$$
\begin{array}{l}
\text { II. 9. Let } x=\cos \theta \text {. Then } \\
f(a, b)=\max _{\theta \in \mathbf{R}}\left\{\left|\cos ^{2} \theta-a \cos \theta-b\right|\right\} \\
=\max _{\theta \in \mathbf{R}}\left\{\frac{1}{2}\left|2 \cos ^{2} \theta-1-2 a \cos \theta-2 b+1\right|\right\} \\
=\max _{\theta \in \mathbf{R}}\left\{\frac{1}{2}|\cos 2 \theta-2 a \cos \theta-2 b+1|\right\} .
\end{array}
$$

Therefore, $f(a, b)$
$$
\begin{array}{l}
\leqslant \frac{1}{2} \max _{\theta \in \mathbf{R}}\{|\cos 2 \theta|+|2 a \cos \theta+2 b-1|\} \\
\leqslant \frac{1}{2} \max _{\theta \in \mathbf{R}}\{1+|2 a \cos \theta+2 b-1|\} .
\end{array}
$$

When $a=0, b=\frac{1}{2}$,
$$
\frac{1}{2} \max _{\theta \in \mathbf{R}}\{1+|2 a \cos \theta+2 b-1|\}
$$

reaches the minimum value $\frac{1}{2}$.
Thus, $f_{\text {min }}(a, b) \leqslant \frac{1}{2}$.
If $f_{\text {min }}(a, b) < \frac{1}{2}$, it leads to a contradiction.
Thus, $f_{\text {min }}(a, b) \geqslant \frac{1}{2}$.
Combining with equation (1), we get that the minimum value of $f(a, b)$ is $\frac{1}{2}$.
9. (16 points) Given the bivariate function
$$
f(a, b)=\max _{x \in[-1,1]}\left\{\left|x^{2}-a x-b\right|\right\} \text {. }
$$

Find the minimum value of $f(a, b)$.