3. Given a function $f(x)$ defined on $\mathbf{R}$ with period $T$ that satisfies $f(1+x)=f(1-x)$ and $f(8+x)=f(8-x)$. Then the maximum value of $T$ is ( ).
(A) 16
(B) 14
(C) 8
(D) 2
3. (B).
$$
\begin{array}{l}
f(x)=f(1+(x-1))=f(1-(x-1)) \\
=f(2-x)=f(8+(-6-x)) \\
=f(8-(-6-x))=f(14+x) .
\end{array}
$$

According to the definition of the period, we know that $T \leqslant 14$. Figure 1 is the graph of a function that meets the conditions with a period of 14. Therefore, the maximum value of $T$ is 14.