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 - Page 259: 35


$\text{a) } (f\circ g)(-3)=-0.5096 \\\\\text{b) } (g\circ f)(-3)=7.6324$

Work Step by Step

$\bf{\text{Solution Outline:}}$ With \begin{array}{l}\require{cancel} f(x)= 0.68x+2.36 \\g(x)= 3.57x+6.49 ,\end{array} to find $ (f\circ g)(-3) ,$ find first $ g(-3) .$ Then substitute the result in $f.$ To find $ (g\circ f)(-3) ,$ find first $ f(-3) .$ Then substitute the result in $g.$ $\bf{\text{Solution Details:}}$ a) Since $(f\circ g)(-3)=f(g(-3)),$ find first $g(-3).$ That is \begin{array}{l}\require{cancel} g(x)=3.57x+6.49 \\\\ g(-3)=3.57(-3)+6.49 \\\\ g(-3)=-10.71+6.49 \\\\ g(-3)=-4.22 .\end{array} Replacing $x$ with the result above in $f$ results to \begin{array}{l}\require{cancel} f(x)=0.68x+2.36 \\\\ f(-4.22)=0.68(-4.22)+2.36 \\\\ f(-4.22)=-2.8696+2.36 \\\\ f(-4.22)=-0.5096 .\end{array} Hence, $ (f\circ g)(-3)=f(g(-3))=-0.5096 .$ b) Since $(g\circ f)(-3)=g(f(-3)),$ find first $f(-3).$ Replacing $x$ with $ -3 $ in $f$ results to \begin{array}{l}\require{cancel} f(x)=0.68x+2.36 \\\\ f(-3)=0.68(-3)+2.36 \\\\ f(-3)=-2.04+2.36 \\\\ f(-3)=0.32 .\end{array} Replacing $x$ with the result above in $g$ results to \begin{array}{l}\require{cancel} g(x)=3.57x+6.49 \\\\ g(0.32)=3.57(0.32)+6.49 \\\\ g(0.32)=1.1424+6.49 \\\\ g(0.32)=7.6324 .\end{array} Hence, $ (g\circ f)(-3)=g(f(-3))=7.6324 .$ Therefore, \begin{array}{l}\require{cancel} \text{a) } (f\circ g)(-3)=-0.5096 \\\\\text{b) } (g\circ f)(-3)=7.6324 .\end{array}
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