Question

1、已知f(x)是一次函数,且f[f(x)]=x+1,求f(x)的解析式.

Answer 1

1,k(kx+b)+b=x+1

k2x+kb+b=x+1

k=1,b=1/2,

y=x+1/2

f(x+y)+f(x-y)=2f(x)f(y)

y=0;2f(x)=2f(x)f(0),f(0)=1

f(π/2+π/2)+f(0)=0

f(π）=-f(0)=-1

f(π+π）+f(0)=2f^2(π）

f(2π）=2-1=1