Three, (20 points) Find the smallest real number $A$, such that for each quadratic trinomial $f(x)$ satisfying the condition $|f(x)| \leqslant 1(0 \leqslant x \leqslant 1)$, the inequality $f^{\prime}(0) \leqslant A$ holds.
Three, let the quadratic trinomial be
$$
f(x)=a x^{2}+b x+c(a \neq 0) \text {. }
$$

From the problem, we know
$$
|f(0)| \leqslant 1,\left|f\left(\frac{1}{2}\right)\right| \leqslant 1,|f(1)| \leqslant 1 .
$$

Notice that $f(0)=c, f\left(\frac{1}{2}\right)=\frac{a}{4}+\frac{b}{2}+c$,
$$
\begin{array}{l}
f(1)=a+b+c, \\
f^{\prime}(0)=b \\
=4\left(\frac{a}{4}+\frac{b}{2}+c\right)-(a+b+c)-3 c .
\end{array}
$$

Then $\left.\left|f^{\prime}(0)\right| \leqslant 4\left|f\left(\frac{1}{2}\right)\right|+|f(1)|+3 \right| f(0)$ |
$$
\leqslant 4+1+3=8 \text {. }
$$

Therefore, $A \leqslant 8$.
$$
\text { Also, } f(x)=-8 x^{2}+8 x-1 \text {, when } 0 \leqslant x \leqslant 1
$$

then, $|f(x)| \leqslant 1, f^{\prime}(x)=-16 x+8, f^{\prime}(0)=8$.
Thus, $A \geqslant 8$.
Therefore, $A=8$.