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Math Help - Image set of a function?

  1. #1
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    Image set of a function?

    How do u find the image set for a function with no domain?
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  2. #2
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    Re: Image set of a function?

    you don't. a function without a domain is an empty function, with a "blank" graph.

    put it this way, how many pairs (x,f(x)) can occur in such a function, when there is no possible "first coordinate"?
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  3. #3
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    Re: Image set of a function?

    I thought it was the lowest possible value for y and the highest possible vaule foe y??
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    Re: Image set of a function?

    Quote Originally Posted by Orlando View Post
    How do u find the image set for a function with no domain?
    Please show us what you mean by "a function with no domain".
    At least, describe it in detail.
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  5. #5
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    Re: Image set of a function?

    Sorry I am tring to find the image set of :

    f(x)=1/4(x-1)^2 -9

    In interval notation.

    There is no domain given so I persume the answear would be

    [-9, to infinity) Sorry don`t know how to us the symbol for infinity

    Or am I completly of the mark
    Last edited by Orlando; November 19th 2011 at 05:40 AM.
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  6. #6
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    Re: Image set of a function?

    Quote Originally Posted by Orlando View Post
    f(x)={1/4(x-1)^2} -9
    In interval notation.
    Assuming that the function is f(x)=\frac{1}{4(x-1)^2}-9 then the domain is all x\ne 1.

    The range is (-9,\infty). Because f(x)\ne -9 for any x.
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  7. #7
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    Re: Image set of a function?

    sorry the function is 1/4*(x-1)^2-9

    sorry for how I have presented it still trying to get use to typing maths in a computer











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  8. #8
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    Re: Image set of a function?

    Then that has "natural domain" of all real numbers. It's graph is a parabola with vertex at (1, 9), vertical axis, opening upward.

    If a function is given without an explicit domain, unless stated to the contrary, the domain is taken to be the "natural domain"- the set of all real numbers for which it is possible to calculate the value. The natural domain of a polynomial is all real numbers. The natural domain of [1/(4(x-1)^2]- 9] is all real number except x= 1. That is because x= 1 would make the denominator of that fraction 0.
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  9. #9
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    Re: Image set of a function?

    Quote Originally Posted by Orlando View Post
    sorry the function is [TEX]f(x)=\frac{1/4}(x-1)^2-9[TEX]
    Is the function f(x)=\frac{1}{4}(x-1)^2-9~???
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  10. #10
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    Re: Image set of a function?

    yes it is. Sorry for messing everyone about still getting my head round La Tex
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  11. #11
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    Re: Image set of a function?

    Quote Originally Posted by Orlando View Post
    yes it is. Sorry for messing everyone about still getting my head round La Tex
    Then the domain is \mathbb{R} and range [-9,\infty)
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  12. #12
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    Re: Image set of a function?

    Thanks for everyones help. Going to learn how to use La Tex looks a great bit of software.
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