9. Given a function $f(x)$ defined on $\mathbf{R}$ with the smallest positive period $T$ that satisfies $f(1+x)=f(1-x)$ and $f(8+x)=f(8-x)$, then the maximum value of $T$ is $\qquad$ .
9. 14. \( f(x) = f[1 + (x - 1)] = f[1 - (x - 1)] = f(2 - x) = f[8 + (-6 - x)] = f[8 - (-6 - x)] = f(14 + x) \).
According to the definition of the period, \( T \leqslant 14 \), Figure 1-7 is the graph of a function that meets the conditions with a period of 14.
Therefore, the maximum value of \( T \) is 14.
Note: This problem uses the method of example verification to show that the maximum value of the period \( T \) is 14. This example method often appears in competitions, which helps to cultivate the originality of thinking.