Find the coordinates of the points on the curve

**Maskawisewin** · August 15th 2012, 06:12 PM

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 $x\rightarrow\infty$ ?

**rainer** · August 15th 2012, 08:58 PM

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?

**Maskawisewin** · August 16th 2012, 09:49 AM

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?

**MaxJasper** · August 16th 2012, 10:25 AM

Your graph has a min and a max point for 0<x<2Pi

**Maskawisewin** · August 17th 2012, 01:40 PM

Thanks I figured it out.

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