解:(1)由-1≤x-c≤1,得g(x)定义域:c-1≤x≤1+c, 由-1≤x-c2≤1,得f(x)定义域:c2-1≤x≤1+c2, 由,得:c+1<c2-1或c2+1<c-1,解得:c<-1或c>2,综上:c的取值范围为{x|c<-1或c>2}。(2)任取x1、x2∈[-1,1],且x1<x2,则,由已知,有,而,∴,∴,∴f(x)在[-1,1]上为增函数。
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