Example 6 Let $f(x)$ be a function defined on $\mathbf{R}$. If $f(0)=2008$, and for any $x \in \mathbf{R}$, it satisfies
$$
\begin{array}{l}
f(x+2)-f(x) \leqslant 3 \times 2^{x}, \\
f(x+6)-f(x) \geqslant 63 \times 2^{x},
\end{array}
$$

then $f(2008)=$ $\qquad$
$(2008$, National High School Mathematics League Competition)
【Analysis】The problem provides two inequalities about $f(x)$. The key to solving the problem lies in appropriately transforming these two inequalities. Since it is a value-finding problem, we should strive to convert the inequality relations into equality relations.
Transform the two inequalities as follows:
$$
\begin{array}{l}
f(x+2)-f(x) \leqslant 3 \times 2^{x} \\
\Leftrightarrow f(x+2)-4 \times 2^{x}-f(x)+2^{x} \leqslant 0 \\
\Leftrightarrow\left(f(x+2)-2^{x+2}\right)-\left(f(x)-2^{x}\right) \leqslant 0 .
\end{array}
$$

Similarly,
$$
\begin{array}{l}
f(x+6)-f(x) \geqslant 63 \times 2^{x} \\
\Leftrightarrow\left(f(x+6)-2^{x+6}\right)-\left(f(x)-2^{x}\right) \geqslant 0 .
\end{array}
$$

By observing and comparing the above two inequalities, construct the function
$$
\begin{array}{l}
g(x)=f(x)-2^{x} . \\
\text { Then } g(x+2)-g(x) \leqslant 0, \\
g(x+6)-g(x) \geqslant 0 . \\
\text { Hence } g(x) \leqslant g(x+6) \leqslant g(x+4) \\
\leqslant g(x+2) \leqslant g(x) .
\end{array}
$$
$$
\begin{array}{l}
\text { Therefore, } g(x) \text { is a periodic function with a period of } 2. \\
\text { Then } f(2008)=g(2008)+2^{2008} \\
=g(0)+2^{2008}=2^{2008}+2007 .
\end{array}
$$