Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 12 - Section 12.1 - The Algebra of Functions - Exercise Set: 5

Answer

$a)$ $(f+g)(x)=\sqrt{x}+x+5$ $b)$ $(f-g)(x)=\sqrt{x}-x-5$ $c)$ $(f\cdot g)(x)=x\sqrt{x}+5\sqrt{x}$ $d)$ $\Big(\dfrac{f}{g}\Big)(x)=\dfrac{\sqrt{x}}{x+5}$, where $x\ne-5$

Work Step by Step

$f(x)=\sqrt{x};$ $g(x)=x+5$ For each case, replace $f(x)$ by $\sqrt{x}$ and $g(x)$ by $x+5$ and simplify if possible: $a)$ $(f+g)(x)$ $f(x)+g(x)=\sqrt{x}+x+5=\sqrt{x}+x+5$ $b)$ $(f-g)(x)$ $f(x)-g(x)=(\sqrt{x})-(x+5)=\sqrt{x}-x-5$ $c)$ $(f\cdot g)(x)$ $f(x)\cdot g(x)=(\sqrt{x})(x+5)=x\sqrt{x}+5\sqrt{x}$ $d)$ $\Big(\dfrac{f}{g}\Big)(x)$ $\dfrac{f(x)}{g(x)}=\dfrac{\sqrt{x}}{x+5}$, where $x\ne-5$
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