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assignment0_sol

# assignment0_sol - MATH1211/10-11(2/Asol0/NKT Department of...

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MATH1211/10-11(2)/Asol0/NKT Department of Mathematics, The University of Hong Kong MATH1211 (2010–11, 2nd semester) Suggested solution to Assignment 0 Note : Some solutions given below are outline only. You may need to give more details in your solution. The parts that are typed in red color are remarks for you to read; they are not a part of the solution. 1. (Easy problem!) (a) f 0 ( x ) = 6 x 2 - 3 2 x - 4 x - 5 . (b) g 0 ( u ) = (sin u + u cos u )(5 + cos u ) + u sin 2 u (5 + cos u ) 2 = u + 5 sin u + 5 u cos u + sin u cos u (5 + cos u ) 2 . (c) h 0 ( t ) = 2 e 2 t - 2 t t 2 + 1 . (d) k 0 ( s ) = 2 s 1 + s 4 . 2. f 0 ( x ) = 6 x 2 + 6 x - 12 = 6( x + 2)( x - 1). Critical points are x = - 2 , 1. At critical points, f ( - 2) = 21 , f (1) = - 6. At end points of domain [ - 3 , 3], f ( - 3) = 10 , f (3) = 46. f 00 ( x ) = 12 x + 6 = 12 ( x + 1 2 ) = f 00 ( x ) < 0 when x < - 1 2 , = 0 when x = - 1 2 , > 0 when x > - 1 2 . (a) f 0 ( - 2) = 0 and f 00 ( - 2) < 0 = f attains relative maximum at x = - 2 with value f ( - 2) = 21. f 0 (1) = 0 and f 00 (1) > 0 = f attains relative minimum at x = 1 with value f (1) = - 6. At left end point x = - 3, f 0 ( - 3) > 0 = f attains relative minimum with value f ( - 3) = 10.

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