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 230: 99


$f$ is restricted to $[0,\infty)$. $f^{-1}(x)=\sqrt{x-1}$ for $x\in[1,\infty)$

Work Step by Step

We are given the function: $f(x)=x^2+1$ Graph the function and use the Horizontal Line Test to check if it has an inverse. We notice that a horizontal line intersects the graph at more than one point, so the function does not have an inverse on the domain $(-\infty,\infty)$. We restrict the domain of $f$ to the interval $[0,\infty)$, on which the function has an inverse. Determine the inverse: $y=x^2+1$ $x=y^2+1$ $y^2=x-1$ $y=\sqrt{x-1}$ as the domain of $f$ equals the range of $f^{-1}$, and the range of $f$ equals the domain of $f^{-1}$.
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