Question

设f(f(f(x)))=27x＋26.求一次函数f(x)的解析式

Answer 1

解设f(x)=kx+b

即f(f(x))=k(kx+b)+b=k^2x+bk+b

故f(f(f(x)))=f(k^2x+bk+b)=k(k^2x+bk+b)+b=k^3x+bk^2+kb+b

又由f(f(f(x)))=27x＋26

知k^3=27

bk^2+kb+b=26

解得k=3,b=2

故f(x)=3x+2