$\begin{array}{l}\text { 9. (16 points) Let } \\ f_{n}(x)=\sin ^{2 n} x+\cos ^{2 n} x(x \in \mathbf{R}, n \in \mathbf{N}) \text {. } \\ \text { Let } S=f_{1}(x)-f_{2}(x)+f_{3}(x)-f_{4}(x)+f_{5}(x) \text {. }\end{array}$
Find the maximum and minimum values of $S$.
$$
\begin{array}{l}
f_{n}(x)=\sin ^{2 n} x+\cos ^{2 n} x \\
=f_{n-1}(x)-\sin ^{2} x \cdot \cos ^{2} x \cdot f_{n-2}(x) .
\end{array}
$$

Let $\sin ^{2} x \cdot \cos ^{2} x=k$. Then
$$
f_{n}(x)=f_{n-1}(x)-k f_{n-2}(x) \text {, }
$$

or $f_{n-1}(x)-f_{n}(x)=k f_{n-2}(x)$.
It is easy to see that $f_{0}(x)=2, f_{1}(x)=1$,
$$
\begin{array}{l}
f_{2}(x)=f_{1}(x)-k f_{0}(x)=1-2 k, \\
f_{3}(x)=f_{2}(x)-k f_{1}(x)=1-3 k . \\
\text { Hence } S=f_{1}(x)-k f_{1}(x)-k f_{3}(x) \\
=3 k^{2}-2 k+1=3\left(k-\frac{1}{3}\right)^{2}+\frac{2}{3} .
\end{array}
$$

Since $k=\sin ^{2} x \cdot \cos ^{2} x=\frac{1}{4} \sin ^{2} 2 x \in\left[0, \frac{1}{4}\right]$, then
when $k=0$, $S_{\max }=1$;
when $k=\frac{1}{4}$, $S_{\text {min }}=\frac{11}{16}$.