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Math Help - Continuity problem

  1. #1
    Member nautica17's Avatar
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    Continuity problem

    I'm not sure how to find these intervals of continuity:

    ln(x-2)

    sin(x^2 - 2)

    I don't understand how I am supposed to find the continuity. I look at the graphs of each and I'm not sure what to look for. I understand functions like tan(x), etc. but with these I'm a little confused. Can anyone point me in the right direction?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by nautica17 View Post
    I'm not sure how to find these intervals of continuity:

    ln(x-2)

    sin(x^2 - 2)

    I don't understand how I am supposed to find the continuity. I look at the graphs of each and I'm not sure what to look for. I understand functions like tan(x), etc. but with these I'm a little confused. Can anyone point me in the right direction?
    these functions are continuous where they are defined. the log function is continuous where its argument is positive: for example, \ln x is continuous where x > 0

    the sine function is continuous everywhere, and so it is continuous wherever its argument is: \sin (f(x)) is continuous where f(x) is continuous.

    so what can you do with that information?
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  3. #3
    Member nautica17's Avatar
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    Quote Originally Posted by Jhevon View Post
    these functions are continuous where they are defined. the log function is continuous where its argument is positive: for example, \ln x is continuous where x > 0

    the sine function is continuous everywhere, and so it is continuous wherever its argument is: \sin (f(x)) is continuous where f(x) is continuous.

    so what can you do with that information?
    Okay, so from what you have said I have come up with this:

    ln(x-2) is continuous when x is greater than 2?

    sin(x^2 - 2) is continuous no matter what and on whatever interval you sort of choose to put it on?
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    Quote Originally Posted by nautica17 View Post
    Okay, so from what you have said I have come up with this:

    ln(x-2) is continuous when x is greater than 2?

    sin(x^2 - 2) is continuous no matter what and on whatever interval you sort of choose to put it on?
    Correct.
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  5. #5
    Member nautica17's Avatar
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    Alright thanks guys.

    Sorry if it's a simple question, but it looked confusing at first.
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  6. #6
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    Quote Originally Posted by Jhevon View Post
    these functions are continuous where they are defined. the log function is continuous where its argument is positive: for example, \ln x is continuous where x > 0

    the sine function is continuous everywhere, and so it is continuous wherever its argument is: \sin (f(x)) is continuous where f(x) is continuous.

    so what can you do with that information?
    hey isn't -1<=sin(x^2-2)<=1
    =>-pi/2<=x^2-2<=pi/2
    =>sqrt(2-pi/2)<=x<=sqrt(pi/2+2)????
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Pulock2009 View Post
    hey isn't -1<=sin(x^2-2)<=1
    =>-pi/2<=x^2-2<=pi/2
    =>sqrt(2-pi/2)<=x<=sqrt(pi/2+2)????
    What?!
    Last edited by Jhevon; January 30th 2010 at 08:15 PM. Reason: fixed over punctuation :p
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  8. #8
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    Quote Originally Posted by Pulock2009 View Post
    hey isn't -1<=sin(x^2-2)<=1
    =>-pi/2<=x^2-2<=pi/2
    =>sqrt(2-pi/2)<=x<=sqrt(pi/2+2)????
    No, that is an interval on which the function is one-to-one and has nothing to do with where it is continous. x^2- 2 is continuous for all x, sin(x) is continous for all x, and the composition of two continuous functions is continuous.


    In a certain, very specific, sense, "almost all" functions are never continuous. But continuous functions are so nice that our ways of writing functions have developed so that functions written as a single "formula" are continuous where ever they are defined.
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