Find the local max and min values of F(x)=x/2-2sin(x/2), 0<x<2pi?

1 Answer

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  • 3 weeks ago

    F'(x) = 1/2 - cos(x/2).

    This will be 0 when cos(x/2) = 1/2,

    so when x/2 = pi/3, 5pi/3, 7pi/3, etc.

    Then x = 2pi/3, 10pi/3, etc.

    But the only such value in the range is 2pi/3.

    At x = 2pi/3, the value of F is pi/3 - sqrt(3).

    This is a local minimum, as F(0) = 0.

    The absolute max occurs at the endpoint x = 2pi, where F(2pi) = pi - 0 = pi. Whether this fits the definition of a "local max" depends what definition you are using!

    • ted s
      Lv 7
      3 weeks agoReport

      1st : F(0) does not exist ; 2nd : F ' ' at the point is > 0 ===> local (here global ) minimum ; 3rd : a global max does not exist on this open interval

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