定义在R上的函数f（x）满足对任意x、y∈R恒有f（xy）=f（x）+f（y），且f（x）不恒为0。
（1）求f（1）和f（-1）的值；
（2）试判断f（x）的奇偶性，并加以证明；
（3）若x≥0时f（x）为增函数，求满足不等式f（x+1）-f（2-x）≤0的x的取值集合。
解：（1）令x=y=1，得f（1）=f（1）+f（1）
∴f（1）=0
令x=y=-1，得f（1）=f（-1）+f（-1）
∴f（-1）=0。
（2）令y=-1，由f（xy）=f（x）+f（y），得 f（-x）=f（x）+f（-1）
又f（-1）=0，
∴f（-x）=f（x），
又f（x）不恒为0，
∴f（x）为偶函数。
（3）由f（x+1）-f（2-x）≤0，知f（x+1）≤f（2-x）
又由（2）知f（x）=f（|x|），
∴f（|x+1|）≤f（|2-x|）
又∵f（x）在[0，+∞）上为增函数，
∴|x+1|≤|2-x|
故x的取值集合为
。
| 完形填空。 | |||
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