Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 2 - Section 2.8 - One-to-One Functions and Their Inverses - 2.8 Exercises - Page 227: 68

Answer

$f^{-1}(x)=(x-2)^{2}-3$

Work Step by Step

$f(x)=2+\sqrt{3+x}$ Rewrite this expression as $y=2+\sqrt{3+x}$ and solve for $x$: $y=2+\sqrt{3+x}$ Take $2$ to the left side: $y-2=\sqrt{3+x}$ Square both sides: $(y-2)^{2}=(\sqrt{3+x})^{2}$ $(y-2)^{2}=3+x$ $3+x=(y-2)^{2}$ Take $3$ to the right side: $x=(y-2)^{2}-3$ Interchange $x$ and $y$: $y=(x-2)^{2}-3$ The inverse of the original function is $f^{-1}(x)=(x-2)^{2}-3$
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