Question

已知定义在R上的奇函数f(x)满足f(x-4)=-f(x),且在区间【0,2】为增函数，比较f(-25),f(11),f(80).

Answer 1

f(x-4)=-f(x) f(x+4)=-f(x) f(x+8)=f(x) f(x+16)=f(x) f(x+24)=f(x)

追问

嫩在详细点解释吗，谢了

追答

f(80)=f(0)=0 f(11)=f(3)=-f(-1) =f(1) f(-25)=f(7)=f(-1)=-f(1)
区间【0,2】为增函数0=f(0)<f(1)
-f(1)<0<f(1) f(-25)<f(80)<f(11)