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    I need to find the domain of : x^2 - 5 /2

    Can't x be anything?
    The exam has got the answer as x>(equal to of greater then) 0
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    It'll probably be that because of the context of the question as they'll probably want you to find the inverse function in a later part of the question and you can't inverse a many to one function
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    Well f^-1 (x) = (x^2 - 5)/2

    The domain should be any real number right but they have it as equal to or greater then 0.

    http://store.aqa.org.uk/qual/gceasa/...W-QP-JUN09.PDF
    http://store.aqa.org.uk/qual/gceasa/...W-MS-JUN09.PDF
    2bii
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    Is the function supposed to be one-one?
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    Did you mean q 2b ii?

    the domain of f^-1(x) is greater than or equal to 0 because it is the same as the range of f(x)
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    Think about it, the domain of f^{-1} (x) will be the same as the range of f (x)
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    (Original post by uxa595)
    Well f^-1 (x) = (x^2 - 5)/2

    The domain should be any real number right but they have it as equal to or greater then 0.

    http://store.aqa.org.uk/qual/gceasa/...W-QP-JUN09.PDF
    http://store.aqa.org.uk/qual/gceasa/...W-MS-JUN09.PDF
    2cii
    If your function were just \dfrac{x^2-5}{2} then your domain would just be \mathbb{R}, but because we're looking at this as the inverse of another function, this won't necessarily be the case. Here, f(x) = \sqrt{2x+5}, which has range f(x) \ge 0.
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    (Original post by nuodai)
    If your function were just \dfrac{x^2-5}{2} then your domain would just be \mathbb{R}, but because we're looking at this as the inverse of another function, this won't necessarily be the case. Here, f(x) = \sqrt{2x+5}, which has range f(x) \ge 0.

    Yes, thats what i initally got, as the range of f(x) which i just got has to be the domain of the inverse.

    But i found it strange as you could have x to be anything if it were on its own.

    so, all your doing is reflecting f in y=x, and looking at that part of the curve?
    So your only looking at one side of the curve which is a flection of f(x)?
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    (Original post by uxa595)
    Yes, thats what i initally got, as the range of f(x) which i just got has to be the domain of the inverse.

    But i found it strange as you could have x to be anything if it were on its own.

    so, all your doing is reflecting f in y=x, and looking at that part of the curve?
    So your only looking at one side of the curve which is a flection of f(x)?
    Yup. The idea is that we could define:
    g(x) = \dfrac{x^2-5}{2},\ 0 \le x < \infty
    h(x) = \dfrac{x^2-5}{2},\ -\infty < x < \infty

    These are different functions, despite being "the same", because they are defined on different domains. Because we're looking at the function that "undoes" f(x) = \sqrt{2x+5}, and because this only takes non-negative values, we must have that the inverse of f is g, and not h.

    The way functions are taught at C3 is a bit weird. When it asks you for the domain of a function, it usually means "the largest possible set of real numbers that can be a domain of" the function. For example, for the function f(x) = x^3, the domain could be \mathbb{R} (which is the largest possible real domain), in which case the range would also be \mathbb{R}; or it could be 0 \le x \le 2, in which case the range would be 0 \le f(x) \le 8. You could even define the domain to be \{3 \}, in which case the range is \{27 \}... but when you're asked the question in C3, it would expect \mathbb{R} as an answer.

    So in this case, although the function \dfrac{x^2-5}{2} can be defined on all the real numbers, because we're looking at it as the inverse of \sqrt{2x+5}, it doesn't make sense for us to allow negative values of x.
 
 
 
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