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Max and min of a function

  1. Aug 8, 2007 #1
    how would i find something like:

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

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

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

    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 + [tex]\sqrt\frac{3}{2}[/tex]

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

    f(pi) = pi/2 - 1

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

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

    critical points are again [tex] x=\frac{\pi}{6} and \frac{5}{6}\pi[/tex], but these both lie outside the range: [tex][-\pi,0][/tex] 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?
  2. jcsd
  3. Aug 8, 2007 #2
    Your results at each of the four points for the first one are a little off, but that could just be latex. cos(pi/6) = sqrt(3)/2, not sqrt(3/2). However, your conclusions are correct.
  4. Aug 8, 2007 #3
    you're right there. the answers are correct as dave said.
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