Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.5 - Continuity - 2.5 Exercises - Page 125: 45


$c = \frac{2}{3}$

Work Step by Step

Clearly the graph is continuous on the intervals $(-\infty,2)$ and $[2,\infty)$ We need to find the value $c$ such that the graph is continuous at the point $x=2$.: $cx^2+2x = x^3-cx$ $c(2)^2+2(2) = (2)^3-c(2)$ $4c+4 = 8-2c$ $6c = 4$ $c = \frac{2}{3}$
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