calculus---helppp me please!

Discussion in 'Math' started by ccc, Jul 14, 2006.

  1. ccc

    Thread Starter New Member

    Jul 14, 2006
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    It asks me to find the value of c for which the function defined below is continuous. What kind of discontinuity is present if c does not have this value?

    ln (x)
    ------ if 0 < x (not=) 1
    x - 1

    c, if x = 1

    Please help me out please!!!:eek:
     
  2. Papabravo

    Expert

    Feb 24, 2006
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    I'm confused, c does not appear in the function, or am I missing something?
     
  3. Dave

    Retired Moderator

    Nov 17, 2003
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    I too am confused. Traditionally the constant c arises due to integration, if so the equation would become:

    [ln(x)/(x-1)] + c

    But then the subsequent question makes little sense. Can you clarify what you mean?

    Dave
     
  4. kinyo

    Member

    Jun 6, 2005
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    when x=1, the function takes the form 0 divided by 0 ... which is considered undefined ... however, the function approaches the value of 1 within the vicinity of x=1 ... so i believe c must have the value of 1 for the function to be continuous
     
  5. Dave

    Retired Moderator

    Nov 17, 2003
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    From my interpretation, the function is continuous for x > 0, x /= 1 (where as you correctly state is an undefined state).

    But that doesn't explain what c is.

    Dave
     
  6. kinyo

    Member

    Jun 6, 2005
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    c is the value of the function that will make it continuous at x=1 ... the function is said to be discontinuous because it is undefined at x=1, where it results to 0/0

    i sound like i'm just repeating the original post ... so let me give another example ... the function (x^2 - 1)/(x-1)

    (x^2 - 1)/(x-1) is discontinuous at x=1 as well because it becomes 0/0, so we ask what is the value of the function (call it c) so that it becomes continuous at x=1? as it turns out, the value of c in this case is 2 .. and its easy to figure that out because (x^2 - 1)/(x-1) is simply x + 1, which will have a value of 2 when x=1

    and to make things clearer, the problem has nothing to do with the constant of integration which incidentally is also called c in possibly all textbooks in calculus

    hth
     
  7. Papabravo

    Expert

    Feb 24, 2006
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    To evaluate the limit of an indeterminate form use L'hopital's rule. Take the derivative of the numerator and the derivative of denominator. Now evaluate that quotient.

    Edit (Thanks Dave)
    Code ( (Unknown Language)):
    1.  
    2. d/dx(ln(x)) / d/dx(x-1) = (1/x) /1  = 1/x evaluated at x = 1 is just 1
    3.  
    so c = 1 is the value of the function which makes the original function continuous.
     
  8. Dave

    Retired Moderator

    Nov 17, 2003
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    143
    To correct the above:

    d/dx(ln(x)) / d/dx(x-1) = (1/x) /1

    However, based this approach it still produces c = 1.

    Dave
     
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