### Video Transcript

Given that π sub one maps values of π₯ which are greater than negative β and less than or equal to two onto the set of real numbers such that π sub one of π₯ is equal to π₯ plus five and π sub two maps numbers from the open interval negative β to negative one onto the set of real numbers such that π sub two of π₯ equals two π₯ squared minus π₯ minus six, find π two over π one of π₯ and state its domain.

Remember, whilst this looks like it might be quite complicated, all itβs really asking us to do is find the quotient of our two functions. Thatβs π sub two of π₯ over π sub one of π₯. Now, we need to be really careful when considering the domain of our function. The domain of the quotient of two functions is the intersection of the domains of the respective functions. But weβll also need to exclude the values of π₯ that make the function in the denominator equal to zero. Now, weβll deal with that in a moment.

π sub two over π sub one of π₯ is two π₯ squared minus π₯ minus six over π₯ plus five. And so now, we move on to considering its domain. Weβre told that the function π sub one maps numbers or values of π₯ greater than negative β and less than or equal to two onto the set of real numbers. And so, since the domain of a function is the input of this function, the domain of π sub one is values of π₯ greater than negative β and less than or equal to two. Similarly, the domain of π sub two is the values of π₯ greater than negative β and less than negative one.

We know that the domain of the quotient of our functions is the intersection or the overlap of the two domains. Well, the values that overlap are values of π₯ less than negative one and greater than negative β. So, our domain is π₯ is an element of the open interval negative β to negative one. But we do need to be really careful here. We know that when working with a rational function or a function thatβs made up of the quotient of two functions, we donβt want their denominators to be equal to zero. We donβt want to divide by zero.

So, π₯ plus five cannot be equal to zero. If we subtract five from both sides of this inequation, we find π₯ cannot be equal to negative five. And so, we include this restriction on our domain. π two over π one of π₯ is two π₯ squared minus π₯ minus six over π₯ plus five. And the domain of this function is π₯ is an element of the open interval negative β to negative one, not including the set which includes the number negative five.