Determine the intervals on which the function is concave up and intervals on which the function is concave down. f(x)=x- 2sin x defined on the interval (0, 3 pi)

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f(x) is concave up when f''(x) > 0 and concave down when f''(x) < 0

f(x) = x-2sin(x)

f'(x) = 1-2cos(x)

f''(x) = 2sin(x)

f''(x) > 0 when sin(x) > 0 and on the interval (0,3pi) this happens on the intervals (0,pi) and (2pi,3pi).

f''(x) < 0 when sin(x) < 0 and this happens on the interval (pi,2pi).  So our answer is

f(x) is concave up on the intervals (0,pi) and (2pi,3pi), and the function is concave down on (pi,2pi).

We can verfy with a graph.

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