Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.3 Exercises - Page 343: 42

Answer

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

Work Step by Step

$$\eqalign{ & f\left( x \right) = \sqrt {{x^2} - 4} ,{\text{ }}x \geqslant 2 \cr & \left( {\text{a}} \right) \cr & {\text{Let }}y = f\left( x \right) \cr & y = \sqrt {{x^2} - 4} \cr & {\text{Interchange }}x{\text{ and }}y \cr & x = \sqrt {{y^2} - 4} \cr & {\text{Solve for }}y \cr & x = \sqrt {{y^2} - 4} \cr & {x^2} = {y^2} - 4 \cr & y = \sqrt {{x^2} + 4} \cr & {\text{Let }}y = {f^{ - 1}}\left( x \right) \cr & {f^{ - 1}}\left( x \right) = \sqrt {{x^2} + 4} ,{\text{ }} \cr & \cr & \left( {\text{b}} \right){\text{ Graph shown below}} \cr & \left( {\text{c}} \right){\text{ The graphs of }}f\left( x \right){\text{ and }}{f^{ - 1}}\left( x \right){\text{ are reflections of each other}}{\text{.}} \cr & \left( {\text{d}} \right){\text{Domain of }}f\left( x \right) = {\text{Range }}{f^{ - 1}}\left( x \right) = \left[ {2,\infty } \right) \cr & {\text{Domain of }}{f^{ - 1}}\left( x \right) = {\text{Range }}f\left( x \right) = \left[ {2,\infty } \right) \cr & \cr & {\text{Graph}} \cr} $$
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