Let f x be an additive function over R, so that F x f x f 2 x if F x F y F x F y , F 1 3

f(1+1)=f(1)+f(1)=6

f(2)=6

So 6 can be replaced with f(2).

f(x)+f(x-2)=f[x(x-2)]; f(x)+f(x-2)>6 can be converted to:

f[x(x-2)]>f(2) ①

The following tries to prove that f(x) is an increasing function;

f '(x)=f '(x)-[f '(2-x)]*1)=f '(x)+f ‘(2-x)

Because f(x) is an increasing function, so.

f '(x)>0; f ’(2-x)>0

So f'(x)>0

So the function f(x) is an increasing function over r, and the equation is equivalent to :

x(x-2)>2

x²-2x-2>0

x-1)²>3

x-1>√3;or X-1<-3

x>1+√3 ;or X-1<-3

Therefore, the value range of x satisfying f(x)+f(x-2)>6 is:

Because f(x+y) = f(x)+f(y).

So f(x)+f(x-2)=f(2x-2), so the inequality is f(2x-2)>6

And because f(4) = f(2+2) = f(2) + f(2) = 3+3 = 6, the inequality is f(2x-2) > f(4).

Because f(x) is an increasing function.

So 2x-2>4

So 2x>6

x>3

It's still not right, otherwise f(1) should be equal to 0, are you confusing f with f?

(1) Proof:

Take baix1>x2, then.

f(x1)-f(x2)=f(x1)-f(2-x1)-f(x2)+f(2-x2)=[f(x1)-f(x2)]+f(2-x2)-f(2-x1)]

x1>x2

2-x2>2-x1

Again, f(x) is a real number.

DAO on the zhi increment function

f(x1)>f(x2), f(x1)-f(x2)>0f(2-x2)>f(2-x1),f(2-x2)-f(2-x1)>0∴f(x1)-f(x2)>0

i.e. f(x1) > f(x2).

f(x) is an increment function on r.

2) Proof of: f(x) = f(x)-f(2-x).

f(1)=f(1)-f(1)=0

f(x) is an increment function on r.

f(x1)+f(x2)＞0=f(1)

If at least one of x1 or x2 is greater than 1, you may wish to set x1>1 (otherwise f(x1)+f(x2)0).

i.e. f(x2)>0=f(1).

x2>1

x1+x2>2

(1) Proof:

Because: f(x) is an increment function.

So: f(-x) is a subtraction function.

Therefore: f(2-x) is also a subtractive function.

Then: -f(2-x) is the increment function.

Then: f(x) is the increment function.

The first question I would ask was: f(x) is an increasing function, 2-x is a decreasing function, so f(2-x) is a decreasing function, so it is an increasing function of -f(2-x), so f(x) is an increasing function.

It's the same increase and the difference is different!

Solution: Take x1>x2 on the defined domain of f(x); then f(x1)-f(x2)=

f(1-x1)-f(3+x1)-[f(1-x2)-f(3+x2)]=f(1-x1)-f(1-x2)+f(3+x2)-f(3+x1).

Because we know that f(x) is an increment function on r and 3+x2<3+x1,1-x1<1-x2 (because x1>x2) is taken

So f(1-x1)-f(1-x2)<0 and f(3+x2)-f(3+x1)<0

So f(x1)-f(x2)=f(1-x1)-f(1-x2)+f(3+x2)-f(3+x1)<0

i.e. f(x) is a decreasing function over r.

fx is an increment function.

Then f(1-x) decreases gradually when x increases, so f(1-x) is a subtraction function.

Then f(x)=f(1-x)+3 is a subtractive function, I hope it can help you!!

f(x) is an increasing function over r, then its derivative f'(x) 0, the derivative of f(x) yields f' (x)=f '(1-x)-f '(3+x)＝－f '(x)－f '(x)＝-2f '(x)＜0

So f(x) is a subtractive function.

From the title, it can be seen that f(x) is an increasing function on r, then f(1-x) is a decreasing function on r, and f(3+x) is an increasing function on r.

