Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 5 - Exponential and Logarithmic Functions - Chapter Review - Review Exercises - Page 345: 1

Answer

(a) $ -26$ (b) $ -241$ (c) $ 16$ (d) $ -1$

Work Step by Step

Given $f(x)=3x-5$ and $g(x)=1-2x^2$, we have (a) $(f\circ g)(2)=f(g(2))=f(1-2(2)^2)=f(-7)=3(-7)-5=-26$ (b) $(g\circ f)(-2)=g(f(-2))=g(3(-2)-5)=g(-11)=1-2(-11)^2=-241$ (c) $(f\circ f)(4)=f(f(4))=f(3(4)-5)=f(7)=3(7)-5=16$ (d) $(g\circ g)(-1)=g(g(-1))=g(1-2(-1)^2)=g(-1)=1-2(-1)^2=-1$
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