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: 54

Answer

The function has an irremovable jump discontinuity at $x=2.$

Work Step by Step

For this piece-wise function, the only possible point for a discontinuity is $x=2:$ $\lim\limits_{x\to2^-}f(x)=-2(2^-)=-4; \lim\limits_{x\to2^+}f(x)=(2^+)^2-4(2^+)+1=-3.$ Since $\lim\limits_{x\to2^-}f(x)\ne\lim\limits_{x\to2^+}f(x)\to\lim\limits_{x\to2}f(x)$ does not exist, the function is not continuous at the point $x=2$. This is non-removable.
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