Determine the inflection point and Concavity for the equation f(x)=2csc(3x/2) on the interval (0,2pi)

Question

The question specifically asks for the point of inflection and where (in interval notation) the graph is concave up and down.

Answer 1

For f(x) = 2csc(3x/2) taking the 1st derivative (utilizing the chain rule) yields:

f'(x) = -2(3/2)csc(3x/2)cot(3x/2) or, simplifying:

f'(x) = -3csc(3x/2)cot(3x/2)

You get a number of critical points, some because it is undefined and others because it equals zero. Bottom line is a local min at x = π/3 and x = 5π/3 and a local max at x = π

To find the 2nd derivative, we need to use the product rule yielding:

f''(x) = - -3(3/2)csc(3x/2)cot(3x/2)cot(3x/2) + - - 3(3/2)csc²(3x/2)csc(3x/2) or simplified:

f''(x) = 9/2csc(3x/2)[cot²(3x/2) + csc²(3x/2)]

To find the points of inflect, set this equal to zero

In this factored form:

1) f'' = 0 when 9/2csc(3x/2) but csc is NEVER equal to zero so there is no solution here.

2) f'' = 0 when cot²(3x/2) + csc²(3x/2) = 0

but, using the Pythagorean Identity cot²(θ) + 1 = csc²(θ) this becomes:

cot²(3x/2) + cot²(3x/2) + 1 = 0

2cot²(3x/2) + 1 = 0

2cot²(3x/2) = -1

cot²(3x/2) = -1/2 but you can't square something and get a negative result so there is no solution here as well.

So there are NO inflection points. In looking at the graph, and in looking at the 2nd derivative test, the section on the interval (0, 2π/3) is concave up while the section on the interval (2π/3, 4π/3) is concave down but the dividing line, at x = 2π/3 is undefined for the function so it cannot be a point of inflection. The same holds true comparing the interval (2π/3, 4π/3) which, as mentioned, is concave down and the interval (4π/3, 2π) is concave up but the dividing line, x = 4π/3 is undefined for the function so it cannot be a point of inflection.

Answer 2

To find critical points, start by taking the 1st derivative and setting it = to 0:

f(x)=2csc(3x/2)

f'(x) = 2 * 3/2 * (-csc(3x/2)) * cot(3x/2)

-3csc(3x/2)cot(3x/2) = 0

the csc of an angle cannot go to 0, so let's just use cot(3x/2) = 0, This happens at 3x/2 = pi/2, x = pi/3, and also at 3x/2 = 3pi/2, x = pi, AND: 3x/2 = 5pi/2, x = 5pi/3

f(x) goes to infinity when 3x/2 = pi, x = 2pi/3 and when 3x/2 = 2pi, x = 4pi/3

So, for 0 ≤ x ≤2pi, we have 4 discontinuities, 2 local minimums, and 1 local maximum.

PLOT the original function to see this....