Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.5 - Continuity - Exercises 2.5 - Page 84: 13


$ y $ is continuous in the domain $$ R-\{2\}$$

Work Step by Step

Given $$ y=\frac{1}{x-2}-3 x $$ Find the zeros of the denominator: $ x-2=0 \Rightarrow x=2$ So, $ y $ is continuous in the domain $ R-\{2\}$ (all values except 2).
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