Question Video: Finding the Domain of Piecewise-Defined Rational Functions Mathematics

If 𝑓 and 𝑔 are two real functions where 𝑓(π‘₯) = 2π‘₯ + 2 if π‘₯ < βˆ’3, 𝑓(π‘₯) = π‘₯ βˆ’ 4 if βˆ’3≀ π‘₯ < 0, and 𝑔(π‘₯) = 5π‘₯ determine the domain of the function (𝑔/𝑓).

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Video Transcript

If 𝑓 and 𝑔 are two real functions where 𝑓 of π‘₯ is the piecewise function defined as two π‘₯ plus two if π‘₯ is less than negative three and π‘₯ minus four if π‘₯ is greater than or equal to negative three and less than zero and 𝑔 of π‘₯ is equal to five π‘₯, determine the domain of the function 𝑔 over 𝑓.

We begin by recalling 𝑔 over 𝑓 of π‘₯ is the quotient of our two functions. It’s 𝑔 of π‘₯ over 𝑓 of π‘₯. So, let’s begin by defining 𝑔 over 𝑓 of π‘₯. Well, since 𝑓 is a piecewise function, we’ll need to define 𝑔 over π‘₯ as shown. It’s five π‘₯ over two π‘₯ plus two if π‘₯ is less than negative three and five π‘₯ over π‘₯ minus four if π‘₯ is greater than or equal to negative three and less than zero. Now, when we’re dealing with the quotient of two functions, its domain is defined as the intersection or the overlap of the domain of the two respective functions.

Well, the domain of a polynomial, unless otherwise defined, is all real numbers. So, the domain of 𝑔 of π‘₯ is the set of real numbers. If we look at both parts of our function, we see that π‘₯ can take values from negative ∞ all the way up to zero. So, the domain of 𝑓, our piecewise function, is the open interval negative ∞ to zero. The intersection or the overlap of the set of real numbers and numbers in the open interval from negative ∞ to zero is numbers in the open interval negative ∞ to zero. So, we can assume that this is the domain of our function 𝑔 over 𝑓 of π‘₯.

However, we do need to be really careful. We’re working with a quotient. So, we need to make sure the denominator is not equal to zero in either case. In other words, we need to ensure that two π‘₯ plus two is not equal to zero and π‘₯ minus four is not equal to zero. If we solve our first inequation, we find π‘₯ cannot be equal to negative one. This is irrelevant though because we were told that this part of our piecewise function only applies when π‘₯ is less than negative three. And so, we don’t need to worry about including this in our domain. If we solve the second equation, we find π‘₯ cannot be equal to four. Well, this is outside of our domain and this part of our piecewise function, so we disregard this bit of information too.

And so, the domain of the function 𝑔 over 𝑓 is the set of numbers in the open interval from negative ∞ to zero.

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