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Maximum and minimum value

  1. May 16, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the maximum and minimum value of:
    (sin a)^4 - (sin a)^2


    I am new to the world of calculus. I have just learned to evaluate the derivatives of functions. (*and nothing else- I am a high school student) For this function the derivative is
    -sin 2a . cos 2a
    Please tell me if I am right.

    While teaching my teacher told me that we can obtain the maximum and minimum values of a function with the help of its derivative. Is it true?
    If yes, please spare some time to tell me how is it possible. I know that the maximum and minimum values can be calculated by coverting this into a perfect square. But I wish to learn it through calculus, if possible.

    Thanks!
    regards,
    Ritwik
     
  2. jcsd
  3. May 16, 2008 #2

    tiny-tim

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    Hi ritwik06! :smile:
    Yes, that's fine! :smile:
    The derivative is the rate at which the function increases.

    At a maximum or minimum, the function has stopped increasing, so the derivative is zero. :smile:
     
  4. May 16, 2008 #3

    ok. thanks a lot tim.

    but the problem now is that if I put:

    - sin 2a . cos 2a=0
    the values of 'a' that are possible are:
    0, 180 ,360, ......
    45, 135, 225.....
    at which value will I get the maximum and at which value the minimum???
     
  5. May 16, 2008 #4
    similarly if I want to fin th maximum and minimum of sin x+cos x

    dy/dx=cos x - sinx

    cos x - sin x=0
    cos x=sin x
    x=45, 225... (till what values of x should I limit myself??)
     
  6. May 16, 2008 #5
    Do I need to check from [0,2*pi]????
    Because after that. the values of the T-ratios repeat. Am I right?
     
  7. May 16, 2008 #6

    HallsofIvy

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    Staff Emeritus
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    The first thing you need to do is rewrite those x values in terms of radians rather than degrees!

    And you had better learn to start thinking in terms of radians. For the functions sin(x) and cos(x), the "x" has to be interpreted in radians, not degrees.
     
  8. May 16, 2008 #7
    Well, you will get an infinite number of local extrema, because the sin and cosine are periodic functions. That being said, half will be local maxima, and half will be local minima.

    You have to use the second derivative test to determine whether each critical point represents a local minimum (second derivative > 0 at the x), a local maximum (second derivative < 0 at the x) or neither (second derivative = 0 at the x).

    But yes, you will have to give your answer like...

    f(x) = sin x

    Maxima: 1/2 PI, 5/2 PI, 9/2 PI, ...
    Minima: 3/2 PI, 7/2 PI, 11/2 PI, ...

    (or)

    Maxima: (4n-3)/2 PI, for all natural numbers n
    Minima: (4n-1)/2 PI, for all natural numbers n

    Hopefully that clarified what was being asked.
     
  9. May 16, 2008 #8

    tiny-tim

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    Welcome to PF!

    Hi csprof2000! Welcome to PF! :smile:

    Don't too much work … best just to give a hint and see if the OP can work out the rest themselves! :wink:
     
  10. Nov 18, 2008 #9
    Thanks A Lot!
     
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