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

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Answer 1 · 2017-10-21T12:58:03

Points of inflection would occur every #pi/2#.

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)#

#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#