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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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6  
Hint: $f(x)$ is a continuous function, and tends to $\infty$ as $x\rightarrow \infty$ and tends to $-\infty$ as $x \rightarrow -\infty$. –  bzprules Dec 14 '12 at 22:09
    
You want to show that $x^3-x=c$ has a real root for all $c$. Why can't it have 3 complex roots? (Hint: Complex roots come in conjugate pairs.) –  Potato Dec 14 '12 at 22:09
    
Hey guys. Sorry, I'm at the very beginning of the calculus semster at the uni, so we haven't gotten to complex roots yet. Don't know how to continue from your hints. –  pie Dec 14 '12 at 22:14
    
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. –  Jonas Meyer Dec 14 '12 at 22:33

2 Answers 2

up vote 8 down vote accepted

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 $f(x_2)<y$. 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.

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