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.



Real root of a cubic equation
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.


