Question

函数周期性题目1,若函数y=f(x)是周期为二的奇函数,切当x∈（0,1）时,f（x）=x+1,求f（π）.2,f(x)

Answer 1

1.f（π）=f（π-4）=- f（4- π）=-（4-π+1）=π - 5

2.f(x+2)[1-f(x)]=1+f(x)

f(x+2)=[1+f(x)]/[1- f(x)]

={1+ [1+f(x-2)]/[1- f(x-2)]}/{1- [1+f(x-2)]/[1- f(x-2)]}

=[1- f(x-2)+1+f(x-2)]/[1-f(x-2)- 1- f(x-2)]

=- 1/f(x-2)

f(x)=- 1/f(x+4)=f(x+8)

f(2008)=f(0+251*8)=f(0)=2008

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

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

上下两个式子相加：f(x-1)+f(x+2)=0

f(x)=- f(x+3)=f(x+6)

f(69)=f(3+11*6)=f(3)=- f(0)=- 6

4.a=f(log½2)=f(-1)=f(1)=f(3)

b=f(-4)=f(4)

c=f(-π)=f(8-π )

因为：3f(8-π)

即：a>b>c