Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 4 - Exponential and Logarithmic Functions - Section 4.1 Composite Functions - 4.1 Assess Your Understanding - Page 280: 45

Answer

(a) $ -\frac{4x-17}{2x-1}$, domain $\{x|x\ne\frac{1}{2},3 \}$. (b) $ -\frac{3x-3}{2x+8}$, domain $\{x|x\ne-4,-1 \}$. (c) $ -\frac{2x+5}{x-2}$, domain $\{x|x\ne-1,2 \}$. (d) $ -\frac{3x-4}{2x-11}$, domain $\{x|x\ne\frac{11}{2},3 \}$.

Work Step by Step

Given $f(x)=\frac{x-5}{x+1}$ and $g(x)=\frac{x+2}{x-3}$, we have: (a) $f\circ g=\frac{\frac{x+2}{x-3}-5}{\frac{x+2}{x-3}+1}=-\frac{4x-17}{2x-1}$, domain $\{x|x\ne\frac{1}{2},3 \}$. (b) $g\circ f=\frac{\frac{x-5}{x+1}+2}{\frac{x-5}{x+1}-3}=-\frac{3x-3}{2x+8}$, domain $\{x|x\ne-4,-1 \}$. (c) $f\circ f=\frac{\frac{x-5}{x+1}-5}{\frac{x-5}{x+1}+1}=-\frac{2x+5}{x-2}$, domain $\{x|x\ne-1,2 \}$. (d) $g\circ g=\frac{\frac{x+2}{x-3}+2}{\frac{x+2}{x-3}-3}=-\frac{3x-4}{2x-11}$, domain $\{x|x\ne\frac{11}{2},3 \}$.
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