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# sole11 - ANALYSIS EXERCISE 11SOLUTIONS 1 By the denition of...

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ANALYSIS EXERCISE 11–SOLUTIONS 1. By the definition of continuity for every x 0 [ a, b ] lim x x 0 f ( x ) = f ( x 0 ) . Using the Heine definition one can compute the above limit along the sequences x n x 0 , x n 6 = x 0 . Choose x n Q . Then f ( x n ) = 0 for all n N , and it is clear that f ( x 0 ) = 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 ψ = f - g . Then ψ is continuous. ψ ( a ) = f ( a ) - g ( a ) < 0, and ψ ( b ) = f ( b ) - g ( b ) > 0. Therefore by the intermediate value theorem ( x ( a, b )) ( ψ ( 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)) ( g ( x ) = 0). 6. (a) Assume for a contradiction that α 6∈ I . There are two possibili- ties: either α < a or α > b . Consider the first case, α < a . Put

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sole11 - ANALYSIS EXERCISE 11SOLUTIONS 1 By the denition of...

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