### Video Transcript

Simplify the function π of π₯ equals π₯ squared plus two π₯ over π₯ squared
minus four and find its domain.

π of π₯ is a rational function and so the expression on the right-hand side is an
algebraic fraction. To simplify this fraction, we need to look for common factors of the numerator
and denominator which we can then cancel out. So our first task is to factor both the
numerator and denominator. Starting with the numerator, we can see that the two terms, π₯ squared and two
π₯, have a common factor of π₯. π₯ squared is π₯ times π₯ and two π₯ is π₯ times two. And so together they are π₯ times π₯ plus two where here we have applied the distributive property.

Now we move on to the denominator which is π₯ squared minus four, and we notice
that that is a difference of two squares. It is π₯ minus two times π₯ plus two. Now that the numerator and denominator are fully factored, we can see that they
have a common factor of π₯ plus two. We can cancel this out. And we see that the simplified form of π of π₯ is π₯ over π₯ minus two and that
we canβt simplify any further.

So we have simplified the function, but now we need to find its domain. The domain of a rational function is the set of values for which its denominator
is nonzero. In other words, it is the set of real numbers minus the set of values for which
the denominator of a rational function is zero. If you look at the simplified function, you might be tempted to think that the
only value of π₯ for which the denominator is zero is π₯ equals two. However, the denominator of the original function, as it was defined, is π₯
squared minus four and not π₯ minus two. And if you look at the factorized form of this denominator, itβs easy to see
there are actually two values of π₯ for which this denominator is zero, two and negative two. The domain is therefore the set of real numbers minus the set containing
negative two and two.

So this is our answer: For every value of π₯ in the domain of a function, π of
π₯ is equal to π₯ over π₯ minus two. However, the domain of the function is the real numbers
minus negative two and two. Had the function originally been defined as just π₯ over π₯ minus two, the
domain wouldβve been bigger. It wouldβve been the real numbers minus just the set of two. If the function had originally been defined as just π₯ over π₯ minus two, then
certainly the domain would be just the real numbers minus the set of two. Had we not excluded this negative two from the domain, then we wouldnβt be
allowed to have cancelled π₯ plus two on the numerator and the denominator because in effect
we wouldβve been dividing by zero on both the top and bottom.

So while itβs possible often to simplify a rational function, the simplification
process doesnβt change the domain of the function. And so when talking about the domain of the function, you should look at the
original statement, the original definition of the function and not the simplified version that
you get after simplification.