已知定义在R上的偶函数f(x)满足f(x+2)=1/f(x)对于x属于R恒成立,且f(x)>0,则f(119)=?

f(x+4)=1/f(x+2)=f(x)
so f(119)=f(120-1)=f(-1)=f(1)
let x=-1 from f(x+2)=1/f(x) we have f(1)=1/f(-1)=1/f(1)
because f(x)>0 so f(119)=f(1)=1