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: 52


The function has no discontinuities.

Work Step by Step

For this piece-wise function, the only possible point of discontinuity is $x=1:$ $\lim\limits_{x\to1^+}f(x)=(1^+)^2=1;\lim\limits_{x\to1^-}f(x)=-2(1^-)+3=1.$ Since $\lim\limits_{x\to1^+}f(x)=\lim\limits_{x\to1^-}f(x)\to\lim\limits_{x\to1}f(x)=1.$ Furthermore the limit is equal to the function's value at the point $x=1$ hence the function is continuous at $x=1$ and has no discontinuities at all.
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