Question

已知fx是定义在r上的函数,且对任意X属于R都有f(x+2)=f(2-x)+4f(2)，若函数y=f(x

Answer 1

1.
函数y=f(x)的图像关于点(-1，0)对称，
=》f(-1+x)+f(-1-x)=0,
=>f((x-1))=-f(-2-(x-1))=>
f(x)=
-f(-2-x)
2.
f(x+2)=f(2-x)+4f(2),
x=0代入，=>
f(2)=f(2)+4f(2)
=>
f(2)=0
3.
由f(2)=0,得
f(x+2)=f(2-x),
f(x+2)=f(4-(x+2))=>f(x)=f(4-x)
4.
1和3得
-f(-2-x)=f(4-x),
=>f(4-x)=-f((4-x)-6)
=>
f(x)=-f(x-6)
5.
f(x)=-f(x-6)=-(-f((x-6)-6))=f(x+12)
6.
f(2017)=f(12*168+1)=f(1)=2

Answer 2

1.
函数y=f(x)的图像关于点(-1，0)对称，
=》f(-1+x)+f(-1-x)=0,
=>f((x-1))=-f(-2-(x-1))=>
f(x)=
-f(-2-x)
2.
f(x+2)=f(2-x)+4f(2),
x=0代入，=>
f(2)=f(2)+4f(2)
=>
f(2)=0
3.
由f(2)=0,得
f(x+2)=f(2-x),
f(x+2)=f(4-(x+2))=>f(x)=f(4-x)
4.
1和3得
-f(-2-x)=f(4-x),
=>f(4-x)=-f((4-x)-6)
=>
f(x)=-f(x-6)
5.
f(x)=-f(x-6)=-(-f((x-6)-6))=f(x+12)
6.
f(2017)=f(12*168+1)=f(1)=2