Calculus of a Single Variable, Chapter 3, 3.1, Section 3.1, Problem 36

Given the function y=tan((pi x)/8) in [0,2].
We have to find the absolute extrema of the function in the closed interval [0,2].

So first we have to take the derivative of the function and equate it to zero.
i.e y'=pi/8 sec^2((pi x)/8)=0
sec^2((pi x)/8)=0
Isolating sec((pi x)/8)=0
sec(x) cannot be equal to 0 because its value ranges between 1 to +infinity and -1 to infinity only.
So the function does not have a critical point.
So we will calculate the absolute extremas at the end points only.
The absolute minima is y= 0 at x=0 and the absolute maxima is y=1 at x=2.