# Question Video: Finding an Unknown in the Sum of Two Rational Functions given Its Domain Mathematics • 10th Grade

Given that the domain of the function π(π₯) = (2π₯/(π₯ β π)(π₯ + 6)) + ((5π₯ + 5)/(π₯ β π)(π₯ + 5)) is β β {β6, β3, 2}, what is the value of π?

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### Video Transcript

Given that the domain of the function π of π₯ equals two π₯ over π₯ minus π multiplied by π₯ plus six plus five π₯ plus five over π₯ minus π multiplied by π₯ plus three is the set of real numbers minus the values negative six, negative three, and two, what is the value of π?

We recall first that the domain of the function is the set of all values on which the function acts. We can think of it as the functionβs input values. When weβre finding the domain of a rational function, as we have here, we need to be really careful to ensure that we arenβt dividing by zero. The domain of a rational function will be the set of real numbers, but weβll need to remove any values which mean weβll be dividing by zero.

Of course, division by zero occurs when the denominator of a quotient is equal to zero. So we need to consider what values of π₯ mean that the denominators of either of these two quotients will be equal to zero. Well, for the first quotient, setting the denominator equal to zero gives the equation π₯ minus π multiplied by π₯ plus six is equal to zero. This product will only be equal to zero when either of the two factors are equal to zero. So we have π₯ minus π equals zero or π₯ plus six equals zero. Each of these equations can be solved in one step.

For the first equation, we add π to each side giving π₯ equals π. And for the second, we subtract six from each side giving π₯ equals negative six. Now, we observe that negative six is one of the values that is excluded from the domain of this function, so that makes sense. We donβt yet know what π is, so letβs consider the denominator of the other quotient. If this were equal to zero, then it follows that either π₯ minus π equals zero, so thatβs the same equation again, or π₯ plus three is equal to zero. This equation can be solved by subtracting three from each side to give π₯ is equal to negative three.

Looking back again at the domain of our function π of π₯, we see that the value of negative three is also excluded. So this makes sense. We havenβt yet found any reason why the value of two canβt be included in the domain of this function. This tells us that two must be the solution to our remaining equation. So π₯ equals two must be the solution to the equation π₯ equals π, and so the value of π must be two. By considering then the set of values of π₯ which would make either of the two denominators equal to zero, and hence the entire function π of π₯ undefined, we found that the value of π is two.