$6 \cdot 82$ 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 $f(x) = ax^2 + bx + c$. When $0 \leq x \leq 1$, it satisfies the inequality
$$
|f(x)| \leq 1.
$$

In particular, the following inequalities should hold:
$$
|f(0)| \leq 1, \left|f\left(\frac{1}{2}\right)\right| \leq 1, |f(1)| \leq 1.
$$

Since
$$
\begin{array}{c}
f(0) = c, f\left(\frac{1}{2}\right) = \frac{a}{4} + \frac{b}{2} + c, f(1) = a + b + c, \text{ and } \\
f'(0) = b = 4\left(\frac{a}{4} + \frac{b}{2} + c\right) - (a + b + c) - 3c,
\end{array}
$$

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

Therefore, $A \leq 8$.
On the other hand, the quadratic trinomial
$$
f(x) = -8x^2 + 8x - 1 = -2(2x - 1)^2 + 1 \text{ satisfies inequality (1).}
$$

In fact, when $0 \leq x \leq 1$, we have $-1 \leq 2x - 1 \leq 1$, so $0 \leq (2x - 1)^2 \leq 1$. Thus, $-2 \leq -2(2x - 1)^2 \leq 0$, and therefore $-1 \leq f(x) \leq 1$. Since $f'(x) = -16x + 8$, we have $f'(0) = 8$. Therefore, $A \geq 8$.
Since $A \geq 8$ and $A \leq 8$, we conclude that $A = 8$.