Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 0 - Before Calculus - 0.2 New Functions From Old - Exercises Set 0.2 - Page 25: 30

Answer

$a. g(5s+2)=\sqrt {(5s+2)}$ $b. g(\sqrt x+2)=\sqrt {(\sqrt x+2)}$ $c.3g(5x)=3\times\sqrt {(5x)}$ $d. 1/g(x)=1/(\sqrt x)=\sqrt x/x$ $e.g(g(x))=g(\sqrt x)=\sqrt {(\sqrt x)}$ $f. (g(x))^2-g(x^2)=(\sqrt x)^2-\sqrt x^2$ $g. g(1/\sqrt x)=1/\sqrt {(\sqrt x)}$ $h. g((x-1)^2)=\sqrt {(x-1)}^2=x-1$ $i. g(x+h)=\sqrt {(x+h)}$

Work Step by Step

$a. g(5s+2)=\sqrt {(5s+2)}$ $b. g(\sqrt x+2)=\sqrt {(\sqrt x+2)}$ $c.3g(5x)=3\times\sqrt {(5x)}$ $d. 1/g(x)=1/(\sqrt x)=\sqrt x/x$ $e.g(g(x))=g(\sqrt x)=\sqrt {(\sqrt x)}$ $f. (g(x))^2-g(x^2)=(\sqrt x)^2-\sqrt x^2$ $g. g(1/\sqrt x)=1/\sqrt {(\sqrt x)}$ $h. g((x-1)^2)=\sqrt {(x-1)}^2=x-1$ $i. g(x+h)=\sqrt {(x+h)}$ $$\boxed{\text{Please see additional work shown in attached image.}}$$
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