Kay, this is gonna be my "help me with all my math questions" 'cause I have a midterm tomorrow. So, yea. Hi Mowicz. Lolol.

Anyways.




Where are each of the following functions discontinuous?

a)

f(x) = (x^2 - x - 2)/(x - 2)

b)

f(x) = { (x^2 - x - 2)/(x - 2) if x=/=2
{ 1 if x=2


Answers:

a) f(2) is not defined, thus f is discontinuous at 2.

b) f(2)=1 is defined and lim as x->2 = (x-2)(x+1) / (x-2) = (x+1) = 3

But lim x->2 f(x) =/= f(2) so f is not continuous at 2.


My question:
Why the hell do you factor b) and not a)?

I think I just don't understand the definition of discontinuity.

Uh, I'm only a first year, and not so great at math, so maybe I don't really know what I'm talking about.

But are you sure the lim as x->0 is DNE?

It seems to me that f is continuous as 0... I don't see why x cannot be equal to 0;

lim as x->0 = 1.

f is indeed discontinuous as 2 though, its a removable discontinuity. I would have definitely factored a, i don't know why they didn't.

Originally Posted by iRetaliate

Uh, I'm only a first year, and not so great at math, so maybe I don't really know what I'm talking about.

But are you sure the lim as x->0 is DNE?

It seems to me that f is continuous as 0... I don't see why x cannot be equal to 0;

lim as x->0 = 1.

f is indeed discontinuous as 2 though, its a removable discontinuity. I would have definitely factored a, i don't know why they didn't.

My bad.
I mixed up the answers. There was no breakline between the answers. D:<

I know you can factor b because the function is 'approaching' 2, thus will never equal two. This means that the part of the function can never be two, therefore you can factor it.

I don't get the first question though... Or at least, what you're asking for.

Originally Posted by mellye

I know you can factor b because the function is 'approaching' 2, thus will never equal two. This means that the part of the function can never be two, therefore you can factor it.

I don't get the first question though... Or at least, what you're asking for.

I'm not asking for anything in regards to a), just copying it straight out of the textbook. It says to find out when the functions are discontinuous.

I just don't see why you would factor one of them and not the other.

Do you have the page number and section please?
If it's a supporting problem I've done it

Originally Posted by goodnews.inc

Do you have the page number and section please?
If it's a supporting problem I've done it

pg 120

example 2, a) and c)

I believe its called holes. I haven't looked at it since gr 11, but basically

something like say a power of the function is greater on the denominator like
(x-2)^2/(x-2)^3 is discontinous because once you simplify, its 1/(x-2)

for a hole, you get something to the same power or greater on top like
(x-2)^2/(x-2)^2 or (x-2)^3/(x-2)^2 once you simplify you either get 1 or (x-2)
You get a hole at x = 2 since you cannot plug in 2 (even if you simplify, since you will be dividing by a 0 in the unsimplified equation), but its continous at all points other than x=2, and allowing you to factor. Not quite sure of the theory behind it, Mocwicz can most likely take care of that, but if the power is the same or greater on the top than the bottom, its factorable and has a hole. Hope that helped, sorry about the bad explanation, I can't really explain anything worth a damn =\

Edit: Mahratta has the right idea I think, I read what you said way wrong t_t.

I think it is because a is asking for you to determine continuity, while b asks for you to determine the limit. For a function to be continuous, it has to be defined at that point - however, regarding a limit, the function doesn't actually have to be defined at that point since the point itself doesn't necessarily concern the limit.
So, for a, even if you factored, you would still get a 'hole' or 'removable discontinuity'. While a limit would exist, the function is not defined at the point, which you would notice without factoring.
B, on the other hand, asks for the limit (at least, I think it does). Since this is the case, the continuity of the function at the point in question doesn't really matter. The function does have a 'hole', so it is not defined, but the limit still exists.

I hope that helps, and I hope I actually was on the right tangent...

