Question

若f(f(x))=4x-1,求一次函数f（x）解析式

Answer 1

解设f（x）=kx+b

则f(f(x))=k（kx+b）+b=k^2x+kb+b

即k^2x+kb+b=4x-1

即k^2=4

kb+b=-1

即解得k=2,b=-1/3或k=-2,b=1

即f（x）=2x-1/3或f（x）=-2x+1