Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.2 The Derivative as a Function - Exercises - Page 116: 75


The function $ f(x)=|x^2-1|$ is not differentailble at $ x=\pm1$.

Work Step by Step

Since $$|x^2-1|=0\Longrightarrow x^2-1=0\Longrightarrow (x+1)(x-1)=0.$$ The function $ f(x)=|x^2-1|$ has corners at $ x=\pm 1$, hence it is not differentailble at $ x=\pm1$.
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