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: 21


$f^{-1}(x)=\frac{1}{3}x^2-\frac{2}{3}x-\frac{1}{3}, x\geq1$

Work Step by Step

$y=1+\sqrt {2+3x}$ $\because \sqrt {2+3x}\geq0, y\geq1+0=1$ $\sqrt {2+3x}=y-1$ $2+3x=(y-1)^2$ $3x=(y-1)^2-2=y^2-2y-1$ $x=\frac{1}{3}y^2-\frac{2}{3}y-\frac{1}{3}$ $x\rightleftharpoons y$ $f^{-1}(x)=\frac{1}{3}x^2-\frac{2}{3}x-\frac{1}{3}$ Domain of inverse: $x\geq1$
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