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Critical Numbers, local extrema, max/min

  1. Nov 21, 2007 #1
    1. The problem statement, all variables and given/known data

    Classify the local extrema of f(x) = x/2 + cos(x), 0 <= x <= 2pi. Give the exact values of the critical numbers and extrema. Find the absolute (or global) maximum and minimum values of the function.

    3. The attempt at a solution

    So i need to find the critical numbers first right? Take the derivative..
    f'(x) = 1/2 - sin(x)
    set equal to zero..
    0 = 1/2 - sin(x) -> 1/2 = sin(x) how do i find x?
     
  2. jcsd
  3. Nov 21, 2007 #2
    arcsin(1/2) = x = pi/6
     
    Last edited: Nov 21, 2007
  4. Nov 21, 2007 #3
    critical numbers: pi/6, 5pi/6
    I drew out this table:
    Interval | Test Value | Sign | Behavior
    (0, pi/6) | pi/10 | + | Inc
    (pi/6, 5pi/6)| pi/2 | - | Dec
    (5pi/6, 2pi) | pi | + | Inc

    f(pi/6) = pi/12 + cos pi/6 = pi/12 + 6sqrt(3)/12 = pi + 6sqrt(3) / 12
    f(5pi/6)= 5pi/12 + [cos 5pi/6 = ???]

    how do i find an exact value for these two?
     
  5. Nov 21, 2007 #4
    The exact value is right as you write it but

    cos(pi/6) = sqrt(3)/2
    and
    cos(5pi/6) = -sqrt(3)/2.

    so

    f(pi/6) = sqrt(3)/2 + pi/12
    f(5pi/6) = 5pi/12 - sqrt(3)/2
     
  6. Nov 21, 2007 #5
    so,

    pi/12 + sqrt(3)/2 is a local max and
    5pi/12 - sqrt(3)/2 is a local min?

    f(0) = 1 is a global min and
    f(2pi) = pi + 1 is a global max?
     
  7. Nov 22, 2007 #6
    When you drew the table, what does it tell you?
     
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