Calculus of a Single Variable, Chapter 2, 2.2, Section 2.2, Problem 61

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)
cos(x)= -1
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:
x=pi
Substitute this angle back to the original function to find the point.
y=x+sin(x)
y=pi+sin(pi)
The value of sine at pi radians or 180 degrees is 0.
y=pi+0= pi
The exact point existent on the given domain would be:
(pi,pi)