11.5. Given a function $f(x)$, satisfying the condition

$$
f(x y+1)=f(x) f(y)-f(y)-x+2
$$

What is $f(2017)$, if it is known that $f(0)=1$?
Answer: 2018

Solution: $f(0 \cdot 0+1)=f(0) f(0)-f(0)-0+2, f(1)=1-1+2=2$,

$$
f(2017 \cdot 0+1)=f(2017) f(0)-f(0)-2017+2
$$

$$
\begin{gathered}
f(1)=f(2017)-1-2017+2=2 \\
f(2017)=2018
\end{gathered}
$$

Criteria: Only for the correct answer - 1 point.