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: 70

Answer

$f^{-1}(x)=\sqrt{4-x^{2}}$

Work Step by Step

$f(x)=\sqrt{4-x^{2}}$ $,$ $0\le x\le2$ Rewrite this expression as $y=\sqrt{4-x^{2}}$ and solve for $x$: $y=\sqrt{4-x^{2}}$ Square both sides: $y^{2}=(\sqrt{4-x^{2}})^{2}$ $y^{2}=4-x^{2}$ Take $-x^{2}$ to the left side: $x^{2}+y^{2}=4$ Take $y^{2}$ to the right side: $x^{2}=4-y^{2}$ Take the square root of both sides: $\sqrt{x^{2}}=\sqrt{4-y^{2}}$ $x=\sqrt{4-y^{2}}$ Interchange $x$ and $y$: $y=\sqrt{4-x^{2}}$ The inverse of the original function is $f^{-1}(x)=\sqrt{4-x^{2}}$
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