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: 32

Answer

$\text{a) } f(g(2))=28 \\\\\text{b) } g(f(2))=-1088$

Work Step by Step

$\bf{\text{Solution Outline:}}$ With \begin{array}{l}\require{cancel} f(x)= 19x+28 \\g(x)= -17x+34 ,\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)=-17x+34 \\\\ g(2)=-17(2)+34 \\\\ g(2)=-34+34 \\\\ g(2)=0 .\end{array} Replacing $x$ with the result above in $f$ results to \begin{array}{l}\require{cancel} f(x)=19x+28 \\\\ f(0)=19(0)+28 \\\\ f(0)=0+28 \\\\ f(0)=28 .\end{array} Hence, $ f(g(2))=28 .$ b) Replacing $x$ with $ 2 $ in $f$ results to \begin{array}{l}\require{cancel} f(x)=19x+28 \\\\ f(2)=19(2)+28 \\\\ f(2)=38+28 \\\\ f(2)=66 .\end{array} Replacing $x$ with the result above in $g$ results to \begin{array}{l}\require{cancel} g(x)=-17x+34 \\\\ g(66)=-17(66)+34 \\\\ g(66)=-1122+34 \\\\ g(66)=-1088 .\end{array} Hence, $ g(f(2))=-1088 .$ Therefore, \begin{array}{l}\require{cancel} \text{a) } f(g(2))=28 \\\\\text{b) } g(f(2))=-1088 .\end{array}
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