f in inertsising, respectively decreasing. 3. Find the intervals on whieh f is cormave up. 1‘espemivt:l_v concave down. 4. Find the local minima. maxima and the inﬂection points of f. s. L's-e apex-ordination of differentials to estimate the value or" f at —l.1 and at 2.1. I3. Find the absolute extrema en the interval [—1.1.2.1:. 1’. Sketch the graph of f based on your computation. Prohlem 5. :‘EU Points] lvlaxinlizc the volume of a right circular cent: in— scribed in a hall of radius R. Setup of the problem only! Sketch the setting. introduce notation1 reduce to one independent variable. state the mathemat— ieal optimization problem, including the interval ot sensible values for the independent variable. [5 points extra credit for a sointion.) 1H, (Ere): a5 ; 1am : <3:ng 1g: 5 (9 +36 a qﬁmwﬁdﬂm} T5 COG-inuumgh [0.2:] . XML This]; 0 4; “13- TM vajrt‘rmLC-‘Jule UQLAL I‘MWMI‘MVL‘W “kﬂd WW 73‘1“ EU” mm“ 0 “ML ‘2; \$6“ 33% (:5; £0.23 m vaL We} :a, (F{1\=.-{~*l+3 '23 2-‘2. I' TI{>‘]:3>ﬁl-£x+3 rpm: -21.: :q. aria] _. ”2— _. 3 Xi: YJ_| __ ____. __ 2- 3: Dgﬁﬁ + 521mg + Li'(3~‘)a : a”, + (&ﬁ+é’)‘re‘ 6‘ 3-52an F (Jim—WV ..r 3‘0?“ v; :: TI" (3;-o‘iﬁww‘") - ahaﬁuk LMEDKIMUM 1n :‘u—L-K—Om ukue ~ |+~SJV€LQ~G 53‘th Qua \ Q V 12‘”; «New few demc-W, : 3%???) E(£F+ ﬂ EYE-fr [2, I:— Fr/QE : I3f§ﬁ*£} : N985) 1’ E gum) t4 # 0 (’an erwaalr U“ 9‘; ‘33? &. =§EQ )=I1~.: ...
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241ex2dsolution - Calculus I(h-‘Iath 241 — Test 2(No...

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