Example 10 Determine all polynomials $f(x)$ that satisfy the following condition:
$$
f(x+1)=\frac{1}{2} f[f(x)]+\frac{3}{2} .
$$
Solution: Let $f(x)=a_{0} x^{n}+a_{1} x^{n-1}+\cdots+a_{n}\left(a_{0} \neq 0\right)$,
substituting into the original functional equation yields
$$
\begin{array}{l}
a_{0}(x+1)^{n}+a_{1}(x+1)^{n-1}+\cdots+a_{n} \\
=\frac{1}{2}\left[a_{0}\left(a_{0} x^{n}+\cdots+a_{n}\right)^{n}+a_{1}\left(a_{0} x^{n}+\cdots+a_{n}\right)^{n-1}+\cdots+a_{n}\right]+\frac{3}{2} .
\end{array}
$$

Comparing the highest power of $x$ on both sides, we get $n^{2}=n$, thus $n=0$ or 1. When $n=0$, equation (*) simplifies to $a_{0}=\frac{1}{2} a_{0}+\frac{3}{2}$, yielding $a_{0}=3$, hence $f(x)=3$. When $n=1$, equation (*) simplifies to
$$
a_{0}(x+1)+a_{1}=\frac{1}{2}\left[a_{0}\left(a_{0} x+a_{1}\right)+a_{1}\right]+\frac{3}{2} \text {. }
$$

Comparing the coefficients of the same powers of $x$ on both sides, we get
$$
\left\{\begin{array}{l}
a_{0}=\frac{1}{2} a_{0}^{2}, \\
a_{0}+a_{1}=\frac{1}{2} a_{0} a_{1}+\frac{1}{2} a_{1}+\frac{3}{2} .
\end{array}\right.
$$

Since $a_{0} \neq 0$, we have $a_{0}=2, a_{1}=1$, thus we deduce $f(x)=2 x+1$.
It is easy to verify that $f(x)=3, f(x)=2 x+1$ both satisfy the original functional equation, hence the solutions to the functional equation are $f(x)=3$ or $f(x)=2 x+1$.