Critical points and Convexity?

Question

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

Answer 1

If $f(x) = 2x+\sin(x), x\in \mathbb{R}$, then $f^\prime(x) = 2+\cos(x)$ is nowhere $=0$, hence there is no critical point. However, $f^{\prime\prime}(x)=-\sin(x)$ changes sign, so $f$ is neither convex nor concave. By scaling you can do that on any interval, as small as you like.

Answer 2

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.