Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 1 - Section 1.5 - Inverse Functions and Logarithms - 1.5 Exercises - Page 67: 24


$y=\frac{1}{2}+\sqrt {x+\frac{1}{4}}$

Work Step by Step

$y=x^2-x, x\geq\frac{1}{2}$ $=(x^2-x+\frac{1}{4})-\frac{1}{4}$ $=(x-\frac{1}{2})^2-\frac{1}{4}$ $(x-\frac{1}{2})^2=y+\frac{1}{4}$ $x-\frac{1}{2}=\pm \sqrt {y+\frac{1}{4}}$ $\because x\geq\frac{1}{2}, x-\frac{1}{2}\geq0$ $\therefore x-\frac{1}{2}=\sqrt {y+\frac{1}{4}}$ $x=\frac{1}{2}+\sqrt {y+\frac{1}{4}}$ $y\rightleftharpoons x$ $f^{-1}(x)=y=\frac{1}{2}+\sqrt {x+\frac{1}{4}}$
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