This is a homework problem so no need to give me an immediate answer.
My general plan has been to try to prove that f(x) cannot have both an upper bound and a lower bound.
One strategy I tried was using induction on the degree of f. The base case for $f(x) = a_0 + a_1x$ is very doable to show that it cannot have both upper & lower bounds. It's when you get to $deg(f) = n$ that things start to get tricky.
The chapter of the book I'm working with, Contemporary Abstract Algebra, 8th edition, has the theorem for the division algorithm for polynomials, and as a corollary we see that f(a) is the remainder of f(x) when dividing by (x - a). So I am assuming they want me to try using that, but I'm not sure.
I feel as though I could brute force this problem by
- Starting with f(b) = lower bound and f(c) = upper bound
- Checking if the leading term has an even/odd power
- Checking if the leading coefficient is positive/negative
Once all those are fixed, I would have to come up with a clever value for $b$ or $c$, and then use that to show that we can't have both an upper bound and a lower bound.
What I would really love is if there is a more elegant solution using the image of the polynomial f(Z) and showing some contradiction if it's finite. But nothing I've come across fits that.