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Domain of (1/3+lnx)

  1. Oct 22, 2011 #1
    Hi just a quick question
    Differentiate f and find the domain of f. (Enter the domain in interval notation.)
    f(x)=(1/(3+lnx))


    Here is my Derive: ((-3+lnx)^-2)(1/x) which is correct

    And for the domain (web assign wants interval notation)
    When I was doing this part, I wasn't sure if it includes 0 or not since ln(0) cannot be defined.
    I tried both (0,∞) and [0,∞) which both were incorrect.

    Did I visualize the graph wrong or something? I even tried to use a graphics calculator

    Please enlighten me ^^
     
  2. jcsd
  3. Oct 22, 2011 #2

    Dick

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    1/(3+ln(x)) is also not defined if ln(x)=(-3), right? For what value of x does that happen?
     
  4. Oct 22, 2011 #3

    SammyS

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    [itex]\displaystyle f(x)=\frac{1}{3+\ln(x)}[/itex]

    What value of x would make the denominator zero ??
     
  5. Oct 22, 2011 #4
    so whenever x=e^-3 this equation will not exist? Since denominator will equal to 0 ^^

    Im still kind of confused how i should input the interval notation...

    will it be (0,e^-3)U(e^-3,∞) if I use interval notation? Should 0 be included since ln(0) is undefined
     
  6. Oct 22, 2011 #5

    SammyS

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    Yes, that's the way to write the domain in interval notation.

    Of course 0 is not included. ln(0) is undefined just as division by zero is undefined.
     
  7. Oct 23, 2011 #6
    Thank you for helping me on solving the question ^^

    From the number of posts, you can see that I'm a new user in this forum and I think this is one really awesome forum
     
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