We observe that this is a rational function in which the numerator is

a polynomial of degree 3 and this polynomial is defined for all real values.

So, this rational function will be undefined at the points where we will have,

denominator = 0

This gives us,

|x - 3| - |x + 1| = 0 ......... (1)

To remove the absolutes in equation (1), we will have following three intervals:

x ≤ -1, -1 ≤ x ≤ 3, x ≥ 3

If x ≤ -1, then using (1), we get

-(x - 3) - (-(x + 1)) = 0

- x + 3 + x + 1 = 0

4 = 0

This is not a true statement.

It means that we have no solution when x ≤ -1.

If -1 ≤ x ≤ 3, then using (1), we get

-(x - 3) - (x + 1) = 0

-x + 3 - x - 1 = 0

-2x + 2 = 0

2x = 2

x = 2/2

x = 1

It means that when -1 ≤ x ≤ 3, then the solution is given by

x = 1

If x ≥ 3, then using (1), we get

(x - 3) - (x + 1) = 0

x - 3 - x - 1 = 0

- 4 = 0

This is not a true statement.

It means that we have no solution when .

Therefore, the solution of the equation (1), is given by x = 1

Hence, the function g(x) is defined for all real values except x = 1.

Therefore, the domain of the function g(x) is (-∞, 1)U(1, ∞).

Domain (-∞, 1)U(1, ∞)