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: 18

Answer

$g'(z) = \frac{-1}{2\sqrt (4-z)}$ $slope= g'(3) = -\frac{1}{2}$ Tangent line: $w = 2 -\frac{1}{2}(z-3)=-\frac{1}{2}z+\frac{7}{2}$

Work Step by Step

$g(z) = 1+\sqrt (4-z)$ $g'(z) = \lim\limits_{h \to 0}\frac{1+\sqrt (4-(z+h))- 1-\sqrt (4-z)}{h}$ $g'(z) = \lim\limits_{h \to 0}\frac{-h}{h(\sqrt (4-(z+h)) + \sqrt (4-z))}$ $g'(z) = \lim\limits_{h \to 0}\frac{-1}{\sqrt (4-(z+h)) + \sqrt (4-z)}$ $g'(z) = \frac{-1}{2\sqrt (4-z)}$ Slope at z = 3: $slope= g'(3) = -\frac{1}{2}$ The equation of the tangent line is: $\frac{w- 2}{z-3} = -\frac{1}{2}$ $w = 2 -\frac{1}{2}(z-3)$
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