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: 96


$x=1$ and $x=-\dfrac{4}{3}$ satisfy the given conditions

Work Step by Step

$f(x)=1-2x$ $,$ $g(x)=3x^{2}+x-1$ and $(f\circ g)(x)=-5$ To find $(f\circ g)(x)$, substitute $x$ by $g(x)$ in $f(x)$ and simplify: $(f\circ g)(x)=1-2(3x^{2}+x-1)=1-6x^{2}-2x+2=...$ $...=-6x^{2}-2x+3$ Set $(f\circ g)(x)$ equal to $-5$: $-6x^{2}-2x+3=-5$ Take $5$ to the left side: $-6x^{2}-2x+3+5=0$ $-6x^{2}-2x+8=0$ Take out common factor $-2$ from the left side and take it to divide the right side: $-2(3x^{2}+x-4)=0$ $3x^{2}+x-4=\dfrac{0}{-2}$ $3x^{2}+x-4=0$ Solve by factoring: $(x-1)(3x+4)=0$ Set both factors equal to $0$ and solve each individual equation for $x$: $x-1=0$ $x=1$ $3x+4=0$ $3x=-4$ $x=-\dfrac{4}{3}$ $x=1$ and $x=-\dfrac{4}{3}$ satisfy the given conditions
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.