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: 37

Answer

$\text{a) } f(g(5))=170 \\\\\text{b) } g(f(5))=490$

Work Step by Step

$\bf{\text{Solution Outline:}}$ With \begin{array}{l}\require{cancel} f(x)= 3x+5 \\g(x)= x^2+4x+10 ,\end{array} to find $ f(g(5)) ,$ find first $ g(5) .$ Then substitute the result in $f.$ To find $ g(f(5)) ,$ find first $ f(5) .$ Then substitute the result in $g.$ $\bf{\text{Solution Details:}}$ a) Replacing $x$ with $ 5 $ in $g$ results to \begin{array}{l}\require{cancel} g(x)=x^2+4x+10 \\\\ g(5)=(5)^2+4(5)+10 \\\\ g(5)=25+20+10 \\\\ g(5)=55 .\end{array} Replacing $x$ with the result above in $f$ results to \begin{array}{l}\require{cancel} f(x)=3x+5 \\\\ f(55)=3(55)+5 \\\\ f(55)=165+5 \\\\ f(55)=170 .\end{array} Hence, $ f(g(5))=170 .$ b) Replacing $x$ with $ 5 $ in $f$ results to \begin{array}{l}\require{cancel} f(x)=3x+5 \\\\ f(5)=3(5)+5 \\\\ f(5)=15+5 \\\\ f(5)=20 .\end{array} Replacing $x$ with the result above in $g$ results to \begin{array}{l}\require{cancel} g(x)=x^2+4x+10 \\\\ g(20)=(20)^2+4(20)+10 \\\\ g(20)=400+80+10 \\\\ g(20)=490 .\end{array} Hence, $ g(f(5))=490 .$ Therefore, \begin{array}{l}\require{cancel} \text{a) } f(g(5))=170 \\\\\text{b) } g(f(5))=490 .\end{array}
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