Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 7: Transcendental Functions - Section 7.1 - Inverse Functions and Their Derivatives - Exercises 7.1 - Page 372: 19


$$ f^{-1}(x)=\sqrt{x-1}$$

Work Step by Step

Given: $ y=x^{2}+1$ We solve for $x$: $\Rightarrow x^{2}=y-1$ $\Rightarrow x=\sqrt{y-1}$ Interchanging $ x $ and $ y $, we get: $y=\sqrt{x-1}=f^{-1}(x)$
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