Calculus: Early Transcendentals, Chapter 4, 4.1, Section 4.1, Problem 58

f(t)=t+cot(t/2)
differentiating,
f'(t)=1-1/2csc^2(t/2)
Now to find the absolute extrema of the function , that is continuous on a closed interval, we have to find the critical numbers that are in the interval and evaluate the function at the endpoints and at the critical numbers.
Now to find the critical numbers, solve for t for f'(t)=0.
1-1/2csc^2(t/2)=0
csc^2(t/2)=2
1/(sin^2(t/2))=2
sin^2(t/2)=1/2
sin(t/2)=+-1/sqrt(2)
sin(t/2)=1/sqrt(2) : t=pi /2 , 3pi /2 in the interval( pi /4 , 7pi /4)
sin(t/2)=-1/ sqrt(2) : No solution
So the critical numbers are pi /2 and 3pi /2
Evaluating the function at the critical points and at the end points of the interval,
f(pi/2)=pi/2+cotpi/4=pi/2+1=2.5708
f(3pi/2)=3pi/2+cot(3pi/4)=3pi/2-1=3.712
f(pi/4)=pi/4+cot(pi/8)=pi/4+1+sqrt(2)=3.19
f((7pi)/4)=(7pi)/4+cot(7pi/8)=(7pi)/4-1-sqrt(2)=3.083
So the absolute maximum=(3pi /2-1) at t=3pi /2
and absolute minimum=(pi /2+1) at t=pi /2