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Homework Help: Find the max and min values

  1. Jul 11, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the maximum and minimum values of the function f(x) = 2sin2x-2sinx on the interval [-pi, pi]]


    2. Relevant equations
    first derivative


    3. The attempt at a solution
    f'(x) = 4sinxcosx-2cosx
    0 = 2cosx (2sinx-1)
    0 = 2sinx-1, sinx = 1/2, x = pi/6, 5pi/6
    0 = 2cosx, cosx = 0, x = pi/2

    I got that MIN = -0.5 (plugged pi/6 and 5pi/6 into the original function) but I can't seem to find where MAX @4 is.

    Thank you!

    Thank you!
     
    Last edited: Jul 11, 2011
  2. jcsd
  3. Jul 11, 2011 #2
    Hi Glissando! :smile:


    You forgot -pi/2 here, no?

    Anyway, that would give you 6 critical values (don't forget to add the boundaries of the interval as critical values!!!!):

    [tex]-\pi,-\frac{\pi}{2},\frac{\pi}{6},\frac{\pi}{2}, \frac{5\pi}{6},\pi[/tex]

    Now you need to calculate f of all these values and find out which one is the greatest!
     
  4. Jul 11, 2011 #3

    Ray Vickson

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    Homework Helper

    Put y = sin(x), to get the function 2*y*(y-1) on the interval -1 <= y <= 1. The min occurs at y = 1/2. You have an "end point" max, where the derivative (with respect to y) need not equal zero.

    RGV
     
    Last edited: Jul 11, 2011
  5. Jul 11, 2011 #4
    Thank you (: Figured it out!
     
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