Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 1 - Section 1.3 - New Functions from Old Functions - 1.3 Exercises: 61


Yes, $m_{1}m_{2}$

Work Step by Step

$f(g(x))=f(m_{2}x+b_{2})=m_{1}(m_{2}x+b_{2})+b_{1}=m_{1}m_{2}x+m_{1}b_{2}+b_{1}$ $f(g(x))$ is still a linear function with gradient $m_{1}m_{2}$
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