3. For any $x \in[0,1]$, we have $|a x+b| \leqslant 1$.

Then the maximum value of $|b x+a|$ is $\qquad$
3. 2 .

Let $f(x)=a x+b$. Then
$$
\begin{array}{l}
b=f(0), a=f(1)-f(0) . \\
\text { Hence }|b x+a|=|f(0) x+f(1)-f(0)| \\
=|f(0)(x-1)+f(1)| \\
\leqslant|f(0)||x-1|+|f(1)| \\
\leqslant 1+1=2 .
\end{array}
$$

When $a=2, b=-1, x=0$, the maximum value 2 is obtained.