## Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole

# Chapter 3 - Exponents, Polynomials and Functions - 3.3 Composing Functions - 3.3 Exercises - Page 258: 31

#### Answer

$\text{a) } f(g(5))=-55 \\\\\text{b) } g(f(5))=-129$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ With \begin{array}{l}\require{cancel} f(x)= 3x+8 \\g(x)= -6x+9 ,\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)=-6x+9 \\\\ g(5)=-6(5)+9 \\\\ g(5)=-30+9 \\\\ g(5)=-21 .\end{array} Replacing $x$ with the result above in $f$ results to \begin{array}{l}\require{cancel} f(x)=3x+8 \\\\ f(-21)=3(-21)+8 \\\\ f(-21)=-63+8 \\\\ f(-21)=-55 .\end{array} Hence, $f(g(5))=-55 .$ b) Replacing $x$ with $5$ in $f$ results to \begin{array}{l}\require{cancel} f(x)=3x+8 \\\\ f(5)=3(5)+8 \\\\ f(5)=15+8 \\\\ f(5)=23 .\end{array} Replacing $x$ with the result above in $g$ results to \begin{array}{l}\require{cancel} g(x)=-6x+9 \\\\ g(23)=-6(23)+9 \\\\ g(23)=-138+9 \\\\ g(23)=-129 .\end{array} Hence, $g(f(5))=-129 .$ Therefore, \begin{array}{l}\require{cancel} \text{a) } f(g(5))=-55 \\\\\text{b) } g(f(5))=-129 .\end{array}

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.