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

Simplify the function 𝑛(π‘₯) = 7π‘₯Β²/(π‘₯ βˆ’ 1) + 3π‘₯/(1 βˆ’ π‘₯), and determine its domain.

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Video Transcript

Simplify the function 𝑛 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 𝑛 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 𝑛 of π‘₯. Well, at first glance, it appears that all real values are valid inputs to the function 𝑛 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 𝑛 of π‘₯, we get an undefined value.

This means that the domain of 𝑛 of π‘₯ is all the real values with the exception of one β€” the real values minus one.

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