4. 3. 11 $\star \star$ Find the maximum and minimum values of the function $y=\sqrt{3 x+4}+\sqrt{3-4 x}$.
Solve the system $\left\{\begin{array}{l}3 x+4 \geqslant 0, \\ 3-4 x \geqslant 0\end{array}\right.$ to get $x \in\left[-\frac{4}{3}, \frac{3}{4}\right]$. Since
$$
\begin{aligned}
y^{2} & =(\sqrt{3 x+4}+\sqrt{3-4 x})^{2}=-x+7+2 \sqrt{(3 x+4)(3-4 x)} \\
& \geqslant-x+7 \geqslant-\frac{3}{4}+7=\frac{25}{4},
\end{aligned}
$$

thus $y \geqslant \frac{5}{2}$, with equality when $x=\frac{3}{4}$. Therefore, the minimum value of $y$ is $\frac{5}{2}$. By the Cauchy-Schwarz inequality, we have
$$
y^{2}=(\sqrt{3 x+4}+\sqrt{3-4 x})^{2} \leqslant(3+4)\left(\left(x+\frac{4}{3}\right)+\left(\frac{3}{4}-x\right)\right)=\frac{175}{12},
$$

thus $y \leqslant \frac{5 \sqrt{21}}{6}$, with equality when $x=-\frac{37}{84} \in\left[-\frac{4}{3}, \frac{3}{4}\right]$. Therefore, the maximum value of $y$ is $\frac{5 \sqrt{21}}{6}$.