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: 33


$\text{a) } f(g(2))=\dfrac{19}{10} \\\\\text{b) } g(f(2))=\dfrac{9}{5}$

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} to find $ f(g(2)) ,$ find first $ g(2) .$ Then substitute the result in $f.$ To find $ g(f(2)) ,$ find first $ f(2) .$ Then substitute the result in $g.$ $\bf{\text{Solution Details:}}$ a) Replacing $x$ with $ 2 $ in $g$ results to \begin{array}{l}\require{cancel} g(x)=2x-\dfrac{3}{5} \\\\ g(2)=2(2)-\dfrac{3}{5} \\\\ g(2)=4-\dfrac{3}{5} \\\\ g(2)=\dfrac{20}{5}-\dfrac{3}{5} \\\\ g(2)=\dfrac{17}{5} .\end{array} Replacing $x$ with the result above in $f$ results to \begin{array}{l}\require{cancel} f(x)=\dfrac{1}{2}x+\dfrac{1}{5} \\\\ f\left( \dfrac{17}{5} \right)=\dfrac{1}{2}\left( \dfrac{17}{5} \right)+\dfrac{1}{5} \\\\ f\left( \dfrac{17}{5} \right)=\dfrac{17}{10}+\dfrac{1}{5} \\\\ f\left( \dfrac{17}{5} \right)=\dfrac{17}{10}+\dfrac{2}{10} \\\\ f\left( \dfrac{17}{5} \right)=\dfrac{19}{10} .\end{array} Hence, $ f(g(2))=\dfrac{19}{10} .$ b) Replacing $x$ with $ 2 $ in $f$ results to \begin{array}{l}\require{cancel} f(x)=\dfrac{1}{2}x+\dfrac{1}{5} \\\\ f(2)=\dfrac{1}{2}(2)+\dfrac{1}{5} \\\\ f(2)=1+\dfrac{1}{5} \\\\ f(2)=\dfrac{5}{5}+\dfrac{1}{5} \\\\ f(2)=\dfrac{6}{5} .\end{array} Replacing $x$ with the result above in $g$ results to \begin{array}{l}\require{cancel} g(x)=2x-\dfrac{3}{5} \\\\ g\left( \dfrac{6}{5} \right)=2\left( \dfrac{6}{5} \right)-\dfrac{3}{5} \\\\ g\left( \dfrac{6}{5} \right)=\dfrac{12}{5}-\dfrac{3}{5} \\\\ g\left( \dfrac{6}{5} \right)=\dfrac{9}{5} .\end{array} Hence, $ g(f(2))=\dfrac{9}{5} .$ Therefore, \begin{array}{l}\require{cancel} \text{a) } f(g(2))=\dfrac{19}{10} \\\\\text{b) } g(f(2))=\dfrac{9}{5} .\end{array}
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