Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 10 - Review - True-False Quiz - Page 719: 1

Answer

False

Work Step by Step

Consider $x=f(t)=(t-1)^{3}$ and $y=g(t)=(t-1)^{2}$ Thus $g'(t)=2(t-1)$ Now put $t=1$ $g'(1)=2(1-1)=0$ which is not equal to $1$ as per the given statement, that is, $g'(1)\ne 1$. Hence , the statement is false.
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