Domain

For l(x) to make sense, it is required that \(\dfrac{x - 1}{x - 3} \in (-1, 1)\).

 

Then, we solve the inequality \(-1 < \dfrac{x - 1}{x - 3} < 1\) to find the domain of f.

I will do one side of the inequality as an example. Let's say we want to solve \(\dfrac{x - 1}{x - 3} < 1\). We can't just multiply (x - 3) on both sides because it is not guaranteed that x - 3 is positive. Instead, we multiply (x - 3)^2 on both sides since it is always nonnegative. Also, note that x cannot be 3 because of the denominator.

 

\(\dfrac{x - 1}{x - 3} \cdot (x - 3)^2 \leq (x - 3)^2, \,x \neq 3\\ (x - 1)(x - 3) \leq (x - 3)^2, \, x \neq 3\\ x^2 - 4x + 3 \leq x^2 - 6x + 9,\, x \neq 3\\ 2x - 6 \leq 0, \, x \neq 3\\ 2x - 6 < 0\\ x < 3 \)

 

Now, you can solve the other inequality \(-1 < \dfrac{x - 1}{x -3}\) to get another inequality for x, combine them, and you get the domain of \(l(x)\).