IVT question involving polynomial with even degree Let $M(x)$ be an even polynomial with a positive leading coefficient, with $a_{2n} > 0, n\ge1 $. Show that there exists a constant $a*\in \mathbb{R}$ such that $M(x)+a = 0$ has a real root if $a<a*$.
I've expressed the polynomial as $M(x) = \sum_{j=0}^{2n} a_jx^j$, not sure where to go from here..
 A: As $M(x)$ has a positive leading coefficient and it is of even degree, $\lim_{x\to\pm \infty}M(x)=+\infty$. This implies in particular there exists a number $A>0$ such that $M(x)>f(0)\,$  if $\,x>A$ or $x<-A$. On another hand, by Weierstrass extreme value theorem, $M(x)$ attains a minimum $m$ value on $[-A,A]$, and $\,m\le f(0)$. Hence $m$ is the minimum value of $M(x)$ on $\mathbf R$, so that $f(\mathbf R)= [m, +\infty)$ by the intermediate value theorem.
Taking $a^*=-m$ answers the question.
A: The first thing I had to do with this problem was to try to figure out what
it was looking for.
I found it helped me to mentally substitute $-b$ for $a$ (that is, to let $b = -a$)
and rewrite the condition accordingly.
With this substitution, we're being asked to show that there is a constant $b^*$
such that whenever $b > b^*$, then $M(x)-b=0$ has a real root.
In other words, there is some $b^*$ such that as $x$ ranges over $\mathbb R$,
$M(x)$ takes on every real value greater than $b^*$.
In terms of the original problem, as $x$ ranges over $\mathbb R$,
$M(x)$ takes on every real value greater than $-a^*$.
So one proof strategy would go like this:
first, identify (somehow) a value $a^*$.
Then let $a$ be an arbitrary real number such that $a<a^*$.
Show that there is an $x_-$ such that $M(x_-) < -a$ (that is, $M(x_-) + a < 0$)
and that there is an  $x_+$ such that $M(x_+) > -a$ (that is, $M(x_+) + a > 0$).
Apply the intermediate value theorem to show the existence of $x$
such that $M(x) + a = 0$.
A hint about choosing $a^*$: if you let $a^*$ be too positive,
then there will be an $a < a^*$ such that $M(x) + a = 0$ has no real root;
in fact there will be no real root of $M(x) + a = 0$ for any $a$
that is "too close" to $a^*$.
(The fact that $M(x)$ has a lower bound has something to do with this.)
So you need to find a value of $a^*$ that is negative enough.
This is also a hint about finding $x_-$, if you think about it.
A hint for finding $x_+$ is: $M(x)$ has no upper bound for $x \in \mathbb R$.
