# Domain of One Variable Function – A rational function inside square root – Exercise 2461

Exercise

Determine the domain of the function:

$$f(x)=\sqrt{\frac{x^2-x-2}{x^2-4x-21}}$$

$$x<-3 \text{ or } x>7 \text{ or } -1

Solution

Let’s find the domain of the function:

$$f(x)=\sqrt{\frac{x^2-x-2}{x^2-4x-21}}$$

Because there is a denominator, the denominator must be different from zero:

$$x^2-4x-21\neq 0$$

Also, there is a square root, so the expression inside the root must be non-negative:

$$\frac{x^2-x-2}{x^2-4x-21}\geq 0$$

Let’s look at the second inequality. When is a fracture greater than or equal to zero? When both the numerator and denominator are positive or when both are negative. Let’s look at the first case – the numerator is positive or zero and the denominator is positive (because denominator cannot be zero):

$$x^2-x-2\geq 0\text{ and }x^2-4x-21> 0$$

We got two quadratic inequalities. To solve them, they must be broken down into factors. We use the quadratic formula for both inequalities and get the factorizations:

$$x^2-x-2=(x+1)(x-2)$$

$$x^2-4x-21=(x+3)(x-7)$$

We solve the first inequality:

$$x^2-x-2=(x+1)(x-2)\geq 0$$

Therefore, its roots are -1, 2 (this means that the graph goes through the x-axis at these points) and the parabola (the graph) in the shape of a bowl (“smiling” parabola), because the coefficient of the square expression is positive. According to the inequality sign, one should check when the graph is on or above the x-axis. It happens when

$$x\leq -1\text{ or }x\geq 2$$

Now, we solve the second inequality:

$$x^2-4x-21=(x+3)(x-7)>0$$

Therefore, its roots are -3, 7 and the parabola in the shape of a bowl (“smiling” parabola), because the coefficient of the square expression is positive. According to the inequality sign, one should check when the graph is on or above the x-axis. It happens when

$$x< -3\text{ or }x>7$$

We intersect (“and”) both results, because we want both a positive numerator and a positive denominator) and get the solution for the first case:

$$x<-3\text{ or } x>7$$

Now let’s look at the second case: both the numerator and the denominator are negative, that is:

$$x^2-x-2=(x+1)(x-2)<0$$

and

$$x^2-4x-21=(x+3)(x-7)<0$$

We solve the first inequality. According to the inequality sign, one should check when the graph is below the x-axis. The graph pass through the x-axis at the points: 1-, 2, so we get the solution:

$$-1

Now, we solve the second inequality. According to the inequality sign, one should check when the graph is below the x-axis. The graph pass through the x-axis at the points: 1-, 2, so we get the solution:

$$-3

We intersect (“and”) both results and get

$$-1

And the final answer is the union (“or” relationship) of both solutions – the solution of the first case and the solution of the second case – because they cannot occur together (either the numerator and denominator are both positive or are negative) and we get

$$x<-3\text{ or } -17$$

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