### Video Transcript

In this video, weβll learn how to
identify the domain of a rational function and the common domain of two or more
rational functions. We know that the domain of a
function is the set of all possible inputs of that function, whilst the range is all
possible outputs after weβve substituted the domain.

Now, for polynomial functions, the
domain is simply the set of real numbers. That means we can substitute any
real value of π₯ into an equation of the form π of π₯ equals π sub zero plus π
sub one π₯ all the way up to π sub ππ₯ to the πth power and the output will be
well defined. There are times though where the
domain of a function would need to be restricted. And this is particularly important
when working with rational functions, that is, a function of the form π of π₯
equals π of π₯ over π of π₯, where π and π are polynomial functions and π of π₯
is not the zero polynomial.

Now, that restriction on π is
important, and this is because dividing by zero is undefined. So we donβt want to be dividing a
polynomial by zero. And that gives us a hint as to how
to find the domain of a rational function. We might recall that when we find
the domain of the quotient of two functions, we find the intersection of the domains
of the respective functions. But we exclude any values of π₯
that make the function on the denominator of the expression equal to zero.

Since the domain of a polynomial is
the set of real numbers and a rational function is the quotient of two polynomials
then, the domain of a rational function is the set of real numbers excluding any
values of π₯ that make the denominator equal to zero. Weβll be using this definition
throughout the remainder of this video. So, with this in mind, letβs find
the domain of a rational function that involves quadratics.

For which values of π₯ is the
function π of π₯ equals π₯ squared minus 25 over π₯ squared minus 12π₯ plus 32 not
defined?

Letβs begin by inspecting the
function π of π₯. π of π₯ is the quotient of a pair
of polynomials. That is, it is a polynomial
function divided by a second polynomial function. In order to identify the values of
π₯ for which the function is not defined, weβll begin by considering the domain of a
rational function. The domain of a rational function,
of course, is the set of values of π₯ for which the function is defined. So, if we consider the set of
values of π₯ for which the function is defined, weβll be able to quickly identify
the values for which it is not defined.

The domain of a rational function
is the set of real numbers, but we must exclude any values of π₯ that make the
denominator of that function equal to zero. This means that our function will
be defined over the set of real numbers excluding the set of numbers that make the
expression on the denominator, π₯ squared minus 12π₯ plus 32, equal to zero.

To find such values of π₯, we will
set the denominator equal to zero and solve, that is, π₯ squared minus 12π₯ plus 32
equals zero. Since we have a quadratic, we can
attempt to solve by first factoring this quadratic expression. We know that we must have an π₯ at
the beginning of each expression because π₯ times π₯ gives us the π₯ squared. Then, we need to find a pair of
numbers whose product is 32 and whose sum is negative 12. Well, negative four times negative
eight is positive 32 as required. But negative four plus negative
eight is indeed negative 12.

So we rewrite our equation as
shown. π₯ minus four times π₯ minus eight
is equal to zero. Then, of course, for the product of
these two expressions to be equal to zero, we know that either one or other of those
expressions must itself be zero. So the solutions to our equation
are given by the solutions to the equations π₯ minus four equals zero and π₯ minus
eight equals zero.

We solve our first equation by
adding four to both sides, so we get π₯ is equal to four. And we solve our second equation by
adding eight to both sides, so π₯ is equal to eight. Remember, if weβre thinking about
the domain of π of π₯, we know itβs the set of real numbers minus the set
containing the numbers that make the denominator equal to zero. So the domain of our function is
the set of real numbers minus the set containing four and eight. Of course, this means our function
is defined over this set. And therefore, it must be undefined
when π₯ is equal to four or π₯ is equal to eight. And so we see that the function π
of π₯ is undefined for the set containing four and eight.

Weβll now consider a second example
which involves finding the domain of a rational function which is the quotient of a
pair of quadratics.

What is the domain of the function
π¦ equals π₯ squared minus one over π₯ squared plus one?

Remember, the domain of a function
is the set of all possible inputs to that function. And if we inspect our function π¦
equals π₯ squared minus one over π₯ squared plus one carefully, we can see itβs a
rational function. That is, itβs the quotient of a
pair of polynomials. So weβll remind ourselves what we
know about the domain of a rational function. The domain of a rational function
is the set of real numbers, but we exclude any values of π₯ that make the
denominator of the function equal to zero. In this case then, the function is
undefined for any values of π₯ which satisfy the equation π₯ squared plus one equals
zero. To establish which values of π₯
this is true for, letβs solve this equation.

We can begin by subtracting one
from both sides, so π₯ squared is equal to negative one. Then, we would look to take both
the positive and negative square root of negative one. But of course the square root of a
negative number is not a real number. And since we said the domain of a
rational function is just the set of real numbers, there are no values of π₯ in this
case which are going to make the denominator equal to zero. We can therefore say that the
domain of our function and the set of numbers that ensure it is well defined is
simply the set of real numbers.

Weβve now considered some examples
where weβve looked at rational functions and calculated the points where theyβre not
defined and, by extension, their domains. We might also encounter problems
where weβre given the domain of a function, and we need to use this to find missing
values. Letβs consider an example that uses
this idea.

Given that the domain of the
function π of π₯ equals 36 over π₯ plus 20 over π₯ plus π is the set of real
numbers minus the set containing negative two, zero, evaluate π of three.

