What are the points of inflection of #f(x)=sin^2x # on the interval #x in [0,2pi]#?

Question

11141 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-28T21:36:43

The points of inflection of #f(x) = sin^2x# in the interval #[0,2pi]# are:

#pi/4, (3pi)/4, (5pi)/4, (7pi)/4#

Evaluate the first and second derivatives of the function:

#d/dx sin^2x = 2sinx cosx = sin2x#

#d^2/dx^2 sin^2x = d/dx sin2x = 2cos2x#

Solve the equation:

#d^2/dx^2 sin^2x = 0#

#2cos2x = 0#

#2x = pi/2+kpi#

#x= pi/4(2k+1)# with #k in ZZ#

Around each of this solutions the second derivative changes sign, so they are all points of inflection. In the interval #[0,2pi]# we have the values:

#pi/4, (3pi)/4, (5pi)/4, (7pi)/4#