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 79: 32

Answer

The function is continuous on the interval $[-3,3]$

Work Step by Step

The function is continuous on the interval because it meets two requirements: 1. $ \lim\limits_{x \to -3^+} 3-\sqrt {9-t^2} = 3 $ and 2. $ \lim\limits_{x \to 3^-} 3-\sqrt {9-t^2} = 3 $ on the closed interval $[-3,3]$
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