9. Let $a, b$ be real numbers, for any real number $x$ satisfying $0 \leqslant x \leqslant 1$ we have $|a x+b| \leqslant 1$. Then the maximum value of $|20 a+14 b|+|20 a-14 b|$ is $\qquad$.
9. 80 .

Let $x=0$, we know $|b| \leqslant 1$;
Let $x=1$, we know $|a+b| \leqslant 1$.
Therefore, $|a|=|a+b-b| \leqslant|a+b|+|b| \leqslant 2$, when $a=2, b=-1$, the equality can be achieved.
If $|20 a| \geqslant|14 b|$, then
$$
\begin{array}{l}
|20 a+14 b|+|20 a-14 b| \\
=2|20 a|=40|a| \leqslant 80 ;
\end{array}
$$

If $|20 a| \leqslant|14 b|$, then
$$
\begin{array}{l}
|20 a+14 b|+|20 a-14 b| \\
=2|14 b|=28|b| \leqslant 28 .
\end{array}
$$

In summary, the maximum value sought is 80, which can be achieved when $a=2, b=-1$.