Question Video: Finding the Quotient of Two Function and Determining Its Domain

Given that 𝑓₁: (βˆ’βˆž, 4) β†’ ℝ such that 𝑓₁(π‘₯) = π‘₯ + 5 and 𝑓₂: (βˆ’8, 6) β†’ ℝ such that 𝑓₂(π‘₯) = 2π‘₯Β² + 13π‘₯ + 15, find (𝑓₂/𝑓₁) (π‘₯) and state its domain.

03:29

Video Transcript

Given that the function 𝑓 sub one maps numbers on the open interval from negative ∞ to four onto the set of real numbers such that 𝑓 sub one of π‘₯ is π‘₯ plus five. And the function 𝑓 sub two maps numbers on the open interval from negative eight to six onto the set of real numbers such that 𝑓 sub two of π‘₯ equals two π‘₯ squared plus 13π‘₯ plus 15. Find 𝑓 sub two over 𝑓 sub one of π‘₯ and state its domain.

Let’s begin by just looking at the notation in this question. Take our function 𝑓 sub one. It takes values of π‘₯ in the open interval from negative ∞ to four. That’s its input. And it gives an output which is a real number. Well, we know that the input to a function is its domain, and the range is its output. So, the domain of our function 𝑓 one is values of π‘₯ in the open interval negative ∞ to four, whereas the range is the set of real numbers.

Similarly, we can say that the domain of 𝑓 sub two is values of π‘₯ is greater than negative eight and less than six. And its range is the set of real numbers. The question wants us to find 𝑓 two over 𝑓 one of π‘₯. And so, we recall that this is really asking us to find the quotient of our functions. So, we’ll begin by finding the quotient of 𝑓 two and 𝑓 one. It’s the function defined by two π‘₯ squared plus 13π‘₯ plus 15 divided by the function π‘₯ plus five.

Now, in fact, we’re able to simplify this fraction a little by factoring the numerator. When we factor the numerator, we get two π‘₯ plus three times π‘₯ plus five. And this is really useful because we now spot that there is a common factor of π‘₯ plus five in our numerator and denominator. We’re therefore going to divide through by π‘₯ plus five. And when we do, we find 𝑓 two over 𝑓 one of π‘₯ is two π‘₯ plus three.

But what’s the domain of this function? Well, when we’re working with two functions, specifically finding the question of the two functions, the domain of the quotient is the intersection of the domains of the respective quotients. And so, we need to find the intersection or the overlap of the domains of 𝑓 one and 𝑓 two. The domain of 𝑓 sub one is values of π‘₯ on the open interval from negative ∞ to four. And the domain of 𝑓 sub two is values of π‘₯ on the open interval from negative eight to six. We see the overlap is values of π‘₯ greater than negative eight and less than four. So, π‘₯ can take values from negative eight to four, not including negative eight and four.

But we need to be really careful. We were working with a quotient of two functions, and we know that when we divide by zero, we get a result that is undefined. So, we need to make sure that the denominator of our fraction is not equal to zero. In other words, π‘₯ plus five is not equal to zero. Let’s solve this inequation to find the values of π‘₯ such that π‘₯ plus five is equal to zero.

We’ll subtract five from both sides. This means that when π‘₯ is equal to negative five, π‘₯ plus five is equal to zero. So, we need π‘₯ to be not equal to negative five. And so, we adjust our domain just a little bit by removing the number negative five from the domain. We can now say that 𝑓 two over 𝑓 one of π‘₯ is two π‘₯ plus three. And the domain is π‘₯ is an element of the open interval negative eight to four, not including the number negative five.

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