已知定义在区间（0，+∞）上的函数f（x）满足f（
（1）求f（1）的值； （2）判断并证明f（x）的单调性； （3）若f（3）=-1，求f（x）在[2，9]上的最小值． |
（1）∵定义在区间（0，+∞）上的函数f（x）满足f（
）=f（x1）-f（x2），x1 x2
∴当x1=x2时，f（1）=O．
（2）f（x）是减函数．
证明：设x1＞x2，则f（x1）-f（x2）=f（
），x1 x2
∵x1＞x2，∴
＞1，x1 x2
∵当x＞1时，f（x）＜0，
∴f（x1）-f（x2）＜0，
∴f（x）在区间（0，+∞）是减函数．
（3）∵f（1）=O f（3）=-1，
∴f（
）=f（1）-f（3）=0-（-1）=1，1 3
∴f（9）=f（3÷
）=f（3）-f（1 3
）=-1-1=-2，1 3
∵f（x）在区间（0，+∞）是减函数，
∴f（x）在[2，9]上的最小值为f（9）=-2．
| 完形填空。 | |||
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