Question

函数f(x)对任何实数x.y都有f(x+y)=f(x)+f(y)-1，且当x<0时，f(x)<1.求f(0)

2.求证：f(x)在R上为增函数
3.若f(4)=7.解不等式f(2x+1)<4

Answer 1

令x=y=0
f(0+0)=f(0)+f(0)-1
f(0)=2f(0)-1
f(o)=1

Answer 2

令x<0
f(x)=f(x+0)=f(x)+f(0)-1
f(0)=1