Question

已知函数f(x)的定义域为R,对任意实数x,y都有f(x+y)=f(x)+f(y)+1/2且f(1/2)=0，当x>1/2，f(x)>0，求单调性

Answer 1

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

= f(x1 - x2) + 1/2

当 x1 - x2 > 1/2时 f(x1 - x2) > 0
所以 f(x1) - f(x2) > 1/2 >0

当 x1 - x2 < 1/2时
f(x1 - x2) + 1/2 = f(x1 - x2) + 0 + 1/2 = f(x1 - x2) + f(1/2) + 1/2 = f(x1 - x2 + 1/2)

因为 x1 -x2 + 1/2 > 1/2
所以 f(x1 - x2) + 1/2 > 0
即f(x1) > f(x2)

综上，f(x)是增函数