1. Determine the minimum value of the function

$$
f(x)=\frac{x^{2}}{x-9}
$$

on the interval $x \in(9, \infty)$.
1. The sought value is the greatest value of the real number $a$ for which the inequality $\frac{x^{2}}{x-9} \geqslant a$ is satisfied on the entire interval $x \in(9, \infty)$. Note that the greatest such value of $a$ is certainly non-negative, since the condition holds for $a=0$. By multiplying by $x-9$ (which we can do because this value is positive) and moving the right side to the left, the considered inequality reduces to $x^{2}-a x+9 a \geqslant 0$. Note that, for $x \leqslant 9$, the last inequality is always satisfied (the left side can be written in the form $x^{2}+a(9-x)$, where both addends are non-negative), so the condition that it be satisfied on the interval $x \in(9, \infty)$ is actually equivalent to the condition that it be satisfied for all $x \in \mathbb{R}$. This last condition reduces to the discriminant being non-positive, i.e., $a^{2}-36 a \leqslant 0$, or $a(36-a) \leqslant 0$. The solution to this inequality is clearly $a \in[0,36]$, so the greatest possible value of the number $a$ is 36.