Calculus

Consider the function on the interval
(0, 2π).

f(x) = x + 2 sin x

relative maximum (x, y) =
relative minimum (x, y) =

From when I worked out using the first derivative test, I ended up with (2π, 0) for the max and (0,0) for the min. Apparently this isn't correct.

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  1. Too bad you didn't show your work...

    f' = 1 + 2cosx
    f'=0 when cosx = -1/2

    x = 2?/3, 4?/3

    f" = -2sinx
    f"(2?/3) < 0, so that is a maximum
    f"(4?/3) > 0, so that is a minimum

    See the graph at

    http://www.wolframalpha.com/input/?i=x%2B2sinx,+for+x+%3D+0..2pi

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