Question

当f(x)在定义R中,且f(x)是奇函数,f(x+2)=-f(x),求f(6)的值

Answer 1

f(x+2)=-f(x)

put x= 0

f(2) = -f(0)

put x =2

f(4) = -f(2)

put x=4

f(6) = -f(4)

therefore

f(6) = f(2) = -f(0)

f(x)是奇函数,=> f(-x) = -f(x)

put x =0

f(-0) = -f(0)

f(0) = -f(0) => f(0) = 0

therefore

f(6) = -f(0) = 0 #