Question

求解题，谢谢！





Answer 1

(1)
f(xy) = f(x)+f(y)

x=y=1
f(1)= f(1) +f(1)
f(1)=0

(2)
x>1, f(x)>0

x>y>0
x = (1+ε) y ; ε>0

f(x) -f(y)
=f((1+ε) y) -f(y)
=f(1+ε) +f(y) -f(y)
=f(1+ε)
>0
f(x) >f(y)
f(x) is increasing on (0, +∞)

(3)
(3)
f(1/3) =-1

-f(1/(x-2)) ≥ 2
f(1/(x-2)) ≤ -2
f(1/(x-2)) ≤ f(1/3) + f(1/3)
f(1/(x-2)) ≤ f(2/3)
=>
x-2>0 and 1/(x-2) ≤ 2/3
x>2 and [3 - 2(x-2)]/[3(x-2)]≤0
x>2 and (2x-7)/[3(x-2)]≥ 0
x>2 and (x≥ 7/2 or x<2 )
x≥ 7/2

追问

这种方法中，“所以。。。”那一步不懂如何得来的？谢谢。

追答

f(xy) = f(x) +f(y)
1/3 = 1 x(1/3)
f(1/3) = f(1) +f(1/3) ≠f(1) -f(3)
这是错误

该是这样！
-f(1/(x-2)) ≥ 2
f(1/(x-2)) ≤ -2
= (-1) + (-1)
f(1/(x-2)) ≤ f(1/3) + f(1/3)
f(1/(x-2)) ≤ f(1/9) ( 初算是出错）
=>
x-2>0 and 1/(x-2) ≤ 1/9
x>2 and [9 - (x-2)]/[9(x-2)]≤0
x>2 and (x-11)/[3(x-2)]≥ 0
x>2 and (x≥ 11 or x<2 )
x≥ 11

Answer 2

(1)令x=y=1，得 f(1)=2f(1) 即 f(1)=0
(2) 设 x1、 x2∈（0，+∞）,且 x2>x1 即 x2/x1>1和x2-x1>0
[f(x2)-f(x1)]/(x2-x1)=[f(x2/x1*x1)-f(x1)]/(x2-x1)
=[f(x2/x1)+f(x1)-f(x1)]/(x2-x1)
=f(x2/x1)/(x2-x1)>0
即f(x)在x x∈（0，+∞）上是单调增
（3）由 f(xy)=f(x)+f(y)
令 x=1/3 y=3
则 f(3)=f(1)-f(1/3)=1
2=f(3)+f(3)=f(9)
-f(1/(x-2)=f(x-2)-f(1)=f(x-2)
则 -f(1/(x02))≥2 转化为：f(x-2)≥f(9)
f(x)为单调增，得
x-2≥9..............(1)
1/(x-2)>0......(2)
解（1）（2），得 x≥11

追问

谢谢，懂了。因为有一朋友先答题了，已经采纳了，谢谢！
有问题再向你请教，再次感谢！