Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 2 - Section 2.8 - One-to-One Functions and Their Inverses - 2.8 Exercises - Page 227: 60

Answer

$f^{-1}(x)=\dfrac{x+3}{8x+4}$

Work Step by Step

$f(x)=\dfrac{3-4x}{8x-1}$ Rewrite this expression as $y=\dfrac{3-4x}{8x-1}$ and solve for $x$: $y=\dfrac{3-4x}{8x-1}$ Take $8x-1$ to multiply the left side: $y(8x-1)=3-4x$ $8xy-y=3-4x$ Take $-4x$ to the left side and $-y$ to the right side: $8xy+4x=y+3$ Take out common factor $x$ from the left side: $x(8y+4)=y+3$ Take $8y+4$ to divide the right side: $x=\dfrac{y+3}{8y+4}$ Interchange $x$ and $y$: $y=\dfrac{x+3}{8x+4}$ The inverse of the original function is $f^{-1}(x)=\dfrac{x+3}{8x+4}$
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