Calculus of a Single Variable, Chapter 3, 3.4, Section 3.4, Problem 28

This function is fully differentiable on a given interval. We can use the second derivative to determine concavity and inflection points. Let's find f'' and where its positive, negative or zero:
f''(x) = -sin(x) - cos(x) = -sqrt(2)*cos(x-pi/4).
This is zero at x-pi/4 = pi/2+k*pi, or x = (3*pi)/4 + k*pi for any integer k. There are two such x at [0, 2pi ], x=(3pi)/4 and x=(7pi)/4.
f''(x) is negative on (0, (3pi)/4), positive on ((3pi)/4, (7pi)/4) and is positive again on ((7pi)/4, 2pi). Therefore f(x) is concave downward on (0, (3pi)/4) and on ((7pi)/4, 2pi), and is concave upward on ((3pi)/4, (7pi)/4).
The points of inflection are (3pi)/4 and (7pi)/4.