Question

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

Answer 1

Have you considered the 1st Derivative?

Answer 2

1st derivative would be -(cot x)
(cosx/-sinx)

how would I get the points from that?

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

horizontal tg = 0(no variation)

Answer 7

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.

Answer 8

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

Answer 9

yes tkhunny that was the original equation

Answer 10

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

Answer 11

gtg sorry.

Answer 12

rapahael can you still help?

Answer 13

yes

Answer 14

so do I need to set the numerator =0

Answer 15

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

Answer 16

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

Answer 17

yes

Answer 18

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?

Answer 19

pi-pi/3,pi+pi/3

Answer 20

@trusev1 get it?

Answer 21

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.

Answer 22

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