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.
