Basic definition of discontinuity

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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)
 

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  • #2
rock.freak667
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f(x)=3/(x2+x-6)

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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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
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?
 
  • #5
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so you are saying that a would be the constant. like in my example if we factor x^2+x-6 we get (x-2)(x+3), so how would that be undefined
 
  • #6
rock.freak667
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so you are saying that a would be the constant. like in my example if we factor x^2+x-6 we get (x-2)(x+3), so how would that be undefined
your f(x) is 3/(x-2)(x+3), as x→2, f(x)→∞, thus f(2) is undefined, similarly for f(-3)
 
  • #7
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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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sorry but i still dont get it, its not like we have a denominator of zero
:confused: but if x=2, don't you have one?

[tex]f(x)=\frac{3}{(x-2)(x+3)}[/tex]
 
  • #9
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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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wat i ment was it dosent cancel out or anything
cancel out with what?

ok then how would you compare that with a function that is not discontinuous like 1/(x+3)
f(x)=1/(x+3) is discontinuous at x=-3.
 
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  • #12
jav
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sorry but i still dont get it, its not like we have a denominator of zero
A function is discontinuous at a point c if the limit as x->c of f(x) does not equal f(c).

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
jav
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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?
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.

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

cancel out with what?



f(x)=1/(x+3) is discontinuous at x=3.
I think you mean x = -3.
 
  • #14
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so does that mean all rational functions are discontiuous.
 
  • #15
jav
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Not necessarily, a rational function just means that it can be written as the ratio of two polynomials. But if, for example, the "denominator polynomial" is a constant, then the function is continuous over its domain.
 
  • #16
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do you mean it would be contiuous if the rational function was a polynomial divied by a constant like for ex. (x+1)/3. but wouldnt that give you a linear function
 
  • #17
jav
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If you choose your numerator to be a first order polynomial, then yes. But what about f(x) = [(x+1)^2]/3
 
  • #18
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would (x)/(x^2-4) be a discontinuity
 
  • #19
jav
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That function would be discontinuous at x=2 and x=-2
 

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