University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.2 - The Derivative as a Function - Exercises: 19

Answer

$s'(t) = -6t$ So, $s'(-1) = 6$

Work Step by Step

$s(t) = 1-3t^{2}$ Therefore, $s'(t) = 0 - 3(2)t^{2-1}$ $s'(t) = -6t$ So, $s'(-1) = 6$
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