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【简答题】
已知二次函数f（x）=ax 2 +bx（a≠0），且f（x+1）为偶函数，定义：满足f（x）=x的实数x称为函数f（x）的“不动点”，若函数f（x）有且仅有一个不动点． （1）求f（x）的解析式； （2）若函数g（x）=f（x）+kx 2 在（0，4）上是增函数，求实数k的取值范围； （3）是否存在区间[m，n]（m＜n），使得f（x）在区间[m，n]上的值域为[3m，3n]？若存在，请求出m，n的值；若不存在，请说明理由．
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