## Task B-2.3.

Let $f(x)$ be a function for which $f\left(\frac{x+2}{x-3}\right)=x^{2}+x+1$ and let the function $g(x)$ be defined by $g(x)=\sqrt{f(x)-1}$. Determine the natural domain of the function $g(x)$.
## Solution.

If we introduce the substitution $\frac{x+2}{x-3}=t$, we get $x=\frac{3 t+2}{t-1}$, and from the equality $f\left(\frac{x+2}{x-3}\right)=x^{2}+x+1$ it follows that $f(t)=\left(\frac{3 t+2}{t-1}\right)^{2}+\frac{3 t+2}{t-1}+1$.

The function $f$ can also be written in terms of $x$. Then the function $g(x)$ is defined for all real numbers $x$ for which $f(x)-1 \geqslant 0$, that is,

$$
\left(\frac{3 x+2}{x-1}\right)^{2}+\frac{3 x+2}{x-1}=\frac{12 x^{2}+11 x+2}{(x-1)^{2}} \geqslant 0
$$

This inequality is equivalent to the inequality $12 x^{2}+11 x+2 \geqslant 0$ under the condition $x \neq 1$.

The solution to the inequality $12 x^{2}+11 x+2 \geqslant 0$ is the interval $\left\langle-\infty,-\frac{2}{3}\right] \cup\left[-\frac{1}{4}, \infty\right\rangle$,

so the natural domain of the function $g$ is the interval $\left\langle-\infty,-\frac{2}{3}\right] \cup\left[-\frac{1}{4}, 1\right\rangle \cup\langle 1, \infty\rangle$.

Note: If a student does not write the condition $x \neq 1$ and does not include it in the domain of the function $g$, but has done everything else correctly, they should be given a total of 4 points.