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 217: 55

Answer

$f\displaystyle \circ g(x)==\frac{2x-1}{2x} \qquad $ domain: $(-\infty,0)\cup(0,\infty)$ $ g\displaystyle \circ f(x)=\frac{x-1}{x+1}; \quad$ domain: $(-\infty,-1)\cup(-1,\infty)$ $ f\displaystyle \circ f(x)=\frac{x}{2x+1}; \quad$ domain: $(-\displaystyle \infty,-1)\cup(-1,-\frac{1}{2})\cup(-\frac{1}{2},\infty)$ $ g\circ g(x)=4x-3; \quad$ domain: $(-\infty,\infty)$

Work Step by Step

f(x) is defined for all x except $-1$ g(x) is defined for all x $ f\displaystyle \circ g(x)=f[g(x)]= \frac{g(x)}{g(x)+1}\quad$on the domain of $g$(x), and $g(x)+1\neq 0$ $=\displaystyle \frac{2x-1}{2x-1+1},\quad$for$ \left\{\begin{array}{l} 2x-1+1\neq 0\\ x\neq 0 \end{array}\right\}$ $=\displaystyle \frac{2x-1}{2x} \qquad $ domain: $(-\infty,0)\cup(0,\infty)$ $g\circ f(x)=g[f(x)]=2f(x)-1 \quad$on the domain of f(x), $=2(\displaystyle \frac{x}{x+1})-1,\quad$for $x\neq-1$ $=\displaystyle \frac{2x-(x+1)}{x+1}$ $=\displaystyle \frac{x-1}{x+1}; \quad$ domain: $(-\infty,-1)\cup(-1,\infty)$ $ f\displaystyle \circ f(x)=f[f(x)]=\frac{f(x)}{f(x)+1},\qquad$on the domain of f(x), and $f(x)+1\neq 0$ $=\displaystyle \frac{\frac{x}{x+1}}{\frac{x}{x+1}+1},\quad $ for $\left\{\begin{array}{ll} x\neq-1 & \frac{x}{x+1}+1\neq 0\\ & \frac{x+x+1}{x+1}\neq 0\\ & 2x+1\neq 0\\ & x\neq-1/2 \end{array}\right\}$ $=\displaystyle \frac{\frac{x}{x+1}}{\frac{x}{x+1}+1}\times\frac{x+1}{x+1}$ $=\displaystyle \frac{x}{x+x+1}$ $=\displaystyle \frac{x}{2x+1}; \quad$ domain: $(-\displaystyle \infty,-1)\cup(-1,-\frac{1}{2})\cup(-\frac{1}{2},\infty)$ $ g\circ g(x)=g[g(x)]=2g(x)-1 \quad$on the domain of $g$(x) $=2(2x-1)-1 \quad$on $(-\infty,\infty)$ $=4x-3; \quad$ domain: $(-\infty,\infty)$
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