2. Find all values of the variable $x$, for each of which both expressions $f(x)=\sqrt{16-x^{2}+6 x}$ and $g(x)=|x-3|$ are defined, and $\min (f(x) ; g(x))>\frac{5-x}{2}$.
Answer: $x \in(-1 ; 1) \cup\left(\frac{11}{3} ; 8\right]$.

Solution. Both functions are defined when $16-x^{2}+6 x \geqslant 0 \Leftrightarrow -2 \leqslant x \leqslant 8$. 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

$$
\left\{\begin{array}{l}
f(x)>\frac{5-x}{2} \\
g(x)>\frac{5-x}{2}
\end{array}\right.
$$

Since the functions $f(x)$ and $g(x)$ are non-negative on the domain of definition, both inequalities of the system are satisfied when $\frac{5-x}{2} < 0$, and considering the domain of definition - when $x \in(5 ; 8]$. If $x \leqslant 5$, then both the left and right sides of both inequalities are non-negative, and squaring them is an equivalent transformation. We get

$\left\{\begin{array}{l}16-x^{2}+6 x>\frac{x^{2}-10 x+25}{4} \\ x^{2}-6 x+9>\frac{x^{2}-10 x+25}{4}\end{array} \Leftrightarrow\left\{\begin{array}{l}5 x^{2}-34 x-39<0 \\ 3 x^{2}-14 x+11>0\end{array} \Leftrightarrow\left\{\begin{array}{l}x \in\left(-1 ; \frac{39}{5}\right), \\ x \in(-\infty ; 1) \cup\left(\frac{11}{3} ;+\infty\right)\end{array} \Leftrightarrow x \in(-1 ; 1) \cup\left(\frac{11}{3} ; \frac{39}{5}\right)\right.\right.\right.$.

Considering the interval $x \in(-\infty ; 5]$, we get $x \in(-1 ; 1) \cup\left(\frac{11}{3} ; 5\right]$. Combining the results, we finally find that $x \in(-1 ; 1) \cup\left(\frac{11}{3} ; 8\right]$.