7. Given real numbers $a, b, c$ satisfy $\left|a x^{2}+b x+c\right|$ has a maximum value of 1 on $x \in[-1,1]$. Then the maximum possible value of $\left|c x^{2}+b x+a\right|$ on $x \in[-1,1]$ is $\qquad$
7. 2 .

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

From $|b x+a+c|$
$$
\begin{array}{l}
\leqslant \max \{|a+b+c|,|a-b+c|\} \leqslant 1 \\
\Rightarrow\left|c x^{2}+b x+a\right|=\left|c x^{2}-c+b x+a+c\right| \\
\leqslant|c|\left|x^{2}-1\right|+|b x+a+c| \\
\leqslant\left|x^{2}-1\right|+1 \leqslant 2 \text {. } \\
\end{array}
$$

When $a=2, b=0, c=-1$, the conditions are satisfied and the equality holds.