6. 82 Find the smallest real number $A$, such that for every quadratic polynomial $f(x)$ satisfying the condition
$$|f(x)| \leqslant 1 \quad(0 \leqslant x \leqslant 1)$$

the inequality $f^{\prime}(0) \leqslant A$ holds.
[Solution] Let the quadratic trinomial be $f(x)=a x^{2}+b x+c$. When $0 \leqslant x \leqslant 1$, it satisfies the inequality
$$|f(x)| \leqslant 1.$$

In particular, there should be the inequalities
$$|f(0)| \leqslant 1,\left|f\left(\frac{1}{2}\right)\right| \leqslant 1,|f(1)| \leqslant 1.$$

Since $f(0)=c, f\left(\frac{1}{2}\right)=\frac{a}{4}+\frac{b}{2}+c, f(1)=a+b+c$, and because
$$f^{\prime}(0)=b=4\left(\frac{a}{4}+\frac{b}{2}+c\right)-(a+b+c)-3 c,$$

we have
$$\begin{aligned}
\left|f^{\prime}(0)\right| & \leqslant 4\left|\frac{a}{4}+\frac{b}{2}+c\right|+|a+b+c|+3|c| \\
& =4\left|f\left(\frac{1}{2}\right)\right|+|f(1)|+3|f(0)| \\
& \leqslant 4+1+3=8.
\end{aligned}$$

Therefore, $A \leqslant 8$.
On the other hand, the quadratic trinomial
$f(x)=-8 x^{2}+8 x-1=-2(2 x-1)^{2}+1$ satisfies the inequality (1). Indeed, when $0 \leqslant x \leqslant 1$, we have $-1 \leqslant 2 x-1 \leqslant 1$, so $0 \leqslant(2 x-1)^{2} \leqslant 1$. Thus, $-2 \leqslant-2(2 x-1)^{2} \leqslant 0$, and therefore $-1 \leqslant f(x) \leqslant 1$. Since $f^{\prime}(x)=-16 x+8$, we have $f^{\prime}(0)=8$. Therefore, $A \geqslant 8$.

Since $A \geqslant 8$ and $A \leqslant 8$, we have $A=8$.