If we know that the fx definition domain is R, f 1 lg3 lg2, f 2 lg3 lg5, f x 2 f x 1 f x, then f 200

f(3)=f(2)-f(1)=lg5+lg2 f(4)=lg2-lg3=-f(1)

f(5)=-lg3-lg5=-f(2) f(6)=-lg5-lg2=-g(3)

f(7)=lg3-lg2=f(1)

According to this, it is found that the period of the function is 6, and f(2009)=f(5)=-lg3-lg5

f(3)=f(2)-f(1)=lg5+lg2=1f(4)=f(3)-f(2)=1-lg3-lg5f(5)=f(4)-f(3)=-lg3-lg5f(6)=f(5)-f(4)=-1

f(7)=f(6)-f(5)=-1+lg3+lg5f(8)=f(7)-f(6)=lg3+lg5f(9)=1

From the above data, it can be concluded that there is a cycle from f(3) to f(8).

f(2009)=f(5)=-lg3-lg5

Let x=0 f(x+2)=f(x+1)-f(x)f(2)=f(1)-f(0).

f(0)=f(1)-f(2)=lg3-lg2-lg3-lg5=-(lg2+lg5)=-lg(2*5)=-1

Next, you may want to push it, for example, f(x+2)=f(x+1)-f(x)f(3)=f(2)-f(1).

This is periodic: f(2009)=f(5)=-(lg3+lg5)=-1

1) Define the domain 2x-1>0

Solve the quarrel x>1 2

2) y=lgx is the increasing function, and if the posture touches lg(1)=0, then f(x)>0 is to find lg(2x-1) > lg(1) then 2x-1>1

x>1 can be traced.

If the function is meaningful, it must be 1-(lgx)2 0, i.e., (lgx)2 1,-1 lgx 1, and the solution is 110 x 10, and the domain of the function is defined as: [110,10] family Qiaodan.

So the answer is: [110,10] Eliminate the town.

<> you refer to let the chaotic mu look at the accompanying auspiciousness! Tansen.

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The function f(x) on r satisfies f(x+y)=f(x)+f(y)+2xy, when the family loses x=0 and y=0.

f(0)=f(0)+f(0)=f(0)=0, when x=1, y=-1, Li Chonghou.

f(0)=f(1)+f(-1)+2*1*(-1) =f(-1)=0 when x=-1 and y=-1.

f(-2)=f(-1)+f(-1)+2*(-1)*(1) =f(-2)=2

When x=-1, y=-2.

f(-3)=f(-1)+f(-2)+2*(-1)*(2) = which spike destroys f(-3)=6

Because the domain of f(x) is (0,4), i.e. in f(lgx).

lgx∈(0,4]

0lg11 so the domain is defined as (1,10000).