Answer
See proof below.
Work Step by Step
Given $I(x)=x$, we have
$f\circ I=f(I(x))=f(x)$
$I\circ f=I(f(x))=f(x)$
$f\circ f^{-1}=f(f^{-1}(x))=x=I(x)$ use fundamental property of inverse functions
$f^{-1}\circ f=f^{-1}(f(x))=f^{-1}(y)=x=I(x)$
Precalculus: Mathematics for Calculus, 7th Edition
by Stewart, James; Redlin, Lothar; Watson, Saleem
Published by
Brooks Cole
ISBN 10:
1305071751
ISBN 13:
978-1-30507-175-9
Chapter 2 - Section 2.8 - One-to-One Functions and Their Inverses - 2.8 Exercises - Page 228: 104
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