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# 2ndsol - Math 119 Week 2 Solutions1 1 Guessing For given >...

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Math 119 Week 2 Solutions 1 1. Guessing δ For given > 0, we want to find δ > 0 such that | (2 x + 3) - 5 | < whenever | x - 1 | < δ ⇒ | 2( x - 1) < whenever | x - 1 | < δ ⇒ | x - 1 | < 2 whenever | x - 1 | < δ Thus, we can choose δ = 2 Showing that this δ works > 0 δ = 2 such that | (2 x + 3) - 5 | < whenever | x - 1 | < δ = 2 | (2 x + 3) - 5 | = | 2 x - 2 | = 2 | x - 1 | < 2 . 2 = 2. Guessing δ For given M , we want to find δ > 0 such that | 1 ( x + 3) 4 | > M whenever | x - ( - 3) | < δ ⇒ | ( x + 3) 4 | < 1 M whenever | x + 3 | < δ ⇒ | ( x + 3) | < 4 r 1 M whenever | x + 3 | < δ so, we can choose δ = 4 r 1 M Showing that this δ works > 0 δ = 4 r 1 M such that | 1 ( x + 3) 4 | > M whenever | x - ( - 3) | < δ = 4 q 1 M | 1 ( x + 3) 4 | > 1 δ 4 = 1 ( 4 q 1 M ) 4 = M 3. We want that f ( x ) is continuous on ( -∞ , ). Obviously f ( x ) is continu- ous on ( -∞ , 3) and (3 , ), because f ( x ) is defined as a polinomial on the ray ( -∞ , 3) or (3 , ). Thus, it is enough that lim x 3 + f ( x ) = lim x 3 - f ( x ) = f (3) to make it continuous. lim x 3 + f ( x ) = lim x 3 + ( cx 2 - 1) = 9 c - 1 lim x 3 - f ( x ) = lim x 3 - ( cx + 1) = 3 c + 1 f (3) = 3 c + 1 = 9 c - 1 = 3 c + 1 c = 1 3 4. [ - 4 , - 2) , ( - 2 , 2) , [2 , 4) , (4 , 6) and (6 , 8) 1 Please email all corrections and suggestions to these solutions to [email protected] All

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2ndsol - Math 119 Week 2 Solutions1 1 Guessing For given >...

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