12. Function f is defined on the real number domain, and satisfies the following conditions: For any real number $\mathrm{x}, \mathrm{f}(2+\mathrm{x})=\mathrm{f}(2-\mathrm{x})$, and $f(7+x)=f(7-x)$. If $x=0$ is a root of $f(x)=0$, find the minimum number of roots of $f(x)=0$ in the interval -1000 $\leqslant x \leqslant 1000$.
Molybdenum First, find the numbers in the domain for which $f$ is periodic from the given equations.
We have
$$
\begin{array}{l} 
f(x)=f(2+(x-2))=f(2-(x- \\
2))=f(4-x),(1) \\
f(4-x)=f(7-(x+3))=f(7+ \\
(x+3))=f(x+10) .(2)
\end{array}
$$

From (1) and (2), we get $f(x+10)=f(x)$

Substituting $x+10$ and $x-10$ for $x$ in (3), we get
$$
f(x+10)=f(x+20) \text { and } f(x-10)
$$
$=f(x)$, and continuing this process, we have
$f(x+10n)=f(x) \quad(n= \pm 1$, $\pm 2, \pm 3, \cdots)$.
Since $f(0)=0$, from (4) we get
$f( \pm 10)=f( \pm 20)=\ldots$
$=f( \pm 1000)=0$.
Thus, we have $f(x)=0$ at 201 roots in the interval $[-1000, 1000]$. Are there any other roots? Setting $x=0$ in (1), we get $f(4)=f(0)$ $=0$, and setting $x=4$ in (4), we get
$\mathrm{f}(4 \pm 10)=\mathrm{f}(4 \pm 20)=\cdots$
$=f(4 \pm 1000)=0$. This gives us another 200 roots of $f(x)=0$ in the interval $[-1000,1000]$, different from the previously found roots. These roots are $\mathrm{x}=-996,-986, \cdots$, $-6,4,14, \cdots, 994$. Since the piecewise linear function shown in the figure satisfies the given conditions and there are no other roots, the solution to the problem is 401.