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

Video Transcript

Simplify the function 𝑛 of 𝑥 equals five 𝑥 over 𝑥 minus four minus 𝑥 plus four over 𝑥 squared minus 16 and determine its domain.

This function 𝑛 of 𝑥 is the difference of two algebraic quotients. In order to simplify this function, we need to combine the two quotients into a single quotient. We see though that they have different denominators. The denominator of the first quotient is 𝑥 minus four. And the denominator of the second is 𝑥 squared minus 16. And we know that in order to add or subtract two quotients, they must have a common denominator.

We should also observe that 𝑥 squared minus 16 is a difference of two squares. 𝑥 squared minus 16 is 𝑥 squared minus four squared. And so it can be factored as 𝑥 minus four multiplied by 𝑥 plus four. So 𝑛 of 𝑥 is equal to five 𝑥 over 𝑥 minus four minus 𝑥 plus four over 𝑥 minus four multiplied by 𝑥 plus four.

Now, remember we said that we need a common denominator in order to add or subtract two fractions. If we observe that, in the second quotient, the factor of 𝑥 plus four is shared between the numerator and denominator and can therefore be canceled out, we see that in fact the two fractions do already have a common denominator of 𝑥 minus four. We have five 𝑥 over 𝑥 minus four minus one over 𝑥 minus four, which simplifies to five 𝑥 minus one over 𝑥 minus four. This can’t be simplified any further as there are no factors shared between the numerator and denominator other than one. And so we found the simplified form of the function 𝑛 of 𝑥.

We’re also asked to determine the domain of this function, which is the set of all 𝑥-values on which the function acts. We can also think of it as the set of all 𝑥-values such that the function 𝑛 of 𝑥 is well defined.

When finding the domain of a rational function, our concern is any 𝑥-values that might make the denominator of any fraction equal to zero. Division by zero is undefined. So any 𝑥-values that lead to this must be excluded from the function’s domain. And we need to consider this for the original form in which the function was specified before we’d carried out any simplification. For the denominator of the first fraction to be equal to zero, it must be the case that 𝑥 minus four is equal to zero. Adding four to each side, we see that this occurs when 𝑥 is equal to four. So the value of four must be excluded from the domain of the function 𝑛 of 𝑥.

For the second fraction, its denominator will be equal to zero when 𝑥 squared minus 16 is equal to zero. Adding 16 to each side, we see that this occurs when 𝑥 squared is equal to 16. Taking the square root of each side, we have 𝑥 is equal to plus or minus the square root of 16, which is equal to positive or negative four. The function 𝑛 of 𝑥 will be well defined for all other real values of 𝑥. So we state that the domain of this function is equal to the set of real numbers minus the set containing the values negative four and four. This means the set of all real numbers with the values negative four and four excluded.

We’ve completed the problem then. The simplified form of the function 𝑛 of 𝑥 is five 𝑥 minus one over 𝑥 minus four. And the domain of this function is the set of all real numbers with the exclusion of the values negative four and positive four.