f(x) = cos(x)
f'(x)=-sin(x) = 0
sin(x) = 0
x=0 + 2nPI, where n is any integer
x= 2PI when n=1
x=PI+2nPI, where n is any integer
Let n=1
x=PI+2PI= 3PI
x=PI,2PI,3PI are solutions
Alternate method of finding extrema: If f(x) is continuous in a closed interval I, then the absolute extrema of f(x) in I occur at the critical points and/or at the endpoints of I.
Therefore, the minimum and maximum occur at critical points PI,2PI,3PI
