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: 48

Answer

Using the inverse function property we have: $f^{-1}(x)=\frac{5+4x}{1-3x}$ $g^{-1}(x)=\frac{x-5}{3x+4}$ Which means that $f$ and $g$ are inverse of each other.

Work Step by Step

To show it we will simply calculate inverse of the both $f$ and $g$ functions. To calculate $f^{−1}(x)$ first we have to write it in terms of $y$ and $x$ and then replace $x$ with $y$ and vice versa. And at last solve the equation for $y$: $f(x)=\frac{x-5}{3x+4}$ $y=\frac{x-5}{3x+4}$ $x=\frac{y-5}{3y+4}$ $x(3y+4)=y-5$ $3xy+4x=y-5$ $3xy-y=-4x-5$ $y(3x-1)=-4x-5$ $y=\frac{-4x-5}{3x-1}$ //Factor out the minus sign $y=\frac{-(4x+5)}{-(1-3x)}$ $y=\frac{5+4x}{1-3x}$ $f^{-1}(x)=\frac{5+4x}{1-3x}$ $g(x)=\frac{5+4x}{1-3x}$ $y=\frac{5+4x}{1-3x}$ $x=\frac{5+4y}{1-3y}$ $x(1-3y)=5+4y$ $x-3xy=5+4y$ $-3xy-4y=5-x$ $y(-3x-4)=5-x$ $y=\frac{5-x}{-3x-4}$ //Factor out the minus sign $y=\frac{-(x-5)}{-(3x+4)}$ $y=\frac{x-5}{3x+4}$ $g^{-1}(x)=\frac{x-5}{3x+4}$
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