# Thread: Find the coordinates of the points on the curve

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

I've encountered the following question:

Find the coordinates of the points on the curve $\displaystyle y=\frac{cos(x)}{2+sin(x)}, 0 \le x < 2\pi$, 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 $\displaystyle f'(x)=\frac{-sin^2(x)-2sin(x)-cos^2(x)}{(2+sin(x))^2}$ however when I check my work with WolframAlpha I'm told I'm wrong with $\displaystyle f'(x)=-\frac{2sin(x)+1}{(2+sin(x))^2}$. Now I can see the identity. However I see -1 and WolframAlpha is saying +1. My train of thought is $\displaystyle -sin^2(x)-cos^2(x)=-1$. 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 $\displaystyle x\rightarrow\infty$?

2. ## Re: Find the coordinates of the points on the curve

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

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?

3. ## 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?

4. ## Re: Find the coordinates of the points on the curve

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

5. ## Re: Find the coordinates of the points on the curve

Thanks I figured it out.