Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 1 - Limits and Their Properties - 1.4 Exercises: 98

Answer

$c=3.$

Work Step by Step

$f(x)$ is continuous for all values of $x$ such that $x\ne1\to f(x)$ is continuous over the interval $(-\infty, 1)$ U $(1, \infty).$ $f(\frac{5}{2})=\dfrac{\frac{5}{2}(\frac{5}{2}+1)}{\frac{5}{2}-1}=\dfrac{35}{6}.$ $f(4)=\dfrac{4(4+1)}{4-1}=\dfrac{20}{3}.$ SInce $f(x)$ is continuous over the interval $[\frac{5}{2}, 4]$ and $\dfrac{35}{6}\lt6\lt\dfrac{20}{3},$ then the Intermediate Value Theorem guarantees the existence of at least one value $c$ such that $f(c)=6.$ $\dfrac{x^2+x}{x-1}=6\to x^2+x=6x-6\to x^2-5x+6=0\to$ $(x-2)(x-3)\to x=2$ or $x=3.$ $c=3$ (the other value is rejected since it is not in the specified domain.
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