Question

已知f（x）是一次函数，且f[f(x)]=4x-1,则f(x)的解析式为

Answer 1

解：设f（x）=kx+b
∵f[f(x)]=4x-1
∴f(kx+b)=k(kx+b)+b=4x-1
k²x+kb+b=4x-1
k²=4，kb+b=-1
∴k1=2，k2=-2
b1=-1/3，b2=1
∴函数的解析式为f（x）=2x-1/3，或者f（x）=-2x+1

Answer 2

f(x) = ax + b
f[f(x)] = a(ax+b) + b = a²x + ab+b = 4x -1
a² = 4, a = ±2
ab+b = (a+1)b = -1
b = -1/(a+1)
1. a = 2, b = -1/3, f(x) = 2x -1/3
2. a = -2, b = 1, f(x) = -2x + 1

Answer 3

设f(x)=ax+b, 则f[f(x)]=f(ax+b)=a^2x+ab+b=4x-1
得a=2, b=-1/3或a=-2, b=1
f(x)=2x-1/3或f(x)=-2x+1