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Math Help - show that f(x) is injective but not surjective

  1. #1
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    show that f(x) is injective but not surjective

    Hi, firstly I've never really understood what injective and surjective means so if someone could give me the gist of that it'd be great!

    secondly the question i'm stuck on at the moment is....

    let A = {x:x (belongs to) [0, (inf))} and B = {x:x(belongs to) [0, inf)} and the function f: A (tends to) B is given by

    f(x) = 1/(2+x^2)

    show that f(x) is injective but not surjective

    Thanks
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  2. #2
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    Injectivity - surjectivity

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  3. #3
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    thanks, a few minutes after posting i learned what they mean, im just unsure of how to prove it :s
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  4. #4
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    Quote Originally Posted by MrMathlete View Post
    Hi, firstly I've never really understood what injective and surjective means so if someone could give me the gist of that it'd be great!

    secondly the question i'm stuck on at the moment is....

    let A = {x:x (belongs to) [0, (inf))} and B = {x:x(belongs to) [0, inf)} and the function f: A (tends to) B is given by

    f(x) = 1/(2+x^2)

    show that f(x) is injective but not surjective

    Thanks
    Injective (one-to-one): If f(x)= f(y) then x= y. If \frac{1}{2+ x^2}= \frac{1}{2+ y^2} show that x= y. It is important here that A consist of non-negative values only.

    Surjective (onto): For any y in B the there exist x in A such that f(x)= y.
    If \frac{1}{2+ x^2}= y, show that is not possible to solve for x for some values of y.
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