题目:已知函数 y=f(x)的图象是自原点出发的一条折线.当 n≤y≤n+1(n=0,1,2,…)时,该函数图象是斜率为 b~n 的线段(其中正常数b≠1),设数列{x_n}由 f(x_n)=n(n=1,2,…)定义.(Ⅰ)求 x_1、x_2和 x_n 的表达式;(Ⅱ)求 f(x)的表达式,并求其定义域;(Ⅲ)证明:y=f(x)的图象与 y=x 的图象没有横坐标大于1的交点.该题解答涉及许多重要的数学思想方法,考 Title: The image of the known function y=f(x) is a polyline starting from the origin. When n≤y≤n+1(n=0,1,2,...), the image of the function is a slope. The line segment of b~n (where the normal number b≠1) sets the sequence {x_n} defined by f(x_n)=n(n=1, 2, ...). (I) Find the expression of x_1, x_2, and x_n; (II) Find the expression of f(x) and find its domain; (III) Prove that the image of y=f(x) and the image of y=x do not have intersection points where the abscissa is greater than 1. Involving many important mathematical thinking methods