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HW2 Solutions

# HW2 Solutions - Math 131A Analysis Summer Session A...

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Math 131A Analysis Summer Session A Homework 2 Solutions 1. Find all solutions to the following: (a) | x - 3 | - | x + 6 | = 5 (b) | x - 3 | ≥ | x + 6 | (c) | x 2 - 8 | = 7 x - 20 Solution. (a) We have 4 cases to check. Case 1: x - 3 0 and x + 6 0 . This case is equivalent to x 3 . 5 = | x - 3 | - | x + 6 | = x - 3 - ( x + 6) = - 9 A contradiction. Therefore no solution exists in this case. Case 2: x - 3 < 0 and x + 6 0 . This case is equivalent to - 6 x < 3 . 5 = | x - 3 | - | x + 6 | = 3 - x - ( x + 6) = - 2 x - 3 Thus x = - 4 is a solution. Case 3: x - 3 < 0 and x + 6 < 0 . This case is equivalent to x < - 6 . 5 = | x - 3 | - | x + 6 | = 3 - x + ( x + 6) = 9 A contradiction. Therefore no solution exists in this case. Case 4: x - 3 0 and x + 6 < 0 This case is impossible. The solution set is {- 4 } . (b) We have the same cases to check: Case 1: x - 3 0 and x + 6 0 . This case is equivalent to x 3 . x - 3 = | x - 3 | ≥ | x + 6 | = x + 6 which implies - 3 6 A contradiction. Therefore no solution exists in this case. Case 2: x - 3 < 0 and x + 6 0 . This case is equivalent to - 6 x < 3 . 3 - x = | x - 3 | ≥ | x + 6 | = x + 6 So 2 x ≤ - 3 or x ≤ - 3 2 . Since in this case, we also have - 6 x < 3 , the values - 6 x ≤ - 3 2 all represent solutions. 1

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Case 3: x - 3 < 0 and x + 6 < 0 . This case is equivalent to x < - 6 . 3 - x = | x - 3 | ≥ | x + 6 | = - x - 6 So 3 ≥ - 6 , which is true for every x in this case. Thus the values x < - 6 all represent solutions. The solution set is { x R : x ≤ - 3 2 } . (c) We have 2 cases to check. Case 1: x 2 - 8 0 x 2 - 8 = | x 2 - 8 | = 7 x - 20 Thus 0 = x 2 - 7 x + 12 = ( x - 3)( x - 4) . So x = 3 and x = 4 are both solutions in this case since 3 2 - 8 = 9 - 8 = 1 > 0 and 4 2 - 8 = 16 - 8 = 8 > 0 . Case 2: x 2 - 8 < 0 8 - x 2 = | x 2 - 8 | = 7 x - 20 Thus 0 = x 2 + 7 x - 28 . The roots are - 7 ± 7 2 - 4( - 28) 2 = - 7 ± 161 2 But - 7 - 161 2 2 > ( - 7 - 12 2 ) 2 > 8 and - 7+ 161 2 2 > ( 5 . 68 2 ) 2 = 8 . 0656 > 8 so these roots cannot be solutions. The solution set is { 3 , 4 } . 2. Let A = {| 3 x 2 - 2 | : - 1 x 1 } , a bounded set. Use Calculus to find inf A and sup A . Solution. Note | 3 x 2 - 2 | = 0 precisely when x = ± 2 3 . Since | 3 x 2 - 2 | ≥ 0 for all x and ± 2 3 [ - 1 , 1] , inf A = min A = 0 . Also, 3 x 2 - 2 0 precisely when x < - 2 3 or x > 2 3 . Thus | 3 x 2 - 2 | = 2 - 3 x 2 when - 2 3 x 2 3 . The function f ( x ) = 2 - 3 x 2 is differentiable on ( - 2 3 , 2 3 ) with derivative f ( x ) = - 6 x . Thus f is decreasing on ( - 2 3 , 0) , increasing on (0 , 2 3 ) and has a local max 2 at x = 0 . On ( - 1 , - 2 3 ) and on ( 2 3 , 1) , f ( x ) = 3 x 2 - 2 has derivative f ( x ) = 6 x . Thus f is decreasing on ( - 1 , - 2 3 ) , increasing on ( 2 3 , 1) , and | 3(1) 2 - 2 | = | 3(( - 1) 2 ) - 2 | = 1 . Thus sup A = max A = 2 .
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