Question Video: Determining the Domain of a Rational Function Mathematics

For which values of π‘₯ is the function 𝑛(π‘₯) = (π‘₯Β² βˆ’ 25)/(π‘₯Β² βˆ’ 12π‘₯ + 32) not defined?

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Video Transcript

For which values of π‘₯ is the function 𝑛 of π‘₯ equals π‘₯ squared minus 25 over π‘₯ squared minus 12π‘₯ plus 32 not defined?

Let’s begin by inspecting the function 𝑛 of π‘₯. 𝑛 of π‘₯ is the quotient of a pair of polynomials. That is, it is a polynomial function divided by a second polynomial function. In order to identify the values of π‘₯ for which the function is not defined, we’ll begin by considering the domain of a rational function. The domain of a rational function, of course, is the set of values of π‘₯ for which the function is defined. So, if we consider the set of values of π‘₯ for which the function is defined, we’ll be able to quickly identify the values for which it is not defined.

The domain of a rational function is the set of real numbers, but we must exclude any values of π‘₯ that make the denominator of that function equal to zero. This means that our function will be defined over the set of real numbers excluding the set of numbers that make the expression on the denominator, π‘₯ squared minus 12π‘₯ plus 32, equal to zero.

To find such values of π‘₯, we will set the denominator equal to zero and solve, that is, π‘₯ squared minus 12π‘₯ plus 32 equals zero. Since we have a quadratic, we can attempt to solve by first factoring this quadratic expression. We know that we must have an π‘₯ at the beginning of each expression because π‘₯ times π‘₯ gives us the π‘₯ squared. Then, we need to find a pair of numbers whose product is 32 and whose sum is negative 12. Well, negative four times negative eight is positive 32 as required. But negative four plus negative eight is indeed negative 12.

So we rewrite our equation as shown. π‘₯ minus four times π‘₯ minus eight is equal to zero. Then, of course, for the product of these two expressions to be equal to zero, we know that either one or other of those expressions must itself be zero. So the solutions to our equation are given by the solutions to the equations π‘₯ minus four equals zero and π‘₯ minus eight equals zero.

We solve our first equation by adding four to both sides, so we get π‘₯ is equal to four. And we solve our second equation by adding eight to both sides, so π‘₯ is equal to eight. Remember, if we’re thinking about the domain of 𝑛 of π‘₯, we know it’s the set of real numbers minus the set containing the numbers that make the denominator equal to zero. So the domain of our function is the set of real numbers minus the set containing four and eight. Of course, this means our function is defined over this set. And therefore, it must be undefined when π‘₯ is equal to four or π‘₯ is equal to eight. And so we see that the function 𝑛 of π‘₯ is undefined for the set containing four and eight.

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