Show that if $p(x)$ is a polynomial, $|p(x)|$ attains its minimum. 
Show that if $p(x)$ is a polynomial, $|p(x)|$ attains its minimum.

Attempt
Let $p(x) = a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$. Then if $|p(x)| =  |a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0|$. If $x > 0,$ then $|p(x)| = |a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0| \leq |a_n|x^n+|a_{n-1}|x^{n-1}+\cdots+|a_1|x+|a_0|.$ Then we can take there derivative of $|p(x)|$ to get a bound to find critical points. I am not sure how to deal with the case of $x \leq 0$.
 A: observe that if the lead coefficient is $c_n x^n$ then for $|x|$ large we have
$$
|p(x)| \geq |c_n/2| |x|^{n}.
$$
So for some large $R$ the value on $|x|>R$ is greater than $|p(0)|.$
This means that the minimum can only be attained and approached on $|x| \leq R.$ So it's enough to work on the set $|x|\leq R.$ However, a continuous function on a compact set always attains its minimum, we are done.
A: Hint: Consider $q(x)=\mid p(x)\mid^2$ it is a polynomial of even degree, its derivative has an odd degree. This implies $lim_{x\rightarrow -\infty}q'(x)=-\infty$ $lim_{x\rightarrow +\infty}q'(x)=+\infty$. So there exists $a<0, b>0, a<b$ such that $q'(x)<0, x<a$ and $q'(x)>0, x>b$. Thus $p^2(x)$ is a decreasing function on $(-\infty,a]$ and an increasing function on $[b,+\infty)$. The minimum of the restriction of $p^2$ to $[a,b]$ exists and its square root is the minimum of $\mid p(x)\mid$.
A: Does $|p(x)|$ always attain $p(x)$'s minimum?
Try using a $p(x)$ such that $a_0<0$ and $a_0$ happens to be its minimum [example: $p(x)=x^2-1$]. Then $|p(x)|$ will never attain $a_0$, but $|a_0|$.
Maybe I'm misunderstanding the question. Try to word it better.
