In Exercises 1–30, find the domain of each function. g(x) = √(x −...

Question

In Exercises 1–30, find the domain of each function. g(x) = √(x −2)/(x-5)

Accepted Answer

Welcome back. I'm so glad you're here. We have this function F of X equals the square root of x minus six, divided by x minus 13. We are asked to find the domain? Well, we have two domain restrictions here. First. This denominator cannot equal zero and second. What's inside the radical here Must be greater than or equal to zero. So let's do each of those step by step. Let's first look at the denominator. Our denominator is X -13. And we know that that can't equal zero because if we have zero in the denominator of a fraction, the fraction, unless our function will be undefined. So in order to figure out what number would make our denominator equal to zero? We set the denominator equal to zero and then we solve for X. So we can add 13 to both sides. And if we have an X that is equal to 13, our denominator will equal zero and it will be undefined. Which is a domain restriction. So therefore X cannot equal 13. There's one of our restrictions. The other one I mentioned, is that what's inside the radical or that X minus six must be greater than or equal to zero. Why? Well, if we have a negative number inside the radical, then we're dealing with imaginary numbers and we only want to deal with real numbers here. So again, we're going to solve for X. So we'll add six to both sides here. And we get that X must be greater than or equal to six. Those are our two domain restrictions. So now how do we express the domain? Well, let's think of this on a number line X can be any numbers from negative infinity. All the way to positive infinity. As long as we keep in mind our domain restrictions. Alright. A zero here in the middle. What are our restrictions? Well, X has to be greater than or equal to six. So if we have six here it can be equal to six. So I'll write this with a filled in circle but it has to be greater than or equal to six. So it's going to be heading to the right toward positive infinity. Now, the only problem is that X can't equal which is in here where we have our line. So I'm going to write an open circle at 13 and that's going to break up our line a little bit. X can be any number from six. All the way up to 13 but not including 13. And then starting just after 13 that it keeps going all the way up to infinity. Now, how do we express this? An interval notation? We're going to start with a bracket because a bracket means including. So it's including six And it's starting at six and going all the way up to 13 but not including 13. So we express not including with the parentheses and that is in union with and starting again with the parentheses at 13 because it can be any number just after 13. Going all the way up to well, it just keeps going all the way up to infinity, and infinity also gets closed with the parentheses because we can never quite obtain infinity. We look at our answer choices and this matches with answer choice C. Well done, we'll catch you on the next one.

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In Exercises 1–30, find the domain of each function. g(x) = √(x −2)/(x-5)

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