Calculus: Early Transcendentals, 8th Edition
Authors: James Stewart
ISBN-13: 978-1285741550
See our solution for Question 22E from Chapter 1.5 from Stewart's Calculus, 8th Edition.
Problem 22E
Chapter:
Problem:
Find a formula for the inverse of the function.
Step-by-Step Solution
Given information
We are given with following function
\[f\left( x \right) = \dfrac{{4x - 1}}{{2x + 3}}\]We have to find the inverse of this function. To find the inverse, we have to find x in terms of f(x), then replace x by $ {f^{ - 1}}\left( x \right)$ and f(x) with x.
Step 1:
Simplify the function and find $x$ in terms of $f(x)$\[\begin{array}{l}\left( {2x + 3} \right)f\left( x \right) = 4x - 1\\2xf\left( x \right) + 3f\left( x \right) = 4x - 1\\\left( {2f\left( x \right) - 4} \right)x = - 1 - 3f\left( x \right)\\x = \dfrac{{ - 1 - 3f\left( x \right)}}{{2f\left( x \right) - 4}}\\x = \dfrac{{1 + 3f\left( x \right)}}{{4 - 2f\left( x \right)}}\end{array}\]
Step 2: Interchange $x$ and $f(x)$
\[{f^{ - 1}}\left( x \right) = \dfrac{{1 + 3x}}{{4 - 2x\,}}\]There for the inverse of the function is: \[{f^{ - 1}}\left( x \right) = \dfrac{{1 + 3x}}{{4 - 2x\,}}\]
We are given with following function
\[f\left( x \right) = \dfrac{{4x - 1}}{{2x + 3}}\]We have to find the inverse of this function. To find the inverse, we have to find x in terms of f(x), then replace x by $ {f^{ - 1}}\left( x \right)$ and f(x) with x.
Step 1:
Simplify the function and find $x$ in terms of $f(x)$\[\begin{array}{l}\left( {2x + 3} \right)f\left( x \right) = 4x - 1\\2xf\left( x \right) + 3f\left( x \right) = 4x - 1\\\left( {2f\left( x \right) - 4} \right)x = - 1 - 3f\left( x \right)\\x = \dfrac{{ - 1 - 3f\left( x \right)}}{{2f\left( x \right) - 4}}\\x = \dfrac{{1 + 3f\left( x \right)}}{{4 - 2f\left( x \right)}}\end{array}\]
Step 2: Interchange $x$ and $f(x)$
\[{f^{ - 1}}\left( x \right) = \dfrac{{1 + 3x}}{{4 - 2x\,}}\]There for the inverse of the function is: \[{f^{ - 1}}\left( x \right) = \dfrac{{1 + 3x}}{{4 - 2x\,}}\]