Calculus (3rd Edition)

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

Chapter 2 - Limits - 2.4 Limits and Continuity - Exercises - Page 67: 57



Work Step by Step

Since $f(x)$ is continuous, then it is continuous at $x=5$; thus we use the property that $$f(5)=\lim_{x\to 5}f(x)$$ where $f(5)=4(5)+2c=20+2c$ and $\lim_{x\to 5}f(x) =5^2-c$ Hence, we get $$25-c=20+2c\Longrightarrow 3c=5.$$ That is, $c=5/3$.
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