Converse: $f'(x)\ge 0 \implies f$ is monotonically increasing? In my textbook the following theorem is stated: If $f'(x)\ge0$ for all $x\in(a,b)$ then $ f $ is monotonically increasing.
Is the converse also true? Intuitively it seems like it is but I know there must be a reason why it's not an if and only if statement.
 A: If $f$ is differentiable and monotone increasing then $f'(x) \geq 0.$
We prove this by contrapositive. if $x < y $ and $f(x) > f(y)$ then the mean-value theorem states, there exists $z \in (x,y)$ such that
$$
f'(z) = \frac{f(y) - f(x)}{y-x} < 0. 
$$
We are done.
A: The converse is true if, for example, you assume $f'$ is continuous away from a discrete set of points. If $f'(x_0)<0$, then continuity implies that $f'(x)<0$ on some interval around $x_0$. Then $f$ is decreasing on that interval.
A: The converse is true if you assume your function is differentiable; this is easily checked by writing down the limit definition of the derivative and using the fact that $f$ is monotone to establish the inequality on the derivative. Probably the reason why they didn't state the converse was that monotone functions need not be differentiable everywhere.
A: To add explicitness to what everyone else is already stating. Take $f(x) = x \cdot \mathbb{1}_{\{x\leq 1\}} + (2x-1) \cdot \mathbb{1}_{\{x>1\}}$ then $f$ is monotonically increasing, yet $f'(1)$ is not defined. 
Now, adding differentiability gives you a completely different story (see any of the other answers).
A: Hint: If $f: [a,b] \to \mathbb  R$ is right differentiable and $f_+'(a) < 0$ then there exists $\delta > 0$ such that $x \in [a,b]$, $$a< x < a+\delta \implies f(x) <f(a)$$  
This comes from 
$$x \in [a,b], a< x< a + \delta \implies \frac{f(x) - f(a)}{x -a } < 0 \implies f(x) < f(a)$$
Just suppose for a moment that $f'(x) < 0$ for some $x \in [a,b]$, and get to a contradiction.
A: claim: differentiable function $f: \Re\rightarrow\Re$ is monotonically increasing $\Rightarrow$ $f'(x)\geq0$ for all x. (note that in here, the premiss have to state that $f$ is differentiable so that this proof can hold)
proof:
let $x_0 \in \Re $ be given.
then $x>x_0 \Rightarrow f(x)-f(x_0) \geq 0 \ $
and $x<x_0 \Rightarrow f(x)-f(x_0) \leq 0 \ $ since f is monotinally increasing.
since $f$ is differentiable, then, in both cases, the limit exists and  $\ \lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}\geq 0 \ $
that is, $f'(x)\geq0$
