问题 解答题
已知定义在区间(0,+∞)上的函数f(x)满足f(
x1
x2
)=f(x1)-f(x2),且当x>1时,f(x)<0.
(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.
答案

(1)∵定义在区间(0,+∞)上的函数f(x)满足f(

x1
x2
)=f(x1)-f(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,∴

x1
x2
>1,

∵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(

1
3
)=f(1)-f(3)=0-(-1)=1,

∴f(9)=f(3÷

1
3
)=f(3)-f(
1
3
)=-1-1=-2,

∵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
完形填空。
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