# Prove that the following polynomial has a minimum in R p(x)=x^4+a_3 x^3+a_2 x^2+a_1x+a_0.

Proof that a polynomial has a minimum in 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 R
$p\left(x\right)={x}^{4}+{a}_{3}{x}^{3}+{a}_{2}{x}^{2}+{a}_{1}x+{a}_{0}$
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
$g\left(x\right)=\left(1+\frac{{a}_{3}}{x}+\frac{{a}_{2}}{{x}^{2}}+\frac{{a}_{1}}{{x}^{3}}+\frac{{a}_{0}}{{x}^{4}}\right)$
So we can now write the polynomial like
$p\left(x\right)={x}^{4}g\left(x\right)$
Since $x,a\in \mathbb{R}$ we can say that $\underset{x\to \mathrm{\infty }}{lim}=\underset{x\to -\mathrm{\infty }}{lim}=1$ (from limit arithmetic).
We know that it is always the case that ${x}^{4}>0$, so the sign of
$p\left(x\right)={x}^{4}g\left(x\right)$
is determined by g(x). For certain values, $g\left(x\right)<0$ so we can say that there is some $a,b\in \mathbb{R}$ such that $g\left(a\right)<0$ and $g\left(b\right)>0$.
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.
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pararevisarii
Step 1
The claim itself is true more generally for any polynomial of even degree with positive leading coefficient.
Step 2
We do not need that $g\left(x\right)<0$ for some x and in fact such x may not exist. What we need is that $\underset{x\to ±\mathrm{\infty }}{lim}g\left(x\right)=1$ implies that there exists $M\in \mathbb{R}$ such that $|x|>M$ implies $|g\left(x\right)-1|<\frac{1}{2}$ (so that $g\left(x\right)>\frac{1}{2}$). Then for $|x|>M$ we conclude $f\left(x\right)>\frac{1}{2}{x}^{4}>\frac{1}{2}{M}^{4}$. We may assume wlog. that $\frac{1}{2}{M}^{4}>{a}_{0}=f\left(0\right)$. Then the minimum the continuous function has on [-M,M] is in fact a global minimum.
###### Did you like this example?
Chaim Ferguson
Step 1
You can't just write $f\left(x\right)={x}^{4}g\left(x\right)$ because f(0) is defined and g(0) isn't. Instead, you can simply prove
$\underset{x\to ±\mathrm{\infty }}{lim}f\left(x\right)=\mathrm{\infty }$
(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
$f\left(x\right)>{a}_{0}\phantom{\rule{2em}{0ex}}\mathrm{\forall }x:|x|>R$
(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 $f{|}_{\left[-R,R\right]}$. This is a continuous function on a closed and bounded set. You should have a theorem giving you that $f{|}_{\left[-R,R\right]}$ attains a minimum.
Since $0\in \left[-R,R\right]$, we know that this minimum is at most $f\left(0\right)={a}_{0}$, so f doesn't attain smaller values outside of [-R,R] by construction of R.
Step 2
This proves
$\underset{x\in \mathbb{R}}{min}f\left(x\right)=\underset{x\in \left[-R,R\right]}{min}f\left(x\right)\in \mathbb{R}$
exists.