2. Find all values of the variable $x$, for each of which both expressions $f(x)=\sqrt{21-x^{2}-4 x}$ and $g(x)=|x+2|$ are defined, and $\min (f(x) ; g(x))>\frac{x+4}{2}$.
Answer: $x \in\left[-7 ;-\frac{8}{3}\right) \cup(0 ; 2)$.

Solution. Both functions are defined when $21-x^{2}-4 x \geqslant 0 \Leftrightarrow -7 \leqslant x \leqslant 3$. Note that the following two statements are equivalent: "the smaller of two numbers is greater than $A$" and "both numbers are greater than $A$". Therefore, the given inequality is equivalent to the system of inequalities

$$
\left\{\begin{array}{l}
f(x)>\frac{x+4}{2} \\
g(x)>\frac{x+4}{2}
\end{array}\right.
$$

Since the functions $f(x)$ and $g(x)$ are non-negative on the domain of definition, both inequalities of the system are satisfied when $\frac{x+4}{2}\frac{x^{2}+8 x+16}{4} \\ x^{2}+4 x+4>\frac{x^{2}+8 x+16}{4}\end{array} \Leftrightarrow\left\{\begin{array}{l}5 x^{2}+24 x-680\end{array} \Leftrightarrow\left\{\begin{array}{l}x \in\left(-\frac{34}{5} ; 2\right), \\ x \in\left(-\infty ;-\frac{8}{3}\right) \cup(0 ;+\infty)\end{array} \Leftrightarrow x \in\left(-\frac{34}{5} ;-\frac{8}{3}\right) \cup(0 ; 2)\right.\right.\right.$.

Considering the interval $x \in[-4 ;+\infty)$, we get $x \in\left[-4 ;-\frac{8}{3}\right) \cup(0 ; 2)$. Combining the results, we finally find that $x \in\left[-7 ;-\frac{8}{3}\right) \cup(0 ; 2)$.