f(domain) = ?

The idea to solve this problem is to find the formula of function $f$ .

First, note that property (i) can actually be generalized as $f(x+k)=f(x)+k$ . This is according to the following fact that:

• $f(x+1)=f(x)+1$

• $f(x+2)=f(x+1)+1=f(x)+2$

• $f(x+3)=f(x+2)+1=f(x+1)+2=f(x)+3$ , etc.

By the generalization above, our next step is to express $\frac {1}{x}$ --from the property (ii)--to be of the form $x+k$ .

$\frac {1}{x}=x+k$

$k=\frac {1}{x}-x=\frac {1-x^2}{x}$

Hence, now we have $\frac {1}{x}=x+\frac {1-x^2}{x}$ . Our generalization is now ready to applied.

$f(\frac {1}{x})=f(x+\frac {1-x^2}{x})=f(x)+\frac {1-x^2}{x}...(*)$

Recall propertiy (ii) and combine it with $(*)$ to get,

$f(\frac {1}{x})=\frac {f(x)}{x^2}=f(x)+\frac {1-x^2}{x}$

Move $f(x)$ from the left-most side to the center side to get,

$\frac {f(x)}{x^2}-f(x)=\frac {1-x^2}{x}$

Put the $f(x)$ out,

$f(x)(\frac {1}{x^2}-1)=\frac {1-x^2}{x}$

Continuing,

$f(x)(\frac {1-x^2}{x^2})=\frac {1-x^2}{x}$

$\Leftrightarrow f(x)=x$

And at last we easily have $f(225)=\boxed{225}$