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Unformatted text preview: ANALYSIS EXERCISE 11SOLUTIONS 1. By the definition of continuity for every x [ a,b ] lim x x f ( x ) = f ( x ) . Using the Heine definition one can compute the above limit along the sequences x n x , x n 6 = x . Choose x n Q . Then f ( x n ) = 0 for all n N , and it is clear that f ( x ) = 0. 2. (a) Let f : R R be defined by f ( x ) = x 3 x + 3. Then f ( 2) = 3 < 0 and f (0) = 3 > 0. By the intermediate value theorem ( x [ 2 , 0])( f ( x ) = 0). (b) Let f : R R be defined by f ( x ) = x 5 + x + 1. Then f ( 2) = 33 < 0 and f (0) = 1 > 0. By the intermediate value theorem ( x [ 2 , 0])( f ( x ) = 0). 3. Let = f g . Then is continuous. ( a ) = f ( a ) g ( a ) < 0, and ( b ) = f ( b ) g ( b ) > 0. Therefore by the intermediate value theorem ( x ( a,b ))( ( x ) = 0). 4. Let f : R R be defined by f ( x ) = x 4 3 x 9. f ( 2) = 13 > 0, f (2) = 1 > 0. But f (0) = 9 < 0. So by the intermediate value theorem there is an x 1 ( 2 , 0) such that f ( x 1 ) = 0, and there is an x 2 (0 , 2) such that f ( x 2 ) = 0. 5. Let g : [0 , 1] R defined by g ( x ) = f ( x ) x . Then g (0) = f (0) 0. If g (0) = 0 then x = 0 is the desired solution. Now assume that g (0) > 0. g (1) = f (1) 1 0. If g (1) = 0 then x = 1 is the desired solution. Next assume that g (1) < 0. Then by the intermediate value theorem ( x (0 , 1))(...
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This note was uploaded on 02/25/2010 for the course MATHEMATIC 11007 taught by Professor Spyros during the Winter '10 term at Bristol Community College.
 Winter '10
 Spyros
 Continuity

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