定义在R上的奇函数f(x)满足f(x+2011)=f(x),则f(2011

问题描述:

定义在R上的奇函数f(x)满足f(x+2011)=f(x),则f(2011

f(0)=0,f(0+2011)=f(0),f(2011)=0

f(x)= -f(-x)
put x=0
=> f(0) =0
f(x+2011) = f(x)
put x=0
f(2011) = f(0) = 0

f(x)是奇函数
所以f(0)=0
f(x+2011)=f(x)
令x=0
所以f(2011)=f(0)=0