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Explain why the function is discontinuous at the …

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Problem 16 Easy Difficulty

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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Video Transcript

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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Video Transcript

James Stewart

Calculus: Early Transcendentals

Catherine Ross

Anna Marie Vagnozzi

Kristen Karbon

Joseph Lentino

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Catherine Ross

Anna Marie Vagnozzi

Kristen Karbon

Joseph Lentino

James StewartCalculus: Early Transcendentals

Catherine Ross

Anna Marie Vagnozzi

Kristen Karbon

Joseph Lentino

James Stewart

Calculus: Early Transcendentals