Question Video: Simplifying and Determining the Domain of Rational Functions Mathematics

Simplify the function 𝑓(π‘₯) = (π‘₯Β² + 2π‘₯)/(π‘₯Β² βˆ’ 4), and find its domain.


Video Transcript

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

𝑓 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 𝑓 of π‘₯ is π‘₯ over π‘₯ minus two and that we can’t 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, it’s 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, 𝑓 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 would’ve been bigger. It would’ve 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 wouldn’t be allowed to have cancelled π‘₯ plus two on the numerator and the denominator because in effect we would’ve been dividing by zero on both the top and bottom.

So while it’s possible often to simplify a rational function, the simplification process doesn’t 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.

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