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Math Help - about ln functions

  1. #1
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    about ln functions

    For ln(x^2+1)

    Here's what I have written:

    Domain: all real numbers
    x intercept: 0
    Asymptote:

    Not sure....I mean, I know that y will never be negative...is that an asymptote? Is there a vertical asymptote?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by theowne View Post
    For ln(x^2+1)

    Here's what I have written:

    Domain: all real numbers
    x intercept: 0
    Asymptote:

    Not sure....I mean, I know that y will never be negative...is that an asymptote? Is there a vertical asymptote?
    x = 0 is a vertical asymptote for y = ln(x). However, since your argument never gets to 0 it does not have one.

    -Dan
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  3. #3
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    However, since your argument never gets to 0 it does not have one.
    What do you mean by this? Do you mean f(x)? Isn't f(x) = 0 when x=0?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by theowne View Post
    What do you mean by this? Do you mean f(x)? Isn't f(x) = 0 when x=0?
    The argument is the quantity inside the function. So the argument of ln(x^2 + 1) is x^2 + 1.

    -Dan
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