Proof that a polynomial has a minimum in $\Bbb R$ I have to prove to following statement and I am having a really hard time here. 
There it is:
Prove that the following polynomial has a minimum in $\Bbb R$
$$p(x)=x^4 + a_3x^3 + a_2x^2 + a_1x + a_0$$
I tried to make the following proof, but I am stuck:
The polynomial can be shown like this:
$$x^4\left(1+\frac{a_3}{x}+\frac{a_2}{x^2}+\frac{a_1}{x^3}+\frac{a_0}{x^4}\right)$$
Let  $$g(x)=\left(1+\frac{a_3}{x}+\frac{a_2}{x^2}+\frac{a_1}{x^3}+\frac{a_0}{x^4}\right)$$
So we can now write the polynomial like
$$p(x)=x^4g(x)$$
Since $x,a \in \Bbb R$ we can say that $\lim _{x \to \infty}=\lim _{x \to -\infty}=1$ (from limit arithmetic).
We know that it is always the case that $x^4>0$, so the sign of $$p(x)=x^4g(x)$$ is determined by $g(x)$. For certain values, $g(x) <0 $ so we can say that there is some $a,b \in \Bbb R$ such that $g(a)<0$ and $g(b)>0$.
$p(x)$ is continouos everywhere, certainly in $[a,b]$, so from the Extreme Value Theorem, the function has a minimum in $[a,b]$.
There are two problems in my proof:


*

*I can't find any value in which $g(x)$ is negative, only if there is some $a$ which is negative. What if $a$ is always positive?

*If the first problem is solved, I managed to prove the statement for some $[a,b]$. Is it just enough? It seems to me that it isn't.
Thanks,
Alan
 A: You can't just write $f(x) = x^4 g(x)$ because $f(0)$ is defined and $g(0)$ isn't. Instead, you can simply prove
$$\lim_{x\to\pm\infty} f(x) = \infty$$
(because the $x^4$ term dominates all others in the limit)
With that settled, by the definition of these limits you can find some $R>0$ such that
$$f(x) > a_0 \qquad \forall x : |x|> R$$
(the $>a_0$ is arbitrary here, any constant works as long as $f$ attains it inside $[-R,R]$; also, $R$ need not be the optimal choice)
Now look at $f|_{[-R,R]}$. This is a continuous function on a closed and bounded set. You should have a theorem giving you that $f|_{[-R,R]}$ attains a minimum.
Since $0\in[-R,R]$, we know that this minimum is at most $f(0) = a_0$, so $f$ doesn't attain smaller values outside of $[-R,R]$ by construction of $R$.
This proves
$$\min_{x\in\mathbb R} f(x) = \min_{x\in[-R,R]} f(x) \in \mathbb R$$
exists.
A: One could also note that the derivative of an even degree polynomial is an odd degree polynomial.  And an odd degree polynomial always has at least one zero.
A: This is true for any polynomial $f(x)$ of even degree with positive coefficient for the highest degree term. Check that as $x\to\pm \infty$ the polynomial values $f(x)\to +\infty$. That means there exist an $N>0$ such that
 for $a\in [-N, N], b\not\in [-N,N]$, we have $f(a)<f(b)$. Now the (global) minimum for $f(x)$ is attained in $[-N,N]$, a compact set.
A: The claim itself is true more generally for any polynomial of even degree with positive leading coefficient.
We do not need that $g(x)<0$ for some $x$ and in fact such $x$ may not exist. What we need is that $\lim_{x\to\pm\infty}g(x)=1$ implies that there exists $M\in \mathbb R$ such that $|x|>M$ implies $|g(x)-1|<\frac12$ (so that $g(x)>\frac12$). Then for $|x|>M$ we conclude $f(x)>\frac12x^4>\frac12M^4$. We may assume wlog. that $\frac12M^4>a_0=f(0)$. Then the minimum the continuous function has on $[-M,M]$ is in fact a global minimum.
