Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 3 - The Derivative - 3.4 Definition of the Derivative - 3.4 Exercises - Page 176: 5

Answer

$2$

Work Step by Step

We need to find the slope of the tangent line at $(5,3)$. We can do this by taking two points - for example $(5,3)$ and $(6,5)$ -, through which the aforesaid line passes through. The slope $m$ is: $m = \frac{y_2-y_1}{x_2-x_1} = \frac{5-3}{6-5} = 2$.
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