### Video Transcript

In this video, weβll learn how to
add, subtract, multiply, or divide two given functions to create a new function and
how to identify the domain of the new function. We are not going to be looking at a
function composition, which is a slightly different concept.

Now, combining functions in this
manner is actually incredibly intuitive. Letβs say we have two functions π
of π₯ and π of π₯. And we have the following
operations and associated notation. The function π plus π of π₯ is
equal to the sum of the functions, π of π₯ plus π of π₯. We can say the same for
subtraction: π minus π of π₯ is equal to the difference of the functions π of π₯
minus π of π₯. Then the function ππ of π₯ is π
of π₯ times π of π₯. And then π over π of π₯ is π of
π₯ over π of π₯. And so these are all fairly
straightforward.

But we do need to be really careful
when combining functions and thinking about the domain of the new function. The domain of the new function will
be the intersection or the overlap of the domain of π and the domain of π. In other words, both functions must
be defined at a given point for the combination of the functions to also be
defined. And, of course, if weβre working
with the division of functions, an additional requirement is that the denominator,
here thatβs π of π₯, cannot be equal to zero. So, now that we have some of these
definitions, letβs see how we might answer a question.

If π and π are two real functions
where π of π₯ is equal to π₯ plus nine over π₯ squared plus 15π₯ plus 54 and π of
π₯ is equal to π₯ plus eight, determine the value of π minus π of negative six if
possible.

Letβs begin by considering what we
mean by the combination function π minus π. π minus π of π₯ is equal to the
difference between the functions π of π₯ and π of π₯; itβs π of π₯ minus π of
π₯. Given that our functions are
defined as π of π₯ equals π₯ plus nine over π₯ squared plus 15π₯ plus 54 and π of
π₯ is π₯ plus eight, we see that π minus π of π₯ here is π₯ plus nine over π₯
squared plus 15π₯ plus 54 minus π₯ plus eight. And we could go ahead and simplify
this, but actually itβs not entirely necessary since weβre looking to work out
whether we can evaluate the function at π₯ equals negative six.

And so we need to consider what the
domain of our function π minus π of π₯ is. Well, the domain of a combination
of functions is equal to the intersection of the domains of the respective
functions. So here, we need to find the domain
of π of π₯ and π of π₯. Firstly, π of π₯ is the function
π₯ plus eight, and thatβs a polynomial. Now we know the domain of a
polynomial is simply the set of real numbers. And so we can say that the domain
of π of π₯ is the set of real numbers; π₯ can be any of these. But what about the function π of
π₯? Well, π of π₯ is the quotient of
two polynomials. And when we think about these
quotients, we must make sure that the denominator is not equal to zero.

And so what weβre going to do is
set the denominator of π of π₯ equal to zero and solve for π₯. And these will tell us the values
of π₯ that we need to exclude from our function since they will give us a result of
zero. So π₯ squared plus 15π₯ plus 54
equals zero. Weβll factor the expression on the
left-hand side. Itβs a quadratic, so weβre going to
have a pair of parentheses with π₯ at the front of each binomial. Then, we want two numbers that
multiply to make 54, they have a product of 54, and they add to make 15. Those numbers are six and nine, so
our expression factors to π₯ plus six times π₯ plus nine.

And now we know that since the
product of these binomials is equal to zero, either one or other of them must itself
be equal to zero. That is, π₯ plus six is equal to
zero or π₯ plus nine is equal to zero. If we solve this first equation by
subtracting six from both sides, we get π₯ equals negative six. And when we solve the second
equation, we get π₯ equals negative nine. And so the values of π₯ that make
the denominator of our function π of π₯ equal to zero are negative six and negative
nine. Now, since π of π₯ is itself the
quotient of two polynomials, we know its domain is the set of real numbers not
including the set containing the elements that make the denominator equal to
zero.

And so, the domain of π of π₯ is
the set of real numbers minus the set including the elements negative six and
negative nine. The domain of our combination
function π minus π of π₯ is the intersection of our two domains, the domain of π
of π₯ and π of π₯. The overlap of the domain of π of
π₯ and π of π₯ is the set of real numbers minus the set including negative six and
negative nine. And so, if we go back to what our
question is asking us, to evaluate π minus π of negative six, we can see that
negative six is not in the domain of this function. And so, we canβt actually evaluate
it. The function π minus π of π₯ is
undefined at that point. And so we say that the value of π
minus π of negative six is undefined.

In our next example, weβll look at
how the inclusion of a root function affects the domain of the sum of two
functions.

If π and π are two real functions
where π of π₯ is π₯ squared minus five π₯ and π of π₯ is equal to the square root
of π₯ plus one, find the domain of the function π plus π.

