Precalculus (6th Edition) Blitzer

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

Chapter 11 - Review Exercises - Page 1179: 48


Function $f\left( x \right)=\left\{ \begin{align} & -1\text{ if }x<0 \\ & \text{ }1\text{ if }x\ge 0 \\ \end{align} \right.$is discontinuous at $x=0$.

Work Step by Step

To find the number at which the provided function is discontinuous, check the continuity of the function at $x=0$. Consider the provided function , $f\left( x \right)=\left\{ \begin{align} & -1\text{ if }x<0 \\ & \text{ }1\text{ if }x\ge 0 \\ \end{align} \right.$ Condition 1: Check the provided function $f\left( x \right)$ is defined at $x=a=0$. Since, $f\left( x \right)=f\left( a \right)=1$ at $x=0$. Hence, the function $f\left( x \right)$ is defined at $a=0$. Condition 2: Check the limit $\underset{x\to a}{\mathop{\lim }}\,f(x)$ exists in the function $f\left( x \right)$. Consider the left-hand limit, $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right),\text{ if }f\left( x \right)=-1$ Therefore, the left-hand limit is $-1$. Consider the right-hand limit, $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right),\text{ if }f\left( x \right)=1$ Therefore, the left-hand limit is $1$. Since the left hand limit and right-hand limit are not equal to each other, the limit does not exist at $x=a=0$. There is no need to check condition (3). Hence, the function is discontinuous at $x=0$.
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