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Continuous Function

  1. Jan 16, 2005 #1
    If you are given:

    [tex] f(x) = \left\{\begin{array}{cc} (3x)/ (x+4),&\mbox{ if } x\leq 4 , x != -4 & \sqrt{x-3}, &\mbox{ if } x > 4[/tex] (3x is numerator and x+4 is denominator)

    find the points of discontinuity of the function. Identify each type. Since [tex] x != 4 [/tex](not equal to 4, don't know how to put not equal sign) , then this is a point of discontinuity. Hence is this a removavble discontinuity (the only discontinuity?) because you get find an extended function?

    Thanks a lot

    PS: Is there any good LaTex tutorial. As you can see I need to work on it!
     
    Last edited: Jan 16, 2005
  2. jcsd
  3. Jan 16, 2005 #2

    dextercioby

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    I think it has 2 points of discontinuity.First is '-4',obviously.

    Find the second.

    Daniel.

    BTW,i think you meant [tex] x\neq -4 [/tex]
     
  4. Jan 16, 2005 #3
    [tex] x = 3 [/tex] the other one because of domain restrictions?
     
  5. Jan 16, 2005 #4

    dextercioby

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    Okay,are u guessing???U have to be sure of your answer...Why is the function not continuous in "3"??

    Daniel.
     
  6. Jan 17, 2005 #5

    HallsofIvy

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    You say "Since x!= 4 (not equal to 4, don't know how to put not equal sign) , then this is a point of discontinuity." but the formula does NOT say x can't be 4, it says x can't be -4. You may be confusing 4 and -4.
     
  7. Jan 18, 2005 #6
    3 things must exist for a function to be continuous at a point

    (a) lim f(x) exists
    x--> a

    (b) lim f(x) = f(a)
    x-->a

    (c) f(a) is defined

    So

    lim f(x) does not exist. Hence 4 is a point discontinuity
    x--> 4

    lim f(x) does not exist. Hence -3 is a point discontinuity.
    x--> -4

    Is this right?

    Thanks
     
    Last edited: Jan 18, 2005
  8. Jan 18, 2005 #7
    any help is appreciated

    thanks
     
  9. Jan 18, 2005 #8

    learningphysics

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    You're correct that 4 is a point of discontinuity since limx->4 does not exist.

    For the second part, I think you meant, hence -4 is a point of discontinuity? You're right that the limx->-4 does not exist, but also f(-4) is undefined. So either reason gives you that there's a discontinuity at x=-4.

    Yes, I find according to the question that there are 2 points of discontinuity, 4 and -4.
     
  10. Jan 18, 2005 #9

    dextercioby

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    Hey,points o discontinuity are in general "x" values.Tell me why the "x=-3" is a point of discontinuity...

    You've already been answered to the problem...

    Daniel.
     
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