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※ 引述《JimCroce (我要下五子棋 ￾N￾ )》之銘言: : Let f be a continuous function and "n" a positive integer. : Show that if 0<= f(x) <=1 for all x (- [0,1] , : then there exists at least one point c in [0,1] for which f(c) = c^n : 抱歉 可以問一下好心人 是要把定義弄熟做這些show的題目才會好上手嘛 Define g(x) = f(x) - x^n for all x in [0,1], take a,b in [0,1] s.t. f(a) = 0 and f(b) = 1 case1. a=0 or b=1 (easy) case2. a≠0 and b≠1 => g(a) < 0 and g(b) > 0, I.V.T. => there is one point c in [a,b] s.t. g(c) = f(c) - c^n = 0 i.e. f(c) = c^n -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 61.219.178.217
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