Suppose we have a function $g$ that is differentiable (and hence continuous) and monotonically increasing on the interval $[P,Q]$.
I know that this alone is not enough to imply that if $a,b\in [P,Q]$ and $a>b$, then $g(a)>g(b)$, because this is a strict order relation and we only know that $g$ is monotonically increasing.
However, it is definitely true that $\forall m,n\in\mathbb{R}$, if $m>n$ then $m^3>n^3$. This strict order relation is definitely true, even though the cubic function $x^3$ is only monotonocally increasing on $(-\infty,\infty)$. My thought is that this is true because the cubic function $x^3$ is strictly increasing except at $x=0$.
This gave me the following thought:
Question
Suppose there is a function $f(x)$ that is differentiable for all $x\in [P,Q]$. Also, $f(x)$ is strictly increasing for $x\in[P,Q]$, except for finitely many values of $x$. And, over the interval $[P,Q], f$ is monotonically increasing.
Is the following statement true?: $\forall m,n\in [P,Q]$, if $m>n$ then $f(m)>f(n)$
To clarify, the function is monotonocally increasing on $[P,Q]$, and it is only stationary at finitely many points. I made this restriction so that the graph of $f(x)$ has no "horizontal lines" in $[P,Q]$, and this should mean that the strict order relation holds.
To me, this is intuitively true, although I haven't been able to come up with a formal proof. Like the function $x^3$, the strict order relations will hold for my function $f$, because the function is only stationary at a finite number of points.
Thanks in advance.
Just prove it by contardiction. If $m>n$ and $f(n)=f(m)$ then $f(m)\geq f(x) \geq f(n)$ for all $x$ between $n$ and $m$ (by monotonicity) which makes $f$ constant througout the interval $(n,m)$. But this contradicts your hypothesis.