Question

Give the value of x where the function f(x)=[abs(x^2-1)]/[(x+1)(x-2)] has a removable discontinuity.

Answer 1

These absolute value questions always trip me up.

Answer 2

Easy - removable dicont. is where technically there is a zero of denominator, but actually the term responsible for zero - is cancelled out....

Answer 3

The absolute value doesn't change how I would go about that at all?

Answer 4

I would automatically go to the answer being x=-1 if it were not for the absolute value bars.

Answer 5

Since, after factoring, I get [(x+1)(x-1)]/(x+1)(x-2)], then, after cancelling (x+1) in both the numerator and the denominator, I'm left with (x-1)/(x-2). But because of the absolute value, isn't there more to be done, or no?

Answer 6

it is -1

Answer 7

0/0 @x=-1

Answer 8

Look abs does NOT change ZEROs and ZERO-locations !

Answer 9

@cuzzin suppose u didn't have abs - where is the removable d. ?

Answer 10

at x=-1

Answer 11

Oookey now - does abs EVER create discontinuity for a continuous function ?

Answer 12

No...right?

Answer 13

Yeap.

Answer 14

Au-contraire - can abs "destroy" discontinuity which is REMOVABLE before its application

Answer 15

Sometimes in class, when dealing with absolute values, the teacher will draw out the peacewise functions, like x<1 and x>1 and so on. Do you only do things like that in certain instances? Is that only when abs is in the denominator?

Answer 16

Concentrate on my questions please

Answer 17

Ok. I'm going to say that abs can destroy discontinuity.

Answer 18

Yes it can destroy - but ONLY if it is NOT , that is if the "actuall value" which "should have been" is zero - abs CANNOT destroy the discont

Answer 19

meant NOT ZERO

Answer 20

Well to cut to the chase : the discontinuity at zero of (x+1) which YOU CORRECTLY identified as x=-1 exists before and AFTER application of abs. This is tye removb. discont

Answer 21

So would I be correct in just removing the absolute value bars and just proceeding with solving the problem as normal?

Answer 22

Nooo

Answer 23

You should point out WHAT actually occurs at x=-1

Answer 24

Which is a removable discontinuity I guess. Plugging (-1) directly in to the problem gives an indeterminant solution.

Answer 25

Argument is necessary why there is NO jump there

Answer 26

Ok. Thanks for the help, I will try and get myself more caught up on all these discontinuities.

Answer 27

So Michael -- what exactly do you write as the reason?

Answer 28

@Mikael