Find the values where the tangent line is horizontal

frosty8688
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1. Find the values for x at which the tangent line is horizontal


2. [itex]f(x) = x + 2sinx[/itex]

3. I found the derivative to be [itex]f'(x) = 1 + 2cosx[/itex] I then set the derivative equal to zero and it came out to be [itex]2cosx = -1, cosx = -\frac{1}{2}[/itex] So the values of the horizontal tangent are 2∏/3 ± 2∏
 
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I think you exchanged some numbers in the result. In addition, there are two different sets of solutions, not just one.
 
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.
 
That is better. "at every 2∏" should get some integer k (or whatever) as arbitrary factor for the 2∏.
 

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