3. A function $f$ is such that $f: \mathbb{R} \rightarrow \mathbb{R}$ where $f(x y+1)=f(x) f(y)-f(y)-x+2$ for all $x, y \in \mathbb{R}$. Find $10 f(2006)+f(0)$.
3. Ans: 20071
$f(x y+1)=f(x) f(y)-f(y)-x+2$, so we have $f(y x+1)=f(y) f(x)-f(x)-y+2$. Subtracting, we have $0=f(x)-f(y)+y-x$ or $f(x)+y=f(y)+x$. Let $y=0$. Then $f(x)=f(0)+x$. Substitute into the given identity and putting $x=y=0$, we get
$$
f(0)+1=f(0) f(0)-f(0)+2, \quad \text { or } \quad(f(0)-1)^{2}=0 .
$$

Thus $f(0)=1$ and $10 f(2006)+f(0)=20071$.