Calculus of a Single Variable, Chapter 3, 3.1, Section 3.1, Problem 35

Given: y=3cos(x),[0,2pi]
First find the critical x values of the function. To find the critical values, set the derivative equal to zero and solve for the x value(s).
f'(x)=-3sin(x)=0
-3sin(x)=0
sin(x)=0
x=0,pi,2pi
Plug in the critical values and the endpoints of the closed interval in to the original function.
f(x)=3cos(x)
f(0)=3cos(0)=3(1)=3
f(pi)=3cos(pi)=3(-1)=-3
f(2pi)=3cos(2pi)=3(1)=3
Examine the f(x) values.
The absolute maximum is at the points (0,3) (2pi,3)
The absolute minimum is at the point (pi,-3)