College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 3 - Review Exercises - Page 268: 5

Answer

(a) $f(2)=0$ (b) $f(-2)=0$ (c) $f(-x)=\dfrac{x^2-4}{x^2}$ (d) $-f(x)=-\dfrac{x^2-4}{x^2}$ (e) $f(x-2)=\dfrac{x^2-4x}{x^2-4x+4}$ (f) $f(2x)=\dfrac{x^2-1}{x^2}$

Work Step by Step

(a) $f(2)=\dfrac{2^2-4}{2^2}=\dfrac{4-4}{4}=\dfrac{0}{4}=0$ (b) $f(-2)=\dfrac{(-2)^2-4}{(-2)^2}=\dfrac{4-4}{4}=\dfrac{0}{4}=0$ (c) $f(-x)=\dfrac{(-x)^2-4}{(-x)^2}=\dfrac{x^2-4}{x^2}$ (d) $-f(x)=-\dfrac{x^2-4}{x^2}$ (e) $f(x-2)=\dfrac{(x-2)^2-4}{(x-2)^2}=\dfrac{x^2-4x+4-4}{x^2-4x+4}=\dfrac{x^2-4x}{x^2-4x+4}$ (f) $f(2x)=\dfrac{(2x)^2-4}{(2x)^2}=\dfrac{4x^2-4}{4x^2}=\dfrac{4(x^2-1)}{4x^2}=\dfrac{x^2-1}{x^2}$
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