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Explain, using Theorems 4, 5, 7, and 9, why the f…

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Problem 27 Easy Difficulty

Explain, using Theorems 4, 5, 7, and 9, why the function is continuous at every number in its domain. State the domain.

$ Q(x) = \dfrac{\sqrt[3]{x - 2}}{x^3 - 2} $

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Video Transcript

Okay. Well given this function and they want us to explain why it's continuous everywhere in its domain and then find its domain. So first we better make sure we know what a domain is. So the domain of of a function is the set of all numbers X. That make a real number come out of the function when you put X in it. So numbers that don't cause a problem in the function. Okay, so instead of this one, let's look at a couple of other functions first. So if I had this F of X equals the square root of X. You know that you can't plug minus four in there Because the square to minus forest to I and two I is not real. So for this function This -4 would not be in the domain. In fact, no negative numbers would be in the domain. The domain of this is all X greater than or equal to zero. Okay. So if there's a square root you have to find out what makes the stuff underneath greater than or equal to zero. I don't know why I start scribbling that out. Okay. I is bad when you're trying to find the domain Can the other bad thing is for example, this let's say we had 20 over X -1. All right, well if you tried to plug one in there, you would get 20/0. Okay, also bad. Okay, because that's undefined. Okay, so you can't let a function to have zero in the denominator. Yeah. Okay, but if you add F of X equals the cube root of X. Thanks. Um You could let X be anything. Okay because you can take the Q. Bert of eight you can take the cube root of zero, you can take the cube root of negative eight. And when you do any of those who get a real number so there's no problem with the Q brew. You can put anything you want in there. So it's domain is all real numbers minus infinity to infinity. I forgot to say the domain of this one. It's All real numbers except minus or accept positive one. So X can't be one. Is its domain. Okay so back to our function here We got a cube root on the top. So there's no problem. You can plug everything you want in there and the next Cube -2. That's a that's not a problem. Except for its in the denominator. Okay so we got to find out what will make that equal to zero and then we have to leave that out. Okay X cubed minus two equals zero. X cubed equals two. X equals the Q. Bert of two. Okay so we cannot plug the Q. Bert of two in there because then we'll get zero on the bottom. Okay so the domain of this all real numbers except X equals the cube root of two. She can block anything you want in there except that. Okay here's a here's another way to write it, keep it up to. So start over here at minus infinity loops over here at minus infinity Until you get to the cube root of two and stop. Leave it out with a round bracket together with that's what the union means. Leave it out again and go on to infinity. Because remember the round brackets says don't include that value. So don't include the cube root of two on this side. Don't include the cube root of two on that side. So here is a way to right the this that I wrote inwards so that either one of those for the domain. Mhm. Okay, now why is it continuous everywhere in its domain? Okay, I don't know what the numbers are in your book. So you'll just have to figure it out, Execute -2. That's polynomial polynomial. Czar continuous, continuous everywhere. Uh X minus two is a polynomial. It continues everywhere, cube roots continues everywhere quotient the quotient. Yeah, the question of one continuous function divided by another is continuous except at the points to make the bottom zero. Okay, so polynomial czar continuous cube roots are continuous. When you divide continuous functions it's continuous. Okay, so all you're really doing, when you're answering this question is you're looking at each individual piece. Okay, this is continuous, This is continuous, add the cube root still continuous, divide by still continuous. Okay, hope

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