It is known that the domain of f x is a non zero real number, and satisfies the analytic formula of

It is known that the domain of f(x) is a non-zero real number.

Since 3f(x)+2f(1 x)=4x

Take x=t and x=1 t, respectively

3f(t)+2f(1 t)=4t

3f(1/t)+2f(t)=4/t

Lianli solution. f(t)=4 5 * (3t-2 t) i.e. f(x)=4/5 *（3x-2/x）.

After verification, the first person turned out to be correct.

3f(x)+2f(1/x)=4x （1）

Let x=1 x, then 1 x=x

So 3f(1 x)+2f(x)=4 x (2)9f(x)-4f(x)=12x-8 x=(12x 2-8) x, so f(x)=(12x 2-8) (5x).

3f(x)+2f(1 x)=4x and 3f(1 x)+2f(x)=4 x are combined to form a system of equations, and the result can be obtained by solving the binary equation!!

This problem is typical, replace the original x with 1 x, get.

2f(1 x)+f(x)=3 x, the two sides of the original formula are called multiplied by 2, and they get.

4f(x)+2f(1 x)=6x, subtract the above equation from this equation to obtain.

3f(x)=6x-3 and bright balance x,f(x)=2x-(1 key grandson x),

Let y=2x, then x=y2

f(y)=f(2x)=2^x=2^(y/2)

So f(x)=2 (x chaotic sock grip 2) is 2 of the celebration x is so big to the 2nd power.

f(1)=0

f(2*1/2)=0=f(2)+f(1/2)f(1/2)=-1

f(x)=-f(1/x)

Take x1>x2>1, then x1*x2>x1, and f(x2)>0f(x1*x2)-f(x1)=f(x2)>0, so f(x) increases monotonically at x>1.

f(1)=0, when x>1, f(x)>f(1) when 0=f(4).

f(x+1)>=f(2x)+f(4)

f(x+1)>=f(8x）

There is x+1>=8x, and x+1>0, 2x>0, and the domain is defined as x>0 and therefore 0

Solution: 1. f(xy)=f(x)+f(y).

f（2）=f(1*2)=f(1)+f(2)∴f(1)=0

f(1)=f(1/2*2)=f(1/2)+f(2)=0∴f(1/2)=-1

2. Let u>v>0then u v>1

and x>1, f(x)>0

f(u/v)>o

i.e. f(u)+f(1 v)>0

f(1)=f(v*1/v)=f(v)+f(1/v)=0f(1/v)=-f(v)

f(u/v)=f(u)-f(v)>0

f(x) is monotonically increasing on (0,+).

3、∵f(1)=f(2x)+f(1/2x)=0∴f(2x)=-f(1/2x)

f(x+1)-f(2x)>=2

i.e. f(x+1)+f(1 2x)>=2

i.e. f[(x+1) 2x]>=2

f(2)=1,2=2f(2)=f(2)+f(2)=f(4)∴f[(x+1)/2x]>=f(4)

f(x) is monotonically increasing on (0,+).

x+1)/2x>=4

Solve the inequality to get 0

Untie. (1) Let x=y=1, then f(1)=f(1)+f(1), i.e., f(1)=0

2) Let 01f(x2)=f(x1*x2 x1)=f(x1)+f(x2 x1)f(x2)-f(x1)=f(x2 x1)>0, i.e. f(x1)0

So the value range of x is (0,2)u[1+ 10,+

This is on the paper, you wait for me, I will copy the answer for you.

Because 3f(x)+2f(1 x)=4x..So 3f(1 x)+2f(x)=4 x..Two-formula one-formula *3-two-formula *2, get.

5f（x)=12x-8/x

then f(x)=(12x-8x) 5

x and 1 x intergeneration:

3f(1 x)+2f(x)=4 x, and the solution with the original synthesis: f(x)=4 5*(3x-2x).

3f（x）+2f（1/x）=4x;

Let x=1 x be substituted to get :

3f（1/x）+2f（x）=4/x（1）;

3f（x）+2f（1/x）=4x（2）；

1), (2) to get a system of equations, subtract f(1 x), that is, f(x) = 12x 5-8 5x

Hope it helps.

(1) Because f(x2)-f(x1) x2-x1>0, there are two categories: aThe numerator and denominator are all greater than 0 b.The numerator and denominator are all less than 0. Then, to use the definition of monotonicity, in either case, the function is an increment.

2) Let x1=x2=1, substitute f(x1x2)=f(x1)+f(x2), f(1)=0

3) According to f(x1x2)=f(x1)+f(x2), then f(x+6)+f(x)=f[(x+6)*x], and because f(4)=1, f(4*4)=f(4)+f(4)=4, i.e., by f(16)=2So the inequality is converted to: f[(x+6)*x]>f(16)

Since the function is an increasing function, [(x+6)*x>16, solve this unary quadratic inequality, and pay attention to the condition of x>0. The final result is x>2

As a small problem, let's use the special function method, let f(x)=logax (the logarithm of x based on a) f(4)=1, so a=4