## anonymous 3 years ago Find all points (i.e. all x-values) between 0 and 2pi where the line tangent to the graph of y =sinx/2+cosx is horizontal

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1. tkhunny

Have you considered the 1st Derivative?

2. anonymous

1st derivative would be -(cot x) (cosx/-sinx) how would I get the points from that?

3. tkhunny

If that is the correct 1st derivative, which I didn't check, you determine where it is zero.

4. anonymous

well the derivative of sinx is cosx and the derivative of any constant is always 0 and the derivative of cosx is -sinx and we know that cotx =cosx/sinx

5. anonymous

I am not sure how I would determine where it would be zero..

6. anonymous

horizontal tg = 0(no variation)

7. tkhunny

No. That will not do. You need the Quotient Rule. I get the 1st Derivative as $$\dfrac{2\cos(x)+1}{(2+\cos(x))^{2}}$$. After that, simply concern yourself with the numerator being zero, unless the denominator happens to be zero in the same place.

8. tkhunny

This assumes your original expression was $$\dfrac{\sin(x)}{2+\cos(x)}$$. Was it?

9. anonymous

|dw:1372217350005:dw|

10. anonymous

yes tkhunny that was the original equation

11. tkhunny

Then I gave the correct 1st Derivative and you should use parentheses to clarify your meaning. Remember your Order of Operations.

12. tkhunny

gtg sorry.

13. anonymous

rapahael can you still help?

14. anonymous

yes

15. anonymous

so do I need to set the numerator =0

16. anonymous

yes, because if the tangent line is horizontal, his angular coefficient is zero

17. anonymous

so then I end up with cosx=-1/2

18. anonymous

yes

19. anonymous

so is that my only x value , -1? ( my reasoning behind that is that cos of 60degrees=1/2) so then y x value is -1?

20. anonymous

|dw:1372217864898:dw|

21. anonymous

pi-pi/3,pi+pi/3

22. anonymous

@trusev1 get it?

23. anonymous

I get that but it is asking for the x-values that are at the line tangent is horizontal. so we got the derivative and even from your picture both those points would have the same x-value.

24. anonymous

no. my figure isnt the graph of your question,it's just the trigonometric circle