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

Exercise

Determine the domain of the function:

f(x)=x2x2x24x21f(x)=\sqrt{\frac{x^2-x-2}{x^2-4x-21}}

Final Answer


x<3  or  x>7  or  1<x<2x<-3 \text{  or  } x>7 \text{  or  } -1<x<2

Solution

Let’s find the domain of the function:

f(x)=x2x2x24x21f(x)=\sqrt{\frac{x^2-x-2}{x^2-4x-21}}

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

x24x210x^2-4x-21\neq 0

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

x2x2x24x210\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):

x2x20  and  x24x21>0x^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:

x2x2=(x+1)(x2)x^2-x-2=(x+1)(x-2)

x24x21=(x+3)(x7)x^2-4x-21=(x+3)(x-7)

We solve the first inequality:

x2x2=(x+1)(x2)0x^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

x1  or  x2x\leq -1\text{  or  }x\geq 2

Now, we solve the second inequality:

x24x21=(x+3)(x7)>0x^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  or  x>7x< -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  or  x>7x<-3\text{  or  } x>7

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

x2x2=(x+1)(x2)<0x^2-x-2=(x+1)(x-2)<0

and

x24x21=(x+3)(x7)<0x^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<x<2-1<x<2

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<x<7-3<x<7

We intersect (“and”) both results and get

1<x<2-1<x<2

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  or  1<x<2  or  x>7x<-3\text{  or  } -1<x<2\text{  or  }x>7

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