Points of inflection are points on the graph at which the concavity changes.

For this function, the concavity changes exactly where the sign of the second derivative changes.

#f(x)=x+sin^2x#

#f'(x) = 1+2sinxcosx = 1+sin(2x)#

#f''(x) = 2cos(2x)#.

To investigate the sign of the second derivative, find the points at which the sign might change. (In general these are zeros and discontinuities, but this #f''# has no discontinuities.)

#2cos(2x) = 0#

#2x = pi/2 + pik# #" "# (#k# an integer.)

#x = pi/4 + pi/2k = pi/4+(2pi)/4k# #" "# (#k# an integer.)

To find values of #x# in #[0,2pi]#, substitute values of #k# until we get #x#'s outside the interval. (There are other methods, but this is the one I am choosing here.)

#{:(k,"|",-1,0,1,2,3,4),(x,"|",-pi/4 <0,pi/4,(3pi)/4,(5pi)/4,(7pi)/4,(9pi)/4 > 2pi):}#

Checking the sign of #f''(x) = 2cos(2x)#.on the intervals defined by these endpoints we see that the sign does indeed change at each opportunity.

#{: (bb "Interval", bb"Sign of "f',bb" Concavity"),
([0,pi/4)," " -" ", " "" Down"),
((pi/4, (3pi)/4), " " +, " " " Up"),
(((3pi)/4 ,(5pi)/4), " " -, " "" Down"),
(((5pi)/4 ,(7pi)/4), " " +, " "" Up"),
(((7pi)/4 ,2pi), " " -, " "" Down")
:}#

Therefore, the inflection points are the points on the graph with

#x = pi/4#, #(3pi)/4#, #(5pi)/4#, and #(7pi)/4#

To find the #y# values use #f(x) = x+sin^2x#