Example 3 Let the function $f(x)=a x^{2}+8 x+3(a<0)$, for a given negative number $a$, there is a maximum positive number $l(a)$, such that on the entire interval $[0, l(a)]$, the inequality $|f(x)| \leqslant 5$ holds. For what value of $a$ is $l(a)$ maximized? Find this maximum value of $l(a)$ and prove your conclusion.
Solution: $f(x)=a\left(x+\frac{4}{a}\right)^{2}+3-\frac{16}{a}$.
Therefore, $\max _{x \in \mathbf{R}} f(x)=3-\frac{16}{a}$.
We discuss in two cases.
(1) When $3-\frac{16}{a}>5$, i.e., $-8-\frac{4}{a} \text {. }$
Thus, $l(a)$ is the larger root of the equation $a x^{2}+8 x+3=-5$. Therefore,
$$
\begin{array}{l}
l(a)=\frac{-8-\sqrt{64-32 a}}{2 a}=\frac{4}{\sqrt{4-2 a}-2} \\
\leqslant \frac{4}{\sqrt{20}-2}=\frac{\sqrt{5}+1}{2} .
\end{array}
$$

The equality holds if and only if $a=-8$.
Since $\frac{\sqrt{5}+1}{2}>\frac{1}{2}$, therefore, $l(a)$ reaches its maximum value $\frac{\sqrt{5}+1}{2}$ if and only if $a=-8$.