Question

@satellite73 need your help really quick, apparently nearly every question we've done together is incorrect.

Answer 1

Give an example of a function with at least one removable and one non-removable discontinuity (that has not been discussed in any lesson). Explain where the discontinuities occur and provide a justification for your reasoning. You may use a graph of the function to help explain your answers.

Answer 2

We stated that f(x)=(x-2)/(x-2)(x+1) is this, x=/=1 and when simplified it is 1/(x+1), leaving you with an asymptote

Answer 3

@satellite73 http://openstudy.com/users/lukecrayonz#/updates/4fc4f52be4b0964abc875a21
My teachers reply to all of those: 1] There are four discontinuities to describe here. For each, say whether it is removable or nonremovable and which condition(s) it fails.

Answer 4

http://screensnapr.com/v/7ChyRB.png

Answer 5

this is two different problems
one is the picture, and one is

"Give an example of a function with at least one removable and one non-removable discontinuity (that has not been discussed in any lesson). Explain where the discontinuities occur and provide a justification for your reasoning. You may use a graph of the function to help explain your answers."

Answer 6

It never says anything about stating which thing it failed..

Answer 7

which problem do you think is incorrect?

Answer 8

Oh sorry, i copied the wrong thing, but we got that one wrong too.

Answer 9

?

Answer 10

http://i.imgur.com/BNGmf.png

Answer 11

Her reply:
1] There are four discontinuities to describe here. For each, say whether it is removable or nonremovable and which condition(s) it fails.
2] Provide me an algebraic equation that contains both a removable and nonremovable discontinuity

Answer 12

I emailed her this about #2:

2] Provide me an algebraic equation that contains both a removable and nonremovable discontinuity

You stated that I got this question wrong, with my answer of f(x)=(x-2)/(x-2)(x+1)

If you put this into a calculator, there is an error at 2 and -1, as the asymptote -1, this forms a removable discontinuity.

with x=/=2, this creates a nonremovable discontinuity. I am not sure why I got this question wrong, or if I am not understanding this completely. Thanks.

Answer 13

@satellite73

Answer 14

does she want you to expand it out instead of leaving it in factored form?

Answer 15

hold on

Answer 16

I have no idea, that's all she said.

Answer 17

\[f(x)=\frac{x-2}{(x-2)(x+1)}\] certainly has a removable discontinuity at \(x=2\)

Answer 18

i think you got it backwards

Answer 19

the discontinuity at \(x=2\) is removable, because you can cancel and get
\[f(x)=\frac{1}{x+1}\] if \(x\neq 2\)

Answer 20

whereas the discontinuity at \(x=-1\) is not removable, because you have a vertical asymptote there

Answer 21

If you put this into a calculator, there is an error at 2 and -1, as the asymptote -1, this forms a nonremovable discontinuity.

with x=/=2, this creates a removable discontinuity. I am not sure why I got this question wrong, or if I am not understanding this completely. Thanks.

Answer 22

But even then, I didn't have to specify what it was, I merely wrote the equation, graphed it, and thats it.

Answer 23

i don't see what the problem is

Answer 24

Neither do I.

Answer 25

http://i.imgur.com/BNGmf.png
For this one, we stated x=/0 0, -2, 2, or 4.

Answer 26

sorry x=/=. But she wants which rule it doesn't satisfy

Answer 27

@satellite73 in number one, why can zero not exist?

Answer 28

ok those are the discontinuities for sure
now why?

Answer 29

http://i.imgur.com/segHB.png

Answer 30

Yes, which rule does it not satisfy from those three?

Answer 31

at \(x=0\) the limit from the left is different from the limit from the right, and therefore the limit itself does not exist

Answer 32

So rule number two.

Answer 33

right, it violates rule 2

Answer 34

And for positive two?

Answer 35

at \(x=-2\) the function does not even exist

Answer 36

so that one violates rule 1

Answer 37

and for 4?

Answer 38

at \(x=2\) the limit exists, but the function does not exist at that point. so that also violates rule 1

Answer 39

and finally, at \(x=4\) the function exists, but the limit does not

Answer 40

So rule two.

Answer 41

yes. the limit must be a number not "going to infinity"

Answer 42

does that clear it up more or less?

Answer 43

Sorry about all the questions, I thought we got all these right and then I was going to finish the course tomorrow but with this it randomly dropped my grade down to a B, so I want to fix it.

Answer 44

hope it is clear now, and you get an A

Answer 45

Okay, i have a piecewise question now. @satellite73

Answer 46

post