Video: Finding and Evaluating the Quotient of Two Functions

Given that 𝑓 and 𝑔 are two real functions where 𝑓(π‘₯) = π‘₯Β² βˆ’ 1 and 𝑔(π‘₯) = √(π‘₯ + 5), find the value of (𝑔/𝑓)( βˆ’2) if possible.

03:21

Video Transcript

Given that 𝑓 and 𝑔 are two real functions where 𝑓 of π‘₯ is equal to π‘₯ squared minus one and 𝑔 of π‘₯ is equal to the square root of π‘₯ plus five, find the value of 𝑔 over 𝑓 of negative two if possible.

Firstly, we recall that 𝑔 over 𝑓 of π‘₯ is simply the quotient of the functions. It’s 𝑔 of π‘₯ over 𝑓 of π‘₯. And the reason this question says to find the value of 𝑔 over 𝑓 of negative two β€œif possible” is because when we think about the domain of our combined function, that’s 𝑔 over 𝑓 of π‘₯, we think about the intersection of the domains of 𝑓 and 𝑔. But we want to make sure the denominator, here that’s 𝑓 of π‘₯, is not equal to zero. So, let’s begin by looking for the domains of our respective functions. 𝑓 of π‘₯ is π‘₯ squared minus one. Now that’s just a polynomial, and the domain of a polynomial is simply the set of all real numbers. So, the domain of 𝑓 of π‘₯ is indeed the set of all real numbers.

𝑔 of π‘₯ is a little more complicated though. We have the square root of π‘₯ plus five. And we know that for the square root of a number to be real, that number must be greater than or equal to zero. So here, we know that π‘₯ plus five must be greater than or equal to zero. Let’s solve this inequality by subtracting five from both sides. And when we do, we find that π‘₯ must be greater than or equal to negative five. And therefore, the domain of our function 𝑔 of π‘₯ is π‘₯ is an element of the left-closed, right-open interval from negative five to ∞. And so we move on to finding the domain of 𝑔 over 𝑓 of π‘₯. It’s the intersection of the domains of 𝑓 of π‘₯ and 𝑔 of π‘₯, which is the left-closed, right-open interval from negative five to ∞.

But we need to exclude the values of π‘₯ that make the denominator equal to zero. In other words, we need to exclude the values of π‘₯ that make the expression π‘₯ squared minus one, that’s the function 𝑓 of π‘₯, equal to zero. This time we’ll solve this equation by adding one to both sides, which gives us π‘₯ squared is equal to one. Next, we take the square root of both sides, remembering, of course, to take both the positive and negative square root of one. And when we do, we find π‘₯ is equal to positive or negative one. And so, this means that the domain of 𝑔 over 𝑓 of π‘₯ is the left-closed, right-open interval from negative five to ∞ minus the set including the elements negative one and one.

So, what does this mean for the value of 𝑔 over 𝑓 of negative two? π‘₯ is equal to negative two is within the domain of the function, and therefore it can be evaluated. And so, we’re now going to move on to finding the function 𝑔 over 𝑓 of π‘₯. We take the function 𝑔 of π‘₯, and we divide it by the function 𝑓 of π‘₯, so that’s the square root of π‘₯ plus five over π‘₯ squared minus one. Then, 𝑔 over 𝑓 of negative two is found by replacing π‘₯ with negative two, so we get the square root of negative two plus five over negative two squared minus one, which simplifies to the square root of three over three. And so, given the functions 𝑓 and 𝑔, we’ve shown that the value of 𝑔 over 𝑓 of negative two is the square root of three over three.

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