Graph Concavity Test I'm studying for my final, and I'm having a problem with one of the questions. Everything before hand has been going fine and is correct, but I'm not understanding this part of the concavity test.
$$f(x) = \frac{2(x+1)}{3x^2}$$
$$f'(x) =-\frac{2(x+2)}{3x^3}$$
$$f''(x) = \frac{4(x+3)}{3x^4}$$
For the increasing and decreasing test I found that the critical point is -2:
$$2(x+2)=0$$
$$x = -2$$
(This part is probably done differently than most of you do this), here's the chart of the I/D
2(x+2), Before -2 you will get a negative number, and after -2 you will get a positive number.
Therefore, f'(x), before -2 will give you a negative number, and after you will get a positive number, so f(x) before -2 will be decreasing and after it will be increasing. Where -2 is your local minimum.
(By this I mean; 2(x+2), any number before -2 (ie. -10) it will give you a negative number.)
As for the concavity test, I did the same thing basically;
$$4(x + 3) = 0$$
$$x = -3$$
However, my textbook says that x = 0 is also a critical point, I don't understand where you get this from. If anyone can explain this I would appreciate it, also if there's a more simple way of doing these tests I would love to hear it, thanks.

If I got your question correctly, you are working on the function $f(x)$ as above and want to know the concavity of it. First of all note that as the first comment above says; $x=0$ is also a critical point for $f$. Remember what is the definition of a critical point for a function. Secondly, you see that $x=0$ cannot get you a local max or local min cause $f$ is undefined at this point. $x=0$ is not an inflection point because when $x<0$ or $x>0$ in a small neighborhood, the sign of $f''$ doesn't change. It is positive so $f$ is concave upward around the origin. 
