f(x) = sin x + x. f'(x) = cos x+1. f"(x)=-sin x. The critical numbers are the solutions of cos x = -1, The first derivative test yields the case {+, +}, and, therefore, we obtain only inflection points at y = (2n+1). x = (2n x= (2n -4-3-2- 4 2T T 1 # 2 3 4 x