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 270: 46


See below:

Work Step by Step

(a) Consider the function: $f\left( x \right)={{x}^{3}}+1$ Follow the steps to determine the inverse ${{f}^{-1}}\left( x \right)$ of the function $f\left( x \right)={{x}^{3}}+1$. Step 1: Replace the function $f\left( x \right)$ with y in the equation of $f\left( x \right)$. $y={{x}^{3}}+1$ Step 2: Interchange the variables x and y. So, the equation is: $x={{y}^{3}}+1$ Step 3: Solve the equation for y. The variable y has to be isolated. So, subtract $1$ to both sides of the equation. So, the equation becomes, $x-1={{y}^{3}}+1-1$ Combine like terms. $x-1={{y}^{3}}$ Take cubic root on both sides of the equation. So, $\begin{align} & {{\left( x-1 \right)}^{{1}/{3}\;}}={{\left( {{y}^{3}} \right)}^{{1}/{3}\;}} \\ & \sqrt[3]{x-1}=y \end{align}$ The equation is $y=\sqrt[3]{x-1}$. 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, $\sqrt[3]{x-1}={{f}^{-1}}\left( x \right)$ Therefore, the inverse function is ${{f}^{-1}}\left( x \right)=\sqrt[3]{x-1}$. (b) Given above. (c) The domain and range of the functions in interval notation are: $\begin{align} & \text{domain of }f=\text{range of }{{f}^{-1}}=\left( -\infty ,\infty \right) \\ & \text{range of }f=\text{domain of }{{f}^{-1}}=\left( -\infty ,\infty \right) \end{align}$
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