Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 3 - The Derivative - 3.1 Limits - 3.1 Exercises - Page 137: 63

Answer

$f$ has a vertical asymptote at $x=1$

Work Step by Step

At $x = 1$, the denominator of the function becomes 0 and therefore $f$ is not defined at $x=1$. Figure $11$ is more deceptive that Figure $10$, since it shows a steep vertical line that supposedly connects the function, but here there should be a vertical asymptote, and x should not be defined at any point at $x = 1$.
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