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关于方程f(x)=f~(-1)(x)的解法,已有好几个杂志对它进行了探讨,这里我们来进一步探讨f(x)(?)af~(-1)(ax+b)+b的解法。定理1 若f(x)在定义域上严格递增,其值域为R,a>0,则 f(x)af~(-1)(ax+b)+b与f(x)■ax+b同解。证明这里仅证f(x)>af~(-1)(ax+b)+b与f(x)>ax+b同解,
The solution to the equation f(x)=f~(-1)(x) has been discussed in several magazines. Here we further discuss f(x)(?)af~(-1)(ax +b) Solution of +b. Theorem 1 If f(x) is strictly increasing in the domain of definition and its range is R, a>0, then f(x)af~(-1)(ax+b)+b and f(x)■ax+ b with the solution. It is proved that only f(x)>af~(-1)(ax+b)+b and f(x)>ax+b have the same solution.