Let $p$ be any real polynomial. Show there exists $c\in\mathbb{R}$ such that $|p(c)| \leq|p(x)|$ for all $x\in\mathbb{R}$.
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I believe you mean p(c),p(x) – ZKe Nov 11 '13 at 07:46
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Let p(x)=x, then there is -no- c such that c=p(c)<=p(x)=x for all real numbers x. – mathematician Nov 11 '13 at 07:48
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@mathematician you are forgetting the absolute value. $c=0$ works for your e.g. – Macavity Nov 11 '13 at 07:50
3 Answers
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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 $|p(c )| \le |f(x)|$.
Macavity
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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!!!
Robert Lewis
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