Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 2 - Limits - 2.8 Intermediate Value Theorem - Exercises - Page 87: 28


$f(x)=\begin{cases} \dfrac{1}{x-2}+1,\text{for }x\in[0,2)\\ 2,\text{for }x=2\\ \dfrac{x}{x-2}-3,\text{for }x\in(2,4] \end{cases}$

Work Step by Step

Consider the function: $f(x)=\begin{cases} \dfrac{1}{x-2}+1,\text{for }x\in[0,2)\\ 2,\text{for }x=2\\ \dfrac{x}{x-2}-3,\text{for }x\in(2,4] \end{cases}$ $\displaystyle\lim_{x\rightarrow 2^{-}} f(x)=-\infty$ $\displaystyle\lim_{x\rightarrow 2^{+}} f(x)=\infty$ Therefore $f$ is not continuous in $x=2$. $f(0.5)f(3.5)=(-1)(2.1)<0$ and there is a zero between $0.5$ and $3.5$.
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