Solution Found!
Each of Exercise gives a formula for a function y = ƒ(x)
Chapter 1, Problem 35E(choose chapter or problem)
Each of Exercise gives a formula for a function \(y=f(x)\). In each case, find \(f^{-1}(x)\) and identify the domain and range of \(f^{-1}\). As a check, show that \(f\left(f^{-1}(x)\right)=f^{-1}(f(x))=x\).
\(f(x)=\frac{x+b}{x-2}, \quad b>-2\) and constant
Questions & Answers
QUESTION:
Each of Exercise gives a formula for a function \(y=f(x)\). In each case, find \(f^{-1}(x)\) and identify the domain and range of \(f^{-1}\). As a check, show that \(f\left(f^{-1}(x)\right)=f^{-1}(f(x))=x\).
\(f(x)=\frac{x+b}{x-2}, \quad b>-2\) and constant
ANSWER:Step 1 of 4
To determine \(f^{-1}\) put \(y=f(x)\) and solve for \(x\) :
\(\begin{array}{c} y=\frac{x+b}{x-2} \\ (x-2) y=x+b \\ x y-x=2 y+b \\ x=\frac{2 y+b}{y-1} \end{array}\)
Now we will interchange x and y to find the formula for \(f^{-1}\).
\(y=\frac{2 x+b}{x-1}\)
which leads us to conclude that the inverse function \(f^{-1}\) is defined as
\(f^{-1}(x)=\frac{2 x+b}{x-1}\)