Question

Help: CONTINUITY and removable discontinuity:
(attached below)

Answer 1

@kropot72

Answer 2

So you have 2 conditions to have a removable discontinuity. The first one is simply that the value must not be defined at that point, which is pretty obvious that is isn't because of the denominator. The second condition is basically the difference between whether or not you have an asymptote at your undefined point or simply a hole. The way to tell whether or not its an asymptote or a whole is if you factored your function, would the factor that is undefined cancel or remain? If the factor on bottom cancels, it is merely a hole, if it does not cancel then it is an asymptote and CANNOT be removed. So I would factor your numerator and see if the denominator cancels. If it does, then simply plug in 4 into your result and you'll get the value you're looking for.

Answer 3

can we work on this together?

Answer 4

i've tried doing this on my own but seem to be getting something wrong.

Answer 5

Sure. So I would factor the numerator and see if x-4 cancels. If it cancels, you have a hole, if it does not cancel, you have an asymptote and it cannot be removed. You know how to factor the numerator, right?

Answer 6

let me try.

Answer 7

Alright, sure.

Answer 8

i got 2(x^2+3x-25) @Psymon

Answer 9

-28you mean? And after that, you still have to further factor the x^2 + 3x - 28.

Answer 10

Find factors of -28 that will add up to positive 3.

Answer 11

yes, ,-28

Answer 12

Right. So When you further factor the numerator, you should have 2(x±__)(x±__). Its just remember how to factor that x^2 + 3x - 28 portion.

Answer 13

2(x+7) (x-4)

Answer 14

Exactly. In turn, that cancels out the denominator, which means the discontinuity is removable and not just an asymptote. Not only that, since the term that would make your function undefined disappeared, you can simply plug in 4 to find out the value you need.

Answer 15

we're left with 2 (x+7)

Answer 16

so we plug in 4?

Answer 17

That's all ya do.

Answer 18

does this apply to every problem with continuity?

Answer 19

Yes. Most of the work is finding a way to eliminate the portion that would give you an undefined answer. Once you manage that, you just plug in the value that the limit is approaching. Of course there are times when you cannot remove the undefined portion, in which that means the limit does not exist and you are merely approaching an asymptote.

Answer 20

thanks:)

Answer 21

Sure :3

Answer 22

ohhh can i practice a little, then ask you for a tricky problem like the one you just mentioned? it would really help.

Answer 23

That works, I have no problem with that.

Answer 24

what about a JUMP DISCONTINUITY? same thing?

Answer 25

That's new terminology for me. I probably know what it is but have not heard it said that way. Could you give me an example?

Answer 26

sure

Answer 27

1 moment

Answer 28

Ah, just one sided limits. We never really gave a name for it before, just that it had to do with one-sided, lol.

Answer 29

oh so how do we begin with this one?
sketch a graph?

Answer 30

i can do graphs but i'd like to do it a simpler way since i won't have enough time when it comes to a final

Answer 31

@Psymon

Answer 32

I apologize, I guess I was making sure what jump kind of meant since I did not have a picture. But yes, all you really need to confirm is that when x = 6 on both functions that they have different limits. It is pretty obvious that they do. As far as the left and right part goes, you need to see which function to just check and see which function is approaching from the left and which one is approaching from the right. Since 8x-8 < 6, that must be the function approaching from the left. Therefore, f(6) of 8x-8 is the value you will use for limit from the left. Since the other function is the one that will becoming from the right, it's value at x = 6 will be the value you use for the right-sided limit. Basically, you just need to see which function to plug the value into.

Answer 33

The delay was graphing and looking at so I knew exactly what to say and to get my brain to work, lol.

Answer 34

it's okay, so how can we work this prob out?

Answer 35

I know you know how to graph, but I wanted the visual up before we move on. So you just want to see which function is from the left and which one is from the right. When it gives you a piece-wise function, it's pretty obvious which one is going to be coming from the left and which one from the right. Once you determine that, just plug in x = 6 into the graph that comes from the left and use that value for lim ->6 - Once you do that, plug x = 6 into the other function, since you know it's the one on the right, and find lim x-> 6 +