Answer
$f^{-1}(x)=\sqrt{x-1},x\geq 1$
Work Step by Step
$f(x)=x^2+1,$ for $x\geq0$
$y= x^2+1$
a.
$x=y^2+1$
$y=\sqrt{x-1}$
$f^{-1}(x)=\sqrt{x-1},x\geq 1$
b.
$f(f^{-1}(x))=(\sqrt {x-1})^2+1=x-1+1=x$
and
$f^{-1}(f(x))=(\sqrt {x^2+1-1})=x$
Intermediate Algebra for College Students (7th Edition)
by Blitzer, Robert F.
Published by
Pearson
ISBN 10:
0-13417-894-7
ISBN 13:
978-0-13417-894-3
Chapter 9 - Section 9.2 - Composite and Inverse Functions - Exercise Set - Page 687: 39
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