14. A function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies
$$
f(x)+f(y)=f(x) f(y)+1-\frac{4}{x y}
$$
for all nonzero real numbers $x$ and $y$. Given that $f(1)>0$, find the value of $f(4)$.
Answer: $\frac{3}{2}$
Solution: Setting $x=y=1$ in the functional equation and letting $c=f(1)>0$, we have $2 c=c^{2}-3$ or $(c+1)(c-3)=0$. Thus $c=f(1)=3$ and setting $y=1$ yields
$$
f(x)+3=3 f(x)+1-\frac{4}{x} \Longrightarrow f(x)=1+\frac{2}{x}
$$
for all nonzero reals $x$. Hence, $f(4)=\frac{3}{2}$.