## Precalculus (6th Edition) Blitzer

Published by Pearson

# Chapter 11 - Section 11.2 - Finding Limits Using Properties of Limits - Exercise Set - Page 1154: 58

#### Answer

The limit $\underset{x\to 5}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)$ is $\frac{-17}{3}$.

#### Work Step by Step

Consider the provided function, $f\left( x \right)=\frac{2x+3}{x+4}$. Follow the steps to find and inverse of the function to get the inverse of $f\left( x \right)=\frac{2x+3}{x+4}$ Step 1 : Replace $f\left( x \right)$ with $y$ $y=\frac{2x+3}{x+4}$ Step 2 : Interchange $y$ and $x$ $x=\frac{2y+3}{y+4}$ Step 3 : Solve for $y$ $x=\frac{2y+3}{y+4}$ Multiply by $y+4$ on both sides: \begin{align} & x\left( y+4 \right)=\frac{2y+3}{y+4}\left( y+4 \right) \\ & xy+4x=2y+3 \end{align} Subtract $2y$ from both sides: \begin{align} & xy+4x-2y=2y+3-2y \\ & xy-2y+4x=3 \end{align} Subtract $4x$ from both sides: \begin{align} & xy-2y+4x-4x=3-4x \\ & xy-2y=3-4x \\ & y\left( x-2 \right)=3-4x \end{align} Divide by $x-2$ on both sides; \begin{align} & \frac{y\left( x-2 \right)}{\left( x-2 \right)}=\frac{3-4x}{\left( x-2 \right)} \\ & y=\frac{3-4x}{\left( x-2 \right)} \\ & =\frac{-4x+3}{\left( x-2 \right)} \end{align} Replace $y$ with ${{f}^{-1}}\left( x \right)$, Thus, ${{f}^{-1}}\left( x \right)=\frac{-4x+3}{x-2}$. Now, solve the limit $\underset{x\to 5}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)$. $\underset{x\to 5}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)=\underset{x\to 5}{\mathop{\lim }}\,\frac{-4x+3}{x-2}$ Use the quotient property of limits: \begin{align} & \underset{x\to 5}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)=\underset{x\to 5}{\mathop{\lim }}\,\frac{-4x+3}{x-2} \\ & =\frac{\underset{x\to 5}{\mathop{\lim }}\,\left( -4x+3 \right)}{\underset{x\to 5}{\mathop{\lim }}\,\left( x-2 \right)} \end{align} Use the limit property $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)-g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)-\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$ in the numerator and $\underset{x\to a}{\mathop{\lim }}\,\left( f\left( x \right)+g\left( x \right) \right)=\underset{x\to a}{\mathop{\lim }}\,f\left( x \right)+\underset{x\to a}{\mathop{\lim }}\,g\left( x \right)$ in the denominator. \begin{align} & \underset{x\to 5}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)=\frac{\underset{x\to 5}{\mathop{\lim }}\,\left( -4x+3 \right)}{\underset{x\to 5}{\mathop{\lim }}\,\left( x-2 \right)} \\ & =\frac{\underset{x\to 5}{\mathop{\lim }}\,\left( -4x \right)+\underset{x\to 5}{\mathop{\lim }}\,3}{\underset{x\to 5}{\mathop{\lim }}\,x-\underset{x\to 5}{\mathop{\lim }}\,2} \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 5}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)=\frac{\underset{x\to 5}{\mathop{\lim }}\,\left( -4x \right)+\underset{x\to 5}{\mathop{\lim }}\,3}{\underset{x\to 5}{\mathop{\lim }}\,x-\underset{x\to 5}{\mathop{\lim }}\,2} \\ & =\frac{\left( -4\times 5 \right)+3}{5-2} \\ & =\frac{-20+3}{3} \\ & =\frac{-17}{3} \end{align} Thus $\underset{x\to 5}{\mathop{\lim }}\,{{f}^{-1}}\left( x \right)=-\frac{17}{3}$

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