- #1
frosty8688
- 126
- 0
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∏
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∏