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 272: 95


The composites ${{\left( f\circ g \right)}^{-1}}\left( x \right)$ and $\left( {{g}^{-1}}\circ {{f}^{-1}} \right)\left( x \right)$ with the functions $f\left( x \right)=3x,g\left( x \right)=x+5$ are $\left( {{g}^{-1}}\circ {{f}^{-1}} \right)\left( x \right)=\frac{x}{3}-5$ and ${{\left( f\circ g \right)}^{-1}}\left( x \right)=\frac{x}{3}-5$.

Work Step by Step

Assume the functions below: $\begin{align} & \left( f\circ g \right)\left( x \right)=f\left( g\left( x \right) \right) \\ & \left( f\circ g \right)\left( x \right)=3\left( x+5 \right) \\ & \left( f\circ g \right)\left( x \right)=3x+15 \end{align}$ To find ${{f}^{-1}}\left( x \right)$: Consider the function $f\left( x \right)=3x$ Put $f\left( x \right)=y$, $y=3x$ Now, put $x\text{ and } y$, $x=3y$ Solve for the value of $y$, $y=\frac{x}{3}$ Now, put $y={{f}^{-1}}\left( x \right)$ Here, ${{f}^{-1}}\left( x \right)=\frac{x}{3}$ To find ${{g}^{-1}}\left( x \right)$ : Let the function below be: $g\left( x \right)=x+5$ So, y = x + 5 Now, put x = y + 5 $y=x-5$ Here, ${{g}^{-1}}\left( x \right)=x-5$ Now, to find ${{\left( f\circ g \right)}^{-1}}\left( x \right)$: The function ${{\left( f\circ g \right)}^{-1}}\left( x \right)$ is the inverse of $\left( f\circ g \right)\left( x \right)$ Therefore, $y=3x+15$ \[x=3y+15\] $y=\frac{x-15}{3}$ Now, the following function: $\begin{align} & {{\left( f\circ g \right)}^{-1}}\left( x \right)=\frac{x-15}{3} \\ & {{\left( f\circ g \right)}^{-1}}\left( x \right)=\frac{x}{3}-5 \\ \end{align}$ Now, to solve for $\left( {{g}^{-1}}\circ {{f}^{-1}} \right)\left( x \right)$ $\begin{align} & \left( {{g}^{-1}}\circ {{f}^{-1}} \right)\left( x \right)={{g}^{-1}}\left( {{f}^{-1}}\left( x \right) \right) \\ & \left( {{g}^{-1}}\circ {{f}^{-1}} \right)\left( x \right)=\frac{x}{3}-5 \\ \end{align}$
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