Let $f_{1}(x)=-\frac{2 x+7}{x+3}, f_{n+1}(x)=f_{1}\left(f_{n}(x)\right)$. Determine the value of $f_{2001}(2002)$.
We will show that $f_{3}(x)=x$.

\[
\begin{aligned}
f_{2}(x) & =f_{1}\left(f_{1}(x)\right)=-\frac{2 f_{1}(x)+7}{f_{1}(x)+3}=-\frac{-2 \frac{2 x+7}{x+3}+7}{-\frac{2 x+7}{x+3}+3}= \\
& =-\frac{-4 x-14+7 x+21}{-2 x-7+3 x+9}=-\frac{3 x+7}{x+2}
\end{aligned}
\]

Similarly, we find that

\[
\begin{aligned}
f_{3}(x) & =f_{1}\left(f_{2}(x)\right)=-\frac{2 f_{2}(x)+7}{f_{2}(x)+3}=-\frac{-2 \frac{3 x+7}{x+2}+7}{-\frac{3 x+7}{x+2}+3}= \\
& =-\frac{-6 x-14+7 x+14}{-3 x-7+3 x+6}=-\frac{x}{-1}=x
\end{aligned}
\]

This means that $f_{6}(x)=f_{3}\left(f_{3}(x)\right)=f_{3}(x)=x$, and by induction, if $3 \mid n$, then $f_{n}(x)=x$. Since $3 \mid 2001$, we have $f_{2001}(x)=x$, and thus $f_{2001}(2002)=2002$.

Remarks. 1. From the solution, it also follows that

\[
f_{n}(x)=\left\{\begin{array}{c}
x \\
-\frac{2 x+7}{x+3}=f_{1}(x) \\
-\frac{3 x+7}{x+2}=f_{2}(x)
\end{array}\right.
\]

depending on whether $n$ modulo 3 is 0, 1, or 2.

2. In the solution, we treated the domain of the functions rather superficially. $f_{1}(x)$ is undefined precisely when $x=-3$. As a composite function, $f_{1}\left(f_{1}(x)\right)$ is defined if the inner function is defined, i.e., $x \neq -3$, and if $f_{1}(x)$ can be substituted, i.e., $f_{1}(x) \neq -3$, which means $x \neq -2$. The denominator of the fraction form of $f_{2}$ excludes exactly this value.

$f_{3}(x)=f_{1}\left(f_{2}(x)\right)$ is defined if $f_{2}(x)$ is defined, i.e., $x \neq -3$ and $x \neq -2$, and if $f_{2}(x)$ can be substituted into $f_{1}(x)$, i.e., $-\frac{3 x+7}{x+2} \neq -3$. Fortunately, this does not impose any new restrictions, as $f_{2}(x)$ does not take the value -3. The precise statement is: if $x \neq -2$ and $x \neq -3$, then $f_{3}(x)=x$, and at these two points, $f_{3}(x)$ is undefined. This means that if $n>3$, the domain of $f_{n}(x)$ does not shrink further, so it is justified to substitute 2002 into $f_{2001}$.

3. The key to the solution was that after three substitutions, we obtained the identity function $x \mapsto x$ (on the domain). This clearly depends on the given coefficients, and it is not obvious at first glance how this happens. The question arises whether we can arrive at this insight only through patient trial and error as a "reward."

## The graph of $f_{1}(x)$

Reconsidering the above domain analysis, we can conjecture and even visually verify the found property without the necessary calculations in the solution.

We saw that in addition to the obvious exclusion of -3, we must also exclude -2, which is $f_{1}^{-1}(-3)$. The crucial observation is that we do not need to exclude any more values, since $f_{1}$ does not take the value -2. The graph shows that this is due to the horizontal asymptote of the curve being $y=-2$.

The asymptotic behavior can also be intuitively described as follows: if $x=-3$, then the function value is "infinitely large," and if $x$ is "infinitely large," then the function value is -2.

If we were to write this down, we could do it as follows:

\[
f_{1}(-3)=\infty, \quad f_{1}(\infty)=-2
\]

If we now add that $f_{1}(-2)=-3$, it is hard not to notice the

\[
-2 \xrightarrow{f_{1}}-3 \xrightarrow{f_{1}} \infty \xrightarrow{f_{1}}-2
\]

"chain": if we "introduce" the $\infty$ value into the real numbers and extend the function $f_{1}$ accordingly, then this three-cycle means that $f_{3}(-2)=-2$, $f_{3}(-3)=-3$, and $f_{3}(\infty)=\infty$.

From this, we can infer that $f_{3}(x)$ will be a "friendly" function, and we can approach the calculations with greater confidence. But this experiment essentially also provides a proof if we consider that $f_{3}$ itself is a rational first-degree function, which is uniquely determined by three substitution values. If $f_{3}(x)$ takes the same value as the identity function $x \mapsto x$ at three points, then it is identical to it over its entire domain. Of course, we need to extend the domain and range of the identity function $x \mapsto x$ to include the $\infty$ value.

4. From the above, it also follows how functions similar to $f_{1}$ can be constructed. On the extended domain, the

\[
f(x)=\frac{a x+b}{c x+d} \quad \text { function has new values } \quad f(\infty)=\frac{a}{c} \quad \text { and } \quad f\left(-\frac{d}{c}\right)=\infty \text {. }
\]

Thus, if we want to achieve the three-cycle $f: \infty \rightarrow u \rightarrow v \rightarrow \infty$ for any $u \neq v$, the function

\[
f(x)=\frac{u(x-v)-(u-v)^{2}}{x-v}
\]

is obtained. This is how the function $f_{1}$ in the problem was constructed with $u=-2$ and $v=-3$.