已知f｛f[f(x)]｝=27x+36求一次函数f(x)的解析式~!

f(x)=3x+m
fff(x)=3*(3*(3x+m))+m)+m=27x+13m
m=36/13

f(x)=3x+36/13

设f(x)=kx+b
则f[f(x)]=k(kx+b)+b=k^2x+(kb+b)
所以f{f[f(x)]}=k[k^2x+(kb+b)]+b
=k^3x+k(kb+b)+b=27x+36
所以k^3=27
k(kb+b)+b=36

k^3=27=3^3
所以k=3
代入k(kb+b)+b=36
3(3b+b)+b=36
12b+b=36
b=36/13

所以f(x)=3x+36/13

设f(x)=kx+b
则f[f(x)]=k(kx+b)+b=k^2x+kb+b
f｛f[f(x)]｝=k(k^2x+kb+b)+b=k^3x+k^2b+kb+b=27x+36
所以k^3=27 k=3
k^2b+kb+b=36 b=36/13
所以f(x)=3x+36/13

很简单
设f(x)=kx+b
则
f[f(x)]=k²x+kb+b
f{f[f(x)]}=k^3*x+k²b+kb+b
根据f{f[f(x)]}=27x+36
对比知道
k^3=27 k²b+kb+b=36
k=3 b=36/13
所以f(x)=3x+36/13

设f(x)=ax+b，则f(f(x))=a(ax+b)+b=a^2x+ab+b
f(f(f(x)))=a(a^2x+ab+b)+b=a^3x+a^2b+ab+b=27x+36
因此a=3，b=36/13