# Video: Simplifying and Determining the Domain of a Sum of Two Rational Functions

Simplify the function π(π₯) = 7π₯Β²/(π₯ β 1) + 3π₯/(1 β π₯), and determine its domain.

03:41

### Video Transcript

Simplify the function π of π₯ equals seven π₯ squared divided by π₯ minus one plus three π₯ divided by one minus π₯ and determine its domain.

In order to simplify the functions, we firstly need to find the lowest common denominator. In this case, we multiply π₯ minus one by one minus π₯. When multiplying fractions, we must do the same to the numerator as we do to the denominator.

Therefore, we need to multiply the numerator of the first fraction seven π₯ squared by one minus π₯ and the numerator of the second fraction three π₯ by π₯ minus one. This gives us a single fraction seven π₯ squared multiplied by one minus π₯ plus three π₯ multiplied by π₯ minus one divided by π₯ minus one multiplied by one minus π₯.

Our next step is to expand the brackets or parentheses using the distributive property. Seven π₯ squared multiplied by one is seven π₯ squared and seven π₯ squared multiplied by negative π₯ is negative seven π₯ cubed.

For the second bracket, three π₯ multiplied by π₯ is three π₯ squared and three π₯ multiplied by negative one is negative three π₯. Grouping the like terms simplifies the numerator to negative seven π₯ cubed plus 10 π₯ squared minus three π₯. Our next step is to get the numerator in its simplest form by factorizing.

Well, firstly, we can see that an π₯ is common. Therefore, factorizing out the π₯ gives us π₯ multiplied by negative seven π₯ squared plus 10π₯ minus three. The quadratic negative seven π₯ squared plus 10π₯ minus three can be factorized into two brackets or parentheses. These two brackets are seven π₯ minus three and one minus π₯.

We could check this by using the FOIL method and expanding the bracket out. As one minus π₯ is a common term, we can divide the numerator and the denominator by one minus π₯. This leaves us π₯ multiplied by seven π₯ minus three divided by π₯ minus one. Therefore, the simplified version of π of π₯ is π₯ multiplied by seven π₯ minus three divided by π₯ minus one.

The second part of the question asked us to determine the domain of the function π of π₯. Well, at first glance, it appears that all real values are valid inputs to the function π of π₯. However, on closer inspection, we can see that there is one value of π₯ that would make the denominator equal to zero. This would give us undefined value.

In order to calculate this value, we need to set the denominator β in this case π₯ minus one β equal to zero. Adding one to both sides of the equation gives us an answer for π₯ equal to one. This means that when we substitute π₯ equals one into the function π of π₯, we get an undefined value.

This means that the domain of π of π₯ is all the real values with the exception of one β the real values minus one.