Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.3 Rules of Differentiation - 3.3 Exercises - Page 151: 41

Answer

a. $x=2; x=-1$ b. $x=4; x=-3$

Work Step by Step

The original function is $f(x)=2x^3-3x^2-12x+4$ therefore, $f'(x)=6x^2-6x-12$ a. Set the derivative function equal to 0 because the slope of a horizontal line is 0: $6x^2-6x-12=0$ $6(x^2-x-2)=0$ $6(x-2)(x+1)=0$ $x=2; x=-1$ b. Set the derivative function equal to 60: $6x^2-6x-12=60$ $6x^2-6x-72=0$ $6(x^2-x-12)=0$ $6(x-4)(x+3)=0$ $x=4; x=-3$
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