Convergence of derivatives near the boundary of an open interval Suppose there are two continuously differentiable functions $a(x_1)$ and $b(x_2)$, where $x_1, x_2\in \mathbb{R}_+$.
I assume that $a'(x_1) >0$ if $x_1<x^o$ and $a'(x_1)<0$ if $x_1>x^o$. It follows that $a'(x^o)=0$ and $x^o$ is the point that gives the maximum of $a(x_1)$.
Similarly, I assume  $b'(x_2) >0$ if $x_2<x^p$ and $b'(x_2)<0$ if $x_2>x^p$. It follows that $b'(x^p)=0$ and $x^p$  is the point that gives the maximum of $b(x_2)$.
I assume $x^p > x^o$ and take the open interval $(x^o, x^p)$.
Suppose $\exists x^q \in (x^o, x^p): |a'(x_1)|>b'(x_2) \forall x_1 \in (x^o, x^p) $ and $\forall x_2 \in (x^q, x^p)$.
I would like to then say that $|a'(x^c)| = b'(x^q)$,  where $x^c$ is a point infinitesmally close to $x^o$ such that $|a'(x^c)| >0$.
The problem is there is always a point closer to $x^o$ than $x^c$.
How would I say that $|a'(x_1)|$ and $b'(x_2)$ converge at $x_1 = x^c$ and $x_2 = x^q$, without negating all the values of $x_1$ that exist between $x^o$ and $x^c$?
 A: There is a subtle but important difference between the title of your question and the body of the post. Your title (and last sentence in the post) talk about convergence, yet the first part of your post is concerned with point-wise statements.
I think introducing the concept of the infimum/supremum will address this issue and your concern about "a point always a little bit closer".
Restating your problem for clarity

*

*We have two continuously differentiable functions $a(x),b(x)$ defined on $x \in [0,\infty)$

*Both functions have unique maxima at $x_a,x_b$, respectively, with $x_a < x_b$

*We assume $\exists x_q \in (x_a,x_b): |a'(x_1)|>b'(x_2),\;\;\forall x_1\in (x_a,x_b),\forall x_2\in(x_q,x_b)$
We want to find an $x_c \in (x_a,x_q):|a'(x_c)|=b'(x_q)$
A couple observations:
Condition (3) above implies the following:
$$0<\sup_{y\in (x_q,x_b)} b'(y) <\inf_{x \in(x_a,x_b)} |a'(x)|=0$$
Therefore, $x_q = x_b$ but $x_b \notin (x_a,x_b) \implies \not\exists x_q \in (x_a,x_b): |a'(x_1)|>b'(x_2),\;\;\forall x_1\in (x_a,x_b),\forall x_2\in(x_q,x_b)\;\;\;\square$
As a corollary, $x_c$ doesn't exist either because it depends on $x_q$ existing.
The overall conclusion is your three conditions are form an inconsistent set of statements.

Discussion
As shown above, there are no points in $(x_a,x_b)$ that satisfy your axioms, and in fact we showed them to be inconsistent. The issue is that, as you correctly pointed out, there is no unique minimum or maximum value in open intervals of the real line ; however, the infimum and supremum exists and is unique. Hence, you correctly pointed out you cannot find $x_c$ that is the point that is closest to $x_a:|a'(x_c)|>0$ -- there is no "closest point".
Now, if we turn to your title and last sentence, we can get to a coherent solution if we switch to convergence and infima/suprema vs min/max, which allows us to use limiting arguments vs point-wise arguments.
First, the infimum of $|a'(x)|$ for $x \in(x_a,x_b)$ exists, and is $0$.
Second, your axiom $(3)$ implies we need to find $x_q:\sup_{a \in(x_q,x_b)} b'(a) = 0$. From the definition of $b(x)$, we know that $\sup_{a \in(x_q,x_b)} b'(a) > 0\;\;\forall x_q<x_b$, so we are again unable to find such an $x_q$.
This is where the idea of convergence comes in.
Instead of $(3)$ we can create $(3')$ with limiting arguments:
$$(3')\quad\quad \;\;\exists x_q \in [x_a,x_b]: \lim_{a \to x_q^-} \sup_{x\in(a,x_b)} b'(x) = 0$$
In this case, we see that $x_q=x_b$ works because $\lim_{x \to x_b^-} b'(x) =0$ (by continuity), hence $\lim_{a \to x_q^-} \sup_{x\in(a,x_b)} b'(x)=0$
Finally, for $x_c$ we can find it via similar limiting arguments:
$$\exists x_c \in (x_a,x_b): \lim_{x \to x_c^+} |a'(x)|=0 \implies x_c = x_a$$
So this is just a restatment of the fact that $\lim_{x \to x^*}f'(x) = \lim_{y \to y^*} g(y)$ where $x^*,y^*$ are extremum points of $f,g$, respectively.
