Example 15 (1998 National High School Competition Question) Let the function $f(x)=a x^{2}+8 x+3(a<0)$, for a given negative number $a$, there is a largest positive number $l(a)$, such that the inequality $|f(x)| \leqslant 5$ holds for the entire interval $[0, l(a)]$. For what value of $a$ is $l(a)$ the largest? Find this largest $l(a)$ and prove your conclusion.
Solution: $f(x)=a\left(x+\frac{4}{a}\right)^{2}+3-\frac{16}{a}$,
$\therefore \max _{x \in \mathbf{R}} f(x)=3-\frac{16}{a}$.
(1) $3-\frac{16}{a}>5$, i.e., $-8-\frac{4}{a}$, and $f(0)=3$, so $l(a)$ is the larger root of the equation $a x^{2}+8 x+3=-5$, i.e., $l(a)=\frac{-8-\sqrt{64-32 a}}{2 a}=$ $\frac{2}{\sqrt{4-2 a}-2} \leqslant \frac{4}{\sqrt{20}-2}=\frac{\sqrt{5}+1}{2}$, with equality holding if and only if $a=$ -8. Since $\frac{\sqrt{5}+1}{2}>\frac{1}{2}$, therefore, $l(a)$ attains its maximum value $\frac{\sqrt{5}+1}{2}$ if and only if $a=-8$.