9. (16 points) The function $f_{n}(x)(n=1,2,3, \cdots)$ is defined as follows:
$$
f_{1}(x)=4\left(x-x^{2}\right) \quad(0 \leqslant x \leqslant 1), \quad f_{n+1}(x)=f_{n}\left(f_{1}(x)\right) \quad(n=1,2, \cdots) .
$$

Let the number of $x$ in $[0,1]$ where $f_{n}(x)$ attains its maximum value be $a_{n}$, and the number of $x$ where $f_{n}(x)$ attains its minimum value be $b_{n}$. Find the expressions for $a_{n}, b_{n}$ in terms of $n$.
9. $y=f_{1}(x)$ reaches its maximum value 1 at $x=\frac{1}{2}$, and its minimum value 0 at $x=0,1$, so $a_{1}=1, b_{1}=2$. For any positive integer $n$, $x=0,1$ are always the minimum value points of $f_{n}(x)$, so the number of maximum value points of $f_{n}(x)$ in $0<x \leqslant 1$ is $a_{n}$. We will use mathematical induction to prove that the number of maximum value points and minimum value points of $f_{n}(x)$ in $0<x \leqslant 1$ are equal. Assuming that the number of minimum value points of $f_{n}(x)$ in $0<x \leqslant 1$ is also $a_{n}$, then the number of maximum value points and minimum value points of $f(x)$ in $0 \leqslant x<1$ are both $a_{n}$. Consider $f_{n+1}(x)=f_{n}\left(f_{1}(x)\right)=f_{n}(y)$, where $y=f_{1}(x)$.

When $0<x \leqslant \frac{1}{2}$, $0<y \leqslant 1$ is monotonically increasing, making $y$ take $a_{n}$ maximum value points and $a_{n}$ minimum value points of $f_{n}(y)$. The $x \in\left(0, \frac{1}{2}\right]$ that make $y$ take these points are also $a_{n}$ each.

When $\frac{1}{2}<x \leqslant 1$, $0 \leqslant y<1$ is monotonically decreasing, making $y$ take $a_{n}$ maximum value points and $a_{n}$ minimum value points of $f_{n}(y)$. The $x \in\left(\frac{1}{2}, 1\right]$ that make $y$ take these points are also $a_{n}$ each.

Therefore, the number of maximum value points and minimum value points of $f_{n+1}(x)$ in $0<x \leqslant 1$ are each $a_{n+1}=2 a_{n}$. Given $a_{1}=1$, it is easy to see that $a_{n}=2^{n-1}$. Thus, $b_{n}=a_{n}+1=2^{n-1}+1$.