## Precalculus (6th Edition) Blitzer

a) ${{f}^{-1}}\left( x \right)=\frac{x+1}{3}$ b) they are inverses
(a) Consider the function, $f\left( x \right)=3x-1$ Step-1-Replace $f\left( x \right)$ with $y$ in $f\left( x \right)$ \begin{align} & f\left( x \right)=3x-1 \\ & y=3x-1 \end{align} Step-2-Interchange $x$ and $y$ \begin{align} & y=3x-1 \\ & x=3y-1 \\ \end{align} Step-3-Solve for $y$. Add $1$ to each side \begin{align} & x=3y-1 \\ & x+1=3y-1+1 \\ & x+1=3y \end{align} Divide each side by $3$ \begin{align} & x+1=3y \\ & \frac{x+1}{3}=\frac{3y}{3} \\ & \frac{x+1}{3}=y \end{align} Step-4-Replace $y$ in step 3 by ${{f}^{-1}}\left( x \right)$ \begin{align} & y=\frac{x+1}{3} \\ & {{f}^{-1}}\left( x \right)=\frac{x+1}{3} \end{align} Therefore, the required inverse of the function is ${{f}^{-1}}\left( x \right)=\frac{x+1}{3}$ (b) Consider the functions: $f\left( x \right)=3x-1$ and ${{f}^{-1}}\left( x \right)=\frac{x+1}{3}$ Replace $x$ with ${{f}^{-1}}\left( x \right)$ \begin{align} & f\left( {{f}^{-1}}\left( x \right) \right)=3{{f}^{-1}}\left( x \right)-1 \\ & =3\left( \frac{x+1}{3} \right)-1 \end{align} Simplify \begin{align} & f\left( g\left( x \right) \right)=3\left( \frac{x+1}{3} \right)-1 \\ & =x+1-1 \\ & =x \end{align} Now, find ${{f}^{-1}}\left( f\left( x \right) \right)$ The equation for ${{f}^{-1}}\left( x \right)$ is given as: ${{f}^{-1}}\left( x \right)=\frac{x+1}{3}$ Replace $x$ with $f\left( x \right)$ \begin{align} & {{f}^{-1}}\left( f\left( x \right) \right)=\frac{f\left( x \right)+1}{3} \\ & =\frac{\left( 3x-1 \right)+1}{3} \end{align} Simplify \begin{align} & {{f}^{-1}}\left( f\left( x \right) \right)=\frac{\left( 3x-1 \right)+1}{3} \\ & =\frac{3x}{3} \\ & =x \end{align} Thus, $f\left( {{f}^{-1}}\left( x \right) \right)=x$ and ${{f}^{-1}}\left( f\left( x \right) \right)=x$ It can be easily observed that ${{f}^{-1}}$ can be expressed in $f$ if we add 3 and divide by 3. Hence, the functions $f$ and ${{f}^{-1}}$ are inverses of each other.