# 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.

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 $$c, then the equation $$(*)$$ obviously has no solution, since the left side always has a value $$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 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

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.)

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