设函数f(x)(x为R)是奇函数,f(1)=1/2,f(x+2)=f(x)+f(2)求f(5)?

f(5) = f(3+2) = f(3)+f(2)
f(3) = f(1+2) = f(1)+f(2)
由於f(x)为奇函数, 所以f(-1) = -f(1)
f(1) = f(-1+2) = f(-1)+f(2)
f(2) = f(1) - f(-1) = f(1) -[-f(1)] = f(1) + f(1) = 1
因此
f(5) = f(3)+f(2) = [f(1)+f(2)] + f(2) = 1/2 + 1 + 1 = 5/2

f(5)=f(3)+f(2)
f(3)=f(1)+f(2)
下面问题转化为求f(2)
由奇函数可得f(-1)=-1/2从而由f(1)=f(-1)+f(2)求得f(2)=1
所以f(5)=2.5