Calculus Concepts: An Informal Approach to the Mathematics of Change 5th Edition

Published by Brooks Cole
ISBN 10: 1-43904-957-2
ISBN 13: 978-1-43904-957-0

Chapter 1 - Ingredients of Change: Functions and Limits - 1.3 Activities - Page 31: 35

Answer

(a) 5 (b)5 (c)5 (d) $f$ is continuous at $x=2$

Work Step by Step

Given $$ f(x)=\left\{\begin{array}{ll}{10 x^{-1}} & {\text { when } x<2} \\ {4 x-3} & {\text { when } x \geq 2}\end{array}\right. $$ (a) Since \begin{align*} \lim _{x \rightarrow2^{-}} f(x)&=\lim _{x \rightarrow2^{-}}10 x^{-1}\\ &=\lim _{x \rightarrow2^{-}}10 (2)^{-1}\\ &=5 \end{align*} (b) Since \begin{align*} \lim _{x \rightarrow2^{+}} f(x)&=\lim _{x \rightarrow2^{+}} (4x-3)\\ &=\lim _{x \rightarrow2^{+}} (8-3)\\ &=5 \end{align*} (c) $f(2)= 4(2)-3=5$ (d) Since $$\lim _{x \rightarrow2^{+}} f(x)=\lim _{x \rightarrow2^{-}} f(x)=f(2)=5$$ Thus, $f$ is continuous at $x=2$
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