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Math Help - What's the domain for the given function?

  1. #1
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    What's the domain for the given function?

    Hey everyone, My friend and I are being asked to find the domain for the following function and it's derivatives:

    From what I can tell, it's R any real number, but my friend insists it's R any real number AND X doesn't equal -1. Who's correct?
    Last edited by theunforgiven; December 4th 2012 at 09:44 AM.
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  2. #2
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    Re: What's the domain for the given function?

    Just do the arithmetic: what is f(1)? What is f(0)?

    Or, perhaps, what is your understanding of the domain of a function?
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  3. #3
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    Re: What's the domain for the given function?

    Quote Originally Posted by HallsofIvy View Post
    Just do the arithmetic: what is f(1)? What is f(0)?

    Or, perhaps, what is your understanding of the domain of a function?
    To answer your question, the domain of a function, as far as I'm concerned, is a set of possible values where the function is defined.
    f(1) is 0/1 which is zero. f(0) is -1/0 which is undefined... Guess that answers my question. It's R any real number expect x=0. We thought the f'(x) and f"(x) had something to do with the domain, but I guess we were wrong.
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    Re: What's the domain for the given function?

    Quote Originally Posted by theunforgiven View Post
    To answer your question, the domain of a function, as far as I'm concerned, is a set of possible values where the function is defined.
    f(1) is 0/1 which is zero. f(0) is -1/0 which is undefined... Guess that answers my question. It's R any real number expect x=0. We thought the f'(x) and f"(x) had something to do with the domain, but I guess we were wrong.
    Given a function f:X→Y, the set X is the domain of f; the set Y is the codomain of f. The domain is the set of 'input' values for which the function is defined. Therefore for each element of the domain, there is a corresponding element in the codomain.

    For the first function it is defined for all x if x does not equal 0.
    Last edited by sjmiller; December 4th 2012 at 04:51 PM.
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