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Proving functions are surjective

  1. Nov 8, 2013 #1

    whether or not the following functions are surjective or injective:
    1) [tex] g: \mathbb{R} \rightarrow \mathbb{R}[/tex] [tex] g(x) = 3x^3 - 2x [/tex]

    2)[tex] g: \mathbb{Z} \rightarrow \mathbb{Z}[/tex] [tex] g(x) = 3x^3 - 2x [/tex]

    my working for 1):

    injective: suppose [tex] g(x') = g(x) [/tex] : [tex] 3x'^3 - 2x' = 3x^3 - 2x [/tex] this does not imply [tex] x = x' [/tex] hence not injective
    surjective: need to show [tex] \forall y \in \mathbb{R} \exists x \in \mathbb{R} s.t. g(x) = y [/tex], [tex] y = 3x^3 - 2x [/tex] as cubics always have one real root then as ## y \in \mathbb{R} ## ## \exists x \in g(x) \in \mathbb{R} ## s.t. ## g(x) = y ## therefore it's surjective

    injective: same as 1:
    surjective: I'm not sure how to phrase this for the integers,

    overall I'm not happy with my proofs, for the injectivity I haven't really shown that x is not equal to x', how would I do it? And for surjectivity I have mainly written it in words, how would I write it out formally for both questions?
  2. jcsd
  3. Nov 8, 2013 #2


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    Why doesn't it imply that it's not injective? You seem to have skipped a step here.
  4. Nov 8, 2013 #3
    I don't really know I just saw it as that :\
  5. Nov 8, 2013 #4


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    Can you find an example which demonstrates that g is not injective?
  6. Nov 8, 2013 #5
    0 and rt(2/3)
  7. Nov 10, 2013 #6


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    Yes, that works. The function's surjectivity can be verified via the intermediate value theorem.
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