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8 - Math 310 hw 8 solutions Friday 6 Nov 2009 18.4 18.6...

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Math 310 - hw 8 solutions 18.4, 18.6, 18.10; Friday, 6. Nov 2009 18.4 Let S R and suppose there is a sequence ( x n ) n in S that converges to a number x 0 / S . Show that there exists an unbounded continuous function on S . We define f : S R by f ( x ) := 1 x - x 0 for x S. Then f is continuous, because it is the composition of the two continuous functions: x 7→ x - x 0 and y 7→ 1 /y . We now verify that f is unbounded. Let M > 0 be given. Then since x n x 0 there is N such that for all n N , n > N implies that 0 < | x n - x 0 | < 1 M . Let n N and suppose that n > N . Then since | f ( x n ) | = 1 | x n - x 0 | > 1 1 /M = M, f is unbounded on S . 18.6 Prove that x = cos x for some x in (0 , π 2 ). We suppose that the function x 7→ cos x is continuous on R . Let f be the R -valued function defined by f ( x ) = x - cos x for x R . Then f is continuous since it is the difference of continuous functions. Observe that f (0) = - 1 < 0 < π 2 = f π 2 . Since f is continuous on [0 , π 2 ] the Intermediate Value Theorem (18.2) applies and we conclude that there is x (0 , π 2 ) such that x - cos x = f ( x ) = 0. Hence, x = cos x as desired. 18.10 Suppose that f is continuous on [0 , 2] and that f (0) = f (2). Prove that there exist x, y in [0 , 2] such that | y - x | = 1 and
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