For example, let the function $f(x) = ax^2 + 8x + 3$ where $a < 0$. For a given negative number $a$, there is a maximum positive number $l(a)$ such that the inequality $|f(x)| \leq 5$ holds for the entire interval $[0, l(a)]$. Question: For what value of $a$ is $l(a)$ maximized? Find this maximum $l(a)$ and prove your conclusion.
(1998 National High School Mathematics Competition Problem)
Solve $f(x)=a\left(x+\frac{4}{a}\right)^{2}+3-\frac{16}{a}, \max _{x \in \mathbf{R}} f(x)=3-\frac{16}{a}$ (since $a5$ , $l(a)$ is the smaller root of the equation $a x^{2}+8 x+3=5$ . $a x^{2}+8 x-2=0$
$$
\begin{aligned}
l(a) & =\frac{1}{2 a}[-8+\sqrt{64+8 a}]=\frac{1}{a}[\sqrt{16+2 a}-4] \\
& =\frac{16+2 a-16}{a(\sqrt{16+2 a}+4)} \\
& =\frac{2}{\sqrt{16+2 a}+4}\frac{1}{2}
\end{aligned}
$$
Thus, the maximum root of $l(a)$ is $\frac{\sqrt{5}+1}{2}$, and at this point $a=-8$ (Figure $4-2$ ).