0 = f( 1)-f( 1)= f( 1/ 1)= f( 1)。 Generally, the special values of F(0)F( 1) are all made by ourselves.

The second question is actually very simple. When you see this kind of problem, you should first check whether the formula given by the problem can be used.

F (x+3)-f (1/x) = f (x 2+3x), not used. What is needed is f (x 2+3x).

Then the title says that the function f(x) is a increasing function whose domain is on (0, positive infinity).

In fact, I want you to find out when f(y) equals 2, such as when y=m equals 2; Just y

F(y) is less than 2, x 2+3x.

In fact, the key is to find M, and it depends on whether we can find this M that makes f(m)=2 through conditions.

If f(6)= 1, find f(m)=2, which is twice the relationship.

Reverse thinking, f(x/y)=f(x)-f(y)

F(x/y)+f(y)=f(x) is deduced.

If x/y=y, it will produce the effect of 2f(y)=f(x).

That is, 2f(y)=f(x) when x = y * y

So when y=6 and x=36, f(36)=2f(6)=2.

x^2+3x-36<； 0

-3/2-√ 153/2 & lt； x & lt-3/2+√ 153/2

Then by definition, fields x+3 and 1/x are both greater than 0.

So the answer is 0.

The idea is comprehensive. Are you satisfied?