Video Transcript
Given that the domain of the
function 𝑛 of 𝑥 is equal to 36 over 𝑥 plus 20 divided by 𝑥 plus 𝑎 is the set of
real numbers excluding negative two and zero, evaluate 𝑛 of three.
In this question, we’re given a
function 𝑛 of 𝑥 which we can see is the sum of two rational functions. And we’re given the domain of our
function 𝑛 of 𝑥 is defined for all real values excluding negative two and
zero. We need to use this information to
determine the value of 𝑛 evaluated at three. Normally, to do this, we would just
substitute 𝑥 is equal to three into our function 𝑛 of 𝑥. However, by doing this, we
immediately run into a problem. We don’t know the value of this
constant 𝑎 in our function 𝑛 of 𝑥. So we can’t directly evaluate 𝑛 at
three; we first need to find the value of 𝑎. And to do this, we’re going to need
to use the fact that we know the domain of our function.
So let’s start by recalling exactly
what we mean by the domain of a function. We recall the domain of a function
is the set of inputs for that function. So this gives us two pieces of
information. First, we can see that negative two
and zero are not in the domain of our function 𝑛 of 𝑥. This means we’re not allowed to use
the value of 𝑥 is equal to zero, and we’re not allowed to use the value of 𝑥 is
equal to negative two. Next, we also see that three is in
the domain of our function. So we are allowed to substitute in
𝑥 is equal to three.
And this can then immediately give
us a question. Why are we not allowed to
substitute in 𝑥 is equal to zero and 𝑥 is equal to negative two? And one way of seeing why these are
not included in the domain of our function 𝑛 of 𝑥 is to try substituting in these
values. Let’s try substituting in 𝑥 is
equal to zero. Substituting in 𝑥 is equal to
zero, we get 𝑛 of zero is equal to 36 over zero plus 20 divided by zero plus
𝑎. And now we can see a problem. When we tried to evaluate 𝑛 at
zero, we can see we’re dividing by zero, so this is not defined. And of course, if a function is not
defined at a value of 𝑥, then it can’t possibly have that value as an input. In other words, zero is not allowed
to be in the domain of our function 𝑛 of 𝑥.
Let’s now try doing exactly the
same thing with 𝑥 is equal to negative two. Substituting 𝑥 is equal to
negative two into our function 𝑛 of 𝑥, we get 𝑛 evaluated at negative two is
equal to 36 divided by negative two plus 20 divided by negative two plus 𝑎. This time, it’s not as immediately
obvious why negative two is not in the domain of our function 𝑛 of 𝑥. If we look at each term separately,
our first term is 36 divided by negative two. We can evaluate this; it’s negative
18. However, in our second term, we can
see we’re dividing by an unknown. We’re dividing by negative two plus
𝑎. Of course, we can divide 20 by any
real number except for zero. This would give us an undefined
value.
Therefore, because negative two is
not in the domain of our function, this implies that the denominator of this term is
going to be equal to zero. So we should have that negative two
plus 𝑎 is equal to zero. Of course, we can solve this for
𝑎. We just add two to both sides of
the equation, and we get that 𝑎 is equal to two. Now, because three is in the domain
of our function 𝑛 of 𝑥, we can just substitute 𝑥 is equal to three into our
function 𝑛 of 𝑥 and use the value of 𝑎 is equal to two.
Substituting 𝑛 is equal to three
into 𝑛 of 𝑥 and using the value of 𝑎 is equal to two, we get 𝑛 evaluated at
three is equal to 36 divided by three plus 20 divided by three plus two. And of course, we can calculate
this expression. 36 over three is equal to 12, and
20 divided by three plus two is 20 over five, which is equal to four. And 12 plus four is equal to 16,
which is our final answer.
Therefore, given the domain of the
function 𝑛 of 𝑥 is equal to 36 over 𝑥 plus 20 divided by 𝑥 plus 𝑎 is all real
numbers excluding negative two and zero, we were able to show that 𝑛 evaluated at
three is equal to 16.
