7. Given a function $f(x)$ defined on $\mathbf{R}$ with period $T$ that satisfies: $f(1+x)=f(1-x)$ and $f(8+x)=f(8-x)$, the maximum value of $T$ is $\qquad$ .
7. 14 .
$$
f(x)=f(1+(x-1))=f(1-(x-1))=f(2-x)=f(8+(-6-x))=f(8-
$$
$(-6-x))=f(14+x)$, then $T \leqslant 14$, a function with a period of 14 that meets the problem's requirements can be constructed.