Question

Find the domain of the function. $$ f(x)=\frac{\sqrt{x-1}}{x^{2}-x-6} $$

Short answer

The domain of the function \(f(x)=\frac{\sqrt{x-1}}{x^{2}-x-6}\) is \(x \in (-\infty, -2) \cup (-2, 1) \cup (1, 3) \cup (3, \infty)\).

Step-by-step solution

Determine the range of the expression inside the square root

The expression inside the square root is \(x-1\). In order for the square root to be defined, the expression inside must be greater than or equal to 0. So we'll solve the inequality: $$x - 1 \geq 0$$

Solve the inequality

To solve for the inequality, add 1 to both sides: $$x \geq 1$$ Now we know that \(x\) must be greater or equal to 1 for the square root to be defined.

Identify the restriction on the denominator

The denominator of the function is \(x^2 - x - 6\). In order for the function to be defined, the denominator cannot be equal to 0. So, we need to find the zeroes of the denominator, as they're not part of the domain. To find the zeroes, set the denominator equal to 0: $$x^2 - x - 6 = 0$$

Solve the quadratic equation

Factor the quadratic equation by searching for two numbers that multiply to -6 and add to -1. In this case, those two numbers are -3 and 2. Therefore, the factored form of the quadratic is: $$(x - 3)(x + 2) = 0$$ Solve for \(x\): $$x = 3, -2$$ These are the two values that make the denominator of the function equal to 0 and cannot be part of the domain.

Combine the two restrictions

Combine the two restrictions to find the domain of the function. Recall that \(x\) must be greater or equal to 1 and cannot be equal to 3 or -2. So the domain is: $$x \in (-\infty, -2) \cup (-2, 1) \cup (1, 3) \cup (3, \infty)$$