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Math Help - Turning points of y = a sin(2x) + 2 cos(x)

  1. #1
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    Turning points of y = a sin(2x) + 2 cos(x)

    1)
    Given function: y=a\sin(2x)+2\cos(x) between 0\leq x \leq \pi
    The function has a stationary point on x=\frac{5}{6}\pi

    A) Calculate a.
    B) Find the vertex of the function in this domain.
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  2. #2
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    Quote Originally Posted by mathwhat View Post
    1)
    Given function: y=a\sin(2x)+2\cos(x) between 0\leq x \leq \pi
    The function has a stationary point on x=\frac{5}{6}\pi

    A) Calculate a.
    B) Find the vertex of the function in this domain.
    Start by finding dy/dx. Can you do that?
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  3. #3
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    Quote Originally Posted by mathwhat View Post
    1)
    Given function: y=a\sin(2x)+2\cos(x) between 0\leq x \leq \pi
    The function has a stationary point on x=\frac{5}{6}\pi

    A) Calculate a.
    B) Find the vertex of the function in this domain.
    y' \left(\frac{5\pi}{6}\right) = 0

    Use the chain rule to differentiate y.
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  4. #4
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    i already found the a, a=1
    my problem is with B.
    i got:
    y'=2\cos(2x)-2\sin(x)=0
    but i dont getting \frac{5}{6}\pi as answer.
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  5. #5
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    Quote Originally Posted by mathwhat View Post
    i already found the a, a=1
    my problem is with B.
    i got:
    y'=2\cos(2x)-2\sin(x)=0
    but i dont getting \frac{5}{6}\pi as answer.
    It saves time and effort if you say in your question what you can already do.

    For B) your job is to find the value of y when x = pi/6. Where are you stuck?
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  6. #6
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    in finding the x from: y'=2\cos(2x)-2\sin(x)=0
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  7. #7
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    Quote Originally Posted by mathwhat View Post
    in finding the x from: y'=2\cos(2x)-2\sin(x)=0
    You asked this question here: http://www.mathhelpforum.com/math-he...tml#post517004

    and got a reply. The reply you were given had a simple error in it that you should have realised if you followed the working closely.

    Thread closed.
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