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Jun10-06, 02:52 PM
P: 229
Quote Quote by mathwonk
then prove (harder) that if f is continuous on a closed bounjded interval [a,b] and f(a) < 0 while f(b) > 0, then f(x) = 0 for some x in (a,b).
Here is a hint for another way to do this problem(a different way than what mathwonk suggested).
First show that for each natural number n, if a_n and b_n are numbers with a_n < b_n and I_(n+1) = [a_(n+1), b_(n+1)] is contained in I_n = [a_n, b_n] for each n and lim n->inf (b_n - a_n )= 0, then there is exactly one point x which belongs to I_n for all n and both of the sequences {a_n} and {b_n} converge this point x. Now recursively define a sequence of nested, closed subintervals of [a,b] whose endpoints converge to a point in [a,b] at which f(x) = 0. This problem is hard I think.