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

Video Transcript

Given that the domain of the function 𝑛 of 𝑥 equals two 𝑥 over 𝑥 minus 𝑎 multiplied by 𝑥 plus six plus five 𝑥 plus five over 𝑥 minus 𝑎 multiplied by 𝑥 plus three is the set of real numbers minus the values negative six, negative three, and two, what is the value of 𝑎?

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 function’s input values. When we’re finding the domain of a rational function, as we have here, we need to be really careful to ensure that we aren’t dividing by zero. The domain of a rational function will be the set of real numbers, but we’ll need to remove any values which mean we’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 𝑥 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 𝑥 minus 𝑎 multiplied by 𝑥 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 𝑥 minus 𝑎 equals zero or 𝑥 plus six equals zero. Each of these equations can be solved in one step.

For the first equation, we add 𝑎 to each side giving 𝑥 equals 𝑎. And for the second, we subtract six from each side giving 𝑥 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 don’t yet know what 𝑎 is, so let’s consider the denominator of the other quotient. If this were equal to zero, then it follows that either 𝑥 minus 𝑎 equals zero, so that’s the same equation again, or 𝑥 plus three is equal to zero. This equation can be solved by subtracting three from each side to give 𝑥 is equal to negative three.

Looking back again at the domain of our function 𝑛 of 𝑥, we see that the value of negative three is also excluded. So this makes sense. We haven’t yet found any reason why the value of two can’t be included in the domain of this function. This tells us that two must be the solution to our remaining equation. So 𝑥 equals two must be the solution to the equation 𝑥 equals 𝑎, and so the value of 𝑎 must be two. By considering then the set of values of 𝑥 which would make either of the two denominators equal to zero, and hence the entire function 𝑛 of 𝑥 undefined, we found that the value of 𝑎 is two.