So, #f(x) = 1/2x - sinx#, is a pretty straightforward function to differentiate.
Recall that #d/dx(sinx) = cosx#, #d/dx(cosx) = -sinx# and #d/dx(kx) = k#, for some #k in RR#.
Hence, #f'(x) = 1/2 - cosx#.
Hence, #f''(x) = sinx#.
Recall that if a curve is 'concave up', #f''(x) > 0#, and if it is 'concave down', #f''(x) <0#. We can solve these equations fairly easily, using our knowledge of the graph of #y =sinx#, which is positive from an 'even' multiple of #pi# to an 'odd' multiple, and negative from an 'even' multiple to an 'odd' multiple.
Hence, #f(x)# is concave up for all #x in (0,pi) uu (2pi, 3pi)#, and concave down for all #x in (pi, 2pi)#.
Generally speaking a curve will have a point of inflection where #f''(x) = 0# (not always - there must be a change in concavity), and solving this equation gives: #x in {0, pi, 2pi, 3pi}#.
We know from part #b# that there are changes in concavity at these points, hence #(0,0), (pi, pi/2), (2pi, pi),# and #(3pi, 3pi/2)# are all points of inflection.
