Proving functions are surjective

1. Nov 8, 2013

synkk

prove

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

2)$$g: \mathbb{Z} \rightarrow \mathbb{Z}$$ $$g(x) = 3x^3 - 2x$$

my working for 1):

injective: suppose $$g(x') = g(x)$$ : $$3x'^3 - 2x' = 3x^3 - 2x$$ this does not imply $$x = x'$$ hence not injective
surjective: need to show $$\forall y \in \mathbb{R} \exists x \in \mathbb{R} s.t. g(x) = y$$, $$y = 3x^3 - 2x$$ 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

2):
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. Nov 8, 2013

FeDeX_LaTeX

Why doesn't it imply that it's not injective? You seem to have skipped a step here.

3. Nov 8, 2013

synkk

I don't really know I just saw it as that :\

4. Nov 8, 2013

FeDeX_LaTeX

Can you find an example which demonstrates that g is not injective?

5. Nov 8, 2013

synkk

0 and rt(2/3)

6. Nov 10, 2013

FeDeX_LaTeX

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