Calculus of a Single Variable, Chapter 3, 3.4, Section 3.4, Problem 26

f(x)=2csc((3x)/2)
differentiating,
f'(x)=-2csc(3x/2)cot(3x/2)*(3/2)
f'(x)=-3csc(3x/2)cot(3x/2)
differentiating again,
f''(x)=-3(csc(3x/2)d/dxcot(3x/2)+cot(3x/2)d/dxcsc(3x/2))
f''(x)=-3(csc(3x/2)(-csc^2(3x/2)*(3/2))+cot(3x/2)(-csc(3x/2)cot(3x/2)*(3/2))
f''(x)=-3(-3/2csc^3(3x/2)-3/2csc(3x/2)cot^2(3x/2))
f''(x)=9/2csc(3x/2)(csc^2(3x/2)+cot^2(3x/2))
f''(x)=9/2csc(3x/2)(1+2cot^2(3x/2))
Now let us find out x for which f''(x)=0,
csc(3x/2)=0 , No value of x exists for which csc(3x/2)=0
1+2cot^2(3x/2)=0
cot^2(3x/2)=-1/2
So no solution exists.
So,f''(x) is undefined for x=0
Now let us test the concavity of the function by taking test value of -pi/2 and pi/2.
f''(-pi/2)=9/2csc(-3pi/4)(1+2cot^2(-3pi/4))
f''(-pi/2)=9/2(-sqrt(2))(1+2)
f''(-pi/2)=-27/2(sqrt(2))
f''(pi/2)=9/2csc(3pi/4)(1+2cot^2(3pi/4))
f''(pi/2)=9/2(sqrt(2)(1+2)
f''(pi/2)=27/2sqrt(2)
Though the concavity of the function is changing, but that does not mean that the inflection point exists at x=0, since the function is not defined at x=0.
Therefore the inflection points do not exist for the function.