Question

已知函数f(x)的定义域为R，对任意实数x、y均有f(x+y)=f(x)+f(y)-1且f(-1/2)=0 当x>-1/2, f(x)>0 证明单调

答案是单调递增

Answer 1

先证明x>0时,f(x)>1
首先,令y=-1/2,有f(x-1/2)=f(x)+f(-1/2)-1=f(x)-1,即f(x)=f(x-1/2)+1
然后令上式的x>0,则x-1/2>-1/2,f(x-1/2)>0,f(x)>1
故x>0时,f(x)>1得证.
最后再证明单调性
对于任意的x1<x2,有x2-x1>0,f(x2-x1)>1
则f(x2)=f(x1+(x2-x1))=f(x1)+f(x2-x1)-1>f(x1)+1-1=f(x1)
故f(x)单调递增.

Answer 2

设x2＞x1
f(x1+x2-x1)=f(x2-x1)+f(x1)-1→f(x2)-f(x1)=f(x2-x1)-1
f(x2-x1-1/2)=f(x2-x1)+f(-1/2)-1
x2-x1-1/2＞-1/2
f(x2-x1-1/2)＞0
f(x2-x1)＞1
所以f(x2)-f(x1)＞0单增