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: 17

Answer

$f'(x) = \frac{-4}{(x-2)^{\frac{3}{2}}}$ Tangent line: $y = 4 -\frac{1}{2}(x-6)=-\frac{1}{2}x+7$

Work Step by Step

$f(x) =\frac{8}{\sqrt (x-2)}$ Using the derivative property: $f'(x) = \lim\limits_{h \to 0} \frac{\frac{8}{\sqrt ((x+h)-2)} -\frac{8}{\sqrt (x-2)} }{h}$ $f'(x) = \lim\limits_{h \to 0}\frac{-8}{(\sqrt ((x+h)-2))(\sqrt (x-2))({\sqrt ((x+h)-2)} +\sqrt (x-2))}$ $f'(x) = \frac{-4}{(x-2)^{\frac{3}{2}}}$ Slope at x = 6: $slope = f'(6) = -\frac{1}{2}$ So, the equation of tangent line is: $\frac{y-4}{x-6} = -\frac{1}{2}$ $y = 4 -\frac{1}{2}(x-6)=-\frac{1}{2}x+7$
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