### Video Transcript

Determine the domain of the function π of π₯ is equal to the square root of π₯ plus seven over the square root of π₯ minus five.

Remember, when weβre thinking about the domain of a function, weβre thinking about the set of possible values that weβre allowed to substitute into that function. More specifically, we need to make sure that when we do substitute those values in, they yield a real output.

And so, looking at the function π of π₯ equals the square root of π₯ plus seven over the square root of π₯ minus five, there are two things we need to take into account. We want to ensure, first of all, that the denominator of the function is not equal to zero. Similarly, we want to make sure that any expressions inside a square root symbol are nonnegative. So letβs deal with the first criteria. That is, we want to make sure the denominator of our function is not equal to zero.

To do so, weβll find any values of π₯ that make the denominator equal to zero. And then weβll disregard those from our domain. So we set the denominator the square root of π₯ minus five equal to zero. We can then square both sides of this equation, so π₯ minus five equals zero. Finally, we add five to both sides. And we get π₯ is equal to five. This is the value of π₯ that makes the denominator equal to zero. So weβre going to disregard this from the domain of our function at the end of the question.

Next, we said that we need to make sure any expressions inside square root symbols are nonnegative. In other words, π₯ plus seven must be greater than or equal to zero, and π₯ minus five must be greater than or equal to zero. If we subtract seven from both sides of our first equation, we find that π₯ must be greater than or equal to negative seven. And if we add five to both sides of our second equation, we find that π₯ must be greater than or equal to five.

So we have three criteria that we need to consider. π₯ must not be equal to five. π₯ must be greater than or equal to negative seven. And π₯ must be greater than or equal to five. Well, we know that if π₯ is greater than or equal to five, then it absolutely must also be greater than or equal to negative seven. And we also said that whilst π₯ must be greater than or equal to five so that the expression inside the bottom square root is nonnegative, π₯ could not be equal to five itself.

So the intersection of all these possible sets is the set of values of π₯ which is simply greater than five. And of course we can use set notation to represent this. The domain of the function is the set of values in the open interval from five to β.