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: 47


See below:

Work Step by Step

(a) Consider the function: $f\left( x \right)={{\left( x+2 \right)}^{3}}$ Follow the steps to determine the inverse ${{f}^{-1}}\left( x \right)$ of the function $f\left( x \right)={{\left( x+2 \right)}^{3}}$. Step 1: Replace the function $f\left( x \right)$ with y in the equation of $f\left( x \right)$. $y={{\left( x+2 \right)}^{3}}$ Step 2: Interchange the variables x and y. So, the equation is: $x={{\left( y+2 \right)}^{3}}$ Step 3: Solve the equation for y. The variable y has to be isolated. As $\sqrt[3]{{{y}^{3}}}=y$ , take cube root on both sides of the equation. So, the equation becomes, ${{\left( x \right)}^{{1}/{3}\;}}={{\left( {{\left( y+2 \right)}^{3}} \right)}^{{1}/{3}\;}}$ Simplify the powers. So, $\sqrt[3]{x}=y+2$ Subtract $2$ from both sides of the equation and simplify. $\begin{align} & \sqrt[3]{x}-2=y+2-2 \\ & \sqrt[3]{x}-2=y \\ \end{align}$ The equation is $y=\sqrt[3]{x}-2$. 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, ${{f}^{-1}}\left( x \right)=\sqrt[3]{x}-2$ Therefore, the inverse function is ${{f}^{-1}}\left( x \right)=\sqrt[3]{x}-2$. (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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