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March 30, 2017

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find the absolute maximum and minimum for f(x) = x+ sin(x) on the closed interval [0, 2pi]

  • calc - ,

    f' = 1 + cos(x)
    f'=0 when cos(x) = -1, or x=pi

    however,
    f''(x) = -sin(x) which is also 0 at x=pi.

    So, (pi,pi) is an inflection point, not a max or min.

    f(2pi) = 2pi

    so that is the absolute max on [0,2pi]

    Note that f'(x) >= 0 for all x, so there is no max or min, which require changing direction.

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