Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 1 - Limits and Their Properties - 1.4 Exercises - Page 80: 45

Answer

No discontinuities; $f(x)$ is continuous for all real numbers.

Work Step by Step

$x^2+1=0\to x^2=-1$ which has no real solutions. Hence the denominator can never equal to $0\to$ no discontinuities.
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