Algebra and Trigonometry 10th Edition

Published by Cengage Learning
ISBN 10: 9781337271172
ISBN 13: 978-1-33727-117-2

Chapter 2 - 2.7 - Inverse Functions - 2.7 Exercises - Page 229: 68


$f^{-1} (x)=x^2+2 $; Domain: $x \geq 0$.

Work Step by Step

When we apply the horizontal test, it has been noticed that the function is one-to-one and verifies the horizontal test. Therefore, the function has an inverse function. To compute the inverse, we will have to interchange $x$ and $y$. $x=\sqrt {y-2} \implies x^2=y-2$ or, $y=x^2+2 $ Replace $y$ with $f^{-1} (x)$. so, $f^{-1} (x)=x^2+2 $ Since the range of the function is $x \geq 0$, the domain for the function is $x \geq 0$.
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