1873_solutions

For limits that f x x 19 given that f r r and that for

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Unformatted text preview: y The function  H is continuous on the compact space K and therefore has a mininimum. Since  H y  0 for every y K there must be a positive number  such that  H y  for every y H. 228 For such a number  we must have d x, y  for all x H and all y K. 12. a. Suppose that S is a compact subspace of a metric space X and that f is a continuous function from X to R. Suppose that c is a point of the set S, that f c  1 and that f x  1 for all x X S. Prove that the function f has a minimum. On the compact space S, the function f must have a minimum. Since f c  1 we know that the minimum value of f on S cannot exceed 1. Since f x  1 whenever x X S, we conclude that the minimum value of f on the subspace S is the minimum value of f on the whole space X. b. Prove that if f is a quartic polynomial of the form f x  x 4  ax 3  bx 2  cx  d for every number x then the function f must have a minimum. Since f x  x 4 1  a  b  c3  d x x2 x x4 for all x 0, it is clear that f x Ý as n Ý. Using this fact we choose a number p  0 such that f x  d whenever |x |  p. p 0 p The minimum value that f has on the interval p, p is the minimum of f on the entire line R. 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 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 229 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 fail...
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