## Precalculus (6th Edition) Blitzer

(a) Consider the function, $f\left( x \right)=\frac{x-7}{x+2}$ Step 1: Replace $f\left( x \right)$ with $y$: $y=\frac{x-7}{x+2}$ Step 2: Interchange $x$ and $y$: $x=\frac{y-7}{y+2}$ Step 3: Now solve for the value of $y$: $xy+2x=y-7$ Or $2x+7=y-xy$ Or $2x+7=y\left( 1-x \right)$ That is, $y=\frac{2x+7}{1-x}$ Step 4: Replace $y$ with ${{f}^{-1}}\left( x \right)$: ${{f}^{-1}}\left( x \right)=\frac{2x+7}{1-x}$ Hence, the inverse function ${{f}^{-1}}\left( x \right)$ of the function $f\left( x \right)=\frac{x-7}{x+2}$ is ${{f}^{-1}}\left( x \right)=\frac{2x+7}{1-x}$. (b) Consider the function, $f\left( {{f}^{-1}}\left( x \right) \right)$ \begin{align} & f\left( {{f}^{-1}}\left( x \right) \right)=f\left( \frac{2x+7}{1-x} \right) \\ & =\frac{\left( \frac{2x+7}{1-x} \right)-7}{\left( \frac{2x+7}{1-x} \right)+2} \\ & =\frac{2x+7-7+7x}{2x+7+2-2x} \\ & =x \end{align} Next consider the function, ${{f}^{-1}}\left( f\left( x \right) \right)$ \begin{align} & {{f}^{-1}}\left( f\left( x \right) \right)={{f}^{-1}}\left( \frac{x-7}{x+2} \right) \\ & =\frac{2\left( \frac{x-7}{x+2} \right)+7}{1-\left( \frac{x-7}{x+2} \right)} \\ & =\frac{2x-14+7x+14}{x+2-x+7} \\ & =x \end{align} Hence, $f\left( {{f}^{-1}}\left( x \right) \right)=x$ and ${{f}^{-1}}\left( f\left( x \right) \right)=x$ for the function $f\left( x \right)=\frac{x-7}{x+2}$.