根據(jù)增函數(shù)的定義求解
設(shè)x1,x2∈R且x1＜x2
f(x2)-f(x1)=[a-2/(2^x2+1)]-[a-2/(2^x1+1)]=2(x2-x1)/[(2^x1+1)(2^x2+1)],
又2^x1＞0,2^x2＞0,故(2^x1+1)(2^x2+1)＞0
所以f(x2)-f(x1)=2(x2-x1)/[(2^x1+1)(2^x2+1)]＞0,根據(jù)增函數(shù)定義知：對(duì)于任意A,f(x)為增函數(shù).