Question

问大家一道比大小的函数题目我要问的是第4道,

Answer 1

∵f(x+1)=-f(x),且是偶函数,

∴f(3)=f(2+1)=-f(2),

f(2)=f(1+1)=-f(1),

f(1)=f(0+1)=-f(0),

f(0)=f(-1+1)=-f(-1)=-f(1),

f(-1)=f(-2+1)=-f(-2)=-f(2),

∴f(3)=f(-1),f(2)=f(0)

f(√2)=f(√2-1+1)=-f(√2-1)=-f(1-√2)=f(2-√2)=f(√2-2),

∵-1