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Homework Help: A derivative

  1. Jan 15, 2010 #1
    1. The problem statement, all variables and given/known data
    I need to use Rolle's theorem to find analitically the maxium and the minimun extremes on the interval [0 , 1/2pi] .

    So I think I must first find the derivative of the fuction and to solve for x

    the function given is f(x) = sin 2x

    2. Relevant equations

    3. The attempt at a solution

    Im not sure about how to start the product rule

    is it like this?

    sin x (2x)

    then the product rule would be

    sin x (2) + 2x (cos x) ?

    I have no idea about how to solve for x.
  2. jcsd
  3. Jan 15, 2010 #2


    User Avatar
    Science Advisor
    Homework Helper

    There is no product rule to be applied.
    sin(2x) is something different than sin(x) * 2x.
    If you call f(x) = sin(x), this is easier to see (f(2x) and 2 x f(x) are completely different things, and the derivative of f(2x) is not 2f(x) + 2x f'(x), or something like that).

    What you are looking for is the chain rule.
  4. Jan 15, 2010 #3
    ok using f ' (g) g'(x) resulted on cos(2x) 2 that is I think 2 cos 2x .

    Now Im not sure about what to do next in order to look for the maximum and minimum extremes on the fucntion.

    On previous examples that were not trigonometric functions I solved for x on the original function and I also solved for x on the derivative and I found the critical points that way.

    How can I find the critical points of f ?
  5. Jan 15, 2010 #4


    Staff: Mentor

    What does Rolle's Theorem say? You have a function, f(x) = sin(2x) for which f(0) = f(pi/2) = 0, and f is a differentiable function (you found its derivative).
  6. Jan 15, 2010 #5
    So the zeros are 0 and pi/2 . The function is continous on [0,pi/2] and differentiable on

    Well I graphed the function and the maximum extreme is obiously 1 but if I evaluate
    2cos(2x) that is the derivtive of f by 0 I get

    2cos(0) = 2

    what am I doing wrong?
  7. Jan 15, 2010 #6


    Staff: Mentor

    What does Rolle's Theorem conclude?
    Why would you want to evaluate f'(0)?
  8. Jan 15, 2010 #7


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    Homework Helper

    You've shown that the derivative of sin(2x) is 2 at x=0. Nothing in math says that the derivative can't be larger than the maximum value of the function.

    Draw out a rough sketch of sin(2x). What's the slope of the tangent line equal to at the highest and lowest points?

    Note that you should be able to solve this question intuitively, without using calculus. What's the maximum and minimum values of any sine function?
  9. Jan 15, 2010 #8
    The derivative of any maximum or minimum equals to 0
    and the pourpose of Rolle's theorem as far as I know is to show that there is a number c in the interval [0,pi/2] such that f '(c) = 0

    Im being asked to find c as well. how can I find c ?

    I think c is called a critical point as well.
  10. Jan 15, 2010 #9
    by the way the answer in my book is 1/4pi I dont understand how that happened.
  11. Jan 15, 2010 #10


    Staff: Mentor

    This is not necessarily true. For example, the minimum value of y = f(x) = |x| is 0, and it occurs at x = 0, but f'(x) is never equal to 0 for this function.
    Close. Rolle's Thm says that there is a number c in the open interval (0, pi/2) such that f'(c) = 0.

    That means that for some number 0 < c < pi/2, 2cos(2c) = 0. Can you solve this equation?
  12. Jan 15, 2010 #11
    Thanks, thats right I think that f '(x) = |x| is undefined at x= 0 .

    no, I cant solve that equation this is kind of new to me. thats the main issue im having with this problem so I dont have a clue. I dont remember ever solving variable equations like that.

    I guess that by guessing it could be factorized like this

    2 (cos (c)) = 0

    but thats about it.
    Last edited: Jan 15, 2010
  13. Jan 15, 2010 #12
    I dont think that this is right but anyway

    2cos(2c) = 0

    2c = 1/2 cos x = c = 1/4 cos x

    but I dont understand where did pi came from in c = 1/4pi which is the original result of the book.

    Can you give me a hint?
  14. Jan 15, 2010 #13


    Staff: Mentor

    No, f(x) = |x| is defined for all values of x, but for this function, f' is not defined at x = 0.
  15. Jan 15, 2010 #14


    Staff: Mentor

    Yes - it's not right.

    2cos(2c) = 0
    ==> cos(2c) = 0

    Now, I'm hopeful that you have seen the graph of y = cos(x).
    For what values of x is cos(x) = 0? One of them is pi/2 and there are lots and lots of values.

    The same values of x for which cos(x) = 0 are the same values of 2c, so what are the values of c?
  16. Jan 15, 2010 #15
    Ok, thanks for your time Mark, take care.
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