Proving a noninvertible function is onto? How do I prove that $f:\mathbb R\longrightarrow\mathbb R, \;f(x) = x^3 - 3x$ is onto?  Is it always necessary to find an inverse function?  I guess there is some way to restrict such a function's domain so it's outside the overlapping parts $x < -\sqrt{3}$ or $x \ge \sqrt{3}$, but how do I express this mathematically?  Or is there some other way to do this?
 A: We want to prove that for any real number $b$, there is an $x$ such that $x^3-3x=b$.
If $x$ is large enough negative, $x^3-3x$ is for sure $<b$.  So there is a number $L$ such that $L^3-3L<b$.  
If $x$ is large enough positive, then $x^3-3x>b$. So there is a number $M>L$ such that $M^3-3M>b$.
Since the function $x^3-3x$ is continuous, by the Intermediate Value Theorem, there is an $x$ such that $L<x<M$ and $x^3-3x=b$.
Or else let $f(x)=x^3-3x-b$. By what was written above, there is an $L$ such that $f(L)<0$, and there is an $M>L$ such that $f(M)>0$. So (again by the Intermediate Value Theorem) there is an $x$ between $L$ and $M$ such that $f(x)=0$, maning that $x^3-3x=b$.  More informally, the function is negative somewhere, positive somewhere, so the graph of $y=f(x)$ must cross the $x$-axis, so by continuity $f(x)$ must be $0$ somewhere.
Comment: There is an old ($16$-th century) complicated but explicit formula, called the Cardano Formula, for the roots of a cubic. Unfortunately, at least for some values of $b$ (those for which $f(x)$ has $3$ distinct real roots) the formula is very difficult to work with. 
A: There is no need to restrict the domain. You need only show that for any $y$ in the range of your function (I'll assume it's $\Bbb R$), that the equation
$$
f(x)=y
$$
has a solution. That, is you need to show that any element in the range is actually an output value of the function.
Here, you'd fix $y\in\Bbb R$ and  consider the equation
$$
x^3-3x=y.
$$
Or
$$
x^3-3x-y=0.
$$
Since this is a polynomial equation of degree 3, it does have a real solution
(by the Fundamental Theorem of Algebra; you could also take a Calculus approach to show this).
If you call this solution $a_y$, then $f(a_y)=y$. Thus $f$ is onto.
