Every real polynomial attains a smallest absolute value Let $p$ be any real polynomial. Show there exists $c\in\mathbb{R}$ such that $|p(c)| \leq|p(x)|$ for all $x\in\mathbb{R}$.
 A: Hints:
1) If $p(x)$ is an odd polynomial, it has a real root (say $c$) and hence $0 = |p(c )| \le |f(x)|$.
2) If $p(x)$ has even degree, WLOG let the leading coefficient be positive.  Then we see that $p(x)$ approaches $\infty$ on both ends of the axis.  This implies that if we pick a number $M$ larger than some element of the range of $p(x)$, we have a closed interval $[a, b]$ where $p(x)>M$ outside this interval.  
As $[a, b]$ is compact and $p$ is continuous, so it attains a minimum at say $c$. Now it is easy to show $|p(c )| \le |f(x)|$.
A: For any polynomial $p(x)$, the function $\vert p(x) \vert$ is continuous and satisfies $\vert p(x) \vert \to \infty$ as $\vert x \vert \to \infty$.  Given that this is the case, it is evident that this question is in fact a special case of this one: problem about continuity and limits, and the solution given there by, of all people, Yours Truly, covers the present question as well.
Whew!  Saved some single-finger 'droid typing!
Hope this helps.  Cheerio,
and as always, 
Fiat Lux!!!
A: Hint:
Every polynomial is either constant or not. If it's not, it must have a certain sort of behavior for arguments with very large absolute values. The local behavior of polynomials also has a useful property.
