`f(x) = x/2 + cos(x)` Consider the function on the interval (0, 2pi). Find the open intervals on which the function is increasing or decreasing, apply first derivative test to identify all relative extrema.
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This functions is infinitely differentiable. To find its extrema we have to solve the equation f'(x)=0:
f'(x)=1/2 - sin(x). This is zero at x=(-1)^k*pi/6+k*pi. At the given interval (0, 2pi) there are two of them: pi/6 and 5pi/6.
f'(x)>0 on (0, pi/6) and (5pi/6, 2pi) and therefore f is increasing on these separate intervals.
f'(x)<0 on (pi/6, 5pi/6) and therefore f is decreasing on that interval.
So pi/6 is a relative maximum and 5pi/6 is a relative minimum.
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