Video: Evaluating the Quotient of Two Functions at Specific Value

Given that 𝑓 : ℝ β†’ ℝ where 𝑓(π‘₯) = βˆ’3π‘₯ βˆ’ 4 and 𝑔 : (1, 7) β†’ ℝ where 𝑔(π‘₯) = βˆ’2π‘₯ βˆ’ 4, find the value of (𝑓/𝑔)(βˆ’1) if possible.

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Video Transcript

Given that the function 𝑓 maps numbers in the set of real numbers onto the set of real numbers, where 𝑓 of π‘₯ is equal to negative three π‘₯ minus four, and the function 𝑔 maps numbers in the open interval from one to seven onto the set of real numbers, where 𝑔 of π‘₯ is negative two π‘₯ minus four, find the value of 𝑓 over 𝑔 of negative one if possible.

We begin by recalling that 𝑓 over 𝑔 of π‘₯ is simply the quotient of the functions 𝑓 and 𝑔, it’s 𝑓 of π‘₯ over 𝑔 of π‘₯, and that when we’re working with two or more functions, we need to be really careful when looking for their domains. The domain of 𝑓 over 𝑔 of π‘₯ will be the intersections of the domains of 𝑓 of π‘₯ and 𝑔 of π‘₯. But when we’re working with the quotient of two functions, we also need to exclude the values of π‘₯ that make the function in our denominator equal to zero. So, what are the domains of our functions? Well, the function 𝑓 maps numbers from the set of real numbers onto the set of real numbers. So, the domain of 𝑓 is simply all real numbers.

The function 𝑔, however, only takes numbers from the open interval from one to seven. So, the domain of 𝑔 is this open interval. We can have values of π‘₯ greater than one and less than seven. The domain of the quotient of our two functions then will be the overlap of these two domains. Well, since the domain of 𝑓 was simply the set of real numbers, the overlap is the open interval from one to seven. And so, let’s evaluate 𝑓 over 𝑔 of π‘₯. It’s negative three π‘₯ minus four over negative two π‘₯ minus four. We’ve said that its domain is the set of numbers in the open interval one to seven. But of course, we need the denominator of our fraction not to be equal to zero because if we divide by zero, we get a result that is undefined.

Let’s add four to both sides and then divide by two. So, we find π‘₯ cannot be equal to negative two. Well, that’s outside of our domain anyway, so we’re not so worried about it. So, are we able to evaluate 𝑓 over 𝑔 of negative one? Well, no, remember, our domain is the open interval from one to seven. In other words, π‘₯ must be greater than one and less than seven. Negative one is less than one. And so, since it’s outside of the domain of our function, we’re unable to substitute π‘₯ equals negative one into our expression for 𝑓 over 𝑔 of π‘₯. And we say that 𝑓 over 𝑔 of negative one is not defined.

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