若函数y=f(x)对于一切实数a,b都满足f(a+b)=f(a)+f(b),且f(1)=8,求f(—1/2).

f(a+b)=f(a)+f(b) 令a=b=0, 则f(0)=f(0)+f(0), ∴f(0)=0 令a=b=1/2, 则f(1)=f(1/2)+f(1/2)=8, ∴f(1/2)=4 令a=1/2, b=-1/2, 则f(0)=f(1/2)+f(-1/2), ∴f(-1/2)=f(0)-f(1/2)=0-4=-4