Question 27 Let $f(x)=\frac{x+6}{x+2}$, try to find $f^{[n]}(x)$.
Solve: Since
$$
f^{[n+1]}(x)=f\left(f^{[n]}(x)\right)=\frac{f^{[n]}(x)+6}{f^{[n]}(x)+2} .
$$

Let $f(x)=x$, solving for the fixed point values gives $x_{1}=2, x_{2}=-3$. Therefore,
$$
\begin{aligned}
\frac{f^{[n]}(x)-2}{f^{[n]}(x)+3} & =\frac{\frac{f^{[n-1]}(x)+6}{f^{[n-1]}(x)+2}-2}{\frac{f^{[n-1]}(x)+6}{f^{[n]}(x)+2}+3}=-\frac{1}{4} \frac{f^{[n-1]}(x)-2}{f^{[n-1]}(x)+3}=\cdots \\
& =\frac{f^{[0]}(x)-2}{f^{[0]}(x)+3}\left(-\frac{1}{4}\right)^{n}=\frac{x-2}{x+3}\left(-\frac{1}{4}\right)^{n} .
\end{aligned}
$$

From this, we solve
$$
f^{[n]}(x)=\frac{\left[2(-4)^{n}+3\right] x+6\left[(-4)^{n}-1\right]}{\left[(-4)^{n}+1\right] x+\left[3(-4)^{n}+2\right]} .
$$

The following problem characterizes the features of functions with fixed points.