已知f(a+b)=f(a)+f(b)-1 当x大于0时 f(x)大于1 ，证f(x)是增函数 若 f(4)=5，f(mm-2)小于8 求m范围

对任意x，y属于R，
若有x<y
y-x>0
=>
f(y-x)>1
f(x + y-x) = f(x) + f(y-x)-1 >f(x)
=>
f(y)-f(x)>0
f(x)为增函数

f(8)=2f(4)-1=9
f(4)=2f(2)-1
=>f(2)=3
f(2)=2f(1)-1
=>f(1)=2
=>
f(8)=f(7)+f(1)-1
=>
f(7)=8
=>
f(mm-2)<f(7)
=>
mm-2<7
=>
-3<m<3