Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 11 - Section 11.3 - Limits and Continuity - Concept and Vocabulary Check - Page 1160: 6


The statement “A piecewise function is always discontinuous at one or more numbers” is false.

Work Step by Step

Consider the piecewise function, $ f\left( x \right)=\left\{ \begin{align} & x-1\text{ if }x\le 0 \\ & {{x}^{2}}+x-1\text{ if }x>0 \end{align} \right.$. Since, $ x-1\text{ and }{{x}^{2}}+x-1$ both are continuous polynomial functions. Therefore, the function can have the point of discontinuity only at $ a=0$. Now check the discontinuity of the function at $ a=0$. Find the value of $ f\left( x \right)$ at $ a=0$, $\begin{align} & f\left( 0 \right)=0-1 \\ & =-1 \end{align}$ The function is defined at the point $ a=0$. Now find the value of $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)$. First check the left hand limit of $\,f\left( x \right)$. $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0-1=-1$ Now find the right hand limit of $\,f\left( x \right)$, $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)={{0}^{2}}+0-1=-1$ Thus the left hand limit and right hand limit are equal, that is $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=-1=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ Thus, $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=-1$ From the above steps, $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=-1=f\left( 0 \right)$ Thus, the function satisfies all the properties of being continuous. Hence, the function $ f\left( x \right)=\left\{ \begin{align} & x-1\text{ if }x\le 0 \\ & {{x}^{2}}+x-1\text{ if }x>0 \end{align} \right.$ is continuous at $0$. Thus, the function $ f\left( x \right)$ is not discontinuous for any number. Hence, the statement is false.
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