Question Video: Evaluating the Sum of Two Linear Functions at a Given Value

If 𝑓: ℝ ⟢ ℝ, where 𝑓(π‘₯) = 4π‘₯ βˆ’ 4, and 𝑔: [βˆ’8, βˆ’2) ⟢ ℝ where 𝑔(π‘₯) = 5π‘₯ + 5, find the value of (𝑓 + 𝑔)(5) if possible.

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Video Transcript

If 𝑓 maps numbers in the set of real numbers onto the set of real numbers, where 𝑓 of π‘₯ is equal to four π‘₯ minus four, and 𝑔 maps values of π‘₯ greater than or equal to negative eight and less than negative two onto the set of real numbers, where 𝑔 of π‘₯ is equal to five π‘₯ plus five, find the value of 𝑓 plus 𝑔 of five if possible.

We’ve been asked to find the value of 𝑓 plus 𝑔 at π‘₯ equals five. Now, this might look quite complicated, but all it’s really asking us is to evaluate 𝑓 of π‘₯ plus 𝑔 of π‘₯ at five. Before we do that, though, let’s look at some of the notation given in our question. We’re told that the function 𝑓 maps numbers in the set of real numbers onto the set of real numbers. This tells us something about the domain of our function 𝑓. Remember, the domain is the set of all possible inputs that will yield real outputs. So, the domain of our function 𝑓 is simply the set of all real numbers.

And what about the function 𝑔? Well, this maps values of π‘₯ greater than or equal to negative eight, that’s what this square bracket means, and less than negative two; that’s what the curly bracket means. And it maps these numbers onto the set of real numbers. And so, the domain of 𝑔 is these values of π‘₯. It’s π‘₯ is greater than or equal to negative eight and less than negative two.

And when finding the sum of two functions, we need to be really careful with the domain. The domain of 𝑓 plus 𝑔 of π‘₯ will be the intersection. Remember, that’s the overlap of the domains of 𝑓 and 𝑔. So, let’s begin by working out what 𝑓 plus 𝑔 of π‘₯ is. 𝑓 of π‘₯ is four π‘₯ minus four and 𝑔 of π‘₯ is five π‘₯ plus five. So, their sum 𝑓 plus 𝑔 of π‘₯ is four π‘₯ minus four plus five π‘₯ plus five. And collecting like terms, we get nine π‘₯ plus one. So, what is the domain of our function?

We saw that the domain of 𝑓 was the set of all real numbers. That’s all numbers up to positive ∞ and all the way down to negative ∞. We see then that the domain of 𝑔 is essentially a subset of the domain of 𝑓; it’s values of π‘₯ greater than or equal to negative eight and less than negative two. The intersection then of the two domains, which we said is the overlap, is the same as the domain of 𝑔. It’s values of π‘₯ greater than or equal to negative eight and less than negative two.

Now, we’re looking to find the value of 𝑓 plus 𝑔 of five. But five is greater than negative two, meaning it’s outside the domain of 𝑓 plus 𝑔. This means we can’t evaluate 𝑓 plus 𝑔 of five; we’re not able to substitute five in. And so, for this reason, we say that the value of 𝑓 plus 𝑔 of five is undefined.

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