Final answer:
The domain of the given functions are determined. The domain of f(x) = 3x - 1 is (-∞, ∞). The domain of f(x) = -1/(x² + 2x - 3) is (-∞, -3) U (-3, 1) U (1, ∞). And the domain of f(x) = √(x - 6) is [6, ∞).
Step-by-step explanation:
1. For the function f(x) = 3x - 1, the domain is all real numbers, since there are no restrictions on the values x can take. In interval notation, the domain is (-∞, ∞).
2. For the function f(x) = -1/(x² + 2x - 3), we need to find the values of x that make the expression inside the denominator equal to zero, because division by zero is undefined. The expression can be factored as (x - 1)(x + 3), so the values of x that make the denominator zero are x = 1 and x = -3. Therefore, the domain is all real numbers except x = 1 and x = -3. In interval notation, the domain is (-∞, -3) U (-3, 1) U (1, ∞).
3. For the function f(x) = √(x - 6), the radicand (x - 6) must be greater than or equal to zero to ensure the square root is defined. Therefore, x - 6 ≥ 0, which implies x ≥ 6. In interval notation, the domain is [6, ∞).