455. Find the domains of the following functions:
1) $y=\sqrt{3+2 x}$; 2) $f(x)=1+\sqrt{9-x^{2}}$;
2) $\varphi(x)=\sqrt{\lg \frac{5 x-x^{2}}{4}}$; 4) $y=\sqrt{3-x}+\arccos \frac{x-2}{3}$.
Solution. 1) The function is defined at those points where $3+2 x \geqslant 0$ or $2 x \geqslant-3$, from which $x \geqslant-\frac{3}{2}$. Thus, the function is defined on the interval $\left[-\frac{3}{2}, \infty\right)$.

2) The function takes real values at those points where the expression under the square root sign will have non-negative values: $9-x^{2} \geqslant 0$ or $x^{2} \leqslant 9$, from which $|x| \leqslant 3$, i.e., $-3 \leqslant x \leqslant 3$ (Fig. $53, a$). Thus, the function is defined on the segment $[-3,3]$.
3) We need to require that $\lg \frac{5 x-x^{2}}{4} \geqslant 0$, from which $\frac{5 x-x^{2}}{4} \geqslant 1$ or $x^{2}-5 x+4 \leqslant 0$. Solving this inequality, we get $1 \leqslant x \leqslant 4$. The function is defined on the segment $[1,4]$.
4) The function $\sqrt{3-x}$ is defined at those points where $3-x \geqslant 0$ or $x \leqslant 3$ (Fig. 53, b).

The function $\arccos \frac{x-2}{3}$ is defined for all values of $x$ satisfying the inequalities

$$
\left|\frac{x-2}{3}\right| \leqslant 1 \text { or }-1 \leqslant \frac{x-2}{3} \leqslant 1,
$$

from which

$$
-3 \leqslant x-2 \leqslant 3 \text { or }-1 \leqslant x \leqslant 5 \text { (Fig. } 53, \text { c). }
$$

Thus, the given function $y$ is defined on the segment $[-1,3]$ (Fig. 53 d).