Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 2 - Chapter Test - Page 235: 5


(a) $-\frac{1}{8}$ (b) $-\frac{1}{28}$ (c) $\frac{\sqrt x}{x^2-18x}~~$, with $x\geq0$

Work Step by Step

$f(x)=\frac{\sqrt {x+9}}{x^2-81}$ (a) $f(7)=\frac{\sqrt {7+9}}{7^2-81}=\frac{4}{-32}=-\frac{1}{8}$ (b) $f(-5)=\frac{\sqrt {-5+9}}{(-5)^2-81}=\frac{2}{-56}=-\frac{1}{28}$ (c) $f(x-9)=\frac{\sqrt {x-9+9}}{(x-9)^2-81}=\frac{\sqrt x}{x^2-18x+81-81}=\frac{\sqrt x}{x^2-18x}~~$, with $x\geq0$
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