Question

若f(f(x))=27x+36,求一次函数f(x)解析式

Answer 1

f(x)=kx+b

则f[f(x)]=k(kx+b)+b

=k²x+b)k+1)

=27x+36

k²=27

b(k+1)=36

k=±3√3

b=36/(k+1)

所以f(x)=-3√3x+18(3√3+1)/13或f(x)=3√3x+18(3√3-1)/13