Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 2 - Section 2.7 - Combining Functions - 2.7 Exercises - Page 216: 15

Answer

$(f+g)(x)=\displaystyle \frac{2}{x}+\frac{4}{x+4}$ , domain: $\{x|x\neq-4,x\neq 0\}.$ $(f-g)(x)=\displaystyle \frac{2}{x}-\frac{4}{x+4}$ , domain: $\{x|x\neq-4,x\neq 0\}.$ $(fg)(x)=\displaystyle \frac{2}{x}\cdot\frac{4}{x+4}$ , domain: $\{x|x\neq-4,x\neq 0\}.$ $(\displaystyle \frac{f}{g})(x)=\frac{x+4}{2x}$ , domain: $\{x|x\neq-4,x\neq 0\}.$ In interval notation, all domains are $(-\infty,-4)\cup(-4,0)\cup(0,+\infty)$

Work Step by Step

The domain of f is $\{x|x\neq 0\}$, because f is defined only for x that yield a nonzero numberin the denominator. The domain of g is $\{x|x\neq-4\}$, for the same reason. Their intersection is $\{x|x\neq-4,x\neq 0\}.$ $(f+g)(x)=\displaystyle \frac{2}{x}+\frac{4}{x+4}$ , domain: $\{x|x\neq-4,x\neq 0\}.$ $(f-g)(x)=\displaystyle \frac{2}{x}-\frac{4}{x+4}$ , domain: $\{x|x\neq-4,x\neq 0\}.$ $(fg)(x)=\displaystyle \frac{2}{x}\cdot\frac{4}{x+4}$ , domain: $\{x|x\neq-4,x\neq 0\}.$ $(\displaystyle \frac{f}{g})(x)=\frac{\frac{2}{x}}{\frac{4}{x+4}}=\frac{2(x+4)}{4x}=\frac{x+4}{2x}$ , domain: $\{x|x\neq-4,x\neq 0\}.$ In interval notation, all domains are $(-\infty,-4)\cup(-4,0)\cup(0,+\infty)$
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