How do you find all points of inflection given y=-sin(2x)?

1 Answer
Oct 21, 2017

Points of inflection would occur every pi/2.

Explanation:

To find points of inflection, we need to find all the points on the graph at which the second derivatives will have a value of 0:

f''(x) = 0

f(x) = -sin(2x)

Using chain rule:

u = 2x

d/(du) (-sin(u)) = -cos(u)

(du)/dx = 2

d/dx = d/(du) * (du)/(dx) = -2cos(u) = -2cos(2x) = f'(x)

Using same principles to differentiate again:

u = 2x

d/(du) (-2cos(u)) = 2sin(u)

(du)/dx = 2

d/dx = d/(du) * (du)/(dx) = 4sin(u) = f''(x)

f''(x) = 4sin(2x)

And now to make f''(x) = 0:

4sin(2x) = 0

sin(2x) = 0

2x = arcsin(0) = 0+npi

x = (0+npi)/2 = npi/2