f(t)=t+cost, f '(t)=1-sint>0,if t≠ π/2 +2πn
f ' (π/2 +2πn)=0. In the left neighborhood of the point π/2 +2πn f(t)<f(π/2 +2πn), in the
right f(t)>f(π/2 +2πn), because in these neighborhood f'(t)>0.Consequently, the points at
which f'(t)=0 are not extrema.
The use of Taylor's formula also confirms the absence of extremum points :
f ''(t)=-cost and f ''( π/2 +2πn)=0 ,f '''(t)=sint and f'''( π/2 +2πn)≠0
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