Question about Spivak proof of "If $n$ is even and $f(x) = x^{n}+a_{n-1}x^{n-1}+\dots+a_{0}$ then there is a number $y$ s.t $f(y) \leq f(x)\ $ The question I have revolves around some of the techniques used in the following proof. I wanted to make sure I'm understanding the reasoning behind it. Here is the proof:

So the idea behind the proof is to split the situation into two cases:

*

*Dealing with the function in the interval $[-b,b]$

*Dealing with the function outside of this interval.

Dealing with case 1, we can refer to a theorem of the existence of a minimum on a continuous function on a closed set. (But I still do have a question on a subtlety)
For case 2 what we end up showing is that our function will always be greater than any value on the closed set from case 1. My question has to do with the "considering the number $f(0)$. What is so special about this point and why do we let "$b > 0$ be a number such that $b^{n} \geq 2f(0)$"? I understand mechanically why $2f(0)$ works it gives us a fraction to relate to the rest the inequalities nicely, but other than that i don't see the "intuition" behind it.
The other small piece I'm trying to understand is at the end of the proof where it is declared "in particular $f(y) \leq f(0)$. Does this declaration have to do with the fact that we showed that outside the interval $[-b,b]$, $f(0) \leq f(x)$, so in theory inside the interval $[-b,b]$, $f(0)$ could be a maximum value and as such the minimum in a closed interval theorem would follow.
 A: You're right in identifying two pieces to the proof. We are trying to analyze the behaviour of the function $f$ in two separate pieces:

*

*first, for $x$ in the closed and bounded interval $[-b,b]$.

*Second, for $x$ outside of this interval.

In the closed interval $[-b,b]$, the continuous function $f$ attains a minimum value: i.e there exists $y \in [-b,b]$ such that for all $x \in [-b,b]$, $f(y) \leq f(x)$. However, this isn't quite what we want. We want to prove that this value of $y$ is in fact a global minimum; i.e for all $x \in \Bbb{R}$ (not just in $[-b,b]$), we want to show $f(y) \leq f(x)$.
This is why Spivak introduces a third "reference number" $f(0)$, which allows us to relate the behavior of the function inside the interval to the behavior outside. Because by construction, $b>0$ was defined to be a number with the property that if $x$ is outside the interval $[-b,b]$ then $f(0) \leq f(x)$. So, for $x \in [-b,b]$, we have by Theorem 7 that $f(y) \leq f(x)$.
\begin{align}
\text{If $x\notin [-b,b]$ then $f(y) \leq f(0) \leq f(x)$}
\end{align}
You see, the $f(0)$ is just there to "bridge the gap in logic" in going from the behavior inside to outside. This is what allows us to prove that for all $x \in \Bbb{R}$, $f(y) \leq f(x)$.

By the way, there is nothing really special about choosing $f(0)$ in particular. The only reason Spivak chose this is out of convenience, because $0$ happens to lie in the interval $[-b,b]$.
For example, if I wanted to, I could also structure the proof like this:

Since $n$ is even, we can show that $\lim_{|x| \to\infty} f(x) = \infty$ (I leave it to you to transcribe precisely what this means. Also, if you notice carefully, this is pretty much what Spivak is trying to do by introducing $M$ and showing that $|x| \geq M \implies x^n/2 \leq f(x)$).
As a result (almost by definition of what the symbol $\lim_{|x| \to \infty}f(x) = \infty$ means), there exists a $r>0$ such that if $|x| \geq r$ then $f(x) \geq f(100)$.  Now, consider the number $b:= r + 100$, so that the number $100$ actually lies inside $[-b,b]$. By Theorem 7, on the interval $[-b,b]$, the function $f$ attains a minimum; i.e there exists $y \in [-b,b]$ such that for all $x \in [-b,b]$, we have
\begin{align}
f(y) \leq f(x)
\end{align}
In particular, we have $f(y) \leq f(100)$. Now, for $|x| \geq b$, we have (by definition of $b$),
\begin{align}
f(y) \leq f(100) \leq f(x).
\end{align}
So, in every case, regardless of what $x$ is, we have that $f(y) \leq f(x)$, which proves the claim.

Again, as you can see, my consideration of the number $f(100)$ is completely arbitrary; the only purpose of it was to relate the behavior of $f$ in the regions inside and outside the interval $[-b,b]$. So, actually, in the first half of the proof, the main idea is to show the following statement:

If $n$ is even and $f(x) = x^n + a_{n-1}x^{n-1} + \dots + a_0$, then there exist $b>0$ and there exist $x_0 \in [-b,b]$ such that for all $x \in \Bbb{R}$:
\begin{align}
\text{if $|x| \geq b$ then $f(x_0) \leq f(x)$}
\end{align}

Or in words, we want to show that $n$ being even implies that outside of some interval $[-b,b]$, the function $f$ is always larger than a particular value $f(x_0)$ which is attained in the interval $[-b,b]$. Most of Spivak's proof is devoted to proving this fact, but for simplicity, Spivak has decided to take $x_0 = 0$, and tries to show the existence of such a $b>0$ accordingly.
