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Use the definition of continuity and the properties of limits to show that the function is continuous on the given interval.

$ g(x) = \frac{x - 1}{3x + 6}, \hspace{5mm} (-\infty, -2) $

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02:51

Daniel Jaimes

06:14

Leon Druch

Calculus 1 / AB

Chapter 2

Limits and Derivatives

Section 5

Continuity

Limits

Derivatives

Missouri State University

Campbell University

University of Michigan - Ann Arbor

Boston College

Lectures

04:40

In mathematics, the limit of a function is the value that the function gets very close to as the input approaches some value. Thus, it is referred to as the function value or output value.

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

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Use the definition of cont…

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Continuity Determine the i…

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Continuity on intervals Us…

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Okay, so we're going to show that um G of X given here as X minus 1/3 X plus six. We want to show that it is a continuous function on the interval given which is minus infinity to minus two. So basically let's review what we need to look for. So we are continuous. Um At Mexico's a at any point on the function, if we need F. Of A to exist, we need F prime F. A. Um Oops, I wanted to I'm jumping ahead of myself. Okay, I want the limit as X goes to A of F of X to exist and I want um So basically the point exists, the limit exists, meaning it is not infinite or it's not like it's not any kind of DNA case, but also then if the limit as X goes to a of F of X equals F. So that's what we're looking to prove and if we can prove that we're good and that's the idea for this. Okay, so let's take a look at each of these parts. Um Alright, so we want G. Of A. Which is a minus 1/3. A plus six. Um This will exist So long as a does not equal to -2 because we don't want the bottom to go to zero, but since our interval does not include -2, we're good. And we have the first category done. Who? Okay. And I did my general form for continuous using ffx but we're just using G F X. Um since they gave us G F X. All right, so now let's check the second one. We want the limit as X goes to a um of my GF x to exist. Well let's see what happens. I get the limit as X goes to a of X -1 over three x plus six. Well now, so long as we're not at any kind of 0/0 case um but then we can simply plug a into X and we will um we'll be able to get our limit. So when we do that we will get a minus 1/3 A plus six And same issue like before so long as a is not -2, then this limit exists. So we have the second one because once again are -2 is not in our interval. And finally let's take a look and double check that our limit equals our point and we can see that in this case I'll rewrite rewrite as X goes to a of G F X equals G of a. We just solved those up ahead. So let's take a look on the left, we get a minus one over three A plus six. And on the right we get also a -1 because we just solved both of those, we got the same values. So this is true. So this is true. So therefore we have shown that our function G of X is continuous at all, points A on our Um interval which does not include -2 few. Okay, hopefully that helped have a wonderful day.

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