Question

已知函数f(x)= (a、b为常数，且a≠0)满足f(2)=1,f(x)=x有唯一解，求函数y=f(x)的解析式和

f［f(－3)］的值.

Answer 1

解：因为f(2)=1,所以1=,

即2a+b=2.                                                                                                              ①

又因为f(x)=x有唯一解，即=x有唯一解，

所以x·=0.解得x₁=0,x₂=.

所以x₁=x₂=0,即=0.

所以b=1.                                                                                                                ②

由①②得

所以f(x)=,

f［f(－3)］=f(6)=.