## 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: 18

#### Answer

$\text{a) } (f\circ g)(x)=-165x+471 \\\text{b) } (g\circ f)(x) =-165x-274$

#### Work Step by Step

$\bf{\text{Solution Outline:}}$ With \begin{array}{l}\require{cancel} f(x)= 11x+21 \\g(x)= -15x+41 ,\end{array} use the definition of function composition to find $(f\circ g)(x)$ and $(g\circ f)(x).$ $\bf{\text{Solution Details:}}$ Using $(f\circ g)(x)=f(g(x)),$ then replace $x$ with $g(x)$ in $f$. Hence, \begin{array}{l}\require{cancel} (f\circ g)(x)=f(g(x)) \\\\ (f\circ g)(x)=f(-15x+41) \\\\ (f\circ g)(x)=11(-15x+41)+21 \\\\ (f\circ g)(x)=-165x+451+21 \\\\ (f\circ g)(x)=-165x+471 .\end{array} Using $(g\circ f)(x) =g(f(x)),$ then replace $x$ with $f(x)$ in $g.$ Hence, \begin{array}{l}\require{cancel} (g\circ f)(x)=g(f(x)) \\\\ (g\circ f)(x) =g(11x+21) \\\\ (g\circ f)(x) =-15(11x+21)+41 \\\\ (g\circ f)(x) =-165x-315+41 \\\\ (g\circ f)(x) =-165x-274 .\end{array} Hence, \begin{array}{l}\require{cancel} \text{a) } (f\circ g)(x)=-165x+471 \\\text{b) } (g\circ f)(x) =-165x-274 .\end{array}

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