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Find Absolute Extrema on a Closed Interval

  1. Dec 10, 2006 #1
    1. The problem statement, all variables and given/known data
    Find the absolute max and min of h(x) = cos(2x) - 2sinx in the closed interval [(pie)/2, 2pie]

    3. The attempt at a solution
    I got
    h(x) = cos(2x) - 2sinx
    h'(x) = -sin(2x)*2 - 2cosx <---chain rule i did :rofl:

    0=-2sin(2x) - 2cosx
    0=[-2sin(2x)/-2] - [2cosx/-2] i divided both sides by -2
    0=cosx(2sinx+1) <---- factored out cosx

    cosx = 0

    and 2sinx+1 = (-1/2)

    The critical numbers i found out where 0, and -0.5

    I dont know if im right or wrong

    The book said the critical numbers are at [3(pie)/2], [7(pie)/2], and [11(pie)/2]
    ^--- i dont know how the book got those answers, can u help me out please
    i understand that all the critical numbers are zero
    Last edited: Dec 11, 2006
  2. jcsd
  3. Dec 10, 2006 #2


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    The critical points are where the derivative is zero, and you correctly found these occur where cosx=0 or 2sinx+1=0 (sinx=-1/2) (which you probably meant for the second one). So where are these points? Then compare the values at these critical points and at the endpoints, and the largest/smallest of these is the max/min on the interval.
  4. Dec 10, 2006 #3


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    You've calculated the expression correctly, and identified that one solution is cosx=0. However, your other solution is incorrect.. it should be 2sinx+1=0.

    Your critcal values are incorrect. The second one is incorrect because you had the sin eqn wrong. However, for the first equation, consider the interval that you give in the question. Is 0 in this integral?

    Incidentally, have you written down the critical values obtained from the book incorrectly?

    EDIT: sorry, the above post wasn't there when I viewed!
  5. Dec 11, 2006 #4
    Then how did the book get the three points

    [3(pie)/2], [7(pie)/2], and [11(pie)/2] <--- critical points that equal zero

    i understand the process of which i got what i got, but then these 3 points came outta right feild, any help please?
  6. Dec 11, 2006 #5


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    Because they have cosx=0.
  7. Dec 11, 2006 #6
    so cosx always equals those points on a closed interval, with sin? Always?
    Last edited: Dec 11, 2006
  8. Dec 11, 2006 #7


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    Please rephrase that so it makes sense.
  9. Dec 11, 2006 #8
    so because i got

    cosx=0 the critical numbers are at these points
    [3(pie)/2], [7(pie)/2], and [11(pie)/2]

    ^-- but why?:yuck:

    is there a chart r table that i can reference from to say that when you get cosx = 0 or whatever else, your points will be at here etc
  10. Dec 11, 2006 #9
    sorry if im not being clear, its just that the book showed no operations for getting

    [3(pie)/2], [7(pie)/2], and [11(pie)/2]

    ^-- cause later on for other parts of the questions i need these numbers, but if i dont know where they came from then well you know
    Last edited: Dec 11, 2006
  11. Dec 11, 2006 #10


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    Are we not looking for x in the interval (pi/2,2pi) (presumably open, since pi/2 is not given by the book as one of the answers!)?

    That's why I asked if the other critical values were copied down incorrectly. Should they not be 7pi/12 and 11pi/12 (corresponding to the solutions of sinx=-1/2 in this region.)
  12. Dec 11, 2006 #11
    and 2sinx+1 = (-1/2)

    meant to equal

    sinx = -(1/2) yes i did copy it down incorrectly, sorry

    where can i find these feilds you talk of?
  13. Dec 11, 2006 #12


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    What do you mean by fields?

    To solve the equation sinx=-1/2, consider first the solutions to sinx=1/2. To solve this, consider an equilateral triangle, of side 2. If you draw a line down the centre you will have a right angled triangle of sides 1, 2 and sqrt3.

    Now, can you find one answer to sinx=1/2? When you have this, sketch the graph of sinx. This should help you find soltions to the required equation (sinx=-1/2) in the given interval
  14. Dec 11, 2006 #13


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    There are some values of sin and cos you should just remember. Like cos(pi/2)=0, and sin(pi/6)=1/2. Also remember the formulas for sin(x+2pi), sin(pi-x). And yes, if you're restricting to the range [pi/2,2pi], then, of the book's answers, only 3pi/2 is in this range (though this isn't the only critical point).
  15. Dec 12, 2006 #14
    so if i got (-1/2) the critical points would be at 7(pi)/6 and 11(pi)/6

    So for example if got (1/2) as my final answer, my critical points would be at (pi)/6, and

    am i right?

    Also too are there 6 different circles for the 6 different trig functions?


    sorry im a idiot
    Last edited: Dec 12, 2006
  16. Dec 12, 2006 #15


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    Is it really so difficult for you to write what you mean?
    "So for example if got (1/2) as my final answer, my critical points would be at (pi)/6, and 5(pi)/6?"
    Your final answer? I assume you mean "if sin(x)= 1/2 then x= pi/6 or x= pi- pi/6= 5pi/6". Yes, that is true.

    I have no idea what you mean by "6 different circles". In the "circle" definition of trig functions (often called "circular functions" when that definition is used), we are given the unit circle in an xy-coordinate system. For any t>= 0, start at the point (1, 0) and measure, counterclockwisem, around the circle a distance t. The coordinates of the resulting point are, by definition, (cos(t), sin(t)). (If t< 0, measure clockwise.) You can then define tan(t)= sin(t)/cos(t), cot(t)= cos(t)/sin(t), sec(t)= 1/cos(t), and csc(t)= 1/tan(t). No, you don't need "6 different circles"!
  17. Dec 12, 2006 #16
    omg sorry, i just wanted to find were the critical points were, but at the time i wasnt familiar with the unit circle all that much

    but now im on the right track thanks to you guys, sorry if i didnt make sense i really didnt understand the 3(pi)/2 or at least didnt know what they meant but now i do

    i will continue to research the unit circle and i will master it

    thank you for you help o:)

    P.S. - sorry
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