Video: Finding the Domain of Rational Functions

Determine the domain of the function 𝑓(π‘₯) = √(π‘₯ + 7)/√(π‘₯ βˆ’ 5).

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Video Transcript

Determine the domain of the function 𝑓 of π‘₯ equals the square root of π‘₯ plus seven over the square root of π‘₯ minus five.

We say that the domain of a function is the complete set of possible values of the independent variable. So here is the set of all possible values of π‘₯, in other words, the input of the function. What we’re looking to do is find the set of all possible values of π‘₯, which makes the function work and will output real 𝑦-values. Now, when dealing with fractional or rational functions, we know that the denominator of the fraction can’t be equal to zero. We also know that the number underneath the square root cannot be negative. It absolutely must be positive.

So if we consider the first definition the denominator cannot be equal to zero, we know that the square root of π‘₯ minus five cannot be equal to zero. Let’s solve this equation for π‘₯ by squaring both sides. And we get π‘₯ minus five equals zero. And then, we add five to both sides. Now, in fact, this value of π‘₯ gives us a dominator that is equal to zero. So part of our definition for the domain of this function will be that π‘₯ cannot be equal to five. But what else are we interested in?

Well, we said that the number under the square root can’t be negative. So we have two numbers under square roots: either π‘₯ plus seven or π‘₯ minus five. So for them not to be negative, we need to say they must be greater than or equal to zero. We will solve the inequality π‘₯ plus seven is greater than or equal to zero by subtracting seven from both sides. And we get π‘₯ is greater than or equal to negative seven. To solve the second inequality, we add five to both sides and we get π‘₯ is greater than or equal to five.

Now, in fact, we’re really interested in the intersection or the overlap between these two inequalities. Well, if π‘₯ is, say, negative six, it satisfies the first inequality, but not the second. The intersection is all values of π‘₯ greater than or equal to five. But of course, we said π‘₯ cannot itself be equal to five. So we represent this using these curved brackets. They say the open interval between five and ∞, in other words, all values of π‘₯ greater than but not equal to five.

And so, we found the domain of the function 𝑓 of π‘₯ equals the square root of π‘₯ plus seven over the square root of π‘₯ minus five. It’s the open interval from five to ∞.

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