文[1]对函数f(x)=∑ni=1aix+bi的最小值进行了研究,得到如下结论:对于函数f(x)=∑ni=1aix+bi(ai∈Q,且ai≠0,bi∈R,i∈N*),总可以写成f(x)=m1[x-x1+x-x2+…+x-xn](x1≤x2≤…≤xn,m,n∈N*)的形式.(1)若n=2k-1(k∈N*),则x=xk时,f(x)取值最小;(2)若n=2k(k∈N*),则x∈[xk,xk+1]时, The paper [1] studied the minimum value of the function f(x)=∑ni=1aix+bi and obtained the following conclusion: For the function f(x)=∑ni=1aix+bi(ai∈Q, and ai≠0 ,bi∈R,i∈N*), always written as f(x)=m1[x-x1+x-x2+...+x-xn] (x1≤x2≤...≤xn,m,n∈N*) In the form of (1) if n = 2k-1 (k∈N*), then x = xk, f(x) takes the smallest value; (2) if n = 2k (k∈N*), then x When [xk,xk+1]