Question Video: Determining the Domain of Quotient of Functions Involving Square Roots | Nagwa Question Video: Determining the Domain of Quotient of Functions Involving Square Roots | Nagwa

Question Video: Determining the Domain of Quotient of Functions Involving Square Roots

Determine the domain of the function 𝑓(𝑥) = √(𝑥 + 1)/(𝑥² − 16).

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Video Transcript

Determine the domain of the function 𝑓 of 𝑥 equals the square root of 𝑥 plus one over 𝑥 squared minus 16.

Remember, the domain of a function is the complete set of possible values of the independent variable. Here, that’s possible values of 𝑥. Essentially, we’re looking to find the set of all possible 𝑥-values, which makes the function work and will output real 𝑦-values. When finding the domain, we need to be really careful with fractional or rational functions. The denominator of these cannot be equal to zero. If it was, we’d have some real number divided by zero, which we know to be undefined.

Similarly, any numbers inside a square root can’t be negative. They absolutely must be greater than or equal to zero. And when finding the domain, we’re actually interested in the intersection of these two criteria. We’ll begin by considering the denominator of our fraction. We know it cannot be equal to zero. So what we’re going to do is we’re going to solve the equation 𝑥 squared minus 16 is equal to zero. And we’ll know that those values of 𝑥 are out of the question.

To solve, we add 16 to both sides. And we see that 𝑥 squared is equal to 16. We then square root both sides of our equation, remembering, of course, we need to take both the positive and negative square root. And we see that 𝑥 is equal to either positive or negative four. And so for our domain, 𝑥 cannot be equal to these values. It cannot be equal to positive or negative four.

The second criteria is that the number under a square root cannot be negative. Well, we have the square root of 𝑥 plus one. So we know that 𝑥 plus one must be greater than or equal to zero. We subtract one from both sides of this equation. And we find that 𝑥 must be greater than or equal to negative one. So the second part of our domain is that 𝑥 must be greater than or equal to negative one. We’re looking for the intersection of these results. Well, for it to be greater than or equal to negative one, we use this sort of square bracket. This says that 𝑥 is greater than or equal to negative one and less than ∞. Remember, it can’t be equal to a positive or negative four. But negative four is less than negative one. So we’re really only interested in positive four. And so, we subtract the set containing the number four.

And so, the domain of our function is as shown. It’s values of 𝑥 greater than negative one, but not including the set containing the number four.

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