Prove that every even degree polynomial function has a maximum or minimum in $\mathbb{R}$ 
Prove that every even degree polynomial function $f$ has maximum or minimum in $\mathbb{R}$. (without direct using of derivative and making $f'$)

The problem seems very easy and obvious but I don't know how to write it in a mathematical way.
For example if the largest coefficient is positive, it seem obvious to me that from a point $x=a$ to $+\infty$ the function must be completely ascending. And from $-\infty$ to a point $x=b$ the function must be completely descending. If it is not like that, its limit will not be $+\infty$ at $\pm\infty$. Now, because it is continuous, it will have a maximum and minimum in $[b,a]$ so it will have a minimum (because every other $f(x)$ where $x$ is outside $[b,a]$ is larger than $f(a)$ or $f(b)$ and we get the minimum that is less than or equal to both of them) . We can also the same for negative coefficient.
But I can't write this in a formal mathematical way.

Let $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$. Let us assume that $a_n>0$ (the case in which $a_n<0$ is similar). Then\begin{align}\lim_{x\to\pm\infty}f(x)&=\lim_{x\to\pm\infty}a_nx^n\left(1+\frac{a_{n-1}}{a_nx}+\frac{a_{n-2}}{a_nx^2}+\cdots+\frac{a_0}{a_nx^n}\right)\\&=+\infty\times1\\&=+\infty.\end{align}Therefore, there is a $R>0$ such that $\lvert x\rvert>R\implies f(x)\geqslant f(0)$. So, consider the restriction of $f$ to $[-R,R]$. Since $f$ is continuous and since $[-R,R]$ is closed and bounded, $f|_{[-R,R]}$ attains a minimum at some point $x_0\in[-R,R]$ and, of course, $f(x_0)\leqslant f(0)$. Since outside $[-R,R]$ you always have $f(x)\geqslant f(0)$, $f$ attains its absolute minimum at $x_0$.
