University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.2 - The Derivative as a Function - Exercises: 14

Answer

$k'(x) = \frac{-1}{(2+x)^{2}}$ $slope = -\frac{1}{16}$ Equation of tangent line: $y = \frac{1}{4}- \frac{1}{16}(x-2)$

Work Step by Step

$k(x) = \frac{1}{2+x}$ $k(2) = \frac{1}{4}$ Therefore, $k'(x) = \lim\limits_{h \to 0}\frac{\frac{1}{2+(x+h)}-\frac{1}{2+x}}{h}$ $k'(x) = \lim\limits_{h \to 0}\frac{-1}{(2+x+h)(2+x)}$ $k'(x) = \frac{-1}{(2+x)^{2}}$ Slope at x = 2: $slope = -\frac{1}{16}$ Equation of tangent line: $\frac{y-\frac{1}{4}}{x - 2} = -\frac{1}{16}$ $y = \frac{1}{4}- \frac{1}{16}(x-2)$
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