Critical points and Convexity? Function $f(x)$ has no critical points in $M$, can we say $f(x)$ is either convex or concave over $M$? 

If $f(x)$ has no critical points, then it's derivative is continuous and either positive definite or negative definite (over the domain $M$).  To be either convex or concave, the second derivative would have to, likewise, by either positive definite or negative definite.  A simple sketch will show that the derivative can be positive definite and yet the second derivative (derivative of the derivative) is not.
As the sketch suggests, you could easily have:
$$
f'(x) = x^2 + C\text{, where } C > 0 \\
f''(x) = 2x \\
f(x) = \frac{x^3}{3} + Cx + D
$$
Clearly, the second derivative produces an inflection point (and thus the concavity changes) yet there are no critical points (no place where $f'(x) = 0$ or $f'(x)$ is discontinuous).
If you make $C=10$ you can make the slope at the inflection point ($x = 0$) a little more dramatic and definitely see that the derivative is never $0$: google graph.