Firstly, we recall that the
combination function π plus π is simply the sum of the functions π and π. Now, weβre looking to find the
domain of this combination function. And so, we recall that the domain
of π plus π, thatβs the set of inputs that will yield real outputs, is the
intersections of the domains of π and π. So, letβs find the domains of π
and π. We begin with the function π of
π₯. Itβs π₯ squared minus five π₯. Itβs simply a polynomial, and we
know that the domain of a polynomial function is the set of real numbers. So, the domain of π of π₯ is the
set of real numbers.

And what about the function π of
π₯? Well, with a root function, we know
that to yield a real output, the number inside the square root must be bigger than
or equal to zero. In π of π₯, we have a function
inside the square root, so π₯ plus one must be bigger than or equal to zero. And this means to find the domain
of π of π₯, we need to solve the inequality π₯ plus one is greater than or equal to
zero. Weβll do this by subtracting one
from both sides. And that tells us that π₯ must be
greater than or equal to negative one. We can use interval notation to
represent the domain of π of π₯. π₯ must be greater than or equal to
negative one, so we say that the domain of π of π₯ is the left-closed, right-open
interval from negative one to β.

Notice that we canβt really define
β, hence why we canβt have a square bracket on the right-hand side of this
interval. And so we know that the domain of
our function π plus π is the intersection, the overlap, between these two
domains. If we consider the domain of π of
π₯ to be a subset of the set of real numbers, then we see that the domain of π plus
π, the overlap, is in fact the left-closed, right-open interval from negative one
to β. And so, that interval, that set of
values for π₯, is the domain of the function π plus π.

Letβs now consider the product of
two polynomials.

Given that π sub one maps numbers
from the set of positive real numbers onto the set of real numbers such that π sub
one of π₯ is equal to π₯ minus four and π sub two maps numbers from the left-open,
right-closed interval from negative nine to one onto the set of real numbers such
that π sub two of π₯ is equal to five π₯ minus two, find π one times π two of π₯
and state its domain.

Firstly, we recall that the product
of π one and π two of π₯ is actually equal to the product of the two
functions. Itβs π one of π₯ times π two of
π₯. And we know that when we combine
functions, the domain of the resulting function will be the intersection of the
domains of the respective functions. The only time we need to take into
account additional information is if weβre finding the quotient and we need to make
sure the denominator is not equal to zero. But of course, thatβs not
applicable here, so letβs begin by finding π one times π two of π₯. π one is π₯ minus four and π sub
two is five π₯ minus two, so the product is π₯ minus four times five π₯ minus
two.

Letβs distribute these parentheses
by using the FOIL method. Weβre going to multiply the first
term in each binomial. π₯ times five π₯ is five π₯
squared. Next, we multiply the outer terms,
which gives us negative two π₯. And we multiply the inner terms,
giving us negative 20π₯. Finally, we multiply the last term
in each binomial. Negative four times negative two is
eight. And so we see π one times π two
of π₯ is five π₯ squared minus 22π₯ plus eight. But what is the domain of this
function?

Well, both π one of π₯ and π sub
two of π₯ are, in fact, polynomials, and usually the domain of a polynomial is the
set of real numbers. But weβre actually told that π sub
one maps numbers from the set of positive real numbers. And so thatβs the domain of π sub
one. Itβs the positive real numbers. Then weβre told that π sub two
maps numbers from the left-open, right-closed interval from negative nine to
one. In other words, π₯ can be greater
than negative nine and less than or equal to one. And so, thatβs the domain of π sub
two. The domain of the product of our
functions will be the intersection, the overlap, of these two domains. And so, letβs use a number line to
establish where this overlap lies.

The domain of π sub one is the set
of positive real numbers. So thatβs any number greater than
zero as shown. The domain of π sub two is values
of π₯ greater than negative nine and less than or equal to one. The overlap lies here. Itβs values of π₯ greater than zero
and less than or equal to one. We can represent this domain using
interval notation as shown. And so we see that π one times π
two of π₯ is five π₯ squared minus 22π₯ plus eight, and the domain is π₯ is an
element of the left-open, right-closed interval from zero to one.

Weβll now look at how we need to
amend our process somewhat when finding the quotient of two functions.

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.

Weβll now move on and establish the
key points from this lesson. In this video, we saw that given
two functions π of π₯ and π of π₯, the combined function π plus π of π₯ is the
sum of the functions. Itβs π of π₯ plus π of π₯. We saw that π minus π of π₯ is π
of π₯ minus π of π₯. And we also saw that π times π of
π₯ is simply π of π₯ times π of π₯. We learned that π over π of π₯ is
π of π₯ over π of π₯. Finally, we learned that the domain
of each of our combinations is the intersection of the domains of π and π of
π₯. In other words, both functions must
be defined at a point for the combination to be defined.

We also saw that an additional
requirement that applies only to the division of functions is that the denominator
cannot be equal to zero. So, the domain of π over π of π₯
is the intersection of the domains of π and π minus the values of π₯ that make the
denominator equal to zero.