Let $f:\mathbb{R} \rightarrow \mathbb{R}$
Is it true that :
$f$ is differentiable $\Leftrightarrow$ we can find $I \subset \mathbb{R}$ such that $f$ is monotonic on $I$ where $I = [a, b]$ ($a \ne b$ and $a, b \in \mathbb{R}$).
If this is true I really don't know how to prove such result... My intuition is saying that this is true because by definition if $f$ is differentiable then the function has a tagent line on every point of $\mathbb{R}$ and hence is monotonic, but I am not sure .
This isn't true. Take $f(x)=\lvert x\rvert$. It is monotone decreasing on $(-\infty,0)$ and monotone increasing on $(0,\infty)$, but yet it isn't differentiable at $x=0$.
To remedy this problem, you might ask that this condition apply pointwise. That is, what if we required that $f$ be monotone on a neighborhood of each point. We could make the following conjecture:
Conjecture: $f$ is differentiable on $\mathbf{R}$ if and only if for all $x\in \mathbf{R}$ there exists an open neighborhood $U$ so that $f$ is monotone on $U$.
This also ends up failing. Take $f(x)=x^2$. This is definitely a smooth function, but at $x=0$, we have a turning point, in the sense that $f'(x)<0$ on $(-\infty, 0)$ and $f'(x)>0$ on $(0,\infty)$. So, on no neighborhood of $0$ is $f$ monotone.
Returning to the original problem, we have the following conjecture:
Conjecture: If $f$ is continuously differentiable on $\mathbf{R}$, then there exists an interval $I=[a,b]\subset \mathbf{R}$ so that $f$ is monotone on $I$.
Proof: If $f$ is continuously differentiable, then either $f'(x)=0$ on all of $\mathbf{R}$ or not. If $f'(x)=0$ everywhere, then we have that $f$ is constant, so that it is indeed monotone everywhere. If $f'(x)\ne 0$ somewhere (say $f'(x_0)>0$) then by continuity, there must exist some open neighborhood $U$ of $x_0$ so that $f'(x)>0$ on $U$. Indeed, then, $f$ is monotone increasing on $U$.
