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101-2008&amp;2009-2-M20-May2009

101-2008&amp;2009-2-M20-May2009 - KUWAIT UNIVERSITY...

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Unformatted text preview: KUWAIT UNIVERSITY Department of Mathematics and Computer Science Math 101 10 May, 2009 Calculus (1) Second Midterm Time: 90 minutes Answer all questions. Calculators and Mobile Phones are not allowed. 1. (3pts) Let f(t) = t2 sint. Find f”(0). 2. (3pts) Suppose y = f(:r) is given by m2 + y2 = 2y. Find the critical numbers of f. 3. (3pts) A car starts at a point A and travels east at 80km/hr. At the same time another car starts at A and goes north at 60km/ hr. At what rate is the distance between them changing one hour after the cars start? 4. (3pts) Find the linear approximation of f(9:) = V2117 + 7 at a = l. in . (313m) Let ﬂan) = 6(2 ~ 3:)4/3 + 5. (a) Show that there is no (2 E (1, 2) such that f’(c) = 0. (b) Does this contradict Rolle’s Theorem for f on [1, 2]? Explain. 6. (lOpts) Let f(x) = x4 —61:2 + 2. (a) Discuss the symmetry of the graph of y = f (x) (b) Find the intervals on which f is increasing and the intervals on which f is decreasing. Find the local extrema of f, if any. ((3) Find the intervals on which f is concave upwards and the intervals on which f is cencave downwards. Find the inﬂection points of f, if any. (d) Find the absolute extrema. of f on [—2,3]. (e) Sketch the graph of f. O. CO Solutions 1. f’ = 2tsint+t2 cost (1pt), f” = 25int-1—4tcost—t2 sint (1pt), so f”(0) = 0(1pt). . 2.7: + 2313/ i 23/ so 3/ —I—y.(1pt) Now, y’ = 0 <2» :3 = 0 and does not exist <:> y : 1 x * ::1(1pt:). Critical points are 0, i1(1pt). Let ac be distance travelled by ﬁrst car, y distance travelled by second car after L hours Distance between them is =x/zz:2 +y§1pt , = TM +2 I 1pt so when t— — 1, 2: =80 y: 60, 32' = d—Dt ’12“)? 80, y’ : 60, dD/dt : 100(1pt)2. - f’ : 2. 51701302“ so L = M) + f’(a)(x — a)(1pt) : 3+ 1(22 ‘1)(1pt). . ’:~63 2—1131/3 1pt :0 4:) 27:2,sonoc61,21pt.Noticethat 3 f(1) = 11 # f(2) : 5 so hypotheses of Rolle not satisﬁed(1pt). (a) (1pt) f(~3:) : f(x). f even, graph symmetric about y—axis. (b) f’ = 43:3 712:5 = 4\$(:c2 — 3). f’< <0f0r,r< ——\/§0rin(0\/_); f’>0f0rz>\/§0r(—in(\/_,0)(1pt)?' Critical points are 0 (legal rriax)(1pt) and :lzx/g, local min (1pt)1(+y’j) ._ .1 o - (c ) f”— — 12(m2 7 1)(1/2pt). f” > 0 for III)! > 1 (so concave up) and f” < 0 on w, "W (— 1 1) so concave down(1pt) Also i1 are points of inﬂection(1/2pt). {(1%) _ -3. (d) f(-2 ) : —6, f(i\/._): —7, f( 0) — 2,f(3 ) : 29(1pt). Absolute max at 3, min at i\/_(1pt). OO 99 ...
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