Real root of a cubic equation I have a function f(x,n) can be expressed as a cubic function of x with coefficients that are functions of n. For example x^3 + (n-2)x^2 + (3n-6)x + n. 
I want to prove that for every positive value of n, there exists a real, positive value of x such that f(x,n)=0.
I know this is true for the function I have in mind but I am not sure how to go about proving it.
 A: For convenience, write the cubic function as
$$f(x) = x^3 + 3ax^2 + 3bx + c,$$
where $a$, $b$, and $c$ are (polynomial?) functions of $n$.  As has been noted in comments, if $c<0$, you're guaranteed a positive real zero $x$, so the only question is what to do for values of $n$ for which $c\ge0$.  
The only way you can have a positive real root when $c\ge0$ is if $f$ has a local minimum at a positive $x$ and takes a non-positive value there.  To check for this, look at the derivative
$$f'(x) = 3(x^2+2ax+b),$$
note that you need $a^2\ge b$ to have a local minimum at all, and then you need $x = \sqrt{a^2-b}-a \gt 0$ to have the local minimum at a positive $x$.  (For example, if $a\gt0$, then you need $b\lt0$.)  You now need only check whether $f(\sqrt{a^2-b}-a)\le0$.  
What's unclear is how easy or hard it is to check the various inequalities for the coefficient functions the OP has in mind.
A: I am assuming you have some different example in mind since as Barry Cipra pointed out your objective is not true in your example (and any example that eventually has all positive coefficients). First of all every cubic has a real root by the intermediate value theorem. Hence if you want to prove the positivity I recommend you start with the general solution to the cubic. It is ugly, but you could perhaps derive positivity (if it is true) in your class of examples. 
