Calculus of a Single Variable, Chapter 3, 3.1, Section 3.1, Problem 29

Hello!
Consider this function for t in [-1, 3] and [3, 5] separately.
At the first interval t<=3 and |t-3| = -(t-3) = 3-t.So y(t) = 3 - (3-t) = t.The maximum is at the maximum t, i.e. at t=3, y(3) = 3.The minimum is at the minimum t, i.e. at t=-1, y(-1) = -1.
At the second interval t>=3 and |t-3| = t-3.So y(t) = 3 - (t-3) = 6 - t.The maximum is at the minimum t, i.e. again at t=3 and with the same value 3.The minimum is at the maximum t, i.e. at t=5, y(5) = 1.
As the whole, the greatest value of y is 3 (at t=3) and the smallest is -1 (at x=-1).
The answer: the global minimum is -1 at t=-1 and the global maximum is 3 at t=3.