## Task B-4.2.

Determine the natural domain of the function $f(x)=\sqrt{\log _{x^{2}-4}\left(x^{2}-4 x+3\right)}$.
## Solution.

Since the base of the logarithm depends on $x$, we need to consider the cases $x^{2}-4>1$ and $0<x^{2}-4<1$.

For the first case, if $x^{2}-4>1$, then $x \in\langle-\infty,-\sqrt{5}\rangle \cup\langle\sqrt{5},+\infty\rangle$. ( $\star$ )

The expression under the root must be non-negative, i.e., $\log _{x^{2}-4}\left(x^{2}-4 x+3\right) \geqslant 0$, or $x^{2}-4 x+2 \geqslant 0$, from which $x \in\langle-\infty, 2-\sqrt{2}] \cup[2+\sqrt{2},+\infty\rangle$. (**)

Finally, the logarithmic function is defined for positive values, so $x^{2}-4 x+3>0$ must hold, but this condition is satisfied as soon as ( $\star \star$ ) holds. Therefore, the solution of the first case is the intersection of conditions $(\star)$ and $(\star \star)$, which is $x \in\langle-\infty,-\sqrt{5}\rangle \cup[2+\sqrt{2},+\infty\rangle$.

Second case.

Now, $0<x^{2}-4<1$, then $x^{2}-4 x+3>0$, i.e., $x \in\langle-\infty, 1\rangle \cup\langle 3,+\infty\rangle .(\triangle \triangle)$

Furthermore, the expression under the root must be non-negative, so $\log _{x^{2}-4}\left(x^{2}-4 x+3\right) \geqslant 0$, i.e.,

$x^{2}-4 x+2 \leqslant 0$. The solution of this inequality is $x \in[2-\sqrt{2}, 2+\sqrt{2}]$. $(\triangle \triangle \triangle)$

The intersection of conditions $(\triangle),(\triangle \triangle)$, and $(\triangle \triangle \triangle)$ is empty, so the solution of the second case is the empty set.

Therefore, the domain of the function is $\mathcal{D}_{f}=\langle-\infty,-\sqrt{5}\rangle \cup[2+\sqrt{2},+\infty\rangle$.