例1已知f(x)=x2+1,x≥0,1,x<0,,则满足不等式f(1-x2)>f(2x)的x的取值范围是.分析粗读题目,很多同学会考虑1-x2与2x的符号,往往需要分x∈(-∞,-1)、x∈[-1,0)、x∈[0,1]和x∈(1,+∞)四种情况来讨论,这其中还要确定端点处是否符合题意,过程比较复杂.随着审题的深入,可以看到x2+1≥1,由f(1-x2)>f(2x),知1-x2>0.因此,只需要考虑x∈(-1,0)和x∈[0,1)这两种情况.x的取值范围缩小了,相应 Example 1 It is known that f (x) = x2 + 1, x≥0, 1, x <0, then the range of x that satisfies the inequality f (1-x2)> f , Many students will consider 1-x2 and 2x symbols, often need to be divided into x∈ (-∞, -1), x∈ [-1,0), x∈ [0,1] and x∈ (1, + ∞ ), We can see that x2 + 1≥1, which is defined by f (1-x2)> f (2x) , We know that 1-x2> 0. Therefore, we only need to consider the two cases of x∈ (-1,0) and x∈ [0,1). The value range of x is reduced,