证明:(1)任取x1<x2,则,∵x1<x2,∴,∴,∴f(x1)<f(x2),即函数f(x)在(﹣∞,+∞)内单调递增.(2)∵,∴m=f﹣1(x)﹣f(x)= =,当1≤x≤2时,,∴,∴m的取值范围是.
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