Example 3 The function $f(x)$ is defined on the set of real numbers, and for all real numbers $x$ it satisfies the equations: $f(2+x)=f(2-x)$ and $f(7+x)=f(7-x)$. Suppose $x=0$ is a root of $f(x)=0$, and let $N$ be the number of roots of $f(x)=0$ in the interval $-1000 \leqslant x \leqslant 1000$. Find the minimum value of $N$.
From $f(2+x)=f(2-x)$, we know that the graph of the function $y=f(x)$ is symmetric about the line $x=2$. Since $x=0$ is a root of $f(x)=0$, it follows that $x=4$ is also a root of $f(x)$. Furthermore, from $f(7+x)=f(7-x)$, we know that the graph of the function $y=f(x)$ is symmetric about the line $x=7$, so $x=10$ is also a root of the equation $f(x)=0$. Thus, there is at least one root $x=4$ in the interval $[2,7]$. If the graph of $y=f(x)$ in the interval $[2,7]$ is a curve segment as shown in Figure 2-1-1, then according to the symmetry of the graph of the function $y=f(x)$ about the lines $x=2$ and $x=7$, we can obtain the graph of the function $y=f(x)$ in the intervals $[-3,2]$ and $[7,12]$. From this, we can further deduce that the graph of the function $y=f(x)$ has periodicity, with a period of 10, and that within one period, the equation $f(x)=0$ has at least 2 roots. The interval $[-1000,+1000]$ is 200 periods of $f(x)$, so in the interval $[-1000, +1000]$, $f(x)=0$ has at least $1+200 \times 2=401$ roots.

Note: This example can also deduce the periodicity of the function $y=f(x)$ based on the given conditions and the definition of periodicity, and then find the minimum number of roots of the equation $f(x)=0$.