(1)f(1/2 +0)=f(1/2)+f(0)-1 得:f(0)=1
f(-1/2 +1/2)=f(-1/2) +f(1/2) 即 1=f(-1/2)+f(1/2);f(-1/2)=-1
(2) 设x1>x2; x1=x2+a(a>0)
f(x1)-f(x2)=f(x2+a)-f(x2)=f(x2)+f(a)-1-f(x2)=f(a)-1 又f(a)=f(a-1/2 + 1/2)=f(a-1/2)+f(1/2)-1=f(c-1/2)+1
因为a>0.故a-1/2>-1/2.即有f(a-1/2)>0
所以
f(x1)-f(x2)>0,在x的定义域内,所以f(x)为单调递增