Show that function is strictly monotonic if certain property holds Let $f:\mathbb R\to\mathbb R$ be differentiable and $f'(x)\geq 0$ for all $x\in\mathbb R$. Show that $f$ is strictly monotonic if there are no two real numbers $a<b$ such that $f'(x)=0$ for all $x\in (a,b)$.
I feel like I'm almost finished with the proof but I am missing a final part: 

If there is no $x_0\in\mathbb R$ with $f'(x_0)=0$ we are done since then $f'(x)\geq 0$ implies $f'(x)>0$. So let's take such an $x_0$. Now, because $f'(x)\equiv 0$ can't happen on open intervals $(a,b)$ we know that the zeros of $f'$ are isolated, meaning that there is an $\epsilon>0$ such that $f'(x)\neq 0$ for all $x\in D_\epsilon (x_0)\backslash \{x_0\}$ and because of $f'(x)\geq 0$ globally this means $f'(x)>0$ on that punctured disk. So for any $a<b$ in that disk the mean value theorem gives me some $c\in D_\epsilon (x_0)\backslash \{x_0\}$ such that $$\frac{f(b)-f(a)}{b-a}=f'(c)>0$$ which means $f(b)-f(a)>0$ and the function is strictly monotonic there. How do I now get away from that disk to some global argument?

 A: I feel like proving the contronominal would be a lot easier. The statement would be: if $f$ is not strictly monotonous, then there exist two real numbers $a<b$ such that the derivative is zero in $(a,b)$. Check it yourself, I often mess up when stating contronominals.
Let's say $f$ is increasing, the non sctrictly-increasing hypothesis tells me $\exists a,b$ such that $f(a)=f(b)$. Now the function is bound to be constant on $[a,b]$ (otherwise the derivative would be negative in a certain interval, contradicting monotonicity), and you're done.
A: From the MWT, $f$ is increasing, but perhaps not strictly. Suppose there are $a<b$ with $f(a)=f(b)$. Then by weakly incrasing, we have $f(a)\le f(x)\le f(b)$ for $a<x<b$, i.e., $f$ is constant on $(a,b)$, which implies that $f'(x)=0$ for $x\in (a,b)$, contrary to the given property..
A: I would do it via contraposition:
Let's assume $a<b$ and that $f'(x)=0, ~\forall ~x \in (a,b)$. If you now simply apply the mean value theorem you get:
$$
\frac{f(b)-f(a)}{b-a}=f'(\xi),
$$
where $\xi \in (a,b)$. This immediately yields $f(a)=f(b)$. So $f$ cannot be strictly monotonic.
