Then f (m)-f (n) = am+(m-2)/(m+1)-an n-(n-2)/(n+1).
=(a^m-a^n)+[(m-2)(n+ 1)-(n-2)(m+ 1)]/(m+ 1)(n+ 1)
=(a^m-a^n)+3(m-n)/(m+ 1)(n+ 1)
a & gt 1,m & ltn
So an m
m & ltn,m-n & lt; 0
m & gt- 1,n & gt- 1,m+ 1 & gt; 0,n+ 1 & gt; 0
So (m+ 1)(n+ 1)>0.
so 3(m-n)/(m+ 1)(n+ 1)< 0
Therefore, (a m-a n)+3 (m n)/(m+1) (n+1) < 0.
That is-1< m <; N times
f(m)& lt; coedna
So f(x) is increasing function at (-1, +∞).