If you factored in part a then f(x) = x+1

which would mean the graph would look like a continous line with a slope of 1 and y-int of 1 as well. However in reality, thats not how the graph is because the original equation is a rational function, which mean it has asymptotes. So by factoring and canceling out (x-2), you will end up missing a key piece of the graph. The original graph will look exactly same as f(x)=x+1 BUT at x=2, it will have a "hole". From my experience in 1st year, you dont wanna simplify if its asking for sumthing like discontinuity.

As for part b, they are factoring because you need a value, and if you were to just plug x=2 into the equation your answer would be undefined. So the only other way would be to somehow simplify the equation and thats why they factored it.

You have to understand that "cancelling" only works when the term is non-zero.

That is (x+1)/(x-2)/(x-2) can only be simplified to x+1 when the cancelled term is non-zero, ie. when x!=2.

you can factor in the b) because you're taking the limit as x approaches 2, it is never equal to 2.

Originally Posted by hmmmcurious

You have to understand that "cancelling" only works when the term is non-zero.

That is (x+1)/(x-2)/(x-2) can only be simplified to x+1 when the cancelled term is non-zero, ie. when x!=2.

you can factor in the b) because you're taking the limit as x approaches 2, it is never equal to 2.

I don't know what you mean by non-zero...

"you can factor in the b) because you're taking the limit as x approaches 2, it is never equal to 2."

It is equal to 2... f(2) = 1.

Why aren't you taking the limit as x approaches 2 in a)?

Originally Posted by lawleypop

I don't know what you mean by non-zero...

"you can factor in the b) because you're taking the limit as x approaches 2, it is never equal to 2."

It is equal to 2... f(2) = 1.

The function defined in b) is a piecewise function. That means the function is defined different depending on the value of x.

For any x that does not equal 2, the function is y = (x^2 - x - 2)/(x - 2)

For x = 2, the function is y = 1

(x-2) will = 0 when x = 2. But when x =2, y=1. Thus, the denominator in y = (x^2 - x - 2)/(x - 2) will never = 0, so you can factor it out.

The def. of continuity states that a function is continuous at a if the limit as x->a exists and is equal to f(a).

In b), f(2)=1, but the limit does not agree. When we take the limit of
{ (x^2 - x - 2)/(x - 2)

as x->2, we are taking values closer and closer to 2 and see what they "tend" towards. We are never plugging in 2 it self, just values very close to 2. Since we are never taking x=2 in the limit, it is ok to factor out the x-2 because it will never be 0 in the limit. Then the limit is simple, lim x->2 (x+1)=2+1=3.
But this does not agree with the def. of f(2)=1. Therefore you will see the a jump in the graph and thus it is discontinous at the point x=2.

Sorry if you dont understand, its somewhat hard to explain in words without using excessive amounts of details. I don't have time to do that :(

Originally Posted by hmmmcurious

The def. of continuity states that a function is continuous at a if the limit as x->a exists and is equal to f(a).

In b), f(2)=1, but the limit does not agree. When we take the limit of
{ (x^2 - x - 2)/(x - 2)

as x->2, we are taking values closer and closer to 2 and see what they "tend" towards. We are never plugging in 2 it self, just values very close to 2. Since we are never taking x=2 in the limit, it is ok to factor out the x-2 because it will never be 0 in the limit. Then the limit is simple, lim x->2 (x+1)=2+1=3.
But this does not agree with the def. of f(2)=1. Therefore you will see the a jump in the graph and thus it is discontinous at the point x=2.

Sorry if you dont understand, its somewhat hard to explain in words without using excessive amounts of details. I don't have time to do that :(

Yeah. Basically, the actual value of a is not of concern in part (b), but is in part (a) Note: when I say 'a' out of brackets, I'm probably not referring to the part, but to the value that x approaches. That's because for f(x) to be continuous at a, f(a) has to be defined. So, even if you factor and find a hole, that just proves that the function is undefined at f(a) in an unneccesarily lengthy manner.
For (b), we only care about the trend of f(x) near a. The function being defined or undefined at x=a does not matter. So, if you factor and find a 'hole', the limit of f(x) as x approaches a can still be determined, and will exist if the one sided limits are valid, etc.