Inflection and concavity How do I find the intervals when the graph below is concave up and when it's concave down as well as its inflection coordinate points just by looking at the graph? 

I'm able to eyeball it here, but I'm confused with a few parts. I'm pretty sure it concaves up in $(0, 2)\cup(5, \infty)$. I'm not sure if the region around $x=4$ should have its concavity considered because the derivative is not defined at this point.
Then I suspect it concaves down in $(2, 3.5)$. Again, I'm not sure if the point after $x=4$ but just before $x=5$ should be considered an interval where it concaves down, but I doubt it. 
Then I think the inflection coordinates are
$(1, 4), (3, 6), (5, 5)$
 A: “Concave up” is like the graph of $x^2$ (any arc of it); “concave down” is like the graph of $-x^2$ (any arc of it).
An inflection point is where the curve has a tangent and is concave up on one side and down on the other side (not “globally”, but in some intervals).
At $(1,4)$ there is no change in concavity, and similarly at $(3,6)$. Indeed, these are a point of minimum and a point of maximum respectively.
The change in concavity happens somewhere in between $1$ and $3$ and the visual symmetry leads to guess that the inflection point is at $(2,5)$.
The point $(5,5)$ is indeed an inflection point (at least if we assume that the picture is “accurate”), because the curve is concave down before it and up past it.
What happens at $(4,3)$? This is a point where there is no tangent, so it has to be analyzed separately. The curve is concave up on either side of it, but it's nonetheless a point of minimum.
The intervals where the graph is concave up are: $[0,2]$, $[5,8]$; it is concave down on the intervals $[2,4]$, $[4,5]$.
Note that you should not say $[2,5]$, because the function is not concave down over the whole interval.
