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: 26

Answer

$\text{a) } (f\circ g)(x)=10x^2-35x+28 \\\text{b) } (g\circ f)(x)=25x^2-55x+24$

Work Step by Step

$\bf{\text{Solution Outline:}}$ With \begin{array}{l}\require{cancel} f(x)= 5x-2 \\g(x)= 2x^2-7x+6 ,\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(2x^2-7x+6) \\\\ (f\circ g)(x)=5(2x^2-7x+6)-2 \\\\ (f\circ g)(x)=10x^2-35x+30-2 \\\\ (f\circ g)(x)=10x^2-35x+28 .\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(5x-2) \\\\ (g\circ f)(x)=(5x-2)^2-7(5x-2)+6 \\\\ (g\circ f)(x)=(25x^2-20x+4)+(-35x+14)+6 \\\\ (g\circ f)(x)=25x^2-55x+24 .\end{array} Hence, \begin{array}{l}\require{cancel} \text{a) } (f\circ g)(x)=10x^2-35x+28 \\\text{b) } (g\circ f)(x)=25x^2-55x+24 .\end{array}
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