Question

设f(X)是R上的奇函数,且满足f(2-x)=f(x).当0《x《1时,f(x)=x^2,求f(2009.5)的值

Answer 1

f(x+4)=f[2-(-x-2)]=f(-x-2)=-f(x+2)

=-f[2-(-x)]=-f(-x)=f(x)

∴周期为4.

∴f(2009.5)=f(4×502+1.5)=f(1.5)=f(2-0.5)

=f(0.5)=0.5²=0.25.