For example, the function $(f x)=\left(1+\frac{x}{2}\right)^{2}-2$, $x \in[-2,+\infty)$, find the solution set of the equation $f(x)=f^{-1}(x)$.
We can first draw the graph, noting that $f(x)$ is a parabola, and on $[-2,+\infty)$, it is intuitively clear that $y:=f(x)$ and $y=x$ intersect at only two points. Applying the proposition we have proven, to find the solution set of $f(x)=f^{-1}(x)$, we can first find the solution set of its subset $x=f(x)$.

$$
\text{It is easy to see } \begin{aligned}
x & =\left(\frac{x}{2}+1\right)^{2}-2 \text{ has solutions } \\
x_{1} & =-2, \quad x_{2}=2 .
\end{aligned}
$$

From the graph, we can see that $f(x)=f^{-1}(x)$ has exactly two roots, so $x_{1}=-2, x_{2}=2$ is the solution set we are looking for.

Please note that in the above problem-solving process, we can only conclude that $\{-2,2\}$ is the solution set of $f(x)=f^{-1}(x)$ if we have intuitively determined that the graphs of $y=f(x)$ and $y=f^{-1}(x)$ intersect at only two points. When solving problems using propositions, it is essential to make the judgment first and then solve the problem. Otherwise, roots may be lost. Sometimes, $M_{1} \neq M_{2}$, $M_{1} \neq M_{3}$, such as $y=-\frac{1}{3} x^{2}+\frac{7}{3} \quad(x \geqslant 0)$ and $y=\sqrt{-3 x+7}$. Note that $P(1,2)$ lies on both $y=-\frac{1}{3} x^{2}+\frac{7}{3}(x \geqslant 0)$ and $y=\sqrt{-3 x+7} \quad\left(x \leqslant \frac{1}{3}\right)$, but $P(1,2)$ does not lie on $y=x$.
(Author is a second-year junior high school student at Tianjin No. 20 Middle School)