`y = x + sin(x), 0<=x<2pi` Determine the point(s) (if any) at which the graph of the function has a horizontal tangent line.
We will need to take the derivative of the function and set the derivative equal to zero.
`y' = 1+cos(x)`
`0 = 1+cos(x)`
The value of `x` is the angle where we have a point on the unit circle with an x coordinate value of -1.
The domain given exists from `[0,2 pi)` .
The only value that the unit circle will have a `(-1,0)` point is when the angle is `pi` radians, or 180 degrees.
We can work our way backward and find that in both radians and degrees ,respectively:
`cos(pi)=cos(180) = -1`
Therefore, the value of `x` in radians where we have a slope of zero is:
Substitute this angle back to the original function to find the point.
The value of sine at `pi` radians or 180 degrees is 0.
The exact point existent on the given domain would be: