Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 1 - Graphs - 2.1 Functions - 2.1 Assess Your Understanding - Page 58: 76

Answer

a. $= \sqrt {x+1} + \frac{2}{x}$ b. $= \sqrt {x+1} - \frac{2}{x}$ c.$= \sqrt {\frac{2}{x}+1} $ d. $\frac{\left(\sqrt{x}+1\right)x}{2}$ e. $4\frac{2}{3}$ f. $\sqrt{5}-\frac{1}{2}$ g. $\sqrt 2$ h. $1$

Work Step by Step

$f(x)= \sqrt {x+1}$ $g(x)=\frac{2}{x}$ a. $(f+g)(x)$ $= \sqrt {x+1} + \frac{2}{x}$ b. $(f-g)(x)$ $= \sqrt {x+1} - \frac{2}{x}$ c. $(f \cdot g)(x)$ $= \sqrt {\frac{2}{x}+1} $ d. $(\frac{f}{g})(x)$ $=\frac{\sqrt {x+1}}{\frac{2}{x}}$ $=\frac{\left(\sqrt{x}+1\right)x}{2}$ e. $(f+g)(3)$ $= \sqrt {x+1} + \frac{2}{x}$ $= \sqrt {3+1} + \frac{2}{3}$ $=4\frac{2}{3}$ f. $(f-g)(4)$ $= \sqrt {4+1} - \frac{2}{4}$ $=\sqrt{5}-\frac{1}{2}$ g. $(f \cdot g)(2)$ $= \sqrt {\frac{2}{x}+1} $ $= \sqrt {\frac{2}{2}+1} $ $=\sqrt{1+1}$ $=\sqrt2$ h. $(\frac{f}{g})(1)$ $=\frac{\sqrt {x+1}}{\frac{2}{x}}$ $=\frac{\left(\sqrt{x}+1\right)x}{2}$ $=\frac{\left(\sqrt{1}+1\right)1}{2}$ $=1$
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