Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 4 - Cumulative Review Exercises - Page 647: 7


$ f^{-1} (x)= x^2+6$

Work Step by Step

Re-arrange as: $ y=\sqrt {x-6}$ $ y^2=(\sqrt {x-6})^2$ or, $ y^2=x-6$ or, $ x=y^2+6$ So, $ f^{-1} (x)= x^2+6$
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