## Precalculus (6th Edition) Blitzer

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