Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole

Chapter 3 - Exponents, Polynomials and Functions - 3.3 Composing Functions - 3.3 Exercises: 19

Answer

$\text{a) } f(g(x))=x-\dfrac{1}{10} \\\\\text{b) } g(f(x))=x+\dfrac{19}{10}$

Work Step by Step

$\bf{\text{Solution Outline:}}$ With \begin{array}{l}\require{cancel} f(x)= \dfrac{1}{2}x+\dfrac{1}{5} \\g(x)= 2x-\dfrac{3}{5} ,\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( 2x-\dfrac{3}{5} \right) \\\\ f(g(x))=\dfrac{1}{2}\left( 2x-\dfrac{3}{5} \right)+\dfrac{1}{5} \\\\ f(g(x))=x-\dfrac{3}{10}+\dfrac{1}{5} \\\\ f(g(x))=x-\dfrac{3}{10}+\dfrac{2}{10} \\\\ f(g(x))=x-\dfrac{1}{10} .\end{array} Replacing $x$ with $f(x)$ in $g.$ Hence, \begin{array}{l}\require{cancel} g(f(x))=g\left( \dfrac{1}{2}x+\dfrac{1}{5} \right) \\\\ g(f(x))=2\left( \dfrac{1}{2}x+\dfrac{1}{5} \right)-\dfrac{3}{5} \\\\ g(f(x))=x+\dfrac{5}{2}-\dfrac{3}{5} \\\\ g(f(x))=x+\dfrac{25}{10}-\dfrac{6}{10} \\\\ g(f(x))=x+\dfrac{19}{10} .\end{array} Hence, \begin{array}{l}\require{cancel} \text{a) } f(g(x))=x-\dfrac{1}{10} \\\\\text{b) } g(f(x))=x+\dfrac{19}{10} .\end{array}

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