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Math Help - Find the coordinates of the points on the curve

  1. #1
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    Question Find the coordinates of the points on the curve

    I've encountered the following question:

    Find the coordinates of the points on the curve y=\frac{cos(x)}{2+sin(x)}, 0 \le x < 2\pi, where the tangent is horizontal.

    My plan was to take the derivative and then try to find where x goes to infinity by taking the limit. Things quickly fell apart.

    I managed to get the derivative of f'(x)=\frac{-sin^2(x)-2sin(x)-cos^2(x)}{(2+sin(x))^2} however when I check my work with WolframAlpha I'm told I'm wrong with f'(x)=-\frac{2sin(x)+1}{(2+sin(x))^2}. Now I can see the identity. However I see -1 and WolframAlpha is saying +1. My train of thought is -sin^2(x)-cos^2(x)=-1. So where is my basic algebra failing me with +1?

    Furthermore, is my thought process correct by thinking I need to find the limit as x\rightarrow\infty?
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  2. #2
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    Re: Find the coordinates of the points on the curve

    Your derivative and Wolfram's agree. Note that in the numerator -\sin^2{x}-\cos^2{x}-2\sin{x}=-(2\sin{x}+1)

    Not sure why you would want to be concerned about the limit of y as x goes to infinity. Think more about the slope of the tangent line. What is the slope of a horizontal tangent line? Now that you have the derivative of the curve, what does the derivative tell you about slope?
    Thanks from Maskawisewin
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  3. #3
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    Re: Find the coordinates of the points on the curve

    I knew it was the same up until that point, I just didn't see the factor and how wolfram got there.

    I think I see what to do now. Set the derivative to zero and solve?
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  4. #4
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    Re: Find the coordinates of the points on the curve

    Your graph has a min and a max point for 0<x<2Pi
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  5. #5
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    Re: Find the coordinates of the points on the curve

    Thanks I figured it out.
    Last edited by Maskawisewin; August 17th 2012 at 02:47 PM. Reason: Solved. I had asked a follow up question but I solved it.
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