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

If 𝑓 and 𝑔 are two real functions where 𝑓(π‘₯) = (π‘₯ + 9)/(π‘₯Β² + 15π‘₯ + 54) and 𝑔(π‘₯) = π‘₯ + 8, determine the value of (𝑓 βˆ’ 𝑔)(βˆ’6) if possible.

04:01

Video Transcript

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

We begin by recalling that 𝑓 minus 𝑔 of π‘₯ is simply the function 𝑓 minus the function 𝑔. So, let’s evaluate 𝑓 minus 𝑔 of π‘₯ first and then consider its domain. We’re given that 𝑓 is the rational function π‘₯ plus nine over π‘₯ squared plus 15π‘₯ plus 54 and 𝑔 is the function π‘₯ plus eight. So, 𝑓 minus 𝑔 of π‘₯ is π‘₯ plus nine over π‘₯ squared plus 15π‘₯ plus 54 minus eight plus π‘₯. We’re actually going to simplify this expression by subtracting the fractions. And to do so, we’re going to write π‘₯ plus eight as π‘₯ plus eight over one then create a common denominator.

To achieve this common denominator, we’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. Let’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, we’re going to subtract every term in the numerator of our second fraction from the numerator of the first, giving us 𝑓 minus 𝑔 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 𝑓 minus 𝑔 of negative six, let’s consider the domain of our function. Remember, the domain is the set of inputs that will yield a real output. And so, whenever we’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 don’t want to be dividing by zero. And so, we’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, we’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. That’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 𝑓 minus 𝑔 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 𝑓 minus 𝑔 of negative six, but we said that negative six is not in the domain of the function 𝑓 minus 𝑔 of π‘₯. And so, we’re, therefore, unable to evaluate 𝑓 minus 𝑔 of negative six. We say that it’s undefined.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.