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: 42


See below:

Work Step by Step

(a) Consider the function: $f\left( x \right)={{x}^{2}}-1$ Follow the steps to determine the inverse ${{f}^{-1}}\left( x \right)$ of the function $f\left( x \right)={{x}^{2}}-1$. Step 1: Replace the function $f\left( x \right)$ with y in the equation of $f\left( x \right)$. $y={{x}^{2}}-1$ Step 2: Interchange the variables x and y. $x={{y}^{2}}-1$ Step 3: Solve the equation for y. The variable y has to be isolated. Add $1$ to both sides of the equation. So, the equation becomes, $x+1={{y}^{2}}-1+1$ Combine like terms. $x+1={{y}^{2}}$ Take square root on both sides of the equation. $\begin{align} & \sqrt{x+1}=\sqrt{{{y}^{2}}} \\ & \sqrt{x+1}=y \\ \end{align}$ 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, ${{f}^{-1}}\left( x \right)=\sqrt{x+1}$ Therefore, the inverse function is ${{f}^{-1}}\left( x \right)=\sqrt{x+1}$. (b) Given above. (c) The domain and range of the functions in interval notation are: \[\begin{align} & \text{domain of }f=\text{range of }{{f}^{-1}}=\left( -\infty ,0 \right] \\ & \text{range of }f=\text{domain of }{{f}^{-1}}=\left[ -1,\infty \right) \end{align}\]
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