π of π₯ is the sum of a pair of
rational functions. Both 36 over π₯ and 20 over π₯ plus
π are the quotient of a pair of polynomials. We also know that we can find the
domain of the sum of a pair of functions by considering the intersection of those
domains. So letβs begin by looking at the
domains of 36 over π₯ and 20 over π₯ plus π to the domain that weβve been given,
the set of real numbers minus the set containing negative two, zero. This will allow us to find the
value of π, which will in turn allow us to evaluate π of three.

Letβs begin by looking at the
expression 36 over π₯. Remember, the domain of a rational
function is the set of real numbers, but we exclude any values of π₯ that make the
denominator equal to zero. In this case, the denominator is
simply π₯. So we set π₯ equal to zero, and we
see that π₯ equals zero is a value that weβre going to exclude from the domain of
this function. The domain π of 36 over π₯ is the
set of real numbers minus the set containing zero. Weβll now consider the second
rational function. We have 20 over π₯ plus π. This time, the domain is going to
be the set of real numbers excluding any values of π₯ that make the denominator π₯
plus π equal to zero.

So letβs set π₯ plus π equal to
zero and solve for π₯. If we do, we find π₯ is equal to
negative π. So we can say that the domain of
this second function is the set of real numbers minus the set containing negative
π. The domain of π of π₯ then is the
intersection of these two domains. The intersection here is the set of
real numbers minus the set containing negative π, zero. Now, if we compare this to the
domain weβve been given, we can see that negative π must be equal to negative
two. And if negative π is equal to
negative two, π itself must be equal to two. So we can rewrite π of π₯ using
the value we have for π. Itβs 36 over π₯ plus 20 over π₯
plus two.

Weβre now ready to evaluate π of
three, and we can do so by substituting three into this equation. When we do, we get 36 over three
plus 20 over three plus two. And that becomes 12 plus four,
which is of course equal to 16. So, given information about the
domain of our function, π of three must be equal to 16.

In this example, we saw that when
we added functions, we had to take into account the domain of both functions. Now, a similar process holds if we
want to find the common domain of a pair or more of functions. In particular, we can take any
number of functions, and the common domain is simply the intersections of the
domains of the respective functions. So all we have to do is work out
the domain of each function individually and identify the regions where they
overlap. And then we can give that in set
notation. Letβs look at an example that
covers this concept.

Find the common domain between the
functions π sub one of π₯ equals negative nine over π₯ plus nine, π sub two of π₯
equals eight over π₯ plus three, and π sub three of π₯ equals seven π₯ over π₯
cubed minus four π₯.

Remember, given any number of
functions, the common domain is the intersection of the domains of the respective
functions. In this case then, we need to find
the domain of π sub one of π₯, π sub two of π₯, and π sub three of π₯. Then, we can find their
intersection. So, next, we remind ourselves how
we find the domain of a rational function. The domain of a rational function
is the set of real numbers, but we exclude any values of π₯ that make the
denominator equal to zero. So letβs take function π sub one
of π₯. Its domain is going to be the set
of real numbers, but we need to exclude values of π₯ that make π₯ plus nine equal to
zero. To solve for π₯, weβll subtract
nine from both sides, and we find the value of π₯ that satisfies this equation is
negative nine. So the domain of π sub one of π₯
is the set of real numbers minus the set containing negative nine.

Letβs now consider π sub two of
π₯. This time, we need to exclude
values of π₯ that make π₯ plus three, the denominator of π sub two of π₯, equal to
zero. The value of π₯ that satisfies this
equation is π₯ equals negative three. And so the domain of this function
is the set of real numbers minus the set containing negative three. Finally, weβll move on to π sub
three of π₯. The denominator here is π₯ cubed
minus four π₯. So we know we need to exclude any
values of π₯ that make this equal to zero. So we have the equation π₯ cubed
minus four π₯ equals zero. And how do we solve it?

Well, we might first look to factor
the expression on the left-hand side. The first step to this is to take
out a common factor of π₯. Then, we can further factor the
expression π₯ squared minus four using the difference of two squares. So π₯ cubed minus four π₯ can be
written as π₯ times π₯ plus two times π₯ minus two.

The first solution to this equation
is when π₯ is equal to zero, so itβs π₯ equals zero. The second solution is found by
setting π₯ plus two equal to zero. And when we solve that equation, we
get π₯ equals negative two. Finally, we solve the equation π₯
minus two equals zero, and we get π₯ equals two. So, finally, we have found that the
domain of π sub three of π₯ is the set of real numbers minus the set containing
these values of π₯. The common domain then is the
intersection of these three domains. So weβre going to have to take the
set of real numbers and exclude the following values of π₯: negative nine, negative
three, negative two, zero, and two. Hence, the common domain between
the three functions weβre given is the set of real numbers minus the set containing
negative nine, negative three, negative two, zero, and two.

Weβre now going to recap the key
points from this lesson. In this lesson, we learnt that a
rational function is of the form π of π₯ over π of π₯. These two functions, π of π₯ and
π of π₯, are themselves polynomials and π of π₯ is not the zero polynomial. With this definition, we can then
identify the domain of a rational function. Itβs the set of real numbers but we
exclude any values of π₯ that make the denominator, π of π₯, equal to zero. Finally, we learnt that we can take
any number of functions, and then their common domain is simply the intersection of
their respective domains.