College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 2 - Functions and Graphs - Exercise Set 2.6 - Page 298: 60

Answer

$a.$ $(f\circ g)(x)=-5x^{2}+20x-7$ $b.$ $(g\circ f)(x)=-25x^{2}+40x-13$ $c.$ $(f\circ g)(2)=13$ $d.$ $(g\circ f)(2)=-33$

Work Step by Step

$f(x)=5x-2$ $,$ $g(x)=-x^{2}+4x-1$ $a.$ $(f\circ g)(x)$ To find $(f\circ g)(x)$, substitute $x$ by $g(x)$ in $f(x)$ and simplify: $(f\circ g)(x)=f(g(x))=5(-x^{2}+4x-1)-2=...$ $...=-5x^{2}+20x-5-2=-5x^{2}+20x-7$ $b.$ $(g\circ f)(x)$ To find $(g\circ f)(x)$, substitute $x$ by $f(x)$ in $g(x)$ and simplify: $(g\circ f)(x)=g(f(x))=-(5x-2)^{2}+4(5x-2)-1=...$ $...=-(25x^{2}-20x+4)+20x-8-1=...$ $...=-25x^{2}+20x-4+20x-8-1=-25x^{2}+40x-13$ $c.$ $(f\circ g)(2)$ Substitute $x$ by $2$ in $(f\circ g)(x)$, which was found in part $a$, and evaluate: $(f\circ g)(2)=f(g(2))=-5(2)^{2}+20(2)-7=-5(4)+40-7=...$ $...=-20+40-7=13$ $d.$ $(g\circ f)(2)$ Substitute $x$ by $2$ in $(g\circ f)(x)$, which was found in part $b$, and evaluate: $(g\circ f)(2)=g(f(2))=-25(2)^{2}+40(2)-13=...$ $...=-25(4)+80-13=-100+80-13=-33$
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