Show that $p(x) = a_nx^n+a...+a_1x+a_0$ either has a root or attains minimum value in R.

Question

Show that if $p(x) = a_nx^n +\dots+ a_1x + a_0$ and $a_n > 0$, then either $p(x) = 0$ has a solution, or else $p(x)$ has attains minimum value on $\mathbb{R}$.

I'm sorry I don't know how to even start the problem. I know that if it is an odd degree polynomial then it has roots.

Thank you. The second part confuses me the most. This is from an undergrad Real Analysis homework.

Answer 1

Hint: Recall that polynomials are continuous on $\Bbb R$; note that $p(x)\to\infty$ as $x\to\infty$; if $p(x)\lt 0$ for some $x=a$, apply IVT to $p$ on $[a,\infty)$; if $p(x)\gt 0$ for all $x$, recall that $\Bbb R$ has the glb property.

Answer 2

If $p$ has no roots, consider $f(x) = |\frac{1}{p(x)}|$. This is well-defined and continuous, and for some $N$, we have that $|x| > N$ implies $f(x) < 1$. Now $f$ assumes a max on the compact $[-N,N]$....

Answer 3

If your polynomial is of an odd degree, then:

$$\lim_{x\to-\infty}p(x)=\lim_{x\to-\infty}a_n\cdot x^n=-\infty$$ $$\lim_{x\to+\infty}p(x)=\lim_{x\to+\infty}a_n\cdot x^n=+\infty$$ So there is a root, or else there is a $x$ where $p(x)=0$.

If your polynomial is of an even degree, then you are not sure if it has a root. But it's first derivative $p'(x)$ is a polynomial of an odd degree($p'(x)=a_n n x^{n-1}+...+a_1$). As I showed before this polynomial has a root, which means there is an $x$ for which $p'(x)=0$, which in turn means that there are some $x$ where $p(x)$ has it's local minimum or local maximum value. Since $a_n>0$ and $p(x)$ tends to $+\infty$ when $x\to-\infty$ or $+\infty$ then at at least one $x$ of the above equation the polynomial has a global minimum value.

So we proved that for any polynomial either it has a root or it has a minimum value.

Answer 4

We recall the well-known fact that polynomials are continuous; thus:

If such a polynomial $p(x)$,

$p(x) = \displaystyle \sum_0^n a_i x^i, \tag 1$

with $a_n > 0$, has no roots, then $n = \deg p(x)$ is even; for if $\deg p(x)$ were odd, then since $a_n > 0$, and the highest degree term $a_n x^n$ dominates for sufficiently large $\vert x \vert$, that is,

$x \to -\infty \Longrightarrow p(x) \to -\infty, \tag 2$

$x \to \infty \Longrightarrow p(x) \to \infty; \tag 3$

it is but a short step, via the intermediate value theorem, to see that $p(x)$ has a zero in $\Bbb R$. So if $p(x)$ has no root, then $\deg p(x)$ is even. But then we have, again since $a_n > 0$, that

$\vert x \vert \to \infty \Longrightarrow p(x) \to \infty; \tag 4$

this means that, given any $B > 0$, there exists $L>0$ such that

$\vert x \vert > L \Longrightarrow p(x) > B; \tag 5$

we also observe that

$p(0) = a_0 > 0; \tag 6$

since $p(x)$ never $0$, it cannot change sign; thus (6) follows from (4)-(5). If we now choose $B > a_0$, and $L$ accordingly large, we affirm that the continuous function $p(x)$ attains its minimum value $p_{min} \le a_0$ at some $x_0 \in [-L, L]$.

And that is indeed that, is it not?