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 ๐‘›(๐‘ฅ) = (2๐‘ฅ/(๐‘ฅ โˆ’ ๐‘Ž)(๐‘ฅ + 6)) + ((5๐‘ฅ + 5)/(๐‘ฅ โˆ’ ๐‘Ž)(๐‘ฅ + 5)) is โ„ โˆ’ {โˆ’6, โˆ’3, 2}, what is the value of ๐‘Ž?

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Video Transcript

Given that the domain of the function ๐‘› of ๐‘ฅ equals two ๐‘ฅ over ๐‘ฅ minus ๐‘Ž multiplied by ๐‘ฅ plus six plus five ๐‘ฅ plus five over ๐‘ฅ minus ๐‘Ž multiplied by ๐‘ฅ plus three is the set of real numbers minus the values negative six, negative three, and two, what is the value of ๐‘Ž?

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 functionโ€™s input values. When weโ€™re finding the domain of a rational function, as we have here, we need to be really careful to ensure that we arenโ€™t dividing by zero. The domain of a rational function will be the set of real numbers, but weโ€™ll need to remove any values which mean weโ€™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 ๐‘ฅ 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 ๐‘ฅ minus ๐‘Ž multiplied by ๐‘ฅ 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 ๐‘ฅ minus ๐‘Ž equals zero or ๐‘ฅ plus six equals zero. Each of these equations can be solved in one step.

For the first equation, we add ๐‘Ž to each side giving ๐‘ฅ equals ๐‘Ž. And for the second, we subtract six from each side giving ๐‘ฅ 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 donโ€™t yet know what ๐‘Ž is, so letโ€™s consider the denominator of the other quotient. If this were equal to zero, then it follows that either ๐‘ฅ minus ๐‘Ž equals zero, so thatโ€™s the same equation again, or ๐‘ฅ plus three is equal to zero. This equation can be solved by subtracting three from each side to give ๐‘ฅ is equal to negative three.

Looking back again at the domain of our function ๐‘› of ๐‘ฅ, we see that the value of negative three is also excluded. So this makes sense. We havenโ€™t yet found any reason why the value of two canโ€™t be included in the domain of this function. This tells us that two must be the solution to our remaining equation. So ๐‘ฅ equals two must be the solution to the equation ๐‘ฅ equals ๐‘Ž, and so the value of ๐‘Ž must be two. By considering then the set of values of ๐‘ฅ which would make either of the two denominators equal to zero, and hence the entire function ๐‘› of ๐‘ฅ undefined, we found that the value of ๐‘Ž is two.

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