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: 1

Answer

$6$ $0$ $-2$

Work Step by Step

$f(x)=4-x^2$, therefore $f'(x)=0-2x$ $f'(-3)=0-2\times(-3)=6$ $f'(0)=0-2\times(0)=0$ $f'(1)=0-2\times(1)=-2$
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