What are the points of inflection, if any, of #f(x)=xsin^2x # on [0,2pi]?

1 Answer

points corresponding to
#f''(x)=0, #
between
#x=0.7 and 0.8#
#x=1. 1and 1.2#
#x=2.3 and 2.4#
#x= 2.5 and 2.6#
#x=3.0 and 3.4#
#x=4.0 and 4.1#
#x=5.4 and 5.5#
#x=5.5 and 5.6#
correct to one decimal place

#x=pi/2#
and
#x=3pi/2#

Explanation:

Given: #f(x)=xsin^2x #
Evaluating the curvature,
#f'(x)=x(2sinx)cosx+sin^2x#
=#2xsinxcosx+sin^2x#
=#xsin2x+sin^2x#
Since, #2sinxcosx=sin2x#
Thus, #f'(x)=xsin2x+sin^2x#
#f''(x)=xcos2x(2)+sin2x(1)+2sinxcosx#
=#2xcos2x+sin2x+sin2x#
Now, #f''(x)=2xcos2x+2sin2x#
#f''(x)=0 #
implies
#2xcos2x+2sin2x=0#
Taking #2cos2x# common
#cos2x(x+tan2x)=0#
Either #cos2z=0#
where, x is clearly #pi/2 and 3pi/2#
or
#x+tan2x=0#
By trial and error,

Inspecting for values between 0 and 2pi in the intervals of 0.1pi
At #x=0, f''(x)=0#
#f''(x)=0, #
between
#x=0.7 and 0.8#
#x=1. 1and 1.2#
#x=2.3 and 2.4#
#x= 2.5 and 2.6#
#x=3.0 and 3.4#
#x=4.0 and 4.1#
#x=5.4 and 5.5#
#x=5.5 and 5.6#
correct to one decimal place
MY thanks for Fleur for reminding