`f(x) = sin(x/2), [0, 4pi]` Find the points of inflection and discuss the concavity of the graph of the function.
This function is defined everywhere and is infinitely differentiable.
At the points of inflection of f the second derivative must be zero:
It is =0 at x/2=k*pi￼, x=2k*￼pi where k is any integer. Because we consider f to be defined on [0, 4pi] there is only one such point inside (0, 2pi), it is x0=2pi.
f'' is negative at the left of x0 and positive at the right. So f is concave downward at (0, 2￼pi) and concave upward at (2pi, 4pi). And 2pi is the point of inflection.