Question

Find the points of inflection;

2csc(3x/2)

(0,2pi)..describe the concavity

Answer 1

\[f(x)=2\frac{1}{\sin(3x/2)}\]
\[f'(x)=2*\frac{0*\sin(3x/2)-3/2*\cos(3x/2)}{\sin^2(3x/2)}=-3\frac{\cos(3x/2)}{\sin^2(3x/2)}\]
\[f''=3*\frac{-3/2*\sin(3x/2)\sin^2(3x/2)-2\sin(3x/2)*\cos(3x/2)*3/2*\cos(3x/2)}{\sin^4(3x/2)}\]
\[f''=3*\frac{3}{2}*\sin(3x/2)*\frac{-\sin^2(3x/2)-\cos^2(3x/2)}{\sin^4(3x/2)}\]
\[f''=\frac{-9}{2}\sin(3x/2)*\frac{1}{\sin^4(3x/2)}\]
to find possible inflection points set \[\sin(3x/2)=0\]

Answer 2

remember sin(0)=0 and sin(pi)=0 and sin(2pi)=0
but you said not include 0 and 2pi since you used () and not [] right?

so anyways sin(pi)=0
so we set 3x/2=pi and solve for x

Answer 3

x=2pi/3

Answer 4

now remember this gives us that we have a possible inflection point at (2pi/3,f(2pi/3))

Answer 5

we need to see if the concavity switches there
if so then we indeed have an inflection point

Answer 6

-------|----------
2pi/3
choose a number btw 0 and 2pi/3 (3x/2=0 => x=0)
choose a number btw 2pi/3 and 4pi/3 (3x/2=2pi=>x=4pi/3)

Answer 7

plug those numbers into f''
a positive number=> concave up
a negative number=> concave down

Answer 8

oops so actually you have one more interval
you need to choose a number btw 4pi/3 and 2pi
actually you will not have any inflection points since f(2pi/3) and f(4pi/3) doesn't exist
but still you use the above numbers to test for concavity