1873_solutions

# X and the fact that v is a neighborhood of f x we

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Unformatted text preview: he function f fails to be closed. We can define f x  x for 0  x  1. 4. Give an example of a function f that is continuous on a bounded set H such that the range f H of the function f fails to be bounded. We can define f x  1 for 0  x  1. x 5. Prove that if a set H of real numbers is unbounded above and f x  x for every number x in H, then f is a continuous function on H and f fails to have a maximum. Since the range of f is the set H which is assumed to be unbounded, the function f must be unbounded above. 6. Prove that if H is a set of real numbers and a number a is close to H but does not belong to H, and if we define 1 fx  |x a | for every x H then f is a continuous function on H but f has no maximum. Solution: We can see that f has no maximum by showing that f is unbounded above. Given any positive number q the inequality fx q says that 1 q |x a | which holds when |x a |  1 q But since a is close to the set H we know that there do indeed exist members x of H for which the inequality |x a |  1 q holds. Therefore there are members x in H for which f x  q and we have shown that f fails to be bounded above. Exercises on Continuity of Functions on Intervals 1. Given that S is a set of positive numbers and that f x  x for all x S, prove that f is a one-one continuous function on S. Prove that S is an interval if and only if the set f S is an interval. Hint: Once you have shown that f is strictly increasing and continuous on S, the fact that S is an interval if and only if f S is an interval will follow from the Bolzano intermediate value theorem and this theorem. 2. Prove that there are three real numbers x satisfying the equation x 3 4x 2  0. Solution: First look at a sketch of the graph y  x 3 199 4x 2 4 2 -2 1x -1 2 0 -2 -4 We now define f x  x 3 4x 2 for every real number x and observe that f 2  0 and f 1  0. Therefore Bolzano’s intermediate theorem guarantees that the equation x 3 4x 2  0 has a solution between 2 and 1. Since f 1  0 and f 0  0 we know that there is a solution of the equation between 1 and 0. Finally, from the fact that f 0  0 and f 3  0 we know that there is a solution of the equation between 0 and 3. 3. Is it true that if a set S of real numbers is not an interval then there must exist a one-one continuous function on S whose inverse function fails to be continuous? Not at all. In fact, we know that if S is closed and bounded then every one-one continuous function S must have a continuous inverse function. Look also at the case in which S is the union of two open intervals that do not intersect with each other. If f is a one-one continuous function on S then the range of f is also the union of two open intervals that do not intersect with each other and the inverse function of f will be continuous. 4. Is it true that if a set S of real numbers is not an interval and is not closed then there must exist a one-one continuous function on S whose inverse function fails to be continuous? Hint: Look at the cas...
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