Precalculus (6th Edition) Blitzer

Published by Pearson

Chapter 4 - Cumulative Review Exercises - Page 647: 7

Answer

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

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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