Question

PLEASE: Determine if the graph of f(x)=x^2+x if ≠0. and 0 if x=0 is continuous for all real numbers, state “no discontinuities” and explain why using the conditions of continuity. If there are any discontinuities, please:



a-Determine what x value the discontinuity occurs
b-Determine if the discontinuity if removable or non-removable
c-If removable, identify the value of the hole. If non-removable, show proof using the conditions of continuity.

Answer 1

I just need a, b and c PLEASEEE

Answer 2

\(y=x^2\) isn't continuous when \(x=0\)

Answer 3

so that is a? @zepp

Answer 4

can you do b and c then?

Answer 5

Oh, \(x^2+x\), my bad, have to check again D:

Answer 6

ok

Answer 7

Ok, a removable discontinuity occurs when the limit from left and right are the same, but from one side, it's undefined.

Answer 8

Yeah

Answer 9

I have been trying forever, can u just please tell me a,b,c? :)

Answer 10

:D

Answer 11

So what would I put??

Answer 12

I understand that, but what I would i put for a b and c @zepp and @apoorvk

Answer 13

Simply verify the limit from 0+ and 0-, and well, when it's undefined, it's at this point the discontinuity occurs.

Answer 14

c'mon, it's on the tray now. I toldya the function is non-continuous at (0,0). So part (a) is staring at your face. But please try to understand the procedure, or you won't be able to solve the other parts. (and am sorry but simply 'answering' is not what we help with, though sure as hell we can toil along with you :))

Answer 15

Ok let me see if I can answer then and tell me if I'm right?

Answer 16

a) x=0 ?

Answer 17

b) removeable and c) 0

Answer 18

@apoorvk @zepp is that right???

Answer 19

You mean (0,0) right for C? yeah they seem okay to me!

Btw just confirming in the first line of the question, what's " ≠0. " ??

Answer 20

does not equal 0

Answer 21

I don't think it's removable D:

Answer 22

is it removable or not? I'm not sure

Answer 23

There's no other point to repair that hole right there.

Answer 24

@apoorvk what do u think? removable or no

Answer 25

this was my answer: Discontinuous at (0,0)!
a) x=0

b) removable

c) (0,0) is the hole.

Answer 26

is that right

Answer 27

OMG. Does the question mean the function is not equal to zero, but is 0 for x=0 only. Does it? @zepp and @schmidtdancer

Answer 28

yes it says ≠..

Answer 29

@_@ I'm confused D:

Answer 30

so is my answer right??

Answer 31

it says ≠

Answer 32

@apoorvk @zepp

Answer 33

I think, thanks to the very stupid language of the question combined with my own stupidity, we may have to solve this again totally. Sorry folks :(

Answer 34

really? so my answer isn't right?!!?

Answer 35

I don't understand this part f(x)=x^2+x if ≠0. and 0 if x=0
If x≠0?...and x=0.. what?

Answer 36

I mean f(x)=x^2+x if x≠0. and 0 if x=0

Answer 37

so @zepp @apoorvk Idk....is my answer ok?

Answer 38

Oh I typed everything and misunderstood it again. Anyways, i'm got him now! :D

So, yeah, when x=0, f(x)=0 as well. right?

Answer 39

Yes! so that makes my answer right, correct????

Answer 40

And, \(x^2+x\) is itself '0' at x=0, right?

Answer 41

Forget the answer, just listen to me for a moment -.-

Answer 42

yes

Answer 43

So, the function graph is actually: