Calculus help: find the points of inflection and discuss the concavity of the graph

[mrevolution#9]

of the function? the equation that is given is : f(x) = 2csc(3x/2), on the interval (0, 2pi)

I really need help on this one..

 

[answering]

It appears that the graph of f(x) = 2csc(3x/2) in the domain (0, 2pi) is discontinuous at 2?/3 and 4?/3; these 2 points divide the domain into 3 same length intervals: (0, 2?/3), ( 2?/3, 4?/3), (4?/3, 2?) where the graph is respectively U ? U: concave up, concave down, concave up. the endpoints of the intervals are vertical asymptotes.

I THINK (but am not positive) x= 2?/3 and x= 4?/3 might be called pts of inflection even tho (and maybe BECAUSE) f(2?/3) and f(4?/3) are undefined. The concavity of the function definitely reverses either side of the pts.