Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 3 - Exponents, Polynomials and Functions - 3.3 Composing Functions - 3.3 Exercises: 20

Answer

$\text{a) } f(g(x))=\dfrac{10}{3}x-\dfrac{5}{3} \\\\\text{b) } g(f(x))=\dfrac{4}{3}x+\dfrac{1}{15}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ With \begin{array}{l}\require{cancel} f(x)= \dfrac{2}{3}x+\dfrac{1}{3} \\g(x)= 5x-3 ,\end{array} replace $x$ with $g(x)$ in $f$ to find $f(g(x)).$ To find $g(f(x)),$ replace $x$ with $f(x)$ in $g.$ $\bf{\text{Solution Details:}}$ Replacing $x$ with $g(x)$ in $f,$ then \begin{array}{l}\require{cancel} f(g(x))=f\left( 5x-3 \right) \\\\ f(g(x))=\dfrac{2}{3}\left( 5x-3 \right)+\dfrac{1}{3} \\\\ f(g(x))=\dfrac{10}{3}x-2+\dfrac{1}{3} \\\\ f(g(x))=\dfrac{10}{3}x-\dfrac{6}{3}+\dfrac{1}{3} \\\\ f(g(x))=\dfrac{10}{3}x-\dfrac{5}{3} .\end{array} Replacing $x$ with $f(x)$ in $g.$ Hence, \begin{array}{l}\require{cancel} g(f(x))=g\left( \dfrac{2}{3}x+\dfrac{1}{3} \right) \\\\ g(f(x))=2\left( \dfrac{2}{3}x+\dfrac{1}{3} \right)-\dfrac{3}{5} \\\\ g(f(x))=\dfrac{4}{3}x+\dfrac{2}{3} -\dfrac{3}{5} \\\\ g(f(x))=\dfrac{4}{3}x+\dfrac{10}{15} -\dfrac{9}{15} \\\\ g(f(x))=\dfrac{4}{3}x+\dfrac{1}{15} .\end{array} Hence, \begin{array}{l}\require{cancel} \text{a) } f(g(x))=\dfrac{10}{3}x-\dfrac{5}{3} \\\\\text{b) } g(f(x))=\dfrac{4}{3}x+\dfrac{1}{15} .\end{array}
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