2. (12 points) The sequence of functions is defined by the formulas:

$$
f_{0}(x)=2 \sqrt{x}, f_{n+1}(x)=\frac{4}{2-f_{n}(x)}, n=0,1,2 \ldots, x \in[4 ; 9]
$$

Find $f_{2023}(9)$.
Answer: -1.

Solution. It is easy to calculate that $f_{3}(x)=f_{0}(x)$, therefore,

$$
f_{2023}(x)=f_{1}(x)=\frac{2}{1-\sqrt{x}}
$$

Then $f_{2023}(9)=-1$.

Remark. One can immediately compute the values of the functions at the given point. The sequence will be $f_{0}(9)=6, f_{1}(9)=-1, f_{2}(9)=\frac{4}{3}, f_{3}(9)=6 \ldots$

Evaluation Criteria. A fully justified solution - 12 points. Calculation errors - minus 1 point. The relation $f_{3}(x)=f_{0}(x)$ found - 7 points, the equality $f_{2023}(x)=f_{1}(x)$ found - another 4 points.