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$f(x)=x^3-x$

The question was the prove $f(x)$ is not injective and is surjective. I didn't manage to prove the latter. Is that a particular technique to prove $y=f(x)$ to show it's surjective or something?

Thanks.

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    Not for purposes of rigorous proof, but if you don't already "see" that this is true, please sketch a graph of the function (or use software to do so), and think about why every possible $y$ value must appear somewhere in the graph, by going far enough left or right. Once you have the intuition, Thomas Andrews indicates a way to make it more precise.2012-12-14

2 Answers 2

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One way is to use the intermediate value theorem.

First, prove that $f(x)$ is continuous.

Then show for any $y$ you can make find $x_1$ so that $f(x_1)>y$ and $x_2$ so that $f(x_2). There there must be an $x_3$ between $x_1$ and $x_2$ such that $f(x_3)=y$.

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Not injective since $x = -1$, $x = 0$, and $x = 1$ all map to $y = 0$. It is sufficient to find two points in the domain that map to the same point in the codomain.

For surjective, this could easily be seen by plotting, but I agree with the previous response.