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: 38

Answer

$\text{a) } f(g(3))=41 \\\\\text{b) } g(f(3))=22$

Work Step by Step

$\bf{\text{Solution Outline:}}$ With \begin{array}{l}\require{cancel} f(x)= 4x-7 \\g(x)= x^2-3x+12 ,\end{array} to find $ f(g(3)) ,$ find first $ g(3) .$ Then substitute the result in $f.$ To find $ g(f(3)) ,$ find first $ f(3) .$ Then substitute the result in $g.$ $\bf{\text{Solution Details:}}$ a) Replacing $x$ with $ 3 $ in $g$ results to \begin{array}{l}\require{cancel} g(x)=x^2-3x+12 \\\\ g(3)=(3)^2-3(3)+12 \\\\ g(3)=9-9+12 \\\\ g(3)=12 .\end{array} Replacing $x$ with the result above in $f$ results to \begin{array}{l}\require{cancel} f(x)=4x-7 \\\\ f(12)=4(12)-7 \\\\ f(12)=48-7 \\\\ f(12)=41 .\end{array} Hence, $ f(g(3))=41 .$ b) Replacing $x$ with $ 3 $ in $f$ results to \begin{array}{l}\require{cancel} f(x)=4x-7 \\\\ f(3)=4(3)-7 \\\\ f(3)=12-7 \\\\ f(3)=5 .\end{array} Replacing $x$ with the result above in $g$ results to \begin{array}{l}\require{cancel} g(x)=x^2-3x+12 \\\\ g(5)=(5)^2-3(5)+12 \\\\ g(5)=25-15+12 \\\\ g(5)=22 .\end{array} Hence, $ g(f(3))=22 .$ Therefore, \begin{array}{l}\require{cancel} \text{a) } f(g(3))=41 \\\\\text{b) } g(f(3))=22 .\end{array}
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