Q:

Concavity on an interval or at a point

I have read the following:

if $f$ differentiable at $(a,b)$ so $f$ is concave up in  $(a,b)$ $\iff$  for all $x_0\in(a,b)$ the graph of the function is above the tangent
if $f$ differentiable at $(a,b)$ so $f$ is concave up in  $(a,b)$ $\iff$ $f'(x)$ is increasing in $(a,b)$
if $f$ is differentiable twice at $(a,b)$ so $f$ is concave up in  $(a,b)$ $\iff$ for all $x_0\in (a,b),  f''(x_0)\geq 0$

So if I know that a function is differentiable I need to test concavity on an interval (or 2 points), and if I know that the function is differentiable twice, I just need a point to determine concavity? 

A:

No.  It's never enough to check only one or two (or any finite number of) points.  What they're saying is that if $f$ is differentiable, then you can choose between the following three ways to check whether it's concave (or confirm that it isn't, if you happen to find a counterexample to one of the following points):

To check that $f$ is concave, for every point $x \in (a, b)$ you need to construct the tangent of $f$ at that point and check that the graph of $f$ is never below the tangent (but may be equal to it) at any points in $(a, b)$.
To check that $f$ is concave, you calculate $f'$, and check that it is always increasing, which is to say for all pairs $x, y \in (a, b)$ with $x< y$ you need to check that $f'(x) \leq f'(y)$.
If $f''$ exists everywhere, then to check that $f$ is concave, you need to verify that for all $x \in (a, b)$ we have $f''(x) \geq 0$.

