(1)
f(x)=(ax+b)/(x^2+1)是奇函数,f(-x)=-f(x)
(-ax+b)/(x^2+1)=- (ax+b)/(x^2+1),
-ax+b=-ax-b,b=-b,
所以b=0.
又f(1/2)=2/5,所以(a/2)/(1/4+1)=2/5,a=1.
∴f(x)=x/(x^2+1).
(2)
设任意-1
(1)
f(x)=(ax+b)/(x^2+1)是奇函数,f(-x)=-f(x)
(-ax+b)/(x^2+1)=- (ax+b)/(x^2+1),
-ax+b=-ax-b,b=-b,
所以b=0.
又f(1/2)=2/5,所以(a/2)/(1/4+1)=2/5,a=1.
∴f(x)=x/(x^2+1).
(2)
设任意-1