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i am looking for a simple definition of discontinuity using the example 3/(x^2+x-6)

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- #1

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i am looking for a simple definition of discontinuity using the example 3/(x^2+x-6)

- #2

rock.freak667

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A simple definition would be that a point of discontinuity on the graph of f(x) occurs at x=a such that f(a) is undefined.

So where would your function be undefined for?

- #3

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at x=2,-3. i understand that but when you say x=a what do u mean by that. can you show me a variable formula (using a, b and c) that would better explain this

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- #4

rock.freak667

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Say your function wasat x=2,-3. i understand that but when you say x=a what do u mean by that. can you show me a variable formula (using a, b and c) that would better explain this

f(x)=1/(x-a)

Wouldn't you agree that [itex]\lim_{x \rightarrow a} = \infty[/itex] and so f(a) is undefined?

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- #6

rock.freak667

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your f(x) is 3/(x-2)(x+3), as x→2, f(x)→∞, thus f(2) is undefined, similarly for f(-3)

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sorry but i still dont get it, its not like we have a denominator of zero

- #8

rock.freak667

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but if x=2, don't you have one?sorry but i still dont get it, its not like we have a denominator of zero

[tex]f(x)=\frac{3}{(x-2)(x+3)}[/tex]

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wat i ment was it dosent cancel out or anything

- #10

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ok then how would you compare that with a function that is not discontinuous like 1/(x+3)

- #11

rock.freak667

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cancel out with what?wat i ment was it dosent cancel out or anything

f(x)=1/(x+3) is discontinuous at x=-3.ok then how would you compare that with a function that is not discontinuous like 1/(x+3)

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- #12

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A function is discontinuous at a point c if the limit as x->c of f(x) does not equal f(c).sorry but i still dont get it, its not like we have a denominator of zero

In this case, consider your c=2. f(c) is not defined. The limit of f(x) as x->2 from the positive side is +infinity. The limit of f(x) as x->2 from the negative side is -infinity.

The first way you can know this function is discontinuous at 2 is that it is not defined at 2. Then no matter what the limit of f(x) is as x->2, it is not equal to f(2) (even if the limit does not exist).

The second way you can know this function is discontinuous at 2 is that the limit of f(x) as x->2 from the positive side does not equal the limit of f(x) as x->2 from the negative side. Therefore the limit does not exist. If the limit does not exist, then it does not equal f(2) (even if f(2) is not defined).

Remember to keep in mind that a function is discontinuous at a point c if the limit as x->c of f(x) does not equal f(c). f(c) could exist, it just happens to not exist in this example.

- #13

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Actually the limit of f(x) as x->a from the left is negative infinity, while the limit of f(x) as x->a from the right is positive infinity.Say your function was

f(x)=1/(x-a)

Wouldn't you agree that [itex]\lim_{x \rightarrow a} = \infty[/itex] and so f(a) is undefined?

Therefore the limit of f(x) as x->a doesn't exist.

I think you mean x = -3.cancel out with what?

f(x)=1/(x+3) is discontinuous at x=3.

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so does that mean all rational functions are discontiuous.

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- #18

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would (x)/(x^2-4) be a discontinuity

- #19

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That function would be discontinuous at x=2 and x=-2

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