若奇函数y=f(x)(x属于R)满足f(2)=1,f(x+2)=f(x)+f(2)则f(5)==

问题描述:

若奇函数y=f(x)(x属于R)满足f(2)=1,f(x+2)=f(x)+f(2)则f(5)==

由题意知:y=f(x)为奇函数,故有:f(x) = -f(-x) (x属于R)当x=-1时,f(-1 + 2) = f(-1) + f(2);即f(1)=f(-1)+1;又f(1)=-f(-1);所以f(1)=-f(1)+1;由此得:f(1)=0.5;当x=1时,f(1+2)=f(1)+f(2),即f(3)=0.5+1=1.5;当x=3时,...