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: 43

Answer

See below:

Work Step by Step

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