### Video Transcript

Given that π and π are two real
functions where π of π₯ is equal to π₯ squared minus one and π of π₯ is equal to
the square root of π₯ plus five, find the value of π over π of negative two if
possible.

Firstly, we recall that π over π
of π₯ is simply the quotient of the functions. Itβs π of π₯ over π of π₯. And the reason this question says
to find the value of π over π of negative two βif possibleβ is because when we
think about the domain of our combined function, thatβs π over π of π₯, we think
about the intersection of the domains of π and π. But we want to make sure the
denominator, here thatβs π of π₯, is not equal to zero. So, letβs begin by looking for the
domains of our respective functions. π of π₯ is π₯ squared minus
one. Now thatβs just a polynomial, and
the domain of a polynomial is simply the set of all real numbers. So, the domain of π of π₯ is
indeed the set of all real numbers.

π of π₯ is a little more
complicated though. We have the square root of π₯ plus
five. And we know that for the square
root of a number to be real, that number must be greater than or equal to zero. So here, we know that π₯ plus five
must be greater than or equal to zero. Letβs solve this inequality by
subtracting five from both sides. And when we do, we find that π₯
must be greater than or equal to negative five. And therefore, the domain of our
function π of π₯ is π₯ is an element of the left-closed, right-open interval from
negative five to β. And so we move on to finding the
domain of π over π of π₯. Itβs the intersection of the
domains of π of π₯ and π of π₯, which is the left-closed, right-open interval from
negative five to β.

But we need to exclude the values
of π₯ that make the denominator equal to zero. In other words, we need to exclude
the values of π₯ that make the expression π₯ squared minus one, thatβs the function
π of π₯, equal to zero. This time weβll solve this equation
by adding one to both sides, which gives us π₯ squared is equal to one. Next, we take the square root of
both sides, remembering, of course, to take both the positive and negative square
root of one. And when we do, we find π₯ is equal
to positive or negative one. And so, this means that the domain
of π over π of π₯ is the left-closed, right-open interval from negative five to β
minus the set including the elements negative one and one.

So, what does this mean for the
value of π over π of negative two? π₯ is equal to negative two is
within the domain of the function, and therefore it can be evaluated. And so, weβre now going to move on
to finding the function π over π of π₯. We take the function π of π₯, and
we divide it by the function π of π₯, so thatβs the square root of π₯ plus five
over π₯ squared minus one. Then, π over π of negative two is
found by replacing π₯ with negative two, so we get the square root of negative two
plus five over negative two squared minus one, which simplifies to the square root
of three over three. And so, given the functions π and
π, weβve shown that the value of π over π of negative two is the square root of
three over three.