calculus – If $f(x) = x^n+a_{n-1}x^{n-1}+cdots+a_1x+a_0$ and $f(y)leq f(x)$ for all $x$, then there is a number $b>y$ such that $f(b)>f(y)$

Chapter 7 of Spivak’s Calculus 3rd edition contains what he calls “Three Hard Theorems” and some of their consequences. The given proofs are clear, but one contains a little detail which bothers me. (Note that all “numbers” are real numbers.)

Theorem 11:
Consider the equation $$(*)quad x^n+a_{n-1}x^{n-1}+cdots+a_1x+a_0=c,$$ and suppose $n$ is even. Then there is a number $m$ such that $(*)$ has a solution for $cgeq m$ an has no solution for $c<m$.

Proof: Let $f(x) = x^n+a_{n-1}x^{n-1}+cdots+a_1x+a_0$. According to Theorem 10 there is a number $y$ such that $f(y)leq f(x)$ for all x. Let $m=f(y)$. If $c<m$, then the equation $(*)$ obviously has no solution, since the left side always has a value $geq m$. If $c=m$, then $(*)$ has $y$ as a solution. Finally, suppose $c>m$. Let $b$ be a number such that $b>y$ and $f(b)>c$. Then $f(y)=m<c<f(b).$ Consequently, by Theorem 4 (Intermediate Value Theorem), there is some number $x$ in $(y,b)$ such that $f(x)=c$, so $x$ is a solution of $(*)$.

Maybe I’m missing something he has mentioned previously in an earlier chapter or problem, but how do I know that there exists a number $b$ such that $b>y$ and $f(b)>c$?

I worked out a proof, analogous to the first part of the proof of Theorem 9 of this chapter, that a number b exists if $f(y)<1$:

Let $M=max(1, frac{n|a_{n-1}|}{1-f(y)}, cdots, frac{n|a_{0}|}{1-f(y)} )$. Then for all $x$ with $|x|>M$ we have $$f(y)<1+frac{a_{n-1}}{x}+ cdots +frac{a_{0}}{x^n},$$
so $$f(y)leq f(y)x^n<left(1+frac{a_{n-1}}{x}+ cdots +frac{a_{0}}{x^n}right)x^n=f(x).$$

Thus, choose $b>M$.

Unfortunately I can’t go any further than that. (The complete proof of Theorem 9 can be found on this site. I can include more details, if needed.)