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Math Help - maximum and minimum of a function

  1. #1
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    maximum and minimum of a function

    how would i find something like:

     max[\frac{x}{2} + cos(x)] where  x\epsilon [0,\pi] and ...

     max[\frac{x}{2} + cos(x)] where  x\epsilon [-\pi,0]
    -------------------------------------------
    here's what i did for  max[\frac{x}{2} + cos(x)] where  x\epsilon [0,\pi]:
    first we need critical points: so f'(x) = 0

    so 1/2 - sin(x) = 0 , so  x=\frac{\pi}{6} and \frac{5}{6}\pi

    maxf(x) = max{f(0), f(pi/6), f(5/6pi), f(pi)}
    f(0) = 1/2(0) + cos(0) = 1
    f(pi/6) = 1/2(pi/6) + cos(pi/6) = pi/12 + \sqrt\frac{3}{2}

    f(5/6pi) = 5/12pi + cos(5/6pi) = 5/12pi - \sqrt\frac{3}{2}

    f(pi) = pi/2 - 1

    so max is at f(pi/6) .. i.e pi/12 + \sqrt\frac{3}{2}

    ---------------------------
    now for  max[\frac{x}{2} + cos(x)] where  x\epsilon [-\pi,0]

    critical points are again  x=\frac{\pi}{6} and \frac{5}{6}\pi, but these both lie outside the range: [-\pi,0] so we only take into consideration -pi and 0:

    so f(-pi) = -pi/2 - 1
    f(0) = 1

    so the max is at point f(0).. i.e 1..

    ........................................
    are these both correct? although seeming a bit obvious.. or do i need to show the final answer differently?
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by physics4life View Post
    how would i find something like:

     max[\frac{x}{2} + cos(x)] where  x\epsilon [0,\pi] and ...

     max[\frac{x}{2} + cos(x)] where  x\epsilon [-\pi,0]
    -------------------------------------------
    here's what i did for  max[\frac{x}{2} + cos(x)] where  x\epsilon [0,\pi]:
    first we need critical points: so f'(x) = 0

    so 1/2 - sin(x) = 0 , so  x=\frac{\pi}{6} and \frac{5}{6}\pi

    maxf(x) = max{f(0), f(pi/6), f(5/6pi), f(pi)}
    f(0) = 1/2(0) + cos(0) = 1
    f(pi/6) = 1/2(pi/6) + cos(pi/6) = pi/12 + \sqrt\frac{3}{2}

    f(5/6pi) = 5/12pi + cos(5/6pi) = 5/12pi - \sqrt\frac{3}{2}

    f(pi) = pi/2 - 1

    so max is at f(pi/6) .. i.e pi/12 + \sqrt\frac{3}{2}

    ---------------------------
    now for  max[\frac{x}{2} + cos(x)] where  x\epsilon [-\pi,0]

    critical points are again  x=\frac{\pi}{6} and \frac{5}{6}\pi, but these both lie outside the range: [-\pi,0] so we only take into consideration -pi and 0:

    so f(-pi) = -pi/2 - 1
    f(0) = 1

    so the max is at point f(0).. i.e 1..

    ........................................
    are these both correct? although seeming a bit obvious.. or do i need to show the final answer differently?
    It looks good to me.

    -Dan
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