### Video Transcript

If π maps numbers in the set of
real numbers onto the set of real numbers, where π of π₯ is equal to four π₯ minus
four, and π maps values of π₯ greater than or equal to negative eight and less than
negative two onto the set of real numbers, where π of π₯ is equal to five π₯ plus
five, find the value of π plus π of five if possible.

Weβve been asked to find the value
of π plus π at π₯ equals five. Now, this might look quite
complicated, but all itβs really asking us is to evaluate π of π₯ plus π of π₯ at
five. Before we do that, though, letβs
look at some of the notation given in our question. Weβre told that the function π
maps numbers in the set of real numbers onto the set of real numbers. This tells us something about the
domain of our function π. Remember, the domain is the set of
all possible inputs that will yield real outputs. So, the domain of our function π
is simply the set of all real numbers.

And what about the function π? Well, this maps values of π₯
greater than or equal to negative eight, thatβs what this square bracket means, and
less than negative two; thatβs what the curly bracket means. And it maps these numbers onto the
set of real numbers. And so, the domain of π is these
values of π₯. Itβs π₯ is greater than or equal to
negative eight and less than negative two.

And when finding the sum of two
functions, we need to be really careful with the domain. The domain of π plus π of π₯ will
be the intersection. Remember, thatβs the overlap of the
domains of π and π. So, letβs begin by working out what
π plus π of π₯ is. π of π₯ is four π₯ minus four and
π of π₯ is five π₯ plus five. So, their sum π plus π of π₯ is
four π₯ minus four plus five π₯ plus five. And collecting like terms, we get
nine π₯ plus one. So, what is the domain of our
function?

We saw that the domain of π was
the set of all real numbers. Thatβs all numbers up to positive β
and all the way down to negative β. We see then that the domain of π
is essentially a subset of the domain of π; itβs values of π₯ greater than or equal
to negative eight and less than negative two. The intersection then of the two
domains, which we said is the overlap, is the same as the domain of π. Itβs values of π₯ greater than or
equal to negative eight and less than negative two.

Now, weβre looking to find the
value of π plus π of five. But five is greater than negative
two, meaning itβs outside the domain of π plus π. This means we canβt evaluate π
plus π of five; weβre not able to substitute five in. And so, for this reason, we say
that the value of π plus π of five is undefined.