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Math Help - Finding stationary points

  1. #1
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    Finding stationary points

    I can't figure this out at all, I've got the derivative

    g'(x)=\frac{1}{2}(4cos^{2}x+4cosx-3)e^{\frac{1}{2}x+cosx}

    and the domain of the function g is [0,2\pi]

    I've factorised to get

    (2cosx-1)(2cosx+3)=0

    But when I put the second bracket as zero to find x for that bracket I get

    x=arccos(\frac{-3}{2})

    which doesn't get me anywhere on my calculator.

    Have I made a mistake somewhere?

    Many thanks.
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  2. #2
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    Re: Finding stationary points

    Why on Earth would you try to find the arc-cosine of -3/2?! The Domain of arccos(x) is [-1,1]. Never do that again.
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  3. #3
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    Re: Finding stationary points

    Sorry but I am still learning.

    I'm pretty sure my factorising is correct, and making the first bracket equal to zero is giving me

    2cosx-1=0

    2cosx=1

    cosx=\frac{1}{2}

    x=arccos(\frac{1}{2})\approx1.047

    which gives me one stationary point, but for the second bracket, like you say, I'm outside of the domain for arccos.

    Any help?
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  4. #4
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    Re: Finding stationary points

    What kind of help do you need? You have correctly determined that the derivative is 0 only when 2 cos(x)- 1= 0 or when 2cos(x)+ 3= 0. Since the range of cosine is -1 to 1, there is NO value of x that makes 2cos(x)+ 3= 0 so the only possible critcal point is where 2 cos(x)- 1= 0.
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  5. #5
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    Re: Finding stationary points

    cos{x} = \frac{1}{2} occurs at two angles on the unit circle ...

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  6. #6
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    Re: Finding stationary points

    Quote Originally Posted by HallsofIvy View Post
    What kind of help do you need? You have correctly determined that the derivative is 0 only when 2 cos(x)- 1= 0 or when 2cos(x)+ 3= 0. Since the range of cosine is -1 to 1, there is NO value of x that makes 2cos(x)+ 3= 0 so the only possible critcal point is where 2 cos(x)- 1= 0.
    Thank you so much,

    All night last night I spent on that thinking I had to find two stationary points.

    Thanks again.
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  7. #7
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    Re: Finding stationary points

    Quote Originally Posted by Srengam View Post
    Thank you so much,

    All night last night I spent on that thinking I had to find two stationary points.

    Thanks again.
    Look at skeeter's post again. There are two stationary points in [0, 2\pi].
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