Precalculus (6th Edition) Blitzer

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

Chapter 1 - Section 1.8 - Inverse Functions - Exercise Set - Page 269: 22


a) ${{f}^{-1}}\left( x \right)=\frac{2}{x}$. b) they are inverses

Work Step by Step

(a) Consider the function: $f\left( x \right)=\frac{2}{x}$ Follow the steps to determine the inverse ${{f}^{-1}}\left( x \right)$ of the function $f\left( x \right)=\frac{2}{x}$. Step 1: Replace the function $f\left( x \right)$ with y in the equation of $f\left( x \right)$. $y=\frac{2}{x}$ Step 2: Interchange the variables x and y. $x=\frac{2}{y}$ Step 3: Solve the equation for y. The variable y has to be isolated. Take the reciprocal of both sides of the equation. So, the equation becomes, $\frac{1}{x}=\frac{y}{2}$ Multiply by $2$ on both the sides of the equation. $\begin{align} & \frac{1}{x}\times 2=\frac{y}{2}\times 2 \\ & \frac{2}{x}=y \end{align}$ Because the obtained equation defines y as a function of x, then the inverse function exists for the function f. Step 4: Replace y by ${{f}^{-1}}\left( x \right)$ in the equation obtained in step 3. Thus, the inverse function ${{f}^{-1}}\left( x \right)$ is obtained as, ${{f}^{-1}}\left( x \right)=\frac{2}{x}$ Therefore, the inverse function is ${{f}^{-1}}\left( x \right)=\frac{2}{x}$. (b) Consider the function: $f\left( x \right)=\frac{2}{x}$. The inverse of the function is obtained as ${{f}^{-1}}\left( x \right)=\frac{2}{x}$ in part (a). The inverse of an equation can be verified by showing that $f\left( {{f}^{-1}}\left( x \right) \right)=x\text{ and }{{f}^{-1}}\left( f\left( x \right) \right)=x\text{ }$. First calculate $f\left( {{f}^{-1}}\left( x \right) \right)$. Replace ${{f}^{-1}}\left( x \right)=\frac{2}{x}$ and evaluate the function f. $\begin{align} & f\left( {{f}^{-1}}\left( x \right) \right)=f\left( \frac{2}{x} \right) \\ & =\frac{2}{\frac{2}{x}} \\ & =2\times \frac{x}{2} \\ & =x \end{align}$ Now, calculate ${{f}^{-1}}\left( f\left( x \right) \right)$. Replace $f\left( x \right)=\frac{2}{x}$ and evaluate the inverse function ${{f}^{-1}}$. $\begin{align} & {{f}^{-1}}\left( f\left( x \right) \right)={{f}^{-1}}\left( \frac{2}{x} \right) \\ & =\frac{2}{\frac{2}{x}} \\ & =2\times \frac{x}{2} \\ & =x \end{align}$ Thus, it verifies that $f\left( {{f}^{-1}}\left( x \right) \right)=x\text{ and }{{f}^{-1}}\left( f\left( x \right) \right)=x\text{ }$. Therefore, the obtained inverse function ${{f}^{-1}}\left( x \right)=\frac{2}{x}$ is correct. Hence, the functions $f$ and ${{f}^{-1}}$ are inverses of each other.
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