6. 81 Function $f$ is defined on the real number domain, and satisfies the following conditions: for any real number $x$, $f(2+x)=f(2-x), f(7+x)=f(7-x)$.

If $x=0$ is a root of $f(x)=0$, then $f(x)=0$ should have at least how many roots in the interval $-1000 \leqslant x \leqslant 1000$?
［Solution］From the given conditions, we have
$$
\begin{array}{l}
f(x)=f(2+x-2)=f[2-(x-2)]=f(4-x), \quad \text { (1) } \\
f(4-x)=f[7-(x+3)]=f[7+(x+3)]=f(10+x),
\end{array}
$$

From (1) and (2), we get $f(x)=f(10+x)$.
Thus, from $f(0)=0$, we have
$$
f(0)=f( \pm 10)=f( \pm 20)=\cdots=f( \pm 1000)=0 .
$$

This gives us 201 roots of $f(x)=0$ in the interval $[-1000,1000]$.
From (1), we also get $f(4)=f(0)=0$.
Then, from (3), we have
$$
\begin{array}{l}
f(4)=f(4 \pm 10)=f(4 \pm 20) \\
=\cdots=f(4 \pm 990) \\
=f(4-1000)=0 .
\end{array}
$$

This gives us another 200 roots of $f(x)=0$ in the interval $[-1000,1000]$.
Therefore, $f(x)=0$ has at least 401 roots in the interval $[-1000,1000]$.