设a是实数.f(x)=a-[2/(2^x+1)] (x∈R).试证明:对于任意a,f(x)在R上为增函数

假设m＞n,m、n∈R
f(m)-f(n)={a-[2/(2^m+1)]}-{a-[2/(2^n+1)]}
=-2[1/(2^m+1)-1/(2^n+1)]
=-2{(2^n-2^m)/[(2^m+1)(2^n+1)]}
=-2{2^n*[1-2^(m-n)]/[(2^m+1)(2^n+1)]}
=负{正*负/正}＞0
所以f(x)在R上是增函数