## Task B-2.5.

The rule of association of the function $f$ is $f(x)=\sqrt{x-2}+\sqrt{x+7-6 \sqrt{x-2}}$. Determine the interval of real numbers on which this function takes a constant value.
## Solution.

The given function can be written in the following form:

$$
f(x)=\sqrt{x-2}+\sqrt{(\sqrt{x-2}-3)^{2}}=\sqrt{x-2}+|\sqrt{x-2}-3|
$$

The domain of the given function is the set $\{x \in \mathbf{R}, x \geqslant 2\}$.

If we introduce the substitution $\sqrt{x-2}=t$, we get the expression

$$
t+|t-3|=\left\{\begin{array}{l}
2 t-3, t \geqslant 3 \\
3, t \leqslant 3
\end{array}\right.
$$

By solving the inequality $\sqrt{x-2} \leqslant 3$, we get $x \leqslant 11$.

The final solution is $\{x \in \mathbf{R}, 2 \leqslant x \leqslant 11\}$.