Show that a polynomial function has a minimum $p:\mathbb{R}\rightarrow\mathbb{R}$ is a polynomial function, which only has positive values. Show that p has a minimum.
This is what I tried to do:
I want to proof this statement using the extreme value theorem.
To be able to use it, the function has to be continuous and bounded.
$p(x) = a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0 > 0$ 
Since the values have to be positive $p(x)>0$
$a_n\neq 0$
$\lim_{x\to +\infty}x^n= \infty $
Since the limit to infinity is not bounded, how do I change my interval so it is bounded and I can use the extreme value theorem?
And how do I make sure that the values are positive?
 A: If $p$ only has positive values $p$ is of even degree, and the coefficient of the highest power term is positive.
$p'$ is of odd degree.
There exists some  $x_1 $ such that for all $x< x_1 \implies p'(x) < 0$
and 
$x_2$ such that $x> x_2 \implies p'(x) > 0$
$p'$ has at least one real root with $p'(x)$ changing sign from negative to positive as $x$ crosses this root.  
This root is a minimum of $p.$
A: Pick some $a \in \mathbb{R}$. Since $\lim_{x\to\infty} p(x) = \lim_{x\to-\infty} p(x) = \infty$, you can choose some $M_1$ such that $x > M_1 \implies p(x) > p(a)$. You can similarly choose $M_2$ such that $x < M_2 \implies p(x) > p(a)$.
Note that $M_2 < a < M_1$. Otherwise, restrict the domain to $[M_2, M_1]$. You can freely reject all other points, as the value of $p$ at $a$ is lesser than on $\mathbb{R} \setminus [M_2, M_1]$.
A: Since $\lim_{x\to\pm\infty}p(x)=+\infty$ (it follows from the assumption that the polynomial has only positive values), there are $a<0$ and $b>0$ such that $p(x)>p(0)$ on $(-\infty,a)$ and $p(x)>p(0)$ on $(b,+\infty)$.
Consider $p$ on $[a,b]$. 
Since all values of $p$ outside this interval are greater than some values of $p$ (e.g. $p(0)$) and $p$ is continuous and bounded on $[a,b]$, the minimum has to be taken on $[a,b]$.
