# Question Video: Finding and Evaluating the Difference of Two Functions Mathematics

If π and π are two real functions where π(π₯) = (π₯ + 9)/(π₯Β² + 15π₯ + 54) and π(π₯) = π₯ + 8, determine the value of (π β π)(β6) if possible.

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### Video Transcript

If π and π are two real functions where π of π₯ is π₯ plus nine over π₯ squared plus 15π₯ plus 54 and π of π₯ is equal to π₯ plus eight, determine the value of π minus π of negative six if possible.

We begin by recalling that π minus π of π₯ is simply the function π minus the function π. So, letβs evaluate π minus π of π₯ first and then consider its domain. Weβre given that π is the rational function π₯ plus nine over π₯ squared plus 15π₯ plus 54 and π is the function π₯ plus eight. So, π minus π of π₯ is π₯ plus nine over π₯ squared plus 15π₯ plus 54 minus eight plus π₯. Weβre actually going to simplify this expression by subtracting the fractions. And to do so, weβre going to write π₯ plus eight as π₯ plus eight over one then create a common denominator.

To achieve this common denominator, weβre going to need to multiply the numerator and denominator of our second fraction by π₯ squared plus 15π₯ plus 54. And when we do, our second fraction becomes π₯ plus eight times π₯ squared plus 15π₯ plus 54. Letβs distribute the parentheses in our second fraction. When we do, the numerator becomes π₯ cubed plus 15π₯ squared plus eight π₯ squared plus 120π₯ plus 54π₯ plus 432, which simplifies to π₯ cubed minus 23π₯ squared plus 174π₯ plus 432.

Now that our denominators are equal, weβre going to subtract every term in the numerator of our second fraction from the numerator of the first, giving us π minus π of π₯ as negative π₯ cubed plus 23π₯ squared minus 173π₯ minus 423 all over π₯ squared plus 15π₯ plus 54.

Now, before we find the value of π minus π of negative six, letβs consider the domain of our function. Remember, the domain is the set of inputs that will yield a real output. And so, whenever weβre calculating the domain of a rational function like this one, we need to consider the fact that the denominator of this fraction cannot be equal to zero. We donβt want to be dividing by zero. And so, weβre just going to begin by calculating the values of π₯ where this expression is equal to zero, where π₯ squared plus 15π₯ plus 54 equals zero.

To do this, weβre going to factor the expression on the left-hand side. We know that the term at the front of each binomial must be π₯ since π₯ times π₯ gives us π₯ squared. Then, we find two numbers whose product is 54 and whose sum is 15. Thatβs nine and six. So, we have π₯ plus nine times π₯ plus six equals zero. Well, for the product of two numbers to be equal to zero, either one of those numbers must themselves be equal to zero. So, π₯ plus nine must be equal to zero or π₯ plus six must be equal to zero.

If we subtract nine from both sides of our first equation, we find π₯ is equal to negative nine. And if we subtract six from both sides of our second equation, π₯ is equal to negative six. And so, we see that the denominator of our fraction will be equal to zero if π₯ is equal to negative six or π₯ is equal to negative nine. So, the domain of our function π minus π of π₯ must be all real numbers not including negative six and negative nine. We can use set notation as shown to demonstrate this.

Now, we were looking to evaluate π minus π of negative six, but we said that negative six is not in the domain of the function π minus π of π₯. And so, weβre, therefore, unable to evaluate π minus π of negative six. We say that itβs undefined.