### Video Transcript

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.