5. The inequality $\left|f^{\prime}(0)\right| \leq A$ holds for all quadratic functions $f(x)$ satisfying $|f(x)| \leq 1(0 \leq x \leq 1)$, then the minimum value of the real number $\boldsymbol{A}$ is $\qquad$
5. 8

Analysis: Since $f(0)=c, f(1)=a+b+c, f(0.5)=0.25 a+0.5 b+c$, we have $f^{\prime}(0)=b=4(0.25 a+0.5 b+c)-(a+b+c)-3 c=4 f(0.5)-f(1)-3 f(0)$.
Noting that $|f(0)| \leq 1,|f(1)| \leq 1,|f(0.5)| \leq 1$, we get
$$
\left|f^{\prime}(0)\right|=|4 f(0.5)-f(1)-3 f(0)| \leq 4|f(0.5)|+|f(1)|+3|f(0)| \leq 8 ;
$$

Taking $a=-8, b=8, c=-1$, the conditions are satisfied, hence $A=8$;