Question 34: Let the function $\mathrm{f}(\mathrm{x})=\mathrm{ax}^{2}+8 \mathrm{x}+3(\mathrm{a}<0)$, for a given negative number $\mathrm{a}$, there is a maximum positive real number $L(a)$, such that the inequality $|f(x)| \leq 5$ holds for the entire interval $[0, L(a)]$. Ask: For what value of $a$ is $L(a)$ maximized? And find this maximum $L(a)$.

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Question 34, Solution: Since $f(x)=a\left(x+\frac{4}{a}\right)^{2}+3-\frac{16}{a}$, it follows that $\max _{x \in R} f(x)=f\left(-\frac{4}{a}\right)=3-\frac{16}{a}$. We will now discuss the cases:
(1) If $3-\frac{16}{a}>5$, i.e., $-8-\frac{4}{a}$, then $L(a)$ is the larger root of $a x^{2}+8 x+3=-5$, which is $L(a)=\frac{-8-\sqrt{64-32 a}}{2 a}=\frac{32 a}{2 a \cdot(\sqrt{64-32 a}-8)}=\frac{4}{\sqrt{4-2 a}-2} \leq \frac{4}{\sqrt{4-2 \times(-8)}-2}=\frac{2}{\sqrt{5}-1}=\frac{\sqrt{5}+1}{2}$. Equality holds if and only if $a=8$.
Since $\frac{\sqrt{5}+1}{2}>\frac{1}{2}$, it follows that $L(a)$ achieves its maximum value of $\frac{\sqrt{5}+1}{2}$ if and only if $a=8$.