`y = 3 - |t - 3|, [-1,5]` Find the absolute extrema of the function on the closed interval.

Textbook Question

Chapter 3, 3.1 - Problem 29 - Calculus of a Single Variable (10th Edition, Ron Larson).
See all solutions for this textbook.

1 Answer | Add Yours

shumbm's profile pic

Borys Shumyatskiy | College Teacher | (Level 3) Associate Educator

Posted on

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.

We’ve answered 318,988 questions. We can answer yours, too.

Ask a question