Show that $p(x) = a_nx^n+a...+a_1x+a_0$ either has a root or attains minimum value in R. 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. 
 A: 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 $[-N,N]$....
A: 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.
A: 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.
A: 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?
