Find the values where the tangent line is horizontal

[frosty8688]

1. Find the values for x at which the tangent line is horizontal


2. f(x) = x + 2sinx

3. I found the derivative to be f'(x) = 1 + 2cosx I then set the derivative equal to zero and it came out to be 2cosx = -1, cosx = -\frac{1}{2} So the values of the horizontal tangent are 2∏/3 ± 2∏

 

[mfb]

I think you exchanged some numbers in the result. In addition, there are two different sets of solutions, not just one.

 

[frosty8688]

The values would be (2n + 1)∏ ± 1/3∏. Since it is -1/2 at 2∏/3 and at 4∏/3 and it occurs at ∏ ± 1/3∏ and at every 2∏ to the left or right.

 

[mfb]

That is better. "at every 2∏" should get some integer k (or whatever) as arbitrary factor for the 2∏.