Relation between bounded first derivative and behaviour of a function Let me begin with an example first:

Consider $\;f:\mathbb R \rightarrow \mathbb R\;$ such that $\;f(x)=\tanh x\;$. Then its plot is given below:


Now, the first derivative of $\;f\;$ is equal to $\;f'(x)=\frac{2}{e^x + e^{-x}} \equiv \frac{1}{\cosh x} \;$ which is a bounded function. Its plot is also given below:



I observe that $\;f\;$ goes from the value $-1$ to the value $+1$ in a "smooth" ,kind of, way.(I'm sorry I lack of the best word to describe the behaviour of $\;f\;$)

I wonder if the above behaviour of $\;f\;$ is related to the fact that its first derivative is bounded. 


If my assumption is true, why does this hold?
If I had a function which first derivative wasn't bounded, then what could I claim about the behaviour of $\;f\;$ in this case?


I'm searching for some examples but I can't find any...

I hope I made my question clear enough. Any help would be valuable.

Thanks in advance!
Take your function to be 

$$
 f (x) =
  \begin{cases}
 0 &amp; \text{if } x&lt;0 \\
 x+1 &amp; \text{if } x \geq 0 \\
  \end{cases}
$$

Then $f$ has a bounded derivative but it is not "smooth" because it is discontinuous at $0$.

If your derivative is not bounded, then you can't claim much about boundness of the original function. For example, $f=\sqrt{x}$ on $[0,1]$ is a bounded function, but it's derivative is not bounded. However, $g(x)=e^x$ is a function that is not bounded and whose derivative is not bounded.
If your 1st derivative is not bounded then either your function is not continuous from both side of domain or it is discontinuous
e.g.
consider these function
\begin{cases}
 \sqrt{x}, &amp; \text{x ∈ [ 0 , 1 ]}\end{cases}
as it has derivative unbounded near zero hence it has not left hand side continuity
and,
\begin{cases}
 1 / x, &amp; \text{x ∈ R}\end{cases}
but in this case the function is not continuous at x = 0
therefore unbounded discontinuity does tells us so much about continuity and much more stuff