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 - Exercise Set - Page 1161: 16

Answer

The function $ f\left( x \right)=\left\{ \begin{align} & x-4\text{ if }x\le 0 \\ & {{x}^{2}}+x-4\text{ if }x>0 \end{align} \right.$ is continuous at $0$.

Work Step by Step

Consider the function $ f\left( x \right)=\left\{ \begin{align} & x-4\text{ if }x\le 0 \\ & {{x}^{2}}+x-4\text{ if }x>0 \end{align} \right.$, Find the value of $ f\left( x \right)$ at $ a=0$, $\begin{align} & f\left( 0 \right)=0-4 \\ & =-4 \end{align}$ 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 find the left hand limit of $\,f\left( x \right)$. The function $ f\left( x \right)=x-4$ when $ x<0$. $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=0-4=-4$ Now find the right hand limit of $\,f\left( x \right)$. The function $ f\left( x \right)={{x}^{2}}+x-4$ when $ x>0$. $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)={{0}^{2}}+0-4=-4$ Since the left hand limit and right hand limit are equal, that is $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=-4=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)$ Thus, $\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=-4$ From the above steps, $\,\underset{x\to 0}{\mathop{\lim }}\,f\left( x \right)=-4=f\left( 0 \right)$ Thus, the function satisfies all the properties of being continuous. Hence, the function $ f\left( x \right)=\left\{ \begin{align} & x-4\text{ if }x\le 0 \\ & {{x}^{2}}+x-4\text{ if }x>0 \end{align} \right.$ is continuous at $0$.
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