### Video Transcript

Given that the domain of the
function π of π₯ equals two π₯ over π₯ minus π multiplied by π₯ plus six plus five
π₯ plus five over π₯ minus π multiplied by π₯ plus three is the set of real numbers
minus the values negative six, negative three, and two, what is the value of π?

We recall first that the domain of
the function is the set of all values on which the function acts. We can think of it as the
functionβs input values. When weβre finding the domain of a
rational function, as we have here, we need to be really careful to ensure that we
arenβt dividing by zero. The domain of a rational function
will be the set of real numbers, but weβll need to remove any values which mean
weβll be dividing by zero.

Of course, division by zero occurs
when the denominator of a quotient is equal to zero. So we need to consider what values
of π₯ mean that the denominators of either of these two quotients will be equal to
zero. Well, for the first quotient,
setting the denominator equal to zero gives the equation π₯ minus π multiplied by
π₯ plus six is equal to zero. This product will only be equal to
zero when either of the two factors are equal to zero. So we have π₯ minus π equals zero
or π₯ plus six equals zero. Each of these equations can be
solved in one step.

For the first equation, we add π
to each side giving π₯ equals π. And for the second, we subtract six
from each side giving π₯ equals negative six. Now, we observe that negative six
is one of the values that is excluded from the domain of this function, so that
makes sense. We donβt yet know what π is, so
letβs consider the denominator of the other quotient. If this were equal to zero, then it
follows that either π₯ minus π equals zero, so thatβs the same equation again, or
π₯ plus three is equal to zero. This equation can be solved by
subtracting three from each side to give π₯ is equal to negative three.

Looking back again at the domain of
our function π of π₯, we see that the value of negative three is also excluded. So this makes sense. We havenβt yet found any reason why
the value of two canβt be included in the domain of this function. This tells us that two must be the
solution to our remaining equation. So π₯ equals two must be the
solution to the equation π₯ equals π, and so the value of π must be two. By considering then the set of
values of π₯ which would make either of the two denominators equal to zero, and
hence the entire function π of π₯ undefined, we found that the value of π is
two.