Calculus of a Single Variable, Chapter 3, 3.1, Section 3.1, Problem 33

Given the function: f(x)=sin(x) in the closed interval [5 pi/6 , 11 pi/6]
We have to find the absolute extrema of the function on the closed interval.
Step 1: Check whether f(x) is a continuous function.
Here f(x) is continuous since all sinusoidal function are continuous functions.
Step 2: Find the critical points.
To find the critical points we have to differentiate the function f(x)=sin(x) and equate it to zero.
So f'(x)=cos(x)=0 implies x= n pi/2
But we need only the critical points that fall inside the closed interval
[5 pi/6 , 11 pi/6]
Therefore the critical points are: pi and 3pi/2 .
Step 3: Now we have to find the values of the function at the two critical points and the end points of the interval.
f(pi)=sin(pi)=0
f((3pi)/2)=sin((3pi)/2)=-1
f((5pi)/6)=sin((5pi)/6)=1/2
f((11pi)/6)=sin((11pi)/6)=-1/2
Absolute extrema are the largest and the smallest the function will ever be. So we can see that the absolute maximum of f(x) is 1/2and it occurs at x= 5 pi/6(end point) and the absolute minimum of f(x) is -1 and it occurs at x= 3 pi/2 (critical point).