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问题 填空题
设f(x)=
x2-4x+6，x≥0
2x+4x＜0
若存在互异的三个实数x1，x2，x3，使f（x1）=f（x2）=f（x3），则x1+x2+x3的取值范围是______．
答案

不妨设x1＜x2＜x3，当x≥0时f（x）=（x-2）2+2，

此时二次函数的对称轴为x=2，最小值为2，

作出函数f（x）的图象如图：

由2x+4=2得x=-1，由f（x）=（x-2）2+2=4时，解得x=2-

2
或x=2+
2
，

所以若f（x1）=f（x2）=f（x3），

则-1＜x1＜0，2-

2
＜x2＜2，2＜x3＜2+
2
，且
x2+x3
2
=2，即x2+x3=4，

所以x1+x2+x3=4+x1，

因为-1＜x1＜0，所以3＜4+x1＜4，

即x1+x2+x3的取值范围是（3，4）．

故答案为：（3，4）．

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