Since the domain of f is R and cos(x) is periodic, then f(x) = arccos(cos(x)) is also a periodic function.
As x increases from 0 to Pi, cos(x) decreases from 1 to 1 and arccos(cos(x)) increases from 0 to Pi. In fact for x in [0 , pi] arccos(cos(x)) = x. As x increases from [Pi , 2pi], cos(x) increases from 1 to 1 and arccos(cos(x)) decreases from Pi to 0.
Since cos(x) has a period of 2Pi, arccos(cos(x)) also has a period of 2Pi. The graph below shows the graphs of arccos(cos(x)) and sin(x) from 0 to 2Pi.
The graph below shows the graphs of arccos(cos(x)) and cos(x) over 3 periods.
Domain of f: (infinity , + infinity)
Range of f: [0 , pi]
Derivative of f(x) = arccos(cos(x))
f(x) is a composite function and the derivative is computed using the chain rule as follows: Let u = cos(x)
Hence f(x) = arccos(u(x))
Apply the chain rule of differentiation
f'(x) = du/dx d(arccos(u))/du = sin(x) * ( 1 / sqrt(1  u^{2}))
= sin(x) * 1 / (1  cos^{2}(x))
= sin(x) / sqrt(sin^{2}(x))
= sin(x) /  sin(x) 
Below is shown arccos(cos(x)) in red and its derivative in blue. Note that the derivative is undefined for values of x for which sin(x) = 0, which means at x = k*pi, where k is an integer. For these same values of x, arccos(cos(x)) has either a maximum value equal to pi or a minimum value equal to 0.
Note that although arccos(cos(x)) is continuous for all values of x, its derivative is undefined at x = k*pi.
More on differentiation and derivatives
