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 1160: 3


Yes, the function $ f\left( x \right)={{x}^{2}}-3x+7$ is continuous at $4$.

Work Step by Step

Consider the function $ f\left( x \right)={{x}^{2}}-3x+7$, First check whether the function is defined at the point $ a $ or not. Find the value of $ f\left( x \right)$ at $ a=4$, $\begin{align} & f\left( 4 \right)={{\left( 4 \right)}^{2}}-3\left( 4 \right)+7 \\ & =16-12+7 \\ & =11 \end{align}$ The function is defined at the point $ a=4$. Now find the value of $\,\underset{x\to 4}{\mathop{\lim }}\,{{x}^{2}}-3x+7$, $\begin{align} & \,\underset{x\to 4}{\mathop{\lim }}\,f\left( x \right)=\,\underset{x\to 4}{\mathop{\lim }}\,\left( {{x}^{2}}-3x+7 \right) \\ & ={{\left( 4 \right)}^{2}}-3\left( 4 \right)+7 \\ & =16-12+7 \\ & =11 \end{align}$ Thus, $\,\underset{x\to 4}{\mathop{\lim }}\,\left( {{x}^{2}}-3x+7 \right)=11$ Now check whether $\,\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ or not. From the above, $\,\underset{x\to 4}{\mathop{\lim }}\,\left( {{x}^{2}}-3x+7 \right)=11\text{ and }f\left( 4 \right)=11$ Therefore, $\,\underset{x\to 4}{\mathop{\lim }}\,f\left( x \right)=f\left( 4 \right)$ Thus, the function satisfies all the properties of being continuous. Hence, the function $ f\left( x \right)={{x}^{2}}-3x+7$ is continuous at $4$.
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