# The Zeros Localization Theorem and the Extreme value Theorem

My intro to analysis book calls the following theorem The Zeros localization theorem:

Theorem: Let $p(x)=x^n + a_{n-1}x^{n-1} +\dots+a_1 +a_0,\ x \in \mathbb R$ , be a polynomial. Then all the zeros of $p$ (if there are any ) lie in the open interval $(-M,M),$ where $M= 1 + max\{ |a_{n-1}| ,\dots,|a_1|,|a_0| \}.$

Here is the problem which I have a solution of but which I DO NOT know if I am correct.

Problem: Using the Zeros Localization theorem and the Extreme Values theorem, prove that every polynomial of EVEN degree $n$ given by $$p(x)= a_nx^n +a_{n-1}x^{n-1}+\dots +a_1x +a_0,\ x \in \mathbb R,$$ where $a_n$ does not equal zero, has a minimum value on $\mathbb R$.

MY PROOF: Note that $$\left(\frac{1}{a_n}\right)p(x)=x^n+\frac{a_{n-1}}{a_n}x^{n-1} +\dots +\frac{a_1}{a_n}x+ \frac{a_0}{a_n}.$$ Hence by the Zeros localization theorem all zeros of $p(x)$ lie in the open interval $(-M,M)$ where $M= 1+ max \{ |\frac{a_{n-1}}{a_n}|,\dots,|\frac{a_1}{a_n}|,|\frac{a_0}{a_n}|\}$.

Also $\frac{1}{a_n}p(x)$ is continuous on $[-M,M]$, and hence by the Extreme Value theorem has a minimum on $[-M,M]$, that is, there exists a point $c \in [-M,M]$ such that $\frac{1}{a_n}p(c) \leq\frac{1}{a_n}p(x)$ for $x \in [-M,M]$. Since $p(x)$ is of even degree, limit of $p(x)$ as $x$ goes to plus or minus infinity equals plus infinity and hence $\frac{1}{a_n} p(c)$ is a minimum of $\frac{1}{a_n} p(x)$ on $\mathbb R ,$ and so $p(c)$ is a minimum value of $p(x)$ on $\mathbb R .$

Is my proof completely correct?

The property, as stated in the Problem, is false.

Here is a counterexample: $p(x)=-x^2$ is a polynomial of even degree ($a_2=-1 \neq 0$) but $p(x)$ does not have a minimum in $\mathbb{R}$.

Your proof must therefore be incorrect. I have detected at least two flaws:

1. The existence of a minimum point $c$ for $\frac{1}{a_n}p(x)$ in $[-M,M]$ does not immediately imply that $\frac{1}{a_n}p(c)$ is a minimum of $\frac{1}{a_n}p(x)$ in $\mathbb{R}$.

2. For $a_n < 0$ (as in the counterexample), if $\frac{1}{a_n}p(x)$ has a (local) minimum at $c$ then $p(x)$ has a (local) maximum at $c$.

• Hi ,do you mean that the problem itself is incorrect ?? – herashefat May 2 '15 at 14:57
• Yes. However, if you replace where $a_n$ does not equal zero by $a_n >0$ it becomes true. – P. Gomes May 2 '15 at 16:22
• Thanks for your response , then how would a correct proof look like?? – herashefat May 2 '15 at 16:36