# Question Video: Evaluating a Product of Two Rational Functions Mathematics • 10th Grade

Given the function π(π₯) = ((π₯ β 6)/(π₯Β² β 15π₯ + 54)) Γ ((π₯Β² β 3π₯ β 28)/(2π₯Β² β 15π₯ + 7)), evaluate π(7), if possible.

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### Video Transcript

Given the function π of π₯ equals π₯ minus six over π₯ squared minus 15π₯ plus 54 times π₯ squared minus three π₯ minus 28 over two π₯ squared minus 15π₯ plus seven, evaluate π of seven, if possible.

Letβs begin by inspecting our function π of π₯. Itβs a combination of two functions. In fact, itβs a product. Specifically, itβs the product of a pair of rational functions. Remember, a rational function is made up of the quotient of two polynomials. And whenever we have rational functions, we know that they may not actually be defined for all values of π₯. So, weβre going to begin by asking ourselves what is the domain of π of π₯, where the domain is the set of possible inputs to the function.

Essentially, we need to check whether the function π of π₯ is actually defined for π₯ equals seven. If it is, then we can then substitute it in. So, how do we find the domain of a combination of functions? Well, the domain of π of π₯ will be the intersection of the domains of each rational function. So, we need to identify the domain of the first fraction, π₯ minus six over π₯ squared minus 15π₯ plus 54, and the domain of the second function, π₯ squared minus three π₯ minus 28 over two π₯ squared minus 15π₯ plus seven.

Now, in fact, when weβre working with rational functions, the domain is actually the set of real numbers. But we exclude any values of π₯ that make the denominator equal to zero. And in fact this makes a lot of sense. We essentially donβt want to put ourselves into a situation where weβre dividing by zero since thatβs undefined. To find the values of π₯ that we exclude from the domain of our first rational function then, letβs set the denominator equal to zero and solve for π₯. We solve for π₯ by factoring. Weβre looking for two numbers that have a product of 54 and sum to make negative 15. Well, thatβs negative nine and negative six. So, when we factor the expression, we get π₯ minus nine times π₯ minus six, and this is equal to zero.

Now, of course, if weβre multiplying a pair of expressions and weβre getting an answer of zero, we know that it makes sense that either one or other of those expressions is itself zero. So, π₯ minus nine equals zero or π₯ minus six equals zero. Solving each of these equations for π₯, and we get π₯ equals nine and π₯ equals six. So, the domain of our first rational function is the set of real numbers minus the set containing six and nine.

So, we repeat this process for our second rational function. We ask ourselves, how do we find the values of π₯ that make its denominator zero? Well, we set it equal to zero and solve by factoring. Since two and seven are both prime numbers, this can be done by inspection. Two π₯ multiplied by negative seven add negative one times π₯ gives us negative 15π₯. And, of course, negative one times seven gives us positive seven. So, factoring gives us two π₯ minus one times π₯ minus seven equals zero.

Then, as we did before, we set each factor individually equal to zero. So, two π₯ minus one equals zero or π₯ minus seven equals zero. By adding one and dividing by two with our first equation, we get π₯ equals one-half. And for our second, we get π₯ equals seven. So, the domain of our second rational function must be the set of real numbers not including these values of π₯.

This means then that the domain of π of π₯, which is the intersection of our two domains, is the set of real numbers excluding all these values of π₯, so the set of real numbers minus the set containing one-half, six, seven, and nine. And this is really interesting because weβre trying to find π of seven. But weβve excluded seven from the domain of our function. Itβs not one of the inputs. Itβs not a value of π₯ that we can substitute in. This means that our function π of π₯ is not well defined when π₯ equals seven, so π of seven is undefined.