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Maxima minina on an interval (calculus+trig)

  1. Jul 25, 2017 #1
    IMG_2066.jpg I need to find the max/min of a function on an interval.
    The function is f(x)=x+cos(x) and the interval is <-PI,2pi>
    There is an attached solution but I do not understand how to arrive at the given solution (see screenshot). I would personally just take the derivative as
    F'(x)=1-sin(x)
    However the solution says this is only half the answer and I do not understand the reasoning (I am trying to do this algebraicly, without thinking of it graphically).
     
  2. jcsd
  3. Jul 25, 2017 #2

    haruspex

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    I think you may be misinterpreting what it says. It is strangely worded. Better would be "f'(x)=1-sin(x), which is ≥0 since sin(x)≤1 for any real x".
     
  4. Jul 25, 2017 #3
    Thanks
     
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