设f(x)为一次函数,且f[f(x)]=2x-1,求f(x)的解析式
问题描述:
设f(x)为一次函数,且f[f(x)]=2x-1,求f(x)的解析式
答
设f(x) = kx + b (k ≠ 0)
则 k(kx + b) + b = 2x -1
k²x + kb + b = 2x - 1
k² = 2 且 kb + b = -1
k = √2 , b = 1 - √2 或 k = -√2 , b = 1 + √2
所以f(x) = √2x + 1 -√2 或 f(x) = -√2x + 1 + √2