Question

已知函数f(x)＝ (a,b为常数，且a≠0)满足f(2)＝1,f(x)＝x有惟一解．求函数y＝f(x)的解析式和f［f(－3)］的值．

Answer 1

答案：
解析：

解：∵f(2)＝1,∴1＝,即2a＋b＝2．                                                                                                               ① 又∵f(x)＝x有惟一解，即＝x有惟一解， ∴x·＝0． 得x1＝0,x2＝,x1＝x2＝0． 得b＝1．                                                                                                                                                                                                           ② 由①、②，得． ∴f(x)＝． 故f［f(－3)］＝f()＝f(6)＝．