2011_Solution_2 - MA 1505 Mathematics I Tutorial 2...

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MA 1505 Mathematics I Tutorial 2 Solutions 1. (a) y = x + 1 x 2 + 1 , x [ - 3 , 3]. y 0 = 2 - ( x + 1) 2 ( x 2 + 1) 2 and y 0 = 0 if x = - 1 ± 2. So critical points are x = - 1 ± 2 and endpoints are x = ± 3. y 0 < 0 if - 3 x < - 1 - 2 , = 0 if x = - 1 - 2 , > 0 if - 1 - 2 < x < - 1 + 2 , = 0 if x = - 1 + 2 , < 0 if - 1 + 2 < x 3 . Hence y is decreasing in [ - 3 , - 1 - 2), increasing in ( - 1 - 2 , - 1 + 2), and decreasing in ( - 1 + 2 , 3]. So local min is: y ( - 1 - 2) = - 1 2( 2 + 1) , y (3) = 2 5 and local max is: y ( - 1 + 2) = 1 2( 2 - 1) , y ( - 3) = - 1 5 . Since - 1 2( 2 + 1) < - 1 5 < 2 5 < 1 2( 2 - 1) , so absolute min. is min x [ - 3 , 3] y = - 1 2( 2 + 1) at x = - 1 - 2 and absolute max. is max x [ - 3 , 3] y = 1 2( 2 - 1) at x = - 1 + 2. (b) y = ( x - 1) 3 x 2 , x ( -∞ , ). y
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This note was uploaded on 01/05/2012 for the course MATHEMATIC MA1505 taught by Professor Freudleong during the Spring '10 term at National University of Singapore.

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2011_Solution_2 - MA 1505 Mathematics I Tutorial 2...

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