## Precalculus (6th Edition) Blitzer

a) ${{f}^{-1}}\left( x \right)=x-5$ b) they are inverses
(a) Consider the function, $f\left( x \right)=x+5$ Step-1-Replace $f\left( x \right)$ with $y$. \begin{align} & f\left( x \right)=x+5 \\ & y=x+5 \end{align} Step-2-Interchange $x$ and $y$ \begin{align} & y=x+5 \\ & x=y+5 \\ \end{align} Step-3-Solve for $y$. Subtract $5$ from each side \begin{align} & x=y+5 \\ & x-5=y+5-5 \\ & x-5=y \end{align} Step-4-Replace $y$ in step 3 by ${{f}^{-1}}\left( x \right)$ \begin{align} & y=x-5 \\ & {{f}^{-1}}\left( x \right)=x-5 \end{align} Therefore, the required inverse of the function is ${{f}^{-1}}\left( x \right)=x-5$ (b) Consider the functions: $f\left( x \right)=x+5$ and ${{f}^{-1}}\left( x \right)=x-5$ Replace $x$ with ${{f}^{-1}}\left( x \right)$ in $f\left( x \right)$ \begin{align} & f\left( {{f}^{-1}}\left( x \right) \right)={{f}^{-1}}\left( x \right)+5 \\ & =\left( x-5 \right)+5 \end{align} Simplify, \begin{align} & f\left( g\left( x \right) \right)=\left( x-5 \right)+5 \\ & =x \end{align} Now, find ${{f}^{-1}}\left( f\left( x \right) \right)$ The equation for ${{f}^{-1}}\left( x \right)$ is given as: ${{f}^{-1}}\left( x \right)=x-5$ Replace $x$ with $f\left( x \right)$ \begin{align} & {{f}^{-1}}\left( f\left( x \right) \right)=f\left( x \right)-5 \\ & =\left( x+5 \right)-5 \end{align} Simplify \begin{align} & {{f}^{-1}}\left( f\left( x \right) \right)=\left( x+5 \right)-5 \\ & =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 subtracted by 5. Hence, the functions $f$ and ${{f}^{-1}}$ are inverses of each other.