Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.3 Inverse, Exponential, and Logarithmic Functions - 1.3 Exercises: 27

Answer

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

Work Step by Step

$f(x)=\sqrt (x+2)$ $y=\sqrt (x+2)$ $y^{2}=x+2$ $y^{2}-2=x$ $f^{-1}(x)=y^{2}-2$
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