摘要：已知f(x)=x2-x+c定义在区间［0.1］上.x1.x2∈［0.1］.且x1≠x2.求证: (1)f(0)=f(1), (2)| f(x2)-f(x1)|＜|x1-x2|, (3)| f(x1)-f(x2)|＜, (4)| f(x1)-f(x2)|≤. 证明:(1)f(0)=c.f(1)=c. ∴f(0)=f(1). (2)| f(x2)-f(x1)|=|x2-x1||x2+x1-1|. ∵0≤x1≤1.∴0≤x2≤1.0＜x1+x2＜2(x1≠x2). ∴-1＜x1+x2-1＜1. ∴| f(x2)-f(x1)|＜|x2-x1|. (3)不妨设x2＞x1.由(2)知 | f(x2)-f(x1)|＜x2-x1. ① 而由f(0)=f(1).从而 | f(x2)-f(x1)|=| f(x2)-f(1)+f(0)-f(x1)| ≤| f(x2)-f(1)|+| f(0)-f(x1)|＜|1-x2|+|x1|＜1-x2+x1. ② ①+②得2| f(x2)-f(x1)|＜1. 即| f(x2)-f(x1)|＜. (4)|f(x2)-f(x1)|≤fmax-fmin=f(0)-f()=.

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