Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 2 - Section 2.4 - Dividing Polynomials; Remainder and Factor Theorems - Exercise Set - Page 366: 81


The inverse of the function $f\left( x \right)$ is ${{f}^{^{_{-1}}}}\left( x \right)=\underline{\frac{10\left( x+1 \right)}{1-x}}$

Work Step by Step

We know that the inverse of a function $f\left( x \right)$ is labeled as ${{f}^{-1}}\left( x \right)$ and it satisfies: $\left( f\circ {{f}^{-1}} \right)\left( x \right)=\left( {{f}^{-1}}\circ f \right)\left( x \right)$ For the inverse of an inverse of any function, find out the expression for x in terms of y. Now, one can solve the function as shown below: $\begin{align} & f\left( x \right)=y \\ & \frac{x-10}{x+10}=y \\ & x-10=y\left( x+10 \right) \\ & x-10=yx+10y \end{align}$ And solve for x as follows: $\begin{align} & x-10=yx+10y \\ & x-yx=10y+10 \\ & x\left( 1-y \right)=10\left( y+1 \right) \\ & x=\frac{10\left( y+1 \right)}{1-y} \end{align}$ Now, replace all the y terms with x terms in the obtained function to get, ${{f}^{-1}}\left( x \right)=\frac{10\left( x+1 \right)}{1-x}$ Thus, the inverse of the function is $\frac{10\left( x+1 \right)}{1-x}$.
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