Precalculus (6th Edition) Blitzer

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

Chapter 11 - Section 11.2 - Finding Limits Using Properties of Limits - Exercise Set - Page 1154: 56

Answer

The limit $\underset{x\to 25}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)$ is 4.

Work Step by Step

Consider the provided function, $ f\left( x \right)={{x}^{2}}+9\,\ \text{ for }\ x\ge 0\text{ }$ Follow the steps to find and inverse of the function to get the inverse of $ f\left( x \right)={{x}^{2}}+9\,\ \text{ for }\ x\ge 0\text{ }$ Step 1: Replace $ f\left( x \right)$ with $ y $. $ y={{x}^{2}}+9\,x\ge 0$ Step 2: Interchange $ y $ and $ x $. $ x={{y}^{2}}+9\,y\ge 0$ Step 3: Solve for $ y $. $\begin{align} & x={{y}^{2}}+9\,y\ge 0 \\ & x-9={{y}^{2}} \\ & \sqrt{x-9}=y\ \end{align}$ Since $ y\ge 0$, so negative root is not considered. Step 4: Replace $ y $ with ${{f}^{-1}}\left( x \right)$. ${{f}^{-1}}\left( x \right)=\sqrt{x-9}$ Thus, ${{f}^{-1}}\left( x \right)=\sqrt{x-9}$. Solve the limit $\underset{x\to 25}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)$ $\underset{x\to 25}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)=\underset{x\to 25}{\mathop{\lim }}\,\sqrt{x-9}$ Use the limit property $\underset{x\to a}{\mathop{\lim }}\,\sqrt{f\left( x \right)}=\sqrt{\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)}$ $\begin{align} & \underset{x\to 25}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)=\underset{x\to 25}{\mathop{\lim }}\,\sqrt{x-9} \\ & =\sqrt{\underset{x\to 25}{\mathop{\lim }}\,\left( x-9 \right)} \end{align}$ Use the limit property $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)-g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)-\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$ $\begin{align} & \underset{x\to 25}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)=\sqrt{\underset{x\to 25}{\mathop{\lim }}\,\left( x-9 \right)} \\ & =\sqrt{\underset{x\to 25}{\mathop{\lim }}\,x-\underset{x\to 25}{\mathop{\lim }}\,9} \end{align}$ Use the limit property $\underset{x\to a}{\mathop{\lim }}\,c=c\ \text{where c is a constant}\text{.}$ $\begin{align} & \underset{x\to 25}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)=\sqrt{\underset{x\to 25}{\mathop{\lim }}\,x-\underset{x\to 25}{\mathop{\lim }}\,9} \\ & =\sqrt{25-9} \\ & =\sqrt{16} \\ & =4 \end{align}$ Thus, the limit $\underset{x\to 25}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)$ is 4.
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