Calculus: Early Transcendentals, Chapter 4, 4.1, Section 4.1, Problem 56

f(t)=root(3)(8-t)
differentiating,
f'(t)=(1/3)(8-t)^(1/3-1)(-1)
f'(t)=-1/(3(8-t)^(2/3))
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/(3(8-t)^(2/3))=0
So, f'(t)=0 is not defined for any value of t.
Now evaluate the function at the end points of the interval.
f(0)=root(3)(8-0)=2
f(8)=root(3)(8-8)=0
So. the absolute maximum=2 at x=0
and absolute minimum=0 at x=8