Are Polynomials one-one functions Given a polynomial of third degree is it possible to show that it is not necessarily one-one in a given Range and Domain. For instance, consider the following problem:-
$$f(x)=2x^3-15x^2+36x+1$$
Here the Domain is $[0,3]$ whereas the range is $[1,29]$.
The question then is whether the function is one-one or not. 
I thought about using the definition of one-one functions, which states that 
$f(x_1)=f(x_2)$ implies that $x_1=x_2$.
But when I tried this approach I was no able to include the constraints offered by the domain or the range. Could anyone provide a clear approach to the above problems..
 A: Visualize the shapes of 3rd order polynomials.  If $f(x_1)=f(x_2)$, then the function must reach a local min or max somewhere between.  So the function is one-to-one in a domain iff df/dx is non-zero for all x in that domain.
A: If you have a hunch, or know that the answer is "no," a natural first approach would be as follows (if it fails, then differentiation and systematic study of $f$ will help). You know that $f(0)=1$, and $f(3)=28$. If you can find any point $c$ in $[0,3]$ such that $f(c)>28$, then the TVI will guarantee that $f$ is not one-to-one. (Can you see why?)
Note that this approach is not systematic per se, as it does not explain how to find $c$ -- it's up to you. But sketching the graph, or trying "obvious guesses" like $f(1)$ and $f(2)$, will usually help a lot -- sometimes give the answer.
A: For the specific example I would try to sketch the polynomial including finding the co-ordinates of any turning points. A polynomial is one to one between turning points. This requires a little calculus (differentiation), if you have that available, and finding the solutions of a quadratic equation.
