anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
tkhunny
  • tkhunny
Have you considered the 1st Derivative?
anonymous
  • anonymous
1st derivative would be -(cot x) (cosx/-sinx) how would I get the points from that?
tkhunny
  • tkhunny
If that is the correct 1st derivative, which I didn't check, you determine where it is zero.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • 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
anonymous
  • anonymous
I am not sure how I would determine where it would be zero..
anonymous
  • anonymous
horizontal tg = 0(no variation)
tkhunny
  • 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.
tkhunny
  • tkhunny
This assumes your original expression was \(\dfrac{\sin(x)}{2+\cos(x)}\). Was it?
anonymous
  • anonymous
|dw:1372217350005:dw|
anonymous
  • anonymous
yes tkhunny that was the original equation
tkhunny
  • tkhunny
Then I gave the correct 1st Derivative and you should use parentheses to clarify your meaning. Remember your Order of Operations.
tkhunny
  • tkhunny
gtg sorry.
anonymous
  • anonymous
rapahael can you still help?
anonymous
  • anonymous
yes
anonymous
  • anonymous
so do I need to set the numerator =0
anonymous
  • anonymous
yes, because if the tangent line is horizontal, his angular coefficient is zero
anonymous
  • anonymous
so then I end up with cosx=-1/2
anonymous
  • anonymous
yes
anonymous
  • 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?
anonymous
  • anonymous
|dw:1372217864898:dw|
anonymous
  • anonymous
pi-pi/3,pi+pi/3
anonymous
  • anonymous
@trusev1 get it?
anonymous
  • 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.
anonymous
  • anonymous
no. my figure isnt the graph of your question,it's just the trigonometric circle

Looking for something else?

Not the answer you are looking for? Search for more explanations.