6. Find the set of values of the function $f(x)=g(2 \sqrt{2.5-g(x)})$, where $g(x)=\frac{3}{|x-2|+1}$.
# Solution:

The function $g(x)=\frac{3}{|x-2|+1}$ is defined on the entire number line and takes all values from

![](https://cdn.mathpix.com/cropped/2024_05_06_331dee27c7cee369cafag-19.jpg?height=545&width=1060&top_left_y=407&top_left_x=178)

the interval $(0 ; 3]$. The function $g(x)$ reaches its maximum value at the point $\quad x=2, \quad g_{\max }=g(2)=3, \quad$ on the interval $(-\infty ; 2)$ the function $g(x)$ is increasing, on the interval $(2 ;+\infty)-$ it is decreasing. $\quad$ The graph of the function $g(x)$ is shown in the figure. The function $\phi(t)=2 \sqrt{2.5-t}$ is defined for $t \in(-\infty ; 2.5] . \quad$ For $t=g(x)$, the function $\phi(t)=2 \sqrt{2.5-t}$ takes its values for $t \in(0 ; 2.5]$, and the set of values of this function is the interval $[0 ; 2 \sqrt{2.5})$. To find the set of values of the function $f(x)$, it is sufficient to find the set of values of the function $g(x)$ on the interval $[0 ; 2 \sqrt{2.5})$. On the specified interval, $g(x)$ takes all values from the set $[1 ; 3]$.

Answer: $E_{f}=[1 ; 3]$.