Question

Calculus Help!!!
f(x)=x-x^2-2 and can someone plz explain how this is Nonremovable discontinuity at x=1 and Removable discontinuity at x=0?????

Answer 1

it isn't unless there is a denominator involved

Answer 2

\(f(x)=x-x^2-2\) is a polynomial, therefore it is continuous everywhere

Answer 3

ok i understand that much but i do not understand the removable and unremovable part and the x=0 either...

Answer 4

a polynomial does not have a discontinuity, removable or otherwise

Answer 5

so the question as is stands is incorrect. this does not have a removable discontinuity at \(x=1\) or anywhere else
is there something missing from this problem?

Answer 6

Not sure where you got this question from. @satellite73 is correct; your function is a polynomial, and polynomials are continuous everywhere. There is no discontinuity here.

Answer 7

o my gosh im sooooooooo sorry the question is x/x^2-x sorry guys :<

Answer 8

if you have a rational function, you will have a discontinuity where the denominator is 0
if you can factor and cancel, then the discontinuity will be removable because you can "remove" it

Answer 9

lol, that makes much more sense

Answer 10

sorry

Answer 11

\(f(x)=\frac{x}{x^2-x}=\frac{x}{x(x-1)}=\frac{1}{x-1}\) so there, we just "removed" the discontinuity at \(x=0\) by cancelling

Answer 12

however there is nothing you can do about the discontinuity at \(x=1\) you are stuck with it

Answer 13

mmmm okay i understand the removed one both actually :}

Answer 14

\(f\) has two discontinuities, at the two zeros of the denominator, namely at \(x=0\) and \(x=1\)
the function is undefined for those to values

Answer 15

but the one at \(x=0\) can be removed, because the two functions \(f(x)=\frac{x}{x^2-x}\) and \(g(x)=\frac{1}{x-1}\) are identical except at \(x=0\) where \(f\) is undefined, but where \(g\) is \(-1\)

Answer 16

okay :] thank u!