(1) Set x1 > x2

f(x1)-f(x2)=f(x1)-f(2-x1)-[f(x2)-f(2-x2)]

The f(x1)-f(x2)+[f(2-x2)-f(2-x1)] function f(x) is an increasing function on the set of real numbers r, f(x1)> f(x2)-x1<-x2, (2-x1)<(2-x2)f(2-x2)>f(2-x1).

f(x1) > f(x2), i.e. the function f(x) is an increasing function on the set of real numbers r.

f(x1) + f(x2)- f(2-x1)- f(2-x2) 0, so f(x1)- f(2-x1) 0

f(x2)- f(2-x2)＞0

So f(x1) f(2-x1) f(x2) f(2-x2) because f(x) is an increasing function.

So x1 2-x1 x2 2-x2

So x1 1 x2 1

So x1+x2 > 2

How to type non-vertical on the computer.

f(x) increase.

then x1-x2

1-x1>1-x2

So f(1-x1)-f(1-x2)>0

x10, so f(1-x1)-f(1+x1)-f(1-x2)+f(1+x2)>0

i.e. x1f (x2).

So it's a subtraction function to choose b

Let x1 x2 then 1-x1 1-x2 1+x1 1+x2

So f(1-x1) f(1-x2) f(1+x1) f(1+x2).

f(x1)-f(x2)=f(1-x1)-f(1+x1)-f(1-x2)+f(1+x2)

f(1-x1)-f(1-x2)]+f(1+x2)-f(1+x1)]＜0

So f(x1)-f(x2) 0

So f(x) is a subtractive function on r

f(x) increases, and f(1-x) decreases. f(1+x) is increasing, so -f(1+x) is decreasing, and the two subtraction functions together are decreasing.

f(x)=f(x)-f(2-x), so there is f(2-x)=-f(x)+f(2-x)=-f(x), so f(x1)+f(x2)=f(x1)-f(2-x2)>0

There is f(x1)> answer is f(2-x2), and there is x1+x2>2 from monotonicity

f(x/y)=f(x)-f(y）

So f(1)=f(1 1)=f(1)-f(1)=0, so f(6)=f(36 6)=f(36)-f(6), so f(36)=2f(6)=2

f(x+5)-f(1 x)=f[(x+5) (1 brother bend x)]=f[(x+5)x].

So f(x+5)-f(1 x)<2 becomes.

f[(x+5)x]0 and talk about dust 1 x>0

i.e. x>0

So the range is 0

Yang with lead Qiaofan knows the happy answer Pei key shouts: f(1 1)=f(1)-f(1),f(1)=,f(36)=2,x 2+5x<36,x (-9,4).

(1) Let x=y=1, then f(1)=f(1)-f(1)=0 let x=1, then f(1 y)=f(1)-f(y)=-f(y) =>f(1 y)=-f(y).

then f(xy) = f(x (1 y)) = f(x)-f(1 y) = f(x) + f(y).

Certification. #2)∵f(2)=1

From (1): f(4) = f(2) + f(2) = 2 from f(x)-f(1 (x-3)) 2

f(x*(x-3))≤f(4)

f(x) is an increment function defined on r+.

x*(x-3)≤4 --

x>0 --

1/(x-3)>0 --

The simultaneous solution yields: 3 Therefore the solution set of the inequality f(x)-f(1 (x-3)) 2 is (3,4].

(1) Proof: f(x y)=f(x)-f(y) is obtained from f(x)=f(x y)+f(y) (let x=ab, y=b substitute (*)

f(ab)=f(ab b)+f(b)=f(a)+f(b) (then let a=x,b=y substitute (**.)

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

2) f(x)-f(1 (x-3))=f[x(x-3)] 2 because f(2)=1, f(2·2)=f(2)+f(2)=2 due to the monotonicity of f(x).

x(x-3)≤4

Solution x -1 (rounded), x 4

i.e. the solution of the inequality is x 4

(1) When x=y=1, f(1 1)=f(1)-f(1)=0 f(1)=0

Let x=1, there is f(1 y)=f(1)-f(y)=-f(y)f(xy)=f[x (1 y)]=f(x)-f(1 y)=f(x)+f(y).

2)f(x)-f(1 (x-3)) 2 x 0,x-3 0,x 3f[x (1 x-3)] 1+1=f(2)+f(2)=f(2*2)=f(4), because f(x) is an additive function defined on r+, so x(x-3) 4 is solved: -1 x 4, and because the definition domain is r+, so 3 x 4