Question

已知f(x)的定义域为R,且满足f(x)=f(x+1)+f(x-1).1求证：f(x)为周期函数.2 若f(3)=2求f

Answer 1

1.f(x)=f(x+1)+f(x-1)

f(x+1)=f(x+2)+f(x)

相加

f(x+2)+f(x-1)=0

f(x+5)+f(x+2)=0

f(x+5)=f(x-1)

周期为6

2.由f(x+2)+f(x-1)=0

f(3)=-f(0)

f(2004)=f(6*334)=f(0)=-2