Calculus: Early Transcendentals 8th Edition

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

Chapter 1 - Section 1.5 - Inverse Functions and Logarithms - 1.5 Exercises: 16


$f^{-1}(3)=1$ $f(f^{-1}(2))=2$

Work Step by Step

Let $f^{-1}(3)=a, f(a)=3,$ $a^3+a^2+a=3$ try $a=1$, LHS=$1^3+1^2+1=3$=RHS $\therefore f^{-1}(3)=1$ Let $f^{-1}(2)=b, f(b)=2,$ $\therefore f(f^{-1}(2))=f(b)=2$
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