已知f(x)=x2-x+c的定义域为(0,1),x1,x2∈(0,1),且x1≠x2,求证:|f(x1)-f(x2)|

已知f(x)=x2-x+c的定义域为(0,1),x1,x2∈(0,1),且x1≠x2,求证:|f(x1)-f(x2)|<|x1-x2|.

答案

证明:∵x1≠x2,所以原不等式等价于||<1,

即要证y=x2-x+c,x∈(0,1)函数图象上任意两点连线的斜率k<1.

∵f′(x)=y′=2x-1,当0<x<1时,-1<2x-1<1,

∴-1<f′(x)<1||<1,即|f(x1)-f(x2)|<|x1-x2|.

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